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Physics for Chemists Ruslan P. Ozerov Department of Physics Faculty of General Technological Sciences D.I. Mendeleev University of Chemical Technology Moscow, Russia The School of Biomedical Biomolecular and Chemical Science University of Western Australia Western Australia Crawly, Australia
and Anatoli A. Vorobyev Department of Physics Faculty of General Technological Sciences D.I. Mendeleev University of Chemical Technology Moscow, Russia
Amsterdam ● Boston ● Heidelberg ● London ● New York ● Oxford Paris ● San Diego ● San Francisco ● Singapore ● Sydney ● Tokyo
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Elsevier Radarweg 29, PO Box 211, 1000 AE Amsterdam, The Netherlands The Boulevard, Langford Lane, Kidlington, Oxford OX5 1GB, UK First edition 2007 Copyright © 2007 Elsevier B.V. All rights reserved No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the publisher Permissions may be sought directly from Elsevier’s Science & Technology Rights Department in Oxford, UK: phone (⫹44) (0) 1865 843830; fax (⫹44) (0) 1865 853333; email:
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Contents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiii Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xv Recommendations to the Solution of the Physical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . xvii
1 Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2.1 Kinematics of a material point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2.2 Kinematics of translational movement of a rigid body . . . . . . . . . . . . . . . . . . 12 1.2.3 Kinematics of the rotational motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 1.3 Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 1.3.1 Newton’s first law of motion: inertial reference systems. . . . . . . . . . . . . . . . . 16 1.3.2 Galileo’s relativity principle: Galileo transformations . . . . . . . . . . . . . . . . . . . 18 1.3.3 Newton’s second law of motion: Momentum . . . . . . . . . . . . . . . . . . . . . . . . . 20 1.3.4 The third Newtonian law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 1.3.5 Forces classification in physics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 1.3.6 Noninertial reference systems. An inertia force: D’Alembert principle . . . . . . 33 1.3.7 A system of material points: internal and external forces . . . . . . . . . . . . . . . . 35 1.3.8 Specification of a material points system . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 1.3.9 The dynamics of rotational motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 1.4 Work, Energy and Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 1.4.1 Elementary work of a force and a torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 1.4.2 Power. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 1.4.3 Kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 1.4.4 A force field. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 1.4.5 Potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 1.5 Conservation Laws in Mechanics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 1.5.1 Conservation law of mechanical energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 1.5.2 Momentum conservation law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 1.5.3 Angular momentum conservation law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 1.5.4 Potential curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 1.5.5 Particle collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 1.6 Einstein’s Special Relativistic Theory (STR) (Short Review) . . . . . . . . . . . . . . . . . . . 90 Problems/ Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 2 Oscillations and Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Definitions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Kinematics of Harmonic Oscillations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Summation of Oscillations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v
105 105 106 113
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2.3.1 Summation of codirectional oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Summing up two codirectional oscillations with slightly different frequencies: beatings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Dynamics of the Harmonic Oscillation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Differential equations of harmonic oscillations. . . . . . . . . . . . . . . . . . . . . . . . 2.4.2 Spring pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.3 The mathematical pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.4 A physical pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.5 Diatomic molecule as a linear harmonic oscillator . . . . . . . . . . . . . . . . . . . . . 2.5 Energy of Harmonic Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Damped Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Forced Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8.1 Introductory remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8.2 An equation of a plane traveling wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8.3 Wave energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8.4 Acoustic Doppler effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9 Summation of Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9.1 Superposition of waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9.2 Standing waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9.3 String oscillations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9.4 Group velocity of waves: wave package . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems/ Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 Molecular Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Kinetic Theory of Ideal Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 Introductory remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.2 Distribution function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.3 An ideal gas model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.4 General equation of an ideal gas. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.5 Absolute temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Distribution of Molecules of an Ideal Gas in a Force Field (Boltzmann Distribution) . . . 3.2.1 An ideal gas in a force field: Boltzmann distribution . . . . . . . . . . . . . . . . . . . 3.2.2 Barometric height formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.3 Centrifugation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.4 Boltzmann factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Distribution of the Kinetic Parameters of an Ideal Gas’ Particles (Maxwell Distribution) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 The Maxwellian distribution of the absolute values of molecule velocities . . . . 3.3.2 The kinetic energies Maxwellian distribution of molecules. . . . . . . . . . . . . . . 3.4 First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.1 Equipartition of energy over degrees of freedom . . . . . . . . . . . . . . . . . . . . . . 3.4.2 First laws of thermodynamics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.3 Heat capacity of an ideal gas: the work of a gas in isoprocesses . . . . . . . . . . . 3.4.4 Heat capacity: theory versus experiment. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 The Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 Heat engines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
113 117 118 118 118 119 121 129 131 133 138 145 145 147 151 154 156 156 157 160 163 165 166
169 169 169 172 174 175 177 178 178 180 183 185 186 186 193 194 194 195 197 204 205 206
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3.5.2 The Carnot cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.3 Refrigerators and heat pumps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.4 Reduced amount of heat: entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.5 Clausius inequality and the change of entropy for nonequilibrium processes . . . 3.5.6 Statistical explanation of the second law of thermodynamics . . . . . . . . . . . . . 3.5.7 Entropy and disorder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 A Real Gas Approximation: van der Waals Equation . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.1 An equation of state of a van der Waals gas . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.2 Internal energy of the van der Waals gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.3 A Joule–Thomson effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Elements of Physical Kinetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.2 Transport processes: relaxation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.3 Transport phenomena in ideal gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.4 A macroscopic representation of a transport coefficient . . . . . . . . . . . . . . . . . 3.7.5 Diffusion in gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.6 Heat transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.7 Viscosity or internal friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.8 A transport phenomena in a vacuum condition . . . . . . . . . . . . . . . . . . . . . . . . Problems/ Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
207 210 211 214 219 220 221 221 226 227 230 230 230 231 233 235 237 238 243 245 247
4 Dielectric Properties of Substances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Electrostatic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.1 General laws of electrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.2 Strength of an electrostatic field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.3 The Gauss law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.4 Work of an electrostatic field force and potential of an electrostatic field . . . . 4.1.5 Electrical field of an electric dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Dielectric Properties of Substances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Conductors and dielectrics: a general view . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Macroscopic (phenomenological) properties of dielectrics . . . . . . . . . . . . . . . 4.2.3 Microscopic properties of dielectrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.4 Three types of polarization mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.5 Dependence of the polarization on an alternative electric field frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.6 A local electric field in dielectrics. Lorentz field . . . . . . . . . . . . . . . . . . . . . . 4.2.7 Clausius–Mossoti formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.8 An experimental determination of the polarization and molecular electric dipole moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems/ Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
251 251 251 252 259 273 276 280 280 282 284 286
5 Magnetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 General Characteristics of the Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1 A permanent (direct) electric current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.2 A magnetic field induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
293 294 296 298 301 303
305 305 305 309
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5.1.3 The law of a total current (ampere law) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.4 Action of the magnetic field on the current, on the moving charge . . . . . . . . . 5.1.5 A magnetic dipole moment in a magnetic field. . . . . . . . . . . . . . . . . . . . . . . . 5.1.6 Electromagnetic induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Magnetic Properties of Chemical Substances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Atomic magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.2 Macroscopic properties of magnetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.3 An internal magnetic field in magnetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.4 Microscopic mechanism of magnetization . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Magnetically Ordered State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.1 Ferromagnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.2 Domains: magnetization of ferromagnetics. . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.3 Antiferro- and ferrimagnetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Displacement Current: Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems/ Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
318 320 327 328 331 332 333 334 336 344 344 347 349 350 358 360
6 Wave Optics and Quantum–Optical Phenomena . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Physics of Electromagnetic Optical Waves. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 An Interference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.1 Superposition of two colinear light waves of the same frequencies . . . . . . . . . 6.2.2 Interference in thin films . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Diffraction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.1 Huygens–Fresnel principle: Fresnel zones . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.2 Diffraction on one rectangular slit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.3 Diffraction grating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.4 Diffraction grating as a spectral instrument . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.5 X-ray diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.1 Polarized light: definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.2 Malus law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.3 Polarization at reflection: Brewster’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.4 Rotation of the polarization plane. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.5 Birefringence: a Nichol prism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Dispersion of Light . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 The Quantum-Optical Phenomena . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6.1 Experimental laws of an ideal black body radiation . . . . . . . . . . . . . . . . . . . . 6.6.2 Theory of radiation of an ideal black body from the point of view of wave theory: Rayleigh–Jeans formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6.3 Planck’s formula: a hypothesis of quanta—intensity of light from wave and quantum points of view . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6.4 Another quantum-optical phenomena . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7 The Bohr Model of a Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems/ Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
361 361 369 369 370 377 378 379 381 383 385 386 386 387 388 389 391 395 398 398 402 404 407 416 421 422
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7 Elements of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Particle-Wave Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.1 De Broglie hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.2 Electron and neutron diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Heisenberg’s Uncertainty Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Wavefunction and the Schrödinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.1 A wavefunction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.2 The Schrödinger equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.3 Standard requirements that the wavefunction should obey . . . . . . . . . . . . . . . 7.4 Most General Problems of a Single-Particle Quantum Mechanics . . . . . . . . . . . . . . . 7.4.1 A free particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.2 A particle in a potential box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.3 A potential step . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.4 A potential barrier: a tunnel effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.5 Tunnel effect in chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 The Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.1 The Shrödinger equation for the hydrogen atom . . . . . . . . . . . . . . . . . . . . . . . 7.5.2 The eigenvalues of the electron angular moment projection Lz . . . . . . . . . . . . 7.5.3 Angular momentum and magnetic moment of a one-electron atom. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.4 A Schrödinger equation for the radial part of the wave function; electron energy quantization. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.5 Spin of an electron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.6 Atomic orbits: hydrogen atom quantum numbers . . . . . . . . . . . . . . . . . . . . . . 7.5.7 Atomic orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.8 A spin–orbit interaction (fine interaction) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 A Many-Electron Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.1 Types of electron’s coupling in many-electron atoms . . . . . . . . . . . . . . . . . . . 7.6.2 Magnetic moments and a vector model of a many-electron atom. The Lande factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.3 The atomic terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.4 Characteristic X-rays: Moseley’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7 An Atom in the Magnetic Field: The Zeeman Effect . . . . . . . . . . . . . . . . . . . . . . . . . 7.8 A Quantum Oscillator and a Quantum Rotator. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8.2 Quantum oscillators: harmonic and anharmonic . . . . . . . . . . . . . . . . . . . . . . . 7.8.3 A rigid quantum rotator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8.4 Principles of molecular spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems/ Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
423 423 423 424 428 432 432 433 434 435 435 436 441 442 445 447 448 451
8 Physical Principles of Resonance Methods in Chemistry. . . . . . . . . . . . . . . . . . . . . . . . 8.1 Selected Atomic Nuclei Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.1 A nucleon model of nuclei . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.2 Nuclear energy levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
497 497 497 499
452 457 460 461 462 467 468 469 470 473 473 477 480 480 481 486 491 494 495
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8.1.3 Nuclear charge and mass distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.4 Nuclear quadrupole electrical moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Intraatomic Electron–Nuclear Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.1 General considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.2 Coulomb Interaction of an electron shell with dimensionless nucleus . . . . . . . 8.2.3 Coulomb Interaction of an electron shell with a nucleus of finite size: the chemical shift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.4 The nuclear quadrupole moment and the electric field gradient interaction . . . 8.2.5 Interaction of a nuclear magnetic moment with an electron shell . . . . . . . . . . 8.2.6 Atomic level energy and the scale of electromagnetic waves . . . . . . . . . . . . . 8.3 -Resonance (Mössbauer) Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.1 Principles of resonance absorption . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.2 Resonance absorption of -rays: Mössbauer effect . . . . . . . . . . . . . . . . . . . . . 8.3.3 -Resonance (Mössbauer) spectroscopy in chemistry . . . . . . . . . . . . . . . . . . . 8.3.4 Superfine interactions of a magnetic nature . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Nuclear Magnetic Resonance (NMR) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.2 Use of nuclear magnetic resonance in chemistry . . . . . . . . . . . . . . . . . . . . . . 8.5 Abilities of Nuclear Quadrupole Resonance. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6 Electron Paramagnetic Resonance (EPR) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems/ Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
500 501 502 502 504 504 506 507 507 508 508 510 513 515 516 516 517 525 526 528 529
9 Solid State Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Crystal Structure, Crystal Lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Electrons in Crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.1 Energy band formation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.2 Elements of quantum statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.3 Band theory of solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Lattice Dynamics and Heat Capacity of Crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.1 The Born–Karman model and dispersion curves. . . . . . . . . . . . . . . . . . . . . . . 9.3.2 The heat capacity of crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Crystal Defects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4.1 Point defects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4.2 Dislocations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 Transport Phenomena in Liquids and Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6 Some Technically Important Electric Properties of Substances. . . . . . . . . . . . . . . . . . Problems/ Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
531 531 537 537 540 544 545 545 550 561 561 563 567 571 577 578
Appendix 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 581 Appendix 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589 Appendix 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591
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Appendix 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 595 Appendix 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 599 Glossary of Symbols and Abbreviations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 605 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 609
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All miracles of nature, no matter how extraordinary they are, have always found their explanation in the laws of physics Jules Verne Journey to the Center of the Earth
Preface In this new century, the development of science, technology, industry, etc., will require new materials and devices, which will in many respects differ from those of the past. Even now, there are many such examples. Certainly, the main foundation of these achievements is science, primarily physics, which enables the solid building of chemical, biological and atomic technologies, etc. In general, this book is a textbook on physics, but takes the above circumstances into account. It is aimed at students and scientists in the field of technology (chemical, biological and other branches of sciences), who will be working in the times ahead. The book differs substantially from standard physics textbooks in its choice of subjects, the manner of its presentation, selection of examples and illustrations as well as problems to be solved by the reader. The book contains problems important for chemists such as the language of potential curves and the essence of the theory of molecular collisions, and a large part is devoted to molecular physics with classical Boltzmann and Maxwell statistics, transport phenomena, etc. In a special part, the dielectric and magnetic properties of molecules are considered from the point of view of their structure. Optics is also covered in order to give the reader some idea of how its laws can be used for molecular structure analysis. Quantum mechanics is presented in an adapted form, aimed at a description of atomic and molecular spectroscopy. A special chapter describes tunneling both as a general phenomenon and as a mechanism of chemical reactions. Special attention is paid, also in an adapted form, to inter-atomic fine and superfine interactions, which are the basis of many modern and productive physical methods in the field of atomic and molecular structural investigations. Solid-state theory is presented on the basis of quantum statistics in order to form a bridge to their properties. A new technological field—nano-scale technology—is touched on here. In our opinion, no other textbook covers the sophisticated modern subjects mentioned above in such an acceptable form. Physics always operates with certain models—simplified representations of real systems. The ideal gas model is one such example. Despite its variety of real gas properties, this simple model assists in understanding the behavior of more complex systems using more complex factors permitted within the model, and it provides numerical results. For example, the introduction of additional interactions leads to van der Waals’s gas and allows further inclusion of virial factors, which in turn make the model more universally applicable to all gases. When using the model, the level of required accuracy has to be defined and on that basis, an appropriate model can be selected. The authors have aimed to make the subject matter easy to master; therefore many theoretical approaches in the text have been presented in an adapted manner, while the more strict proofs are given separately throughout each chapter in the Examples, in the Problems/Tasks at the end of each chapter, and particularly in the Appendices. A set of important constants is given in Appendix A1 to facilitate the solution of the problems. Units of measurement of physical values are also listed there. xiii
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The text is further enhanced by the illustrative material, including selected drawings, graphs and tables, which are an inseparable part of the book and greatly simplify understanding of the text. The book will be particularly useful for students, not only in the narrow area of their future profession but also in allowing them a broader glimpse of the surrounding world. In our opinion, this is necessary to encourage in the young people of the twenty-first century a firm perception of the world as an objective reality. The book pays a good deal of attention to the laws they need to learn in order to acquire new knowledge and to use if to expand human possibilities both in industrial and spiritual spheres of activity. The book’s nine chapters provide a description of the main laws of mechanics, statistical physics, thermodynamics, physics of dielectrics and magnetics, wave optics, quantum mechanics and physics of electronic shell of atoms, solid-state physics, physics of electromagnetic waves and physical methods of investigation. It contains a large amount of comprehensive information, useful for everybody in all stages of tutorial, practical and scientific activity. For successfully understanding the book, the reader should have a knowledge of the mathematical laws of, and some experience regarding, operation with vectors, differentiation and integration of elementary functions and others. Mathematics is the language of physics: the faults in mathematics must be considered as the faults in physics. The book is the fruit of our long experience at the Mendeleev University of Chemical Technology (MUChT) in Moscow. The results we have achieved have had a great influence on the content of the book and the problems chosen. One of us (RO) participated in publishing a textbook on physics under the auspices of the “State Program of Education and Science Integration” (“Physics in Chemical Technology” in collaboration with Professor E.F. Makarov from the Institute of Chemical Physics of Russian Academy of Sciences). Although it appeared in a very limited edition, which meant that the book wasn’t available for purchase, it has nevertheless greatly influenced the publication of this new work. One of the authors (RO) has spent a relatively long time at The University of Western Australia (UWA) in Perth, and this too has had a significant influence on the content and style of the book. R.P. Ozerov A.A. Vorobjev
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Acknowledgments It happens that both co-authors of this book, learning together in the same institute (Moscow Institute of Physical Engineering) and working thereafter for a very long period of time in the same department of the Moscow University of Chemical Technology have now been separated by long distance, corresponding only by telephone and e-mail. Therefore this part of the acknowledgment is expressed mostly by one of us, namely RPO. Firstly I would like to give my thanks the late Professor Edvard N. (Ted) Maslen, my colleague on the commission on Electron Density of the International Union of Crystallography; he strongly supported an acceptance of my previous student Dr. Victor A. Streltsov to a postdoctoral position at the Crystallography Centre of the University of Western Australia, which has resulted in an profitable collaboration with the D.I. Mendeleev Institute of Chemical Technology in Moscow. Ted and Professor Sid Hall made provision for a very favorable and productive investigation into X-ray crystallography. I greatly appreciate the initiative that Professor Sid Hall showed when he undertook on the position to assist me, especially considering the unusual circumstances. Dr. Alexandre N. Sobolev jointed the project a little latter; though the skill he has acquired in UWA made him a high-class specialist in X-ray crystallography. His skill and determination are extremely valued and his endless commitment enabled me to complete this book. I have never met a more kind and generous specialist like Professor Alan White. He was always forthcoming with valuable advice and support. I would also like to acknowledge Professor Brian N. Figgis who helped me immensely with my previous book “Electron density and chemical bonding in crystals.” Dr. Lindsay Byrne—a rare find as an NMR guru—is a devoted man with much enthusiasm and was always willing to help. I appreciate a lot Professor Ian McArthur for his attention and professional help. AV and me would like to express our gratitude to Professor Vitalii I. Khromov for his help and valuable advice. I would dearly like to thank the head of the School of Biomedical, Biomolecular and Chemical Sciences, The University of Western Australia, Professor Geoff Stewart and Mrs Leigh Swan for the pleasant and prosperous working environment they provided. Without this comfort and assistance I would not have been able to finish this project. I express my gratitude to my friend Mrs Suzanne Collins—the extra class specialist in English teaching—for her valuable help and advice. I am greatly obliged to all member of my family who helped me at all stages of my work.
Professor Ruslan P. Ozerov Professor Anatoli Vorobyev
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Recommendations to the Solution of the Physical Problems Readers are recommended to begin solving the problems by writing down the given data in the normal way. On the left-hand side, all the data should be written in a column as accurately as possible. At this point, it is useful to translate all the given date to the SI system of units in order to avoid confusion at the end. In the majority of cases, it is also very useful to make a competent analysis of the task, choose a reference system, and make the drawing indicating all the details correctly; with the proper indication of all details; it can be said that a reliable drawing is 50% of the solution. The solution should be carried out, as a rule, in a general form, i.e., without intermediate numerical calculations down to the final answer; this means that the symbol for the physical value sought should remain on the left-hand side of the answer, while the symbols for the physical values of the given conditions and the necessary physical constants should be on the right-hand side. The values of physical constants are listed in Appendix 1. It is also useful to accompany the decision with a brief explanation, both physical and mathematical, to state the physical ideas behind the solution. Moreover, the mathematical treatment should also be explained: if the definite integral is treated, the limits should be explained. In square root calculations it is desirable to explain whether both roots should proceed or one root should be rejected and why? The final answer in a general form should be marked by any way. We especially want to emphasize that a general solution is of greater significance and value: it means that not just the particular problem has been solved, but a real task. The results can then always be used to solve similar problems without starting the treatment again from the very beginning. The analysis of the dimension of the result is one of the important stages of the solution; it permits one to be confident of the result. It is necessary to carry out calculations keeping the desired accuracy in mind; more often three significant figures will be sufficient. The numerical answer should be written down as a number increased to the proper power of 10 or using multiple prefixes, e.g., A = 2.56⫻107 J and/or 25.6 MJ. It is important that one should be certain that a reasonable result has been achieved, i.e., the speed of a body does not exceed the speed of light, or its size does not exceed the size of the universe, etc. We wish our readers all success!
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–1– Mechanics
1.1
INTRODUCTION
An enormous number of physical events and phenomena are taking place around us all the time: the movement of all types of transport (bicycles, cars, trains, airplanes, etc.), building activity, athletes in competition, rain falling, wind blowing, water flowing, earthquakes and a wide range of other phenomena. All of these are performed at speeds much smaller than the velocity of light (c ⬇ 3 108 m/sec) and at scales much greater than atomic scales (⬃1010 m). All are described by classical mechanics, based mainly on Newton’s laws. This does not, of course, exclude the existence of other phenomena described by other physical branches. Quantum mechanics deals with the world of atoms and molecules, their transformations and accompanying changes in property. The overwhelming majority of them are invisible to the naked eye, but experience shows the following to be true: all materials, though differing in their characteristics, consist of a limited number of various particles— atoms and molecules. This is the world of so-called quantum mechanics. We can indirectly observe these phenomena manifest themselves, but for their investigation and understanding, a special knowledge is needed. To continue this analysis, we can mention one more branch of phenomena that manifest themselves at velocities close to the velocity of light; this is the more exotic area of classical and quantum relativistic physics.
1.2
KINEMATICS
Kinematics is the branch of mechanics that explores the motion of material bodies from the standpoint of their space–time relationships, disregarding their masses and the forces acting on them. 1.2.1
Kinematics of a material point
For a description of a point’s motion in space and time, a reference system should be chosen. The reference system is a collection of instruments: the time-measuring device (e.g., a watch) and the bodies conditionally considered as being fixed in space with respect 1
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to which the motion is considered. Time, a continuously changing scalar value, is measured by a watch, and cannot be negative. In problems of kinematics time is usually taken as an independent variable (or argument), the rest of the parameters being considered as functions of time. For different problems the reference system can be chosen either in the form of Cartesian coordinates, or as a cylindrical or spherical coordinates system. A moving point describes a certain continuous line in space that is referred to as a trajectory. In a number of problems the path itself will define the motion (for instance, its rails will dictate the motion of a railway carriage). At a certain instant, corresponding to a certain body motion, tangent unit vectors—principle normal and binormal vectors—are taken as natural axes. In the following we will consider only plane motion, so there is no need for a binormal vector. The principle normal is perpendicular to the tangent and is directed to the center of curvature. The direction of the tangent and normal unit vectors will be denoted as and n. Let us recall some information about the line curvature (trajectory). The tangent lines assigned by vectors 1 and 2 at two adjoining points A and B of the plane form an angle (Figure 1.1) to be drawn, which is referred to as the angle of contingence. If we then make the distance AB shorter, an arc AB l aspires to zero. At the limit /艎, it gives the trajectory curvature K in a given point: lim
K ᐉ ᐉ
at
AB 0.
The reciprocal value 1/K is the curvature radius in point A. In fact, a circle’s curvature is equal to its radius; the curvature radius of a straight line is infinity. The simplest object in mechanics is called a material point (MP); this implies a body whose size in the framework of a given problem can be considered to be negligibly small. Another definition of an MP is that it is a point that possesses a mass. Different objects in different problems can be considered differently: the molecules acting on a vessel’s wall can be imagined as an MP, the earth moving around the sun may, in some instances, also be treated as an MP. However, the same objects in different problems cannot be considered
1 B ∆
A
2
n
Figure 1.1. The trajectory curvature.
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as MPs: e.g., molecules in molecular spectroscopy rotating around their center of mass (CM), the earth rotating around its geographical axis, etc. An important task in kinematics is to assign an equation of motion, i.e., to construct the necessary mathematical equations that are sufficient to determine the MP’s position in space at any instant of time. In the Cartesian coordinate system such an equation is the time dependence of the radius vector r(t); three scalar equations x(t), y(t) and z(t) correspond to one vector equation. If a point in a time interval t moves from point A to B along an arc l (Figure 1.2), the vector r r2 r1 is referred to as the displacement, whereas the length of the arc AcB is the distance travelled. If one takes one’s car in the morning, travels some distance during the day and then returns the car to the garage, the overall day displacement is equal to zero, whereas the distance travelled is the non-zero speedometer indication. The distance travelled and the displacement can coincide in two cases: when the movement occurs along a straight line or at t → 0. The equation 冓冔
r . t
(1.2.1)
allows us to calculate the average speed at a time interval t. The instant velocity is given by the equation lim
t0
r dr r(t ). t dt
(1.2.2)
V
c A
B
∆r
r1
r2
O
Figure 1.2. A displacement vector ∆r and distance travelled AcB.
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The velocity at a given point is a physical value, numerically equal to the time derivative from the radius vector of the MP in the reference system under consideration. Remember that for brevity of writing, the time derivative function is denoted by a point above the letter, expressing a given function. Where the direction of the vector is concerned, in the limit of the movement of point B to point A the secant will coincide with the tangent to the trajectory in point A. Consequently, an instant velocity vector is directed along the tangent to the trajectory, and the modulus is the time derivative from the function, expressing the law of point movement. As usual, the point radius vector r(t) can be decomposed upon the orts r(t ) x(t )i y(t ) j z(t )k,
(1.2.3)
(t ) r(t ) ix (t ) jy (t ) kz(t ).
(1.2.4)
ix jy kz ,
(1.2.5)
and therefore,
The velocity vector is
where x, y and z are its projections onto the coordinate axes: x x (t ), y y (t ), z z(t )
(1.2.6)
The modulus of the velocity vector is the square root sum of their projections’ squares: x2 y2 z2
(1.2.7)
Acceleration is the change of the velocity vector in time. If, in the time interval t, an MP displaced along the trajectory and a change in velocity and its direction had taken place then 2 1. The mean acceleration in the t interval is then 冓 a冔
. t
(1.2.8)
The acceleration at a given time instant (instantaneous acceleration) is the limit of the ratio (/t) at t → 0.
a lim
d (t ). t dt
(1.2.9)
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or a ix jy kz ,
(1.2.10)
(because orts are in this case independent of time). In another form a ia x ia y kaz ,
(1.2.11)
where ax, ay and az are projections of the a vector onto the coordinate axes. Comparison of eqs. (1.2.6) and (1.2.11) gives a x x (t ) x(t ); a y y (t ) y(t ); az z (t ) z (t )
(1.2.12)
and correspondingly a a x2 a y2 az2 .
(I.2.13)
At curvilinear movement the velocity vector is the product , where is a tangent ort. Because of the fact that the point is moving along a curvilinear trajectory and “draws” the unit vector behind, its position is also dependent on time. In this case:
a
d d d . dt dt dt
(1.2.14)
Expression (1.2.15) shows that acceleration is the sum of the two vectors: the first is directed along the tangent and is equal to the first derivative of velocity and the second term depends on the change of in time. To determine the magnitude and direction of the second term, we need to find the meaning of the derivative d/dt. Let the direction of the velocity vector at two adjacent positions separated by time interval t be specified by orts 1 and 2 (Figure 1.3). Then the change of the vector in the time interval t can be expressed by vector 2 1. We shall consider the derivative d/dt as a limit of a ratio / for t ; 0. We find the value of vector magnitude from the triangle (/2) ACD: sin , then sin , at t → 0 numerically t → , since 2 2 2 the unit-vector magnitude is unity and sin(/2) ⬇ /2 at ^ 1. Then d lim lim t 0 t 0 dt t t
(1.2.15)
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/t by the arc Multiplying both the numerator and denominator of the function length l we obtain d ᐉ ᐉ lim lim lim dt t0 t ᐉ t0 ᐉ t0 t Let us consider both these limits. Since an angle is the angle of contiguity, the lim(/l) K is equal to the curvature of a curve at a given point, i.e., to curvature radius . The second limit is the velocity magnitude lim
to
ᐉ d ᐉ . t dt
Thus, d K . dt
(1.2.16)
To determine the direction of the vector (d/dt) we shall draw a straight line from point A parallel to and examine the value of an angle ⬔CAE at the limit t → 0. As can be seen from Figure 1.3, an angle ⬔CAE ⬔CAF /2 (/2) /2). At t → 0 the → 0 whereas ⬔CAE → ( /2). Therefore, the vector (d/dt) contiguity angle lim(/t) will be directed along the normal to the center of curvature at point A; it can be presented as d n. dt
(1.2.17)
C 1 ∆ A
2
∆ D
F Ε
n ∆/2
derivative. Figure 1.3. Calculation of the d dt
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Returning to expression (1.2.14), we can write a
d 2 n dt
(1.2.18)
Therefore, the total acceleration a in curvilinear movement can be separated into two parts: the first is the tangent acceleration a
d dt
(1.2.19)
an
2 .
(1.2.20)
and the second is
The tangent part influences the absolute velocity magnitude whereas the normal part changes the direction of the velocity vector. The square of the total acceleration can be written as 2
⎛ 2 ⎞ ⎛ d ⎞ ⎜ ⎟ ⎜ ⎟ . ⎝ dt ⎠ ⎝ ⎠ 2
a
2
a2 an2
(1.2.21)
The expressions derived are valid for the general movement of the MP along the curve with an arbitrary regime of velocity change. Let us consider some particular cases: Rectilinear movement: , an 0, a a Uniform movement along a circle: const., a 0, a n(2/R) Nonuniform movement along a circle: const., a 苷 0, an 苷 0. In the case of uniformly alternating movement a const. 苷 0, an 0; it is possible to derive the general expressions. Integration of an expression (d/dt) a over time gives (t ) ∫ a(t )dt C at 0 ,
(1.2.22)
where 0 is the initial speed (at t 0). The distance travelled can be derived from (1.2.22) by repeated integration as
x(t ) x0 0 t where x0 is the initial coordinate (at t 0).
at 2 , 2
(1.2.23)
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EXAMPLE E1.1 A car moving uniformly covers a distance of 100 km at a speed 60 km/h, although on the way back it travels at a speed of 40 km/h. Determine the average speed of the car. Solution: At first glance the answer is simple: 50 km/h. However, this is incorrect. The average speed is the total distance travelled (200 km) divided by the total time spent (100/60) (100/40) 4.16 h. Therefore, the average speed is (200/4.16) ⬇ 44 km/h.
EXAMPLE E1.2 The movement of a MP along an x-axis is described by the equation x A Bt CT 3, where A 4 m, B 2 m/sec, C –0.5 m/sec3. For the instance of time t1 2 sec determine: (1) the MP coordinate x1, (2) an instant velocity 1, and (3) an instant acceleration a1. Solution: (1) To find the point coordinate one should substitute time t for the instant time t1x1 A Bt1 Ct13. Inserting the given values we obtain: x1 4 2.2 0.5 23 4 m. (2) To find an instant speed at any time we should differentiate a coordinate on time: (dx/dt) B 3Ct12. Introducing B, C and t1 we obtain: 1 4 m/sec. The sign shows that at that very moment the point moves in a negative direction on the x-axis. (3) To find acceleration as a function of time we should take the second time derivative from coordinate: a (d2x/dt2) (dx/dt) 6Ct. To find the instant acceleration at t1 we should introduce the given data and obtain the result. a1 – 6.0 5.2 –6 m/sec2; the sign shows that the movement is decelerative.
EXAMPLE E1.3 The movement of an MP along an x-axis is described by equation x A Bt Ct 2, where A 5 m, B 4m/sec, C –1.0 m/sec3. Draw a graph of x(t) and the distance travelled S(t). Solution: For the drawing of the graph of the point coordinate time dependence x(t), we find characteristic values of movement: initial and maximum coordinates The initial coordinate corresponds to the moment t 0, its value equals x(0) A 5 m. The point reaches maximum height corresponding to the moment when the point starts to move back (speed changes sign). We can find this moment having equated to zero the time first derivative from coordinate: (dx/dt) B 2Ct 0, wherefrom t –(B/2C) 2 sec. The maximum coordinate xmax x(2) 9 m. The time instant t when x 0 can be found from equation x ABtCt2 0. Solving the quadratic equation we obtain t (2 3) sec. The negative value does not satisfy the problem. Therefore t 5 sec. Using the data obtained we can draw the
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graph of coordinates’ dependence on time. The distance travelled and the coordinate coincide until the point stops; from this time the point goes in opposite direction and its coordinate diminishes; however the distance travelled continues to grow (Figure E1.3).
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2
4
6
t
–5
EXAMPLE E1.4 A mortar is installed on a hill at a height of H 60.0 m above ground level. It fires a missile at an initial angle of 60° to the horizon. The missile’s initial velocity is 0 80 m/sec. Derive: (1) the kinematical equations of the missile’s flight x(t) and y(t); (2) the equation of trajectory y(x); (3) the expression for projections x(t) and y(t) on the coordinate axes x and y and the time velocity dependence (t); (4) the velocity dependence on time (t) on absolute value and direction; (5) the absolute values of tangential 冟a冟 and normal 冟an冟 acceleration dependence on time (derive a corresponding formula and execute calculations): (6) the maximum height ymax of flight; (7) the time of the missile’s flight ; (8) the range L of missile; (9) the missile’s velocity (on modulus and direction) at the moment of falling on the ground; (10) the curvature radii trajectory 1 and 2 at the moment of falling and at the highest point of flight, respectively. Solution: To solve this problem we have to begin with the choice of reference frame. The motion of the missile is subject to a constant acceleration g directed downward. Therefore, the flight trajectory is a plane (two-dimensional). Choose a Cartesian system xOy in such a way that the x-axis is horizontal and the y-axis is vertical; the flight will occur in this plane. The origin is superposed with the earth
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surface, axis x directed horizontally and y vertically upward. Accept the missile as an MP. The movement in this case can be separated into two independent components: along axes Ox and Oy. A movement along axis Ox is uniform with a speed x 0 cos ; however along axis Oy it uniformly accelerates with initial coordinate y0 H and initial speed oy 0 sin and acceleration ay g (see inset in Figure E1.4).
y
x(τ)=L an H
a g
x()=L x
O an
a g
Thus, (1) Kinematical equations for mine movement projected on axes x and y can be written:
x(t ) 0 t cos and
y(t ) H 0 (sin )t
gt 2 . 2
(2) An equation of the trajectory can be obtained by excluding time from the kinematical equation for the missile’s movement: sin x , then y( x ) H 0 x 0 cos 0 cos g g 2 x 2 H (tan ) x 2 x2 . 2 20 cos 20 cos2
since t
(3) The velocity (t) projection on coordinate axis can be found by time differentiation of x(t) and y(t) as x (t )
dx dy Ox 0 cos const., y (t ) 0 sin gt. dt dt
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(4) The dependence of velocity (t) on time, in vector form, can be obtained in the form (t) ix(t) jy(t). Then the velocity modulus is 冷 (t ) 冨 x2 y2 02 2 gt0 sin g 2 t 2 . The velocity (t) direction can be determined by an angle between this vector and the axis Ox. It can be seen from a Figure E1.4 that tan (t )
y (t ) x (t )
⎛ sin gt ⎞ 0 sin gt , therefore (t ) arctan ⎜ 0 . 0 cos ⎝ 0 cos ⎟⎠
(5) Since the total acceleration is constant (i.e., g) the moduli of tangential 冟a冟 and normal 冟an冟 components (as can be seen from Figure E1.3) will be equal to: a g sin and an g cos , where sin (y /x) and cos (x /y ) or a (g(oy gt) / (2o 2gtoy g2t2)1/2 ) and an n(goy /(2o 2gtoy g2t2)). (6) The highest point of the flight ymax can be found from the kinematical equations y(t) and y(t). From the first dependence one can find the time of ascent tasc, from the second—the maximal ascent ymax y(tmax). In the upper trajectory point y 0, therefore Oy gtasc 0, then tasc (Oy /g). Therefore: ymax H
2 2 2 2 Oy Oy gOy gtasc H 0 sin 2 H 245m 2g 2 g 2g
(7) The missile’s total flight time can be found from the fact that in the moment of its drop y(t) y() 0, i.e., H 0 sin α (g2/2) 0 and 2 (20 sin /g) (2/g) H 0. Solving the quadratic equation regarding τ we can arrive at 1,2 12
20 sin 2 sin 2 2 gH 0 2 2 , g g g
1g 冢0 sin 02 sin 2 2 gH 冣
(Since the time cannot be negative we should accept sign “”). Executing calculations we obtain 14.9 sec. (8) The missile’s range L can be found by inserting τ into the x(t): L x() Ox 596m. (9) The modulus of velocity at the moment it hits the ground can be found using the equation given in point (4) substituting a running time on as 2 () 兹 苶苶 苶苶0苶g苶si 苶n苶苶 苶 苶g2苶苶2 87 m/sec 0 2
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or, converting this expression to a form 兹苶 02苶 苶苶2g苶H 苶, one can obtain the same result. Since tan (y /x), then () arctan 62.5°; a minus sign shows that the velocity vector makes with the x-axis an angle , counts off in a negative direction, i.e., clockwise (Figure E 1.4, insertion). (10) To find curvature radius one can use expressions n (2/) wherefrom ( 2/an), and an g cos , and is the speed at the moment of hitting the ground: (τ). Therefore,
1
02 20 g g 2 2 02 2 gH 1.67 m, g cos g cos
however, at the highest point 2 2Ox/g 163 m. Note that the curvature at the maximum height is approximately 100 times less than at point of hitting the ground. By solving a problem in this way, we can then use all these particular equations in future.
1.2.2
Kinematics of translational movement of a rigid body
A body in which the distance between two arbitrary points remains at constant temperature, unchanged by any motions or interactions, is referred to as an ideally rigid body (IRB). During the translational motion of the IRB, any segment inside of it remains parallel to itself at any time. With such motion the displacement, velocity and acceleration of any point of the IRB are the same at any time. Therefore, many characteristics of the IRB’s translational movement can be described by the motion of a single body’s point with a mass equal to the mass of the whole body moving with velocity (acceleration) in any point of the body. The best point to choose is the centre of mass (CM) (see below). 1.2.3
Kinematics of the rotational motion
Rotational movement is widespread in nature, no less (but can be even more) than translation motion. Indeed, the motion of electrons around the nucleus (within the Bohr atomic model) and the earth around the sun, the rotation of a gyroscope, the rotation of numerous details and assemblies in technology and industry, the rotation of a wheel (this genius invention of mankind)—all of these are examples of rotational motion. The rotational motion of the IRB around a motionless axis Oz in which all points of the body are moving in parallel planes, making circles with their centers lying on a single straight line coinciding with the z-axis, is referred to as the rotational motion of the IRB. When rotating, all points of the IRB have linear velocities differing in size and direction, depending on the point distance from the axis of rotation. So, for a description of rotational motion we should introduce angular kinematic features unique to the whole body: angular displacement, angular velocity and angular acceleration. Let us restrict ourselves to the case of IRB rotation around an axis whose space position does not change in time.
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Angular displacement Consider a body revolving around axis Oz. Select in the body a point, not lying on the axis of rotation (point A in Figure 1.4; the body itself is not shown in the figure). In accordance with the definition of rotational motion, this point while moving describes a circle with a radius R, the center of which (O) is lying on an axis Oz. While rotating, vertical planes drawn through the axis of rotation and any body point turn on the same angle. Let the plane (and the body) turn on an angle d. This angle is referred to as the angular displacement. The angular displacement is a vector, coinciding with the axis of rotation, whose direction is defined by the right-handed system. Remember that this rule concludes that if the right screw reconciles with the axis of rotation and turns it in a direction complying with the rotating body, the translational direction of the screw movement along the axis of rotation . complies with the direction of the vector d Vectors whose directions are aligned with the rotation direction are called axial vectors. is an axial vector, the modulus of which is equal to the ratio of Angular displacement d arc dS and radius R and the direction of which coincides with the rotation axis in accordance with the right screw rule. d
dS k, R
(1.2.24)
k being the ort of rotation axis Oz. The value to which the limt → 0 tends is called the angular velocity :
d (t ). dt
(1.2.25)
z
d
k 0 R
d A
dl
. Figure 1.4. Elementary angular displacement vector d
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∆ > 0
∆ < 0
(a)
(b)
Figure 1.5. Relationship between angular velocity and angular acceleration.
Angular velocity is a first time derivative from the vector of angular displacement. It shows the speed of angular displacement changing with time. Angular velocity is also an axial vector, which coincides in direction with the angular displacement vector d. The value to which the limit (/t) tends is called angular acceleration: d k k, t0 t dt
lim
(1.2.26)
Angular acceleration is also an axial vector. The different mutual orientations of angular velocity and angular acceleration are presented in Figure 1.5: when the angular velocity is rising (d 0), then the direction of the angular acceleration vector coincides with the former (both are directed along the axis of rotation); if the angular velocity decreases (d 0), then the direction of the angular acceleration vector is opposite to the angular velocity. If the axis of a body rotation changes its orientation in the course of time, some interesting effects appear which are unfortunately beyond the scope of our consideration here. There exists a linear relationship between angular and translation features. It can be seen in Figure 1.4 that dl Rd, i.e. d (dl/). Time derivation (d/dt) (1/R)(dl/dt) (1/R), i.e., R.
(1.2.27)
The relationship between angular acceleration and linear tangent acceleration a can also be obtained. The modulus of angular acceleration is (d/dt), where (/R); then (d/Rdt). Since (d/dt) is a point linear (tangential) acceleration a then (a/R) or a R.
(1.2.28)
The last formula connect the linear and angular characteristics (Figure 1.6). They can be given in vector form: [r ],
(1.2.29)
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z
ω
k
O r
υ
Figure 1.6. Relationship between vectors of angular and linear velocity.
and [a r ].
(1.2.30)
Using eqs. (1.2.27) and (1.2.28) we can obtain by integration the dependencies of (t) and (t) like (1.2.22) and (1.2.23) valid for uniformly accelerated rotation 0 t ,
(1.2.31)
and
0 0 t
t2 . 2
(1.2.32)
where 0 and 0 are angular characteristics at the initial instant of time. The structure of these expressions is equivalent to those obtained for linear motion (eqs. (1.2.22) and (1.2.23)). However, rotational motion is distinct from linear because of the fact that it is periodical. In rotation, the values of are repeated in a certain time interval. If these intervals are constant (uniform rotation, const.), the period T, a duration of one full turn (on 360°), and accordingly rotating frequencies, i.e., numbers of full repetition in the unit time (v n 1/T ), can be used. Bearing in mind that one turn corresponds to an angular displacement equal to 2 radian, we can introduce an angular velocity 2v 2/T. The difference between angular vector features of revolution and the corresponding features of linear motion lies in the fact that angular vectors are directed not along the linear motion of each point of the IRB, but along the axis of rotation (perpendicular to their planes of motion). Many remarkable characteristics of rotational motion are bound up with this circumstance (refer to Section 1.3.9, Figure 1.17 and Appendix 2).
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In general, the arbitrary motion of the IRB can be presented as a combination of translation motion of an MP with a mass equal to the mass of the whole IRB and located in the center of inertia (refer to Section 1.3.7, see below), and rotation of the body’s points around the center of inertia. EXAMPLE E1.5 A disc of a radius R 10 cm starts to rotate with angular acceleration 0.1 rad/sec2 around a motionless axis, perpendicular to the disc’s plane passing its geometrical center. Determine at the time instant 12 sec after the beginning of the disc rotation: (1) an angle of the disk turns (an angular displacement); (2) the number of complete revolution Ntot; (3) the net turn angle ; (4) the distance traveled by any point A of the disk crown S along an arc; (5) the angular speed value and (6) a frequency of rotation n at this moment. Solution: (1) For a uniformly accelerated rotation a kinematical equation is (1.2.32), where (t) is a turning angle for time instance t, 0 and 0 are the initial angle and angular speed; in our case 0 0 and 0 0. Therefore (t) ( t2/2). Introducing the numerical values and execute calculations for the time instant we obtain () (0.1122)/2 rad 7.20 rad (413°). (2) We can find the number of revolutions N by dividing the previous result by 2, i.e. N (/2) 7.20/(2 3.14) 1.15 revolutions. Since the number of revolutions is an integer then Ntot 1. (3) The net turn angle can be found as a difference between the final turn angle minus the 2 integer value: (720 2 1) rad 0.917 rad (52.6°). (4) The total distance S traveled by point A along an arc can be found multiplying the turning angle by the radius R:S R 7.20 0.1 m 0.72 m. (5) To determine the disk angular speed at the time instance one first should take the time derivative of an angular displacement
d d ⎛ t 2 ⎞ ⎜ ⎟ t . dt dt ⎝ 2 ⎠
For t we obtain () . Execute the calculations 0.1 12 1.20 rad/sec. (6) The instant frequency of rotation n() can be obtained as n (/2) (1.20/2) 0.19 sec1. 1.3 DYNAMICS Dynamics deals with the study of a body’s motion with definite mass under the action of applied forces. 1.3.1
Newton’s first law of motion: inertial reference systems
Generally speaking, the same physical events can be described differently in different reference systems. Undoubtedly, we would wish to find a reference system in which the laws
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of different physical phenomena have the simplest expression. On the other hand, it would be interesting to find in the surrounding world a system that would be at absolute rest so that any motion could be considered with respect to this system. Is it possible to find such a system? To answer this question we shall analyze the simplest form of motion—the motion of a free body. A body is called free if it is at such a distance from all other bodies that their effect on it is negligible. (Such a body and such a motion is actually a physical abstraction since it cannot be fully realized. Nevertheless this model has played a very important role in the development of physics, from Aristotle to Galileo and Newton). So, experiencing no external effect, the free body must move rectilinearly and uniformly. Such a motion cannot be achieved in any reference system but only in the so-called inertial one. A reference system is referred to as inertial if the free body moves in relation to it with constant velocity—in magnitude and direction. Besides, if one of the reference systems moves relative to another, additional effects can appear. All these questions are the subjects of different theories of relativity, realizing the relationships between physical laws in reference systems moving relative each other (refer to Section 1.6). The classical theory of relativity is based on Galileo’s and Newton’s hypotheses. Their main feature is separation and independence the space and time and their independence. The laws of the classical theory of relativity appear from mankind’s everyday experience of isotropic and uniform space (all directions are equivalent and space metric is constant everywhere) and independence of time intervals from the reference system (an interval in Moscow is the same as in London). These laws prove to be perfectly justified in the case of motion of a material body with velocities << c. Experiencing no external influences, a free body must, consequently, move rectilinearly and uniformly. However, this cannot be achieved in all reference systems, but only in those that are referred to as inertial systems. A reference system in which a free body of constant mass proceeds with constant velocity is called an inertial reference system. The existence of an inertial reference system is a sequence of definite characteristics of space and time: the uniformity and isotropy of space and uniformity of time. The uniformity of space and time means the equivalence of all positions of free bodies in space at all instants of time, and space isotropy means the equivalence of different directions. Therefore, it is possible to give another definition of an inertial reference system: as a system relative to which space is homogeneous and isotropic and time is uniform. The statement describing inertial reference systems as systems in which the motion of a free body is rectilinear and uniform, forms the essential part of inertia law: a free body preserves a state of the rest or the uniform rectilinear motion until another body makes it leave this state. The idea of an inertial system was incorporated by Newton into his system of the main laws of dynamics: it is referred as the first law of dynamics or Newton’s first law. Having found a single, accidental inertial system it would be imagined that the unique motionless system is found, relative to which any motion of bodies in the Universe should be considered. However, this is not so, because there exist countless inertial reference systems. We can illustrate this quite simply. Let there be two reference systems moving towards each other uniformly and rectilinearly; one of them is known to be inertial. Confirm that the other system will also be inertial. In fact, a body that is in a state of uniform and rectilinear motion with regard to the first reference system (which we know is
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inertial), will move uniformly and rectilinearly towards the second reference system (though, with different velocity); however this (second) reference system will then also be inertial. Thereby, any reference system, moving rectilinearly and uniformly relative to any system (which we know to be inertial), is also an inertial one. Hence, there are countless sets of equivalent inertial systems. Equivalence of all inertial systems is equivalent to the statement that laws of mechanics are invariant with respect to the Galileo's transforms. The term “invariant” signifies that laws of mechanics and their mathematical writings are similar in all inertial systems. 1.3.2
Galileo’s relativity principle: Galileo transformations
So, there are countless sets of equivalent inertial systems. Moreover, it has been proved to be physically impossible to distinguish one inertial system from another—all inertial systems are equivalent. This last statement is Galileo’s mechanical relativity principle: there are no mechanical experiments that can be carried out within a given closed inertial reference system that can distinguish whether this reference system moves rectilinearly and uniformly or is at rest. Being, for instance, in the windowless sheep’s hold uniformly and rectilinear moving without heaving (example have been taken from Galileo’s book) or in the cabin of plane with closed windows, an explorer arranging any possible mechanical experiments (balls collision, throwing any subjects in different directions, free fall of bodies, etc), can not find, whether moves this system uniformly and rectilinear or is motionless. (We also refer the reader to Jules Verne’s novel “From the Cannon to the Moon”, in which passengers discuss a missile flying to the moon.) The mechanical principle of relativity, coupled with the suggestion of uniformity of time flow in all inertial reference systems, is referred to as Galileo’s principle of relativity. Let us find a correlation that would allow us to transmit from one inertial system to another. Suppose that there are two inertial reference system K(x,y,z,t) and K (x ,y ,z ,t ), a second system (K ) being moved with regard to the first (K) with a constant velocity V0 so that axes x and x coincide (Figure 1.7). If, at the initial instant both coordinate systems coincide, at moment t, the coordinates x and x will be bonded by the correlation x x VOxt. For three-dimensional movement a similar correlation appears between all coordinates so the correlation system will look like x x V0x t , y y V0y t , z z V0z t.
(1.3.1)
In general form this system can be written as r r V0 t.
(1.3.2)
Expressions (1.3.1 and 1.3.2) together with the independency of the time flowing in both systems (t t ) are called Galileo transitions. They permit one to go from one inertia system to another. The equivalence of all inertial reference systems is similar to the statement that the laws of mechanics are invariant with respect to Galileo’s transforms. The phrase “are invariant”
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z
z'
K
K'
V0t
x' x
x' x
y
y'
Figure 1.7. Galileo’s transforms.
signifies that the laws of mechanics and their mathematical writings are similar in all inertial systems. However, each particular physical value can differ when turning from one system to another. Having differentiated both right- and left-hand sides of expression (1.3.2) in respect to time, we can find a known law of velocities summation: dr dr
V0 dt dt
(1.3.3)
V V0 V ,
(1.3.4)
or
where V and V are velocities of a MP in inertial systems K and K . It can be seen that velocity is noninvariant regarding Galileo’s transforms. The second derivative gives d 2 r d 2 r
2 , dt 2 dt
(1.3.5)
a a ,
(1.3.6)
or
i.e., acceleration is invariant regarding Galileo’s transforms. An important conclusion can be derived from this consideration: all the velocity’s dependent physical values (for instance, momentum, kinetic energy, etc.) are noninvariant; however, all the acceleration-dependent physical values (for instance, force, etc.) are invariant regarding Galileo’s transforms. Other examples of invariants are mass, distance, temperature, time, etc.
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Einstein has generalized Galileo’s principle of relativity. According to Einstein’s principle of relativity, it is impossible by either mechanical or by physical experiment (in particular, electrical, magnetic or optical) conducted in an inertial system, to distinguish if this system is at rest or in rectilinear uniform motion. This statement is the basis of the special relativistic theory (see Section 1.6). Now we have to solve a problem: of the many real reference systems we usually deal with, in practice, those that can be considered to be inertial. Many problems of mechanics are considered in a laboratory reference system, strictly bound to the surface of the earth. Is this reference system an inertial one? Strictly speaking, the answer is no, since the earth rotates daily, the points on the terrestrial surface (excluding the poles) possess different acceleration, perpendicular to the axis of the earth’s rotation. However, in comparison with free-falling acceleration, this acceleration is very small, and for practical problems connected with the earth in a laboratory system, it is possible to consider it to be inertial. 1.3.3
Newton’s second law of motion: Momentum
Any change in the movement of an MP or in its state is caused by the action of other bodies or force fields. The quantitative measure of mechanical action of bodies on another body is called a force. This action can be exhibited as a change in the velocity of a body, or as its deformation. It can be measured in both cases, so the force can be quantitatively evaluated experimentally. Newton’s second law gives a relationship between the change in motion velocity and the force F causing this change. Generalizing a large number of experimental facts, Newton suggested that acceleration is proportional to force. The law can be given as follows: the body’s velocity change is proportional to the applied force and results in the direction of the line along which the force is acting. With the action of the same force on different bodies, their accelerations can be different, depending upon the body’s mass. In this instance, mass acts as a proportionality factor between the force, acting on the body and its acceleration. Such mass is identified as the inert mass (unlike the gravitational, which will be discussed below). The mass is a measure of the body’s inertial property relative to its translational motion. Accordingly, the mathematical expression of Newton’s second law can be written in vector form: F ma m
d , dt
(1.3.7)
or in coordinate form: mx Fx , my Fz , mz Fz . Equations (1.3.8) are called the differential equations of an MP’s movement.
(1.3.8)
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If the mass is constant, eq. (1.3.7) will have the form d (m) d p F. dt dt
(1.3.9)
This equation is the most general mathematical expression for Newton’s second law valid in many cases (for constant and varying masses, relativistic and quantum mechanics). Vector p here equals the product of the body’s mass and its velocity and is referred to as the momentum: p m.
(1.3.10)
Vectors of momentum and velocity coincide in direction. In Cartesian reference frames the momentum vector can be expressed as: p m imx jmy kmz ipx jpy kpz ,
(1.3.11)
where px, py and pz are projections of p on the coordinate axis. Using expression (1.3.9), one can find the relation between the force and the momentum increment, which is produced by the force: the rate of change of the momentum of a particle is proportional to the net force acting on the particle and is in the direction of that force d p Fdt.
(1.3.12)
It can be seen that the elementary body’s momentum change is a product of the force acting on the body and the time of its action. The product Fdt is called the elementary force impulse.
dp
p
F
Figure 1.8. Second Newtonian law: relation between the direction of force and the direction of momentum increment. A material point trajectory is shown.
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Hence, Newton’s second law can be reformulated: the change of the body’s momentum is proportional to the force applied and coincides with the direction of the force action (Figure 1.8). Independence of force action principle When discussing the force action we implied a single force only. In many cases, however, a body exerts the action of several forces. In this case the principle of independence of force action is taking place: if there are several forces acting on a body the acceleration exerted by the body under the action of each force is independent of whether or not other forces exist. Let forces F1, F2, F3, … simultaneously act on a body. The i-th force imparts to a body an acceleration ai Fim. The simultaneous action of all forces will impart to the body an acceleration equal to the sum of all accelerations:
∑ Fi a i1 . N
i
m
(1.3.13)
This is the generalization of the second Newtonian law to the case of simultaneous action of N forces. The geometrical sum of forces is called the resultant force applied to a body. In the general case, the resultant force imparts acceleration to a body, its direction coinciding with the direction of the resultant force. If N forces are applied, the body’s motion can be written in the coordinate form as
m m m
d 2x dt 2 d 2y dt 2 d 2 z dt 2
N
∑ Fxi , i1 N
∑ Fyi , i1 N
(1.3.14)
∑ Fzi . i1
EXAMPLE E1.6 An acting force is a function of a body displacement (deformation) F(x) x (such force is characteristic of elastic forces, e.g., stretching a string). Find the law of a body’s motion. Solution: Let the initial conditions be: at t 0, 0 and x0 0. The differential equation of the motion has the form: mx x. Reduce the derivation order m(dx /dt) x. The number of variables is three; therefore one should try to go over from
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variable t to variable x. This can be done by multiplying and dividing the left-hand side of the equation by dx: (dx /dx) (dx/dt) dx /dx). Simple transformations give: x dx
x2 x 2 c1 . 2 2m
xdx; and then m
0 The initial conditions allow us to find a constant c1 : c1 . Then 2 x2 x 2 2 0 2 m 2 2
or x 02
2 x . m
Since x (dx/dt), we obtain dx dt. 2 0 x2 m
冪莦莦冢莦莦冣莦莦
Integration gives:
( m) x t c2 , 0 m
arcsin
At t 0 the coordinate is x0 0, therefore c2 0. Then x 0 兹(m 苶/ 苶)苶sin兹( 苶/m 苶)t; i.e., under an elastic force a system acquires periodical movement according to sin (or cos) law (this case will be considered in more detail in Section 2.4.2).
EXAMPLE E1.7 A body of mass m 80 kg falls from a motionless airborne helicopter. Besides the gravity force mg, the force of resistance of the air Fc k (k 10 kg /sec) operates on the body. Find the time dependence of speed (t). Find also the body’s steady speed. Solution: Direct an axis y vertically downwards. Two forces will act on a falling body: gravity mg and force of air resistance Fc (Figure E1.7a). According to the second Newtonian law the equation of movement in the vector form looks like ma mg Fc. Since a d/dt and Fc k, hence m(d/dt ) mg k. In projection onto axis Oy an equation of movement in a scalar form is m(d/dt ) mg k. This is a differential equation of the first order with separatable variables. Therefore, (d/mg k) (dt/m). Integration gives:
t
d 1 ∫ mg k m ∫ dt, 0 0
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i.e., 1 t ln 冷 mg k 冨 0 k m
and then 1 mg k t ln k mg m
or
ln
mg k k t. mg m
Exponentiation gives mg k e( k m )t . mg
From this the dependence (t) can be derived:
(t )
mg (1 e( k m )t ). k
The graph of (t) is presented in Figure E1.7(b). One can see that the body’s speed asymptotically reaches the speed of steady movement st, i.e.:
st lim
mg (1 e( k m )t ). k
At t → ∞, st → mg/k. After introduction of the figures one arrives at: st
80.9 81 m sec 78.5m sec. 10
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O
v
Fc= – k =
mg k
mg
t z (a)
(b)
EXAMPLE E1.8. A high-speed boat moving on calm water achieves a speed of 90 km/h. The total mass (m) of the boat with a man on board is 400 kg. The force of water resistance to the boat’s movement changes under law Fc k. Draw a graph of the time dependence of the speed (t) and calculate by how much the speed of the boat changes () in time t 1 min after the man has switched off the engine. The resistance force constant k equals 0.8 kg/m. Solution: Two vertical forces operate on the boat in the lake: the gravitational (mg) and extruded Archimedes force Fa; both of them equalize each other, therefore we do not need to consider them. We shall connect the inertial system of reference with the earth. We shall direct the coordinate axes horizontal. The equation of the Newtonian second law in aprojection to an axis x will be written as m
d Fc k2 dt
Separating the variables we obtain (d/ 2) (k/m)dt. In order to draw a graph it is more convenient to take an indefinite integral ∫ (d 2 ) (k m)∫ dt C.
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After taking an integral we obtain (1/(t)) (k/m)tC. We can find the constant using the initial conditions: at t 0, (0) 0 and then C –(1/ 0). Therefore, 1 k 1 k t m t 0 . m 0 m0 We should solve an equation relative (t): (t )
m0 . k0 t m
The graph is a hyperbola (Figure E1.8). To determine we should find a difference (t) 0 ∆ 0(1(m/kt 0 m)). To obtain the answer we should express all values in brackets in terms of SI. The speed value before a bracket can be left in km/h (then the answer will turn out to be in the same unit): 400 ⎡ ⎤ 90 ⎢1 ⎥ km h 22.5 km h. ⎣ (400 400 0.8 25 60) ⎦
(t)
0
t
Motion of a body with a variable mass In some problems of mechanics a body’s mass, in the process of motion, does not stay constant. Examples can be found in modern technology (e.g., refuelling airborne aircraft, etc.), particularly in systems whose movement is based on the combustion of fuel accumulated entirely in the moving system. Let us derive the main laws of such a motion using a space rocket as an example (Figure 1.9). As a simple example, we can consider a rocket fired in an open space where no gravitational force or air resistance exists. All of the fuel stored is eventually burned and ejected from the nozzle of the rocket’s engine; the rocket mass is
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F
R dm
C
u x
dm
Figure 1.9. Principle of rocket propulsion.
time dependent in this case. It is impossible to apply Newton’s second law to the rocket alone; however taking the rocket and its ejected combustion products together allows us to consider such a combined system to be a closed one and to apply this law. During the flight time the mass of the rocket and its velocity become time-dependent, m(t) and (t). For a time accretion dt the rocket mass and velocity increments are dm and d, the value of dm being negative. The momentum of the rocket will become (m dm) ( d), and the momentum of gas exhausted will be dmgug, where dmg is the mass of the combustion gas and ug is the exhausted gas speed. The momentum increment for the time dt can be obtained by subtracting from the written momentum its initial value. According to eq. (1.3.12), this difference is a forward impulse Fdt. In projections on an axis along which the motion proceeds it is (m dm)( d) dmg g m Fdt
(1.3.15)
Opening brackets and taking into account that dm dmg and rel g (rel is the velocity of the gas exhausted speed relative to the rocket), neglecting the term of the lower infinitesimal order (dm.d) one can arrive at the expression: md rel dm Fdt or m
d dm rel F dt dt
Since we consider the movement in open space without the influence of any gravitation or resistance, we assume F 0. Therefore, m
d dm rel dt dt
The right-hand term rel (dm/dt) is equidimensional to a force; it presents the reactive propulsive force or thrust of the rocket R rel
dm R dt
(1.3.16)
According to its form, eq. (1.3.16) corresponds to the second Newtonian law eq. (1.3.7). However, here the mass is not constant: it is decreasing with time because of the fuel
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combustion; the faster the combustion rate and exhausted gas speed, the larger the rocket engine thrust. This equation was derived for the first time by I.V. Mescherski, and is referred to as the first rocket equation. In the absence of external force, F 0 (the motion in outer space), only the propulsive force is acting on the rocket. Finally, if no external force F acts on the rocket, with an initial condition m(0) m0 and under constant combustion rate ((dm/dt) const.), the velocity (t) can be found:
(t ) 0 r ln
m0 m(t )
(1.3.17)
This equation is referred to as the second rocket equation. Actually, from the expression m(d/dt) (dm/dt)rel, it follows that d (dm/m)rel. By integration, the expression presented above can be derived. The relation of the rocket velocity and its mass change can be derived from expression (1.3.17) (at initial velocity 0 0): ⎛ (t ) ⎞ m(t ) m0 exp ⎜ . ⎝ rel ⎟⎠ This expression is referred to as K.E. Tsiolkovsky’s law (derived in the 1920s).
EXAMPLE E1.9. A spacecraft of mass m 10 T is at a great distance from the earth and the other planets of the solar system. To change its speed to ∆ 0.8 km/h a jet engine is turned on for a time . Determine this time if the fuel combustion rate is 100 kg/sec. The exhaust speed u is 2 km/sec. Solution: Accept that the spacecraft moves with some speed in an inertial reference system relative to the Sun (heliocentric reference system). Its great distance from all planets allows us to neglect the gravitational forces acting on it. Therefore, on turning on the jet engine the single force acting on the spacecraft is the jet thrust R –u. Using the first rocket equation for 0, we shall obtain m(t) d/dt –u. The mass of the spacecraft is continuously decreasing (owing to the fuel burning and the exhaust from the engine nozzle) according to the equation m(t) (m – t), where t is the fuel mass burned up in time t. Let the coordinate axis Ox be codirected with vector . We shall write down the equation of movement in coordinate form as (m – t) d/dt u. After variable separation the equation looks like d dt .
u mc t
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Time integration in the limits from 0 to and, consequently, from to gives 1
u
v v
∫ v
1 dv ln 冷 m t 冨0 .
Deleting on and introducing the limits we obtain
v m
⎞ ⎛ ln ln ⎜ 1 ⎟ . ⎝ u m m⎠
After exponentiation, this equation reduces to 1 (/m) e( /u). The time sought can be extracted from this equation:
m⎡ ⎛ v ⎞ ⎤ 1 exp ⎜ ⎟ ⎥ . ⎝ u ⎠⎦
⎢⎣
Calculations give the time (104/102)(1e0.8/2) 33 sec. 1.3.4
The third Newtonian law
Till now we have discussed the question of how other bodies act on a given body. Quantitatively, this action is defined by force. Experience shows that such an action cannot be unilateral, any action has the nature of interaction. “Actions” and “counteraction” are equal and indistinguishable. They simultaneously appear and simultaneously disappear, but are attributed to different bodies. All these facts form the essence of the third dynamics law or Newton’s third law: the forces with which bodies act on each other are always equal in magnitude and oppositely directed. This signifies that at the interaction of two bodies a force F12, with which the first body acts on the second, is of the absolute value and oppositely directed to the force F21, with which the second body acts on the first. That is, F12 F21 . 1.3.5
(1.3.18)
Forces classification in physics
All the known forces of interaction existing in nature can be reduced to a small number of main types. Belonging to the first type are the gravitational and electromagnetic forces; belonging to the last type are the forces of interatomic and intermolecular interaction pertaining to which macroscopic manifestation are elasticity forces. (Outside the scope of this book are the short-range nuclear forces, bonded nucleons in nuclei, and weak interactions, revealed in the decay of elementary particles.)
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The forces of gravitation are weaker than all the others. At the same time their action is realized through a gravitational field onto great distances. The expression for gravitational interaction between two point masses M and m is defined by the law of Newtonian attraction
F G
Mm r , r2 r
(1.3.19)
where G is the gravitational constant, r is the radius vector drawn from one MP to another being equal in absolute value to the distance between them. These forces govern the interaction between the heavenly bodies. A body on the Earth’s surface (r R, i.e., R is the distance between the center of the earth and the body) experiences the attraction F mg, or in another form (according to eq. (1.3.19)), ⱍFⱍⱍG(mM/R2)ⱍ; it then follows that mg G(mM/R2), or g G(M/R2). (This relationship can be used in order to simplify the solution to some problems.) We meet here for the second time the notion of “mass.” In this respect the mass is called gravitational. Generally speaking, this coefficient can be different from that appearing in the second Newtonian law. However, practice shows that, fortunately, inertial mass is just the same as gravitational mass; i.e., the mass is the objective characteristic of a body exhibiting equally both inertial and gravitational laws. It is worth mentioning here the difference between the terms “weight” and “mass.” Mass is an inherent property of a body, whereas weight is a measure of an action of the body on a support or suspension. A reaction from the support to the gravitational force exists. When a body lies on a motionless bench, two forces—gravitational and support reaction—compensate each other (according to the third Newtonian law); however, the first is applied to the body and the second to the supports. If the “bench” falls with an acceleration g no reaction appears at all, the weight diminishes to zero and a state of weightlessness occurs. Therefore, the mass is a property of the body, but the weight depends on its motion (on acceleration).
EXAMPLE E1.10A A body of mass m 10 kg is resting on spring scales in an elevator. The elevator moves with an acceleration of a 2 m/sec2. Determine the readings of the spring scales in two cases: when the elevator’s acceleration is directed vertically upwards and then vertically downwards (Figure E1.10). Solution: To determine scale readings means to find the weight of the body mg (a), i.e., the force with which the body acts on a spring. This force, under the third Newtonian law in the inertial system connected to the earth, is equal on the modulo and is opposite in direction to the force of elasticity (force of a support reaction) from which the spring cup of the scales operates on a body N, P being the scale reading, that is mg –N or in scalar form P N.
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ai
N
N
N m
a
mg
m
a
mg
mg
ai (a)
(b)
(c)
Hence, the problem of the scale readings is reduced to finding the reaction of the support N. There are two forces acting on a body: the gravity force mg and the support reaction N. Let first the acceleration a be directed upwards. (We can descend the index z because both forces are collinear; the direction will be marked by signs.) The second Newton’s law equation can be written as N – mg ma, whereas N mg ma m(g a). Since N m(g a). The sign of acceleration should be accounted: at a 0 N 10(9.81 2) kg m/sec2 118 N (b) whereas at a 0 N 10(9.81 – 2) kg m/sec2 78 N (c). Electromagnetic forces retain electrons in atoms, keep atoms in molecules and crystals, and define the interaction of molecules between themselves, etc. Electromagnetic forces are long term (i.e., similarly to gravitational forces they decrease with distance as ∼(1/r2)). In practice, one usually deals with gravity forces, elasticity and friction (resistances). These forces reduce to those already mentioned: gravity forces are a result of gravitational interactions, elasticity forces and friction are manifestations of electromagnetic interactions of atoms and molecules both inside the bodies and between them. Examples of macroscopic elastic forces are forces acting on the suspension to which a body is attached. Under gravitational attraction the mass m will act with the force mg to the support, and, as a reaction, elastic forces appear in the support. The body’s weight is the force with which this body acts on the support or stretches a suspension, all of them being unmovable relative to the earth. The elasticity force appearing in suspension is called tension. The force acting on a body from the side of the support is also an elastic force. In this case elastic forces appear as a result of the support’s deformation. Such forces are called a reaction. Under small deformations, the elastic strength linearly depends on the deformation value, i.e., it follows a linear Hooke’s law: the value of deformation is proportional to the deforming force and is opposite in sign: Felast x,
(1.3.20)
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where is the coefficient of proportionality (coefficient of elasticity or rigidity), x is a value of deformation. The term “quasi-elastic force” is often used in physics though the forces are not of mechanical tension; it implies that a force is proportional to deformation. Under large deformations the elastic force can depend on deformation nonlinearly: F x x 2 x 3 ...
(1.3.21)
The coefficients are called unharmonicity coefficients. Friction forces of sliding appear under the direct contact of surfaces at the relative motion of one body upon the other. They are stipulated by the phenomena occurring in the shallow layers of the surfaces of the contacting bodies. Such friction, which acts between surfaces of different bodies, is called external friction. The friction force P in this case is proportional to the normal pressure force N P fN ,
(1.3.22)
where f is the friction factor. This factor depends on the material, conditions of the surfaces, the presence of lubrication and others. Friction force is always directed against the direction of motion and lies in planes of contiguity. Friction can appear as a result of the interaction of different parts of one and the same material, for instance, between the different layers of a moving liquid or gas. Such friction is referred to as internal and the phenomena itself as viscosity. When a solid body moves in liquids or gas, internal friction can appear. This is a result of the molecules sticking to the solid body surface and moving together with it. The phenomena are referred to as internal (viscous) friction; under small velocities such friction force can have a linear dependence on velocity: F k,
(1.3.23)
where k is a coefficient dependent on the medium property and the dimension and form of a body. The nonlinearity can appear as the velocity increase. In some velocity interval the force of friction can appear to be proportional to the second power of velocity Ffr k12 .
(1.3.24)
The k1 coefficient is also dependent on both the medium nature and the body size and form. The information presented on the nature of forces in physics does not, however, give any hope of receiving an answer to the simple question: what keeps atoms in molecules. Moreover, it is proved that forces of only electrostatic nature are unable to ensure the molecules’ stability. We should accept that within the framework of classical physics the
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answer to this question is absent in general. Quantum laws, namely the identity of electrons, bring about the so-called exchange interactions, creating additional forces and stabilizing molecules. The nature of these forces however still remains electrostatic. We can say that the exchange forces are the result of using usual classical presentations in the field of, and according to, quantum physics. 1.3.6
Noninertial reference systems. An inertia force: D’Alembert principle
So far we have considered the motion of a body in inertial reference systems. However, there exist many problems where it is necessary to use noninertial reference systems such as, for example, the motion of molecules in a centrifuge along a circular path or accelerating motion in the rocket. In noninertial reference systems expressions (1.3.14) are not fulfilled. Reference systems in which the motion of a free body is not rectilinear and uniform are referred to as noninertial systems. Consequently, any reference system moving with acceleration relative to any inertial reference system is a noninertial one. The acceleration can be both translational (aτ ≠ 0) or rotational (an ≠ 0). In the general case, the acceleration of different points of a moving body can be different. This means that the space connected with the noninertial reference systems is neither uniform, nor isotropic. The equation of a MP motion regarding the noninertial system looks different from an inertial one. Consider the specific example of a noninertial reference system K (x , y , z ) moving with an acceleration a0 comparative to a certain inertial system K(x, y, z). Suppose then that a MP in this inertial system moves with an acceleration a, and this acceleration is caused by the action of forces F1, F2, …, FN. According to eq. (1.3.13) we can write N
ma ∑ Fi . i1
(1.3.25)
In the system K the same MP will have an acceleration a (relative acceleration), which is the sum of a and the acceleration of the noninertial reference system a0, i.e., a a a0. We can multiply the right- and left-hand sides of this expression by the mass of an MP (m): N
ma ∑ Fi ma0
(1.3.26)
i1
The expression obtained differs from the equation of motion in the inertial reference system (1.3.7) by the term ma0. The noncompliance with the second Newton law is caused by the appearance of that additional term. Moreover, if the geometric sum of acting forces is equal to zero, then a a0, whereas according to the second Newton law it also has to be zero. The product of the body’s mass and the acceleration of the noninertial reference system taken with the opposite sign is called the force of inertia. Fi ma0 .
(1.3.27)
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Inertia forces are the uncommon forces that disobey the laws of classical Newton mechanics. Indeed, in a noninertia reference system we are unable to indicate a body whose action can explain the appearance of inertia forces. This signifies that Newtonian laws are not executed in noninertial reference systems. Figuratively speaking, there exists a force of “actions” (the force of inertia), but no force of “counteraction.” In noninertial reference systems, these particularities of inertia forces do not allow the selection of a closed system of bodies (refer to 1.3.7), since for any body in a noninertial system the inertia forces are the internal ones. Thus, in the noninertial reference system the conservation laws of energy and momentum, which will be considered below (see Section 1.5), are not valid. The importance of introducing the forces of inertia consists of the fact that with the provision of these forces it is possible to use Newtonians laws, as they would occur in an inertial system. With provision for eq. (1.3.26) Newton’s second law (1.3.7) takes the form: N
ma ∑ Fi Finert .
(1.3.28)
k1
If a body is resting in the noninertial system (a 0), eq. (1.3.28) simplifies to: N
∑ Fi Finert 0 i1
(1.3.29)
and the problem of dynamics (in a noninertial reference system) is reduced to the equality to zero as the result of all acting on the body forces and the force of inertia, i.e., to the problem of the static in the inertial reference system. The statement that problems of dynamics can be reduced to problems of static by the addition of the usual forces acting on the body, the force of inertia, is called the D’Alembert principle. It must be remembered that the tasks of static can be solved more easily than the problems of dynamics.
EXAMPLE E1.10B This task can be solved in a noninertial system. In this case a reference system can be connected to the elevator. It means that to all forces the D’Alambert force (Fi mai) should be added (ai is the acceleration of the elevator movement relative to the earth). Three forces are acting on a body in this case: gravitational force mg, elasticity force N and the inertia force Fi. In the reference system connected with the elevator the body is at rest. Therefore, the sum of the forces is zero: mg N Fi 0. After projection on the z-axis the equation is transformed to N g ma 0, whereas the support reaction is N mg ma m(g a); we arrive at the same equation as in the Part A of this example. Of particular interest is a specific case of the manifestation of inertia forces. This case concerns the uniform motion of an MP along the circle trajectory. An MP participating in
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such a motion possesses normal (centripetal) acceleration an. Then, obviously, the reference system connected with this point (!) will be a noninertial one. In this noninertial reference system to the MP of mass m, aside from the others, an inertia force Fi ma0 is applied; it is called the centrifugal inertia force. This force is attached not to the MP but to the bonds retaining this MP on the circle trajectory; it is directed from the center along the radius. It is important not to confuse them!
1.3.7
A system of material points: internal and external forces
Any set of MPs (or bodies) is called a material points system. Each system’s points can interact both with bodies of the same system and with bodies not belonging to it. Forces acting between the system’s MP (bodies) are referred to as internal forces. Forces acting on the system’s points from outside are referred to as external forces. A system is called closed (or isolated) if it comprises all interacting bodies. Thus, in the closed system only internal forces are acting. They compensate each other according to the third Newtonian law. Strictly speaking, there are no closed systems in nature. However, it is almost always possible to define a task neglecting external forces in the limit of accuracy. The choice of the border surface is a person’s own prerogative and can be chosen by the researcher on the basis of an analysis of internal and external forces. One and the same system can be considered as closed or open in different situation, depending on the statement of the problem and pre-given accuracy. All processes are described more easily and decided more clearly in the closed system. This is what in physics is called the physical model. The these two notion are of the some importance and should be given accordingly of a system plays an enormous role not only in art (architecture, painting, ornaments, composing, etc.), but to no less a degree in science. Let us first provide some definitions. Any operation superposing an object with itself is referred to as an operation of symmetry. Atoms and molecules, plants, animals and people, building materials, etc. are symmetrical. In crystallography, chemistry and quantum chemistry translations (t), mirror planes (m) and axes of symmetry (Ln) are mainly used. Endless checked paper (e.g., graph paper) allows one to illustrate an infinite translation: under certain translations (refer to Section 9.1), this sheet being shifted on a certain vector will coincide with itself. A plane perpendicular to the benzene ring going through two opposite atoms of carbon is a mirror symmetry plane. An axis going through the oxygen atom in the water molecule along the bisector of the valence angle is an axis of symmetry of the second order L2. (The numeral 2 appears because of the fact that the symmetry rotation here is (360°/2 180°)). The combination of an L2 axis and a perpendicular mirror plane generates the center of symmetry. There are also other combinations: 360°/n. At n 3 (L3 is a designation of an axis of symmetry of the third order, for instance, a molecule of ozone O3), at n 4 (axis of symmetry is of the fourth order, square molecules), at n 6 (axis of symmetry is of the sixth order, for instance, the benzene ring with the axis L6, passing perpendicularly through the center of the carbon ring). A very important element is the center of symmetry, the “reflection in the point” (the benzene ring has a symmetry center, but the molecule of water does not).
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In many cases, analysis of a problem’s symmetry simplifies its solution. We use this in many sections of this book. Center of mass Let us introduce a very important notion, a centre of mass (CM) (or the center of inertia). The CM of a system is the point at which the system’s mass can be assumed to be concentrated. It can be described with a radius vector
∑ mi ri R c iN1 ∑ i1 mi N
(1.3.30)
The total mass of the system is further denoted by m. If one places the origin into the CM (point C) then Rc 0 and
∑ mi ri 0.
(1.3.31)
i1
Another definition can be drawn from this equation: the CM of a mechanical system is the point for which the sum of products of the masses of all MPs comprising the system, on their radius vectors, drawn from this point, are zero. In Figure 1.10 all these positions are illustrated using the example of a system consisting of two MPs. In the Cartesian reference system, the radius vector Rc has coordinates XC, YC, ZC:
∑ mi xi , i1 N
XC
m
∑ mi yi , i1 N
YC
m
∑ mi zi . i1 N
ZC
m
x2
x1 C
m1 Rc
r2
m2
r1
O
Figure 1.10. The center of a two-mass system.
(1.3.32)
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So far we have considered the set of discrete MPs. The question is now: how to find the CM in a continuous body? It is quite natural to move from the sum to integrals:
RC
1
m V∫
rdm
(1.3.33)
and in coordinate form
XC
∫V xdm ;
YC
m
∫V ydm ; m
ZC
∫V zdm . m
(1.3.34)
It is easy to guess that for bodies having a plane of symmetry, the CM is located on this plane. If a body possesses a symmetry axis, the CM certainly must lie on this axis. If a body possesses a center of symmetry, it is not necessary to determine the CM position: they have to coincide with each other. In order to show the importance of the CM point, let us determine how it moves. Let us write the expressions (1.3.32) in the form N
N
N
i1
i1
i1
∑ mi xi mXc , ∑ mi yi mYc , ∑ mi yi mZc . The time second derivation of them gives: N
N
N
i1
i1
i1
∑ mi xi mXC , ∑ mi yi mYC , ∑ mi zi mZc Comparing these equalities with eq. (1.3.8) one can find that mXC ∑ Fxi , mYC ∑ Fyi , mZC ∑ Fzi . or N
F . mR ∑ i C
(1.3.35)
i1
These equations, called the differential equations of the CM movement, coincide in their structure with the differential equations of MP movement. Therefore, one can conclude: the CM of the mechanical system moves as MP, the mass of which is equal to the total system’s mass and to which all the acting forces are applied.
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If the system is free from the external forces (i.e., closed) (Fi 0 or ΣFi 0), then V const., R C C
(1.3.36)
and, hence, the velocity of the CM is constant (i.e., is preserved). The internal forces do not influence its movement. If at some time in some reference system the CM’s point of a closed system is at rest it means that it will rest further. Many problems of mechanics can be solved in the easiest way in the coordinate system connected with the mass center.
EXAMPLE E1.11 The ends of a half hoop are connected by a straight weightless wire. The radius of the half hoop is R. Find the position of the CM of this figure. Solution: The half hoop has a symmetry axis of a second order; it divides the figure into two equal parts (see Figure E1.11). The center of inertia should definitely lie on this axis, direct a z-axis along this axis; therefore we have to find only one coordinate. Use the formula for the ZC coordinate (eq. (1.3.32)). In our case it is Zc ( 冕dmz) 冒M, where M is an unknown mass of a half hoop. Allocate a segment dl Rd (see figure), write down the evident relations dl dm, (z /R) cos , M/R . Substituting these equations into the expression for ZC we arrive at
1 1 M d ᐉR cos ∫ dR R cos ∫ M M R 2 2R R cos d . ∫ 2
ZC
0 d R Zc
dl
z
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1.3.8 Specification of a material points system To assign a system state means that one can describe a system’s configuration (i.e., to know the position of all the elements) at the initial instance of time, being able to calculate it at any other time. The space position of an MP (Figure 1.11) can be assigned by radius vector r, which in turn is assigned by three coordinates (for instance, Cartesian coordinates x, y, z). How many numbers does one have to know to be able to find a system position at any following points of time t dt? For the time dt a point displaces from position 1 to position 2: r(t dt ) r(t ) d r r(t ) V(t )dt.
(1.3.37)
This implies that for the determination r(t dt) one needs to know r(t) and (t), i.e., six numbers in total. If a system consists of N particles, the number of parameters is 6N. For the assignment of a system state in modern physics, a so-called configuration space is introduced. The dimensionality of this space is the number of parameters defining the state of a system at a point of time, the change of this state being defined by the set of points, i.e., the line. The element of the configuration space volume d in analogy with a three-dimensional case is written as N
d ∏ dxi dyi dzi dpxi dpyi dpzi .
(1.3.38)
i1
We will use this representation in Chapters 3 and 9.
z
dr = Vdt r(t) r(t+dt)
y x
Figure 1.11. A material point movement description.
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The dynamics of rotational motion
When considering the kinematics of the rotational movement of an IRB, we had to introduce some new characteristics: elemental angular displacement, angular velocity and angular acceleration. These values are identical to the whole rotating body, whereas the linear characteristics for all the body’s MPs differ. In rotational motion other dynamic characteristics are also required, such as a force moment (torque) with regard to a motionless axis, a moment of inertia (MI) and an angular momentum, being in some respect analogous to the characteristics of linear motion (mass, force, momentum). These models do not exhaust the whole description of problems concerning rotational motion, but cover a wide range of phenomena with which the chemist may be confronted. Ignoring the logic of physics for the sake of simplification, we will start from the description of the rotation of an IRB relative to a motionless axis. We will then generalize the results obtained and apply them to the motion of an MP around a pole. The dynamics of rotation of an IRB around a motionless axis Suppose that force F is arbitrarily applied to a body’s point (in Figure 1.12 the body itself is again not shown). Divide the force vector into two components: one parallel to the axis of rotation F⏐⏐, and the other lying in the plane perpendicular to the rotation axis F⊥. Only one of them (F⊥) influences the rotation, whereas F⏐⏐ exerts pressure on the bearings in which an axis is fixed. The force momentum (torque) M in respect to an axis Oz is the value M F⬜ R sin F⬜ h,
(1.3.39)
where is an angle between the radius of circular point path (R) and the force arm (h). In turn, Rsin h, h being the shortest distance from the point O to the line along which the force component F⊥ is acting (the force arm). Note that the torque can be zero though the force itself is not, if the force line action crosses the axis of rotation or is parallel to it. z Lz z
F
Fll
O h
R F1 r
Figure 1.12. A force moment; a work of this moment at a body rotation.
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Since all the body’s points are at different distances from the axis of rotation their linear velocities and, correspondingly, their momentums are different. In order to find a body’s rotation characteristics let us draw at a moving point, a mass element dm dV, where is the body’s density. The value dLz R dp R dm,
(1.3.40)
is referred to as the angular momentum of a mass element dm relative to the axis Oz, R being the distance of the dm element from the axis and the linear velocity of this element. Referring to eq. (1.2.27) we can arrive at dLz z R 2 dm z dI z .
(1.3.41)
dI z R 2 dm
(1.3.42)
A value
is referred to as a moment of inertia (MI) of the mass’s element dm relative to the axis Oz. Integrating it over the body’s volume V, one obtains Lz z ∫ dI z z ∫ R 2 dm. V
V
In this expression I Z ∫ R 2 dm
(1.3.43)
V
is referred to as the body’s momentum of inertia relative to an axis Oz. Then one can arrive at LZ I Z Z .
(1.3.44)
That is, the angular momentum of a body relative to a motionless axis is the product of the moment of inertia and the angular velocity of a body’s rotation relative to the same axis. (This definition according to its “structure” is equivalent to the definition “momentum of the translational movement is a product of its mass and their velocity.”) Moments of inertia of some symmetric bodies Consider moments of inertia of some symmetric figures relative to their central axes, i.e., axes passing through their CM (i.e., the symmetry axes). Thin rod. Let us derive the MI of a thin rod (i.e., a rod of mass m whose transverse linear dimensions are much less than its length l ) with regard to the axis Oz passing perpendicularly to the rod passing through its CM (Figure 1.13). Choose an elementary fragment dx, remote
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from the axis at distance x from the axis Oz. The mass of this fragment is dm (m/l)dx. According to the formula (1.3.43) we will arrive at l 2
Iz
m ml 2 . x 2 dx ∫ 12 l l 2
(1.3.45)
Hoop. Consider a hoop of mass m and radius R. The material cross-section is negligibly small (Figure 1.14). The ring MI regarding the axis Oz drawn through its center perpendicular to the ring plane is 2 R
Iz
∫
R2
0
m dl mR 2 2R
(1.3.46)
where dl is the length of an arc of dm (m/2R)dl. z
x
x, x'
dx
Figure 1.13. A moment of inertia of a thin rod. dl z R
O
Figure 1.14. A moment of inertia of a thin hoop.
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R
0
r dr
z h
Figure 1.15. A moment of inertia of a disc (cylinder).
Disc (cylinder). Let us select in the solid-bulk disc of mass m and radius R (Figure 1.15) an elementary volume dV in the form of a coreless cylinder of radius r, height h and thickness of walls dr. Its mass dm is dm dV. As m/V m/R2h then dm (m/R2 h)2rdrh R 2 2 3 (2m/R )r dr, h and Iz makes up Iz (2m/R ) 冕 r dr (2mR4 / R24), or finally 0
Iz
mR 2 . 2
(1.3.47)
In many cases it is necessary to calculate the MI with regard to an axis z parallel to the symmetry axis z but shifted from it to the distance d remaining parallel to the first (Figure 1.16). The MI IZ regarding the central axis z is taken as known. The Iz value should be calculated. In the shifted reference system the coordinate of mass element dm is x′ the last being equal to x′ d x. The MI Iz can be written as l 2
I z
∫
l 2
(d x )2 dm
l 2
∫
l 2
d 2 dm 2 d
l 2
∫
l 2
xdm
l 2
∫
x 2 dm
l 2
However, l 2
∫
l 2
l 2
d 2 dm d 2
∫
l 2
dm d 2 m, 2 d
l 2
∫
xdm 0
l 2 l/2
(according the definition of the CM (1.3.31)). The term d 2 冕 dm d 2m is IZ; therefore we l/2 arrive at
I Z I Z md 2 .
(1.3.48)
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z'
z
x' x
x,x'
dx
d
Figure 1.16. A parallel axis theorem.
This expression is referred to the parallel axis theorem. Notice that the value md 2 is always positive; it means that the MI relative to the symmetry axis IZ has a minimum value. There exists one more method of simplifying the MI calculations. We mean calculations of the MIs of planar figures, i.e., figures that slightly differ from a two-dimensional form (have low thickness). A method is derived and presented in Example E1.12. The main essence of this method is the relationship between MI regarding three orthogonal axes Iz Ix Iy .
(1.3.49)
Sometimes this method is referred to as the theorem of orthogonal axis. EXAMPLE E1.12 Calculate the MI of a hoop relative to its diameter (a y-axis, Figure E1.12); R is its radius and m its mass (an axis y lies in the hoop’s plane passing its center). y
x
dl
d R z
x
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Solution: Let us solve this problem by the first method. On the hoop, allocate an elementary segment dl with mass dm (m/2R)dᐉ. Find the MI of a chosen element considering it as MP: dIy x2dm x2(m/2R)dᐉ. Note that x (the shortest distance to axis Oz) can be expressed through an angle : x Rcos. In its turn, dᐉ/R d and a MI of a selected element is dIy R2cos2(m/2)d. By integration in the limits from 0 to 2 we arrive at: Iy
mR 2 2
2
∫ 0
cos2
mR 2 1 2 2
2
∫ (1 cos )d 0
mR 2 mR 2 2 4 2
Now consider the second method. Consider the same plain figure: a hoop R in diameter and mass M: the x- and y-axes coincide with the diameters; the z-axis is directed perpendicular to the plane. Allocate an elementary area (as MP) of a mass dm with coordinates x and y (z 0). Then the Iy value relative to the y-axis is: dIy x2 dm; dIx y2 dm and dIz (x2 y2)dm x2 dm y2 dm; then integration over the whole figure gives Iz Ix Iy Note that we already know the Iz of a thin ring relative to an axis perpendicular to the ring’s plane and passing through its center: Iz mR2 (see eq. (1.3.46)). Symmetry consideration gives Iy Ix, then according to (1.3.49) we can write
1 mR 2 I z I x I y 2 I y and I y I z 2 2 It can be seen that the second method in some cases is less troublesome than the direct solution. This approach is referred to as the perpendicular (orthogonal) axis theorem. EXAMPLE E1.13 Calculate the MI of a H2O molecule relative three mutually orthogonal axes x, y, z passing through the molecule’s CM. Interatomic distances O–H are d 95.7 pm, valence angle 104.5° (see Figure E1.13). Relative nuclear masses of atoms: Ar,1 1 and oxygen Ar,2 16. O
z'
y'
a = xC
z
y
C
H
H
x
x'
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Solution: Let’s arrange a molecule as it is represented in Figure E1.13. A water molecule consists of three MPs (nuclei of atoms) with a total mass m 2m1 m2, where m1 and m2 are masses of hydrogen and oxygen atoms, respectively. Calculate first the MI Iz of the water molecule relative to an axis z which passes through oxygen atom perpendicular to a molecule plane. The origin (intersection of three axes C) is superimposed with the molecule CM. We direct an axis z upwards perpendicular to a molecule planes. To find the molecule MI we shall take advantage of the theorem on parallel axis (see (1.3.48)): Iz Iz,C ma2, where Iz is the MI of the molecule relative axis z , passing through an oxygen atom and parallel to the axis z. The required MI is Iz,C Iz ma2*, where a is a distance between two parallel z axes. One can find the MI relative the axis z as a sum of two MI of MP being at the distance d from the z axis Iz 2m1d 2. The distance a is just the coordinate xC of the oxygen atom laying on y axis; it can be found according eq. (1.3.34), i.e., a xC
∑ mi xi 2m1 x1 m2 x2 . m
m
Therefore, a (2m1x 1 m2x 2)/m. Taking into account that x'1 d cos(/2) and x 2 0, we can obtain a 2 m1d cos(/2)/m. Substitute Iz , m and a into the star* equation we obtain
⎛ 2 m1 ⎞ I z,C 2 m1d 2 ⎜ 1 cos2 ⎟ 2 m1 m2 2⎠ ⎝ The Iy can be obtained accordingly: Iy 2m1d2 sin2(/2). Now we can use the advantage of the previous correlation of MI’s of the plain figures (see E1.12) (Iz Ix Iy) and found
2m 2 m1 ⎞ ⎞ ⎛ ⎛ I x 2 m1d 2 ⎜ 1 1 cos2 sin 2 ⎟ 2 m1d 2 cos2 ⎜ 1 ⎝ m 2 2⎠ 2⎝ 2 m1 m2 ⎟⎠ Calculations can be simplified a bit taking into account that expression in bracket are dimensionless. The mass m1 and d should be translated into SI, keeping in mind that 1 a.m.u. 1.66 1027 kg and 1 pm 1012 m, then m1 A r,1 1.66 1027 kg and d 95.7 1010 m. Substitute all these values into the final formula and executing calculation gives ⎡ 2.1 ⎛ 104.5 ⎞ ⎤ 2 I z,C 2 1.66 1027 (0.957 1010 )2 ⎢1 cos2 ⎜ ⎟⎠ ⎥ kg m ⎝ 2.1 16 2 ⎣ ⎦ 2.911047 kg m 2 , I y,C 2 1.66 1027 (0.957 1010 )2 1.888 1047 kg m 2 , I x,C 2 1.66 1027 (0.957 1010 )2 1.02 1047 kg m 2
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Such kinds of information can be found from the optical molecular spectroscopy and from experiments on inelastic scattering of neutrons. Diatomic molecule as a rigid rotator A system rotating around a motionless axis is referred to as a rotator. If the intermolecular distance is constant, the rotator is called a rigid one. Consider a diatomic molecule, rotated around an axis passing through the CM (Figure 1.17). The problem is to express the MI of this molecule and the corresponding kinetic energy of rotation through its parameters, which are supposed to be known (for instance, from the reference literature). In the figure, the masses of two atoms of the molecule are marked by letters m1 and m2, and the letter d denotes an interatomic distance (d x1 x2). The point C is a mass center. Bearing in mind the characteristic of the CM (1.3.31) we can derive x1m1 x2m2. Therefore, x1 x2(m1/m2), x2 x1(m2/m1) further x1 d m2 /(m1 m2) and x2 d m1/(m1 m2). It is no trouble to calculate the MI: I Z m1 x12 m2 x22
m1m2 2 d d 2 , m1 m2
(1.3.50)
where m1m2 m1 m2 /(m1m2) is referred to as the reduced mass of the molecule and Iz as the reduced molecular MI. It can be seen from the equation derived, that the rotation of diatomic molecule relative to the axis passing through the CM can be reduced to a single mass () rotation around the same axis at the distance from axis being d. Note that eq. (1.3.50) contains only tabulated molecule characteristics which can easily be found in the reference books. The kinetic energy of a rotating body can be given by the equation
x
I z 2 L2 z 2 2Iz
(1.3.51)
z
x1
x2
m1
C
m2
d
d
Figure 1.17. A diatomic molecule as a rigid rotator.
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The equations derived are used in molecular spectroscopy. Let us now analyze projections of the dynamic characteristics of a body rotating around the motionless axis. The time derivation equation (1.3.44) gives dLz dz d ( I z z ) I z I z z . dt dt dt
(1.3.52)
(with IZ not changing in time). If one acts on the mass element dm by the force F along the tangent to the circular trajectory, the mass element will receive acceleration aτ ( ZR): dF aτdm R Zdm. Multiplying the left- and right-hand sides of this equation by R, we receive dMZ dIZ Z, where dMZ is a force momentum with regard to the axis Oz. Integration over the whole body gives MZ I Z Z .
(1.3.53)
(This equation is equivalent to the second Newtonian law for translation motion.) Comparing eqs. (1.3.52) and (1.3.53) we arrive at dLz Mz . dt
(1.3.54)
Applying this equation for the case of N forces (with corresponding momentums), N dLz ∑ ( M z )i . dt i1
(1.3.55)
This equation presents another form of Newton’s second law for body rotation relative to the axis Oz: the change of the angular momentum projection onto the axis Oz is equal to the sum of projections of all the force momentums applied to the body relative to the same axis. The planar motion of the material point relative to a pole Let an MP of mass m move along the planar orbit around the pole O with a velocity (and, consequently, with momentum p m). The angular momentum of a MP relative to the pole O is a vector product L [r p ],
(1.3.56)
where r is a radius vector of the MP (Figure 1.18). According to the definition, the vector L is perpendicular to the plane in which both vectors r and p are lying (i.e. to the orbit plane) and is directed in such a manner that r, p and L produce the right-hand rule system.
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Z Lz
L r e r
Figure 1.18. A force moment relative to a pole.
Calculating the time derivation of eq. (1.3.56) gives dL ⎡ d r ⎤ ⎡ d p ⎤ p r . dt ⎢⎣ dt ⎥⎦ ⎢⎣ dt ⎥⎦
Hence dr/dt and p m, the first term is then zero since both vectors are collinear. Taking into account that dp/dt F, the second term can be written as [r.F]. This vector is called the momentum of the force (torque) F with respect to pole O. M M 0 (F) [r F].
(1.3.57)
dL M, dt
(1.3.58)
Therefore,
i.e., the time derivative of the angular momentum of an MP is equal to the applied force momentum. This is the basic law of the dynamics of rotational motion or Newton’s second law for planar MP revolution. The projection of the equation onto the axis Oz gives dLz /dt MZ and we arrive at the known eq. (1.3.54).
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It is very useful to apply eq. (1.3.58) to the explanation of the effect known as a gyroscopic effect. Using this example, one can see how the rules of rotational motion are distinguished from our usual beliefs about the mechanics of motion and what effects they can bring about. In Figure 1.19 a sufficiently sophisticated situation is shown. Imagine a body (in Figure 1.19 it is arbitrarily drawn in the form of a disc) which rotates around axis z. The angular momentum of this body L is directed along the same axis. Apply to the axis of the rotating body a force F acting along the x axis. The torque of this force M is a vector, directed along axis y. According to eq. (1.3.58) the increment dL is directed not along the line of the force action but along the torque mentioned, i.e., along axis y. (The indexes x and y in the figure indicate the fact that force F is directed along axis x, but vector dL is along axis y). Thereby, a force directed along axis x creates a rotation of vector L perpendicular to the force Fx. z Lz
dL y
Fx
My
x
y
Figure 1.19. A gyroscopic effect. Table 1.1 The relationships between the characteristics of translational and rotational motion Translational motion (along z axis)
Rotational motion (relative to z axis)
M
Body’s mass
Iz
Fx
Force projection
Mz
px mx
Momentum projection
Lz Izω
dpx Fxdt
dLx Mzdt
dA Fxdx
Basic law of dynamics of the translational motion Elementary force work
W Fxυx
Power
W Mzz
dA Mzd
Moment of inertia relative the rotation axis Force moment relative the rotation axis Angular momentum relative the rotation axis Basic law of rotational motion relative an axis Elementary work of force momentum Power
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It is reasonable here to look at Figure 1.8 where an increment dp in the translation motion, according to Newton’s second law, is directed along the acting force line but is by no means perpendicular to it. As an example, a precession of an unbalanced gyroscope (a top) in a uniform gravity field is considered in Appendix 2. Many physical events, such as diamagnetism, precession of magnetic moments (atomic and nuclear) in the external magnetic field and others are based on gyromagnetic effects (refer to Chapter 8 and Appendix 2). In conclusion, it is useful to note that the structure of the formulas of the kinematics and dynamics of rotational motion relative to a fixed axis have the same “structure” as formulas of translation motion. One has only to substitute all translational characteristics with rotational ones. This analogy can be seen in Table 1.1.
1.4 1.4.1
WORK, ENERGY AND POWER
Elementary work of a force and a torque
Let an MP, under the action of a force F, undertake an elementary displacement dl (Figure 1.20). The elementary work dA of the force F is the scalar product of the force and the elementary displacement of a point of force application: dA (F dl )
(1.4.1)
dA F dl cos Fl dl.
(1.4.2)
or
Depending on the magnitude of the angle α, the elementary work can be positive or negative: at 0 π⁄2 the work is positive and at π/2 π it is negative. When π/2 the work is zero. The work on the finite displacement L is equal to A ∫ (F dl ).
(1.4.3)
L
Fl dl A
F
Figure 1.20. Force work F over a displacement dl.
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z Mz
d
F
R dl
Figure 1.21. Torque work on an angular displacement d.
In a particular case in which the force is constant and the application point is moving along straight line l, eq. (1.4.3) simplifies to: A Fl cos Fl l
(1.4.4)
where Fl is the force projection on the direction of motion. A force F being applied to a point A of a rotating body performs a work dA Fdl (Figure 1.21). Since Rdl is an angle displacement d then dA M Z ( F⬜ )d,
(1.4.5)
where MZ(F) is the torque of the force F regarding the Oz axis. The work done on a finite angle is
A ∫ M Z (F) d.
(1.4.6)
A M Z (F).
(1.4.7)
If MZ(F) const. one has
Let us consider some examples. Elastic force work. Elastic force depends on deformation (displacement of the force application point at a body) according to the linear law F(x) – x (refer to Section 1.3.5). The simplest example of elastic force is the small deformation of a spring. Let us superpose the origin with the point of the force application when the spring is in a nondeformed state and an external force is not acting (Figure 1.22, point O). Suppose that a body can move without friction (!) along the horizontal x-axis. After application of an external force, two forces will act on the body: the external force F2 and the elastic one F1. At any position they are in balance: F1 –F2. When the body returns to the initial position both forces perform equal work but of opposite sign. The elementary work of the elastic force on a displacement dx is dA F ( x )dx x dx;
(1.4.8)
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x O F1
F2
x x F(x) F=−x x A
Figure 1.22. A work of elastic force F1.
The total work of the external force on the displacement x is x
A ∫ x dx 0
x 2 , 2
(1.4.9)
whereas the elastic (internal) force produces the positive work:
A
x 2 . 2
(1.4.10)
The external force work is equal to the internal one in absolute value but is opposite to it in sign. 1.4.2
Power
When work is produced during some time interval, the question arises of how fast the work is made. This leads to the notion of power. For time ∆t, let work ∆A be accomplished. Then an averaged power P for a given interval ∆t is 冓 P冔
A t
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Under ∆t → 0 one can obtain a power made by a force in a given time instant, i.e., the instant power A dA . t0 t dt
P lim
(1.4.11)
Keeping in mind eq. (1.4.1), one can write P
Fdl F dt
(1.4.12)
The power is the scalar product of the force and the velocity of point of its application. When rotational work is produced by a force moment MZ(F) applied to the body the power is P
1.4.3
M z d M Z . dt
(1.4.13)
Kinetic energy
The kinetic energy of an MP is a scalar measure of its mechanical motion, equal to half of the MP’s mass and the square of its velocity
x
m2 . 2
(1.4.14)
As can be seen, the kinetic energy is always positive. Because of the motion, an MP possesses a certain stock of mechanical energy, referred to as kinetic energy. Reduction of velocity means a loss of kinetic energy; velocity increase leads to an accumulation of kinetic energy. Changing the energy goes to accomplishing a work. The kinetic energy is assigned in a certain inertial reference system. Turning to another inertial system the kinetic energy value (in consequence of a velocity change) will be different; the kinetic energy (in the same way as the velocity) is noninvariant with respect to Galileo’s transformations though the mathematic expressions are the same and the physical values differ only by a constant. The total kinetic energy of a mechanical system consisting of a set of N material points is the sum of the kinetic energy of all the system’s elements:
mi i2 . i1 2 N
x ∑
Consider a value of kinetic energy for some types of motion.
(1.4.15)
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Translational motion. Taking into account that under translational motion of an IRB the velocities of all the body’s points are the same, we arrive at
mi i2 2 2 i1 2 N
x ∑
N
∑ mi i1
m2 , 2
(1.4.16)
where m is the body’s total mass and is the velocity of any point of the body. Rotational motion. For the rotational motion of a body around a fixed axis, the velocity of an arbitrary point is i Ri where is the angular velocity of a body, Ri is the distance of a corresponding point from the axis of rotation. Then N
x ∑ mi i1
2 Ri2 2 2 2
N
∑ mi Ri2 .
(1.4.17)
i1
When the last sum is a moment of inertia Iz of a given body relative to the axis of rotation then
x
I z 2 L2 z . 2 2Iz
(1.4.18)
General case of a body’s (a system’s) energy. For an arbitrary system a theorem is equitable: the kinetic energy of a system of MPs is the sum of the kinetic energy of the system’s mass as a whole, imaginary concentrated in the CM and moving together with it, and the kinetic energy of all the system’s elements with respect to the CM. Apply the theorem to an arbitrary body motion. In this case the motion can be divided into the translational part (at the velocity of the CM) and the rotational part relative to the axis, passing through the CM. According to the theorem, the total kinetic energy in this case is the sum of the kinetic energy of a translational motion (with the velocity of its CM, VC) and the rotational motion relative to the axis, passing through the CM (with the angular velocity ): x
mVc2 I c 2 . 2 2
(1.4.19)
Theorem on the kinetic energy change. Let an MP of a mass m under the action of a force F gain an increment dl. Projecting the vectors of equation ma F onto the tangent to the trajectory in the point of force application we obtain ma F, where a is the absolute value of the point’s tangent acceleration of an MP, and Fτ is the corresponding force projection. Taking into account that a d/dt (d/dl) (dl/dt) d/dl we obtain: md F dl. Considering the right-hand side of this equation as the force’s elementary work dA and the left-hand side as a differential of the body’s kinetic energy we arrive at d(m2/2) dA, or dK dA,
(1.4.20)
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i.e., the elementary change of the kinetic energy is equal to the elementary force work. In integral form, this equation can be presented as x 2 x 1 A,
(1.4.21)
i.e., the change of the MP’s kinetic energy is equal to the work of all forces applied to it. No limitation has been applied to the nature of the force in eq. (1.4.21). Therefore it is valid for any system and has to take into account all kinds of forces: internal and external, potential and dissipative. The theorem discussed is also valid for a system of MPs.
EXAMPLE E1.14 A uniform hoop and/or disc begins to roll, without either friction nor sliding, along an inclined plate with an angle of slope α 10° from the height h 40 cm (Figure E1.14). Determine: (1) linear velocities of CM Vc1 (hoop) and Vc2 (cylinder) of both bodies at the end of the plane; (2) time of their rolling τ1 and τ2.
Oc d
O'
dŠ H
Solution: (1) In consequence of the absence of the dissipative forces we can take advantage of the law of the mechanical energy conservation E1 E2, where E is the total mechanical energy at the top and bottom position. This energy can be written as the sum of kinetic and potential energies £ 1 U1 £ 2 U2. In the initial and final positions both bodies have identical potential energies. Their kinetic energies at the initial position is zero, both bodies are at rest. In the final position, according to the energy conservation theorem and taking into account the theorem on total energy, the total energy will be equal to the sum of kinetic energies of the translational motion of the body’s CM and rotational motions, £ 2 (m2C/2) (Iz2/2), where C is the CM’s speed of bodies, and MI is the MI concerning a
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horizontal axis z passing through the CM of bodies parallel to the inclined plane and is the angular speed of rotation of bodies concerning these axes. The point of contact of rolled bodies and the inclined plane is the instant center of speeds, in all time instants it is at rest; movement can be considered as instant rotation of the bodies concerning this point. Therefore,
V dt V d dS C C. dt R dt R dt R
In order not to solve the same problem twice we designate the MI of bodies IZ mR2 (for the hoop 1, for the disc 1/2). Therefore x 2
mVC2 mR 2 (VC R)2 mVC2 mVC2 2 2 2 2
2
and finally, £ 2 (mVC /2)(1) In this case the rolling kinetic energy can be written down for both the bodies and for any other cylindrically symmetric body, one should substitute only a proper value. In order to find the velocities one should use the conservation law. mgh (mVC2 2)(1 ). Reducing expression of masses m we can arrive at
VC
2 gh . 1
Carrying out calculations for speed of the CM we shall obtain (see below). We can find the time of the bodies’ slope run by taking the formula of the kinematics of uniformly accelerating movements l/<>, where l is the inclined plane length and 〈〉 is an average speed of uniformly accelerating motion 〈〉 VC /2 (because the initial speed was equal to 0). We can then arrive at: for the hoop ( 1) VC1 gh 1.98 m sec and 1
for the disk ( 1 2) VC2
4
3 gh
2h 2.33sec VC1 sin
2.29 m sec and 2
2h 2.01sec. VC2 sin
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1. Mechanics
A force field
A physical field is a particular form of matter that links together material particles and transmits an influence from one body to another (with finite velocity). Each type of interaction has its own special corresponding force field. The force field is an area of space in which a force acts on any material particle placed in this space point, depending on coordinates and time. A force field is called a stationary one if the acting forces do not depend on time. A force field at any point of which the acting force has one and the same value (on modulus and direction) is referred to as a uniform one. It is possible to characterize a force field by force lines. In this case, the tangent to force lines defines the direction of a force and the line’s density is proportional to the force value. The force field is referred to as a central one if every force line passes through one particular point, called the center of forces (Figure 1.23). The magnitude of force F, acting on a MP in such a field, depends only on distance r from the center of forces, i.e., r F(r ) F (r ) , r
(1.4.22)
(r/r is a unity vector in the direction of r). All the force lines pass through a single point (pole) O; the force momentum in this case is identically equal to zero (M0(F) ≡ 0). Gravitational and Coulomb forces are related to the central ones. An example of a uniform force field is depicted in Figure 1.24: in every point the lines of force action are the same both in direction and magnitude, i.e., F(r ) F const.
O
r
Figure 1.23. Force lines of a central force field lines.
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In Figure 1.25 an example of a nonuniform force field is given. In this case F ( xyz ) const. and
F ( xyz ) F ( xyz ) F ( xyz ) , ,
x
y
z
(i.e., all partial derivatives) deviate from zero.
y
x
Figure 1.24. A schematic representation of a uniform force field lines. y
x
Figure 1.25. Force lines representation of a nonuniform force field.
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All the mechanical forces can be divided into two groups: conservative forces (acting in potential fields) and nonconservative forces (or dissipative). The forces are referred to as conservative (or potential) ones if their work depends neither on the trajectory form, nor on the path length, and is defined only by the position of the point of the force application in the initial and final positions. The field of conservative forces is called the potential (conservative) one. Let us see that the work of the conservative forces along a closed contour is zero (Figure 1.26). Arbitrarily divide the contour into two parts: 1a2 and 1b2. Since the force is conservative A1a2 A1b2. On the other hand it is obvious that A1b2 –A2b1. Then A1a2b1 A1a2 A2b1 A1a2 – A1b2 0, which has to be proven. The inverse statement is fair: if the work of a force on the closed contour is zero, the forces are conservative and the field is potential. This condition can be written as a contour integral (circulation of a vector along a closed contour):
∫ Fdl 0.
(1.4.23)
L
i.e., the circulation of vector F along the closed contour L is zero. The work of the nonconservative (dissipative) forces in the general case depends both on the form of the paths travelled and the lengths of the way. Examples of nonconservative forces can be given by friction and resistance forces. In both cases the mechanical energy transforms into another type of energy. The central forces are referred to as the conservative forces (Figure 1.27). In fact, if a force F is a central one, the work of this force dA can be presented as dA Fdl F(r)(r/r)dl and dA F(r)cos dl F(r)dr (because dl cos dr). Then the work is r2
A12 ∫ F (r )dr f (r2 ) f (r1 ),
(1.4.24)
r1
where f(r) are the antiderivative functions. It can be seen from this equation that the work A12 of the central force depends only on the form of the f-function and on the positions of the initial and final points (r1 and r2) but not on the path length. This statement is just the indication of a conservative field (conservative force). a
2 b 1
Figure 1.26. Proof of the work of a conservative force to be equal to zero.
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F dr dl 2 1
r
r2
r1
O
Figure 1.27. Proof of the conservative nature of a central force (a central field).
The proof given is common to any central force and, consequently, covers the abovementioned types of forces—gravitational and electrostatic ones. 1.4.5
Potential energy
In a potential system, the notion of potential energy can be introduced as a function of a point coordinate. In a system, first choose a state that we can arbitrarily admit as a point with zero potential energy (position U0 0). Further, suppose that we need to find an MP potential energy in another point of the system, which we assign as position 1 (i.e., find the value U1). The potential energy of a system in position 1 is taken to be numerically equal to the work of the field force on transferring the system from position 1 to that position where the potential energy is chosen as zero: A10 U1 U 0 U1 .
(1.4.25)
If the field is potential, the work A10 does not depend on the pathway 1–0. It characterizes the system in point 1 with respect to point 0. If one needs to define the potential energy in position 2, the work of the field force should be measured. Obviously, A20 U2 and A12 U2 U1. Because A21 A12, the work of the force is A12 U1 U 2 U ,
(1.4.26)
i.e., the work of the internal force (force of the field) is equal to the decrease of the potential energy. On the other hand, the work of the external force, acting against the field force, brings about the potential energy growth. A21 U U 2 U1 .
(1.4.27)
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Position 0 was chosen arbitrarily; any point of the system can be accepted as the zero point. This signifies that the defined potential energy is accurate to a constant value C. This “arbitrariness” is not essential, since in the calculations of the difference of energy (refer, for instance, to eq. (1.4.25–27)) constants C are mutually canceled out. Also, the presence of the constant in the equation does not affect the derivative of the potential energy function in respect to the coordinates. The correlation obtained shows how one can determine the potential energy of a system at a certain position. There is no universal formula for such a calculation (as for the kinetic energy). Correlation (1.4.25) shows a way of determining the system’s potential energy by calculating the force work which leads a system to the given zero point. Below are some important examples.
EXAMPLE E1.15 Calculate the potential energy of a deformed spring using the potential energy definition. Solution: In Figure 1.22 the scheme of the spring, originally in a nondeformed state, is presented: the left end of the spring is rigidly fixed, the other end, under the action of an external force, can move along an axis x. The spring is also stretched under the action of external force F2. On the movable end of the spring in an arbitrary state two oppositely directed forces operate: external force F2 and force of elasticity F1: F1 F2. For the zero position (with zero potential energy) we choose the spring to be in a nondeformed state (x 0). According to eq. (1.4.25) dA xdx, i.e., 0
0
X
X
U ( x ) ∫ dA ∫ xdx
x 2 . 2
The straight line in Figure 1.22 represents Hooke’s law whereas the potential energy is shown by a hatched area.
EXAMPLE E1.16 Determine the potential energy U(r) of the body in the gravitational field of the earth. The earth’s mass is M and the distance from the center (!) of the earth to the body of mass m is r. Solution: According to Newton’s gravitation law (1.3.19) we have F(r) G(Mm/r2). Assume as a position with zero potential energy an infinite body’s remoteness from the earth (U(∞) 0). By definition, the potential energy of the
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body in the given (arbitrary) space point r is numerically equal to the work of the gravitational force when carrying the body from a position r to r ∞
U (r ) GMm ∫ r
r dᐉ dr Mm GMm ∫ 2 G r r3 r r
r
Therefore, one arrives at U(r) G(Mm/r) The potential of the gravitational field (U/m) is numerically equal to the potential energy of the MP of a unit mass placed into a given point r; therefore, (r) G(M/r). Note that r is the distance to the body from the center of the earth; this formula is valid for any spherically symmetric force’s field. The relationship is also useful: U(r) m(r). This is the most general expression. Relationship between the force and potential (work-energy theorem). In order to establish a relationship between energy and work we shall consider a body in a stationary field. Acquire an elementary body’s displacement dl. The internal forces of the field will produce the work dA equal to the potential energy change (dA –dU). Therefore, Fl
dU . dl
(1.4.28)
This is the relation sought. The dl quantity is a displacement. In a one-dimensional problem, consider dl ≡ dx. Therefore: F ( x )
dU ( x ) . dx
(1.4.29)
Considering further a motion in the central force field dl ≡ dr F (r )
dU (r ) . dr
(1.4.30)
In a three-dimensional case ⎛ U
U
U ⎞ F(r ) ⎜ i j k gradU ( xyz ), ⎝ x
y
z ⎟⎠
(1.4.31)
i.e., the force is equal to the gradient of the potential energy taken with an opposite sign. That is, the potential energy is determined within an accuracy to a constant component; this does not influence the result: calculating the differences or derivations using these formulas (1.3.28–1.3.31) the arbitrary component makes no contribution to the result.
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EXAMPLE E1.17 Calculate the potential energy U(x) of an elasticity force F(x) using the relation of force and potential energy. Solution: Since F(x) − x and F(x) −(dU/dx), then x (dU(x)/dx); therefore dU(x) xdx. Integration gives U(x) (x2/2) C. Assigning U(0) 0 we can obtain C 0. Then U(x) (x2/2). The work to deform a spring from x1 to x2 can be expressed by the equation: A21 U U 2 U1 12 ( x22 x12 )
EXAMPLE E1.18 Derive an expression for the potential energy U(h) for a body raised on h, not high above the Earth’s surface. Solution: We should start with the precise relation (see example E1.16). According to the definition, the potential energy is numerically equal to the gravitation force work
A(r ) U (r ) GMm ∫
dr r2
or U (r ) G
Mm C. r
We can determine an integration constant C in two ways: firstly, having accepted the point where the potential energy of a body removed to an infinite distance has zero potential energy, i.e., U(∞) 0. Then C 0. Hence, U(r) G(Mm/r)*. Secondly, we can assign the potential energy to be equal to zero when the body is resting on the surface of the earth, i.e. U(R) 0; that is to measure the potential energy from ground level. Then the equation * can be rewritten as U ( R) G
Mm Mm C 0. Therefore, C G R R
In this case the potential energy of a body at height h above ground level will be: U (h ) G
Mm Mm GMmh G . Rh R R( R h )
At h^R we have U(h) ⬇ G(Mm/R2)h; since G(M/R2) g and U(h) mgh. This is a well-known formula describing the potential energy of a body lifted above the Earth’s surface to a moderate height (h^R) (a straight line U(h) in Figure 1.28).
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Generally, the gravitational field of the earth, as for any central field, is nonuniform (the force lines of the field are not parallel to each other). However, if we limit the volume of the field by a definite dimension h, we can always calculate an accuracy within which we can use either a precise eq. (1.3.19) or an approximate formula (1.4.37). We shall note that eq. (1.4.37) is only a special case, though very important in normal engineering practice. The precise formula is given in Example E1.16. In Figure 1.28 the graph of the function U(r) is presented: hyperbolic dependence U(r) ∼(−1/r) lies in the areas of negative values of potential energy. We can see that U(r) → 0 at r → ∞. If we take another border condition, U(R) 0, then the origin in the graph should be shifted by G(Mm/R) down the U-axis and by R along the r-axis. At (h^R) dependence U(h) is represented by a straight line. The place where the curve U(r) and linear line U(h) at small h practically coincide can be seen in Figure 1.28. In general, the gravitational field of the earth, as with any other central field, is nonuniform (force lines of a field are not parallel to each other). However, if a volume of the field is limited by a definite dimension, it is always possible to calculate to an accuracy within which either a precise or an approximate formula can be used. Figure 1.28 presents a graph of the function U(r): hyperbolic function U(r) ∼(−1/r) lies in the areas of negative values of potential energy. It can be seen that U(r) → 0 at r → ∞. If we accept that U(R) 0 then the origin in the graph should be shifted on G(Mm/R) down along the U-axis and on R along the r-axis. At h^R, dependence U(h) is represented by a straight line. It can be seen in Figure 1.28 that the curve U(r) and linear dependence U(h) at small h practically coincide. (All these aspects are analyzed in Examples E1.16 and E1.18 in detail.) Simple calculations show that to within 1% the equation U mgh can be used up to a height of 64 km (see Example E1.19). In fact, the gravitational field of the earth up to a certain accuracy can be counted as homogeneous. The zero position of the potential energy is not necessarily connecting with a level of the Earth’s surface; potential energy can be counted from any level near the earth.
U(r) 0
r
R U=mgh ~−
−
1 r
GMm R
Figure 1.28. Work of a gravitational force.
h
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EXAMPLE E1.19 Find in some respect a less-precise expression for the potential energy of a body in the Earth’s gravitational field at large distances. Evaluate the approximations. The concept of potential energy of the gravitational interaction permits some arbitrariness: everything depends on the choice of from which level to count this energy. There are different options, among which are exact and approximate ones. In the latter case it is always possible to estimate an error arising when using the given approximations. Estimate this error using the formula (1.4.37): in what limits of a body lifting will the error not exceed 1% in comparison to a general approach (1.4.28). Solution: The equation in Example E1.16 is definitely the most general. It has already been noticed that the most important thing is not the absolute value of the potential energy but the difference in their two states. There is one more approach to the gravitational potential energy evaluation U (h ) U ( R h ) U ( R). Then mM ⎛ mM ⎞ 1 ⎞ ⎛1 ⎜G ⎟ GmM ⎜⎝ ⎟ Rh ⎝ R ⎠ R Rh ⎠ R mM R . Gh mgh Rh R( R h ) R
U (h ) G
In a very good approximation we can consider the gravitational force on the earth level: mg G(mM/R2). Then U(h) mgh(R/(Rh)). This expression is precise enough and can be used sometimes at any h. In order to obtain an answer to our main question we should calculate the uncertainty ratio
U U (h )
R R h R h 1 h . R R R mgh Rh
mgh mgh
To find the uncertainty 1% error we equate the last prescribed accuracy with 0.01R: h 0.01 R 0.01 6.4 103 m 64 km. Therefore, we can use the mgh formula in the limit of 1% while climbing Everest, but not in planning the spacecraft flights. Note that the potential energy of a body (e.g., a satellite) is negative; it is because of the fact that we chose the value U 0 at infinite interval from the earth. The potential energy of a body lifted above the earth is always positive.
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Because in the framework of Galileo’s transforms the distance is invariant, the potential energy is also invariant (remember that kinetic energy is noninvariant in relation to Galileo’s transformations because of the dependence of a body’s velocity on the choice of the coordinate system). The work of the gravitational forces at a body displacement from some point 1 to a point 2 will generally be expressed as follows:
⎛ 1 1⎞ A12 U U1 U 2 GMm ⎜ ⎟ m(1 2 ), ⎝ r1 r2 ⎠ and in the case of a uniform gravitational field: A12 mg(h2 h1 ).
1.5
CONSERVATION LAWS IN MECHANICS
Conservation laws are the most general, fundamental laws of nature. They have an enormous scientific value. Their importance is defined by the fact that the solution to many kinds of problem can be achieved with their help and without detailed analysis of specific circumstances and details. 1.5.1
Conservation law of mechanical energy
Let 1 and 2 be the two positions of an MP in a potential-force field; U1 and U2 are their potential energies at these points. According to the theorem of kinetic energy change (refer to (1.4.3)) x 2 x 1
m y 22 m y12 A12 , 2 2
(1.5.1)
where A12 is the work of all forces, external and internal, conservative and dissipative applied to the MP on displacement from point 1 to point 2. If we restrict ourselves to conservative (potential) systems and remove the external forces (according to a criterion of the closed system), then according to (1.4.26) A12 U1 U 2 . Combining the two last equations, we obtain x 2 U 2 x 1U1 .
(1.5.2)
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Because two arbitrary positions were chosen, the general equation for the total mechanical energy can be written as: xU E const.,
(1.5.3)
which is valid for any body of the system and for the system as a whole. So, the total mechanical energy of the closed potential system is conserved. The given determination comprises certain conditions, the removal of any of which breaks the law of conservation. In fact, if one takes into consideration the dissipate forces (for instance, friction), part of the mechanical energy transforms into heat, and the law of conservation of mechanical energy becomes invalid. On the other hand, if one takes into consideration other forms of energy (heat, electric and others), the condition of system conservation becomes an excessive condition: the total energy of any closed systems is preserved. In differential form, the law of mechanical energy conservation can be denominated by equation, dUd£ 0 or dU d x.
(1.5.4)
This means that in the resting system the kinetic energy can appear only as a consequence of the reduction of potential energy. But if the potential energy in the given state has a minimal value, the motion simply cannot appear. Consequently, the system is in a state of equilibrium when the potential energy has a minimum value. As we know, the potential energy is a function of a body’s coordinate; the kinetic energy, however, depends on its velocity (or momentum). E (r, p ) U (r ) x (p ) U ( xyz ) x ( px , py , pz ) As was mentioned in Section 1.3.8, in order to specify a system state one has to know the coordinates and momentums of all points, i.e., the same parameters that define the energy. In this sense it can be said that the total mechanical energy is a function of a system’s state. When changing a system’s state its energy changes as well. The work in this case is presented as a measure of changing a system’s energy. This concerns the physical sense of work. The law of total mechanical energy conservation is invariant with respect to Galileo’s transforms. This does not mean, however, invariance of the total energy with respect to Galileo’s transforms literally, since the kinetic energy with respect to different reference systems has different values. So, the constant characterizing the total energy in each case can be different, though the principle in general is the same.
EXAMPLE E1.20 In a gravitational field of the earth, a body with a weight m moves from a point 1 to a point 2 (Figure E1.20). Define the speed of the body at point 2 if its speed at point 1 was 1 兹苶g苶 R 7.9 km/sec. Assume the acceleration of free fall at all points to be equal g.
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R 2R
3R
2
1
Solution: An earth–body system is conservative and closed. Therefore, we can use the energy conservation law. We connect the origin to the earth’s centre. We can write: E1–E2 i.e. U1 £ 1 U2 £ 2, where U and £ are the potential and kinetic energies at points 1 and 2. Because the reference frame is connected with the earth, the relative body energy does not depend on its movement. Then
x 1
m y12 my 2 Mm Mm ; U1 G ; x 2 2 ; U 2 G . 2 3R 2 2R
Substituting these expressions in the energy conservation law we obtain: m12 Mm m22 Mm G G . 2 3R 2 2R
Substituting GM by gR2 and executing elementary simplifications we obtain: 22 12
gR gR and then 2 12 . 3 3
Because 21 gR then 2 兹4苶g苶R 苶/3 苶 1兹4苶/3 苶. Executing calculations we arrive at 2 9.12 km/sec. 1.5.2
Momentum conservation law
Returning to expression (1.3.9), consider a case when there are no forces acting on an MP, i.e., when the right-hand part of the equation is zero. Then d(m) 0 and, hence, p m const.,
(1.5.5)
i.e., in the absence of external forces the momentum of an MP remains constant. We can apply the result obtained for a mechanical system. If the system is closed, external forces are absent, hence C const. (According to Newton’s third law, the mutual action of all bodies of a given system is counterbalanced and when calculating the total
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momentum of the system, the internal forces make no contribution to the momentum of the system.) At constant system mass the total system momentum mC pc also remains constant. p c const.,
(1.5.6)
That is, the total momentum of a closed system pc is conserved. Note that the law of total momentum is conserved in any closed system regardless of whether it is conservative and/or dissipative. Several consequences arising from the law of momentum conservation can be mentioned: 1. 2.
A reference system associated with the CM of a closed mechanical system is inertial. In fact, if pc const. the constant speed of the CM υC should be constant too. The center of inertia of the closed system is preserved as a resting state or a state of rectilinear uniform movement despite any mutual displacements (under the action of internal forces) of any of the system’s elements. From this it is clear that it is impossible to change the position of the mass’ center of the closed system by only internal forces.
Some special cases can be noted in this respect. A system is not closed, but the result of forces is zero. In this case the total momentum of the system is preserved. A system is not closed but one of the projections of external forces (e.g., FX) is zero. Then ∆pX 0 and px const. Hence, the law of conservation is valid only to the given motion direction. This case corresponds, for example, to the motion without friction in a field of a gravity along a horizontal axis with non-zero initial velocity (refer to Example E1.4).
EXAMPLE E1.21 A carriage L 3 m in length and mass m1 120 kg rests on a smooth horizontal surface. At one end of the carriage is a man whose weight is m2 80 kg. Define the displacement L (modulo and direction) if the man walks from one end of the carriage to another. Neglect friction. Solution: Choose an inertial reference system, having connected with the earth. Since projections of the external forces acting on the system “man-carriage” from the earth are perpendicular to the Earth’s surface this system can be counted as closed (regarding a horizontal axis); a momentum conservation law (see Section 1.5.2) can be used. In the initial state the man and the carriage are motionless and, hence, the system’s total momentum is zero. We shall assume that the man moves along the carriage uniformly at a speed u, thus the carriage also begins to move uniformly at a speed 2 relative to the earth. As the relative speed u 1 – 2 (where 1 is the speed of the man relative the earth), v1 u v2
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Then the momentum conservation law for the case considered can be written as m1 1 m2 2 0 or m1( 1 2) m 2 0, whence after disclosing brackets and regrouping terms we obtain 2(m1 m2) m1u 0 or v 2
m1 u. m1 m2
Let us multiply this equality on both sides by the time taken by the man to move from one side of the carriage to the other; we obtain
⎡ m1 ⎤ v2 ⎢ ⎥ u, ⎣ m1 m2 ⎦ where v2 L is the carriage displacement relative to the Earth’s surface and u is the man’s displacement relative to the carriage (i.e., L). Therefore, L
m1 L m1 m2
Introducing the values given above, we obtain the carriage displacement. L 2 m. The negative sign indicates that the man and carriage displacements are in opposite directions to each other. EXAMPLE E1.22 A nucleus decays into two fractions m1 1.6 10−25 kg and m2 2.4 10−25 kg. Determine the kinetic energy £ 2 of the second fragment if the kinetic energy of the first is £ 1 18 nJ. Solution: According to the momentum conservation law, the momentums of fragments after decay are the same p1 p2*. Express the momentums through their kinetic energies p mv, p2 m2 2, £ m 2/2 and, 2m£ m2 2. From these expressions find
苶苶 £ Substituting this equality into * we obtain 兹2苶m 苶苶 苶m 苶1苶 £ 2苶 whence p: p 兹2苶m 1£ 1苶 兹2 we can find £ 2:£ 2 (m1/m2) £ 1 1.2108 J 12 nJ. 1.5.3
Angular momentum conservation law
We will start from the general law of the dynamics of a rotational motion (1.3.58). In a closed system (at MF(F) 0) no change of angular momentum is observed, dL 0 and, hence, L const.,
(1.5.7)
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z
L=const
r F
m m
Figure 1.29. The movement of a body in a central field.
That is, in an absence of torques of external forces the angular momentum of a system remains constant. This statement concerns an MP, an MP system and an IRB. In other words, the angular momentum of a closed system is conserved. Apply this law to the analysis of the motion of an MP under the action of a central force (Figure 1.29). Let an MP of mass m be under the action of an external force so that in all its positions the line of force action passes through one point (through the center of a circle). Then MF(F) ≡ 0, accordingly (dL/dt) 0 and L const. It can be seen that if movement takes place under the action of the central force, vector L is fixed, therefore vectors p and L are fixed as well (as [r⋅p] L). It, in turn, fixes a plane, in which vectors r and p lie. Hence, under the action of the central force the MP (a body) moves along a flat trajectory (circular, elliptic or hyperbolic) so that [r⋅p] const. (Again it is appropriate to recollect the conversations of Jules Verne’s heroes in the projectile in which they tried to reach the moon.) Examples of such movement are the motion of the planets around the Sun (according Kepler’s laws) and the electron motion in atoms (within the framework of the Bohr model, refer to Chapter 6, Section 6.7). EXAMPLE E1.23 A person of mass m2 75 kg stands at the edge of a platform which is in the form of a homogeneous disc of a radius R and mass m1 200 kg. The platform can rotate freely around a vertical axis that passes through its center. Define the angle of turn of the platform if the person, moving along its edge, returns to the initial point of the platform. Solution: (Before solving this problem it is useful to remember the example on the law of momentum conservation, Example E1.21). An inertial reference system is useful to relate to the earth. Only gravitational forces and bearing reactions act on the system, all of them being parallel to the rotation axis; their torques being zero. It is therefore possible to take advantage of the angular
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momentum conservation law. Since, at the initial moment, the “person–platform” system is at rest, the total angular momentum is zero. We shall consider that the person starts to move uniformly on the platform. Since the total angular momentum should remain zero, the platform should move in the opposite direction with an angular speed axis 2. We shall denote the angular velocity of the person concerning the earth as 1. Then, according to the conservation law of angular momentum, in a projection to axis Oz: Iz11 Iz22 0*, where Iz,1 is the MI of the person relative to the Oz axis and Iz,2 is the MI of the platform relative to the same axis. Note that the angular velocity ωrel of the person relative to the platform is determined by the equality rel 1– 2. Wherefrom 1 rel 2. Then the star equation becomes I z,1 ( rel 2 ) I z,2 2 0 Having removed the brackets and rearranged the terms we obtain (Iz,1 Iz,2)2 –Iz,1rel. Let us multiply both parts of the equality by the time taken by the person to return to the initial point: ( I z,1 I z,2 )2 I z,1 rel Here 2 is the turning angle of the platform relative to the earth and rel 2 is the angle over which the person travels relative to the platform. If we consider the person as an MP, then Iz,1 m1R2. The MI of a disc can be calculated according to eqn (1.3.47): Iz,2 (1/2) m2R2. Having substituted these values in the formula obtained and made simple transformations, we arrive at –4 [m1 /(m2 2m1)] Substitution of numerical values gives the final result:
6 2.69 rad 154 7
EXAMPLE E1.24 A disc-shaped wheel of mass m 50 kg and radius r 20 cm is twisted to promptness n1 480 min1 and released to rotate freely. It then stops because of friction. Find the torques for the following two cases: (1) the wheel stops in t 50 sec; (2) the wheel makes N 200 revolutions before stopping. Solution: (1) According to Newton’s second law (applied to rotation motion) we can write M t I2 I1 ,
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where I is the wheel’s MI and is the angular speed. Since 2 0 and t is t then Mt I1 and therefore M (I1/t)*. The disk’s MI is I (mr 2/2). Substituting this equation into * we obtain M mr 2 1 2t. Since 2n we obtain M –1 Nm. (2) We can find the torques from its work A (J21/2) M (see Section 1.4.1). Then
M
mr 2 12 . 4
The total number of revolutions until disc shutdown (i.e., its total angular displacement) can be found: 2πN 1256 pad. Therefore, we arrive at the same torque M 1 Nm. The sign shows that the torque has damped the movement. 1.5.4
Potential curves
We will consider two problems in the framework of the conservation laws: the potential curves principle and the theory of collisions in areas of science close to chemistry. One of the main problems in chemistry is the investigation of interactions between particles. One of the most accepted methods for describing such interaction is the language of potential curves. Potential curve U(r) is a graphic picture of the potential energy between interacting particles as dependent on the interparticle distance. If the origin is kept in one of the particles (for instance, in particle number 1), then r is the distance between particles 1 and 2. As was mentioned earlier, when describing the interaction due to the force field, the potential energy is usually taken to be zero when particles are at infinite distance (i.e., at r -). In spite of the fact that curve U(r) reflects only the potential energy change, with its imaging we can also find the value of the kinetic energy in each interparticle point (at given total energy E). The potential energy of the gravitational interaction Let two MPs having mass m1 and m2 be at a distance r from each other; then the potential energy of their interaction, as was shown above, (U/(r) G(m1m2/r)), can have a hyperbolic form. In the case of attraction (Figure 1.30a) the potential energy increases as far as the points move further from each other. At infinity, the potential energy of the interaction reaches a maximum. On the accepted condition this maximum will correspond to zero potential energy. If the maximum is zero, all the potential energy values, in the case of attraction force on finite distances, will be negative.
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U(r)
U(r)
U(r) > 0
r
r
U(r) < 0
(b)
(a)
Figure 1.30. The potential curves for forces of (a) attraction and (b) repulsion.
Potential energy of electrostatic interaction It describes the interaction between two resting charges depending on the intercharge distance r; it is expressed by the formula U (r )
q1q2 , 4 0 r
where q1 and q2 are values of point charges, 0 is the electrical constant (refer to Section 4.1). The sign of potential energy is taken into account “automatically” due to signs of interacting charges. In the case of different charges, the potential energy of interaction (attractions) is negative. The form of the graph is qualitatively just the same as in the gravitational interaction provided the charges are of different sign. In the case of similar charge signs, the potential energy is positive going to zero in infinity (Figure 1.30b). Potential energy of elastic interaction In Section 1.4.5 the particularities of elastic interaction energy have been considered. Note that elastic forces manifest themselves equally successfully both in the macro- and the micro-world. When deforming a body or “stretching” an intermolecular bond, an external force produces work. The magnitude of this work is equal to the change of the potential energy of the elastic interaction. As mentioned in Section 1.4.5, the potential energy in this case is U (1/2)x2, where x (in this instance) is the deformation magnitude. The graph of this function is described by a parabolic curve, symmetrical relative to the U-axis (Figure 1.31). Since the elastic forces are conservative, the total energy E is constant (according to the laws of conservation of mechanical energy). This is depicted in the graph
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U
E=Umax
E=Umax K=0
K=0
K(a) E
Kmax
U(a)
−x0
a
F
0
F
x0
x
Figure 1.31. Potential curve for a harmonic oscillator.
by the straight line E const. In the arbitrary point x a the potential and kinetic energy values are expressed by lengths of segments U(a) and £ (a). The elastic force is always directed to the origin. Approaching the origin, the particle’s potential energy U decreases. Conversely, at the origin (x 0) kinetic energy reaches its maximum (£ max E, since U(0) 0). Continuing motion, a particle loses its kinetic energy until it vanishes to zero. At this point, potential energy reaches its maximum and is equal to the total energy, 1 U max E x02 , 2 where x0 is a maximum displacement. Thereby, the maximum point displacement x0 兹苶2苶 E苶 / is defined by its total energy. The particle performs an oscillation relative to the origin (Chapter 2). More complex type of interaction is the interplay of molecules Without looking deeply into the nature of these interactions, we can say that both attraction and repulsion forces act simultaneously between molecules. Because attraction forces decrease with distance slower than repulsion forces, attraction forces dominate at longer distances and at shorter distances repulsion forces dominate. One of the popular approximations is a power dependency of potential energy upon distance r in the form
U (r )
a b 12 . 6 r r
(1.5.8)
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U(r) U = − a6 + b r r12
Urep = b12 r r
0 Uattr = − a6 r
Figure 1.32. “6–12” (Lennard–Jones) potential.
This is the so-called Lennard–Jones potential “6–12,” offered for the interaction description of nonpolar molecules. Values a and b for different molecules are different. The first term expresses the potential energy of attraction whereas the second term expresses the potential energy of repulsion. In Figure 1.32 both curves are represented by dotted lines and the solid line is a resulting curve (eq. (1.5.8)). The formula of the Lennard–Jones potential can be written in another form as ⎡⎛ ⎞ 12 ⎛ ⎞ 6 ⎤ U (r ) 4 ⎢⎜ ⎟ ⎜ ⎟ ⎥ , ⎝r⎠ ⎥ ⎢⎣⎝ r ⎠ ⎦ where and are again different for different molecules. Let us look at the change of the force acting on particle 2 as it approaches the origin, where particle 1 is resting. For this purpose it is effective to use the connection of the central force and the potential energy in the form (1.4.30) and carry out a graphic derivation (Figure 1.33). It can be seen that at great distances the derivative is small (the magnitude of angle of the slope of the tangent line £ £ 1 to abscissa is small) and is positive (angle is sharp). It means that the force F acting on particle 2 is negative. This signifies that force F is directed to the negative direction of the r axis, i.e., to the origin. As particle 2 approaches particle 1 the derivative increases and in the inflexion point a reaches to its maximum (angle of the tangent slope has at this point its highest value). At this point the force F is minimum (the maximum value of attraction force). Then the angle begins to decrease and at point r0 vanishes to zero (the tangent line to the curve U(r) is parallel to the abscissa). At this point
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U(r)
F 1
αs
F r0
2
dma
K1 α
r
K b
a
F(r) REPULSION 1 (F>0)
F=− 2
dU dr r
ATTRACTION (F<0)
Figure 1.33. Potential energy and force as a function of interparticle distance.
the attraction and repulsion forces are in balance. Distance r0 is referred to as the equilibrium distance. At distances r r0 an angle is obtuse (for instance, at point b); in this area the tangent changes its sign (dU/dr 0). This signifies that the force F changes its direction, becoming repulsive (F 0) and then increasing rapidly as particle 2 approaches the origin. A graph of the dependence F(r) is given in the lower part of Figure 1.33. Considering curve U(r) (Figures 1.30, 1.32 and 1.33) one can be assured that only by being at a point with the minimum potential energy, a particle does not feel any force action at all. When the particle displaces to the right or to the left from the position of equilibrium, in which the potential energy is minimal, the force appears, directed to the position of balance. Therefore, it is seen that the particle will always tend to occupy a position with minimum potential energy (however the force of inertia does not allow it, assuming that the dissipation of energy is absent). Let us now trace the change of kinetic and potential energy, provided the total energy is known, when molecules approach each other. Let, at infinite distance, the molecules have a total energy E 0. As we have already agreed, the potential energy of infinitely separated molecules is zero. Consequently, kinetic energy is equal to the total energy E (Figure 1.33). When molecules are approaching, their kinetic energy increases and at the point of balance (r0, refer to Figure 1.34) reaches a maximum. While approaching further (r r0), the kinetic energy of interaction decreases rapidly until it equals zero. According to the conservation law of mechanical energy, a total energy at this point is admitted a maximum potential energy E Umax. The closest approach , to which under a given total energy molecules can be reached, is referred to as an efficient molecule diameter. After an instant stop, all events will run in the inverse sequence and the molecules will return to the initial position (to infinity), possessing kinetic energy, which is the same as before the rapprochement. In this case, a moving particle is unlimited in space. Such spatially unrestricted motion is referred to as an infinite one. Particles whose total energy turns out to be negative behave otherwise (U 0, and ⱍUⱍ£ , then E 0). In this case, a particle turns out to be “locked” in the potential well
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U(r)
E
E 0
K r
U<0
Figure 1.34. A particle with energy over a potential well (E 0).
(Figure 1.34). Particles are not able to approach more than to distance rmin, and cannot retreat more than distance rmax. Such spatially restricted motion is referred to as a finite one. The particle will perform an oscillatory motion. The two-particle state is a bonded one (i.e., particles are not able to separate) (Figure 1.35). From this qualitative analysis one can draw an important conclusion. If the total energy of the two-particle system is positive (E 0), no bound state of particles occurs. If the total energy of the particles is negative (E 0), the bound state is possible. We will meet examples of this in the book.
1.5.5
Particle collisions
In this section we will apply the laws of mechanical energy and the momentum conservation principle to processes of particle collisions. By collision, we mean any short interaction between particles. Unlike the collision of macroscopic bodies, where the direct contact of bodies and their deformation on impact can be observed (e.g., the action of a boot-kicking a ball can be captured by a rapid photograph), interaction of atomic particles is realized by means of force fields and cannot be directly observed. In this respect, the collision of electrical charges, which interact through their electric fields or the collisions of neutrons (where the interaction takes place via nuclear field), etc. can be mentioned. In principle, within the framework of classical macroscopic mechanics one could consider the detailed process of collision and draw a conclusion about a body’s deformation and interaction forces (their changes during the process of collision), taking into account the results of collisions, analyzing direction and velocities of collided particles during and after the collision. However, such detailed consideration even for classical objects is very difficult, but for quantum objects it is absolutely impossible. So, for instance, it is impossible to
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U(r)
rmin
rmax
r
r
0 U<0
E<0
E=U; K=0
E=U; K=0 K>0
Figure 1.35. A particle in a potential well (E 0).
describe in detail the process of the collision of neutrons because no one knows the exact laws of the nuclear forces’ properties. Nothing can be said about the mechanism of the collision of neutrons with phonons (refer to Chapter 9), -quantum with an electron, etc. Moreover, a detailed description of the process of collision of microparticles in general is impossible because of the uncertainties principles (refer to Section 7.2). Nevertheless much of our information about atomic and molecular particles is obtained experimentally by observing the effects of collisions between them. For instance, it was just such collision experiments that allowed Rutherford to suggest a planetary model of atoms. Fortunately, it appears that in many cases such detailed consideration of interactions is not needed. For the determination of velocities of particles after the collision it is sufficient to know an initial state and use the conservation laws. Collisions are not limited to cases in which two bodies come into contact in the usual sense. However, we will restrict ourselves to only pair collisions. Collisions of particles can be of two types: elastic and inelastic. Elastic particles collisions A collision is called elastic if the total particle kinetic energy is preserved. Only the kinetic energy repartition between collided particles takes place, whereas the inner-particle state remains unchanged. Consider in general form an elastic collision of two particles noninteracting at distance (i.e., their potential energy before and after collision is zero, U 0). The collision is called a head-on impact (or frontal one), if vectors of their velocities (of the CMs of collided bodies) before the impact are directed along one and the same direction (say, axis x). If this condition is not valid, the collision is called a glancing one.
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m2 m1
1
2 x
Figure 1.36. A head-on collision.
Consider a frontal collision of particles moving translationally (without rotation) (Figure 1.36) along the x-axis. Momentums of particles are directed along this line, and therefore we can change the vector writings to a scalar form (in projections on the axis x) (the sign x for simplicity is deleted). The velocities of the particles after collision are denoted by u. Consider a system of two colliding particles being a closed and a conservative one; it is possible to apply to collision both conservation laws: the energy and momentum conservation laws in mechanics. The energy conservation law looks as follows: m112 m222 m1u12 m2 u22 , 2 2 2 2
(1.5.9)
and the momentum conservation law m11 m22 m1u1 m2 u2 .
(1.5.10)
Expressions (1.5.9) and (1.5.10) can be considered as a system of two equations with two unknowns u1 and u2. To solve this system one can transform it to the form: m1 (12 u12 ) m2 (22 u22 ) and m1 (1 u1 ) m2 (2 u2 ), and then divide the upper equation with the lower one. The result is: 1 u1 2 u2 . The first equation divided by m1 gives 1 u1
m2 m u2 2 2 . m1 m1
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We arrive at the new system consisting of two linear equations with two unknowns. In order to find u2 one can sum up these two equations:
⎛ ⎛ m ⎞ m ⎞ 21 u2 ⎜ 1 2 ⎟ 2 ⎜ 1 2 ⎟ , m1 ⎠ m1 ⎠ ⎝ ⎝ from which it follows: ⎛ m ⎞ 21 2 ⎜ 1 2 ⎟ m1 ⎠ ⎝ . u2 m 1 2 m1
(1.5.11)
The velocity of the first particle can be found analogously: ⎛ m ⎞ 22 2 ⎜ 1 2 ⎟ m1 ⎠ ⎝ u1 . m2 1 m1
(1.5.12)
(This equation can be obtained from (1.5.11) by changing all indexes 1 ↔ 2 keeping the mathematical symmetry in mind.) With a little effort, one can obtain an analogous expression in the form: u1 1 2
m11 m22 m1 m2
(1.5.13)
u2 2 2
m11 m22 m1 m2
(1.5.14)
and
The fractions in (1.5.13) and (1.5.14) present the velocity of the two particles’ mass center motion. Then u1 1 2Vc and u2 2 2Vc ,
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where Vc
m11 m22 , m1 m2
underlining once more that 1 and 2 are the velocities of each particle and Vc is the CM velocity. (Knowing m1, m2, 1, 2, i.e., all the initial data, it is easy to find Vc; refer to eq. (1.3.32) and Figure 1.13). If one transfers now according to the Galileo principle to the coordinate system connected to the mass center (i.e., Vc 0), the equations can be simplified significantly: u1 1 , u2 2
(1.5.15)
After collision in this coordinate system, the colliding particles change their direction of motion in such a way that the absolute magnitudes remain unchanged. Some particular cases are of general interests and importance. 1. If the second particle remains at rest in the laboratory system (2 0). Then ⎛ m ⎞ 1 ⎜ 1 2 ⎟ m1 ⎠ ⎝ m m1 u1 2 1 m m2 m1 1 2 m1
(1.5.16)
and u2
2 m1 1 . m1 m2
(1.5.17)
It can be seen from eq. (1.5.17) that, after the collision, the initially resting particle 2 acquire the velocity u2, the direction of which always coincides with the initial velocity of the first particle before collision. However, the velocity direction of the first particle after collision depends upon the mass ratio of both particles. If m1 m2 the first particle changes the direction of flight to the opposite one (as can be seen from eq. (1.5.16)). If m1 m2, the direction of motion of both particles after collision is the same. If the masses of both particles are the same (m2/m1 1), then u1
22 (11) 2 (11) 2 , u2 1 1 , 11 11
i.e., in the case of the central elastic collision of particles of equal mass, a simple exchange of velocities and, consequently, kinetic energies occurs. The following takes place: the moving particle stops (after the collision u1 2 0); however, the resting particle begins
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to move with a velocity u2 1, i.e., at the velocity of the first particle. In this case, the moving particle will completely transfer all its kinetic energy to the resting particle. 2. Consider now the collision of particles that differ noticeably in their mass, say m1 / m2 ^ 1. Let the second particle with the larger mass be at rest (2 0). One can neglect the fraction m1/m2 in eq. (1.5.16) in comparison with 1. One obtains u1 ⬇1 and u2
21 ⬇ 0. m 1 2 m1
Thus, the particle with the smaller mass after collision changes its velocity to the opposite without loss of kinetic energy. This effect appears to be very important in chemical kinetics. For example, in exothermic reactions with the production of atomic hydrogen as an intermediate product (which is very light in comparison with another reagents), its kinetic energy can significantly exceed the quasi-equilibrium one. This phenomenon in kinetics is called the “effect of hot atoms”. The slow energy transfer from the small particle to the large one is also determined in the relaxation processes in lasers with compounds of halogens with hydrogen as a working medium.
p p
p' ∆p
p
p' ∆p
p x
Figure 1.37. An impact of a particle onto a wall.
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The result obtained is fulfilled with even greater accuracy when a particle collides with a wall (for instance, when a molecule collides with a vessel wall, Figure 1.37). In the latter case a molecule moving perpendicularly to the wall reflects from it and proceeds with the previous velocity backwards u1 1 (Figure 1.37). Determine the momentum change ∆p in this case. According to the definition p p 1p1, where p1 m11 and p 1 m1u1, then p m1(u1 1). Since u1 1, then p 2m11 or p 2 p1 .
(1.5.18)
The negative sign shows the direction of momentum changing, which complies with the direction of “elastic force” acting on the particle from the wall during the collision. If the particle falls to the wall under an angle to a normal, the component of velocity parallel to the surface of the wall remains unchanged, and only the component normal to the wall (along the x axis) takes part in the momentum transfer to the wall (i.e., ∆p –2p1cosα) (see Figure 1.37). As it follows from the results obtained, the particle does not change its kinetic energy on collision with the wall, but the wall receives from the particle a momentum equal to 2p1. (Herewith the wall “velocity” is zero; however the value of transferred momentum is not zero). One can calculate (this we will leave to the reader), that on direct central collision with a resting particle of mass m2 the colliding particle of mass m1 will transfer part of its kinetic energy: K 2 冷 K1 冨
4 m1m2 K1 . (m1 m2 )2
Then the relative loss of the particle kinetic energy will be: 冷 K1 冨 4 m1m2 . K1 (m1 m2 )2
(1.5.19)
It is easy to see that at the equality of mass (m1 m2) the first particle loses all its energy, i.e., K1/K1 1, and stops. On collision with a wall m1/m2 → 0 and then K1/K1→0, its kinetic energy remains unchanged. The process of deceleration of neutrons in atomic reactors is based on the phenomena of a “fast” particle transferring energy to a resting particle. This is because in the elementary act of chain nuclear reactions only fast neutrons are produced. For the realization of the next act of a chain reaction—capture of a neutron by a nucleus of a uranium-235 atom—it is necessary to slow neutrons down until their energy becomes commensurate with the energy of the thermal motion of the molecules. This occurs when neutrons collide with the atomic nuclei of the moderator material. Judging by the formulas given, the best
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moderator is a hydrogen-containing material: during a single front collision with protons, the neutrons immediately lose the whole of their kinetic energy (u1 0). However, protons easily participate in the reaction with neutrons; protons capture a neutron (with a deuteron formation) and remove neutrons from the chain reaction process. Therefore, the heavy water D2O is more often used (then K1/K1 0.9) or graphite (K1/K1 0.28). Note also, that in the use of tungsten nuclei this ratio is 0.02, the impact of the neutron to the tungsten nucleus is closer to the case of the collision with a wall, with a negligible loss of kinetic energy.
EXAMPLE E1.25 How many times k will the neutron kinetic energy decrease after N consecutive collisions with atomic nuclei which practically do not capture neutrons. Consider the collisions to be elastic and central with atoms: deuterium 2H, carbon 13C and tungsten 84W. Let N be 3. Solution: The problem with the single collision of a moving and motionless particle is solved in the text where formulas for the speed of each of the particles after collision (see formulas (1.5.16) and (1.5.17)) are given. Using these results it is possible to obtain expressions for neutron kinetic energy after collision £ 1 in relation to its initial energy £ 1: 2
x 1
2
m1u12 m1 ⎛ m1 m2 ⎞ 2 ⎛ m1 m2 ⎞ ⎜ v1 ⎜ x1 2 2 ⎝ m1 m2 ⎟⎠ ⎝ m1 m2 ⎟⎠
After the second collision the kinetic energy £ 1 should be treated as initial. Then the neutron kinetic energy £ 1(2) is 2
2
⎛ m m2 ⎞ ⎛ m1 m2 ⎞ ⎛ m m2 ⎞ x 1(2) ⎜ 1 x 1 ⎜ 1 ⎟ ⎜ ⎟ ⎝ m1 m2 ⎠ ⎝ m1 m2 ⎠ ⎝ m1 m2 ⎟⎠
22
x1
It is easy to guess that after N collisions x 1(N ) (m1 m2 ) (m1 m2 )2 N x 1 Then the ratio after N collisions is k £ 1/£
(N) 1
(m1m2 )/(m1m2)2N.
EXAMPLE E1.26 An argon atom collides with one N atom of a resting molecule N2 perpendicular to the N–N bond (Figure E1.26). The velocity of the argon atom velocity is 1 400 m/sec. The impact is elastic. Atoms can be represented as MP and the N2 molecule considered as a rigid rotator. Relative masses are ArA,r 40 and Ar,N 14, the interatomic distance N–N is 0.109 nm. For the nitrogen molecule after collision, determine: (1) the
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velocity Vc of its mass center; (2) the angular momentum Lz acquired by N2 molecule relative to the z-axis (perpendicular to the drawing plane and passing through its CM, point C in the figure); (3) the angular velocity of the molecule rotation relative to the z-axis (Figure E1.26).
Ar
N
C
N
Solution: Assume that collision of Ar and N(1) atoms is head-on and elastic. Using the formulas of Section 1.5.5 (eq. (1.5.16)) we obtain u1
m1 m2 1 . m1 m2
To determine the CM’s velocity Vc we can use only the momentum conservation law because part of the kinetic energy will go to the kinetic energy of the molecule rotation. Therefore, this part of the interaction should be considered as inelastic and the energy of rotation as internal. Then m11 2 m2VC m1u1 , therefore VC
m1 (1 u1 ) 2 m2
.
Substituting the velocity u1 in this expression we obtain
VC
m m2 ⎞ m1 m ⎛ 1 1 1 ⎟ 1 . ⎜ 2m ⎝ m1 m2 ⎠ m1 m2
Executing calculations we obtain VC (40/(40 14)) 400 296 m/sec. (2) Since there are no external force actions for angular momentum calculations we can use the angular momentum conservation law relative to the z-axis and passing the CM: Lz,1 L z,1 L z,C , where Lz,1 m11(d/2) is the angular momentum of the argon atom before the impact,
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L z,1 m1u1(d/2) after the impact and L z,C is the angular momentum of the nitrogen molecule after the collision relative to the Oz axis, wherefrom
⎛ m m2 ⎞ d mm d L z,C m1 (1 u1 ) m1 ⎜ 1 1 1 1 2 1d 2 m1 m2 ⎟⎠ 2 m1 m2 ⎝ Executing calculations we arrive at L z,C
40 14 1.66 1027 400 1.09 1010 7.511034 kg m 2 ssec 40 14
(3) We can find the angular velocity of the molecule rotation knowing the angular momentum value after impact; writing the angular momentum according to eq. (1.3.44): L z,C I z z d 2 z , where is a reduced molecular mass (eq. (1.3.50)) (in our case m1 m2 and m2/2). Then
2 Lz,C m2 d 2
mm 2 m1 2 1 2 1d or 1 m1 m2 d m2 d 2 m1 m2
Executing calculation we arrive at
2 40 400 5.44 1012 rad sec 40 14 1.09 1010
Inelastic collision of particles As indicated above, the collision of particles is called inelastic if the total kinetic energy of the particles before the collision is not equal to the total kinetic energy after the collision. The kinetic energy partly or totally transforms to the internal energy (i.e., particles change their energy state, for instance, temperature). A limiting case of inelastic collision is absolute inelastic collision, under which a maximum loss of kinetic energy occurs. After the collision, both particles do not move away, but move together as a single particle. Under inelastic collision, the law of mechanical energy conservation is not executed, since the system is a dissipative one (mechanical energy transforms into another type of energy). The general law of total energy conservation is certainly executed, but we are unable to use it, since we do not know a priori, what part of the kinetic energy transforms into internal energy. It is therefore possible to use only the law of momentum conservation: p1 p 2 p,
(1.5.20)
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where p is the total momentum of particles after an absolute inelastic collision. Under frontal inelastic collision all velocity vectors are directed along a single line, passing through the center of the particle masses. In projections onto axis x, complying with the direction of motion, the law of momentum conservation can be written as (sign x is omitted) m1u1 m2 u2 (m1 m2 )u, where u is the velocity of particles after the collision. From this expression, we can find the velocity of the particles after collision
u
m11 m22 . m1 m2
(1.5.21)
Now it is possible to define that part of the total kinetic energy, which at collision transforms into the internal energy of the particles. If we denote U as the internal energy of particles, then U £ 1 £ 2 £ , where £ is the kinetic energy of both particles after collision. In the comprehensive form the above formula can be rewritten as
U
m112 m222 (m1 m2 )u2 . 2 2 2
(1.5.22)
Formula (1.5.22) can be written in another form: 1 U (1 2 )2 , 2
(1.5.23)
where is the reduced mass of the system of two particles:
m1m2 . m1 m2
(1.5.24)
We can see that if particles of similar masses and similar velocities move to meet each other, their velocity after absolutely inelastic collision is zero, and, consequently, the whole kinetic energy transforms into internal energy. Chemistry students should note that the above consideration of collision processes cannot be directly extended to the case of atomic–molecular collisions even in classical physics. This is, in particular, because in a real atomic–molecular system a process of interaction significantly depends on the so-called adiabatic factor, i.e., on the correlation of the collision time and the intrinsic molecular frequencies.
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EINSTEIN’S SPECIAL RELATIVISTIC THEORY (STR) (SHORT REVIEW)
Until now, our discussions have been based on the classical Galileo–Newton representations. Moving on to quantum-optical phenomena and others, these ideas appear to be insufficient and it is necessary to consider some more general notions, in particular, Einstein’s special theory of the relativity (STR, 1905). There are a number of theories on the relationships between descriptions of the same physical phenomena in various systems moving relative to each other. They are referred to as theories of relativity (e.g., Galileo’s principle, refer to Section 1.3.2). Another more general theory is Einstein’s STR. This theory is based on a reliably established experimental fact: the speed of light propagation is independent of the speed of it source. This fact has been proved in the well-known Mickelson–Morley experiments: they had coordinated the parts of their large-scale devices on the earth that could determine the speed of light sent along the trajectory of the earth’s movement around the sun and opposite to it with very high accuracy. Experiment has shown that this speed does not change whether the movements are directed in parallel or oppositely. The speed of light (i.e., electromagnetic waves) appeared to be independent of the speed of a source. This greatly contradicted Galileo’s principle of speed additions (eq. (1.3.4)). Moreover, this result appeared contradictory to the theory of Maxwell electrodynamics. In order to preserve the principles mentioned above, it was necessary to proceed further than Galileo’s transformations (1.3.1) and (1.3.2). These transformations have been replaced by the mathematical Lorentz equations already known in physics. Lorentz transformations replace Galileo’s transforms. When a coordinate system £ (x , y , z , t ) moves relative to the system £ (x, y, z, t) in x direction (this restriction is made for simplicity) with a speed u (ux u) they look as follows: x
x ut
x
x ut
1 1 2 y y, y y , z z, z z , ux ux t 2 t 2 c c t t 2 1 1 2 2
(1.6.1)
where u/c; an expression is often used 1/ 兹1苶 苶 苶2苶. These equations were derived by Lorenz in 1904 as mathematical expressions of coordinates and time transformation as an attempt to preserve the system of Maxwell’s theory in all inertial coordinate systems. The same transformations were obtained by Einstein in 1905; he proceeded from Newton’s postulate on the equality of all inertial coordinate systems and experimental fact on the independence of light speed upon the source speed. Analysis of eq. (1.6.1) shows that time is incorporated in space and movement. Thus, Einstein considered it necessary to change Newton’s representation about time and space (refer to Section 1.3.1). Moreover, in order to make Newton’s transformations be invariant
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to Lorentz’s equations it was necessary to accept that the mass of a particle is speeddependent, being described by the expression m
m0 1 2
,
(1.6.2)
where m0 is the particle mass at rest (at 0). Some conclusions follow from the STR: 1. Relativistic speed’s summation law. If an MP m moves with a speed ′ ( x , 0, 0) in an inertial system of coordinates K which, in turn, moves relative to another inertial coordinate system K with a speed u (ux, 0, 0), then the speed (x, 0, 0) of that MP in a system K (according to Lorentz’s transformation), can be presented as:
u
. u
1 2 c
(1.6.3)
The formula (1.6.3) refers to relativistic law of speed’s summation. Obviously, the resulting speed is less than the sum of the two speeds u and ′. From (1.6.2) and (1.6.3) it follows that: firstly, the speed of light is the same in any inertial system as at c for any u c, and, secondly, the body with non-zero m0 cannot reach in any inertial system the speed of light in vacuum or exceed it, as ′ c and u c it follows that c. In the theory of relativity the expressions ct and ut have the dimension of length; it behaves as the fourth spatial coordinate. From (1.6.3) it follows that values ct and x can mix depending on the speed of the observer. 2. Relativistic shortening of length. Let a rod of length L lie along the x axis and move with a speed u in an inertial system K′; its length L′ as measured from the resting system K will be shorter than the length L: L L′ L 1 2 L .
(1.6.4)
Various observers (being in different inertial systems) consider the same rod to have a different length. From a physical point of view this discrepancy can be explained by the concepts of simultaneity, i.e., events for one observer are not simultaneous with those for another.
EXAMPLE E1.27 Determine a relativistic electron’s momentum p and its kinetic energy £ if the electron moves at a speed 0.9c.
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Solution: The relativistic momentum is p m0c (/ 兹1苶 苶 苶2苶). Executing all calcu−22 lations we arrive at p 5.6 10 kg m/sec. The kinetic energy in the relativistic mechanics is the difference between total energy E mc2 and energy E0 m0c2 at rest. Therefore, K
m0 c 2 1 2
m0 c 2 ,
or finally ⎛ ⎞ 1 K m0 c 2 ⎜ 1⎟ . ⎜⎝ 1 2 ⎟⎠
Executing calculations we arrive at £ 1.06 × 1013 J 0.66 MeV
EXAMPLE E1.28 A spacecraft moves with a velocity 0.9c to the center of the earth. What distance l does it cover in the earth reference framework K in a time interval ∆t0 1 sec, measured by the spacecraft clock (system K′). Ignore diurnal rotation and the earth’s movement around the Sun. Solution: We can determine the distance l that the spacecraft can cover in the K′ system according to the formula l t, where t is the time interval measured in the K system. This interval in the K system and that measured in the K′ system are related by the formula t
t0 1 2
.
We should substitute this value into the previous relation:
l
t0 1 2
.
Executing all calculations we arrive at l 6.19 × 108 m. A body such as a physical pendulum consists of a rod of l 1 m in length and of mass m1 1 kg. A disc of mass m2 0.5m1 is fixed on one side of the rod. Find the MI
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l/3
of such a body relative to an axis passed perpendicularly to the picture through a point O (Figure E1.29).
l
a1
O
l/4
a2
l/2
C1
R C2
Solution: The MI of the combined body can be calculated according to the formulas presented in Section 1.3. Choose an x-axis directed along the rod with its origin at point O. The overall MI of the whole body relative to the x-axis is the sum of the composite details: Iz Iz1 Iz2, where part 1 is the rod and 2 is the disc. In order to find Iz1 and Iz2 we should use the theorem on parallel axis (1.3.48). The rod’s MI can be given by the expression I z1
m1ᐉ2 m1a12 12
(where a1 OC1 ).
We can see in Figure E1.29 that a1 l/6. Therefore, I z1
m1ᐉ2 m ᐉ2 m1 (l 6)a 2 1 0.111 m1l 2 . 12 9
The disc’s MI is Iz2 (m2R2/2)m2a 22, where R l/4 is the disc radius and m2a 22 is the addition of parallel axis transfer. A distance OC2 a2 is equal to l(2/31/4) l(11/12). Therefore, 2
J z2
2
m2 ᐉ2 ⎛ 1 ⎞ 2 ⎛ 11 ⎞ 2 2 ⎜ ⎟ m2 ᐉ ⎜⎝ ⎟⎠ m2 l (0.0312 0.840) 0.871 m2 l . 2 ⎝ 16 ⎠ 12
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Summing up the results for two parts, we arrive at Jz 0.111m1l 2 0.871m2l 2 (0.111m10.871m2)l 2 0.547m1l 2 0.547 kgm2. (We used here the condition that m2 0.5m1). EXAMPLE E1.29 How many times does the density of a rod in the laboratory reference system (system K) change if its speed relative to this system equals 0.8c (c – speed of light in vacuum). Solution: It is clear that
m0 m m 1 1 0 2 2 ᐉS ᐉ S 2 1 0 ᐉ 0 1 S 1
Therefore 1 . 0 1 2 Executing calculations we arrive at 1 1 1 2.78 2 2 0 1 0.36 1 0.8 3. Dilation of time. Let the inertial system of coordinates K move regarding another inertial coordinate system K with a speed u(u,0,0). If in a moving system K at the origin two events A and B with a time interval t′ take place the observer in motionless system K will find that the time interval between these events t is shorter than t′: t
t
1 2
(1.6.5)
That is, from the point of view of the motionless observer time in the moving system flows slower. For example, for u 0.5c the interval t′ 1 sec will correspond to an interval t 1.15 sec. As time dilation in moving systems is a property of time, not only the moving watch but all physical processes (including the ratio of chemical reactions) is taking place as well. This means that the ageing of organisms also slows down. However, the real speed of a spacecraft is still much less than the speed of light and the effect of dilation on the ageing of an astronaut is very small. 4. A relativity of simultaneity. If two events A and B occur in moving inertial system K′ at different points of space at the same instant of time t A t B, for example, in points with
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x′ 0.5a and x −0.5a, the same events are not simultaneous for the observer in a motionless inertial system K: t t
1 ua 2 c 1 2
(1.6.6)
The effect is very small: at a 1000 km calculation gives ∆t 0.023 sec. In SRT the concept of simultaneity is meaningful for events in one and the same coordinate system only. 5. Relativistic speed and acceleration. According to Newton’s physics, the constant force F acting on some constant mass m accelerates it. If t 0, 0 0, m m0, one can find speed (t) and acceleration a(t) of a mass m. The expression for F according to Newton’s second law has in SRT the form:
F
m0 a d p d ⎡ m0 ⎤ ⎥ ⎢ dt dt ⎢ 1 2 ⎥ (1 2 ) 3 2 ⎣ ⎦
(1.6.7)
It can be seen from eq. (1.6.7) that it is impossible to accelerate a body with a non-zero resting mass by a finite force F to a speed equal to the light speed in vacuum. In order to explain this circumstance Einstein had to introduce the speed dependence of the particle mass (1.6.2). From the expression (1.6.7) we can obtain: (t )
Ft m0
1 ⎛ Ft ⎞ 1 ⎜ ⎟ ⎝ m0 c ⎠
2
, a (t ) =
F m0
1 ⎡ ⎛ Ft ⎞ 2 ⎤ ⎢1 ⎜ ⎟ ⎥ ⎢⎣ ⎝ mo c ⎠ ⎥⎦
3
. (1.6.8)
In the motion beginning for small time intervals the speed is still small and corresponds to classic mechanics: (t) Ft/m0, a(t) F/m0. If the time increases (t → ∞), at constant acting force the speed asymptotically approaches the light speed value c and the acceleration decreases to zero: 1 2 3 2 ⎡ 1 ⎛ m c⎞ 2 ⎤ ⎛ m ⎞ ⎛ c⎞ c ⎢1 ⎜ 0 ⎟ ⎥ c at a ⎜ 0 ⎟ ⎜ ⎟ 0. ⎝ F ⎠ ⎝t⎠ ⎢⎣ 2 ⎝ Ft ⎠ ⎥⎦
EXAMPLE E1.30 Determine the relativistic momentum of an electron and its relativistic kinetic energy if the electron is moving at a speed of 0.9c.
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Solution: The relativistic momentum can be determined according to equation p m0c (/兹1苶 苶苶2苶). Calculation gives p 5.6 × 1022 kgm/sec. The kinetic energy is determined as the difference between the total energy and the resting energy: £ EE0, where E mc2 and E0 m0c2. We then obtain x
⎛ ⎞ 1 m0 c 2 and finally x m0 c 2 ⎜ 1⎟ . ⎜⎝ 1 2 ⎟⎠ 1 2
m0 c 2
Substituting all values we arrive at £ 1.061013J. Sometimes the particle energy is expressed in terms of energy: the electron energy at rest is m0c2 0.51 MeV. Then £ 0.66 MeV. 6. Relativistic momentum. As with many other properties the relativistic momentum in STR is p m, p
m0 1
2
m0 c
1 2
.
(1.6.9)
Unlike the classical definition, the relativistic expression permits the momentum p to approach infinitely large values as the particle speed approaches the speed of light. 7. Relativistic energy. The relativistic expression of the total energy E is E mc 2 m0 c 2 x ;
(1.6.10)
E0 m0c2 is the total energy at rest. The kinetic energy is ⎛ ⎞ 1 x E0 ⎜ 1⎟ . ⎜⎝ 1 2 ⎟⎠
(1.6.11)
The total energy E is therefore expressed as follows: E mc 2 mc 2 x
(1.6.12)
8. Relation of energy and momentum. From the above relations it follows that E 2 ( pc)2 (mc 2 )2 ,
(1.6.13)
pc 2 x ( x2 E0 ).
(1.6.14)
and
Equation (1.6.10) allows a different treatment. There is a treatment that explains this equation as a measure of possible energy release confined in mass m. From the other point of view there are no equivalence of “mass” and “energy” in such a simplified statement. We can explain this expression as a kind of energy conservation law: at certain conditions a body can release some amount of energy in the form of -radiation which, being absorbed by other atoms, increases its kinetic energy, i.e., heating it; the energy of -quanta initiates the heat of the environment. A simple calculation shows that the general law of energy conservation is
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valid in this case too; however part of the inert mass transforms into the -ray’s mass, i.e., into the mass of particles which do not possess the resting mass. Both explanations are nearly equivalent; however, the latter accepts the physical meaning but not a scholastic statement on energy and mass equivalence. From the equations presented it follows also that the photon energy h corresponds to its mass m h/c2. It should be noted that in 1911 Einstein had expanded his theoretical consideration of noninertial systems and had suggested the general relativistic theory of gravitation. On the basis of this theory Einstein postulated the principle of equivalence: the action of a gravitational field is equivalent to the action of accelerated motion of a system. Corresponding mathematical expressions can be interpreted that any mass perturbs the environmental space; therefore all bodies will move on the trajectories curved in a vicinity of the disturbing mass while approaching it. Obviously, all relativistic expressions transform into classic ones at speeds that are small in comparison with the speed of light in vacuum. Therefore the principles presented by Einstein do not contradict the general statements of the Galileo relativity. Chemists only meet the relativistic approach occasionally, e.g., energy of inner electrons of heavy atoms, some details of physical methods of investigations and some others. For physical objects and reference frames moving with speeds ^c or u^c, Einstein’s theory led to the results of classical nonrelativistic theory: Lorentz transformations changed into Galileo’s transforms and the Einstein relativity principle into Galileo’s relativity principle.
PROBLEMS/TASKS 1.1. Two direct roads cross at a corner 60°. From the crossroads two cars start simultaneously along these roads with speed v1 60km/h and v2 80 km/h. Determine the rate at which the cars move away from each other. Consider two variants. 1.2. For the four cases presented in Figure T1.2 calculate: (1) kinematical equations of movement x(t) and y(t); and (2) the trajectory equation y(x). y
y
O
x
h
O
x v0
h g
v0
A
y
y A
g
A
v0
A g
h O
x s
h
v0 O
g
x s
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1.3. A body covers the first half of its journey at t1 2 sec and the second part at t1 8 sec. Determine the body’s average speed 〈υ〉 if the distance travelled is s 20 m. 1.4. From what height (H) has a body fallen if it moves the last meter of its drop (s 1 m) in time t 0.1 sec? 1.5. A car covers three quarters of its journey at 1 60 km/h and the remainder at a speed of 2 80 km/h. What is the average speed 〈υ〉 for the whole journey? 1.6. An MP moves along a circle with a radius R 4 m. The initial point speed is 0 3 m/sec, tangential acceleration aτ 1 m/sec. For time interval t 2 sec determine: (1) the distance travelled by the MP; (2) the displacement modulus |∆r|; (3). the average speed 〈υ〉, (4) the modulo of an average velocity |〈〉|. 1.7. Determine the relationship of the free fall acceleration g with a distance from the Earth’s center r. Assume that the earth density is a constant independent of the earth point. Draw a graph g(r). Assume the earth’s radius R to be constant. 1.8. A thin stick with a length of l 25 cm staying vertically on a horizontal surface begins to fall. Determine the angular velocity and linear speed v of (1) the middle point of stick; and (2) the upper end of the stick. The friction is so great that the stick’s lower end does not slip. 1.9. A stone is thrown upwards at an initial speed of 0 20 m/sec. In 1 sec another stone is thrown in the same direction at the same speed. At what height h do they meet? 1.10. The movement of an MP is set by equation r(t) A(i cost j sint) where A 0.5 m/sec, 5 rad/sec. Draw the MP’s trajectory. Determine the speed modulo of |v| and | an|. 1.11. An aircraft flying at a height H 2940 m at a speed of v 360 km/h has to reset a bomb. At what time (t) before passing above the target and at what distance (s) from it should the plane reset the bomb to strike the target? Neglect any air resistance. 1.12. A projectile is fired at an angle 30° to the horizon. It twice reached the same height at t1 10 sec and t2 50 sec after the shot. Determine the initial speed v0 and height h. 1.13. A tank l 4 m in length filled with water moves with acceleration a 0.5 m/sec2. Find the difference of the water levels (h) in the front and the rear of the tank. 1.14. A helicopter with a mass m 3.5 T and propeller vane length d 18 m hangs motionless in the air. Determine the velocity v with which the propeller throws down the air-blast. Assume that the air-blast diameter is equal to the rotor diameter. 1.15. Initially at rest, a disc with a radius r 10 cm begins to revolve with constant angular acceleration 0.5 rad/sec2. Find the tangent (aτ), normal (an) and total (a) acceleration of the points on the wheel crown at the end of 2 sec of movement. 1.16. A spacecraft of mass m 3500 kg begins to reorient in space. The exhausted gas speed is 800 m/sec; fuel consumption is Qm 0.2 kg/sec. Find the reactive trust (R) and acceleration of the craft. 1.17. A rocket of mass M 6 T is launched upwards. The thrust of the engine is F 500 kN. Determine the rocket’s acceleration (a) and the tension force of a free hanging cable (T) in a section one quarter of the total cable length distant from the fixation point. 1.18. A disc-shaped wheel of mass m1 48 kg and radius R 40 cm can rotate freely around a horizontal axis. One end of a thin nonstretched rope is fixed to the rim of the wheel. A weight m2 0.2 kg is fastened to the other end of the rope. The weight
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1.19.
1.20. 1.21.
1.22.
1.23.
1.24.
1.25.
1.26.
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is lifted to h 2 m and then let drop. The rope tightens and the wheel begins to rotate. Calculate the angular velocity of the wheel. Two skaters of mass m1 80 kg and m2 50 kg hold two ends of a stretched rope and stay motionless on the ice. One of them begins to extend the rope at a speed 1 m/sec. Find the speeds u1 and u2 at which the skaters move on the ice. Ignore any friction force. Find the distance (x) of the CM of the system earth–moon (Mearth 6 1024 kg, mmoon 7.33 × 1022 kg, d 3.84 × 108 m ) from the center of the earth. A molecule decays into two atoms. The mass of one piece is N 3 times bigger than the other. Neglecting the initial molecule kinetic energy, determine their kinetic energies £ 1 and £ 2 if the total kinetic energy is £ 0.032 nJ. A projectile with a mass m 10 kg has at the upper point of its motion a velocity of 200 m/sec. It explodes at this point into two parts. The smallest part m1 3 kg acquires a velocity u1 400 m/sec in the previous direction. Find the velocity of the second part after the explosion. The movement of an MP along a curvilinear trajectory is given by equations x A1t3 and y A2 t, where A1 1 m/sec3 A2 2 m/sec. Find the MP trajectory equation, MP speed and the total acceleration a at the moment t 0.8 sec. A bullet of mass m 10 g moves horizontally at a speed v 800 m/sec rotating around a longitudinal axis with a frequency n 3000 sec−1. Assuming the bullet to be a cylinder with diameter d 8 mm determine the bullet’s total kinetic energy K. Two balls of masses m1 2 kg and m2 3 kg move at velocities 1 8 m/sec and 2 4 m/sec. Find the change of the inner energy ∆U after their inelastic collision in the following two cases: (1) when the smallest ball overtakes the other; and (2) when the balls move towards each other. A uniform thin stick of mass m1 0.2 kg and length l 1 m can oscillate freely around an axis passing a point O (Figure T1.26). A sticky ball with a mass m2 10 g, moves horizontally at a speed of 10 m/sec and get stuck at point A on the stick. Determine both angular and linear velocity u of the lower point of the stick in the initial instant of time. Carry out the calculation for the (1) a l/2; (2) l/3; and (3) l/4. A a
l
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1.27. A uniform disc of mass m1 0.2 kg and radius R 20 cm can rotate freely around a horizontal axis Oz, which passes through point O. Onto a point A on the disc rim a small sticky ball of mass m2 10 g moving horizontally at a velocity 10 m/sec, strikes the disc and cleaves to it. Find the disc’s angular velocity and linear velocity u of point B at the instant of the blow provided the ball hits the rod at point A on the disc rim. Carry out a calculation for (1) a b R; (2) a R/2, b R; (3) a 2R/3, b R/2; and (4) a R/3, b 2R/3 (Figure T1.27).
B b O a v
A
1.28. A block is suspended to spring scales. A cord is crossed over the block. Masses m1 1.5 kg and m2 3 kg are attached to the ends of the cord. Find the scale readings at which the weights will move. Ignore block mass, the cord and the friction in the block. 1.29. A hammer of mass m 1000 kg falls from a height h 2 m onto an anvil. Impact duration is 0.01 sec. Determine the average value of the impact 〈F〉 force. 1.30. A ball of mass m 300 g collides with a wall and rebounds from it. Find the momentum p1 obtained by the wall if, in the last moment before the impact, the ball has a velocity of 0 10 m/sec directed at an angle of 30° to the wall’s surface. The impact is perfectly elastic. 1.31. A rocket of mass mc 2 T leaves surface of the moon. After time it reaches the first space (moon) velocity 1 1.68 km/sec. Determine the mass fuel consumption if the nozzle is 4 km/h. Ignore the gravitation of the moon. 1.32. The ratio of a rocket’s fuel mass to total starting rocket mass is ¾. Determine the velocity of the rocket after total consumption of the fuel if the fuel consumption from the nozzle u is 2 km/sec. Ignore the air resistance. 1.33. Determine the maximum part (w) of kinetic energy £ 1 which a particle of mass m1 2 10−22 g can transmit to particle m2 6 10−22 g through an elastic collision. The second particle is at rest before the collision. 1.34. A bullet of mass m 10 g moving at a speed 600 m/sec hits a ballistic pendulum of mass M 5 kg and lodges in it. Determine the maximum height h of the pendulum’s lift (Figure T1.34).
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M
v
m
h
1.35. A ball of mass m1 2 kg collides with another ball m2 8 kg. The momentum p1 of the first ball is 10 kg m/sec. The impact is direct and elastic. Determine just after collision: (1) the momentum of the first p1 and of the second p2 ball; (2) the change in momentum of the first ball ∆p1; (3) the kinetic energies of the first £ 1 and the second £ 2 ball; (4) the change in kinetic energy of the first ball £ 1 ; and (5) the portion w of the kinetic energy transferred from the first ball to the second. 1.36. A ball of mass m1 6 kg collides with another, rested ball m2 4 kg. The momentum p1 of the first ball is 5 kg m/sec. The impact is direct and inelastic. Determine just after collision: (1) the momentum of the first p1 and the second p2 ball; (2) the change of momentum of the first ball ∆p1; (3) the kinetic energies of the first £ 1 and the second £ 2 ball; (4) the change in kinetic energy of the first ball £ 1 ; (5) the portion w1 of the kinetic energy transferred from the first ball to the second, and the portion w2 of the residual kinetic energy of the first ball; (6) the change of the inner energy ∆U of the balls; (7) the portion w of the kinetic energy of the first ball transferred into the inner energy of the balls. 1.37. The kinetic energy of a rotating wheel is £ 1 kJ. Under the action of a constant retarding torque it begins to rotate, uniformly retarded. After making N 80 revolutions it stops. Find the retarding torque M.
ANSWERS 1.1. v′ 122 km/h; v″ 72 km/h. 1.2. (a) x 0t, y −h−(gt2/2); y h gx2/220. gx2 . (b) x 0tcos, y h 0t sinα (gt2/2); y h xtg 20cos2 g(xs)2 (c) x s 0t, y h−(gt2/2); y h . 20 g(xs)2 . (d) x s 0tcos, y h – 0tsin –(gt2/2); y h (xs)tg 202cos2
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1.3. 〈〉 s/(t1 t2) 2 m/sec. 1.4. H (2s gt2)/(8gt2) 5.61 m. 1.5. 〈〉 64 km/h. 1.6. (1) s 8 m; (2) |∆r| 6.73 m; (3) 〈υ〉 4 m/sec, (4) |〈〉| 3.36 m/sec. 1.7. g(r) (4/3)πGr at r^R and g(r) 4 GR3/(3r2) at r≥R (Figure T1.7).
g(r) (4/3) GR
~r
~r −2
R
r
1.8. (1) 1 14 rad/sec, u1 1.05 m/sec; (2) 2 14 rad/sec, u2 2.1 m/sec. 1.9. h 19.2 m. 1.10. |v| 2.5 m/sec, |an| 12.5 m/sec2. 1.11. t 24.5 sec, s 2.45 km. g(t1t2) 1.12. 0 588 m/sec; h g t1t2 2.45 km. 2sin 1.13. h 20.4 cm. 1.14. v (1/d)[(4m)/π]1/2 10.2 m/sec ( is the air density). 1.15. aτ 5 cm/sec2, an 10 cm/sec2 , |a| 11 cm/sec2. 1.16. R – Qmv –160 N, a – Qmv/m 4.57 cm/sec2. m F 3 1.17. a0 g 73.5 m/sec2; T F 625 N. Mm 4 Mm
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2m2兹苶2苶g苶h 1.18. 0.129 rad/sec. (m1 m2)R 1.19. u1 0.385 m/sec; u2 –0.615 m/sec. 1.20. x 4.69 × 106 m (radius of earth is R 6.4 × 106 m). 1.21. K1 nK/(n 1) 24 pJ. 1.22. u2 114 m/sec. 1.23. y3 – 8x 0; 2.77 m/sec, a 4.8 m/sec2. 1.24. K (m/4)(2v2 π2n2d2) 3.21/kJ. m1m2 (1 2)2 1.25. U 9.6 J and 86.4 J. 2(m1m2) 1.26. (1)
6m2 3m2 1 2.61 rad/sec. u1 1.30 m/sec. (3m2 m1)ᐉ (3m2 m1)
3m2 2m2 (2) 2 1.43 rad/sec. u2 0.952 m/sec. (m2 m1)ᐉ (m2 m1) 4m2 3m2 0.629 m/sec. (3) 3 0.839 rad/sec. u 3 (m2 (7/3)m1)ᐉ m2 (7/3)m1 1.27. (1) 1 4.55 rad/sec, u1 0.909 m/sec. (2) 2 2.27 rad/sec, u2 0.454 m/sec. (3) 3 3.03 rad/sec, u3 0.303 m/sec. (4) 4 1.52 rad/sec, u4 0.202 m/sec. 4m1m2 1.28. F g. m1 m2 1.29. 〈F〉 (m/) 兹2苶g苶h苶 626 kN. 1.30. p1 2m0sin 3 N sec. 1 mc 1.31. 1exp 1.68 km/sec. u
冢
冣
1 1.32. u ln 2.77 km/sec. 1
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4m1m2 1.33. w (m1m2)2 0.75. 1.34. h m22/[2g(m M)2] 7.32 cm. 1.35. (1)
2m2 m1 m2 p 1 6 kg.m/sec, p 2 16 kgm/sec. m1 m2 m1m2
(2) p1 p 2 16 kg m/sec. 2m2 p21 m1m2 (3) £ 1 9 J, K 2 p21 16 J. m 2m m1m2 1 m2 (4) 冟£ 1冟 K 2 16 J. 4m1m2 £ 1 (5) w 2 0.64. (m1 m2) £ 1 m1 p2 m2 p1 1.36. (1) p 1 3 kg m/sec, p 2 2 kg m/sec. m1 m2 m1 m2 (2) p 1 p 2 2 kgm/sec. m1 p12 m2 p12 (3) £ 1 2 0.75 J, K 2 2 0.5 J. 2(m1 m2) 2(m1 m2) (4) £
1
m2(2m1m2) p12 1.33 J. 2(m1 m2) 2m1
£ 2 m21 m1m2 £ 1 (5) w1 0.24, w 0.36. 2 £ 1 (m1 m2)2 (m1 m2)2 £ 1 m2 p12 (6) U 0.833 J. 2m1(m1 m2) m2 U (7) w m1 m2 0.4. £ 1 £ 1.37. M 1.99 Nm. 2N
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–2– Oscillations and Waves
2.1
DEFINITIONS
Along with translational and rotational motions, oscillations play an important role in the macro- and micro-world. We can distinguish both chaotic and periodic oscillations. Periodic oscillations are characterized by the time of repetitions: through certain periods of time a system passes one and the same position, running in one and the same direction. An example is given in Figure 2.1, which shows the man’s cardiogram in the form of a graph of electrical signals of the heart’s oscillations. It is possible to select the period of time of one complete oscillation T, thus presenting a periodical process. However, periodicity is not a unique feature of oscillations: a rotational movement can also be characterized by periodicity. The presence of an equilibrium position is a second particularity of oscillatory motion (rotation is characterized only by the so-called indifferent balance: a well-balanced raised car wheel, being rotated, stops in any position with equal probability). Thirdly, any deflection force tends to return an oscillatory system to its initial equilibrium position; i.e., the restoring force is always directed to the position of equilibrium. The presence of all these three signs distinguishes oscillations from other types of motion. Consider several specific examples of oscillatory motion. Clamp one end of a steel straightedge in a vice and let the other end move freely. The returning force will try to draw the free end of the straightedge toward the equilibrium position. Passing by this position, the straightedge will have a certain velocity and a certain stock of kinetic energy. The inertia forces will not permit the straightedge to stop in the position of equilibrium and will work against the internal elastic force to decrease the kinetic energy. This will bring about an increase in potential energy. When the kinetic energy is completely exhausted, the potential energy will reach a maximum. The forces of elasticity will also reach a maximum and will be directed to the position of equilibrium. All these features were described in detail in Sections 1.3.5 (eq. (1.3.20)), 1.4.1 (eq. (1.4.9)) and 1.5.4 (Figure 1.33) in the language of potential curves. Oscillation will repeat until the total mechanical system energy disappears into the surrounding space. Another well-known example is that of pendulum oscillation. We have seen this example in Chapter 1 (refer to Figure 1.33) and will often come across it in different aspects.
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Figure 2.1 An example of a periodic oscillation process: the cardiogram of a human being.
Oscillation can be not just mechanical. So, for instance, one can consider the oscillations of an electric current in an oscillatory circuit or a magnetic field strength in a dynamo, etc. These can be described by an equation similar to the one that defines mechanical displacements from a position of equilibrium. In spite of this fact, we will mostly analyze mechanical oscillations, keeping in mind their applicability to other types of oscillation. The time in which a system accomplishes one complete oscillation is called the oscillation period T. A value inverse to the period expressing a number of full oscillations in the time unit is referred to as the oscillation frequency (i.e., completely identical to rotation)
1 T
(2.1.1)
Let us begin the analysis of oscillatory processes with the simplest case of a one-dimensional oscillation, i.e., of a system with one degree of freedom. The degree of freedom described a number of independent variables required for the complete description of the positions of all parts of a given system. If, for instance, pendulum oscillations are limited by one plane and the thread of the pendulum is not stretched, it is sufficient to assign either an angle of deflection of the thread from a vertical line or any other value of displacements from the position of equilibrium. Each of them is enough to define the position of the pendulum in full. In this case the system considered possesses one degree of freedom. The same pendulum, if it can occupy any position on a section of a spherical surface, possesses two degrees of freedom.
2.2
KINEMATICS OF HARMONIC OSCILLATIONS
From the entire variety of periodic oscillations we will select first of all the so-called harmonic oscillations. Interest in harmonic oscillations is due to the following reasons: firstly, it is relatively simple to describe harmonic oscillations mathematically, and, secondly, any periodic oscillations can be presented as a superposition of harmonic oscillations. This latter circumstance is very important, and we will return to it in Section 2.3.2. Harmonic oscillations are an abstraction, since they have to continue for an infinitely long period (t), according to certain laws, without any changes, which is not
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the case in the real macroscopic world. We consider harmonic oscillations to be one of the common physical models. Oscillations are referred to as harmonic if the changes in time of some physical values occur under the sine or cosine law (t ) A sin(t 1 ) (t ) A cos(t 2 ),
(2.2.1)
where (t) is the time dependence of a displacement. By the term “displacement” we understand the magnitude of the displacement of any physical value at a given time instance t. In particular, when considering the simplest mechanical oscillations under displacement we shall understand a deflection of varying point from the position of equilibrium. The maximum value of the displacement is called the oscillation amplitude A; it is always taken as a positive value. The expression in parentheses is the phase of oscillations, is the initial phase of oscillations (i.e., the phase of oscillation at moment t 0) and is the angular (or circular) oscillation frequency. The choice of sine or cosine for describing the harmonic oscillations as well as the initial phase is rather arbitrary and is chosen for convenience. A cosine form is preferable in most cases as will be seen later. The transition from one form to another is easily realized by a corresponding change of the initial phase. So, for instance, if harmonic oscillation is described by an expression (t) A sin(t 1), it is also possible to present it in the form (t) A cos(t1 /2) A cos(t2), where 2 1 /2. Let us make an interconnection between the angular frequency , the frequency v and the period T. Sine and cosine are periodical functions with period 2. This means that after the time interval T a system returns to its initial state and the phase is changed to 2, i.e., [(t T )][t ] 2. Thereby, the angular frequency is connected with period T and frequency v by the expressions
2 T
(2.2.2)
or 2v.
(2.2.3)
Let us examine the change in the speed and acceleration of an oscillating point. If the displacement of an oscillating MP is expressed by eq. (2.2.1), its velocity and acceleration can be found by the first and second time derivative of displacement as d(t ) (t ) A sin(t ), dt
(2.2.4)
d 2(t ) (t ) A2 cos(t ). dt 2
(2.2.5)
(t ) a (t )
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Comparing these expressions, we arrive at a(t ) 2(t ).
(2.2.6)
This means that the acceleration of harmonic oscillations is always proportional to the displacement and is in the opposite direction. A graphical diagram of displacement (a), velocity (b) and acceleration (c) depending on the phase t is presented in Figure 2.2. It can be seen from these curves that the phase of velocity differs from that of displacement by /2; however, the phase of acceleration shifted by in comparison to the displacement phase. This can be summarized as follows: the velocity phase leaves behind the displacement phase by /2; however the acceleration is in antiphase to the displacement. In other words, when the displacement is maximum ( A) the velocity is equal to zero; however acceleration reaches its maximum value (amax 02A). On the other hand, while the MP passes the equilibrium position the velocity is maximum (max 0 A), and acceleration at this moment is zero. It is instructive to consider the relationship between a simple harmonic motion along a line and uniform circular motion. In this respect, the harmonic oscillation can be presented by uniform rotation of the radius vector (or amplitude vector). Let us imagine a segment with a length numerically equal to the amplitude value A uniformly rotated around one of its ends (Figure 2.3) with an angular frequency . We denote as an angle with an abscissa axis . At the instant of time t this angle is t . Projection of a point B onto the -axis will increase in time and be described mathematically in just the same manner as the harmonic oscillation (t) A cos(t ). Therefore, the radius vector accomplishes a rotational motion, whereas its projection on the -axis oscillates according the harmonic law. 2
ξ
a
ξ0 0 ϕ
1− 2
3− 2
2
5− 2
ωt
.
ξ 0 υ0
b 1− 2
3− 2
2
5− 2
ωt
..
ξ
0
c 1− 2
3− 2
2
5− 2
ωt
a0
Figure 2.2 Kinematics of .harmonic oscillations. Initial values of displacement, velocity and accel.. eration 0 (0), 0 (0) and a0 (0), respectively, are shown.
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The angular velocity of rotation can be found by time derivation [d(t )/dt] . So, the harmonic oscillation can be formally presented by the rotating radius vector A with angular velocity , the phase being found by an angle of the radius vector OB with the -axis. Here, the initial phase is an angle the radius vector forms with the -axis in the instant of time t 0. By describing harmonic oscillation in this way, we can accept that the phase is a more exhausting characteristic of harmonic oscillations than displacement. This can be better seen in Figure 2.4. The advantage of phase to displacement is that the former uniquely
B A
0
Figure 2.3 A vector diagram of harmonic oscillations.
A 1
2
0 2
1
Figure 2.4 Ambiguity of oscillations with displacement: two phases 1 and 2 correspond to one and the same displacement , whereas velocities 1 and 2 have opposite directions.
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describes the oscillation, whereas in the latter case one should choose the real phase of two (two phases 1 and 2 lead to the same displacement; in the successive moments the oscillating points will scatter in different directions with velocities 1 and 2). Besides comparing two harmonic oscillations with the same frequency but with different and unknown amplitudes, values of displacements cannot give a valuable result; whereas a knowledge of phases can provide a clear picture. For instance, if phases coincide, both harmonic oscillations reach maximum displacement simultaneously (oscillations are in-phase) and vice versa. When changing a phase by 2n harmonic oscillation returns to its initial state (n 0, 1, 2, ...). If phases differ by , 3, 5, etc. (in the general case (2n 1)), such oscillations are accepted to be in opposite phases or in antiphase. Figure 2.5 shows a vector diagram with displacement, velocity and acceleration. It can be seen that velocity is ahead of displacement by an angle /2 and acceleration is antiphase to displacement. Here, the vector diagram of the harmonic oscillation enables us to use representations of complex numbers. In some cases it allows us to avoid bulky trigonometric transformations and essentially simplifies mathematical calculations and physical view. Let us consider a complex number plane. On the abscissa we shall put the real part of a complex number and on the ordinates its imaginary part (Figure 2.6). Then any complex number can be written as Z i.
(2.2.7)
Its modulus is equal to A Z 2 2
(2.2.8)
A B
C
0 OC = . OB = .. OA =
Figure 2.5 A vector diagram of displacements, velocities and acceleration. The frequency is chosen as unity.
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B
A 0
Figure 2.6 Graph of a complex number Z = i.
Real () and imaginary () parts of complex number i can be expressed through the modulus A and argument by formulas A cos , A sin .
(2.2.9)
Therefore, any complex number can also be expressed as Z A(cos i sin ).
(2.2.10)
At uniform rotation of a radius vector A as an argument we can take t . Then the real part (Re) of a complex number Z will change under the harmonic law: Re ( Z ) A [cos(t )],
(2.2.11)
where the modulus of the complex number is equal to the amplitude of the harmonic oscillations represented by this complex number. The Euler formula plays an important role in interpretation of oscillations. Accordingly, the real part of the complex number (written in exponential form) Z A exp[i(t )] changes in time under the harmonic law: Re[ A exp i(t )] A cos(t ).
(2.2.12)
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In all these cases, displacement of harmonic oscillations can be presented as Re( Z )
(2.2.13)
Henceforth, we shall designate complex number Z by the same letter as displacement (), meaning every time that harmonic oscillation is described by the real part of this complex number. Hence, the time dependence of displacement at harmonic oscillations can be written as follows: A cos(t ).
(2.2.14)
A exp(i)exp(it ).
(2.2.15)
This expression can be rewritten as
A exp(i) defines the length and direction of the radius vector (A) at the initial instant of time and is referred to as complex amplitude (which we shall designate a) a A exp(i).
(2.2.16)
Then harmonious oscillations can be written even easier: a exp(it ).
(2.2.17)
The sign in an exponent shows the direction of the radius vector A rotation. In physics, factor ei sometimes stands for an operator of rotation. In fact, multiplication by this factor is equivalent to turning the vector A counterclockwise at an angle . EXAMPLE E2.1 A small weight is suspended on a long, nonextendable string. Prior to oscillating, it was removed from the position of equilibrium to the utmost left-hand side and then set off. Write down the equation of oscillation and find the initial phase. Solution: First of all, we have to choose the form of the answer (sin or cos, with their signs). Let it be cos. The sign depends on a positive direction of the axis chosen; let it be from left to right. If the weight was released at its negative utmost deviation the displacement at t 0 should be A. Therefore, the equation is (0) A cos(t ); is just the initial phase. If we choose function (sin) the initial phase would be 3/2. It is expedient to draw the vector diagram for this case.
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EXAMPLE E2.2 An MP oscillates with simple harmonic motion according to the equation x(t) = A cos(t ), amplitude A being equal to 2 cm. Find the initial phase if x(0) 3 cm and x.(0) 0. Draw a vector diagram for the zero instance of time (t 0). Solution: Express a displacement at t 0 via initial phase: x(0) A cos . The initial phase is arcos [x(0) /A] and further arcos(3 / 2). Two angles correspond to these phases 1 (5/6) and 2 (7/6). To find for a certain phase we have to use the condition x.(0)0. Keeping in mind that x.(t) A sin(t ) we can substitute numerical values 1 and 2 and find x.1(0) A/ 2 and x.2(0) A/ 2. Since A 0 and 0, only the first value satisfies the condition x.(0)0. Hence, the value sought is 1 (5/6). The results can be seen in Figure E2.2. y
5/6 A x 0
x(0) (0)
2.3 2.3.1
SUMMATION OF OSCILLATIONS
Summation of codirectional oscillations
Let us begin with the simplest case: summation of two oscillations with the same frequencies: 1 (t ) A1 cos(t 1 ) and 2 (t ) A2 cos(t 2 ).
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The resulting displacement (t) can be found as an algebraic sum of oscillations (t) 1(t) 2(t) cos(t ), where A and are the amplitude and initial phase sought of the final oscillation. Such a summation can be realized both graphically and analytically, although the graphical method is more visual. Each oscillation in the same time instance (say, t 0) can be presented as vectors A1 and A2 (Figure 2.7), plotted from the abscissa under angles 1 and 2, correspondingly. Since the frequencies of both oscillations are the same, the mutual positions of both vectors A1 and A2 remain unchanged during their rotation with the same angular velocity. Consequently, the total oscillation can be represented by vector A, which is the vector sum of A1 and A2. This is the oscillation accomplished with the same cyclic frequency (the complex parallelogram is rotated with this angular velocity). The amplitude A can be determined according to the cosine theorem taking into account that instead of angle 180° (2 1), in the following expression, we use angle 2 1 that influences the sign in radicand A A12 A22 2 A1 A2 cos(2 1 ).
(2.3.1)
Let us analyze the result. The phase difference of the summing oscillation remains constant; at any time instance it is the difference of the initial phases, i.e.,
(t 2 ) (t 1 ) 2 1 .
B
A
A2
2
G
H
A1
2 0
C
1 2
D
1
Figure 2.7 Summation of two codirectional oscillations with same frequency.
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As can be seen from expression (2.3.1), the resulting amplitude depends on the phase difference of the summed oscillation. In particular, if the phase difference satisfies the condition 2n,
(2.3.2)
where n 0, 1, 2,..., then cos 1, and the resulting amplitude will have the maximum value Amax A12 A22 2 A1 A2 or Amax A1 A2 .
(2.3.3)
(2n 1),
(2.3.4)
If the phase difference is
then cos 1 and the resulting amplitude will have the minimum value Amin A12 A22 2 A1 A2 or
Amin A1 A2 .
(2.3.5)
The modulus is used here because the amplitude must be positive. It is clearly seen that oscillation will not take place at all in this case: having equal amplitudes and oscillating in antiphase, they have cancelled each other out. Determine now an initial phase . From Figure 2.7, tg ( BD/OD) can be derived; however, BD A1 sin 1 A2 sin 2 and OD A1 cos 1 A2 cos 2. Therefore
tg
A1 sin 1 A2 sin 2 , A1 cos 1 A2 cos 2
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and the phase is
⎛ A sin 1 A2 sin 2 ⎞ arctan ⎜ 1 ⎝ A1 cos 1 A2 cos 2 ⎟⎠
(2.3.6)
Hence, summing up two harmonic oscillations, we also obtain the harmonic oscillation with the same cyclic frequency and amplitudes given by expression (2.3.1) and the initial phase given above (2.3.6).
EXAMPLE E2.3 Two oscillations take place along one and the same direction. They are expressed by equations x1 A1 cos(t 1) and x2 A2 cos (t 2), where A1 1 cm, A2 2 cm, 1 1/6 sec, 2 1/2 sec, sec1. Determine (1) initial phases 1 and 2 of the component oscillation; and (2) the amplitude and initial phase of the resulting oscillation. Write down the equation of the resulting oscillation and draw the corresponding vector diagram.
y
A A2
A1 2
1 x
0
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Solution: (1) In general form the equation of oscillations is x A cos(t 1). Rewriting the given equation for both oscillations: x1 A1 cos(t 1) and x2 A2 cos(t 2). From the conditions given above we can find 1 w 1 (/6) rad and 2 2 (/2) rad. (2) Keeping in mind the vector diagram (Figures 2.4 2 2 and 2.7), we can find A A A 2 A co s . We can find from the data 1 2 1A 2 given above; (/3) rad, and further, tan (A1 sin 1 A2 sin 2) / (A1cos 1 A2 cos 2). Executing the calculations we arrive at arctan (5/ 3) 70.9° 0.394 rad. 2.3.2 Summing up two codirectional oscillations with slightly different frequencies: beatings For simplicity, consider that 1 2, so (1 2) and A1 A2 A. This summation can be made analytically without difficulty. However, we will use here the summation method of oscillations based on the vector diagram as we did earlier in this chapter to define the qualitative nature of the result. Turn first to Figure 2.7. In this case vectors A1 and A2 rotate with slightly different angular velocities. This signifies that at some point in time both vectors can be in antiphase and the resulting oscillation amplitude is zero. Hereon, vector A2 runs after A1 and at another time instant “catches” it up, the phases of both vectors coinciding. The resulting oscillation amplitude will become 2A, whereupon vector A2 will overrun vector A1 and the total oscillation amplitude will gradually decrease. At least vector A2 overruns A1 by : once again the oscillations will disappear in a moment. Then the whole process will be repeated again. It is easy to see that oscillations like those depicted in Figure 2.8 are composed oscillations. These types of oscillations are called “beatings” and are easily observable. When two sources of oscillations (for instance, two engines in an aircraft) rotate at close frequencies,
T=
2
t
T=
2 ∆
Figure 2.8 Beatings.
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a resulting sound can be heard: low-frequency fluctuations on a background of the highfrequency roar of the engines.
2.4 2.4.1
DYNAMICS OF THE HARMONIC OSCILLATION
Differential equations of harmonic oscillations
It was shown earlier (refer to eq. (2.2.6)) that there exists a simple correlation between displacement and acceleration of an oscillating MP: (t ) 2(t ) or (t ) 2(t ) 0.
(2.4.1)
This equation is referred to as the differential equation of harmonic oscillations. By analyzing this equation, we can arrive at an important conclusion: when solving a problem and arriving at an equation like that presented above, it means that the problem can be reduced to harmonic oscillations and the coefficient before the displacement function is the square of its cyclic frequency. A general equation is a sum a cos t b sin t
(2.4.2)
or, in exponential form A exp(it ).
This is proved by substitution of any of the proposed solutions into eq. (2.4.1). 2.4.2
Spring pendulum
A system consisting of a body with a mass m, which, moving without friction (!), can oscillate under the action of elastic force (weightless springs) with the rigidity coefficient , is called a spring pendulum. One end of the spring is attached to the weight and the other end is fastened to a rigid wall (Figure 1.22). (This system has already been considered in Sections 1.4.1 (Figure 1.22) and 1.5.4 (Figure 1.31).) The starting
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position is that of a nondeformed spring. According to the second law of dynamics, we can write m ,
(2.4.3)
where is the value of the shift from the origin. In this case, the elastic force is a single force acting on the body (since the projection of the gravitational force to the abscissa x is zero and friction is neglected). A negative sign is stipulated by the fact that the force acting on the weight is always directed toward the origin. Transferring both terms of eq. (2.4.3) to the left-hand side and subdividing them by m, we arrive at
0. m
(2.4.4)
Compare the expression obtained with the general type of differential equation of harmonic oscillation (eq. (2.4.1)). From the fact that both equations have a similar form, it can be stated that the weight makes a harmonic oscillation. Thereof, the other definition of a harmonic oscillation is an oscillation that occurs under the action of an elastic force. By equating the multipliers in the similar terms of the equation, we can derive an expression for the cyclic frequency of the spring pendulum:
. m
(2.4.5)
The same expression can be obtained for the cyclic frequency of free-oscillating bodies appearing by the action of any quasi-elastic force, which is linearly proportional to displacement and directed opposite to it. According to eq. (2.2.2), the period T of harmonic oscillations produced by the action of elastic and quasi-elastic forces is given by the formula
T 2
2.4.3
m .
(2.4.6)
The mathematical pendulum
Another example of harmonic oscillations is that of a mathematical pendulum. An MP suspended on a weightless, nonstretched and ideally flexible thread, is referred to as a mathematical pendulum. Consider small displacements of a pendulum from the equilibrium position, i.e., l, where l is the length of the mathematical pendulum. At a certain instant of time, let the pendulum occupy the position depicted in Figure 2.9. Using the second Newtonian law, equation of motion can be written as m mg sin
(2.4.7)
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O
ΟΑ = l0 T
F
A
mg
Figure 2.9 A mathematical pendulum.
At small pendulum’s angle deflection (/l 1), sin and the returning force F (mg/l) can be considered as a quasi-elastic one. The coefficient characterizing the “rigidity” of the quasi-elastic force for the mathematical pendulum is mg/l . Introducing this expression for the rigidity of a quasi-elastic force into eqs. (2.4.5) and (2.4.6), we can obtain an expression for the cyclic frequency and period of small oscillations of a mathematical pendulum:
g ᐉ
(2.4.8)
and
T 2
ᐉ . g
(2.4.9)
These formulas are valid only for small displacements ( l), under which approximation sin is also valid. This approximate equality will be executed if angle 1. So, for instance, at 5° ( 0.1 rad) replacing sin by brings about an inaccuracy of the order 0.2%. On reducing angle this inaccuracy quickly decreases: at 1°, it reaches an
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insignificantly small value 0.005%. By contrast, at greater amplitudes it is impossible to consider oscillations to be harmonic and their period will depend on amplitude. 2.4.4
A physical pendulum
Any body having the possibility to oscillate freely under a gravitational force around a horizontal axis, not passing through the body’s CM, is referred to as a physical pendulum. In this case, all points of a rigid body move along an arc of concentric circles. Consequently, for the description of a physical pendulum’s oscillations, the rotational laws of dynamics should be applied. Let an axis of rotation z pass horizontally through point O (Figure 2.10) perpendicular to the plane of drawing. Also, let the physical pendulum be deflected from the position of equilibrium by angle , which, as previously, is considered to be small. Then, the main law of dynamics of rotational motion can be written as I z M z ,
(2.4.10)
.. where Iz is the moment of inertia of the physical pendulum regarding axis Oz, is a time second derivative of and Mz is the moment of external force with respect to axis z returning the pendulum to the position of equilibrium. In a given case this moment is stipulated by gravitational force mg, attached to the CM of the physical pendulum. In Figure 2.10, the
O
C
O1 OC = lC
mg
Figure 2.10 A physical pendulum.
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physical pendulum CM is marked by the letter C whose distance from the oscillation axis O is marked by the letter lc. From Figure 2.10 it can be seen that for small angular displacements M z mg ᐉc sin mg ᐉc .
(2.4.11)
Sign “” corresponds to the accepted sign rule for the returning force moment of the Oz axis. Thereby, the differential equation for small physical pendulum oscillations according to eqs. (2.4.10) and (2.4.11) can be written as
mgᐉc 0. Iz
(2.4.12)
Comparing this expression with eq. (2.4.1) we can conclude that the physical pendulum makes harmonic oscillations with cyclic frequency
mgᐉc Iz
(2.4.13)
and the period of small oscillations is
T 2
Iz . mgᐉc
(2.4.14)
The length of such a mathematical pendulum, which is equal to the physical pendulum’s oscillation period, is called the reduced length of a physical pendulum. An expression for the reduced length of a physical pendulum can be found by comparing eqs. (2.4.9) and (2.4.14): OO1 L
Iz . mᐉ c
(2.4.15)
Point O1 on the line OC (Figure 2.10) at a distance L from the axis of rotation z is called the center of swing of the physical pendulum. It is noteworthy that if a pendulum is turned over and hung up on the horizontal axis passing through the point O1 the period of its oscillation does not change, point O being the new center of oscillation. We will leave the proof of this property as an exercise for the reader.
EXAMPLE E2.4 On the ends of a thin rod of weight m3 and length l, small-sized balls of weights m1 and m2 are fixed. The rod makes small oscillations about a horizontal axis perpendicular to
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the rod and passing through its middle point. Define the period T and frequency of the oscillation of the pendulum. Solve the problem for l 1 m, m1 200 g, m2 300 g and m3 400 g. Solution: The frequency of the physical pendulum’s oscillation and the period of its oscillations can be found according to eqs. (2.4.14) and (2.4.13). There are three values to define: total weight m, the MI of the pendulum relative to the axis of oscillation Iz and the distance from the axis of oscillations up to the center of weights lc. Oscillation axis z passes through the center of the rod perpendicular to the plane of drawing: let us direct an axis x vertically downward (parallel to the rod) and super-
m1
l m3
O lC C
m2
pose its origin with an oscillation axis. The coordinates of all bodies included in the system can be determined from Figure E2.4. We can find the distance lc from eq. (1.3.32):
2
lc
∑ mi xi 1
m
m1 (l 2) m2 (l 2) (m2 m1 ) 5.55 cm. 2m m
Remember that the oscillation axis coincides with the position of the oscillation axis, i.e., the coordinate of the CM numerically coincides with lc. The weight of a physical pendulum is equal to m m1 m2 m3 (0.9 kg).
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We shall find the MI of a physical pendulum relative to oscillation axis as the sum of the moments of inertia of three bodies: 2
2
m ᐉ2 ᐉ2 (3m1 3m2 m3 ) ⎛ ᐉ⎞ ⎛ ᐉ⎞ 0.158 kg m 2 . I I1 I 2 I 3 m1 ⎜ ⎟ m2 ⎜ ⎟ 3 ⎝ 2⎠ ⎝ 2⎠ 12 12
Substituting the data into the general eqs. (2.4.13) and (2.4.14) we arrive at T 2.17 sec and 2.89 rad/sec.
EXAMPLE E2.5 A physical pendulum consists of a rod and a hoop of masses mrod 3m1 and mhoop m1; the length of the rod is l 1 m. The horizontal axis of oscillation Oz is perpendicular to the rod and passes it at its center O. Determine the oscillation period of such a pendulum. The rest of the definitions are given in Figure E2.5. I Solution: The period of oscillation is expressed by eq. (2.4.14), T 2 .
mgl c
3m1
l O lc C l/2 a
l/4
m1
To find the period, we must first choose a reference frame (axis x), mark a zero position on it (see Figure E2.5) and find the MI of parts of the pendulum, Iz,1 is the MI
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of the rod and Iz,2 is the MI of the hoop and the total MI is I. It is also necessary to find the distance lc between the oscillation axis (point O) and the CM. The MI of the physical pendulum relative to the oscillation axis is the sum of I1 (the MI of a rod) and I2 (the MI of the hoop both relative to the same axis I I1 I2). The rod MI is I1 m1l 2/4 (because the rod mass is 3m1 and the axis passes through the center of the rod). The MI of the hoop is the sum of the MI of the hoop itself (first item) and the addition from the parallel axis theorem (second item): 2
2
5m ᐉ ⎛ 3ᐉ ⎞ ⎛ ᐉ⎞ I 2 m1 ⎜ ⎟ m1 ⎜ ⎟ 1 . ⎝ 4⎠ ⎝ 4⎠ 8 2
2
2
2
m1l 5m1l 7m1l The total MI of the pendulum is the sum I . 4
8
8
The distance ᐉc
∑ mi xi 3m1 0 m1 (3ᐉ 4) 34 m1ᐉ 3 ᐉ. 3m1 m1 4 m1 16 ∑ mi
(In order to simplify calculation of lc it is useful to mark zero on axis x at the same level as point O; in this case the CM coordinate is simultaneously lc). Obtaining these preliminary results we can place all the values under the square root:
T 2
7 8
m1ᐉc2
4 m1 g 163 ᐉc
2
7ᐉc , 6g
therefore, T 2.17 sec. EXAMPLE E2.6 A small weight of mass m 5 g performs harmonic oscillations under the action of a gravitational force with frequency v 0.5 Hz. Amplitude is A 3 cm. Find (1) the velocity of the weight at a time instant when x 1.5 cm; (2) the maximum force F acting on the weight; and (3) the total energy of the oscillator. Solution: (1) The oscillation equation is x(t) A cos(t ), whereas the expression for velocity can be obtained by time derivation of (t): (d/dt) A sin(t )*. In order to find the relation between the velocity and , we should exclude the time from the last two equations. For this it is necessary to square both equations, divide the first by A2, the second by A22 and sum the results: (x2 /A2)( 2/A22) 1. Solving the equation relative to we can arrive at 2v A 2 x2 (keeping in mind that 2v). Executing all calculations, we obtain 8.2 cm/sec.
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The sign “” is valid when the velocity is directed along the positive direction of x-axis and vice versa. The same result can be obtained if the sin function is used instead of cos. (2) The force value can be found according to the Newton’s second law. The time first and second derivative can be seen in eqs. (2.2.4) and .. (2.2.5): a x (d/dt) A2 cos(t ) or a 42v2Acos(t ). Using acceleration amplitude in any of these equations can yield Fmax 2mA or 1.49 mN. (3) The total energy is the sum of the kinetic and potential energies. Therefore, the total energy can be calculated at any position; for instance, in the lower position the kinetic energy is maximum at this point. Therefore using the equation marked as * (see above) and taking sin(t ) 1 we obtain E Kmax (m2max /2). Finally, E 22v2A2; executing the calculation we arrive at 22.1 106 J or 22.1 J. EXAMPLE E2.7 An areometer (densitometer) consisting of a long tube of diameter d 1 cm weighing m 50 g freely floats vertically in still water. It is submerged a little and then released; it begins to oscillate up and down. Neglecting the water viscosity find the period of its oscillations. Solution. Choose an axis vertically and denote an origin 0 at the areometer tube prior to its oscillation (Figure E2.7,a). In this state its gravity and Archimedes force are equalized. Oscillations will be accomplished by the periodically changing Archimedes force because the gravity remains constant. The value of areometer’s
a
b
0
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immersion can be arranged as (Figure E2.7,b) which will determine the Archimedes buoyancy force (d 2/4) g ..; this is the restoring force .. in our case. According the second Newtonian law, m (d2/4) g and then (d2/4m) g 0. Firstly, we obtained the confirmation that our system accomplishes harmonic oscillation (compare this equation with eq. (2.4.1)), and secondly can easily extract the 2 value: (2/T ) (d/4) ( g/ m) whence T (4/d)( m / g) 1.6 sec. EXAMPLE E2.8 A neon atom (Ne) collides with an oxygen atom of a molecule (O2) along its bond direction (Figure E2.8). The kinetic energy of the Ne atom is K1 6. 1021 J. The oxygen bond rigidity coefficient is 1.18. 103 N/m. Relative masses of the Ne and O atoms equal Ar,Ne 20; Ar,O 16. Consider the collision to be elastic and the oscillations of the O2 molecule after impact to be harmonic. Determine: (1) the translational kinetic energy Kz,tr of the oxygen molecule after impact; (2) the oscillation energy E2,osc of the oxygen molecule acquired by the impact (suppose that the O2 molecule did not oscillate before the collision); (3) the average values of kinetic K2,osc and potential U2,osc energies; (4) amplitude A of the harmonic oscillation; and (5) the angular oscillation frequency . Ne
O
O
x
Solution: (1) Assume that impact is elastic and head-on impact takes place. Using eq. (1.5.16) we can obtain u1
m1 m2 y1 , m1 m2
where 1 is the velocity of the Ne atom prior to impact. After collision its kinetic energy is 2
K1
m u2 ⎛ m m2 ⎞ 1 1 ⎜ 1 K1 . 2 ⎝ m1 m2 ⎟⎠
The CM velocity can be found using only the momentum conservation law because part of the kinetic energy is taken for the inner (oscillation) energy of the O2 molecule
Vc
m m2 ⎞ m1 m ⎛ y1 1 y1 y1 . 2 m ⎜⎝ m1 m2 ⎟⎠ m1 m2
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Then the kinetic energy of the translational movement of the O2 molecule K2,trs is
K 2,trs
2 m2Vc2 (m1 m2 )
2
2 m1m2 K1 . (m1 m2 )2
Executing calculations we arrive at K 2,trs
2 20 16 6 1021 2.96 1021 J 20 16
(2) The energy of oscillation can be found using the energy conservation law: Therefore K1 K1 K1,trs E2,osc
E2 K1 K 2 E2,trs , or 2
⎛ m m2 ⎞ 2 m1m2 E2,osc K1 ⎜ 1 K1 K1 ⎟ ⎝ m1 m2 ⎠ (m1 m2 ) 2 m1m2 K1 Since E2,osc K 2,trs , 2 (m1 m2 ) therefore, E2,osc 2.96 1021 J. (3) The average values of oscillation kinetic and potential energies can be found from eqs. (2.5.7) and (2.5.8). It follows that
K 2,osc U 2,osc
m1m2 K1 K 1.48 1021 J. m1 m2
(4) The amplitude of the harmonic oscillations can be calculated from the expression E2,osc ( 12 ) 2 A 2 The rigidity coefficient 2, therefore E2,osc (1/2) A2 and then A
2 E2,osc
.
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Using the oscillation energy we can execute all operations
A
2 2.96 1021 2.25 1012 m. 1.18 103
(5) The angular frequency can be found from equation 2. Since (m2/2),
2
. m2
And finally we arrive at
2.4.5
2 1.18 103 2.98 1014 sec1 . 16 1.66 1027
Diatomic molecule as a linear harmonic oscillator
The diatomic molecule is an example of a linear harmonic oscillator provided that the interatomic force is an elastic one. Consider a molecule to be close to an isolated system. This signifies that two atoms of a molecule make oscillations relative to their CM, so that such oscillation can be reduced to an oscillation of a single body (with the mass equal to the reduced mass system) regarding the motionless fixed point under the action of the same interatomic force. Superpose the origin into point C, which is the CM of two points with masses m1 and m2 (Figure 2.11). Then their coordinates x1 and x2 determine the equilibrium positions of both. For this case, we can write m1x1 m2x2 (refer to Section 1.3.7). Considering the oscillations to be symmetrical, for any instant of time it is fair to say that m1 ( x1 1 ) m2 ( x2 2 ). d m2
m1 C
1
x1 x1+1
x2
2 x2+2
Figure 2.11 A diatomic molecule as a linear rigid harmonic oscillator.
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Simplifying, m11 m22 . Assume for the linear oscillator the interatomic force is an elastic one. This corresponds to the problem in harmonic approximation. Then, a force acts on any atom that is out of its equilibrium position: F (1 2 ). Use Newton’s second law for each atom m1 1 (1 2 ) and m2 2 (1 2 ). Express in the first equation 2 through 1 and in the second equation 1 through 2:
⎛ ⎛m ⎞ m ⎞ m11 ⎜ 1 1 1 ⎟ , m2 2 ⎜ 2 2 2 ⎟ . m2 ⎠ ⎝ ⎝ m1 ⎠ Rewriting these expression as m1m2 1 m1 m2
and
m1m2 2 m1 m2
and summing both expressions, we arrive at m1m2 (1 2 ) (1 2 ). m1 m2 Because 1 2 , where is the displacement .. of one .. atom .. relative to the other, we can write the expression for relative acceleration as 1 2 . Value m1m2/ (m1 m2) is the reduced mass of the molecule, which is denoted by (Section 1.3.9). Then the above equation corresponds to the harmonic oscillation of a single material point under the action of an elastic force :
0,
(2.4.16)
where
.
(2.4.17)
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The equation derived is similar to that describing the oscillations of a spring pendulum. Here the interatomic force plays the role of a spring. This shows that the problem of the atom vibrating in the molecule is reduced to the problem of harmonic material point oscillations with the mass, equal to the reduced molecular mass. A physical system, making oscillations, is referred to as an oscillator. If it complies with eq. (2.4.6), it is referred to as a one-dimensional harmonic oscillator. In the first approximation any molecule can be considered as a classical one-dimensional harmonic oscillator; this is the simplest physical model explaining some (but by no means all) particularities of atom vibrations in the molecule. In Chapter 7 it will be shown that in quantum mechanics a much better approximation is given by a model of a quantum linear harmonic oscillator. However, the next best approximation is a nonharmonic (nonlinear) model (this model is more complicated; it describes atomic vibrations in more detail and introduces new phenomena).
2.5
ENERGY OF HARMONIC OSCILLATIONS
An oscillating body possesses both potential energy U and kinetic energy K. Its total energy E is the sum: E U K. First an expression for the potential energy of the oscillating body is found. When displacing from its equilibrium position, an elastic force acts on the body. The potential energy of the body in this case was determined in eq. (1.5.4): U 12 2 .
(2.5.1)
The time dependence of displacement is expressed by formula (2.2.1). Then the potential energy is equal to U 12 A2 cos2 (t ). The kinetic energy is equal to K
1 2
(2.5.2)
m 2. Since A sin(t),
K 12 A2 2 m sin 2 (t ).
(2.5.3)
K 12 A2 sin 2 (t ).
(2.5.4)
Substituting m2 ,
Now the total mechanical energy combining both U (2.5.2) and K (2.5.3) is E 12 A2 [cos2 (t ) sin 2 (t )]
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or E 12 A2 .
(2.5.5)
It can now be seen that total mechanical energy is defined by the coefficient characterizing the rigidity of the system, and by the oscillation amplitude squared. Therefore, the total mechanical energy is equal to E 12 m2 A2 .
(2.5.6)
It is important to note that the total energy of harmonic oscillations is proportional to the square of amplitude: E A2. Obviously, the system considered is conservative and its total energy is conserved, i.e., only the transfer of kinetic energy into potential energy and back again is taking place (refer to eq. (1.5.4)). The potential energy reaches a maximum at the largest (amplitude) displacement, whereas kinetic energy is at its highest possible when the system crosses the origin. Note that expressions (2.5.2) and (2.5.4) can be presented in the form
U 14 A2 [1 cos2 (t )],
(2.5.7)
K 14 A2 [1 cos2 (t )].
(2.5.8)
and
It can be seen that the oscillation of kinetic and potential energies accomplishes with the doubled frequency 2 in comparison with the initial one. Graphs of functions (t), K(t) and U(t) are presented in the Figure 2.12. The averaged values of kinetic and potential energies of harmonic oscillations for the period are equal to half the total energy, i.e.,
K U 12 E
(2.5.9)
This equation is the consequence of the fact that the average values of sin and cos functions for the period are equal to
T
sin 2 t
T
1 1 sin 2 t dt and cos2 t ∫ cos2 t dt. ∫ T 0 T 0
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A 1T 2
0
t
T
U.K.E E K t
0 U
Figure 2.12 Energy graphs of harmonic oscillations.
2.6
DAMPED OSCILLATIONS
Hitherto, we have considered harmonic oscillations appearing under the action of a single (restoring) force in a system. Such oscillations are called free or natural ones, their frequencies being designated by 0. Strictly speaking, there are no such systems in surrounding macroscopic nature. In real systems, there are forces other than elastic forces; they are distinguished in nature from quasi-elastic forces and appear when an oscillation system interacts with its surroundings. The final result of these interactions is the transformation of the mechanical energy of the moving bodies into heat. In other words, a dissipation of the mechanical energy occurs. The process of energy dissipation is not purely mechanical and for its description another section of physics is required. Within the framework of mechanics we can describe this process by introducing forces of friction or resistance. As a result of mechanical energy dissipation, the oscillation amplitude decreases. The damping oscillations are no longer harmonic ones, since oscillation amplitude changes. Oscillations that, in consequence of energy dissipation, have a continuously decreasing amplitude are referred to as damped oscillations. Consider nearly free damped oscillation with a small resistance. At small oscillation amplitudes, the velocity of the body will also be small; under small velocities the force of resistance is often proportional to the velocity value (refer to eq. (1.3.5))
F r y r,
(2.6.1)
· where is the velocity of a body’s motion and r is the proportionality factor, called the resistance coefficient. The minus sign in the expression of resistance force is stipulated by the fact that its direction is opposite to the velocity of the moving body.
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· Taking the quasi-elastic ( ) and resistance forces (r ) into account, we can write an equation of motion of damping oscillations as m r.
(2.6.2)
Substituting the coefficient by m02 in this expression and dividing both sides by m, we have r 20 0. m
(2.6.3)
Suppose that for damping oscillations the expression sought has the same form as previously discussed: A0 exp(it ),
(2.6.4)
Here is as yet an unknown value. The initial phase is taken as zero, i.e., we begin to measure time when the phase crosses the zero position. To find this quality we can substitute the form (2.6.4) into equation (2.6.3) together with their first and second derivatives: d (t ) A0 ie it , dt
d (t ) 2 A0 i 2 2 e it A0 2 e it . dt 2
Substitute these equations into (2.6.3):
A0 2 e it
r A0 ie it 20 A0 e it 0. m
After reducing on A0eit and changing signs we obtain
2 i
r 20 0 m
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This quadratic equation relative has two roots:
r2 ir 20 2 . 2m 4m
(2.6.5)
We can find the time dependence of displacement by introducing (2.6.5) into (2.6.4). For the sake of convenience introduce two more values:
r 2m
(2.6.6)
and 20 2 .
(2.6.7)
2 Since i 2 i the displacement in damping oscillation can be pre0 sented in the form:
(t ) A0 e i(i )t A0 et eit .
(2.6.8)
The choice of the sign corresponds to the phase shift on . We shall write the solution with a “” sign. Then eq. (2.6.8) can be rewritten as (t ) A 0 et e it .
(2.6.9)
This is the expression sought for damped oscillation. In trigonometric form it can be given as A 0 et cos t .
(2.6.10)
In all the sequent expressions is the frequency of damped oscillations. It is always lower than the frequency 0. The time dependence of the amplitude can be given as A(t ) A0 exp(t ),
(2.6.11)
where A0 is the amplitude in the initial instant of time (at t 0). The constant (refer to (2.6.6)), equal to the ratio of the resistance coefficient r to the doubled system mass, is usually called the coefficient of the damped oscillations.
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Find the time during which the amplitude of the damped oscillations reduces e times, i.e., A(t)/A(t ) e.Using eq. (2.6.11) we can obtain A0 exp(t ) e A0 exp[(t )] or exp( ) e,wherefrom 1.
(2.6.12)
Hence, the damping coefficient is the value reciprocal to the time of amplitude reduction into e times. In order to characterize the process of attenuation, the so-called logarithmic decrement of damping (attenuation) is also used. It is accepted to be equal to the natural logarithm of a ratio of two oscillation amplitudes, separated from each other by an oscillation period: A(t ) . A(t T )
(2.6.13)
A0 exp(t ) ln exp[T ] T . A0 exp[(t T )]
(2.6.14)
ln
Using expression (2.6.11), we obtain ln
Find a physical sense of logarithmic decrement of fading. Let the oscillation amplitude decrease e times after N oscillations. The time , for which a system makes N oscillations, can be expressed in periods: NT. Having substituted this value in (2.6.12), we obtain NT 1. As far as T , we can get N 1 or
1 . N
(2.6.15)
Consequently, the logarithmic decrement of fading is a value inverse to the number of oscillations after which the amplitude decreases e times. In a number of cases, it is suitable to express the dependency of oscillation amplitude from time through the logarithmic decrement of fading. The degree t in expression (2.6.11) can be written according to (2.6.14) as follows:
t
. T
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Then expression (2.6.11) takes the form ⎞ ⎛ A A0 exp ⎜ ⎟ , ⎝ T⎠
(2.6.16)
where /T is the magnitude, which shows the number of oscillation a system accomplishes for time . The approximate values (orders of magnitude) of the logarithmic damping decrements of some oscillation systems are plotted in Table 2.1. Let us now analyze the influence of resistance force on oscillation frequency. When displacing a body from its position of equilibrium and allowing it to return to its initial position, the resisting force will act all the time. This signifies that a body will cover the same distance in a longer time. It means that the period of damped oscillation will be larger than that of free oscillations. From expression (2.6.7) it can be seen that the difference in oscillation frequency becomes larger the larger the damping factor . Under greater resistance force, oscillations degenerate into the aperiodic process. Figure 2.13 shows a graph of the time dependencies (t) and A(t) (at the initial phase 0). The dashed line expresses a change of oscillation amplitude (2.6.11) in the course of Table 2.1 Magnitudes of the logarithmic decrements of damping of some systems System
Decimal logarithm of the oscillating system
Acoustic waves in gases Electric oscillating systems Tuning fork Quartz plate
1 2 3 5
A0e−δt
A0
0
T
A(t+T)
A(t)
T
2T
3T
4T
Figure 2.13 Damping oscillations.
5T
t
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time. If friction in the system increases sufficiently to be commensurate with the magnitude of the free oscillation frequency 0, the oscillations become less and less. Finally, when 0, oscillations become completely impossible (this corresponds to an imaginary magnitude of frequency, refer to (2.6.7)). Then the system becomes a damping one.
2.7
FORCED OSCILLATIONS
If a periodical force is acting on a system in addition to restoring and damping forces, the so-called forced oscillations will take place. Consider the simplest case of forced oscillations when the driven force changes according to the periodical law F F0 cos t ,
(2.7.1)
where is the frequency of the driven force and F0 is its amplitude. In addition to the driven force, a quasi-elastic ( ) and a resistance force (r) are acting on the system as discussed before. The equation of motion can be written in the following way: m r F0 cos t.
(2.7.2)
Let us divide the above equation by mass m: F
r 0 cos t. m m m
(2.7.3)
We will introduce the definitions used in the previous section:..02 / m and . r / 2m; let us denote F0 / m by f0. The equation will then take the form 02 2 f0 cos t. For convenience we present the driven force in the complex form F F0eit. The equation will then be as follows: 2 20 f0 e it . The steady forced oscillations, i.e., oscillations occurring after a time long enough to ensure their stability, will be accomplished with a frequency equal to the frequency of the driven force . Because of the system’s inertia, the displacement will be detained upon the phase . Hence, the solution of the above equation can be rewritten as (t ) Ae it ,
(2.7.4)
where A is the complex amplitude containing the phase multiplier ei. Therefore, the solution of eq. (2.7.2) is now known with accuracy to the phase multiplier and to the magnitude of the oscillation amplitude. Our task is just to find these values.
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. .. Introduce into eq. (2.7.2) the expressions iAeit and A2eit. Therefore, A2eit 2iAeit 02Aeit f0eit, Divide the equation by a common multiplier and find from it the amplitude as a function of the frequency of the driven force:
A()
f0 2 0 2 2i
The complex quantity Z(022) i2 is presented in the denominator whose modulus is Z ( 2 2)2 42 2. This complex number can be presented in the form i
Z Z e , where arctan(2/022). The expression for the amplitude receives the form A(, )
f0 (20 2 )2 42 2
ei .
As can be seen from this expression, amplitude depends both upon and . We are not interested in the dependence on . Determine only the A() dependence. It takes the form A()
f0 (20 2 )2 42 2
.
(2.7.5)
Consider first the case of the absence of a resisting force ( 0). Then eq. (2.7.5) simplifies to A()
f0 2 0 2
.
(2.7.6)
Let us analyze this expression. If 0, then A f0/0 F0/m2 or A f0/2. The damped oscillation amplitude in this case turns out to be practically equal to the displacement, caused by constant force F0 in the system, characterized by the quasi-elastic coefficient . If the frequency is increased (0 0), the oscillation amplitude increases, as can be seen from expression (2.7.6), and under 0 will go to infinity. With a further increase in frequency ( 0) the amplitude decreases and at 0 will practically not depend on the elastic characteristics since in this case it is possible to neglect the natural oscillation frequency 0 in comparison with . Then A f0/2 and at the forced oscillation amplitude will go to zero. Figure 2.14 presents the dependence of forced oscillation amplitude on the force frequency. In the absence of damping at 0 the amplitude reaches infinity. This result is practically unrealistic, since it means an infinitely great energy of oscillatory motion. In fact, due to the resistance force, the oscillation amplitude remains finite. Its magnitude is
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A
F0 0
Figure 2.14 Amplitude of forced oscillations versus a frequency of external driving force in the absence of damping.
determined by the formula (2.7.5). (This expression at 0 turns over to (2.7.6), considered above.) Thereby, the amplitude of the given oscillating system depends on damping coefficient value and correlations of natural and damped oscillation frequencies. Consider now how the amplitude of forced oscillation will behave when damping takes place. With a driven force frequency equal to zero (oscillation will not take place, but will become steady-state displacement under the action of force F0): ACT
f0 20
f0 m
2
F0 .
This coincides with the case of the absence of resistance. The amplitude increases with the force frequency increase because the denominator in eq. (2.7.5) decreases. When w approaches 0, the amplitude of forced oscillation increases, reaches a maximum and then decreases at 0. The increase in amplitude of forced oscillation at the approach of the frequency of the driven force to that of natural frequency is called resonance. In order to determine the resonance frequency in the presence of resistance, we must use the well-known principle for finding extrema A/ 0. Since the denominator in expression (2.7.5) is constant, it is easier to find the extremum of the radicand in the denominator. We obtain 2(20 2res )(2 res ) 82 res 0,
from which we can arrive at the resonance frequency res res 20 22 .
(2.7.7)
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Then the resonance amplitude of forced oscillations will be Ares
f0 2 20 2
.
(2.7.8)
At small damping ( 0) Ares
f0 . 20
(2.7.9)
The dependence of the forced oscillation amplitude A on the driving force frequency under different damping factors is depicted in Figure 2.15. The following conclusion can be made from consideration of this graph: the less the resistance force (i.e., is small), the sharper the resonance peak and the closer the resonance frequency res to the natural frequency 0. On the contrary, at significant resistance the resonance peak is smoothed and shifted to the low frequency region. At small resistance the resonance amplitude is given by eq. (2.7.7), whereas at static shift it is Ast f0/0. Now the ratio of the amplitudes is Ares f 20 0 0 . Ast 2 f0 20
(2.7.10)
=0
A
1
2> 1>0
2 F0 r r2 r1
Figure 2.15 Amplitude of forced oscillations versus a frequency of external driving force in the presence of damping: the smaller the damping the higher the oscillation amplitude; r are the resonance cyclic frequencies.
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Obviously, at 0 the ratio will be very large. This explains the great significance of resonance phenomena in physics and technology. It is widely used when a weak effect is to be measured. The best example of this is radio broadcasting. On the other hand, if the resonance phenomena can lead to some destruction it must be avoided. The resonance technique is widely used nowadays in methods of chemical structure investigations (refer to Chapter 8). Such a property of an instrument as a figure of merit is its critical property. It characterizes the sharpness of the spectral lines in which the main information of the subject studied is included. Let us begin with eq. (2.7.5). Dividing both nominator and denominator by 0 we arrive at
A()
f0 02 ⎛ 2 ⎞ ⎛ ⎞ 1 ⎜ 2 2 ⎟ ⎜⎝ 2⎟ ⎠ ⎝ ⎠ 0
2
. (2.7.11)
0
Value 0 /2 G is called the figure of merit. The above equation can be rewritten using this definition:
A()
f0 20 ⎛ 2 ⎞ ⎛ 1 ⎞ 1 ⎜⎝ 20 ⎟⎠ ⎜⎝ G 0 ⎟⎠
2
.
(2.7.12)
The time of merit characterizes the rate of the energy loss by an oscillating system (oscillator). At small damping E E(t) E(t T ) (T is the period of damping oscillations, which in this case is equal to the period of natural oscillations). Then E(t) E0exp(2t/ ) E0e (0/G)t (because 2/ 2(0/G)). Therefore, ⎡ 2T 2 ⎛ 2T ⎞ ⎤ E E (t ) ⎢1 exp ⎜ ⎟ ⎥ E (t ) E (t ) . ⎠ ⎝ ⎦ G ⎣
The relative energy loss for the period is E 2 . E G
(2.7.13)
Regarding electrical oscillations, the value E/E is called the figure of merit of a vibration contour. The width of the spectral line at half of its height at 0 (its half-width) is called the figure of merit. The values of resonance characteristics in some systems are presented in Table 2.2.
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Table 2.2 Values (up to the order of magnitude) of the figures of merits of some systems System
Decimal logarithm of the natural oscillation frequencies (Hz)
Decimal logarithm of the figures of merit
Decimal logarithm E of the ratio E
Oscillation counter Resonator in a quartz watch Optical spectral line CO2 laser -Radiation of atomic nuclei in Mössbauer effect (refer to Chapter 8).
4 5 1415 13 19
2 4 57 9 915
2 4 57 9 915
EXAMPLE E2.9 A body of weight m 10 g makes a damping oscillation in a viscous media. The resistance coefficient is 2 4 104 kg/sec. Determine the ratio E /E0 of energy loss by the body in time 1 min. Solution. Two forces operate on an oscillating body in the presence of damping. One of them is a quasi-elastic force F , where is the coefficient of the quasi-elastic force and is deformation. The other force is the . force of resistance dependent on the speed of the moving body Fc r r . According to the second Newtonian..law, the equation .. of movement in a projection on an x-axis can be written as m r . Dividing both sides of the equality by mass m, making some replacements r / m 2 and / m 02 and .. . .. rear-. ranging the terms, we obtain (r / m) ( / m) 0 or in compact form 2 02 0 (refer to Section 2.6). Solving this differential equation we obtain the time dependence of the amplitude of damping oscillations A(t) A0 exp(t)*, where A0 is the initial amplitude of oscillations. The total mechanical energy is dependent on amplitude E(t) ½ A2(t) (refer to eq. (2.5.5)) where A(t) is substituted by the equation *. We shall obtain E(t) ½ A02 exp( 2t), where ½ A02 is the initial energy E0 . Therefore, the energy of the damping oscillation’s time dependence can be expressed as E(t) E0 exp( 2t). The lost energy ratio at damping oscillation in time can be found by dividing E E0 E( ) by the initial energy E0:
E E E ( ) E ( ) 0 1 E0 E0 E0
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or E 1 exp(2 ). E0 Returning to the initial definitions, we substitute into r/m and arrive at the final expression E ⎛ r ⎞ 1 exp ⎜ ⎟ . ⎝ m ⎠ E0
After all calculations, we obtain ⎛ 2 104 ⎞ E 1 exp ⎜ 60⎟ 1 exp(1.2) 1 0.301 0.699. 2 E0 10 ⎝ ⎠ That is, the loss of energy is approximately 70% of the initial value.
EXAMPLE E2.10 A body of weight m 0.2 kg is attached to one end of a spring with rigidity
2 N/cm 200 N/m. The body can move along a horizontal pivot without friction. The other end of the spring is fixed (refer to Figure 1.22 and Section 2.7). The oscillations occur in viscous media. An external harmonic variable force operates on the body: F(t) F0 cos t where F0 is the force amplitude value (F0 3 N) and is its angular frequency. For this system, define the resonant frequency res and resonant amplitude Ares. Make calculations for two values of the resistance coefficients r1 0.5 kg/sec and r2 5 kg/sec. Solution: Let us consider the forces working on the body. There are several forces acting in a vertical direction; however, they all mutually compensate each other (according the third Newtonian law) and are therefore excluded from our consideration. Operating along the horizontal direction are: (1) a periodically changing with frequency external force F(t) F0 cos t, (2) an elasticity force Fel , (3) the velocity-dependent force of resistance Fr rv (see Section ..2.6, eq. (2.6.1)). . Therefore, the equation of the body’s movement can be written as: m r F0 2 cos t. This equation was solved in Section 2.7 and the results are res 22 * 0 2 2 characterizes and amplitude Ares f0 / 2 **. There are two frequencies: 0 the driven force and 0 is the natural frequency of the free system. For visualization we first calculate separately the natural frequency 0 / m. 200 / 0.2 31.62 sec1. The fourth significant digit is useful here because it ensures that we do not lose the accuracy at intermediate calculations. 1 r1 / 2m
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0.5 / (2 0.2) 1.25 sec1. The same is valid for 2 12.5 sec1. In the first case 212 02 (3.12 1000), and hence res 0 31.62 sec1. In fact, calculations according to the precise formula* give 31.57 sec1, so the coincidence is up to the third significant digit. In the second case 222 312 sec2 and 02 1000 sec2, and therefore calculations should be executed mainly according the precise formula *: res 1000 312 s1 26.2 sec1. Let us next calculate the resonance amplitude. In the first case of 1 we can use the insignificance of 12 in comparison with 02 (1.56 1000). An approximate formula gives Ares,1
f0 15 0.19 m 19 cm 210 2 1.25 31.6
In the second case (2 12.5 sec1) the approximate formula is invalid (22 156 sec2 is commensurable with 02 1000 sec2) and calculations have to be made according to precise formula **:
A res,2
f0 2 2
20 2
15 2 12.5 1000156
0.0207 m or A res,2 2.07 cm.
In conclusion we can present the static displacement under the action of the force F0(max F0 / ) (Hooke’s law), i.e., max (3/200) m 1.5 cm. One can see that in the first case at small damping the resonance phenomena are more pronounced than at high damping: Ares / max 29 / 1.5 12.7. 2.8 2.8.1
WAVES
Introductory remarks
Oscillations originating from any source propagate further in space. The propagating oscillations are referred to as waves. It was noted in the preceding sections that the mathematical descriptions of different kinds of oscillations are similar, thus allowing a general mathematical description to be made regardless of the type oscillation. Different waves exist (mechanical, electromagnetic, acoustic, etc.), depending on what physical value is “propagated”; herewith their mathematical description will once again be the same. Thus, we shall mostly consider mechanical waves, bearing in mind the possibility of applying the results to other kinds of waves, e.g., electromagnetic waves. Mechanical waves can propagate only in an elastic media. If particle vibrations are agitated in a region of an elastic medium (solid, liquid or gaseous), as a consequence of the
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interaction between particles, this disturbance will be transmitted to surrounding particles, which in turn, will distribute excitation further. In this manner, the wave appears. The process is not instantaneous; a wave propagates with a speed , which depends on the properties of the medium. However, it must be noted that no transportation of the medium’s particles take place, particles oscillate around their permanent equilibrium positions. We can distinguish different kinds of waves by considering how the motion of the constituent particles is related to the direction of propagation. Distinction is made between two kinds of waves. A wave is called longitudinal if the direction of particle oscillations coincides with the direction of wave propagation (Figure 2.16). Longitudinal waves can be agitated in a medium that is elastic in terms of compression and stretching. All media—solid, liquid and gaseous—possess these properties. A wave is called transverse if the medium particles oscillate in a direction perpendicular to the direction of propagation (Figure 2.17). It follows from this definition that the transverse
1
t=0
2
3 4
5
6 7
8
9 10 11 12 13 14 15 16 17
x
wave propagation direction 1 2
t=1/4T
3
4
5
x
t=1/2T 2
1
3
4
4
5
5
9
x
t=3/4T 2
3
1 2
3
9
13
x
1
t=T
4
5
13
9
17
x
Figure 2.16 Generation of a longitudinal wave. t=0
1
2
1
3
2
4
3
t=1/4T
t=1/2T
1
2
3
t=T
2 3
7
8
9 10 11 12 13 14 15 16 17
x
wave propagation direction 4 4
t=3/4T 1
6
5
4
5
x
5 6
5
7
6 7
8 8
9 9
x 10 11 12
13
x
13 9
1 2
3 4
5
Figure 2.17 Generation of a transverse wave.
17
x
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wave can propagate in media that possess the property of elastic deformation of slip. Only solid media possess this property; only in solid media can the transverse wave propagate. Hence, only in solid media can both transverse and longitudinal waves be propagated. The space in which the wave propagation occurs is called the wavefield. The geometrical locus of points that, at a given instant, the propagation process reaches, is called the wavefront. For a periodic wave one can draw a surface through the points that oscillate in phase. This surface is called a wave surface. The extreme wave surface is the wavefront. The direction of propagation in isotropic media is always perpendicular to the wavefront. A line perpendicular to the wavefront is called a ray. From the wave’s viewpoint a ray is an imaginary line along the direction of travel of the wave. A bundle of parallel rays form a beam. In isotropic elastic media all waves propagate at the same speed. Therefore, if the source of waves is tightened down to a point the wavefront is spherical and the wave is also spherical. If the wave front is a plane, a plane wave is produced. If the initiating oscillation is harmonic, the wave produced in isotropic media is also harmonic. 2.8.2
An equation of a plane traveling wave
(x0,t)
For the majority of problems it is important to know the dependence of oscillations of different points of media at a given instant. This dependence can be considered as determined if the amplitudes and phases of oscillation are known. For transverse waves it is also necessary to know the polarization. For a plane one-dimensional polarized wave it is sufficient to have an expression defining the displacement of any wave point of (x,t) with the coordinate x in the instant of time t. Such an expression is called an equation of wave. Consider a so-called traveling wave, i.e., a wave propagated in one direction. Direct the x-axis along the wave propagation. In this case the wavefront is perpendicular to axis x. Let particles of media, just verging on the source of plane waves, accomplish harmonic oscillations according to the harmonic law (0,t) A cos t (Figure 2.18). In Figure 2.18a the T A t (a)
(x0,t)
0 T
A
t (b)
0
(x,t0)
x= A 0
= T x (c)
Figure 2.18 Schematic representation of a traveling wave: (a) oscillations at the oscillation source origin (0,t), (b) oscillations at the distance x from the origin (x , being the delay relative to the source oscillations), (c) deviation of the wave’s particles from their equilibrium positions (x,t0) at a time instant t0. The periods T and wavelength are shown.
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displacement of particles at x 0 is presented. Zero time has been chosen to ensure the initial phase is also zero. Particles lying in this plane are all oscillating in the same phase. Find the expression (x,t) for the displacement of particles that are at distance x from the wave’s source (origin). The wavefront covered this distance in time x/v. This means that the vibrations at point (plane) x will be behind by the time from that in origin. These points will also accomplish the harmonic oscillations but with propagation delay . In the absence of damping the oscillation amplitude is constant. Therefore, ( x, t ) A cos (t ).
(2.8.1)
This is the plane traveling wave equation. As already noted, the equation allows us to define a displacement of the media particles with coordinate x at the instant of time t. The phase of oscillation [t (x / )] depends on two variables: particle coordinate x and time t. At a given fixed instant of time the phases of the different particles will, generally speaking, be different. However, it is possible to select particles oscillating in the same phase (in-phase). The phase difference is 2m (where m 1, 2, 3). The shortest distance between two particles of a traveling wave, oscillating in phase, is called the wavelength . Find the relationship of wavelength with other values, characterizing the wave propagation in a definite media. In accordance with the definition of the wavelength we can write x ⎞ x⎞ ⎛ ⎛ ⎜t ⎟⎠ ⎜⎝ t ⎟⎠ 2 ⎝
or, after cancellations, / 2. Since 2 / T, then T .
(2.8.2)
This expression allows another definition of the wavelength: wavelength is the distance a wave can propagate for a time equal to the period of oscillation. The traveling wave equation develops thereby a double periodicity: on the coordinate x and on time t. One can, for instance, fix a particle coordinate (x const.) and consider its displacement as a function of time. Alternatively, one can fix a moment of time (t const.) and consider particle displacement as a function of coordinates. So, standing on a pier one can take a picture of the surface of the sea at time instant t, or having thrown an object into the sea (i.e., having fixed a coordinate x), one can check its oscillation in time. Both these cases are given as graphs in Figure 2.18. Equation of wave (2.8.1) can be written in another way: ⎞ ⎛ ( x, t ) A cos ⎜ t x ⎟ . ⎝ ⎠
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The ratio / is usually defined by letter k, referred to as the wavenumber k
.
(2.8.3)
2 .
(2.8.4)
Since (2/T) and (2/T) (2/), k
Hence, the wavenumber shows how many wavelengths can be placed in a length of 2 units. One can now rewrite the equation for the traveling wave in the most popular form ( x, t ) A cos(t kx ).
(2.8.5)
Find the relation of the oscillation phase difference of two particles with a difference of their coordinates x x2 x1. Using eq. (2.8.5) can be written as (t kx1 ) (t kx2 ) k ( x2 x1 ). Then kx or, according to (2.8.4),
2 x.
(2.8.6)
The plane wave propagating in an arbitrary direction can be expressed as ( x, t ) A exp[i(t kr )], where r is a radius vector, drawn from the origin to a point where a particle occurs, k is the wave vector equal modulo as eq. (2.8.4) and coinciding with the direction of propagation (or to the normal to the wave surface). The exponential form is also appropriate for wave description. So, in the case of a plane wave propagating along x-axis ( x, t ) A exp[i(t kx )]
(2.8.7)
and in the general case of an arbitrary direction ( x, t ) A exp[i(tkr )].
(2.8.8)
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The traveling wave equation can be obtained as the solution of a differential wave equation referred to as the wave equation. Knowing the solution in the forms (2.8.5) and/or (2.8.7) we can find the wave equation itself. Differentiation of the equation of a plane wave (x, t) twice upon the time and upon the coordinate gives 2 k 2 A exp[i(tkx )]k 2 x 2 2 2 t 2
and equating the value from both equations, we can reduce it to 2 1 2 1 2. 2 2 2 k x t
Taking eq. (2.8.3) into account, we can write 2 1 2 x 2 2 t 2
(2.8.9)
This is the one-dimensional wave equation. In the general case this equation looks as follows:
2 2 2 1 2 x 2 y 2 z 2 2 t 2
(2.8.10)
or
1 2 , 2 t 2
where is the Laplace operator ⎛ 2 2 2 ⎞ ⎜ 2 2 2 ⎟. y z ⎠ ⎝ x
(2.8.11)
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The speed of propagation of the constant phase point is referred to as the phase speed. In other words, this is the propagation speed of the wave crest or hollow as well as any other wave point preserving the constant phase. The wavefront is the surface of the constant phase. Consequently, the speed of the wavefront is just the phase speed. Therefore, the speed, determined by eq. (2.8.3),
, k
(2.8.12)
is also the wave phase speed. The same results can be obtained by finding the speed of propagation of the wave points with constant phase, t kx const. Finding the dependence of coordinates upon the time x 1 (t const.) we can derive the speed of points with a constant phase dx / dt k
/k, that coincides with eq. (2.8.12). The equation of the plane traveling wave in the opposite direction is written as (x, t) A cos(t kx). The phase speed in this case is negative dx . dt k The phase velocity in a given media depends both on the properties of the medium and the frequency of the source of oscillations. This relationship is called dispersion, whereas the media relates to dispersive media. (One should not think that expression (2.8.12) is the dependence discussed. The point is that in the absence of dispersion the wavenumber k is directly proportional to the frequency , therefore, ( / k) const. Dispersion takes place when (k) is nonlinear (refer to (2.9.4).) The plane traveling wave is monochromatic if the source of oscillations is harmonic. Eqs. (2.8.5) and (2.8.7) describe monochromatic waves. In linear media this corresponds to the fixed wavelength. It must be emphasized that the amplitude A and frequency of a monochromatic wave is accepted as being independent of time. This means that a monochromatic wave must be infinite in time and space, i.e., is an idealized model. Any real wave is not monochromatic from a formal point of view (refer to Section 7.2). However, the longer the time such a wave is maintained, the nearer the wave is to being monochromatic. In practice, a wave is considered as monochromatic if it lasts sufficiently long.
2.8.3
Wave energy
As mentioned earlier, there is no macroscopic transfer of matter accompanying a wave. This, however, does not mean that there is no energy transfer with wave propagation. On the contrary, forcing every particle to oscillate, a wave carries that energy which is
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consumed in its creation. This energy can be easily calculated. To make this calculation, the total kinetic energy of all particles participating in oscillation must be counted:
WN
2
2 mmax max V . 2 2
(2.8.13)
Here, W is the total energy of all particles in the volume V, N and m are their number and mass, respectively, and is the media density. To estimate the max value, we can proceed from expression (2.2.4): max A, as the trigonometric term is equal to unity. Substituting this equation into eq. (2.8.4) we can find the total energy contained in the volume V of the oscillating media:
W
2 A2 V 2
or
W
2 A2 w, 2 V
(2.8.14)
where w (W /V) is the volumetric energy density. Find the wave energy flux , i.e., the energy carried by a wave through the area S perpendicular to wave propagation. So it is defined as a scalar flux averaged upon the area S. It can, however, be different both at different points of the area and in different directions. In order to characterize the energy flux locally the value of the flux density j is introduced: j (r )
d , dS⬜
(2.8.15)
which is equal to the energy flux through the unit area perpendicular to the propagation direction. The energy flux density can depend on the direction. Therefore, it has to be defined as a vector numerically equal to d/dS⊥. Therefore, j (r )
d , dS⬜
where / is the unit vector of the wave propagation direction.
(2.8.16)
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Now find connection of the energy flux density with the wave phase velocity. Choose the area S⊥ perpendicular to the unit vector / and calculate the energy carried through this area in the unit time interval t (Figure 2.19). In the time interval t through the area defined above there will be energy E will carried through; it contains in the cylindrical volume with a height t and a base area S⊥. This energy can be expressed as the product of the volumetric energy density w (2.8.14) and volume V. The energy carried will be E wS⊥t and the energy flux E/t wS⊥. The scalar energy flux density is j /S⊥ w. Taking the vector characters j and into account we obtain j w.
(2.8.17)
The total energy flux d (2.8.15) can be determined as d jdS⬜ .
(2.8.18)
In order to obtain a more general significance of the last expression, attach to the elementary area dS the vector character by taking into account the different orientations of the area dS regarding the vector field. Figure 2.20 shows the disposition of the area dS⊥ as part of the more general area dS projected onto a plane perpendicular to the vector j. Attribute to the scalar value dS a vector character by multiplying dS by the normal unit vector: dS dSn,or dS dS cos ,
∆S⊥
(2.8.19)
n
∆t
jn
Figure 2.19 An elementary flow d of a vector through the area dS normal to .
dS⊥
dS n
n
j
Figure 2.20 An elementary flow d of a vector j through the area dS at their arbitrary orientation.
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where is an angle between vectors n and j. At 0 the equation dS = dS takes place. Now expression (2.8.18) can be given in another form: d ( j dS),
(2.8.20)
i.e., elementary flux d of the energy flux density j through the elementary area dS is the scalar product of vectors j and dS. The total flux through the surface S can be written more generally as ∫ j dS.
(2.8.21)
∫ jn dS or ∫ j dSn . s s
(2.8.22)
S
and
Expression (2.8.21) is the particular case of the vector flux calculations through the surface. It can also be used in evaluation of liquid mass consumption in pipes, electrostatic and magnetic strength vectors, etc. (refer to Chapters 4 and 5).
2.8.4
Acoustic Doppler effect
Hitherto, we have considered the propagation of waves in a media in a coordinate system in which both source and receiver are unmovable. Let us examine further how a wave frequency will change when turning from one inertial coordinate system to another. In other words, what frequency will a wave detector measure if it moves with respect to the source in the same direction? The relationship between frequency of the relative motion of the wave source and detector, which was originally established by P. Doppler, is well known both in acoustics and optics; in spite of the fact that they are related to different principles, the resulting mathematical expressions are mainly the same. Note, however, that we deal with nonrelativistic cases. We will restrict ourselves only to acoustic waves. In the resting reference system K, let a plane wave be generated propagating in direction x. Denote the frequency by . Determine what frequency will the wave detector perceive in system K moving with velocity V0 along the same direction x relative to the system K (Figure 2.21). Apply the wave equation to the system K, i.e., (x,t) aexp{it ikx}. Turning to another coordinate system K we should use the Galileo transforms (1.3.1). In the other system, the equation takes the form ( x, t) a exp{it ik ( xV0 t)} a exp{i( kV0 )t ikx}.
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155 s
d
V
s
d
V
Figure 2.21 An acoustic Doppler effect; V is the mutual speed of source and detector, s and d are emitted and measured frequencies, d and s are detector and source.
Comparing these two equations, it can be seen that the detector in system K will perceive a wave number process with a frequency equal to kV0 and the same wave k k. Hence, k /, where is the propagation wave speed in system K; then ⎛ V ⎞ ⎜ 1 0 ⎟ . ⎝ ⎠
(2.8.23)
Note that there is no contradiction between the change of frequency and stability of wavenumber: In fact, according to eq. (1.3.1) the velocity of wave propagation in system K is equal to V0. Therefore, k
1 (V0 t) k. V0 V0
We use here the relation (2.8.23). This relation allows us to determine how much the frequency measured by the detector det differs from that emitted by the wave source sor provided they both move relative to each other. Suppose now that the wave source is resting and the detector moves away from it at a speed V0 (Figure 2.21). Then in expression (2.8.23) sor and det. Therefore, ⎛ ⎞ det sor ⎜ 1 ⎟ sor . ⎝ V0 ⎠
(2.8.24)
Changing the motion direction, the relative sign of the velocity also changes. Hence, in this case ⎞ ⎛ det sor ⎜ 1 ⎟ sor . ⎝ V⎠
(2.8.25)
An analogous result is obtained when the detector is unmovable but the source is moving.
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The Doppler effect is widely used, since it gives the possibility to change wave frequency, changing, for instance, velocity of source motion. This is exactly the way to obtain the resonance absorption of -rays in MÖssbauer effect in -resonance spectroscopy (refer to Chapter 8). 2.9 2.9.1
SUMMATION OF WAVES
Superposition of waves
Now consider a situation in which, instead of one source, there are several sources of waves (oscillators). In a certain area of space, these waves can interact with each other. Here again we come across the very important physical phenomena – the principle of superposition. It relates equally to waves and to many other types of excitations. Its essence is exceedingly simple. Suppose that instead of a single source of excitation there are a number of sources in space (it can be mechanical vibrations, oscillation of electrical charges, electrical currents, etc.). What result will the measuring instrument register, accepting simultaneously all excitements from all sources? If each component does not influence the others, the total result will simply be the sum of the separate excitations. This precisely implies the superposition principle. This principle is common for many phenomena, but its mathematical expression can sometimes be different depending on the nature of the events considered, e.g., vector or scalar. The principle of wave superposition is not valid for all cases but only for the so-called harmonic sources and linear media. A medium can be considered linear if its particles are under the action of quasi-elastic forces. Otherwise, a medium is nonlinear. Very unusual and important phenomena can appear in the latter case, e.g., the propagation of ultrasound and/or laser rays in nonlinear media. Extremely interesting and technically important phenomena can appear. Scientific and technical investigations dealing with nonlinear phenomena are referred to as nonlinear acoustics and optics. Although nonlinear effects are of great importance in certain modern devices, we will only consider linear effects further. When applied to waves, the principle of superposition affirms that each wave is propagated regardless of the presence in the given media of other sources of waves. This can be mathematically expressed as N N i(t ki x ) ( x, t ) ∑ i ( x, t ) ∑ Ai e , i1
i1
(2.9.1)
where N is the number of wave sources, (x, t) is the total wave. Consider superposition of two waves generated by two sources 1(x, t) A1 cos(t k1x) and 2(x, t) A2 cos(t k2x). Fix an arbitrary point M and examine the result of the superposition at this point. Fixing the point, we transform a wave into oscillations: 1,M(t) A1 cos(t k1xM) and 2,M(t) A2 cos(t k2xM), since a product k1xM can be considered as a phase. In order to find the resulting oscillation process (t) we should sum 1 and 2 at the point M: (t) 1(t) 2(t). Such a problem was solved in Section 2.3.1
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(refer to Figure 2.7 and eq. (2.3.1)). Correspondingly, we can write a total amplitude A through partial amplitudes A1 and A2: AM A12 A22 1 2 A1 A2 cos(2 2 1 ).
(2.9.2)
The value AM depends on the difference of oscillation phases 21. In Section 2.3.1 the situation was analyzed in detail. In particular, the total amplitude AM can be changed from zero to 2A provided A A1 A2 and the phase difference remains stable in time. In order to observe interference, the phase difference should be constant. This can be obtained if the wave’s sources are coherent and, vice versa, two sources are referred to as coherent if the phase difference remains constant. If waves are coherent the sources are coherent as well. The methods of experimental interference phenomena will be considered in more detail in Chapter 6, which is devoted to wave optics. 2.9.2
Standing waves
A wave that appears as the result of the superposition of two similar waves coming from opposite directions is referred to as a standing wave. Find the equation of the standing wave. Let us assume that a flat traveling wave with amplitude A and angular frequency , extending along the positive direction of an axis x, meets a counter wave of the same amplitude and frequency. Equations of these primary waves can be written in trigonometric form as follows: 1 A cos(t kx) and 2 A cos(t kx), where 1 and 2 are the displacement of the medium’s points caused by two running waves. According to the superposition principle, in an arbitrary point with coordinate x in time instance t the displacement is the sum 1 2 or A cos(t kx) A cos(t kx). Using a wellknown trigonometric equation
cos cos 2 cos
cos , 2 2
we can obtain (t ) 2 A cos kx cos t.
(2.9.3)
There are two trigonometric terms in this expression. The first term, cos kx, is a function of a point’s coordinates only and can be considered as a variable amplitude of the standing wave changing from point to point, i.e., Ast 2 A cos kx .
(2.9.4)
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(Since the oscillation amplitude is essentially positive, the sign on the modulus is written.) The second term, cos t, depends only on time and describes harmonic oscillation of the point with the fixed coordinate x. Thus, all wave points make harmonic oscillations with different (dependent on x) amplitudes. It is clear from eq. (2.9.4) that the amplitude of a standing wave depends on x and changes from zero up to 2A. Points in which the amplitudes of oscillations are maximum are referred to as antinodes of the standing wave. The points with permanently zero displacements are referred to as nodes of a standing wave (Figure 2.22). Find coordinates of the standing wave nodes. Write the obvious expression 2A cos kx 0 whence cos kx 0. Therefore, kx ( 2n 1)/2, where n 0, 1, 2,…. Having replaced wavenumber k by its expression k 2/ we obtain (2 )x ( 2n 1) (/2). So we find the nodes’ coordinates xknt (2n 1) , n 0,1,2,... 2
(2.9.5)
The antinodes’ coordinates can be found from eq. (2.9.4). Indeed, the point’s coordinate in which oscillation acquires the maximum displacement satisfies equation kx n with n 0, 1, 2, … So we can obtain the antinodes’ coordinates: xank n , n 0,1,2,... 2
antinode
node
(2.9.6)
node
t=0 =0 t=1/4T =0 t=1/2T =0 t=3/4T =0 t=T =0
st
Figure 2.22 Standing transverse waves. Open circles represent a position of oscillating particles, whereas arrows show a direction and the magnitudes of their speed.
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The distance between adjacent nodes (or antinodes) is referred to as the standing wave wavelength. It can be seen from eqs. (2.9.5) and (2.9.6) that this length is /2, i.e., st . 2
(2.9.7)
Nodes and antinodes are shifted relative to each other on a quarter of the original wavelength. In Figure 2.22 an origin (x 0) is imposed with the node point at n 0 (2.9.6). For t 0 deviations of all the wave’s points pass through equilibrium positions and thus the wave is degenerated into a straight line. This instant of time is taken as zero. However, at this very instant each point (except for the nodes) possesses a certain speed specified in the figure by arrows. Displacements achieve a maximum at t T/4, the wave is represented by sine, but the speed of each point of the wave becomes equal to zero. The instant t T/2 again corresponds to the straight line but the speeds of all points are directed to the opposite side, and so on. Comparison of traveling and standing waves reveals the following difference. In a plane traveling wave all media points oscillate with identical amplitude, but their phases are different and repeat in t T. In a standing wave all points (from node to node) make oscillations in one phase, but the amplitudes of their oscillations are different. The points of the wave shared by a node oscillate in antiphase. In fact, standing waves do not transfer energy. An example of a standing wave is the oscillation of a flexible cord fixed rigidly at one end, with the other end in the hands of the experimenter; the latter generates oscillations. When an antinode reaches the fixed end (Figure 2.23a) the wave affects a fastening. Under Newton’s third law the fastening produces a reciprocal influence on the cord, equal in magnitude and oppositely directed. It generates a return wave; the displacement of the cord’s
node antinode
1 2
x st (a)
st =
1 2
0 1 4
x
(b)
0
Figure 2.23 A model of a wave reflection in the case of a more dense (a) and less dense (b) medium; the solid curves represent a cord vibrations and a dashed lines show the “less dense” media.
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point at a border is in the opposite direction to the displacement of the “incident” wave. As a result, both waves are in an antiphase; therefore, the loss of half a wavelength occurs at the border point. An example of the mobile (less dense) border is a thin weightless braid connecting the cord end with the fastening (Figure 2.23b). The cord end generates free oscillations running along the cord wave in the opposite direction. Analysis of the wave reflections in these two cases shows that at reflection from the more dense border (Figure 2.23a) loss of half the wavelength occurs, and there is a phase shift of . Reflection from the less dense border is not accompanied by a phase change; therefore, at the junction point of the cord and the braid there will always be an antinode (Figure 2.23b). 2.9.3
String oscillations
The oscillations of a tightly stretched string can be considered as a special case of standing waves: on both fixed ends of the string there is the reflection of a running wave resulting in the formation of standing waves. This steady picture of standing waves will take place at the string fastening in any of the nodes; they do not seem to participate in oscillations. Hence, the integer of half wavelength m can be confined on the length of string between the fastenings: L m , m 1,2,3. 2 The definite wavelength corresponds to these oscillations
m
2L . m
(2.9.8)
We can transform these wavelengths into frequencies:
m
m. m 2L
(2.9.9)
The vibrations of a string with the smallest frequency when only one antinode is confined corresponds to the basic tone in a sounding string (Figure 2.24, above). Other vibrations with multiple frequencies are overtones. The rules presented operate not only in many musical instruments and the formation of standing waves in them (in the resonator of a guitar, for instance) but are also used in models of an ideal black body, in quantum mechanics, in physics of solid-state properties, etc. All of these are discussed in other sections of the book.
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L 1=2L 1=
2L
2=1/2 1 2=21
3=1/3 1 3=31 4=1/4 1 4=41
Figure 2.24 String oscillations.
EXAMPLE E2.11 A transverse wave runs along an elastic cord at a speed 15 m/sec. The period of oscillations of the cord points is T 1.2 sec, amplitude A .2 cm. Define (1) wavelength, (2) phase of oscillation , displacement , velocity and acceleration .. of the point at a distance x = 45 m from the source of waves at time instant t 4 sec and (3) phase difference of the oscillation of two points lying on the cord at a distance x1 20 cm and x2 30 cm from the source of waves. Solution: (1) According to definition, wavelength is T. Substituting the values we obtain 18 m. (2) The wave equation is A cos (t (x/))*. The phase of the point oscillations with coordinate x in the time instant t stays under cos sign (t (x/)) or (2/T ) (t (x/)). Calculation gives 5.24 rad or 300°. We can find the displacement from the equation * substituting the A and values: appears equal to 1 cm. The velocity of the point can be. found by the time derivation of : substituting all the values derived we arrive at 9 cm/sec. The acceleration is the second time derivative on the displacement
x ⎞ 4 2 A ⎛ A2 cos ⎜ t ⎟ 2 cos . ⎝ ⎠ T Therefore, ¨ 27.4 cm/sec2. (3) The phase difference is related to the distance between the points x by equation (2/)x. This gives 3.49 rad or 200°.
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EXAMPLE E2.12 A wall MN is located perpendicular to the wave at a distance l 4 m from the source of a plane wave with frequency 440 Hz. Define the distance from the source of the wave to the points in which the first three nodes and first three antinodes of the wave arising as a result of the superposition of two waves, running to and reflected from the wall, occur. Assume the wave’s velocity to be 440 m/sec (Figure E2.12). Solution: Let us choose an axis x directed perpendicularly to the wall and the origin at a distance l from the reflected wave source. Then the equation of the wave will be written as A cos(t kx )ⴱ . As in a point with coordinate x the reflected wave will come back covering twice the distance (l x) and the reflected wave will lose a phase /2 at reflection; the reflected wave equation can be written as
M
O
x'2
x'1
x'0
x0 x
x2
x1
N l
A cos t k ( x 2(2l x )) and further
A cos t k (2l x ))ⴱⴱ . . Summing two equations * and ** we can find the standing wave:
1 2 A cos(t kx ) A cos(t k (2l x )).
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According to trigonometry we can obtain
2 A sin k (l x )sin (t kl ). Since A sin k(l x) does not depend on time it can be considered as the standing wave amplitude (being taken as modulus 2A sin k(l x)). From this equation we can obtain the nodes’ and antinodes’ coordinates. Nodes occur when 2A sin k(l x) 0 and consequently k(l xn) n; since , /v, k 2v/. Therefore, 2v(l xn) n. The nodes’ coordinate can be obtained from the last equation: xn l (n /2v). The first three nodes’ coordinates are x0 4 m, x1 3.61 m, x2 3.23 m. Correspondingly the antinodes appeared when k (l xn) (2n 1)(/2); therefore, 4vxn 4vl (2n 1) and the antinode’s coordinates are xn l(((2n 1) )/4v). The first three antinodes will appear at x0 3.81 m, x1 3.42 m, x2 3.04 m. The results are depicted in Figure E2.12. 2.9.4
Group velocity of waves: wave package
By definition the monochromatic wave is boundless in space. The real wave is always limited in space and is emitted during a limited interval of time, which is why it cannot be monochromatic in full measure. However, any real wave can be considered as a result of the superposition of a large number of strictly monochromatic flat waves. As a result of interference, in one part of space these waves strengthen each other, and in other parts extinguish. Such waves have some features that can be discovered using a simple model of superposition of two plane monochromatic waves. Let two plane cross-sectional polarized monochromatic waves with equal amplitudes be distributed along an axis x. Such waves are described by equations: 1 A cos(1t k1x) and 2 A cos(1t k2x). Because of the superposition principle a combined wave can be represented as 1 2 A cos(1t k1x) A cos(1t k2x), or
k k1 ⎛ 1 2 A cos ⎜ 2 t 2 ⎝ 2 2
k k1 ⎞ ⎛ 1 x ⎟ cos ⎜ 2 t 2 ⎠ ⎝ 2 2
⎞ x⎟ ⎠
Suppose now that the angular frequencies 1 and 2 and wave vectors k1 and k2 differ only slightly, i.e., 2 1 ( 1), k2 k1 k (k k1). Therefore, we can write
1 2 1 2
and
k1 k2 k1 . 2
(2.9.10)
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Then k ⎛ 2 A cos ⎜ t ⎝ 2 2
⎞ x ⎟ cos(1t k1 x ). ⎠
(2.9.11)
We can see that the superposition of two monochromatic waves with equal amplitudes with slightly different frequencies and wavenumbers produces a new wave with variable in space and time amplitude k ⎛ 2 A cos ⎜ t ⎝ 2 2
⎞ x⎟ . ⎠
(2.9.12)
Since and k are small in comparison with 1 and k1 a change of amplitude will take place comparatively slowly. Such a wave is called a wave with modulated amplitudes. Let us determine the rate of crest displacement of such a composed wave. In order to solve this problem we can repeat the method used in Section 2.8.2 when we evaluate the phase velocity rate. The crest corresponds to the constant phase in eq. (2.9.11), i.e., k t x const., since the model package velocity dx/dt g x / t appears to be 2 2
equal to /k. If now we make the model more realistic and take not just two but a continuous set of waves with k lying in narrow interval k, and dependence (k), close to linear, eq. (2.9.11) becomes more complicated. The expression for the crest velocity distribution of the wave package turns out to be obtained from eq. (2.8.13) at k 0 (i.e., replacement of final increments by differentials). The corresponding group of waves (or a wave package) is depicted schematically in Figure 2.25. The group of waves transfers energy and momentum at a velocity that is generally different from the phase velocity, and is characterized by group velocity g. The group velocity is
2a∆k sin
2a∆k
2a∆k
−3π
−2π
−π
0
π
2 3π
2a∆k
2π v=
dω t - x ∆k dx t=const
Figure 2.25 A wave packet.
2 5π
3π
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1
2
k
Figure 2.26 An example of a dispersion curve.
the speed of movement of a point corresponding to any fixed wave amplitude. Remember that the phase velocity of a monochromatic wave is /k, whereas the group velocity is g d / dk. In the absence of dispersion (frequency is linearly dependent on wavenumber) the group and phase velocities coincide. At the presence of dispersion, group and phase velocities are different. An example of function (k) is given in Figure 2.26. In region 1 the frequency linearly depends on k; here the phase and group velocities coincide. In region 2 the function (k) is nonlinear; the group and phase velocities differ. PROBLEMS/TASKS 2.1. A point accomplishes oscillations according to the formula x A cos(t ), . where A 4 cm. Determine the initial phase if x(0) 2 cm and x (0)0. . .. 2.2. Determine the maximum values of speed x max and acceleration x max if a point accomplishes harmonic oscillation with amplitude A 3 cm and angular velocity /2 sec1. 2.3. A physical pendulum consists of a rod of mass m and length l 1 m and of two small balls of masses m and 2m fixed to the rod at lengths l/2 and l, respectively. The pendulum makes small oscillations relative to a horizontal axis passing perpendicularly to the rod through the middle of the rod. Determine the frequency of the harmonic pendulum oscillations. 2.4. A hoop of mass m made of thin metal is suspended on a long nail hammered into a wall. It makes harmonic oscillations in a plane parallel to the wall. The radius of the hoop is R = 30 cm. Calculate the oscillation period T. 2.5. A point oscillates according the law: x A cos(t ). In some instant of time . .. the displacement is 5.0 cm, its velocity is x 20 cm/sec and acceleration x 80 cm/sec2. Find the amplitude A, angular velocity and phase (t ) for this time instant. 2.6. Two similar harmonic oscillations with the same periods T1 T2 1.5 sec and amplitudes A1 A2 are summarized. The initial phases are 1 /2 and 2 /3.
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2.7. 2.8.
2.9. 2.10.
2.11.
2.12. 2.13. 2.14. 2.15.
2.16.
2.17.
2.18.
2. Oscillations and Waves
Determine the amplitude A and initial phase of the resulting oscillations. Draw a vector diagram of the superposition in a scale chosen. Two tuning forks sound simultaneously. Their frequencies are v1 440 Hz and v2 440.5 Hz. Determine the resulting beatings period T. An MP oscillates according to equation x A cost, where A 8 cm, /6 sec1. At the instant of time when a force F reaches a value 5 mN for the first time, the potential energy U of the MP becomes equal to 100 J. Find this instant of time t and its corresponding phase t. A small weight is suspended on a spring, which has been expanded by x 9 cm. Find the period T of the weight when it will oscillate freely. A mathematical pendulum of l1 40 cm in length and a physical pendulum in the form of a thin straight rod of length l2 60 cm oscillate around a common horizontal axis. Find the distance a between the rod CM and the oscillation axis. An oscillation period of a pendulum l 1 m in length during the time interval t 10 min was lowered two times. Determine the logarithmic decrement of damping . A body of mass m 5 g accomplishes a damping oscillation. After t 50 sec the system loses 60% of its energy. Determine the friction coefficient. Find the number N of total oscillations of a system during which the system energy decreases by n 2 times. The logarithmic decrement of damping 0.01. Under the weight of an electromotor, a bar on which it is fixed deflects on h 1 mm. At what frequency of the motor rotation n can the danger of resonance appear? A carriage of m 80 ton has 4 bow springs. The rigidity of each spring is k = 500 kN/m. At what speed will the carriage begin to intensively swing on the rail’s conjunction if the rail is L 12.8 m in length? The amplitudes of forced harmonic oscillations at the frequencies v1 400 Hz and v2 600 Hz are equal to each other. Determine the resonance frequency vrez. Neglect damping. A swollen log whose section is constant along the whole length, has plunged vertically into the water so that only a small part (in comparison with length) remains above water. The period of oscillation of the log is equal to T 5 sec. Determine the whole length L of the log. Mercury with a mass of 200 g is poured quickly into a U-shaped tube with a cross section of S 0.4 cm2, open at both ends. Determine the period of oscillation of the mercury in the tube. Neglect the viscosity of the mercury. The density of the mercury is 13.6 g/cm3. ANSWERS
2.1. /3 rad. . .. 2.2. x max 4.71 cm/sec, x max 7.40 cm/sec2. 2.3. v (1/) (3 /7)( g/l) 0.652 Hz. 2.4. T 2 2R /g 1.55 sec.
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167
.. .2 2 2 2.5. x /x 4 sec1, (2/) 1.57 sec, A x x 7.07 cm, (t ) arcos(x/A) (/4) rad. 2.6. A 3.68 cm, 0.417 rad, x A cos(t ), where (2/T) 4.19 sec1. 2.7. T 2 sec. 2.8. t 2 sec, t /3. 2.9. T 0.6 sec. 2.10. a1,2
2.11
; 10 and 30 cm.
1 2 l1 l12 (1/3 )l 2 2
2 ᐉ A1 ln 2.31 103. ᐉ g A2
2.12. r 9.16 105 kg/sec. 2.13. N 35. 2.14. n
1 g 16 sec1 . 2 h
2.15. y
L
k 10.2 m sec. m
2.16. vrez 510 Hz. 2.17. L 6.21 m. 2.18. T 2
m 0.86 sec. 2 gS
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–3– Molecular Physics
3.1 KINETIC THEORY OF IDEAL GASES 3.1.1
Introductory remarks
So far, we have considered the behavior of small amounts of particles and can follow almost each of them. We come now to a description of physical systems consisting of enormous numbers of particles, moving chaotically (i.e., randomly) in a limited space. Such objects are referred to as statistical systems, related to which are gases, liquids and real (nonideal) solids. Here, special approaches are necessary. By a thermodynamic system, we understand a set of macroscopic bodies that can accomplish an exchange of energy both among themselves and with bodies that are outside the system. These laws are referred to as statistical ones. In fact, there is a difference between these and Newton’s laws that have an exact character; they cannot give any information on a particular particle, but describe the general properties of a system that has a statistical character. Statistical physics, however, operates both by integrated values and allows one to evaluate uncertainties (fluctuations) as well. As a rule, in molecular physics, the latter are negligibly low, therefore we will not pay much attention to them. These systems as a whole are described by a restricted set of parameters unified by corresponding equations of state. Such systems can be in equilibrium states if their parameters do not depend on time, remaining unchanged for an infinitely long time. We will deal mainly with such systems. Very important examples are systems that are in thermal equilibrium with other systems and the environment. The parameters describing such a system, for gases in particular, comprise mass (M ), pressure ( p), density (), temperature (T ), and, in some respects, the volume (V ) that a system can occupy. Any statistical system can be in various states described by a set of parameters. These parameters do not always have certain values. If, for example, the pressure (or temperature) in different elementary volumes of gas is different, it is impossible to attribute certain general values to the whole gas. Such a state is referred to as nonequilibrium. If such a nonequilibrium closed system is left alone for a period of time, the temperature and/or pressure inside the system will equalize in due course. The system will have passed into a state of equilibrium. From this point of view equilibrium systems possess maximum disorder, whereas nonequilibrium systems have a smaller disorder parameter. 169
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A state in which the parameters are identical at any point of the system is referred to as an equilibrium state; in the absence of external excitement, it remains unchanged in time. From all possible statistical systems the equilibrium closed systems are allocated. For such systems the physical theory, referred to as equilibrium thermodynamics or simply thermodynamics, is well developed. Thermodynamics is the phenomenological doctrine of heat. Classical thermodynamics asserts that “the isolated thermodynamic system cannot spontaneously change it’s state”. This statement is sometimes referred to as the zeroth law of thermodynamics, another assertion of which is that “If two thermodynamic systems are in thermodynamic equilibrium with some third body they are in thermodynamic equilibrium with each other”. The transition of a system from one state to another is called the process. The process consists of a sequence of nonequilibrium stages. If the relaxation time of the system is much less than the time of the external disturbing effects, the system will have enough time to reach equilibrium; such a process consists of a sequence of equilibrium states (strictly speaking, it should be an indefinitely slow process). The process consisting of a continuous sequence of equilibrium states is referred to as the equilibrium or quasiequilibrium process. Thus only an indefinitely slow process can be equilibrium, although at a slow enough transformation, a real process can come nearer to equilibrium than a closed.
EXAMPLE E3.1 Determine the average distance l between the centers of two neighboring molecules in water vapor at normal conditions and compare it with the diameters of water molecules. Solution: Suppose that all molecules of water vapor are at equal distances to each other; this distance will coincide with an average distance l. Each molecule in this case will have a volume l3. If N is the number of molecules then VNl3 is the volume occupied by N molecules. Dividing this equality by V, we obtain (N/V)l31; 3 this is the concentration n of water molecules and we can obtain 具l典l1/兹n苶. If the water vapor is under normal conditions (p1.013105 Pa and T300 K) the molecule’s concentration is known (nnL2.69.1025 m3 is the Loschmidt number.) Executing the calculations we arrive at
冓l 冔
1 3
nL
1 3.33 109 m. 2.69 1025
The same formula is valid for the estimation of the molecule’s diameter in liquid water since in this case, the molecule centers are at distances practically equal to 3 the molecule’s diameter d1/兹n苶; in this case n is the molecule’s concentration in the water. In order to evaluate the diameter of the water molecule one should determine
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the molecule concentration in the liquid phase. Since n(NA/M) where is density of the water then,
d
1 3
n
3
M . N A
Substituting the constants and executing the calculations, we arrive at ⎛ 18 103 ⎞ d ⎜ ⎝ 6 1023 103 ⎟⎠
1/3
⬇ 3.11010 m.
Thus, under evaporation the average intermolecular distance increases by approximately 11 times. EXAMPLE E3.2 Chlorine, m 0.8 g in weight and at a temperature 2 K is placed in a vessel of capacity V. Find the pressure of the gas if it is in part dissociated. The degree of dissociation 0.4 (i.e., 40% of gas is dissociated). Solution: Partly dissociated gas can be considered as a mixture of two gases: monatomic mdis m and a mole mass M/2 and nondissociated mass mnondis m(1) and mole mass M. According to Dalton’s law, the pressure is an additive value, i.e., each of the gas components yields “its own pressure”: monoatomic pdis gas and diatomic gas pnondis. Their sum is the total pressure of a mixture of the two component gas ppdispnondis. Find these two values using the ideal gas equation pV
m RT . M
Then, 1 m RT and V M 2 1 (1 )m pnondis. RT and V M mRT (1 ). p MV pdis
Transform all values into the SI system and execute calculations: p
8 104 8.31 2.103 Pa 2.62 105 Pa. 35.5 103 2 103
p 262 kPa or P 2.59 atm (1 atm 1.013. 105 Pa).
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3. Molecular Physics
Distribution function
For the definition of the average value of any characteristic of a particle system, statistical physics offers a special function: the so-called function of distribution of particles on a given physical property. We shall consider the notion of the distribution function in more detail. In a multiparticle system some particles may have different characteristics. To find the distribution of particles of a given system on any characteristic means to find the relative number of particles for which the given physical value has a numerical magnitude lying in the given interval of values. This distribution is nonuniform; for example, the concentration of particles with low speeds is much less than the concentration of particles with average speeds. We shall designate by x a value of distribution upon which a system is examined; this can be speed, energy, etc. We shall designate dN(x) as the number of particles, the parameter x of which lies in a limit from x up to x dx. Certainly, dN(x) increases as wider the interval and greater the number of particles in the system; that is dN(x) ⬃ Ndx or dN(x) ANdx, where A is the proportionality coefficient. This coefficient should be a function of x since in identical intervals at different x intervals, the number of particles can be different, i.e., A f(x). Then dN(x) f(x)Ndx, or dN ( x ) f ( x )dx. N
(3.1.1)
The magnitude dN(x)/(N) is a share (or relative amount) of particles, the parameter x of which lies in an interval from x to x dx. This quantity can be treated in the probability sense w. If one managed to measure the property of any specified particle then it is a probability dN(x)/N, that the particular particle has a property lying in the interval x–x dx. Therefore, dN ( x ) dw( x ) f ( x )dx. N
(3.1.2)
The value, f ( x)
dw( x ) dN ( x ) , dx Ndx
(3.1.3)
is referred to as the distribution function density. Figure 3.1 presents an example of a distribution function. In the graph, the hatched area is dw(x) f(x)dx. We also know definitely that a particular particle is in a vessel, therefore,
∫ f ( x)dx ∫ dw( x) 1. This is the normalization condition.
(3.1.4)
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f(x) dw(x)=f(x)dx f(x)
x
Figure 3.1 A model of a distribution function.
Find out how we can determine the average value. Begin first with a case where the values x can accept any integer data. Then, 冓 x冔
x1 x2 x3 xN 1 N N
N
∑ xi . i1
If the value xi has ki particles and the overall measurement is N, then, 冓 x冔
k1 x1 k2 x2 k3 x3 1 k1 k2 k3 N
N
N
i1
i1
k
N
∑ ki xi ∑ Ni xi ∑ wi xi . i1
In the last expression, the fraction ki /N under the sum sign expresses the share of particles, which with probability wi determines the value xi. If the property x can accept a continuous set of values, the summation should be changed to integration: 冓 x 冔 ∫ xdw( x ) ∫ xf ( x )dx.
(3.1.5)
Integration is accomplished over all values of x. If one has to obtain the averaged value of a function, the prescription is the same: 冓冔 ∫ ( x ) f ( x )dx,
(3.1.6)
in the case of normalized f(x) function, and
冓( x )冔 ∫
if it is not normalized.
( x ) f ( x )dx
∫
f ( x )dx
,
(3.1.6 )
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Thus we have obtained a procedure for calculating an averaged value starting with the distribution function, which is supposed to be known. 3.1.3
An ideal gas model
The simplest gas model is the model of an ideal gas, the essence of which is as follows: 1.
2. 3.
4. 5.
A gas is represented by a set of an enormous number of molecules. It allows one to obtain average values of its physical parameters and to reduce deviations (fluctuations) from them. The thermodynamic balance in the ideal gas is established only due to an exchange of energy between molecules at their collisions. Molecules are represented by very small spherical particles, the total volume of which is negligibly small in comparison to the volume that the gas occupies. It is accepted that molecules do not interact at intermolecular distances larger than their effective diameter. The potential curve for the model of the ideal gas looks like that presented in Figure 3.2. If we imagine molecules as “incompressible balls” with radius ro, the potential energy of their interaction is equal to zero at distances larger than their diameter 2r0, and increases sharply at r 2r0. It is supposed that the collisions of molecules occur under the laws of absolute elastic collision (Section 1.5.5). It is supposed also that there is no additional physical influence on the system as a whole.
In this approximation, it is possible to obtain an exact equation (within the framework of statistical laws) of the ideal gas state. A few words should be added. There are no ideal gases in nature: neither hydrogen gas nor uranium hexafluoride is ideal. There exists only the notion of an ideal gas state. Each gas can be both in the state of an ideal gas or behave as a real gas. All depends on the temperature and pressure. The model of the ideal gas is very convenient for understanding many gas state peculiarities.
U(r)
def
r
Figure 3.2 Potential curve U(r) of an ideal gas molecules interaction.
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175
General equation of an ideal gas
One of the basic features of gas is its property to exert pressure on the walls of the vessel in which it is contained. Let us determine this pressure for a state of ideal gas. On the way, we shall obtain a number of other important equations. In spite of the fact that we will be using a considerably simplified model, we will obtain results conterminous to our exact consideration. In addition to those simplifications that we have already made when introducing the ideal gas model, we will also make some additional assumptions that will help us to simplify our calculations in order not to lose generality. In particular, we will: (1) count all molecules as identical, spherically symmetric and moving at the same speeds (let this speed be the average root square velocity rms兹具 苶2苶); 典 (2) consider that all molecules move with equal probability along three coordinate axes, i.e., 1/6 part of all molecules moves along each direction; one of the axes is chosen to be perpendicular to the vessel wall. Choose a cylindrical element with the cross-section S and a side l in the vessel close to the wall. Each molecule hitting the wall obtains from it a momentum p 2m (refer to eq. (1.5.18) and Figure 1.37). In turn, the molecule transfers the momentum p 2m to the vessel’s wall. Assuming a molecule-wall interaction time , we can obtain the force produced by this impact F 2mrms /, and acting on the wall by a single molecule. In order to find the total force, we need to find the number of molecules that hit the wall in time . If we choose a cylinder of length equal to l rms (Figure 3.3), then all molecules in the cylinder moving toward the wall hit the wall; their number is (1/6)n.2mrms. Then the force exerted upon the wall from the molecules in the cylinder will be Fp(1/6)nSrms(2mrms/)(1/3)nS 2rmsm, and the pressure will be
p
2 F 2 m y rms n . S 3 2
(3.1.7)
x X
S
l
Figure 3.3 Model of gas pressure on a vessel’s wall.
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Taking into account the fact that the actual speeds of all the molecules vary, we should take the average values of the speed squared, 冓 冔
my 2 , 2
(3.1.8)
and the pressure, p 23 n冓冔.
(3.1.9)
Therefore, the pressure acquired by the ideal gas on the vessel’s walls is equal to 2/3 of the volumetric density of the translational kinetic energy of the molecules.
2 my 2 p n . 3 2
(3.1.10)
Since n具典N具典/VK/Vw, where w is the volume averaged kinetic energy density. Therefore, the pressure acquired by the ideal gas on the vessel’s walls is equal to 2/3 of the volumetric density of the translational kinetic energy. Another form of this equation can be obtained by its multiplication with volume in moles pVm 冢 23 冣nVm 冓冔.
(3.1.11)
Since pVmRT and nVmNA, then RT(2/3)NA具典. The ratio (R/NA) is referred to as the Boltzmann constant (1.381023 J/K). Then, 冓冔 23 T .
(3.1.12)
i.e., the average kinetic energy of the translational energy of the molecules of the ideal gas is equal to 2/3 of the product of the Boltzmann constant and absolute temperature. Therefore, product T is a measure of the thermal movement of molecules. It follows from the last formula that absolute temperature T is a measure of the average kinetic energy of the translational motion of molecules. Two important conclusions follow from this: at zero absolute temperature the translational kinetic energy of each molecule is zero; their motion stops. The second conclusion is that the ideal gas temperature can be measured in energy units. In reality, at a room temperature of 300 K, the averaged value of energy 具典 is 1.381023300 J41021 J25 meV; it is equivalent to saying “the gas energy is 300 K” or “the gas temperature is 25 meV”. There exists a relation between the pressure and absolute temperature. It can be obtained from comparison of eqs. (3.1.11) and (3.1.12). We can obtain p n T .
(3.1.13)
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Multiplying this equation by molecule mass m we can obtain the expression, T
3.1.5
mp
(3.1.14)
Absolute temperature
Let us determine the absolute temperature included in expression (3.1.14). Then we will establish the relation between various temperature scales. It is known, for example, that at water freezing temperatures, the density of air (basically, air consists of nitrogen) at sea level (pressure equals 105 Pa) is equal to 1.255 kg/m3. The mass of a molecule of nitrogen is 4.681026 kg. To what absolute temperature in Kelvin’s scale (K) does the temperature of water freezing (namely 0 °C) correspond? From eq. (3.1.14), it follows that °C corresponds to zero.
T ( C)
4.68 1026 1.01105 273.15K. 1.38 1023 1.255
Thus, according to definition, water freezes at 273.15 K. According to the formula (3.1.12) at T 0 K, any translational motion of molecules stops. Such a temperature is referred to as absolute zero. It corresponds to 273.15 °C. Both in degrees Kelvin and Celsius, the difference between water freezing and boiling is equal to 100 degrees; therefore, in both of them the measurement of one degree is identical. Therefore between Celsius and Kelvin scale there is a ratio: T (K) T ( C) 273.15. Experiments show that – 273.15°C (T 0 K) is the lowest possible temperature limit. In practice, different thermometer approaches are used: at very high temperatures one uses the laws of thermal radiation (refer to Chapter 6.6); at average temperatures—the thermal expansion of hydrogen or helium at constant pressure and at low temperatures— the characteristics of solids according to the laws of quantum physical statistics (refer to Chapter 9.3) are used. There are other scales of temperature also, for example, some countries use the Réaumur scale (°R), and in the English-speaking countries the Fahrenheit scale (°F) is used. These scales are connected to the Celsius scale by the equations: T ( R) 冢 45 冣 T ( C) T ( F) 冢 95 冣 T ( C) 32,
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where 0( C) 0( R) 32( F) and 100( C) 80( R) 212( F).
3.2
DISTRIBUTION OF MOLECULES OF AN IDEAL GAS IN A FORCE FIELD (BOLTZMANN DISTRIBUTION)
In an ideal gas, a large number of molecules are considered as noninteracting particles, i.e., the potential energy does not appear in this case. However, in an external force field this situation varies and molecules have a potential energy due to the influence of external forces. The question arises as to how molecules are distributed in this external field. In the absence of any external influence, because of chaotic thermal movement, the gas uniformly fills the given volume. If, however, an ideal gas is in an external force field, the distribution of molecules changes; it becomes dependent upon the potential energy of the molecules in this field. 3.2.1
An ideal gas in a force field: Boltzmann distribution
Let us first find the distribution of a molecule concentration of an ideal gas in a uniform one-dimensional external force field. Consider the field acting on the molecules to be conservative and uniform. Choose a unique z-axis coinciding with the direction of a force vector, along which we shall examine the molecules’ distribution. In the gas, imagine there to be two identical plates with an area S that dispose perpendicularly to the z-axis with a distance dz between them (Figure 3.4). Because of the action of external forces, the pressure upon the upper plate will be higher than on the lower plate.
z
p+dp
p
U-dU
dz
U
Figure 3.4 Gas in an external force field U(z). Boltzmann distribution function.
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The pressure difference dp is equal to the total force working on all molecules being of volume S dz acting on the plates’ areas: dp
F ( z )n( z )S dz F ( z )n( z ) dz S
(3.2.1)
where n(z) is the molecule concentration at height z and F(z) is the force acting on a single molecule in a layer with coordinate z. According to the problem condition, the force is conservative, which means that the field is potential. In this case, there exists a relation between the force F(z) and the potential energy U(z) (1.4.32) dF –dU(z)/dz. Therefore,
dp n( z )
dU ( z ) dz n( z ) dU ( z ). dz
(3.2.2)
Assume that the gas is ideal and so its pressure and concentration are proportional to each other (eq. (3.1.13) below). The gas is considered to be uniformly spread, therefore, the temperature is the same at all points. From (3.1.13) one obtains dp T dn.
(3.2.3)
Changing the pressure to concentration in eq. (3.2.2), we obtain T dn–n dU. Separating the variables we obtain dn/n(dU/T). Integration gives ln n (U/T ) ln C. We can rewrite this expression ln (n(z)/C)(U(z)/T ). And further, n(z)C exp((U(z)/T )). Assume that at a z-level the concentration is n0. Then the constant C is equal to n0. Therefore, finally ⎛ U (z) ⎞ n( z ) no exp ⎜ . ⎝ T ⎟⎠
(3.2.4)
The expression obtained relates the molecular concentration of the ideal gas in a force field n(z) as well as p(z) with the potential energy of a single molecule U(z) at uniform temperature; the law is referred to as Boltzmann’s law (or the Boltzmann distribution). A graph is presented in Figure 3.5a from which it is apparent that the greatest concentration of molecules takes place in those points of a field in which the potential energy is minimum; we accept this potential energy to be the zero level. With an increase in the potential energy, the concentration of molecules decreases. At UT, the molecule concentration will be e time less than at zero level. The distribution presented in the figure corresponds to some certain temperature. Figure 3.6 shows the family of curves corresponding to different temperatures. At high enough temperatures, the influence of a force field on the molecule distribution decreases; the energy of chaotic molecular movement also increases. That is, at high temperatures, the concentration of molecules will be equating; gas will regularly fill the whole given
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n(U) p(U)
U
(a) n p
h
(b)
Figure 3.5 (a) Bolzmann distribution n(U); (b) barometric height distribution function.
volume. On the contrary, a decrease of temperature will lead to a sharp change of concentration with z. The force field becomes more effective. As the concentration of molecules and their pressure are proportional to each other, we can similarly write for pressure: ⎛ U (z) ⎞ p( z ) po exp ⎜ . ⎝ T ⎟⎠
(3.2.5)
In this formula p0 and p(z) are the pressure in the points where the potential energy equals zero and U(z), respectively. 3.2.2
Barometric height formula
Consider a particular example: the distribution of molecular concentration (or pressure) in a field of gravity not very high above the earth’s surface. In this case, the gravitational field
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n(z) n0 1
181
T=∞
T1 T2 < T1 T→0 K U
Figure 3.6 Boltzmann height distribution function at different temperatures.
can be counted as being homogeneous (see Section 1.4.5). Then the potential energy of the gas molecules is described by the formula U(h) mgh where m is the molecules’ mass, and h is the height above the earth’s surface. Substituting this expression for potential energy in the general formula we obtain ⎛ mgh ⎞ p(h ) po exp ⎜ . ⎝ T ⎟⎠
(3.2.6)
This is the barometric height formula. It allows us to calculate the atmospheric pressure above the earth’s surface, a graph of which (Figure 3.5b) is analogous to that presented in Figure 3.5a. In fact, it is assumed that the atmospheric temperature is constant and convection absent. The barometric formula can be rewritten in a different form. Remember that m/ (mNA/TNA) (M/R) (M is a mole mass and R is the universal gas constant). Therefore, ⎛ Mgh ⎞ p(h ) po exp ⎜ . ⎝ RT ⎟⎠
(3.2.7)
If a gas represents a mixture of different gases (like air), every component should be considered independently. Applying the barometric formula to the partial pressure of every component of the mixture, we can see that the pressure (concentration) of gases with greater mole mass decreases faster than light gases. This should lead to an enrichment of the higher layers of the atmosphere by light gases, for example, hydrogen. This, however, is not the case: tests of air at different heights reveal approximately identical percentages of light and heavy gases. This shows that our approximation is incorrect: the temperature of the atmosphere at different heights is different. Moreover, it is impossible to ignore macroscopic streams of air (convection). If there was no thermal movement of molecules, they would all fall to the ground under the action of gravitational force, and all the air would concentrate as a layer, several meters in thickness, on the surface of the ground. If there were no gravitational force field, the molecules would scatter throughout open space. The atmosphere is
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obliged for its existence to the competition of the thermal chaotic motion of molecules and gravitational forces. Thus, in atmosphere, a quite definite distribution of molecules is established. Notice that the earth’s gravitational field is completely unable to keep gas in a current balance. According to Boltzmann’s formula, a nonzero (let it be very small) concentration of molecules is present at any height. In fact, it is possible that as a result of manifold casual collisions a molecule can gain a speed greater than the second space speed ( 11.2 km/sec) and can leave the earth. Therefore, the earth’s atmosphere continuously, but very slowly, loses its light component. The moon, in particular, has already lost its atmosphere as the gravitational field of the moon is much weaker than that of the earth’s.
EXAMPLE E3.3 In the cabin of an airborne plane, the barometer shows an identical pressure of p 79 kPa all the time because of what the pilot counts as the flying-height h1 constant. However, the temperature of the outside air changes from T1 5°C to T2 1°C. What uncertainty h in the definition of height has the pilot supposed? Assume pressure p0 at the ground to be normal. Solution: To find the solution we have to use the barometric height (bh) formula (refer to Section 3.2.2). The barometer can show constant pressure p at various temperatures T1 to T2 onboard only in the case where the plane is not at height h (which the pilot counts constant), but at some other height h2. Write the barometric height formula (3.2.7) for these two heights:
⎛ Mgh1 ⎞ ⎛ Mgh2 ⎞ . p po exp ⎜ and p po exp ⎜ ⎟ ⎝ RT1 ⎠ ⎝ RT2 ⎟⎠ Take the logarithm from both sides of equations:
ln
⎛ Mgh2 ⎞ po ⎛ Mgh1 ⎞ p . ⎜ and ln o ⎜ ⎟ p ⎝ RT1 ⎠ p ⎝ RT2 ⎟⎠
Find the difference of heights h : h h2 h1
R ln( po p) (T2 T1 ) . Mg
The test of the measurement units gives the correct values. Substitution of the numerical values give the results h 28.5 m. The sign shows that the plane is at a height 28.5 m lower than the height supposed.
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183
Centrifugation
One of the practical applications of the Boltzmann distribution is centrifugation, which is used in technology mainly for mixture separation. The basic part of a centrifuge is the cylindrical rotor rotating with high angular velocity relative to the axis of the cylinder; this rotation creates a centrifugal inertia force field. The rotor is filled with mixtures of gases to be processed, suspensions, isotopes mixes, etc., and rotation is started. The particles with the greater weight will concentrate near the rotor walls whereas, the lighter ones concentrate in the center. In this way, the division of the fractions of the mixture is achieved. There are many types of centrifuge; all are subdivided, first of all, according to the separation factor Fr, the ratio of centripetal acceleration near the rotor’s wall to the acceleration of free fall, Fr an/g. It can be seen that the separation factor is dimensionless and for normal industrial centrifuges, it is approximately 3500. Usually centrifuges are used for the separation of disperse systems (particles with sizes of 10–50 m). In the socalled supercentrifuges, Fr is about 3500; these are used for separating emulsions and highly disperse (less that10 m) suspensions. In laboratory practice, ultracentrifuges have Fr reaching 1.2106 min1. These are applied mainly in the research of fibers, viruses and pigments. Ultracentrifuges are used for isotopes separations. The distribution of molecules along the rotor radius depends on the relative mass of the isotopes. By selecting fractions in the rotor center and at the wall, and repeating the process of centrifugation, it is possible to achieve significant enrichment of gas by a given isotope. The efficiency of division depends on the difference between the masses of two isotopes; therefore, the separation of heavy isotopes is no more complex than separation of light ones. Let us find the concentration distribution of different particles in a centrifuge depending on the distance of a cylindrical layer from a rotation axis. Choose the reference system connected with the rotor; direct an axis r along the rotor radius. Such a system is not inertial. Each particle of mass m at distance r feels the action of centrifugal force F (r ) m2 r.
(3.2.8)
Thus, all particles are in centrifugal forces (of inertia). Such a field is not uniform. Nevertheless one can use the concept of potential energy caused by the presence of forces of inertia. Therefore, the relation between centrifugal force and potential energy can be used (see eq. (1.4.32)): dU (r ) F (r )dr. After integration we obtain
U (r )
m2 r 2 C. 2
(3.2.9)
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n(r) M2 M1
n0
=0 1 2
0
R
r
Figure 3.7 Molecule distributions in a centripetal force field.
This expression can be substituted into eq. (3.2.4). We obtain ⎛ m2 r 2 ⎞ ⎛ m2 r 2 C1 ⎞ ⎛ C ⎞ n(r ) C exp ⎜ ⎟ C exp ⎜ 1 ⎟ exp ⎜ . ⎝ T ⎠ T ⎠ ⎝ 2T ⎟⎠ ⎝ 2T Denote the molecule concentration at r 0 as n(0), i.e. n(0) C exp(C1/T). Then we arrive at ⎛ m2 r 2 ⎞ n(r ) n(0) exp ⎜ , ⎝ 2T ⎟⎠
(3.2.10)
⎛ M 2 r 2 ⎞ n(r ) n(0) exp ⎜ . ⎝ 2 RT ⎟⎠
(3.2.11)
and
The particle concentration increases with increasing r. This expression confirms that the heavy particles concentrate near the walls (Figure 3.7).
EXAMPLE E3.4 Radon (M 0.222 kg/mole) is placed in a centrifuge’s rotor with radius R 0.2 m at temperature T 300 K. Determine the ratio of concentration n(R)/n(0)) of radon atoms near the rotor’s walls to that on the centrifuge axis if frequency of the rotor rotation is equal to n 12103 min1. Solution: The most rational way to solve this problem is to use a noninertial system of coordinates in the framework of a D’Alambert principle. We shall connect the coordinates system with the rotating rotor. Radon atoms, in this case, will
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possess the so-called centrifugal potential energy that is caused by centrifugal forces of inertia. In this case, a Boltzmann distribution can be given in the form n(r) n(0)e(Ucf (r)/T), where Ucf is the centrifugal potential energy. First find the expression for the centrifugal force Fcf. It was shown in Section 1.2.1 that centripetal acceleration is 2/r and correspondingly, the centripetal force is Fcf –man m(2/r) m(2r2/r) m2r. So, centrifugal force is directed along the radius from the rotation axis. According to eq. (1.4.34), the centrifugal force is F cf –(dU cf /dr), therefore dU cf –F cf dr. Therefore, after integration U(r)–(1/2)m2r2C. Accepting U(0) 0 C, we obtain U(r) –(1/2)m2r2. The Boltzmann equation in the centrifugal reference system is now n(r) 2 2
e2(m r /T). Because of the fact that we know the rotation frequency n but not 2 2 2 (2 n) and (m/) (M/R) we finally arrive at n(r)/e(2 Mn r /T). Keeping in mind that n 12103 min–1 200 s1 and executing calculations we obtain a numerical result ⎡ 2 2 0.222 (200)2 (0.2)2 ⎤ n( R) 2.81 exp ⎢ ⎥ e 16.6. n(0) 8.31 300 ⎣ ⎦ 3.2.4
Boltzmann factor
We have started considering the behavior of those systems consisting of an enormous number of particles, for some special cases. In particular, we have obtained the one-dimensional Boltzmann distribution. It is easy to generalize this approach to a system in three-dimensional space. Let dN be the number of particles near a point x, y, z in a volume dV dxdydz. Keeping in mind eq. (3.2.4), it is possible to write dN ( x, y, z ) n( x, y, z )dV or, ⎛ U ( x, y, z ) ⎞ dN ( x, y, z ) n0 exp ⎜ ⎟ dx dy dz. ⎝ T ⎠
(3.2.12)
This expression describes the relationship of molecules of ideal gas concentration and their coordinates in an external three-dimensional force field. It can be seen that the probability for the gas particles to be in volume dV around a point x, y, z is proportional to the exponential multiplier ⎛ U ( x, y, z ) ⎞ exp ⎜ ⎟, ⎝ T ⎠
i.e., the higher the potential energy in a given point of a field, the less the probability of finding molecules there. This multiplier is referred to as the Boltzmann factor (multiplier).
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The significance of the Boltzmann factor consists of the fact that it establishes an interrelation between two competing tendencies: the aspiration of the molecules to their ordering under the action of a force field and the tendency to disordering under the action of molecular chaotic thermal motion with average energy T. So, if there was no thermal motion (T 0), all molecules would concentrate near a point with U 0; if there was no ordering action of a force field, all molecules would scatter uniformly all over space. This particular result is described by the Boltzmann factor. This factor is common in deciding many questions in various areas of physics. The Boltzmann factor plays an analogous and important role in chemical kinetics. Here, it describes the distribution of molecules on their internal energy. As for the majority of endothermic reactions, overcoming a potential barrier occurs basically due to the energy concentrated on internal degrees of freedom. The Boltzmann factor defines their kinetics. In Figure 7.14, a qualitative example of a potential energy curve of interatomic interactions in a chemically reacting system is presented. Only those molecules that have a sufficient stock of energy are capable of overtaking a potential barrier Q in height (so-called activation energy) at temperature T. The Boltzmann probability for a molecule to have the necessary stock of energy at temperature T is defined by factor exp
Q . T
(3.2.13)
This defines its importance in chemical kinetics. The Boltzmann factor determines the exponential (temperature) part of Arrhenius’ formula for the ratios of the majority of chemical reactions. This factor is common to the solution of many problems in various areas of physics and chemistry.
3.3
DISTRIBUTION OF THE KINETIC PARAMETERS OF AN IDEAL GAS’ PARTICLES (MAXWELL DISTRIBUTION)
3.3.1 The Maxwellian distribution of the absolute values of molecule velocities The distribution of molecules based on their potential energy is described by eq. (3.2.4). Compare it to the distribution function definition (Section 3.1.2). These two expressions are almost identical with one exception: in Section 3.1.2, the total number of particles is N whereas, in eq. (3.2.4) the constant multiplier is n0 in a point U 0. Hence, they differ by a constant term. Therefore, one can write ⎛ U ( x, y , z ) ⎞ f ( x, y, z ) C exp ⎜ ⎟. ⎝ T ⎠
(3.3.1)
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i.e., the distribution function of molecules in a force field on coordinates is proportional to the Boltzmann factor. With sufficient evidence, we can assert that the distribution of molecules on the full mechanical energy is described by an analogous expression ⎡ U ( x, y, z ) K ( y x , y y , y z ) ⎤ ⎛ E⎞ f ( x, y, z, y x , y y , y z ) C exp ⎜ ⎟ C exp ⎢ ⎥ (3.3.2) ⎝ T ⎠ T ⎣ ⎦ Then, dw( x, y, z, y x , y y , y z )
dN ( x, y, z, y x , y y , y z ) N
⎡ U ( x, y, z ) K ( y x , y y , y z ) ⎤ C exp ⎢ ⎥ d . T ⎣ ⎦ (3.3.3)
The formula requires an explanation. Here, there is the full mechanical energy U K E, being a function of the system’s state, i.e., dependent on particle coordinates and components of molecular speeds (or momentums). The factor d dxdydzdxdydz is an element of a configuration space (eq. 1.3.38). Constant C is not yet determined, but can be found from normalization. This is the Maxwell–Boltzmann distribution. The coordinate distribution function has already been obtained (Section 3.2.1). Therefore, this part of the equation can now be withdrawn from consideration, leaving only those terms that depend on the components of the molecules’ speed
dw( y x , y y , y z )
dN ( y x , y y , y z ) N
⎡ K (y x , y y , y z ) ⎤ C exp ⎢ ⎥ dy x dy ydyz . T ⎣ ⎦
(3.3.4)
What is the physical meaning of an elementary volume d? To uncover this, let us draw an imaginary coordinate framework (in a velocity’s space) and allocate the elementary volume in a configuration space (Figure 3.8a). The physical meaning of dV dx dy dz z
dy
z
dz dx
d
0 x (a)
y
z y x
0
y
x (b)
Figure 3.8 Maxwell distribution: the transition from the Cartesian (a) to the spherical symmetric coordinate (b) in the velocities space.
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consists of the fact that volume d takes into account only those molecules whose velocity vectors begin in the origin and ends in an elementary volume dV; i.e., these vectors are fixed both in length and in direction. Remember our problem: we have aimed to derive the distribution of molecules on kinetic energy. The last is defined by the velocity squares, i.e., they are not dependent on the direction of molecular movement. Thus, we are dealing with distribution of the speeds regardless of the direction of molecular movement. This means that from an orthogonal elementary volume we should pass to an elementary volume in the form of a thin spherical layer which takes into account only the vector length dV 4 2d. Now we can understand that the function of distribution should look like ⎛ my 2 ⎞ f (u)du C exp ⎜ 4 y 2 d y. ⎝ 2T ⎟⎠
(3.3.5)
We now have to calculate the constant C . Keeping in mind a table integral
∫x
2
exp(ax 2 )dx
0
1 , 2 a3
and normalization condition (Section 3.1.5), we obtain C (m/2 T)3/2. Finally we obtain 4 ⎛ m ⎞ ⎟ ⎜ ⎝ 2 T ⎠
32
f ( y )d y
⎛ my 2 ⎞ y 2 exp ⎜ dy ⎝ 2T ⎟⎠
4 ⎛ m ⎞ ⎟ ⎜ ⎝ 2 T ⎠
32
f (y )
⎛ my 2 ⎞ y 2 exp ⎜ . ⎝ 2T ⎟⎠
and (3.3.6)
This function describes the molecular speeds distribution of the ideal gas. The resultant curve is presented in Figure 3.9 together with the constituent functions. Let us examine the relationship obtained. In the area of small speeds, the exponential multiplier is close to unity. When the speed increases sufficiently the power function increases slowly in comparison with the drop of the exponent. The curve has a welldefined maximum. There are three characteristic speeds of this distribution: the most probable speed p, the average molecule’s speed 具典 and the root mean square rms speed. The value of the most probable speed can be found by calculating an extremum of the distribution function: yp
2T 2 RT . m M
(3.3.7)
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f() 2 2 e
Figure 3.9 A Maxwellian distribution f() on molecule’s speed. f()
p<>rms
Figure 3.10 Maxwell distribution and characteristic speeds: the most probable p, the averaged 具典 and root mean square rms velocities.
The specified average molecule’s speed 具典 and the root mean square speed rms can be obtained using table integrals 冓y冔
y rms
8T 8 RT m M
冓y 2 冔
3T 3 RT . m M
(3.3.7 )
(3.3.7 )
All these values are different, and are presented in Figure 3.10. The geometrical sense of average speed consists of the fact that it divides the area under a curve f() into two equal parts. The root mean square rms characterizes the molecular energy and was used in Section 3.1.3. For instance, we consider the distribution of molecules of nitrogen on their speeds at room temperature. The most probable speed is approximately 420 m/sec. The division of all molecules into groups is shown in Table 3.1. It can be seen from the table that 60% of all molecules are grouped near the central area. The relative number of slow and fast molecules is low. At the same time, at any temperature, there are always some molecules with speeds considerably exceeding the most probable value.
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Table 3.1 Relative amounts of nitrogen molecules having their speed in given intervals (at room temperature) Velocity intervals (in m/sec)
Relative amounts of molecules (in%) having their speed in a given interval
0–100 100–200 200–300 300–400 400–500 500–600 600–700 700–800 800–900 900–1000 Higher than 1000
0.6 5.0 7.0 10.0 20.0 18.0 10.0 9.0 8.0 7.0 5.4
f() T1 T2 T3
Figure 3.11 Maxwell distribution at different temperatures.
Temperature has an essential influence on the kinetics. From eq. (3.3.7) it can be seen that the maximum of the Maxwellian curve shifts toward the higher speeds (Figure 3.11). Although the curves flatten, the area under every curve (i.e., the overall number of particles) remains the same. When gas is heated, the number of molecules with high speeds increases though their total amounts remain constant. EXAMPLE E3.5 Knowing the molecules’ speed distribution find (1) the most probable p, (2) the average speed 具典 and (3) the root-mean square rms value of speeds. Solution: (1) The molecules’ speed distribution is described by eq. (3.3.6). The tabular integrals used are:
3 2 ∫ x exp(ax )dx 0
1 3 and ∫ x 4 exp(ax 2 )dx (5 2) . 2 2a 8a 0
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In order to find the most probable p speed one should determine the extremum of f()-function, i.e., to calculate the first derivative of that function on and equate it to zero. This gives df ( y ) ⎛ m ⎞ 4 ⎜ ⎝ 2 T ⎟⎠ dy
32
⎡ ⎛ my 2 ⎞ ⎤ ⎛ my 2 ⎞ 2 my y exp ⎢2 y exp ⎜ ⎜⎝ 2T ⎟⎠ ⎥ 0. T ⎝ 2T ⎟⎠ ⎢⎣ ⎥⎦
This equation corresponds to three roots: 0, and p兹2苶 苶T 苶/m 苶. The first two of them are trivial, the third is the one sought. (2). To find 具典, we should use eq. (3.1.5) and calculate an integral 具典f()d. 0 Therefore, 冓y冔
2 ⎛ me2 ⎞ 2m3 8T 1 2 m 3 ⎛ 2T ⎞ 3 d y y . exp ⎜ ⎟ ∫ ⎜⎝ 2T ⎟⎠ m 2 (T )3 ⎝ m ⎠ (T )3 0
(3) To find the root mean square speed
y rms
冓y 2 冔
2m3 (T )3
⎛ my 2 ⎞ 3T 4 y exp ∫ ⎜⎝ 2T ⎟⎠ d y m . 0
The relationship between the values are y p : 冓 y 冔 : y rms 2 : 8 : 3 1.41 : 1.59 : 1.71.
EXAMPLE E3.6 An ideal gas is in normal conditions and occupies a volume V10 mL. Determine the number of molecules whose speed are in a limit from 1 0 to 2 0.01p (p兹2苶 苶T 苶/m 苶, see example E3.5) Solution: The number of molecules whose speeds lie in a narrow interval –d can be determined according to an approximate equation dNdN f()d* where N is the total number of molecules in the given volume, f() is the Maxwell distribution function (3.3.6). In order to determine the amount of molecules sought, we have to substitute the distribution function into the expression * and calculate a corresponding integral. Therefore, p
N
∫ 0
4N ⎛ m ⎞ f ()d ⎟ ⎜ ⎝ 2T ⎠
3
⎛
m2 ⎞
∫ exp ⎜⎝ 2T ⎟⎠ d. 2
0
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It is impossible to take this integral analytically; however, we can make some simplifications. In particular, since the speed interval lies in very low speed values, we can substitute exponential multiplier by unity (accurate calculations give its magnitude as 1.0001).Besides, the multiplier in brackets can be rewritten as: ⎛ m ⎞ ⎟ ⎜⎝ 2T ⎠
32
1 . y p3
Then the integral can be rewritten as: 4 N N yp
0.01y p
∫
y 2 d y.
0
After integration we obtain
N
4
N 3 6 y p 10 . 3 y p3
The particle number can be expressed knowing the Loschmidt value (the number of gas particles in 1cm3 at normal condition), nL2.691019 cm3. Substituting all intermediate values we arrive at N4/(3兹 苶)2.691019101066.071014 (molecules). EXAMPLE E3.7 Find the share N/N of molecules of the ideal gas, speed of which differs from the (½)p by not more than 1%. Solution: In order to solve this problem, it is convenient to write the Maxwell distribution using the relative molecular speed u u y yp
y 2T m
.
The distribution on u can be found as f(u) (4/兹 苶)u2eu 2 The number sought can be written as: N/N f(u)u (4/兹 苶)u2eu u. Notice that u 1 at p. The magnitude of the interval between u-limits is u 0.01. The interval to be taken is narrow; therefore, the integral can be approximated accordingly 具u典 f(u)du ⬇ f u, i.e., 2
u
N 4 ⎛ 1 ⎞ (1 2)2 e1 4 ⎛ 1⎞ f ⎜ ⎟ u e 0.01 0.01. ⎜ ⎟ ⎝ 2⎠ N ⎝ 2⎠ 2
Executing calculations we arrive at N/N (e1/4 / 兹 苶)0.01 4.39103
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Distribution of the Kinetic Parameters of an Ideal Gas’ Particles
3.3.2
193
The kinetic energies Maxwellian distribution of molecules
We will now find the distribution function of the values of molecule kinetic energies. For this purpose we shall express the relative number of those molecules having speeds in interval –d using the kinetic energies expression corresponding to these speeds. Express the equal number of molecules on velocities and energies f ( y )d y f ()d.
(3.3.8)
Make a substitution of variables in expression (3.3.6). As m2/2 then, 2 2/m and 兹2苶/m 苶 兹苶. Taking into account eq. (3.3.8.), expression (3.3.6) will then be as follows: 4 ⎛ m ⎞ f () ⎟ ⎜ ⎝ 2T ⎠
32
2 1 ⎛ ⎞ exp ⎜ ⎟ d. ⎝ T ⎠ 2 m m
Finally, we arrive at f ()
2 (T )
3
⎛ ⎞ exp ⎜ ⎟ . ⎝ T ⎠
(3.3.9)
The distribution function f() is presented in Figure 3.12. At small kinetic energy ( T), an exponential term is close to 1. Therefore f() ⬃ 兹苶 (i.e., runs up steeply). At large T,
f()
Q
Figure 3.12 Maxwell molecular energy distribution.
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the exponent exp((/T)) mostly influences the function. Hence, curve f() leaves the origin as 兹苶, passes over a maximum and is exponentially going down toward zero. The distribution obtained allows us to estimate the concentration of chemically active molecules, which can overcome the activation barrier Q and get in touch with another molecule to produce a new product. To find the relative concentration of molecules with kinetic energy Q, we should calculate an integral
N dN ∫ ∫ f ()d. N N Q Q
(3.3.10)
There are no such integrals in the mathematical tables. We can see that at large , the main contribution to the result is made by the exponent; the multiplier 兹苶 contributes vaguely in comparison with the exponent and can be neglected. Therefore, the situation is simplified and calculation of the integral results in 2 Q N N (T )3 2
⎛
⎞
∫ exp ⎜⎝ T ⎟⎠ d
Q
2
Q ⎛ Q⎞ exp ⎜ ⎟ . ⎝ T ⎠ T
(3.3.11)
The number of chemically active particles is not N, but much less; in Figure 3.12, it is shown by the hatched area. The proportion of such particles can be defined according to eq. (3.3.11); they can overcome activation barrier Q.
3.4 3.4.1
FIRST LAW OF THERMODYNAMICS
Equipartition of energy over degrees of freedom
In order to analyze the energy that molecules can accept, we should first find the number of its degrees of freedom i, i.e., the number of independent coordinates that describes the position in space of all the molecules’ atoms. Remember, the position of atoms in a biatomic fixed material MP (i.e., atom) is defined by three independent coordinates (i 3). The position of a biatomic molecule is described by six degrees of freedom; however, the presence of one interatomic chemical bond reduces this number to five: each bond reduces the i-number per unit, i.e., i 3 × 2 – 1 5 (three translational and two rotary degrees of freedom). For a bent three-atomic molecule i 3 × 3 – 3 6 (i.e., there are three translational and three rotary degrees of freedom). For any other multinuclear molecule, the numbers of translational and rotary degrees of freedom are all the same (six), because of the presence of chemical bonds. Apart from translational and rotary movements, intramolecular oscillations of atoms can also exist; there are oscillatory degrees of freedom with doubled i for the kinetic and potential part of oscillatory energy. However, at not so high temperatures (not appreciably higher than room temperature) intramolecular oscillations are not excited.
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On the basis of this simplified model, it would appear that monoatomic molecules, which are essentially point like and have practically zero MI, can store energy only in their translational motion. Maxwell suggested a theorem of equipartition of energy among the degrees of freedom: every degree of freedom can store an equal amount of energy; it is associated with an energy of (1/2)T per atom or (1/2)RT per mole (refer to Section 3.1.4). 3.4.2
First law of thermodynamics
Thermodynamics is a section of physics in which physical properties of macroscopic systems are studied on the basis of macroscopic energy transformations, mainly without accounting for their microscopic structure. When investigating the macroscopic properties of thermodynamic systems in gaseous, liquid, solid or plasma states, we are dealing with enormous numbers of particles. Statistical methods consist of averaging the physical characteristics of constituent particles. As a result of such averaging, there appear a small number of parameters describing the state of the system as a whole, these parameters being pressure p, volume V, thermodynamic (absolute) temperature T, and total number of particles N, or amount of substance . Such an approach to the description of thermodynamic systems does not examine a particular system’s models and its internal structure. Thermodynamics operates with such concepts as internal energy U, amount of heat Q and work A of a thermodynamic system. In particular, a very important characteristic of a system is its internal energy U. Internal energy is the total energy of each system component minus its macroscopic kinetic energy as the whole and potential energy in an external force field: N
U ∑ i i1
1 N N ∑ ∑ Uij (i j ). 2 i j
(3.4.1)
Internal energy is a function of the system’s state. It means that every time the system appears in a given state, its energy takes values corresponding to this state. At the transition of a system from state 1 (characterized by internal energy U1) to state 2 (characterized by internal energy U2), the change of internal energy is defined by the difference of these energies and will not depend on the process by which the given transition is taking place; i.e., U U2 – U1. In an ideal gas state, interaction between molecules is absent; hence there is no total N potential energy in eq. (3.4.1). Then U冱i iN具典 where N is the total number of particles. Since N(m/M)NA and 具典3/2T (see eq. (3.1.12)) then,
U
m 3 m 3 T N A RT . M2 M2
(3.4.2)
So, the internal energy of a particular amount of ideal gas depends only on its temperature T.
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There are two possible ways of changing the internal energy of a thermodynamic system: (1) by the production of macroscopic work under the action of external forces applied to the system; and (2) by heat exchange. The process of the heat exchange of contacting bodies not accompanied by the production of macroscopic work is referred to as heat exchange. The energy transferred to a body by another body during heat exchange is referred to as the amount of heat or, simply, heat. These two methods of energy exchange between macroscopic bodies are qualitatively various and nonequivalent. In fact, the accomplishment of the work of a system can lead to the change of any other kind of energy (potential, internal, etc.), whereas, the energy transferred to a system as heat contributes directly to the increase of the chaotic motion of the system’s particles, i.e., it influences the internal system energy only. Both methods of energy transfer accompany each other; i.e., the heat transferred to a system can contribute to a change of internal energy as well as the system’s work against external forces in the framework of the law of energy conservation. The law of energy conservation is referred to as the first principle of thermodynamics. This law states that: the amount of heat transferred into the system, Q, goes to the change of internal energy dU and the work accomplished by the system against external forces A: Q U A.
(3.4.3)
All values are expressed in the same energy units. Only dU in this equation is an exact differential because of the fact that U is an exact function of the system’s state; here, Q and dA are not the functions of state and mean only a very small quantity. Eq. (3.4.3) enables us to define the internal energy of the system from a thermodynamic position only: internal energy is a body property, the change of which is equal to the difference of heat brought into the system and the work produced against external forces. The work produced by a system concerns gas compression and/or expansion. For this purpose, we can imagine cylinder with diameter S with a piston under which there is a gas (Figure 3.13). There is a manometer to measure pressure p. The cylinder has a ruler on its side to measure piston position (i.e., volume). We shall act on the piston by a force F. Under the action of this force the piston, having compressed gas, will move on dx. The work of the force under gas we shall define as dA Fdx. Since pressure is p F/S then F pS and dA pSdl is the change of volume under the piston dl S dV. The work of external force dA is equal to the increase of the system’s potential energy dA pdV .
(3.4.4)
The same consideration is valid also for the work of gas against external forces. In the coordinate system p(V) an elementary equation pdV is represented in Figure 3.14 as a hatched strip; however, in macroscopic processes integration is often needed to determine the total work (Figures 3.14a and b).
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197
F S
V p
Figure 3.13 An experimental device. p
(a)
p
∆V
V
V1
V2
V
(b)
Figure 3.14 Work done in processes: (a) elementary volume change dV; (b) in a finite process V.
3.4.3
Heat capacity of an ideal gas: the work of a gas in isoprocesses
Heat capacity is a very important characteristic of a system because it enables the calculation of the amount of heat needed for initiation and maintenance of chemical processes. Consider a case when the amount of particles remains constant. The heat capacity is an amount of thermal energy (of heat) that raises a body’s temperature by 1 K. Therefore, c
dQ . dT
(3.4.5)
In this form the heat capacity depends of the amount of substance. For uniform systems, a specific heat capacity is used: the heat capacity of a mass unit. Therefore, Cs
dQ c . m dT m
(3.4.6)
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The molar heat capacity relates to one mole of a substance C
dQ M dQ M c.
dT m dT m
(3.4.7)
where m is the mass of the substance and is the number of moles. An intercomparison gives c mCs ,
(3.4.8)
C MCs ,
(3.4.9)
m C . M
(3.4.10)
c
Consider the particular forms for the work of a gas in isoprocesses on the basis of the first law of thermodynamics. 1. Isochoric process (V const.). In this process gas does not produce any work because V const and correspondingly dV 0. This means that all the heat consumed from outside will increase the internal energy: Q dU и C dU/dT. However, for the closed system of an ideal gas, the change of internal energy means the change of total kinetic energy dU dK. Therefore, dU (C )V dT
m 3 RdT . M2
(3.4.11)
where, (C )V
m 3 3 R R. M 2 2
(3.4.12)
This is the expression for mole heat capacity CV at constant volume. For the isochoric heating of moles of an ideal gas for T, the thermal energy should obtained Q 3 R T (C )V T . 2
(3.4.13)
2. Isobaric process (p const.). In this case the heat consumed goes toward increasing the internal gas energy dU and to producing the work pdV: Q dU pdV d (U pV ).
(3.4.14)
In thermodynamics, the sum U pV H is referred to as the enthalpy (the heat content) of a system: the enthalpy is a thermodynamic quality, the change of which in an isobaric process is equal to the heat consumed by the heat of the system. Since U depends only on
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199
temperature, for a molar dU (C)VdT and Q dU pdV dU RdT. Dividing the equation by dT we obtain ⎛ Q ⎞ ⎟ (C )V R. ⎜⎝ dT ⎠ P
(3.4.15)
(C )P (C )V R.
(3.4.16)
or
The work produced by a system in an isobaric process is dApdV RdT or A RT(m/M)RT. In order to heat isobarically an ideal-gas system, one should provide for T the heat Q 冢 25 冣 (C )P T 25 R T . 4. Adiabatic process. This process occurs without heat exchange to the environment. Hence, at adiabatic process Q 0 and, according to the first law of thermodynamics, dA –dU. The work of the system against external forces in an adiabatic process is at the expense of internal gas energy change. In practice, the adiabatic process is carried out with a sufficiently fast gas expansion or compression that heat exchange with the environment has no time to take place. Use both expressions dApdV and dU(m/M)dT. Therefore, pdV(m/M)dT(C )v. Using the ideal gas equation, we can write (m/M)RdT d(pV) pdV Vdp. Combining the two last equations we can write as follows: pdV ((C )v/R) (pdVVdp). Using eq. (3.4.16) we can write (C )V (C )V 1 . (C )P R (C ) p (C )V 1 (C )V The ratio (C)P/(C)V is usually designated as the adiabatic index (C )P (C )V
(3.4.17)
and referred to as the ratio of specific heats. Then pdV(1/(1))(pdVVdp) or pdV – pdV –pdV – Vdp. therefore pdV Vdp 0. Dividing this equation by pV, we obtain (dV/V) dp/p 0. Integration gives ln V ln p ln C. After exponentiation we arrive at pV const. This equation is referred to as the Poisson equation.
(3.4.18)
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The adiabatic graph falls noticeably more steeply than the isothermal curve (Figure 3.15). For two states with parameters p1V1 and p2V2 we obtain p1V1 and p2V2. Using these equations and equation we can obtain TV1const. and Tp(1/)const. and for volumes: T2 ⎛ V1 ⎞ T1 ⎜⎝ V2 ⎟⎠
1
(3.4.19)
and for pressures: T2 ⎛ p1 ⎞ T1 ⎜⎝ p2 ⎟⎠
1
(3.4.20)
Let us find the equations for the work in an adiabatic process. Since dApdV and pdV(m/M)(C )VdT, then dA(m/M)(C )VdT. After integration in the limits T1–T2, we obtain A(m/M)(C )V(T1T2). Taking initial temperature T1 out of the brackets, we obtain a slightly different expression: A(m/M)(C )VT1(1(T2/T1)). Using one of eqs. (3.4.19), we can obtain an expression for the work as a function of the system’s initial and final state; for volumes: A
⎡ ⎛ V ⎞ 1 ⎤ m (C )V T ⎢1 ⎜ 1 ⎟ ⎥ M ⎢⎣ ⎝ V2 ⎠ ⎥⎦
(3.4.21)
and for pressures: 1 ⎡ ⎤ ⎛ p1 ⎞ ⎥ m ⎢ A (C )V T1 ⎢1 ⎜ ⎟ ⎥ M ⎝ p2 ⎠ ⎢⎣ ⎥⎦
p
T = const adiabat
isotherm
V
Figure 3.15 Isotherm and adiabatic curve.
(3.4.22)
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Since, in an adiabatic process, the system’s temperature changes without consumption of external heat, the heat capacity in this process is equal to zero (dQ 0 at dT 0). The important property of the heat capacity is its additivity. This means that heat spent for heating a mixture is equal to the sum of the amounts of heat necessary for heating these system’s parts separately (i.e., heating to the same temperature). For two gases 1 and 2 this statement has the following mathematical expression: Qs.mix mmix T (Q1.s m1 Q2.s m2 )T
(3.4.23)
or further Qs,mix
Q1,s m1 Q2,s m2 mmix
.
(3.4.24)
The same is true for molar quantities. EXAMPLE E3.8 Calculate the specific heat capacity Cv,mix of a mixture of two gases: m1 6 g of helium and m2 10 g of nitrogen at constant volume. Solution: The gas-specific heat capacity is determined by eq. (3.4.6). The heat capacity of a mixture of two gases can be found due to the additivity properties of specific heat capacity: this means that the heat required to warm each gas separately is equal to the heat required to warm the mixture to the same temperature m1CV1tm2CV2 t Cv,mix(m1 m2)t. Solving this equation relative to Cv,mix and reducing by t we arrive at the final expression
Cv,mix
m1CV,1 m2CV,2 m1 m2
i1 m1 i2 m2 2 M1 2 M 2 R. m1 m2
Notice that the gases are different: helium is a monoatomic gas, but nitrogen is diatomic, therefore i1 is 3 (M1 4 103 kg/mole) and i2 is 5 (M2 28 103 kg/mole) (see Section 3.4.1). Executing all operations, we arrive at a numerical result: Cv,mix 1.63 kJ/(kg K) EXAMPLE E3.9 The degree of dissociation of a diatomic gas is equal to 0.3. Determine the adiabatic rate (i.e., Cp/Cv) of such a partly dissociated gas. Solution: The degree of dissociation shows the ratio of a number of decomposed molecules Ndis to the initial number of molecules N, i.e. (Ndis/N).
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Multiplying both the numerator and denominator by the mass of a single molecule m1 we obtain
m1 N dis mdis , m1 N m
i.e., we obtain the ratio of mass of dissociated molecules to the initial mass of molecules. Then, knowing the degree of dissociation, we can find the mass of dissociated mdis (mdism) and mass of nondissociated mnondis (mnondismmm(1) molecules. Therefore, we have in fact a mixture of two gases: monoatomic with i 3 and mole mass M/2 (M is the mole mass of nondissociated diatomic gas) and the rest of the diatomic gas with i 5 and mole mass M. Taking into account the fact that the heat capacity is additive, we can write: Cv,mix
mdis i1 m i R nondis 2 R. ( M 2) 2 M 2
Substituting in this expression all numbers we already know, we obtain R ⎛ 2m 3 (1 )m 5 ⎞ R m (5 ) . Cv,mix ⎜ ⎟ ⎠2 M ⎝ M 2 M
In an analogous way, we can find heat capacities at constant p Cv,mix
mdis i1 2 m i 2 m R R nondis 2 R (7 3) ( M 2) 2 2 2 M M
Therefore, the adiabatic rate can be written
Cp,mix Cv,mix
7 3 7 3 0.3 7.9 1.49. 5 5 0.3 5.3
EXAMPLE E3.10 Determine the heat amount absorbed by hydrogen m 0.2 g in mass on being heated from temperature T1 0°C to T2 100°C. Find the change of the gas external energy U and work done A. Solution: For isobaric heating the amount of heat required is Q mCpT* (see Section 3.4.2). Substitute the heat capacity Cp((i2)/i)(R/M) into equation *: Q((i2)/i)(R/M)mT. Execute the calculations and arrive at the result Q 291 kJ. The change of external energy can be found according to the formula U (i/2)(m/M)RT. Calculations give U 208 kJ. The work done at the gas expansion is A Q–U 83 kJ.
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EXAMPLE E3.11 A uniform ideal gas is in a state which is characterized by parameters V1 1 L and p1 2 atm. At a rapid contraction to V2 0.52 L, the pressure rises to 5 atm. Determine: (1) the number of degrees of molecular freedom, (2) the change of internal energy U, and (3) the change of enthalpy H. Solution: (1) It was mentioned that the process is rapid, therefore, it can be considered an adiabatic one. The equation for an adiabatic process (Poisson equation) is as follows: p1V1 p2V2 ,
where is the adiabatic rate: Cp Cv
i 2 R i 2 2 , i 2
R 2
(here is an amount of substance, i.e., number of moles). From the last equation, it is possible to find i: i 2/(1). Rewrite the adiabatic equation and find :
p2 ⎛ V1 ⎞ , p1 ⎜⎝ V2 ⎟⎠ therefore,
i
2 p2 ln p1 1 V1 ln V2
⎛V ⎞ 2ln ⎜ 1 ⎟ ⎝ V2 ⎠ . p2V2 ln P1V1
Execute calculations and arrive at: i 4.99 ⬇ 5.0. This means that the gas is biatomic. (2) The internal energy of an ideal gas is equal by equation: U (m/M) Cv, mT. The internal energy change U U2 U1 (m/M)CV, mT2 – (m/M) CV, mT1. That is U (m/M)CV, m(T2 T1). The temperatures T1 and T2 will be found from the ideal gas equation T1
冢
p1V1 pV and T2 2 2 . m m R R M M
冣
Then, U (m M )(i 2) R ( p2V2 p1V1 ) (m M ) R .
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After some simplifications, we arrive at U (i/2)(p2V2 p1V1) (the answer). Using the SI system: p1 2 atm ⬇ 2.026 105 Pa; V1 1 L ⬇ 103 m3; p2 5 atm ⬇5.065 105 Pa; V2 0.52 L ⬇ 0.52 103 m3. Execute calculations: U (5/2)(5.065 105 0.52 103 – 2.026 105 1 103) J 2.5 (2.634 – 2.026) 102 J 152 J We have to carry out the calculations in SI. U 2.5 (5.052 21) L atm 1.50 L atm. Since l atm 101.3 105 Pa, then U 1.50 101.3 J 152 J. (3) Enthalpy H is H U pV. The change of enthalpy can be expressed as H (U 2 p2V2 ) (U1 p1V1 ) or H U ( p2V2 p1V1 ). Since U (i/2)(p2V2p1V1) then, H (i/2)(p2V2p1V1) (p2V2p1V1) that is, H (i2)/2 (p2V2p1V1). Using the calculations for U presented above we arrive at H 3.5 (2.634 – 2.026) J 213 J 3.4.4
Heat capacity: theory versus experiment
The expressions obtained earlier for average kinetic energy show that each degree of freedom can realize only the definite energy. Therefore, the heat capacity of gases significantly depends on several factors. Firstly, their properties depend on the degree to which they approach the state of ideal gases; i.e., the nearer the gas is to an ideal gas, the better the theory describes its thermodynamic properties. Secondly, the simple theory often only
CV
J/mole.K
7R/2 25.0 20.0
5R/2
18.0 3R/2
10.0
R/2
5.0 0
25
50 75 100
250
1000
2500
5000
Figure 3.16 The experimental heat capacity values of hydrogen.
T,K
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takes into account the molecular translational energy, though the molecules’ rotation and vibration should also be accounted for. Thirdly, only quantum mechanics can explain the thermodynamic properties of complex gases. Therefore, comparison with experiment in the framework of used approximation gives only approximate agreement. We will return to this question in Chapters 7 and 9. Figure 3.16 imagines the validity of the classic theory. At very low temperatures, quantum processes make the picture more complex. At 50–100 K, the (3/2)T law works relatively well. However, at higher temperatures, the heat capacity becomes temperature-dependent and, so far, no theory exists that explains this curve in detail.
3.5
THE SECOND LAW OF THERMODYNAMICS
The second law of thermodynamics is a fundamental law of nature, covering numerous phenomena of the world around us and having deep practical and philosophical consequences. However, compared with the first law that has a simple and obvious physical sense as it is based on the well-known general law of conservation energy, the second law and its physical essence are less obvious and demand a more profound understanding of the physical processes occurring in nature. Before formulating the main ideas, it is necessary to make a number of remarks. Any thermal machine works under a closed cycle creating a circular process. The process by which a thermodynamic system passes through a number of states and comes back to the initial point is referred to as circular process or a cycle. Figure 3.17 depicts the cycle made by an ideal gas in coordinates p-V. Along line 1a2, the extending gas performs mechanical work A1a2 0. At compression along a line 2b1, the work is taken over by external forces; therefore the mechanical work performed by the gas is A2b1 0. As the work performed by an ideal gas on the graph in coordinates pV is numerically equal to the area of a curvilinear trapeze, the difference between these areas, (the shaded area) is numerically equal to the work accomplished by the gas for a cycle. It is accepted that the cycle going clockwise is direct 冢ADC 养 pdV0冣 and the cycle is reversed if it goes in the opposite direction (Figure 3.17), 1a2b1 i.e., ADC 养 pdV0. 1a2b1
p
p 1
1 a
a
b A>0
(a)
b 2
2 A<0
V
(b)
V
Figure 3.17 Thermal cycles: (a) represent a direct thermodynamic cycle 1-a-2-b, (b) the reversed cycle 1-b-2-a-1 is depicted.
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heater, TH
QH
engine
W
QC
cooler, TC
Figure 3.18 A heat engine. The engine is represented by arrows pointing in a clockwise direction around a central block: heat extracted from a heat reservoir is converted partly into work W.
The thermodynamic process is referred to as reversible if a thermodynamic system passes through the same intermediate equilibrium states in both the forward and in the opposite direction but in the reverse sequence; additionally, there will be no change in the surrounding bodies. It follows from this definition that any equilibrium process is reversible. Any process that does not meet these requirements is irreversible. 3.5.1
Heat engines
Heat engines (machines) are intended for the production of useful work due to the heat received from the combustion of fuel or other energy sources. A device that changes heat into work while operating in a cycle is referred to as a heat engine. The following components are necessary for the cyclic work of such an engine: a heater, a working body and a cooler (refrigerator). As a rule, one uses the environment as a cooler; in the design of a thermal engine a cooler as the component of the thermal machine is therefore absent. The basic scheme of a thermal engine is given in Figure 3.18. The heater at temperature T1 transfers an amount of thermal energy Q1 to a working body; as a result, the temperature of the working body rises. Due to expansion, it carries
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out mechanical work A, making any mechanical devices rotate (e.g., a turbine) or move (e.g., a piston in a cylinder). To return the system to its initial state, a cooler with temperature T2 is used to which the working body transfers an amount of heat Q2. Notice that the presence of a cooler is an obligatory condition, otherwise the periodic (cyclic) work of the thermal engine is impossible: in fact, in this case the working body will eventually come into thermal balance with the heater and the thermal stream from the heater to the working body stops. According to the first law of thermodynamics, after the return of the working body to its initial state, its internal energy U for a cycle does not change. Therefore the difference of amounts of heat Q1 and Q2 will be equal to useful mechanical work A, accomplished by the thermal engine for a cycle; this is the thermal efficiency e. The thermal efficiency of the engine cycle e is the ratio of useful work A to the amount of heat absorbed from a heater Q1. e
A Q1 Q2 Q1 Q1
or e 1
Q2 . Q1
(3.5.1)
As for all cyclically working thermal engines, quantity Q2 should necessarily be transferred to the cooler (otherwise the return of the working body to an initial state is impossible) so at Q2 0, the thermal efficiency e is always less than unity. 3.5.2
The Carnot cycle
The development of thermodynamics is a direct result of the need for the description and calculation of thermal engines. The low thermal efficiency of the engines existing in S. Carnot’s time prompted him to investigate the work of heat engines from a thermodynamic point of view. He aimed to find a thermal cycle, the efficiency of which would be maximal. It was obvious from this requirement that the cycle should consist only of reversible processes, excluding thermal losses. There can be only two processes: an isothermic process, which should be carried out infinitely slowly as only then will the working body (e.g., gas or vapor) pass a number of quasi-stationary states (without thermal losses since they are reversible); and an adiabatic process which, as a matter of fact, proceeds reversibly without heat exchange with the environment, i.e., it is carried out instantly. An ideal gas is used as a working body. Thus, in the search for the optimum process, the cycle should contain two isotherms corresponding to temperatures T1 (the temperature of the heater) and T2 (the temperature of the cooler). These two isotherms should be cut by two adiabatic curves. Such a cycle is represented in Figure 3.19. The line 1–2 corresponds to the isothermal gas expansion resulting from obtaining the heat Q1 from the heater. This is
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p pa
1
TH QH
2
pb W
pd
4 QC TH 3
pc
TC 0
Va
Vd
Vb
Vc
TC V
Figure 3.19 Carnot cycle.
in accordance with the first thermodynamic law Q1 U A1.2 (where in our case U 0) and leads to the expression Q1 A1.2 RT1ln
V2 V1
(3.5.2)
where is the amount of gas moles. V1 and V2 are the initial and final gas volumes. The line 2–3 corresponds to adiabatic gas expansion (Q 0) with no contact with the heat reservoir. Then U23 A23 0. Therefore, the work done by gas along this line is A23 CVm (T2 T1 ) CVm (T1 T2 )
(3.5.3)
When it reaches temperature T2 (at the end of the adiabatic expansion), the gas transfers to the refrigerator heat Q2 A34 RT2 ln
V V4 RT2 ln 3 V3 V4
(3.5.4)
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Then along the line 4–1, the gas is compressed so that its temperature acquires the initial temperature T1. Since the heat exchange along this line is absent, the work of the external forces A 4–1 equals the increment of the internal energy U41 CV.m (T1T2). Then the work carried out by the gas is A41 –A 4–1 or A41 CVm (T1 T2 ).
(3.5.5)
The total work performed by gas in one cycle is the sum of work produced on each segment A A12 A23 A34 A41. or A RT1ln(V2/V1) CV.m(T1T2) – RT2ln(V3/V4) – CVm(T1–T2). After some transformation, the expression adopts the form:
⎛V ⎞ ⎛V ⎞ ⎛ V ⎞ V A RT1ln ⎜ 2 ⎟ RT2 ln ⎜ 3 ⎟ R ⎜ T1ln 2 T2 ln 3 ⎟ . V1 V4 ⎠ ⎝ V1 ⎠ ⎝ V4 ⎠ ⎝
(3.5.6)
This work is numerically equal to the W area in the graph. We will execute some transformations that will be useful to us. Recall the equitable expression for adiabatic process TV1 const. Then some relationships for lines 23 and 41 can be written T1V21 T2V31 and T1V11 T2V41 or T2 / T1(V2 / V3)1 and T2 / T1(V1 / V4)1. The volume relationships can be obtained from these expressions V2/V3V1/V4 or V2/V1V3/V4. Therefore, in the expression (3.5.6), the transformation of the logarithmic terms as 1n(V2/V1)1n(V3/V4) allows us to write A R(T1 T2 )ln
V2 . V1
(3.5.7)
Since Q1 RT11n(V2/V1), the maximum thermal efficiency of the ideal thermal engine is e id
A R(T1 T2 )ln(V2 V1 ) , Q1
RT1ln(V2 V1 )
(3.5.8)
T1 T2 T 1 2 . T1 T1
(3.5.9)
that is, e id
The results of these thermodynamic investigations of heat engine processing have resulted in Carnot’s theorem (1850): the thermodynamic efficiency e of the ideal engine working on the Carnot cycle depends only on the heater T1 and cooler T2 temperatures and does not depend on the engine construction and on the kind of working body used.
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Note that any real engine has an apparently smaller thermal efficiency eid than the ideal one ereal: ereal eid : Carnot’s work permits problems to be approached from thermodynamic positions. It seems that there exist only two ways of increasing the thermal efficiency of engines: either increase the heater temperature T1 or lower the cooler temperature T2. However, the second way is economically unjustified because of its complexity. Therefore, there is only one way of increasing engine thermal efficiency, i.e., raise the heater temperature. However, there are also difficulties inherent in method. For example, if water vapor is used as the working body the problem of corrosion arises which inevitably leads to the deterioration of the engine. Thus, in a real heat engine the transformation of heat into work inevitably causes the transfer of a certain amount of heat to the environment. This results in the engine’s thermal reservoir cooling and the cooler (i.e., the environment) heating; this is in agreement with the second law of thermodynamics. At the same time, an infringement of this law would mean an opportunity to create a socalled perpetuum mobile of the second kind, i.e., one executing mechanical work due to the internal energy of the thermal reservoir but not changing the thermodynamic state of the environment. This allows us to suggest one more rule of the second law of thermodynamics: it is impossible to create a perpetuum mobile of the second kind. 3.5.3
Refrigerators and heat pumps
Another rule of the second law of thermodynamics is as follows: heat cannot flow from one body to another body at a higher temperature without other changes being involved. The transfer of heat from a cooler body to a warmer one can take place only because of work A done under the action of external forces on the working body. The thermodynamic cycle initiated in the reverse direction is used in refrigerators and thermal pumps. The removal of an amount of heat Q2 from a cool body and its transfer to a more heated one (the thermal reservoir) is the task of the refrigerating machine (refrigerator). The thermal coefficient of performance K is defined by the ratio of the amount of heat Q2 removed from a cool body and given to the warm body to the external mechanical forces work A K
Q2 Q2 . A Q1 Q2
(3.5.10)
Depending on the design of the refrigerator, K can be either less or more than unity. Notice that the refrigerator permits the cooling of bodies that are inside it, while simultaneously heating the surrounding air. The idea of using the heat machine’s backwards cycle to carry an amount of heat from the cooler environmental air to heat the housing is achieved. Such a device is referred to as a thermal pump. The efficiency of the thermal pump is defined by the ratio of the amount of heat Q1 received by the heated body to the mechanical work spent A produced by external forces
K th.p.
Q1 Q 1 , A Q1 Q2
(3.5.11)
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where Q2 is the amount of heat taken from the cool body; it is always distinct from zero therefore, the efficiency of the thermal pump is always higher than unity. The basic difference between a thermal pump and a refrigerator is that in the thermal pump the amount of heat Q1 is given to the heated body, and in the refrigerator Q2 is removed from the cooled body. Heat pumps require less power expense for the work than electro-heating devices. 3.5.4
Reduced amount of heat: entropy
Write two expressions for the thermal efficiency of the ideal thermal engine eid1(Q1/Q2) and eid 1 (T1/T2). Since these two equations are equal then, Q1 / Q2 T1 / T2 or Q1 / Q2 Q2 / T2. Consider an amount of Q1 obtained by a working body from a heater positive (Q1 0) and the heat amount yielded by the working body to the cooler heat negative (Q2 0). Then the equation can be written as Q1 / T1 (Q2 / T2) or, for the Carnot cycle, Q1 Q2 0. T1 T2
(3.5.12)
The ratio of the amount of heat to the temperature is referred to as the reduced amount of heat. So for the Carnot heat engine, the sum of the reduced amount of heat is zero. Since the Carnot cycle is reversible, any reversible cycle can be presented as a sequence of elementary (Carnot) cycles. Ten such cycles consisting of isotherms crossed by adiabats are depicted in Figure 3.20.
i-th reversible Carnot mini-cycle
adiabats
isotherms
Figure 3.20 Carnot microcycles.
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The larger the amount of such microcycles used, the better the degree of approximation. For each microcycle the eq. (3.5.12) is valid. Therefore, for any reversible cycle, we can write ⎛ Q1i Q2i ⎞ ⬇ 0. T1i T2i ⎟⎠ i1 N
∑ ⎜⎝
(3.5.13)
Notice that the amount of heat Q1i obtained by a working body in the ith cycle is approximately equal to the amount of heat Q2.I, which is yielded by a working body in the preceding cycle (i–1). The precise equality can be reached at an infinitely large amount of cycles with infinitely thin Carnot microcycles. Therefore, all the heat flows into the system along the internal parts of cycles compensating each other; instead of a sum we can write an integral
∫
Q 0 T
(3.5.14)
(for reversible cycles) where Q is the small amount of heat obtained or given off by a working body. This equation is referred to as a Clausius equation. Since the integral is taken over a closed counter, it can be taken from any point and in any direction (since the cycle is reversible) (Figure 3.21). This allows one to present it as a sum of two integrals 1a2 Q/T2b1 Q/T0. Thus the integral from 12 Q/T0 between any two points of 1 and 2 equilibrium states does not depend on the form of a trajectory along which a thermodynamic reversible process is going on. As the choice of points 1 and 2 is arbitrary, we can assert that the integral describing the change of value (designated by the letter S) is a function of a system’s state. Suggested by Clausius, this value is referred to as entropy (measured in J/K units). For a reversible cycle 养 S0. From the expression obtained, it follows that for reversible processes 12 Q/T does not depend on the form of a line (trajectory) representing this process on the graph; it is
p
a 2 b 1
0
V
Figure 3.21 The cycle and the work done; entropy.
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defined only by the points describing initial and final equilibrium states. Thus, the elementary reduced amount of heat is the full differential of the S-function dependent only on the system’s state, i.e., dS
Q T
(3.5.15)
Therefore, the entropy change S at the transition from the first equilibrium state (1) to another equilibrium state and (2) at reversible process is determined by equality S S2 S1
Q . T 12
∫
(3.5.16)
Thus, entropy is only a function of an equilibrium state of a thermodynamic system. It does not depend, in this case, on particular processes leading the system into a given equilibrium state. Entropy, as well as potential energy, can be defined only to some constant. This is a consequence of the fact that the formula (3.5.16) does not allow us to define the absolute entropy value but only its difference. For the determination of entropy change at quasi-equilibrium, processes one can use a property of its additivity, i.e., N
S ∑ Si
(3.5.17)
i1
where Si is the entropy change at the i-th quasi-equilibrium process. N is the number of such processes. The entropy change will vary depending on the process if the initial and final positions of the thermodynamic system do not coincide. At quasi-equilibrium (sufficiently slow) heating of the system from temperature T1 to temperature T2, we obtain the following. (1) For an isochoric process: Q CvmdT (where Cvm is a molar heat capacity at constant volume): 2
T
2 T Q dT Cvm ∫ Cvm ln 2 . T T T1 1 T
Sv ∫
1
(2) For an isobaric process: Q CpmdT (where Cpm is a molar heat capacity at con2 T2 stant pressure). Sp1dQ/T T dT/T Cpm1n(T2/T1) (for reversible process). Since 1 CpmCvm, then SpSv. If the thermodynamic system is an ideal gas in which simultaneously the quasiequilibrium changes of several parameters take place, we can use the first thermodynamic
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law (3.4.3) in the form Q Cvm pdV then,
Q dT dV nCvm R . T T V
Since for ideal gas pdV RT
dV V
then, T
2
V
2 2 Q dT dV Cvm ∫ R ∫ . T T V 1 T V
S = ∫
1
1
Executing the integration we obtain: S Cvm ln
T2 V Rln 2 . T1 V1
(3.5.18)
3.5.5 Clausius inequality and the change of entropy for nonequilibrium processes Remember that the thermal efficiency of the thermal Carnot heat engine is given by equations e
Q1 Q2 T1 T2 , Q1 T1
therefore, Q1 Q2 . T1 T2
(3.5.19)
As noted earlier, the work of any heat engine is always accompanied by a thermal loss. Therefore, the thermodynamic cycles of real engines are irreversible and their thermal efficiency is less than that of ideal and reversible Carnot cycles ereal (non-rev.)eideal (rev.) However, the mathematical equations e(Q1Q2)/Q1 (real, irreversible) and e(T1T2)T1 (ideal, reversible) are still valid. That is, (Q1Q2)/Q1(T1T2)/T1 or 1Q2/Q11T2/T1. It follows then Q2 T2 Q1 T1
(3.5.20)
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or Q2 Q1 . T2 T1
(3.5.21)
From comparison of these two equations, it follows Q2 Q1 . T2 T1
(3.5.22)
The sign of equality relates to reversible processes whereas the sign of non-equality relates to irreversible processes. Let us consider the heat obtained by a working body from a heater positive and the obtained from a working body to a cooler negative. Then the inequality will be Q2 Q1 0 T2 T1 and, in the more general form, Q
∑ Ti 0. i
(3.5.23)
i
At the limit (for minicycles), the sum can be substituted by integral
∫
Q 0. T
(3.5.24)
This expression is referred to as the Clausius inequality. Let us change this inequality by incorporating the entropy notion. Use a cycle like that presented in Figure 3.21. Let us consider also that part of the cycle (1a2) is reversible and the other (1b2) is irreversible. From the Clausius inequality it then follows that rever.,1a2( Q/T)irrev.,1b2( Q/T)0, and further, rever.,1a2( Q/T)irrev.,1b2( Q/T), i.e., the integral from ( Q/T) over the irreversible path is always lower than that over the irreversible path between the same states. Since the integral from the right is the change of 2 the entropy S, therefore, S S2S11( Q/T) for the irreversible cycle. Then for the irreversible cycle (approximately) 养( Q/T)S Applying this equation to a minicycle we obtain dS
Q . T
(3.5.25)
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Combining expressions (3.5.15) and (3.5.25) we obtain dS
Q , T
or dS 0.
(3.5.26)
The equality sign relates to reversible processes and the inequality sign to irreversible processes. This form of the second law of thermodynamics possesses the greatest generality as microprocesses are practically elementary and on the basis of such mini-processes, other probable processes can be composed, both noncircular and circular (cyclic). Thus, using the entropy concept, we can give one more rule of the second thermodynamic law: the entropy of an isolated thermodynamic system can only increase and, after reaching the maximum value, remain constant. This rule is also referred to as the law of entropy increase. We can be convinced of this by a simple example. Let there be two bodies representing an isolated thermodynamic system. One body (hot) is at temperature T1 another (cold) at temperature T2. We know that the first body cools down whereas the second is heated up; this process proceeds spontaneously. How then will entropy change? An amount of heat will be transferred from the hot body to the second body, warming it up. The entropy change of the first body will be equal to the reduced amount of heat ( Q/T1) taken with the negative sign: S1(Q/T1). This amount of heat is obtained by the second body. Therefore, the reduced amount of heat (Q/T2) is positive. The change of the second entropy can be similarly determined: S2
Q . T2
We can find the entropy change of the whole system according to the additivity principle as the sum of S1 and S2: SS1S2 or S(Q/T1)(Q/T2)Q ((T1T2) /T1T2). As T1T2, therefore the entropy at a spontaneous irreversible process increases: this is in accordance with the rule of the second law of thermodynamics given above. Notice that in open systems entropy can decrease as well. In this case, however, there will always be changes in surrounding bodies so the total entropy of the system and the surrounding bodies will increase. The first and second laws of thermodynamics can be combined. For this purpose, in the expression of the second law (dS( Q/T ), a small amount of heat is presented as the sum dUpdV as is required by the first law. Then dS(dUpdV) / T or TdS dU pdV
(3.5.27)
This expression is referred to as the basic thermodynamic equation. The equality sign relates to reversible processes; the inequality sign to irreversible processes.
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EXAMPLE E3.12 An ideal diatomic gas in an amount of 1 mole is under a pressure p1 250 kPa and occupies volume V1 10 L. The gas is heated to T2 400 K and further isothermically expanded to initial pressure. After that, the gas returns to its initial state by isobaric compression. Define the cycle thermal efficiency e. Solution: For visualization, we shall draw the whole cycle in coordinates p-V with characteristic points 1, 2, 3. The thermal efficiency of any cycle is given by expression (3.5.9) where Q1 is the amount of heat obtained by the gas from the heater, Q2 is the heat returned by the gas to a cooler. The heat difference is the work done by the gas in one cycle (marked W in Figure 3.19 in the graph). The working body obtains the heat Q1 at two lines: 1–2 (isochoric process) and 2–3 (isothermal process); therefore Q1 Q1–2 Q2–3. The amount of heat obtained at isochoric process is Q12 Cv (T2T1). T1 can be found from the ideal gas equation p 2
p2
p1
3
1
0
V1
V2
T1
V
p1V1 .
R
Executing calculations we arrive at T1 300 K. The amount of heat obtained by a gas in an isothermal process is Q23RT2ln (V2 / V1), where V1 is the volume at T2 and pressure p1 (point 3 in the graph). On the line 3-1, gas returned the heat Q2 Q2 Q31 Cp (T2 T1 ). Consequently the thermal efficiency is E 1
Cp (T2 T1 ) V
CV (T2 T1 ) RT2 ln 2 V1
.
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We can change the ratio of volumes to that of temperatures V2/V1T2/T1 and express the heat capacities according to their temperatures (see Section 3.4.2). Making substitutions we arrive at e 1
(i 2)(T2 T1 ) i(T2 T1 ) 2T2 ln
T2 T1
.
Substituting corresponding values and executing calculations, we arrive at e 0.041 4.1%. EXAMPLE E3.13 There is cold (T1 10°C) and hot (T2 90°C) water in two vessels of masses m1 7 kg and m2 3 kg, respectively. Determine the entropy change S on mixing these portions of water. Ignore the thermal capacity of the vessels. Consider the specific thermal capacity of water to be independent of temperature and equal to Cs 4.18 kJ/kg K. Solution: We must first find the temperature of the mixture produced. We shall take advantage of the equation of thermal balance Q1 Q2 where Q1 is the heat obtained by the cold portion of water and Q2 is the heat given by the hot water (we ignore the thermal capacity of the vessels.) Therefore Cm1( – T1) C m2(T2 – ). Wherefrom we can obtain
m1T1 m2T2 m1 m2
In order to avoid complication, we calculate the value of :
(710 3 90)/ (73) 34°C or 307 K. The entropy is additive. Its change S consists of two parts: part S1 describes the change of entropy as a result of heating the cold water from T1 to due to the cooling of the hot water, and S2 is equal to the change of entropy of hot water at its cooling from T2 to the average temperature . Hence S S1 – S2. The change of 2 entropy is determined by the integral S1(dQ/T). Here limits 1 and 2 symbolize the initial and final states of the thermodynamic system. Apply to the given task
S1 ∫
T1
dQ1 dQ2 and S2 ∫ T T T 2
where dQ1 Cm1dT and dQ2 Cm2dT. Substituting these equations into integrals and executing integrations, we obtain S1Cm1ln(/T1) and S2Cm2ln(/T2). Therefore the entropy change is SC(m11n(/T1)m2ln(/T2)). All values
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required for calculation have already been mentioned. The final calculation is S 4.18 103 (7ln (307/283) 3.ln (307/363)) J/K 4.18 103 (0.5698 0.5027) J/K 280 J/K. We know that the spontaneously running process goes with increasing of the entropy and our result corresponds to this regularity S 0. 3.5.6
Statistical explanation of the second law of thermodynamics
The physical significance of entropy becomes clearer in a statistical interpretation offered by Boltzmann (1875). As already mentioned, thermodynamics does not consider the microscopic structure of a substance; it establishes a connection between the macroscopic properties of a thermodynamic system in various states of its existence. These states are defined by a small number of thermodynamic parameters (p, T, V, etc.). The states of a system characterized by such thermodynamic parameters are referred to as macrostates. The microscopic structure of a substance is taken into account in molecular-kinetic theory using statistical descriptions. On the basis of the statistical laws of an ideal gas, the assumption of the independence of coordinate and speed of any single molecule from the speeds and coordinates of all other molecules was adopted. Therefore, the most detailed description of a gas state would be the enumeration of the six parameters for each particle (see Section 1.3.8); for a monoatomic gas consisting of N molecules there will be 6N parameters. Such a representation of a system’s state is referred to as a microstate approach. As the gas molecules are in continuous motion accompanied by the exchange of energy at collisions, the microstates are changing almost continuously; however, the macrostate remains constant. This means that any macrostate can be realized through an enormous number of microstates. Assuming that all microstates are equiprobable (ergodic hypothesis), one can find the probability of any macrostate. The so-called thermodynamic probability of a macrostate’s statistical weights is used in statistical physics. The thermodynamic probability W of a system macrostate is determined by a number of various microstates that can assure the given macrostate. Contrary to mathematical probability that cannot be more than a unit, thermodynamic probability is, as a rule, expressed by a giant number. Boltzmann showed that a system’s entropy is proportional to the natural logarithm of the number of possible microstates corresponding to the given macrostate, i.e., to the thermodynamic probability. The proportionality factor is , a value later called the Boltzmann constant. Therefore, entropy can be expressed as S ln W
(3.5.28)
Notice that, under constant external conditions, the spontaneous transition of a system from one state to another is possible if the second state appears thermodynamically more probable. However, large entropy corresponds to a greater thermodynamic probability. As a transition is spontaneous, a system can pass only from a state of nonequilibrium state to a state of equilibrium; this means that entropy increases in irreversible processes. The equilibrium condition is achieved at the maximal value of thermodynamic probability and, hence, at the maximum entropy value compatible with the given external conditions. Thus
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the Boltzmann formula explains entropy increase in irreversible processes, which follows from the second law of thermodynamics. Using the notion of entropy, the second law of thermodynamics can be interpreted as follows: any irreversible processes in a closed system occurs when the system’s entropy increases. Notice that the system should be closed because in an open system entropy can behave in any way, i.e., it can decrease, increase or remain constant. In closed systems entropy remains constant only in reversible processes. In irreversible processes entropy always increases. 3.5.7
Entropy and disorder
The energy of thermal motion is characterized by the irregular chaotic motion of molecules. It differs from the mechanical energy of a moving body where all molecules of the body participate both in the ordered motion together with the whole body, and also in chaotic motion. If the moving body were to stop sharply, it would heat up: the mechanical energy of the translational motion would transform to the kinetic energy of the thermal movement of the molecules, i.e., there would be a transformation of ordered into disordered motion. It is much more troublesome to carry out the return process. Another example in shown in Figure 3.22. There are ordered balls, black and white, in a box (Figure 3.22a). When the box is shaken, the balls get mixed up (Figure 3.22b); this takes place almost spontaneously. However, all attempts to return the balls to their initial ordered state by shaking or stirring them will be unsuccessful. This simulated inconvertibility can be seen even more dramatically in any molecular system with an incommensurably large number of “balls”. One more example can be taken from a real problem of either the chemical or isotope separation of molecules. Let two different gases be divided by a partition. If the partition is removed, both gases will spontaneously become mixed. However, the return process of separation will not take place; separation demands a huge expenditure of energy and effort. These examples show that any process aspires spontaneously to proceed to a state of greater disorder. This corresponds to the aspiration of a system to proceed to the state that has the greater entropy. The irreversibility of thermal processes corresponds to the irreversibility of order and disorder. Thus, entropy is a measure of the system’s disorder. Despite its generality, the second law of thermodynamics has no absolute character. Deviations from it due to fluctuations are quite natural; the fewer the number of particles, the greater the probability of deviations. Infringement of the second law is shown at small concentrations. The first and second laws of thermodynamic allow us to judge the behavior of thermodynamic systems near to absolute zero (0 K). This problem was formulated by Nernst
(a)
(b)
Figure 3.22 Irreversibility of the order (a) and disorder (b).
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in 1906. On the basis of the generalization of experimental data, Nernst formulated a theorem which was subsequently referred to as the third law of thermodynamics: at the aspiration of any equilibrium thermodynamic system to absolute zero, the entropy aspires to some universal constant, the value of which does not depend on any thermodynamic parameters of the system and can be taken as equal to zero lim T0 S 0 and lim S 0. T0 Simultaneously the following relation will also be fulfilled ⎛ S ⎞ lim ⎜ ⎟ 0, ⎝ Z ⎠ T
(3.5.29)
where Z is any thermodynamic parameter (pressure, volume, etc.). Index T means that differentiation is executed at constant temperature. Nernst’s theorem is applicable only to equilibrium systems. From the third law of thermodynamics, it follows that absolute zero is unattainable because, according to eq. (3.5.27), if near a temperature of absolute zero, a small amount of heat is taken off a system (T 0), a large enough (in a limit infinite) entropy change will take place; this contradicts Nernst’s theorem. Notice that at the aspiration of the temperature to absolute zero, the thermal heat capacities Cv and Cp will also aspire to zero.
3.6 3.6.1
A REAL GAS APPROXIMATION: VAN DER WAALS EQUATION
An equation of state of a van der Waals gas
In order to come nearer to the state of an ideal gas, the conditions given in Section 3.1.3 must be satisfied: the volume of gas must be increased and the pressure reduced. In fact, these suggestions cannot be accepted by engineers. It is therefore important to improve the model (or make it more complex) by removing some of the restrictions and making them positively accounted. Plenty of the equations aimed at developing an ideal gas model have been suggested (up to 150). Many of them, when applied in practice to certain classes of chemical substances and processes, give good agreement with experiment in limited intervals of temperatures and pressure. The most successful approach has been worked out by van der Waals. The van der Waals equation contains two new parameters a and b, which takes into account intermolecular interactions (a) and the total volume of molecules (b). It supposes that the amount of interaction between molecules would affect the pressure value P, and that the amount of molecule volume will lead to a reduction of the volume accessible to free their movement. For one mole of van der Waals’ gas, the following equation can be written: ⎛ pa ⎞ ⎜ V 2 ⎟ (Vm b) RT . ⎝ m ⎠
(3.6.1)
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2
Here, terms (a/Vm ) and b describe the deviations of gas from ideality. The value (a/Vm ) corresponds in its dimension to pressure; it results from the molecules’ interaction and represents a so-called “internal (cohesive) pressure” pi. The term b takes into account the total volume of all gas molecules. Therefore (Vmb) is the volume free to molecular motion. Having removed the brackets the van der Waals equation can be rewritten as: ⎛ RT ⎞ a ab b⎟ Vm 0. Vm3 Vm2 ⎜ ⎝ p ⎠ p p
(3.6.2)
This equation is cubic relative to Vm; at a definite set of temperatures, it should have either one real root or three roots, two of them being imaginary. (Further on, we shall omit index m so as not to block up the formula). Solution of van der Waals by means of a mathematical treatment gives the isotherms depicted in Figure 3.23, from which it can be seen that there is a value of the T parameter (T TC) that divides qualitatively various types of isotherms into two parts. At T TK curves p(V) monotonously falls down; this corresponds to the presence of one real equation root: only one value V corresponds to each value p. At T T, the gas behaves approximately as ideal (exact conformity is absent only at T , i.e., when it is possible to p
pc
C
Tc
Vm, c
Vm
Figure 3.23 Mathematical solution (isotherms) for a van der Waals gas.
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neglect the interaction between molecules and count them as MP). At lower temperatures (T T) one p value corresponds to three values V; the form of isotherms essentially changes. At critical temperature T TC, the van der Waals’s isotherm has one specified point (one solution). To this point corresponds pC (critical pressure) and VC (critical volume). This point C corresponds to the new substance state referred to as critical; as experiments show that in this state a substance is no longer either a gas or a liquid. To obtain the real experimental isotherms, one can use the device presented in Figure 3.13. Reduction of volume V at T TC leads to the pressure increasing according to the theoretical curve (Figure 3.23). This occurs up to point N. Further reduction of the volume does not lead to an increase in pressure; part of the gas undergoes a transition from a vapor state to a liquid state (Figure 3.24). Reduction of the volume from point N to point M does not produce a pressure change but leads to a change in the amount of liquid and gaseous phases. The pressure, referred to as saturated vapor pressure, remains unchanged, whereas the amounts of the liquid and vapor phases change at this constant pressure (on Figure 3.24 this is marked by the letters svp—saturated vapor pressure). In a point of M all vapor volume liquefies. With a further reduction of volume, the isotherm sharply rises upwards; this corresponds to the sharp reduction in the compressibility of a liquid in comparison with vapor. When temperature increases, the length of segment MN decreases; at T TC, it collapses into a critical point C. A locus of N and M points form a bell-like curve separating the diphase area (under a bell-like curve) from single-phases—the vapor phase (at large V) or the liquid phase (at small V) (Figure 3.25). The gas cannot liquefy at any temperatures higher than TC. (One can use this circumstance to distinguish gas and vapor—vapor can be compressed to liquid, but gas cannot be.)
p
C
L M' psv
N'
M
N TC O V
Figure 3.24 Real isotherms of a van der Waals gas.
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p
liquid
C
gas
vapor+liquid vapor
V
Figure 3.25 A phase diagram of a van der Waals gas.
In accurately adjusted experiments, metastable states characterized by segments MO and NL (Figure. 3.24) can be obtained. These states are supercooled vapor (line MO) and a superheated liquid (line NL). The supercooled vapor is in a state such that, according to the parameters it should be a liquid, but its properties continue to follow the behavior of a gaseous state–it aspires to extend, for example, as the volume increase. On the other hand, a superheated liquid is in a state such that, according to the parameters, it should be a vapor, but its properties remain liquid. Both these states are metastable: at small external influences they pass into a two-phase state. A line OL corresponds to a negative factor of compression; it is unstable and cannot be realized at all. Constants a and b are independent of temperature and are different for different gases. The van der Waals equation can be modified; moreover being modified it describes the behavior of any gases provided that they are described by expression (3.6.1). For this purpose, we shall find the relation between constants a and b, as well as critical parameters: pC , VC and TC. Eq. (3.6.1) can be rewritten in another way:
p(V )
RT a 2. V b V
(3.6.3)
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We can then use the property of the critical point; in this point value (dp/dV) is zero since its tangent inherits a horizontal lines sv, and (d2p/dV2) is zero because it is a point of inflection. For the critical point it is fair to say that pC
RTC a VC b VC2
(3.6.4)
RTC 2a ⎛ dp ⎞ 3 0 ⎜⎝ ⎟ 2 dV ⎠ C VC (VC b)
(3.6.5)
⎛ d2 p ⎞ 2 RTC 6a ⎜⎝ dV 2 ⎟⎠ (V b)3 V 4 0. C C C
(3.6.6)
All these equations are valid for critical point C. The solution gives: VC 3b, pC
a 8a , TC , 2 27 Rb (27b)
(3.6.5)
and consequently,
b
8p V VC , a 3VC2 pC , R C C . 3 3TC
(3.6.7)
Let us now substitute these parameters into the initial equation: ⎛ pCVC2 ⎞ ⎛ V ⎞ 8 T . p V C ⎟ pC VC ⎜ ⎜ ⎟ ⎝ ⎠ 3 3 VC ⎠ TC ⎝ After dividing the equation by pCVC and expressing the parameters in a reduced form ( p/pC, V/VC, T/TC), eq. (3.5.1) became very simple: 3 ⎞ ⎛ 2 ⎜⎝ 2 ⎟⎠ (3 1) 8. V
(3.6.8)
This is the van der Waals equation in reduced parameters, universal for all van der Waals gases (i.e., the gases submitting to eq. (3.6.1)). Eq. (3.6.8) allows us to come to the law of corresponding states: if for two different gases, two of three ( , , ) reduced parameters coincide, the third parameter should also coincide. These substances are in corresponding states. Writing the van der Waals equation in the form (3.6.3) allows us to apply the representations given to the case of any other gases that do not originally satisfy the van der Waals
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equation. Notice that eq. (3.6.3) reminds us of a decomposition series of the function p(V) over a Vn term; it is only desirable to continue the series
p(V )
RT A1 (T ) A2 (T ) A3 (T ) 2 3 4 V V V V
(3.6.9)
Factors An(T ) are called virial factors. At an infinite number of terms in this decomposition, it could precisely describe the state of any gas. Notice that factors An(T ) are temperature-dependent functions. In different technological processes, various models for their calculation are used. There is also a theoretical estimation: what is the amount at which members of this decomposition term reach the desired accuracy. 3.6.2
Internal energy of the van der Waals gas
The internal energy of the van der Waals gas should contain the potential energy of the molecules’ interaction as well as the kinetic energy part. Kinetic energy represents the total kinetic energy of all the gas’s molecules; however, for calculating the potential energy, it is necessary to determine the work of the internal forces while carrying a system from a given state to a state with “zero” potential energy (see Section 1.4.5). It has been shown above (see Section 3.6.1) that the additional (internal) pressure appears because of intermolecular interactions in van der Waals gas; it is defined by the expression (a/V 2). Then the elementary work of a gas expansion is equal to dA(a/V 2)dV. An integration results: Aa(dV/V 2)(a/V ). Thus, the internal energy of one mole of a V gas is U(i/2)RT(a/V )CVT(a/V ). From this equation it follows that adiabatic expansion brings about gas cooling. In fact, dUdQ(i/2)RdT(a/V 2)dV0. Therefore, dT(a/(CV)V 2)dV and dT ⬃ dV. Indeed a volume increase leads to the decrease of the gas temperature since all factors in the equation are positive. However, this simple estimation becomes more complicated at a qualitative level.
EXAMPLE E3.14 Oxygen with density 120 kg/m3 under a pressure of 10 MPa is in a pressure vessel. Considering oxygen under such pressure to be a van der Waals gas, find its temperature T and compare it with analogous calculations in the framework of an ideal gas. The van der Waals constants are: (a) 0.136 N m4/mole2, (b) 3.17.105 m3/mole. Solution: Write the van der Waals equation for any amount of the substance
a ⎞ ⎛V ⎞ ⎛ ⎜⎝ p 2 ⎟⎠ ⎜⎝ b⎟⎠ RTvdW ⴱ
V
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Here TvdW is the temperature of the van der Waals gas. Making some apparent transformation ( m/M, m V and (V/M) and substituting them in equation *, we obtain ⎞ ⎛ 2 a ⎞ ⎛M b⎟ RTvdW . ⎜⎝ p ⎜ 2⎟ ⎠ ⎝ ⎠ M
Therefore, ⎞ a ⎞⎛M ⎛ 1⎞⎛ TvdW ⎜ ⎟ ⎜ p 2 2 ⎟ ⎜ b⎟ . ⎝ R⎠ ⎝ ⎠ M ⎠⎝
Executing the calculations, we arrive at TvdW
⎞ 1 ⎛ 7 (120)2 0.136 ⎞ ⎛ 32 103 10 3.17 105 ⎟ 337 K. ⎜ 3 2 ⎟ ⎜ 8.31 ⎝ (32 10 ) ⎠ ⎝ 120 ⎠
Considering the oxygen in the pressure vessel as an ideal gas, we can obtain from the ideal gas equation Tid
pV MP . m R R M
Thus, Tid
32 103 107 321K. 120 8.31
Although the difference in temperatures is not so high (16 K), it can, nevertheless, be significant at precise experiments. 3.6.3
A Joule–Thomson effect
The potential energy of molecules interaction U depends on the average distance between molecules for the van der Waals gas and, consequently, it changes with the change in the volume occupied by gas. When there is no exchange of energy between the gas and its environment, the internal energy of gas should remain constant. Hence, if one part of the energy changes, there should be a respective alteration in the other part: dU –dK. If we take into account only a molecule’s attraction at the gas’s expansion, its potential energy grows since the average distance between molecules increases. The increase in potential energy under these conditions causes reduction of the gas’s average kinetic energy and, consequently, reduction of its absolute temperature. Therefore, if a real gas is allowed to extend adiabatically in vacuum (without producing any work against external forces) its temperature should go down. This phenomenon is called the Joule–Thomson effect. The
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reduction in temperature at gas expansion is referred to as a positive Joule–Thomson effect. In some cases, heating of gas can occur and then the effect is referred to as being negative. For van der Waals gases the sign on the effect depends on which correction terms in the equation play the larger role. The potential curves of molecule interaction are qualitatively presented in Figure 3.26. Consider first the gas for which it is possible to neglect the first correction term in the equation with attraction forces between molecules. Then we should take into account only repulsion forces described by the second correction member b (Figure 3.26b). In this case, the average potential energy decreases with the increase in intermolecular distance. As with the reduction of pressure (expansion of gas in vacuum), the average distance between molecules is significantly enlarged; the average potential energy of the gas decreases. Reduction of the potential energy causes an increase in the gas’s kinetic energy and, hence, leads to an increase in its temperature. Thus, gas in which one can neglect the action of the attraction forces, but in which the correction parameter b is essential, brings about gas heating at expansion—a negative Joule–Thomson effect occurs. Now let us consider the gas for which forces of repulsion between molecules can be neglected. This case concerns a gas in which an inherent volume of molecules can be ignored (Figure 3.26a). We should then only account for attraction forces. Thus, the potential energy of molecule interaction is negative and its numerical value decreases with an increase in the intermolecular distance. Because with the reduction in the pressure of the gas the average distances between molecules increases, the change of its potential energy will be positive growing with the reduction of the gas pressure at its expansion in vacuum. Therefore, the average kinetic energy of the molecules of such a gas decreases and therefore its temperature goes down. Hence, in a gas in which attraction forces play an essential role and the coefficient b can be neglected, expansion brings about its cooling; the Joule–Thomson effect will be positive.
U(r)
U(r) r2
r1
Z ∆U
a is small
∆U b is small
Z r1 ∆U > 0
∆K > 0
U+K=const
∆U < 0
∆K > 0
therefore ∆T > 0
therefore ∆T< 0 (a)
r2
(b)
Figure 3.26 (a) and (b) An explanation of the Joule–Thomson effect.
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For any gas, the sign of the Joule–Thomson effect depends on temperature and pressure. The positive effect for each gas is observed only in the limited interval of temperatures and pressures. For each gas there are values of temperature and pressure at which the Joule–Thomson effect is equal to zero (no temperature changes occur at gas expansion in vacuum). These points (Ti, pi) are called points of inversion. At these points, the influence of forces of attraction is completely compensated for by the influence of repulsion forces; consequently the gas temperature does not change. The set of inversion points forms an inversion curve in a p–T diagram. Figure 3.27 presents the inversion curve for nitrogen. It can be seen that, to a given value p, two points of inversion can occur. The curve of inversion outlines two points of inversion for which a positive Joule–Thomson effect is observed. Values for the upper and lower inversion points for some gases at various pressures are given in Table 3.2. For the majority of gases, the upper point of inversion lies above room temperature. Hydrogen and helium are an exception. The Joule–Thomson effect finds important practical applications in the techniques of gas fluidization, when the gas is throttling over a wide range of pressures from 2×107 to 105 Pa. Successive repetition results in reduction of the gas temperature down to its boiling point. If this procedure is applied to air, first oxygen gas is condensed (90 K) and then nitrogen (77 K). The boiling point of hydrogen is 20 K and helium 4.2 K. All these gases are widely used in various technical equipments, scientific explorations and in various crystallographic technologies.
p(10 MPa)
3
2
1
0
73
573 T,K
Figure 3.27 An inversion curve of the Joule–Thomson effect. Table 3.2 Upper and lower temperature inversion points for some gases (in K) Gas
p (in 105 Pa)
Upper
Lower
He H2 Air CO2
1 113 150 18–100
23.6 192.7 583 2050
140 249
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3.7 3.7.1
ELEMENTS OF PHYSICAL KINETIC
Introduction
In the previous sections we have considered ideal gases in a state of thermodynamic equilibrium. Such systems are stationary, i.e., their parameters do not change in time. In this section we shall consider macroscopic systems removed from an equilibrium state and aspiring to return to it. The area of physics that deals with this process is referred to as physical kinetics. To describe quantitatively, such a process is possible only within the framework of the model of an ideal gas. The application of the results of this research to real gases and even to liquids can be provided on a qualitative or semiqualitative level. Although such analysis is carried out with such approximation, it can be usefully applied to nonideal systems. In physical kinetics, there are two approaches to the study of physical phenomena: empirical phenomenological and microscopic at the molecular level. In the first of these, problems are investigated from a macroscopic point of view without considering the detailed atomic mechanism. In the second method the behavior of systems is investigated from microscopic standpoint on the basis of molecular representations. Both methods should yield the same results as the description of the same phenomena. A system can be removed from a condition of thermodynamic equilibrium by external influence. For example, one can inject another gas into a certain point of the predominant gas and, due to thermal (chaotic) molecular movement the concentration of the second component will tend to spread over the whole volume. Sooner or later it will be equal. One can heat gas locally in one area of the volume and the gas temperature will also start to equalize over the whole volume due to molecular chaotic movement. From the resulting examples it is clear that the thermal movement of molecules plays an active role in reaching equilibrium. We shall examine below the phenomena of alignment caused exclusively by this factor—chaotic movement of molecules, i.e., at a molecular level. This does not mean that there are no other mechanisms of alignment; however, we will avoid them at the moment. 3.7.2
Transport processes: relaxation
Transport phenomena appear in different substances, mostly gases, because of chaotic molecular movement within them and permanent collisions with an interchange of their kinetic properties. During chaotic molecular movement and collisions, a transfer of energy and momentum and masses diffusion can take place and there is hence a gradual alignment. In this way a system can reach a state of equilibrium. The process of a system returning back to equilibrium is called relaxation. Let us evaluate the speed of the relaxation process from a mathematical point of view. We shall suppose that the displacement rate (d/dt) of the return to equilibrium is proportional to the displacement itself and is opposite to it in sign: (d/dt)k, where k is a coefficient of proportionality. Integration of this equation gives ln (t ) kt ln C
(3.7.1)
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or, (t ) C exp(kt ).
(3.7.2)
Constant C can be found from the initial displacement. If at t 0 the maximum displacement 0 is taking place C 0 and eq. (3.7.2) takes the final form: (t ) 0 exp(kt ).
(3.7.3)
The time for which the displacement will decrease in e time is called the time of relaxation. This concept can be applied to a number of phenomena: radioactive fusion, damped oscillation, etc. Let us find the physical sense of factor k. If we write eq. (3.7.3) for the time , the displacement becomes: () 0 exp(k). Therefore, (()/(0)) exp(k) e1. Therefore, k 1 or k (1/). Hence, ⎛ ⎞ (t ) 0 exp ⎜ ⎟ . ⎝ t⎠
(3.7.4)
This equation is valid for many relaxation phenomena, not only for transport properties in gases. We must emphasize that the transport phenomena can take place either in a closed system (where the system is removed from an equilibrium state and recovered coming back to it due to chaotic molecular movement), or in open stationary system where nonequilibrium can be induced permanently due to an external influence. In both cases, the relaxation consists of the redistribution of the uniformity over the whole system. 3.7.3
Transport phenomena in ideal gases
We define an ideal gas (see Section 3.1.3) as a gas whose molecules move from collision to collision without interaction. As a first rough approximation, we shall represent molecules as solid spheres with a certain diameter d. Enter an important characteristic of the gas, namely the free path length, i.e., the average distance , which a molecule moves freely from one collision to another. We can calculate the average time between molecule collisions as t /, where is the average speed of the thermal movement of the molecules (to keep things simple in this section, we shall not write the sign on averages). The average number of collisions of a molecule for 1 sec will be written as: 1 y
. t
(3.7.5)
The shortest distance between the centers of the molecules on impact is referred to as the effective diameter of molecules d (see Section 1.5.4, Figure 1.34). During its
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movement, a molecule (which we consider as a solid spheres) collides with those molecules whose centers lie at distances smaller than d from the line of its movement (Figure 3.28 inset). The value d2 is the target area in which the center of the molecule should lie in order to guarantee the collision. This area is called a collision cross section. On collision, each molecule performs a zigzag path. If one rectifies the molecule’s path one can assert that in the unit of time, the molecule describes the cylinder volume with a height equal to the speed of the molecule and with a cross-section d2. The collision will take place with those molecules whose centers lie inside this cylinder. We shall consider that all molecules are at rest except those, which we follow up. Then the number of collisions in a unit of time will make
d 2 yn,
(3.7.6)
where n is the molecule concentration. Taking into account the movement of other molecules leads to an additional term 兹2苶. Then expression (3.6.6) will have the form:
2 d 2 yn,
2 d 2 yn.
(3.7.7)
From comparison of eqs. (3.7.5) and (3.7.7) it can be deduced that
1 2n
1 2 d 2 n
.
(3.7.8)
Consider the relation of the free path length on the parameters of an ideal gas state. In an isothermal process p nT then ⬃ 1/p. So for nitrogen under normal conditions 107 m, at pressure 1 Pa 102 m, at pressure 104 Pa a molecule on average runs a long distance without collision (about 103 m). If n const., (isochoric process) should
def = πd 2
d
d
Figure 3.28 A path traveled by a molecule.
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not depend on temperature. However, practice shows that increases a little if the temperature rises; this dependence is described by empirical law:
T T C
(3.7.9)
where is the free path length at T . 3.7.4
A macroscopic representation of a transport coefficient
Molecules in a closed volume can differ in their characteristics (mass, momentum, energy, etc.); assume that the spatial distribution of property values initially is nonuniform. Due to the thermal movement of the molecules and their collision, a process of restoration of uniformity in the volume can take place. We began a general analysis of these phenomena, then at particular processes: diffusion, heat conductivity and internal viscosity. Denote G a transferable property and let it be distributed along an x-axis nonuniformly (Figure 3.29). We shall fix a point x0 on the x-axis and an area S normal to axis x. In time
G(x)
G2 −G1≈− dG 2 dX
x0−
x0+ x0
υ0∆t
x
υ0∆t
Figure 3.29 Macroscopic consideration of transport phenomena.
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t, due to thermal motion, a number of molecules will pass through this area from right to left and from left to right; both numbers being identical. However, each molecule will transfer that “quantity” of a property G, which corresponds to the last collision of the molecule. Due to nonuniformity of distribution G(x), it will cause resulting transport G through the given area S which we shall designate q. We shall consider q 0 if the property G is transferred to the positive direction of the x-axis and vice versa. At the same time it is easy to understand that the sign q should be opposite to the sign on the gradient of the G-function, i.e., to the sign of derivative dG/dx. Thus, from the most general macroscopic consideration, it follows that q ⬃
dG( x ) S t . dx
Let us introduce a macroscopic (phenomenological) coefficient , which we denote as a transport coefficient. Then q
dG( x ) St . dx
(3.7.10)
The expression (q/t)(1/S) is a flow of the G property. Let us determine the microscopic meaning of coefficient . We will take two layers of the gas, parallel to the area S at distances from both sides of point x0. Let G1 and G2 be the values of G in these layers. All molecules allocated in parallelepipeds (see Figure 3.29) and moving in the definite direction (to area S ) at a speed and in time t will cross the area S in both directions. As the molecules move chaotically, the number of molecules moving in each side is 1/6 of their total number. Their number is equal, but they carry different amounts of property G. All molecules moving in a positive x-direction carry property G1whereas in a negative direction they carry property G2. The total transport of the G-value through the area S is q 1 1 ny (G1 G2 ). t S 6
(3.7.11)
Assuming that the gradient of the G-value is small, we can write (G1G2)(dG/dx)2. Hence, q 1 1 dG ny. t S 3 dx
(3.7.12)
Comparing eqs. (3.7.10) and (3.7.12) we arrive at 13 yn.
(3.7.13)
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The significance of this coefficient consists of the fact that it connects the macroscopic (technical) value with microscopic one ( and ). Notice that this result is obtained from the comparison of macroscopic (phenomenological) and microscopic approaches.
3.7.5
Diffusion in gases
Imagine a gas in which a certain constant difference of concentration exists in the various fixed points (a nonequilibrium, but stationary system), whereas the temperature at any point in this system is the same and remains constant. Thus, there is a resulting flux of gas molecules referred to as stationary diffusion. We would like to emphasize that this flux is caused only by the chaotic movement of molecules and not for any other reasons. Another situation arises when the difference of concentration is not constant; the system aspires to equalize concentration; this will be nonstationary diffusion. In any case, the driving force of molecular diffusion is the concentration gradient. We shall restrict ourselves to self-diffusion, i.e., diffusion of similar molecules distinguished by any insignificant (for diffusion) property (e.g., labeled by radioactive mark) in the environment of the same nonradioactive molecules. In fact, the diffusion of carbon oxide in nitrogen can be considered as self-diffusion because the molecules of these gases differ insignificantly with regard to their size and mass. Our task is to find a relationship between the macroscopic and microscopic characteristics of diffusion. We shall consider a simple one-dimensional case: imagine a vessel in the form of a long thin cylinder (an x-axis is directed along one axis of the cylinder) containing a mixture of two gases; the concentration gradient of one of the gases is artificially maintained constant. Assume that the masses and the sizes of the molecules of both gases are identical: (m1 m2 m). Therefore, the molecules can acquire identical average speed u; free path length can be taken as equal to 1/(兹2苶n), where n n1 n2 is the total average concentration of molecules in the volume. A comparison of what we obtained from a macroscopic consideration and that of microscopic one looks like
j
q G D , St x
(3.7.14)
where j is the mass diffusion flux and ⬅ D is the diffusion coefficient. In this form, this equation is referred to as Fick’s law. The transport value G(x) in this case is the relative concentration of the chosen molecules n1, that is G(x) (n1(x))/n. Comparison with the eq. (3.7.13) shows that
j
n ( x) q ⎛ 1⎞ ⎜ ⎟ ny 1 . ⎝ 3⎠ St n
(3.7.15)
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By comparing eqs. (3.7.13) and (3.7.14), we can further derive the microscopic expression for the macroscopic diffusion coefficient D: D 冢 13 冣 y.
(3.7.16)
It follows from this expression that the physical sense of the diffusion coefficient D consists of the fact that it shows the number of molecules that diffuse through a unit area in a unit time and at a unit gradient of relative concentration. Data on key parameters of diffusion as well as other transport phenomena, i.e., of heat conductivity and viscosity, are shown in Table 3.3. If molecules differ considerably in their masses and the dimensions mentioned above, the calculations demand specification. More detailed examination shows that the process of diffusion is determined by the speed of the fastest (smallest) molecules, whereas for effective cross section determination, it is the larger molecules. For the nonstationary diffusion, it is possible to estimate the time t for which there is an alignment of concentration (reduced in e times) from a dimension consideration. In fact, is defined only by the character of distribution of molecular masses in the initial instant of time and gas property. The initial state is defined by the dimension of heterogeneity area L. There is only one combination from D and L that has a dimension of time, namely Table 3.3 Transport phenomena characteristics Phenomenon
Main laws Transferred value
Driving force
Diffusion
q dc( x ) D St dx Fick’s law Mass
dc( x ) dn1 ( x ) dx dx Concentration gradient
Transfer coefficients
1 D y 3
Heat conductivity
q dT ( x ) St dx Fourier’s law Energy
dT ( x ) dx Temperature gradient
1 y(C )sp 3
Internal friction
q du( x ) St dx Newton’s law Momentum
du( x ) dx Velocity gradient
1 y 3
Coefficient’s value Units Gases Liquids Resin, glass Crystals
m2/sec
W/(m)
104106 1091010
104105
10161018
100 102
Non-SI units: *Poise (P) 0.1(N sec/m2); **Stokes (St) 104(m2/sec).
1 y 3
N sec/m2* m2/sec** 103105 ⬃102 ⬃106 ⬃1015
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⬃L2/D. The time in gases can be estimated under the order of value: at D⬃104 m2/sec and L ⬃ 0.1 m, the order of value is 103 sec, i.e., diffusion is a slow process even in gases and in liquids it is slower by some orders. 3.7.6
Heat transfer
If, at different points of any gas in a closed volume for a short time, a different temperature is created and the gas is then left to itself, the temperature at all gas points is equalized owing to the process of heat transfer. From a macroscopic point of view, the phenomenon of heat transfer in gases consists of temperature transport from hotter to colder places. Within the framework of the molecular kinetic theory, the process of heat transfer consists of the fact that molecules from a heated site of the gas, where they have large kinetic energy, transfer the energy to cooler areas via collisions; a flow of heat is thus created. In reality, in gases and liquids this phenomenon is usually accompanied by heat transport by the steam of a gas or a liquid initiated by their density difference, i.e., so-called convection. However, we will now consider heat transfer exclusively from the point of view of molecular kinetics. By the way, it is very difficult to subdivide these two processes. The transfer of heat caused by the thermal (chaotic) movement of microparticles is referred to as a heat transfer phenomenon. Thus, in the general transport formula (3.7.10), q is the flux of heat. The transferable value in this case is the amount of heat Q or G(x) CVT(x) where CV is the heat capacity of a substance. This heat is produced by the total kinetic energy of the molecules. Hence the heat transport description can be given as: dq 1 dT ( x ) , dt S dx
(3.7.17)
where is the heat transfer coefficient. In this form, this equation is referred to as Fourier’s law. The minus sign shows that the heat transfer is directed against the temperature increase (against the temperature gradient). From a microscopic point of view, the driving force is the molecular averaged kinetic energy gradient: d具典/dx. Taking into account that 具典(i/2)T, we obtain: d 冓冔 i dT i dT N A iRm dT dT (CV ) m . sp dx dx 2 dx 2 dx N A 2 N A m dx
(3.7.18)
where sp index specify the specific heat capacity. From the other side a value of the particle’s flux is equal to (dq/dt)(1/S)(1/3)n (d具典) / (dx). Substituting here the value d具典/(dx) from eq. (3.7.18), we obtain: (dq/dt) (1/S) (1/3)(CV)spmn(d具T典) / (dx). From this equation the heat transfer coefficient is derived 13 y(Cv )sp ,
(3.7.19)
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where mn is the density of the substance. The physical meaning of the factor of heat transfer consists of the fact that it defines the amount of heat transferred in a unit of time through a unit of perpendicular area S⬜ and at a unit of temperature gradient. Analyzing expression (3.7.19), we can see that the heat transfer coefficient depends on the structure of the gas’s molecules (as depends on the molecules’ number of degrees of freedom). In isothermal processes, does not depend on gas concentration and pressure ( ⬃ n and ⬃ 1/n). In the isochoric process depends on temperature as 兹T 苶. The independence of on pressure seems strange at first sight: in fact it is well known that for manufacturing high quality Dewar vessels (thermoses), the presence of air in the space between the walls reduces the quality of the vessel (outflow of heat): the less pressure between the walls of the vessel, the higher the Dewar quality. However, the result obtained seems to contradict this statement. In fact, there are no contradictions in the result, there is incorrectness in its interpretation: at low pressure when becomes commensurable with the space between the walls, dependence of on (1/n) disappears and becomes independent on n (and on p) as will be shown in Section 3.7.8. 3.7.7
Viscosity or internal friction
Viscosity is a property of gases, liquids and solids to resist a flow under the action of external forces. We shall consider in more detail the viscosity of gases. In Figure 3.30, the laminar flow of gas or a liquid is schematically shown; the division of the overall flow into layers is also shown. Due to viscosity, the speed of movement of various gas layers is different: because of its interaction with the unmovable wall’s border, the edge layer is zero and due to the external pressure increases to a central line. The microscopic description of internal viscosity of gases is based on the fact that all molecules in the limit of its layer participate in the macroscopic motion with the momentum, where u is the macroscopic velocity of a molecule with the layer. However, owing to their chaotic movement, molecules jumping over one layer to another transfer the macroscopic momentum m u. Such jumping can influence the macroscopic speed of layers: molecules from faster layers accelerate slower layers and vice-versa. It looks like one layer renders a friction on its neighbor. Thus, viscosity in gases is the phenomenon of the chaotic momentum transfer of the macroscopic movement from layer to layer in a flowing gas. Consider now the law to which the phenomenon of viscosity submits. For this purpose, we shall consider the behavior of a gas between two flat parallel plates (Figure 3.30b). Let one of the layer be resting and the other moves with a constant speed u parallel to the plane of the plates. If there is a viscous medium between the plates in order to move this plate with constant speed u, it is necessary to apply some constant force (directed along the speed) since the medium will exhibit resistance to such movement. Accordingly there appear tangent forces between the separate layers in the medium. Experience shows that this force F is proportional to the speed of plate u and to the area of plates S and is inversely proportional to the distance between the plates x. In a limit at x 0 ⎛ du ⎞ F ⎜ ⎟ S, ⎝ dx ⎠
(3.7.20)
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X F dx S
(a) X
(b)
u(x)
Figure 3.30 The viscous flow: (a) streamline motion of gases and liquids at a laminar flow, (b) friction between adjacent layers: the Newton’s law of internal friction.
i.e., force F, which needs to be applied to move the two layers of gas over each other, is proportional to the area of contact of layers S and to the gradient of relative speed (du/dx) perpendicular to the moving layers. This is Newton’s law of internal friction, is the coefficient of dynamic viscosity of the media in which the movement takes place. To clarify its physical meaning, we shall increase the left- and right-hand parts of eq. (3.7.20) by t. In this case, we have Ft (du/dx)St. On the left-hand side we substitute Ft, which is equal to p (eq. 1.3.12), i.e., ⎛ du ⎞ p ⎜ ⎟ S t , ⎝ dx ⎠
(3.7.21)
where p is the change in momentum of a flux element due to the change in the speed of movement. The coefficient of dynamic viscosity is numerically equal to the momentum of macroscopic movement, which is transferred in a time unit between adjacent layers of flux at a speed gradient along the x-direction equal to unity. In the phenomena of viscosity,
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the transferred quantity is the momentum of the macroscopic movement of the molecules’ layer G(x) mu(x). Substituting this value in eq. (3.7.10) we obtain: q 1 du( x ) , nym St 3 dx
(3.7.22)
冢 13 冣 y.
(3.7.23)
therefore,
For dynamic viscosity in SI units, the coefficient of viscosity of that media is that at a speed gradient equal to unity through area S in 1 m2, the total molecule momentum 1 N sec / m2 kg m / sec is transferred. Thus the unit of viscosity in SI units is N sec/m2 kg/(m sec). Another widely used unit of viscosity is g/(cm.sec) (poiseuille or poise (P)) (in honor of J.L.M. Poiseuille). In tables, viscosity is usually expressed in centipoise (cP). The ratio between units is 1 kg/(m sec)10 P. Besides the coefficient of the dynamic viscosity , the coefficient of the kinematical viscosity is used in technology; there is a relation between the two coefficients: the kinematical coefficient is the ratio of the dynamic viscosity to the medium density /. The coefficient of the kinematical viscosity is measured in stokes (St): 1 St 1 cm2/sec. In SI the unit of kinematical viscosity is m2/sec (1 m2/sec 104 CT). Some viscosity factors are given in Table 3.3.
EXAMPLE E3.15 A free path flight of CO2 gas at normal conditions is 40 nm. Determine the molecule’s average speed 具典 and the number of impacts z a molecule undergoes in 1 sec. Solution: The average molecule speed can be found according to expression (3.3.7´). Substituting the known and given values we obtain 具典 362 m/sec. The average number of impacts can be found according to equation z / (the averaging sign is omitted here). Substituting the given values in this equation, we arrive at z 9.05 109 sec1.
EXAMPLE E3.16 Two horizontal disks with identical radii R 20 cm settle one over another so that their z-axes coincide. The distance between the disks is d 0.5 cm. The lower disk rotates around the common axis (refer to Figures E3.16a and b) with frequency n 10 cm1, but the upper disk is kept motionless. Find the torque M working on the upper disk. Both disks are in air, the coefficient of the dynamical viscosity of air being 17.2 Pa sec.
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z Driven disk
R d
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(a)
R
dr r
(b)
Solution: Due to the viscosity of the air that is between the disks, the rotational moment of forces is transferred from the bottom to the upper disk. Together with rotation of the bottom disk, the adjoining air layer comes in movement. In order to find the moment of friction forces working on the upper disk, we shall allocate an elementary thin layer of air with thickness dr. The bottom air layer rotates together with the bottom disk with angular velocity . Accordingly, the linear speed of air particles is u r. We shall take advantage of Newton’s law of external friction (see Section 3.6.4, eq. (3.7.20) and Figure 3.30) and make some simplifications (proceeding from a condition of the problem). We shall take the velocity gradient to be equal to (du/dz) ⬇ (u/d) (u/d) (since u2 is equal to 0). We shall express the force of viscous friction by an elementary tangent force dF and the small area
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of air layer touching as dS 2 rdr. Accordingly, Newton’s law for our particular problem will be u dF 2 rdr. d We can find the elementary moment of friction forces relative to the z-axis working on the part of the allocated air layer having increased a tangent force dF on a shoulder r, i.e., ⎛ n⎞ dM dF r 4 2 ⎜ ⎟ r 3 dr. ⎝ d⎠ To find the total force moment, we have to integrate dM over the whole disk area, i.e., find the overall contributions from all elementary layers dr, that is to integrate the elementary force moment upon the radius in the limits from 0 to R. We obtain R
n R4 R4 ⎛ n⎞ M 4 2 ⎜ ⎟ ∫ r 3 dr 4 2 n . ⎝ d⎠ 0 d 4 d Checking the dimension of the result, we obtain mN m, which corresponds to the torque dimension. Let us express all values in SI units and execute calculations: MF3.1421.72105 (0.2)4/510—3 mN m 0.543 mN m (milli-Newton m). EXAMPLE E3.17 Define how many times the coefficient of diffusion of gaseous hydrogen differs from that of gaseous oxygen; both gases are under identical conditions. The effective diameters of the hydrogen and oxygen molecules are 0.27 and 0.36 nm, respectively; the molar masses M of oxygen and hydrogen molecules are 16 and 1, respectively. Solution: The coefficient of diffusion D is expressed through physical characteristics of molecules by the formula (3.7.16): D (1/3)具典, where 具典 is the average molecules speed (3.3.7 ) and the mean free path. Therefore, we can write the diffusion coefficients for oxygen and hydrogen (to an order of 1/3): D1
1 8RT 3 M1
1 2 d12 n
and D2
1 8RT 3 M 2
1 2 d22 n
.
In the equations, we took into account that the conditions (n and T ) of both molecules are identical. Therefore the ratio of coefficients is as follows: 2
D1 M 2 ⎛ d2 ⎞ . D2 M1 ⎜⎝ d1 ⎟⎠
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Substituting the given values we arrive at: 2
D1 32 ⎛ 0.36 ⎞ ⎜ ⎟ 7.11 D2 2 ⎝ 0.27 ⎠ 3.7.8
A transport phenomena in a vacuum condition
If the free path length of the gas molecules becomes commensurate (or more) with the linear size of the vessel, the condition of the gas is referred to as a vacuum. This means that the process of intermolecular collisions, which usually plays a predominant role in establishing a balance is no longer of any importance; the interaction between the molecules and the vessel’s walls is more important in a vacuum. During its last collision with the surface of a hot wall, a molecule acquires additional kinetic energy and carries it, without sharing it with other molecules, to the surface of another wall to heat it. This kind of heat transfer is characterized by an absence of temperature gradient inside the gas (as molecules move the whole distance at the same speed); all occasions we have discussed so far have vanished. In order to estimate heat transfer in a vacuum, we shall consider a simple experiment. Imagine a chamber divided in two by a partition wall with a hole of area S in this dividing wall (Figure 3.31). The temperature of the left wall is T1 and of the right is T2. Let T1 T2. On impact with wall 1 a molecule obtains energy iT1/2. Moving toward wall 2 along an axis x and reaching it the molecule gives the acquired energy to wall 2. The amount of energy will make: ⎛ 1⎞ ⎛ i ⎞ q ⎜ ⎟ ny ⎜ ⎟ (T1 T2 )S. ⎝ 6⎠ ⎝ 2⎠
(3.7.24)
As the factor similar to the heat transfer coefficient in the given expression appears as (1/6)n(i/2); it is proportional to concentration (and hence to pressure, contrary to what we obtained earlier in Section 3.7.6). This can be expected as this fact is well-known
T1
S
T2 X
Figure 3.31 A scheme of vacuum experiments. Effusion.
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from human experience. In fact, the quality of Dewar vessels (thermoses) depends significantly on the gas concentration between walls (residual molecule concentration): the less molecule concentration, the better the thermal protection. Notice that this phenomenon starts to operate only when becomes commensurate (or more) with the distance between the walls. At a distance of 1 cm, the pressure should be no higher than 0.10 Pa. A further reduction in pressure reduces the heat transfer between the walls under the linear law. This distinction from the usual transport phenomena is not the only one for rarefied gases. The flow of molecules at outflow differs significantly too (it is referred to as Knudsen’s flow.) Viscosity or internal friction in high vacuum is absent as there is no collision of molecules. In spite of the fact that pressure does not influence the individual molecular movement, it does influence the current. Let us consider a gas outflow through an aperture under these conditions; i.e., molecular effusion (effusion is the slow outflow of gases through small apertures.) We shall restrict ourselves to the case of isothermal effusion. Refer to the scheme presented in Figure 3.31: two vessels where gas is at different pressures p1 and p2, The effusion flow will take place through the aperture. If p1 p2, the resulting flow will be directed from left to right. In the time unit the number of molecules crossing from left to right can be estimated as N1 n1S/6. The opposite stream is less as concentration is less: N2 n2S/6. Hence, the resulting stream is N N1 N 2
yS (n1 n2 ) ⎛ 1 ⎞ 8 T 2 ⎛ 1⎞ ⎜ S p⎜ ⎟ S p. ⎟ ⎝ 3 ⎠ Tm ⎝ 6T ⎠ m 6
(3.7.25)
One can see from this equation that effusion depends on molecular mass. This circumstance is used for separation of gas isotope mixes. At transmission of a gas mix stream through a porous partition with fine pores, there is an enrichment of the mixture by molecules to lighter isotope. Repeating the process results in a mixture enriched mainly by a definite isotope. With different temperatures in both chambers, the ratio of concentrations n1/n2 will be other than at normal pressure. In fact, in normal conditions, it follows from the formula p nT that n1/n2 T2/T1. Another ratio is observed in vacuum. Equating the number of particles moving in the time unit in two sides, we can obtain: N1N2n11Sn22S and finally arrive at the ratio (n1/n2)(2/1)(兹T 苶2苶 / 兹T 苶1苶), which differs from normal conditions. Notice that the laws marked in this section are relevant only to ideal gases. For gases that cannot be considered as ideal, or for liquids, the laws obtained should be used carefully. The method of measuring small pressures is based on the effusion phenomenon (104–105 Hg mm or 0.1–0.01 Pa), if the diameter of the aperture is small in comparison with the free path length. Otherwise, the gas outflow in the unit time occurs under the laws of hydrodynamics; the gas volume that is penetrated in the time unit is ⬃(1/兹 苶), where is the gas density. This allows one to determine the density of gases measuring time of outflow through small (0.1–0.01 mm) apertures.
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PROBLEMS/TASKS 3.1. How many atoms contain in 1 g weight each of gases (1) helium, (2) carbon, (3) fluorine, (4) polonium? 3.2. The cylinder contains gas at temperature t1 100°. To what temperature t2, is it necessary to heat gas so that its pressure p has increased twice? 3.3. Assume that dry air consists only of oxygen and nitrogen. In this approximation air mass fractions of both gases are w10.232 and w20.768, respectively. Define the relative molar weight Mr of air M(air)rM/k (M is air molar weight; k103 kg/mol). 3.4. A centrifuge rotor with a radius a 0.2 m is filled with atomic chlorine at T 3000 K. The chlorine consists of two isotopes 37C and 35C. The share of isotope 37 C atoms make is w1 0.25. Determine shares w1 and w2 of atoms of both isotopes near the centrifuge’s rotor if it is rotated at an angular speed of 104 rad/sec. 3.5. Knowing the Maxwell distribution of a molecule’s speed (3.3.6) derive function f(u) where u is the molecule’s relative speed u(/prob), i.e., expressed in dimensionless units. Apply this distribution to find that the particular molecule has the probability w of having a speed that differs from half the most probable speed prob/2 by not more than 1%. 3.6. Determine the relative number N/N of ideal gas molecules that have a speed in the limits from zero to 0.1 of the most probable speed prob 3.7. Find the relative number of ideal gas molecules whose kinetic energy differs from the average energy 具典 by not more than 1%. 3.8. How many times will the maximum of the f() function change if the gas temperature T increases two times? Illustrate the result graphically. 3.9. Find a number of all collisions N per t 1 sec between all hydrogen atoms in a volume of V 1 mm3 at normal conditions. 3.10. Find the average time 具典 of the free path flight of oxygen molecules at T 250 K and pressure p 100 Pa. 3.11. Calculate the diffusion coefficient of an ideal gas at (1) normal conditions, (2) p 100 Pa and temperature T 300 K. 3.12. Calculate the dynamic viscosity coefficient of oxygen at normal conditions. 3.13. Oxygen of mass m 160 g was heated for T 12 K, and expended an amount of heat Q 1.76 kJ. By which process did the heating take place: at constant V or at constant p? 3.14. Water vapor is expanded at constant pressure p. Find the work of expansion A, if the amount of heat consumed is Q 4 kJ. 3.15. Nitrogen is heated at constant pressure by an amount of heat Q 21 kJ. Determine the gas work A and the change of its internal energy U. 3.16. An ideal gas performs a Carnot cycle. The cooler temperature T2 is 290 K. By how much is the cycle efficiency e increased if the heater temperature increases from T 1 400 K to T 2 600 K? 3.17. An ideal gas performs a Carnot cycle. The work of isothermal expansion of gas is A1 5 J. Define the work A2 of isothermal compression if thermal efficiency e is 0.2.
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3.18. As a result of isochoric heating of hydrogen in weight m 1 d, the gas pressure increased by two times. Determine the change of gas entropy S. 3.19. A piece of ice m 200 g in weight taken at t1 10°C is heated to t2 0°C and fused; the water formed is heated to t 10°C. Define the entropy change at these processes. 3.20. Water in weight m 36 g is at boiling temperature at normal atmospheric pressure. Determine change U of the internal energy at its full evaporation. Specific heat of evaporation is L 2.26 J/kg. Ignore the volume of liquid water. 3.21. A gas mixture contains 1 2 mole of helium and 2 3 mole hydrogen. Determine the specific heat capacity Cp,sp of the mixture. M14103 kg/mol and M2 2103 kg/mol. 3.22. Determine the adiabatic index of partly dissociated nitrogen gas if its dissociation ratio is 0.2. 3.23. A mixture of gases consists of argon of mass m18 g and nitrogen of mass m24 g. Determine the specific heat capacity CV,sp of the mixture. (M140103 kg/mole, M228103 kg/mol). 3.24. Determine the dissociation rate of oxygen gas if its specific heat capacity is CV,sp 727 J/(kg K). (Moxg 32103 kg/mol). 3.25. A gas mixture consists of equal masses of neon and oxygen. Determine the specific heat capacity of this mixture. (M120103 kg/mol—neon, M232103 kg/mol—oxygen). 3.26. A gas mixture consists of an equal number of moles of argon and oxygen. Determine the adiabatic index . 3.27. A gas mixture of helium and hydrogen are at equal conditions. Find the adiabatic index of the mixture consisting of V1 4 L of helium and V2 3 L of hydrogen. 3.28. Equal volumes of gas neon and oxygen are at equal conditions. Determine the specific heat capacity CP,sp of the mixture. (M120103 kg/mol, M232103 kg/mol). 3.29. Determine the dissociation rate of the partly dissociated chlorine gas if the adiabatic index is (CP /CV) 1.48. 3.30. Identical particles of dust of mass m41019 g are in air as a suspension. The temperature of the air is T 284 K. Define the relative change of concentration (兩n兩/no) of particles of dust at levels with a height difference of h 1 m. 3.31. Air is in a uniform field of gravity. Assume that the temperature of the air is uniform and equals T 290 K. Define the relative change in air pressure (兩p兩/po) at levels with a height difference of h 8.5 km. Assume the molar mass of air to be 29 103 kg/mol. 3.32. Oxygen gas is in a uniform field of gravity. Assume the temperature of oxygen is identical in all layers and is T 300 K. Define the relative change in gas pressure (兩p兩/po) at levels with a height difference of h 80 m. Molar mass of the oxygen is 32103 kg/mol. 3.33. The rotor of a centrifuge is filled with air and finely dispersed particles of weight m 1019 g. The temperature of the air is T 286 K. Find the ratio of the concentration of particles (n/no) at the walls of the rotor and in its center if the rotor revolves with a frequency 8 sec1 and its radius is r 25 cm. 3.34. A concentration no of nitrogen molecules near to the ground surface is equal to 2 1019 cm3. What is the change in concentration 兩n兩 of the molecules at height
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Answers
3.35.
3.36.
3.37.
3.38.
3.39.
3.40.
3.41. 3.42.
3.43. 3.44.
3.45.
247
h 10 m? The temperature of the air is T 280 K. The molar weight of nitrogen is 28 103 kg/mole. Identical particles of dust are suspended in air. The temperature of the air is T 300 K. Under the action of a gravitational force, the relative change of the concentration of suspension particles (兩n兩/no) in levels with a height difference h 3 cm reaches a value of 0.01. Determine the particles’ mass m. Identical particles of dust of mass m 1 1018 g are suspended in air. The temperature of the air is T 300 K. At what distance h are the layers in which the ratio (no/n) of the concentration of the particles is equal to 4? Carbonic acid is in a uniform field of gravity. The temperature of the gas is T 300 K. At what height h does the relative change of the molecule concentration (兩n兩/no) of gas reach 1%? The molar weight of the carbonic gas is 44 103 kg/mol. The rotor of a centrifuge is filled with nitrogen at temperature T 300 K. The rotor radius r is 20 cm. At what frequency of the rotor rotation n (sec1) will the gas pressure on its wall exceed the gas pressure in the center of the rotor by 2%? The molar weight of nitrogen is 28 103 kg/mol. The rotor of an ultracentrifuge is filled with radon at temperature T 305 K. The radius of the rotor is r 20 cm. At what frequency of the rotor rotation n (sec1) will the pressure on its walls be times higher than the pressure in the center of the rotor? The molar weight of radon is equal 0.222 kg/mol. Determine the molecular thermal conduction coefficient of saturated water vapor at temperature T 373 K (100 ºC). The effective molecular diameter d is 0.30 nm, molar mass M 18 103 kg/mol. Find the dependence of the diffusion coefficient on molar mass M at p constant and T constant. Find the average time between oxygen molecule collisions at p 1 Pa and T 300 K. The effective diameter of oxygen molecule d is 0.36 nm, mole mass M 32 103 kg/mol. Determine the ratio (1/2) of the viscosity coefficients of oxygen and hydrogen at similar conditions if the ratio of effective diameters of their molecules is (d1/d2) (4/3). Helium of mass m 0.5 g is in a high-pressure vessel with volume V 20 L. Find the average path length of the helium atoms. The effective diameter d of a helium atom is 0.22 nm, molar mass is M 4.0 103 kg/mole. Oxygen is at a temperature of T 300 K. At what pressure p will the average molecule free path distance be N 103 times bigger that its effective diameter d 0.36 nm? ANSWERS
3.1. (1) 1.50 1023; (2) 5.02 1022; (3) 3.17 1022; (4) 2.87 1021. 3.2. t2
p2 (t1 To ) To 473 C(To 273 C). p1
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3.3. Mair
1 ⎛ w1 w2 ⎞ ⎜⎝ M M ⎟⎠ 1 2
28.9 kg mol.
3.4. 28% and 72%. 3.5. f (u) du 3.6. 3.7. 3.8. 3.9. 3.10. 3.11. 3.12. 3.13. 3.14. 3.15. 3.16. 3.17. 3.18. 3.19.
4
exp(u2 )u2 du; w 4.39 103.
N/N 7.52 107. w N/N 9.3 103. Decreases three times. N 1.57 1021. 具典 147 ns. (1) 90 105 m2/sec, (2) 0.061 m2/sec. 18 Pa sec. At p constant. A 1 kJ. A 6 kJ. U 15 kJ. e 1.88 e. A2 4 J. S 7.2 J/K. S 291 J/K.
3.20. U
m ( ML RTk ) 75.2 J. M
3.21. C p,sp 3.22.
5 1 7 2 R 9.20 kJ (kg K).
1 M1 2 M 2 2
7 3 1.46. 5
m1 m 5 2 M1 M2 R 3.23. CV,sp 455 J (kg K). m1 m2 2 3
3.24.
2M CV,sp 5 0.6. R
⎛ 3 5 ⎞R 636 J (kg K). 3.25. CV,sp ⎜ ⎝ M1 M 2 ⎟⎠ 4 3.26.
(i1 2) (i2 2) (12 8) 1.5. i1 i2
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3.27.
249
5V1 7V2 1.52. 3V1 5V2
3.28. CP,sp 3.29.
i1 i2 4 R 6 R 959 J (kg K). M1 M 2 2 M1 M 2
5 7 0.263. 3
3.30.
冷 n 冨 1 exp(gh T ) 1 e1 0.632. no
3.31.
冷 p 冨 1 exp(gh RT ) 1 e1 0.632. po
3.32.
冷 p 冨 (gh RT) 0.01. po
3.33.
n 2 m 2 r 2 exp e 2 7.39. no T
3.34. 冷 n 冨
gh no 2.36 1016 cm3 . RT
3.35. m (T gh )
冨 n 冨 1.411022 kg. no
3.36. h(T/mg) 1n (no/n)0.588 m. 3.37. h ( RT g )
冷 n 冨 57.8 m. no
3.38. n
1 r
冷 p 冨 RT 47.5s1 . 2 po
3.39. n
1 r
RT ln 100 s1 . 2
3.40.
i 3 d 2
3.41. D( M ) ⬃
RT 22.9 mW mK. M 1 M
.
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3. Molecular Physics
3.42.
3.43.
Page 250
MRT 4 d 2 pN A
1 d22 2 d12
3.44.
3.45. p
16.1s.
M1 2.25. M2 MV
2 d 2 mN A T 2 d 3 N
1.24 m.
20 kPa.
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–4– Dielectric Properties of Substances
Electric and magnetic interactions play an enormous role in chemistry and chemical technology; they govern the processes taking place in atoms and molecules, crystals, electrolysis, surfaces of solids electrolyzing of dielectric polymer materials and others. Because the electric field in molecular systems has a very complex structure, for the convenience of the reader, we will give the nomenclature of electric fields at the beginning of the chapter. Let us start by briefly presenting the main information on electric charges and the characteristics of the fields they create.
4.1 4.1.1
ELECTROSTATIC FIELD
General laws of electrostatics
The main electric charge carriers are charged particles—electrons and protons. They carry a charge |e| 1.6 1019 C, electrons being negative and protons positive. In neutral atoms and molecules, negative electron charge is compensated for by a positive nuclear charge. By removing one or more electrons from an atom, one can make a positive (monoor multicharged) ion and, conversely, by adding electrons, it is possible to create a negative ion from an originally neutral particle. In such a way, positive and negative carriers of charge are created (in an electrolyte, for instance). When an excess or lack of electrons is created in a body, the body becomes charged, carrying a charge Q. The value Q is always proportional to |e|. However, at large Q (in comparison with |e|), this discontinuity is not exhibited, so it is possible to consider the charge changing continuously. The distribution of charge in a body can be described by the function (r) such that (r)dV dq, where dq is the charge, comprised in the volume dV. Function (r) is called the volume charge density. Charge distribution on a surface is described by surface charge density (r)dS dq. Charge distribution along a line gives a linear charge density (l) so that dq (l)dl. (In the first two cases, r is a radius vector of the elements dV and dS, and l is a coordinate of a point measured along the charged line.) Electric charge is invariant with respect to Galileo transformations. If an electric system is closed, its total algebraic charged is preserved. 251
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The force of interaction of two point charges q and Q in a vacuum is defined by the Coulomb’s law: F
1 Qq r , 4 0 r 2 r
(4.1.1)
where r is the radius vector, drawn from one charge to another (|r| is the distance between charges) and 0 is the electric constant. Factor 1/40 is included in the Coulomb’s law in order to correspond to the International System: then q is measured in coulomb, r in meter and F in newton; Maxwell equations (Chapter 5.4) also appear very rational in this case. Coulomb’s law has much in common with Newton’s law of gravity: in both laws, the same functional dependence on the distance (1/r2) is present. There is a profound physical meaning in this fact which we will consider below. There exists however an essential difference: attraction between dissimilar charges and repulsion between similar charges are automatically taken into account in the Coulomb’s law (i.e., Q and q are taken with their signs); however, in Newton’s law, the negative sign of force always corresponds to the masses’ attraction. Comparing these two laws, we notice that Coulomb interaction is billions of times greater than Newton’s law. If we calculate, for instance, the gravitational attraction of electrons and protons with their electric interaction, we would arrive at the enormous value of an order 1040 times. Therefore, in atomic systems gravitational interaction is not taken into consideration. Note that eq. (4.1.1) is applicable only to the interaction of point charges; an exception is the interaction of two spherically symmetric balls uniformly charged over their volumes and/or surfaces (see below). Eq. (4.1.1) cannot be applied directly to the calculation of interaction of nonregularly charged bodies. In this case the extensive charges have to be subdivided into elementary volumes dV; knowing (r), find dq, then find the Coulomb interaction dF between elementary charges and then integrate these elementary forces over the volumes of both charges. The exception is the case when charges are spread over symmetrical bodies; then it is useful to apply mathematical methods to this problem, e.g., the Gauss theorem (refer to Section 4.1.3). 4.1.2
Strength of an electrostatic field
An electrostatic field is created by motionless (in the given system of coordinates) charges. If a charge q is located at any point near another charge Q, the Coulomb force (eq. (4.1.1)) is acting, and it can be accepted that charge Q creates an electrostatic field (see eq. (1.4.4)), in which a charge q is situated. If Q is a spherically symmetric body, the field also possesses spherical symmetry, i.e., is central (Figure 1.23). It is convenient to describe a field by the force characteristic —strength of an electrostatic field E. At a point A, the vector magnitude E is numerically equal to force, acting on the positive dimensionless (point) unit charge.
E
F 1 Qr ; q 40 r 2 r
(4.1.2)
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E A
Figure 4.1 Force lines of a nonuniform electrostatic field and orientation of an electric field strength vector at a point A.
The force F(r) can be substituted by the strength E(r): F(r ) qE(r ),
(4.1.3)
with which the field acts on the point charge q. The electrostatic field can be described by the so-called force lines (Figure 4.1): at a given point A the force is directed along the tangent to a line and the vector magnitude 冟F冟 being numerically equal to force, acting on the unit positive point charge placed in this point of space. If q is equal to unity, E is equal to F. This permits us to investigate a field configuration by placing a probe point unit charge q at different points of the field and, measuring vector force F, to find the distribution of E. Consequently, the electrostatic field can be depicted by the force lines. Taking into account the general law, these lines are drawn in such a way that the tangent to them at any point gives the direction of vector E at that point, and the density of lines per cross-sectional area gives the magnitude of E. Looking at a picture of force lines, it is easy to judge the configuration of a particular field, the direction and the magnitude of vector E at every point of space, gradient E, etc. In Figures 4.2a and b, the central fields of positive and negative charges are depicted, and in Figure 4.2c, a uniform field (the plain capacitor with plates of infinite extent) is represented. Electrostatic fields obey the general physical principle of superposition (first given in Section 2.9.1): if the electrostatic field is created by several charges, the field of every charge is created irrespective of the presence of other charges. The mathematical expression for this case is reduced to the total field E as the geometric sum of strength of fields created by every point charge N
E ∑ Ei , i1
where N is the total number of charges.
(4.1.4)
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+
+
–
(a)
(b)
–
(c)
Figure 4.2 A central field of the point charges (a and b) and a uniform field of a plane condenser (c).
The field created by an extended body can be obtained by integration over the whole volume of a body V: E ∫ dE. V
The integral from vector quantity can be calculated, having expanded the vector onto components along the unit orts vectors E i∫ dE x j∫ dE y k ∫ dEz .
(4.1.5)
It is very important that the vector E (as a polar vector) has its own axis of symmetry directed along it (vector E can be rotated around itself at any angle, i.e., it possesses an axis of symmetry of an “infinite” order). This means that, if the charged body creating an electrostatic field has an axis of symmetry, the vector of the field strength E is directed necessarily along this axis. Hence, when dealing with the problem of calculating the strength of electrostatic field E it is strongly recommended first to discover the charged body symmetry: if the charged body has a symmetry axis, the overall strength E must by all means be directed along this axis! There is then no need to derive Ey and Ez: they automatically become equal to zero. From the formula (4.1.5), only one item of sum remains
E i∫ dE x .
(4.1.6)
There is no problem calculating the integral from the scalar function. EXAMPLE E4.1 The electric field is created by two point charges Q1 and Q2. The distance between the charges is d. Determine the strength of electrostatic field E at a point which is removed at a distance r1 from the first charge and r2 from the second. The numerical values are Q1 30 nC, Q2 10 nC, r1 15 cm and r2 10 cm. Solution: According to the superposition principle of electric fields, each charge creates a field, irrespective of the presence of other charges. Therefore, the intensity
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of a field at a specified point can be found as the vector sum of the fields, created by two charges separately, E E1 E2. The strength of the electrostatic field can be determined by eq. (4.1.2). The vector relations can be seen in Figure E4.1. The E1
A
E
−
r1
E2 r2 −Q2
+Q1
partial field strength created by each charge can be determined according to the Coulomb’s law; they are E1
| Q1 | 4 0 r12
and E2
| Q2 | 4 0 r22
.
The sum can be found according to the cosine theorem E E12 E22 2 E1 E2 cos, whereas cos can be found from triangle (see figure)
cos
E
1 4 0
d 2 r12 r22 0.25. 2r1r2
Q12 r14
Q22 r24
2
| Q1 | | Q2 | r12 r22
cos .
Substituting all data and executing calculations, we arrive at E 1.67 104 16.7 kV/m.
EXAMPLE E4.2 Three identical positive charges Q1 Q2 Q3 1 nCb are located at the vertex of a flat equilateral triangle. What negative charge should be placed in the center of the triangle that would counterbalance the forces of mutual repulsion of the positive charges in the vertexes (Figure E4.2)?
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4. Dielectric Properties of Substances
+ Q2
− Q4
r1 F4
F3
+
F
Q1
+ r
Q3
F2
Solution: All three charges located in the vertexes of a triangle are in identical states. Therefore, for the solution of the problem, it is enough to find out what charge is necessary to place in the center of the triangle so that one of the charges, for example, Q1, would be in equilibrium. According to the principle of superposition each charge is exhibited irrespective of the presence of the others. Therefore, the charges will be in equilibrium if the vector sum of forces acting on them is equal to zero, F2 F3 F4 0 F F4 where F2, F3 and F4 being the forces with which charges Q2, Q3 and Q4 act on charge Q1; F is the resultant of forces F2 and F3. Since forces F and F4 are codirectional they can be substituted by scalar sum F F4 0 or F4 F. The force F can be presented as a sum F2 F3, and as F2 F3 we can write F4 F2兹苶2苶 (1苶 苶 co苶 s 苶苶 )*. Applying Coulomb’s law and taking into account that Q2 Q3 Q4, one can find Q1Q4/40r 12 Q12/40r 2兹苶2苶 (1苶 苶 co苶 s 苶苶), wherefrom Q4
Q1r12 r2
2(1 cos )ⴱ .
From geometrical considerations, we can obtain r1 r/(2 cos 30°) r/ 兹3苶. Since cos cos 60° 0.5, the formula * takes the form Q4 Q1/ 兹3苶. Taking into account that Q1 is 1.0 nCb, the charge Q4 is 0.58 nCb. Note that such electrostatic equilibrium is unstable.
EXAMPLE E4.3 The field is created by a uniformly charged thin rod of length l with a linear charge density . Define the strength of an electrostatic field at a point A lying on the line of the charged rod at a distance d from its end. Numerical values are: 1.2 C/m, l 10 cm and d 5 cm. Solution: Let us draw an x-axis along the rod, and let the origin be placed at the beginning of the rod (Figure E4.3). Because the problem has an element of symmetry (an axis of infinite order collinear to the direction of the rod), the vector of the field strength must certainly lie on this axis.
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y dx
A
E
X
X
a
l
Since the electrostatic field is created by a continuous body (rod), it is not possible to apply Coulomb’s law directly to the solution. Therefore, we have to allocate an elementary charged segment on the rod, to calculate intensity created by it and to integrate this with the length of the whole rod. The element of the rod dx carries charge dx. The elementary field strength is calculated using eq. (4.1.2) dE
dx 1 . 4 0 (ᐉ a x )2
Integration gives ᐉ
ᐉ
E
dx 1 ⎛ ᐉ ⎞ . 4 0 ∫0 (ᐉ a x )2 4 0 (ᐉ a x ) 0 4 0 ⎜⎝ a(a ᐉ) ⎟⎠
It is easy to prove that the dimension corresponds to the field strength dimension. There are other ways to choose the origin, though all choices will give the same result.
EXAMPLE E4.4 One-third of the circle’s circumference of radius R 20 cm made of a dielectric material is charged uniformly with linear charge density 1 106 C/m. Determine the strength of electrostatic field E created by this piece of charged arc in its center—point O (Figure E4.4).
y dE
dEy j
d dl
i dEx x
O
R
/3
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Solution: It is impossible to apply Coulomb’s law directly to the whole part of the ring because it is extensive (not a point). We must divide an arc body into pieces and apply the law mentioned to each elementary piece. Notice that the whole problem has an axis of symmetry of the second order passing through the middle of the arc and the arc center O. This means that the resulting electric strength vector is directed along this axis (why?). Therefore, we direct an axis y along an axis of symmetry. Axis x is of no significance; let it be perpendicular to the y-axis. We allocate an elementary piece dl, bearing charge dq, dq dl. This charge creates an elementary field strength dE at the point O. This vector should be projected onto the y-axis in order to be able to make further integration over the whole charged arc. The projection dE on y-axis is . dE y dᐉ cos 4 0 R 2 d cos 4 0 R The component dEx can be ignored, because we definitely know that after integration the x-component should give zero (“why?” check up by integration). We have to express two variables l and through one, let it be : d dl/R. Hence 3
E
∫
3
3
dE y 2 ∫ dE y 0
2 4 0 R
3
∫ cos d 20 R 0
(pay attention to the integration limits, to coefficient 2 and to the way of measuring the value). To substitute the given and the physical constants, we arrive at the value E 兹苶3/20R; executing the calculations we obtain E 2.18 kV/m.
EXAMPLE E4.5 Along a ring of radius R 1 m, a charge Q 1 C is uniformly distributed. Find an electric field strength E(h) in points lying on the axis that pass through the center of the circle perpendicular to the plane of the ring. Solution: As the field strength E is a polar vector, therefore in this problem it must certainly coincide with the coordinate axis z. The linear charge density is Q/2R. Allocate an elementary vector dl at the ring. This segment carries the charge dq dl Qdl/2R; at point A it creates a field 冟dE冟 dq/40r2, where r is the distance from dl to the point A. Vector dE is directed along the line joining the element dl with the point A. Since we a priori know the general direction of the electric field strength E (along the axis z), we have to project the dE vector on the z-axis and integrate projections (since an integral over perpendicular components must be equal to zero). It can be seen from Figure E4.5 that dEO NdEN cos , cos being equal to h/ 兹苶h2苶 苶 R2苶. Therefore dE
h d ᐉ. 4 0 (h R 2 )3 2 2
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Z
dE dEz A
h
r
O R dl
The integration dE over the whole ring gives:
E
Q h . 2 4 0 (h R 2 )3 2
This general expression permits us to calculate the electric strength in any point of x-axis. In particular, it is very useful to use this example to investigate the extreme conditions. In fact, we know that the field in the center of the ring is zero. This corresponds to the limit h → 0: indeed at h 0 the relation obtained gives E 0. From the other side, at h R, or at R → 0, E ⬃ Q/h2 which corresponds to Coulomb’s law.
EXAMPLE E4.6 A thin half-ring carries a charge uniformly distributed along the ring with a linear charge density 10 nC/m. Determine the electric potential in the center of ring O. Solution: A ring element dl carries a charge dl. This charge produces an electric potential equal to d dq/40R dl/40R in the point O. After integration of the scalar values over the whole half-ring (/2 /2), we arrive at /40 282 V. Note that potential calculation is less troublesome than the field strength!
4.1.3
The Gauss law
A vector flux through an area and vector circulation along a closed contour is the basic characteristic of a vector field in vector algebra. The application of these concepts to an electrostatic field appears extremely fruitful.
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We apply this concept, which we first met in Section 2.8.3, to a flux and density of flux. Accordingly, the elementary flux d of the electrostatic field strength E through an elementary area dS is d E dS
(4.1.7)
(see eq. (2.8.18)). Having written down an expression of scalar product and then attributing cos first to E and then to dS, we obtain: d E dS cos En dS E dSn ,
(4.1.8)
where En is the projection of the vector E onto a normal n to the elementary area dS, and dSn is the projection of dS on a plane, perpendicular to E (refer to Figure 2.24). It can be seen that the flux d is subject to change not only due to E magnitude, but also due to the mutual position of vectors E and dS (due to the change of cos ) from E dS up to E dS, i.e., positive and negative. If the field is produced by a point charge, the flux d of the field strength is proportional to a solid angle d . The part of space confined by a conical surface (Figure 4.3) is referred to as a solid angle. A measure of a flat angle d is the ratio of the length of an arc dl of a circle, drawn by an arbitrary radius r about a point O, to the value of the radius mentioned, i.e., d
dᐉ ; r
(4.1.9)
accordingly a measure of a solid angle d is the ratio of an elementary area of a spherical surface dS, drawn by any radius r around a point O, to the square of the radius r2 mentioned (Figure 4.3):
d =
dS dSn = 2 . r2 r
(4.1.10)
dl d 0 dS
r
r
0 dΩ
Figure 4.3 A measure of a solid angle.
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(Replacement of dS by dSn is permissible because the area dS is always perpendicular to the radius.) When dSn is numerically equal to r2, the solid angle is equal to 1 sr (steradian). 2 Similar to the way in which the total flat angle is equal to dl/r 2r/r 2 rad, the 0 total solid angle is equal to
冕
养
dSn 4r 2 2 4 sr. r2 r
(4.1.11)
Gauss law asserts that: the flux of an electrostatic field strength vector through any closed surface S is equal to the algebraic sum of the charges confined by surface S divided by 0. We shall call this virtual (imaginary) surface the Gauss surface. Note that we ourselves choose the form and size of a Gauss surface! Let us prove the theorem taking a single point charge Q as an example. We place it in a point O and cover this charge by a closed spherical Gauss surface S with the center in the charge Q position (Figure 4.3). Let us find the value of the elementary flux d of the vector E through the elementary surface dS: d E dSn. The magnitude E is defined by expression (4.1.2) and therefore
d
1 Q 1 Q 2 dS r d 40 r 2 n 40 r 2
(4.1.12)
(we used here the expression (4.1.10)). After integrating over the total solid angle, we arrive at
养 En dS 养 S
S
Q 1 Q d 4 0 4 0
养 d S
1 Q, 0
or, keeping in mind expressions (4.1.7) and (4.1.8), 养 EdS s
1 Q. 0
(4.1.13)
Let us comment on the expression derived. In this expression the point charge Q is comprised of a Gauss surface S. It was taken in spherical form, but it can also be a irregular shaped surface. On the right-hand side the charge Q can be seen which we encompassed by the Gauss surface (Figure 4.4). If the charge is outside, the flux through the Gauss surface is equal to zero (Figure 4.5) since part of the flux is negative and the other part is positive (because of various orientations of a normal n to the surface relative to vector E). This is the case for all those charges which lie outside the Gauss surface; they can be excluded from consideration.
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S
E dS n dΩ
r
Q >0
Figure 4.4 The Gauss theorem: Q is a charge, S is a closed Gauss surface, dS is an elementary area and n is a unit vector to dS. n E
O Q>0
E
n
Figure 4.5 The zero total flux of the electric field strength for a particular case when an electric charge is outside the Gauss surface.
Values Q can be determined if distribution of the charge in a given problem is known. If the field is produced by a sum of N individual charges, then according to the superposition principle 养 E dS s
1 0
N
∑ Qi .
(4.1.14)
i1
If the field is set up by a volume charged body, the expression can be modified as follows: 养 E dS s
1 (r ) dV , 0 V∫
(4.1.15)
where (r) is a charge density distribution function. If the field is set up by a charged surface with the surface charge density (r), the Gauss law looks like: 养 E dS s
1 (r ) dS. 0 S∫ⴕ
(4.1.16)
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If the source of field is a charged line, the Gauss law can be given as 养 E dS s
1 0
∫ (l ) dl,
(4.1.17)
L
where (l) is the linear charge density distribution function and l is the point coordinate along the charged line. The body volume V, charged surface S and the length L of the charged line correspond to those parts of the charged physical bodies that are encompassed by the Gauss surface. Repeat once more that in Gauss law the total charge laying inside of the Gauss surface is present on the right side which is equal being divided by o to the electrostatic field strength flux on the left side. The Gauss law is valid for all forms of charged bodies and any Gauss surface. However, its most fruitful application is in calculating the field when the problem possesses some symmetry. In this case the skilful choice of the Gauss surface form permits one to achieve essential simplification: to provide that E En const., to take E out of an integral and to integrate only upon a surface, or, in general, to arrange the field strength being En 0 onto a part of a surface. Examples of application of the Gauss law given below will show calculations of fields of the charged physical body possessing some symmetry. These problems could be solved using only the principle of electric fields’ superposition. However, this way is troublesome; it demands integration upon a volume. Application of the Gauss law allows many problems to be solved “in a single line”. EXAMPLE E4.7 An electrostatic field is created by two parallel, infinitely large, charged plates with surface charge density 1 and 2. Define the electrostatic field strength created by these two plates between and behind them. Numerical values are 1 0.4 C/m2 and 2 0.1 C/m2. Solution: According to the principle of superposition of electric fields, each charge creates a field irrespective of the presence of other charges. In this case it concerns the charged planes. Therefore, the intensity of a field in specified areas E can be found as the vector sum of the fields, created by two planes separately: E E1 E2. The absolute values are in this case: E1 1/20 and E2 2/20, E1 being higher than E2. The sign of each term depends on two peculiarities: the sign of the plate charge (objective characteristic) and the choice of the axis direction (subjective characteristic). Let us divide the problem into three parts (I, II and III) and choose an x-axis direction, we recommend that it is usually accepted from left to right (Figure E4.7). The superposition principle offers us the choice of the direction and absolute values of particular fields. Therefore, we recommend beginning the solution of the problem with the determination of E(I) and E(III) E (I) E (III)
1 (I) (III) . 0 2
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E
1
2
x
Between the plates the field is directed oppositely from both plates and therefore E(II)
1 | 1 2 | . 0 2
The situation presented in this example can rarely be met in practice. Much more popular is the plane condenser, which differs from that given here by the fact that 1 2; the electric field inside the condenser is E /0, there is no field outside the condenser to say nothing on the edge effects (see Figure 4.11).
EXAMPLE E4.8 Two concentric spheres of radius R1 6 cm and R2 10 cm, respectively, carry charges Q1 1 nC and Q2 0.5 nC. Find the electric field strength at points r1 5 cm, r2 9 cm and r3 =15 cm (Figure E4.8). Draw a graph of E(r). | E | = En
E
E
n n
Q2
R2
E,V/cm
I
2500
Q1 R1 III
II
I
II
O 900 III
450 (a)
(b)
0
R1
R2
r
Solution: The three points mentioned are correspondingly disposed just inside the inner sphere (domain I, r1 < R1), between the two spheres (domain II, R1 < r2 < R2) and outside the spheres (domain III). The two spheres really exist carrying a definite charge. Apply the Gauss theorem to solve the problem. First, we have to solve the
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situation inside the smallest (internal) sphere. One should choose an imaginary sphere with r1 < R1 and write down the Gauss theorem for this domain: wEn dS 0. Because En on the sphere is constant, En 冟E冟 E, then Ew dS E4r 2 0; since r 2 is not zero it means that E1 ⬅ 0*. In domain II, R1 r2 R2, the Gauss surface should be chosen having R1 R R2. In this case the whole first real sphere occurred inside the Gauss surface. Therefore, wEndS Q1/0, E4r2 Q1/0 and E (Q1/40)(1/r2)**. (Sometimes the sphere charge is given through surface charge density and the whole charge of the sphere is 4R2. Note that R is a number, but r is a variable.) In order to find the E(r) function in the domain III, the Gauss surface radius should be larger than R2; it incorporates both spheres. Therefore, EwdS (Q1Q1)/4r 32 ***. The values of E’s presented in the graph can be obtained from the marked functions E2 and E3. Three marked equations permit us to draw the graph. Try to find yourself the values presented in Figures E4.8a and b.
EXAMPLE E4.9 An electron with zero initial speed has passed an electrical potential difference between a cathode and an anode U0 10 kV and entered a space between horizontal plates of the flat condenser on line AB parallel to plates (Figure E4.9); the condenser is charged up to potential difference U1 100 V. A distance d between plates is equal 2 cm. The length L1 of the condenser plates in direction AB is equal 20 cm. Determine distance BC (BD DC) between fluorescent screen spots at the distance from the condenser’s end to a screen is L2 1m. L1 Cathode A
–
L2
0
B 0
M + + + + + + + 1
Anode
l1 D
l2
C Screen
Solution: An electron movement inside the condenser consists of the two components: (1) by inertia, along the line AB with the constant speed 0 acquired earlier by action of accelerating potential difference between anode and cathode U0 which an * m02 electron has passed up to the condenser eU0 and (2) a uniform acceleration 2 movement inside the condenser in a vertical direction due to an action of constant electric field. After leaving the condenser at point M, an electron moves uniformly and rectilinearly.
冢
冣
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One can see from the figure that the sought distance consists of two values BC l1 l2, one of them l1 being a distance on which the electron is displaced in a vertical direction during movement thought the condenser, the other one l2 between a point on the screen DC (see figure). Let’s estimate separately l1 and l2. The value l1 can be found using the formula of travel distance in the uniform acceleration movement l1 at2 /2 **, where a is the acceleration acquired by the electron under the action of the condenser’s constant electric field and t is the time of its action. According to the Newtonian second law, acceleration is a F / m (the force F which acts on the electron in condenser, m being its mass). In its turn, F eE eU1 /d. The time of the electron flight inside the condenser l1 we shall find by the formula of a uniform movement t L1 / 0. Consequently, substituting values from other expressions into the formula * we can obtain l1 U1L12 / (4dU0). We can find the length of a segment l2 from corresponding triangles MDC and then build on vectors 0, 1 and assuming that the deviation from AB-direction is small, l2 (1L2 / 0)**, where 1 is the electron speed in a vertical direction at point M; and L 2 is a distance from the condenser end up to the screen. We can find the speed 1 using the formula 1 at which in view of expressions for a, F, and t will become
1
eU1 L 1 . dm0
Having submitted the expression derived into the formula **, we obtain l2 eU1L1L2 / (d m02) or, having replaced 0 from equation **, we found l2 U1L1L2 / 2 dU0. For the required distance BC l1 l2, we shall finally arrive at
BC
U1 L 12 U1 L1L 2 U1 L1 ⎛ L1 ⎞ L 2⎟ . ⎠ 4 dU 0 2 dU 0 2 dU 0 ⎜⎝ 2
Substituting all given values into the last expression and having made calculations, we arrive at BC 5.5 cm. Let us start with calculation of a field set up by a single charge Q. The field of the point charge depends only on the distance from a charge to a point of observation and does not depend on the direction, i.e., the field is spherically symmetric. Choose an arbitrary point A (arbitrary means that it is typical and not distinguished from any other). We shall carry out a spherical Gauss surface through this point at the radius r with the center in charge Q (Figure 4.6). We shall write down for this sphere S the Gauss law in the form sEn dS Q/0. In this expression, En is the projection of vector E on the normal vector n to the sphere surface. Any line passing through point O is the axis of symmetry because any turn around it superposes a system with itself.
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E, r A n
∝1/r
E =En
Q
E ∝1/r 2 r
Figure 4.6 The electric field of a point charge. Point A is a point of observation, the dashed line represent a Gauss surface.
Therefore, the vector E coincides with the symmetry axis. It means that |E| En. Besides, as a consequence of the symmetry of the problem, E const. at any point of the Gauss sphere. Therefore, wsEdS EwsdS 40r 2 . The right-hand side of the equation is equal to Q/0. So, E4r2 Q/0 or E Q/40r 2. This coincides with eq. (4.1.2). This coincidence is quite natural because while deriving the Gauss law, we use for simplicity the point electric charge. It is worth mentioning here that the Gauss law is valid for all fields whose strength falls as 1/r2. Obviously, only in this case the r2 terms in the nominators and denominators of the expression (4.1.12) are cancelled. We know two such fields: electrostatic and gravitation; in these two cases the Gauss theorem is valid. Let us now apply the Gauss law to consideration of a field set up by both uniformly charged spherically symmetric bodies (sphere) and a ball uniformly charged upon the surface, the radius of spheres being R in both cases. This means that there is no charge inside the body. The charge surface density can be denoted as Q/S, where S is the surface area. In both cases we have to signify two domains: domain I is the region inside the bodies (rA < R, where rA is the distance from the center of spheres to an arbitrary chosen point A) and domain II corresponds to outer space (Figure 4.7). As was valid above, any line passing through the center of the spheres is the symmetry axis and |E| En const. In the first domain 0 r R, the Gauss integral is sEn dS 0. As En |E|, then EwsdS E4r2 0 and since r 2 苷 0, then E ⬅ 0. In domain II, r > R, the arbitrary point A has to be chosen outside the spherical bodies. The total charge Q will here be fully encompassed by the Gauss surface. Therefore
E ∫S dS E 4r 2
Q 0
and
E=
Q 40r
2
⎛ 1 ⎞ ⎛ 4R 2 ⎞ ⎝ 4 0 ⎟⎠ ⎜⎝ r 2 ⎟⎠
=⎜
=
R2 . 0r 2
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r O R A
E(r)
(r)
E(r)
R
r
Figure 4.7 The electric field strength E(r) and potential (r) of the surface-charged sphere.
The result is depicted in Figure 4.7. It is apparent that the strength values on the internal and external sides of the surface are different, i.e., the field strength on the charged surface experienced a break. In a similar way it is possible to obtain the values of the field strength of several spherical surfaces with the same center (e.g., for a spherical condenser). If the sphere is charged homogeneously ( Q/V const.), the field inside is not equal to zero. As before, the problem has a spherical symmetry; therefore, the Gauss sphere should also be chosen in spherical form. With such a choice, |E| En const. at Gauss sphere. Accordingly, the arbitrary point A is chosen in two domains: inside the sphere (r < R) and outside it (r > R). To find the electrostatic field strength dependence at the distance r from the charged sphere center inside it, we shall take advantage of eq. (4.1.15). We shall find the charge density by dividing overall charge by the volume Q/(4/3)R3 3Q/4R3. Then 养 En dS E 养 dS E 4r 2 s
s
3Q 4 3 r , 40 R3 3
wherefrom
E (r )
Q 40 R3
r.
Pay attention that E ⬃ r! This graph is given in Figure 4.8.
(4.1.18)
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r1 R
O
A
E
E~r E~1/r 2 E~r/
r
R
Figure 4.8 Field of a uniformly charged ball.
Calculation of the field outside the sphere is easier: the Gauss surface encompasses the whole sphere. Then
养 En dS E 4r 2 s
Q 0
and
E
Q 40r 2
.
As previously seen E is proportional to 1/r2. If there is a material with definite (see below), the dependence E(r) should contain term 1/ (Figure 4.8). Let us generalize the result. In all three cases of spherical symmetry, a function E(r) at r R is described by dependence E ~ 1/r2. It means that an observer who is outside the sphere at any point r, knowing the fragment of the measured dependence E(r) cannot judge whether the field is created by the point charge, a sphere
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uniformly charged over its surface or the uniformly charged solid ball. In all three cases dependence E(r) is identical! Moreover, in the case of spherical symmetry nothing prevents “swiping” the sphere or ball to a point and to consider a field as if it is created by point charge! Sometimes it essentially facilitates a problem. Remember that we have already done this in Chapter 1: in studying the force and potential energy of a gravitational field (a field of the earth), we dealt with formula F(r) GMm/r2 and U(r) GMm/r, where r is the distance from the center of the earth. We swiped the whole mass of earth to a point (see Section 1.4.5). The possibility of the generality of the results is in identical dependence on Coulomb and Newton forces from distance r! Calculate now a field created by a body with cylindrical symmetry, for example, infinite and uniformly charged cylinder (for simplicity—over a surface) with linear charge density dQ/dl const. (Note that there are no such cylinders or infinite planes in nature. This physical problem is equivalent to the condition where the cylinder has finite length L, however, we consider a field near to the charged surface, i.e., at r L . Then it is possible to neglect the edge effects and solve the problem for an infinite cylinder.) Let us analyze a problem that has cylindrical symmetry, i.e., the axis of symmetry coincides with the axis of the cylinder; this means that in calculating the field strength, we deal with E (not with E), it depends only on the distance from a cylinder axis to a point of observation r (but not from r!). Furthermore, any straight line perpendicular to the cylinder axis and crossing it is an axis of symmetry of the second order (the turning of the infinite cylinder around the axis on superposes it with itself). This means that vector E should be directed along such a straight line. Choose accordingly a closed Gauss surface in the form of a coaxial cylinder of length l (cyl) and end surfaces (end) perpendicular to the cylinder axes (Figure 4.9). Then the integral over the Gauss surface will be separated into three parts:
养 En dS s
∫ En dS 2 ∫
cyl
En dS.
end
The last item is equal to zero because En 0 at end surfaces (E ⊥ n). Over all cylindrical surface, E ⱍⱍ n; therefore
∫ En dS ∫ En dS 2r ᐉE. S
cyl
At r < R (not depicted in the scheme), E 0 (according to the same consideration why the field inside an empty sphere is absent). At r R, E2rl Q/0 l/0. Thus, E(r) /20r. A graph of this dependence is presented in Figure 4.9.
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R n E r
l
E
n A
E
~1/r r
R
~-ln r r
Figure 4.9 Field and potential of a system with cylindrical symmetry.
Let us now calculate a field created by an infinite plane (i.e., with linear sizes much larger than the distance from the plane to an observation point), the plane being uniformly charged ( = dQ/dS = const.). Any straight line perpendicular to the infinite plane is the axis of symmetry (because any turn around it will impose a plane on itself). Therefore, E in any point should be directed along this axis, i.e., perpendicular to the plane. In this case, it is expedient to choose the Gauss surface to be a cylinder with the generatrix perpendicular to the plane, as is shown in Figure 4.10. Therefore, the electric strength flux through a side of the cylinder surface is zero, whereas the flux through two ends is
养 En dS 2 s
∫
En dS 2 ESend .
Sbot
According to the Gauss theorem, this expression is equal to the charge inside the Gauss surface, i.e., Send. So 2ESend = Send/0, i.e., E /20 const (see Figure 4.10)
E
. 2 0
(4.1.19)
It can be seen that the field strength near to plane does not depend on distance (!). Negative value of E to the left of the plane means that vector E is directed opposite to the axis x.
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n
+
E n
n
E x S
S
S E +/ 2 0
–/2ε 0 x
Figure 4.10 Field of a uniformly charged plate.
+
–
+
–
d
d
Figure 4.11 A Uniform condenser field.
The system of two similar planes carrying an equal charge with an equal charge density represents a plane condenser, provided they carry charges of opposite signs. According to the superposition principle, the field in the condenser is an algebraic
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dm=υtρdS υt
S dS
Figure 4.12 A Gauss theorem and a liquid’s flow.
sum of the fields created by each plane separately (Figure 4.11). So, the field inside the plane condenser is
E 2 E
; 0
(4.1.20)
however, the field outside it is zero. The plane condenser is a source of a uniform electric field. In summary, we make two more remarks on the Gauss theorem. The first concerns the physical nature of the equation. Because in eq. (4.1.13) an electric charge is presented on the right-hand side, the theorem asserts that the source of an electrostatic field is electric charge. Another concerns the general meaning of the theorem, wider than only electrostatics. Imagine, for instance, a flow of a liquid in a pipe (Figure 4.12). Each particle of the liquid moves with speed u. The mass of the liquid will cross a surface dS in time t, making dm ut dS. The mass of the liquid m s t dS t s dS will pass through surface S in time t, and if S and dS are directed parallel to each other, m t dS. On one hand, these equations allow us to calculate the mass of the liquid at its flow in pipes (at known distribution of speed, at turbulent or laminar current), and, on the other hand, to formulate a criterion for liquid incompressibility, 养sdS constant. If there is a source of liquid inside the Gauss surface, 养sdS j, where j m/t is a source power. 4.1.4 field
Work of an electrostatic field force and potential of an electrostatic
Coulomb forces are central and, consequently, conservative; the field of these forces is potential (refer to Section 1.4.4, Figure 1.29 and eq. (1.4.24)). Indeed, by definition, an elementary work dA of a force F on a displacement dl is determined as: dA F dl qE dl qE dl cos, where E Q/40r 2 and dlcos dr. Then dAqQdr/40r2 and A12(qQ/40) r r21 dr/r 2. Therefore,
A12
qQ ⎛ 1 1 ⎞ , 40 ⎜⎝ r1 r2 ⎟⎠
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i.e., the work in fact does not depend on the distance traveled but on r1 and r2, the radius vectors of the initial and final points. If a charge starting from the initial point returns to it, the work of electrostatic forces is equal to zero: Aq LEl dl 0, where El E cos is a projection of vector E on the displacement vector dl. Then 养LEl dl 0: circulation of the electrostatic field strength along the closed trajectory is equal to zero. This is the definition of the potential character of a force field. From equation 养LEl dl 0 it follows that the force lines of an electrostatic field cannot be closed (otherwise with a positive detour the integral would be essentially positive). They start from the positive charges and end at negative charges (or pass into infinity). The work of electrostatic field forces can be presented as a potential energy decrease, A U1 U2 U. From a comparison of this equation with that given above, the potential energy of a charge q in a field of another charge Q can be written as U(qQ/40r)C. If one accepts that at r ∞ the potential energy U(∞) is equal to zero, then C 0 and the potential energy is finally
U
qQ . 40r
(4.1.21)
The electrostatic potential is a value numerically equal to the potential energy of a unit positive point charge placed into the given field point. Then U = q .
(4.1.22)
The point charge potential is therefore
Q . 4 0 r
(4.1.23)
Inasmuch U q , then
A12 U1 U2 q( 1 2 ) q .
(4.1.24)
If one assumes (∞) 0, then A1∞ q 1; therefore, 1 A1∞ /q. This means that the potential is numerically equal to the force work done on the displacement of the unit positive point charge from the given point to infinity. The field potential is the scalar power characteristic of an electrostatic field. The potential of a field having been created by a system of motionless charges obeys the superposition principle, i.e., each charge creates a field irrespective of the presence of other charges in the space. As the potential is a scalar value, the potential in any point of the field, created by the system of charges, is equal to the algebraic sum of potentials of fields created by each charge individually: N
(r ) ∑ i (r ), i1
where N is the total number of charges.
(4.1.25)
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From the point of view of the potential distribution, the electrostatic field can be graphically characterized by surfaces of equal potential, i.e., by equipotential surfaces with constant (Figure 4.13). The force lines are always normal to them. Because electric field strength E(r) and potential (r) are functions of a point radius vector r, there should be an interrelation between them. Let us find it. The elementary work of a field force can be determined as either dA F dl qE dl or dA dU q d . Comparing them, we can arrive at El d /dl. If the field is created by a spherically symmetric charge, E and are both r-dependent; therefore
Er (r )
d (r ) . dr
(4.1.26)
Ex ( x)
d ( x) . dx
(4.1.27)
In the case of linear dependence
In the general case, r(x, y, z), E(r) and (r) are interconnected by ⎛ d
E(r) ⎜
⎝ dx
i
d d ⎞ j k⎟ grad (r), dy dz ⎠
(4.1.28)
i.e., the electrostatic field strength is equal to the electric field potential gradient taken with opposite sign. One can see that the field strengths’ vector is always normal to equipotential surfaces (Figure 4.13). It follows from eqs. (4.1.26)–(4.1.28) that the function (r) cannot be a discontinuous function. In fact, if a function undergoes a break (for instance, in the point x0), then, according to eq. (4.1.27),
E ( x0 ) lim , x0 x since in this point x → 0, but tends to a finite difference; this means that E(x0) is equal to infinity. This cannot be the case. = const
Figure 4.13 Force lines and equipotential surfaces.
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The relation established between the electrostatic field strength and potential allows us to find dependence (r) knowing the function E(r) for all examples considered above. In fact, it follows from eqs. (4.1.26) and (4.1.27) that (r ) ∫ E (r )dr C and ( x ) ∫ E ( x )dx C. Substituting here the already known functions E(r) and E(x), we can calculate functions (r) and (x). The constant C reflects the fact that the function , as well as potential energy, is known to be within a constant; it can be found from boundary conditions. The choice of C can provide a continuity of function . Thus, the potential for the problems considered can be obtained. Corresponding graphs are depicted in Figures 4.6, 4.7, 4.9–4.11. In order to calculate the potential difference , integration should be carried out in certain limits. So, for instance, the potential difference in a plane condenser is d
d
0
0
∫ E (x )dx E ∫ dx Ed , where d is the distance between the plates. The reverse problem (knowing , calculate E) can be obtained as well. This method will be used, for example, when we calculate an electric dipole field (because potential calculation is easier than calculating the field strength since the former is a scalar value, whereas the latter is of vector quality). 4.1.5
Electrical field of an electric dipole
A system consisting of two charges, equal in absolute value and opposite in sign, is referred to as an electric dipole. The vector drawn from a negative charge to a positive one l is called a dipole arm. The vector p equal to a product of the charge and arm is referred to as electric dipole moment (Figure 4.14): p ql.
(4.1.29)
It can be seen that the unit of dipole moment measurement is C m. This is a very large value and therefore a significantly smaller one is used in chemistry, namely 1 Debye (D): 1 D is equal to 3.33 1030 C m. When a dipole field is studied at a distance much larger than the dipole arm, it is called the point dipole. In spite of the fact that a dipole seemed to be an electrically neutral particle, it produces an electric field; this field has different properties compared to a field of
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point charges. In order to describe a dipole field, we use the superposition principle: , i.e., ( q 4 0 r ) ( q 4 0 r ) q (r r ) 4 0 rr , We consider the point dipole (l r) (Figure 4.14); therefore, rr ⬇ r2, r r ⬇ l cos which means that
(r , q )
qᐉ 4 0 r
2
cos
p cos . 4 0 r 2
(4.1.30)
In order to determine the strength of the dipole field at point A, the relationship between strength and potential can be used. We use a polar coordinate r and with a polar axis coinciding with the dipole arm direction (Figure 4.15). Component Er (projection E on the radius vector r) is Er
2p d cos . dr 4 0 r 3
A r
−
l + +q
p
–q
Figure 4.14 An electric dipole model.
Er
Eθ
A r+
r– r
l cos −
p
+
l
Figure 4.15 Dipole electric field calculation.
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The E component perpendicular to r is equal to E
d p sin , r d 4 0 r 3
where r d is the length of an arc with radius r. Then, the total electric field strength at point A is E Er2 E2
p 13 cos2 . 4 0 r 3
(4.1.31)
The potential and strength of the dipole field are defined not by a value of charge q but by the dipole electric moment p |q|l. With distance, the dipole potential and strength decrease faster ( ~ 1/r2, E ~ 1/r3) than those of a field of a point charge (1/r and 1/r2, respectively). A configuration of the dipole electric field is shown schematically in Figure 4.16. In a uniform electric field a torque MF aspires to turn the dipole moment p to be oriented along the field E. A force couple MF is equal to the product of force F |q|E and the force arm l sin (Figure 4.17), i.e., M |q|El sin pE sin or, in vector form, M F [ pE].
(4.1.32)
The vector M is directed perpendicular to the plane of drawing from a spectator. The resulting force in the uniform electric field is zero. In order to calculate the potential energy of a dipole in the uniform electric field we use the recipe given in Section 1.4.5. To calculate the potential energy we should first z E
p
Figure 4.16 A dipole electric field configuration.
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find the elementary work of the force couple (dA dU, where dA MF d is the force couple work and dU is the increment of the potential energy). Then
U ∫ M F d ∫ pE sin d, after integration we can obtain U() pE cos C. If we accepts U(/2) 0, then C 0 and U pE cos or in vector form U (pE).
(4.1.33)
A graph of dependence U() is given in Figure 4.18. When the vector p coincides with vector E, the dipole potential energy acquires the lowest value ( 0, Umin pE): the dipole is in a stable equilibrium; the force couple is also equal to zero. When the vector p is oriented perpendicular to the vector E, the dipole energy is equal to zero ( /2, U 0). When the vector p is directed opposite to the vector E, the dipole energy is maximum ( , Umax pE), the dipole is in an unstable equilibrium: any deviation from this state leads to turn-off of the dipole. In a nonuniform electric field, forces acting on dipole charges are not equal in absolute value and direction. In Figure 4.19, a graph of a nonuniform field (dE/dx < 0) with a dipole p in this field is depicted. Both the force couple MF and resulting force F will act on the dipole.
F+ P
l.sin
F–
Figure 4.17 Electric dipole in a uniform electric field.
U +pE
0
/2
–pE
Figure 4.18 A dipole potential energy in a uniform electric field.
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F+ +
x
p F–
–
Figure 4.19 An electric dipole in a nonuniform electric field.
To calculate a force projection on x-axis, we can use eq. (1.4.33) where the relation of the force projection with the field nonuniformity is given (Fx ∂U/∂x); in our instance, U pE cos and then Fx p
E cos . x
(4.1.34)
If a dipole in the electric field is in a stable state (cos 1), then the force will draw the dipole into an area of stronger field (for Figure 4.19, Fx 0). The ratio (eq. (4.1.34)) shows that an interaction force can exist even between neutral molecules, though having the dipole moments. In fact, if an electrically neutral polar molecule creates the nonuniform field described by expression (4.1.31) and another molecule with a dipole moment p is in this field, a dipole–dipole interaction appears between them.
4.2 4.2.1
DIELECTRIC PROPERTIES OF SUBSTANCES
Conductors and dielectrics: a general view
From the point of view of electric properties, all substances can be divided into two main classes—conducting and nonconducting an electric current. Metals, their alloys and a small number of chemical compounds with metal character of interatomic interactions relate to the class of conductors. The second class includes other substances and represents the overwhelming majority. Conductivity is defined by the presence of free charge carriers in a substance; their absence determines dielectric properties. So, dielectrics are substances in which there are no free charges capable of covering long distances in the substance (in comparison with molecular sizes). Depending on their molecular structure, all dielectrics can, in turn, be divided into two large groups—polar and nonpolar. In polar dielectrics, molecules themselves represent the electric dipoles with the electric moment p; it appears due to displacements of electric charges from positions of their equilibrium in free atoms as a result of chemical bonding (e.g., H2O, HCl and NH3). The molecular dipoles of polar dielectrics participate in thermal motion; this can be translational motion (in gases and liquids), oscillation
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about equilibrium positions (solids and liquids) and rotation around the center of mass. As a result, the dipole electric moments p are chaotically distributed along directions N (Figure 4.20a). For the whole dielectric, the vector sum i1pi is equal to zero. Therefore, in spite of the fact that each molecule possesses a dipole moment, the whole sample does not. Molecules of nonpolar dielectrics do not possess a dielectric moment. This means that the positions of the positive and negative charge centers in the molecules coincide. Examples of nonpolar molecules are H2, CCl4, C6H6, CH4, etc. In this case, the macroscopic dielectric sample does not possess a dipole macroscopic moment either. However, when placed in an external electric field E0, all dielectrics regardless of their molecular properties are polarized, i.e., pi becomes nonzero, the dielectric acquires a macroscopic dipole moment (Figure 4.20b). For the quantitative description of dielectrics, the notion of the polarization (degree) or the polarization vector is introduced; numerically, it is equal to the electric dipole moment of a unit volume. Accordingly
1 V
N
∑ pi .
(4.2.1)
i1
In this expression V is a so-called physically infinitesimal volume, i.e., a volume containing enough dipoles that the vector sum in eq. (4.2.1) adequately reflects the macroscopic dielectric state of this volume, but, simultaneously, small enough that within the limits of this volume the polarization degree can be considered as uniform. It is important to notice once more that the sum in eq. (4.2.1) is of a vector character. However, in the case when all N (identical) moments are directed along one and the same axis (we shall choose an x-axis), expression (4.2.1) can be written as x
∑ px Npx np V
V
x
np,
(4.2.2)
where p px (because all vectors p are focused along axis x).
E0 = 0
(a)
E0 > 0
(b)
Figure 4.20 Mutual orientation of a dielectric’s molecules in the absence (a) (random orientation) and in a presence of an external electric field (b).
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There are two approaches to the description of dielectric properties: phenomenological which includes the description of macroscopic dielectric characteristics (subject to direct measurement) and microscopic which includes the analysis of dielectric behavior taking place at an atomic and molecular level. We shall begin by considering dielectric behavior in an external electric field from the macroscopic point of view. 4.2.2
Macroscopic (phenomenological) properties of dielectrics
Consider now what occurs with a dielectric if it is placed in an external field E0 (Figure 4.21), created, for example, by a plane condenser. As was mentioned above, the dielectric polarizes. All charges inside a volume remain mutually compensated; however, there is no compensation near surfaces. Bound charges with density are created at the edges of a body. These charges cannot be taken from the dielectric surfaces; therefore, they are referred to as bounded charges (opposite to those free charges which are on capacitor plates and form a field E0 (E0 free /0) (4.1.19)). The surface-bounded charges produce another charged condenser-like “plate” which creates the electric field E1 directed opposite to the external field E0 (Figure 4.21). This field is referred to as a depolarized field or an electric field of bounded charges. The electric field E inside dielectrics can be treated as a superposition of two fields: the external field E0 and the depolarized field E1, i.e., E E0 E1, or in scalar form E E0 E1 .
(4.2.3)
Thus, the resulting average field E in the dielectric body is always less than external field E0. This field is referred to as an average macroscopic field E in the dielectric; the average field represents the result of superposition of an external field and electric fields of bounded charges. The value equal to a ratio of the strength E0 of an external electric field to the strength E of an average field in the dielectric is referred to as dielectric permeability of the dielectric medium.
– '
− − −+ − − − − − − − − − −
−+
E0 . E
−+
E
+ + −+ −+ + + + + + + + + E1 + +
(4.2.4)
'
E0
Figure 4.21 Dielectric in an external electric field E0: formation of surface–bounded charges with surface density is shown.
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Namely this value has to be introduced into the Coulomb’s law (4.1.1) in order to determine a force acting between two point charges in a medium with dielectric permeability . F
1 Qq r . 4 0 r 2 r
(4.2.5)
There is a certain correlation between the polarization and the surface-bounded charges’ density. To establish this, we must imagine that we have cut in a flat dielectric plate an elementary volume (e.g., in the form of a cylindrical rod) along the field force lines perpendicular to the dielectric’s surfaces (Figure 4.22). Assume the face area to be S, the cylinder length l and the bounded charges density . Therefore qS and the induced electric dipole moment p ql Sl. Then the polarization of the allocated dielectric rod shall be found relating the electric moment p of the rod to its volume V, i.e., p/V. As an elementary volume is V Sl (see Figure 4.22) and p Sl, then by definition (see (4.2.1))
ⴕ Sᐉ ⴕ. Sᐉ
So, the surface density of the bounded charges is numerically equal to the dielectric polarization. ⴕ.
(4.2.6)
As experiments show, the isotropic dielectric polarization is proportional to the electric field strength E and coincides with it directionally ( ~ E). In the SI, the relation between and E is written as 0 E,
(4.2.7)
where is dielectric susceptibility. Note that E in this expression is the strength of the average electric field (4.2.3).
E0 +'
–'
+∆q
–∆q
∆S
l
Figure 4.22 Relationship of the bounded charges’ field and a body polarization.
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Dielectric susceptibility characterizes the dielectric from the point of view of its ability to be polarized in the electric field. The dielectric susceptibility in the SI is numerically equal to dielectric polarization divided by 0 of the dielectric polarization in the field of the unit electrostatic field strength. Thus, the ability of a dielectric to be polarized is characterized by two values: dielectric permeability and dielectric susceptibility . There must be an interconnection between them. Indeed, the average field in a dielectric, according to (4.2.3), is E E0 E1. From the other side, E0 E and E1 /0. Then E E /0. Taking into account (4.2.6), we can write E E /0 E 0E/0, and therefore 1 .
(4.2.8)
The numerical values of dielectric permeability of various substances change over a wide range and depend on the frequency of the external electric field. We shall discuss this question in more detail below. For the majority of nonpolar liquids, dielectric permeability is 2–2.5; for polar liquids, it is significantly higher at 10–80. For the majority of solids, 1.5–2.5; however, for ferroelectrics (see Section 9.6), achieves a value of 104. For gases, differs a bit from unity. In order to characterize an electric field inward the dielectric, it is convenient to use one more value similar to electric field strength, namely a vector of an electric induction (or a vector of electric displacement) D. In isotropic dielectrics, D 0E. If does not depend on E, values D and E are proportional to each other. From the above-mentioned relationships, it follows: D 0 E . If E depends on whether dielectrics are present in the field and what their dielectric permeability is, D does not depend on these circumstances and does not change its value at the transition from vacuum to dielectrics (remember that this concerns isotropic dielectrics). If depends on E, as it sometimes takes place (e.g., in ferroelectrics, see Section 9.6), then D nonlinearly depends on E, remaining independent of the presence in space of other dielectrics. 4.2.3
Microscopic properties of dielectrics
Among the microscopic properties of a dielectric, the basic place is occupied by the molecular dipole moment p; this depends, in turn, on atomic structure and chemical bonding. In Figure 4.23 examples of molecules are given and both directions and values of the dipole moments (in D) are specified. Note that the concept of molecular dipole moments is one of the most important in chemistry. Some main principles connecting a molecule structure with its dipole moment can be noted. Firstly, the polarity of multiatomic molecules depends on the polarity of a given chemical bond and their mutual arrangement. On a simplified level, the electric dipole moment of a multiatomic molecule can be considered as a result of geometrical summation of the individual moments of each bond. Such an approach is based on the additive property of the total dipole moment: each chemical bond being considered as a dipole
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– p =1.84 D
H2O +
+ –
HCl
p=1.03 D + –
NH3
+
+
p=1. 48 D
+
p= 0
+ CO2
+
–
Figure 4.23 Molecular models and dielectric properties of some molecules.
moment, the total moment is obtained as a vector sum of the moments of a particular bond. The transferability of chemical bonds polarity is used here as well: it is accepted that the dipole moment is the specific property irrespective of in which particular molecule these bonds participate. At the same time, if a molecule possesses the center of symmetry (e.g., C6H6, CO2, etc.), its dipole moment a priori is equal to zero. This statement follows from the fact that the polar dipole moment vector is incompatible with the center of symmetry because any vector itself does not possess that center. Secondly, the molecular dipole moment strongly depends on the charge transfer from one atom to another. So, the diatomic molecules consisting of identical atoms, due to the symmetry of the electron pair arrangement, do not possess polarity at all; their electric dipole moment is equal to zero. The diatomic molecules consisting of different atoms are, in most cases, polar. The polarity of the bond is determined by the electron affinity of the constituent atoms. The greater the difference in the electron density, the more polar is the bond. The polarity reaches the highest value at a purely ionic bond. Thus, the electric dipole moment characterizes the degree of ionicity of a chemical bond. For example, the dipole moment of halogen–hydrogen bonds increases from HI to HF according to increase of electronegativity of the halogen atoms (see Table 4.1). Molecular dipole moments can be calculated by modern methods of quantum chemistry from the first principles (for small molecules). Experimentally dipole moments can be derived from X-ray diffraction experiments using a function of electron density distribution (r). Every dielectric at microlevel represents a discrete structure in which molecules or atoms are distributed in ordered (in crystals) or chaotic (in gases and liquids) manner. Therefore, the electric field strength at any point of the dielectric is the superposition of an external field and the fields of all neighboring molecules. The structure of an electric field in a dielectric is highly nonuniform; the value of an average field obtained above represents a highly averaged picture. It means that the electric field strength really acting on a given molecule is not equal to the averaged field: we should consider some effective field, referred to as a local field. The strength of the local field Eloc is defined as the geometrical sum of the strengths of an external field E0 and the total field of all the dielectrics’ dipoles except for the dipole being considered.
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Table 4.1 Electric dipole moments of some of halogen–hydrogen molecules Molecule
HF HCl HBr HI
Dipole moments 10−30 C m
D
6.37 4.33 2.63 1.30
1.91 1.03 0.79 0.39
If an isotropic nonpolar dielectric is placed in an external electric field, the local field will act on each molecule. Displacement of charges of different signs will take place and a dipole will be created. Such a dipole is called an induced dipole. Its value is proportional to the strength of the local field. In SI units, this dependence is p 0 Eloc .
(4.2.9)
The value characterizes the ability of a molecule to be polarized in an electric field, and is referred to as polarizability. Formula (4.2.9) permits us to rewrite the expression for dielectric polarization. Starting from (4.2.2), we arrive at np n 0 E loc .
(4.2.10)
We can compare the expression for polarization with that derived earlier (see eq. (4.2.7)) on the basis of macropresentations. Analysis shows that they are identical when E ⬇ Eloc, i.e., when the depolarizing field E1 is rather small. This corresponds in particular to a low molecule concentration. Then n.
(4.2.11)
Of course, this is valid for low dielectric polarization. The connection between and n at significant polarization is given in Section 4.2.7. 4.2.4
Three types of polarization mechanisms
There are several mechanisms of dielectric polarization, each being directly dependent on the molecules’ structure. Let us distinguish a deformation and an orientation polarization. In the first case the molecule deformation takes place when imposing it in the external field. Deformation can touch both the single atom and the whole molecule. We shall consider each of these mechanisms separately.
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Electronic polarization Relates to the deformation type of polarization and appears as a result of the displacement of an atomic electron shell relative to the nucleus which, accordingly, leads to an induced dipole moment. Certainly, such a polarization mechanism takes place in all systems containing atoms, i.e., in all dielectrics; however, in a pure state it can be seen in the case of nonpolar homo-atomic molecules and atoms. Let us estimate the value of electronic polarization. As an example, take a simplified model of an atom. Assume that the electron charge Ze is uniformly distributed around the nucleus in a sphere of radius R, i.e., with constant electron density Q/((4/3)R3)ZNeN/((4/3)R3). In the absence of an external electric field, the nucleus is in the center of a spherically symmetric electronic cloud, the centers of positive and negative charges coincide; the atom does not possess a dipole moment. We can impose an origin with negative charge center. Let us place an atom in an external electric field. Under the action of this field, the centers of charges of both signs will shift away from each other and an induced dipole moment will appear (Figure 4.24) p Z | e | l,
(4.2.12)
where Z|e| is a nuclear charge and that of the electron cloud as well and l is the displacement value of charges; the direction of dipole moment p coincides with the direction of external field Eloc. A nucleus displaced from its center will be under the action of two competitive forces: the action of a local electric field, Fⴕ Z e Eloc ,
(4.2.13)
and an internal atomic electric field, Fⴖ Z e
Z |e | l, rZ e 3 0 4 0 R 3
(4.2.14)
where l is the dipole arm; the latter force F tries to remove the nucleus in the initial central position. Because the forces are balanced, F F and ElocZ|e| Z |e|(Z|e|/40 R 3)l, wherefrom, after cancellation on Z |e|, it follows that Z⏐e⏐l 40 R3Eloc. The expression for
R
F" F' O l
+
E
Figure 4.24 A model of the electron polarization.
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the dipole moment is written on the left-hand side. Therefore, p 40 R3E loc. Comparing this expression with eq. (4.2.9), we can conclude that the electron polarization coefficient is el 4R 3 .
(4.2.15)
It can be seen that the atomic electron’s ability to be polarized is proportional to its volume. Although the result has been obtained at a certain level of simplification, it is qualitatively correct in describing a real picture of electron polarization. It can be seen from Table 4.2 that a cube of atomic radius increase and an electronic polarizability change in parallel. Let us estimate to an order of magnitude the value of electronic polarizability. We estimate 3 1010 m for atomic radius, then ⬃ 4R3 ⬃ 10 1030 m3 ⬃ 1029 m3 which coincides with the values of electronic polarizability of middle-sized atoms. The electron polarization is determined by the atomic electron shell. Therefore, it does not depend on temperature. Note that it consists of a shift of a light part of the atom regarding heavy nuclei and therefore possesses low persistence. Atomic (ionic) polarization Is observed in heteroatomic molecules in which atoms, due to different electronegativity, endure a redistribution of electron density. Therefore, each atom appears carrying an excessive (positive and/or negative) charge (Table 4.3). This charge is referred to as effective atomic charge and is defined in terms of an electron charge, i.e., q/|e|. Certainly, the sum of all over the whole molecule is equal to zero. Depending on the effective charges and interatomic distance, the hetero-atomic molecules exhibit a permanent dipole moment even in the absence of an external electric field Table 4.2 Electron polarizability of some atoms Atom
Polarizability coefficient, el (10−30 m3)
He Ne Ar Kr Xe
2.3 4.7 16 25 41
Cube of atomic radius, R3 (10−30 m3) 0.78 1.4 3.6 4.8 6.9
Table 4.3 The effective charges of atoms in selected molecules Substance
Atom
Substance
Atom
Na2O MgO Al2O3 P2O5 SO3 Cl2O7
O O O O O O
−0.81 −0.35 −0.31 −0.13 −0.06 −0.01
NaF NaBr NaCl MgO MgBr2 MgCl2
Na Na Na Mg Mg Mg
0.58 0.83 0.92 1.01 1.38 1.5
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+ +
− – l
–
−
+ F=qE +
F=− ∆l
E ∆l
Figure 4.25 Upper (without the external field) and lower (a molecule in the external field).
p |q|l. However, under the action of an electric field, there is additional displacement of the atoms relative to each other. Therefore, the interatomic distance is enlarged and the dipole moment increased. An additional (induced) electric dipole moment appears which causes the additional polarization. In order to estimate the atomic polarization, we can use a model: consider a molecule consisting of two atoms with effective charges and (Figure 4.25) bounded by a quasi-elastic force with rigidity coefficient . In the absence of the external electric field, its dipole moment is p |e|l (Figure 4.25, upper). Under the external field action, the interatomic distance l undergoes an enlargement l (Figure 4.25, lower). Two acting forces (|q|E and l) are balanced in absolute value: l |qef|Eloc. We can find the interatomic distance enlargement l (|qef|/)E. Due to the action of the local electric field, the molecules acquire an additional dipole moment p q l
qef2 Eloc .
(4.2.16)
Denote the polarization coefficient by at. Then the additional dipole moment can be rewritten as p at0E, wherefrom at p/0Eloc. Substituting ∆p according to (4.2.16), we arrive at
at
qef2 2 e 2 . 0 0
(4.2.17)
Evaluation of at, for a HCl molecule, for instance, gives: at 0.2|e|, 10 N/m, the coefficient of atomic polarization is at((0.2 1.6)2 1038)/(1011 102)⬇1029 m3. This is a bit lower than el for the same molecule. At moderately high temperatures when electronic density and, accordingly, interatomic forces can be considered as constants, the atomic polarization does not depend on temperature. The nuclear subsystem participates in atomic polarization and, therefore, atomic polarization possesses greater persistence than the electronic one. Orientation polarization Consider a polar dielectric, each molecule of which possesses an inherent, permanent dipole moment p. In the absence of an external field, dipoles are oriented chaotically due to the
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molecules’ thermal motion; this means that the dielectric’s polarization (4.2.1) is zero (Figure 4.21). In an external field, due to the action of force pairs (refer to Section 4.1.5 and formulas (4.1.32) and (4.1.33)), each dipole will acquire a tendency to be oriented by the field, but the thermal motion will prevent it. The general problem of the behavior of an ensemble of permanent dipoles (magnetic and electric) in an external permanent field was solved by the French scientist P. Langevin. Here, we shall take advantage of his unsophisticated dielectric model which arrived, nevertheless, at the correct result. Consider a molecular system consisting of the permanent moments p with concentration n. Accept a model in which dipoles are focused with equal probability along three coordinate axes, so 0. In the absence of an external field in any direction (positive or negative) of each coordinate axis, n/6 of all molecules will be directed. Assume now that operating on each molecule is a field whose direction coincides with the positive direction of the axis y (Figure 4.26). This means that each molecule will obtain a potential energy U pE pE cos , where is the angle between vectors p and E. For molecules whose dipole moments are directed along the field ( 0, cos 1), U pE. For molecules whose dipole moments are directed oppositely ( , cos 1), U pE. Assume also that dipoles directed along axes x and z do not participate in polarization. The competition between ordering tendency of an external field (in our case, this action is described by the value of potential energy U) and disordering tendency caused by thermal chaotic molecule movements (with energy T ) is described by the Boltzmann factor (see Section 3.2.4). This competition results in the fact that the concentration of molecules with dipoles oriented along the field (n) and oppositely (n) will be different and equal n
n ⎛ pE ⎞ exp ⎜ loc ⎟ ⎝ T ⎠ 6
and n
n ⎛ pE ⎞ exp ⎜ loc ⎟ . ⎝ T ⎠ 6
Since the concentration of molecules whose dipoles are oriented in a positive and negative y-axis direction will be different, the resulting dipole moment appears in a positive direction. According to (4.2.2), it is equal, y (n n)p. We will restrict our consideration to weak fields for which pEloc T (the potential energy of dipole interaction with an external field U is significantly less than the energy
z
E
n–p x
n+p
y
∆np
Figure 4.26 To the model of a orientational polarization.
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of the molecules’ thermal motion T). In this case, an exponent function can be decomposed in a series: we will limit ourselves to only the first two terms: n⬇ (n/6)(1(pEloc/T)) and n⬇ (n/6)(1(pEloc/T)). Then an excess of dipole concentration directed along the electric field will be n nn (n/3)(pEloc/T)) and the polarization will be
n pEloc np2 np2 p Eloc 0 Eloc . 3 0T 3 T 3T
(4.2.18)
The averaged effective dipole moment counting on a single molecule can be obtained by dividing by n: ef
p2 Eloc . n 3T
The orientation polarization can be characterized by the coefficient of orientation polarization op. Then p op0Eloc, wherefrom
op
p2 . 3 0T
(4.2.19)
Evaluation of or for an HCl molecule gives p 1 D 3.33 1030 C m and T 300 K
or
((1 3) 1029 )2 1028 m 3 , (3 1011 ) 1023 300
which is higher than el. The value of or is inversely proportional to absolute temperature and is proportional to the square of the molecular dipole moments. In extreme conditions (in very high fields or at very low temperatures when pE T— something that is technically very difficult to achieve), all dipoles are built along an external field; further increase does not essentially change polarization, the polarization reaches saturation. The polarization value in this condition depends only on dipole moment value p and concentration n. In Figure 4.27 a graph of polarization versus external electric field strength is given. At pET this function is linear (eq. (4.2.18)), the higher the molecular dipole moment and the lower the temperature, the steeper is the linear function, i.e., tan d/dE. In the other limiting case (pE
T), polarization is constant ( constant, saturation state). A smooth curve connects these two extremes. It is obvious that in polar dielectrics all three mechanisms of polarization (two of deformation and one of orientation) are exhibiting simultaneously. The total polarization of isotropic polar dielectrics el, atomic at and orientation or is a sum: el at or. The total polarization can also be given as the sum (el at or )n 0 E loc ,
(4.2.20)
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4. Dielectric Properties of Substances np
tg =
p2 30κT E
Figure 4.27 A relationship of a dielectric polarization and the external electric field strength E at orientational polarization.
a molecule can be characterized by a total averaged polarization coefficient el at or .
(4.2.21)
In different molecules each term can exhibit differently depending on structure and electron density distribution as well as syntheses, temperature, experimental methods, etc.
EXAMPLE E4.10 An electric dipole with a moment p 2 nC m is in a uniform electric field E 30 kV/m; vector p is oriented at 60° to E. Determine the work A of external forces to rotate the dipole at an angle 30°. Solution: From the initial position the dipole moment can be turned on /6 30° in two directions: (1) clockwise at an angle 1 0 /3 /6 /6 or (2) anticlockwise at 2 0 /3 /6 /2. In the first case, the rotation occurs under the action of inner forces, therefore the work is negative; in the second case, only external forces can do the turn and correspondingly the work is positive. The work can be determined by two methods: (1) direct integration and (2) using the relation between work and potential energy change. The first method comes to the integration dA M d pE sin d in the limits from 0 to
0
0
A ∫ pE sin d pE ∫ sin d; executing the calculation we obtain A pE(cos 0 cos ). Accordingly, the clockwise rotation gives A pE(cos 0 cos 1) 21.9 J and in anticlockwise rotation, A pE(cos 0 cos 2) 30 J. We think that the second method is preferable. In fact, A U2 U1 (see Section 1.4.5) and A pE(cos 0 cos ). This coincides with the previous expression.
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EXAMPLE E4.11 In an iodine atom at a distance r =1 nm from an -particle, an induction dipole moment p 1.5 1032 C m appears. Find the polarization coefficient of the iodine atom. Solution: The polarization coefficient can be found according to eq. (4.2.9), p/0Eloc, where Eloc is the field in which a given atom occurs. In this case, the field created by an -particle is a local field. Therefore, it is equal to Eloc E 2冟e冟/40r2. Combining the last two expressions, we arrive at 2r2p/ 冟e冟. Executing the calculation, we obtain 5.9 1030 m3.
EXAMPLE E4.12 Krypton is under a pressure p 10 MPa at a temperature T 200 K. Find (1) the dielectric permittivity of krypton and (2) its polarization in an external field E0 1 MV/m. Krypton polarizability is 4.5 1029 m3. Solution: (1) Expression (4.2.26) is suitable for solving this problem (1)/ (2) n/3, where n is krypton concentration. Find the dielectric polarizability from this equation: (1(2/3)n)/(1(1/3)n). The concentration is equal to n p/T; therefore, (3T2p)/(3Tp). Substituting all data in the expression, we arrive at 1.17. (2) In the uniform electric field krypton polarizability can be given by eqs. (4.2.2) and (4.2.10) np and p 0nEloc. To rewrite the local field via the external field for nonpolar substances (Eloc((2)/3)E0) and np/T 3.61027 m3, we obtain p 0 ((2)/3)nE0. Using all data obtained we arrive at p 1.30 106 C/m2. 4.2.5 Dependence of the polarization on an alternative electric field frequency Polarization is a measure of a dielectric’s “response” to the action of an external electric field. In a static electric field all the molecular dipoles align along the external field (we will not take the thermal motion into account at the moment). To measure total polarization in the static field is easy: having placed a dielectric sample in a condenser, its capacitance depends on the dielectric permeability of the material between the plates (C C0, C0 being the capacitance of an “empty” condenser, remember the ratio between susceptibility and permittivity ). In an alternating electric field the molecular dipoles should have time to turn as a whole when the electric field changes its direction. While field frequency is not so high, the dipoles have time to reorient, following the change of field direction. In a field of high frequencies when ⬃ 106–107 Hz due to the molecules’ inertia, they begin to delay, being unable to follow the electric field reorientation. The higher the frequency, the greater is the delay. At very high frequencies the molecules will not be able to reorient and the polarization comes to naught.
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We can conclude that in static electric field, all molecules regardless of their origin participate in overall polarization; therefore, such measurements give the sum of three polarization mechanisms (refer to (4.2.21) and Figure 4.28 in which a range of frequencies is given). When the electric field frequency reaches a value of 1011 Hz, the molecules are unable to turn over because of their great inertia; therefore, the orientation polarization is switched out. Atomic polarization does not need the molecules’ reorientation; in a high frequency field, the induced oscillations of atoms along the field direction take place. This will be preserved up to a frequency of approximately 1012 Hz. At this value, atomic oscillations also vanish and only electron oscillations (the lightest part of atoms) remain. At ⬃ 1013–1015 Hz, only electrons can accomplish their oscillations. This frequency corresponds to oscillations of light vectors, magnetic and, mainly, electric. At such frequencies, only the electron part remains; at higher frequencies, even the electron part disappears. The frequency dependence explains why the dielectric permeability of water measured in a static electric field is 81, but at optic frequencies is only 1.77: in the first case all polarization mechanisms are participating in the polarization but in the latter case only electron polarization takes place. Otherwise the polarization is exhibited in solids. The ability of a molecule to change orientation or oscillate depends essentially on its geometrical form and interaction forces with its neighborhood. If the form of the molecule is close to spherical and its electric moment is not high, it can rather easily change its orientation (e.g., molecular group CH4). Molecules HCl and H2O are less symmetric; in solids they have some steady orientations and rather slowly pass from one to another. The average time of such a transition is referred to as relaxation time. The value reciprocal to the relaxation time is referred to as relaxation frequency. When the frequency of an external alternative electric field is higher than the relaxation frequency, the system will no longer react to the action of such a field. The relaxation time depends on temperature and the aggregate substance state. So for water (t 200 °C), 3 1011c; however, for ice (t 20 °C), 103c. 4.2.6
A local electric field in dielectrics. Lorentz field
A local or microscopic electric field is a field acting on the given dipole in a dielectric. The strength of a local field is the geometrical sum of external field strength E0 and the or at el 109
1010
1011
1012
1013
1014
1015
1016
Figure 4.28 Dielectric polarization in the alternating field of different frequencies.
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complex field ∑Ei created by all dipoles, except that on which this field operates (Eloc E0 ∑Ei). This field strongly changes within the limits of intermolecular distances and, owing to thermal molecular motion, changes as well in time. In fact, it is impossible to calculate Eloc according to the formula presented because of the enormous number of dipoles. However, if the dielectric is polarized homogeneously, i.e., the polarization at any dielectric point is the same in magnitude and direction, it is possible, at a certain approximation, to calculate its value. Without carrying out a detailed calculation (because of its complexity), we shall, nevertheless, note the general remarks that allowed H.A. Lorentz to solve this problem. His method is as follows. Remember that the further the dipole from the point of consideration, the less is its contribution to the local field (the dipole field falls down as E ~ 1/r3, eq. (4.1.31)) but the larger their number. This allows him to replace summation upon individual dipoles onto integration, i.e., calculate a local field macroscopically. Lorentz suggested allocating in a homogeneously polarized dielectric a sphere of rather small radius, inside which there exist a large but, nevertheless, limited number of dipoles. The center of the Lorentz sphere should coincide with the point of observation (Figure 4.29). In an external field, polarization will take place, including the allocated sphere. We shall mentally remove the originally allocated sphere from the dielectric body. There appears a spherical cavity with charges distributed on its internal surface. The strength of the local field can now be given by four terms Eloc E0 E1 E2 E3, where E0 is the external field, E1 is the depolarizing field created by bounded charges distributed on an external dielectric surfaces, E2 is the field created by the bounded charges on the internal surface of the spherical cavity (Lorentz field) and E3 represents the field of the nearest neighbors; in isotropic dielectrics (in gases, liquids and isotropic crystals), this field is equal to zero. The values E0 and E1 have already been considered above, and calculation of the strength of the field E2 is the subject of our consideration. Integration over the internal cavity surface gives a field E2 which is referred to as a Lorentz field:
E2
1 . 3 0
(4.2.22)
Correspondingly, the local field strength can be written as: E loc E
–
. 3 0
(4.2.23)
+
0
E0
Figure 4.29 A dielectric’s body polarization.
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In scalar form, it can be given as Eloc E
1 . 3 0
(4.2.24)
This field should be presented in all expressions in which a local field appears. 4.2.7
Clausius–Mossoti formula
For rarefied dielectrics, eq. (4.2.11) connecting the macroscopic characteristic with the microscopic one , which in turn provides access to the analysis of the molecule properties, was given. In more complex cases of the more condensed matter at n noticeably larger than unity, the given simple ratio is not fair. In order to find the proper ratio in more dense substances we should substitute in the expression (4.2.10) the local field Eloc by eq. (4.2.24): ⎛ 1 ⎞ n n 0 Eo n 0 ⎜ E n 0 E . ⎟ 3 0 ⎠ 3 ⎝ Solving this equation relative to , we obtain
n 1 ( n 3)
0 E.
If we accept n 1, it goes to (4.2.10), i.e., n. However, if this is not the case, then we have to compare it with eq. (4.2.7). The comparison gives
n 3n , 1−(n 3) 3n
or n . 3 3
(4.2.25)
Substituting here (4.2.8) accordingly, we arrive at 1 1 n . 3 2
(4.2.26)
This is one of the forms of the Clausius–Mossoti law. It connects the macroscopic value of the susceptibility with polarizability of molecules.
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One can express a molecule concentration n in eq. (4.2.26) via the Avogadro constants NA and a molar volume M/: n NA/(M/) (/M)NA. Then the Clausius–Mossoti formula can be written as M 1 1 NA . 2 3
(4.2.27)
Clausius and Mossoti independently obtained eq. (4.2.27) for nonpolar dielectrics in the middle of the 19th century. In 1912, using Langevin theory for magnetization of paramagnetic substances, Debye obtained for polar dielectrics a connection between polarization and electric dipole moments. He suggested a concept of orientation polarizability of polar molecules and has generalized the Clausius–Mossoti equation for the case of polar dielectrics. This Debye–Langevin formula is usually written as M 1 1 ⎛ p2 ⎞ NA, ⎜ def 2 3 ⎝ 3 0T ⎟⎠
(4.2.28)
where def is the deformational polarization (see Section 4.2.4) or the polarization of elastic displacement, which is the sum of electronic el and atomic at polarizations. Taking into account all three types of polarization mechanisms, the Debye–Langevin formula can be written as: p2 ⎞ M 1 1 ⎛ ⎜ el at NA . 2 3 ⎝ 3 0 KT ⎟⎠
(4.2.29)
The Debye–Langevin formula is applicable to polar dielectrics at definite restrictions. It achieves good fulfillment for gases and vapors at low pressure, and for highly dissolved solutions of polar liquids in nonpolar solvents. This formula is of great importance in the interpretation of molecular structures. Being written as M 1 1 N , 2 3 ( el at ) A it successfully describes nonpolar gases at low and average pressures (500 kPa and lower), can be applied approximately for nonpolar gases at elevated pressures (above 500 kPa) and nonpolar liquids, and is good enough for crystals with face-centered lattice if atoms possess only electronic polarization and approximate for ionic crystals with a cubic lattice (see Section 9.6). As was already mentioned in Section 4.2.5, the polarizability of molecules depends on the frequency of the alternative electric field, especially at high frequencies. In the Maxwell electromagnetic theory, the ratio between a refraction index n and the dielectric permeability of substances is given. For low-magnetic substances, n 兹苶 . If in eq. (4.2.27) we substitute by n2 and take into account that at optical frequencies
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( ⬃ 10151016 Hz) all polarizability mechanisms, except the electronic one, are suppressed, it is possible to write M n 2 1 1 el N A . n2 2 3
(4.2.30)
This is the H.A. Lorentz–L. Lorentz formula. The right-hand side of this equation is a molar refraction R. It is fair for gases, nonpolar liquids and isotropic crystals (with cubic lattices). It is approximately applicable to nonpolar liquids at relatively high frequencies when the orientation polarization is not exhibiting.
EXAMPLE E4.13 The density of liquid benzene is 899 kg/m3 and its refraction index n 1.50. Determine the benzene electron polarizability el. Solution: The Lorentz–Lorentz formula (4.2.30) can serve us in solving this problem. Using it, we can solve it relative to el
el
3 M (n2 1) . N A (n2 2)
In this equation all the entries are known except for mole mass M. Since the atomic composition of benzene is C6H6, the benzene relative mole mass is 78. Therefore, the molar mass is M 78 10−3 kg/mol. Substituting all the values, we arrive at el 1.27 10−28 m3. 4.2.8 An experimental determination of the polarization and molecular electric dipole moments Experimental determination of microscopic characteristics is based on the Debye–Langevin (eq. (4.2.28)) and Lorentz–Lorentz (eq. (4.2.30)) equations. They connect macroscopic molecular characteristics, measured directly in physical experiment, with microscopic ones. Measuring dielectric permeability in an electrostatic field, the molar polarization can be found: 1 (el at or )N A , 3 from which the total polarizability of a molecule can be obtained 3
NA
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or 3
M 1 . N A 2
(4.2.31)
Measurements of a refraction index n in an optical range of frequencies allows us to find the molar refraction R, and consequently, the molecule electron polarizability using eq. (4.2.30) (R elNA/3) or
el 3
M n 2 1 . N A n2 2
(4.2.32)
The difference between and el gives the sum: el or el 3
R . NA
(4.2.33)
There are two ways of separating atomic and orientation polarizabilities: either taking advantage of the frequency dependence of the polarizability or using the temperature dependence of the molar polarization (or ~ 1/T ). In the latter case the temperature dependence of molar polarization is removed and a graph (1/T ) (Figure 4.30) is drawn. The segment under the horizontal dashed line gives that part of the molar polarization which does not depend on temperature 1 el at (el at ) N A . 3
(4.2.34)
The value el can be determined by refractometric experiments (4.2.33) and at can be found according to formula (4.2.34) at 3
el at el . NA
(4.2.35)
∆=or def =def +
p2 30
.
1 T 1/T
1/T
Figure 4.30 A relationship of a molar dielectric polarization versus reciprocal temperature.
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The orientation polarization op, corresponding to a given temperature, can also be found from the graph: the angle coefficient tan is determined by op. Indeed, tan
1 or or N A T , (1T ) 1T 3
from which the orientation polarizability can be found or
3 tan . N AT
(4.2.36)
The tan value is determined by expression / (1/T) m3 C/mol. Therefore, the slope has the same dimension. Note that op decreases with decrease in temperature. Knowing or, we can determine the molecular dipole moment. According to eq. (4.2.19), p230Tor; substituting further op according to eq. (4.2.36), we can obtain p2(9/NA)0 tan , from which ⎛ ⎞ p 3 ⎜ 0 tan ⎟ ⎝ NA ⎠
1 2
(4.2.37)
.
We would like to underline once more that dielcometry is a relatively simple though rather powerful, method of chemical structure investigation.
EXAMPLE E4.14. Thin semi-infinite rod is charged uniformly with linear density 1 C/m. At a distance r0 20 cm from the end of the rod perpendicular to it a point O is located (Figure E4.15). Calculate the electric field strength created by the charged rod in point O. y K
dE
dEy
N dl M
O dEx
d r0 K N dl M
L
x
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Solution: Allocate a piece of the rod with a charge dQ dl in an arbitrary point (1dl / 40) of the rod. This charge create a field in the point OdE , r being the r2 distance MO from the elementary charge to the point mentioned. Denoting angles MOL and, consequently MON, as we have r r0 / cos and dl r d / cos 2 . Substitute these equations into formula for E, we can obtain 冷dE冨 d . 40r0 Decompose the vector 冷dE冨 into two components: dEy dE cos and dEx dE sin . Proceeding further, we can find dEy (cos / 40r0 )d and dEx (sin / 40r0 )d . Integration over the limits 1 0 and 2 /2 (both ends of the semi-infinite rod) gives 2
Ey
∫ 0
cos and E x d 4 0 r0 4 0 r0
2
∫ 0
sin . d 4 0 r0 4 0 r0
兹苶2 After summing these two vector components, we arrive at the final value E . 40r0 1069109兹苶 2 Substituting the numerical values we obtain E 7.05 kV/m 20102 1 (keeping in mind that numerically 9109 ). 4πε0
PROBLEMS/TASKS 4.1. There are two similar small balls of mass 1g each. Find the electric charge q which should be given to the balls in order to compensate for the force of their mutual Newtonian attraction. 4.2. In the semiclassical theory of the hydrogen atom, an electron is supposed to travel around a proton along a circular orbit. Determine its speed if the orbit radius is r 53 pm and the frequency of the electron’s revolution. 4.3. In vertexes of an ideal hexagon with a side length a 10 cm, charges Q, 2Q, 3Q, 4Q, 5Q and 6Q (Q 0.1C) are situated. Find the force F, which acts on a point charge Q placed in the center of the hexagon. 4.4. A thin rod of l =10 cm in length is uniformly charged with =103 nC/m. On an extension of it at a 20cm from its end, a point charge Q 100 nC is placed. Determine the interaction force between the rod and the charge. 4.5. A thin ring of R 10 cm radius carries a uniformly distributed charge Q 102 nC. On a plane perpendicular to the ring, just over its center, is a point charge Q1 =10 nC at heights of (1) l1 20 m and (2) l2 2m. Determine the interaction force of the ring and the point charge.
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4.6. There are two coaxial cylinders made of thin metal foil of radius R and 2R. They carry uniformly distributed charges with density and − ( 60 nC/m2). (1) Using the Gauss theorem find the electric field distribution inside both cylinders and between them (areas I and II) and outside them (area III). (2) Calculate the field strength at the point E(3R). Draw a graph E(r) in all areas and calculate the characteristic (border) values. 4.7. There are two concentric sphere made of thin metal foil of radius R and 2R. They carry uniformly distributed charges −2 and ( 0.1C/m2). Using the Gauss theorem find the electric field distribution inside both cylinders (areas I and II) and outside them (area III). Calculate the field strength at the point E(3R). Draw a graph E(r) in all areas and calculate the characteristic values. 4.8. In an area extending between two half-circled rings of radii R and 2R (R 10 cm), a charge Q 20nC is uniformly distributed. Find in the central point O (the center of the rings) the field E(O). 4.9. An infinite thin wire is charged with a linear density 0.2 µC/m. The wire is bent at right angle. Determine point A is denoted on a continuation of one of the wire side at a distance ro15 cm from the corner. Determine at this point the field E. 4.10. An electric dipole p 0.4 C m is in a uniform electric field of strength E 25 kV/m at an angle 1 /6. Find the work of external forces at dipole reorientation (2 7/6). 4.11. An electric dipole p 200 nC m is in a uniform electric field of strength E 50 kV/m at an angle /3. Find the change of its potential energy U at its rotation anticlockwise at the angle 2π/3. 4.12. An electric dipole p 0.2 C m is in a nonuniform electric field with dE/dx −10 kV/m2 oriented against the electric field E and electric field gradient. Find the force direction and calculate its Fx value. 4.13. Two HCl molecules with the same orientation of their electric dipole moments p 1.91 D in value are at a distance r 5 nm from each other. Considering the molecules as point ones, determine the potential U energy of their interaction. 4.14. Two polar molecules SO2 (p 1.60 D) are at a distance r 8 nm from each other. Considering them as a point dipole, determine the force of their interaction. 4.15. Argon is under normal conditions in an electric filed E 30 kV/m. Determine the shoulder l of the induced dipole moment of an Ar atom if the dielectric permeability at the same state is 1.000554. 4.16. The dielectric susceptibility of the gas Ar under normal conditions is 5.54 10−4. Find the dielectric permeabilities 1 and 2 of liquid (1 1.40 g/cm3) and solid (2 1.65 g/cm3) argon. 4.17. What minimum velocity min should a proton possess in order to reach the surface of a metallic sphere charged up to 400 V? 4.18. From point 1 on the surface of an infinite negatively charged cylinder ( 20 nC/m) of radius R, an electron starts with zero velocity. Find the electron kinetic energy K at point 2 which is at a distance of 9R from the cylinder surface. 4.19. Knowing the electric dipole moment p1 of a chlorobenzene (phenyl chloride) molecule (C6H5Cl) (p1 1.59 D), find the dipole moment of ortho-dichlorobenzene (orthodi-phenyl chloride) p2.
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4.20. A xenon atom (the polarizability of which is 5.2 10−29 m3) is at a distance r 1 nm from a proton. Determine the xenon atom’s induced electric moment p. ANSWERS 4.1. Q 2m兹 苶苶 0G 苶 86.7 1015C (G is a gravitational constant). 4.2.
e 4 0 mr
2.19 106 m/sec; n (2r) 6.58 1015sec1 .
4.3. F
6Q 2 54 mN. 4 0 a 2
4.4. F
Qᐉ 1.5 mN. 4 0 (ᐉ a )a
4.5. (1) F1
QQ1ᐉ 1 4 0 ( R 2 ᐉ21 )3 2
15.7 kN; (2) F2
EA
2.26 KVm. 3 0
4.7. (1) E I 0(r R); EII (r )
(2) EA
2.25 N.
R R ( R r 2 R); EIII (r ) (r 2 R). 0r 0r
4.6. (1) E I 0 (r R); EII (r )
(2)
QQ1 4 0 ᐉ22
2R 2 2R 2 ( R r 2 R ); E ( r ) (r 2 R). III 0r 2 0r 2
2 2.51 kVm. 9 0
4.8. E
Q ln2 5.29 kVm. 3 2 0 R 2
4.9. E
5 26.8 kVm. 4 0 r0
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4.10. A12 2pE cos 17.3 mJ. 4.11. U pE[cos − cos( )] 15mJ. 4.12. Fx p(dE/dx)cos 2 mN. 4.13. U 4.14. F
p2 5.82 1024 J ( 36.4 eV). 2 0 r 3
3 p2 3.74 1016 N. 2 0 r 4
4.15. ᐉ
( 1) 0 E 1.9 1018 m. nnorm e Z
4.16.
3 M 2Vom ; 1 1.51; 2 1.61. 3 M Vom
4.17. min 4.18. K
3e
e 2 o
2m
0.24 106 msec.
ln10 828 eV.
4.19. p2 2 p1 cos . 6 4.20. p 6.6 10−31 C/m.
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–5– Magnetics
Before describing the magnetic properties of a chemical substance we will briefly discuss some of the properties of a magnetic field itself. Whereas the sources of an electrostatic field are motionless electric charges, the sources of a magnetic field are moving charges, i.e., an electric current. Let us consider some characteristics of a permanent electric current and the conditions of its maintenance.
5.1 5.1.1
GENERAL CHARACTERISTICS OF THE MAGNETIC FIELD
A permanent (direct) electric current
An electric current can run in a substance in which free charges (current carriers) are present. They can be either electrons (in metals, for example) or ions (in liquids or solid electrolytes). Such substances are referred to as the conductors of an electric current. However, if a conductor is brought into an external electrostatic field there will be only an instant displacement of charges according to electrostatic laws; this will lead to the creation of an internal electrostatic field in a conductor directed opposite to the external electrostatic field and numerically equal to it; the current instantly stops. Therefore, inside a conductor the electric field is always zero. This means that additional conditions to support the current flow are necessary. We shall consider these a little later, but first we shall introduce some ideas about the electric current. The ordered motion of electric charges is referred to as an electric current (a current of conductivity). The scalar value I determined by a total charge dQ having run through a cross-section of the conductor in a unit time interval dt, i.e., I
dQ , dt
(5.1.1)
is referred to as a current strength or simply current. The motion of positive charges is accepted as the current direction, the current flow generally being opposite to the electron movement. If the current magnitude does not change with time it is referred to as a permanent (constant, direct) current. If the current strength changes with time it is referred to as an alternating current. 305
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The current I is a macroscopic characteristic of a particular conductor. For the distribution of an electric current throughout the conductor section a vector of current density j is introduced. The vector of current density j is directed along the carrier’s motion and is numerically equal to the electric charge current, which in the unit time crosses a unit conductor area perpendicular to the carrier velocity (see Figures 2.19 and 2.20). Thus, the modulus of a j-vector is equal to the ratio of the current strength dI through an elementary area dS located in a given conductor point perpendicular to the direction of the ordered carrier motion: j dI/dS⊥. To attach to the current density a vector character, we can multiply j to the unit vector of a direction of the carrier motion /; therefore, .
(5.1.2)
I ∫ j dS,
(5.1.3)
j
dI dS⬜
The integral current strength is therefore
S
where S is the conductor cross-section. Let us establish correlation between the microscopic current characteristics, the density of current j, the concentration of current carriers n and the average velocity of carrier motion (current speed of drifts). Let charge Q be transferred in time t through the cross-section S, perpendicular to the ordered carrier movement (Figure 5.1). By definition, the density of a current j is numerically equal to the ratio I/S. The charge Q is equal to the product of a single carrier charge and their full number in the volume V S⬜ l. Therefore j q nlS⬜/tS⬜ q n. Because current density vector j and the velocity of positive curriers are codirectional, hence j qn冬 冭.
(5.1.4)
Taking into account the possible movement of both positive and negative charges (for instance in an electrolyte) the total current density is j qn 冬 冭 qn 冬 冭.
q +
S⊥
j ∆l = t
Figure 5.1 Relationship of a current density j, carrier concentration n and drift velocity .
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Now derive Ohm’s law in differential form. At a fragment of a uniform conductor in an electric circuit, the current I is proportional to voltage loss U and inversely proportional to its resistance R: I U/R. Consider an elementary cylindrical volume in the vicinity of an arbitrary point inside the conductor (Figure 5.2) with the cylinder generatrix parallel to current density vector j. A current jdS flows through the cross-section of the cylinder. The voltage loss U on the cylinder ends is equal to Edl, where E is the strength of the electric field in the vicinity of the given point. This means that dI jdS
Edl Edl dS, R dl
where is the specific resistance and R(dl/dS). Since the charge carriers at every point move parallel to vector E, the Ohm’s law in differential form acquires the form j E/ E, where 1/ is the specific electroconductivity. Therefore, the current density at any current point j(r) is equal to the product of specific electroconductivity and the electric field strength E(r): j(r ) E(r ).
(5.1.5)
This is Ohm’s law in differential form. Another characteristic of a current is the mobility of the current carrier b, which is the average speed acquired by the current carrier in a field of unit electric field strength. If charges have average speed in a field E then, by definition, their mobility is b /E. The mobility b can be expressed through the specific electroconductivity and carrier concentration n. As the current density is j nq and j E, therefore, E nq. Having divided both parts of equality by E we shall obtain: nqb or b
. nq
(5.1.6)
Let us now consider the conditions that can maintain the electric current in a closed electric circuit. In fact, the current in a circuit can exist only if external forces maintain at the conductor ends a constant voltage difference (to say nothing about superconductivity). Therefore, in the closed circuit, along with the parts in which positive charges move along the decreasing potential , there should be parts where the positive charges move against
j dS E dl
Figure 5.2 Derivation of the Ohm law in differential form.
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the potential loss, i.e., against the forces of an electrostatic field. Figure 5.3 shows a closed circuit with a part where the electrostatic field is acting (1-a-2) and a part (2-b-1), where the so-called extraneous (outside) forces operate. The extraneous forces are of a different, nonelectrostatic nature; they can be of chemical (galvanic cells), induction (electrogenerators), or of thermal origin, etc. These extraneous forces are capable of maintaining the ordered movement of carriers against Coulomb forces. There exists a field of extraneous forces, characterized by the extraneous force Fext having field strength Eext. By analogy with eq. (4.1.2) we can write E ext
Fext . Q
(5.1.7)
The extraneous forces produce the work on charges moving along a circuit. The physical value equal to the work of the extraneous forces produced in moving a positive unit charge Q from point 1 to point 2 is referred to as electromotive force (EMF) Thus, 2
12
Aext ∫ E ext dl. Q 1
Thus, for the closed circuit A Q ∫ E,ext dl. Therefore, the work of the EMF along the closed circuit is Q养El dl. Both extraneous and Coulomb forces act on the charge moving along the closed circuit. The work produced is AQ养E dl, where E symbolizes the sum of the extraneous Eext and the Coulomb Ecol field strengths, i.e., AQ养(EextEcou)dl. Since 养Ecoudl0, hence AQ养Eext dlAext. Therefore A/Q, i.e., the work on the unit positive charge along the closed circuit, i.e., the EMF force is ∫ E dl,
dA = Eextra⋅dl 2
b
1 a
dA = Ecol⋅dl
Figure 5.3 A closed circuit with extraneous forces.
(5.1.8)
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where E is the combined strength of both fields. If we consider the charge motion in a limit 2 2 from point 1 to point 2, then A12Q1 EextdlQ1Ecoudl, and consequently A12 Q12 Q(1 2 ).
(5.1.9)
The value U12 is numerically equal to the work of both fields, electrostatic and extraneous, at the displacement of a unit positive charge from point 1 to point 2, referred to as voltage loss (voltage) at a given section of circuit U12 IR12 12 (1 2).
(5.1.10)
If the circuit is open ended (I 0, U12 0) then 12 (2 1), i.e., the EMF is equal to the potential difference on the clumps of a current source. 5.1.2 A magnetic field induction The electric current (moving charges) creates a magnetic field in the surrounding space. This field affects the charges (currents) moving in it. Thus, the interaction of two currents has an electromagnetic character. Ampere’s experiments showed that two parallel infinite currents I1 and I2 running in one direction attract each other, whereas currents directed oppositely repelled (Figure 5.4). The interaction force falling at a unit conductor length, f, is inversely proportional to distance b between them, i.e., f⬃(I1I2/b). The given statement is the essence of Ampere’s law. In SI this law takes the form
f
0 2 I1 I 2 , 4 b
(5.1.11)
where 0 is the permeability constant (refer to Appendix 1).
F
F
I1
F I2
(a)
F I1
I2
(b)
Figure 5.4 Ampère’s law: interactions of currents, (a) currents are codirectional and (b) antidirectional.
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It can be imagined that one conductor creates the magnetic field and the other in this field acquires its action. Consider first how an electric current creates a magnetic field and then how the magnetic field acts on the conductor with a current. When we analyzed an electrostatic field it was convenient to use a unit point positive charge (probe charge), which we imagined to be placed in each point of the field and measured the force acting on it; this was the electric field strength E. It is impossible to apply such a procedure in the case of a magnetic field because there are no magnetic charges (monopoles) in nature. It is necessary to use a multipole of the next order, i.e., a dipole, placing it at different points in the magnetic field and measuring the torque acting on it (refer to Appendix 3). In order to carry out such a magnetic field analysis, we need to use a model. Such a model is a magnetic dipole moment, which is a small (ideally, a point) flat contour with a current (a probe contour). Orientation of the probe contour in space is determined by a vector n normal to the contour (Figure 5.5). Placing the probe contour into every point one can see that in a given point it accepts a definite position: the normal vector n is oriented in a strictly determined direction. Experience also shows that the action of the magnetic field on the contour is associated with the value IS (where I is the current flowing around an area S). This contour M IS is called the magnetic moment of a contour. The magnetic moment of a contour is the vector value directionally coinciding with the normal n to the contour and determined by the contour current by a right-hand system rule, so MMn ISn.
(5.1.12)
Accept the direction of the probe contour normal n freely oriented in a given point of the magnetic field as a direction of the force line. The force characteristic of the magnetic field is a vector B in the direction of the normal vector n; its absolute value is determined by the maximum torque acting on a contour in this point. Then,
B
M max . M
(5.1.13)
Vector B Bn is referred to as a magnetic field induction vector.
n M
I
Figure 5.5 An elementary magnetic moment.
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The magnetic field is schematically presented in full analogy to the electric field: the B vectors at any point are parallel to the tangent to force lines and their density is proportional to the 兩B兩 value. For the description of the magnetic field we require one more characteristic—field strength H. In vacuum, the two characteristics differ only by a constant, namely, B 0H; however inside of a body magnetic characteristics mentioned differ noticeably (refer to Section 5.4). Remember that the two characteristics appear also in the description of the electric field properties (see Section 4.2.2). The magnetic field obeys the general principle of superposition (see Section 2.9.1). With reference to the case given it can be formulated as follows: the magnetic field created at a given point of space by any current does not depend on whether there are other sources of a field (other currents) in this space or not. Owing to the vector character of the magnetic field, the total induction of a system of currents is equal to the vector sum of particular field inductions, which are created by each current separately: N
B∑B , i
(5.1.14)
i1
and when current is distributed continuously B ∫ d B,
(5.1.15)
L
where dB is the induction created by the elementary current Idl. An empirical method of calculating the induction of a magnetic field at some point in space if the distribution of currents is known was suggested by Biot and Savart; a corresponding law relates a current element Idl (I is a scalar current running in the conductor element dl) to an induction dB at a point A, the latter being assigned by the radius vector r drawn from the element dl to point A (Figure 5.6): dB
0 I [dl r ] . 4 r 3
A multiplier 0/4 entered into the law formula in the SI.
I
dl r
A + dB
Figure 5.6 Biot–Savart law.
(5.1.16)
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We can illustrate the application of the Biot–Savart law with two important examples. Calculate a field that creates a circular current (Figure 5.7) in its center point O. The magnetic induction dB at the point O is created by an element of current Idl accordingly. The vector product in (5.1.16) assign elementary vector dB: it is directed perpendicular to a plane of a drawing and is related to the current element by the right-handed system rule. The vector dB is directed along the same axis (a symmetry axis of a ring) independently of where on the circuit the current element is chosen; therefore, integration (5.1.15) can be executed in a scalar form. Proceeding from the general formula (5.1.16), keeping in mind that the angle between vectors dl and r remains /2, we arrive at dB( 0/4 )(Idl/R2), where r R is the ring radius. The circular current integration gives B 0 ∫ dB L
0 I 4 R
2 R
∫
dl
0
0 I . 2R
(5.1.17)
Next, calculate a field created by the current I running along a rectilinear section of the conductor. In Figure 5.8 conductor MN and an observation point A at a distance b from it are presented alongside the current element dl and a corresponding vector r. Wherever the element dl is chosen, the vector dB is directed along the same direction (perpendicular to the plane of the drawing). Therefore, integration (5.1.15) can be executed in a scalar form. Applying (5.1.16), we obtain
dB
0 Idl sin . 4
r2
There are several variables in this expression; they should be expressed by a single variable. We shall choose for the integration variable an angle (see Figure 5.8). Expressing r and dl through the distance b and an angle : r
b rd bd , dl 2 . sin sin sin
Having substituted these expressions in the formula for dB, we obtain dB(rd /4 )(I/b) sin d , and consequently
B
0 I 2 I sin d 0 (cos 2 cos 1 ). ∫ 4 b 4 b
(5.1.18)
1
Angles 1 and 2 are defined by the extreme points M and N.
dB 0 r
dl
Figure 5.7 Application of Biot–Savart law to the calculation of the magnetic field of a circular current.
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313
A + dB
b α
d r
r d
dl
N
Figure 5.8 Application of Biot–Savart law to the calculation of the magnetic field of a direct current.
Figure 5.9 Magnetic force lines of a direct current in planes perpendicular to the current.
For the infinite direct conductor wire (where 1 0 and 2 and the difference cos0 cos is 2): I I (5.1.19) B 0 2 0 . 4 b 2 b The magnetic field of a direct current in a plane, perpendicular to a conductor, is depicted in Figure 5.9 by the force lines. They appear as concentric circles; their density decreases as the distance increases. EXAMPLE E5.1 Along two parallel indefinitely long wires identical currents I 60 A flow in the same direction. The wires are located at distance d 10 cm from each other. Define the magnetic induction B at a point A (Figure E5.1) at distance of r1 5 cm from one conductor and r2 12 cm from another. B B2 A
r1 D + l
B1 r2 d
+ C l
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Solution: In order to find the magnetic induction at the specified point A we should define the directions of induction vectors B1 and B2 created by each conductor separately and then combine two vectors B B1 B2. We can find the module of total induction according to the cosine theorem B B12 B22 2 B1 B2 cos ⴱ . Induction values B1 and B2 are expressed accordingly by the current strength I and distances r1 and r2: B1( 0I/2 r1) and B2( 0I/2 r2). Substituting these values in * we obtain B
0 I 2
1 1 2 cos . r12 r22 r1r2
Find cos using the cosine theorem
cos
r12 r22 d 2 0.576. 2r1r2
Substituting all constants and given values we arrive at 286 T.
EXAMPLE E5.2 Determine the magnetic field induction B produced by a section of infinitely long wire at point A at a distance r0 20 cm from the center of the wire segment (Figure E5.2). The flowing current I 30 A, the segment length l 60 cm.
dl
r d
1 d
r
l
r0 I
2
A
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Solution: The magnetic field induction B can be calculated according to eq. (5.1.18). The scalar form of this equation is dB( 0I sin d/4 r2). There are three variables in this equation: , r and l. It is more convenient to integrate over an angle provided all the variables are expressed through this angle. So drd /sin . Substituting this ratio into the first equation we obtain 0 sin rd 0Id dB ; 4 r2 sin 4 r besides r is also a variable and should be expressed through r (r0/sin ) and therefore dB 0I sin d /4 r0. This expression should be integrated over variable . dB
0 I 4 r0
2
0 I
∫ sin d 4 r0 (cos 1 cos 2 ) . ⴱ
1
Note that in the case of the symmetrical position of wire cos 1 cos 2, therefore the formula * is ( 0I/2 r0) cos 1. We need to define angle . It can be seen from Figure E5.2 that cos
2
r02 冢 2 冣2
4r02 2
.
Therefore, B
0 I I cos 1 0 2 r0 2 r0
4r02 2
.
Substituting all data into the formula obtained we arrive at B 2.49 105 NA m 24.9 T.
EXAMPLE E5.3 A long wire with a current I 50 A is bent at an angle 2 /3. Determine induction B at a point A (Figure E5.3). The distance d is equal to d 5 cm. I 1 B + A
I
0 r0 π−
2
d
2
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Solution: A bent wire can be considered as consisting of two semiinfinite pieces. According to the principle of magnetic fields superposition, the magnetic induction B at a point A is equal to the geometrical sum of the magnetic inductions B1 and B2, i.e., the fields created by two wire pieces 1 and 2, B B1 B2. The field magnetic induction B2 is zero: it follows from the Biot–Savart law, according to which in points lying on an axis of a conductor, dB 0 ([dlr] 0). Therefore, we need to find only B1. In Section 5.1.2 this problem was considered in detail and eq. (5.1.18), i.e., B( 0 I/4 r0) (cos 1cos 2) was derived, where r0 is the shortest distance from the wire to point A (the length of the perpendicular descended from point A on the wire). In the case considered 1 → 0 (the wire is infinite, cos 1 1), 2 2 /3 (cos 2 2 /3 1/2). The distance is r0 d sin( ) d sin ( /3) d(兹苶3/2). Correspondingly, 2 0 I ⎛ 3 0 I 1⎞ . ⎜⎝ 1 ⎟⎠ 2 4
d 4 d 3
B1
Executing calculations we obtain B 34.6 T. Vector B is directed perpendicular to the drawing in downward direction.
EXAMPLE E5.4 A wire in the form of a thin half ring of radius R 10 cm is in a uniform magnetic field (B 50 mT) perpendicular to magnetic force lines. A current I 10 A flows along the wire. Find the force acting on the half ring. y
B X
jdFy
I
dF
dl
id Fx
R j
d
0
x
i
Solution: (See first the Section 5.1.4). (Let us arrange the wire in a plane of the drawing and direct the coordinate axes as is represented in Figure E5.4. On the wire allocate an elementary section with a current Idl. On this area an Ampere force dF I[dl B] (5.1.24) operates. Let us divide the elementary force into two components dF i dFx j dFy. The force acting on the whole conductor can be found by the integration F i ∫ dFx j∫ dFy , L
L
where the symbol L indicates that the integration is taken over the whole half ring length.
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Using symmetry of the problem we can a priory write L dFx0. Therefore the whole force depends only on y-component dFy dF cos , where dF is the modulus of the force dF, dF IBdl. Expressing dl through R and (dl Rd ) we can write dFy IBR cos d . Integration over a quarter of ring (multiplied by 2) gives
/2 Fj IRB2 0 cos d and |F | 2IBR. It can be seen that the force is directed along the y-axis. Executing calculations gives F 0.1 N. EXAMPLE E5.5 A current I 80 A is flowing along a thin conducting ring of radius R 10 cm. Find the magnetic induction B at the z-axis perpendicular to the circle crossing the ring center at an arbitrary point of the axis. Then find the magnetic induction at a point r 20 cm (Figure E5.5). z
dBz dB dB⊥
r
dl
Solution: Since the z-axis is perpendicular to the ring plane and passes the center of the ring, it is a symmetry axis L . This means that the induction vector certainly must be codirectional to the z-axis and only the Bz component gives contribution to the field induction B. First, derive the general expression for B(z). Use the Biot–Savart law dB( 0/4 )(I[dlr]/r3) to determine the dBz component: dBz 冟dB冟 sin dB (R / 兹苶 R2苶 苶 Z2苶). The vector product [dl.r] is perpendicular to the plane fixed by vectors dl and r. Then dBz
0 I dl R , 2 2 2 4 R z ( R z 2 )1 2
where sin R/(R2z2)1/2. The only variable is dl. The integration over the whole circle gives
Bz
0 I 2 R R 0 IR 2 . 2 2 32 2 4 ( z R ) 2( z R 2 )3 2
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This is the general result. To make sure of the result we can use the border conditions B(0) and B( ). We know that B(0) is equal to ( 0I/2R) (see eq. (5.1.17)); by assuming z 0 we arrive at the correct result. The magnetic field diminishes at z → , which is also in agreement with our result. In order to find the field at an equidistant point r we can use the result obtained, accept r2 x2 R2 and substitute it into the general result; therefore B( 0IR2/2r3). Executing calculations, we arrive at 62.8 105 T or 628 T.
5.1.3
The law of a total current (Ampere law)
The sign of the potential character of a force field is the equality to zero of the circulation of the field intensity vector along any closed contour. Let us see whether the magnetic field is potential, i.e., whether the integral 养LBdl is equal to zero or not. Consider the simplest case when a magnetic field is created by a linear conductor with current I. The magnetic force lines in this case are the concentric circles lying in parallel planes, perpendicular to the conductor, with their centers on the linear conductor (Figure 5.9). Choose for simplicity contour L coinciding with one circular force line of any radius R (Figure 5.10). Then the circulation of the vector B along the contour L will be equal to
∫ B dl L
0 I I Rd 0 ∫ 2 R L 2 R
2 R
∫
dl.
0
Therefore, by using eq. (5.1.19):
∫ B dl 0 I .
(5.1.20)
L
Since circulation of the magnetic field induction is not zero the magnetic field is not potential. (Notice that in the above integral dl is an element of a contour L but not current.) To obtain this ratio for a noncircular contour of any form is not a difficult task. Expression (5.1.20) is the essence of Ampere’s law: circulation of the induction vector along a closed contour L is equal to the current multiplied by 0 comprised by this contour.
L I B dϕ
dl
Figure 5.10 An Ampere law consideration.
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(This means that if the current passes outside the contour chosen, this particular current does not contribute to the total current.) A field satisfying the condition (5.1.20) is referred to as a nonpotential field. Expression (5.1.20) is also referred to as Ampere’s law to emphasize the unity of the phenomena of interaction of currents with each other and with the magnetic fields. If contour L comprised N currents then, according to the principle of superposition, the circulation of a vector B is equal to their algebraic sum N
Ii , ∫ B dl 0 ∑ i1
(5.1.21)
L
the current is considered positive if it corresponds to the clockwise rule, otherwise it is considered negative. If the current is distributed nonuniformly across the conductor this law can be rewritten as
∫ B dl 0 ∫ j(r)dS, L
(5.1.22)
S
where a surface S is resting on contour L (Figure 5.11).
S
L Figure 5.11 A surface rested on a contour loop.
Let us apply Ampere’s law to calculate the induction of a magnetic field created by a solenoid. Remember that a coil that has been reeled up by thin wires without misses on the cylinder (Figure 5.12) is referred to as a solenoid. We shall choose a rectangular contour 1–2–3–4, depicted in the figure. Then circulation along the whole contour can be divided into four integrals:
∫ B dl ∫ Bl dl ∫ L
L
12
Bl dl
∫
Bl dl
23
∫
Bl dl
34
∫
Bl dl.
41
It can be seen that the integrals along segments 2 3 and 41 are zero since the angle between B and dl is /2. The integral on the segment 34 is also zero because this segment can be chosen far enough from the solenoid where B 0. Therefore,
∫ Bl dl ∫ L
12
Bl dl 0 ∑ I i 0 nlI , 12
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1
2
4
3
Figure 5.12 Calculation of a solenoid magnetic field strength.
where l is the length of segment 1–2, n is a number of turns over a unit solenoid length, I is the current in the solenoid. Therefore, B 0 nI ,
(5.1.23)
i.e., the induction inside an infinitely long solenoid is proportional to the overall current running onto the length unit. The magnetic field inside the solenoid is uniform. In this respect the solenoid plays the same role as a plate condenser plays in electrostatics. 5.1.4
Action of the magnetic field on the current, on the moving charge
Comparing the expression of the Ampere law (5.1.11) with the expression for a field created by an infinite conductor (5.1.19), one can imagine that current I1 (Figure 5.4) is creating the magnetic field B1 which in turn acts on a neighboring conductor with current I2; then f12 B1I2. Accordingly, f21 B2I1. If the conductor is not rectilinear, we should consider a conductor element dl and then f(dF/dl)IB or, finally, in the vector form d F I [ dl B ].
(5.1.24)
Figure 5.13 shows the arrangement of the vectors describing the field B, the conductor element dl and force acting on the conductor dF. Since the differential expression (5.1.24) is obtained starting from the Ampere law (5.1.11) the force dF is accordingly referred to as Ampere force. Another manifestation of the Ampere force is the action of a magnetic field on a moving charge. Moreover, from the point of view of electromagnetic dynamics all macroscopic (ponderomotive) forces can be reduced finally to the forces applied to the electric charges included in this body. We shall reduce force FA, which operates on the whole conductor, to a force that
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I dl B dF
Figure 5.13 Vector disposition according an Ampere law in the differential form.
operates on each single charge moving inside the conductor. For this purpose allocate the elementary cylinder of a volume dV dlS inside a conductor so that its generatrix is parallel to the direction of the carriers’ motion (Figure 5.14). According to eq. (5.1.24) force dFA, acting on this cylinder, is equal to dFA I [dl.B] jS [dl.B]. On the other hand, the current density j can be submitted according to eq. (5.1.4). Taking into account that j is codirected with u and, accordingly, with dl, the force dFA can be rewritten as dFAqnSdl[ B]qdN[ B], where dN is the number of electric carriers in the allocated cylinder. If the overall force is divided by the number of carriers (dFA/dN), force acting on a single charge is obtained Flor
d FA dN
q[ B].
(5.1.25)
The force acting on a single charge moving with a speed in a magnetic field B is referred to as a Lorentz force. The sign (direction) of the force depends on the sign of the moving charge, i.e., from sign q in eq. (5.1.25). From the expression defined for the Lorentz force Flor it can be seen that it is always perpendicular to the particle velocity. Therefore, a Lorentz force does not produce work. It follows that it is impossible to accelerate the particles by means of a Lorentz force, i.e., an electric field is required to do so. (Nevertheless, the Lorentz force is used in accelerators to make a motion cyclic.) The Lorentz force defines the movement of charged particles in a magnetic field. If the particles enter the magnetic field in a plane perpendicular to the induction B, the Lorentz force will act perpendicular to both vectors and B. In the absence of any other force, the Lorentz force is centripetal, and a circular movement will occur (Figure 5.15). Write the equation of Newton’s second law for this case qBmanm(2/R). The radius of a circle R can be derived from this expression R
1 , B ( q m )
(5.1.26)
where q/m is the specific particle charge. The period T of the circular particle motion turns out to be
T
2 R 1 2
. B(q m)
The period does not depend either on the particle radius R or on the speed.
(5.1.27)
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dl
dF
Figure 5.14 To a Lorentz force derivation.
+
+ Fπop
B +
+
Figure 5.15 Movement of a charge in a plane perpendicular to a magnetic field induction.
If the particle velocity is directed at an angle to a vector B (Figure 5.16, where B is directed along the z-axis), vector has to be projected in two directions: perpendicular ⊥ and parallel
to the induction vector B. Accordingly, the component ⊥ defines the circular motion of the particle and another component
determines its uniform motion along axis z since the Lorentz force for this component is zero. This results in the particle’s spiral movement. The basic spiral increment h is defined as h ||T 2
cos , B(q m)
(5.1.28)
whereas the circus radius R depending upon the perpendicular constituent ⬜ sin is R
sin . B(q m)
(5.1.29)
Therefore, the particle entering the magnetic field moves with a winding-up movement on the magnetic force lines. The charged particle movement described above permitted the development of a powerful instrument – the mass spectrometer: a device for “sorting” ions on their specific charge q/m. Such an opportunity is extremely tempting for modern chemistry for the analysis and synthesis of new substances and for many other problems. The basic scheme of a mass spectrometer is shown in Figure 5.17. The gas to be analyzed enters a vacuum chamber at a point S and is ionized (by any method, for example, by an electron beam impact). Between points S and A the potential difference is applied and ions are accelerated by an electric field. Passing an aperture (at point A) all ions possess identical energy, but not speed. To select ions with identical speed from a beam a speed filter is used in which both forces, Coulomb’ (Fcou qE) and Lorentz’ (Flor qB1), operate perpendicular to
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z
h
B
Fxop
r
Figure 5.16 Movement of a charge at an arbitrary orientation of a field induction B and a charge velocity vector. C
S
υ
A
+ B1 × + E −
∅
∅
qυB1 + + × υ × × υ ×
−
× −
×
qE
×
×
× R2 A1 R1 × × K
× × B2
× υ ×
×
×
×
×
Figure 5.17 Scheme of a mass spectrometer.
each other; the filter allows the singly ionized particles with speed E/B1 to pass through. This filtered beam goes to another point A1 in the chamber where the ions start to move in a circular trajectory with a radius depending on the q/m ratio. In a collector (or a film, or special detector device) located on line A1C ions with different specific charges q/m fall in different points. In Figure 5.18 the mass spectrum of the air is shown, where along abscissa axis values A/q (A is a mass number, q is a particle charge) are plotted, whereas along the ordinate a relative number of the given molecules in the object under investigation are shown. The construction of the modern mass spectrometer differs significantly from the one described above, although it is based on the same principle.
EXAMPLE E5.6 An electron is projected into a uniform field of induction B (B 30 mT) with its velocity vector ( 2 106 m/sec) making an angle of 30o with B. It begins
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x3
324
A+
(N216)+ x3
x100
(N14N15)+
x30
(arbitrary units)
x300
N (OTH. eΠ)
x10
O2+
44
40
x3
OH+
H2O+
34 32
29 28
20
x3
A2+
x3
CO2
x3
x3
(O16O18)+
+
N+ (N22+)
O+ (O22+)
18 17 16
14
A/q
Figure 5.18 Air mass spectrum (every peak relates to a definite ion; numbers over intense peaks show how many times its height was curtailed in order to present all peaks in one spectrogram).
to move along the helix (refer to Figure 5.16). Find the radius R of the helix and its pitch h. Solution: A Lorentz force is acting on the electron changing the direction of travel of the electron. Let us divide electron’s velocity into two components || and ⬜ relative to vector B: ⱍⱍ sin and ⬜ cos . The helix radius can be found from the second Newton law: F man, therefore F 冟e冟 ⬜B and an ⬜2/R. Then 冟e冟⬜B m(⬜2/R) and further R(m⬜/冟e冟B)(m sin /冏e冏B). Substituting the given values we obtain R 0.19 mm. The helix pitch is h ⱍⱍ T, where T 2 R/⬜. Substituting this value into an expression for h we obtain ⎛ 2 R|| ⎞ ⎛ 2 R cos ⎞ h⎜ or h ⎜ 2 R tan . ⎟ ⎝ sin ⎟⎠ ⎝ ⬜ ⎠ Calculation shows that the pitch is h 2.06 mm.
EXAMPLE E5.7 An electron enters a uniform magnetic field of induction B 0.03 T and begins to move along a circle of radius R 10 cm. Determine the electron’s speed . Solution: The second Newton law can be applied to the movement of the electron along a circle (m2/r)冏e冏B; its momentum can be found from the expression p m 冏e冏Br. However, relativistic laws should be used in this case (as we will
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see in the end). In this case p(m0c /兹1 苶 苶 苶2苶) (the electron’s velocity is included in the value ). Solve the last expression relative to :
p
m0 c
2
⎛ p ⎞ 1 ⎜ ⎝ m0 c ⎟⎠
2
冷 e 冨 Br m0 c ⎛ 冷 eBr 冨 ⎞ 1 ⎜ ⎝ m c ⎟⎠
2
.
0
Our solution can be simplified; calculating separately the value appears equal to (冏e冏Br/m0c) 1.76. Therefore, the value can be calculated ( 0.871) and then the velocity found c 2.61 108 m/sec. The electron moving with such a velocity is the relativistic one. EXAMPLE E5.8 An alpha particle runs through the accelerating potential difference 104 V and projects into two fields crossing at right angles – magnetic B 0.1 T and electrostatic E 10 kV/m. Find the charge/mass ratio of this particle if, when traveling in these fields perpendicular to both the fields, it does not diverge from a rectangular motion. Solution: In order to find the ratio it is useful to apply the relationship between the electrostatic forces work qU and the change of its kinetic energy m2/2. From this equality it follows that (q/m)(2/2)*. The special arrangement of the Coulomb and Lorentz fields provides the equality of their action and straight-line motion (Figure E5.8). Therefore, qE qB. This equation permits us to find particle speed as E/B. Inserting this equality into the * fraction, we arrive at (q/m)(E2/2B2). Executing calculations, we obtain (q/m)4.81107 C/kg. Let us check the dimension of the result: (E2/B2)((1B/m)2/1BT2) (1B A2/1B N2)1J C/(1N sec)21C m/(1N sec)21C/kg. Dimension is just the specific charge. z B
Flor
E O
υ x
Fc
y
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EXAMPLE E5.9 In one plane with an infinite direct wire is a frame with sizes as shown in Figure E5.9. Along the wire flows a current I 50 A. Find the magnetic induction flux d through the frame in square S al . x
dx
I
l + B a
a
Solution: In full analogy with the electrostatic strength flux the induction of the magnetic field is d B dS. In our problem induction of the magnetic field depends on the distance from the wire x, therefore the flux is
冕
2a
d B(x)dx. Since B(x)( 0 I/2x) the flux is equal to ( 0lI/2)ln 2. Executing a
the calculation we arrive at 4.5 Wb.
EXAMPLE E5.10 An infinite wire is bent as is shown in Figure E5.10. The circle radius is R 10 cm and the current flowing along it is I 80 A. Determine the magnetic induction B at a point O. 3
O
1
2
π
x B = B1 + B2 + B3
2
l
1
Solution: Divide the current wire into three pieces: 1, 2 and 3. The magnetic field induction is the vector sum B B1 B2 B3. The first segment does not produce a magnetic field at point O since dB 0 ([dl r] 0), according to Biot–Savart law along the whole segment piece. Two components remain both giving codirectional
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induction perpendicular to the drawing plane; they can be summed as scalar values. The half-ring induction B2 can be found according to eq. (5.1.17) taking only half of it: B2( 0I/4R). Induction B3 can be found according to eq. (5.1.18) taking into account that 1 /2 and 2 ; therefore B3 ( 0 I / 4 R). Then B B2 B3
5.1.5
0 I 0 I 0 I ( 1). Therefore B 3.31104 T 331 T. 4 R 4 R 4 R
A magnetic dipole moment in a magnetic field
Just as an electrostatic field acts on an electric dipole moment (refer to Section 4.1.5), a magnetic field acts on a magnetic dipole moment. At a significant difference of working forces the results are very similar. Let us first consider a homogeneous magnetic field; we accept a magnetic dipole moment as a rectangular hoop (Figure 5.19). If the contour is oriented so that vector B is parallel to its plane, the sides having length b will not fill any action of the Ampere force because the vector product in eq. (5.1.24) is zero. The forces acting on the side a are F IaB (refer to eq. (5.1.24)), the force couple renders the torque rotating moment of the contour M F Fb IaBb ISB M B or in vector form: M F [M B ].
(5.1.30)
This expression corresponds to eq. (4.1.32). The torque MF aspires to turn the contour so that the magnetic moment M as a vector turns along the field. On two sides having length b the Ampere forces act oppositely and will stretch (or compress) the contour but not rotate it. It is also possible to show that formula (5.1.30) is valid for a contour of any form and, hence, can be used regardless of the form of the magnetic moment M. The potential energy of a magnetic moment in a magnetic field can be calculated according to the recipe given in Section 1.4.5. Taking into account that in this case MF is the moment of external forces, we obtain dU M F d M B sin d
MF a
B
F n
F b
M
Figure 5.19 A frame with a current in a uniform magnetic field.
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and after integration U ∫ M F d M B cos C. Accept C to be zero at 0. Thus, finally in vector form U (M B ).
(5.1.31)
The graph is just the same as for an electric analog (Figure 4.18). In a nonuniform field besides the torque, a force F will act on the magnetic dipole moment (providing the moment is turned into a stable position with 0 and cos 1). To calculate the force we can take the advantage of eq. (1.4.32) describing it as dependent on potential energy. Then (for a one-dimensional case, B B(x)) Fx
U dB M cos . x dx
(5.1.32)
It follows from this equation that the magnetic moment M can be dragged in or pushed out of the magnetic field depending on the angle (cos 1) and the sign of magnetic field gradient. 5.1.6.
Electromagnetic induction
The concept of a vector flux d through a surface dS has been given in Sections (2.8.3) and (4.1.3). Being a particular case of a more general theory of a vector field, the same concepts can be applied to a magnetic field as well. An elementary flux d of a magnetic field induction vector B through the surface dS is equal to a scalar product of B and dS d (B dS).
(5.1.33)
Depending on the angle between a normal n to the surface dS (Figure 2.20) and the induction vector B, a flux d can vary in limits BdS. In general, flux through surface S is defined by the integration ∫ (B dS ) ∫ Bn dS. S
(5.1.34)
S
In 1812 the English physicist Michael Faraday made a discovery, which has significantly influenced the development of all mankind. Having made a real conducting contour, confined to a surface S, he established that by changing the flux (5.1.34) an electric current appeared in the circuit. By numerous experiments, Faraday established that current value does not depend on the way the flux changes but on the speed of this change. The mathematical form of Faraday’s law is extraordinary simple: i
d . dt
(5.1.35)
The minus sign in this expression corresponds to the general physical law of inertia: the induction current in a contour is always directed in a way that opposes the reason of its appearance. This statement is referred to as the Lenz rule.
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To exemplify Faraday’s law, let us imagine a simple experiment. We shall create a closed electric contour with motionless rails and a metal axis with wheels moving on them (Figure 5.20). On this “construction,” impose a magnetic field perpendicular to the plane of the image. Move the axis at a uniform speed . At the same time and with the same speed begin to move electric carriers creating a current in the metal axis and in the circuit in general. The force F directed along the axis and equal to q[ . B] will operate on the carriers. The action of this force is equivalent to the action of the electric field E [ . B]. This field is not electrostatic because it has been created in a different way – the movement of charges in a magnetic field. The circulation of the electric field strength E along the contour will produce EMF in the contour (see eq. (5.1.8)): i ∫ E dl ∫ [ B ]dl. L
(5.1.36)
L
Only the movable part of the contour creates EMF, therefore 2
i ∫ E dl ∫ [ B ]dl. L
1
Supposing that both the axis movement and the magnetics field are uniform, we can obtain 2
i B∫ d B. 1
Multiplying and dividing this intermediate expression by dt we derive
i
B dt . dt
1 i
E' B +
υ
l
dl 2
Figure 5.20 A modeling of a Faraday magnetic induction law.
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Taking into account that ldt dS and BdS d , we obtain i(d /dt), which coincides with eq. (5.1.35). The sign appeared in this expression after analysis of the vector disposition and accounting for the negative sign of electron. The integral 养LE dl is not zero since E is not an electrostatic field. It represents quite different, solenoidal (or vertex or curl) electric field. The acting force F is also nonCoulomb in origin; in contrast, it can be related to the extraneous one. Therefore, we can proceed by adding to the nonzero integral 养LE dl the zero’s addition 养LEdl and write the expression ⴱ i ∫ E d.
(5.1.37)
L
Here, however, E* is the strength of both the conservative and (potential and nonpotential) electric fields. Furthermore, if the left-hand part of eq. (5.1.35) is written in the form (5.1.37) and using eq. (5.1.34), we can arrive at an important expression which is referred to as a Maxwellian equation ⎛ B ⎞
d
∫ E d dt ∫ Bn dS ∫ ⎜⎝ t ⎟⎠ n dS. ⴱ
L
S
(5.1.38)
S
At the left-hand side the integral is taken on any contour L, whereas the surface S is fixed by the already chosen contour L: the surface S rests on the contour L (see Figure 5.11). From this Maxwellian equation it follows that any change of a magnetic field (the righthand side of the equation) generates an electric field (the left-hand side of the equation). If a conducting wire is drawn along the contour L an electric current would occur in it. If contour L is in vacuum, along it the electric field would be excited. Once again we would like to emphasize that an induction electric field is not electrostatic. In fact, the source of an electrostatic field is motionless electric charge whereas in producing a nonpotential electric field the source of the field is an alternating magnetic field. This field, induction by origin, is solenoidal (i.e., is not potential) and certainly possesses other than electrostatic field properties. Faraday’s law is one of the general laws of electrodynamics. In an alternating magnetic field, induction leads to the excitement of EMF. It defines the mutual induction of one conductor onto another. However, even if there is only one conductor with an alternative current a force appears that renders back the current value state. A magnetic flux that penetrates its own contour with current I and generates current variation, is referred to as its own, intrinsic magnetic flux and is designated as S. This flux is not influenced by a change of counter orientation; it is firmly connected with a contour. Since, according to Biot–Savart law, B is proportional to I, therefore it is also proportional to c LI .
(5.1.39)
The coefficient L is called the self-inductance of the counter. It describes the relation between an alternating current and the intrinsic magnetic field produced by it.
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If the counter consisted of N windings and the magnetic flux penetrated all of them without omission, then the total magnetic flux linkage is equal to N c or LIN L I.
(5.1.40)
Excitation of an induction in a closed counter when a current change is taking place is referred to as self-induction. It is equal to the speed of flux linkage change taken with the sign minus: S
d dt
or for solenoids without a ferromagnetic core: S L
dI . dt
(5.1.41)
In order to calculate the inductance of a long solenoid, we can use an expression for the total magnetic flux (Ampere law) LI and N c nlBS (where n is the number of solenoid windings on a unit length and l is the total solenoid length; the stroke at L is omitted). Since the solenoid magnetic induction is B 0nI (see eq. (5.1.22)), hence 0 n2 lSI . Therefore, L 0 n 2V .
(5.1.42)
It can be seen that the solenoid inductance depends quadratically upon the number of windings on a unit length and is proportional to the solenoid volume. In the absence of a ferromagnetic core the inductance is constant and is not dependent on the current strength.
5.2
MAGNETIC PROPERTIES OF CHEMICAL SUBSTANCES
From the point of view of their reaction to an external magnetic field, all substances are referred to as magnetic. Its chemical structure defines the magnetic properties of a substance. All magnetic materials can be divided mainly into three main classes: diamagnetic, paramagnetic and magnetically ordered substances. Diamagnetics are pushed out of an external nonuniform magnetic field; they consist of those molecules that do not possess their own (i.e., “intrinsic”) magnetic dipole moments. Paramagnetics are drawn into the external nonuniform magnetic field; they consist of molecules possessing inherent magnetic dipole moments in the absence of an internal magnetic field. Among the magnetically ordered substances are ferromagnetics and ferrimagnetics, highly reacting on an external magnetic field. There is also a class of antiferromagnetic substances, weakly reacting on an external field. Further we shall describe in more detail the nature of all these substances and consider those characteristics that describe macro- and microproperties of magnetics.
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Moreover, we shall follow approximately the same logic we used for the description of dielectrics. In particular, we should establish where the magnetic properties originated from. It is logical to connect magnetic properties with magnetic dipole moments of atoms and molecules. 5.2.1
Atomic magnetism
In semiclassical Bohr theory an atom is represented as consisting of a nucleus and electrons traveling along circular stationary orbits. This motion can be characterized by an orbital angular momentum: the orbital angular (mechanical) moment Ll (see Section 1.3.9). An intrinsic electron angular momentum is also considered; sometimes it is described by electron rotation around its own axis (see Chapter 6.7 and Section 7.5.5); it is referred to as electron spin with its own angular momentum Ls. Orbital and spin states are sometimes imagined as circular electric currents. These currents create orbital and spin magnetic dipole moments. More advanced notions have been developed in quantum mechanics, but these semiclassical representations are very distinct and useful at this point. There are three sources of the magnetic properties of substances: (1) electron spin; (2) orbital electron motion; (3) change of the electron orbital angular momentum at the imposition of an external magnetic field. The first two can explain paramagnetism, and the third can be used in considering diamagnetism. A nuclear magnetic moment is very weak in comparison with orbital and spin electron magnetic moments; thus it can be temporarily neglected here. However, nuclear magnetism will be closely considered in Chapter 8 because the nuclei participate strongly in resonant methods of investigations in chemistry. We shall take advantage of the semiclassical Bohr theory as it provides an elementary model for understanding the physical essence of the phenomenon. We will begin by calculating a inasmuch as the Bohr theory permits us to do it very easily. The gyromagnetic ratio is the ratio of the magnetic and mechanical moments is referred to as giromagnetic ratio (Figure 5.21). Remember that the angular momentum L of an MP (an electron, in our case) relative to an origin (nucleus) is defined as a vector product L [rp], where r is the electron radius vector relative to the nucleus, and p the linear electron momentum (p m). On traveling along a circular orbit the linear velocity is perpendicular to r ( ⬜ r), then Ll ⱍLlⱍ mr. The direction of the vector Ll is defined by the rule of vector product (the right-hand screw rule).
Ll
I
e
− +
r M
Figure 5.21 A gyromagnetic ratio for orbital electron movement in an atom.
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The circular current caused by the electron traveling along the orbit produces an orbital magnetic moment Ml, the direction of which is also defined by the right-hand screw rule (Figure 5.21). The module of the orbital magnetic moment is Ml IS (ⱍeⱍ r2/T ), where T is the period of revolution. Thus, the gyromagnetic ratio is Ml 冷 e 冨 r 2 Ll T m yr
Canceling r and taking into account that (2 r/T ) we arrive at Ml 1 冷 e 冨 . 2 Ll m
(5.2.1)
Ml 冷e冨 , g 12 Ll m
(5.2.2)
Eq. (5.2.1) can be rewritten as
where g is the gyromagnetic ratio in the unit 12 (ⱍeⱍ/m). In this unit the orbital gyromagnetic ratio is equal to 1 (gorb 1). In their outstanding experiments, Einstein and de Haas showed that for spin the gyromagnetic ratio is equal to 2 (gsp 2) and consequently (Ms /Ls)2 12 (ⱍeⱍ/m). This gyromagnetic ratio anomaly is a source of some of the most interesting and important phenomena and is used, in particular, in many physical methods of chemical substances’ research. 5.2.2
Macroscopic properties of magnetics
The magnetization of a substance is quantitatively characterized by magnetization (, which is numerically equal to the magnetic moment of a volume unit (
∑ Mi . V
(5.2.3)
Alongside with magnetization of a unit volume, a specific and mole magnetization are considered as well. The specific magnetization (magnetization of a mass unit) is equal to: ( sp
1 ∑ Mi , m
(5.2.4)
where m is a mass of the physically infinitesimal volume V (refer to Section 4.2.1). Having replaced m on V where is a substance density we shall obtain ( sp 1 (.
(5.2.5)
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Mole magnetization (magnetization of one mole) is (M
1 N ∑ Mi , v i1
(5.2.6)
where is the number of moles in a physically infinitesimal volume m/M, M is a molar mass. Alongside with a magnetic induction B one more value characterizes the magnetic field H, a strength of the magnetic field, is used; in isotropic magnetic the magnetization ( is proportional to magnetic field strength H, that is ( H
(5.2.7)
where is the scalar value referred to as a magnetic susceptibility. The magnetic susceptibility characterizes ability of substance to be magnetized in a magnetic field. As ( and H have identical dimension is dimensionless value. For diamagnetic materials 0, its value is ⬃105107 , for paramagnetic materials 0, its value is ⬃103106; for ferromagnetic ⬃ 103105. One can see that diamagnetic weakly and oppositely magnetized in an inner magnetic field and therefore pushed out from it. Paramagnetic weakly magnetized too but positively. Ferromagnetic magnetized very strong and intensively drawn in magnetic field. Alongside with magnetic susceptibility of the unit volume a specific magnetic susceptibility sp is often used in practice sp((sp/H ) the relation being exist sp(1/). The same is for molar magnetic susceptibility M which is equal to M
(M H
M
M .
(5.2.8)
and
5.2.3
An internal magnetic field in magnetics
One more magnetic characteristic, a magnetic permeability , is usually introduced; it shows how much the magnetic induction in a magnetic B is larger then that of the external magnetic field B0: B B0 .
(5.2.9)
This means that this value should be substituted into the Bio-Savare law (5.1.16) dB =
0 I [dl r ] 4 r 3
(5.2.10)
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B0 l (
Figure 5.22 A magnetic in an external magnetic field; a circular micro-molecular current and surface currents can be seen.
Consider now a sample of a cylinder form placed in an external magnetic field with induction B0. Let the cylinder by a cross section S and length l is oriented along the external field force lines (Figure 5.22). Under an action of a field all of molecular currents (vectors of the magnetic dipole moments) will be ordered in the field (along or opposite, it does not matter at the moment). At averaging, inside of the magnetic cylinder molecular currents will be mutually canceled. Not compensated micro-currents will be left on the cylinder’s surface. The picture remained a solenoid and can be considered as a certain total macroscopic current, flowing the cylinder over (with a magnetization current Jm). We can introduce a value of current linear density ℘ being equal to ℘Jm / l. The situation is really just like the solenoid with the magnetic field induction B1 inside. The field inside the solenoid is defined by eqn (5.1.23). Considering the magnetized cylinder as the solenoid we can calculated the induction B1 superimposed on the external field B1 0 nI 0℘
(5.2.11)
Find the relation of the magnetic magnetization ( and the surface density of the current ℘. According to eqn (5.2.3), the magnetization is the magnetic moment of the unit volume. Therefore for this case ( = ∑ Mi V IS (Sl ) ℘,
(5.2.12)
i.e. the magnetization of a piece of the magnetic is numerically equal to the linear density of a surface current. The expressions obtained allow one to find a ratio outside and inside of the magnetic and to establish the ratio between a magnetic susceptibility and magnetic permeability. The macroscopic field in a substance is characterized by a magnetic induction which is the geometrical sum of magnetic inductions of the external B0 and internal B1 fields, i.e. B B0 B1 .
(5.2.13)
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Having replaced in this expression B0 and B1 according to (5.2.9) and (5.2.11), and taking also into account a collinear arrangement of all three vectors, we obtain B = 0 H 0℘ 0 H 0 (.
(5.2.14)
It shows that the strength of the magnetic field inside of the magnetic differs from that of an external field on the value of magnetization. Writing further the magnetization according to (5.2.7) one can obtain B = 0 H 0 H 0 (1 ) H .
(5.2.15)
On the other hand, according to (5.2.9), B = B0. Therefore 1
(5.2.16)
and the induction inside of isotropic magnetic is B 0 H.
(5.2.17)
This expression corresponds to the definition of the magnetic susceptibility (5.2.9). It follows that H
B ( 0
Let’s note here a distinction of the induction and the strength notions. The induction of a magnetic field according to (5.2.9) depend on substance property, however the strength of the field outside and that of inside of the magnetic is the same. Moreover, the strength doesn’t depend at all on the sample magnetic properties (i.e. on ) (H B/ 0 0 H/ 0 H). At the same time the induction varies at transmitting from one magnetic to another. Therefore at calculations of magnetic circuits to use the strength H is more convenient.
5.2.4
Microscopic mechanism of magnetization
It has already been mentioned that from the point of view of their magnetic properties, we can distinguish three main classes of substance: diamagnetic, paramagnetic and magnetically ordered substances. We shall now consider the same question from a microscopic point of view, i.e., which processes cause magnetic properties and how these properties are related to their chemical structure. We shall start with a diamagnetic.
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Diamagnetics magnetize opposite to an external magnetic field and are pushed out of it. The magnetic susceptibility of a diamagnetic is thus negative and depends neither on temperature, nor on the strength of the magnetic field. Diamagnetic properties are defined by the electron atomic orbit. It is easier to begin with one-electron atom. In a magnetic field the electron orbit precesses in the same way as a spinning top in a gravitational field (refer to Appendix 2). This precession arises because an atom possesses both magnetic and angular (mechanic) momentums (refer to 1.3.57 and Figure 1.19). Find the angular velocity of orbit precession. Let the atom possess an angular momentum L and magnetic moment , directed opposite to each other (see Figure 5.23). In an external magnetic field B, excited along an axis z, a torque MF will operate MF [B], directed perpendicular to vectors and B (Figure 5.24). Under the action of this torque, vector L in time dt will acquire an increment dL MF dt and, accordingly, L (t dt) L(t) dL. The vector dL is perpendicular to vector L and therefore ⱍL ⱍⱍ ⱍLⱍⱍ. Thus, the action of torque MF changes the direction of vector L, but not its length. Thereof, the plane in which the axis z and vector L lie, will turn by an angle d
d
dL Mdt . L sin L sin
Since M |MF| MB sin , hence
d
M B sin M dt Bdt. L sin L
z B
d L sin
L dL
L'
Figure 5.23 An electron orbit precession.
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B
M
−e
r′ d dL
−e
r′
I′ s=πr′2
Figure 5.24 Origin of a diamagnetic moment.
Dividing both sides by dt, we obtain the angular velocity of the electron orbit precession L
d M B dt L
Using the gyromagnetic ratio the orbit angular velocity can be found:
L g
1 冷e冨 B 2 m
(5.2.18)
This expression can be applied both to orbit and spin precession taking the g-factor into account. The angular velocity is named after J. Larmor and the magnetic precession is also referred to as Larmor precession. In particular, for orbit precession, the angular velocity is 1冷e冨 L B. (5.2.19) 2 m Additional electron movement caused by orbit precession leads to the excitation of an equivalent circular current I (Figure 5.24). This current induces the magnetic moment M, which is the diamagnetic moment. Irrespective of the direction of the
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torque vector MF or of the direction of induction of an external magnetic field B, the induced diamagnetic moment M is always directed against B. This is the origin of the diamagnetic effect. Therefore, the minus sign is present in the expressions for the diamagnetic moment. The additional electron movement occurs along a circle smaller than r (the electron radius designated in Figure 5.24 as r ). A circular electric current corresponds to this movement
I
e L e2 B. 2
4 m
Strictly speaking, the radius (r ) is not the radius of an electron orbit; besides, it depends on the inclination angle . As an electron orbit can be inclined to the induction direction (to an external field), averaging over all these parameters allows us to obtain 冬(r )2 冭 2 冬(r )2 冭, 3 where 具(r)2典 is the average value of the square of the electron distance from the z-axis. Having substituted this value in the expression by M, we shall obtain for one-electron atom
M
e2 2 冬r 冭 B. 6m
(5.2.20)
Summing up the expression obtained for all electrons in a multielectron atom, we shall find its induced moment as M
e2 kZ 2 B ∑ 冬rk 冭, 6 m k1
(5.2.21)
where Z is the number of electrons in an atom. If we now increase this value by Avogadro number NA we shall obtain the value of the mole magnetizations
( MN A
N A e2 kZ 2 B ∑ 冬rk 冭. 6m k1
Multiplying numerator and denominator by 0 and comparing the expression obtained with eq. (5.2.7), we shall find the molar magnetic susceptibility of multielectron atoms
M 0
N A e2 kZ 2 B ∑ 冬 rk 冭. 6m k1
(5.2.22)
It can be seen that the more electrons in an atom and the larger the radius of electron orbits, the greater is the diamagnetic susceptibility. Substituting here values of fundamental physical values and accepting the radius of atoms ⬃1010 m, we obtain M ⬃ 107108 m3/mol which corresponds well to experiment for molar susceptibilities of diamagnetics.
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The molar susceptibility of a number of compounds is presented in Table 5.1. Notice that diamagnetism is inherent in all substances without exception. Paramagnetics magnetize in the direction of an external magnetic field and are drawn into it. The magnetic susceptibility of a paramagnetic is positive, depending on the temperature and value of the magnetic field strength. Paramagnetism is inherent in substances whose molecules possess permanent magnetic moments regardless of magnetic field. We shall connect all these experimental facts with the microscopic properties of paramagnetics. In the absence of an external magnetic field, the magnetic moments of paramagnetic molecules are disordered in space by the action of chaotic thermal movement (Figure 5.25a). This means that the vector sum for magnetization (5.2.3) is zero; hence, the magnetization ( is also zero. When a paramagnetic substance is brought into the magnetic field, each magnetic moment aspires to be guided in the field’s direction; however, the molecule’s thermal movement prevents it from doing so. A balance is established (Figure 5.25b); as a result the vector sum in eq. (5.2.3) becomes distinct from zero, the substance is magnetized. A factor which should be taken into account when describing the competition of the ordering action of a magnetic field (whose energy is U( )) and the disordering tendency due to chaotic movement (with averaged energy T ) is the Boltzmann factor: exp(U( )/T). It is necessary to take all these circumstances into account when considering the magnetization process.
Table 5.1 Values of molar susceptibilities of some diamagnetic compounds Substance
M, 1011 m3/mol
Substance
Helium (He) Neon (Ne) Argon (Ar) Krypton Xenon (Xe)
2.4 8.2 25 40 84
Silver (Ag) Bismuth (Bi) Glass (SiO2) Methane (CH4) Naphthalene (C10H8)
(a)
M, 1011m3/mol 27 350 50 76 240 (perpendicular to the molecule plane)
(b)
Figure 5.25 A paramagnetic (a) outside a magnetic field (Mi 0) and (b) inside a magnetic field (Mi 0).
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An analysis of all these processes was carried out by the French physicist P. Langevin. He considered the behavior of a system of magnetic dipole moments in an external magnetic field. (An identical problem exists in dielectric physics when describing the orientation polarization, see Section 4.2.4.) The essence of both phenomena consists of the competition between the two processes: the aspiration of a field to direct the dipole moments along the field direction and the action of chaotic thermal movement interfering with it. At each temperature a compromise is achieved. Finding this compromise is the essence of the Langevin theorem. (The Langevin theorem is presented in full in Appendix 4.) The result of this theorem is an expression for mole magnetic susceptibility of a paramagnetic substance. At ( B/T) 1 the mole paramagnetic susceptibility has been found to be
M
0 N A 2 3 T
(5.2.23)
(see eq. (A4.10) in Appendix 4). It can be seen from this formula that M is inversely proportional to temperature: M
C T
(5.2.24)
This dependence is referred to as Curie’s law. Comparing eqs. (5.2.23) and (5.2.24), we can find a constant C, which is also named after Curie: C=
0 NA M2 . 3
(5.2.25)
Substituting into formula (5.2.23) the values of fundamental physical constants and the value of the spin magnetic moment, we can obtain M ⬇ 108 m3/mol, which is compatible with most experimental data (see Table 5.2 below). In very highly magnetic fields and/or at very low temperatures ( B T ) the field can orient all the magnetic moments of all molecules in parallel; this results in saturation, i.e., further increase of the field intensity cannot appreciably change the magnetization of the sample since all moments are already parallel along the field ( M N A M.
(5.2.26)
Table 5.2 The molar magnetic susceptibility of some paramagnetics Substance
M, 1011 m3/mol
Substance
M, 1011 m3/mol
Sodium (11Na) Aluminum (13Al) Lithium (3Li) Vanadium (23V) Oxygen (O2)
2.0 2.1 3.1 37 430
MnSO4 Fe2O3 NiSO4 FeCl2 Dysprosium (66Dy)
1.7 4.8 5.0 16 150
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In an intermediate area where B⬇T one dependence (m(H) smoothly passes to another (Figure 5.26). The curve repeats for orientation polarization (Figure 4.28). Consider now a number of examples that will allow us to see how the structure of chemical substances influences their magnetic properties. In the creation of diamagnetic properties the main role is played by the electron orbit. This means that diamagnetism is inherent to all substances without any exception. All atoms possess diamagnetic properties but they cannot always be measured because, as a rule, diamagnetism is masked by a larger paramagnetic effect. Accordingly, diamagnetism becomes apparent in substances consisting of atoms with completely compensated magnetic moments. This takes place in noble gases (He, Ne, Ar, Xe and Rn). As an example consider a neon atom: it has 10 electrons in its electron shell (1s22s22p6). Figure 5.27 shows the electron distribution among quantum cells. Notice that for s-electrons the orbital magnetic moment is zero (l 0). The total magnetic moment of s-shells is also zero because all quantum cell are occupied by pairs; the p-shell is also filled completely. So the total magnetic moment of the neon atom is zero. In this case diamagnetic properties can be exhibited and measured. Ions Na and Cl whose electron configurations coincide with Ne and Ar will also be diamagnetic. On the other hand, the neutral atoms Na and Cl possess magnetic moments as there is one noncoupled 3s electron in the Na atom and a 3p electron in the Cl electron shell. While forming a chemical compound, the Na atom’s 3s electron passes to the Cl atom and a NaCl molecule with ionic bond is formed. Since Na and Cl magnetic moments do not possess noncoupled spins, the molecule NaCl is diamagnetic. Note that, in general, the majority of chemical compounds are diamagnetic. In particular, this is true for the ionic compounds of the type considered and covalence compounds with nonsaturated bonds. By way of example, we can consider a CCl4 molecule. Upon formation of this molecule, the carbon atom, having two noncoupled 2p electrons, is excited (Figure 5.28); 2s electrons are dicoupled and, as a result, sp3-hybridizations are generated; four equivalent sp3-hybrid orbits arise, with two electrons on each bond. These orbits form chemical bonds with the Cl atom. Thus, in the molecule CCl4 there are no free noncoupled electrons and consequently it is diamagnetic. mB0>>T
( (sat mB0<<T
H
0
Figure 5.26 Paramagnetic magnetization versus magnetic field strength H.
s 2 1
p Ne
Figure 5.27 Neon atom’s electron configuration.
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In addition to the substances already listed, metals Pb, Zn, Hg, Sb, Ge, elementary sulfur, water and others also exhibit diamagnetic properties. (See Table 5.1 above for values of the molar susceptibilities for some of diamagnetic substances.) The origin of paramagnetism is mainly due to the electron spins. Therefore, an analysis of paramagnetism should begin with a consideration of quantum cell filling by spins. A lithium atom, for example, has two coupled electrons in 1s-state and one 2s electron (Figure 5.29). The spin magnetic moments of the coupled electrons are zero. Therefore, the paramagnetic moment of a lithium atom is defined exclusively by the single 2s-electron spin. A free carbon atom is also paramagnetic as it possesses a significant magnetic moment due to the two noncoupled 2p2 electrons. However, diamond and graphite are diamagnetic; the diamond’s diamagnetism is caused by four equivalent saturated sp3-hybrid covalence bonds participating in the crystal formation. The diamagnetism of graphite with layered structure is of the other type. The layer-forming carbon atoms are bonded by sp2-hybrid bonds; whereas between the layers, the forces are of the van der Waals type. Within each layer probably free electron’s movement is possible along the closed orbits of the large ring radius; this leads to strong diamagnetic effect. Oxygen paramagnetism is defined by the presence of a noncoupled electron pair on the antibonding molecular orbits. Many rare earth elements together with their alloys and compounds clearly exhibit paramagnetic properties. The atoms of these elements have incompletely filled deeply laying 4f-shells; they are shielded by outer electrons. So, for example, a dysprosium atom 66Dy has four noncoupled (according to the Hund rule) 4f electrons (Figure 5.30). They define
s
p
p
s
2 1
2 1
C(1s22s2p2)
C+(1s22s12p3)
Figure 5.28 A carbon atomic configuration in sp3-hybridization state. s 2 1
Li (1s22s)
Figure 5.29 Lithium atom’s electron configuration.
6 5 4 3 2 1
s p 6s2 d 5s2 5p6 2 6 10 4s 4p 4d M L K
f
Dy
Figure 5.30 Dysprosium atom’s electron configuration (K, L and M are closed and omitted).
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the magnetic properties of this atom. At the same time, the outer electron shell (5s25p66s2) takes part in bond formation and very poorly influences the magnetism. Therefore, the magnetic molar susceptibility of the rare earth element compounds depends only insignificantly on particular compounds and on what state the rare earth atom is in. Compounds of elements with incompletely filled d-shells exhibit strong magnetic properties, especially 3d-metals. However, in this case some of the 3d electrons take part both in magnetism and in chemical bonding. Therefore, the magnetic properties of compounds of the same d-element can behave differently in different chemical compounds. Analysis of the participation of orbital state in the formation of atomic magnetic moments shows that, for a variety of reasons, its contribution can frequently be ignored; it is either too small in comparison with spin, or is suppressed (“frozen”) by a crystal field. In Table 5.2 values of molar susceptibilities of some paramagnetic substances are given. There are, however, exceptions to this general law (for example, 29Cu, 47Ag, 83Bi with odd numbers of electrons), exhibiting diamagnetic properties. Due to the large number of electrons (Z in eq. (5.2.22)) diamagnetism is great in these substances and suppresses paramagnetic effects. If the number of electrons is even the orbital and spin magnetic moments can be compensated in pairs.
5.3
MAGNETICALLY ORDERED STATE
A magnetic state is magnetically ordered if the atomic magnetic moments in the absence of an external magnetic field in a macroscopic crystal volume are orderly directed to each other. From this definition it is clear that a magnetically ordered state is mainly typical for a crystalline state. There are two basic magnetically ordered structures, which can be distinguished from each other by the presence or absence of permanent macroscopic magnetization, namely ferromagnetic and antiferrimagnetic. In the ferromagnetic-ordered state the atomic magnetic moments are spontaneously oriented parallel to each other (Figure 5.31a), and in antiferromagnetics they are oriented mutually antiparallel (Figure 5.31b). A ferrimagnetic state is referred to as an ordered state and arises when the antiferromagnetic structure consists of atoms (ions) with different magnetic moments. The magnetic moments in this case do not completely compensate each other (the moments in positions A in Figure 5.31c differ from that in position B). Macroscopically such a state appears to be ferromagnetic; however, the magnetic ordering is nearer to an incompletely compensated antiferromagnetic state.
5.3.1
Ferromagnetism
Ferromagnetics are intensively macroscopically magnetized in an external field and are strongly drawn into it. They have the greatest application techniques. Ferromagnetic
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(a)
(b)
(c)
Figure 5.31 Magnetic ordering in (a) ferro, (b) antiferro and (c) ferrimagnetic crystals.
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susceptibility is 1 and, consequently, magnetic permeability is nearly equal to its susceptibility ( ⬇ ). The ferromagnetic state is also characterized by the following features: •
• •
the existence of spontaneous magnetization in macroscopic volumes that are referred to as domains; magnetic hysteresis, i.e., ambiguous dependence of the magnetization on the strength of an external magnetic field; pronounced dependence of ferromagnetic susceptibility on temperature and on the intensity of an external magnetic field; the presence of a Curie point, i.e., temperatures above which the ferromagnetic loses its specific properties and transforms into a paramagnetic state (for example, the Curie point for Fe is 1043 K, for Ni is 631 K, etc.).
Einstein and de Haas carried out an experiment to determine the gyromagnetic ratio (see Section 5.2.1) for ferromagnetic materials; later their results were repeatedly confirmed. Experience has inevitably led to the gyromagnetic ratio g 2 instead of the expected g 1 (as for orbital magnetic moments). Goudsmit and Uleneck explained this result suggesting an inherent electron magnetic moment (spin). Thus, it was established that ferromagnetism is defined by electron spin. A typical example of a ferromagnetic is iron. Figure 5.32 shows a scheme of the electron distribution on the quantum cells in an iron atom. The d-shell has four noncoupled electrons. They define the atomic magnetic moment. In an isolated iron atom an orbital electron movement creates some orbital moment. In crystals, however, the orbital moments are “frozen” by the action of intracrystalline fields and do not contribute to the atomic magnetic moment. The reason for the “freezing” of the orbital moments is not yet completely understood. However, experience clearly shows that the orbital moments do not participate in the ferromagnetism of transition d-elements. Moreover, in the transition metals (iron, for instance) the distribution of the valence electrons on the quantum cells is influenced by many factors; so it is impossible to predict, a priori, in which state outer 3d and 4s electrons are. From saturation magnetization measurements (see formula (5.2.35)), it is known only that the atomic magnetic moment of iron is equal to 2.86 B (Bohr magnetons, see Section 7.5.5). In iron compounds the iron atom can be either in a di- or a trivalent state; in this case the electron distribution on quantum cells is certainly known (Figure 5.33). In an Fe2 ion there are four noncoupled spins and in Fe3 there are five such electrons. Accordingly, the magnetic moment of ion Fe2 is 4 B and ion Fe3 is 5 B. Fe
s 4 3 3s2 3p6
p
d
Figure 5.32 Electronic configuration of an iron atom. HOH Fe2+
s 4 3 3s2 3p6
p
d
s 4 3 3s2 3p6
p
HOH Fe3+ d
Figure 5.33 Iron’s ions Fe2 and Fe3 electron configuration.
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In addition to transition d-elements (Fe, Co, Ni), magnetic ordering (including ferromagnetic) is found in rare earth metals: gadolinium, terbium, dysprosium, etc., as well as other elements of the yttrium group, their compounds and alloys. In spite of the fact that the number of magnetically active atoms (ions) is limited, the number of their combinations in alloys and chemical compounds is very high. The presence of noncoupled d- and f-electrons in atoms does not yet completely provide the conditions necessary for ferromagnetism to occur. For example, such elements as Cr, Mn, Pt, etc., also have noncoupled electrons, but they are not ferromagnetic. In some cases they possess antiferromagnetic behavior. The phenomenon of magnetic ordering is collective; magnetic atoms themselves, being isolated from each other, do not show any ferromagnetic properties. The nature of magnetic ordering has a quantum mechanical origin, a description of which is beyond the scope this book. 5.3.2
Domains: magnetization of ferromagnetics
At temperatures lower than Curie point Tc the magnetic moments of a ferromagnetic are ordered, i.e., they build in parallel to each others. If such ordering is established in the whole sample (as in Figure 5.31a) it would be saturated even in the absence of an external magnetic field. However this is not the case. Below Tc the ferromagnetic sample splits up into small volumes referred as domains. There is nearly perfect magnetic ordering in each domain however all of them are oriented disorderly, i.e., a sample exhibits no magnetic moment. A scheme of the ferromagnetic sample splitting into domains is shown in Figure 5.34a-d; in Figure 5.34e the final splitting is depicted. The sample splitting into domains is due to the fact that a single-domain sample would posses too big energy because of a presence on a sample edges the similar poles (Figure 5.34a). The sample splitting into domains with opposite moments directions leads to reduction of energy (Figure 5.34b-d). Partitioning the sample body into domains is accomplished by creation of the large number of domain walls to what the energy is spent. Splitting comes to the end when the gain as a result of splitting into domains is equal to expense of the domain walls production. In Figure 5.35 the scheme of such domain wall between two neighboring domains is shown. In Figure 5.36 a curve of a preliminary ferromagnetic magnetization, i.e., dependence of a magnetization ( on the magnetic field strength H is given. Figure 5.37 describes graphically a respective alteration of the sample domain structure. N
S (a)
N
S
S
N (b)
(c)
(d)
(e)
Figure 5.34 Scheme of the ferromagnetic sample splitting into domains.
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Domain’s wall
Figure 5.35 A domain wall. (
final turn of domains
(c)
domains rotation
(b)
elastic domain’s (a) displacement H
Figure 5.36 At stages of a primary ferromagnetic magnetization process: (a) elastic domain walls displacement, (b) turn of the domain magnetization directions, (c) final turn of domains directions.
The magnetization process can be conditionally divided into some stages. In the beginning (at low H) volume of those domains which are energetically more favorable (from the point of view of their orientation in the magnetic field) increases. This increase is accomplished by the displacement of the domain walls. The volume of “unfavorable” domain decreases in their total amount and vise versa. When process of domain walls comes to the end an additional turn of an already single domain to the direction of the external field takes place. As a result almost all atomic magnetic moments of the whole sample drew up in the direction of the external field; the sample comes nearly to saturation. There is only a small part of the moments disoriented due to the thermal vibration. Finally, all magnetic moments are aligned, there comes the saturation. The subsequent reduction of intensity of magnetic field strength does not occur in the same way as discussed earlier. The magnetization decrease follows a curve 1–2 (Figure 5.38). When the external field disappears, magnetization is kept at a nonzero level (r, referred to as residual magnetization. In fact, this is the magnetization that we all experienced working with permanent magnets. To demagnetize the sample, an opposite field (a part of a curve 2–3) must be applied. The field strength corresponding to the sample demagnetization Hc is referred to as the coercive force. To continue increasing an opposite field, the sample magnetization will follow along curve 3–4; to then change the field direction the magnetization process will follow curve 4–5–6–1. The closed curve of the ferromagnetic magnetization (Figure 5.38) is called a magnetic hysteresis loop. Modern techniques have certain requirements of various magnetic materials in respect of their saturation magnetization value, residual magnetization, coercive force, etc. Thin
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H
H
H
H>0
H=0 (a)
H
(b)
(c)
(d)
(e)
H
Figure 5.37 Change of a ferromagnetic domain structure at magnetization: (a) the domain structure of a sample out of the magnetic field, (b) the growth of the “favorable” domain in the external magnetic field, (c) the single domain crystal, (d) the turn of the whole domain along the external magnetic field to say nothing about the single magnetic moments, (e) saturation. ( (sat
1
(r 2 Hc
6 3
H
5
4
Figure 5.38 A hysteresis loop. x
xmax
x0 0
H
Figure 5.39 A ferromagnetic susceptibility in external magnetic field H.
and thick, high- and flat-loop materials are required and these can be achieved by the synthesis of new magnetic compounds and alloys. Differentiation of the primary magnetization curve (Figure 5.36) shows that the magnetic susceptibility (H) of a ferromagnetic depends on the strength of the external field (Figure 5.39). 5.3.3
Antiferro- and ferrimagnetics
The magnetic structure of antiferromagnetics can be imagined as two identical ferromagnetic sublattices inserted into each other with opposite moment directions. As an
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Figure 5.40 An antiferromagnetic structure of MnO.
example, Figure 5.40 shows the magnetic structure of MnO (diamagnetic oxygen atoms are not shown). It can also be visualized by ferromagnetic atomic layers with antiparallel orientation of the magnetic moments in every following layer. Naturally, such crystal does not show ferromagnetic properties, all the magnetic moments are mutually compensated. The study of antiferromagnetics is nevertheless of interest exclusively from the point of view of the theory of magnetic interactions in magnetic materials. Also, antiferromagnetism exists in a limited range of temperature. The transition of an antiferromagnetic (as well as a ferrimagnetic) to a paramagnetic state occurs at certain temperature points, referred to as Neel points. There is technical interest in ferrimagnetics. The majority of them are antiferromagnetics, the structure of which includes two (or more) ions whose magnetic moments differ, and therefore sublattices do not completely compensate each other in the crystal. The macroscopic magnetic moment arises. In many respects, a ferrimagnetic crystal behaves as a ferromagnetic one. The domain structure and, consequently, a hysteresis loop, also exhibit. Magnetite Fe3O4, having the structure of a noble spinel (CaAl2O4) is a good example of a ferrimagnetic. The magnetite’s chemical formula does not, however, reflect a valence state of atoms in this compound. It would be more correct to present the chemical magnetite formula as Fe23Fe2O4. As we have already mentioned, the ionic magnetic moments of iron Fe3 and Fe2 are different (Figure 5.33). They can occupy (in ordered or disordered manner) different crystallographic positions in the structure. In every case crystals can possess different properties. Moreover, a lot of isomorphic substitutions are possible in the crystal structure. They are all referred to as ferrites. Each particular compound can have different magnetic properties (saturation magnetization, coercive force, residual magnetization, etc.). They are dielectrics and do not lose energy for induction currents and Joule heat release, which is why they are widely used in portable radio sets. The creation of special materials with preset properties is one of the tasks of the chemist.
5.4
DISPLACEMENT CURRENT: MAXWELL’S EQUATIONS
To end this chapter it is useful to generalize all the aspects discussed; they compose a system of equations of classical Maxwellian electrodynamics from which all the laws of electricity and magnetism can be derived, including electromagnetic radiation. Nearly all the equations are already known to the readers, so we can concentrate mainly on the physical conclusions. Let us start with those equations that describe stationary phenomena. One of the equations is the Gauss law (Section 4.1.3). Its physical sense concerns the statement: the sources
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of an electrostatic field are motionless (in the given reference frames) electric charges. In fact, if there are no electric charges in a system the right-hand parts of all specified equations are equal to zero; this also leads to the absence of an electrostatic strength flux. The significance of Coulomb and Gauss laws has already been mentioned in Section 4.1.3. The Ampere law from which the nonpotential character of the magnetic field follows is the next Maxwell equation. It follows from this law that the source of the magnetic field is the electric current and that it is of a nonpotential character:
∫ H dl ∫ j dS L
H, dl , d S, j
(5.4.1)
S
The potential character of an electrostatic field follows from equality to zero of circulation of the strength vector E of the electrostatic field. Since there are no magnetic charges (monopoles) in nature, the flux of the magnetic induction through the closed surface is zero (养SBdS0). The next equation in the Maxwellian series is the Faraday’s law (5.1.38) ⎛ B ⎞
∫ E dl ∫ ⎜⎝ t ⎟⎠ n dS, L
(5.4.2)
S
a change of the magnetic field causes the appearance of an electric vortex field (being not of an electrostatic nature). Comparison of eqs. (5.4.1) and (5.4.2) allows us to find an infringement of symmetry in the magnetic and electric laws. In fact, if the source of a vertex electric field is an alternating magnetic field, it can be expected that the alternating electric field should cause the occurrence of the magnetic field. From the equations presented above, this does not follow. So Maxwell came analytically to the symmetry brake phenomenon, which later originated the new notion of displacement current. Let us consider an electric contour with a condenser as its component (Figure 5.41). If the contour is connected to a permanent electric source, the current in the circuit does not flow. If, however, the source generates an alternating voltage, there appears an alternating current in the circuit, which depends on the frequency of the electric generator and on condenser capacity. Charges will periodically appear on the condenser clamps creating a variable electric field in the condenser. At the same time lines of electric current density seem to terminate on the surface of the condenser plates. Moreover, the alternative current in the circuit creates a magnetic field around it, though such a field is unlikely to be present around the condenser. Maxwell solved this uncertainty by introducing a current inside the condenser (Figure 5.41), which closed the electric current lines. This current is referred to as a displacement current.
I
Figure 5.41 Lines of a displacement current.
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Consider these phenomena in more detail. According to the definition the conductivity current I in an electric circuit is dq/dt. It can also be written as jcond S dq/dt, from which follows jcond dq/(Sdt) d/dt (since dq/S is a surface charge density d). On the other hand, the electric strength of the field inside the condenser is /0. Therefore, ⎛ dE ⎞ ⎛ dD ⎞ ⎛ d ⎞ ⎟ D. ⎜⎝ ⎟⎠ 0 ⎜⎝ ⎟⎠ ⎜⎝ dt dt dt ⎠
(5.4.3)
The value (dD/dt)D is the displacement current. Denote it as jdisp. Notice that this value is of a vector nature as far as the jcond is concerned; then we can write (dD/dt)D . The displacement current force lines enclose the conduction current forces lines, besides, the displacement current produces a magnetic field around. Therefore the current density is the sum j jcond jdisp jcond D .
(5.4.4)
∫ H dl ∫ ( jcond D )dS.
(5.4.5)
Eq. (5.4.1) can be rewritten as
L
S
Two more equations are included in the Maxwellian system, which connect the fields in a vacuum to the fields inside a medium (refer to Sections 4.2 and 5.2). B 0 H,
(5.4.6)
D 0 E.
(5.4.7)
j(r ) E(r )
(5.4.8)
And the last equation
relates to the local characteristics of the medium. Remember that this is the Ohm’s equation in differential form (refer to eq. (5.1.5)). Let us come back to eqs. (5.4.2) and (5.4.5). There are no conductivity currents in a vacuum; therefore the last equation can, for this particular case, be rewritten as ⎛ B ⎞
∫ E dl ∫ ⎜⎝ t ⎟⎠ dS, L
⎛ D ⎞
∫ H dl ∫ ⎜⎝ t ⎟⎠ dS. L
(5.4.9)
S
S
The symmetry between the two is clearly seen.
(5.4.10)
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It follows from them that the change in time of a magnetic field is the source of an electric field, whereas a time-varying electric field produces a magnetic field. Together they explain the existence of an electromagnetic wave. Notice that the excitation of electromagnetic waves needs only one field: the variable electric field is easier to create. Consider in brief the principle of the creation of an electromagnetic wave. This can be produced by a dipole antenna. It is well known that in an electric contour containing inductance L and capacitance C (insertion in Figure 5.42), oscillations can be excited with a period T depending on the values of L and C. Turn the condenser plates to align them as a simple dipole antenna (Figure 5.42). Certainly, the value of the condenser’s capacitance will change, but this does not interest us at the moment. Suppose, in a given instant of time, the left side of the antenna is charged positive and the right side is charged negative. In a plane xy there will be an electric field, and an excitement created immediately begins to move away from the antenna along the x-axis with a high but finite speed. This speed c in vacuum is equal to c ⬃ 3 1010 m/sec (see Appendix 5). After a quarter of the period, the charge on the sides of the antenna becomes zero, the force lines of the already created electric field become closed, but excitement continues the movement. After another quarter of the period, the antenna will have an opposite charge to the initial one. Again, there will be an electric field with oppositely directed force lines. After three quarters of the period, the antenna again becomes uncharged and the force lines of the second “ringlet” become isolated (closed), but in the opposite direction. All these “ringlets” will follow each other and never be caught because they move at the same speed. In a period of time, all the events will be repeated. According to eq. (5.4.10) the change in time of an electric field generates a variable magnetic field in a plane perpendicular to the “ringlets” of an electric field. There will be similar “ringlets” of a magnetic field moving together with the first. Sometimes this picture is represented as “couplings” as shown in Figure 5.43a. However, a more realistic picture is given in Figure 5.43b: in the Cartesian coordinates two mutually perpendicular planes result in which oscillations of the vectors of the strengths of electric and magnetic fields take place. The wave is propagated with a speed c. It was shown in Section 2.8.3 that the energy transferred by an electromagnetic wave is proportional to the square of the wave amplitude E. Without proof, we state that the energy density carried by a wave is defined by the vector product [E.B] S; vector S being referred to as the Poynting vector. The mechanism of electromagnetic radiation emission in electronic and nuclear subsystems of atoms differs significantly from that discussed above. However, all characteristics
C
x
x
x
L +++
−− −
−− −
+++ y
Figure 5.42 A work of a dipole antenna; inset (top left): an oscillation circuit.
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of the radiation remain the same. A more detailed consideration of the Maxwellian equation system is given in Appendix 5. Table 5.3 gives the range of known electromagnetic waves.
l
B
B
B
E
E (a)
y
x
E
V
B
z (b)
Figure 5.43 Electromagnetic wave: (a) schematic representation of electromagnetic bundles and (b) oscillation of the field inductions vectors E and B in an electromagnetic wave. Table 5.3 Electromagnetic wave scales and their peculiarities
−21
−2
12
−3
−20 −19 −18
−1 0 +1
−17 −16 −15
+2 +3 +4
13 14 15 16 17 18
−4 −5 −6 −7 −8 −9
−14
+5
19
−13
+6
20
Radiation type
Short Micro waves Infra red
Emitter
Radio waves
Medium
Wavelength (γ/1m) 6 5 4 3 2 1 0 −1 −2
Long
Energy Frequency (s−1) lg(E/1J) lg(E/1eV) 3 −11 −30 −29 −10 4 −28 −9 5 −27 6 −8 −7 7 −26 −25 −6 8 −24 −5 9 −4 10 −23 −3 −22 11
Oscillatory circuit
Detector
Radio devices NMR, NQR
Klystron, magnetron
Crystal diodes
Heated body
Visible light.
Eye, photo
Ultra violet
Exited molecules
−10
x − rays
Valence electrons
−11
gamma rays
nuclei
Counter of nuclear particles
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EXAMPLE E5.11 A square frame with a side length of a 2 cm consisting of N 100 winds of a thin wire is suspended by an elastic thread. The thread elasticity C is equal to C 10 N/m grad. The frame plane coincides with an external magnetic field induction. Determine the field induction if at a current I 1 A the frame plane is turned at angle 60° (Figure E5.11). M2 a I
I
pm
B MI
Solution: The frame is in equilibrium if two angular momentums, the action of the magnetic field M1 pm Bsin and the resistance of the elastic thread M2 C, are equal: M1 M2 and, consequently, pm Bsin C. We know that pm ISN Ia2N (where S a2); we can rewrite the previous equation as NIa2 Bsin C. Therefore, B (C/Na2 sin ). In this particular case ( /2) (refer to Figure E5.8); therefore sin cos . Finally we can find B (C/Na2 cos) and substituting all known values arrive at B 0.03 T 30 mT. EXAMPLE E5.12 A frame of area S 150 cm2 consisting of N 1000 winds is rotating in a magnetic field of induction B 0.1 T with frequency n 10 sec1. Find the instant EMF i in the magnetic field with induction B 0.1 T which corresponds to the turn angle 30o. Solution: An instant value of i can be determined according to eq. (5.1.35): i (d/dt), where N is a magnetic linkage. Therefore, i N(d/dt). At revolution changes according to the induction law BScost, where 2 n is the angular frequency of rotation. Substituting the last equation into the previous equation we obtain i 2 nNBS sin t. Executing the calculations we arrive at i 47.1 V. EXAMPLE E5.13 Determine the bismuth magnetic susceptibility and its molar susceptibility M if specific bismuth susceptibility sp 1.3 109 m3/kg. The bismuth density is 9.8 103 kg/m3.
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Solution: All the equations in use are given in Section 5.2.2. In order to solve the problem we will use the equation sp: –1.3 105 (dimensionless quantity). To determine the molar susceptibility M we have to multiply sp by the molar mass M: M Msp. Executing the calculations we arrive at M ⬇2.7 1010 m3/mol.
EXAMPLE E5.14 A bismuth ball of radius R 1 cm is in a uniform magnetic field with B 0.5 T. Determine the magnetic moment of the ball pm acquired by it if the bismuth magnetic susceptibility is 1.5 104. 1 N Solution: Use the expression for magnetization ( (5.2.3) ( v i1i , where i is the magnetic moment of an ith bismuth atom and N is the number of atoms in the volume V. The vector qualities stay under the sum sign; however, bismuth is a very strong diamagnetic. Therefore all induced atomic moments are codirectional and equal on modulus. Therefore, we can rewrite the expression in scalar form as ((1/ V) JN. The product JN is the total magnetic moment of the unit volume; therefore the ball magnetic moment is pm (V. Since the magnetization ( relates to the magnetic field strength H as (H (B / 0) and the ball volume is V(4/3) R3, hence pm 4 (B/ 0)R3. Express all values in SI and execute calculations: 3
0.5 4 3 4 2 2 pm 3 (1.5104) 4 107 (0.01) 2.510 Am 250 A m . The minus sign shows that the acquired ball magnetic moment is directed opposite to the external magnetic field. EXAMPLE E5.15 A square frame with a side b 10 cm is made of a thin wire. It consists of N 255 winds. The frame is in a uniform magnetic field with induction B 0.25 T; the frame can rotate around an axis which passes through the middle of its opposite sides and is perpendicular to the magnetic induction force lines. Determine the maximum amplitude value of EMF induction Ei,max arising in the frame winding with frequency n 1800 min1. Solution: The change of a magnetic flux crossing the frame occurs at the rotation of the frame. Instant value of magnetic flux (t) is defined by the expression (t) BS cos , where S is the frame area (S b2); is the angle between the normal vectors n to the frame plane and that of a magnetic induction B. At uniform rotation this angle linearly changes with t, i.e., t where is the angular velocity of the frame rotation. According to law of Faraday electromagnetic induction in which Ei is defined by the time derivative from the flux Ei – (d/dt) and using flux linkage notion Ei is Ei –(d/dt) or Ei –N(d/dt). Executing differentiation we obtain Ei NBb2 sin t. The maximum instant EMF is at maximum sin t, i.e., sin t 1.
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The Ei,max NBb2 or Ei,max 2 NBb2n. Express all values in the SI: N 255; B 0.25 T; b 10 cm 0.1 m; n 1800 min1 30 sec1. Substitute them in the expression obtained: Ei,max 2 255 0.25(0.1)2 30 120 V. EXAMPLE E5.16 Two identical codirectional point magnetic moments pm 5 Am2 are at a distance r 1 m from each other. Determine: (1) the potential energy U of their interaction; (2) the force F of their interaction. Solution: In order to solve this problem it is convenient to use the result of Example E5.5, where an expression for the magnetic field induction on the perpendicular axes B(z) was obtained. In the case of point dipole r R and the expression takes the form B
0 IR 2 2r 3
ⴱ.
The magnetic moment pm of the ring with electric current I is pm nIS, where n is a normal unit vector to the plane of the ring and S is an area of the ring S R2. It is easy to obtain from the equation *: B( 0I( R2)/2 r3) and further B ( 0pm/2 r3)**. (1) Assume that one dipole creates a field and the other is experiencing its action. The potential energy of the magnetic dipole in the magnetic field is defined by an expression U pm Bcos (see Section 5.1.5), where is the angle between vectors pm and B. In our case 0 and cos 1. Therefore, U –pmB. Substituting B by the expression ** we obtain U( 0 p2m /2 r 3 ). Expressing all values in SI units and executing calculations we obtain U
4 107 (5 103 )2 6.25 1013 J 0.625 pJ. 2 23
(2) In order to determine the interaction force we can use the relation (1.4.28) ⎛ U ⎞ F ⎜ . ⎝ r ⎟⎠
Using the potential energy expression obtained after differentiation one has ⎛ 3 p2 ⎞ F ⎜ 0 4m ⎟ . ⎝ 2 r ⎠ Substituting all the values into this expression (in SI) we arrive at F
3 4 107 (5 103 )2 N 9.38 1013 N 0.938 nN. 4 2 2
The minus sign in both final expressions shows that the force is attractive.
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PROBLEMS/TASKS 5.1. A thin tape l 40 cm in width is folded into a cylinder of radius R 30 cm (Figures T5.1). An electric current I 200 A flows, uniformly distributed, across the tape. Define the magnetic induction B on the axis of the cylinder at two points: at a central point of the cylinder B1 and at a point at the end of the cylinder B2. z I
l
1
2
R
5.2. Entering the uniform magnetic field with B 0.1 T an electron began to move along a circle of radius R 5 cm. Find the equivalent current I produced by this movement and associated magnetic moment . 5.3. Assume that in metallic state an iron atom contains four noncoupled 3d electrons. Determine the theoretical value of the corresponding saturation magnetization (sat of the metallic iron. 5.4. An infinitely long straight wire is bent at a right angle at point O. An electric current flows through the wire (I 100 A). Calculate the magnetic field induction B at the points on the line of bisection of the right angle at distances a 10 cm from point O to both sides of it. 5.5. An electric current I 60 A flows through a thin wire bent as a rectangle. The dimensions of the rectangle are 30 and 40 cm. Calculate the magnetic field induction (B) at the point of intersection of the diagonals. 5.6. An electric current I 1 kA flows through a straight wire passing through a uniform magnetic field that is perpendicular to the lines of the field induction. With what force F/l does the magnetic field act on a l 1 m long wire. The magnetic field induction is equal to B 1 T. 5.7. A square wire frame is placed in the same plane as a long straight wire; the wire is parallel to the two opposite sides of the square frame. The same current (I 1 kA) flows along all wires. Determine the force F applied to the whole wire frame if the distance of the nearest frame side from the direct wire is equal to the length of the frame side. 5.8. A current of I 10 A flows along two similar wire rings with a radius R 10 cm. Determine the force F interaction between the rings if they are on the same plane parallel to each other; the distance (d) between the rings is 1 mm.
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359
5.9. A thin ring with a radius R 10 cm is carrying a charge Q 10 nC. The ring is rotating at n 10 sec1 around its symmetry axis, i.e., perpendicular to the ring plane. Find: (1) the magnetic moment pm of the ring; (2) the relationship between the magnetic moment and the angular momentum (pm/L), if the weight of the ring is 10 g. 5.10. Two ions with the same charge but different masses move into a uniform magnetic field. The first ion moves along a circular path with a radius R1 5 cm and the second ion with a radius R2 2.5 cm. Find the ratio between the masses (m1/m2) of the two ions if they initially pass through the same accelerating potential difference. 5.11. An electron enters a uniform magnetic field of strength H 16 kA/m with a velocity 8 106 m/sec. The velocity vector is oriented at an angle 60° relative to the force field lines. Determine the spiral radius R and pitch h. 5.12. A rod of length l 10 cm is spinning in a uniform magnetic field with an induction B 0.4 T in the plane perpendicular to the induction force lines. The axis of rotation passes through one end of the rod. Determine the potential difference inducted at ends of the rod if the rod rotates with frequency n 16 sec1. 5.13. Determine the nonuniformity of the magnetic field (dB/dx) if the maximum force acting on the point magnetic dipole is Fmax 1 mN. The dipole moment of the point dipole is m 2 mA m2. 5.14. Find the electric current I flowing along the thin ring of a radius R 0.2 m if the magnetic field induction in the equidistant point at r 0.3 m on the z-axis from the ring is B 20 T (Figure T5.14).
A
z
r
O
R
5.15. A rectangular copper bail (Figure T5.15) of a wire section S a2 2 mm2 is in a uniform magnetic field of B 10 mT directed vertically downwards. The bail can freely rotate around a line OO . A current I 20 A is flowing along the wire. Find the angle on which the bail will turn as a result of Ampere interaction. The copper density is 8.9 103 kg/m3.
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5. Magnetics
O' B a O
a
5.16. A long solenoid is wound closely with wire d 5 mm in diameter. What is the strength H of the magnetic field inside the solenoid at electric current I 4 A?
ANSWERS 5.1. B1 349 T, B2 251T. 5.2. Ief 1.1 mA, H 10 MA/m. Be2 R2/(2m) 14.08pA m2. 5.3. (sat 3.13MA/m. I I 5.4. B1 0 ( 2 1) 482 T; B2 0 ( 2 1) 82.8 T. 2 a 2 a 2 2 5.5. B 2 0 I a b 200 T.
ab 5.6. (F/l) 1 kN/m.
5.7. F
0 I 2 0.1 N. 4
0 I 2r 12.6 mN. d (1) m qnR2 3.14 nAm; (2) (m/L) 500nC/kg. (m1/m2) 4. R 1.96 mm, h 7.1mm. l2Bn 201mV. (dB/dx) (Fmax/m) 0.5 T/m.
5.8. F 5.9. 5.10. 5.11. 5.12. 5.13.
5.14. I
2 Br 3 21.5 A. 0 R2
5.15. arctan
IB 29.8. 2gS
5.16. H 8 kA/m.
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–6– Optical Wave Optics and Quantum Phenomena
6.1
PHYSICS OF ELECTROMAGNETIC OPTICAL WAVES
Wave optics deals with the propagation of light waves and their interaction with matter. A schematic drawing of electromagnetic waves was given at the end of Chapter 5 (Figure 5.43b) and their main characteristics were summarized (Table 5.3). The frequencies (the wavelength) corresponding to the range of waves visible to the naked eye are presented in Figure 6.1: relative eye sensitivity S is depicted which reflects the property of the human eye. Two areas adjoin the optical range: infrared (IR) and ultraviolet (UV) radiation (see Table 5.3). Remember also the ratio c/n between light velocity in a vacuum c and in some isotropic media . n being the refraction index, n . A useful scheme of the wavelengths of visible light is also presented.
S
1.0
0.5
0.40 | violet
0.60 , m | | green yellow
0.50
0.70 | red
Figure 6.1 The relative eye sensitivity S of the standard observer to different wavelengths, S(), for a normal level of illumination. 361
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All sets of optical phenomena comprise three general parts: geometrical, wave and physical optics. In this chapter, substantial attention will be given to wave optics and its transition to quantum optics, in which light exhibits as a flow of particles. To make an atom emit an electromagnetic wave it is necessary to excite the atom, i.e., to transfer it to a higher energy state with a short lifetime. In the ground state, an atom can exist for an indefinitely long time without changing its state. However, an atom can be transferred from the ground state to some other state with higher energy, i.e., to the socalled excited state with a limited lifetime (in many cases approximately 108 sec; see Section 7.2 and Section 8). For this time, an atom emits a continuous wave referred to as a wavetrain or zug. The wavetrain length l can be estimated as follows: time (~108 sec) multiplied by the light velocity (3.108 m/sec) gives value of the order of the wavetrain length l ~ 1 m. We can describe a simplified diagram of a wavetrain as a single wave with definite properties. Groups of such wavetrains form a bunch of waves or a light beam. Electromagnetic waves do not require a medium to propagate. Different kinds of electromagnetic waves are produced by different emitters. As a rule, an emitter does not emit a single wave, but a bundle of waves (we don’t touch coherent laser radiation here). An emitter produces waves by changing its own state (for instance, at the transition from one energy level to another, the energy difference contributes to the wave’s energy). Each specified wave is characterized by some properties. So, a beam consists of an enormous amount of particular waves having a definite length (Section 7.2). Electric interactions are stronger than magnetic ones. In other words, all physiological, photochemical, photoelectric, etc., actions of light are caused by the operation of an electric field. Therefore, the vector E is referred to as the light vector and the plane of its oscillation is called the plane of oscillation. Each light beam consists of many wavetrains with particular orientation of the plane of oscillation, i.e., polarizations. Figure 6.2 shows the polarization for a number of cases, viewed along the wave propagation direction. In Figure 6.2a, a natural beam of transverse electromagnetic waves is depicted; a cylindrical symmetry can be seen in the beam; the direction of the symmetry axis coincides with the beam propagation direction (axis x in the picture). The beam as a whole is completely unpolarized. In Figure 6.2c, the electric field vector E (i.e., the single orientation of the oscillation plane) is shown; a beam is totally polarized. Such a wave is called a plane-polarized wave. However, in many cases the beam consists of waves with
(a)
(b)
(c)
Figure 6.2 Polarization of electromagnetic wave beams: (a) completely polarized beam, (b) partly polarized beam, (c) natural unpolarized beam.
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partly ordered planes of oscillations; a partly polarized beam is formed (Figure 6.2b). In solids, in the given propagation direction two perpendicular oscillation planes can exist; they are denoted by the letter s, which can be equal to 1 and 2; in addition, a third, longitudinal, polarization can exist too (s 3). The waves in solids can thus have three polarizations (s → 1, 2, 3), but in liquids and gases only one (s 3, longitudinal) polarization exists. Different orientations of the planes of oscillation are possible in a beam. However, many planes of vibrations take place in a beam, they can all always be reduced to two mutually perpendicular planes. The equations of running waves were derived in Chapter 2 (eqs. (2.8.5) and (2.8.7)). In these equations, (x, t) symbolizes the displacement of any point from its equilibrium position x in time instance t. In this chapter we have passed from mechanical to electromagnetic waves; by displacement we shall now mean oscillations of the electric field strength E and those of the magnetic field H in mutually perpendicular planes (Figure 5.45). The cross line of these planes coincides with the axis of wave propagation, which coincides with a wave vector k (refer to Section 2.8.2). A plane that is determined by the vector k and the plane of E vector oscillations is a plane of oscillation. Let us present some fundamental laws of optics; later we shall use them from the point of view of the wave nature of light. Consider a beam of light falling on the border of two media 1 and 2 with absolute refraction indexes n1 and n2 (Figure 6.3). The medium with the higher refraction index is referred to as the optically denser one (Figure 6.3, n1 < n2) and vice versa. At a point O, every wavetrain splits: part of it reflects (reflected beam OB) and the other part refracts (refracted beam OO). The law of reflection–refraction asserts that all beams, namely, initial AO, reflected OB and refracted OO together with a normal KL to the border plane in a point O, lie in one plane. The angle of incidence is equal to the reflection angle . The ratio of the sine of the angle of incidence to the sine of the refraction angle is equal to the ratio of the absolute refraction indexes of two media sin n2 n2,1 , sin n1
(6.1.1)
where n2,1 is the relative refraction index. This equation is referred to as Snell’s law. Since the air refraction index is near to 1, the relative refraction index on the border with air is practically equal to the absolute index.
K A
n1 B
' O
K' O'
L
n2 n1
L'
Figure 6.3 Reflection and refraction at an air–glass interface (n2 n1).
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If, having undergone a number of reflections and refractions, a beam has passed along a path; another beam directed along the same path, but in the opposite direction, will pass along the path; in exactly the same way, having repeated all features. This statement is referred to as the law of reversibility. Taking advantage of this law, if we follow the path of the return beam at its propagation from optically more dense medium 2 and on the border of the media contact, then sin n21sin. Since n21 1, at some angle , the refracted beam will propagate along the border without leaving medium 2 (the refraction angle becomes /2; Figure 6.4). The angle C being in this very case an incident angle is referred to as the critical angle of total internal reflection. At greater values of ( C) the total reflection is preserved. The phenomenon of total internal reflection plays a basic role in the construction of fiber light guides, i.e., thin fibers made up of transparent materials (Figure 6.4): if an optic ray enters the guide, it will be unable to abandon it, propagating inside to its end because any incident angle for it will be larger C because of the small fiber diameter. Consider now a path of beams in a prism in air with a vertex angle of a material with a refraction index n (Figure 6.5). At small angles a deviation is determined by the refraction index n and angle as: (n 1),
(6.1.2)
i.e., the angle of deviation linearly depends on the refraction index n of the material that the prism was made of. Since the refraction index depends on the wavelength (see Section 6.5), the prism is the simplest device for decomposition of light in a spectrum on the wavelength. A rainbow in the sky after rain is a widely known example of decomposition of white light into a spectrum by water drops in air. In Chapter 2 (Section 2.9.2 and Figure 2.27), the effects arising from the reflection of waves from the borders were considered. There, a very simple model of a traveling wave
n1 < n2 n1
=/2
n2
c
S
Figure 6.4 A total internal reflection of light from a source S at different incident angles: larger and smaller than the critical angle C.
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n
red
A
violet
Figure 6.5 Light ray paths in a prism.
Table 6.1 Refraction indexes of some materials Media
Refraction index (n)
Light speed in the medium (in the light speed unit)
Absolute vacuum Air Water Glass Diamond Silicon
1 1.0003 1.33 1.4 1.8 2.4 3.5
1 0.9997 0.75 0.56 0.71 0.42 0.29
propagated in some media with a reflection index n1 was used. The reflection was examined for two cases: when the reflecting wave on an interface has a node or antinode. This is of importance since it influences whether or not the wave at reflection loses a halfwavelength. It was shown that if the reflection takes place from the denser media with a refraction index n2 n1, the wave loses a half-wavelength and, in contrast, if n2 n1 it does not. In Table 6.1, the refraction indexes of some substances are presented.
EXAMPLE E6.1 From a source of monochromatic light with wavelength into a screen point A, there come two beams (Figure E6.1): one beam S1A comes directly from a source horizontally perpendicular to the vertical screen, and the second beam S1BA is reflected in a point B from a horizontally located mirror. The distance from the source to the screen l1 is 1 m, the distance from the horizontal beam to the mirror surface is h 2 mm. Determine (1) what will be observed in point A of the screen—strengthening or
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lowering of intensity? (2) how will the intensity in point A be changed if a planeparallel glass plate (n 1.55) of thickness d 6 m is placed perpendicular to the path of the horizontal beam? Screen
l1
l1 S1
B
d
S1 A
h
A H l2
Mirror S2
Solution: Let us find the position of an imaginary image S2 of the source S1 in the mirror. Sources are coherent; therefore, at superposition of the waves coming from these sources on the screen point A there will be an interference. Strengthening or weakening of intensity depends on the difference of the length of the optical path of two beams; in other words, from the number of half-wavelengths stacked on an optical path difference : m /(/2)*; if is a whole even number, intensity will be maximal, if a whole odd number then intensity is minimal. (1) The optical path difference 1 will consist of both the geometrical difference l2l1 (both beams go in air) and the additional difference in /2 appearing on reflection from the mirror. Therefore,
1 l2 l1
2
ⴱⴱ
.
Since , l2 艎2 H 2, 2 2 ⎡ ⎤ ⎛ H⎞ ⎛ H⎞ ᐉ 2 ᐉ 1 ᐉ 1 1 ⎜ ⎟ ᐉ 1 ᐉ 1 ⎢ 1 ⎜ ⎟ 1⎥ . ⎢ ⎥ ⎝ ᐉ1 ⎠ ⎝ ᐉ1 ⎠ ⎣ ⎦
Since (H/l1) ^ 1, therefore, for the root calculation we can use the approximation a 1 a 1 %. 2 Thus, ⎡ 1 ⎛ H ⎞2 ⎤ H2 ᐉ 2 ᐉ 1 ᐉ 1 ⎢1 . ⎥ 1 ⎜ ⎟ ⎢⎣ 2 ⎝ ᐉ1 ⎠ ⎥⎦ 2ᐉ 1
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After substituting this expression into **, we obtain
1
H2 . 2ᐉ 1 2
Then we have to decide commensurability of the last expression with (/2): H2 2ᐉ 2 H2 1. m1 1 ᐉ 1 2 Since H 2h, then
m1 4
h2 1. ᐉ 1
Executing calculations, we have m1 33. Therefore, in point A there will be a minimum of intensity. (2) The glass plate of thickness d introduced into the path of the horizontal beam will change the optical path length. The optical path length will be sum of the geometrical path length and optical path length of the beam in the plate itself: L (ᐉ 1 d ) nd ᐉ 1 (n 1)d. The optical path length difference is now
2 l2 L l2 [l1 (n 1)d ] , 2 2 or using *
m2
2
(n 1)d d (n 1) 1 m1 2 . 2 2
Executing calculations, we obtain m2 19.8. Since this number is nearer to 20 (even) than to 19 (odd) in point A there will be a partial increase of intensity.
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EXAMPLE E6.2 On a thick glass plate with refraction index n3 1.5 covered with a very thin film, the refraction index of which is equal n2 1.4, a parallel beam of monochromatic light ( 0.6 m) falls nearly along a normal. The reflected light, owing to interference, is maximally weakened. Determine the thickness of the film. p S1
S n1
1
D S2 C
A
d n2 > n1
Film
2' B
n3 > n2
Solution: Let us allocate a narrow light beam from the light falling on the film. The path of this beam in the case when the angle of incidence is is shown in Figure E6.2. At point A, the beam is in part reflected and refracted. The reflected beams AS1 and ABCS2 fall on a convex lens S1S2 and interfere at point F. As the parameter of refraction of air n1 1 is less than the refraction index of the film, which in turn is less than the glass refraction index, in both rays reflection occurs in total without phase change. Since the light is maximally weakening the optical path length l2n2 – l1n1 (AB BCn2 – ADn1 should be equal to an odd number of halfwavelengths (AB BCn2 – ADn1(2k 1) (/2). If the incidence angle tends to zero, AD → 0 and AB BC → 2d and we obtain 2dn2 (2k 1) (/2). Then the film thickness is d [(2k1)]/4n. Taking k 0, 1, 2, 3, we arrive at a number of possible values for d: d0
3 0.11 m, d1 3, d0 0.33 m,etc. 4 n2 4 n2 EXAMPLE E6.3
A monochromatic light of wavelength falls onto a nearly parallel glass wedge (two plate) with a very small wedge angle , normal to its sides. An interference picture appears. It consists of a sequence of light and dark strings (see Figure 6.9). On a length of wedge l 1 cm, 10 strips are observed. Define the refracting angle of the wedge (Figure E6.3).
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dk+10 −dk
6.2
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l
dk
k + 10
k
k+9
k+1
2
l
Solution: A parallel beam of light falling normally onto the wedge reflects from both sides of the wedge, upper and lower. Both beams are coherent and practically parallel; though an optical path length difference is created, therefore an interference picture is observed. Dark strips are visible on those sites of the wedge for which is multiple to an odd number of half-wavelengths: (2k 1)(/2) with k 0, 1, 2…. The geometrical difference of paths length is 2dn cos; losses of /2 appeared at the reflection from the upper side of the wedge and should be added. Therefore, for dark strips we have (2k 1)(/2) 2dkn cos (/2), where n (n 1.5) is refractive index of glass, dk is the thickness of the glass wedge in a point of dark strip. The incident angle is assumed to be zero. Therefore after simplification, we obtain 2dkn k*. Let the thickness for any dark strip be dk and the thickness of glass in the point k 10 is dk 10, the distance between them being l. Then ( expressed in radians)
dk10 dk . ᐉ
Calculating values of d’s from expression * and substituting them into the last expression we arrive at 5/(nl). The angle sought is then 2 104 rad. This angle in degrees is 2 104 2.06 105 41.2 .
6.2 6.2.1
AN INTERFERENCE
Superposition of two colinear light waves of the same frequencies
In Section 2.3.1 a summation of two oscillations of the same frequencies and propagating in the same direction was considered analytically and using a vector diagram. It was also shown in Section 2.8.3 that the oscillation intensity is proportional to the square of amplitude. Apply now the conclusions mentioned to the calculation of the light intensity at waves imposed in any fixed point of space; let it be x0. At the fixed coordinate the equation of a running wave
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E (x0, t) E0 cos ( t–kx0) transforms into the equation of oscillation E(t) E0 cos ( t ), where product kx0 is included as a certain number into the phase . A splintered wavetrain traveled in two parts, 1 and 2; then they superpose (Figure 6.6) enhancement or weakening of the light intensity. The expression for the square of total amplitude (2.3.1) is 2 2 E02 E0,1 E0,2 2 E0,1 E0,2 cos( 1 2 ).
(6.2.1)
Consider a number of cases of the addition of two light waves with equal amplitude (i.e., at E0,1 E0,2) at various phase differences ( 1 2) . If light waves from two different independent sources are summed, the average value of cos is equal to 0. This 2 means that if in expression (6.2.1), E 20 2E1,2 , i.e., the intensity of light will increase twice, there will be an enhancement of intensities. If remains constant the two waves are coherent (refer to Section 2.9.1). At (2m 1) (m is an integer), cos 1 and the resulting light amplitude and intensity in the given point are equal to zero. At 2m, cos 1 and the resulting light amplitude is twice as high as the amplitudes of each of the initial waves, i.e., an enhancement of amplitudes takes place; intensity thus grows four times. The phenomenon of redistribution of intensity of light in space on imposing two or several coherent waves is referred to as the interference. Interference phenomena (including diffraction phenomena) are the direct consequence, and proof of, the wave nature of light. 6.2.2
Interference in thin films
The practical realization of two coherent light sources is very difficult (it can be achieved, for instance, with the use of lasers). However, there is a relatively simple way to carry out an interference. It consists of splitting a single light beam into two components by reflection from a pair of mirrors and then superposing them in a single point; they will interfere, thus P1
n1 Ο
M
n2
P2
Figure 6.6 Reflection and refraction ray paths of two split parts of a single wavetrain in two media: 1 and 2; P1 and P2 are mirrors.
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a splintered wave “interferes with itself.” The basic scheme of such an experiment is submitted in Figure 6.6. At point O, on the border of the two media with the refraction indexes n1 and n2, a wave is splintered in two parts. With two mirrors P1 and P2 both parts go to point M at which they interfere. The speed of propagation of the two beams, due to different media properties, is c/n1 and c/n2. At point M, the two parts of the wave will superpose with each other with constant shift in time equal to S1/1 S2/2, where S1 and S2 are geometrical lengths of the path traveled by the two parts of the wave. The oscillations of electric field strength at point M will be E0,1 cos (t – S1/ 1) and E0,2 cos (t– S2/2). The square of the resulting oscillation amplitude at point M is ⎡⎛ S ⎞ ⎛ S ⎞ ⎤ 2 2 E0 E0,1 E0,2 2 E0,1 E0,2 cos ⎢⎜ 1 ⎟ ⎜ 2 ⎟ ⎥ . ⎣⎝ y1 ⎠ ⎝ y 2 ⎠ ⎦
(6.2.2)
Since 2/T and c/n, the expression in square brackets is equal to (2/cT )(S2n2–S1n1) (2/o)(S2n2–S1n1). The product of the path traveled S and refraction index n is referred to as optical path length denoted by . Keeping in mind that cT 0 (0 being the wavelength in vacuum),
2
. 0
(6.2.3)
This expression joins the phase differences and the optical length traveled in the splintered wave. defines the interference effects. Indeed, cos 1 corresponds to the maximum intensity since (2/0) 2m. From this, the condition of the intensity maximum can be derived:
m
(6.2.4)
The largest diminishing of the light intensity corresponds to cos 1, i.e., (2m 1). Then (2m 1) (2 /0) or
(2m 1)
0
2 .
(6.2.5)
It is easy to see that the summation of waves described above with fourfold enhancement of intensity corresponds to the displacement of the two “parts” of the splintered wave from each other by the difference in lengths equal to the integer wavelengths (or, accordingly, to the phase difference 2m). The complete extinction of the wave’s intensity is observed at the displacement of the two wave parts on the wavelength half (on an odd number of the wavelength half, i.e., (2m 1)). Consider as an example the interference of light at the reflection from thin films (or from a thin plane-parallel plate; Figure 6.7). The direction of a beam falling on the film is shown in the figure by an arrow. Splitting of the wavetrains occurs in this case at partial reflection
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of each part of it on the upper (point A) and the lower surfaces (point B) of the film. We shall consider that the light beam goes from air and leaves after a point B into air (with air refraction index equal to unity) whereas the parameter of this film material is equal to n. Every wave of the beam falling at an angle at point A is split into two parts: one of them is reflected (beam AD) and the other refracts (beam AB). At point B every wave of the refracted beam is split again: part is reflected from the lower film surface and part refracts leaving the film. At point C the wave is again split, but we will follow that wave part which leaves the film at the same angle as beam AD. The two reflected beams are gathered by a lens (not shown in the picture) at one point. Being parts of the same primary wave the beams are coherent and can participate in the interference, the intensity being dependent on the difference of their optical traveled lengths (or differences in phases). The phase difference in waves 1 and 2 is accumulated in traveling along path lengths AD and ABC. The optical path length is (AB BC)n – AD, where AB BC 2d/cos and AD 2d sin sin /cos . Remembering that sin n sin , then (2dn/cos ) (1 – sin2) or 2dncos . Since angle is usually given in problems but not , it is more convenient to present the value in the form
2 d n2 sin 2 .
(6.2.6)
When defining the conditions of light intensity (maximum and/or minimum), it was necessary to equate the value to the integer or half integer to the number of wavelengths (eqs. (6.2.4) and (6.2.5)). However, as well as estimating the optical path length difference
, it is also necessary to analyze the opportunity of the loss of half a wavelength during reflection. This depends on a specific condition, namely, whether the media from which the reflection occurs is more or less dense. So, if the film with n 1 is surrounded with air with n 1, the loss of half-wavelength occurs at point A (Figure 6.7). If the film is on a surface of a medium whose reflection index is higher than for the film material, the loss
D
A
C
d B
Figure 6.7
Ray paths in a thin film.
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of half a wavelength occurs at two points—A and B; as in this case the whole wavelength is accumulated, this effect should not be taken into account at all. It follows, that specific tasks demand individual consideration. The main principle consists in finding the whole difference of optical lengths , to consider the possible loss of half a wavelength at reflection, if necessary, to add (or subtract, it does not matter) it to (or from) and to bring it into correlation with the conditions presented. In the case of a film in air represented in Figure 6.7, the condition of maximum interference looks like
2 d (n2 sin 2 )1 2
0 m 0 . 2
(6.2.7)
Since the refraction index depends on the wavelength (see Section 6.5), the interference conditions are qualitatively different. Therefore, the film will decompose falling light in a spectrum, i.e., in falling white light the thin film always looks as if it has been painted. We all have met examples of this: observing multicolored soap bubbles or an oil stain on the surface of water. Consider now the example of a thin air wedge. This wedge is opposite to the thin film picture (Figure 6.8). A plate with well-polished surfaces lies on another, similarly perfect plate. At a definite place between the two plates a thin subject (e.g., a thin wire) is introduced, so an air wedge is formed. Consider a beam of light falling normally onto the upper plate. We shall accept that there is no divergence at surface points on reflection and refraction, keeping in mind that the wedge angle is very small. Admit that A is a point on the lower plate where the optical path length between plates is equal to integer m of wavelengths plus /2 (due to reflection from the optically more dense lower plate); two reflected waves are nearly parallel to each other. Suppose that there is the condition of maximum interference intensity at this point. An equation describes this condition (factor 2 appears because the beam runs the distance twice): 1⎞ ⎛ 2 m ⎜ m ⎟ . ⎝ 2 2⎠
(6.2.8)
If we look at this picture from above (for this purpose a simple optical system is required), it is possible to see geometrical strips in which, at certain m, light (or dark)
A
B
Figure 6.8 An air wedge.
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Figure 6.9 Lines of constant thickness.
strips are formed. Along this strip condition (6.2.8) is fulfilled, i.e., along it the backlash of air has the same thickness. Such strips referred to as strips of equal thickness. Provided that the plates are made carefully, the strips of equal thickness are represented by parallel straight lines. If, however, there are defects in the plates, the appearance of the strips changes appreciably, and the position and form of the defects develop clearly. Fringes of equal thickness are shown in Figure 6.9: in an air wedge a narrow stream of warm air is produced, the density of which and, accordingly, the refraction index, differ from the values for cold air. The curvature of the lines of constant thickness is visible. If a convex lens touches a perfect flat plate, at a favorable ratio of the lens curvature radius, light wavelength and the presence of an optical magnifying system, so-called Newton’s rings can be observed. They represent the fringes of equal thickness in the form of concentric circles.
EXAMPLE E6.4 A vivid example of strips of equal thickness is Newton rings. They appear when a lens of a radius of curvature R lying on a carefully processed glass plate is irradiated with monochromatic light in the wavelength . Determine the radius of the mth ring.
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incident beam R reflected rays 1
2
R−dm
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lens
dm glass (a)
(b)
Solution: A geometrical diagram and the interference picture are presented in Figure E6.4a. The condition of interference maxima is 1⎞ ⎛ dm ⎜ m ⎟ . ⎝ 2⎠ 2
The air refraction index is assumed to be unity. In Figure 4a, it can be seen that
dm R R 2 rm2 R R 1
rm2 R2
,
where r is the radius of the mth ring. Taking into account the fact that the radius of curvature R is much larger than the size of the interference picture, we can expand the last equation into binominal series limiting ourselves by two terms ⎡ 1 r2 ⎤ r2 d R R ⎢1 . ⎥ 2 ⎣ 2R ⎦ 2R
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The ring’s radii r can be deduced as 1⎞ ⎛ r ⎜ m ⎟ R ⴱ. ⎝ 2⎠ The character of rings—light or dark—depends on the loss of /2 at the reflection from the more dense matter: in each particular case this needs to be derived separately. In our problem, we use a light falling from above, therefore, there is only one reflection from a denser media (from a glass plate). Therefore, for the light ring 1⎞ ⎛ rm ⎜ m ⎟ R . ⎝ 2⎠ 2 An overall picture of Newton’s rings is depicted in Figure E6.4b. If the light is directed from below (and the results are observed from above), there will be no loss of a half-wavelength because the reflection from the denser media takes place twice: from the lens and from the glass plate).
EXAMPLE E6.5 Find the radii of the second r2 dark, and fifth r5 light Newton rings in a monochromatic light 0 0.56 m provided the lens radius is R 1.2 m. Solution: Using the star equations from the previous example, keeping in mind that the air refraction index is 1 and taking into account that we are first searching for the radius of the dark ring, we can write (2m1)(/2) (rm2 /R)(0 /2), there7 m 1.16 mm. fore, rm m R. Therefore, at m 2, r2 2 1 .2 5 .6 10 For the light ring, m0 (rm2 /R) (0/2) and rm (2 m 1)R ( /2 ). Executing 7 .5 .2 (5 .6 /2 )1 0 1.74 mm. calculations, we obtain r5 (2 1)1 The phenomena of interference find wide application in chemistry and the chemical industry. In particular, they are used in interferometry in defining the refraction indexes of substances in their three states: solid, liquid and gaseous. There is a large number of various interferometers which differ by their assignment. Let us illustrate the determination of the refraction index of substances by a simple interferometer intended for the measurement of the refraction indexes of liquid and gaseous substances (Figure 6.10). Two completely identical thick plane-parallel glass plates A and B are fixed in parallel to each other. The light from source S falls onto the surface of plate A at an angle close to 45°. As a result of its reflection from both sides
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S
l 1 A
K1
K2
B 2 L
P
Figure 6.10 The diagram of a simple interferometer.
of plate A, two parallel beams 1 and 2 are produced. Running through two identical glass cells K1 and K2, these beams fall onto plate B and are again reflected from both its sides and are gathered at a point P by a lens L. At this point, they interfere, and the interference strips are examined with an ocular, which is not shown in the figure. If one of the cells (e.g., K1) is filled by gas with a known absolute refraction index n1 and the second with a substance with a measured refraction index n2, the optical path length difference between plates will be equal to (n1n2) l, where l is the cell length. With the help of a special device, the displacement of the interference strips concerning their position with empty cells can be observed. Displacement is proportional to the difference (n1n2), which allows one to determine one parameter knowing another. We note that while there are rather low requirements as to the accuracy of the measurement of the strips’ position, the relative accuracy in defining the refraction indexes can achieve values of 106–107. This accuracy enables the study of small impurities in gases and liquids, measurement of the different dependences of the refraction indexes on temperature, pressure, humidity, etc. There are still many other designs of interferometer construction, intended for various physical measurements. In particular, using a specially designed interferometer, Michelson and Morley in 1881 established the independence of light speed from the speed of its source (refer to Section 1.6). Einstein took this fact as a principle of his Special Theory of Relativity.
6.3
DIFFRACTION
Diffraction is a set of phenomena arising from the propagation of light in a media with pronounced heterogeneity and consisting of light deviations from the laws of geometrical
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optics. Diffraction leads to light deviating from rectilinear distribution, bending around opaque obstacles and penetrating into an area of geometrical shadow. 6.3.1
Huygens–Fresnel principle: Fresnel zones
Taking into account experiments in which light exhibited its wave nature C. Huygens assumed that each point of a primary light wavefront serves in its turn as a source of secondary spherical wavelets. The new position of the wavefront will be the enveloping surface of these secondary waves (Figure 6.11); in turn, each point of the secondary wavefront is again the source of the next generation of waves and so on. In Figure 6.12, this principle is illustrated with an example where a light wave is passed through an aperture; it can be seen that, due to secondary waves, light can penetrate into the area of geometrical shadow. These phenomena are only exhibited in an appreciable measure
envelop of secondary waves at time instance t+dt
wavefront at instant of time t
r =c∆t wavefront at instant of time t+dt
Figure 6.11 Huygens principle (the wavefront at time instants t and t t is shown, each point is the source of the second waves).
diffracted beams
areas of geometric shadow
Figure 6.12 Diagram of penetration of the diffraction radiation into the area of geometrical shadow.
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when the diameter of the aperture is commensurable with the light wavelength: only in this case do the angles of diffraction appear to be measurable (see below). A.J. Fresnel enlarged this principle by assuming the same laws of interference which had already been developed for primary waves were also applicable to secondary waves. It is rather difficult to calculate the distribution of intensity in a diffraction picture. However, a method allowing an essentially simplified calculation of the diffraction effects, at least at a qualitative level, has been offered: Fresnel has suggested mentally breaking a wave surface into zones, the distance from respective points of which up to the sighting point differs from the previous one by /2. In this case light waves from the adjacent zones are in an antiphase (because of the shift by /2); this leads to the mutual cancellation of such waves; in other words, the adjacent zones extinguish each other. The method has been successfully used to solve different problems of wave optics, in particular, in the explanation of rectilinear distribution of light. We shall take advantage of this principle by considering the diffraction on a slit. Two kinds of diffraction can be distinguished: diffraction in parallel light rays from a plane front wave (referred to as Fraunhofer diffraction), and diffraction in converging beams (Fresnel diffraction). Here we will consider only the Fraunhofer diffractions. The scheme of this diffraction is presented in Figure 6.13: point S marks the light source, a condenser lens K provides a parallel light beam, and lens L with a focal length f concentrates the result of the diffraction at an angle at a screen point P. A central ray O and axis sin along which the figure is expanded are shown. 6.3.2
Diffraction on one rectangular slit
We use the Fresnel zone principle for qualitative consideration of the Fraunhofer diffraction on a single rectangular slit. By definition, each following zone extinguishes the previous one. This means that if, at the slit width d and wavelength at an angle , an even number
S
K
L f
screen P
0
sin
Figure 6.13 Diffraction in parallel rays (Fraunhofer diffraction).
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of Fresnel zones opens, all zones extinguish each other; in this direction the intensity of diffraction becomes equal to zero. On the contrary, if at another angle , an odd number of zones open, there should be a maximum in the spectrum. It can be seen in Figure 6.14 that if the condition is even, the zone number corresponds to the length of a segment MN dsin on which the integer of wavelengths is stacked (the even half-wavelengths). This is an indication of the minimum intensity. Mathematically, it looks like d sin ,
(6.3.1)
where k is an integer which shows the diffraction order. The maximum intensity appears when an odd number of half-wavelengths stack up in the segment MN d sin (2 1) . 2
(6.3.2)
In Figure 6.15, an experimental diffraction spectrum on a single slit is schematically depicted: at 0 in a direct beam the maximum is seen because only one zone is opened
d
N
M
Figure 6.14 Fraunhofer diffraction from a single slit (the angle corresponds to eight Fresnel zones half of them faintly marked in the figure).
I
− 2 d
− d
0
d
2 d
sin
Figure 6.15 Light intensity distribution after single slit diffraction.
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(the whole slit width). On changing in both directions the maximums and minimums alternate in an orderly way. The quantitative solution of this plot shows that ⎡ ⎤ A0 sin ⎢ d sin ⎥ ⎣ ⎦, A ⎡ ⎤ ⎢⎣ d sin ⎥⎦ the square of which gives the so-called interference function I (sin ) A 2 .
(6.3.3)
This function describes the intensity distribution at the diffraction on a single slit (Figure 6.15). At the diffraction on a single slit the intensity of diffraction rapidly decreases with angle. 6.3.3
Diffraction grating
A diffraction grating is composed of a large number of identical, regularly distributed alternating transparent strips on an opaque flat carrier. A constant of the diffraction grating b is the distance between corresponding points of two adjacent strips (Figure 6.16).
N N
M
Figure 6.16 Diffraction from a diffraction grating: e is a grating constant, N is the number of slits, eN is the total width of the grating, is the diffraction angle; the faintly marked zones in Figure 6.14 are now completely nontransparent.
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For simplicity we shall consider that the grating consists of transparent and opaque strips of identical width. We shall designate N as the number of transparent and opaque pairs. Then the general width of the grating will be bN. Compare the condition of minimum intensity for one slit (eq. 6.3.1) with that for diffraction grating. Imagine that we could close every second Fresnel zone at the diffraction on a single slit. Therefore all open zones, having no “antagonists,” make a full contribution to the diffraction spectrum. Figure 6.16 presents a scheme to illustrate this idea. What was the slit width d, is now bN, the minimum condition (6.3.1) transforms into the maximum condition eN sin or esin k/N , where k and N are integers. There can be two cases, the most important is when is divisible by N, i.e., when (/N) m, where m is a simple integer. In this case, a so-called main maximum of order m is obtained; it corresponds to diffraction maximum when all transparent slits are “in a phase.” The main maximum condition can be written as: b sin m.
(6.3.4)
The second case is when in expression (/N) both numbers are integers, but are not divisible by each other. This gives the so-called subsidiary maxima of small intensity, which are obtained due to diffraction only on a single-grating slit. As a result, the spectrum consists of a rear strong main and many weak subsidiary maxima, as shown in Figure 6.17a. The intensity of the diffraction maxima (eq. (6.3.2) and (6.3.3)) increases N2 times in comparison with one slit, and the maxima width decreases by 1/N. The condition of the main maximum (6.3.4) is of primary importance. It shows that for a given diffraction grating (at fixed b), a different wavelength gives maxima at different points of the spectrum. This is the basis of the use of diffraction gratings in optical spectroscopy.
m=−3
m=−2
m=−1
m=0
m=1
m=2
m=3
a
m=4
m=−4 m=−3 m=−2 m=−1
b
m=2 m=1
m=3
m=0 −60
−40
−20
0
20
40
60
Figure 6.17 Spectrum of white light as viewed in a grating instrument. The different orders of spectra identified by the order number m are shown separated vertically. The central line in each order corresponds to 0.55 m.
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If white light falls on a diffraction grating each of the main maxima is broadened. The width of a white spectrum is defined by boundary wavelengths 0.4–0.7 m of visible light. Spectrum of adjacent orders can sometimes overlap. This overlapping is shown in Figure 6.17b where, for clarity, spectra of different orders are given on a different vertical level. It can be seen that spectra of zero-, first- and second-orders exist separately, whereas spectra of the third- and fourth-orders are partially overlapped. Notice that a diffraction grating can also be used in a reflecting position. 6.3.4
Diffraction grating as a spectral instrument
Spectroscopy is the method of studying the composition and structure of a substance or the control of technological processes (refer to Chapter 7, Section 7.8). The main stage in spectroscopy is the decomposition of electromagnetic radiation in a spectrum on the wavelength or frequency. Optical spectroscopy deals with the optical range of electromagnetic radiation, including UV and IR. The basic units of optical spectrometers are either a prism or a diffraction grating. The most important characteristics of the quality of a spectral device are dispersion D and resolution R. Distinguish an angular and linear dispersion. The value numerically equal to the ratio of the angular distance between spectral lines to the difference of wavelengths of these spectral lines is referred to as angular dispersion. The angular dispersion is equal D
,
(6.3.5)
In order to obtain an expression for angular dispersion, we should find a derivative d /d from eq. (6.3.4) and change further the differentials into finite increments (neglecting the minus sign). At small angles, eq. (6.3.4) can be rewritten as b ≈ m and then m . b
D
(6.3.6)
Linear dispersion Dl is the value numerically equal to the ratio of the linear distance in the spectrum between the spectral lines l to the difference corresponding to those lines Dl
l ;
(6.3.7)
m , e
(6.3.8)
and at small Dl f
where f is the focal length of lens L (Figure 6.13). From the above formulas, it can be seen that the dispersion (both angular and linear) is larger for higher order of the spectrum m.
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Another device that permits the decomposition of incident radiation in the spectrum is a prism. Decomposition in the spectrum by a prism is due to the dependence of the angle of refraction on the wavelength. The corresponding formula for the prism can be obtained using expression (6.1.2). Notice that the sign of the derivative / for the diffraction grating and for the prism is different. On wavelength resolution (resolving power) of the spectral device, there is a minimal distance at which two close spectral lines are accepted as being two instead of seeing them as one single widened line (Figure 6.18). Rayleigh has offered a criterion by which the spectral lines are considered as resolved if the middle of the maximum position of one line coincides with the edge of the adjacent line (Figure 6.18, center). The resolving power R of a spectral device is a dimensionless reversed value of the wavelength difference of the resolved neighboring lines to the wavelength of one of them: the value is R
.
(6.3.9)
Using the Rayleigh criterion, we arrive at the expression R mN .
(6.3.10)
It can be seen that the resolving power is larger when a longer grating length and higher order reflections are used. In Figure 6.19, two spectral lines obtained with three different diffraction gratings are presented. Gratings I and II are characterized by identical resolution (lines have identical half-widths) but provide a different dispersion, whereas gratings II and III have different resolution (maxima have different half-widths at identical dispersion).
Rayleigh’s criterion
0 (a)
0 (b)
0 (c)
Figure 6.18 Image of two distant point objects formed by a converging lens; (a) the angular separation of the objects is so small that the images are not resolved, (b) the objects are farther apart and the images obey Rayleigh’s criterion of resolution, (c) the objects are still farther apart and the images are well resolved.
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385 1
2 I
1
2 II
1
2 III
Figure 6.19 The intensity patterns of light with wavelength 1 and 2 incident on the different gratings; grating II has the same resolution as I however higher dispersion, grating III – the same dispersion but lower resolution.
6.3.5
X-ray diffraction
X-rays, discovered by W.K. Röntgen in 1895, as well as visible light are both electromagnetic waves; however the X-ray wavelength is 103–104 times shorter (about 1010 m, i.e., 0.1 nm). This circumstance defines their high penetrating ability, which the great majority of mankind has experienced during medical inspections. Our interest here is in X-ray diffraction (XRD) in crystals. For experimental observation of diffraction, the radiation wavelength should be of the same order of magnitude as the diffraction grating period. This follows from eq. (6.3.4): in order to measure diffraction angle the ratio mb/ should have the order of unity, i.e., b should be commensurable to . Therefore, to observe XRD using diffraction gratings is extremely difficult in practice. At the same time, a diffraction grating with a period of about 1 Å has the nature that the interatomic distances in crystals are about this size. As the interatomic distances are approximately 1010 m and the size of even the smallest crystal is 107 m (repetition is 103 in the majority of cases), the crystal can be considered infinite. If a beam of X-rays falls on a crystal, under the action of an electromagnetic wave the atoms’ electrons begin to oscillate and scatter secondary radiation of the same wavelength in all directions (compare with Compton-effect, Section 6.6). As the atoms in a crystal are ordered, these secondary waves are coherent and interfere; this defines the diffraction effect. The diffraction problem of X-rays in crystals “on transmission” has been solved by M. Laue. However, a more evident picture has been given by W.L. Bragg and also independently by G.V. Wulf. Formalizing the picture described above, they reduced the scattering of secondary waves to the X-ray reflection from so-called crystallographic planes (see Section 9.1). (These are planes drawn through the nodes of a crystal lattice.)
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(a)
(b) 1 2
A
C
B
d
AC+CB= 2d sin
Figure 6.20 The Bragg’s law: (a) the ordered atom’s array and two arbitrary crystallographic planes, (b) an incident X-ray beam scattered by the entire family of crystallographic planes, the X-ray’s paths difference ACB is marked.
The atoms of a crystal and two most rational crystallographic planes are shown in Figure 6.20a. An X-ray beam falls on a crystallographic plane at an incident angle . Because of its high penetrating ability, the X-ray radiation passes into the crystal without refraction (the refraction index n 1). Therefore, the difference in the lengths traveled by waves 1 and 2 can be easily counted, making 2d sin , where d is the distance between the nearest parallel planes (interplanar distance). The maximum of intensity will be observed if this difference is equal to an integer of wavelengths (refer to Section 6.2): 2 d sin m,
(6.3.11)
where m is the reflection order. This formula is referred to as the Bragg formula. Knowing the arrangement of atoms in a crystal, it is easy to calculate the intensity of X-ray reflection. More difficult, however, is the problem of calculating the arrangement of atoms in a crystal from an experimentally measured diffraction picture. This problem is the essence of modern X-ray crystal structure analysis for which M. von Laue (1914) and W.L. and W.H. Bragg (1915) were awarded Nobel Prizes.
6.4 6.4.1
POLARIZATION
Polarized light: definitions
An important feature of a wave beam is its polarization. A wavetrain has two mutually perpendicular planes in which oscillations of the vectors of E and H takes place (Figure 5.45). It has already been mentioned that the action of an electromagnetic wave is defined mainly by the vector E. Therefore, vector H in many cases is neglected in drawings, whereas the plane of E vector oscillation is referred to as the plane of oscillations. The wavetrain is, therefore, linearly polarized, i.e., it possesses a single plane of oscillation (Section 2.8.1; Figures 6.2a and 6.21b and c).
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Because all atoms of a source emit electromagnetic waves independently, the beam consists of large numbers of independent wavetrains; their planes of oscillations are not correlated, such light being referred to as nonpolarized or natural. In this case, axial symmetry of the oscillation planes disposition takes place (Figure 6.2b and 6.21b and c). The direction of the axis of symmetry coincides with the direction of the wave propagation. If there is a partial infringement of the axial symmetry, the light beam is partly polarized (Figure 6.2b and 6.21b and c). The following designations are accepted in the schematic representation of light polarization in physics literature (Figure 6.21). The plane of oscillations of the vector E is set by arrows. The polarized beam is represented accordingly by a number of parallel arrows. If the plane of oscillations is perpendicular to the drawing plane, arrows are projected in points. A nonpolarized beam is represented by alternate points and arrows. 6.4.2
Malus law
There are devices called polarizers, sensitive to polarization of a light beam. These devices freely transmit the incident electromagnetic waves with a plane of oscillation parallel to the plane of the polarizer, and completely absorb oscillations perpendicular to this plane. Hence, behind the polarizer the natural light becomes polarized with the plane of oscillation parallel to the plane of the polarizer. If, in the way of this secondary beam, a second polarizer is installed with a plane perpendicular to the first, it will detain the first polarized beam completely. This second polarizer is in the position of an analyzer; it is sensitive to the degree of polarization of the light beam. What will happen if the plane of oscillations in the beam makes an angle with the plane of the polarizer? Let the plane polarized beam falls on the analyzer with the oscillation plane oriented at an angle relative to this plane and the plane of the polarizer. Separate the E0 vector into two components: parallel and perpendicular to the polarizer planes (Figure 6.22). The perpendicular component will be completely absorbed by the polarizer, whereas the component of the electric field in the parallel position will be equal to E0 cos , and the corresponding intensity will be I ( ) E 20 cos2 I 0 cos2 .
(6.4.1)
(a) Natural light
(b) Partly polarized light
(c) Plane polarized light
Figure 6.21 Schematic representation of light polarization: (a) natural light, (b) partly polarized light and (c) completely polarized light.
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2 I ( ) E0
1 I0
Figure 6.22 Illustration of Malus’ law: (1) the polarization device and (2) the plane of polarization.
This equation is called Malus’ law. It can be seen that if the wave’s plane of oscillation is parallel to the polarizer axis, the beam will pass through with no intensity loss ( 0, cos 1). In contrast, at /2 cos 0 and light will be absorbed completely. If natural light falls on a polarizer the intensity of the passed light is proportional to average value of cos2 ; since in an interval 0 /2 the value cos2 is equal to 1/2, Ipol. Inat/2: intensity of light passed through the polarizer is a half of that of the incident natural light.
6.4.3
Polarization at reflection: Brewster’s law
If natural light falls on the border surface of two media the reflected and refracted beams are partly polarized. This occurs because of the fact that from a dielectric surface only the component of the E vector which is parallel to the border surface (perpendicular to the incidence plane) is reflected. Then, in the reflected light the oscillations perpendicular to the plane of incidence will predominate, whereas in the refracted beam the oscillations parallel to the plane of incidence will prevail. It has been experimentally established that when reflected from a dielectric surface light is completely polarized if there is a certain relationship between the incidence angle and the refraction index: tan B n.
(6.4.2)
Here the angle B is referred to as the Brewster angle and the given reflection is known as Brewster’s law. However, the refracted beam is polarized only in part. When the beam falls on the two-phase border at the Brewster angle, the angle between the reflecting and refracting beams is equal to /2 (Figure 6.23). Indeed, as tan (sin /cos ) and (sin /sin ) n to satisfy Brewster’s law (tan B n) it turns out that cos B sin , which is possible only when B /2.
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Incident natural ray
Reflected planepolarized ray
B =
Refracted partly-polarized ray
Figure 6.23 Illustration of Brewster’s law.
EXAMPLE E6.6 A beam of natural light falls on a polished glass surface plate submerged in a liquid. The beam of light reflected from the plate is at an angle of 97° to the incident beam. Define the refraction index of the liquid if the reflected light is completely polarized (Figure E6.6).
1
1'
n1
n2
2'
Solution: According to Brewster’s law, when reflected from a dielectric light is completely polarized if the tangent of the incidence angle is equal to tan 1 (n2/n1) n21 where n21 is the relative index of the second body (glass) relative the first (liquid). The relative refractive index n21 is the ratio of absolute indexes, i.e., tan 12 (n2/n1). The reflected beam makes an angle of 2 and, consequently, tan ( /2) (n2/n1). Therefore, n1 n2/tan ( /2) and the refraction index n1 is 1.33.
6.4.4
Rotation of the polarization plane
When passing plane-polarized light through some substances, the plane of polarization can change its position in space, namely, it rotates around the light wave vector k. Substances
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possessing such properties are referred to as being optically active. Among optically active substances there are many anisotropic crystal (i.e., whose structure does not relate to cubic and hexagonal systems, see Section 9.1) and liquids (e.g., turpentine, nicotine, solutions of many organic and inorganic substances in inactive solvents, etc.). Experience shows that the angle of rotation of the polarization plane around the wave vector k in optically active media is proportional to the length l, traveled by a beam in a sample l
(6.4.3)
The coefficient , generally dependent on the wavelength, is referred to a rotation constant and is expressed in angular degrees on millimeters of distance run. In solutions of optically active substances the angle of polarization plane rotation is proportional to the length traveled l and concentration of the active substance c: [] c l,
(6.4.4)
where [] is the specific rotation constant. So, knowing [] and having measured l, it is possible to define the concentration of an active substance in a solution. The direction of rotation of the polarization plane depends on the substance: if the plane of polarization turns clockwise in relation to k, the substance is referred to as a right-hand (or dextrorotatory); if it turns anticlockwise, the substance is a left-hand (or laevorotatory) substance. Thus the direction (the wave vector k) and the beam direction of rotation in a dextrorotatory substance forms a left-hand system, and in a laevorotatory substance forms a right-hand system. For an explanation of the rotation of the polarization plane it is supposed that plane polarization in inactive substances is the superposition of two oppositely directed circular polarizations with identical amplitude and angular velocity. In Figure 6.24, a scheme explaining this supposition is given. On the left, vectors E1 and E2 rotate around the k vector in opposite directions with equal angular velocities, therefore the total vector E oscillates in the vertical plane. If the angular velocities differ, the plane of oscillations turns around k (Figure 6.24, on the right). The angular velocity’s characteristic in optically active substances is caused by an asymmetric arrangement of atoms in molecules and crystals. In Figure 6.25, an example of a hypothetical tetrahedron in two various enantiomorphous forms is depicted. In the center of the tetrahedron is an atom (e.g., carbon as a complexation atom, not shown in the picture), and in the vertexes various atoms are arranged A, X, Y and Z. If the tetrahedron is looked at from above at a detour alternation XYZ (Figure 6.25a), a clockwise motion takes place. The tetrahedron in Figure 6.25b is a mirror image of the one Figure 6.25a. Such molecules are referred to as enantiomorphic. Therefore, if a substance with tetrahedrons of a-type in the structure is, for instance, dextrorotatory, an isomer with tetrahedrons of b-type is a left-handed isomer. Research into the effects of rotation of the polarization plane is one of the methods of structural chemistry.
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P'
P E E1
P
E E2
E2 E
Figure 6.24 The polarization plane rotation.
A
A
Y
X
Y
X (b)
(a) Z
Z
Figure 6.25 Enantiomorphic atomic arrangement.
6.4.5
Birefringence: a Nichol prism
Let us consider now a phenomenon known as double refraction in anisotropic crystals. In the XVIIth century, Huygens discovered that light passing through some crystals is split into two beams (Figure 6.26). One passes through the crystal in strict conformity with the laws of geometrical optics and is referred to as an ordinary beam (marked on the figure by the letter “o”). The other beam is called an extraordinary beam (marked on figure by the letter “e”); it passes the crystal’s surfaces with an infringement of the law of refraction: i.e., it cannot lie in one plane with an incident beam and a normal to an interface. The important thing is that both beams are completely polarized in mutually perpendicular planes. This is the basic, and most practically important, property of birefringent crystals. In the crystals described there are one or two directions along which the double refraction does not occur. These directions are referred to as the optical axes of a crystal (in Figure 6.26 and further defined by line MN). Certainly, they are determined by the atomic structure of a crystal. If the crystal has one such direction it is referred to as a single-axis crystal; there are also biaxial crystals with two such directions. Any plane which runs through the crystal’s optical axis is referred to as the main section or the main plane. Most interesting is the main section containing the light beam. The plane of the vector E oscillations in an ordinary beam is perpendicular to the main section and in extraordinary beam lies in the main plane.
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The properties of isotropic media (including optical properties) are identical in all directions. Nearly all gases and liquids and highly symmetric crystals with cubic and, in part, hexagonal structure (see Chapter 9.1) are referred to as isotropic ones. With anisotropic crystals light interacts differently than with isotropic media. Remember that the refraction index n defines the light speed in a medium c/n. For weakly magnetic substances (M ≈ 1), it is connected to the dielectric permeability of the medium by an equation n . In anisotropic crystals the dielectric permeability depends on direction. In particular, in the optic single-axis crystals the dielectric permeability in the direction of the optical axis and perpendicular to it have different values, || and ⊥ respectively. In other directions has intermediate values. If we draw a sketch of values in a single-axial crystal for different directions by segments from an origin, the ends of these segments form a rotation ellipsoid. Its axis of symmetry will coincide with the crystal optical axis. In Figure 6.27, an ellipsoid of the dielectric permeability of a single-axis crystal is presented. Because the light speed in a substance depends on the dielectric permeability , the given scheme also represents a diagram of the dependence of on the crystal direction. In this case this figure is called an indicatrix of speeds. Since the light speed does not depend on direction in isotropic media, the indicatrix is represented by a spherical surface. In anisotropic crystals, the properties of which depend on direction, the indicatrix differs from a spherical one. Moreover, they can differ for ordinary and extraordinary
M e 0
N
Figure 6.26 The birefraction of natural light by a single crystal of spar CaCO3: MN is the optical axis, o is the ordinary ray (o-ray), e is the extraordinary ray (e-ray). y g
S
x x
z
Figure 6.27 Huygens wave surfaces generated by a point source S embedded in calcite.
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beams. Therefore, two indicatrixes exist: spherical for ordinary beams and as rotational ellipsoids for extraordinary beams. They appear to be “inserted” in each other. Both indicatrixes touch in the direction of an optical axis because in this direction they have an identical light velocity. In a perpendicular direction, both indicatrixes differ maximally. In single-axis crystals there are two opportunities: in optically positive crystals the velocity of extraordinary beam e is less than that of ordinary 0, in optically negative crystals e 0 (Figure 6.28). By taking into account the difference in the optical properties of the crystal, it is possible to find the refraction of all rays in all directions graphically. Nichol suggested a relatively simple method of making a completely polarized beam; the method is based on arranging two split pieces of a crystal of Iceland spar (CaCO3) in such a way that the beam transmitted through it is polarized. Such a device is now referred to as a Nichol (Figure 6.29). In order to obtain such a polarizer the single crystal of Iceland spar should be cut first into two pieces of proper orientation and then be stuck together by a special glue substance. This substance should have the refraction index n, lying in an interval between indexes n0 and ne of an initial crystal (n0 n ne); it is the Canadian balm. The angle of fall onto the plane of pasting is selected in order to make the ordinary beam undergo a total internal reflection (refer to Section 6.1) and then be absorbed by the frame of the prism. The extraordinary beam freely passes through the thin layer of balm and leaves the Nichol. Accordingly, it is completely polarized and can be used in optical measurements. υe
υo υo
υe
M
N
O
M
(a)
N
O
(b)
Figure 6.28 Ellipsoid index of the light wave velocity in a mono-axis crystal: (a) single axis positive and (b) negative. MN is the optical axis.
M 48°
B
C
22
68°
S
c A N
76.5°
o D
Figure 6.29 Schematic presentation of a Nichol. BD—the Canadian balm layer.
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A similar Nichol can be used as an analyzer. If two Nichols are one after another, one as polarizer and the other as analyzer, their rotation around the beam axis essentially influences the transmission of light through this double system. If both Nichols are installed identically the light transmission has maximal intensity, the rotation of the second Nichol (the analyzer) around the beam at a right angle (crossed Nichol) completely extinguishes light. If, however, an optically active substance is placed between the Nichol, the sample becomes visible in crossed Nichols. To determine the angle of polarization plane rotation it is necessary to turn the Nichol-analyzer at a certain angle to achieve extinction again. The angle of rotation will be equal to angle (Section 6.4.2). This fact relates to optical methods of substances research.
EXAMPLE E6.7 A plate of quartz of thickness d1 1 mm cut out perpendicularly to the optical axis of a crystal, turns a polarization plane of monochromatic light of a certain wavelength at an angle of 20°. Define: (1) what the thickness of the quartz plate placed between two parallel Nichols should be in order to extinguish the light completely; (2) what lengths a tube l with a solution of sugar of mass concentration C 0.4 kg/l should be placed between Nichols in order to obtain the same effect? The specific rotation of the sugar solution is 0.665°/(m.kg.m3) (refer to Section 6.6.4). Solution: (1) An optically active medium rotates a polarization plane at an angle d*. Therefore we can present the thickness of the quartz plate as d2 ( 2/)**, where 2 is an angle totally extinguishing the light ( 2 90°). The rotation constants can be found from * formula ( 1/d1). Substituting this expression into **, we obtain d2 ( 2/ 1) d1. Executing the calculations, we obtain d2 4.5 mm. (2) The length of the tube with the sugar solution can be found from the expression 2 []Cd which defines the sugar solution turning angle of the polarization plane 艎 2[]C. Substituting all known data we obtain l 0.38 m.
EXAMPLE E6.8 A parallel beam of light with a wavelength 0.5 m falls normal to a diffraction grating. A lens with a focal length l 1 m is behind the diffraction grating to project a diffraction picture on a screen (Figure E6.8). The distance between two first-order maxima is l 20.2 cm. Determine (1) the diffraction grating constant, (2) the specific number of grating grooves (on 1 cm), (3) the limiting diffraction maxima and their total amount N and (4) the angle max corresponding to this maximum.
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Diffraction grating
L
II
I
0 l
I
II
Screen
Solution: (1) The diffraction grating formula is d sin k. In our case k 1. sin ≈ tan (because L p l) and 2 tan l/L. Therefore, d·艎/2L . From this equation it follows that d 2L/艎. Executing calculations, d 4.95 m. (2) The specific number (on 1cm) of grating grooves is n (1/d) 2.02 103 cm1. (3) In order to find the number of diffraction maxima we need to calculate the kmax corresponding to kmax d/(sin max)*. kmax corresponds to max sin . Therefore, kmax 9.9; however for this number sin is larger than 1, therefore, kmax is 9. This value allows us to find the general number of diffraction maximums N. The obvious relation exists N 2kmax 1. Therefore, N 19. (4) max can be found from star relation: max arcsin(kmax/d) 65.4°. 6.5
DISPERSION OF LIGHT
All phenomena caused by the dependence of the refraction index on frequency (or on wavelength) are united under the name light dispersion. Dispersion is referred to as normal if the refraction index steadily falls with an increase of wavelength (dn/d 0) or, grows with an increase of frequency ((dn/d ) 0); otherwise the dispersion is referred to as being anomalous ((dn/d ) 0). A typical picture of dependence n() for a normal dispersion is given in Figure 6.30.
n
Figure 6.30 Normal dispersion: the relation between refraction index and .
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For a better understanding of the presence of abnormalities in the dependence n(), it is useful to consider the process of light interaction with atoms in a substance. In Section 4.2.4 in the analysis of electronic types of dielectric polarization, it was shown that the internal electric field in an atom linearly depends on displacement and creates a force that returns the charge to its initial position. This means that an electron, being forced out of its equilibrium position, begins to make harmonic oscillations with a frequency of its natural frequency 0. The strength of an electric field in a light wave acts with its own frequency . Under the action of an electric field, an electron starts to oscillate with frequency causing secondary radiation. The process of forced oscillation was considered in Chapter 2.7. There it was shown that the amplitude of the forced oscillations of a system A( ) depends inversely on the difference of squares of natural frequency and the frequency of the driving force ( 02 – 2). When 2 approaches 02, a sharp increase in oscillation amplitude can be observed (eq. (2.7.4) and Figure 2.15). Accordingly, the resonance is accompanied by an additional absorption of the incident wave; which is expressed in more or less sharp lines in spectra of absorption and emission. The refraction index n is expressed through dielectric permeability and magnetic susceptibility by an equation n ()1/2. In Section 4.2.5, it was mentioned that only electron polarization will be exhibited in light with a frequency ≈ 1015 sec1. Accordingly for substances with ≈ 1 (which is characteristic of the overwhelming number of chemical compounds) n2 . Taking into account that 1 and ℜ/0E (refer to Section 4.2.2, eq. (4.2.8)), we obtain n2 1(ℜ/0E). For dielectrics in this frequency range the polarization ℜ can be presented as: ℜ np (eq. (4.2.2)). In its turn an induced electric moment p, according to definition, is the product of the charge ⏐e⏐ and shoulder x of the induced dipole; then
n2 1 e
nx o E
(6.5.1)
The displacement value x can be found by solving a differential equation of forced oscillations. In this case the electric field strength is acting as the force: Ffrc ⏐e⏐E(t) ⏐e⏐E0cos t, a restoring force can be considered as an elastic force Frst x m 20x. Therefore, the differential equation for x determination has the form: ⎛ d2 x ⎞ m ⎜ 2 ⎟ m 20 x e E0 cos t. ⎝ dt ⎠
(6.5.2)
The solution of such an equation is given in Chapter 2.7 where it was shown that:
x
e E0 cos t m( 20 2 )
e E (t ) m( 20 2 )
.
(6.5.3)
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Substituting the result into eq. (6.5.1) we obtain
n2 ( ) 1
ne2 . o m( 20 2 )
(6.5.4)
At ^ 0 or p 0, the value n2 does not depend on and is near to 1. If the light frequency approaches 0, resonance phenomena appear and the refraction index rises significantly (Figure 6.31). In reality the picture, n2( ) is like that presented in Figure 6.32. It is important that the position of abnormal dispersion ((dn/d ) 0) coincides with the light absorption line in a substance.
n2
1
0
Figure 6.31 Graphic representation of the relationship of the square refraction index and the light frequency (as follows from eq. (6.5.4)). n2
1
0
Figure 6.32 The experimentally measured dependence of n2 versus the incident light angular frequency . The position of the anomaly dispersion maximum coincides with the absorption line position.
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The figure at which n 1, the light speed in a substance becomes larger than the speed of light c, should not confuse the reader. In this case the phase speed is considered, i.e., the speed of distribution of the given, constant phase. The wave energy and momentum are transferred by group speed which cannot be more than the speed of light (see Section 2.9.4).
6.6 6.6.1
THE QUANTUM-OPTICAL PHENOMENA
Experimental laws of an ideal black body radiation
The radiation of electromagnetic waves by a heated body refers to as thermal radiation. Any body radiates at any temperature, however at medium temperature range its intensity cannot be measured by an ordinary device at sure. For the quantitative characteristic of thermal radiation the concept of an emittance R is used: the emittance is referred as to the energy that is emitted by a unit surface of a heated body in all directions within a solid angle 2 (a half of full solid angle, i.e., one side of a plane) in a unit time in a whole interval of frequencies (wavelengths). Thermal radiation basically contains waves of all frequencies. When allocate an interval of frequencies d at temperature T, part of emittance dR corresponds to it: the wider d the higher dR. However the ratio between them is not linear, it depends on frequency (wavelength ). The value r, connecting dR with d depends also on radiation frequency and referred to as spectral density of emittance, i.e. dR(T ) r ( , T )d , or dR( , T ) r ( , T ). d
(6.6.1)
The total emittance R can be obtained by integration of function r( ,T ) over the whole interval of frequencies, therefore R does not depend on frequency and is entirely defined by temperature:
0
0
R(T ) ∫ dR( , T ) ∫ r ( , T )d .
(6.6.2)
Function r( ,T ) also describes of a body’s ability to emit thermal radiation. Basically, the curve of dependence r ( ) at the fixed temperature T can be experimentally measured (Figure 6.33). The figure shows: a muffle furnace with a radiating body inside, aperture in the furnace door and in a screen cutting out the desirable stream of
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heated body
screen
spectral unit
intensity final curve wavelength
Figure 6.33 The experimental scheme of the measurement of the spectral composition of heat radiation; furnace to heat a body, screen protector, spectral device to measure the radiation intensity; the experimental graph coordinates are shown.
the thermal energy directed further to a spectral device. A purpose of the spectral device is to decompose the whole falling radiation into a spectrum on frequency. Consequently the spectrum of thermal radiation is obtained which should be investigated and explained. To measure such curve represents significant difficulty as beams of different wavelength demand various techniques. In a result, the various parts of a general curve are “tailored”. Figure 6.34 schematically depicts the emission spectra of bodies heated to various temperatures, from almost room temperature to temperature of the sun surface. In the same figure the curve of luminosity presents (see also Figure 6.1) which enables one to know radiation at what emitter temperature is radiation perceived by eye as light. Part of a spectrum of thermal radiation can be felt as heat on the skin of the human – this is mainly IR radiation. UV radiation can also be felt by the skin – this we think of as sunburn. All this is an insignificant part of the general thermal radiation. Perhaps, only the sun radiation contains in its spectrum all the wavelengths that humans perceives with almost all their sensory organs. The curves can be schemed in coordinates r( ) and r() keeping in mind the relation 2c ,
(6.6.3)
In order to transfer from r() to r( ) one should to compare equivalent peaces of the graphs areas of both functions r ( )d r ( )d .
(6.6.4)
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r pick of 180 cm height
6000 K
3000 K , m
2000 K 1000 K
B
r
vertical scale drawn up in 250 times
V R
1000 K 800 K 400 K , m
VR
Figure 6.34 Spectral intensity distribution of the heat radiation at different emitter temperature: boiling water (400 K), electric heater (800 K), red incandescence (1000 K), blowtorch (2000 K), voltage arc (3000 K) and the sun radiation (6000 K).
Derivation of the eq. (6.6.3) gives d
2c 2 d d . 2c 2
(6.6.5)
(The minus sign in this expression specifies only that with increase one value another one decreases. Therefore the sign minus further will be omitted). Changing in the equation (6.6.4) d on d according to (6.6.5) we can obtain r ( )d r ()(2πc/ 2)d r()(2/2c)d , or, finally: r ( ) r ( )
2c 2 r ( ) . 2c 2
(6.6.6)
The radiation power flux d ( ) is the thermal radiation energy emitted in the unit of time from the surface dS of the radiated body d ( ) R( )dS.
(6.6.7)
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Falling on any surface, this flux will be partly absorbed (this part we will call the absorbed flux of energy d ( )), and partly reflected. The dimensionless value
a( , T )
d ( , T ) d ( , T )
(6.6.8)
is referred to as the absorbing capacity of a body a. In general cases, absorbing capacity a depends both on frequency and on temperature of the radiator. For a body that completely absorbs all the thermal radiation falling on it in all ranges of frequencies and at any temperature, the value a is constant and equal to 1; such a body is called a perfect black body (PBB). If the body’s absorbing capacity depends on frequency and/or on temperature, but is constant (less than 1), the body is called gray. The majority of bodies, however, are not PBB or gray, their absorbing capacity is less than 1 and depends on frequency and temperature; these are arbitrary bodies. There is a certain connection between the emitting and absorbing capacity of a body namely the Kirchhoff law: the ratio of the body’s emitting and absorbing capacity does not depend on the nature of the body but is identical, i.e., universal, for all bodies’ functions of frequency and temperatures (function f( ,T ) r ( , T ) f ( , T ). a( , T )
(6.6.9)
This function is referred to as Kirchhoff function. The Kirchhoff law defines one of the most important properties of thermal radiation, distinguishing it from other types of radiation (fluorescence, luminescent, etc.): thermal radiation is an equilibrium one. From eq. (6.6.9), it follows that the more a body absorbs, the more it radiates. Hence, in an isolated system of bodies their temperature will eventually be equalized, becoming identical. If a body absorbs more, it also radiates more. The values r( ,T) and a( ,T) can differ, but their ratio is identical. The analysis of the curves, similar to those presented in Figure 6.34, allows one to understand and formulate some laws of thermal radiation. Thus it has been experimentally established that emittance is proportional to the fourth degree of absolute temperature: R(T ) T 4 . For a gray body the given ratio can be rewritten: R(T ) aT 4 .
(6.6.10)
This is the Stefan–Boltzmann law. The value of constant in this law is experimentally established: 5.7 108 W/(m2 K4). The expression (6.6.2) shows that the area under the curves in Figure. 6.34 is proportional to the fourth degree of absolute temperature.
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From the same figure it can be seen that, depending on temperature, the radiation spectrum is shifted to the shorter wavelengths. This law can be written as max
b T
(6.6.11)
where max represents the wavelength corresponding to the maximum of radiation spectral distribution. This dependence is referred to as Wien’s law of displacement, constant b is equal to 2.90 103 mK. All the above-mentioned experimental facts and laws require a theoretical explanation. 6.6.2 Theory of radiation of an ideal black body from the point of view of wave theory: Rayleigh–Jeans formula Before examining the theory of thermal radiation, one should suggest the model of an ideal black body (IBB), i.e., a body that absorbs all falling radiation. The most simple, yet successful, is a model representing an almost completely closed cavity with a small aperture (Figure 6.35): all beams that get inside the cavity lose their intensity after consecutive reflections, and do not leave the cavity. Because the heated-up walls of the cavity are a source of thermal radiation and only an insignificant part of it leaves, a certain equilibrium density of radiation is established in the cavity. Standing waves (such as in a string (see Section 2.9.3) are produced in the cavity, the wavelength of which is defined by eq. (2.9.8). As for standing waves in a string, the maximum wavelength of a standing wave is determined by the size of the cavity. Also, the minimum length of a wave is determined by the discrete character of the material from which the walls are made. Therefore, the density of standing waves in the cavity is finite. After detailed consideration, one can obtain 2/42c2. Within the framework of classical physics, an oscillator with certain frequency can be put in conformity to each standing wave. According to the law of uniform distribution of energy on degrees of freedom (see Chapter 3), to every oscillator an average energy T can be attributed (because two (½)T goes to the oscillation degree of
Figure 6.35 Model of an ideal black body (IBB).
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freedom connected to kinetic and potential energy). Hence, the total energy of radiation of an IBB is r ( , T )
2 T . 4 2 c 2
(6.6.12)
or in terms of wavelengths
r ( )
2c T . 4
(6.6.12)
This formula was first suggested by Rayleigh and Jeans, and was irreproachable from the point of view of the wave nature of light. It equally concerns experimentally measured dependence of heat radiation of an ideal black body. At the same time, the theory and experiment are in a glaring contradiction with one another, mainly in the area of small wavelengths (large frequencies). From Figure 6.36b, it can be seen that at small wavelengths (or at increase of frequency, in other figures) the theoretical curve soars sharply upwards, whereas the curve achieved by experiment (Figures 6.34 and 6.36a) goes downwards. This also leads to the incorrect conclusion that the luminosity R (6.6.2) becomes senselessly infinite. Because of the area where the divergence takes place, the discrepancy between theory and experiment has been called an “UV accident,” thus recognizing the inability of the theory to explain the laws of radiation within the framework of wave theory as it existed at that time. A completely different approach to the theory of radiation was, therefore, necessary. One was proposed by W. Wien (awarded the Nobel Prize in 1911) who, on the basis of the laws of thermodynamics, obtained a “bell-shaped” theoretical curve. A revolutionary approach was suggested by Planck (1900), which resulted in the full agreement between theory and experiment.
r
r
r
h too small
"catastrophe"
λ
λ (a)
h=6.64×10−34 J.sec
(b)
h too big
λ
(c)
Figure 6.36 The spectral density of heat radiation r versus the wavelength : (a) experimental curve, (b) “ultraviolet catastrophe,” (c) the correspondence of the theory and experiment in the framework of Planck theory.
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EXAMPLE E6.9 The maximum of spectral density of the sun’s radiation emission corresponds to the wavelength 500 nm. Assuming that the sun radiates like an IBB, determine: (1) emittance of sun R* (R* means the emittance of IBB), (2) energy flux of sun radiation , (3) total mass m of electromagnetic radiation irradiated by the sun in one sec (refer to Section 6.6.1). Solution: (1) The Stefan–Boltzmann law describes the radiation emittance R* of an IBB R* T 4, where equals to 5.67 108 W/(m2K4). The sun’s surface temperature can be determined using the Wien’s law max b/dT, where b 2.90 103 mK. Combining these two formulas, we obtain 4
⎛ b ⎞ Rⴱ ⎜ . ⎝ max ⎟⎠ Executing calculations, we arrive at R* 6.4 107 W/m2. (2) The energy flux radiated by the sun is the product of radiation emittance and the sun’s surface area RS or 4r2R* where r is the sun’s radius. Substituting all data into the last formula, we arrive at 3.9 1026 W. (3) The total mass of electromagnetic radiation emitted by the sun in 1 sec can be determined using the correspondence mass and energy E mc2. The energy of the electromagnetic radiation in the time t is equal to the product of the flux and time E t. Therefore t mc2 and further m (t/c2). Executing calculations, we arrive at m 4.3 109 kg. 6.6.3 Planck’s formula: a hypothesis of quanta—intensity of light from wave and quantum points of view Analysis of the state of theory and experiment concerning IBB radiation and the mathematical description of the phenomenon led Planck to recall Newton’s hypothesis that light is a stream of particles (corpuscles); this had been rejected on the basis of successive works on interference and diffraction. Planck suggested a revolutionary idea: that each particle of radiation is a corpuscle or quantum, i.e., a particle bears a portion of energy h where h is a certain constant, and h/2. Then the total radiation energy flux should be expressed by the total number of quanta, , i.e., E Ni1ni i, where is the energy of a single quantum, ni is the amount of such quanta and N is their total number. The distribution of quanta on energy is set by the Boltzmann factor. According to the statistical method of average values calculation (see eq. (3.2.11)), we can write a similar expression for the average energy of a quantum oscillator having replaced integrals by sums. Therefore, following Planck, we arrive at: N
⎡ n ⎤ ⎥⎦ [exp(x ) 2 exp(2 x ) ] , 1 exp(x ) exp(2 x ) ⎡ n ⎤ exp ⎢ ⎣ T ⎥⎦
∑ n0 n exp ⎢⎣ T ∑ n0 N
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where x ( /T). In the last expression a fraction can be rewritten as d log[1 exp x exp 2 x ] dx d 1 log dx 1 exp x exp(x ) 1
from which the sought value is derived:
. ⎛ ⎞ exp ⎜ 1 ⎝ T ⎟⎠
(6.6.13)
Leaving unchanged the part of the calculation in which the oscillation density in the cavity of an IBB was counted, and attributing to every oscillator the above-mentioned average energy , Planck came to the formula of function r( ,T) for an IBB
r ( , T )
2 4 2 c 2
. ⎛ ⎞ exp ⎜ 1 ⎝ T ⎟⎠
(6.6.14)
The formula obtained not only correctly reflected an agreement between the theory and experiment, but also allowed the determination of the h value; in Figure 6.36c the theoretical results are “adjusted” to the experimental, from which the value of the constant h ( 6.626 1034 J sec) has been determined. This value was later named after M. Planck; it is typical that Planck’s constant has the dimension of the momentum or quantum of action (i.e., the product of energy and time of its action). Proceeding further to corresponding functions from wavelength ((6.6.1) and (6.6.6)), it is possible to obtain the dependences r and f on :
r (, T ) f (, T )
4 2 c 2 5
1 exp ⎛⎜⎝ 2c ⎞⎟⎠ 1 T
.
(6.6.15)
Planck’s formula well describes limiting transitions. So, at ^ T the exhibitor in the denominator of function (6.6.14) can be decomposed in a series and be limited by two terms. This leads to Rayleigh–Jeans formula which describes very well experiment in this area of frequencies. In contrast, at p T the unit in the denominator can be neglected and functions r and f fall according to the exponent that is found by experiment.
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The Planck formula suggests how to find numerical values of constants in Stefan– Boltzmann and Wien laws. In particular on integration of Kirchhoff’s law on the whole frequency range one can arrive at the Stefan–Boltzmann formula. The constant in Wien’s law b can be found by derivation of the Kirchhoff’s function on frequency and equalizing it to zero. We hope that readers can carry out these calculations themselves. Thus in the phenomena described, radiation is represented by a flow of corpuscles, quanta of energy which have been called photons. The energy of each photon is defined by product . Though for a long time it was known that in other experiments the same radiation manifests itself as a flow of waves. There is a problem, which has occupied the minds of many physicists, which is now known as particle-wave dualism and to which we shall pay more attention. Wave and quantum theory lead to completely different representations of the intensity of light. We should remember that intensity is understood as energy falling normally on a unit of area in a unit of time. Within the framework of wave theory, the intensity of a monochromatic beam of light is defined by the square of the wave amplitude and does not depend on frequency, i.e., I ~ A2. In quantum theory at a fixed wavelength (and, correspondingly, frequency) intensity is defined by the number of quanta, i.e., I ~ N . As will be shown below, a number of experiments have excellently confirmed Planck’s quantum hypothesis. Let us emphasize once again that the theory of thermal radiation became the starting point for quantum mechanics, which has subsequently received confirmation in many areas of physics. Laws of thermal radiation are widely used in technology to initiate and support chemical processes and to measure the temperature of bodies, contact with which is either impossible or complicated (e.g., measuring the temperature of the stars, the heated up gases during the launch of missiles, etc). These laws form the basis of optical pyrometry. It can be seen in Figure 6.34 that measurement of an integral of luminosity R (the area under the curves) can be a measure of body’s temperature. There are several kinds of pyrometry: one is based on color and another on brightness. The first is based on the position of the curve maximum and the second on the ordinate at fixed wavelength. Certainly, all the laws used here are only fair for IBB, however there are ways to account for the uncertainty arising in experiments. The stated theory of thermal radiation also allows an explanation of a phenomenon that has an influence on life on earth. This is the so-called green-house effect. The sun’s radiation (the spectrum is depicted in Figure 6.34), passes through open space, and reaches the external layers of the earth’s atmosphere, naturally with a loss of intensity, but without a special change of spectral composition. In the atmosphere there is selective absorption of the sun’s radiation by natural and industrial gases. This selectivity is defined by the structure of molecules, by their concentration and properties. It is natural also, that absorption of radiation depends on humidity, dust content and other properties of the atmospheric layers close to the surface of the earth. The sun’s radiation reaches the surface of the earth and heats it up and, together with the internal heat of the planet, defines the temperature of its parts (depending on geographical place). At the same time, the earth’s surface also radiates thermal energy. The temperature of the “radiator” in this case is essentially less than the sun’s temperature, accordingly the entire spectrum, under Wien’s law, is shifted to the long-wavelength region area (see Figure 6.34).
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This radiation is directed from the earth surface and should again penetrate through the atmosphere in the opposite direction. However, being long wavelength the earth’s radiation is absorbed by the atmosphere differently than the radiation from the sun. The transmission ability of long-wavelength radiation is less than solar radiation, the earth’s radiation is appreciably “absorbed” in the atmosphere, heating it up. This is the green-house effect. Taking billions of years to establish, the thermal equilibrium in the solar system defines life on earth. Every large-scale action can affect the established balance to some extent, displacing it in one way or another. In particular, the industrial activity of mankind leads to a change in the chemical composition of the atmosphere, increasing the concentration of industrial waste products. This change influences the absorption of the radiation falling to earth and leaving it. However, to a much greater degree, it concerns the radiation of the earth rather than that of the sun. All these events cause “over-warming” of the atmosphere and disturbance of the equilibrium. The effect is probably not so large: it is estimated at approximately 1–2°. The results, however, can be catastrophic. One example is the appreciable effect on the people living in those European countries that are below sea-level and protected from the sea by dams (e.g., Denmark, The Netherlands). The increase in atmospheric temperature can melt much more ice than would normally maintain the existing balance. The consequences are dangerous for large cities such as Venice, Saint Petersburg and will affect the climate of Florida and many other pearls of human civilization. In this connection it is also worth mentioning the so-called “ozone holes”—the local destruction of a centuries-old balance in the composition of the atmosphere resulting from the products of industrial activity (e.g., chlorofluorocarbons—freons) which create areas (holes) in the atmosphere that are transparent to short-wave UV radiations. These holes in the ozone layer are making the affected areas dangerous to live in because of the excess of UV radiation, which is harmful to life on earth. It is also probable that short wavelengths can cause undesirable mutations in living organisms. 6.6.4
Another quantum-optical phenomena
Planck’s hypothesis was confirmed and developed by Einstein’s theory of an external photo-effect. The photo-effect consists of knocking electrons out of the surface layers of some metals and oxides on their irradiation by quanta of electromagnetic radiation. The scheme of an experiment on the photo-effect is presented in Figure 6.37. The main part of the experimental equipment is a vacuum bulb with two electrodes C and A with a window allowing the irradiation of electrode C. The interaction of electromagnetic radiation of definite frequency and amplitude A results in knocking the electrons out of the surface of electrode C. A voltage difference is applied to the electrodes (Figure 6.38). First a negative pole is applied to K accelerating the kicked-out electrons. As the voltage increases, all the electrons reach electrode A, and saturation takes place (the horizontal line in the scheme). However, a device permits the signs of the electrodes to be changed; when a decelerating voltage difference is applied, part of the electrons are not able to reach electrode A. When the difference is zero only those electrons that have their own high enough kinetic energy
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can reach A. Furthermore, even at a coercive force some of the energetically active electrons can still reach A. Only a locked-out voltage can stop the current through the bulb. If the light intensity (i.e., the amplitude of the incident light, A) is increased and the same is kept, the saturation increases but the locked-in voltage remains the same. This means that it is not the light electric field amplitude (intensity) but the frequency which is responsible for kicking electrons out from the electrode and locking them. In fact, if the light frequency is increased, the locking in voltage also increases. The result obtained in the experiment is in agreement with the supposition that, in a given phenomenon in the photo-effect, light behaves as a flux of particles (photons). The results obtained regarding the volt–ampere characteristic shown in Figure 6.38 are deceptive. These results can be explained as follows. The energy of the falling quantum is transferred to an electron in photocathode producing the work A. Part of the energy is expended for the work A1 of moving the electron from the deep layers of the photocathode up to its surface, then in overcoming electron binding to the photocathode body A2; the remaining energy is left to the
Light
-
C
A µA V
R
−
+
B1 −
0
B2 −
+
+
Figure 6.37 Diagram of a device for the photo-effect measurements: an electric device permits to change an electric field polarity.
I
I
I
1
saturation
2
1= 2
1= 2 Locking voltage (lv)
V (a)
lv
V (b)
lv2 lv1
V (c)
Figure 6.38 The V–A characteristics of the photo-effect at different light frequencies: (a) typical V–A characteristic, (b) the same frequency and different light intensity, (c) the same intensity of light however different frequencies.
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photoelectron as its kinetic energy. According to the law of energy conservation we can write:
A1 A2
my 2 . 2
(6.6.16)
In this expression, work A1 is undetermined since it is not known at precisely what point of the cathode body the collision took place. The work A2 is a characteristic of every metal and oxide and is referred to as work function. If we exclude the unknown term from this line, i.e., remove the term A1, the equation become valid only for those electrons which, at the moment of collision, were on the surface of electrode A; they have the highest possible kinetic energy. Therefore:
A2
2 m y max . 2
(6.6.17)
It is possible to determine the kinetic energy of the photoelectron using the experimental value of the locking out voltage. Then we can write:
eU loc
2 m y max . 2
(6.6.18)
It follows from this equation that there is a limiting frequency k below which the photo-effect in a given photocathode disappears completely. In fact, the quantum k does not have enough energy to tear an electron out of the surface of electrode C. This happens when is lower than the work function. The so-called photo-electric threshold takes place at k A2. In addition to the theory of an IBB, it has been proved in quite another experiment that, in some circumstances, light behaves as a particle flux rather than a wave. Albert Einstein was awarded a Nobel Prize in 1921 for his outstanding contribution to physical science in general and especially for the photo-effect theory, which belonged mainly to Einstein and provided convincing confirmation of Planck’s hypothesis energy quanta not only in the theory of heat radiation but also in some other physical events. Einstein was also responsible for the concept of a photon which is widely used in modern physics. A short-wave border of X-ray radiation is another phenomenon, which supports the quantum idea. Discovered by W.C. Röntgen in 1895 and referred to as X-rays, this is the electromagnetic radiation with a wavelength of the order 1010 m (see Chapter 5, Table 5.3), arising on the electron transition in atoms and also on electron movement with acceleration (linear or centripetal). In the majority of countries this radiation is referred to as X-rays as it was called by Röntgen himself, but in Germany and Russia the term “Röntgen rays” is used. The generator of such radiation is the X-ray tube, the principle scheme of which is given in Figure 6.39. There are two electrodes in vacuum glass cylinder. The cathode represents a heated string and the anode is a massive metal cylinder, compulsorily cooled from inside by
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flowing water. The cathode’s task is to emit electrons. A potential difference from tens to hundreds of keV (and more) is applied between the cathode and the anode. This electric field accelerates all emitted electrons up to an energy of 10–100 keV. The main part of the electron energy, allocated in the anode as heat, is taken away by flowing water. The remaining energy is used in the excitation of X-rays. Two kinds of X-ray radiation are known. Characteristic radiation results from the return electron’ transitions from excited to ground state levels in atoms. This radiation has a linear spectrum and is widely used in modern science and technology for the analysis of chemical structures (refer to 7.6.4). Bremschtralung radiation arises at the instant of the electron stopping in the anode substance. According to classical theory, the distribution of a frequency (wavelength), arising due to electron stopping X-ray radiation should cover a wide range of spectrum from zero to infinity (as in thermal radiation spectra). Experimental results contradict this supposition: in Figure 6.40 the X-ray intensity versus their wavelength is plotted, the curves sharply terminating at the shortest wavelengths. An explanation of this fact can be found in the quantum theory of radiation. The law of energy conservation in this case can be written as: N
eU ∑ i .
(6.6.19)
i1
The electron energy before impact with an anode is written on the left-hand side of this expression; on right-hand side is the sum of all the photons’ energies, which appeared on collision. Since the process of electron braking is uncontrollable in this process, photons of all energies are produced, and the spectrum contains all wavelengths (so-called “white” spectrum). However, a limiting case exists when an electron gives all its energy to produce only one single unique photon. In this experiment this photon possesses the largest energy. It defines the boundary value of the wavelengths; photons of larger energy (smaller wavelength) in the bremschtralung X-ray spectrum cannot appear. For such a photon, expression (6.6.19) becomes simpler eU max
2c , min
(6.6.20)
cathode anode cooling water device to heat cathode X-rays high voltage transformer ground
Figure 6.39 Scheme of an X-ray tube source.
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10 50×8 8 40×8 I
6
4 30×8 2 20×8
r 0.02
0.04
0.06 .Å
0.08
0.10
Figure 6.40 The relation of the X-ray bremschtralung short-range limit versus wavelength .
whence min
hc . eU
(6.6.21)
The expression excellently coincides with experiment. The presence of a short-wave limit of X-ray radiation in the X-ray tube spectrum is a fact that cannot be explained by wave theory; in this experimental arrangement, the photon with larger energy cannot appear under the law of energy conservation: all the electron energy has already been given to the single X-ray quantum, the photon with smaller wavelength (greater energy) simply has no electron energy to appear; The Compton effect is another phenomenon contradicting classical wave theory. This effect arises on X-rays scattering by electrons weakly bonded to atoms. The scheme of the experiment is given in Figure 6.41. A beam of monochromatic X-rays (with wavelength 0) falls through a collimator onto a sample and is scattered. A special device investigates the intensities of both incident and secondary radiation scattered at an angle . Proceeding from wave theory, it follows that the scattered radiation should contain only one wavelength: the one that falls on the sample, i.e., 0. In fact, the electric field of an electromagnetic wave in the X-ray range should oscillate the electrons, which in turn should radiate secondary waves of the same wavelength. However, in the scattered radiation, experiment reveals that besides one unshifted 0, there is one more component, referred to as a shifted component with wavelength greater than 0 (Figure 6.42). It is experimentally established that the value of displacement (shift) 0 does not depend on the sample material and that this displacement is greater, the larger the scattering angle , namely,
0 (1 cos ) 2 0 sin 2 . 2
(6.6.22)
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Pe X-ray source
ricoil electron
sample
hk
diafragms spectral unit hk'
Figure 6.41 The Compton scattering experiment.
It is possible to explain all the features of the Compton effect if we consider them as a process of elastic collisions of X-ray photons with peripheral atomic electrons (in terms of the theory of particle collisions, see Section 1.4.5). We recall that, in elastic collision, both conservation laws (kinetic energy and momentum) are valid. Since a feedback electron can have a speed commensurable with the speed of light, it is more appropriate to use the relativistic theory (refer to Chapter 1.6) for the analysis. At an initially rested electron, weakly bonded to an atom (with its kinetic energy and momentum practically equal to zero), the photon falls with energy and a momentum k. In this case, the above-mentioned conservation laws in this case look like: h h (m m0 )c 2
(6.6.23)
⎛ ⎞ hc hc 1 m0 c 2 ⎜ 1⎟ . ⎜⎝ 1 2 ⎟⎠
(6.6.24)
the energy conservation law, and
the momentum conservation law, /c. For the feedback electron, having lost its bonding to the atom, the momentum relativistic expression pe can be given as: pe
m0 y 1 2
,
(6.6.25)
where is its speed. Using a vector diagram (Figure 6.43), projecting electron and photon momentums on x-axis, we obtain h h cos pe cos ,
(6.6.26)
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= 0.7 Å
= 90°
= 0°
ns
ns
sh
= 35° ns
= 135°
sh
ns
sh
Figure 6.42 Results of the measurement of the Compton effect versus the scattering angle : ns, nonshifted component; sh, shifted component.
and on the y-axis
0
h sin pe sin .
(6.6.27)
Exclude from the last expressions the electron parameters and we arrive at the final expression:
h (1 cos ) (1 cos ). m0 c
(6.6.28)
The expression h/m0c is referred to as Compton wavelength and is denoted as Λ. The same results can be obtained in the framework of nonrelativistic physics though the Compton effect belongs to the relativistic case. Both arrive at the same result, but the nonrelativistic derivation is simpler. However, when examining the feedback electron, it is necessary to use relativistic theory. For his discovery and explanation of the effect, A. Compton was awarded the Nobel Prize in 1927. At present this effect is used for the study of atomic valence electrons in the structure of the chemical compounds.
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m
hk θ hk'
Figure 6.43 Vector diagram of Compton scattering.
EXAMPLE E6.10 An electron runs accelerating voltage 104 V in an X-ray tube. Determine the wavelength corresponding to the short-wave limit of the bremschtralung spectrum of the X-ray radiation. Solution: A simplified construction of an X-ray generating device is given in Chapter 6.6. The accelerating electrons in the X-ray tube knock on an anode, X-rays being emitted in this process. In this example, a continuous spectrum is of interest. The shortest wavelength limit in this bremschtralung spectrum appears. It corresponds to the case when the whole electron energy transfers to a single X-ray quantum. Therefore, for this point the energy conservation law is valid eU hc/. Therefore, min
hc . eU
Substituting all the values we arrive at
min
o 6.63 1034 3 108 1.24 A. 19 4 1.6 10 10
EXAMPLE E6.11 Determine the cesium photoelectric threshold 0 if at its surface irradiation by violet light 400 nm a maximal speed of photoelectrons max is equal to 0.65 106 m/sec (refer to Section 6.6.4). Solution: The threshold corresponds to the situation where both the speed and energy of photoelectrons are equal to zero. Therefore, Einstein’s equation is
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A where A is the photoelectric work function and is the electron kinetic energy. We can obtain the expression 2c 2c A or 0 = ⴱ. 0 A The photoelectric work function can be determined using Einstein’s equation
A K max
2 2c m y max . 2
To execute calculations we should express all the values in the SI system: 1.05 1034 J sec, c 3 108 m/sec, 400 nm 4 107 m, m 9.11 1031 kg, max 6.5 105 m/sec. Calculations give us A 3.05 1019 J. To define photoelectric threshold 0, we should substitute the already known data and obtain 0 651 nm. EXAMPLE E6.12 A photon of energy 0.75 MeV is scattered by a nearly free electron at an angle 60°. Assuming that the electron’s kinetic energy and momentum before the collision were negligible small, define: (1) the energy ′ of the scattered phonon, (1) the kinetic energy K of the recoil electron and (3) the direction of its movement. Solution: (1) According to the Compton formula (refer to Section 6.6.4):
2 (1 cos ), mo c
we can express and ′ using energy of photons and ′: 2c 2c 2 (1 cos ) or m0 c 1 1 (1 cos ) . m0 c 2 Solving this equation regarding we obtain:
(1 cos ) 1 m0 c 2
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Executing calculations we arrive at 0.43 MeV. (2) The kinetic energy can be found from the energy conservation law K – 0.32 MeV. (3) The direction of the electron recoil motion can be found by applying the momentum conservation law (see Figure 6.43) p p m0. From a triangle OCD we can find CD
CA sin or OD OA CA cos sin p sin si n . tan p p p cos cos cos p tan
.0 (We use here the general equation p /c). Let us express (1 cos ) 1 m0 c 2
tan through the given data; therefore we find the ratio (1 cos ) 1. mo c 2 Hence tan
sin . ⎛ ⎞ 1 (1 cos ) ⎜⎝ m c 2 ⎟⎠ o
Taking into account some trigonometric relation we arrive at 2 . tan 1 m0 c 2 c tan
Executing calculations we obtain tan 0.701 and correspondingly 35. 6.7
THE BOHR MODEL OF A HYDROGEN ATOM
In previous sections, experiments which do not have explanations within the framework of Newtonian physics have been described. It is also known that reliably measured linear
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spectra of atoms, primarily hydrogen, also required the development of a completely new approach. This approach was suggested by Bohr, who formulated the theory of the hydrogen atom (Nobel Prize, 1922). Today it is considered to be semi-quantitative although it has not lost its significance. If we accept that atoms radiate electromagnetic waves with an energy h, it is necessary to establish where this energy originates from. We can equate energy h to the loss of energy E2 - E1 but it is then necessary to explain the nature of these energies. It is tempting to accept Rutherford’s planetary atomic model, but this seems impossible since it is known that the movement of a charged particle on a curvilinear trajectory undergoes a continuous loss of energy and the electron will inevitably fall onto a nucleus. However, in order to connect energy E2 and E1 with orbital movement, it is necessary to understand the stability of their orbits. The answers to all these questions were given by Niels Bohr in his theory of hydrogen atom. Bohr’s planetary model of the atom states that electrons in a hydrogen atom move in a circular orbit of radius r around a proton. The proton is so heavy in comparison with the electron that the center of mass of this system coincides with the position of the nucleus. Following Bohr, calculate the total electron energy. According to Newton’s second law 2 e2 , m r 4 0 r 2
(6.7.1)
where a Coulomb force electron–proton interaction equates to ma, a being centripetal acceleration. The kinetic energy K can be derived from this equation: K
m2 e2 , 2 8 0 r
(6.7.2)
whereas the potential energy of a negative charge in a field of the positive nucleus can also be found U (e)
e2 . 4 0 r
(6.7.3)
where is the nucleus electrostatic potential at distance r (refer to eqs. (4.1.21) and (4.1.22)). The total energy is then
E K U
e2 . 8 0 r
(6.7.4)
Since the orbit radius can apparently take on any value, so can the energy E. The problem of E quantization reduces to quantization of r.
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All the values listed above are unequivocally connected with r, other values are expressed with its help as well. In particular, the linear electron speed can be expressed as e2 . 4 0 mr
(6.7.5)
The frequency is
e2 . 2r 163 0 mr 2
(6.7.6)
The linear momentum p is p my
me2 . 4 0 r
(6.7.7)
L pr
me2 r . 4 0
(6.7.8)
And the angular momentum L is
Thus if r is known the orbit parameters K, U, E, , , p and L are also known. If any one of them are quantized all others must be also. Up to this point only classical physics has been used. Here, Bohr suggested that the necessary quantization of the orbit’s parameters shows up most simply when applied to the angular momentum and that, specifically, L can take on only values given by L n n
h , 2
(6.7.9)
where n can accept integer values 1, 2, 3, etc. The Planck constant appears again in a fundamental way. Combining eqs. (6.7.8) and (6.7.9) leads to r
h2o me2
n2
(6.7.10)
and E
i.e., to the energy quantization.
me 4 1 8 20 h 2 n2
(6.7.11)
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If we accept n 1, the well-known radius of the first orbit in H-atom can be obtained r0
h2o me2
.
(6.7.12)
Further, the atom radiates or absorbs energy only when an electron passes from one stationary orbit to another. This portion of radiation has been referred to as a quantum (of energy). As the energy is connected to the frequency of the quantum and, accordingly, to wavelength (in vacuum) the frequency of the quantum can be expressed through the quantum numbers corresponding to two orbits (j and k being their quantum numbers) (compare with (7.5.33)):
1⎞ me 4 ⎛ 1 2⎟. 2 3 ⎜ 2 8 0 h ⎝ j k ⎠
This is the famous serial formula which allows calculation of all the spectral lines in the hydrogen atom spectrum. An expression me4/802h3 is referred to as a Rydberg constant. The quantum mechanical theory of the hydrogen atom is given below (see Chapter 7.5) The Bohr model of the hydrogen atom is a transition from purely classical presentations to quantum mechanical ones: the motion of electrons along the orbits is accepted; however not all orbits are permitted, the angular momentum is accepted, though its values and orientations are subject to strict limitation. One can consider the Bohr model as the transition from classical mechanics to quantum mechanics with the preservation of many its attributes. As a result, many of the ideas of the Bohr model will often be met in order to simplify the students’ understanding. A typical quantum mechanical object such as an atom possesses some classical characteristics unexplainable within the framework of generally accepted presentations (no orbital motion, yet the existence of angular momentum; no rotation of an electron around its own axis, yet intrinsic angular and magnetic moments, i.e., spin, etc.). As a result, these terms are used irrespective of their classical sense.
EXAMPLE E6.13 In the framework of the Bohr model of the hydrogen atom (refer to Section 6.6.7) calculate the radius of the first electron orbit r and the linear electron speed . Solution: From Section 6.6.7, we know that r and are united in equation mr n (in our case n 1). In order to find two values one more relation is needed. For the rotation of an electron around a nucleus we can write the Newton’s second law equation
m
1 e2 y2 . 4 0 r 2 r
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or
my 2
1 e2 . 4 o r
Solving all equation we can find
r
4 o 2 me2
.
Substituting all known values, we arrive at r1 ao 5.29 1011 m and /mr 2.18 106 m/sec.
EXAMPLE E6.14 It is measured experimentally that the CuK X-ray beam ( 1.542 Å) being diffracted by a corundum single crystal deviates from its initial direction at an angle 41.66° (compare to Figure 6.20). The diffraction takes place from the crystallographic plane (600). Find the deviation angle from the same crystallographic plane using the MoK radiation ( 0.710 Å).
d
A
B
C
Solution: Bragg’s law should be used to solve this problem (refer to Section 6.3.5 and eq. (6.3.11), Figure 6.20). In the equation mentioned, is the incident angle and is the deflection angle. The picture is repeated here to make the situation clearer (Figure E6.14; an atomic arrangement is not depicted substituted by two reflected planes with interplanar distance d600). It is seen in the picture that 2. The crystallographic plane’s index is (600), in fact there is no such plane in the crystal; this should be understood as a 6th order of reflection from the plane (100) (i.e., at distance
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ACB in Figure 6.20 six wavelengths stack). Therefore, sin( /2) 6/2d600; this equation is valid for both wavelengths. Dividing two equations for two wavelengths we obtain (sin( Mo/2) sin( Cu/2)(Mo/Cu). Executing calculations we arrive at sin( Mo/2) 0.164, therefore, Mo is equal to 4.72°.
PROBLEMS/TASKS 6.1. The surfaces of a glass wedge form an angle 0.2. On the wedge perpendicular to its surfaces a beam of monochromatic light with wavelength 0.55 m falls. Determine the width of interference strips b (the distance between the adjacent maxima). 6.2. The diameters of two light Newton rings are d1 4.0 and d2 4.8 mm. It is known that three light rings settle between the two measured rings. The rings were observed in reflected light. Find the curvature radius R of the plane-convex lens. 6.3. In the experiment with Newton rings, a liquid oil was poured between a lens and a sample stage table, with its refraction index less than that of glass. The radius of the eighth dark ring is d8 2 mm ( 700 nm) whereas the radius R of the planeconvex lens is 1 m. Find the refraction index n of oil. 6.4. On illuminating diffraction gratings by a white light, the spectra of second- and third-orders partly overlap. On what wavelength in the second-order spectrum, does ultraviolet (UV) of the third order ( 0.4 m) fall. 6.5. A monochromatic light with wavelength 600 nm falls on diffraction gratings with the period d 10 m at an angle 30°. Find the diffraction angle corresponding to the second main maximum. 6.6. The energy flux radiated through a muffle’s sight hole is 34 W. Assume that the muffle radiates as an IBB and find its temperature T if its area is S 6 cm2. 6.7. Assume that the sun radiates as an IBB; calculate its emittance R and surface temperature T. The solar constant (the energy radiated by the sun per unit area measured on the outer surface of earth’s atmosphere) is C 1.4 kJ/m2 s. Assume the sun–earth distance to be d 1.49 1011 m. 6.8. Because of the change of temperature of an IBB, the position of the maximum spectral density emittance shifts from 1 2.4 m to 2 0.8 µm. How many times are the emittance R and the maximum spectral density of emittance r changed? 6.9. The temperature T of an IBB is T 2000 K. Calculate: (1) the spectral density of emittance r() for the wavelength 600 nm, (2) emittance R in an interval of wavelength from 1 590 nm to 2 610 nm. Assume that the averaged spectral density of emittance in this interval is equal to that for 600 nm. 6.10. Compton scattering of X-ray 55.8 pm occurs by graphite plate. Find the wavelength ′ of light scattered at an angle 60° 6.11. A photon ( 1 pm) is scattered by a free electron at an angle 90°. What part of its energy W does the photon transmit to the electron? 6.12. The spectral density maximum rmax of the emittance () of the bright star Arcturus corresponds to 580 nm. Assuming that the surface of this star emits as an IBB, determine its temperature T.
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6.13. Find the wavelength of a 1 MeV photon. Compare it with resting electron mass m0. 6.14. A photon’s wavelength is equal to Compton length C. Determine the photon energy and momentum p.
ANSWERS
6.1. 6.2. 6.3. 6.4. 6.5. 6.6. 6.7. 6.8. 6.9. 6.10. 6.11. 6.12. 6.13. 6.14.
b /(2n) 3.15 mm. R 880 mm. n 1.4. 0.6 m. arcsin (sin m/d) 38.3°. T 1 kK. R 64.7 MW/m2; T 5.8 kK. Increases R 81 and r 243 times. r() 30 MW/m2 mm, R 600 W/m2 . 57 nm. 0.70. T 4.98 kK. 1.24 pm, mph 1.8 1030 kg, pph 5.3.10 22 kg m/sec, mph 2moe. 0.511 MeV, p 2.7 1022 kg m/sec.
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7.1 7.1.1
PARTICLE-WAVE DUALITY
De Broglie hypothesis
After the experiments and theoretical explanations that led to the revival of the corpuscular theory of light outlined in Section 6.6, it was logical to take the following step: restoring symmetry and assigning wave characteristics to those micro-objects that are known to us as particles. This step was taken in 1924 by the French theoretical physicist L. de Broglie (Nobel Prize, 1929). His idea consisted of the following. If quanta of electromagnetic radiation—photons—possess corpuscular characteristics, i.e., possess a mass (m E/c2), momentum (p = hk) and energy (E = h), the converse must also be true, i.e., a wave process must be related to a moving particle. De Broglie suggested a formula for the wavelength of such a process:
h , m
(7.1.1)
where h is the Planck’s constant, m is the particle mass and is its velocity. These waves are called “de Broglie waves.” It must be said that no one can imagine what lies behind this statement. We will see here a case that is quite typical in quantum mechanics: our normal imagination is not adequate to conceive of the main principles of this science. There is certainly an explanation for this effect. Indeed, let us imagine a 10 g bullet moving at a speed of 1000 m/sec. According the formula (7.1.1), the corresponding wave process (de Broglie wave) has the wavelength ~ (1034/(102 103)) ~ 1035 m . In order to reveal the character of such a wave, we must find a grating with a comparable period. Certainly, there are no such gratings. Generalizing, it can be said that in the everyday world there is no way to prove that de Broglie waves corresponding to a moving body can be visualized. Therefore, it is purely academic. This is why the referee of de Broglie’s thesis, the well-known French physicist P. Langevin, was not able to evaluate de Broglie’s formula and stated “de Broglie’s idea is either the fruit of his morbid imagination or, conversely, a brilliant idea, though in both cases it deserves high recognition.” The Nobel Prize was the relevant award. 423
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Quite a different situation occurs in the micro-world. Here, because of small particle masses, the de Broglie wavelength is commensurate with the interatomic distances in crystals; therefore, the corresponding experiment can be realized: a crystal can serve as a diffraction grating (refer to Section 6.3.5). In fact, according to formula (7.1.1) and the conservation law we can obtain
h h 150 ⬃ 1010 m, 1 2 m (2 meU ) E
(7.1.2)
i.e., a voltage of 150 V is needed to produce an electron beam with a wavelength of 1010 m and to observe diffraction. It is quite natural to ask the question: what do the microparticles (electrons, neutrons, etc.), on one side, and the phonons on the other represent? The modern development of science enables it to be stated that all micro-objects simultaneously possess a set of properties, among which both wave and corpuscular properties are equally present, and they reveal them depending on the conditions to which an experimenter subjects them. This statement is the essence of corpuscular-wave dualism, which in turn is the basis of quantum mechanics. 7.1.2
Electron and neutron diffraction
Only 2 years were needed to prove electron diffraction experimentally. In 1927, simultaneously in several laboratories in different countries, attempts were made to observe electron diffraction. Davisson and Germer in America, Thomson in England and, several years later, Tartakovsky in Russia observed electron diffraction after the interaction of electrons with single crystals (C.J. Davisson and G.P. Thomson, Nobel Prize 1937) and polycrystalline films. In a simplified representation, the principles of an electron diffraction device are depicted in Figure 7.1. The main part of the instrument is the vacuum column, in which all the elements of the device are contained. The heating filament 1 (cathode) emits electrons by thermoemission, which then speed in an electrostatic field with the potential difference U. Passing through the diaphragm, monochromatic electrons (i.e., electrons with constant wavelength, refer to formula (7.1.2)) fall onto the polycrystalline film sample 3. The polycrystalline sample contains an enormous number of small microcrystals, absolutely chaotically oriented in space. From the whole set of microcrystals, some are oriented with their crystallographic planes p1, with interplanar distance d1 with respect to the incident beam of electrons at an angle 1, for which the Bragg condition is met (refer to 6.3.5). Planes p1 of this crystal will give a reflected beam with an angle 21 with respect to the direction of the primary electron beam. In a polycrystalline sample, an ensemble of other microcrystals can always be found which have the planes p1 identically oriented with the same reflecting angle ; all the p1 planes look like they are rotated around the primary beam direction. Therefore, all the reflected beams are situated on a cone around the direction of the primary beam with a cone opening angle 41. Each electron colliding with the fluorescence screen 5 produces a flash; all the flashes create a diffraction ring.
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P1
1 2
Θ1
3 2Θ1
4Θ1
4 5
Figure 7.1 An electron diffraction investigation scheme.
Under known instrument parameters (distance from the sample to the screen L, applied voltage U, etc.), the diffraction circle diameter D is connected with the interplanar distance d (refer to Sections 6.3.5 and 9.1) by the formula
nh 2 meU
2 d sin ⬵ 2 d
dD , 2L
and, accordingly, d
2 Lh D 2 meU
,
(7.1.3)
where m and e are the mass and charge of the electron. There are other crystallographic planes p2, p3 and p4 in the polycrystalline sample too, with other diffraction cone opening angles 42, 43 and 44, respectively. As a result, the typical electron diffraction pattern looks like a set of concentric circles, all circles having a different intensity. The set of interplanar distances di and the corresponding intensity Ii of each circle characterize the material analyzed. The diffraction pictures of MgO films obtained by means of the diffraction of X-rays (usually accepted as waves) and electrons (usually considered as particles) are compared in Figure 7.2; since they were taken using different wavelengths, for the best comparison both pictures were brought to the same scale. (Spots on the electron diffraction rings are caused by the presence of partial crystalline order in the MgO film.) The similarity of the diffraction images made with electrons and X-rays can clearly be seen in this picture. In 1932, the English scientist J. Chadwick discovered a neutral elementary particle—the neutron with a mass practically equal to the proton mass (Nobel Prize, 1935). Experiments to prove neutron diffraction immediately followed. In 1936, such experiments gave a positive result. However, in the modern view as it is currently used, neutron diffraction as a method of structure investigation became possible only after nuclear reactors were set up. Neutron diffraction became one of the most powerful methods for the analysis of
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Figure 7.2 A MgO powder diffraction patterns obtained by X-ray (left) and electrons (right).
solids. In 1995, the American physicist C. Shull, who contributed most to the development of the method of neutron diffraction in physics and chemistry, was awarded a Nobel Prize. Along with neutron diffraction, experiments of inelastic neutron scattering (where the neutron behaves as a material particle interacting with phonons (refer to Chapter 7.3), according to collision theory (refer to Section 1.5.5)), were also realized. For the discovery of inelastic neutron scattering and establishing a new method of solid state investigations, the Canadian physicist B. Brockhause was awarded the Nobel Prize (1995). Thereby, the de Broglie hypothesis has been widely accepted as a scientific law. To reiterate, particle-wave dualism is one of the cornerstones of quantum mechanics.
EXAMPLE E7.1 An electron acquires a kinetic energy K under an accelerating voltage U. Determine its de Broglie wavelength for two cases U1 51 V and U2 510 kV. Solution: The electron wavelength depends on its momentum p: (2/p)*. In the case of classic mechanics the momentum depends on energy p 兹2苶m 苶苶 0K ; however, for a relativistic case this relationship is p (1/c)(兹(2 E K )K ). The rela苶苶0苶苶苶苶 tion * for two cases mentioned can accept the forms 2/(兹2苶m 苶苶 0K ), and correspondingly 2/冢(1/c) 兹苶2苶 E苶 苶 K 苶 )K 苶 冣. 0 Compare the electron’s kinetic energy for two cases and compare them with the electron’s rest mass; the kinetic energy is 51 eV and 0.51 MeV; the electron’s rest mass is just 0.51 MeV. Therefore, the second case corresponds to a relativistic one. The wavelengths for these cases are: 1 2/eU1 172 and 2 2/ 兹苶3m0c 1.4 pm.
EXAMPLE E7.2 Assume that on a narrow slit of a width a 1 m, a plane-parallel electron beam of very low intensity, having speed 36.5 106 m/sec, is directed (Figure E7.2,
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refer also to Figure 6.15). Taking into account the electron wave properties, define the distance between two first minima of intensity of the first order (that is width of a zero maximum) in a diffraction spectrum. Assume the distance from the slit to the screen to be L 10 cm.
a
L
Ltg x
Solution: Assume that according to de Broglie hypothesis, a wave of wavelength 2/m* is attributed to a particle of mass m moving with a speed . Therefore, diffraction should be observed. The diffraction minimum in the spectrum can be found according to the relation a sin (2k1)(/2) (refer to Section 6.3.2), where k indicates the diffraction order (in our case k 0; Figure E7.2). Besides diffraction angle is small, sin ⬇ tan⬇ and therefore, a (3/2). The sought distance x on the screen is . x 2 L tan ⬇ 2 L (3 2)(a ) 3L a. Using the equation * we obtain x
6L a1 . m
Executing calculations, we obtain x 6 .105 m. This means that a spectator can find a particular electron in this range, i.e., the uncertainty to locate the electron’s position is x.
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EXAMPLE E7.3 A plane-parallel incident electron beam falls on a crystallographic plane of a nickel single crystal at an angle of 64°. Diffraction takes place. Accepting the interplanar distance of the given crystallographic planes as d 200 pm, define the de Broglie electron wavelength and their speed . Solution: Diffraction of X-rays is described in Section 6.6.4; electrons exhibit the same property. The Bragg’s equation is applicable to this case 2d sin n, n being the diffraction order, n 1 in our case. From this equation we can find the wavelength 2d sin 360 pm. From the de Broglie relation the electron’s speed can be found:
2
7.2
2 Mms. m
HEISENBERG’S UNCERTAINTY PRINCIPLE
Particle-wave duality entails important consequences. The question is, can a micro-object simultaneously possess precise values of its coordinate and momentum? Indeed, a certain internal self-contradiction exists between the characteristics of a material particle that can be localized in space with arbitrary accuracy and the monochromatic de Broglie wave, which according to its nature is extended from ∞ to ∞ and is completely delocalized in space. However, it is exactly the latter that possesses a certain, exact wavelength and, accordingly, fixed momentum. Quantitative analysis allowed W. Heisenberg in 1927 to suggest the principle (Nobel Prize, 1932), which nowadays is given as follows: there exists no state in which the coordinates of a microparticle and its momentum have precise values. If a micro-object is traveling along the x-axis, one can characterize the uncertainty of the coordinate and component of momentum by values x and px; then the Heisenberg principle (for coordinate and momentum) has the form x p x h,
(7.2.1)
i.e., the product of the uncertainties in the coordinate and corresponding component of momentum cannot be less than h. It is possible to apply another interpretation to the Heisenberg uncertainty principle. It is well-known that a wave can be characterized by the precise value of the wavelength, when it spreads from ∞ to ∞. It is also known that such a wave is a mathematical abstraction; any wave has an origin and an end. Accordingly, this model corresponds to the precise values of the wavelength (and wave vector k) and, consequently, momentum p. It means that in this case, uncertainty in the momentum px is zero (Figure 7.3, on the left). As a result, we are unable to assign any position to the particle; its uncertainty is equal to infinity. Certainly, the product of x and px can be less than h. Such values of uncertainties save an equitable correlation of Heisenberg uncertainties.
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∆px ∆px −∞
∆x
p +∞
∆x
Figure 7.3. An uncertainty principle for x and px: the higher the particle localization, the lower the determination of its momentum.
However, if we try to reduce the uncertainty in the particle’s position and put it in a state when x becomes any finite quantity, say a value L (Figure 7.3, middle), the product L 0 cannot be larger than h. In other words, this will bring about the appearance of an uncertainty in the momentum, which is displayed in Figure 7.3 (on the right) in the form of a curve, the maximum width of which is x (which can be evaluated as h/L). The Heisenberg uncertainty principle imposes essential restrictions on some laws of classical mechanics. In particular, these affect a very important notion such as trajectory. As an example, let us consider a hydrogen atom within the framework of the Bohr model: an electron revolves around a proton along a circular orbit. From the known electron mass and charge, as well as the electrical constant, within the framework of classical electrodynamics one can define (at the order of magnitude) the linear electron velocity, which turns out to be approximately 106 m/sec. Then the uncertainty in the coordinate x is x
h h 1034 30 ⬇ 1010 m, p m 10 106
i.e., it coincides with the atomic size. One can conclude, therefore, that the notion “trajectory” in this case (and in quantum mechanics, in general) loses it’s meaning: the uncertainty in the electron coordinate is comparable with the object size. It is clear that a new approach to describing micro-objects is necessary. The principle of uncertainty itself allows one, in some cases, to arrive at the decision not to solve the problem exactly. As an example, one can consider the state of a particle limited in its motion in space (i.e., existing in a potential well, refer to Section 1.5.4) by the width L. Let us pose a question: can the particle energy accept any values or an undetermined one in this case? Can a particle “settle down to the bottom” (i.e., possess exact (zero) energy and, accordingly, exactly determined momentum)? In order to decide, let us choose an uncertainty in the momentum: let this uncertainty be equal to 100%, i.e., it will accept pp. Bearing in mind the relationship of energy E with the momentum, we can write: pp 兹苶2苶 m苶 E. The uncertainty in the coordinate x in this case is the width of well L: we know that the particle is in the potential well, but do not know precisely at what point. As a result, the uncertainty principle looks like x px (兹苶2苶 m苶 E) L h, whence E
h2 2mL2
.
(7.2.2)
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This shows that the answer to the question above is as follows: a particle cannot occupy a position on the bottom of a potential well of finite width, and the expression derived presents the lowest permitted value of energy. This solution has been reached only by the uncertainty principle, without using the main attributes of quantum mechanics. As we will see below, this conclusion complies well with the result of the exact solution of this problem. The uncertainty principle also offers the energy E of a micro-object and the lifetime of a system in this state: the product of uncertainty in energy E and in the time the system exists in this state cannot be less than h (7.2.3)
E h.
For the ground state of a micro-object that can exist in this state infinitely long (t∞), the uncertainty in energy E is zero, i.e., the energy of the ground state can have an absolutely precise value. Though for excited states with a limited lifetime of, say 108 sec, the uncertainty in energy is: E ⬇ 1034/108 1026 J⬇107eV. This is a very small value: however, in some cases it plays an important role in the physical phenomenon. In Figure 7.4, the illustration of broadening of a spectral line on account of uncertainty principle (for energy and time) is depicted. The linewidth caused solely by the energy level broadening because of the Heisenberg uncertainty principle (i.e., not subjected to the influence of an instrumental imperfection), is called a natural spectral linewidth. It is necessary to note that the uncertainty principle does not impose any restrictions on the possibility of the simultaneous existence of precise values of coordinates and momentums along different coordinate axis. In other words, products y px and x py can be equal to zero, i.e., the corresponding values of coordinates and component of momentums can be determined precisely.
E1
E1
E0
E0
I Γ = 2∆E/h
Figure 7.4 An uncertainty principle for energy and time of life. Two energy levels are presented (left) without uncertainty principle accounting (the both are infinitely narrow and a spectral line is infinitely narrow too); taking the principle mentioned into account leads to the broadening of the width of both an excited energy level E1 and spectral line ( is the natural width of a spectral line).
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EXAMPLE E7.4 An electron’s kinetic energy in a hydrogen atom is of the order 10 eV. Using the Heisenberg uncertainty principle, determine the linear size of the hydrogen atom. Solution: The momentum and coordinate uncertainties are related by eq. (7.2.1). Let the atom have a linear size l, then the electron will be somewhere in the limits xl/2. The uncertainty principle can then be written as:
2 . p
Suppose that the physical reasonableness p should be less than p: p p. Therefore, we can assume p ⬇ 2 mK
and min
2 2 mK
.
Substituting the known and given values, we arrive at lmin 124 pm, i.e., commensurable to the atom’s size.
EXAMPLE E7.5 Using the uncertainty principle of energy and time, define the natural width of a spectral line in the excited state and at its transition from the excited state to the ground state. Define also the corresponding . Assume equal to 108 sec and wavelength 600 nm. Solution: The natural width of the energy level E and the time of its life are related by the Heisenberg ratio . Therefore the natural width is / , i.e., ⬇
1034 1026 J. 108
Since the photon energy and the wavelength are related as
2c ,
the uncertainty relation can be found by
2c 2
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(the minus sign is ignored). Hereafter,
2 . 2c
Substituting the given value, we arrive at 2 1014 m
7.3 7.3.1
WAVEFUNCTION AND THE SCHRÖDINGER EQUATION
A wavefunction
What we have discussed above makes it necessary to develop another approach to the description of micro-objects, different from that in classical mechanics. This was done by the Austrian physicist E. Schrödinger (Nobel Prize, 1933), who (together with P. Dirac and W.K. Heisenberg) suggested the idea of the wavefunction ψ(x, y, z, t) as well as, an equation that this wavefunction should obey. Accordingly, a microparticle state in quantum mechanics is defined by the wavefunction ψ(x, y, z, t). Knowing the wavefunction, for instance for electrons in an atom, one can define their behavior when changing the chemical bonding between atoms, the probability of forming one or other molecular structures, judging the strength of interatomic bonding in molecules, etc. Thereby, the wavefunction of electrons is a key to deciding many principle problems of chemistry. However, function itself has no physical sense; it cannot be measured itself. The wavefunction presents a certain mathematical expression by means of which it is possible to find the probability of one or other real physical features of electrons in atoms and molecules. In order to calculate these probabilities, we need to use a value *, where * is a complex conjugate with , since in general can be a complex function. In a particular case, when a wavefunction is real, * 2. We will restrict ourselves to some limitations: we will only deal with those problems that do not depend on time; an electron spin will be incorporated further, see Section 7.5.5; besides, we have considered so far only nonrelativistic problems. The physical meaning of ⱍ ⱍ2 consists in the following: ⱍ (x, y, z)ⱍ2 is proportional to the probability dw(x, y, z) of finding a particle in the elementary volume dV in the vicinity of a point with coordinates x, y, z: 2
2
dw( x, y, z ) ( x, y, z ) , dV ( x, y, z ) dxdydz.
(7.3.1)
The square of the wavefunction | (x, y, z)|2 is hence the probability density (i.e., the probability related to the unit volume) of finding a particle in a point x, y, z: dw( xyz ) 冷 ( x, y, z ) 冨2 dV
(7.3.2)
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Certainly, wavefunction can be dependent on time, because physics knows many problems that are time dependent. However, according to our task, we will consider only stationary processes, not depending on time: the force fields in which particles move are stationary (refer to (1.4.4), i.e., they do not depend on time). It is possible to show that in this case the wavefunction (x, y, z, t) disintegrates into two factors, one of which depends on coordinates only and the other cyclically depends on time: ( x, y, z, t ) ei t ( x, y, z ) e(i ) Et ( x, y, z ),
(7.3.3)
where is frequency, and E is the total particle energy. In this case ( x, y, z, t ) ⴱ ei t e i t ⴱ ( x, y, z ) , 2
2
(7.3.4)
i.e., the wavefunction square is also time independent. 7.3.2
The Schrödinger equation
The wavefunction is the solution of a certain equation that was introduced by E. Schrödinger—the Schrödinger equation. This is the main equation in quantum mechanics. In general, it cannot be derived theoretically; however, its validity is proved in practice: the results obtained in solving this equation are confirmed in numerous experiments. Here it plays the same role as the second Newtonian law in classical physics (refer to Section 1.3.3). For stationary nonrelativistic problems, the Schrödinger equation can be written as follows:
( x, y, z )
2m [ E U ( x, y, z )] ( x, y, z ) 0. 2
(7.3.5)
Besides the well-known notions, a new one is presented in the equation: the Laplacian operator that is the sum of the second partial coordinate derivatives acting on the function that follows it (in our case on wavefunction (x, y, z)): ⎛ 2
2
2 ⎞ ⎜ 2 2 2⎟.
y
z ⎠ ⎝ x
(7.3.6)
If a one-dimensional problem is being solved, only the first term of eq. (7.3.6) is used. The term U(x, y, z) is the particular particle potential energy in the force field; the information on the particular type of problem is concentrated exactly in this term. In order to solve a quantum mechanical problem, we should substitute an analytical expression for the particle potential energy in the given force field, U(x, y, z), in eq. (7.3.5); find the values of parameter E at which the Schrödinger equation allows solutions and calculate the wavefunction (x, y, z) as a solution of this equation. The parameter E stands out as the particle energy.
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Standard requirements that the wavefunction should obey
In all cases, the wavefunction (x, y, z) as the solution of the Schrödinger equation should possess some properties that themselves play an important role in quantum mechanics. Before describing the methods for solving Schrödinger equation, we have to analyze the standard requirements (conditions) that a wavefunction should obey; these requirements result from the type of equation and the physical sense of the functions. This especially concerns cases when the problem should be divided into pieces because of the complexity of the potential distribution. So, the wavefunction (x, y, z), in any case, must be a single-defined (univocal) function of coordinates, finite and continuous in the whole range of variable coordinates x, y, z, including infinity. In the case of the potential energy partition, the most important is that (x, y, z) must join smoothly at the boundaries. The wavefunction must be finite, i.e., we require that (x)t0 at xt -. As far as the Schrödinger equation contains the coordinate second derivatives, the first coordinate derivative (x, y, z) should be continuous also. All of these are obvious from physical and mathematical considerations. Certainly, | (x, y, z)|2, and, consequently, (x, y, z) , must have no more than a single value under fixed coordinates x, y and z or under such operations which return a particle into the former point of space (requirement of univocacy). Otherwise the solution would be ambiguous, which would make no sense. Furthermore, as we know that the micro-object really exists in the whole region, the wavefunction (x, y, z) should obey the normalization condition. The total probability of finding a particle in the whole range of variable must be unity, i.e.,
∫ dw ∫∫∫ dw ∫∫∫ ( x, y, z)
2
dxdydz 1.
(7.3.7)
The properties mentioned are called the standard requirements (conditions) that the wavefunction should obey. They result from the type of equation and physical sense of these functions. The requirements imposed on the wavefunction have the result that the Schrödinger equation, as a second-order differential equation in partial derivatives under the given U(x, y, z), can have solutions satisfying them only at definite values of E, values which play the role of parameters in the equation. Values of E, under which the Schrödinger equation has solutions satisfying the standard conditions, are called eigenvalues of energy. The set of eigenvalues E forms an energy spectrum. If eigenvalues form a set of definite values E1, E2, E3, etc., the energy spectrum is called a discontinuous one. If energy E can accept any values, the energy spectrum is called a continuous one. Below we will see that a discrete energy spectrum appears when a particle motion is restricted in space (motion is finite). In the case of infinite motion, the energy spectrum is continuous (energy can change continuously). The wavefunctions N that correspond to parameters EN are called the eigenfunctions.
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MOST GENERAL PROBLEMS OF A SINGLE-PARTICLE QUANTUM MECHANICS A free particle
Let us consider first the simplest case: free particle motion. Free motion means the case with U ⬅ 0, i.e., the motion of a particle on which no irrelevant forces are exerted. Such a particle is moving uniformly and in a straight line. Let the x-axis be directed along the particle motion. The Schrödinger equation in this case has the form 2m d2 ( x ) 2 E ( x ) 0. 2 dx
(7.4.1)
As here depends on only one variable, the partial derivatives are replaced onto the full derivative. The energy in this case is kinetic energy, because potential energy for free particles is zero. Therefore, E
p2 , 2m
(7.4.2)
where p is particle momentum. Let us introduce the definition
2 mE k2. 2
(7.4.3)
( x ) k 2 ( x ) 0.
(7.4.4)
Then the equation accepts the form
The solution of such an equation is well-known: ( x ) A exp(ikx ),
(7.4.5)
where k p/ (refer to Section 1.6) and, correspondingly, k 2/ (eq. (2.8.4)). Because of the fact that we are dealing with the uniform Schrödinger equation and the derivation from the wavefunction is taken on the coordinate x, this solution can be multiplied by any coefficient not depending on the coordinate. Let this coefficient be ei t. Then the solution accepts the form (x,t) A exp(ikx) exp(i t) A exp(i tikx). The real part of it is therefore the wave running in x direction ( x, t ) Acos(i t ikx ),
(7.4.6)
(refer to 2.8.2). It is of use to express this expression via the energy and momentum 1 ( x, t ) A cos ( px x Et ).
(7.4.7)
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The calculation of the probability density to find a particle on any point of the x-axis gives dw( x ) ⴱ A exp(i t ikx ) A exp(i t ikx ) A2 , dx
(7.4.8)
i.e., the probability density is uniform along the whole x-axis. Thereby, if a particle momentum is strictly defined (as in our case), its position in space is indefinite: there is no priority to find it on any point of the x-axis, the particle is as if it is smashed along the x-axis. It is worth noting that this result completely corresponds to the uncertainty principle (refer to Section 7.2). Correlation (7.4.6) allows us to express carefully the statement (not attempting any of the serious generalization), that the wavefunction (x, t) in the form of waves can be called the wave of probability, although this statement certainly does not make the physical picture clearer. The real physical sense has only * 2. The solution obtained satisfies a standard condition: the energy E and particle momentum can accept any magnitude (the spectrum of energy is continuous). Motion is infinite and energy, as in classical physics, can accept any value. We must pay attention to one more important property of the Schrödinger equation (7.4.1), apparent from its uniformity: any superposition of solutions is also a solution of this equation. Such a superposition property often leads to what in chemistry is called hybridization.
7.4.2
A particle in a potential box
Let us consider now the state of a particle placed in a one-dimensional square potential box. This problem is not a far-fetched abstraction, but presents a model of a bound molecule particle. In Section 1.5.4, the Lennard-Jones potential was considered (and below the Morse potential will be used). It was shown that under the total energy E lower than the depths of the potential well, a particle makes an oscillatory motion (whether it is harmonic or anharmonic plays no essential role at present) near its equilibrium position. Exactly such a potential function has to be substituted in the Schrödinger equation to decide this problem. To solve such an equation in analytical form, however, is impossible at the moment. Therefore, the potential of a complicated profile is approximated by a model, the consideration of which, certainly, does not solve the problem precisely, but nevertheless allows U
r
Figure 7.5 A Lennard-Jones’ and a model’s potentials.
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U
I
II 0
III L
x
Figure 7.6 A one-dimensional infinitely deep potential box.
us to make a number of important conclusions. In Figure 7.5, the Lennard-Jones potential is approximated by a rectangular, one-dimensional potential box. In Figure 7.6, this box is provided in the form in which it will be used below. First consider an infinitely deep potential box of the width L. Beyond the borders of the box, potential energy U is infinitely large, hence the particle cannot exist there. Inside, the potential U can be taken as zero. As a result a particle in the potential box is considered, the latter being assigned as follows (Figure 7.6): U(x) ∞ at x 0 and at x L, U(x) 0 at 0 x L. Accordingly, three areas are presented in the scheme: I, II and III. In areas I and III, the particle cannot exist, the wavefunctions there are zero: I (x) III(x) 0. In area II the potential energy is zero (U 0) and the Schrödinger equation gains the type II ( x ) kII2 II ( x ) 0, similar to eq.(7.4.4). Solution of the equation in this area gains the form II ( x ) A sin kx B cos kx.
(7.4.9)
From standard conditions for wavefunction continuity, it follows that at the point x 0, the wavefunctions on the left and on the right sides are zero: I(0) II(0). It follows that 0 A 0 B 1, hence B 0. At the point x L the same condition requires A sin kL to be zero. This signifies that kIIL n, where n is an integer. Taking the expression (7.4.3) into account this equality can be rewritten as En
2 2 2 n . 2 mL2
(7.4.10)
The fundamental result has been obtained: the particle energy in an infinitely deep potential well can accept only discrete values! In other words the bound particle energy is quantized. The integer n, incorporated earlier as an arbitrary integer, now stands out as a certain parameter that determines the value of energy. This integer is called the quantum number. The position of the energy levels of a particle in the infinitely deep potential well is given in Figure 7.7.
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E
0
n
n2
5
25
4
16
3
9
2 1
4 1
Figure 7.7 The derived energy levels of a particle in a one-dimensional infinitely deep potential box.
It is worth noting that the quantization of energy in this instance has resulted not from mathematical deduction but by using one of the standard conditions of wavefunction continuity. A result of the same significance (7.2.2) was obtained earlier in the analysis of the problem by means of the uncertainty principle; the particular expression differs from eq. (7.4.10) by only a constant multiplier nearer to unity. We want to emphasize this logical unity and intercoupling of different approaches in quantum mechanics, including the standard conditions the wavefunction must obey. Each of these aspects can be used in different cases when deciding a particular problem. In order to obtain the total final expression for the particle wavefunction in the infinitely deep potential box, one has to define the normalizing multiplier. The normalizing condition (7.3.7) can be used here. In this instance it looks like L
∫ 0
2
L
( x ) dx A2 ∫ sin 2 kx dx = 1.
(7.4.11)
0
Replacing the integrand according to the trigonometry formula sin2kx (1/2)(1/2) cos2kx, we can arrive at A2 2/L. Finally, the expression for the wavefunction of a particle inside the rectangular infinitely deep potential box is given as II ( x )
2 n sin x. L L
(7.4.12)
The graphs of the wavefunctions (a) and their squares (b) of the particle in the infinitely deep potential well in different quantum states are depicted in Figure 7.8. As a matter of convenience, they are distributed along the ordinate axis corresponding to the quantum numbers. The specific question is, can n be equal to zero? This is equivalent to the question already discussed in Section 7.2: whether a particle can “lie” on the bottom of a well; to which we have already obtained a negative answer from the uncertainty principle.
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U
U
439
n=3
n=3
*
n=2
n=2
n=1 n=1
0
x
L
0
x
L
Figure 7.8 Wavefunctions of a particle in a one-dimensional infinitely deep potential box (a) and their squares (b).
Graphs of n(x) are extremely reminiscent of graphs of standing waves and, particularly, the string oscillations (refer to 2.9.3). There is nothing amazing in this fact: moving in the potential box, particles are reflected from potential barriers on the right and on the left so that their wavefunctions interfere, forming standing waves. Half of the standing wavelength is equal to L/n (refer to formula (2.9.8)). Only in such a case is a state stable! This example is suitable to illustrate the Bohr correspondence principle. It states that all regularities of quantum mechanics turn into the regularities of classical mechanics under the increasing quantum numbers. It is well known that the different levels of physical approximations are characteristic to certain areas of this science. Transformation from one area to another occurs not abruptly, but gradually. So, Newtonian mechanics becomes less and less exact when the velocity of particle motion increases, transforming into the relativistic one. We are interested here in the transition from quantum mechanics (in which quantization plays a fundamental role) to classical (in which the energy levels discontinuity is not observed). Let us start from the formula for energy (7.4.10). From this formula and Figure 7.7, one can see that the distance between adjacent levels increases with increasing the quantum number n. However, if one analyzes not the absolute but the relative value of energy, this fraction with increasing n decreases. In fact, (En1En)/En E/E and E (n 1)2 n2 2 n 1 1 2 ⬇ 0, E n n2 n
(7.4.13)
manifestation of quantization decreases rapidly. Consider now (on the qualitative level) what will occur if the depth of a potential box becomes finite; let it be denoted U0 (Figure 7.9). Suppose herewith, that the total particle
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U
U0 (x)
ekx
sin kx
e−kx
E
0
L
x
Figure 7.9 A wavefunction of a particle in a one-dimensional potential box of a definite depth.
energy E is less than U0, i.e., the particle remains bounded. In this case the potential U in the Schrödinger equation (7.3.5) will not be equal to infinity and the wavefunction in areas I and II will not be zero. Moreover, the expression in brackets (E – U) is negative. In order to avoid the operation with complex qualities let E and U change their places in the brackets ( x ) [U 0 ] ( x ) ( x ) k 2 ( x ) 0.
(7.4.14)
This problem can be solved exactly, but we will restrict ourselves to qualitative consideration. From the given equation, one can see that the second derivative from the wavefunctions on the coordinate in areas I and III must have the same sign as the function itself. An exponent AIexp( kIx) satisfies this condition. In the area I (∞ x 0), the requirement for the wavefunction to be finite corresponds to a sign “” in the exponent. Therefore, the solution for I(x) must be of the form AI exp(kIx). In area III (L x ∞), for the preservation of the wavefunction to be finite, the sign in the exponent must be negative: III(x) AIII exp(–kIIIx). Constants AI and AIII are to be determined from boundary conditions I(0) II(0) and II(L) III(L). The solution for wavefunction expression and conditions of quantization are, certainly, changed, though not in principle. The graph of relationship 2(x) for n 1 is schematically depicted in Figure 7.9. The most important point is that the particle wavefunction is not zero even in those areas of space where E U0, i.e., where the total energy is less than the potential or where the kinetic energy is formally negative. In classical physics, this cannot be the case: total energy must always be higher than potential since kinetic energy cannot be negative; the particle cannot exist where this condition is not fulfilled. The explanation of this apparent confusion is contained in the uncertainty principle! The fact is, potential energy depends on the particle positions, but kinetic energy is a function of momentum. Therefore, a particle in quantum mechanics cannot simultaneously possess precise values of potential and kinetic
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energy. Quantum mechanics operates mainly with total and potential energy, and only in some cases considers kinetic energy when a system does not possess potential energy at all.
EXAMPLE E7.6 An electron is in a one-dimensional potential box of infinite depth and width of l. Calculate the probability to find the electron in the excited state (n 2) in the middle third of the box (see Section 7.4.2) and Figure E7.6).
n=2
Solution: The probability to find a particle in an interval x1 x x2 is defined by an integral x2
w ∫ n ( x ) dx, 2
x1
where (x) is normalized wavefunction corresponding to a given state. In our case this function is 2 n n ( x ) sin x (n 2) (eq. (7.4.12)). Substitute all given values into this equation, we obtain x
w
2 2 2 2 sin xdx. x∫ 1
In our case x1 (l/3) and x2 (2l/3). Using a trigonometric equation, sin x 2
2 1⎛ 4 ⎞ x ⎜ 1 cos x 2⎝ ⎟⎠
we arrive at w 0.195. 7.4.3
A potential step
The results we obtained in preceding sections can help us to analyze quantitatively some other quantum mechanical problems, not having analogues in classical physics. We will consider further a potential step and a potential barrier.
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U(x) |(x)|2 U0
A2I
A2Iexp(-2k IIx) E
I
II
x
Figure 7.10 A rectilinear potential step.
The graph of potential energy as a function on the distance (along the axis of particle motion x) in quantum mechanics is referred to as a potential step: U(x) 0 at x 0 and U(x) U0 at x 0 (Figure 7.10). This problem can be solved at two total energy E: higher and lower the heights of the potential step U0. The latter case E U0 corresponds to an infinite particle motion; consequently, the energy E can accept a continuous spectrum of values. We will move our attention to another case with E U0. The problem can be solved analytically in the framework of Schrödinger’s equation and the standard conditions for (x), in particular using the requirement to be continuous at the step boundaries (at x 0). As in preceding cases, divide the problem into two parts: x 0 (area I) and x 0 (area II). In area I the particle motion is infinite; it can be described by the periodic wavefunction I(x) (7.4.2) with constant probability of finding a particle in any point of this area. Let us denote the amplitude of the de Broglie wave AI, then ⏐ I(x)⏐2 AI2. In area II the energy E U0, therefore all that has been said in Section 7.4.2 about the wavefunction behavior outside a potential box of finite depth (see above) is applicable to the particle falling onto the potential step (Section 7.4.2). The Schrödinger equation is the same as above (eq. 7.4.14). The solution of this equation is analogous to the case of exponential decrease of the probability of finding the particle while moving away from the boundary, i.e., AII exp(-kIIx). This wavefunction corresponds to the probability reduction of finding a particle under the step at x 0. From the condition of the wavefunctions continuity, it follows that AI AII. In Figure 7.10 a graph of the wavefunction⏐ II(x)⏐2 AI2 exp(–2kIIx) is presented. It can be seen that the probability of finding a particle under the potential step is exponentially decreased when moving away from the border. Considering this problem, we ignore the possibility of a wave reflection from the potential step boundary: this does not change the essence of the solution and is not important for us at the moment. 7.4.4
A potential barrier: a tunnel effect
Let us apply the results obtained to the consideration of the problem of a potential barrier depicted in Figure 7.11. Consider the case when the total particle energy E is lower than
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U ||2 A2Ie−2kIIx
U0 A2I
A2Ie−2kII d E x I
0 II d
III
Figure 7.11 A rectilinear potential barrier.
the height of the barrier U0. As previously, we subdivide the problem into three areas: area I (−∞ x), area II (0 x d) and area III (d x). The Schrödinger equation may be written as follows: ( x ) k 2 ( x ) 0,
(7.4.15)
for areas I and III with k2I,III (2mE/2), and ( x ) k 2 ( x ) 0
(7.4.16)
for area II with kII2
2m(U 0 E ) 2
Solutions should be sought in the following form: In area I: I(x) AI exp(–ikIx) BI exp(ikI x); In area II: II(x) AII exp (–kIIx) BII exp(kIIx); (7.4.17) In area III: III(x) AIII exp(ikIIIx). In these expressions, the amplitudes marked by the letter A correspond to the wavefunction propagation from left to right and amplitudes B describe a reflection from external and internal edges of the potential barrier. We shall not take reflected waves into account and, accordingly, shall not consider the waves containing the amplitudes marked by B (the exact solution using the boundary conditions and the wavefunction’s properties does not present a particular difficulty though they have no special interest). We denote the incident wave amplitude as AI and will express all other qualities using this value. A graph of the wavefunction’s amplitudes is given in Figure 7.11. In area I, the value of the square of the wavefunction amplitude AI2 is depicted. As was shown earlier for free particles, this value does not depend on coordinates and, within the given area, is constant. In area II the solution is not periodic but exponential (as was obtained when solving a potential
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step problem). The boundary conditions consideration gives: AI(0) AII(0). So the solution for the second area is given by the expression: II ( x ) AII exp(kII x ), which is also depicted in Figure 7.11. In the third area a motion once again is infinite and, accordingly, wavefunction is presented by periodic function and constant probability density. Its absolute magnitude is defined by one of the boundary conditions (condition of continuity). It links AIII with AI by the correlation 2 | III ( x ) |2 AIII AI2 exp(2 kII d ).
(7.4.18)
Since kI kIII, the de Broglie wavelengths in areas I and III are the same, whereas the amplitudes are different. The result also shows that, regardless of the energy of the falling particles, they can penetrate the barrier from area I to area III. Let us estimate what part of the particles falling onto the barrier penetrates the barrier or, what amounts to the same thing, how probable is it that a single particle shall penetrate it. We can call this value the barrier transparency and denote it by the letter D. Because of the fact that squares of amplitudes describe the “intensity” of de Broglie waves in a particular area, the barrier transparency can be defined as the ratio of their “intensities,” or
D
III (d ) I (0 )
2
2
AI2 exp(2 kII d ) AI2
⎛ 2d ⎞ 2 m(U 0 E) ⎟ . exp(2 kII d ) exp ⎜ ⎝ ⎠ (7.4.19)
One can see that the transparency of the rectangular potential barrier depends exponentially upon several factors: on the particle mass, on the barrier width and on the difference (U0 – E). The table below illustrates the barrier transparency for electrons upon the barrier width. d (10–10 m) D
1.0 0.1
1.5 0.03
2.0 0.008
2.5 10–7
If the barrier is not rectangular and its form is described by the function U(x) (Figure 7.12), the barrier transparency can be given by the expression
⎛ 2b ⎞ D exp ⎜ ∫ 2 m [U ( x ) E ] dx ⎟ . ⎝ a ⎠
(7.4.20)
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U
E
0
x
L
Figure 7.12 A potential barrier of an arbitrary form.
EXAMPLE E7.7 An electron with an energy E 4.9 eV moves in the positive direction of the x-axis (see Figure 7.11). The height of the potential barrier is U0 is 5 eV. At what width of the barrier will the probability to penetrate it be w 0.2? Solution: The penetration probability (the barrier transparency) is given by the eq. (7.4.19) ⎛ 2d ⎞ w⎜ 2 m (U 0 E )⎟ ⎝ ⎠
and hereafter, nw
2d 2 m(U 0 E ).
It is possible now to find the main equation:
d
n(1w) 2 2 m(U 0 E )
.
Calculation gives d 4.95 1010 m 0.495 nm. 7.4.5
Tunnel effect in chemistry
An approximate formula for the temperature dependence of the reaction rate coefficient K is given by the Arrhenius equation: ⎛ Q⎞ K exp ⎜ ⎟ , ⎝ T ⎠
(7.4.21)
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where Q is an activation energy and is the Boltzmann constant. If the activation energy is constant, a plot of ln K versus 1/T gives a straight line. This equation is often given in so-called Arrhenius coordinate (Figure 7.13): ln K
Q . T
(7.4.22)
In order to give a demonstrative presentation of the activation energy, let us imagine some group of atoms having two equilibrium states in two potential wells. Let them be characterized by two potential curves as depicted in Figure 7.14 (refer to Section 1.5.4). Such a state can be characteristic, for instance, for two atoms creating a molecule after their collision, or for two independent molecules forming a more complex molecule, or monomer molecules joining in polymer chains, and others. Certainly, different states are characterized by potential wells of a different form and depth: the more stable states have a deeper potential well. Suppose that the energy of an initially stable (ground) state in one potential well is characterized by the energy level noted in Figure 7.14 by a horizontal line. In order for the reaction to occur, the system should overcome the potential barrier Q, noted in the same scheme. This can occur either if the system has enough energy and can overcome the potential barrier Q (concentration of such active particles was calculated in eq. (3.3.7), Figure 3.6) or by means of penetration of the potential barrier. The overwhelming majority of chemical reactions really complies with the Arrhenius law and, in the logarithmic scale, is schematically expressed by a straight line (Figure 7.13). However, not long ago, the effect of the limitation of chemical reactions’ ratio at low temperatures was observed, the ratio became constant (Figure.7.13), in contradiction to Arrhenius theory. An explanation of this behavior can be given within the framework of the tunnel mechanism. In Figure 7.14, it can be seen that a system can overcome the potential barrier, not by climbing over the barrier, but having penetrated the barrier from one potential well into another one (in the manner of using an underpass in a mountainous area). From eq. (7.4.19)
ln K
l/T
Figure. 7.13 A chemical reaction ratio K versus reciprocal temperature 1/T.
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U
Q
Figure 7.14 A Scheme of activation (over a potential barrier with activation energy Q) and tunnel (penetration of a barrier) mechanisms of chemical reactions.
it can be seen that the barrier transparency indeed does not depend on temperatures. The penetration of the barrier is purely a quantum mechanical effect. As an example, we will describe the polymerization reaction of formaldehyde. In Figure 7.15, the potential curves of formaldehyde both in monomer and polymer states are shown. From a comparison of the curves it can be seen that the polymeric state is more stable than the mixture of monomers. The scheme above shows chain linkage in monomers and in polymer. In order for a monomer molecule to join the polymer chain, an electron from the double CO bond should abandon it and “penetrate” into the space between the atoms of carbon and oxygen of the adjacent molecule with the formation of a single bond C–O (this is shown in the middle of the scheme above). Certainly, herewith a redistribution of electron density occurs, developing and changing the interatomic distances: the density of polymer is higher than of the monomer mixture (the corresponding distances df, di and d are shown in the scheme). Since the polymerization process brings about greater energy of bonding (the potential well is deeper for the polymer) and the allocation of heat. The time of joining of the next monomer molecule to the already created polymeric chain, measured by very sensitive equipment, proves to be 102 sec, whereas computed from the Arrhenius equation gives a value near 1030 years. The effect of tunneling enables us to take quite a new look at some physical, chemical and, particularly, biological processes in respect of the organization and behavior of biologically active systems, accompanying all natural processes—from the formation of planets to the most complex particularities of biosyntheses. The scales of these phenomena are inconceivable though it is presently possible to assert that “tunneling effects” play a very important role in many processes of vital activity.
7.5
THE HYDROGEN ATOM
Consider now a more complex problem, which is very important for chemistry: the motion of a charged particle in a spherically symmetric electric field. In this case a particle’s
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polymer O df
C H
H
monomer O
O C
C H
H
H
O
O ∆d
H H
U(d) 2
di
C H
O
C H
C H
H
H
-CH2-Opolymer unit
1
CH2=O monomer molecule Q
df di
∆d
distance between adjacent molecules d
Figure 7.15 The tunnel effect of formaldehyde polymerization (after V.I. Goldanski et al.).
energy depends only on the distance from the center but not on direction (refer to expression (1.4.22)). The simplest problem of this kind is the motion of an electron in the field of a positively charged, dimensionless, heavy (in comparison with the electron) nucleus (proton), i.e., the problem of a hydrogen atom (and a singly ionized helium atom). The solution to this problem has an exceptionally important role in quantum mechanics and especially in quantum chemistry. Firstly, this is a problem that can be solved analytically (though some special mathematical functions must be used). Secondly, the solution is of great importance for chemistry where the electronic orbits arise from the solution; moreover the theory of the chemical bond has been worked out using the results. Thirdly, an empirically modified hydrogen atom’s orbits are widely used generally for heavier atoms because there are no other ways of achieving results. 7.5.1
The Schrödinger equation for the hydrogen atom
Let us choose the nucleus (proton) as the origin. Assume the nucleus is point-like. Because the proton’s electric field is spherically symmetric, we have to change from a Cartesian (x, y, z) to a spherical (r, , ) coordinate system. Figure 7.16 illustrates this transform: x r sin sin , y r sin cos , z r cos .
(7.5.1)
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z
r
y
x
Figure 7.16 Spherical coordinate system.
Note, that the z-axis is formally distinguished from the other axes. It can be said that this axis is selected: it is specially distinguished only geometrically, though it is always specially distinguished if there is an outer influence (electric, magnetic, etc.). It is said that z is the distinguished axis. The Schrödinger equation for hydrogen atom has the form (refer to (7.3.5))
(x,y,z )
2m [E U (x,y,z )] (x,y,z ) 0, 2
(7.5.2)
in which U(x, y, z) is the electron’s potential energy in the proton’s Coulomb field
U ( x, y, z ) U (r )
1 e2 , 4 0 r
(7.5.3)
(refer to Section 4.1.4), where r 兹苶 x2 苶苶 y2苶 苶z2苶. This expression has to be substituted into the Schrödinger equation and solved. As was noted earlier, this problem can be solved analytically; however, it takes much effort and a long time. Therefore, we will use a reasonable mathematical simplification paying more attention to the physical meaning. The Laplace operator (7.3.6) in spherical coordinates has the form: (r , , )
1 r2
⎧ ⎛ 2 ⎞ 1 ⎛
⎞ 1
2 ⎫ sin ⎨ ⎜⎝ r ⎬. ⎟ ⎜ ⎟
r ⎠ sin ⎝
⎠ sin 2 2 ⎭ ⎩ r
(7.5.4)
The Schrödinger equation in this case can be expressed as ⎧1 ⎡ 2
⎡
1 ⎤ ⎤ sin (r , , ) ⎥ r (r , , ) ⎥ 2 ⎨ 2 ⎢ ⎢
⎦ r sin ⎣ ⎦ ⎩ r r ⎣ r 2 1
⎪⎫ 2 m 2 2 (r , , )⎬ + 2 [E U (r )] (r , , ) 0. 2 r sin ∂ ⎭⎪
(7.5.5)
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Let us represent the wavefunction (r, , ) as the product of three independent functions where each function depends only on one argument: either on a radial coordinate or on the angles and : (r , q, ) R(r )()() R(r ) Y (, ),
(7.5.6)
Y (, ) ()()
(7.5.7)
where, is the so-called angular part of the wavefunction, and the function R(r) is the radial part. Consider first the electron’s motion along a spherical surface at fixed radial coordinate. This corresponds to the expressions r const. and Y/ r 0. This permits us to find the electron’s probability distribution on a sphere of fixed radius. The first term in eq. (7.5.5) becomes zero and therefore, the equation for the angular part of the wavefunction will be of the form 1 r2
⎧⎪ 1 ⎡ ⎫⎪ 1
2 ⎤ sin Y ( , ) Y (, ) ⎬ ⎨ ⎢ ⎥ 2 2
⎦ sin ⎩⎪ sin ⎣ ⎭⎪ 2 2r + 2 [E U (r )]Y (, ) 0.
(7.5.8)
The problem has been reduced to the investigation of the motion of a body along a sphere with fixed radius. Such a problem is called the rigid rotator; the classic version of which was considered in Section 1.3.9 (Figure 1.17). Recall that the electron’s angular momentum L in the framework of the Bohr model is L [rp] m[r].
(7.5.9)
It is perpendicular to the two vectors r and p m. If the angle between the two vectors r and p is 90°, the magnitude of L is L mr
(7.5.10)
It was shown in Section 1.3.9 that the rotation of two masses with distance d between them around a motionless center of mass O can be represented by the planar rotation of one mass (m1m2)/(m1m2)(the reduced mass) around the axis that passes through the center of mass. The result relates both to rotation of a diatomic molecule relative to its center of mass and to the rotation of the electron relative to the nucleus. The moment of inertia is called the reduced moment of inertia Ie I d 2 ,
(7.5.11)
where d being the interparticle distance. It is worth to notice that the reduced mass for hydrogen atom is half of electron mass whereas for hydrogen molecule is half of hydrogen atom mass. The kinetic energy of rotation is K
L2 L2 . 2 I 2d 2
(7.5.12)
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One can see that the same approach applies both to diatomic molecular rotation and to the hydrogen atom. 7.5.2.
The eigenvalues of the electron angular moment projection L z
The kinetic energy of a free particle forms a continuous spectrum, i.e., can possess any value. Within the framework of quantum mechanics we can ask: is the spectrum of values of the kinetic energy of a molecule’s free rotation (within the framework of the rigid rotator model) as well as an electron in the hydrogen atom either discrete or continuous? The answer is not a priori obvious. To answer this question we need first to solve the Schrödinger equation (7.5.8) with boundary conditions imposed on the wavefunction. For the rotation of the electron in the hydrogen atom, it is necessary to find a solution for the function Φ(). With reference to Figure 7.16, it will be noticed that this function describes the electron’s rotation in a xy plane, which is perpendicular to the z-axis. Consequently, it describes the behavior of the projection of the angular moment L onto the z-axis, i.e., Lz. It was shown above that free particle translational motion is described by the wavefunction (x) a exp(ipxx/) (time dependence is not important here). If we remember that the formulas of translational and rotational motion have the same structure (Section 1.3.9, Table 1.1), it is easy to write down the wavefunction for rotation by replacing px on Lz and x by . Therefore, we can write ⎛i ⎞ () a exp ⎜ Lz ⎟ . ⎝ ⎠
(7.5.13)
The condition of a single-valued solution has the form () ( 2).
(7.5.14)
Certainly, one revolution brings a system back to the initial state. Combining eqs. (7.5.13) and (7.5.14), and using the Euler formula we obtain: ⎛i ⎞ ⎛i ⎞ ⎛i ⎞ a exp ⎜ Lz ⎟ a exp ⎜ Lz ⎟ exp ⎜ Lz 2⎟ , ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ and further, ⎛ L 2 ⎞ ⎛ L 2 ⎞ cos ⎜ z ⎟ i sin ⎜ z ⎟ 1. ⎝ ⎠ ⎝ ⎠
It follows from this equation that (LZ/)2 m(2), where m is integer. And finally, Lz m.
(7.5.15)
Thus, without solving the Schrödinger equation, starting only from the boundary condition of uniqueness of the wavefunction, an important conclusion has been obtained: the
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projection of the angular momentum onto the distinguished z-axis can take only integer values of . The integer m, equal to 0, ±1, ±2, ±3, etc., is a quantum number that defines the value of Lz projection. The wavefunction Φ() is, consequently, expressed as m () a exp( im).
(7.5.16)
The magnitude a can be found from the normalization condition: 2
2
2 im im d a 2 2 1, ∫ () d a ∫ e e 2
0
0
i.e., a 2 (1 2) and finally ()
1 im e . 2
(7.5.17)
(7.5.18)
Thus the angular part Φ() of the wavefunction has been specified. 7.5.3.
Angular momentum and magnetic moment of a one-electron atom
Eq.(7.5.8) is modified by taking into account that E – U is the kinetic energy K because of the fact that in this particular case the potential energy of free rotation is zero (K E), and m is the reduced mass of the rigid rotator. Since the angular part of the wavefunction is the product of two functions Y(, ) Φ()Θ(), we can obtain () ⎡
() ⎤ () 2() 2r 2 sin 2 ( E U )()() 0. (7.5.19) ⎢ sin ⎣
⎥⎦ sin 2 2
Multiplying the equation by sin2/Φ()Θ() we can obtain 2 sin ⎡
() ⎤ 1 2() 2 2r sin sin E 0. () ⎢⎣
⎥⎦ () 2 2
(7.5.20)
Note that in the last equation each term depends either on or on . They can be grouped in the following manner: 2 sin ⎡
() ⎤ 1 2() 2 2r sin sin E . Θ() ⎢⎣
⎥⎦ () 2 2
(7.5.21)
A function Θ() is on the left-hand side and a function Φ() is on the right-hand side. Therefore, this equality can be satisfied only if both sides are equal to a constant value. Since the function Φ() (7.5.18) is already known, we can find that constant. In fact, this function is equal to:
1 2() m2 . () 2
(7.5.22)
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It means that 2
sin ⎡
() ⎤ 2r sin sin 2 2 E m 2 . () ⎢⎣
⎥⎦
(7.5.23)
This illustrates the previous statement that the wavefunction of the hydrogen atom can be presented as a product (7.5.6). The eq. (7.5.23) is a Legendre equation. It follows from the theory of Legendre equations that they have solution only if the coefficient of expression at sin2 is equal to 2r 2 E ( 1), 2
(7.5.24)
m .
(7.5.25)
where l is an integer with
Taking into account eq. (7.5.12) one can write E
2 ( 1). 2I
(7.5.26)
Therefore, we have found the energy eigenvalues of hydrogen atom and rigid rotator energy. The angular eigenvalues will be given later in Section 7.5.7. Thus the answer to the question that we posed earlier is: the rotational energy can take only definite discrete values (i.e., it is quantized). Using eq. (7.5.12), we can obtain also L2 2l(l1) or L ( 1),
(7.5.27)
i.e., the quantization of the kinetic energy of the rigid rotator originates from the quantization of the absolute value of angular momentum. The expression (7.5.27) and a gyromagnetic ratio (see Section 5.2.1) allow us to determine an orbital magnetic moment of the one-electron atom. Because the gyromagnetic ratio is /L ge/2m (where g 1), the magnetic moment is as:
e 2m
( 1) B ( 1).
(7.5.28)
The value B (e)/(2m) is known as the Bohr magneton. The general solution of eq. (7.5.23) for the function Θ() can be written with Legendre polynomials. The expressions for Y(, ) for the quantum number l 0,1,2 are given below in Table 7.1. Two important conclusions can be drawn from the solution obtained. One concerns the quantization of the rigid rotator energy; the other describes the properties of angular momentum in quantum mechanics. We will use the first one later in Section 7.8.2 in the description of rotational spectroscopy; here, we consider the properties of the angular momentums in quantum mechanics. As has already been noted, if one has to select a particular axis it is usually identified with the z-axis. It follows from the results described by eqs. (7.5.15), (7.5.25) and (7.5.27), that the
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Table 7.1 Wavefunctions for s-, p- and d-electrons n
l
ml
1
0
0
1s
0
⎛Z⎞ 2s 4 2 ⎜⎝ a0 ⎟⎠
2
2
2
3
3
3
3
0
1
1
0
1
1
2
1 ⎛Z⎞ ⎜ ⎟ ⎝ a0 ⎠ 1
3 2
e 32
(2 )e( 2 ) 32
2pz
⎛Z⎞ 4 2 ⎜⎝ a0 ⎟⎠
32
2p x
⎛Z⎞ 4 2 ⎜⎝ a0 ⎟⎠
32
2p y
⎛Z⎞ 4 2 ⎜⎝ a0 ⎟⎠
32
0
⎛Z⎞ 3s 81 3 ⎜⎝ a0 ⎟⎠
32
0
3pz
2 ⎛Z⎞ 81 ⎜⎝ a0 ⎟⎠
32
3p x
2 ⎛Z⎞ 81 ⎜⎝ a0 ⎟⎠
32
3p x
2 ⎛Z⎞ 81 ⎜⎝ a0 ⎟⎠
0
±1
±1
0
1
e( 2 ) cos
1
e( 2 ) sin cos
1
1
e( 2 ) sin sin
(27 18 2 2 )e( 3)
(6 − )e( 3) cos
(6 )e( 3) sin cos
(6 )e( 3) sin sin
⎛Z⎞ 3d z 2 81 6 ⎜⎝ a0 ⎟⎠ 1
3d xz
2 ⎛Z⎞ 81 ⎜⎝ a0 ⎟⎠
32
2 e( 3) (3cos2 1)
32
2 e( 3) sin cos cos
(Continued )
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Table 7.1 (Continued ) n
l
ml
2 ⎛ Z⎞ 81 ⎜⎝ 0 ⎟⎠
32
3
2
±1
3d yz
3
2
±2
3d 2
3d xy
⎛Z⎞ 81 2 ⎜⎝ a0 ⎟⎠
x y2
2 e( 3) sin cos sin
⎛Z⎞ 81 2 ⎜⎝ a0 ⎟⎠ 1
1
32
2 e( 3) sin 2 cos2
32
2 e( 3) sin 2 sin2
Note: The following units are accepted in the table: (Z/a0)兹2苶r苶; a0 (h2/4e2)
absolute value of angular momentum and its projection on the z-axis satisfy certain conditions; in particular they signify that the absolute value of the angular momentum vector is always longer than the length of any (even the longest) of its projections (Figure 7.17). Indeed, mmax l and l 兹苶 l(苶 l苶 1) 兹苶 l 2苶 苶l and , therefore Lz is always smaller than |L| (with one exception when l 0). Therefore, the vector L is fixed in the space within angle . It follows also from the results presented that the vector L can never lie on the z-axis. This law concerning angular momentum vectors is called spatial quantization. Figure 7.18 shows the spatial quantization for several l values. Moreover, the theory gives no way of finding the other projections (Lx and Ly); they are completely nondeterminated. This fact can conditionally be added to uncertainty principles. Note that this is not a lack of theory but is the law of nature. Generally speaking, there is no need to give visual evidence of a given fact: this is the nature of things. However, attempts to present a situation as a precession of the vector L around the axis z are sometimes met. According to our understanding, this interpretation is unlikely to be correct: with precession a certain frequency and, further, energy must be bound though no one has observed such an additional energy. More acceptable is a model of “uniform smearing” of the L vector upon a conical surface at a fixed angle, however this model also does not give any reliable ideas for further development. At a given quantum number the quantum number ml can accept 2l1 values. Therefore, the angular momentum vector can be situated on one of 2l1 cones (Figure 7.18). The opening angle can be found as: cos
LZ m L ( 1)
(7.5.29)
where is an angle between the vector L and z-axis and m is one of the quantum numbers connected to l. Complications appear also when we have to sum up the angular momentums vectors of several rotators (electron orbits). All restrictions described above are valid in this case too.
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z LZ |L |
0
Figure 7.17 A space quantization of the angular momentum.
Suppose, for example, that it is necessary to sum up two vectors L1 and L2 in order to obtain the third (resulting) vector L. One has to remember that the vectors being added cannot be exactly parallel or antiparallel to each other. Really, since 兹苶 (苶 )(苶 1苶 苶 2苶 苶 1) 兹苶 苶2苶 (苶 1苶 )苶1兹苶苶 苶 1苶), the length 1 1苶 2(苶 2苶 of the resulting vector L is always less than the sum of the absolute values of composite vectors L1 and L2 (兩L兩 兩L1兩兩L2兩). Formally, one can write L L1 L2, where L1 兹苶 1苶 (苶 1) and L2 兹苶苶 苶 1苶), 1苶 2(苶 2苶 the resulting vector having magnitude L 兹 苶 1). All the vectors have z-projections 苶(苶苶 which are quantized: L1z m1, L2z m2, Lz m, where mmax for all three vectors. Since projections are scalar values, projection of the resulting vector will be found as an algebraic sum of the projections of the individual vectors. The maximum value of the projection of the vector L is Lz , where l2, whereas the shortest length of vector L will be if projections have opposite signs, L l2. Thereby, when adding the angular momentum vectors, the resulting vector L can take all values from maximum to minimum with quantum numbers , running over all values from 1 2 to 1 2 in integer steps. Figure 7.19, illustrate this summation. One should remember that the resulting vector L in quantized as well relative the new axis z. Since the magnetic moment vector is tightly connected with the angular momentum, everything that has been said applies to them both. The main conclusion from the whole consideration is that the summation of angular momentum vectors in quantum mechanics is accomplished according to a certain scheme. The procedure L1 L2 L must be understood as a summation of two quantum mechanical angular momentum vectors giving the third vector, complying to the same rules. An attempt to give a primitive image is presented in Figure 7.19. However, it appears that there is no need to carry out such a huge drawing especially in the case of several vectors, reducing the geometric summation to the combination of quantum numbers. Sometimes, the prescription given above for vector summation is reduced to the symbolic sum l 2, suggesting combining quantum numbers 1 and 2 according to the general rule presented above. After introducing electron spin, we have to deal with the set of quantum numbers related to orbit and spin.
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/2
+½ (a) B −½
) ×2
(1 (1 ×2
×3
)×
l=½ 2l + 1 = 2
3/
2
1× 2
+1
1×2
(b) B
2 1×
l=1 2l + 1 = 3
−1
(3/
+½
2)× (5/
2)
+
(c) B
(3/ 2)× (
5/2
)
l= 2l + 1 = 4
2)
(5/ 2)×
(3/ −½
)
5/2 )×( (3/2
−
Figure 7.18 Permitted projections of the angular momentum vector. (a) 1/2, (b) 1, (c) 3/2.
7.5.4 A Schrödinger equation for the radial part of the wavefunction; electron energy quantization Assume that the values of angles and are constant. Thus we set all partial derivatives of these arguments to zero and look at the behavior of the radial part of the wavefunction. The Schrödinger equation (7.5.5) in these circumstances will take the form
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z′ L
L2
L1
Figure 7.19 Summation of two angular momentum vectors.
1 ⎡ 2 R(r ) ⎤ 2 m r [E U (r )]R(r ) 0,
r ⎥⎦ 2 r 2 r ⎢⎣
(7.5.30)
where instead of the reduced mass the electron mass m is again substituted. This equation has solutions at the definite values of the eigenvalues E, and the wavefunctions R(r) are expressed via special functions, the adjoined Laguerre functions. The eigenvalues of energy can be expressed in terms of the fundamental physical constants: En
e4 m 1 13.56 2 eV. 2 2 2 n 32 0 n
(7.5.31)
In this expression the energy E is given in eV in order to avoid a source of errors caused by the fact that in different systems of units, the constants are written differently. The integer n in this expression is the principal quantum number; it determines the electron energy E in the hydrogen atom. It can take any positive nonzero value. Eq. (7.5.31) describes the discrete spectrum of electron energies (Figure 7.20). From this drawing, one can see that value –13.56 eV is the energy of the ground state of the hydrogen atom. Figure 7.20 also depicts the values of some excited states with n 1–7. Excitation of an electron up to the energy E 0 transfers it to a state with a continuous energy spectrum and infinite motion. This corresponds to the removal of the electron from the atom or, in other words, to the ionization of the atom. The first ionization potential is the lowest energy that should be given to an atom from outside to tear the ground state electron away. Consequently, the first ionization potential of a hydrogen atom is 13.56 eV. If the atom is already excited, the ionization energy is less. Accordingly, one can distinguish the second (3.4 eV), third (1.5 eV), etc., ionization potentials of the hydrogen atom. The transitions between electron energy levels, associated with absorption or emission of a quantum of electromagnetic radiation, are not limited by selection rules, all transitions are permitted. The main equation is: En En ,
(7.5.32)
where n and n are principal quantum numbers corresponding to the initial and final energy levels. Equation (7.5.32) corresponds to serial formula (6.7.13) (see Section 6.7).
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E(eV) 0
n ∞
7 6
5
4 3 2
−13.56
1
∞
Figure 7.20 The hydrogen atom energy levels; the visible part of hydrogen atom radiation (Balmer series) is presented below: wavelengths —656, —486, γ—434 and δ—410 m. A spectrum border is 365 m.
Different series of spectral lines are associated with the quantum number n. According to the quantum number n, the so-called series is distinguished: Lyman series (n 1), Balmer series (n 2), Pashen series (n 3), etc. The Balmer series lies in the visible region of the spectrum. The eigenfunctions R(r) are presented in Section 7.5.7 and in Figure 7.22. As has already been mentioned above, the Schrödinger equation can be solved exactly for several particular problems only, including the hydrogen atom and the helium ion He. Even for the neutral helium atom with its two s-electrons, the equation is rather complicated because the potential energy U must take into account not only the interaction of each electron with the nucleus, but the electrons’ interaction with each other as well. The expression for the total potential energy takes the form
U (r1 , r2 ) ⬃
1 ⎡1 1 1 ⎤ , 4 0 ⎢⎣ r1 r2 r12 ⎥⎦
(7.5.33)
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r1 and r2 being the radial coordinates of each electron, and the last term takes into account the mutual interaction of the electrons. One can imagine to what extent the Schrödinger equation becomes complicated. This equation can be solved only approximately, though modern methods of quantum chemistry enable a solution with good enough accuracy to be obtained. However, cases exist when the solution for the hydrogen atom can be successfully used for more complicated atoms. These are the so-called hydrogen-like atoms, in which a particular electron is rather more distant from the nucleus than the other electrons. The latter form a closed shell of inner electrons, and the residual electron is situated in the complex electric field. In this case the changes in the energy are taken into account by atomic number Z:
E
13.56 2 Z eV. n2
(7.5.34)
It must be emphasized however, that the value Z in this case is not an exact atomic number because an external, valence electron is presented in the effective field, created both by the nucleus and by inner electrons. This effect is termed “screening” because the inner electrons make a negative shield around the positive nucleus. Therefore, the effective Z value differs from the ideal atomic number and is often called the effective atomic number. 7.5.5
Spin of an electron
Besides an orbital angular momentum, the electron has an inherent angular momentum that has been called spin. This is often explained as the rotation of the electron around its own axis. However, neither orbital nor spin “motions” have been considered as motion, but as a state, characterized by definite quantum numbers. It was found by Einstein and deHaas (1915) and Stern and Gerlach (1922) in classic experiments that the number of possible projections of the electron angular momentum vector on the selected axis is two. So in accordance with eq. (7.5.3) one can obtain s ½ (because 2 2s1), s being the spin quantum number. This signifies that the length of electron intrinsic angular momentum vector LS according to the general rule for the magnitude of the spin angular momentum vector is 兹苶 s(苶 s苶1苶), therefore, LS
3 3 . 4 2
(7.5.35)
According to the quantization rule this means that the spin projections onto the selected z-axis can have two values, namely, 1 L m z s 2 where mS is a quantum number of spin projection.
(7.5.36)
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The quantum number mS does not influence the electron wavefunction significantly; however, it influences the electron distribution among the energy levels and quantum cells to a great extent. The Pauli principle is a consequence of these quantum mechanical rules. Because of the fact that not the absolute value of the angular momentums but their projections are measurable quantities, it can often be said that the electron spin is (1/2) or simply (1/2). The spin spatial quantization is depicted in Figure 7.21. In full analogy with the orbital state, the spin magnetic moment can be determined through the gyromagnetic ratio g (Section 5.2.1). For spin g 2 (in e/2 m units), the absolute value of spin magnetic moment is therefore equal to M 2
1 1 3 e ( 1) 2 B , 2m 2 2 4
(7.5.37)
The projection of the magnetic moment on the z-axis is M z B .
(7.5.38)
When chemists look at atomic and/or ionic electron states, and distribute electrons over the energy levels and/or fill in corresponding quantum cells by arrows in accordance with the Pauli exclusion principle (not more that two electrons into one quantum cell) and the Hund’s rule, they bear in mind exactly these two projections of spin magnetic moment. Note that arrows are represented by schematically signs and , since projections themselves are not vector but algebraic values. 7.5.6
Atomic orbits: hydrogen atom quantum numbers
Now it is desirable to generalize the information that we have obtained following the analyses of the Schrödinger equation for the hydrogen atom and introducing the electron spin. We mean the systematization of the values characterizing an atom’s state. Herewith
+½
)
1½
½(
B −½
½(
1½
)
Figure 7.21 Spin space quantization.
S=½ 2S + 1 = 2
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we will try to analyze the approaches of physicists and chemists, which in some respect look different. A typical quantum mechanical object such as an atom possesses some classic characteristics that are unexplainable within the framework of generally accepted presentations (refer to Section 6.7) (no orbital motion, yet the existence of angular momentum; no rotation of an electron around its own axis, yet intrinsic angular and magnetic moments, i.e., spin, etc.). As a result, these terms are used irrespective of their classical sense. Moreover, the Bohr model is a transition from the purely classical presentations to the quantum mechanical ones: the motion of electrons along the orbits is accepted, but not all orbits are permitted; the angular momentum is accepted though its values, and orientations are the subject of strict limitation. It is possible to consider the Bohr model as a transition from the classical mechanics to quantum, with the preservation of many of its attributes. As a result, many of Bohr model notions will often be met in order to simplify the students’ understanding. Quantum mechanical angular momentums and its projections, as well as the electron energy, are defined now by quantum numbers: the electron energy is defined by the principal quantum number n, the angular momentum vector length is defined by quantum number l, its projection on the selected z-axis is given by a quantum number ml. Each of the quantum numbers enumerated is included into a particular wavefunction, and their product gives total wavefunction describing an electron distribution in the language of probabilities. Such an image of wavefunctions and their squares brings about the atomic orbits. For chemists, it is precisely the most valuable result. The quantum number mS is characterized by the z projection of the electron spin angular momentum. In first approximation, it does not influence the energy and wavefunction shape, but influences significantly the electron distribution among the energy levels. Consequently, the total wavefunction is the product of all three parts (refer to eq. 7.5.6): (r , , ) R(r )()() R(r )Y (, ),
(7.5.39)
the R(r) function depends on n and Y(, ) on l and ml. 7.5.7
Atomic orbits
The general expressions for R(r), Θ() and Φ() can be written using the special mathematical functions. They are presented in special literature on mathematics, quantum mechanics and quantum chemistry. We restrict ourselves to giving here the description of the main physical concepts of electron orbits as the basis for the theory of chemical bonding. The state of an electron in an atom is much more sophisticated than can be expected from the Bohr theory. Quantum mechanics shows that an electron can be found in any point of space, but the probability of its presence changes from point to point. The notion of an electron orbit appears more productive than electron clouds. Under the electron orbit, the physicist often understands the mathematical expression of the wavefunction itself corresponding to definite quantum numbers. In chemistry, the orbit is understood as a set of electron positions around a nucleus taking the probabilities into account. This probability is defined by wavefunctions R, Θ, Φ.
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Expressions for s-, p- and d-electrons in the analytical form in the spherical coordinate system (Figure 7.16) are given in Table 7.1. In the radial part of the wavefunction R(r) argument r is given in atomic units, i.e., in the unit of the first Bohr atomic radius a0 0.5292 1010m. Graphs of the R(r) functions (a) and the probability density curves, i.e., the probability density of finding an electron in the spherical layer with a thickness dr(dw/dr 4r2R2(r))(b) as dependent on r are depicted in Figure 7.22. It should be noted that R(r) at the point r 0 (i.e., on the nucleus) has the maximum value. However, this does not contradict common sense because the probability of finding an electron at point r 0 (Figure 7.22b) is equal to zero. A scheme for drawing graphs of the angular part of the wavefunction Y(, ) and its square Y2(, ) (particularly for pz orbit) is given in Figure 7.23a and b, respectively. The value of Y for the given is proportional to a line OM. It is worth noting that the function Y() is presented as spheres, whereas the Y 2() is presented as the elongated dumbbells more popular in chemistry. The wavefunctions in Table 7.1 (above) are presented for n 1, 2 and 3. In the first line the data for the 1s-state is given, in this case R(r) has a maximum at r 0 and falls down with increasing r. The Y(, ) function depends neither on , nor on , therefore the ||2 graph is spherically symmetric. The same is true for the 2s- and 3s-states. The analytical expression for n 2, and 1, ml 0 and 1 are given in the next three lines. It can be seen that the pz-orbit solution has the simplest look than the two others (px and py); such inequality is the result of the spherical coordinate system; moreover, the last ones have imaginary form. In order to obtain the expression in a real form, one should compose a linear combination of particular solutions, i.e., carry out the hybridization of orbits (since any combination of the solutions of the Schrödinger equation is also the equitable solution). We must use here the Euler formula and compose the linear combination of Y1,1 and Y1,−1 orbits: 1 1 3 Ypx (Y1,1 Y1,1 ) sin [cos i sin cos i sin ] 2 2 8 3 sin cos 8
(7.5.40)
and, Yp y
=
1 ⎛ 1⎞ (Y1,1 Y1,1 ) ⎜ ⎟ ⎝ 2i ⎠ 2i 3 sin sin . 8
3 sin [cos i sin cos i sin ] 8 (7.5.41)
The angular parts in the real form for electron d-states were also obtained in this way. In this form, they are widely used in chemistry. Having determined all parts of the wavefunction in any point, the r(r, , ), the total wavefunction can be derived by multiplication of all its parts. In the abstract case of the absence of any external influence when there are no arguments for the choice of quantization of axis z, all solutions of the Schrödinger equation and
0.8 4 0.4 n = 1, l = 0 0 0.8
2 1
0.4
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n = 1, l =0 0 1
4πr 2[R(r)] 2
0 n = 2, l =0
0 1
n = 3, l =0 0
0.4
n = 3, l = 0
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0 0.8
1
0.4 n = 2, l = 1
0 1
0 0.4
n = 3, l =1
0
n = 3, l = 1
0 0
1
2
3
4
5 (a) r
6
7
8
9
10
0
1
2
3
4 5 (b) r
6
7
8
Figure 7.22 A wavefunction’s radial parts R (a) and corresponding values 4πr2R2 (b) for some electron states.
9
10
7. Elements of Quantum Mechanics
n = 2, l =1
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z
z M
M
0
(a)
0
(b)
Figure 7.23 How to draw the wavefunction’s angular parts Y(, ) (a) and Y2(, ) (b).
all their linear combinations can virtually exist. However, there is no physical possibility of finding either all or some of them, because any attempt to select a quantization axis destroys the atom itself. This fact reveals a property of quantum mechanics: an instrument of investigation destroys the state of an object. If the atom considered falls into the orbit of other atoms, the mutual chemical influence makes essential changes to their state. In different circumstances, it can appear that other linear combinations can be more advantageous, for example, the well-known s-p and s-p-d hybrid orbits are created (see Tables 7.1). Note that the probability density of finding an electron in different points of an orbit graph is different. To depict the total electron distribution in three-dimensional form and to make them understandable to the reader is a very difficult task. Nevertheless, some efforts have been made to create a suitable image. The most recent is the presentation of electron density as a graph of the charge distribution dq(r, , ) |e|||2(r, , )dV as a spatial pattern in the form of isolines and three-dimensional pictures. The results of X-ray diffraction investigation are at present given in such a form. Graphs in the form of closed surfaces are often used in chemistry; inside a closed volume, a definite amount of atomic electrons are contained (more often 90%). Such a picture is presented in Figure 7.24 showing the orbits of different electron states in the hydrogen atom. Note that the orbits do not touch the origin (the nucleus). This is because of the fact that in this area the probability density of finding an electron is very small due to the radial part of the wavefunction (the argument r is too small). Therefore the total density is also small. It turns out that, even for hydrogen-like atoms, atomic orbits appear vastly more complex. Regrettably, it is impossible to obtain the exact solution even for these atoms. So in quantum chemistry different kinds of approximation are used, more or less successfully, to describe one system or another, and one atomic area or another. For instance, the factor is introduced as a multiplier in the exponent order of the radial wavefunctions to describe an orbit’s compression-expansion (Slater multiplier). Sometimes, not one but two, or even several, multipliers are used, each of which is better for describing the electron density near the nucleus or far from it. These empirical modifications for different atoms are given in quantum chemistry.
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z
1s
+
x y z
z
z
+
−
2px
−
+
x
y
2pz
2py
x
x
+
y
y z
z
+
−
+
−
+
−
x
x +
y
3dx2-y2
y
−
3dz2
z
z −
+ −
−
+
x
+
z −
−
x
+ y
y
y
+
+
3dxy
3dxz
x
− 3dyz
Figure 7.24 Representation of a wavefunction’s angular parts in different electron states.
EXAMPLE 7.8 A hydrogen atom is in a ground 1s state. Determine the probability of finding an electron w in a sphere of radius r 0.1a. Solution: The probability of finding an electron in a spherically symmetric 1s state is given by the normalized wavefunction 100(r)(1/兹苶 苶 a3)exp((r/a)) the volume ele2 ment being dV 4r dr. Therefore
dw
2
4 ⎛r⎞ ⎛ 2r ⎞ exp ⎜ ⎟ 4r 2 dr 3 exp ⎜ ⎟ r 2 dr. 3 ⎝ a⎠ ⎝ a⎠ a a 1
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It is convenient to move to the atomic units further r/a; therefore r2 2a2, dr ad and dw 4 exp(2)2d. The probability can be found by integrating dw in limits r1 0 to r2 0.1a (or from 1 0 to 2 0.1) 0.1
w 4 ∫ 2 exp(2)d . 0
Decompose the exponential into a MacLoren series exp(2) ⬇12 .. and limit by two terms; we can present the integral as 0.1
0.1
0.1
0
0
w 4 ∫ (1 2)2 d 4 ∫ 2 d 8 ∫ 3 d 0
4 3 3
4 0.1 0 8 4
4 −3 −3 0.1 0 10 0.2 10 . 3
Therefore the final result is w 1.53.103. It is useful to compare this result with Figure 7.22.
7.5.8
A spin–orbit interaction (fine interaction)
The above scheme of the Schrödinger equation solution did not take two circumstances into account: firstly, relativistic effects and, secondly, electron spin. The relativistic effects appear when a particle possesses high energy and consequently moves at a speed close to the light velocity. In an atomic planetary model, the inner electrons are nearest to a nucleus (E is negative and great in the absolute value, refer to eq. (7.5.34) at Z 30), are precisely the relativistic particle. In this case, one should take relativism into account; we will not, however, discuss this further. Consideration of electron spin raises the question of the so-called spin–orbital interaction influencing the electron energy values. The mechanism of this interaction consists of the fact that the electron orbital “movement” produces the magnetic field acting on its own spin. In order to make the picture more understandable, let us place the origin in an electron, the nucleus in this case appears moving along the circular orbit around the electron under consideration, creating a magnetic field at its position. Interaction of the magnetic moment of the electron spin with the magnetic field produced by a nucleus is called the spin–orbit interaction. The energy of this interaction must be taken into account when estimating the atomic total electron energy. Fortunately, such a complex procedure can be considerably simplified using the procedure of substitution of the vectorial summation by the combination of quantum numbers (see Section 7.5.3). To account for the spin–orbit interaction, the additional quantum number j is introduced, j being the sum of quantum numbers l and s j s 12
(7.5.42)
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Instead of the single energy level, characterized by quantum number n, two levels appear with quantum number l equal to l (1/2). (At l0 quantum number j accepts only one value j(1/2).). Accounting for the spin–orbit interaction, the electron energy becomes dependent not only on the principal quantum number n but on the quantum number j as well. Exact calculations give ⎛ ⎞ 13.56 2 13.56 Z 4 2 ⎜ 1 3⎟ En, j 2 Z ⎜ 1 4n ⎟ n n3 ⎜ j ⎟ ⎝ ⎠ 2
(7.5.43)
(as in expression (7.5.30), the energy is given here in eV units). Estimation of En,j shows that it is in the order 104 eV. The dimensionless value (e2/c) with good accuracy is equal to 1/137; it is called the constant of the fine structure. Its rational value has been the reason for a great deal of speculation about the rational correlations between atomic characteristics and, accordingly, the creation of the united theory of elementary particles. However, all of these came to nothing and were personal tragedies for a number of scientists. In spite of the fact that the energy of fine interactions is small, analysis of the last two formulas shows that because of fine interaction all energy levels with l0 split into two. This is developed, for instance, in the splitting of some spectral lines of atomic electron spectrums. So the well-known transition p→s in alkaline metal atoms instead of one spectral line contains two reliably measurable lines.
EXAMPLE E7.9 An excited electron is in a 3p-state. Find the change in the orbital magnetic moment of the atom’s transformation to a ground state. Solution: We have to find the difference MlMl2Ml1. The orbital magnetic moment depends only on orbital quantum number: Ml B 兹苶( 苶 苶1苶)苶. Therefore, in the ground state l 0 and l2 0. In the excited state, l 1 and Ml1 B兹2苶. Hence the difference is Ml B兹2苶. Execute the calculation remember that B 0.927 1023 J/ T, we arrive at the numerical value Ml –1.31 1023 J/ T.
7.6
A MANY-ELECTRON ATOM
As we already know, it is impossible to solve analytically the Schrödinger equation for atoms with two and more electrons. This prompted the development of new quantum mechanical methods of approximate solutions or modified solutions, equitable for the hydrogen atom, by introducing empirical adjustments. In this chapter we will consider intra-atomic interactions between electrons and the complication that this interaction causes. (Here we must emphasize that any complication in the theory forces the development of more sophisticated experimental methods of investigation,
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opening new possibilities of studying more precise questions relating to the chemical structure.) The wavefunctions of many-electron atoms will not be studied in this book; these are the subject of quantum chemistry. 7.6.1
Types of electron’s coupling in many-electron atoms
If there are several electrons in an atom, the total angular momentum (sometimes it is called a mechanical moment unlike magnetic) is obtained as a sum of mechanical moments of atomic electrons. Depending on the electron interaction energies there exists two ways to combine all mechanical moments (spin and orbital) into the total one: either to take into account the fine spin-electron coupling first (refer to Section 7.5.8), obtain the quantum numbers j’s and corresponding LJ or primarily unite all mechanical moments of all electrons, orbital and spin separately, obtain total L and Ls and finally obtain LJ. Such coupling is referred as Russell-Saunders binding. Consequently, to obtain the total angular momentum of a manyelectron atom one should first to add to each other the orbital moments of all electrons LL, then sum all spin moments into Ls and only afterwards to sum them both into the total one ( and s being orbital and spin quantum numbers of a single electron and L and S the quantum numbers of the total angular momentum of the whole atom). Herewith there is no need to combine vectors with provision for rules of the angular momentum summation in the quantum mechanics, it is easier do this by combination of the quantum numbers. Let us do the summation of the two electrons with quantum numbers l1 and l2, s1 and s2. Summation of orbital moments gives L 1 2 , 1 2 1, 冨 1 2 冨 .
(7.6.1)
Consequently, the total angular momentum LL, can accept as much values as many terms contain the series (7.6.1). Their values can be determined according the general rule (refer to 7.5.3) L L L ( L 1).
(7.6.2)
We can operate quite analogously with spin angular momentums. As s 1/2 and N is the number of electrons, the total spin angular momentums S can accept the values S N 12 , N 12 1, ..., 0,
(7.6.3)
S N 12 , N 12 1, ... , 12 ,
(7.6.4)
when N is even and
when N is odd. The total spin angular momentum is expressed by analogy with the previous expression L S S (S 1).
(7.6.5)
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In order to obtain the total (atomic) angular momentum (mechanical moment), we must put the quantum numbers together J L S, L S 1, ... , L S ,
(7.6.6)
where J is the total atomic quantum number; it determines the atomic mechanical moments L J J ( J 1). Its z-projections are LJ,z mJ where
(7.6.7) (7.6.8)
mJ J , J 1, J 2, ... ,J ,
(7.6.9)
overall 2J 1 values. The graphic summation will be given below. When the individual electron spin–orbit interaction energy prevails over the spin–orbit interactions of different electrons, we first have to sum the orbital and spin moments of every electron (to obtain all j’), then sum these j’ and further obtain the total atomic angular momentum quantum numbers and finally the overall quantum number J, characterizing the total angular momentum of the atom. This type of binding is called a j–j bond. Since the spin–orbit interaction is proportional to Z 4, this type of binding occurs mainly in heavy atoms. We will not consider further this type of coupling. Sometimes all states with the same quantum number J have the same energy, because different mI’ does not often influence it. In these cases, states are called degenerated ones. However, there are often cases when the energy with definite J and different mI’ have different energy. These are cases without degeneration. Sometimes, this only partly concerns a state. For instance, in a state with J 2 corresponding to 2J 1 values of mJ (2, 1, 0, –1, –2), all of them can possess the same energy. Such states are called fivefold degenerate ones. An external influence (for instance, a crystal field) can split one degenerated fivefold electron level into twofold and triple fold split levels (this situation is often encountered in the d-state of d-element complexes). Such a process is called lifting (or removing, or waiving,) the degeneration. In this case the degeneration is removed only partly. Which quantum number the degeneration lifting relates to is usually indicated. 7.6.2 Magnetic moments and a vector model of a many-electron atom. The Lande factor In Section 5.2.1 the gyromagnetic ratio g, the ratio of the magnetic moment to the angular momentum, was introduced. This ratio is given in the unit (e/2m) and allows the determination of the magnetic moment knowing the angular (mechanical) moment. The g value for the orbit (g 1) can be calculated theoretically, whereas for spin (g 2) it is determined experimentally. In this section, we shall consider the problem of what to do if, in the many-electron atom with a Russell-Saunders coupling scheme, both orbit and spin participate in producing the magnetic moment. The vector model of an atom will help us.
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Formally, one can write the expression for the magnetic moment MJ
g B J ( J 1),
(7.6.10)
where B is the Bohr magneton and J is the total quantum number of atomic mechanical moments. Our task is to determine the g-value when both orbit (quantum number L) and spin (quantum number S) participate in magnetic moment creation. The mutual disposition of all total atomic mechanical and magnetic moments is presented in Figure 7.25. The vectors LL and LS, obtained according to the rule described above for Russell–Saunders coupling can be seen in the picture. Their sum gives the general mechanical moment LJ (all of them in the scale of ). The magnetic moments (in the scale B) correspond to each mechanical moment; all of them are directed oppositely to their counterparts because of the negative electron charge. The length of magnetic moments takes into account the corresponding g-factor: g 1 for orbit and g 2 for spin. Therefore, their sum J* is directed not strictly antiparallel to its counterpart J , but at some angle (denoted as η) to the line of LJ. (In an atom, as in any other rotating system, the main vector is the angular (mechanical) momentum vector; it specifies the main system axis). Therefore, the total atomic magnetic moment is not J*, but J that is a projection of the first on an axis specified by the LJ vector. It is seen in Figure 7.25 that M J M J
ⴱ
cos
(7.6.11)
The total vector J length is a sum M J M L cos M S cos .
(7.6.12)
As ML 1(e/2m)兹苶 L苶 (L苶 苶1苶) 1 B兹苶 L苶 (L苶 苶1苶) and MS 2(e/2m)兹苶S苶 (S苶 苶1苶)2B 兹苶S苶 (S苶 苶1苶) we have to determine cos α and cos . Let us consider a triangle with sides LL, LS and LJ. According the cosine theorem: L2S L2L L2J 2 L L L J cos and L2L L2S L2J 2 L S L J cos . Then, cos
L2L L2J L2S L ( L 1) J ( J 1) S (S 1) 2L L L J 2 L ( L1) J ( J 1)
and cos
L2S L2J L2L S (S1)J ( J1) L ( L +1) . 2L S L J 2 S (S +1) J ( J +1)
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B,z
LJ
LS
LL
-
ML
MJZ MS
MJ
MJ*
Figure 7.25 The vector diagram of a multielectron atom.
Substituting these values into eq. (7.6.12), we arrive at ⎛
M J B ⎜ L ( L 1)
L ( L 1) J ( J 1) S (S 1)
2 L ( L 1) J ( J 1) S (S 1) J ( J 1) L ( L 1) ⎞ 2 S (S 1) ⎟. 2 S (S 1) J ( J 1) ⎠ ⎝
This expression can be re-written as: ⎛ ⎝
M J B ⎜ 1
J ( J 1) S (S 1) L ( L 1) ⎞
2 J ( J 1)
⎟⎠ J ( J + 1).
(7.6.13)
Comparing expressions (7.8.10) and (7.8.13), we obtain g 1
J ( J 1) S (S 1) L ( L 1) . 2 J ( J 1)
(7.6.14)
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Expression (7.6.16) determines the so-called Lande factor (multiplier). It permits us to write the expression of the magnetic moment of the multi-electron Russell–Saunders atom in a compact form: MJ B g
J ( J 1)
(7.6.15)
In the particular cases at S 0, J L and the Lande factor is, certainly, equal to 1; at L 0, J S g 2, which correspond to purely spin magnetic moment. Expressions that determine the magnetic moments of multielectron atom are obtained in full. 7.6.3
The atomic terms
It is obvious now that an atomic state has to be characterized not only by the electron distribution between energy levels (with Pauli exclusion principle validity), but also by moments: total LJ, orbital LL and spin LS, their mutual positions being taken into account. The range of quantum numbers J, L and S comprise the so-called atomic term. The symbolic presentation of the atomic term is given as: βAJ : A is characterized by the quantum number L according to the following scheme: L A
0 S
1 P
2 D
3 F
The number is called the term multiplicity; it characterizes the number of acceptable states at fixed quantum number L. The multiplicity is equal to 2S 1 at S < L and 2L 1 at S L. For example, for states with L 1 and S 21 multiplicity is equal 2 12 12 (term 2P1/2, 2P3/2); for states with L 1 and S 2, 2 1 1 3 (terms 3P3 , 3P2 and 3P1), etc. The definite energy corresponds to each term. However, systems can often be degenerated. The external action can remove the degeneration, i.e., make each term have its own energy. 7.6.4
Characteristic X-rays: Moseley’s law
As we saw in Section 6.6.4, an electro-magnetic radiation with a wavelength of 10–10–10–11 m is regarded as an X-ray. Two types of X-rays are known; both of them are produced in an X-ray tube as the result of the interaction of primary radiation (X-rays or electrons) with an anode material. X-rays of the first type are produced as a result of intraatomic energy transitions and are called characteristic X-rays; their line spectrum is specified by the anode material’s atoms. The second are bremsstrahlung X-rays emitted while an electron moves with high deceleration in the surface layers of the anode material. This spectrum is white, i.e., it contains a continuous range of wavelengths (see Section 6.6.4). In both cases a great amount of heat is emitted and then removed by a special water flow inside the anode.
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White radiation is used in material X-ray radiography (translucence, human body). Its properties were mentioned in Section 6.6.4. We will restrict ourselves in this section to a consideration of characteristic X-rays. The primary electrons in the X-ray tube (see Figure 6.39) are emitted by a heated cathode by thermo-emission and accelerated by an electrostatic field between anode and cathode with the potential difference 103–105 V. In the nonexcited atoms of the anode, all lower levels are occupied in accordance with the Pauli exclusion principle. Rapidly moving electrons strike a metal target (refer to Section 6.6.4). Colliding with the atom, the accelerated electron kicks the inner, lower-level electron out producing free places (holes) in the low-lying (ground) levels. An upper-level electron drops down to fill the hole. The transition is accompanied by photon emission, its wavelength being determined by the energy difference: 2c/(E – E), where (E – E) is the energy difference. In Section 7.5.4, the spectrum of the hydrogen atom was discussed. Equation (7.5.34) shows that the energy levels, at least for the hydrogen-like atoms, go down (in a negative region) proportional to Z 2: the potential well deepens though the relative positions of the levels do not change significantly. At the above-mentioned voltage difference and Z~10–40, the emitted photons enter the X-ray range. In other types of atoms, the mutual positions of levels are changed: now the energy level depends not on the principal quantum number n only but on the orbit L, spin S and total J numbers as well. A spectral fine structure appears (Figure 7.26). It can be seen that the X-ray spectrum has a line feature with a relatively small (in comparison with optic molecular spectra) number of spectral lines. The X-ray tube radiation usually contains low intensity white radiation background and strong lines of characteristic spectral lines rise above it (Figure 7.27). The most important fact is that the appearance
2D s/2 2
D3/2
2P
M-shell
3/2
2P
1/2
2S
1/2
M series
2P
3/2
2P
L-shell
1/2
2
S1/2
L series 1 2 1 2 1 2 2S
K-shell
1/2
K Series
Figure 7.26 Electron levels of a multielectron atom.
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I K
K
0.4
0.6
0.8
(Å)
Figure 7.27 The tube’s X-ray spectrum; high-intensity characteristic spectral lines over the background of a white bremsstrahlung can be seen.
of the spectrum is very specific: each element has its own spectrum and it can be used to reveal the material that has emitted from the spectrum. The X-rays resulting from electron transition from any higher level to a lower level with the principal quantum number n 1 are known as K-series; transition to a level with n 2 are known as L-series; and series M, N... describe with n 3, 4..., etc. Because of the fact that every level after K is split, the transition L → K in particular produces three closely set lines, the most intense being denoted as Kα1. A Z-dependence of an emitted X-ray radiation frequency ν is given by Moseley’s law (1913). The British scientist H.G.J. Moseley found that the most intense short-wavelength line in the characteristic K-series X-ray spectrum from a particular target (anode) element varied evenly with the element atomic number Z (Figure 7.26). He also found that this relationship could be expressed in terms of X-ray frequency by a simple formula v 2.48( Z 1)2 1015 Hz
(7.6.17)
The Z1 value is the so-called effective atomic number. The point is that one electron kicked out of the K level left a second one untouched. The latter takes part in the screening of the positive nuclear charge by the negative electron charge. The above formula only applies to an L→K transition. An analogous expression describes the other series. An illustration of Moseley’s law for K-series lines of different elements is given in Figure 7.28. The X-ray spectra of all elements are tabulated. Moseley’s discovery is closely connected with the Bohr atomic model, both being announced in the same year (1913). The spectra discussed are produced by inner transitions and are not affected by the valence state and chemically bonded atoms. Therefore the characteristic spectra are the probe of the chemical composition of the material investigated. Analysis of the specimen
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24 Zr
√·10-s (Hz1/2)
Y Cu
16
Cr Cr Fe
Ti Cl 8
Al
Ni
Zn
V K
Si
0 1
8
16
24
32
40
Z
Figure 7.28 A graph of versus periodic number of a chemical element Z (Moseley law).
chemical composition on the basis of Moseley’s law is the essence of the chemical X-ray fluorescent method. For excitation, a corresponding device can use either the primary X-ray beam or primary electron beam described above (refer to Section 7.2). In the latter case, special magnetic coils (lenses) can squeeze the electron beam down to microns and carry the experiment out to the chemical element distribution point by point. Such devices are called quantometers. In order to see the composition of secondary X-ray radiation, X-ray diffraction analyzers are used (refer to Section 6.3.5): an analyzing crystal with a known structure and, hence, with known crystal interplanar distance enables the measuring of the reflecting angle and then the wavelength of the secondary radiation. By consulting tables of the characteristic X-rays, one can find the sample chemical composition in every point, in crystallites and intercrystallite junctions, and in amorphous samples, etc. As an example, the characteristic spectrum of a BaTiO3 ceramic is presented in Figure 7.29. The wavelength is plotted along the abscissa and the relative intensity along the ordinate. The numbers below are the tabulated data of corresponding elements and series of Ti, Ba and O. The line intensity is proportional to the relative amount of a particular element in the sample (at the point irradiated). The spectral lines of Ti-Kα and Ba-Lα are located near each other, whereas the O-Kα line is at a great distance from the first ones. Moreover, such a wavelength is the subject of intense absorption by the air molecules. Therefore, a high vacuum device is needed for such experiments. The method is successfully used with relatively heavy substances (from Be up to U). The analysis of light element (C, S, O, etc.) is rather troublesome because their fluorescent radiation is of a long wavelength and is strongly absorbed by air. Therefore, the precision and sensitivity increases with the increase in the atomic mass of the analyzing materials; however, the possibility does exist to analyze elements with Z 2. Its application ranges from on-line industrial analysis and in-field inspection of geological samples to ultra-trace analysis of semiconductor surfaces.
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I
TiK BaL
27.48 27.75
OK
235.7
(Å)
Figure 7.29 Results of X-ray investigation of a chemical composition of the ferroelectric ceramic BaTiO3: in an arbitrary scale as a graph of the X-ray intensities versus the wavelength is presented (after Yu. Ya. Tomashpolsky et al.).
7.7
AN ATOM IN THE MAGNETIC FIELD: THE ZEEMAN EFFECT
The Zeeman effect consists of atomic energy level splitting and, accordingly, the splitting of the spectral lines of a sample when an external magnetic field is imposed on a sample. If a multi-electron atom is placed in an external magnetic field, depending on the magnitude of the magnetic field induction B, two cases can occur. In the first case, the weak magnetic field is unable to tear out the Russell–Saunders coupling; this is the case of a weak magnetic field or anomalous Zeeman effect. In this particular case, Larmor precession definitely takes place, indeed all cones in Figure 7.25 are precession cones: the whole “construction” takes part in the precession around the direction of this field (z-axis) and mechanic vectors. In the second case, corresponding to a strong magnetic field, the Russell–Saunders coupling is thrown out, and the orbital and spin angular momentums participate in precession independently (normal Zeeman effect). Consider first the case of a weak magnetic field when the bonding of the orbit and spin angular momentums remains unbroken. It was shown in Section 5.1.5 (eq. (5.1.35)) that magnetic moment in the magnetic field with induction B depending on the angle α between vectors M and B acquires additional energy (MB) MB cos . This energy can be expressed in the form E (M J B) M J B cos( ) M J B cos ,
(7.7.1)
(refer to Figure 7.25). ( − ) is an angle between vectors M and B. The product M J cos is the projection of the magnetic moment onto the z-axis, eq. (7.6.15), i.e., M J*. Then, E Mⴱj B g B m J B.
(7.7.2)
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The value E is, accordingly, that energy which an atom admits by the interaction of its magnetic moment with the external magnetic field (with induction B) and which is added to the main electronic energy levels. As a consequence of ∆E depending on the quantum number mJ, and taking 2J 1 values, the subsidiary energy also takes such values. This signifies that the main electron energy level splits into 2J 1 sublevels. If, before the imposition of the magnetic field, the states with different mJ had the same energy, the magnetic field has brought about the levels splitting, i.e., lifting the degeneration on this quantum number mJ. The number of sublevels depends on the quantum numbers, i.e., splitting depends on the atomic term. In Figure 7.30 an example of the splitting of the atomic S- and P-levels in an external magnetic field is presented. Before the field was imposed, each term was degenerated, the S-level was double, and the P-level was sixfold degenerated; both levels can be regarded as singular (Figure 7.30b), i.e., the energy of all sublevels is the same. The energy transition between them is defined by the energy difference EP - ES E, which corresponds to the quantum emission with the frequency 0
E .
(7.7.3)
At the same time, the spin–orbit interaction, even in the absence of an external field (B 0), brings about the splitting of the P-level into two (one of them with J 3/2 and the other with J 1/2, Figure 7.30c). This corresponds to the lifting of the degeneration on the quantum number J. The S-level remains singular. The value of splitting of the P-level is defined by the expression (7.7.2). In the experimentally measured spectrum, provided the resolution is sufficient, instead of one line with the frequency ω0, two lines (a doublet) appear, whereas the central line disappears (as shown in the scheme below). The shift of the lines from the position of the central (absent) line ω0 is defined by the value of energy gB divided by .
mJ 3/2 1/2 −1/2 −3/2 1/2 −1/2
B=O +1
B=O
2
P
0
P3/2
B=O
B=O
2
P1/2
−1
2
S1/2
S
0
gmJ 2 2/3 −2/3 −2 1/3 −1/3
1/2 1 1/2 −1
0
0
0
0
0
(a)
(b)
(c)
(d)
(e)
Figure 7.30 Zeeman effect: (a) normal Zeeman effect, (b) energy levels before imposing a magnetic field, (c) transition P→S (without the magnetic field), splitting because of spin–orbit interaction, (d) and (e) an anomalous Zeeman effect.
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Further splitting occurs when the external field is imposed (B 0, Figures 7.30d and e). The degeneration on the quantum number mJ is lifted according to eq. (7.7.2): at J 3/4 four levels appear, at J 1/2 there are two. In Figures 7.30d and e, the splitting scheme is presented specified by values mJ and gmJ for all sublevels. Transitions can occur between levels (and sublevels). Selection rules limit their number. Strictly speaking, selection rules follow from the analysis of quantum mechanical transition probabilities, however, a qualitative explanation of these rules can be suggested; the law of spin conservation can be attracted. The point is that an electromagnetic radiation quantum (photon) carries away from a system (from an atom) the spin equal to its own spin. The latter equals l . The spin angular momentum can be oriented in a triple way regarding the photon wave vector: perpendicular to it (upward ( ) and downward ( )) and along it (0). Accordingly, mJ can accept three values mJ 0, 1
(7.7.4)
Transitions with mJ 0 are referred to as and with mJ 1 as σ transition; photons are differently polarized, correspondingly. Thus, at transitions P1/2→S1/2, four spectral lines occur whereas at transition P3/2→S1/2 there are six of them (Figures 7.30d and e). The line’s shift relative ω0 is defined as:
B B (mJ g − mJ g),
(7.7.5)
where two strokes characterize the top and one stroke characterizes the bottom levels. This description of splitting of the spectral lines, corresponding to an abnormal Zeeman effect, is given only by quantum physics and cannot be explained within the framework of classical physics. Quite a different picture appears when a high magnetic field is imposed; the Russell– Saunders coupling is torn out, and the orbital and spin moments participate separately in
Z B LL LS
Figure 7.31 A Russell–Saunders coupling breakdown in a high magnetic field.
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precession around the direction of an external field (Figure 7.31). The additional energy E consists of the contribution of both the orbit and spin accounting for the gyromagnetic ratio anomaly: ∆E B BmL 2 B BmS B B(mL 2 mS ).
(7.7.6)
Because transitions with a change of spin quantum number are forbidden (as a quantum emission is connected only with the orbital atomic state change), the spin term in the sum (7.7.6) is excluded and only the degeneration lifting on mL is considered. The S-level remains single- and doublefold degenerated, whereas the P-level (with L 1) splits into three doublefold degenerated levels (Figure 7.30a) (i.e., the degeneration on mL is lifted). The spectral triplet turns out to be exhibited. The shift of each spectral line from the central one is defined as
B B ,
(7.7.7)
In spite of the fact that D-, F- and the other subsequent levels split into a greater number of sublevels because the g-factor is equal to unity in all level splitting, and the selection rule is still valid here as well, the normal Zeeman effect is exhibited as a spectral triplet regardless of the transitions. The exhibiting of a triplet is referred to as a normal Zeeman effect. A normal Zeeman effect can be explained within the framework of classical physics. Displacement of both the energy levels and the spectral line (7.7.7) corresponds to the Larmor precession frequency (refer to eq. (5.2.19))
eB , 2m
(7.7.8)
This coincides with the previous statement that it is possible to treat the additional energy obtained by the magnetic moment in the external magnetic field either as a shift of the energy level on the value (M B), or as the energy of the Larmor precession (7.7.8). In conclusion, we have established that the precession gives a gain of additional energy. Therefore, to treat spatial quantization of the angular momentum (7.5.3) as precession is incorrect from our point of view, because this effect does not produce any specific energy.
7.8 7.8.1
A QUANTUM OSCILLATOR AND A QUANTUM ROTATOR
Definitions
Like translational motion, all molecular motions accomplish rotational and oscillatory motions, too. In Section 2.4.5, an example was given of a diatomic molecule with an interatomic distance d rotating around a stationary axis z passing through the molecular center of inertia. It was shown that rotation of these two masses can be substituted by the rotation
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of one reduced mass accomplishing rotation around the same axis at the same distance d. Remember that any molecule rotated around a stationary axis is called a rotator; if the molecular intra-atomic distance remains constant the rotator is referred to as a rigid one. The expression for the reduced moment of inertia Ie d2 and kinetic energy KIe 2/2 was also given there. No requirement on the value of rotational kinetic energy is imposed in Newtonian physics. Since, in this case, the potential energy is accepted as zero, the kinetic energy is the total one. In Section 1.5.4, a potential curve for classic harmonic oscillations in parabolic form (refer to Figure 1.33), as well as a potential curve describing anharmonic oscillations (Lennard-Jones potential “6–12”, Figure 1.31) were presented. In both cases, the total particle energy in a potential well can have a continuous range of values. Also remember that any oscillating system is called an oscillator. The quantum mechanical consideration of molecular rotation and intramolecular oscillation is of significant interest since it is the basis of optical methods of investigation, both scientific and technological.
7.8.2
Quantum oscillators: harmonic and anharmonic
The potential energy of a system accomplishing small oscillations around an equilibrium position was considered in Section 2.4. The condition “small” signifies that the restoring force is linearly dependent on displacements; oscillations then behave according to the harmonic law (sine or cosine). For classical harmonic oscillation, alongside the continuous spectrum of energy, is the distinctive fact that the probability of finding a system beyond the amplitude displacements is zero. Because we consider here a conservative process, the total mechanical energy is preserved (remember that in a conservative system the energy does not dissipate). In accordance with Section 7.3.2, in order to solve the quantum harmonic oscillator problem, we have to write the potential energy expression in the form x2/2, to substitute it into the Schrödinger equation (7.3.5), find the wavefunction satisfying the standard condition and then find the spectrum of the energy whether it is continuous or discrete. This particular problem can be solved in analytical form. However, while not solving the Schrödinger equations, we will give here the essence of the results. Firstly, the equation for the quantum harmonic oscillator shows that the energy can accept only definite values of energy equal to E 冸 v 12 冹,
(7.8.1)
where v is the oscillation quantum number. The corresponding spectrum of allowed energies is depicted in Figure 7.32 on the background of the classical potential curve. It consists of the equidistant levels with the interlevel distance E ⎡⎣冸 v 1 12 冹 冸 v 12 冹 ⎤⎦ h .
(7.8.2)
Secondly, the number of levels is not limited; the quantum number v can accept any value.
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Thirdly, calculations of the quantum mechanical probability of the energy transitions showed that the quantum number can be changed in increments of 1 only (v 1); the transitions are allowed only between the adjacent levels. In other words, since the levels are equidistant, only one spectral line can be emitted (or absorbed) regardless of which levels the transition takes place between. The emitted/absorbed quantum energy is in any case, always ω. Fourthly, at T 0 K (v 0), the oscillations do not come to an end. The so-called zero oscillations are preserved even at absolute zero temperature. The zero point oscillation energy is equal to ω/2. Finally, the eigenvalues of the wavefunctions can be expressed by Hermite special polynomials. Some of these, together with heir squares, are depicted in Figure 7.32. The principle difference from the classic case is that the probability of finding the oscillator beyond the amplitude values is not zero; it vanishes as the deviation is increased. Of course, we are interested mostly in the oscillation of atoms in molecules. For the sake of simplicity we will deal with the diatomic molecule. As a rule they are the anharmonic ones, i.e., described by a nonsymmetrical potential curve. One with the prevailing potential of such a type in physical chemistry is the Morse potential presented in Figure 7.33: U (r ) U 0 [1 e(rr0 ) ]2 .
(7.8.3)
The potential curve U(r) is equal to zero at r r0; this is an equilibrium interatomic distance [U(r) U(r0) 0]. In the case of small oscillations near r r0, the exponent can be decomposed into a series 1–α(r–r0)±… and we arrive at the quadratic harmonic law
U ψ(x)
X (a)
|ψ(x)|2
V=3
V=3
V=2
V=2
V=1
V=1
V=0
V=0 (b)
(c)
Figure 7.32 A linear harmonic quantum oscillator: (a) classic potential energy curve, (b) wavefunctions and (c) their squares.
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U(r) ⬇ 2(r r0)2. At large values of r the exponent approaches zero and U(∞) → U0. Therefore U0 is the depth of the potential well. Notice that the Lennard-Jones and Morse potentials resemble each other; they differ in the choice of origin of the ordinate axis (one is lifted relative to another by a value of U0; this is insignificant because the potential energy is determined accurate to the constant, U0 in this case; refer to Section 1.5.4). To solve the quantum mechanical problem of anharmonic atomic oscillations in a diatomic molecule it is necessary to accomplish the standard procedure: substitute the Morse potential expression into the Schrödinger equation and solve it, the eigenvalues of energy and eigenwave functions can be found. The given problem cannot be solved in analytical form. We will use only eigenvalues of energy E(v). This is given by the approximate expression E(v) [(v 12 ) (v 12 )2 ].
(7.8.4)
In this expression is the anharmonicity coefficient that characterizes the peculiarities of the interatomic interactions. For some simple diatomic molecules lies in the limit 0.01–0.07; it can be found in the reference books for many molecules. The Morse potential curve is depicted in Figure 7.34 with energy levels of anharmonic oscillations. At small v the second term in eq. (7.8.4) is small and the energy levels are approximately the same as for the harmonic oscillator. While increasing the quantum number v, the negative contribution of the second term increases. As a result, the curve E(v) has a maximum and further increase of the v value brings about a decrease of energy (Figure 7.35): this reduction has no physical sense. So the number of levels of the inharmonic oscillator is limited, i.e., there exists a maximum value of v (v vmax). There are two ways to find the value of this limiting quantum number: either find the extreme value of the functions E(v) from an equation dE/dx0, or equate ∆E to zero at v vmax. In the latter case one obtains 0 E(vmax ) ⎡⎣冸 vmax 1 12 冹 v冸 vmax 1 12 冹2 冸 vmax 12 冹 v冸 vmax 12 冹2 ⎤⎦ , (7.8.5) U U0
r0
r
Figures 7.33 Potential Morse for unharmonic oscillator.
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U U0 3 2
D
1 0 r0
r
Figure 7.34 Potential Morse, the energy levels and the dissociation energy D.
therefore, [1 (vmax 1)] 0,
(7.8.6)
and further, vmax
1 1. 2
(7.8.7)
1 . 2
(7.8.8)
As far as 1, one can obtain vmax ⬇
Substituting this value into the general expression for energy (7.8.4), we obtain the maximum energy and, consequently, the depth of the potential well Emax U 0
. 4
(7.8.9)
In order to tear out the interatomic bond, one has to excite a molecule into the state with E U0, i.e., into the state when an infinite motion of one atom relative to another is realized. If the molecule is in the lower energy quantum state, this energy is E U 0 E0
. 4 2
The energy D required for tearing out the chemical bond of the molecule being in the lower state with v 0 is called the dissociation energy. Accordingly, it is equal to ∆E, i.e., D
(1 2). 4 2 4
(7.8.10)
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E Emax
max
Figure 7.35 Energy of unharmonic oscillator versus the oscillation quantum number v.
The coefficient can be derived from the last expression:
. 4D2
(7.8.11)
As can be seen, the dissociation process is tightly connected with the molecule’s anharmonic oscillation. Within the framework of the harmonic model, it is impossible to explain the dissociation process. The frequency , appearing in all the last equations, is the frequency of natural vibrations, i.e., a frequency corresponding to the lower molecule energy level E /2; the anharmonic term brings a negligible contribution to this energy. When a molecule turns from one energy level into another, emission or absorption of energy in the form of quantum of the electromagnetic radiation occurs. The energy of such quanta is equal to the energy difference. The quantum mechanical calculations show that all transitions are allowed, i.e., there are no selection rules in this case. The frequency of a photon emitted at transitions between adjacent levels is equal: ph 1(Ev1Ev). The corresponding wavelength is
2c . Ev1 Ev
(7.8.12)
An arbitrary quantum transition has been shown in Figure 7.34. EXAMPLE E7.10 The natural frequency of HCl molecule vibration is 5.63 1014 sec1, its anharmonicity coefficient is 0.0201. Find (1) the energy ∆E2,1 molecule transition from the second to the first energy levels (in eV) of the oscillatory spectra; (2) the highest quantum number vmax; (3) the maximum oscillatory energy Emax; (4) the dissociating energy D. Solution: (The theory of molecular anharmonic oscillations is presented in Section 7.8.2 (eq. (7.8.4))).
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U
U0
3 2 =2h
D 1 0
r
r0
(1) The energy difference at transition is Ev+1,v [12(v1)] the quantum number v being the lowest level. Executing calculations, we obtain E2,1 1.09 1019 J 0,682 eV. (2) The vmax can be found according to eq. (7.8.8) Vmax 1/(2)23. (3) The Emax can be obtained if we substitute vmax into the expression for oscillation energy (7.8.4). We can obtain (Emax /4). Substituting all values already obtained we arrive at Emax U0 7.38 1019 J 4.61 eV. (4) The dissociation energy is given by eq. (7.8.10) D U0(1-2). Calculation shows that D 4.43 eV. The corresponding drawings is presented in Figure E7.10. 7.8.3
A rigid quantum rotator
The rotation of micro-objects around a motionless axis was analyzed in Section 7.5.3. The orbital motion of an electron was used as an example. It was found that in this case the rotational energy can accept only discrete values defined by eq. (7.5.26). Since potential energy in free rotation is accepted to be equal to zero, the total energy is kinetic. One of the important characteristics of such movement is the rotational constant B:
B
2 . 2I e
(7.8.13)
The Ie value is called the reduced moment of inertia d2. The quantum number in the optical rotational spectroscopy is defined by another letter (letter j instead of l). Therefore the rotational energy can be written as E Bj ( j 1),
(7.8.14)
(compare with eq. (7.5.26)). No restrictions in quantum number are present here. Besides, j can accept the value j 0, corresponding to zero rotational energy. The rotational energy levels E(j) are depicted in Figure 7.36. The selection rule in the case of rotational spectra
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j
1
22
33
j( j+1)
3
12
2
6
1
2
0
0
44
Figure 7.36 The rigid quantum rotator energy levels.
is the same as in harmonic oscillations, the quantum number difference can accept values j 1, i.e., transitions between adjacent levels only are allowed. It is easy to calculate the energy distance between rotational levels. Naturally, E E j1 E j B[( j 1)( j 2) j ( j 1)] 2 B( j 1).
(7.8.15)
The absolute distance between levels increases at increasing j; however, the relative values, E/E vice versa, decrease. This corresponds to the Bohr correspondence principle.
EXAMPLE E7.11 For an HCl molecule determine: (1) the moment of inertia; interatomic distance is d 91.7 pm, relative atomic masses are correspondingly 1 and 19 a.m.u.; (2) the rotational constant; (3) the energy necessary to excite the molecule from the ground state to the first rotational level (j 1) (see Section 1.3.9, Figure 1.17 and Section 7.8.2, Figure 7.36.) Solution: (1). The moment of inertia can be found according to eq. (1.3.50) I 1.33 10–47 kg m2. (2) The rotational constant is given by eq. (7.8.13). Calculation gives B 4.37 10–23 J 2.73 meV. (3) The energy in question can be calculated according to eq. (7.8.15) E1,0 2B 5.46 meV.
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EXAMPLE E7.12 A multielectron atom is in a state 3P. For this term draw a vector diagram for the maximal atomic orbital angular momentum L (Figure E7.12) and determine the angle between this vector and the orbital angular momentum LL (refer to Figure 7.19)
α LL
LJ LS
Solution: The atomic term 3P is characterized by a set of quantum numbers (refer to Sections 7.5 and 7.6.3): L 1 and S 1. One can draw three vector diagrams like that depicted in Figure 7.19 for different J (which can take on values from L S to |L - S|, i.e., 2, 1, 0. It was mentioned in the problem situation that the maximum value of L is supposed (J 2). Hence the vector diagram is presented in Figure E7.12 according to LS兹苶2, LL兹苶2 and LJ兹苶6 (see eqs. (7.6.5), (7.6.2) and (7.6.7)). The angle sought can be found according to cosine law (Figure 7.25), angle cos
L ( L 1) J ( J 1) S (S 1) 2 L ( L 1) J ( J +1)
3 . 2
Hence arccos 兹3苶/2(/6)30°. EXAMPLE E7.13 A beam of neutrons is thermolyzed by a room temperature moderator (T 300 K) in a nuclear reactor. It then passes a special collimator in a hole in the reactor wall and falls on a graphite single crystal. A diffracted plane of the first order n 1 from the graphite base with interplanar spacing d 33.5 pm was measured at an angle 2 25.5o. Find neutron wavelength , its velocity υ and mass m. Solution: Produced in chain nuclear reactions, fast neutrons should be slowed down in order to participate in further reaction (refer to Section 1.5.5 and Example E1.25). Besides, slow neutrons are widely used in crystal structure investigation and solid state physics (refer Section 7.1.2). Therefore, this example has a practical interest. According to Bragg’s law (eq. (6.3.11) and Figure 6.20) at n 1 we can write 2dsin . Knowing d and angle 2 we can find the neutron wavelength . 2 3.35 1010 sin 12.75 1.46 1010 m.
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The neutron mass can be found from de Broglie formula (7.7.1) and rms thermolyzed neutron velocity (eq. (3.3.7″/)) m
h2 43.96 1027 1.66 1027 m. 3T 2.13 9 1.38 2
Dimension checking gives
J 2 s2 m. m (JK)K 2
EXAMPLE E7.14 Define into how many sublevels will energy levels 2S, 3P, 4D, 9F split in atoms with Russell-Saunders bonds mode owing to spin–orbit interaction and write down the terms describing each sublevel. Solution. In light atoms, a Russell-Saunders type of electron interaction usually take place. In Section 7.6 the method is described in detail. The atomic energy state, in this case, is characterized by a set of quantum numbers: the total orbital quantum number L, total spin quantum number S and total internal quantum number J. Thus quantum number J can accept values from Jmax L S up to Lmin |L –- S |, changing on unit. A certain energy state of atom (spin–orbital interaction) corresponds to each value J, i.e., an energy sublevel appears. The number of sublevels or number of possible mutual orientations of vectors LL and LS at is defined by the ratio 2S 1 referred as multiplicity. At L < S the number of sublevels is defined by another ratio 2L 1 (see Section 7.6.3). Accordingly, for a term 2S we have: multiplicity 2S 1 2, since S ½; L 0 corresponds to symbol S (to not be confused with S-state). Hence, condition L < S and the number of sublevels determined by expression 2L 1 1 is valid. This means that the given energy level is not split and is characterized by quantum number J L S ½ and accordingly term 2S1/2 is created. For term 3P multiplicity is 2S 1 3, therefore S 1; to symbol P corresponds L 1 and L S; the number of sublevels determined by expression 2S 1 3 coincides with term multiplicity, i.e., it is equal 3. Thus, to each sublevel there corresponds the term with various J values. As Jmax L S 2 and Jmin |L – S | 0. J accepts values 0, 1, 2 and then corresponding terms will be as 3P0, 3P1, 3P2. The term 4D corresponds to S 3/2, since multiplicity is 2S 1 4, L 2 (D state). As condition L S is valid J accepts values from L S 2 3/27/2 up to (L – S) (2 – 3/2) 1/2, i.e.,1/2, 3/2, 5/2, 7/2. For this set of sublevels the quantum numbers is equal to multiplicity (i.e., 4); the corresponding terms are 4D1/2, 4 D3/2, 4D5/2, 4D7/2. The term 9F has S 4 (i.e. 2S 1 9); therefore, the F-state meets L 3. Thus, inequality L S and the number of possible mutual orientations of vectors LL and LS and consequently the number of sublevels is defined by a ratio 2L 1 23 1 7 is fulfilled. Thus quantum number J will accept the following values; 1, 2, 3, 4, 5, 6, 7 and each sublevel will be characterized by a term: 9F1, 9F2, 9F3, 9F4, 9F5, 9F6, 9F7.
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EXAMPLE E7.15 In the Stern–Gerlach experiment a narrow beam of sodium atoms in the ground state was passed through a highly nonuniform magnetic field (Figure E7.15). What nonuniformity of the magnetic field ∂B/∂z should be provided so that the distance between the components of the sodium atoms split beam fixed on the screen will be equal to 6 mm? The installation dimensions are l1 10 cm, l2 15 cm, speed of atoms is equal to υ0 400 m/sec. The whole installation is placed in a hermetic vacuum shield. z sample
S
S = 2( 1+ 2) υ0
heater N
N
υ0
υz
1 2
x screen
l2
l1 z
(a)
(b)
Solution. Figure E7.15a shows the main way of producing a highly nonuniform magnetic field due to tailor-made poles and Figure E7.15b describes the path the atom travels. We have to consider some details. The basis for the decision of this problem is formula (5.1.32): a force working on a magnetic dipole depends on both the nonuniformity of the magnetic field and the orientation of the atomic magnetic moments relative to the quantization axis z. If the atomic magnetic moments submit to laws of classical physics, there could be any values of angles and the experiment would result in a fuzzy maximum. However, in quantum mechanics this is not a case! The force working on an atom depends on atomic state, degree of nonuniformity of the magnetic field in the device and a set of magnetic quantum numbers. Eq. (5.1.32) is still valid in quantum physics though the angle depends here on the set of quantum numbers mentioned: M z /Mcos mJ/兹苶 J(苶 J苶1苶). The quantum number mJ can accept an amount of the discrete values dependent on quantum number J, that is in all 2J 1 values. Since to each value mJ corresponds the force as in eq. (5.1.32), the atomic beam is split on some component; there will be a system of strips on the screen. To each strip there corresponds a certain number of mJ . The sodium atom in the ground state has one 3s electron in M shell. Therefore, the set of quantum numbers are: L 0, S1/2 and J1/2, this corresponds to mJ 1/2. So we know that there should be two strips on the screen. In order to calculate the distance between them and find the magnetic field nonuniformity, we have to solve the problem qualitatively. There is no need to calculate g-factor Lande since we know that only spin contributes to the sodium atomic magnetic moment, i.e., g 2 (see Section 7.6.2). In Figure E7.15, a beam of atoms from a heater pass some of the slits shown. Entering the nonuniform magnetic field, a beam splits into two parts, which move
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along curvilinear (parabolic) trajectories. The acceleration inside the magnet can be calculated according to the second Newtonian law azFz/m. Then the shift is azt12 az12 1 where t1 1 and 1 . The movement along l2 is assumes 0 2 202 uniform. In time t2 the atom acquires the velocity υz azt2. The shift gained is 12 2zt2azt1t2az . The total relative shift is then 2(12) and 02 az1 ( 22). Taking into consideration all calculations we obtain 02 1
⎛ B ⎞ g B ⎜ ⎟ mJ ⎝ z ⎠ m02
1 ( 1 2 2 ).
Therefore, we arrive at
A02
B
z B 1 ( 1 2 2 ) N A (keeping in mind that g|mJ| 1). Executing calculations we obtain
B 6 103 23 103 4002 J T
z 9.27 1024 0.1 (0.1 2 0.15) 6.02 1023 98.9 JT ⬇ 1 Tcm. 7.8.4
Principles of molecular spectroscopy
The electromagnetic radiation emitted or absorbed by the substances under investigation is the subject of the oscillation–rotation (molecular) spectroscopy. In this section, the main attention is concentrated on molecular spectroscopy near and within the optical range of frequencies. Emission of the radiation quanta or their absorption is defined by energy transitions. Therefore, the basis of all spectroscopy methods is the discontinuity of energy spectrum and the mutual position of energy levels. Of course, this is only a general treatment: many factors playing an important role in spectroscopy remain outside our discussion. Naturally, the best effect in the interaction of radiation with matter can be reached in the case when the energy of radiation quanta used coincides in order of values with energy transitions. Therefore, we will consider that range of quanta energy complies with that of energy transition on the basis of the data given in Table 5.3. The total molecule energy can be written as a sum: E En Ev E j ,
(7.8.16)
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where quantum numbers n, v and j denote electronic, oscillatory and rotational contributions. Let us evaluate each part separately. The electronic energy levels are assigned by eqs. (7.5.31) and (7.5.34). Depending on the nuclear charge Z the ground state energy varies from 10 eV up to 100 keV. Therefore, only transitions among the high lying atomic levels enter the optical frequency range (for example, the Balmer series with v 2). In order to evaluate the oscillatory energy levels, it is necessary to know their intrinsic frequencies. Experiments show that they are 10121014 Hz. Correspondingly, the emission frequency belongs to the IR radiation region and the wavelengths are 104–106 m. The purely rotational frequencies lie in the frequency range dependent on the molecule’s rotational constant (7.8.13). Estimations show that B ⬇ (1068/1047) ⬇ 1021 J or 102103 eV. This corresponds to the wavelength ⬇ 103 m, which corresponds to a microwave frequency far from the optical range. Important conclusions follow from these consideration. First, emission and/or absorption due to electron transition, lying far from the optical range in the short wavelength side, are not considered in molecular spectroscopy. Second, the same applies to purely rotationary spectra, though the shift is in the long wavelength region. The main interest in optical molecular spectroscopy concerns oscillatory-rotationary transitions, presented by the last two terms of the sum (7.8.16). The energy transitions in this case correspond precisely to the visible and adjoining wavelength regions (IR and UV). An arrangement of the electronic, oscilationary and rotationary levels of energy of a hypothetical molecule is submitted in Figure 7.37. The distances between oscillatory levels only, determined by (eq. (7.8.2)), are larger than distances in the rotationary levels (eq. (7.8.14)). Therefore, rotary levels settle down between oscillatory levels (according to the sum (7.8.16)), however, the number of rotary levels between the next oscillatory levels is limited to the distance between the last. If E(v) accepts a certain value, the number of rotationary levels is defined by their highest value (in an accepted interval E(v)). In other words, the difference E Bjmax ( jmax 1) should be less than or equal to E. Transitions are carried out between levels with two sets of quantum numbers (v, j). Let us consider in general the scheme of a spectroscopic experiment. We usually distinguish emission spectroscopy (Figure 7.38a) and adsorption spectroscopy (Figure 7.38b). In the first case, the sample under investigation is excited in order to force it to emit quanta of radiation (e.g., by heating the sample). Radiation is directed onto the spectral device with a prism or diffraction grating (see 6.3.4), decomposing the radiation in a spectrum along wavelengths. The investigator writes down the results on a paper (screen) as the dependence of intensity on wavelength (wavenumber). In adsorption spectroscopy the light source emits a “white” spectrum (containing the entire wavelength in a certain interval), which then goes through the sample. The falling radiation is absorbed by the substance that cuts out a definite characteristic wavelength from the incident beam. The corresponding spectral device gives the results as strips with the “cut out” frequencies. As an example, the absorption spectrum of chloroform is presented in Figure 7.39. Traditionally in molecular spectroscopy energy is measured in wavenumbers 1/, i.e., it is expressed in cm1. It is simple to derive the relation between wavenumbers and energy:
2c hc. v
(7.8.17)
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E n
V
j
2
−3
1
−2 −1 0
0 En" 2
1
−3
0
−2 −1 0
En'
Figures 7.37 The overall electronic, oscillator and rotator electron levels (not to scale).
I K Sp S 1/ (a)
I I
K
1/
S Sp
E (b)
1/
Figure 7.38 Molecular spectroscopy schemes: (a) emission, (b) absorption; are samples, K collimation slits, Sp spectral recorder, E source of radiation, I(1/) type of specter’s measured. Incertion: corresponding source spectra.
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Cl C
H
Cl
Cl 3
5
C-Cl 7
9
11
13 16
25 ν
Figure 7.39 Chloroform absorption optical spectrum.
In conclusion, we should note that when translating from the SI system to reciprocal centimeters (cm–1), it is necessary to divide the result by 100.
PROBLEMS/TASKS 7.1. A narrow beam of sodium atoms in the ground state passes through a Stern-Gerlach device with a nonuniformly magnetic field ( B/ z1 T/cm) and l 10 cm in length. Determine the distance between the components of a split beam at the outlet of the magnet. The sodium atom’s speed is 300 m/sec (refer to Figure E7.15). 7.2 An atomic state is characterized by the two spectral terms 1D and 1P. Find all possible quantum numbers J for these terms and draw a scheme of level splitting in a weak magnetic field. 7.3. Determine the minimum energy of ⏐E⏐min (in meV) of an atom in the state 2F in a uniform magnetic field B 0.8 T. 7.4. An atom is in a state characterized by the term 4D. Calculate the minimum value of the total angular momentum LJ and draw the corresponding vector model (refer to Figure 7.19). Determine the angle between the spin LS and total LJ angular momentums. 7.5. An electron of energy E 0.5 U0 moves in the positive direction of an axis x. Estimate the probability that the electron will penetrate through the potential barrier of height U0 10 eV and width d 0.1 nm. It is useful to make a corresponding drawing. 7.6. An electron of energy E 9 eV moves in the positive direction of an axis x. At what width of potential barrier d will the transparency factor be equal to D 0.1 if the height of the barrier U0 is 10 eV? It is useful to make a corresponding drawing. 7.7. A monochromatic electron beam meets in its motion a potential barrier d 1.5 nm in width. At what energy difference (U0 – E) (in eV) will the barrier penetrate 0.001 of incident electrons. It is useful to make a corresponding drawing. 7.8. An electron meets in its motion a potential barrier of the height U0 10 eV and in width d 0.2 nm. At what electron energy E (in eV) will the barrier transmission probability w be w 0.01. It is useful to make a corresponding drawing.
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7.9. A particle is in an infinitely deep potential box of width L. The particle state corresponds to wavenumber k (/L). What is the probability of finding the particle in an area (L/2) x (L/4)? It is useful to make a corresponding drawing. 7.10. A particle is in an infinitely deep potential box of width L. The particle state corresponds to the quantum number n 2. What is the probability of finding the particle in an area (L/3) x (2L/3)? It is useful to make a corresponding drawing. 7.11. A vanadium atom in the state 4F3/2 passes through a Stern-Gerlach device (refer to Figure E7.15b). The atom’s velocity is 400 m/sec. Determine the distance between the upper and lower component of the split beams spots if l1 l2 10 cm and B/ z3 T/cm. 7.12. A silver atom’s beam is passed through a Stern-Gerlach device (refer to Figure E7.15b). The atom’s velocity is 300 m/sec. Determine the magnetic field gradient B/ z if the distance between the ends component of the split beam is 2 mm, l1 10 cm and l2 0. Silver atoms are in the ground state (L 0, J 1/2, g 2). 7.13. Find the number N of vibration energy levels for HBr molecule if the anharmonicity coefficient is 0.0201. 7.14. Knowing the natural angular frequency of a CO molecule (ω 4.08 .1014 sec–1), find its rigidity (quasi-elastic) coefficient . 7.15. Determine the dissociation energy D (in eV) of a CO molecule if the natural frequency is ω 4.08 1014 sec–1 and the anharmonicity coefficient is 5.83 10–3. 7.16. For an O2 molecule, find (1) the reduced mass , (2) the internuclear distance d, if the rotational constant B 0.178 meV, (3) the angular velocity ω if the molecule is on the first rotational energy level. 7.17. Find the angular momentum of an O2 molecule if its rotational energy EJ is 2.16 meV. 7.18. Can monochromatic electromagnetic radiation with wavelength 3 µm excite vibrational and rotational energies of the HF molecules if it is in a ground state? 7.19. Determine the multiplicity of the energy levels of a diatomic molecule with quantum number J. 7.20. Calculate the internuclear distance d in CH molecules, if the V interval in the purely rotational emission spectrum is 29 cm1.
ANSWERS 7.1. B gJl ( B z ) 2.70 mm. m2 2
7.2. For the term 1P J 1 and mJ 0, 1; for the term 1D J 2 and mJ 0, 1, 2. 7.3. ⏐E⏐min Bg ⏐mJ⏐min, B 0.0199 meV (g (6/7) at J (5/2); ⏐mJ⏐min (1/2). 7.4.
LJ ,min
3 ; cos 0.447; 166.5 2
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⎡ 2d ⎤ 2 m(U 0 E ) ⎥ 0.1 7.5. w exp ⎢ ⎣ ⎦ 7.6.
d
n(1D) 2 2 m(U 0 E )
0.22 nm. 2
7.7.
⎡ n 冸1w 冹 ⎤⎦ (U 0 E ) ⎣ 0.20 eV. 8md 2 2
7.8.
E U0
[n(1w)] 5.0 eV. 8md 2
1 1 7.9. w 0.409. 4 2 1 3 7.10. w 0.196. 3 4
7.11.
7.12.
3 B Jgl 2 m2
B
z 3.7 mm.
B m2 116 Tm.
z 3 B gJ 2
7.13. N Vmax (1/2) ⬇ 24. 7.14. 2 1.91 kN/m. 7.15. D
1 2 11.4 eV. 4
7.16. (1) 1.33 .10–26 kg , (2) d 121 pm, (3) ω 7.61 .1011 sec–1. 7.17. L 3.66 .10–34 J sec. 7.18. Only rotational motion occurs. 7.19. 2J 1. 7.20. d 112 pm.
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8.1
SELECTED ATOMIC NUCLEI CHARACTERISTICS
So far we have considered the atomic shell and the point (dimensionless) nucleus with charge Ze interacting with electrons by Coulomb forces according to the laws of quantum mechanics. The account of the nuclei characteristics with a definite dimension raises many important and interesting phenomena. They are the basis of modern methods of investigation of atomic and molecular structure of chemical substances. The aim of the present chapter is to give relevant information on the atomic nucleus. Taking into account the general aim of the book, we will only touch on those nuclei properties, which have a direct bearing on the description of methods, so this information cannot be considered as comprehensive. Among the modern chemical and physical methods of investigation, spectroscopic methods are widely used (they are often called instrumental methods). These methods allow quantitative information on composition and structure to be obtained effectively and rapidly, and also enable analytical and other problems to be solved without the destruction of the materials tested. Further, these methods are in the process of development and improvement. Very popular among these methods are the nuclear resonance (spectroscopic) methods, which are based on the fundamental discoveries made in quantum mechanics and radioengineering in the second part of the 20th century, namely, nuclear magnetic resonance (NMR), nuclear quadrupole resonance (NQR), -resonance (Mössbauer) spectroscopy (RS), electron paramagnetic resonance (EPR), and others. As these resonance methods are based on electron–nuclear interactions, we must provide some information on the nuclear physics required to understand the subject. 8.1.1
A nucleon model of nuclei
A nucleus possesses an internal structure. It consists of many elementary particles— nucleons—the main of which are protons and neutrons. They move in the field of nuclear forces and continuously change their state from that of charged particle (proton) to neutral particle (neutron) and back. However, on an average in a given nucleus, a certain number 497
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of protons, Z (which defines the given element’s number in the periodic system) and number of neutrons, N (which defines the particular isotope of a given element) are always present. Since the mass of protons and neutrons are approximately 2000 times greater than the mass of electrons, the total mass of an atom is practically comprised in its nucleus. The total number of nucleons, A, is called the mass number, i.e., A⫽N⫹Z. The symbol AZN is used to identify a nucleus, where Z is the chemical symbol of a given element. Mass number A defines the mass of a nucleus. Thus, there are 92 protons and 143 neutrons in the uranium nucleus 235U92. The number of nucleons Z is often omitted because the relevant information is contained in its chemical symbol. Thus, 235U is the usual symbol. In the text below, a proton mass is marked by mp. Mass can be measured in kilograms (mp⫽1673⫻10⫺27 kg), or can be denominated in atomic units of mass (1 a.m.u.⫽ 1.66⫻10⫺27 kg). It is sometimes also expressed in energy units (mp⫽938.2 MeV). Each nucleon possesses its own angular momentum, i.e., spin. The quantum numbers of neutron and proton spins are the same as an electron spin, i.e., 1/2. In general, a complex nucleus also possesses a spin angular momentum. As a result of the partial or complete compensation of nucleonic spins, the total spin can have values of 0, 1/2, 1, 3/2, etc. Herewith, in full analogy with the electronic shell, the angular moment (mechanical moment) of a nucleus is: LI ⫽ I ( I ⫹1),
(8.1.1)
where I is the spin quantum number of a particular nucleus. Analogously with electron characteristics, the projection LI on the selected axis z can have the values: Lz ⫽ mI ,
(8.1.2)
where quantum number mI ⫽ I, I⫺1, … 0, … ⫺I can have 2I ⫹ 1 values. From eqs. (8.1.1) and (8.1.2), it can be seen that the value of electron and nucleon angular momentums have the same order (⬃). Because the nucleus spin is stipulated by the behavior of both neutrons and protons, the existence of nuclear magnetic momentums can be expected as well. The vectors of mechanical and magnetic moments are tightly bonded to each other. Nuclear physics was developed much later than atomic physics and this fact has influenced the approach to the description of nuclear properties. So, eq. (5.2.2), describing the electron gyromagnetic ratio of magnetic to mechanical moments, is applied to the nuclei as well. As a result, the gyromagnetic ratio for a spin-possessing nuclei was formally transcribed as: 冏e冟 M ⫽ gN LI 2 mp
(8.1.3)
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(in SI), however, instead of the mass of electron me in the denominator, the proton’s mass mp is present. Hence, the magnetic moment of a nucleus can be derived as:
M I ⫽ gN
冷e冨 I ( I ⫹1) 2 mp
(8.1.4)
The value e ⫽ N 2 mp
(8.1.5)
is referred to as the nuclear magneton and denoted as N (N ⫽ 5.05 ⫻ 10⫺27 J/T). Since in nuclear physics the mutual orientation of the angular and magnetic moments can be both parallel or antiparallel, the gyromagnetic ratio can also be positive (parallel orientation of these two vectors) and negative (antiparallel orientation as for orbital electron state). Therefore, the nuclear g-factor (gN) (in units e/2mp) can have both positive and negative signs. They cannot be calculated but are defined from experiment only. Therefore, one can see that the gN sign is historically casual. For proton gp is ⫹5.5851, for neutron gN⫽⫺1.9103. It can be seen from eqs. (8.1.4) and (8.1.5) that the magnitudes of the nuclear magnetic moment are approximately 2000 times lower than the corresponding electron values. Therefore, special instruments are necessary for observing them. The z-projection of MI is: M Iz ⫽ gN N mI
(8.1.6)
Because MI and LI are tightly bonded, nuclei precession in the magnetic field also takes place (refer to Section 5.2.4). The magnetic field created by electron shell can in this case be regarded as external to the nucleus. It is necessary to note that the nature of the nucleon’s magnetism is still not sufficiently clear. For instance, it is not understood why such an electrically neutral particle as a neutron nevertheless possesses a magnetic moment. 8.1.2
Nuclear energy levels
One of the distinctive signs of any quantum mechanical system, including nuclei, is the discontinuity of its energy states. The permitted values of nuclear energy depend on its particular structure, the nature of the nuclear forces, and more. Unfortunately, at present, nuclear physics does not allow us to calculate theoretically the arrangement of energy levels; these are experimentally found values. Nuclear ground states E0 are stable, not changing their energy in the course of time. However, by one or other external influence, they can be excited to states E1, E2, etc. All
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excited states are unstable: in the course of time, nuclei spontaneously transform to the lower or ground state, emitting in the surrounding space an excess of energy in the form of rays. The time during which the number of originally existing nuclei decreases in e times is referred to as the nuclear lifetime . For a given nucleus in different excited states, can have widely different values, however, a quantum mechanical relationship always exists between the value and uncertainty of energy E in the given energy state (refer to Section 7.2). E ⱖ h.
(8.1.7)
For instance, if ⫽⬁ (ground state), energy E0 can have a precise value (E⫽0). If ⬃10⫺8 sec, then E⬃10⫺7 eV. Usually, relative to E0, the first excited energy levels of many nuclei are E1⫺E0⬃104 eV. Therefore, the ratio E/(E1⫺E0) is of 10⫺11 in the order of value and seemingly can be regarded as negligible. However, as we will see below, this is not so. 8.1.3
Nuclear charge and mass distribution
The distribution of neutrons and protons in a nucleus can be different. In the majority of experiments, a nucleus behaves as a system possessing a center of symmetry and axis of symmetry of an endless order. Consequently, a nucleus also possesses a mirror plane of symmetry perpendicular to the symmetry axis passing through the center of symmetry. The charge distribution can be described by the function of the charge density (x,y,z). The normalization of the charge to the total nuclear charge gives:
∫ (r )d ⫽ Ze,
(8.1.8)
V
The distinctive features can be mentioned in this description. Firstly, the average density function can differ for charge and mass distribution. If, in particular, protons are distributed at the nucleus periphery and neutrons are concentrated nearer the origin, the characteristic radii of the two can be different. They can differ even for the two energy states of one and the same nucleus. Secondly, the nuclear forces are of short-range nature. Therefore, nuclei are sharply outlined in space. There is an empirical expression for spheri cal nuclei size: rЯ ⬇ 1.25 ⫻10⫺15 3 A m,
(8.1.9)
where A is the mass number. Thirdly, the nucleus can deviate from the spherical symmetry toward the ellipsoid of revolution with three mutually perpendicular second-order symmetry axes. Such an ellipsoid is of two radii: a is directed along z-axis and b is perpendicular to it. The ellipsoids of revolution can be flattened (a⬍b) or elongated (a⬎b) (Figure 8.1).
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z a z
z
x
b y
y
y
x
x (a)
(b)
(c)
Figure 8.1 Models of different nuclear forms: (a) spherical, (b) elongated and (c) flattened.
8.1.4
Nuclear quadrupole electrical moment
Point charge and dipole electrostatic fields have been described in Chapter 4. There are more complex electrical systems created by irregularly distributed charges (refer to Appendix 3). It is shown in Appendix 3 that such a system can produce an electrostatic field, which can be described by the sum of terms differently dependent on the distance r (Table 8.1). Being charged, the nucleus has nonzero monopole contribution. Since nuclei possess the definite symmetry elements listed above, it cannot possess a dipole moment; a dipole moment as a vector cannot coexist with the symmetry operations mentioned. This signifies that the nucleus does not possess a dipole moment. However, this does not prohibit a nucleus from having an electrical multipole moment of higher order: quadrupole, in particular. In other words, the deflection of the nucleus from a spherical form brings about the existence of the quadrupole moment eQ (refer to Appendix 3). Value eQ is measured in C.m2 units (in SI). The mathematical description of the quadrupole moment can be given as:
∫∫∫ ( x, y, z)(3z
2
⫺ r 2 )dV ⫽ eQ
(8.1.10)
V
where r2⫽x2⫹y2⫹z2, dV ⫽ dx dy dz. Integration is subjected over the whole nucleus volume V. Bearing in mind that in the reference system connected with the nucleus (refer to Figure 7.16), this expression can be presented in trigonometric form: eQ ⫽ ∫∫∫ ( x, y, z )r 2 (3 cos2 ⫺1)dV .
(8.1.11)
V
As shown in Appendix 3, the charged system elongated up to the z axes possesses a positive quadrupole moment, though for the flattened nucleus it is negative. The deflection of the nucleus from a spherical form and the nucleus spin are interrelated. Therefore, one can expect that the experimental manifestation of a quadrupole moment and nuclear spin are also correlated. This is exhibited in the following. The uncertainty in the angular momentum orientation (see Figure 7.17) is present in the quadrupole moment
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Table 8.1 Properties of a number of electric systems Potential as dependent on distance r
Charges system
Electric field E as dependent on distance r
Energy in an external electric field
Monopole
Ec ⬃
q r2
c ⬃
q r
U ⬃ q
Dipole
Ed ⬃
p r3
q ⬃
p r2
U⬃p
Quadrupole
Eq ⬃
qa 2 r4
q ⬃
qa 2 r3
U ⬃ qa 2
d dr d 2 dr 2
too. The measured value is not the eQ itself but the quantity eQz that can be expressed through the nuclear spin I as: eQz ⫽
I ⫺ (1Ⲑ 2) eQ. I ⫹1
(8.1.12)
This means that at I⫽0, the quadrupole term is meaningless; at I⫽1/2, it cannot be measured and it is reliably measured at I⬎1/2. 8.2 8.2.1
INTRAATOMIC ELECTRON–NUCLEAR INTERACTIONS
General considerations
If we consider a nucleus being not a point but a volumetric nucleus, additional effects in the electron–nucleus interaction appear. They play a very important role in the physical description of an atom. These additional effects are exceedingly small in comparison with the main Coulomb and even with fine interactions (refer to Section 7.5.4). So, they refer to the number of intraatomic “superfine interactions.” Remember that the energy of the electrostatic interaction of two electric charges is defined by product q (refer to Section 4.1.4) in which is a potential created by the electron’s charge in the point where the proton’s charge is located. Therefore, the additional interaction energy appears. (Certainly one can consider that the field is created by a volumetric nucleus and an electron is in this field; this effect will also be considered further.) The most general expression for the electrostatic (marked by letter E) interaction energy can be written as:
EE ⫽ ∫∫∫ ( x1 , x2 , x3 )( x1 , x2 , x3 )dx1 dx2 dx3 . V
(8.2.1)
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For convenience in this expression, instead of Cartesian coordinates x,y,z we can choose another form of coordinates x1;x, x2;y, x3;z. The function that describes a charge distribution in a nucleus is (x1,x2,x3), the potential created by the electron shell at the nucleus is denoted (x1,x2,x3), and the integral is taken upon the whole nucleus volume V. Decomposing the function (x1,x2, x3) in the MacLoren series near the origin: 3 ⎛ ⎞ x ( x1 , x2 , x3 ) ⫽ (0,0,0) ⫹ ∑ ⎜ x ⎟⎠ 0 ⫽1 ⎝
⫹
1 3 ⎛ 2 ⎞ 2 1 3 3 ⎛ 2 ⎞ x ⫹ ∑ ∑ ⎜ x x ∑ 2 ⫽1 ⎜⎝ x 2 ⎟⎠ 0 2 ⫽1 ⫽1 ⎝ x x ⎟⎠ 0
(8.2.2)
where (⭸/⭸x)0 are electric field components along the axis x, (⭸2/⭸x2)0 are the electric potential gradient along the same axis. Because of the small dimension of nuclei, it is possible to limit the number of terms in expression (8.2.2) by the second order. The electric field gradient possesses the symmetry axis along the x3 axis. Therefore, all crossed terms of the type (⭸2)/(⭸x⭸x) with ⫽ are equal to zero. Substituting the expression (8.2.2) into the electric energy (8.2.1) we have: 3 ⎧⎪ ⎛ ⎞ 1 3 ⎛ 2 ⎞ 2 ⎫⎪ EE ⫽ ∫∫∫ ( x1 , x2 , x3 ) ⎨(0,0,0) ⫹ ∑ ⎜ x ⫹ ⎟ 2 ∑ ⎜⎝ x 2 ⎟⎠ x ⎬ dx1 dx2 dx3 V ⫽1 ⎝ x ⎠ 0 ⫽1 0 ⎩⎪ ⎭⎪
(8.2.3) Opening the brackets and carrying out elementary transformation we obtain: 3 ⎛ ⎞ EE ⫽ ∫ ( x1 , x2 , x3 )(0, 0, 0)dV ⫹ ∑ ⎜ ⎟ ⭈ ∫ ( x1 , x2 , x3 ) x dV ⫽1 ⎝ x ⎠ 0
⫹
1 3 ⎛ 2 ⎞ ⭈ ( x1 , x2 , x3 ) x2 dV ⫹... ∑ 2 ⫽1 ⎜⎝ x2 ⎟⎠ 0 ∫
(8.2.4)
For our convenience, let us write every term separately presenting eq. (8.2.4) as a sum EE⫽EE1⫹EE2⫹EE3 and then: EE1 ⫽ ∫ ( x1 , x2 , x3 )(0, 0, 0)dV Because (0,0,0) is the electric potential at the origin, it is the number and can be taken out from the integral. The rest is the nucleus charge, Ze. Therefore, EE1 ⫽ eZ (0, 0, 0),
(8.2.5)
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The first term in eq. (8.2.5) EE1 is the interaction energy of atomic electrons with the nucleus, the charge of which is contracted to point; the nucleus becomes dimensionless. Integrals of the type EE2⫽ (x1,x2,x3)xdV in (8.2.4) are equal to zero because of the fact that according to their mathematical structure, they describe nuclear electric dipole moments though we already know that the nucleus cannnot have one. Therefore, eq. (8.2.4) reduces to two terms, the first and the third: EE ⫽ eZ (0, 0, 0) ⫹
1 3 ⎛ 2 ⎞ ( x1 , x2 , x3 ) x2 dV , ∑ 2 ⫽1 ⎜⎝ x2 ⎟⎠ 0 ∫
(8.2.6)
By some transformations, the term EE3 can be transformed into the sum: EE′ 2 ⫽ 23 e2 Z 冷 (0) 冨2 冬r 2 冭
(8.2.7)
1 3m 2 ⫺ I ( I ⫹1) . EE′′3 ⫽ zz [eQ] 2 4 3I ⫺ I ( I ⫹1)
(8.2.8)
and
Consider all terms separately. We will refer here to the energy level scheme in Figure 8.2. 8.2.2 Coulomb interaction of an electron shell with dimensionless nucleus The first term in (8.2.4) (EE1 in eq. (8.2.5)) describes the electrostatic interaction of an electron shell with the point nucleus charge at the origin (Figure 8.2a). This term corresponds fully to the Coulomb energy, which was discussed in Section 7.5; hence, the quantization of the energy is just the same as was accepted above. In other words, E1 is the energy that describes the electron shell state in the absence of superfine interactions. Evaluation of the energy has already been made in Section 7.5.4. It shows that depending on Z, the energy EE1 is 101⫺104 eV. Remember that the addition to EE1, the energy of spin–orbit fine interactions is Enj⬃10⫺1/10⫺3 eV. Other types of interactions have made their own particular contribution to the energy levels. 8.2.3 Coulomb interaction of an electron shell with a nucleus of finite size: the chemical shift Let us consider the next term eq. (8.2.7). Two multipliers exist in this term: 冓r2冔 is the mean-square nuclear charge radius (determined according to the general rule (eq. (3.1.6⬘)) 冓r2冔⫽( (x1,x2,x3)r2dv)/( (x1,x2,x3)dv) and 兩 (0)兩2 is the probability density of finding an electron at the origin. Notice that this term accounts for the nonzero nuclear size.
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(a)
(b)
505
(d)
(c)
(e) +3/2
∆2
+3/2
3/4 1/4
+1/2
+1/2 –3/2 –1/2
–1/2 –3/2
+1/2 1/4
∆1
–1/2
∆ 0
0
∆ 0
0
Figure 8.2 The particular effects influences on the energy state of an atom: (a) a stripped nucleus, (b) the chemical shift (refer to Section 8.2.3), (c) the fine electron-nucleus interaction (refer to Section 7.5.8), (d) and (e) an atom in an external magnetic field (in (d) influence of both quadrupole and magnetic splitting is very complicate), (e) the influence of the magnetic field produced by the electron shell on the nucleus’s magnetic moment position (nuclear Zeeman effect, see Section 7.7). The shift of the spectral line relative 0 is presented in the lower part of the Figure. The energy transitions scale is not followed. The long arrows represent R transitions, the double arrow in the figure represent the NMR transitions.
It follows from eq. (8.2.7) that this energy contribution characterizes the electron cloud interaction with the nuclear charge. This type influences the shifting of the nuclear energy levels but does not split them. The energy shift depends on a product: the electronic parameters [e兩 (0)兩2] and the nuclear properties (Ze冓r2冔) (Figure 8.2b). Because nucleus excitation can lead to a change of its mean-square radius 冓r2冔, this value can differ for two levels—the ground and the excited ones. Therefore, the energy difference can be changed. This phenomenon is referred to as the chemical shift. The experimental spectral lines are schematically shown under the energy transitions. Analysis of the electron radial curves (Figure 7.22 and Table 7.1) shows that only s-electrons can influence the 兩 (0)兩2 value. In the isolated atom, the energy E⬘E2 could change the line position only a little. This is schematically shown in Figure 8.2b. The E⬘E2 value is often 10⫺4 eV and has a positive sign. The difference between the energy of two levels is significantly smaller than the energy itself and is usually 10⫺7–10⫺8 eV. As a result, the line position on the experimental spectrum is shifted as well by , as shown in the figure. For a given atom in different atomic groups and/or molecules, the E⬘E2 value can be changed. When an atom enters the chemical bonding with other atoms, it produces new molecules and the whole electron system undergoes a change (s-electrons including). This last influences the E⬘E2 value. Thus, the chemical shift is a very sensitive measure of the chemical bonding.
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It is impossible at present to calculate the E⬘E2 value theoretically. However, the change in the energy transition can be found experimentally. In order to give this value a quantitative character, measurements are carried out relative to a specially chosen reference substance. In this case, one can measure the quality: EE′ 2 ⫽ Ze2 冬r 2 冭冦冷 (0, 0, 0) 冨2 ⫺ 冷 (0, 0, 0) 冨2refr 冧,
(8.2.9)
where 冓r2冔⫽{冓r2excited冔⫺冓r2ground冔} is the change of the mean-square nuclear charge radius in the transition from the excited to the ground state. As soon as the reference substance is chosen (the second term in the difference in the figure bracket), the E⬘E2 quality is called the chemical shift relative to a definite compound. In modern chemistry, the chemical shift is one of the main, widely used characteristics of an atom in a molecule. Measurement of the chemical shift can be achieved by means of NMR and R. It should be noticed, however, that the specificity and individual development of these two methods are the reason that different data and different reference substances determine the difference in results, although the physical content of both methods remains the same. 8.2.4 The nuclear quadrupole moment and the electric field gradient interaction A second term in eq. (8.2.6) corresponds to the so-called quadrupole effects related to electron–nuclear superfine interactions. Unlike the preceding, it brings about the splitting of nuclear energy levels into sublevels, lying both above and below the main level because the energy of this interaction can have both positive and negative signs. As well as chemical shift, the quadrupole interaction expression consists of two terms: one describes the nucleus feature, i.e., the nuclear quadrupole moment, and the others present the electrical field gradient in the point where the nucleus is located (⭸2/⭸x2)0⫽(⭸E/⭸z)0 The nuclear quadrupole moment was considered in the previous chapter (refer to formulas (8.1.10) and (8.1.11)). It depends on the quantum numbers I and mI: depending on these numbers, the originally singular level splits into several sublevels (degeneration on the quantum number mI is partly lifted). The scheme of quadrupole splitting with the nuclear spin I⫽3/2 is given in Figure 8.2c. Since in the expression for the energy, the square of quantum number mI is entered, the four values of mI (⫾1/2 and ⫾3/2) correspond only to two energy levels. When a system turns from the upper (split) onto the ground (singular) level, two lines appear (as shown in Figure 8.2c). The distance between sublevels depends on the electrical field gradient at the nucleus. These are called quadrupole splitting. In order for quadrupole splitting to be developed two conditions must be fulfilled. First, the nucleus must possess a quadrupole electric moment (refer to eq. 8.1.10). Second, it is necessary for the nucleus be in the point where the electric field gradient exists. This appears in those cases where an atom is in an asymmetric environment. Thus, a relationship of the quadrupole splitting and the chemical structure appears. Unlike the chemical shift, which depends only on s-electrons, the gradient of an electrical field can be created by p-, d- and f -and other electrons.
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The evaluation of the value developed in this splitting is usually in the order of 10⫺7⫺10⫺8 eV. It is possible to observe nuclear quadrupole splitting using R and NQR methods. 8.2.5
Interaction of a nuclear magnetic moment with an electron shell
In addition to the electrostatic interactions of a nuclear charge with an atomic electron shell, there is also the interaction of a nuclear magnetic moment with a magnetic field created by an electronic shell at the point of nucleus location. Two phenomena appear simultaneously: the electron shell creates a magnetic field at the point of nucleus location, and, on the other hand, the nuclear magnetic moment creates a magnetic field influencing the electron cloud. The interaction of the nuclear magnetic moment with the electron magnetic field can be described by the Zeeman effect (refer to Section 7.7). The Zeeman splitting in this instance is: EM ⫽ gN mI N B(0, 0, 0),
(8.2.10)
where B(0,0,0) is the magnetic field induction produced by electrons at the origin. In magnetically disordered materials, as a result of the thermal chaotic motion of atoms, the averaged nuclear magnetic field is zero and the effect is nonobservable. In magnetically ordered materials, the chaotic motion does not occur in full measure and a nuclear Zeeman effect takes place. In this case, the degeneration upon mI is lifted (refer to Section 7.7), i.e., the nuclear energy level with a certain I is split into 2I⫹1 sublevels. Figure 8.2e presents such spectral lines splitting with I⫽3/2: this level will form six equidistant superfine magnetic splitting lines. The splitting energy is defined by an expression close to eq. (7.7.5). It can be seen from expression (8.2.10) that the additional energy of the magnetic interaction depends on the magnetic field imposed on the nucleus. Because the nucleus magnetism is 1840 times lower than the electronic, the Zeeman nuclear splitting is significantly less than the electronic. However, owing to the large magnitude of the magnetic field created by electron shell at the nucleus position, this splitting can be experimentally observed, for instance, by the -resonance method. The same happens when the nuclear magnetic moment acts on the atomic electron’s shell. This makes it possible to investigate the changes in the shell caused by the chemical interactions. 8.2.6
Atomic level energy and the scale of electromagnetic waves
Let us analyze the results. A general scheme of nuclear energy levels with consideration of all electronic–nucleus interactions is given in Figure 8.2. A scheme corresponding to idealized models of stripped nucleus (without electronic shell) is given in Figure 8.2a. As an example, the energy states—ground and excited—were accepted with quantum numbers I ⫽ 1/2 and 3/2, accordingly. The excitation energy for the majority of cases is of the order 104 eV.
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When a nucleus undergoes transformation from an excited to a ground state, a -quantum is emitted, the frequency of which is ⫽(E3/2⫺E1/2)/. For stripped nucleus, we denote this frequency as 0. If we put the originally taken nucleus of nonzero size into the usual electronic shell and observe the interactions, we should take the Coulomb forces of the nucleus with electrons into account (eq. (8.2.5)) (Figure 8.2b). This event shifts both levels. The displacement value is in the order of 10⫺4 eV. (Attention should be paid to the fact that it is impossible to draw the levels and splitting in Figure 8.2 to the right scale: on the background of main transition 104 eV, we must depict details of 10⫺7⫺10⫺8 eV, the difference is 11 orders of value less.) However, it is important that the nucleus in ground and excited states can have different radii (different root mean square radii 冓r2冔) and displacements of those two levels are different (however, 兩(0,0,0)兩2) in both cases is the same). Measurements show that this difference is of the order 10⫺7 eV. So, transition frequency decreases by the value ⬃10⫺7/10⫺15⬃108 Hz (in comparison with 1019 Hz). The next step is an account of quadrupole interaction (eq. 8.2.8) provided the nucleus possesses a quadrupole moment and is in a nonuniform electric field. If both these conditions are fulfilled, the energy levels split according to eq. (8.2.8) (see Section 8.2.4). When quantum number mI present in this expression is in square, the top level splits into two sublevels (mI2 ⫽ 9/4 and 1/4) and the lower level does not split at all (mI2 ⫽ 1/4). Therefore, one existing spectral line splits in two. The experimental evaluations show that the usual value of quadrupole splitting is about 10⫺7 eV. If one takes into consideration the magnetic field created by the electronic shell at the nucleus position (Figure 8.2d), the scheme of energy transitions becomes more complicated (see Section 8.2.5). The degeneration on mI is lifted. The value of splitting is defined by the particular magnitudes of the terms in eq. (8.2.4). If the magnetic and quadrupole contributions are presented simultaneously, the spectral picture gets more and more complex (not shown in Figure 8.2d). If we consider magnetic splitting alone, the spectral picture gets comparatively simple (Figure 8.2e). Here, we should keep in mind that the selection rules permit only the transition for which quantum number mI is changed to unity or is not changed at all. If we take I⫽(3/2), the spectrum will split into six lines. The magnitude of splitting depends on the nuclear g-factor and on magnetic field induction at the nucleus position. In addition to the transition between sublevels of excited and ground states, other transitions are also possible, in particular between two sublevels of a single ground state (the double arrow in Figure 8.2e). The single arrows correspond to RS and the double arrows describe the NMR. 8.3 8.3.1
-RESONANCE (MÖSSBAUER) SPECTROSCOPY
Principles of resonance absorption
In certain circumstances, a substance is capable of absorbing electromagnetic energy, particularly, powerfully and selectively. This happens when the energy of a falling quantum coincides with the electron and/or nuclear energy level difference. Then, so-called resonance absorption of the electromagnetic radiation occurs.
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Suppose that the sample contain atoms with two intrinsic energy levels E1 and E0. The resonance condition can be written as: E1 ⫺ E0 ⫽ res ⫽ gN N B,
(8.3.1)
(remember that the change in quantum number mI is ⫾1). Here, B is the magnetic induction at the nucleus position. An experimental scheme is depicted in Figure 8.3. In Figure 8.3a, the main parts of the experimental equipment are shown: an emitter (1), a sample (2) and a detector (3). The resonance is observed when condition (8.3.1) is met. In Figure 8.3b, the absorption curve is
1
2 –V
3 V (a)
Emitter
Detector
Sample
Absorption
Absorption
(b)
Intensity
Ι
(c)
Γ
Derivation
dI d
(d)
Figure 8.3 An experimental arrangement of the -resonance experiment: (a) experimental arrangement, (b) absorption curve, (c) transmission curve and (d) dI/d curve. is the width of a bell-like curve on the half-height (half-width).
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presented, whereas in (c) the transmission is depicted. Only a single frequency can be found in experiments. In order to measure the whole curve and reveal reliably resonance absorption, a displacement from the resonance frequency should be accomplished. The frequency can be changed although it is easier to move the emitter (or sample) relative to each other using the Doppler effect. In some cases, the derivative curve is measured (Figure 8.3d). The point is that if both emitter and detector are motionless, the device would register quanta with just the emitted frequency. If either the emitter or detector begins to move along the connecting straight line with relative velocity V, the harmony is destroyed. Depending on V, the detector will register another frequency ⬘. The difference
⫺ ⬘⫽ can be found according to the Doppler effect (refer to Section 2.8.4 and to eq. (2.8.23)), namely, V ⎛ V⎞
′ ⬇ 0 ⎜ 1⫹ ⎟ ⫽ 0 ⫹ 0 , ⎝ ⎠ c c
(8.3.2)
The resonance curve is characterized by several parameters: position, width on half of the height (half-width) and intensity (height of maximum or absorption “depth”). 8.3.2
Resonance absorption of -rays: Mössbauer effect
Assume that there are two samples with exactly the same structure and nuclei. This signifies that the positions of two energy levels in both samples are exactly equal. Assume further that there is a way to initiate the nuclei of the first sample to emit quanta. The emitted spectral line E1⫺E0⫽E will exhibit a frequency, . One can evaluate the value of the natural width of this spectral line (i.e., the width as defined by quantum physics (refer to Chapter 7.2) but not by imperfection of the experimental equipment). The uncertainty principle gives: 10⫺15 c ⫽ ⫽ ⫺7 ⫽ 10⫺8 eV 10 c
(8.3.3)
where is the natural half-width of the spectral line and is the average time-of-life of a nucleus in an excited state with the energy E1. The ratio of the natural half-width to the absolute energy value will then be /(E1⫺E0)⬇(10⫺8 eV/10⫺4 eV)⬇10⫺12. It can be seen that this relative width is very narrow. If this radiation is directed onto a second quite similar sample, resonance should occur, i.e., the resonance absorption should be observed. Indeed, the -quantum energy is precisely equal to the energy difference E1⫺E0. The absorption of the electromagnetic waves with -quantum energy just equal to the energy difference in a sample is caused by the resonance absorption, i.e.:
res ⫽
E1 ⫺ E0
(8.3.4)
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If it is possible to change smoothly the electromagnetic radiation frequency, then the resonance radiation absorption can be observed in the background of the nonresonance processes. Such an experimental scheme is depicted in general form in Figure 8.3. However, there are at least two factors that destroy the resonance. The first factor is recoil, which obtains free nucleus when emitting the quantum. Really, the law of momentum conservation requires that the total system momentum, being equal to zero before emitting the quantum (0⫽p⫹pn), should be kept at zero after disintegration, so 0 ⫽ p ⫹ pn
or
p ⫽ ⫺pn ,
(8.3.5)
where p and pn are the p⫺ and pn⫺nuclear momentums. The law of energy conservation in this case has the form E2⫺E1⫽E⫹En, where: E ⫽ ( E2 ⫺ E1 ) ⫺ En
(8.3.6)
It can be seen that the -quantum energy E is less than the ideal value E1⫺E0, on the value of nucleus recoil energy En (denoted usually as R). From eqs. (8.3.5) and (8.3.6), we can find values EnR. The energy of a photon is correlated with its momentum by the equation (see Chapter 1.6): E ⫽ p c
(8.3.7)
where c is the speed of light. Energy of recoil, R, from the above given equations is R⫽
p2 E2 pN2 ⬃ 10⫺3 ⫺10⫺2 eV ⫽ ⫽ 2 mN 2 mN 2 mN c 2
(8.3.8)
This signifies that the emitting line is displaced along the energy axis on R to a side of its reduction. This shift itself is small, particularly in comparison with the quantum energy (104 eV); however, in comparison with the natural width of a spectral line (10⫺8 eV), it is large. Similarly, the spectral absorption line will be displaced as well, but in the opposite direction. Two lines, the natural width of which are 10⫺8 eV, move apart by 10⫺3 eV (Figure 8.4). Thereby, no overlap of the spectral lines takes place, and, consequently, no resonance absorption occurs. The second factor is chaotic thermal atomic motion: different nuclei are emitted in different chaotic states of movement. As a result of the Doppler effect, the broadening D of both spectral lines, emitting and absorbing, occurs (Figure 8.4); moreover, at room temperature, this broadening is of many orders of magnitude larger than the natural line width. As a result, only the tails of the spectral lines overlap; absorption will reach a miserable value from the expected effect. Quite another picture is observed if both nuclei are confined in a crystal. In this case, the whole crystal needs to be considered as a closed system. The theory of this effect (at energies
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I D
E R
R
Figure 8.4 An infringement of the resonance.
of quanta smaller than the binding energy of atoms in the crystal) shows that, while emitting, two possibilities can arise. The first is the creation of an elastic wave in the crystal, i.e., phonon (refer to Section 9.3.1), which would carry away the excess of the energy of a quantum. This is inelastically scattered nonresonance quantum. The other possibility is an elastically scattered quantum, when the recoil energy is transmitted to the whole crystal. Herewith, in the formula for the determination of R, instead of the nuclear mass, mN, we should substitute the macroscopic crystal mass; then the recoil energy reduces practically to zero, and the -quantum energy becomes almost precisely equal to the energy difference E1⫺E0. The Doppler broadening from the thermal motion also vanishes. As a result, the emitting and absorbing lines narrow down to the natural width and coincide with each other. The resonance requirement becomes satisfied. The theory of this effect was worked out by R. Mössbauer (Nobel Prize, 1961) and the effect itself is referred to as the Mössbauer effect. It consists of the recoilless emission and resonance absorption of quanta by nuclei. The ratio of the number of elastically emitted quanta to their total number is called the Mössbauer effect probability f and is proportional to: ⎧⎪ 冬 x 2 冭 2 ⎫⎪ f ⬃ exp ⎨⫺ ⎬, c 2 ⎪⎭ ⎪⎩
(8.3.9)
where 冓x2冔 is a mean-square displacement of nuclei from their equilibrium position in a crystal as a result of their thermal motion (in the direction of quantum emission). The probability is proportional to the atomic mobility in a crystal. How can -radiation resonance absorption be seen and measured? Assume that the emitter and absorber materials are quite similar and are in the same state. The maximum value of resonance absorption must be observed, when emitter and absorber are motionless (V⫽0) (eq. 8.3.2). When one begins to move, e.g., the emitter regarding the absorber, this resonance absorption is destroyed and the experiment reveals an absorption curve; a very small relative velocity is needed to “separate” the spectral lines (emitted and absorbed). V ⬃ ⬃ 10⫺12 E c
(8.3.10)
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The relative velocity needed to destroy the resonance can be found from this equation. A striking result appears: despite the fact that light travels very fast, the speed needed to destroy the resonance is very small V⫽c⫻10⫺12⫽10⫺4 m/sec! Therefore, the simplest way to observe resonance absorption is to measure it as a function of the relative speed, V. A schematic diagram of a R experiment is presented in Figure 8.5a. 8.3.3
-Resonance (Mössbauer) spectroscopy in chemistry
As far as the volume of information that can be obtained on chemical and physical properties of chemical compounds, RS occupies a leading position. Nuclear transitions correspond to energy difference 104 eV but the effect to be measured relates to 10⫺7–10⫺8 eV; however, Mössbauer spectroscopy resolution is so high that it successfully solves such problems. In addition, this method can find nearly all the effects of nuclear interactions with an electronic shell; it carries extremely valuable information on molecular or crystal structure. One of the main parameters of a resonance curve is its position on the velocity axis. In Figure 8.5a, a resonance curve is presented schematically when the emitter and absorber are one and the same motionless substance: the maximum of absorption falls on the zero value of relative velocity. Intensity of absorption (depth of curve minimum) is proportional to the probability of the resonance absorption, f. So, such measurements enable the measurement of a change in the atomic mobility in a sample under different treatments
(a)
(b)
(c)
(d)
0
∆ 0
∆
0
2E/h
0
Figure 8.5 Exhibiting different effects in R: (a) primary position of the absorption spectral line on the velocity axis by a stripped nucleus 0, (b) chemical shift (c) quadrupole splitting and (d) magnetic SF interaction development.
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connected with such chemical problems as catalysts, chemisorption, polymerization and others. If the source substance and absorber are different, the spectral absorption line is shifted along the velocity axis; that is caused by the additional energy EE2 (8.2.9). The energy EE2, denominated in units of velocity, is called in RS the chemical shift (Figure 8.5b). The sodium nitroprusside Na2[Fe(CN)(NO)5]⭈2H2O is mostly used as the standard (reference) substance in Mössbauer spectroscopy. Indication of the reference substance in scientific publications is obligatory. If some atoms in a substance are in an asymmetric environment and their nuclei possess a quadrupole moment, quadrupole splitting can occur (Figure 8.5c). Knowing spin I, the splitting allows the electric field gradient value of the nuclei that is dependent on the structure’s peculiarities to be found. If the substance under investigation is in a magnetically ordered state (ferro-, antiferro- or ferri-magnetic, refer to Section 5.3), a nuclear Zeeman effect takes place: energy levels split according to eq. (8.2.10). In Figure 8.5d, the case of a nucleus with I⫽3/2 in symmetrical surroundings in a magnetically ordered substance (e.g., antiferromagnetic, -Fe2O3) is given. The number of lines allows the nuclear spin I to be found, whereas the distance between lines enables us to find the magnetic field induction (strength), created by the electronic shell, at the nucleus position B(0 ,0 ,0) (or H(0,0,0)). The latter is a feature of electronic shell and differs for different atomic states. Let us look at some examples. A particular interest for chemical investigations is the relative change of the chemical shift in a number of various isostructural absorbers at a fixed (motionless) source. For example, one can judge the valence state of a Mössbauer atom in various compounds by measuring the chemical shifts. So, all bivalent tin compounds have positive chemical shifts relative to grey tin (the source of the radiation moves to the absorber), whereas all tetravalent compounds have a negative shift relative to grey tin (the source of the radiation goes from the absorber). The chemical shift value is proportional to the ionicity of a Mössbauer atom in the compound under investigation. In Figure 8.6, the spectra of a series of isostructural tin tetra-halogenides are presented with tetrahedrons SnI4, SnBr4 and SnCl4, confirming this belief. However, from the same figure, it can be seen that there are no tetrahedrons with equivalent Sn–F bonds in the SnF4 compound; a couple of resonance lines are present. Theoretical analysis of the spectrum obtained shows that there is a quadrupole electric super-fine interaction in this compound; this is explained by the reticular coordination polymer structure with nonequivalent Sn-F bonds (Figure. 8.6). If a nucleus with energy E possesses a magnetic moment M, which is in a magnetic field B, the energy of the nuclear state (8.2.10) will change by a magnitude E. In R experiments, the transitions between two different nuclear states are observed. As EM is proportional to the first order of mI, the (2I⫹1)-fold degeneration is lifted in the magnetic field: to every mI, there corresponds its own value EM. Distinction is measured in RS effect at the transitions between the sublevels of two various nuclear states (Figure 8.2c). Thus, the number of lines of radiation/absorption spectra is defined by selection rules mI⫽0, ⫾1. For example, for nucleus 57Fe and 119Sn (Iex⫽3/2, Igr ⫽ 1/2) from eight formally possible transitions, only six are realized because of the selection rule.
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0 0.5
515
SnI4
v
2Γ
1.0
F
Absorption
v
SnBr4
SnF4
2Γ
F v
SnCl4 2Γ
v
SnF4 2Γ
∆
2Γ
Figure 8.6 Results of a RS investigation of tin tetrahalogenides. Quadrupole splitting is seen for SnF4 (after E.F. Makarov et al.).
For chemical researchers studying magnetic fields acting on the nucleus of Mössbauer atoms, this is of essential interest as the magnetic field on the nucleus is caused by spin, radial and angular distributions of electronic density in an atom. 8.3.4
Superfine interactions of a magnetic nature
If an atom with a nucleus with a magnetic moment M appears in an inner field with magnetic induction B, the nuclear energy E (8.2.20) increases: EM ⫽⫺(M B),
(8.3.11)
The transition between the two nuclear states is measured. As EM is proportional to the first degree of mI, (2I⫹1)-fold degeneration is lifted, and a separate value of the energy level EM corresponds to each quantum number mI. This value is measured in RS as the energy of the transitions between the sublevels of the two different nuclear states (Figure 8.2e). The number of spectral lines is the subject of limitation by selection rules mI⫽0, ⫾1 and depends on the number of energy lines in both ground and excited states. It can be seen from eq. (8.3.31) that experimentally measured EM allows calculation of the magnetic field strength H (or induction B), which electrons of the same atom create on the nucleus. We will not discuss this event in more detail, but would like to underline that these fields appeared to be rather high (up to 105 A/m and even higher). In Figure 8.7, the
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temperature dependence of RS in ferromagnetic iron metal measurements is depicted: at elevated temperatures, all the SF spectral lines collapse into a single one; the crystal undergoes transformation to a paramagnetic state. The interline distance in spectra gives the magnetic field mentioned above.
8.4 8.4.1
NUCLEAR MAGNETIC RESONANCE (NMR)
Introduction
NMR consists of the resonance absorption of electromagnetic radiation by a system of nuclear magnetic moments of the substance under investigation. NMR was discovered in 1946 by E.M. Parcel and F. Bloch (Nobel Prize, 1952). At present, the method is widely used in chemistry because of its remarkable ability to solve many chemical problems. There are several approaches to nuclear resonance descriptions. In this section, we will try to develop the energy approach, which is simpler for understanding. In Chapter 7.7 the behavior of a system of atomic magnetic moments in an external magnetic field was considered. Additional energy appeared; being applied to nuclei, this energy is E⫽gNNmIB0, where gN denotes the nuclear g-factor, N is the nuclear magneton, mI is
Absorption, arbitrary units
773°
744°
444°
22°
V mm/s
Figure 8.7 Magnetic R experimental curves of iron heated to above TC (see Section 5.3.1): all the SF lines collapse at T ⬎ TC (after E.F. Makarov et al.).
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the nuclear quantum number that determines the projection of the magnetic moment vector onto the quantization axis and B0 is the induction of the permanent external magnetic field applied to the substance. The splitting of the ground state of the nuclear energy levels is the result of the nuclear Zeeman effect (Figure 8.2). This effect lifted the degeneracy of the ground state energy making it dependent on the quantum number, mI. This means that the system that previously had one degenerated level, has 2I⫹1 nondegenerate sublevels in the external magnetic field. For proton I⫽1/2, the number of sublevels is two with mI ⫽⫾(1/2). Therefore, the ground state level will admit either positive or negative energy; in other words, it will be split, the splitting gap being E ⫽ N gN B0
(8.4.1)
(the quantum number mI is absent in this equation because the transition difference corresponds to mI⫽⫾1). This is depicted by a grey double arrow in Figure 8.2e. An alternating electric field with definite frequency is additionally applied onto the sample. This alternating field will be absorbed more intensively by the sample if the following resonance condition is achieved: res ⫽ E ⫽ gN B B0
(8.4.2)
There is a special coil to measure the absorption of an electromagnetic field. There are many types of NMR apparatus available. Figures 8.8 depicted two main schemes: an old one (a) and greatly advanced one (b). Since a very stable magnetic field is needed, the superconducting magnetic coil (SMC) is constructed, merged into liquid helium. The special devices permit one to change samples without warming the main vessel. Moreover, they enable one to change the frequency range as well. It can be seen from eq. (8.4.1) that the resonance condition can be met either by changing the frequency of the secondary alternating magnetic field at constant magnetic field B, or the induction variation B at constant frequency. Both possibilities can be realized in modern NMR apparatus. Sometimes it is easier to change the field (by means of sweep coils “c” in Figure 8.8a) and use the constant frequency. The majority of devices work according to these principles; actually, results do not depend on the choice of measurements. The frequency ranges used are presented in Table 5.3. So, the nuclear paramagnetic resonance NMR is the selective absorption of either the alternating electromagnetic energy in the presence of a permanent magnetic field by a substance containing atoms whose nuclei have nonzero spins, or vice versa, the absorption of magnetic induction B at constant frequency of the alternating field. 8.4.2
Use of nuclear magnetic resonance in chemistry
In order to realize the resonance absorption of an alternating magnetic field, one should have a sample with nuclei that possess nonzero magnetic moments, i.e., nonzero spin quantum number. The nuclei 1H (protons), 11B, 13C, 19F and some others have such nonzero
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(a)
(b)
z −1 a x
N2
b
N2 He
He
b
SCC
d S
N
y
−2
SCC 4 −3 CM
g c
c f
h
probe
e
Figures 8.8 Schemes of NMR spectrometers. (a) An original old spectrometer scheme of the continuous wave (a: sample tube; b: magnet; c: sweep coil; d: receiver coil; e: transmitter; f: amplifier; g: register unit; h: recorder), the electromagnet poles are also seen. (b) Section of a more advanced NMR spectrometer, the nitrogen and helium vessels with the superconducting magnetic coil are depicted. A sample is introduced from above; however, the device (probe), inserted from below, realizes the connection of sample measurement coils with the register system. CM: position of the sample investigated.
spins. It can be seen that many atoms dealt with in modern chemistry are present in this list. Proton magnetic resonance (PMR) is most popular in chemistry, but resonance on 19F and 13C is also often used. However, the information quality is so great that using the spectra of these atoms, one can find the behavior of other atoms of the mother molecule. Consider the behavior of hydrogen nuclei (protons) in atoms in a certain molecule in the external magnetic field, B0. The proton spin quantum number is 1/2, therefore, its lower energy level is split (Figure 8.2e). The purpose of the constant magnetic field is to split the single level in two due to the nuclear Zeeman effect. The alternative electromagnetic field B( ) facilitates transitions between sublevels. The third coil is to measure absorption. The principle difference between NMR and RS is as follows: on account of its enormous resolution, resonance allows one to define the superfine structure investigating the transition between different nuclei states (levels) with different I, whereas NMR allows one to measure the transition
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between sublevels of one and the same ground level. So, in spite of the difference between the methods, the results obtained often have the same nature. Consider the PMR ability using the example of the ethyl alcohol molecule CH3–CH2–OH. It contains six protons in three different atomic groups. However, only hydrogen protons give a contribution to the spectrum in the given frequency range. The PMR spectrum of ethyl alcohol obtained by a spectrometer with a very low resolution would contain only one spectral line to which all the protons contribute simultaneously. The information that can be obtained from such a spectrum is restricted by the statement that in fact there are hydrogen atoms in the sample. Substitution of hydrogen by, say, fluorine atoms would remove the signal completely. By improving the spectrometer resolution, the amount of information can be significantly increased. Consequently, it can be found that protons of different atomic groups absorb an electromagnetic wave under a different external magnetic field (Figure 8.9a). The amount of information in such a spectrum increases: the presence of three maxima shows that there are three types of hydrogen atoms in a compound, the resonance for different functional groups CH3, CH2 and OH takes place at the three different experimental conditions. This is because of the fact that, as well as the outer magnetic field, an additional magnetic field acts on each proton. The origin of these additional fields is the diamagnetic response of the given atom to the external field. Every particular atom has its
(a) Thu Jan 12 15:47:14 2006: (untitled) W1: 1H Axis = ppm Scale = 25.39 Hz/cm
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(b) Thu Jan 12 15:49:45 2006: (untitled) W1: 1H Axis = ppm Scale = 25.39 Hz/cm
5.500
5.000
4.500
4.000
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Figure 8.9 The NMR spectra of ethyl alcohol obtained at intermediate (a) and higher (b) resolution. Zero position signal corresponds to TMS.
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own electron cloud subject to some individual changes under the influence of the chemical bonding. This diamagnetic field is local, i.e., different for different atomic groups. How small the differences in the electron shell of hydrogen atoms in different positions, this difference is exhibited in the diamagnetic screening of the given nucleus. This signifies that not only will the external magnetic field with the induction B0 act on the given proton, but an additional local field as well, specific only to this particular atom. It was shown in Section 5.2 that the diamagnetic field is proportional to the external magnetic field and is in the opposite direction to it. Thereby, BA ⫽ B0 ⫺ A B0 ⫽ B0 (1⫺ A ), BB ⫽ B0 ⫺ B B0 ⫽ B0 (1⫺ B ), BC ⫽ B0 ⫺ C B0 ⫽ B0 (1⫺ C ),
(8.4.3)
where A, B and C show the structure-nonequivalent atoms of hydrogen in the molecule. Values are screening constants; they depend upon the electron structure, which, in turn, depends on more distant effects of the influence of the neighboring atoms and atomic groups. Dimensionless values correspond to atomic diamagnetic susceptibility; they are of the order 10⫺6 (this value’s order dictated the high requirement of magnetic field stability). On the background of unity, this is a very small value; nevertheless, it is reliably measured in an experiment. This is because of the fact that being in the outer magnetic field B0, protons of the hydrogen atoms “fill” not that field but fields BA, BB, BC, etc. Consequently, resonance occurs under some field other than B0 (at a fixed frequency). Thereby, the NMR spectrum consists of as many spectral lines as the structure-nonequivalent hydrogen atoms present in the molecule: for each of them, resonance will be observed under the different induction of the magnetic field. Such displacement of the position of the NMR signal along the magnetic induction axis depending on the atomic electron shell properties is referred to as the chemical shift. NMR spectrum of the same ethyl alcohol molecules is given in Figure 8.9b measured by a high-resolution NMR spectrometer. It can be seen that the proton resonance of the different functional groups CH3, CH2 and OH takes place at different values of the magnetic field (on account of the difference in ). There follows, in particular, an important conclusion that the number of NMR signals on the spectrum is equal to the number of structurenonequivalent protons in the molecule. (For instance, in benzene there will be one signal, in mono-substituted benzene there will be three, in ortho-di-fluorine-benzene—one, in para-di-fluorine-benzine—one, in meta-di-fluorine-benzine—two, etc). It would be useful to impart a quantitative meaning to the notion of the “chemical shift” in NMR. For this purpose, for instance, it was possible to obtain an NMR signal for a “stripped” proton and measure the shift from its signal. However, it is too complicated to obtain a signal from a “necked” proton, particularly for every sample. It was agreed, therefore, to choose a standard “reference” substance. At present, tetramethilsilan Si(CH3)4 (TMS) has been chosen as the most widespread standard substance. This substance has many advantages. Firstly, there is a relatively large amount of hydrogen in this compound.
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Secondly, all protons of methyl groups are structure-equivalent, so TMS exhibits itself by only one NMR signal. Thirdly, in solutions the CH3 groups revolve around the Si–C bonds and, therefore, the magnetic fields are averaged and the TMS signal is very narrow. Fourthly, the screening constant of the methyl group is minimal with respect to the majority of other hydrogen-containing substances. Consequently, the TMS reference line lies nearly always on the right side on all other signals on the axis of the magnetic field induction. From this line, the chemical shift is always positive (measured to the left side). To compare chemical shifts obtained on different instruments with different working frequencies, it was accepted to express the chemical shift quantitatively in relative units (in millionth units, or in m.u.) (not dependent neither on B0 nor 0): i ⫽
B0 i ⫺ B0 ( TMS) B0 ( TMS)
6 ⫻ 10 ,
(8.4.4)
where B0i and B0 (TMS) are the induction of external magnetic fields corresponding to the resonance absorption at the principle frequency of the group of equivalent nuclei (i) of molecules under investigation and in TMS, correspondingly. Thereby, i is defined by the relative displacement of the NMR signal of a given molecule on the spectrum to the left from the TMS lines. The evaluation shows that, for the hydrogen atom, removing one selectron brings about a chemical shift of 20⫻10⫺6 or in 20 m.u. Therefore, i for protons in the majority of different atomic groups have values from 0 to 10 m.u. There exist other, less often used, scales of chemical shifts. The generalization of NMR spectra allows experimental results on chemical shifts to be exhibited in the form depicted in Figure 8.10. Since the protons in similar atomic groups have somewhat different , the chemical shift values are plotted on the abscissa axis in the form of segments (not points) of i (in m.u.); the corresponding functional atomic groups are plotted on different levels on the ordinate axis. The black strips correspond to those cases, which are met with more often. 14
12
10
8
6
4
2
0 TMS
– COOH C(O)H C–NOH N = CH ArH RC(O)NHR(H) ArOH C = CH
C = C – NH C = OH RNH SH CH CH2 CH3
Figure 8.10 Generalization of the proton chemical shift in different atomic groups.
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Having obtained a PMR spectrum, the information can be used to gain a representation of whatever substance is being dealt with. Similar tables are accumulated for other nuclei as well. The picture described, however, is equitable only in some respects. There actually exist other finer effects influencing the spectrum. The mutual interaction of the proton magnetic moments of different functional groups can be mentioned in the first order. This interaction turns out to be stronger than was assumed, keeping in mind the interaction between two nuclear magnetic moments (magnetic dipole–dipole interaction). This is because magnetic moments of protons superpose magnetization onto electrons, participating in the chemical bonding in the molecule (particularly, electrons participating in chemical coupling), which, in turn, creates a magnetic field on the neighboring proton. This additional chemical bond magnetization noticeably enlarges the hydrogen proton chemical shift, as well as enlarging the interaction between proton containing groups, and does its more long-range acting. Such a type of interaction is called a spin–spin interaction. Spin–spin interaction is exhibited in the high-resolution spectrum. The spectrum gets more complicated whereas the information on the molecular structure increases. Let us consider this phenomenon using the example of the same ethyl alcohol, measured on the high-resolution spectrometer (Figure 8.9b). As has already been noted, three types of hydrogen groups manifest themselves in three line series, belonging to protons of different functional groups. The one right utmost line is a mark of resonance in TMS (⫽0). The group of lines at ⬇1.2 m.u. corresponds to protons of the CH3, the group of lines at ⬇3.5 m.u. corresponds to the group CH2, and at ⬇4.5 m.u. to the OH group. Let us look at the spin–spin interaction between protons of three functional groups in the molecule of ethyl alcohol. The proton can occupy two positions in the external field with projections defined by quantum numbers mI⫽⫾(1/2). Denote the state with mI⫽⫹1/2 by the letter and with mI⫽⫺1/2 by the letter . Analyze first the action of protons of group CH2 on the nearest groups, CH3 and OH. Two protons of CH2-groups can accept three total quantum numbers, S: ᎑ spin ⫽⫹1 ᎑ and ᎑ spin ⫽ 0 ᎑ spin ⫽ −1
S ⫽ 1, S ⫽ 0, S ⫽⫺1,
the total number of states with S⫽0 being twice as large as the other S values. Correspondingly, the neighboring lines of the CH3 and OH groups split into three with relative intensity 1:2:1 (Figure 8.9b). Evaluate now the action of CH3 on the protons of their neighbors. The number of states of the group of three protons can have four values of the total S: ᎑ spin ⫽ 23 , , ᎑ spin ⫽ 12 , , ᎑ spin ⫽⫺ 12 ᎑ spin ⫽ ⫺ 23
S ⫽ 23 , S ⫽ 12 , S ⫽⫺ 12 , S ⫽⫺ 23 ,
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with the relative number of states 1:3:3:1, accordingly. The SF dipole–dipole splitting is depicted in Figure 8.11 (without following the scale on). Consequently, the influence of group CH3 on group CH2 will be exhibited in the splitting of the signal. The action of the hydroxyl group in a favorite case obeys the general rule herewith each of the four lines must else be split into two. So, the high-resolution signal of the group CH2 in the ethyl alcohol is usually split into eight components. The general rule for nuclei with I⫽1/2 corresponds to this example: the number of spin–spin splitting caused by spin–spin interaction, is n⫹1, where n is the number of protons in the adjacent atomic group whose influence is considered (in the general case with I⬎½, this number is 2nI⫹1). This phenomenon can be described quantitatively by the spin–spin interaction constant J, proportional to the distance between the lines of the spin–spin splitting in the spectrum. Unlike the chemical shift, J does not depend on inductions of magnetic field B0 and so it is expressed right in Hz. For many atomic groups, J is well known and is reported in reference books. If, in the molecule in the main atomic group alongside the protons, there are other atoms with magnetic nuclei they can participate in spin–spin interaction not exhibiting their own magnetic moments. In this case, the splitting character is more complicated, though the amount of information on the substance increases as well. The above-considered example of the PMR spectrum of ethyl alcohol is equitable only for very pure material. If a small admixture of water is present in the sample, the OH group will not be exhibited in the spectrum. This is because the proton of the hydroxyl group
Hz
, m.u.
4.0
200
3.0
Figure 8.11 The scheme of a SF nuclear dipole–dipole splitting (scale not followed).
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sometimes participates in the rapid exchange between the alcohol and water molecules. The time of exchange influences the width of the PMR signal. Depending on the rate of this exchange, a signal can be significantly broadened or disappear completely. This also illustrates the possibilities for PMR in chemical applications. In the last decade, some methods have been offered that are more complex and significantly more informative. In particular, a number of methods have greatly increased the ability of NMR (R.R. Ernst, Nobel Prize, 1991). Ernst found that the ability of NMR, previously limited by the small number of nuclei, could be greatly extended by substituting a slow variation of the magnetic field by strong short pulses, decomposed by means of Fourier transformations in a certain number of harmonics. Thereby, for one pulse, a field runs the whole spectrum of frequencies. The inverse Fourier summations permit a normal NMR spectrum to be obtained. The method allows greatly enlarged sensitivity of the NMR spectroscopy and involves a greater number of nuclei. The method allows influence upon the processes of spin–spin interaction, enlarging or reducing their intensity. A method of two-dimensional scan has also been worked out with the participation of R. Ernst. This approach permits one to influence the spin–spin interactions, enlarging and/or suppressing them. This enables one to untangle complex spectra of polymeric and biological objects. A mechanism of spin–lattice and spin–spin relaxations was also worked out at the same time. The point is that the relative levels occupancy N/N, between which a transition occurs, is defined by the Boltzmann factor. The probability of transition between two levels is greater the larger the distance between them and, consequently, the more N/N differs from 1. In -gamma resonance, the energy difference is of the order of 104 eV, so the upper level is always nearly free and resonance absorption of quantum is nearly always possible. In NMR resonance, the difference E⫺E is extraordinarily small and factor N/N is close to 1. Under such conditions, the occupancy of both levels is the same and the probability of transition should be very small; and, basically, paramagnetic resonance must not be observed. However, it turns out that another mechanism of removing the excitation from the upper level exists. Two such processes are the excitation of either thermal crystal lattice oscillations (spin–lattice relaxation) and/or excitement of oscillation in the nuclear magnetic system (spin–spin relaxation). Both processes are characterized by constants T1 and T 2: T1 is the spin–lattice relaxation time and T 2 is the spin–spin relaxation time. Values of T1 and T2 render an essential influence upon the form of resonance curves. On the other hand, the relaxation processes depend on the mobility of one atomic group or another in the substance. So, experimental measurement of the relaxation processes serves as a method of measurement of the dynamics of chemical conversions depending on time, temperature, chemical conversion particularity and other factors. New, rather complex instruments—tomographs—have been constructed on the basis of NMR. They can be successfully applied in the medical diagnosis of normal and pathological human tissues. The general principle of tomography consists of the fact that the amplitude of the NMR response is proportional to the proton concentration in the sample investigated. In the human body, hydrogen is present in water and in adipose tissue; in the latter, its percentage being higher than in water. So, in picture imaging the distribution of hydrogen, one can see a distinct contrast between adipose and other tissue. To increase contrast, another characteristic of the NMR signal can be used: the strong dependence of the relaxation times on the chemical composition of the medium.
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Being practically harmless, NMR is used for the diagnosis of pathology in the human body, particularly for the hard-to-reach parts of the body such as the heart, liver and brain.
8.5
ABILITIES OF NUCLEAR QUADRUPOLE RESONANCE
NQR is a method based on the resonance absorption of electromagnetic energy in a substance owing to quantum transitions between nuclear energy levels, created by a nuclear quadrupole moment interaction with the gradient of the intracrystal electric field (Section 8.2.4, Figure 8.2c). NQR has been developed on account of the interaction of the electrical field gradient of the shell electrons with the gradient of electrical field appearing because of nonspherical electron density in atoms (refer to Appendix 3). Quadrupole interaction brings about a change in the energy states, corresponding to nuclear spin different spatial orientation regarding the crystallographic axes (formula (8.2.10)). An alternating electromagnetic field causes a magnetic dipole transition between sublevels of the main level, which is found as resonance absorption of electromagnetic energy. Since the energy of quadrupole interaction changes are in rather broad limits depending on the nuclear characteristics and on crystal structures, so the NQR frequencies lie in a range from hundreds kHz to thousands MHz. At present, NQR is not used as widely in science and technology as NMR. Nevertheless, many questions can be solved more quickly and easily with NQR than by NMR, not least because this method does not require an external magnetic field and availability of single crystals. NQR measurements are characterized by a high spectral resolution, selectivity and rate. In particular, of current importance is the remote control of air on the presence of different nitrogen compounds by means of NQR. Nuclear charge quadrupole moment eQ interacts with the gradient of electrical field, e2Qq, defined by an asymmetry parameter, ; it contains information on the molecular structure. These parameters can be determined from experiment. The NQR method can also be used in solids and, predominantly, crystals. There are four areas of successful investigation: electron density studies in molecules, in particular, changes in the orbital occupancy at complexation and substitutions, study of molecular dynamics, in particular, reorientation, rotation of atomic groups, hindered rotation, phase transformation study, revealing and studying defects and mixed crystals investigation. The NQR method is used in nuclear physics for the determination of nuclear quadrupole moments. In crystal chemistry, the method gives information on symmetry and structure, macromolecules ordering degree, their motion and nature of chemical bonding. The method permits the definition of a number of structure-nonequivalent atoms even with nonzero quadrupole moments. While with the NMR method, the structure of a molecule manifests itself only as broadening and splitting of lines; with NQR measurements, the crystal structure is defined by resonance frequencies themselves. Since in NQR, there is a distinctively strong dependence of the spectral line width from defect concentration in crystals; the line-width measurement allows the study of internal stresses in the crystal, the presence of admixtures and ordering processes. The parameter of asymmetry strongly depends on structures; so, in particular, in an ion NC2H2⫹, when substituting two hydrogen atoms on carbon, it is changed from 0 to 1, which is easy to observe in an experiment.
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EXAMPLE E8.1 Calculate the relative emitter/absorber velocity V requires to compensate recoil energy and observe resonance on a free 197Au isotope (wavelength is ⫽0.016 nm). Solution. In order to destroy the resonance, one can move the resonance line on its width (eq. (8.3.10)) taking into account the fact that the recoil process relates both to emitter and to absorber. Therefore, to compensate for this effect, it is necessary to move one of them with velocity V, which can be determined according the equation:
V ⫽ 2 R. c
Therefore, V⫽
( )2 c 2 Rc 2 h . ⫽2 ⫽2 ⫽ 2 2 mN mN 2 mN c
Substituting all known values into this equation, we arrive at V ⫽ 126 m/sec. This value is not small, however, in comparison with quantum speed (V/ c)⬃(0.1/108) ⫽ 10⫺9, which seems negligible.
EXAMPLE E8.2 The spin quantum number of a nucleus of 14N is I⫽1. Therefore, the nucleus has three energy levels with frequencies 0, ⫹ and ⫹ (for mI⫽⫹1, 0, ⫺1, correspondingly). Two frequencies of 14N in sodium nitrate (NaNO2) are ⫹⫽4.640 MHz and –⫽3.600 MHz. Calculate (1) NQR frequency 0, (2) asymmetry parameter and (3) quadrupole coupling constant e2qzzQ/h. Solution. (1) A simple average v0 can be easily calculated v0⫽v⫹⫺v⫺⫽1.040 MHz. (2) The quadrupole moment is v⫹⫽3/4(e2qzzQ/h)(1⫹/3), v⫺⫽3/4 (e2qzzQ/h) (1⫺/3): the solution of these two equations give: ⫽3(v⫹⫺v⫺)(v⫹⫹v⫺)⫽0.38. (3) The quadrupole moment of 14N is: {(e2qzzQ/h)⫽2(v⫹⫺v⫺)/]⫽5.474 MHz
8.6
ELECTRON PARAMAGNETIC RESONANCE (EPR)
The effect of resonant absorption of electromagnetic waves by electron paramagnetic centers in a substance in a permanent magnetic field is referred to as electronic paramagnetic resonance (EPR). In the EPR method, the resonance condition for electrons looks like that for nuclei (refer to eq. (8.3.1): (8.6.1) E1 ⫺ E0 ⫽ res ⫽ gE B B,
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though instead of nuclear, electron characteristics are included. A comparison of resonance in nuclear and electronic subsystems (see Section 8.3.1) shows that the frequency of the EPR-resonance used is higher than that in NMR, i.e., it is approximately 109 Hz, which corresponds to the wavelength 10⫺1 m (several centimeters) lying in the microwave range (see Table 5.3). In EPR, an external permanent magnetic field influences electron spin energy level split; however, a high-frequency field throws the spins from one state to another. In other words, the external permanent magnetic field removes degeneration from electron energy states on quantum number mS. The basic EPR circuit is similar to that for NMR though the technique of EPR measurements differs significantly from that of NMR. Intraatomic superfine interaction is described differently depending on the aggregative state of the substance investigated. In particular, it has the same nature in liquids as that described in Section 8.3 and is referred to as contact interaction; it is described by the term:
a⬵
8 M I 冷 (0, 0, 0) 冨2 3 I
(8.6.2)
It can be seen that contact interaction is caused mostly by s-electrons, because only for s-electrons is the wave function on the nucleus distinct from zero. However, contact interaction of the same character can be observed in some cases in the absence of a noncoupled s-electron; it can arise to the account of -electron of aromatic hydrocarbons belonging, for example, to anion-radicals: contact interaction of a nucleus with s-electrons is rather sensitive even to small electron excitation at hybridization. The interaction of noncoupled electron spins with the nucleus can cause SF EPR line splitting EPR signal. The character of the splitting, or SF splitting of the EPR signal, is defined by a nucleus spin interaction with the number of specific atomic configuration in the nearest neighborhood. For a proton, the rule n⫹1 (where n is the number of nonequivalent atoms) is valid in this case as well, since I⫽(1/2), like electron spin. The simplest case is atomic hydrogen where SF splitting is defined by electron interaction with a nucleus (proton) belonging to the same atom. Notice that the reorientation time of a nucleus in an alternating magnetic field is much longer than the corresponding electron time. This means that the nuclear spin assigns the quantization axis and electron spin reorients alone (transition from mS⫽⫹1/2 to mS⫽–1/2). Quantum numbers for an electron spin and proton are identical and equal to 1/2. This means that the number of possible states is equal to two (spin of an electron and nucleus are parallel, the coupled spin is 1; or antiparallel, the coupled spin is zero). Therefore, the EPR spectrum of an atomic hydrogen consists of two SF signals (Figure 8.12a) split on 502 Oe (remember that the size of splitting does not depend either on the external field or on frequency). A deuteron’s spin quantum number is equal to 1. Therefore, there are three possible mutual orientations of the deuteron and electron spins resulting in 1, 0 and –1. Hence, the EPR signal of atomic deuterium consists of a three-line SF structure divided by a field 78 Oe (Figure 8.12b). The splitting is proportional to a; this allows one to determine the magnitude 兩 (0,0,0)兩2 for hydrogen (deuterium) atoms. Notice that the rule n⫹1 for a deuteron is no longer valid, since I⬎1/2.
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(a) 502 oe
(b)
78 oe
Figure 8.12 EPR spectra of atomic H (a) and D (b).
The EPR spectrum of each atom, ion or radical, has a specific shape (due to SF electron–nuclear interaction) well known to experts and given in the specialist reference literature. The most successful fields of application of the EPR method are: atoms and radicals (stable and/or unstable) with odd number of electrons, short-lived particles as elements of intermediate stages of the chemical reactions, ions of transition elements with partly builtup electron shells, crystal defects, etc. Integral signal intensity is proportional to paramagnetic center number; hence, the EPR spectrum contains important characteristics of a substance, such as the concentration of free radicals, number of defects of a crystal lattice, etc. The position of a signal allows one to determine the g-factor of a paramagnetic atom. Therefore, one can find the state of the paramagnetic center (refer to Section 7.6.2). Notice that, in view of relativistic effects, the spin g-factor is not precisely to 2, but is 2.0023. PROBLEMS/TASKS 8.1 A free 40K nucleus emits a -photon with the energy ⫽30 keV. Find the relative displacement / of the spectral line due to nucleus recoil. 8.2 A 67Zn nucleus with the excitement energy E⫽93 keV emits a -photon transferring from the excited to the ground state. Determine the relative change of (/) of the -quantum energy because of the recoil of the initial nucleus.
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8.3 Calculate an atom’s recoil energy R after emission photon in (1) visible range energies (⫽500 nm), (2) in X-ray range (⫽0.5 nm) and (3) in range (⫽5⫻10⫺3 nm). Consider one and the same nucleus of mass m⫽100 a.u.m. 8.4. The atomic bonding energy Eb in a crystal is Eb ⫽ 20 eV; atom mass is 20 a.m.u. Find the minimum quantum’s energy, which kicked the atom out of the crystal. 8.5. Simulate the R spectrum of ilmenite (FeTiO3) if iron’s chemical shift EI relative to stainless steel is 1.3 mm/sec and the quadrupole splitting EQ⫽0.5 mm/sec. The Fe excitation energy is 14.4 keV. Express EI and EQ in eV units. ANSWERS 8.1. ⫽1.7⫻10⫺17; (/)⫽(/2mNc2)⫽4⫻10⫺7 m. 8.2. /⫽(/2mNc2)⫽7.45⫻10⫺7. 2 ⎛ ⎞ ; (1) 33 peV; (2) 33 meV; and (3) 0.33 eV. m A ⎜⎝ ⎟⎠
8.3.
R⫽
8.4.
⫽ c 2 mN Eb ⫽ 8.6 ⫻10 −5 eV .
8.5. EI ⫽ 6.24 ⫻ 10⫺8 eV, EQ ⫽ 2.4 ⫻ 10⫺8 eV.
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–9– Solid State Physics
Solids are mostly subdivided into crystal and amorphous substances. In this chapter the crystalline state is predominantly considered although some features of the amorphous and liquid states will also be briefly touched on.
9.1
CRYSTAL STRUCTURE, CRYSTAL LATTICE
A crystal is characterized by a three-dimensional, regular, periodic array of particles— atoms and/or molecules. Modern experimental techniques allow us to see the structure of large molecules in crystals using an electron microscope. Figure 9.1 is an electron photograph of a crystal of tobacco mosaic virus, showing the regular packing of the molecules in a crystal. Remember that by “molecule” we usually mean the smallest part of a substance that can exist alone and retain the characteristics of that substance. Imagine that we can divide a crystal until it becomes the “brick,” which retains the main (but, certainly, not all) characteristics of the whole crystal. The contents and form of this “brick” defines the crystal structure: the relative amount and mutual disposition of atoms (molecules), the chemical composition of crystalline material, the interatomic distances and valence angles (i.e., chemical bonding), etc. The smallest part of the crystal that retains the specified characteristics (the “brick”) is referred to as a unit cell. Using the property of periodicity one can build the whole crystal by regular repetition of unit cells along coordinate axes, as shown in Figure 9.2. A crystal can be characterized both by the unit cell (the carrier of the chemical composition and atomic structure), and by three translation vectors a, b and c; the latter can be used to build the whole crystal from original cells. Each of these three vectors corresponds to a symmetry operation, since these operations superpose a crystal with itself: each atom (supposed to be a point) moves over to a similar atom in the nearby unit cell. Any point (atom, including) can be transferred to another crystal point, identical to the first, by operation of a translation t, which is t manb pc,
531
(9.1.1)
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Figure 9.1 Image of the tobacco mosaic crystal as seen in the electron microscope (size of molecules is approximately 25 nm).
z
y
>
c
>
>
(a,b) = (a,c)= (b,c)=
b
x 0
a
Figure 9.2 Crystal lattice: a unit cell based on a, b, c vectors and the crystal build-up.
where m, n and p are integers which number the unit cell to which the translation is taking place. Unlike the symmetry elements considered in Section 1.3.7, translation is an attribute of the crystalline state, because it translates a given atom to a similar one in the neighboring unit cell, but does not combine an atom with itself. It is important to note that the crystal
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is here considered to be of infinite size; this is a good approximation, since crystal size is usually many orders of value larger than the unit cell. The system of translations forms a so-called crystal lattice; this presents a mathematical abstraction: describing the translational characteristics of a given crystal. In Figure 9.2 this position is illustrated. Vectors a, b and c represent a set of three translations in three dimensions, which together with angles , and , define the form and size of the unit cell. The crystallographic axes x, y and z are usually directed along the main translations. Note that the choice of three main translations is ambiguous, there exist certain rules to make this choice but this is not important here. The cross points of the axes are called lattice nodes. It should once again be emphasized that while the crystal structure is the arrangement of atoms in the crystal, the crystal lattice is a system of translations, describing the translational properties of the crystal. The form of cells, the location of atoms in them and, accordingly, the crystal’s physical properties are defined by the symmetry laws. All these comprise the subject of crystal physics and crystal chemistry. Correlations between the translation vector length and the angles between them define possible crystal classes or syngony, resulting from the unit cell form and the crystal lattice. There are seven crystal classes (or syngony), plotted in Table 9.1. The position of origin is chosen from considerations of rationality. So, in Figure 9.3 the atomic structure and crystal lattice of a Cl2 crystal is depicted: the origin is chosen not in the position of an atom, but in the CM of the Cl2 molecule. (In this instance a crystal node is in “emptiness.”) Planes drawn through the nodes (but not in general through atoms) are called crystallographic planes (see Section 6.3.5). A lattice node is defined by its coordinates (in the units of the vector length), which are placed in double square parentheses (for instance, [[001]]). All parallel directions in a crystal are equivalent. So a straight crystal line (or simply line) is conducted through the origin and its indexes are defined by the coordinate of the first node, lying on this line; a line’s indexes are enclosed in brackets (so, direction [001] complies with the direction to axis z, since the first node on it has the coordinates [[001]]).
Table 9.1 The crystal classes (syngony) Syngony
Relations between lattice periods
Relations between angles
Triclinic Monoclinic Orthorhombic Tetragonal Trigonal (rhombohedral) Hexagonal Cubic
abc ab c abc abc abc abc abc
90° 90° 90° 90° 90°, 120° 90°
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Figure 9.3 Crystal unit cell of Cl2: there is an arbitrary rule in the choice on the unit cell.
z 2
1 0
1
2 y
1 2 3 x
Figure 9.4 An order of the Miller indexes determination.
The family of parallel crystallographic planes is defined by three indexes h, k, l called the Miller indexes. The indexes are inversely proportional to the segments cut by the given plane on the axes, provided it is the nearest plane to the origin; they are bracketed in parentheses. Let a plane cut definite segments on the axes. In Figure 9.4 such lengths (in units a, b and c) are 3,2,2. (These lengths in unit cell periods can be whole or fractional numbers.) They must be “turned over” and multiplied by a single whole number (here the number is 6) to make them whole numbers
冤冢
冣
冥
1 1 1 → (2,3,3) . The three whole numbers obtained (2,3,3) are 3,2,2
the Miller indexes of this plane. If a plane cuts an axis on the negative side of the origin, the corresponding index is negative indicated by placing a minus sign above the corresponding index (e.g., h ,k, l). In Figure 9.5 an example of some crystallographic planes in a cubic lattice is presented. Index zero amongst the Miller indexes corresponds to the fact that a plane cuts on the corresponding axis a segment equal to infinity, i.e., this plane is parallel to a given axis. If there are two zeroes amongst the indexes it means that a given plane is parallel to the corresponding unit cell edge.
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z
535
z
(110)
z
(100)
y x
y
x z
y x
z
z
y x
y
x (111)
(110)
y
x (100)
(200)
Figure 9.5 Selected crystallographic planes in a cubic crystal.
The correlation presented describes the so-called primitive cells in which there is only one atom (each atom in the nodes belongs simultaneously to eight cells, which gives for one unit cell 8 (1/8) 1 atom). The unit cell of each crystallographic class is characterized by a definite symmetry corresponding to the crystal space symmetry. It is possible to place an additional atom in a primitive cell in such a manner that the cell symmetry does not change. For instance, in the cubic primitive cell one can introduce an additional atom into the cell’s center
冤冤 12 , 12 , 21 冥冥 without destroying its symmetry. A body-centered
lattice (BCC) with two atoms to the cell is obtained. Iron crystals, for instance, possess such a structure. It is possible to place atoms at the centers of the edges of a primitive cell
冤冤 2 2 0冥冥, 冤冤 2 0 2 冥冥 and 冤冤0 2 2 冥冥 preserving the cubic symmetry. In a similar manner, 1 1
1
1
1 1
the so-called face-centered cubic (FCC) with four atoms in the unit cell is obtained. Copper, for instance, has a FCC structure. Such cells are called cells with basis (or Bravais lattices). Crystallographic directions are characterized by identity distances (periods), i.e., the shortest distance between identical atoms in a given direction. Parallel crystallographic planes are also identical to each other. The distance between two adjacent planes with indexes h, k, l, or what amounts to the same things, the distance from the origin to the nearest crystallographic plane with the same indexes are called interplanar spacing dh,k,l. Exactly this value falls into the Bragg equation (refer to Sections 6.3.5). Periods of lattices a, b, c together with angles , , , plane indexes (h, k, l) and interplanar spacing d are bound by a so-called quadratic form. For the simplest case of a cubic crystal, the quadratic form is presented by the equation 1 h2 k 2 l 2 d2 a2
.
(9.1.2)
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Structure and crystal lattice define the positions of atomic CM in the unit cell. In addition, atoms in crystals perform thermal oscillations, significantly influencing the crystal’s physical properties. These oscillations are characterized by root mean square displacements. The main method of crystal structure determination is X-ray diffraction analysis (refer to Section 6.3.5). Experiments that use small single crystals and/or polycrystalline samples allow one to determine the mutual location of atoms in crystals of rather complex chemical compounds (for instance, a thousand or more atoms in the unit cell) together with the nature and parameters of atomic thermal vibrations (their root mean square displacement values). For investigation of special questions, methods of neutron and electron diffraction are also used (refer to Chapter 7.2). EXAMPLE E9.1 Knowing the density of a Ca single crystal 1.55 103 kg/m3 determine: (1) the lattice period a; and (2) the closest interatomic distance d. The Ca crystal is of the FCC structure type. Solution: The unit cell volume V of a cubic crystal can be bounded with the lattice period a by a simple expression V a3. On the other side, it can be expressed as a ratio of the molar volume to the number of unit cells in one mole of the crystal a3 Vm/Zm*. The molar volume Vm is Vm M/. The number of the unit cells in one mole is Zm NA/n where n is the number of atoms in the unit cell. Substituting these 3 values into * we obtain a3 nM / NA wherefrom a 兹苶 (n苶 M苶/苶 苶 N苶 A). Taking into account the number of atoms in the FCC unit cell (Chapter 9.1) we arrive at a 556 pm. (2). The closest interatomic distance in the FCC lattice is the face diagonal d a兹2苶 / 2. Therefore, d can be found d 393 pm, where the atomic radius can be calculated from r (d/2) 1.81 Å. EXAMPLE E9.2 Write down the indexes of crystallographic directions presented in Figure E9.2 by a bold line (refer to Chapter 9.1).
z
y x
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Solution: The crystallographic direction does not cross the origin (axes symbols without asterisks). However, we know that all the nodes of a crystal lattice are equivalent. Therefore, we can shift the origin in the point [[100]] in the figure. This point is the new origin. Then the line will cross the node [[ 101]]. These are the indexes of the direction [ 101].
EXAMPLE E9.3 In Figure E9.3 a crystal lattice is presented. Write down indexes of the crystallographic plane crossing three nodes indicated in the figure.
x
y
z
Solution: It can be seen from the Figure E9.3 that points cut on coordinate axes (in periods units) are x 3, y 1 and z 2. According the rule we should obtain 1 3
冢
1 2
冣
the reciprocal values; they are ,1, . Multiply the three value on 6 we arrive at
( 2, 6, 3 ). Note that indexes can be multiplied by a constant value, including negative. All planes obtained will be identical.
9.2 9.2.1
ELECTRONS IN CRYSTALS
Energy band formation
In Chapter 7 the electron structure of free atoms, i.e., not subjected to any external influences, was considered. In a crystalline state the distance between the atoms is comparable with their size; so each of them appears strongly influenced by its neighbors. The interatomic interaction and periodic character of a crystal field render an extremely strong
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influence on the atomic ensemble. As a result, crystals possess a complex of the properties strongly distinguishing them from ensembles of free atoms. Let us consider the essence of the process using the example of the hydrogen-like atom of sodium with 11 electrons in its shell in a free state. In Figure 9.6 (above) a potential electron energy curve of two neighboring atoms in a field of their nuclei is presented. All levels of energy up to 2p are occupied completely though there is one noncoupled electron in the outer 3s level. The potential curves go up to U 0 but are not overlapping. Each electron is fixed near its nucleus. The same picture is presented in Figure 9.6 (below); however, here the distance between atoms corresponds to the shortest interatomic distance in sodium metal (a 4.3 Å). Firstly, the potential curves in a crystal form a unique periodic potential with the maxima located appreciably below the zero level of energy. Secondly, an upper occupied 3s-electron level has risen above the potential barriers and a corresponding electron appears capable of moving along the whole crystal. It is usually said that collectivization of valence electrons has occurred. Such collectivized electrons form an ensemble of quasi-particles that has lost part of their initial properties (see Section 9.2.2). The overlapping of the outer 3s orbits has occurred. On approach, changes also touch the inner electrons. The mutual influence of atoms transforms the very “thin” electron’s levels into a band of final width. In Figure 9.7a scheme of such level expansion is given for several crystals. In Figure 9.7a the 1s level of a lithium atom is expanded insignificantly whereas the 2s level has formed a rather wide
E
E
+ r
r
E
E E=0
Figure 9.6 Interaction of two sodium atoms. above – the initial state, the potential curves of the two free atoms not interacting to each other, below – overlapping of the upper electron shells at the interatomic distance became closer reduced to that in metallic sodium: this give rise to Bloch’s free electron formation.
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E
539
E
E
E
E
2p 2s
2s
1s
0
(a)
Ec
0
P S
Ev
1s
r
a
E
r
r
a
0 (b)
a (c)
Figure 9.7 Broadening of the electron’s energy levels and formation of energy bands: (a) lithium crystal (in the 2s band part of the level is occupied, this is shaded in a section, whereas other part remains free), (b) bands in beryllium, (c) bands in diamond-like crystals.
2s band. In a beryllium crystal (Figure 9.7b), 2s and 2p levels have formed an overlapping band. This mixed band is called a hybrid band. In diamond-like crystals with strongly pronounced covalency the bands were split in two (Figure 9.7c); each of them contains four states for one atom: one s-state and three p-states. These bands are subdivided by the forbidden energy band. The lower occupied band is referred to as a valence band, and the upper band as a conductivity band. The picture described corresponds to the so-called strong-binding approximation. In a weak-binding approximation the electron behavior in a periodic crystal field is considered. For the analysis of the latter case it is necessary to solve the Schrödinger equation for a particle moving in a periodic field F. Bloch has offered a function consisting of the product of two functions: U ( x ) u( x ) e ikx ,
(9.2.1)
The function u(x) describes the electronic potential of a single atom and the other (exponential) ensures the periodicity u(x a) u(x). This function is called the Bloch function. Such potential should be substituted into the Schrödinger equation and the allowed values of energy can be derived. Instead of a smooth parabolic function E (h ¯ 2 / 2m)k 2 a curve with breaks is derived. Some representation of the physical origin of such breaks can be given using diffraction electron properties (see Section 7.1): breaks occur at values k, corresponding to back reflection according to Bragg’s equation (6.3.11). Back reflection in this case can take place when an angle is 90° and, accordingly, n 2d (n is an integer designating the order of reflection). Remembering that 2/k the equation can be presented as
k
n , d
(9.2.2)
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E
2 − a
− a
a
0
2 a
k
Figure 9.8 Electron energy E as a function of the wavenumber k. The same curve for a free atom is presented by a broken curve.
where d is the interplanar distance. If the planes (h00) are considered, the interplanar distance is a and eq. (9.2.2) appears in the form
k
n . a
(9.2.3)
As a result the curve E(k) looks like that given in Figure 9.8: on a background of classical parabolic dependence in the places determined by expression (9.2.3) the curve has breaks. The forbidden energy gap, which we met above, is formed. We are unable now to call the particles as electrons; we must call them quasi-particles (quasi-electrons). The properties of a quasi-particle with a wavenumber k close to n/a (at the bottom or near the top of a band), differ appreciably from the properties of really free electrons. In particular, their effective mass m* differs at the “top” and “bottom” of the energy band. It follows from this fact that m* is defined by the second derivative of an energy E on wave vector k, i.e., m* (d2E/dk2); from the curve it can be seen that m* depends on k and can even change sign. 9.2.2
Elements of quantum statistics
Because of mutual influence, some atomic electrons are generalized forming a “gas” of quasi-electrons in crystals. These electrons preserve some properties of free electrons (e.g., each of them possesses a classical momentum) but, at the same time, also possess properties that distinguish them from really free particles (e.g., they have a mass different from the classical electron mass). Some well-known electric properties of metals are caused by this “gas.” However, it appears that in a model of free “electron gas” theoretical calculations strongly overestimate experimentally known characteristics; only a small part of the generalized electrons can take part in the formation of these properties.
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Are these electrons indeed the gas of free electrons? The answer to this question has already been obtained: no, there are no free electrons in metals; electrons form energy bands that can either be overlapping, or be separated by the forbidden energy gap. Besides, moving in a periodic crystal field, electrons are symmetry-dependent. It is necessary to answer one more question: how are electrons distributed among these bands and how are they distributed inside the band? In classical Newtonian physics the elementar volume of a configurational space cell is infinitesimal (it looks like Planck’s constant ¯h is accepted to be zero); the electron distribution upon their energy is given by Maxwell–Boltzmann statistics: there are large amounts of particles, all of which tend to occupy the state with the lowest energy, though chaotic temperature motion, on the other hand, scatters them on different energies. This process is described by the Boltzmann factor. In quantum physics the volume of a phase space cell is no smaller than h3 because of the uncertainty principle; the number of cells is limited accordingly and this leads to the limited amount of energy levels in the bands. The problem of particle distribution among cells (and energy levels) is not apparent. Regulation is given by quantum statistics. The energy spectrum is presented by energy bands with limited “capacities.” Consider the energy structure of a valence band. The example of sodium atom was given above; every atom from its 11 electrons delivers only 1 to the valence band; besides, every atom “brings” one level to each energy band; levels regularly fill the band, they cannot overlap since the uncertainty principle and exclusion principle do not permit overlapping. If we assume that sodium crystal comprises N atoms, the conduction band is half-filled because of the spin degeneration. How are these electrons distributed on the levels? Figure 9.9 presents such a half-filled band. The upper occupied level at 0 K is referred to as the Fermi-level (EF); energy counts up from the bottom of the band. The same is presented in Figure 10a; the graph f(E) forms a step, all levels under EF are occupied whereas all levels over EF are empty. It is suggested in quantum statistics that particles with half-integer spin obey Fermi–Dirac statistics; such particles are called fermions. Electrons have a spin quantum number equal to 1/2 and therefore are fermions and must obey Fermi–Dirac statistics. The mathematical description of Fermi–Dirac statistics is given by the Fermi distribution function f ( E ) 冤e( EEF ) (T ) 1冥1 .
(9.2.4)
The magnitude of this function depends on the sign of difference E EF. At 0 K and E EF the difference is negative and f(E) 1. At E EF the difference is positive and f(E) 0. This corresponds to the fact that up to Fermi energy, all levels are occupied by electrons according to Pauli’s exclusion principle. When the temperature increases, only those electrons close to the EF level transmit to higher levels (Figures 9.9 and 9.10b) and take part in excitation. The number of such electrons is proportional to T, as is shown in the figure. The point is that only such electrons can participate in electric properties. At T EF function f(E) corresponds to nondegenerated gas and Fermi–Dirac distribution transforms to Boltzmann distribution. However, this can happen only at very high temperatures when the crystal hardly exists.
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At absolute zero temperature all levels are occupied by electrons up to EF. When temperature increases some of the electrons are excited and occupy free levels of energy. These electrons define the electron conductivity of metals. The average energy width of this “excitation band” is equal to T, as is marked in Figures 9.9 and 9.10.The opposite case is presented by particles with integer spin, in particular photons with a spin quantum number 1. Such particles are referred to as bozons (Bose particle). They obey Bose–Einstein statistics. A distribution function f(E) for bozons is expressed as f ( E ) 冤 e ( E ) (T ) 1冥1 .
(9.2.5)
The value in this equation is the so-called chemical potential, i.e., the thermodynamic function of a system, which is determined by the system energy change at the change of the particle number by 1. A feature of photon gas is the fact that, at first, photon gas cannot exist by itself: a source of photons as heated body surface is necessary for its maintenance. Further, all photons in vacuum move at an identical speed equal to the speed of light c and possess energy ¯h , momentum ¯h /c and mass E/c2 (refer to Section 1.6). They cannot stop; their resting mass is zero. Moreover, photons do not collide with each other and their equilibrium is achieved only on interaction with heated surfaces; therefore their quantity is not fixed, it is established by the equilibrium with the heated body.
E
0
{
T
E EF
zone bottom
Figure 9.9 An energy band structure. E is the total depth of the band, EF is the Fermi energy count from the band bottom. Schematically the dark field presents the fully occupied by electrons band at zero temperature, non-occupied levels space is white, a variable part of a band of T in width is active, i.e., can be exited and participate in conductivity. The energy levels are very close to each other and therefore are not shown in the diagram.
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f 1 T=0
(a)
EF
E
f 1 T>0
(b)
E
T f 1 T
(c)
∞
E
Figure 9.10 The graph of Fermi-Dirac distribution: (a) at 0 K it looks like a step, (b) at intermediate temperature, the border near Fermi energy is fuzzy. (c) at T → ⴥ the distribution transforms to Boltzmann law.
An electron with half-integer spin is a fermion, i.e., it has to obey Fermi–Dirac statistics. All the usual metal electroconductivity properties are in agreement with these statistics. However, in some cases two electrons can produce the so-called Cooper’s pairs with compensated spin, the spin of the Cooper pairs is zero. Therefore, such pairs transform electrons from fermions to bozons. There are no Pauli’s exclusion principles for bozons; they can be “condensed,” i.e., they occupy all the levels, and nearly all participate in electroconductivity. Superconductivity can then take place. This suggestion theoretically explains the metal and intermetallic compound superconductivity phenomenon (J. Bardin, L.N. Cooper and R. Schrieffer, Nobel Prize 1972), discovered earlier by H. Kamerling-Onnes (Nobel Prize, 1913). This phenomenon occurs only at very low temperatures (⬇20 K). However, superconductivity has been discovered in nonmetallic, oxide-type chemical compounds with critical points of superconductivity up to ⬇140 K (in liquid nitrogen region) (so-called high-temperature super conductors, HTSC) (J.G. Bednorz and K.A. Muller, Nobel Prize, 1987). Intensive attempts to synthesize new materials of this kind are in progress.
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Band theory of solids
Band theory relates to the electrical properties of solids and the character of their energy bands. The corresponding relation is given in Figure 9.11. The case of metallic sodium already examined above is presented in Figure 9.11a. Electrons fully occupy the 2p band, but the conducting 3s band is half-filled. Therefore, 2p electrons do not participate in conductivity but the electrons lying near EF in the 3s band can easily be excited by an external electrical field and take part in an electrical current. Consequently, sodium, like all other alkali metals, is a conductor. This situation is shown in Figure 9.11b, where a scheme of another case of a conductivity band created both by 3p- and 3s-electrons is depicted. Another picture of energy band filling is shown in Figure 9.11c. It corresponds to full occupation of a valence band (EF equals the band top energy) with the presence of the forbidden energy gap Eg between the valence band and the band of conductivity. Electrons of the valence band cannot be so easily excited now by the action of an inner field to take part in conductivity (for this, electrons have to be transmitted to the conduction band, overcoming a broad energy gap); the corresponding materials possess a dielectric property. If the gap is narrow in comparison to T, the material is an intrinsic semiconductor (Figure 9.11d). In addition, there exist two great classes of semiconductors characterized by the presence of local energy levels (to the account of different admixtures) in the forbidden energy gap. If these levels lie close to the top of the valence band (Figure 9.12a) (this is called the acceptor level), electrons of the valence band occupy them and release some levels in this band. The so-called hole conduction appears (on account of vacancies near the top of the valence band). Such materials are semiconductors of p-type. Germanium crystals with indium admixture can be numbered amongst them. If occupied local electron energy levels are near the bottom of an empty conduction band they can be excited directly to this band (Figure 9.12b). Such levels are called donor levels and the material is a semiconductor of n-type (germanium with the admixture of arsenic). In intrinsic semiconductors (for instance, pure germanium and silicon) the local levels are absent; however electrical conductivity appears because of the narrowness of the forbidden gap and at moderately high temperature an excitation of electrons from the valence band directly to the conduction band is possible (Figure 9.11d).
3s
3p
2p
3p
Eg
3s
Eg
Eg
Eg
2p
2p
(a)
(b)
2s (c)
3s (d)
Figure 9.11 Band structure of different crystal types: (a) and (b) conductors, (c) dielectric, (d) semiconductors.
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545
Conduction zone EA
Ec
Ec
Eg
Eg
Donor levels
Ea
Accepter levels
(a) Valence zone
(b) Valence zone
Ev
Figure 9.12 A band structure of doped semiconductors: (a) n-type, (b) p-type.
9.3 9.3.1
LATTICE DYNAMICS AND HEAT CAPACITY OF CRYSTALS
The Born–Karman model and dispersion curves
By crystal dynamics one usually understands the theory describing atomic oscillations around their equilibrium positions and those features of the properties that depend on these oscillations. Only harmonious oscillations occurring because of the action of quasi-elastic forces will be considered here (refer to Chapters 2.2 and 2.4). Being independent, oscillations of all atoms can be described by a system of running waves with certain frequencies, polarization and wave vectors (see Section 2.9.5). According to the general principles of particle-wave duality (refer to Chapter 7.1), each wave associates with a particle called a phonon with energy ¯h S, polarization s (s 1, 2, 3) and momentum p ¯h k. Since phonons exist only within a crystal they are related to the category of quasi-particles. Nevertheless, interaction of phonons with other particles (electrons, neutrons, etc.) occurs according to classical laws of collision (refer to Section 1.4.5). Consider the simplest model of a crystal as a one-dimensional chain of identical atoms. Direct the chain along an axis x. Denote a period along the chain by the letter a, and let l be the number of the atom counted from an arbitrary chosen atom (Figure 9.13). Value xl represents instant coordinate of atom l. In this model one more simplification is entered, namely, the nearest neighboring atoms are considered to be interacting among themselves only. Such a model is called the Born–Karman model. The equation of movement (Newton’s second law) can be written for the atom l as ⎤ ⎡ m l [( l l1 ) ( l1 l )] 2 ⎢ l l1 l1 ⎥ , 2 ⎣ ⎦
(9.3.1)
where is a force constant and the expression in square brackets is the shift of the atom l from its equilibrium position. This system of equations can be solved, the solution being sought in the form of a running wave l 0 exp[i( t kla)].
(9.3.2)
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a
a
1
2
3
Figure 9.13 A one-dimensional chain of similar atoms with inter-atomic distance a ( is atom’s deviation from the equilibrium position), is a force constant of the inter-atomic bonds.
The connection between and k is 2 m (e ika eika 2) (e( ika 2 ) e( ika 2 ) )2 4 sin 2
ka , 2
(9.3.3)
wherefrom
4 ka sin . 2 m
(9.3.4)
A dispersion curve is a curve of allowed -values as a function of k. Within the framework of the Born–Karman model this dependence is represented by eq. (9.3.4) (Figure 9.14). Let us look at its main points. Firstly, this equation is nonlinear; the dispersion of waves (refer to Section 2.9.4) is taking place. Secondly, this dependence is periodic; in space of wave vectors an independent (repeating) part is lasting from zero to /a. This area is referred to as the Brillouin zone and a value /a is the border (in this case –one-dimensional border) of this zone: outside this zone the function (k) repeats periodically. Besides, the number of waves with various wavelengths is limited: from below by the period of a chain a, from above by the length of all chain L. At small k((ka / 2) ^ 1) sin(ka/2) in eq. (9.3.4) can be substituted by its argument. Then
m
ka
(9.3.5)
and the dependence (k) is getting linear. The wave phase speed can be derived from eq. (9.3.5) (refer to eq. (2.8.12)); this is an acoustic dispersion wave curve. The derivative (d / dk) is zero at the zone boundary. The dispersion curve is referred to as an acoustic branch. A more complex model is a diatomic chain consisting of alternating atoms of different masses m1 and m2 which are located at equal distances from each other (which still equal a) and between which the same elastic forces operate (Figure 9.15). “The unit cell” of such a “crystal” is twice as large as the previous one. Two equations for movement should be written and solved together. The solution in the form of running
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ka
( / m)
547
–π/a
π/a
0
k
Figure 9.14 Dispersion curve (k) for Borman–Karman model. a
1
a
m1
m2
2
Figure 9.15 A one-dimensional model consisted of two different atoms with the same inter-atomic distance. A period is equal to 2a.
waves has the same appearance, but frequency depends on two masses and has two solutions: 2
2
2 ⎛ 1 ⎛ m m2 ⎞ 1 ⎞ sin ka ⎜ ⎟ ⎜ 1 . 4 ⎟ m1 m2 ⎝ m1 m2 ⎠ ⎝ m1m2 ⎠
(9.3.6)
Accordingly, a graphic representation of the (k) function has two branches (Figure 9.16). It is typical that one solution ( ) behaves in the same way as in the previous case. However another solution ( ) essentially differs: firstly, at k 0 frequency aspires to a nonzero value, secondly, at k ( / 2a) the curve comes to different point on the Brillouin zone border. (The factor 2 in the denominator appeared because the identity period in the model presented in Figure 9.15 has changed.) The difference between frequencies on the Brillouin zone border is proportional to the distinction of masses of the two kinds of atoms. This second branch is referred to as an optical branch. In a real three-dimensional crystal there are similar waves extending in all directions. Taking into account that in the crystal the existence of waves of three polarizations is possible (one being longitudinal L and two transverse wave T1 and T2, refer to Section 2.8.2),
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A
–/2a
/2a k
0
Figure 9.16 Graphical representation of dispersion curves (k) for the two-atomic chains. Acoustic (A) and optic (O) branches are presented.
L
T1
T2
0
k
/2
Figure 9.17 A general case of dispersion curves of three different kinds: L—longitudinal wave, T—two transverse waves.
three branches (k) can occur (Figure 9.17). In some crystals their degeneration is probable: two or all three branches can merge into one. Similarly to the one-dimensional chain, the borders within the framework of which function (k) is independent are outlined. Such borders form the three-dimensional Brillouin zone. The picture of the dispersion curves dependent on direction and from wave polarization, becomes complex and, frequently, confusing. Figure 9.18 shows an example of experimentally measured dispersion curves in an aluminum crystal for the different directions specified in the figure; Figure 9.19 presents an example of the oscillation frequency distribution function g( ). Note that the acoustic branch corresponds to the oscillation of the crystal unit cells relative to each other, whereas the optical branches describe the oscillation of atoms relative to each other within the volume of a single unit cell. Since the elastic waves in crystals are caused by atomic oscillations and are interconnected, waves are developed as collective excitations. Hence, in modern terminology,
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[111]
arb. units
[100]
549
0
1.0
0
0.5
k/k max
Figure 9.18 Dispersion curves for different crystallographic directions (shown in the scheme) in aluminum obtained by the neutron inelastic scattering method.
g
2
1
0
3
6 /2, TΓµ
D
Figure 9.19 The experimental frequency density distribution g( ) for aluminum ( D is the Debye frequency).
lattice heat capacity is a development of collective excitation in solids, i.e., the heat energy goes to phonon excitation (see below). Participation of electron excitations in heat capacity is limited by quantum statistics laws (refer to Section 9.2.2). As only a small part of the electrons takes part in excitation, the electronic heat capacity usually makes a very small addition in comparison with the lattice one.
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In a polycrystalline sample consisting of a large number of microcrystals, ideally— randomly dispersed in space—the resulting picture is averaged. As a result, the function g( ), presents the projection of all dispersion branches to the axis of frequencies (phonon spectrum). It describes the distribution of oscillations on frequencies: product g( )d
gives the number of oscillations dz falling on an interval d . In Figure 9.19 the experimentally measured function g( ) for aluminum is given as an example. The fluctuation distribution function determines the internal energy of a crystal:
U ∫ g( )d ,
(9.3.7)
where is an averaged oscillation energy. The dispersion curves and g( ) function can be obtained by neutron scattering methods. 9.3.2
The heat capacity of crystals
Heat capacity was introduced in Section 3.4.3 as the heat energy capable of heating a body by 1 grad. For an ideal gas in molecular kinetic theory the mole heat capacity appeared to be equal to CV
ief i 2 R and CP ef R, 2 2
where ief is the effective number of a gas molecule’s degrees of freedom. In this case ief turns out to be the sum of translational, rotational and oscillation movements; in the latter ief, osc 2 because oscillation simultaneously possesses potential and kinetic energy. Apply these representations to a solid. For this purpose we should first analyze the character of possible movements of the molecules in it. Clearly, translational motion is excluded in this case. The rotation of molecules in a crystal is basically possible. For example, in crystal NH4Cl a group NH4 at different temperatures can make rotations around axes of different symmetry and sometimes exhibit a free rotation. However, the contribution of rotation to the thermal capacity of solids is appreciably less than oscillatory degrees of freedom. As a result only oscillations of atoms basically define the crystal thermal capacity. Furthermore, the three-dimensional character of a crystal should be taken into account. The molar internal energy of a crystal U will then be given as a product
U 3
ief,osc RT 3RT . 2
(9.3.8)
The derivative from internal energy on temperature gives a mole thermal crystal heat capacity C. As the volume of a crystal does not vary appreciably with temperature, we exclude index V. We shall then obtain C 3 R,
(9.3.9)
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i.e., the molar heat capacity of a crystal does not depend on temperature and is numerically equal to 3R (⬇ 25 J/(mol K)). This result is referred to as the Dulong–Petit law. Experiment shows that at moderate temperatures this law is carried out well enough for crystals with rather simple structure (Na, Al, Fe, Cu, Sn, etc.). However, crystals with more complex structures fail the Dulong–Petit law. So the atomic thermal capacity of a boron crystal is 14.2 and diamond is 5.7 J/(mol K). It has also been experimentally established that in the region of low temperatures the heat capacity falls as T 3, coming nearer to zero at T → 0 K. The experimental facts therefore show that the simple classical theory considered is applicable only in a narrow temperature interval and for crystals with a simple structure. The first quantum theory of heat capacity was suggested by Einstein. Einstein’s model assumed that each atom in a crystal is an independent quantum oscillator (refer to Section 7.8.1). The energy spectrum of such an oscillator is presented in Figure 7.32. Because of the selection rule, the spectrum of absorption/emission contains only one frequency. A level’s population is defined by the Boltzmann factor E ⬃ exp(E/T ), i.e., the population gets less when the energy increases. In Section 6.6.3 it was shown that the average oscillator’s energy is (h ¯ ) / (exp (h ¯ / T ) 1) (eq. (6.6.13)). Accordingly, the internal energy U of one mole of a substance is
U = E N A = N A
+U , exp(
T ) -1 0
(9.3.10)
where U0 is the zero oscillation energy. The latter is of no significance in the heat capacity of crystals. The heat capacity C is obtained as 2
C
exp(
T ) dU ⎛ ⎞ NA ⎜ . ⎝ T ⎟⎠ (exp(
T ) 1)2 dT
(9.3.11)
Denoting the ratio (h ¯ /) by , the last expression can be rewritten as 2
exp( T ) ⎛ ⎞ . C R⎜ ⎟ ⎝ T ⎠ (exp( T ) 1)2
(9.3.12)
The value is called the characteristic temperature. If one derives a function of the heat capacity upon T/ for a series of simple substances, it will be represented by a single curve (Figure 9.20). It appears that Einstein’s characteristic temperature defines the border behind which an essential deviation from the Dulong–Petit law takes place for every element. If we substitute the frequency 兹 苶苶/m 苶 (see eq. (2.4.5)) into the expression for the value, we can arrive at
, m
(9.3.13)
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6
C/n
5 4
Ag Al C(grafite) Al2O3 KCl
3 2 1 0
0
0.5
1.0 T/
1.5
2.0
Figure 9.20 A reduced temperature dependence of heat capacity for some simple substances.
i.e., the value is inversely proportional to the square root from atomics’ mass. Let us consider some extreme cases. At T the exponent can be decomposed into a series (ex ⬇ 1x ...), and we can limit ourselves to the first two terms. For a system of one-dimensional oscillators we can arrive at ⎛ ⎞ ⎛ ⎞ 1 冢 T 冣 C ⬵ R⎜ ⎟ R ⎜ 1 ⎟ ⬇ R ⎝ T⎠ ⎝ T ⎠ 冢 2 T 2 冣 2
or for a three-dimensional crystal C 3R, i.e., the Dulong–Petit law. At T the exponent in eq. (9.3.12) is much larger than 1. It gives C ⬵ 3R ( / T )2 ( / T ) e , i.e., heat capacity is decreases with the decrease in temperature (because the exponent influence dominates over the term ( / T)2). Apparently, Einstein’s model gives good conformity with Dulong–Petit at high temperature and explains the decrease of thermal capacity at low temperatures. However, it contradicts the law of approaching absolute zero: according to experiment, the corresponding curve should look like a cubic parabola T 3, whereas according to Einstein this dependence is described by the law e(1/T). P. Debye modified the Einstein’s model by introduction of inter-atomic forces in a crystal model. This is equivalent as to take phonons into account (refer to Section 9.3.1). To each elastic wave (phonon) the Born-Karman atomic chain was attracted spread out in a three-dimensional array (Figure 9.13 and 9.15). As a result of reflection from external crystal borders, standing waves with various values and k (refer to Section 2.9.2 and 2.9.3) are formed. There is the certain relationship between the wavelength of standing waves and the size of the crystal L expressed by the eqn (2.9.8). Phase speed of running
, therefore elastic wave Ph is related to and k by expression (2.8.3) Ph k
kPh
2Ph T Ph n, L
(9.3.14)
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where n is the number of the harmonic wave. At the crystal volume V a large but limited number of standing waves are formed. The wavelengths of these standing waves are limited by the period of the crystal lattice from the short wavelength size ( a) and by the crystallite size L ( L) from the large one. Accordingly,
min
2ph max
2ph 2L
ph L
(9.3.15)
and
max
2ph min
ph a
.
(9.3.16)
These values limit the range of frequencies. The number of standing waves dz is proportional to the elementary volume 4 2d . For a cubic crystal with volume V L3 the number of harmonics (normal oscillations) in the limits from 0 to is equal to z (2L /n)3 8 (V / n3) or more precisely
z
4V V 2 3 2 . 3n 2 ph
(9.3.17)
Differentiating z by we can obtain the frequency distribution g( ) as
g( )
dz 3V
2 . 3 d 2 2ph
(9.3.18)
The Debye function g( ) is depicted in Figure 9.21.
g
0
Figure 9.21 The frequency density distribution g( ) in the framework of the Debye model ( D is the Debye frequency).
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The maximum spectrum frequency is called the Debye characteristic frequency D. It can be determined from the normalization to the total number of normal frequencies (the number of standing waves).
D
V 3
∫ g( )d 22D3
3N ,
ph
0
from which
⎛ 6N ⎞
D ph ⎜ ⎝ V ⎟⎠
1 3
.
The notion of “characteristic temperature” acquires a new meaning:
D . T
(9.3.19)
According to eq. (9.3.7) the general expression for U in a given case is:
D
U
∫ 0
D
g( )d
∫ 0
9N 3 3 d . exp(
T 1) max
(9.3.20)
Differentiating over temperature, we can obtain the crystal heat capacity
C
⎡ ⎛ T ⎞ 3 T ⎛ x 3 dx dU 3( T ) ⎞ ⎤ 3 R ⎢12 ⎜ ⎟ ∫ ⎜ x x ⎟ dx ⎥ . dT ⎢⎣ ⎝ ⎠ 0 ⎝ e 1 e ( T ) 1⎠ ⎥⎦
(9.3.21)
The expression in square brackets is the Debye function. This function is tabulated. We shall find extreme values of heat capacity: at T p D the Debye function aspires to 1 and C → 3R; at T ^ D the integral in the Debye function is close to 1 and C → (T / )3. Good consent with experiment is achieved. Figure 9.22 shows the results of comparative approaches for three models of crystal dynamic properties. In conclusion, we give in Table 9.2, the Debye temperature for some substances. Despite the success of the theory, it is necessary to ascertain whether its use in practice meets many difficulties. This is visible even from a comparison of the curves of dependence (k), obtained theoretically and experimentally (with the help of inelastic scattering of thermal neutrons) (Figures 9.19 and 9.21). At concurrence of some features of the curves (square dependence at small phonon wave vectors k, sharp decrease at certain values of frequencies, etc.) a definite difference can also be found. The theory allows one to
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D
B-K
E
/a
k
Figure 9.22 Comparison of the g( ) function for different models: Born–Karman (B–K), Einstein (E) and Debye (D). Table 9.2 Debye temperature values for some simple substances Substance
Debye temperature (K)
C (diamond) Be Si Cr Pb
1910 1160 658 420 95
obtain only estimated or comparative values of heat capacity; therefore experimental data are often used in serious work. A promising direction in chemical technology developments in the last 10 years is nanotechnology, i.e., the syntheses of materials consisting of microscopically small particles (clusters), small in number (N ⬃ 103 atoms), being comprised on the “nanoscale” L ⬃ 109 m. This technology gives us a good assurance that this chapter is of use in the book. In practice, this border is somehow degraded, because in technology by “nanoparticle” one usually implies an atomic or molecular object, the internal energy of which complies in order of magnitude with its surface energy. More important is the fact that the mechanical, electrical, magnetic and other characteristics of nanoparticles are vastly distinct from similar characteristics of the same bulk material. Some of thesxe characteristics are directly connected with the nanoparticle’s size, so this opens the way to adjust them to a goal-directed image. This fact permits one to produce new materials with predetermined properties, e.g., semiconductors that are required for creating more reliable generations of computers.
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Let us consider as an example the well-known Debye theory of the heat capacity of crystals. The internal energy of a bulk crystal minus zero energy is given in this theory by an equation 9 N T 4 U 3D
D T
∫ 0
x3 dx, e x 1
where x (h ¯ / T ) A nanoparticle is distinguished from a bulk crystal since its phonon spectrum is truncated not only from the site of high frequencies but from low frequencies as well (refer to Example E9.5). Let us consider the averaged solid nanoparticle model with number of atoms N, crystal lattice period a, primitive cubic lattice with the unit cell volume V and linear size L. The physicochemical features, however, and in particular the nanoparticle’s heat capacity in the implicit description can depend on their forms, for instance, by the way it fastens to the substrate and the nature of interaction with it, i.e., from border conditions. A detailed consideration of all these factors forms the subject for specialized future studies. Therefore, the further evaluations are based on the assumption of the independence of the nanoparticle’s thermal characteristics from the particular type of border conditions (similar to that of the classical theories of Einstein and Debye). The longest wavelength corresponds to the lowest frequency of the nanoparticle that can be put in correspondence to the temperature, i.e., N
N 2 , N
therefore,
N
2 ⬇ . L 2 N 1 3 a N 1 3 a
Since the Debye temperature is assumed to be D (h ¯. /a), the ratio (N/D) N 1/3 3 or (N / D) ⬇1/10 with N ⬇ 10 . Therefore, nanoparticles have two characteristic temperatures that distinguish them from that of the bulk materials: the usual Debye temperature D and the temperature N both depend on the particle size. For further simplification, let us restrict ourselves to the limit that corresponds to the condition D T, because of the fact that this condition corresponds to the lowest value of x (h ¯ / T ) (N T ) 1 . This permits us to simplify the integral expression, i.e., assume ex p 1. In this case the internal energy of the nanoparticle in the low temperature approximation is equal to
UN
9 N T 4 3D
∫
N T
x3ex dx
9 N T 4 3 ( x 3 x 2 6 x 6) 3D
N T
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For diamond dust (D 2230 K), L ⬃ 10a, N ⬃ 103, N (D /10) ⬃ 223 K, i.e., the condition (N / T ) 2 is fulfilled at temperatures T 111 K. At ex p 1 the expression for the nanoparticle heat capacity is changed as well. The Debye equation at T ^ D, i.e., ⎛T ⎞ C 9 Nk ⎜ ⎟ ⎝ D ⎠
3
x 4 e x dx
∫ (e x 1)2 0
can be simplified by decomposition to
⎛T ⎞ CN 9 Nk ⎜ ⎟ ⎝ D ⎠
3
∫
N T
3
⎛T ⎞ x e dx 9 Nk ⎜ ⎟ ( x 4 4 x312 x 2 24 x24)ex ⎝ D ⎠ 4 x
N T
.
The well-known classical Debye law for bulk crystals Cⴥ⬃ T 3 is disobeyed, and, as the calculations show, rather significantly. Once more, this confirms the fact that nanoparticles really are new materials, with physicochemical characteristics greatly different from the characteristics of materials with the same chemical composition but in the bulk state.
EXAMPLE E9.4 Determine the amount of heat Q for NaCl of mass m 20 g heated at T 2 K in two cases (1) from T1 D and (2) T2 2 K. Accept the characteristic Debye temperature D to be equal to 320 K. Solution: In general, the amount of heat Q needed to heat a system from 1 to
冕
2 can be calculated according to the expression Q
2
1
CdT where C is the heat
capacity of the system. The heat capacity of a body is related to molar heat capacity C (m/M)Cm where m is body mass and M is molar mass(refer to Section 3.4.3). m
Substituting into the integral gives Q
冕
2
1
CMd T *. In a general case, CM
depends on T and therefore it is not allowed to take it out of the integral sign. However, in the first case we can neglect the CM change and consider it to be constant, throughout the interval T, equal to CM(T1). Therefore, the integral takes the form Q (m/M)CM(T1)T**. The molar heat capacity in Debye theory is given by eq. (9.3.21). In the first case calculations give CM 2.87 R. Substituting this result into ** we obtain Q 2.87(m/M)RT and executing calculations we arrive at Q 16.3 J. In the second case (T ^ D) calculation of Q is simplified by the fact that we can use the limited property of the Debye law where the heat capacity is
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proportional to T 3: we cannot use the previous approximation but have a mathematical expression for 3
12 4 ⎛ T ⎞ CM R⎜ ⎟ . 5 ⎝ D ⎠
Q
12 4 m R 5 M 3D
T2 T
∫
T 3 dT .
T2
Executing integration we obtain
Q
12 4 m R ⎡ (T2 T )4 T24 ⎤ ⎥. ⎢ 5 M 3D ⎣ 4 4 ⎦
Since T2 T 2T2 this equation takes the form
Q
3 4 m R m R 4 15T24 or 9 4 T2 . 5 M 3D M 3D
Substituting all known values and executing calculations we obtain Q 1.22 mJ.
EXAMPLE E9.5 How much does the amount of heat required differ for heating a nanoparticle consisting of N atoms from temperature T1 0 K to T2 D/50, compared to the amount of heat required to heat the same amount of particles in the form of a bulk crystal. Accept the nanoparticle’s size L 10a, a being the period of the crystal lattice; D is the Debye temperature).
L
a
k
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Solution: The frequency of a nanoparticle’s dependence on the wavenumber k is depicted in the Figure E9.5. From the problem’s conditions, we have D 1 . N L 10 The internal energy U of a bulk crystal consisting of N atoms at T1 T2 is equal to
UN
9 N T 3D
D 4 T2
∫ 0
x3 dx . e x 1
The upper limit in the task is (D / T2) 50 1, therefore one can accept it, as usual, to be equal to ⴥ. The tabular point D 50 1, T2
x3 ∫ e x 1 15 6.5. 0
For T T2 the internal energy is
UN
9 N T24 3D
∫
x3ex dx,
N
T2
where N 50 5. T2 10 The zero energy as independent of temperature cannot be taken into account. The integral in the last expression for UN can be calculated in parts:
∫x e
3 x
dx ( x 3 3 x 2 6 x 6) ex
5
1.57
5
Then the ratio (Uⴥ) /(UN) (6.5 /1.57) ⬇ 4.15, i.e., the internal energy difference for a bulk crystal and a nanoparticle with equal amounts of atoms is really very significant.
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EXAMPLE E9.6 Using a free electron model calculate for potassium at 0 K: (1) Fermi energy F, (2) Fermi temperature TF, (3) the ratio of averaged potential energy 具U典 of the two adjacent electrons in the electron gas to their averaged kinetic energy 具典. Assume the potassium density 860 kg/m3 and atomic mass A = 39.1 103 kg/mol. Solution. Although in general in the main text we can come to the conclusion that the free-electron model is applicable to crystals, we can use in some cases, with certain limitations, the classical approach to the evaluation of electron gas properties. (1). At T 0 K, the Fermi energy (refer to Section 9.2.2) depends only on the electron concentration n F (h ¯ 2 / 2me)(32n)2/3. Assume that every potassium atom gives one electron to the free electron gas state. Therefore, the concentration of free electrons is equal to the potassium atom concentration, namely n nat (/A)NA. In order to avoid very large expressions and calculation we will deviate from our recommendation and calculate each quantity separately. Thus, we can calculate the electron concentration accordingly:
n
860 6.02 1023 m3 1.32 1028 m3 39.1103
We can calculate the Fermi energy according to the mentioned formula as F
(1.05 1034 )2 冸3 2 1.32 1028 冹2 3 J 3.23 1019 J or 2.02 eV 2. 9.111031
(2) The Fermi temperature can be calculated from the equation F T: T F /. Executing calculations we obtain: TF
3.23 1019 2.34 10 4 K 23.4 K 1.38 1023
(3) In order to find the zero temperature electron gas pressure we can use the expression p (2/3)n具典. Here we will assume 具典 the averaged value of kinetic energy at T 0 K. Find this expression using the distribution of the free electron upon energies dn() (1/22)(h ¯ 2/2me )3/2 1/2 d, where dn () is the number of electrons in a unit volume, the energy of which lie in the small interval from to d. Then dn() C1/2d, where C is a constant. We can find the averaged value 具典 for free electrons by dividing the total energy in an unit volume by their concentration, i.e., F
具典
∫ dn 0 F
∫ dn 0
F
C ∫ 3 / 2 d 0 F
C ∫ 1 / 2 d 0
(2 5) F
5 2
(2 3) F
3 2
3 F 5
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Substituting this result into the expression for free electron gas pressure p (2/3) n (3/5) F we obtain p (2/5)1.32 1028 3.23 1019 1.71 109 Pa 1.68 104 atm. (4) The averaged potential energy of the electrostatic interaction of two point charges (refer to eq. (4.1.21)) 具U典 (e2)/(40具l典) where 具l典 is an averaged distance between two adjacent electrons of the degenerated electron gas (refer to Example E3.1) 具l典 n(1/3). Then 具U典 (e2n1/3) / (40) . We already know the 具典 (see above) therefore
5e2 n1 3 具U 典 e2 n1 3 5 . 40 3 F 12 0 F 具典 Substituting the numerical values into this expression we arrive at 具U 典 (1.32 1028 )1 3 5(1.6)2 2.81 具典 12 8.85 1012 3.23 1019
9.4
CRYSTAL DEFECTS
A crystal represents a complex quantum mechanical system with an enormous amount of particles with a strong interaction between them. If all the particles are located in space strictly ordered with the formation of an ideal three-dimensional crystal structure, then such a system possesses minimal free energy. In a real crystal, however, the ideal periodicity is often broken due to inevitable thermal fluctuations: some atoms break periodic array, abandon their ideal position and produce a defect. The frequency and amount of fluctuations are defined by the Boltzmann factor, i.e., they depend on temperature and binding strength or, in other words, on the depth of the potential well corresponding to the regular position of atoms. The role of defects in crystal properties is very high. On the other hand, new and improved old methods of defect crystal structure studies have appeared. This leads to the fact that, in recent decades, there has been extensive investigation into the physics of solids from the point of view of their deflection from perfection. Natural and new synthetic materials are being investigated. Let us distinguish point (zero-dimensional) and extended defects (dislocations). 9.4.1
Point defects
Point defects can appear in a small area of the crystal, not exceeding several internuclear distances. Due to thermal fluctuations, a single atom can abandon its ideal position and occupy a position in the so-called interstitials (a position, though having a potential well, is of small depth). The regular position in the lattice remains empty, and is called a vacancy. Such pair defects are referred to as Frenkel’s defects, they contain a vacancy in a regular array and interstitial atoms (Figure 9.23).
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+ Sh
–
+
–
+
–
+
–
–
+
–
+
+
–
+
–
F
+
–
–
+
–
+
Figure 9.23 The Frenkel (F) and Schotky (Sh) types of point defect in a crystal.
If an atom comes out onto a crystal surface and leaves a vacancy in its regular position, then such a single defect is referred to as a Schotky defect. The probability of a Schotky defect formation is proportional to the Boltzmann factor P⬃ exp(Esh/T), where Esh is the energy required for an atom’s removal from its regular position to the crystal surface. If N is the total amount of atoms and n is the vacancy number then P n/(Nn) exp (Esh/T ) or, at n N, n ⎛ Esh ⎞ ⬇ exp ⎜ . ⎝ T ⎟⎠ N
(9.4.1)
Estimations show that at T ⬇ 1000 K and vacancy activation energy E ⬇ 1 eV, the relative vacancy concentration is (n/N) ⬇ e12 ⬇ 105. Note that the relative vacancy concentration is proportional to exp(1/T). The number of Frenkel pair defects n depends on the number of regular positions in the crystal N and the number of interstitials N ⎛ EF ⎞ n ⬃ ( N N) exp ⎜ , ⎝ T ⎟⎠ where EF is the energy required for an atom’s disposition from the regular site into the interstitial. The vacancies in ionic crystals are mostly Frenkel defects. Crystal defects are not static. They wander around the crystal: a neighboring atom occupies the vacant position; the interstitial atom can occupy a regular site, etc. Only their relative number is determined by the Boltzmann factor.
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Along with such elementary defects, more complex ones can also appear; for example, divacancy, the formation of which is energetically more favorable in comparison with two individual vacancies. Besides, the mobility of a divacancy is higher than that of individual vacancies. The congestion of many vacancies leads to the formation of the so-called clusters. Vacancies strongly influence diffusion and related processes. Elastic displacements in the area surrounding a point defect decrease proportionally to 1/r3, where r is the distance from the defect. This shows that distortions in the neighborhood of a defect can be rather significant, but quickly falls down at distance. Possessing mobility, the point defects can interact with each other and with other defects. On meeting, the vacancy and interstitial atom can annihilate each other. An important role in vacancy formation is played by impurity atoms. They can replace regular atoms, and also be in interstitials. In particular, in the crystal structure KCl the replacement of K by Ca atoms leads to the occurrence of vacancies. Small amounts of impurities considerably change the electric properties of semiconductor crystals. 9.4.2
Dislocations
Apart from point defects, there also exist extended, in particular, one-dimensional edge, dislocations: a failure of a crystal lattice having in one direction a long (even macroscopic) dimension but small sizes in the other directions (one or several internuclear distances). Such defects play a very important role in the formation of many properties of crystals, in particular, mechanical properties. Quantitatively, the extended failure of the crystal lattice is described by the so-called Burgers vector. Let it be described in a simple cubic crystal. In Figure 9.24 a plane (001) of such a crystal is presented. In an ideal nondeformed lattice we shall choose a closed contour and number the atoms included in this contour (Figure 9.24a). Such a contour is called Burger’s contour. Travelling toward the east over the contour for two periods, south for two periods, then
0 0
1
1
2
0
0
2 7
7
3
1
2
b 7
w
6
5
3
3 4 6
5 (a)
4
4 6
5 (b)
(c)
Figure 9.24 Burger's vector in (a) an ideal, (b) an elastically deformed and (c) a crystal with edge dislocation.
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west and, finally north, we will arrive at the initial atom. In a numerical expression, the travel over the contour is 0–1–2–3–4–5–6–7–0. If a crystal is elastically deformed the atoms, being close to their ideal positions, are displaced by small distances (Figure 9.24b): the contour is deformed but still remains closed. Now create a defect in the crystal as a half-plane (i.e., an infinite plane in two directions limited by a line in the third side) and introduce it into the crystal parallel to similar crystallographic planes (Figure 9.24c). Now, traveling over Burger’s contour we will not come to the initial atom but to another point. To close the contour, it is necessary to draw a closing vector b, which is referred to as Burger’s vector. Note that Burger’s vector is perpendicular to the plane introduced. Dislocation is a linear crystal defect for which Burger’s vector is not zero. This definition is common for rather large numbers of a different kind of linear defect. The considered dislocation is referred to as an edge disposition; the extreme line (edge) of the induced plane is referred to as the line of dislocation (in Figure 9.24c this line in a point w is perpendicular to the plane of drawing). For such a dislocation, Burger’s vector is perpendicular to the dislocation line. A model of edge dislocation made of balls is presented in Figure 9.25. In Figure 9.26 a three-dimensional image of the edge disposition with Burger’s vector perpendicular to the plane is shown. The crystal above the dislocation edge is stretched and in its lower part is compressed. Now consider another kind of disposition, namely, a screw disposition. Let us imagine a pile of sheets of paper. Using a sharp knife, cut it from the center of the pile to its edge. Shift the free paper edges perpendicular to the sheets by the thickness of one sheet. Sticking the edges of the new planes to the edges of the old ones will result in a spiral. This spiral gives a graphic representation of screw dislocation. Its crystallographic image is given in Figure 9.27. Dislocations can move in crystals as do point defects; the edge dislocation is presented in Figure 9.28. Under the action of an external pressure the shift of the upper part of the
Figure 9.25 A ball model of the edge dislocation.
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Figure 9.26 A three-dimensional image of the edge dislocation.
Figure 9.27 A model of a screw dislocation.
crystal has forced the dislocation to be switched to the next atomic plane and, finally, to appear on the crystal surface. Similar movements can make screw dislocation too. There exist bigger crystal defects called small-angle boundaries. The scheme of such a boundary is depicted in Figure 9.29. It can be imaged by a set of number of edge dislocations. The presence of imperfections essentially influences the mechanical properties of crystals. Experience shows that theoretically calculated strength properties appear, as a rule, to be appreciably higher than experimental ones; this is caused by neglecting crystal imperfections. Table 9.3 lists the physical methods that allow the investigation of the real structure of crystals.
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Figure 9.28 A scheme of an edge dislocation movement.
b b
Figure 9.29 A low-angle edge. Table 9.3 Methods of investigation of crystal defects Method
Object thickness
Width of image
Maximal defects density (cm2)
Electron microscopy X-rays’ transmission X-rays’ reflection Metal sputtering onto surfacea Etching pitsb
100 nm 0.11.0 mm 2/50 m ⬇10 m
⬇10 nm 5 m 0.5 m 0.5 m
10111012 104105 106107 2 ·107
No limits
0.5 m
4 108
a
Sputtering of thin layers onto a surface and observation of dislocation by different methods. Surface etching and observation of etching pits by different methods.
b
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567
TRANSPORT PHENOMENA IN LIQUIDS AND SOLIDS
The laws of transport phenomena in states other than gases differ among themselves in many respects. This distinction is defined mainly by the difference in their atomic structure. This will be explained using the example of a monoatomic crystal with a cubic structure (refer to Table 9.1). The function of radial distribution G(r) has great value in this respect. To explain the physical sense of this function, we shall define one more concept: the coordination sphere, perhaps known to readers from chemistry. For this purpose we choose in the crystal any atom and superpose it with the coordinate system origin. Allocate around it a spherical layer of radius r and thickness r (r r); the layer volume is V 4r2r. The corresponding scheme is shown in Figure 9.30; for simplicity a two-dimensional section of a cubic lattice is used. If r r1 no atom falls onto the sphere. When r is less than the shortest interatomic distance a no atom falls into the spherical layer either. At r a all atoms which meet at this distance fall into the spherical layer (in a threedimensional cubic crystal there are six such atoms). With a further increase of r in an interval a r 兹苶2a , again no atoms fall in the layer, but at r 兹苶2a atoms do fall again in the layer; their number in the three-dimensional array will be 12. At r 兹苶3a in the layer eight atoms will fall. This procedure can be continued further. These spheres are referred to as coordination spheres and the numbers of atoms falling onto them as coordination numbers. Count the number of atoms N enclosed in the spherical layer 4r2r. Write this number as: N 4r 2 r n0 G(r ). Here n0 is the average over the whole crystal volume atomic concentration, and G(r) is the radial distribution function. Wherefrom: G(r )
N N n(r ) , 2 Vn0 4 r r n0 n0
(9.5.1)
where n(r) is the local atomic concentration in the spherical layer with radius r.
1 3
2
4
Figure 9.30 Coordination spheres in a crystal. A two-dimensional section of a cubic crystal is presented. Numbers denote the first, second, third and fourth coordination spheres.
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It can be seen from the last expression that when the atomic concentration in the allocated spherical layer is equal to average concentration n0, G (r) 1. The function of radial distribution is distinct from unity only in the case when the concentration of atoms in the allocated layer is different from n0. Hence G(r) characterizes the concentration deviation in a specified layer from the average value. The functions G(r) for gases, liquids and crystals are presented in Figure 9.31. In crystals atoms are arranged in the ordered lattice; they are at certain distances from each other. The G(r) function is different from zero when r is equal to one of the possible interatomic distances, i.e., when r touches the coordination sphere. In the resultant radial distribution function G(r) in a crystal is represented by a system of discrete peaks (Figure 9.31a). The regular arrangement of peaks in crystalline solids specifies the presence in a crystal of the long-range order. If atoms are at rest in their ideal positions, function G(r) is represented by a system of discrete, narrow spectral lines. Because of the thermal vibrations of atoms around their regular positions the lines are widened. In gases G(r) ⬅ 1; this means that there is a total disorder in gases. However, because gas atoms cannot approach each other closer than r d Эф in an area r def function G(r) falls to zero (Figure 9.31c). All this corresponds to an absence of even short-range order in gases; the full disorder in an arrangement of atoms. It has been experimentally established that function G(r) in liquids looks like that presented in Figure 9.31b: there are dim maxima at r (13) a, and further G(r) aspires to 1. This result indicates some order in an arrangement of atoms in the liquids at specified distances and the absence of order at large r. Such a situation corresponds to the shortrange order. It can be imagined as a presence in liquids of very small crystallites, completely disordered to each other. Due to the thermal motion of atoms these crystallites
G (c) 1
(b) 1
(a)
a
2a
r
Figure 9.31 A function of the atomic density radial distribution in (a) crystal, (b) liquids and (c) gas.
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continuously collapse and new ones are created instead. Each atom in a liquid spends some time (referred to as time of settled life) in a regular position and makes fluctuations around it. At this time the atoms of a liquid behave like atoms in a crystal. Later, the atom abandons its regular position and jumps to a new position remote from the first by a distance approximately equal to the interatomic distance in the liquid. During the jumping process, the atom of a liquid assimilates to a gas atom. If the temperature decreases, the time of settled life is increased; near the crystallization point this time increases more and the properties of the liquid come nearer to the properties of solids. On the contrary, if the temperature rises, coming nearer to the boiling point, the time of settled life decreases; and the liquid’s properties become similar to those of gas. On melting there is an increase in the specific volume of a substance, on an average by 3%. It is reasonable to assume that the increase in volume is caused by an increase in interatomic distances in the liquid in comparison with that in crystal. However, the compressibility dependence on pressure does not confirm this supposition. Since there is a short-range order in a liquid the local potential curve U(x) in its nearest surroundings resembles a periodic character (Figure 9.32). Atoms settle down in points of potential energy minimum. In a settled state an atom oscillates with an amplitude significantly smaller than the interatomic distance . It was found experimentally that the frequency of these fluctuations have the same order as in solids, namely v0 ⬇ 1013 sec1, and the oscillation period T0 (1/v0) ⬇ 10 13 sec. In order to leave the pseudo-equilibrium position an atom should overtake the potential barrier u. The probability of this process is proportional to the Boltzmann factor e(u / T ). In 1 sec, an atom makes v0 fluctuations; hence the frequency of jump over to the new position is v v0 exp((u/T )). Thus, the time of settled life can be obtained by (1 / v ) 0 exp(u / T ) F
u
2a
Figure 9.32 Atomic potential curve in a liquid.
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When a particle leaves a temporary equilibrium position and occupies a new one, it covers a distance . The particle’s average speed of wandering (chaotic motion) in liquids is then ⎛ u ⎞ ⬇ exp ⎜ ⎟ . ⎝ T ⎠ 0
Suppose that there is a concentration gradient in the liquid. In that case the chaotic wandering creates an excessive atom flow in the direction opposite to the concentration gradient; i.e., diffusion flow will take place. The diffusion coefficient for ideal gases is described by the formula D (1/3) , where is the molecular free path length, and is the average speed of thermal motion (refer to Section 3.3.8). In a liquid the role of free path length is played by the displacement and the role of average speed is (/). Therefore the diffusion coefficient in the liquid can be evaluated as
D
1 2 ⎛ u ⎞ exp ⎜ ⎟ . ⎝ T ⎠ 3 0
(9.5.2)
It can be seen from this expression that diffusion in a liquid sharply increases with temperature according to the exponential law. In gases the diffusion coefficient also rises with temperature, but not as fast: under the power D.T1/2 if heating is isochoric, and as D. T3/2 if heating is isobaric (see Section 3.7). Although the expressions obtained are approximate, they usually correctly estimate the order of diffusion coefficient provided that ⬇ 1010 m, v0 ⬇ 1013 sec1 and u (the activation diffusion energy) is approximately equal to the latent heat of melting. The data on diffusion parameters in liquids is given in Table 3.3. Theoretical treatment of a viscosity phenomenon of liquids can also be carried out proceeding from the same representations. Leaving aside the treatment itself, we write down the final result concerning the temperature dependence of the viscosity as Ae(u T ) ,
(9.5.3)
where A is a pre-exponential factor dependent on the parameters of the liquid and weakly dependent (in comparison with the exponent) on temperature. It can be seen that liquid viscosity falls very quickly when temperature increases. The result is not unexpected; such behavior is well known from our everyday experience. It essentially differs from the conclusion that was derived for ideal gas. The discussion on liquid properties can be used when considering solids. Diffusion in crystals is also defined by the speed of the atom’s wanderings, though it is greatly determined by a crystal’s imperfections. However, time of settled life in crystals is higher by some order of values than in a liquid because of the fact that practically all crystallographic positions are occupied (see Section 9.4). Atomic hopping (i.e., diffusion) is carried out on those points that are most preferable for this purpose: on vacancies, interstitials, along dislocations and other
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places in which the atomic order is destroyed. As a result, the intensity of diffusion in solids is much less than that in liquids. At the same time, the diffusion coefficient also increases exponentially with temperature and in some cases diffusion plays an appreciable role. Viscosity of solids is extremely high (see Table 3.3). For solids of different nature it differs essentially. It also depends on the structure of substances: degrees of crystallinity, the nature of chemical bonding, etc. Heat conductivity of solids also changes over a wide range. Similar to gases, the heat conductivity of liquids and crystals depends on their structure, especially of those particles that play the role of energy carriers. In crystals it can be phonons (refer to Section 9.3, in all solids) and electrons (electron gas in metals). As a result, the heat conductivity of solids changes over a wide range (see Table 3.3). We hope that readers can apply the representations given here to amorphous materials; it should be borne in mind that they have much in common with liquids, but their time characteristics are closer to solids. Unfortunately, there is no general theory allowing calculation of transport properties of condensed systems. This means that when deciding technical problems based on the use of Fick, Fourier and Newton laws, it is also necessary to use empirical laws and factors. However, it should be remembered that the essence of the phenomena and their physical sense remain the same, as for the description of the elementary model of an ideal gas.
9.6
SOME TECHNICALLY IMPORTANT ELECTRIC PROPERTIES OF SUBSTANCES
Ionic polarization The mechanisms of polarization are caused by the structure of molecules (previously considered in Chapter 4). There is a type of crystal polarization that has much in common with atomic polarization (refer to Section 4.2.4). In crystals, atoms are in an ordered array. As a result of interaction between neighboring atoms (i.e., chemical bond formation) redistribution of electron density occurs and atoms acquire an effective charge, positive and/or negative. The presence of the two kinds of atomic charge allows one to consider a crystal lattice consisting of two sublattices inserted into one another. Consider for example a crystal structure of cesium chloride (CsCl) (Figure 9.33). Ions Cs and Cl form two simple cubic sublattices shifted from each other along a spatial diagonal by a distance equal to half of its length. The total crystal electric dipole moment is zero; the crystal is not polarized. When an electric field is imposed, each sublattice is displaced from the other. The crystal then gets an uncompensated electric dipole moment, i.e., polarization of the whole crystal takes place. Such polarization is referred to as ionic polarization. Elasticity forces will be opposite to the sublattices’ displacement; these forces will compensate the action of the electric field. Thus, ionic polarization arises at elastic displacement due to the action of an electric field on positive and negative ions, shifting them from their equilibrium positions. By analogy with atomic polarizability, one can introduce an ionic polarizability with coefficient ion being attributed to each pair of oppositely charged ions. Under the order of value ion coincides with the value of at, i.e., ion1030 m3.
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ⴚ
ⴚ ⴚ
ⴚ ⴙ
ⴙ ⴙ
ⴙ
ⴚ
ⴚ ⴚ
ⴚ ⴙ
ⴙ ⴙ
ⴙ
Figure 9.33 Crystal structure of CsCl.
As a result of sublattice displacement, the polarization of an ionic crystal is accompanied by its deformation—the crystal is extended in the direction of the field. This crystal lengthening does not depend on the direction of the field: if the field direction is changed to the opposite direction, it will still lead to the same lengthening of the crystal. Such a change of crystal sizes by the action of the electric field is referred to as electrostriction. The relative crystal deformation l/l depends quadratic on the electric field strength (l/l ⬃ E2). The value of the sample length change is very small; for example, for SiO2 in a field strength E ⬃ 104 V/m the relative lengthening of a sample is 109. Currently, normal electrostriction has no practical application. However, there are crystals (e.g., Rochelle salt) in which the positive and negative sublattices are found to be asymmetrically disposed in the crystal; therefore these crystals are polarized even in the absence of an external field. Such polarization is referred to as spontaneous (see below). Charges are formed on the edges of spontaneously polarized crystals, however to measure it one should have a fresh chip. In due course ions of an opposite sign from the surrounding atmosphere neutralize the surface charges and prevent charges being found. When the crystal temperature changes, crystal deformation takes place. Consider an edge of a crystal on which the bounded positive charge exists; they are compensated by adsorbed negative ions. With an increase in temperature, crystal polarization will also increase leading to an increase in the density of the bounded charges, and adsorbed negative ions will no longer compensate them. Hence, a positive charge will be spread on this surface. Similarly, the negative charge will be concentrated on the opposite edge. Thus, the potential difference appears at heating between crystal sides. The phenomenon of the occurrence of electric charges on the edges of a crystal with a change in its temperature is referred to as a pyroelectric effect. A piezoelectric effect This effect exhibits the presence of spontaneous polarization of a crystal. G. Curie and P. Curie found that at mechanical compression or stretching of some crystals, the electric charges appeared on their edges. This phenomenon is referred to as a direct piezoelectric
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effect. Later, the opposite effect was also found experimentally, i.e., a reversed piezoelectric effect that consists of crystal deformation by applying an external electric field. At reversed piezoelectric effect, unlike electrostriction, the relative deformation of a crystal l/l depends linearly on the intensity of an electric field (l/l ⬃ E); consequently, if the crystal is extended in some direction of the electric field, the crystal will be compressed in the opposite field direction. In this case, the relative deformation of the crystal is by some order of values larger than at electrostriction. The piezoelectric effect is observed in noncentosymmetric crystals and is especially great in quartz, tourmaline, Rochelle salt, barium titanate, sugar, blende, and in some others. The most important at present is quartz, which has found wide application in practice, for example, in the widely known lighters. Consider the nature of the piezoelectric properties of quartz, SiO2. In this crystal the silicon atom bears a positive charge and the oxygen atom a negative charge. A freely grown quartz crystal represents a hexahedron prism topped with many-sided prisms (Figure 9.34). The axis of the prism Oz is an optical crystal axis. Directions Ox1, Ox2 and Ox3 are electric axes of the crystal; apparently all of them are equivalent. If one cuts out a quartz plate perpendicularly to the optical axis and compresses it along one of the electric axes, bounded charges will appear on the surface of the plate. On stretching, the signs of the charges change to its opposite. On compression or stretching along the optical axis Oz, a piezoelectric effect usually does not appear. Let us find the mechanism by which charges occur on crystal surfaces on deformation. The hexagonal atomic crystal structure is shown in Figure 9.35a where silicon atoms are z
x3
x1 x2
Figure 9.34 Crystal forms of SiO2 single crystal.
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x1 + 6
x1 1
+ −
+
+ x3
6
2
5
4− (a)
3
1
−
−
+
2
x2
x3
+ 5
4−
3
x2
(b)
Figure 9.35 A scheme of disposition of positive (Si) and negative (O) atoms in: (a) a free crystal; (b) a forced crystal.
designated by a “” sign and oxygen atoms are designated by a “” sign. If the crystal is compressed in the direction of one of the electric axes (Figure 9.35b) the Si1 ion will be wedged between ions O2 and O6 and the O4 ion will be wedged between Si3 and Si5 ions. Since in the absence of electric voltage, all charges compensate each other, the introduction of a positive charge of the Si1 ion creates an excessive negative charge on the edge of the crystal, and the displacement of the O4 ion creates an excessive positive charge. Hence, a negative-bounded charge appears on the top surface of the crystal and a positive charge appears on the bottom. On stretching, the signs of the charges will be opposite. The reverse piezoelectric effect can be similarly explained. On bringing a quartz plate into an electric field, it will be deformed due to displacement of the charges; the sign on deformation changes to the opposite when the direction of the field is changed. Direct and reverse piezoelectric effects have found very wide application in practice: for measuring pressure in rapidly proceeding processes, for transformation of electric vibrations in mechanical and in acoustoelectronics, etc. Ferroelectricity There are substances among dielectrics whose dielectric permeability in a narrow temperature interval is extremely high (⬃103 104) and depends nonlinearly on the electric field strength E. Such crystals are referred to as ferroelectrics; the Rochelle salt NaKC4H4O6.4H2O was the first to be discovered in the series by I.V. Kurchatov in the last century. A condenser with such dielectrics between plates has various capacitances, depending on the potential difference applied. The dependence of the polarization ≠ from the value of the electric field strength is given in Figure 9.36. Before application of an electric field, the ferroelectric is not polarized, i.e., ≠ 0 (point O). During the process of increasing the electric strength, the polarization changes according to curve ABC. If we now begin to reduce the strength E, the polarization decreases not along the previous curve CBOA, but along curve CBE. When the field strength falls to zero, the polarization appears to be ≠r.This polarization value is referred to as residual ferroelectric polarization. If the field is changed to the opposite direction and increased, it reduces polarization and at E EF the ferroelectric polarization will vanish to zero. The value of the residual field that removes ferroelectric polarization EF is referred to as a coercive force. With a further
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ᑬ B
ᑬHac ᑬr
C
E A
EF
D F
O
E
G
H
Figure 9.36 A hysteresis loop in ferroelectrics.
field increase the polarization changes along the curve FGH. If after point H, the field strength E again starts increasing in a positive direction to axis E, the polarization will change along the curve HDC. Thus, Figure 9.36 shows that the ferroelectric polarization change detains from the change of the polarizing field. This phenomenon is referred to as hysteresis and the whole closed curve is a hysteresis loop (similar to ferromagnetics). In general polarization, decreases with temperature, and ferroelectricity absolutely disappears at a temperature called the Curie point TC; the crystal transforms to a paraelectric state. The phenomena of magnetization/polarization in ferromagnetics and ferroelectrics have much in common. In the Russian literature, however, another definition is used, i.e., segnetoelectrics, in accordance with the name of the first-discovered compound of that type. Rochelle salt in Russian is called segnetova salt. The phenomenon was discovered by I.V. Kurchatov in 1932. There are many ferroelectric crystals with the common formula ABO3, natural and synthesized, on the basis of the first-discovered perovskite mineral CaTiO3 (Figure 9.37). All constituent atoms can be isomorphic substituted, both in hetero- or homovalence manner, the resulting compound possessing ferroelectric properties. In the paraelectric phase of these crystals, ions of small size (type B, black balls) are in the centers of ideal oxygen octahedrons. At temperatures lower than TC, due to specificity of interatomic interaction and features of atomic thermal vibration, some ions are displaced from their ideal positions, the structure becoming noncentrosymmetric. The ion of type B (or A) can displace along one of three symmetry axes—the second, third and fourth orders. Any of these displacements leads to the occurrence of a spontaneous dipole moment. This takes place simultaneously in a certain area of the crystal. Such areas with a certain macroscopic volume of spontaneous polarization are called domains. If the ferroelectric is not polarized as a whole, the distribution of the domains is random. When an external electric field is imposed the domains turn gradually along the field. Saturation is reached at large fields: all dipole moments of all domains are turned in one direction. Indeed, the picture is quite similar to that described in Section 5.3, which is devoted to ferromagnetism.
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Figure 9.37 A crystal type of perovskite CaTiO3.
All ferroelectrics are also piezoelectrics; therefore they have found wide application in sound and ultrasonic generators, microphones, ceramic condensers, etc. Electrets There are some dielectrics which, in the absence of an external electric field, can preserve the polarized state for a long time. These are referred to as electrets. An example of such a compound is wax, which consists of long molecules with a permanent electric dipole moment. If a small amount of wax is melted and, while it is not yet hardened, is placed in a strong electric field, the wax molecules will partly be oriented by the guidance given along the field; the wax remains in polarized state after it has hardened. Organic dielectrics possess the same property as well (naphthalene, paraffin, ebonite, mica, nylon, etc) and inorganic dielectrics (sulfur, boric glass, titanates of alkali earth materials of the perovskite type, etc.). Electrets are analogs to permanent magnets. They can be produced both as described above (thermoelectrets), or by the action of light on a photoconducting dielectric by the action of a strong electric field (photoelectrets). All electrets have a stable surface charge ,⬃ 100 C/m2. The time of an electret’s life, i.e., the time for which the surface charge will decrease in e time, lasts from several days to many years. Electrets are used as sources of a permanent electric field. They are also used in radio engineering (microphones and phones), in dose metering (electret dosimeters), in measuring techniques (electrostatic voltmeters), in computer facilities (memory elements), etc. Liquid crystals In modern techniques the so-called liquid crystals are gaining increasing importance. The liquid crystal state is characterized as being an intermediate between isotropic liquids and anisotropic solids: in some temperature interval they preserve some properties of a liquid (e.g., fluidity) and some properties of a crystal (e.g., anisotropy). The liquid crystal state is realized more often in substances whose molecules consist of a long flat atomic structure (rods) with included benzene rings (e.g., 4-met-oksi-benziliden-4-butil-aniline). Properties of liquid crystals are defined by different features of molecular packing in
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domains: they are built usually by the long axis along a single direction; whereas azimuthal orientation in different substances can behave differently. Depending on the molecular properties (e.g., the presence of dipole moments and orientation), the existence of the short- or long-range ordering in a molecule arrangement (refer to Section 9.5) and specificities of intermolecular interaction liquid crystals possess the whole scale of macroscopic properties. Practically all liquid crystals are dielectrics with strong anisotropy of electric properties, i.e., their dielectric permeability depends on direction (there are at least two value of : ⊥and 储). In an external electric field the liquid crystal molecules are oriented in such a way that the direction of the maximum dielectric permeability coincides with the external field direction. Almost all liquid crystals are diamagnetic. An exception is made with those substances whose molecules incorporate free radicals with permanent magnetic moments. Anisotropy of electric and optical properties of the liquid crystals, closely connected with anisotropy of their molecular structure, causes variety of their electrooptical properties. The external electric field can render a very strong influence on optical properties of liquid crystals, changing, for instance, their transparency and color. Therefore they have found very wide application as electronic indicator panel in microelectronics (calculators, watches and other devices). Such types of devices require very low voltage (parts of volts) and power ( W). In some cases optical properties of liquid crystals depend very strongly on temperature. Substances possessing such properties find application, for example, in industry and in medicine (for diagnostics of deviation from normal of the thermal mode of body parts).
PROBLEMS/TASKS 9.1. A crystal lattice is depicted in Figure T9.1. Find the Miller indexes of the crystallographic direction given in bold. z
z'
y
y' x x'
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9. Solid State Physics
9.2. A crystal lattice is depicted in Figure 9.5e. Find its Miller indexes. 9.3. Using the free electron model, calculate the maximum electron’s velocity in a potassium metal at T 0 K assuming the Fermi energy F 2.02 eV (see Example E9.5) and corresponding electron’s momentum pF. 9.4. Two nanoparticles with linear particle dimensions L1 10a and L2 20a (a is the lattice period) are at the same temperature T D/100, where D is the Debye temperature. Calculate the ratio of the internal energy U2/U1 for the case of the same amount of atoms (the zero energy can be neglected). 9.5. Two nanoparticles with linear particle dimensions L 10a and 20a (a is the lattice period) are at the same temperature T D/100, where D is the Debye temperature. Calculate the ratio of heat capacities C2(T)/C1(T) for the case of the same amount of atoms (the zero energy can be neglected). 9.6. A nanoparticle with linear particle dimensions L 10 a (a is the lattice period) is heated from temperature T D/100 up to T2 D/50, where D is the Debye temperature. Calculate the ratio of the internal energies U2(T2)/U1(T1) and compare the ratio with that of bulk crystal Uⴥ(T2)/Uⴥ(T1). (The zero energy can be neglected.) 9.7. Nanoparticles with linear dimensions L 10a and 20a (a is the lattice period) are heated (in average) from temperature T1 D/100 to T2 D/50 (D is the Debye temperature). Determine how much its heat capacity will change C(T2)/C(T1). Compare the result obtained with the analogous ratio Cⴥ(T2)/Cⴥ(T1) for bulk crystal. 9.8. Determine how much a nanoparticle’s heat capacity at temperature T D/100 (D is the Debye temperature) with linear dimensions L 10a (a is the lattice period) differs from that of Cⴥ of the same bulk crystal after converting to an equal number of atoms. 9.9. There are nanoparticles with linear dimensions L1 10a (a is the lattice period) and L2 20a of the same composition and temperature (T D/100). Estimate the ratio of heat transfer coefficients 1/2 of nanomaterials ( (1/3)C, CV being the heat capacity and is an averaged sound speed; assume that the free path in the nanoparticles is limited by particle dimensions L. 9.10. How many times larger is the ratio of free electrons to one metal atom at T 0 K in aluminum than in copper. The Fermi energy of these two metals is F,Al 11.7 eV and F,Cu 7.0 eV. 9.11. A nanoparticle with linear dimensions L 10a (a is the lattice period) is heated from a temperature T1 D/100 up to T2 D / 50, where D is the Debye temperature. Calculate the internal energies ratio U2(T2) / U1(T1) and compare the ratio with that of bulk crystal (the zero energy can be neglected).
ANSWERS 9.1. Indexes are [1,0,1]. 9.2. Indexes are (1,2,1). 9.3. max 兹2苶苶 F苶/苶 m 8.42 105 m/sec2 and pF 9.11 10318.42105 7.67 25 10 kg m/sec.
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9.4. (U2/U1) 26.2. 9.5. (C2/C1) 16.2. 9.6. [U2N(T2)/U1N(T1)] 405 and [Uⴥ(T2)/Uⴥ(T1)] 8. 9.7. [C(T2)/C(T1)] 130, [Cⴥ(T2)/Cⴥ(T1)] 8.
/
9.8. (Cⴥ CN) 41.5. 9.9. (1/2) 0.03. 9.10. Three times. 9.11. [U2(T2)/U1(T1)]N 405, [U2(T2)/U1(T1)]ⴥ 16.
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Appendix 1 System of Units Used in the Book
Units of physical properties measurements, based on the International System of units (SI) are described below. According to this system the basic units in mechanics are: meter (m), second (sec), kilogram (kg); an additional unit is the radian—a unit of measurement of a flat angle (rad) and that of a solid angle steradian (sr). All other units of the SI system are derived and can be obtained with the help of corresponding transformations: Area unit—square meter (area of a square with a 1 m side) [S] 1 m2 . –Volume unit—cubic meter (volume of a cube with a 1 m side) [V ] 1 m 3 . –Velocity unit—meter per second (velocity of uniform straight line motion; 1 m per 1 sec) [ y ] 1 msec. –Acceleration unit—meter per square second (acceleration when the uniform straight line motion velocity change is 1 m/sec) [ a ] 1 msec 2 . –Frequency unit—second to the power of minus one (at revolution) and hertz (Hz) (unit of oscillation frequency; the frequency when a single periodic process is accomplished in 1 sec) [] 1 sec1 . 581
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–Angular velocity unit—1 radian per second (angular velocity when a uniform rotating body turns by 1 rad in 1 sec) [] 1 radsec. –Angular acceleration unit—1 radian per 1 sec in the second power (angular acceleration when angular velocity changes by 1 rad/sec in 1 sec) [] 1 radsec 2 . –Force unit—newton (N) (the amount of force required to give a 1 kg mass body an acceleration of 1 m/sec2) [ F ] 1 N 1 kg msec 2 . –Density unit—kilogram per cubic meter (the density of a uniform substance whose mass per 1 m3 is equal to 1 kg) [] 1 kgm 3 . –Pressure unit—pascal (Pa) (pressure produced by a force of 1 N acting on an area of 1 m2) [ P ] 1 Pa 1 Nm 2 . –Momentum unit—kg m/sec (a body of mass 1 kg moving translational with a velocity of 1 m/sec) [ p] 1 kg msec. –Force impulse—newton second (a force impulse produced by a force of 1 N for 1 sec) [ F t ] 1 N sec. –Work (energy) unit—joule (J) (the amount of work done when an applied force of 1 N moves in the direction of the force through a distance of 1 m) [ A] 1 J 1 N m. –Power unit—watt (W) (a watt is used to measure power or the rate of doing work; 1 W is a power of 1 J per second). [W ] 1 W(Jsec).
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–Torque unit—newton meter (moment of force produced by a force of 1 N relative to a point which is at distance of 1 m away from the force action line) [ M ] 1 N m. –Moment of inertia unit—kilogram square meter (moment of inertia of a material point 1 kg in mass relative to a rotation axis 1 m away) [ I ] 1 kg m 2 . –Angular momentum unit—kilogram square meter per second (angular momentum of a body with a moment of inertia in 1 kg m2 rotated with angular velocity 1 rad/sec) [ L ] 1 kg m 2 sec. In addition to the units presented above in molecular physics following units are also used. Unit of heat energy or heat—calorie; 1 cal 4.1868 J [Q] 1 cal. Heat capacity unit—calorie per kelvin (amount of heat to warm a body by 1 K) [C ] 1 calK. –Specific heat capacity unit—calorie per kilogram kelvin (amount of heat to warm 1 kg of a substance by 1 K) [Csp ] 1 calkg K. Mole heat capacity unit—calorie per mole kelvin (amount of heat to warm one mole of substance by 1 K) [Cmole ] 1 calmole K. In addition to the mechanical units of measurements, in the sections describing electricity and magnetism, one basic unit—the ampere (A) and a number of derivative units are used. In the SI system 1 A is defined as the force of that current which produces a specific force between two parallel infinitely long conductors which are 1 m apart, in 2 107 H/m.
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System of Units Used in the Book
Charge unit—coulomb (C) (the charge that runs through a cross-section of a conductor when a current of 1 ampere is flowing) [q ] 1 A 1 sec (Asec) (then an electric current I in 1 A corresponds to 1 coulomb transfer in 1 sec) [I ]
1C 1 sec
(A).
–Current density unit—ampere per square meter (electric current of 1 A per 1 m2 of crosssection of a conductor) [ j]
1A 1 m2
(Am 2 ).
–Electric field strength unit—volt/meter (electric field strength, acting on a point charge 1 C with a force 1 N) [E]
1N 1V 1C 1m
(kg mA sec 3 ).
–Electric displacement (induction) unit—coulomb/meter2 (electric field strength multiplied by 0) [ D] 0 E (A secm 2 ). –Electric field potential unit—volt (V) (electric field potential in which the charge of 1 C possesses potential energy 1 J) [] 1 J1 C (kg m 2 A sec 3 ). –Electric dipole moment unit—coulomb meter (dipole electric moment of a pair of opposite charges equal in value and being 1 m apart) [ p] 1 C 1 m(A sec m). –Electric quadrupole moment—coulomb meter2 (quadrupole electric moment of a system of two pairs of opposite charge equal in value of 1 C displaced alternately in square corners at side length 1 m) [Q ]1 C1 m 2 (A sec m 2 ).
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585
Eelectric linear density unit—coulomb/meter (charge of 1 C uniformly distributed along a line of 1 m) [ ] 1 C1 m (A secm). Electric surface charge density unit—coulomb/meter2 (charge of 1 C, uniformly distributed over an area of 1 m2) [ ]
1C 1 m2
(A secm 2 ).
Electric volume charge density unit—coulomb/meter3 (charge, uniformly distributed in a volume of 1 m3) [ ]
1C 1 m3
(A secm 3 ).
–Dielectric polarization unit—coulomb/meter2 (a dielectric’s volumetric dipole moment) [ ¬]
1C 1 m2
(A sec m 2 ).
Dielectric susceptibility unit—dimensionless (polarization of isotropic dielectric in a unit field strength divided by 0) [ ]. Dielectric permeability unit—dimensionless (a value indicating by how much an averaged macroscopic field in a dielectric is less than an external field) [ ]
E0 . E
Polarization of a molecule unit—meter3 (a molecular dipole moment in a field of a unit strength divided by 0) [ ]
p 0 E
(m 3 ).
Electronic capacitance unit—farad (F) (capacitance of conductor, which is charged to potential 1 V receiving a charge of 1 C) [C ]
1C 1V
(A 2 sec 4 (kg m 2 ));
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System of Units Used in the Book
–Magnetic moment unit—ampere meter2 (electric current of 1 A flowing around an area of 1 m2) [M] 1 A 1 m 2 (A m 2 ). –Off-system unit—Bohr magneton (1 = 0.927 10–23 A m2). Magnetic field induction unit—tesla (T) (maximal magnetic force moment, acting on a unit magnetic moment) [ B]
1N 1 A m2
(kg (A sec 2 )).
Strength of magnetic field unit—ampere/meter (magnetic field induction, divided by 0) [H]
1A m
(Am).
Magnetization unit—amper/meter (moment of an unit volume moment) [ (]
1A (Am). 1m
Magnetic susceptibility unit—dimensionless (magnetization of an unit volume of a magnetic in an unit strength field) [ ]. –Specific magnetic susceptibility unit—meter3/kilogram (magnetization of a unit mass of a magnetic in a field of unit strength) [ sp ]
(m 3 kg).
Mole magnetic susceptibility unit—meter3/mole (magnetization of one mole of magnetics in a field of unit strength) [ M ]
M
(m 3 mole).
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587
Magnetic permeability unit—dimensionless (shows how many times greater is the magnetic field than an external magnetic field) []
B . B0
Magnetic flux unit—weber (Wb) (a magnetic field induction flux in 1 T through the surface of unit area) [ ] 1 T 1 m 2 1 Wb; (kg m 2 (A sec 2 )). Inductance unit—henry (H) (inductance of a conductor in which at a current of 1 A appears a total magnetic leakage of 1 Wb) [ L ] 1 Wb1 A; (kg m 2 (A 2 sec 2 )). Some important physical constants: Acceleration of free falling Gravitational constant Avogadro constant Mole gas constant Molar volume at normal conditions Boltzmann constant Elementar charge Electron mass Specific electron charge Light velocity in vacuum Stefan–Boltzmann constant Win shift constant Planck’s constant Rydberg constant First Bohr orbit radius Compton wavelength Bohr magneton Ionization energy of hydrogen atom Atomic unit of mass Nuclear magneton Electric constant Magnetic constant
g 9.81 m/sec2 G 6.67 1011 m3/(kg/sec2) NA 6.02 1023 mol1 R 8.31 J/(K mol) Vm 22.4 103 m3/mol
1.38 1023 J/K ⱍeⱍ 1.60 1019 C me 9.11 1031 kg (e/m) 1.76 1011 C/kg c 3.00 108 m/sec 5.67 108 W/m2 K4 C 2.90 103 m K h 6.63 1034 J sec; h/2 1.051034 J sec R 1.097 107 m1 a 5.29 1011 m C 2.43 1012 m B 9.27 1024 J/T Ei 2.16 1018; J 13.56 eV 1 a.u.m 1.66 1027 kg N 5.05 1027 J/T 0 0.885 1011 F/m 0 1.26 106 H/m
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Appendix 2 Gyroscope Precession in a Gravity Field
A symmetric body with a single motionless point and able to rotate at high angular velocity around an axis z⬘ passing through this motionless point is called a gyroscope. There are two types: balanced (the motionless point coincides with the center of inertia) and unbalanced (where this condition is not fulfilled). A child’s spinning top is a primitive example of an unbalanced gyroscope. Figure A2.1 shows an unbalanced gyroscope acquiring a rotation (precession) in a gravitational field. The pivot point O is a unique motionless point and the axis of rotation z⬘ passes through it. The gravity force is directed vertically downwards along axis z. The angle between axes z and z⬘ is denoted by and is assumed to be small. In the figure, for simplicity, the angular momentum vector L terminates in the center of mass C, the distance OC being lC. According to the basic equation of rotational motion dynamics, we can write dL ⫽ Mdt (refer to (1.3.57)). The force momentum (torque) of the gravitational force relative to point C is M = mglC sin. z z' d
g =
L
d dt
C
dL
mg O
The rotation of the gyroscope’s axis z⬘ relative to the vertical axis z is referred to as gyroscope precession. Under the action of the gravity force momentum the vector of the angular momentum L of the unbalanced gyroscope obtains an increment dL directed along 589
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Gyroscope Precession in a Gravity Field
the vector M (see (1.3.57) and Figure 1.19). Since dL is perpendicular to L the modulus of L is constant and only precession takes place. The gyroscope precession angular velocity g , as seen in Figure A 2.1, can be found as g ⫽
d mgᐉ C sin mgᐉ C ⫽ ⫽ . dt I z⬘ sin I z⬘
(A2.1)
Note that the angle is small and I and regarding both axes, z and z⬘, are approximately same. It can be seen that the precession angular velocity g is higher, the lower the gyroscope’s moment of inertia I and angular velocity z⬘; g does not depend on the angle between axis z and z⬘. Anyone can conduct an experiment using a child’s spinning top.
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Appendix 3 An Electrostatic Field of an Arbitrary Distributed Charge
Among the real problems the chemist can come across in practice, a simple situation with a discrete set of point charges is rarely seen. Any molecule consists of positively charged nuclei encircled by negative electrons, each particle being vibrated around positions of equilibrium. Therefore, the overall charge distribution is described in this case by the distribution function (r): dq ⫽ (r )dV .
(A3.1)
The charge density distribution (r) is of great importance because it permits the calculation of a wide number of molecular and crystal properties and enable us to follow the paths of chemical reactions. Consider a field created by the electric charge system described by the function (r) (refer to Figure A3.1). Our task is to calculate the electrostatic field created by this system in a certain point A. Direct an axis z of the Cartesian coordinate system in such a way that
z A
r
R
dV θ r' y x
591
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An Electrostatic Field of an Arbitrary Distributed Charge
it crosses point A. The electrostatic potential in point A is the superposition of contributions from all elements dq. (r⬘)dV ⬘ (r⬘)dV ⬘ ⫽∫ , R 冷 r ⫺ r⬘ 冨 V V
(r ) ⫽ ∫
(A3.2)
where r is the z component of the radius-vector of point A, r⬘ is the argument of the function (r⬘), ⱍRⱍ = ⱍr–r⬘ⱍ is the distance from the element dV to point A. The integration is over the coordinate r⬘ over the whole charge containing space. Denoting the angle between vectors r and r⬘ as and using the cosine theorem; we obtain R ⫽ (r2 ⫹ r⬘2 ⫺ 2rr⬘ cos)1/2. Then the integral can be rewritten as A ⫽ 冕(r⬘)dV⬘(r2⫹r⬘2 ⫺2rr⬘ cos)⫺1/2. If we calculate the field far from the origin (i.e., r⬘^r) the expression ⫺ 21
2 ⎞⎤ r⬘ 1 1 1 ⎡ ⎛ r⬘ ⫽ 2 ⫽ ⫹ ⫺ 2 cos ⎟ ⎥ 1 ⎢ ⎜ 2 1Ⲑ 2 2 R (r ⫹r⬘ ⫺2rr⬘ cos ) r ⎣⎢ ⎝ r r ⎠ ⎥⎦
can be decomposed into a series and can be expanded over the r⬘ orders
1 3 (1⫹ )⫺1Ⲑ 2 ⫽ 1⫺ ⫹ 2 ⫹ , 2 8 where ⎡ ⎛ r⬘ ⎞ 2 r⬘ ⎤ ⫽ ⎢⎜ ⎟ ⫺2 cos ⎥ . ⎝ ⎠ r ⎢⎣ r ⎥⎦ Summing up all the terms with the same order of r⬘/r and neglecting the terms of higher orders than quadratic, we obtain the expression 2 ⎤ 1 1 ⎡ r⬘ 1 ⫽ ⎢1⫹ cos ⫹ ⎛⎜ r⬘⎞⎟ ⫻ (3 cos2 ⫺1) ⎥ . ⎠ ⎝ R r⎣ r 2 r ⎦
Introducing this expression into eq. (A3.2) and taking into account that integration is accomplished over r⬘, we can obtain for A the sum 1 1 (r⬘)dV ⬘⫹ 2 ∫ r⬘ cos(r⬘)dV ⬘ ∫ r r 1 1 2 ⫹ 3 ∫ r⬘ ⫻ (3 cos2 ⫺1)dV ⬘. 2 r
A ⫽
(A3.3)
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593
The magnitude of each of these integrals depends only on the properties of the electron density function (r⬘). Being calculated for the given system they can be expressed as numbers k0, k1 and k2 correspondingly. The dependence of A on ⱍrⱍ will be expressed by the sum
A =
k0 k1 k2 ⫹ ⫹ . r r2 r3
(A3.4)
The kn are referred to as the electric moments of the system. Let us analyze each term of this sum. The k0 value is expressed by an integral
k0 ⫽ ∫ (r⬘)dV ⬘.
(A3.5)
and is the total charge of the system concentrated in origin. It is referred to as a monopole moment or simply a monopole. For a neutral system k0 ⫽ 0. The coefficients k1 and k2, unlike k0, depend on charge distribution. The coefficient k1 characterizes an electric dipole moment
k1 ⫽ ∫ r⬘ cos (r⬘)dV ⬘.
(A3.6)
Since the value r⬘cos is z-coordinate of element dV⬘, this term characterizes the relative displacements of the positive and negative charges (r⬘)dV⬘ along this axis. Indeed, if one imagines a system consisting of two dissimilar charges q in points (0,0,z) and (0,0,⫺z) with z ⫽ 1/(2l), then a value r⬘cos⫽⫾(1\2)ᐉ can be factorized from the integral. The resultant expression 冕 (r⬘)dV⬘ will be equal to q and the whole coefficient k1, which is now equal to lq ⫽ p, composes the electric dipole moment oriented along the z-axis (see Section 4.1.5 and eq. (4.1.29)). The coefficient k2 1 2
k2 ⫽ ∫ r⬘2 (r⬘) (3 cos2 ⫺1)dV ⬘.
(A3.7)
is a so-called quadrupole moment. Try to understand what electron density distribution is described by such a factor. For spherically symmetric electron distribution k2 ⫽ 0. It follows from a specific type of k2 factor: keeping in mind that r⬘2 ⫽ x2 ⫹ y2 ⫹ z2 for the specified symmetry all three coordinates are equivalent, therefore x2 ⫹ y2 ⫹ z2 ⫽ 3z2 and, consequently, 3z2 ⫺ r⬘2 ⫽ 0. A flattened out electron density model is the charge q rotating around an z-axis at a level z ⫽ 0 at a distance r0 from the axis. Then ⫽ /2 and the expression in brackets becomes negative. Since r ⫽ const., then k2 is equal to ⫺r 02 q for a positive charge and ⫹r 02 q for a negative charge. It is reasonable to assume that as for
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every “flattened out” distribution the quadrupole moment has such signs. It is easy to show that for the “extended” model distribution the signs will be the opposite. Expression (A3.4) shows that in the electrostatic field created by a particular system, the electric potential falls differently with distance (refer to Table 8.1) the higher the order of the moment, the sharper the potential falling down. Even neutral systems (atoms, molecules) create an electric field by means of which these systems interact with each other. Accordingly, the higher the order of the moment, the lower the energy of interaction; for example, dipole–dipole interaction is appreciably weaker than the interaction of monopoles (Coulomb interaction). All this information is useful in Chapter 8.
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Appendix 4 Langevin Theorem
Consider a system consisting of identical molecules weakly interacting with each other, characterized by a magnetic dipole moment M. An external magnetic field B acts on the system trying to orient all magnetic moments along z axis contrary to chaotic thermal motion. Allocate in this system a spherical volume V of a radius R. Suppose that there are a large enough number of molecules in this volume with all possible orientations of magnetic moments. Among all N molecules in the volume V we shall denote as dN() such molecules whose magnetic moments form an angle from up to d with a direction of vector B (Figure A4.1).
d R sinθ B z Rd
These dN() molecules account for a physically infinitesimal elementary volume dV(). Then the concentration of such molecules is n
dN . dV
(A4.1)
An equilibrium distribution of noninteracting particles in an internal force field is described by the Boltzmann formula n C exp(U/T ), where C is a normalizing coefficient. Since the potential energy of the dipoles in the external field is determined by a ratio (eq. (5.1.31)) (U MBcos), the equilibrium distribution of dipoles upon potential energies (i.e., on various orientations of the magnetic moments) in field B can be written as: n C exp
冸 T cos 冹 . MB
595
(A4.2)
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Langevin Theorem
From the last two equations we can derive ⎛
⎞
dN () C exp ⎜⎝ M B cos ⎟⎠ dV (). T
(A4.3)
Express an elementary volume dV(), occupied by a molecule the magnetic moment of which is directed at an angle , through the angles and d. Since dV() (V/4)d, where d is an elementary solid angle obtained by two coaxial cones with a common vertex at point 0 and openings of 2 and 2 d, then from the solid angle definition (4.1.10) and Figure A4.1, it follows that
d
dS 2( R sin )( R d) 2 sin d. R2 R2
Then dV() (1/2) V sind and (A4.3) could be rewritten as 1 ⎞ ⎛ MB dN CV exp ⎜ cos ⎟ sin d. ⎠ ⎝ T 2
(A4.4)
The constant C can be determined from the normalizing procedure: the integral from dN() over all possible orientation of angles (from 0 to ) must equal the total number of molecules N in volume V, that is:
∫ dN () N . 0
(A4.5)
To simplify further calculations mark (MB/T) as a and cos as x; then sin d dx. This change brings about the change in limits in (A4.5): instead of the inferior limit there will be 1 (cos 1), and the superior limit will be 1 (cos 1). After such transformation and change of a sign before an integral one arrives at 1
1 CV ∫ e ax dx N . 2 1 After these transformations we obtain
1 e a ea CV N, 2 a
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597
hence
C
2 aN (e a ea )V
2a n, (e a ea )
or C
where n N/V is the total molecules concentration. Then dN() can be rewritten as
dN()
a nVe a cos sin d. e ea a
(A4.6)
These dN() dipoles make a contribution of d(() to the general magnetization (. Taking into account the magnetization definition (5.2.3) d(() we can obtain
d (()
M cos dN ()
V
.
Changing in this expression dN() according to (A4.4), we can obtain
d (()
anM a cos e cos sin d. e ea a
The total magnetization can be found by integration 1
(
anM xe ax a e ea ∫1
dx.
Integration by parts (u x, dv eax dx) can give 1
e ax x e dx x ∫ a 1
+1
1
ax
-1
1 e ax dx. a ∫1
After this integration and substituting the limits: 1
∫
1
xeax dx
e a ea e a ea . a a2
(A4.7)
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Langevin Theorem
Magnetization ( can now be presented as
(
anM ⎛ e a ea e a ea ⎞ ⎟ e ea ⎜⎝ a a2 ⎠ a
⎛ e a ea 1 ⎞ or ( nM⎜ a . ⎝ e ea a ⎟⎠
The expression in brackets is referred to as the Langevin function L(a). Thus L
e a ea 1 . e a ea a
(A4.8)
Using the function L(a) we can finally write the magnetization ( as ( nM L (a )
(A4.9)
At limiting values of a (MB/T ), the L(a) function can be decomposed into the MacLoren series. At small a values we can have 1 1 2 5 L (a) a a 3 a . 3 45 945 This series is alternating-signed and therefore it diminishes rather quickly. If we limit ourselves to the first term, then L(a) (1/3)a. The expression already given can be obtained
(
nM2 B. 3T
The paramagnetic molar magnetic susceptibility at (MB/T ) ^ 1 is obtained M
0 N A M2 . 3T
(A4.10)
This expression coincides with eq. (5.2.23) derived from the very simple suppositions. The volumetric and specific susceptibilities can be calculated according to the formulas given above (refer to Section 5.2.2).
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Appendix 5 Maxwell Equations in Differential Form: Electromagnetic Wave Propagation in Vacuum
Vector algebra provides us with a good opportunity to write Maxwell equations in differential form, i.e., to characterize an electromagnetic field in a point. It allows us to see most clearly the physical sense of the equations and their importance for understanding the laws of electrodynamics. Let us start with the equation of a Gauss law, which in the integral form looks like: 养 DdS ∫ (r )dV. S
V
(A5.1)
Remember that in differential form, the divergence of a vector D in a point r is the limit to which the left-hand side of this equation tends under a contraction S (and V) to a point:
lim
V → 0
1 养 DdS div D(r ). V
(A5.2)
The symbol div means the sum of first particular derivatives. Therefore, ⎛ ⎞ div D(r ) ⎜ ⎟ D( xyz) D(r ). ⎝ x y z ⎠
Here an operator is introduced, well known in mathematics ⎛⎞ ⎛ ⎞ ⎛⎞ i ⎜ ⎟ j⎜ ⎟ k ⎜ ⎟ . ⎝ x⎠ ⎝ z ⎠ ⎝ y ⎠
Correspondingly, divD(r) is the scalar product of an operator and a vector D(r). In fact, the divergence is the flow of D vector “outflows” from a point r. Integrating the divergence 599
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over the whole volume V, we arrive at the total power of the source, i.e., the flow through the closed surface S comprising the volume V.
∫ div D(r )dV 养S DdS.
V
This equation is referred in mathematics as the Ostrogradski theorem, from which an important expression originates: div D(r ) (r ).
(A5.3)
The divergence of a vector of an electric displacement in a point r (x, y, z) is equal to the density of an electric charge (that is the source power of the electrostatic field) in this point. This is the Causs theorem in differential form. It follows that the force lines of an electrostatic field proceed from a positive electric charge (a source), come to an end in a negative electric point charge (a drain) or go to infinity; at the drain divergency has a negative sign. Previous consideration (see Chapter 5) shows that there are no magnetic charges in nature, therefore one can write div B(r ) 0,
(A.5.4)
This expression is also a Maxwell equation. Hence, the magnetic field force lines are closed. The next Maxwellian equation is the law of electromagnetic induction. This equation describes the nature of producing the electric field E by variation of the magnetic field induction B ⎛ B ⎞
∫ E dl∫ ⎜⎝ t ⎟⎠ n dS. L
S
(A5.5)
Note that a rotor of vector E in a point r is the limit of the ratio of the electric field circulation E over the closed contour L, comprises an area S, to the area S while aspiring tightening contour L (and area S) to zero (see Figure 5.11), that is lim
S0
1 ∫ Edl div E(r). S L
(A5.6)
Integrating rot E upon surface S, we obtain circulation of vector E along a contour that comprises this area 养 Ed ᐉ ∫ rot EdS. L
S
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In mathematics this equation is referred to as Stokes’ theorem. Comparing this expression with (A5.5), rot E (r )
B(r ) B(r ), t
(A5.7)
that is, the rotor of the electric field strength in point r is equal to the time derivative from the magnetic induction in the same point. This implies that the induction electric field is a curling (vortical) field in contrast to the electrostatic potential field. Using the previous notions, we can write instead of rot E the vector product and E, that is (r). E] [ In the same way the next Maxwellian equation can be derived which connects the circulation of magnetic field strength and currents. It has the form rot H(r) D (r) jcond(r). There are no electric currents in vacuum ( jcond 0), therefore, the equation simplifies to rot H(r ) D (r )
(A5.8)
This means that the source of the magnetic field in the point r is the time changeable electric field (in the same point r). It is expedient to put all equations analyzed above together. div D(r ) (r ), div B(r ) 0
rot E(r ) B (r ) rot H(r ) D (r )
(A5.9)
To these equations it is expedient to add two, which connect the strength of both fields in vacuum and fields in an isotropic medium B 0 H,
D 0 E.
(A5.10)
The last equations are equitable only for isotropic media. In anisotropic media they have a tensor character. The last equation in the Maxwellian system is the relation between the strength of an electric field in a point r with the current density in the same point (Ohm’s law in differential form) j E.
(A5.11)
In these equations all electrodynamics is described! Try to estimate an electromagnetic wave’s propagation speed based on the Maxwellian treatment of electrodynamics. As usual, let us make the task simpler, i.e., we shall analyze a
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certain physical model and look to see how far it corresponds with accepted representations. In this case our problem is the definition of the speed of propagation of an electromagnetic wave in vacuum. The Maxwell equation can be written in the form: ⎛ ⎞ 养 Eᐉ d ᐉ ⎜ B ⎟ , ⎝ t ⎠ L
(A5.12)
⎛ E ⎞ ⎛ D ⎞ ⎛ E ⎞ 0 ∫ ⎜ ⎟ dS 0 0 ∫ ⎜ ⎟ dS. ⎟ ⎝ ⎠ ⎝ t ⎠ n ⎠ t t S s
∫ Bᐉ dᐉ ⎜⎝ L
(A5.13)
Let us imaging an electromagnetic wave as successive steps of “constant” electric and magnetic fields (with vectors E and B fixed in their magnitudes, perpendicular to each other), running along an x-axis with planes of vectors oscillation E in x0y and B in x0z (see Figure A5.1(a)) with a “wave” front motion speed, which we should define. Consider that in some instances, the front of the “wave” reaches line 1–2. Allocate to planes x0y an imaginary rectangular contour defg and estimate the vector B flux through an area limited by the contour (Section 5.1.6, eq. (5.1.38)). In time dt the line 1–2 displaces to positions 4–3. The area A limited by the specified contour is A = c.t h, where h is the length of segment 1–2 and ct is the distance run by the front of the “wave” in time t. As the vector B everywhere is perpendicular to the plane x0y, d B ⎛ dA ⎞ Bhc dt Bhc B⎜ ⎟ ⎝ dt ⎠ dt dt 2 y
c
4
E
(A5.14)
c∆t e
d
2 E
∆A 1
c
h
x
d g
e
f
3
B
w g
x
(a) z
4
c∆t
(b)
B
1
3
f ∆A
In contrast, the circulation of the E vector along the contour 1–2–4–3, i.e., the left-hand side of eqs. (5.1.38, 5.4.6, 5.4.7 and A5.12), is equal to Eh. Therefore
养 Eᐉ d ᐉ Eh, L
(A5.15)
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since E = 0 on the segments 2–4, 4–3 and 3–1 (the wavefront has not yet reached them). From a comparison of eqs. (A5.14), (A5.15) and (A5.12), we can derive the ratio between E and B: (A5.16)
E cB.
Continuing to consider the same model, look what occurs in the plane x0z (Figure A5.1(b)). Again we shall allocate a rectangular contour (into planes x0z) and we shall make the same calculations as in the previous case. Using eq. (5.1.2), circulation of the vector B (A5.13) gives Bw (w is the length of segment 2–4); the vector E flux through area A (that is E) gives 00wct. Time derivation and successive cancellation on w gives B 0 0 Ec.
(A5.17)
If we substitute in this equation the E value from (A5.17) and canceling by B we obtain 00c2 = 1. Whence
c
1 00
.
(A5.18)
Three basic constants of electrodynamics appear connected with each other. Substituting values 0 and 0, we obtain for light speed in vacuum c 2.998 108 km/sec. The agreement of this value with that obtained experimentally was a triumph of Maxwell’ theory. A similar result can be obtained with a more exact model. It is also not so difficult to show that propagation of electromagnetic waves in a medium with dielectric susceptibility and magnetic susceptibility occurs with the speed
y
1 0 0
c
(A5.19)
Since the refraction index is the ratio of the light propagation in a medium to that in a vacuum n
1
(A5.20)
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This glossary is intended to free the main text from multiple repetition of the explanation of the notation used. As a rule, an explanation is given on its first occurrence in the text and occasionally elsewhere. Since the vocabulary of physics is very broad, as well as the whole range of Greek letters, the same roman letters have been used several times, partly in different fonts. Vector values are given in bold. Roman letters: A a, a a, b, c B C Csp CM D D d, dhkl d苹 E e, ⏐e⏐ e e E EF F f(υ) f(ε) g g H h h,k,l I Iz i i i, j, k
force work acceleration lattice periods vector of magnetic field induction heat capacity of a body, system specific heat capacity molar heat capacity vector of electric field displacement (induction) diffusion coefficient crystallographic interplanar spacing molecule’s effective diameter vector of electric field strength absolute value of the electron charge (elementary charge) logarithmic natural base (exp) thermal efficiency energy, total mechanical energy Fermi energy vector of force Maxwell molecular velocity distribution molecule kinetic energy distribution free fall acceleration vector gyromagnetic ratio, Lande factor vector of magnetic field strength Planck’s constant Miller indexes electric current, moment of inertia (MI), intensity moment of inertia relative to z-axis number of degrees of freedom imaginary unit unit vectors (orts) of Cartesian coordinate 605
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j k £
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vector of energy density wavevector (|k|⫽k⫽2/) kinetic energy, degrees of absolute thermodynamic temperature, performance factor Boltzmann constant L angular momentum’s vector Lz angular momentum relative to axis z l, ᐍ length, distance M molar mass M vector of force moment (torque) Mz force moment relative to axis z M vector of the magnetic dipole moment, electron atomic orbit or spin moment, nuclear magnetic moment m mass of a body, atom, molecule, total system’s mass n concentration n a unit vector of a normal NA Avogadro’s number a molar polarization p vector of the electric dipole moment p vector of momentum, pressure P power Q heat, activation energy Q, q electric charges R(r) a radial part of atomic wavefunction R molar refraction, heat emittance r(x, y, z) Cartesian radius vector r(r, , ) radius vector in a spherical coordinate system S area S entropy s wave polarization index T absolute thermodynamic temperature t, time T period U potential energy U internal energy of a molecular system V volume VM molar volume 〈〉 average speed of a particle rms ⫽ 兹冓苶2苶冔 root mean square of a particle’s speed prob most probable value of a particle’s speed velocity vector W thermodynamic probability Y(,)⫽ ()() angular part of the atomic wave function Z statistical sum
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607
Greek letters:
⫽/c
〈〉 0 〈osc〉 〈rot〉 o B N (x,y,z)
angle, molecular polarizability angle, a force constant the relative speed coefficient in special relativistic theory coefficient of anharmonicity; adiabatic index attenuation coefficient dielectric susceptibility, a micro-object energy average kinetic molecular energy angular acceleration vector, thermal process efficiency electric constant average energy of atomic oscillations in molecules average energy of molecule rotation coefficient of dynamical viscosity, coefficient of kinematic viscosity dielectric permeability, coefficient of molecular heat transfer wavelength, logarithmic attenuation decrement, mean free path length magnetic permeability, reduced molecular mass magnetic constant Bohr magneton nuclear magneton coefficient of kinematic viscosity, frequency ( ⫽ n speed of rotation) displacement from equilibrium position density of matter, electron density surface charge density unit vector of a tangent system’s time of life, relaxation time, linear charge density angular velocity, = 2 time-independent wave function magnetic flux linkage/time-dependant wave fuction flux angle, electric field potential magnetic susceptibility time-dependent wave function solid angle
Others fonts: ( ℜ ℘
magnetization polarizability (polarization vector) surface density of electrical current
Quantum numbers: n, ᐉ, mᐉ, s, ms quantum numbers of one-electron atom L, mL, J, mJ , S, ms quantum numbers of multielectron atom I, mI nuclear quantum numbers s, p, d ... one-electron states S, P, D ... many-electron atoms states
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v j
a molecule vibrational quantum number a molecule rotational quantum number
Abbreviations: MP CM MI IRB IBB SF
material point center of mass moment of inertia ideal rigid body ideal black body superfine
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Index A A, see Ampere Absolute zero temperature 177 Absorption spectrum 492 Acceleration angular 12, 14, 40 average 4 center of mass 12 centripetal 25, 183 constant 9 due to gravity 20, 30, 587 in electric field in simple harmonic motion 108 instant 4 radial (normal) 7 tangential 5, 7, 14 Activation energy 186, 194, 446 Adiabatic process adiabatic index 199 with ideal gas 199 Alpha particle 325 Alternating current 305 Ampere (unit) definition 583 Ampere’s law 318 Ampere–Maxwell law 351 Amplitude 107, 112 Angle of incidence 363 of polarization 390 of reflection 363 of refraction 363 Angular acceleration 13, 14 displacement 12–14 force moment 51 torque 51, 52 frequency 107, 157 momentum 40, 41, 48–50 conservation of 71
orbital 332 spin 460, 498 vector form velocity 15 velocity 13, 14, 41 Antinode 158, 159 Approximations 76, 439 Archimedes’s buoyant force 127 Area, units of 581 Atmosphere (unit) 171 Atmosphere of earth 182, 406 Atom 332 Atomic mass unit 498 Atomic number 460 Avogadro constant 297, 339, 587 B B, see Magnetic field induction Barometric height distribution function 181–182 Beatings 117 Binding energy 512 Biot–Savart law 311 Birefringence 391 Bloch function 539 Bohr atom model 416–419 Bohr magneton 454, 586 Boltzmann constant 176, 219, 587 Boltzmann distribution at different temperatures 179, 181 Boltzmann factor 185–186, 551 Boson 542 Bragg’s law 386 Brewster’s law 388–389 C C, see coulomb Conservation law of charge 251 Capacitor displacement current in 254 parallel-plate 253, 254 609
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Carnot cycle 207–210 engine 211 microcycles 211 Celsius temperature 177 Center of mass acceleration of 12 of two particles 36 rigid body 47 symmetry and 37 Centripetal acceleration 35, 183, 185 Centripetal force 184, 185 Charge interaction force 75 unit 584 Chemical potential 542 Chemical shift 504–506 Chemical shift in ΓRS 514 Chemical shift in NMR 520 Circuit, electrical 307, 308 Critical angle total internal reflection 364 Clausius inequality entropy change 215 in nonequilibrium processes 214–216 Coherence 170 Collision center of mass 80 conservation of 79 conservation of energy in 81 elastic 80 head to head 80, 81 inelastic 88 momentum in 81 one-dimensional 81 Complex variable 110, 111 Conductivity electrical 280, 352 thermal 233, 236 Configurational space 541 Conservation energy law in thermodynamics 196 Conservation laws in collisions 79 of angular momentum 71 of energy 74 of linear momentum 69 of mechanical energy 67 Conservative force 60
Index
Constant-pressure process 198 Constant-volume process 198 Convection 181, 237 Cooper’s pairs 543 Coordinates, space-time 39 Coordination number 567 sphere 567 Coulomb 584 Coulomb’s law 252 and Gauss’ law 259–263 Crystal class 533 Crystal lattice 531–536 periods 556 Crystal structure 531–536 Crystallographic direction 535 Crystallographic plane 533–535 interplanar spacing 535 Curie point 345, 575 Curie temperature 346, 347, 575 Current alternating 305 charge carrier of 251, 280 displacement 350–354 Current density 306, 307 Current drift speed 306 Current loop as magnetic dipole 319 as magnetic moment 348 Cycles direct 205 reversed 205 Cyclic process nonreversible 214–216 reversible, as Carnot cycles 206, 207 D Damped harmonic motion energy 131–133 Damped oscillations 133–138 Debye unit 276 Deceleration 85, 473 Degree of freedom 106 Density electron density 285, 287, 288 linear 251, 263 surface 251, 262, 270, 283 volume, 251
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Deuteron 86, 527 Diamagnetics 331, 337 Dielectric hysteresis loop 575 Dielectrics 280–282 Diffraction by grating 381–383 by single slit 379–381 Fraunhofer 379, 380 Huygens–Fresnel principle 378–379 of electrons 424–426 of neutrons 424–426 types of 379 X-ray 385–386 Diffraction grating 381–383 Diffusion coefficient D 235 Fick’s law 235 Dipole, electric, see Electric dipole Dipole, magnetic, see Magnetic dipole Dipole–dipole interaction 280 Dispersion of grating 383, 384 of light 395–398 Dispersion curves 545–550 acoustic branch 546, 548 optic branch 546, 548 Displacement of simple harmonic motion 108 Displacement current in capacitor 351 Distribution function calculation average value 172–174 Domains, ferroelectric 575 Domains, ferromagnetic 347–349 Doppler effect acoustic 154–156 for light 154 Drift speed of currier 306 E E, see Electric field Earth mass 62 Earth–moon system 99 Efficiency of engine 207, 209, 210 Elastic collision 80, 412 Electrets 576 Electric charge 251, 273, 305
611
Electric current 305–309 Electric charge in dielectrics 281 Electric dipole moment 276 potential energy 278–279 torque on 278 Electric displacement 284, 584 Electric field equipotential surfaces 275 flux of 261 induced 353 induced magnetic field 353 lines of force 253, 254 of electric quadrupole 501 of electromagnetic wave 33 of finite rod 256–257 of infinite cylinder 270 of infinite line 256–257 of infinite plate 263–264 of point charge 253, 254 of ring 258–259 of semi-infinite rod 300–301 of spherically symmetric charge 252, 267, 275 potential 29 Electric field potential capacitor 276 dipole 276–288 distribution of charge 251 of system of charges 274 point charge 274 ring 259 Electric potential superposition principle 274 Electric quadrupole moment 501–502 Electric strength flux 261 Electromagnetic force 29, 31 Electromagnetic induction 328–331 Electromagnetic radiation 353 Electromagnetic spectrum 354 Electromagnetic waves amplitudes of fields 353 energy density in 353 from dipole 353 Poynting vector 353 scale 354 sources of 351 speed of 353
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Index
wave equations 352–353 See also Light Electromotive force 308 Electron charge to mass ratio 325 magnetic dipole moment 332 orbital dipole moment 332 spin angular momentum 332 Electrons as quasi-particles 540 Electron paramagnetic resonance (EPR) 526–529 Electrons in crystals 537–545 Electrostatics 251–252 Electrostriction 572 Elevator, weight in 30 EMF, see Electromotive force Energy activation 186, 446, 447 binding 512 conservation 196 quantization of 457 quantized level 437–438 relationship with mass 97 rest 96 SI units of 582 simple harmonic motion 105 Energy density of electromagnetic radiation 353 Engine efficiency of 207 Entropy changes in 213 irreversible process 214–216, 220 reversible process 211, 212, 216, 220 statistical definition 212 Equation of state, ideal gas 175 Equilibrium in thermodynamics 169, 170, 230 Equipotential surface 275 Equipartitioning of energy on degree of freedom 194–195 Extraneous force 308
Fermi level 541 Fermion 541 Ferroelectrics (segnetoelectrics) 574, 575 Ferromagnetism 344–347 Fiber light guides 364 Fine interaction 467–468 Fine structure constant 468 First law of thermodynamics 194–205 in various processes 196, 197 Flux of electric field strength 261 magnetic field induction 328, 329 vector field 259, 328 Force as vector 20, 40 centripetal 185, 321 conservative 60 derived from potential energy 61, 63 dissipative 60 external 28, 52, 53 internal 53 non-conservative 60 unit of 582 Force constant 545, 546 Force fields 58–61 Forced oscillations 138–145 Fraunhofer diffraction 379, 380 Free expansion 196 Free fall acceleration 18, 20, 587 Free path length 235 Frequency and period 107 angular 107 of string vibration 160–161 resonant 140 simple harmonic motion 108 units of 581 Friction non-conservative nature 32, 238 Fresnel diffraction 378–379 Fuel combustion 26–28, 206
F
G
F, see Farad Fahrenheit temperature 177 Farad 585 Faraday 328–330, 587 Faraday’s law, see Electromagnetic induction
Galilean transformation 18–20 Gamma-resonance (Mössbauer effect) 510–513 Gas, ideal 174 Gas constant R, universal 181, 587
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Gauss’ law applied to charge distributions 259–263 Gauss surface 261, 262 General theory of relativity 90–91 Grating, diffraction dispersion of 383 principal maxima of 382 resolving power 383, 384 secondary maxima 382 transmission 385 Gravitational constant 30, 587 Gravitational field gravitational mass 30 gravitational potential energy 61–62, 66 Gravity, acceleration of free fall 18, 20, 587 Gravity of earth 65, 66, 182 moon 182 Ground state 362 Group speed 398 H H, see Henry Harmonic motion damped 133–138 forced simple 138–145 See also Simple harmonic motion Harmonic oscillator 76, 129–131 Heat engine 206–207 pump 210–211 Heat capacity of crystals Debye model 552 Debye frequency 553, 554 Debye function 553, 554 Debye temperature 55, 554 Dulong–Petit law 551 Einstein model 551 Heat capacity of ideal gas in dissociation 202 versus experiment 204–205 Heat transport coefficient 237 Henry 587 Hertz 581 Hooke’s law 31 and potential energy 31, 62 elastic force work 52
613
Huygens’ principle diffraction 378–379 Hydrogen atom 416–419 Hysteresis loop in ferroelectrics 575 in ferromagnetics 348, 349 Hz, see Hertz I Ideal gas adiabatic process 199 and temperature 197 at constant pressure 213 at constant volume 213 average energy 176 general equation 175 heat capacities 197–204 in force field 178–180 internal energy 195 isotherms 200, 207, 208, 211 model 174 pressure 175 Impulse of force 21 Incidence, plane of 388 Index of refraction 297, 361, 363–365, 388, 392, 395, 396 total internal reflection 364, 393 Induced electric field, see Electromagnetic induction Induced EMF 329, 330 Induced magnetic fields 309–313 Inductance mutual 330 of solenoid 331 self 330 units 587 Induction, Faraday’s law of, see Electromagnetic induction Inelastic collision 88–89 Inertia, law of 17 Inertia, rotational 40 Inertial mass 30 Initial conditions, in harmonic motion 107, 108 Intensity in single-slit diffraction 380, 381 wave amplitude 406 Interference, from thin films 370–374
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Internal energy 88, 89, 195, 198 of ideal gas 186, 195, 196 of van der Waals’s gas 226 Internal forces 35, 38, 53, 61, 63, 70 International System of Units, see SI Invariance 68 Invariant quantities 18, 19 Ionic polarization 571–572 Irreversible process entropy change of 214–216, 220 Isotherm 200, 207, 208, 211, 222, 223 Isotope 183, 220, 244, 498 Isotropic material 147, 284, 286, 291, 295, 361, 392, 576 J Joule–Lenz law 328 Joule–Thomson effect 227–229
Index
Laue diffraction 385 Length, relativity 91 Lenz’s law 328 Light energy quanta 408, 409 momentum 412 polarization 386–395 polarized 386 quantization and emission of 387–388 sensitivity of eye to 361 speed of 392 unpolarized 362 visible 361 Linear density 251 Linear motion, with constant acceleration 7 Linear oscillator 129–131 Lines of force, electric field 275 Liquid crystals 576–577 Lorentz transformation 90
K M Kinematics 1–16 Kinetic energy center of mass 55 in simple harmonic motion 131 in transverse wave 152 of rotation 47, 55 relativistic 96 Kinetic theory pressure, ideal gas 175 temperature, ideal gas 177 Kirchhoff’s law 401 Knudsen flow 244 L Lattice defects clusters 563 dislocation 563–566 Burgers vector 564 edge 564, 565 linear 564 screw 564, 565 small angle boundaries 565 interstitials 561, 563 point 561–563 Frenkel 561, 562 Schotky 562 vacancies 561, 563
m, see meter Magnetic dipole 327–328 Magnetic splitting 347 Magnetic dipole moment of iron atom 347 of iron ions 347 Magnetic domain 347–349 Magnetic field circulating charges 305 electric current 305–309 electromagnetic wave 353, 354 flux 336 induction 309–313 induced electric fields 330 strength 307, 308, 310, 311, 320, 334, 342, 347, 348 Magnetic flux 328, 330, 331 Magnetic force between parallel wires 313 on moving charge 320–327 on wire with current 313 Magnetic monopole 310, 351 Magnetically ordered state 344–350 Magnetization 333, 336–344 Magneton Bohr 346, 453, 586, 587 nuclear 499, 516, 587
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Magnets 348 Magnetic field induction 309–318, 586 Mass and energy 96, 97 and weight 30 atomic unit of 587 center of 36–38 equivalence principle 96, 97 gravitational 30 inertial 30 in relativity 91 Mass number 323, 498 Mass spectrometry mass spectrometer 322, 323 Maxwell’s equations 599–603 Maxwell energy distribution function 193–194 Maxwell velocity distribution function average 188, 189 most probable 188, 189 on molecular energy 189 root square mean 188, 189 Maxwell’s law 186–190, 195, 599–603 Mean free path 242 Mechanical energy kinetic 54 potential 61 potential energy curves 74–79 Miller indexes 534 Molar heat capacity at constant pressure 198 at constant volume 198 of ideal gases 198, 213 Molecular speeds 187, 188 Molecular mass 88, 131 Moment of inertia 40 –43 Momentum and Newton’s second law 20–29 angular, see Angular momentum conservation of 69, 71 kinetic energy 68 relativistic 91 velocity of center of mass 38 Moon 182 Mössbauer effect 510–513 Mössbauer spectroscopy 508–516 Mount Everest, potential energy 66 Mutual inductance units of 87 Mutual induction 330
615
N N, see Newton Nanoparticles 555–557 Natural frequency 140, 141, 396 Natural width of spectral line 510 Negative charge 253, 288 Neutral matter 251, 280, 342, 426, 497, 499 Neutron 425, 489, 497, 499 Newton 582 Newton rings 374 Newton’s first law 16–18 Newton’s law of gravity 30 Newton’s second law angular form 21 for particle 21 for system 22 in relativity 21, 95 momentum form 21 simple harmonic motion 119 Newton’s third law 29 Node 158, 159 Nonconservative forces 60 Nuclear forces 497, 499, 500 Nuclear magnetic resonance (NMR) 516–525 proton magnetic resonance (PMR) 518, 521 Nuclear magnetism 516–525 Nuclear physics nucleon model of nucleus 497–499 Nuclear quadrupole resonance (NQR) 525 Nucleon energy levels 499–500 form 500, 501 magnetic moment 499 mass 498 nuclear gyromagnetic ratio 498 nuclear magneton 499 quadrupole moment 501–502 size charge distribution radius 500–501 mass distribution radius 500–501 symmetry 500 Nucleus 332, 497–499 O Ohm’s law 307, 352 Orbital magnetic dipole moment 333, 342, 453
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Order long range 568 short range 568 Oscillation, center of 122 Oscillations damped 133–138 forced 138–148 simple harmonic 106–113 P Pa, see Pascal Parallel axis theorem 43–44 Paramagnetism 332, 340, 343 Pascal 582 Path independence and conservative forces 60 Pendulum physical 121–122 simple, mathematical 119–121 spring 118–119 Performance, thermal coefficient of 210 Period and frequency 107 of linear oscillator 119, 120, 122 of simple harmonic motion 132 Permanent magnetism 348 Permeability constant 309–310 Permittivity constant 293 Perovskite type crystal 575, 576 Perpendicular axis theorem 45 Phase of simple harmonic motion 107 Phase changes, on reflection 160, 368 Phase difference, in interference 370 Phase speed 151, 398, 546 Photo-effect external Einstein photo-effect law 407, 408 Physical kinetics collision cross-section 232 effective diameter at collisions 231 relaxation process, time of 230–231 Physical model 35 Physical pendulum 121–122 Physics, quantum 33, 479, 490, 510, 541 Piezoelectric effect 572–574 Piezoelectrics 572–574 Planck constant 404, 587 Planes, mirror 35 Plane motion 2 Plane of incidence 388
Index
Plane polarization 390 Polar material 280, 284, 291, 297 Polarization dielectric 286–292 reflection 388–389 Polarization of electromagnetic waves 362, 386–394 Polarized light plane 387, 389 refection 388–389 Polarizer 387–388 Polarizing angle 390 Postulate relativity 90 speed of light 90 Potential, electric, see Electric field potential Potential energy and work 61 electric 75 electric dipole 278 force 63 gravitational 62, 74 magnetic dipole 327 simple harmonic oscillator 131 Potential–energy curve 74–79, 186, 482 Potential difference 276 Power average 53 instantaneous 54 units of 582 Poynting vector 353 Precession 51, 337, 338, 477, 589–590 Pressure, kinetic theory of 175, 176 Prism 364, 365, 384, 391 Process adiabatic 199 cyclic 205 constant volume 198 irreversible 206 isobaric 198 reversible 206 Projectile motion 9 Proton 497–499 Pyroelectric effect 572 Q Quadrupole moment 501, 584, 593 Quadrupole resonance 525 Quadrupole splitting 506
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Quantization at emission of light 512 electron energy in atom 457–460 of angular momentum 455, 456 Quantum mechanics 423–496 Quartz crystal oscillator 137 R R (universal gas constant) 181, 487 Radial distribution function 567 Radian 581 Radiation, blackbody 398–402 Ratio of specific heats for gases 199 Ray 147 Rayleigh’s criterion 384 Real gas approximation corresponding states law 225 critical point 223 internal energy of 226 Joule–Thomson effect 227–229 inversion curve 229 inversion point 229 phase diagram of state 223, 224 van der Waals gas 221–226 real isotherms of 223 van der Waals equation 221, 222 in reduced parameters 225 Recoil 511, 512 Reduced amount of heat: entropy change in isobaric process 213 in isochoric process 213 Reference frames 21, 97 Reference system (frame) 1 Inertial 16 Noninertial 33 Reflection law of 363 phase change 368 polarization 388–389 thin film 371, 372 total internal 364 Refraction Huygens’ principle 378–379 index of 361, 363, 365 law of 363 Refrigerator 210–211 Relativity motion at high speed 90 speed summation law 90
617
Shortening of length 91 dilation of time 94 simultaneity 94 Resolving power of grating 384 Resonance 140, 497–529 Resonance absorption 508–510 Rest mass 426 Reversible process, entropy change in 213 Rigid body angular momentum and velocity 41 Rocket, propulsion 26, 27 Rolling motion, energy of 57 Root-mean-square molecular speed 188, 189 Rotation angular momentum 41 analogy with translation motion 50, 51 with constant acceleration 15 Rotational motion analogies to translational motion 50 kinematics 12 dynamics 16 Rotation of plane of 389–391 Rotator, rigid 47, 450 S Saturation ferromagnetic 349 paramagnetic 341 Second law of motion, see Newton’s second law Second law of thermodynamics 205–221 entropy 211–214 Selection rules 479, 480 Self-induction 331, see also Inductance Semiconductors acceptor type 544 donor type 544 intrinsic type 544 n-type 544, 545 p-type 544, 545 SI 581 Simple harmonic motion acceleration 108 amplitude 107 angular frequency 107 displacement 107
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equation of motion for 107 kinetic energy of 131 period of 106 phase 107 Simple harmonic oscillator period of 131 Single-slit diffraction 379–381 Snell’s law 363 Solenoid, magnetic field of 319, 320 Solid-state dynamics Born–Karman chain 545 Special relativistic theory 90–97 Specific heat capacity 197, 198, 583 Speed angular 57, 73 average 3, 8, 307, 570 molecular average 188 distribution of 186–190 most probable 188 root-mean-square 188 radial acceleration 14 speed and acceleration 95 Speed of light 603 Spherical symmetry 252, 268, 269, 270, 500 Spin, quantum number 460, 469, 480, 498, 517, 518, 527, 541, 542 Spring scale 30 Spring, force law 52 Spring, potential energy 62 Standing waves 157–160 Statistical thermodynamics 219–220 Statistics Bose–Einstein 542 classic 541 distribution function 541 Fermi–Dirac 541 quantum 540–543 Steradian 261, 581 String, waves on 160–161 Superconductivity 543 Superfine dipole–dipole interaction 523 Superfine interaction 515–516 Superposition principle 156, 157, 274, 277 Surface charge 576 Surface current 335 Symmetry operation 531 Syngony 533
Index
T T, see Tesla Tangent, unit vector of 2 Tangential acceleration 14 Temperature in kinetic theory 177–178 Temperature scale Celsius 177 Fahrenheit 177 Kelvin 177 Réaumer 177 Tesla 586 Thermal conductivity 233, 236 Thermal energy 197, 198, 206, 406 Thermal equilibrium 169, 407 Thermodynamics first law of 194–205 second law of 205–221 zeroth law of 170 Thermodynamic process 170, 206 Third law of motion 29 Thrust, of rocket 27, 28 Time of settled life 568 Time dilation 94 Torque and angular acceleration 48 and angular momentum 49 on electric dipole 278 on magnetic dipole 327 Total internal reflection 364 Trajectory of projectile 72 Transformation Galilean 18–20 Lorentz 90 space time 90 velocity 91 Translation 531, 532 Translational motion and rotational motion 50, 51 Transport phenomena in ideal gases 233–235 flow of G property 234 macroscopic representations 233–235 U Unit cell body-centered cell (BCC) 535
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face-centered cell (FCC) 535 primitive unit cell 535 Unit vector 2 V V, see Volt Vacuum heat transfer in vacuum 243–244 Velocity angular 13, 14 average 3 in simple harmonic motion 107 linear 14, 41 transformation 18, 19 Velocity space 187 Viscosity (internal friction) dynamical coefficient 239 kinematical coefficient 240 Volt 584 Voltage, see Electromotive force Volume, units of 581 W Water molecule 3, 45, 46, 170 Waves electromagnetic, see Electromagnetic waves interference 369–377 longitudinal 146 mechanical 145 sinusoidal 159 standing 157–160 string 160–163 transverse 146 transverse simple harmonic, energy of 151–154 transverse standing equation of 158 traveling, equation of 147–151
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Wave equations 148, 150 Wave number 149, 492 Wave speed 155 Wavelength 148 Wavelength of light 354 Wavetrain 362 Weber 587 Weight 30 Wire magnetic field of 313 magnetic force on 316 Work–energy theorem 63 Work of gas in adiabatic process 199 in elementary processes 226 in isobaric process 198 in isochoric process 198 in isoprocesses 197–201 in isothermic process 200, 207 X X-rays bremschtralung 410 characteristic 410 X-ray diffraction 385–386 Z Zeeman effect 477–480 anomalous 477, 478 normal 477, 478, 480 nuclear 478 Zeroth law of thermodynamics 170 Zone theory conductivity band 539 energy bands 539 forbidden energy gap 539 hybrid band 539 valence energy band 539
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