Methods for Electromagnetic Field Analysis (IEEE Press Series on Electromagnetic Wave Theory)

METHODS FOR ELECTROMAGNETIC FIELD ANALYSIS

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METHODS FOR ELECTROMAGNETIC FIELD ANALYSIS

!sMO

V.

LINDELL

HELsINKI UNIVERSITY OFTECHNOLOGY

IEEE PRESS

The Institute of ElectricalandElectronics Engineers, Inc., NewYork

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© Ismo V. Lindell, 1992. All rights reserved No part ofthis book may be reproduced in anyform, nor may it be stored in a retrieval system or transmitted in any form, without written permission from the publisher.

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ISBN 0-7803-6039-7 IEEE Order No. PP5625

Library of Congress Cataloging-in-Publication Data

Lindell, Ismo v. Methods for electromagnetic field analysis I Ismo V. Lindell. p. em. - (IEEE Press series on electromagnetic wave theory) Originally published: New York: Oxford University Press, 1992. IEEE order no. PP5625-T.p. verso. Includes bibliographical references and index. ISBN 0-7803-6039-7 1. Electromagnetic fields-Mathematics. I. Title. II. Series. QC665.FA IA6 2000 530.14'1--dc21

00-047149

Preface The present monograph discusses a number of mathematical and conceptual methods applicable in the analysis of electromagnetic fields. The leading tone is dyadic algebra. It is applied in the form originated by J.W. Gibbs more than one hundred years ago, with new powerful identities added, making coordinate-dependent operations in electromagnetics all but obsolete. The chapters on complex vectors and dyadics are independent of the rest of the book, actually independent of electromagnetics, so they can be applied in other branches of physics as well. It is claimed that by rnemorizing about five basic dyadic identities (similar to the well-known bac-cab rule in vector algebra), a working knowledge of dyadic algebra is obtained. To save the memory, a collection of these basic dyadic identities, together with their most important special cases, is given as an appendix. In different chapters the dyadics are seen in action. It is shown how simply different properties can be expressed in terms of dyadics: boundary and interface conditions, medium equations, solving Green functions, generalizing circuit theory to vector field problems with dyadic impedances, finding transformations between field problems and, finally, working on multipole and image sources for different problems. Dyadic algebra is seen especially to aid in solving electromagnetic problems involving different linear media. In recent years, the chiral medium with its wide range of potential applications has directed theoretical interest to new materials. The most general isotropic medium, the bi-isotropic rnedium, has made electromagnetic theory a fresh subject again, with new phenomena being looked for. The rnedium aspect is carried along in this

text. What is normally analysed in isotropic media is done here for biisotropic or sometimes for bianisotropic media, if possible. Especially new is the duality transformation, which actually exists as a pair of transformations. It is seen to shed new light on the plus and minus field decomposition, which has proved useful for analysing fields in chiral media, by showing that they are nothing more than self-dual fields with respect to each of the two transformations. In Chapter 5, Green dyadics for different kinds of media are discussed and a systematic method for their solution, without applying the Fourier transformation, is given. In Chapter 6, source equivalence and its relation to non-radiating sources is discussed, together with certain equivalent

vi

PREFACE

sources: point sources (multipoles) and surface sources (Huygens' sources). Everywhere in the text the main emphasis is not on specific results but methods of analysis. The final chapter gives a summary of the work done by this author and colleagues on the EIT, exact image theory. This is a general method for solving problems involving layered media by replacing them by image sources which are located in complex space. The EIT is presented here for the first time in book form. The contents of this monograph reflect some of the work done and courses given by this author during the last two decades. The results should be of interest to scientists doing research work in electromagnetics, as well as to graduate students. For classroom use, there are numerous possibilities for homework problems requiring the student to fill in steps which have been omitted to keep the size of this monograph within certain limits. The EIT can also be studied independently and additional material, not found in this text, exists in print (see referece lists at section ends of Chapter 7). The text has been typed and figures drawn by the author alone, leaving no-one else to blame. On the other hand, during graduate courses given on the material, many students have helped in checking a great number of equations. Also, the material of Chapters 1 and 2 has been given earlier as a laboratory report and a few misprints have been pointed out by some international readers. For all these I am thankful. The rest of the errors and misprints are still there to be found. This book is dedicated to my wife Liisa. A wise man is recognized for having a wife wiser than himself. I have the pleasure to consider myself a wise man. Helsinki July 1991.

LV.L.

For the second edition, the main text has remained unchanged except for a small number of misprints, which have been corrected. To assist in classroom use of the book, three new appendices have been added. Appendix A contains a collection of problems on the topics of Chapters 1 - 6 and Appendix B a set of solutions for most of these problems. Appendix C gives a collection of most useful formulas in vector analysis for convenience. Appendix A (Dyadic identities) of the first edition is now relabeled as Appendix D. Helsinki

March 1995.

LV.L.

Contents 1

2

Complex vectors

1

1.1 Notation

1

1.2 Complex vector identities

4

1.3 Parallel and perpendicular vectors

5

1.4 Axial representation

8

1.5 Polarization vectors 1.6 Complex vector bases

10

References

15

Dyadics

17

2.1 Notation 2.1.1 Dyads and polyads 2.1.2 Symmetric and antisymmetric dyadics

17 17 19

2.2 Dyadics as linear mappings

20

2.3 Products of dyadics 2.3.1 Dot-product algebra 2.3.2 Double-dot product 2.3.3 Double-cross product

22

2.4 Invariants and inverses

28

2.5 Solving dyadic equations 2.5.1 Linear equations 2.5.2 Quadratic equations 2.5.3 Shearers

31 33 35

2.6 The eigenvalue problem

36

2.7 Hermitian and positive definite dyadics 2.7.1 Hermitian dyadics 2.7.2 Positive definite dyadics

39 39 40

2.8 Special dyadics 2.8.1 Rotation dyadics 2.8.2 Reflection dyadics 2.8.3 Uniaxial dyadics 2.8.4 Gyrotropic dyadics

41 41

14

22 24

25 30

43

44 45

CONTENTS

viii

3

2.9 Two-dimensional dyadics 2.9.1 Eigendyadics 2.9.2 Base dyadics 2.9.3 The inverse dyadic 2.9.4 Dyadic square roots

50 50

References

51

Field equations

53

3.1 The Maxwell equations 3.1.1 Operator equations 3.1.2 Medium equations 3.1.3 Wave equations References

53 55

3.2 Fourier transformations 3.2.1 Fourier transformation in time 3.2.2 Fourier transformation in space References 3.3 Electromagnetic potentials 3.3.1 Vector and scalar potentials 3.3.2 The Hertz vector 3.3.3 Scalar Hertz potentials References 3.4 Boundary, interface and sheet conditions 3.4.1 Discontinuities in fields, sources and media 3.4.2 Boundary conditions 3.4.3 Interface conditions 3.4.4 Sheet conditions 3.4.5 Boundary and sheet impedance operators References 3.5 Uniqueness 3.5.1 Electrostatic problem 3.5.2 Scalar electromagnetic problem 3.5.3 Vector electromagnetic problem References 3.6 Conditions for medium parameters 3.6.1 Energy conditions 3.6.2 Reciprocity conditions References

47 48

49

56 57 58 59 59 62 63

63 64

66 67 68 69 69 71

74 77

81 83 84

85

86 88

88

89 89 93 94

CONTENTS

4

5

Field transformations

ix

4.1 Reversal transformations 4.1.1 Polarity reversal 4.1.2 Time reversal 4.1.3 Space inversion 4.1.4 Transformations of power and impedance References 4.2 Duality transformations 4.2.1 Simple duality 4.2.2 Duality transformations for isotropic media 4.2.3 Left-hand and right-hand transformations 4.2.4 Application of the duality transformations 4.2.5 Self-dual problems 4.2.6 Self-dual field decomposition 4.2.7 Duality transformations for bi-isotropic media References 4.3 Affine transformations 4.3.1 Transformation of fields and sources 4.3.2 Transformation of media 4.3.3 Involutory affine transformations

97 97 97 98 99 99 99 100 100 100 101 103 105 107 110 112 112 113 115 116

4.4 Reflection transformations 4.4.1 Invariance of media 4.4.2 Electric and magnetic reflections 4.4.3 The mirror image principle 4.4.4 Images in parallel planes 4.4.5 Babinet's principle References

117 117 118 119 121 122 124

Electromagnetic field solutions

125

5.1 The Green function 5.1.1 Green dyadics of polynomial operators 5.1.2 Examples of operators References

125 126 128 129

5.2 Green functions for homogeneous media 5.2.1 Isotropic medium 5.2.2 Bi-isotropic medium 5.2.3 Anisotropic medium References

129 129 132 133 137

CONTENTS

x

6

5.3 Special Green functions 5.3.1 Two-dimensional Green function 5.3.2 One-dimensional Green function 5.3.3 Half-space Green function References 5.4 Singularity of the Green dyadic 5.4.1 Constant volume current 5.4.2 Constant planar current sheet 5.4.3 Singularity for a volume source 5.4.4 Singularity for a surface source References 5.5 Complex source point Green function 5.5.1 Complex distance function 5.5.2 Point source in complex space 5.5.3 Green function References 5.6 Plane waves 5.6.1 Dispersion equations 5.6.2 Isotropic medium 5.6.3 Bi-isotropic medium 5.6.4 Anisotropic medium References

138 138 140 141 141 142 142 144 146 147 148 148 149 150 151 153 154 154 156 156 157 163

Source equivalence

165 165 165 168 169 170 170 171 173 174

6.1 Non-radiating sources 6.1.1 Electric sources in isotropic medium 6.1.2 Sources in bianisotropic media References 6.2 Equivalent electric and magnetic sources 6.2.1 Bianisotropic medium 6.2.2 Isotropic medium 6.2.3 Chiral medium References 6.3 Multipole sources 6.3.1 Delta expansions 6.3.2 Multipole expansion 6.3.3 Dipole approximation of a multipole source 6.3.4 Electric and magnetic dipole approximation References

174

175 176 178 181 181

CONTENTS

xi

6.4 Huygens' sources 182 6.4.1 The truncated problem 182 6.4.2 Huygens' principle 183 6.4.3 Consequences of Huygens' principle 185 6.4.4 Surface integral equations 187 6.4.5 Integral equations for bodies with impedance surface 189 6.4.6 Integral equations for material bodies 191 References 191 6.5 TE/TM decomposition of sources 192 6.5.1 Decomposition identity 193 6.5.2 Line source decomposition 194 6.5.3 Plane source decomposition 196 197 6.5.4 Point source decomposition References 197

7

Exact image theory

199

7.1 General formulation for layered media 7.1.1 Fourier transformations 7.1.2 Image functions for reflection fields 7.1.3 Image functions for transmission fields 7.1.4 Green functions References

199 199 205 214

218 223

7.2 Surface problems 7.2.1 Impedance surface 7.2.2 Impedance sheet 7.2.3 Wire grid References

228

7.3 The Sommerfeld half-space problem 7.3.1 Reflection coefficients 7.3.2 Reflection image functions 7.3.3 Antenna above the ground 7.3.4 Transmission coefficients 7.3.5 Transmission image functions 7.3.6 Radiation from a loop antenna into the ground 7.3.7 Scattering from an object in front of an interface References

235 236 238 247 249 250 255 257 259

7.4 Microstrip geometry

260 260 267

7.4.1 7.4.2

Reflection coefficients and image functions Fields at the interface

224 224 231

235

xii

CONTENTS

7.4.3 The guided modes 7.4.4 Properties of the Green functions References 7.5 Anisotropic half space References

271 272 274 275 279

Appendix A Problems

281

Appendix B Solutions

291

Appendix C Vector formulas

307

Appendix D Dyadic identities

311

Author index

315

Subject index

317

Chapter 1

Complex vectors Complex vectors are vectors whose components can be complex numbers. They were introduced by the famous American physicist J. WILLARD GIBBS, sometimes called the 'Maxwell of America', at about the same period in the 1880's as the real vector algebra, in a privately printed but widely circulated pamphlet Elements of vector analysis. Gibbs called these complex extensions of vectors 'bivectors' and they were needed, for example, in his analysis of time-harmonic optical fields in crystals. In a later book compiled by Gibbs's student WILSON in 1909, the text reappeared in extended form, but with only few new ideas (GIBBS and WILSON 1909). Thenceforth, complex vectors have been treated mainly in books on electromagnetics in the context of time-harmonic fields. Instead of a full application of complex vector algebra, the analyses, however, mostly made use of trigonometric function calculations. As will be seen in this chapter, complex vector algebra offers a simple method for the analysis of time-harmonic fields. In fact, it is possible to use many of the rules known from real vector algebra, although not all the conclusions. Properties of the ellipse of time-harmonic vectors can be seen to be directly obtainable through operations on complex vectors.

1.1

Notation

As mentioned above, complex vector formalism is applied in electromagnetics when dealing with time-harmonic field quantities. A time-harmonic field vector F( t), or 'sinusoidal field' is any real vector function of time t that satisfies the differential equation (1.1) A general solution can be expressed in terms of two constant real vectors F 1 and F 2 in the form

F(t)

= F 1 coswt + F 2 sinwt.

(1.2)

2

CHAPTER 1. COMPLEX VECTORS

The complex vector formalism can be used to replace the time-harmonic vectors provided the angular frequency w is constant. There are certain advantages to this change in notation and, of course, the disadvantage that some new concepts and formulas must be learned. The main bulk of formulas, however, is the same as for real vectors. As an advantage, in using complex vector algebra, work with trigonometric formulas can be avoided, and the formulas look much simpler. The complex vector f is defined as a combination of two real vectors, f re the real part, and fim the imaginary part of f: (1.3) The subscripts re and im can be conceived as operators, giving the real and, respectively, the imaginary parts of a complex vector. The essential point in the complex vector formalism lies in the oneto-one correspondence with the time-harmonic vectors f f-+ F(t). In fact, there are two mappings which give a unique time-harmonic vector for a given complex vector and vice versa. They are:

f ~ F(t): F(t) = ~{feiwt} = fre coswt - fim sinwt, F(t)

--+

f:

f

= F(O) - jF(1r/2w) = F 1 - jF 2 •

(1.4) (1.5)

Thus, for the two representations (1.2) and (1.3) we can see the correspondences fre = F 1 and fim = -F 2 . The mappings (1.4), (1.5) are each other's inverses, as is easy to show. For example, let us insert (1.4) into (1.5): (1.6)

which results in the identity f = f. It is important to note that there always exists a time-harmonic counterpart to a complex vector whatever its origin. In fact, in analysis, there arise complex vectors, which do not represent a time-harmonic field quantity, for example the wave vector k or the Poynting vector P. We can, however, always define a time-harmonic vector through (1.4), maybe lacking physical content but helpful in forming a mental picture. A time-harmonic vector F(t) = F 1 coswt + F 2 sinwt traces an ellipse in space, which may reduce to a line segment or a circle. This is seen from the following reasoning. • If F 1 x F 2 = 0, the vectors are parallel or at least one of them is a null vector. Hence, F(t) is either a null vector or moves along a line and is called linearly polarized (LP).

1.1. NOTATION

3

• If F l x F 2 =1= 0, the vectors define a plane, in which the vector F(t) rotates. Forming the auxiliary vectors b = F l X (F l x F2) and c = F 2 X (F 1 x F 2 ) , we can easily see that the equation [b- F(t))2 + (c· F(t))2 = IF l X F 2 14 is satisfied. This is a second order equation, whose solution F(t) is obviously finite for all t, whence the curve it traces is an ellipse. • The special case of a circularly polarized (CP) vector is obtained, when IF(t)12 = Fjcos2wt+F~sin2wt+Fl·F2sin2wt is constant for all t. Taking t = 0 and t = 1r/2w gives F~ = F~, which leads to the second condition F l · F 2 = O. Thus, to every complex vector f there corresponds an ellipse just as for every real vector there corresponds an arrow in space. The real and imaginary parts fre, fim both lie on the ellipse. fre equals the time origin value and is called the phase vector of the ellipse. The direction of rotation of F(t) on the ellipse equals that of fim turned towards fre in the shortest way. A complex vector which is not linearly polarized (NLP) has a handedness of rotation, which depends on the direction of aspect. The rotation is right handed when looked at in a direction u (a real vector) such that fim X fre · U is a positive number and, conversely, left handed if it is negative. An LP vector must be represented by a double-headed arrow (infinitely thin ellipse), which is in distinction with the one-headed arrow representation of real vectors. The difference is of course due to the fact that the time-harmonic vector (1.2) oscillates between its two extremities. The complex conjugate of a complex vector f, denoted by f*, is defined by (1.7) From (1.4) we can see that f* corresponds to the time-dependent vector F( -t), or it rotates in the opposite direction along the same ellipse as f(t). The complex vector f is LP if and only if fre x fim = O. This is equivalent with the following condition: f is LP

¢:}

f x f* = O.

(1.8)

The corresponding condition for the CP vector is f is CP

¢:}

f ·f

= O.

(1.9)

In fact, (1.9) implies that f:e = f?m and fre · fim = 0, which is equivalent with the CP property of the corresponding time-harmonic vector, as was seen above.

4

CHAPTER 1. COMPLEX VECTORS

Every LP vector can be written as a multiple of a real unit vector u in the form f = au. Every CP vector f can be written in terms of two orthogonal real unit vectors u, v in the form f = a(u + jv). In these expressions a is a complex scalar, in general.

1.2

Complex vector identities

The algebra of complex vectors obeys many of the rules known from the real vector algebra, but not all. For example, the implication a-a = 0 =} a = 0 is not valid for complex vectors. To be more confident in using identities of real vector algebra, the following theorem appears useful:

all multilinear identities valid for real vectors are also valid for complex vectors. A multilinear function F of vector arguments aI, a2, ... is a function which is linear in every argument, or the following is valid for i = 1 ... n: F(al' a2, ..., (aa~

+ ,Ba~/), ..., an)

=

aF(al, ..., a~, ..., an) + ,BF(al' ..., a~/, ...a n).

(1.10)

A multilinear identity is of the form

F(al' ..., an) = 0 for all ai, i = 1...n.

(1.11)

Now, if the identity is valid for real vectors a, and the function does not involve a conjugation operation, from the linearity property (1.10) we can show that it must be valid for complex vectors a, as well. In fact, taking Q = 1,,B = i, the identity is obviously valid if the real vector a, is replaced by the complex vector a~ + ja~/. This can be repeated for every i and, thus, all vectors a, can be complex in the identity (1.11). As an example of a trilinear identity we might write

a x (b x c) - (a· c)b + (a· b)c = 0 for all a, b, c.

(1.12)

Also, all non-linear identities which can be derived from multilinear identities are valid for complex vectors, like a x a = 0 for all vectors a. The conjugation operation can be introduced by inserting conjugated complex vectors in multilinear identities. Thus, the identity (1.13)

can be obtained from the real quadrilinear identity (a x b) · (c x d) = (a· c)(b· d) - (a· d)(b· c),

(1.14)

1.3.

PARALLEL AND PERPENDICULAR VECTORS

5

with the substitution c = a", d = b*. The absolute value for a complex vector is defined by (1.15) lal2 = a· a". All irnplications that can be derived from identities are valid for complex vectors if they are valid for real vectors. The basis for these is the null vector property: [a] = 0 ¢:> a = 0, (1.16) 2 which can be shown to be valid by expanding lal = la + jainll2. re

Two important, although simple looking theorems can be obtained from vector identities: a x b = 0 and a =j:. 0

a .b

=0

and a

=1=

:::} 30, b

0 :::} 3c, b

= oa,

=c

(1.17)

x a.

(1.18)

(1.17) follows from the identity a" x (a x b) = (a* · b)a - (a· a*)b,

(1.19)

from which b can be solved. Correspondingly, (1.18) is obtained from

a x (a* x b) = (a· b)a* - (a· a*)b.

(1.20)

As a consequence, from (1.17), (1.18) we see that the theorem a xb

=0

and a· b

=0

:::} a = 0 or b

= 0,

(1.21)

valid for real vectors, is not valid for complex vectors. In fact, assuming [a] =1= 0 gives us either b = 0 or a is CP, which implies b is CPo A theorem corresponding to (1.21) for complex vectors is the following one:

a x b = 0 and a· b* = 0 :::} a

=0

or b

= 0,

(1.22)

as can be readily verified from (1.17), (1.18).

1.3

Parallel and perpendicular vectors

For real vectors a, b there exist the geometrical concepts of parallel vectors for a x b = 0 and perpendicularity for a . b = O. Although the geometrical content will be different, it is helpful to define parallelity and perpendicularity with these same equations for complex vectors. This leads, however, to the existence of vectors which are perpendicular to themselves, namely the CP vectors.

CHAPTER 1. COMPLEX VECTORS

6

A complex vector b is parallel to a non-null vector a if there exists a complex scalar 0: such that b = oa, as implied by (1.17). Let us denote a = Aej 8 with real A,O, and A > O. It is easy to see from the definition (1.4) that if we have the correspondence a

~

A(t),

(1.23)

then we also have ~

()

AA(t+-). w

(1.24)

Thus, the magnitude of the ellipse is multiplied by the factor A and the phase of the ellipse is shifted by () /w. The form of the ellipse as well as its axial directions and sense of rotation are the same for parallel vectors. Parallel vectors are said to have the same polarization. The geometrical content of perpendicular complex vectors is more difficult to express. Let us find the most general vector b perpendicular to a given non-null vector a. Obviously, if a is LP, or parallel to a real unit vector u: a = au, b may be any vector in the plane perpendicular to u, or of the form u x c. For an NLP vector a there exists a real unit vector n satisfying n- a = 0, which is normal to the ellipse of a. Writing b = ,Bn+b a with b, in the plane of R, we see that {3 may have an arbitrary complex value. The problem is to find b.,, which must also be perpendicular to a. From the identity b; x (n x a) = n(b a • a) - a(b a . n), whose right-hand side vanishes, and (1.17) we see that there must exist a scalar a such that b., = an x a. Thus, the most general vector b perpendicular to a can be written as b

= ,Bn + an x a,

with n· a =

o.

(1.25)

The corresponding time-harmonic vector B(t) is easily seen to be a sum of an LP vector along n plus an NLP vector in the plane of a, which is obtained from a vector parallel to a rotated by 1r /2 in its plane. The most general b vector can be seen to lie on an elliptic cylinder, whose cross section is the ellipse of b a . It is also easy to see that there exist vectors orthogonal to a given vector a of any ellipticity, because we can obtain ellipses with every axial ratio by cutting the elliptic cylinder with planes of different orientations. In particular, the LP vector b is a multiple of n, whereas the two CP vectors b±

= a(n x a±jya:an),

(1.26)

1.3. PARALLEL AND PERPENDICULAR VECTORS

are obtained from (1.25) with the CP condition b· b

7

= o.

b

b 8(0)

(b)

(a)

Fig. 1.1 (a) Parallel complex vectors a and b. (b) Elliptic cylinder construction of a vector b perpendicular to a given vector a. The projection vector b; is parallel to n x a, where n is a real unit vector normal to a.

Any vector b can be written as

b

= b· aa _ a·a

a x (a x b). a·a

(1.27)

This gives a decomposition of a vector b into parts parallel and perpendicular to a vector a, and it is used in real vector algebra. Although (1.27) is also valid for complex vectors, it fails when a is CPo A more practical decomposition theorem is the following one:

b

=

b· a a" _ a x (a* x b), a : a* a a*

(1.28)

which splits the vector b into vectors parallel to a" and perpendicular to a. This decomposition has the property of power orthogonality of its parts. In fact, writing (1.28) b = beD + b er as respective terms, where beD is the co-polarized part and b er the cross-polarized part of the vector b with respect to the vector a, we can write (1.29)

The co-polarized component is thus parallel not to a but to a". The definition is needed, for example, in antenna theory when reception of an incoming wave with the field vector E is considered with the polarization match factor

p(h, E)

=

[h X E*1 2 (h. h*)(E . E*) = 1 - (h. h*)(E . E*) ,

Ih· El2

(1.30)

which tells us how well the polarization of the incoming field can be received by an antenna with the effective length vector h. It is seen that only the

CHAPTER 1. COMPLEX VECTORS

8

co-polarized component of E with respect to h contributes to the value of p(h, E) and complete polarization match p(h, E) == 1 is obtained for h x E* == 0, or when hand E* are parallel vectors. On the other hand, there is a total mismatch p(h, E) == 0 for' perpendicular vectors h, E, or when the incoming field is cross polarized with respect to the antenna vector h. As an example of the polarization match factor, let us consider radar reflection from an orthogonal plane. The far field of an antenna has the same polarization as its effective length vector h. Reflection from the surface does not change the polarization of the field (but its handedness is changed!), whence the polarization of the field coming back to the antenna is also that of h. Because the polarization match factor is independent of the magnitude of the field, it equals p(h, h) in this case. It is seen that for an LP antenna h x h" == 0 and p(h, h) == 1, or there is no polarization mismatch between the antenna and the incoming field. On the other hand, if h is CP, we have p(h, h) == 0, or there is complete mismatch. A CP radar does .not see reflections from an orthogonal plane, or other circularly symmetric obstacles, wheras an LP antenna receives the best possible signal.

1.4

Axial representation

Polarization properties of a complex vector such as an electromagnetic field are often needed. For example, given a complex vector, how can we determine its axial directions and magnitudes? Usually, in books working first with complex vectors, the notation is suddenly changed to time dependent representation and the quantities needed are obtained through trigonometric function analysis. This is, however, unnecessary, because the same can be written down in complex vector notation quite simply. The procedure is based on the following simple facts.

(i) The vector b ==

ejBa

has the same ellipse as a for real 8.

(ii) There exists () real such that b., . binI == O. (iii) If b., . binI == 0, b re and b im lie on the axes of the ellipse of b. (i) was demonstrated above and (ii) defines an equation for 8, which obviously has solutions. (iii) can be shown to be true through a consideration of the corresponding time-harmonic vector B(t), because IB(t)12 ==

1.4. AXIAL REPRESENTATION

8

9

1m

Fig. 1.2 Construction of the axis vectors of a complex vector a.

We are now ready to write an axial decomposition for any NCP vector a, from which the axes of its ellipse can be obtained. Obviously, a CP vector does not have any preferred axes, so it can be excluded. For an NCP vector, the scalar is non-zero and we can define another complex vector b through a (1.31) b =

va:a

,va:-a,--. va:a

The factor multiplying a is obviously of the form e- j8 , whence b is of the form (i) above implying that b and a have the same axial vectors. There is no need to solve for B. Because b- b = [a-a] is real, (ii) is also satisfied, and (iii) is valid, whence the real and imaginary parts of b are the axis vectors. Since b . b is positive and equals b~e - b~m' the real part of the vector lies on the major axis and the imaginary part of the vector on the minor axis of the ellipse of b and, hence, a. The axial representation of complex vectors was probably first given by MULLER (1969) in his monograph on electromagnetic theory. The axial decomposition (1.31) giving the major axis vectors defined by b re =

1~1!R{J:.a}'

(1.32)

and the minor axis vectors by bim

= lva:-al~{

J:.a}'

(1.33)

can easily be memorized and applied to simplify the analysis. For example, we can write a· a* = b· b* = Ibrel2 + Ibiml2 , to obtain a geometrical interpretation for the magnitude lal = Va· a* of a complex vector a, as the hypotenuse of the right triangle defined by the vectors b re and bim in Fig. 1.2.

10

CHAPTER 1. COMPLEX VECTORS

1.5

Polarization vectors

As stated above, two parallel vectors have the same polarization. Thus, the polarization of a vector a consists of all its properties that are not changed when multiplied by a complex scalar a. Because this operation changes the magnitude and phase of the ellipse, the following are left as properties of polarization: • plane of the ellipse (can be defined by its normal vector n), • direction of rotation on the plane (right hand in the direction n), • e, the axial ratio of the ellipse, which defines its form,

• axial directions on the plane of the ellipse (major axis along axis along U2).

Ul,

minor

Because the complex vectors are defined by 3+3=6 real parameters and complex scalars by 2 real parameters, the definition of the polarization concept requires 4 real parameters. For example, the unit vector n takes 2 parameters to define, the axial ratio e one, and the direction of the major axis Ul on the plane one more parameter (angle on the plane). The minor axis direction is then obtained as U2 = n x uj , The polarization of a complex vector is very often of more interest than the complex vector itself. As one example, in certain microwave ferrite devices, a piece of ferrite material should be positioned in the spot where the magnetic field is circularly polarized, whatever the magnitude of the field may be. Polarization can be represented most naturally in terms of a normalized complex vector U with a=au.

(1.34)

For NCP vectors, a can be defined as ~, whence u is a complex unit vector satisfying u · u = 1. For CP vectors, however, this breaks down. As another possibility we could try to define a as the real and positive number .;a:-a*, whence u is another complex unit vector satisfying u· u" = 1, but this u is no longer a representation of polarization because it contains the phase information of a.

p vector representation A very useful way to present the polarization is by two real vectors p and q to be described next. p is defined as the following non-linear real vector function of a:

a x a* p(a) = - . - . Ja· a*

(1.35)

1.5. POLARIZArrION VECTORS

11

This vector has the following properties. 1. [p(a)]* = pea), or it is a real vector. 2. p(a*)

== -p(a), or a change in the direction of rotation changes the

direction of p.

== pea), or p is independent of the magnitude and phase of a. However, for a = 0 the p vector is indeterminate.

3. p(aa)

4. Ip(a)1 == 2e/(e 2+1), where e is the ellipticity (axial ratio) of a. Hence, pea) == 0 {:} a is LP and Ip(a)( == 1 {:} a is CPo Otherwise the length of p is between 0 and 1.

5. pea) = 2ainl X are/a· a", whence for NLP vectors pea) points in the positive normal direction of the a ellipse. The rotation of a is right handed when looking in the direction pea).

+ n x asinO) = pea) for n == p(a)/lp(a)1 and 0 real. This makes sense for NLP vectors only and means that the ellipse may be rotated in its plane by any angle () without changing its p vector. Thus, it is not sufficient to represent the polarization by the p vector only.

6. p(acosO

Although the real vector function pea) does not carryall the polarization information of a, it is useful in analysing elliptic polarizations. In fact, it gives us the following information about a: • whether it is an LP vector or not; • for NLP vectors, the plane of polarization, sense of rotation and ellipticity. It does not give the following information: • direction, magnitude or phase of an LP vector; • for NLP vectors, the magnitude or phase or axial directions on the plane of polarization. It is seen that the p vector provides least information for LP vectors and most information for CP vectors, for which in fact the polarization is totally

12

CHAPTER 1. COMPLEX VECTORS

known. p(a)

~-t (a) a

.q;E (b)

>

~a)

a

Fig. 1.3 Polarization vectors p(a) and q(a) of a complex vector a. q vector representation

The complementary representation is the q(a) vector defined as follows:

q(a) = la· al ~{a/va:a} a· a* 1~{a/va:a}I·

(1.36)

In fact, (1.36) defines a pair of real vectors because of the two branches of the square-root function. Hence, we may depict it as a double-headed arrow. The vector function q(a) has the following properties. 1. [q(a)]* = q(a), or it is a real function. 2. q(a*)

=:

q(a), or the sense of rotation has no effect on q.

3. q(aa) = q(a), or the magnitude and phase of a have no effect on the q vector. For a null vector q is indeterminate.

4. Iq(a)1 = (1 - e2)/(1 + e2 ) , whence p2 + q2 = 1 and p, q is a complementary pair of vectors. q(a) --+ 0 for a approaching CP and Iq(a)1 = 1 for a LP. Otherwise, the magnitude of q is between and 1.

°

5. Ifa = ei 8b with b-b > 0, q(a) = ±brelq(a)I/lbrel, or q(a) is directed along the major axis of a. 6. For a

=:

aUt

+ j(3u2

with real a, (3, uj , U2 and lit . U2

= 0, we have

q(a) = ±lq(a)IUt, or the direction of the minor axis of the ellipse does not affect on the q vector.

1.5. POLARIZATION VECTORS

13

Unit vectorrepresentation Because real vectors have three parameters, it is not possible to represent polarization, requiring four parameters, by either of the p and q vectors alone. Some information (like the ellipticity of the complex vector) is shared by both vectors, in other respects they are complementary. It is possible to form a pair of real unit vectors as a combination of the two real vectors:

u±(a) = p(a) ± q(a),

(1.37)

which together exactly represent the polarization of the complex vector a. The subscript + or - is not essential, because the pair ±q is not ordered. The properties of u±(a) can be listed as follows. • For a LP we have u., = -u+. Thus, a pair of opposite unit vectors gives us the polarization of the LP vector. • For a CP, u., = u+. Coinciding unit vectors give the plane and sense of polarization of the CP vector. • In the general case, there is an angle 1/J between the unit vectors. From u., + u., = 2p(a) the plane and sense of polarization as well as the ellipticity are obtained, whereas u., - u_+ = 2q(a) gives us the direction of the major axis. The ellipticity and the angle 1/J have the relation e = cos(1/J/2)/(1 + sin'l/;/2)) = tan[(1r -1/J)/4]. (8)

u2

u1 (

-q(a)

)

a

q(a)

a -q(a)

q(a)

(c)

a Fig. 1.4 Unit vector pair ui = u+(a), U2 = u-(a) representation of the complex vector a in (a) linearly polarized, (b) elliptically polarized and (c) circular ly polarized cases.

In fact, any NCP complex vector can be written in the form (1.38)

14

CHAPTER 1. COMPLEX VECTORS

or in the equivalent form

a = 20[(1 + Iq(a)l)q(a) + jq(a) x p(a)], where

(1.39)

a·a

a

= ((1 + Iql)2 _

p2) q2·

(1.40)

These expressions are not valid for CP vectors a, for which u., = u., and q = O. Equation (1.39) can, however, be extended to CP vectors if we let q ~ 0 so that oq = c is finite, whence a = c

+ jc x p.

(1.41)

If only vectors a on a certain plane orthogonal to n are considered, the direction of p(a) is fixed on the line ±n. Then, one single unit vector is enough to represent the polarization of a, since the other one can be obtained from u., + u., = 2p. The unit vector pair corresponds to two points on a unit sphere. This is closely related to the well-known Poincare sphere representation of plane polarized vectors, where only a single point is used. In fact, if a direction on.the plane is chosen, from which the angle

1.6

Complex vector bases

In many practical cases it is necessary to expand a given vector, real or complex, in a base of complex vectors. For example, a wave travelling in the ionosphere is split into characteristic waves with different polarizations, whose propagation can be simply calculated. The propagation of a wave with general polarization must then be written in terms of these characteristic polarizations, after which it is easily computed. The characteristic polarizations are complex so that a decomposition theorem for an arbitrary vector d in terms of three given complex vectors a, b, c is needed, such as the following Gibbs' identity:

(a· b x c)d = (d· a x b)c + (d· b x c)a + (d. c x a)b.

(1.42)

Being a tetralinear identity, (1.42) is valid for all real as well as complex vectors a, b, c, d. It can be derived by expanding the expression (a x b) x

1.6. COMPLEX VECTOR BASES

15

(c X d) in two ways and equating the results. The vector triple a, b, c is called a base if a . b x c # 0, in which case (1.42) gives the decomposition theorem (1.43) d = d· a'a + d· b'b + d· c'c, with the reciprocal base vectors defined by a' == b x c] J, h' = ex a/ J, c' = a x hi J, with J = a x b . c. As an example of longitudinal ionospheric propagation in the direction u (real unit vector), a base corresponding to characteristic polarizations can be formed with two CP vectors a, a* both orthogonal to u, satisfying u x a == ja and u x a" == -ja*. Because p(a) == p(u x a) = u, a has right-hand and a" left-hand polarization with respect to the direction of propagation u. The vector triple is a base, since u . a x a" == ja . a" f:. O. Hence, any field vector E can be expanded as

a*· E a·a

* a· E a·a

E == a--* +a --* +uu·E.

(1.44)

A similar base can be generated from any NLP vector a as the triple

a, a", p(a), because obviously a x a" . p(a) == j(a· a*)p(a) . p(a), which is nonzero for p(a) # 0. An interesting and natural base can be generated from any NCP vector a through the following eigenvalue problem: a x v = Xv,

(1.45)

The eigenvalue A can be easily seen to have the values 0, jva:a., -jva:a. corresponding to the respective eigenvectors v = a, v +, v _, defined by V±

= O:'±

p(a)) va:a. '

.a x

(

pea) =f J

(1.46)

with arbitrary coefficients o±. The vectors defined by (1.46) are easily seen to be CP vectors. If a is CP, the base does not exist. In fact, all three vectors tend to the same vector a as it approaches circular polarization. Also, for LP vectors a (1.46) does not seem to work because p(a) = o. However, the limit exists as a approaches linear polarization, if the product a±p( a) is kept finite.

References DESCHAMPS, G.A. (1951). Geometrical representation of the polarization of a plane electromagnetic wave. Proceedings of the IRE, 39, (5), 540-4.

16

CHAPTER 1. COMPLEX VECTORS

DESCHAMPS, G.A. (1972). Complex vectors in electromagnetics. Unpublished lecture notes, University of Illinois, Urbana, IL. GIBBS, J.W. (1881,1884). Elements of vector analysis. Privately printed in two parts, 1881 and 1884, New Haven. Reprint in The scientific papers of J. Willard Gibbs, vol. 2, pp. 84-90, Dover, New York, 1961. GIBBS, J.W. and WILSON, E.B. (1909). Vector Analysis, pp. 426-36. Scribner, New York. Reprint, Dover, New York, 1960. LINDELL, I.V. (1983). Complex vector algebra in electromagnetics. International Journal of Electrical Engineering Education, 20, (1), 33-47. MULLER, C. (1969). Foundations of the mathematical theory of electromagnetic waves, pp. 339-41. Springer, New York.

Chapter 2

Dyadics Dyadics are linear functions of vectors. In real vector space they can be visualized through their operation on vectors, which for real vectors consists of turning and stretching the vector arrow. In complex vector space they correspondingly rotate and deform ellipses. Dyadic notation was introduced by GIBBS in the same pamphlet as the original vector algebra, in 1884, containing 30 pages of basic operations on dyadics. Double products of dyadics, which give the notation much of its power, were introduced by him in scientific journals (GIBBS 1886, 1891). Gibbs's work on dyadic algebra was compiled from his lectures by WILSON and printed a book Vector analysis containing 150 pages of dyadics (GIBBS and WILSON 1909). Of course, not all the formulas given by Gibbs were invented by Gibbs, quite a number of properties of linear vector functions were introduced earlier by Hamilton in his famous book on quaternions. In electromagnetics literature, dyadics and matrices are often used simultaneously. It is well recognized that the dyadic notation is best matched to the vector notation. Nevertheless, often the vector notation is suddenly changed to matrices, for example when inverse dyadics should be constructed, because the corresponding dyadic operations are unknown. The purpose of this section is to introduce the dyadic formalism, and subsequent chapters demonstrate some of its power. The contents of the present chapter are largely based on work given earlier by this author in report form (LINDELL 1968, 1973a, 1981).

2.1 2.1.1

Notation Dyads and polyads

The dyadic product of two vectors a, b (complex in general) is denoted without any multiplication sign by ab and the result is called a dyad. The order of dyadic multiplication is essential, ab is in general different from bat A polyad is a string of vectors multiplying each other by dyadic products and denoted by ala2a3 ...an' For n = 1 we have a vector, n = 2 a dyad,

CHAPTER 2. DYADICS

18

n = 3 a triad and, in general, an n-ad. Polyads of the same rank n generate a linear space, whose mernbers are polynomials of n-ads. Thus, all polynomials of dyads, or dyadic polynomials, or in short dyadics, are of the form a, b l + a2 b 2 +... + ak b k . Similarly, n-adic polynomials form a linear space of n-adics. A sum of two n-adics is an n-adic. _Here we concentrate on the case n = 2, or dyadics, which are denoted by A, B, C, etc. Dyadics and other polyadics arise in a natural manner in expressions of vector algebra, where a linear operator is separated from the quantity that is being operated upon. For exarnple, projection of a vector a onto a line which has the direction of the unit vector u can be written as u(u· a). Here, the vectors u represent the operation on the vector a. Separating these from each other by moving the brackets of vector notation, gives rise to the dyad uu in the expression u(u . a) = (uu) . a. A dyad is bilinear in its vector multiplicants: (alaI

+ (}2 a 2)b

= (}1(aIb)

+ (}2(a2b),

(2.1)

+ ,82 b2)

== ,81(ab 1 )

+ ,82(ab2).

(2.2)

a(,81 b l

This means that the same dyad or dyadic can be written in infinitely many different polynomial forms, just like a vector can be written as a sum of different vectors. Whether two forms in fact represent the same dyadic (the same element in dyadic space), can be asserted if one of them can be obtained fro~the other through these bilinear operations. A dyadic A can be multiplied by a vector c in many ways Taking one dyad ab of the dyadic, the following multiplications are possible: 0

Co

(ab) == (c v ajb,

(2.3)

c x (ab) == (c x a)b,

(2.4)

(ab)oc==a(boc),

(2.5)

(ab) x c == a(b x c).

(2.6)

In dot multiplication of a dyad by a vector, the result is a vector, in cross multiplication, a dyad. Likewise, double multiplications of a dyad ab by another dyad cd are defined as follows: (2.7) (ab) : (cd) == (a· c)(b· d), (ab)~(cd) == (a x

c)(b x d),

(2.8)

(ab}" (cd) == (a x c)(b· d),

(2.9)

(ab); (cd) == (a· c)(b x d).

(2.10)

2.1. NOTATION

19

These express~ns~a~b~g~eralized.!9 corresponding double products between dyadics, A : B, A~B, A~ B and A~B, when dyads are replaced by dyadic polynomials and multiplication is made term by term. The double dot product produces a scalar, the double cross product, a dyadic, and the mixed products, a vector. These products, especially the double dot and double cross products, give more power to the dyadic notation. Their application requires, however, a knowledge of some identities, which are not in common use in the literature. These identities will be introduced later and they are also listed in Appendix A of this book. The linear space of dyadics contains all polynomials of dyads as its elements. The representation of a dyadic by a dyadic polynomial is, however, not unique. Two polynomials correspond to the same dyadic if their difference can be reduced to the null dyadic by bilinear operations. Because of Gibbs' identity (1.42), any dyadic can be written as a sum of three dyads. In fact, taking three base vectors a, b, c with their reciprocal base vectors a', b', c', any dyadic polynomial can be written as n

n

n

n

i=l

i=l

L a.b, == a L(a' . ai)b + b L(b' . ai)b i + c L(c' .ai)bi. i

i=l

i=l

(2.11)

This is of the trinomial form ae + bf + cg, which is the most general form of dyadic in the three-dimensional vector space. If we can prove a theorem for the general dyadic trinomial, the theorem is valid for any dyadic. A sum sign L without index limit values in this text denotes a sum from 1 to 3. 2.1.2

Symmetric and antisymmetric dyadics

The transpose operation for dyadics changes the order in all dyadic products: (2.12)

Because (AT)T == A, the eigenvalue pr~lem AT == ,\A has the~igenvalues ,\ == ±1 corresponding to symmetric As and antisymmetric Aa dyadics, which satisfy

A; == As, AaT

== -Aa.

(2.13)

Any dyadic can be uniquely decomposed into a symmetric and an antisymmetric part: =

1=

=T

A == 2(A + A )

1=

+ 2(A -

=T

A ).

(2.14)

Every symmetric dyadic can be written as a polynomial of symmetric dyads:

CHAPTER 2. DYADICS

20

=

As

1

= Laibi = "2 L(aibi + b.a.) =

"2 L ((ai + bi)(ai'+ b.) - a.a, 1

b.b.},

(2.15)

The number of terms in this polynomial is, however, in general higher than 3. A single dyad cannot be antisymmetric, because from the condition ab == -ba, by multiplying by a*· and dividing by a· a" we see that b must be of the form oa, whence ab = oaa == -aaa == O. Instead, any antisymmetric dyadic can be expressed in terms of two dyads in the form ab - ba. The vectors a, b are not unique. This is seen from the following expansion with orthonormal unit vectors Uj:

(2.16)

with (2.17)

Thus, the general antisymmetric dyadic can be expressed in terms of a single vector c. If we write c == -a x b, going (2.16) backwards we see that any antisymmetric dyadic can be expressed as ab - ba. The choice of orthonormal basis vectors does not affect this conclusion. The linear space of dyadics is nine dimensional, in which the antisymmetric dyadics form a three-dimensional and the symmetric dyadics a sixdimensional subspace.

2.2

Dyadics as linear mappings

A dyadic serves as a linear mapping from a vector to another: a -+ b == D . a. Conversely, any such linear mapping can be expressed in terms of a dyadic. This can be seen by expanding in terms of orthonormal basis vectors u, and applying the property of linearity of the vector function

rea):

rea) = I : u.u, . f(I: UjUj . a) = j

(I: I : u, . f(Uj)UiUj) . a. i

(2.18)

2.2. DYADICS AS LINEAR MAPPINGS

21

The quantity in brackets is of the dyadic form and it corresponds to the linear function f(a). _ _ The unit dyadic 7 corresponds to the identity mapping 7 · a = a for any vector a. From Gibbs' identity (1.42) we see that for any base of three vectors a, b, c with the reciprocal base a', b', e', the unit dyadic can be written as (2.19) 1 = aa' + bb' + ee'. Taking an orthonormal base u, with form

u~ =

u., the unit dyadic takes the (2.20)

The unit dyadic is symmetric and satisfies 1· D = D·l = D for any dyadic D. This and (2.19) can be applied to demonstrate the relation between matrix and dyadic notations by writing (2.21) or any dyadic can be written in terms of nine scalars D i j . These scalars can be conceived as matrix components of the dyadic w~h respect to the base {ail. The matrix components of the unit dyadic 1 are {6i j } in all bases. From (2.16) it can be seen that the most general antisymmetric dyadic can be written as

A a ~ ab - ba = (b x a) x I =

I

x (b x a),

(2.22)

as is_seen_if (2.20) is substituted in (2.22). Thus, dyadics of the form e x 1 = 1 x e are antisymmetric. Th~vector e corresponding to the antisymmetric part of a general dyadic D can be obtained through the following operation: 1-c(D) = 2.1~ D.

(2.23)

For an antisymmetric dyadic, (2.23) can be easily verified from (2.22). For a symmetric dyadic, e(D) = 0 is also easily shown to be valid. All dyadics can be classified in terms of their mapping properties. • Complete dyadics D define a linear mapping with an inverse, which is represented by an inverse dyadic D-l. Thus, any vector b can be reached by mapping a suitable vector a by D · a = b.

22

CHAP1'ER 2. DYADICS

• Planar dyadics map all vectors in a two-dimensional subspace. If we take a base [a.}, the vectors {D . a.] do not form a base, because they are linearly dependent and satisfy (D·al)·(D·a2) x (D·a3) == o. Writing the general D in the trinomial form ab + cd + ef, we can show that the vector triple a, c, e is linearly dependent and one of these vectors can be expressed in terms of other two. Thus, the most general planar dyadic can be written as a dyadic binomial ab + cd.

• Strictly planar dyadics are planar dyadics which cannot be written as a single dyad. • Linear dyadics map a!!- vectors in a one-dimensional subspace, l.e. parallel to a vector c: D· a == oc. Thus, D must obviously be of the form cb. Linear dyadics can be written as a single dyad. Finally, we can distinguish between strictly linear dyadics and the null dyadic. As examples, we note that the unit dyadic I is complete, whereas an antisymmetric dyadic is either strictly planar or the null dyadic. The ina.b, can be written quite verse dyadic of a given complete dyadic D == straightforwardly in trinomial form. First, to be complete, the vector triples [a.}, {bi} must be bases because from linear dependence of either base, a planar dyadic would result. Hence, there exist reciprocal bases {a~}, {b~}, with which we can write

E

(2.24) That (2.24) satisfies D . n- 1 == D- 1 . D

2.3

== I, can be easily verified.

Products of dyadics

Different products of dyadics playa role similar to dot and cross products of vectors, which introduce the operational power to the vector notation. The products of dyadics obey certain rules which are governed by certain identities summarized in Appendix A. 2.3.1

Dot-product algebra

The dot product between two dyadics has already been mentioned above and is defined in an obvious manner: (2.25)

2.3. PRODUCTS OF DYADICS

23

With this dot product, the dyadics form an algebra, where the unit dyadic, null dyadic and inverse dyadics are defined as above. This algebra corresponds to the matrix algebra, because the matrix (with respect to a given base) of A . B can be shown to equal the matrix product of the matrices of each dyadic. Thus properties known from matrix algebra are valid to dot-product algebra: associativity

A. (B· C) = (A. B)· C, and (in general) non-commutativity,

A· B

=1=

(2.26)

B . A. Further, we have

(A . B)T = B T . AT, (A. B)-1 = B- 1.

(2.27)

A-I.

(2.28)

Powers of dyadics, both positive and negative, are defined through the dot product (negative powers only for comp~te dyad~cs) in an obvious manner. For example, the antisymmetric dyadic u, satisfies for all n > o.

A = u x 7 with an

A4n = 1=A4n +l --

NCP unit vector

uu,

(2.29)

A,

(2.30)

A4n +2 - -1 + uu,

(2.31) (2.32)

Because for re~l u,. A can be interpreted as a rotation by

1r /2

around u,

the powers of A can be easily understood as multiples of that rotation. Two dyadics do not commute in general in the dot product. It is easy to see that two antisymmetric dyadics only commute when one can be written as a multiple of the other. This is evident if we expand the dot product of two general antisymmetric dyadics:

(a x 1) . (b x 1)

= ba -

(a· b)1.

(2.33)

If this is r~quired to be symmetric in a and b, we should have ab = ba or (a x b) x I = 0, which implies a x b = 0 or a and b are parallel vectors. A dyadic commutes with an antisymmetric dyadic only if its symmetric and ~ntisymmetric parts commute separately. In fact, writing D = D s + d x I in terms of its symmetric and antisymmetric parts, we can write

D . (a x

I) - (a x I) . D =

(D s x a)

+ (D s

x a)T - (a x d) x

I.

(2.34)

CHAPTER 2. DYADICS

24

Equating the antisymmetric and symmetric parts <>!. (2.34) to zero, shows us t~at the symmetric and antisymmetric parts of D must commute with a x separately. Thus, the antisymmetric part of D must be a multiple of a x 1. The symmetr~ part of D must be such that D 8 X a is antisymmetric, i.e. of the form b x 1. Multiplying this by -a gives zero, whence b must be a multiple ~f 8. It is easy to show that the symmetric dyadic must be of the form 0./ + {3aa. Thus, the_most general dyadic, which commutes with the antisymmetric dyadic 8 x I is necessarily of the form

J.

D =

QJ + {3aa + ,a x I.

(2.35)

A dyadic of this special form is called gyrotropic with axis 8, which may also be a complex vector. From this, it is easy to show that if a dyadic commutes with its transpose, it must be either symmetric or gyrotropic.

2.3.2

Double-dot product

The double-dot product of two dyadics A scalar

= L a.b., B = L cjdj gives the (2.36)

This is symmetric in both dyadics and satisfies

(2.37)

I) : (b x I) = 2a . b, A : I = 2)ai · hi) = trA. (a x

(2.38) (2.39)

The last operation gives a scalar which can be called the trace of A because it gives the trace of the matrix of A in any base [c.]. In fact, writing

= , . A = L AijCiCj gives

us

As special cases we have I :I = 3 and A : B = (If . BT) : I = (B . AT) : I. A dyadic whose trace is ze.!o is called trace free. Any dyadic can be written as a sum of a multiple of I and a trace-free dyadic: -

1- --

-

1---

D = -(D : 1)1 + (D - -(D : /)/). 3 3

(2.41)

2.3. PRODUCTS OF DYADICS

25

Antisymmetric dyadics are trace free. In fact, more generally, if symmetric and A antisymmetric, ~e have from (2.37)

S is

(2.42)

The scalar D : D* is a non-negative real number for any complex dyadic D. It is zero only for D = 0, which can be shown from the following: = =T =* ~ = . u.] 2 . D : =* D = (D . D ): = I == L...J u, . =r D . =* D . u, == ~ L...J ID

(2.43)

Here, {Ui} is a real orthonormal base. (2.43) is seen t~give a non-negative number and vanish only if all D . u, vanish, whence D == L D . u.u, == O. We can define the norm of D as

II DII = JD:D*. 2.3.3

(2.44)

Double-cross product

The double-cross product of two dyadics produces a third dyadic. Thus, it defines a double-cross algebra. Unlike the dot-product algebra, the doublecross algebra is commutative:

=x= =x= AxB = BxA,

(2.45)

and non-associative, because A~(B~C) =/; (A~B)~C in general. The commutative property follows directly from the anticommutativity of the cross product: a x b == -b x a, as is easy to see. It is also easy to show that there does not exist a unit element in this algebra. A most useful formula for the expansion of dyadic expressions can be obtained from the following evaluation with dyads: (ab)~[(cd)~(ef)]

= [a x (c x e)][b x (d

x f)] ==

[ca e - a · ce][db . f - b · df] = (ab : cd)ef + (ab : ef)cd - ef· (ab)T . cd - cd· (ab)T . ef.

(2.46)

This expression is a trilinear identity for dyads. Thus, every dyad can be replaced by any dyadic polynomial because of linearity, whence (2.46) may be written for general dyadics: (2.47)

CHAPTER 2. DYADICS

26

Use of this dyadic identity adds more power to the dyadic calculus. Its memorizing is aided by the fact that, because of the commutative property (2.45), Band C are symmetrical in (2.47). The method used above to obtain a dyadic identity from vector identities can be generalized by the following procedure. 1. A multilinear dyadic expression (linear in every dyadic) is written in terms of dyads, i.e. every dyadic is replaced by a dyad. 2. Vector identities are applied to change the expression into another form. 3. The result is grouped in such a way that the original dyads are formed.

4. The dyads are replaced by the original dyadics. _ To demonstrate this procedure, ~t us expand the dyadic expression (A~B) : I, which is linear in A and B. Hence, we start by replacing them by ab andcd, respectively, and applying the well-known vector identity (ab~cd) : I = (a x c) · (b x d) = (a· b)(c · d) - [a- d)(b· c). This can be grouped as (ab : I)(cd : I) - (ab) : (cd)?'. Finally, going back to A and B leaves us with the dyadic identity ==

=

(A~B) : I

== = (A=:= I)(B : I) -

=nr

A :B ,

(2.48)

or trace of A~B equals trA trB - tr(A · B). New dyadic identities can also be obtained from old identities. As an example, let us write (2.48) in the form (2.49) To obtain this, we have applied the invariance in any permutation of the = = = = = =x= = triple scalar product of dyadics, A~B : C = A~C : B = BxA : C = · ... (Of course, the double-cross product must always be performed first.) Because (2.49) is valid for any dyadic B, the bracketed dyadic must be the null dyadic, and the following identity is obtained:

=

= == 7T AxI=(A:I)I-A.

=x=

(2.50)

That D : B = 0 for all B implies D = 0, is easily seen by taking B = UiUj from an orthonormal base [u.}, whence all matrix coefficients Dij of D can be shown to vanish. The identity (2.48) is obtained from (2.50) as a special case by operating it by : B.

2.3. PRODUCTS OF DYAD/CS

27

An important identity f~ the double-cross product can be obtained by expanding the expression (A~B)~(C~D) twice through (2.47) by considering one of the bracketed dy~dics as a single dyadic, and equating the expressions. Setting C = D = / we obtain

=x= AxB= This identity could be also conceived as the definition of the double-cross product in terms of single-dot and double-dot products and the transpose ~p~atio~ It is easily seen that (2.50) is a special case of (2.5~. _ Also, /~1 = 2/ is obtained as a further special case. The operation A~/ is in fact a mapping from dyadic to dyadic. Its properties can be examined through the following dyadic eigenproblem: =x=

Ax!

= = ..\A.

(2.52)

Taking t~ trace of (2.52) leaves us with (2 - "\)A : / = 0, whence either ,.\ = 2 or A is trace free. Substituting (2.50) in (2.52) gives us the following different solutions: • ,.\ = 2 and

A=

a/ where a is any scalar;

• ,.\ = 1 and A is antisymmetric; • ,.\ = -1 and

it is symmetric and trace free.

Any dyajic can be written uniquely as a sum of three components: a multiple of 1, an antisymmetric dyadic and a trace-free symmetric dyadic,

(2.53) respectively. It is a simple matter to check that the right-hand side of (2.53), each term multiplied by the corresponding eigen~alue ,.\, gives the sam~ ~sult as (2.50). T~r~e~sts a~ inverse maR.ping to 1~ which, denoted

by I(A) and satisfying I(A~/) = A for every of inverse eigenvalues, or in the simple form

A, can

be written in terms

(2.54)

28

CHAPTER 2. DYADICS

Finally, let us ~onsider the double-cross mapping ~hrough an antisymmetric dyadic d x 1. Every antisymmetric dyadic a x 1 is mapped onto the planar symmetric dyadic

(a x Every symmetric dyadic

1) ~ (d x I) = ad + da.

(2.55 )

S is mapped onto the antisymmetric dyadic: S~(dXI)=(S.d)XI.

(2.56)

The mapping of the general dyadic is obtained when it is decomposed into a symmetric and an ~ntisymmetric part. There does not exist an i~verse to the mapping ~ (d x I), because all symmetric dyadics satisfying S . d == 0 are mapped onto the null dyadic.

2.4

Invariants and inverses

In matrix algebra, invariants are such functions of a matrix that are independent of the basis in which the matrix is formed. Thus, they can be expressed as functions ~ th! corresponding dyadic. The trace was defined in (2.39) as the scalar A : I, corresponding to the sum of diagonal terms in the matrix. The second invariant can be denoted, and expanded from (2.48), as -

spmA ==

1- - 1 - - - -A~A: 1 = -[(A: 1)2A: AT],

(2.57) 2 2 and its counterpart in matrix algebra is the sum of principal minors. The third invariant corresponds to the determinant of the matrix and can be expressed as -

1- -

-

detA = 6A~A; A.

(2.58)

Finally, the cross-product square of the dyadic is defined as

A(2) = !A~A. 2

(2.59)

t

As an example, 1(2) = can be easily verified. Writing the dyadic A in the form Laibi, from (2.59) we can find the following vector expression for the cross-product square:

(2.60)

2.4. INVARIANTS AND INVERSES

29

where {a'l and {hi} are reciprocal to the bases {a}, {b}. If A is not complete, the last expression in (2.60) is not meaningful, since the reciprocal bases do not both exist. Further, we can write for the determinant, (2.61) Combining (2.24), (2.60) and (2.61) gives us the formula for the inverse of a dyadic: =

A-I

A(2)T (AXA)T x __ = ---== 3__ - - -, detA

(2.62)

A~A: A

which implies (2.63) Because detI = 1, the in~rse of I is easily seen to be I, as expected. The inverse exists only if detA ~ 0 is satisfied, which serves as a definition of the complete dyadic. In fact, the classification of dyadics can be written in the form • complete dyadic if detA =f.

o.

• strictly planar dyadic if detA = 0 and A~A ~ • linear dyadic if A~A =

o.

o.

There exists a relation between the cross-product square of a dyadic and its dot-product square. In fact, from (2.51) we can write (2.64) After transposing (2.64), dot-multiplying by if ~nd applying (2.62) we end up with a relation between powers of a dyadic if (2.65) which is called the Cayley-Hamilt~equation, but actually it is an identity, since it is satisfied by all dyadics A. As other properties of the cross-product square we can easily prove that (2.66)

30

CHAPTER 2. DYAD/CS

which tells ~ t~t if is planar, A~A is a linear dyadic. In fact, a dyadic of the form A~A cannot be st:!.ctly planaE! it is either complete or linear. To prove the property det{An) = (detA)n known from matrix algebra, we need the identity

A

(2.67) which can be derived with the method described above. The following is a special case: (2.68) which can also be written as

A(2) . B(2) = (A . B)(2).

(2.69)

Applying this and (2.63) we have det{A· B)

= (detA){detB)

(2.70)

after some algebra. This in fact shows that the dot product of any number of dyadics has a determinant function which equals the product of determinant functions of individual dyadics. Thus, the product of dyadics is planar if anyone of the dyadics is a planar dyadic.

2.5

Solving dyadic equations

In this section, some principles for and results of solving dyadic equations are considered. The presentation does not cover all the kin~ of equations ~at may occur. By a dyadic equation we mean the form A = B, where A and B ~ay ~ontain unkn~n quantities. In the following cases, the equation A = B reduces to A = 0 and B = 0, or is split up into two equations which are usually easier to solve than the original one.

• 11 =

aC where C is a complete dyadic, and B is known to be planar. (Taking the determinant function of each side gives a = 0.) If a =I 0, there is no solution at all.

• A = 0:1 and B is trace free. (Taking the trace operation of each side gives us a = 0.)

• 11 is symmetric and B

is antisymmetric.

• A is antisymmetric and B is linear dyadic. (Because an antisymmetric dyadic is either strictly planar or null dyadic, both are null dyadics.)

2.5. SOLVING DYADIC EQUATIONS

31

For example, we might encounter an equation of the type

=x= = AxA=axI.

(2.71)

Because A~A is either complete, linear or null dyadic and the antisymmetric dyadic a ~1 either strictly planar or null dyadic, we have a = 0 and A~A = 0, or A may be any linear dyadic.

2.5.1

Linear equations

Let us study the linear dyadic equation (2.72) If A is complete, the solution X = A-i. B is defined by (2.62). Problems arise when A is planar. If in this case B is complete, (2.72) does not possess a solution at all. Thus, for (2.72) to have a solution when 11 is planar, B must also be planar, which does not warrant a solution, however. If there is a solution, it is not unique unless we restrict it somehow. Let u~study the problem (2.72) where A is strictly planar, and try to solve for x. Because A is strictly planar, we have detA = 0 and A (2) i= o. In fact, the cross-product square of A is a strictly linear dyadic and can be written in theJorm A(2) = ab =1= o. From (2.63) we have A(2)T . A = A· A(2)T = 0, or a . A = A · b = o. Thus, A defines a class A of planar dyadics which are orthogonal to a from the left and to b from the right. This class can be used to define a unique solution to the equation (2.72). From (2.72) it is seen that there does not exist a solution unless B satisfies A(2)T . B = 0, (2.73) or a . B = O. B need not belong to class A. Let us look for the solution of (2.72) in the form X = D·B and find a dyadic D, whose conjugate transpose is in class A. Such a dyadic is called the planar inverse or generalized inverse of A and is simply denoted by D = A -1. Because the inverse in the strict sense (2.62) does not exist for planar dyadics, there should not be any place for misinterpretation due to the notation. The planar inverse can be written in the form =

A- 1

=

(A x lI* (2»)T . X

A(2) : A*(2)

(2.74)

32

CHAPTER 2. DYADICS

This is a solution whose conjugated transpose is in class A, as can be easily checked. To verify (2.74), we expand using (2.47) and (2.64):

A· (A~(A*~A*))T = 2[tr(A. A*T)(A. A*T) -

(A. A*T)2] =

2[spm(A . A*T)I - (A . A*T)(2)T],

(2.75)

whence we can write = (A~A*(2»)T = (A* . AT )(2) = A· _ _ =1_ _ =lp •

spm(A* . AT)

spm(A* . AT)

(2.76)

Here, the dyadic I p can be~alled the planar unit dyadic of class A. Multiplying (2.76) by ·B from the right and taking (2.69), (2.73) and the identity (2.77) into account, we see that X =_A- 1 . B is really a solution of (2.72). (2.~) is called the planar inverse of A. The denominator in (2.74) equals spm(A· A*T) and is never zero for strictly planar dyadics The most general solution to (2.72) can be written in the form

A.

x = A-I. B + x ;

(2.78)

where X 0 is any solution to the homogeneous equation

A·Xo=O.

(2.79)

!

Because is strictly planar, X 0 must be a linear dyadic, as is easily seen writing A = al hI + a2 h 2 . In fact, we obviously must have

= X;

= =(2)T A .= C,

(2.80)

as t.!:e most general solution with an arbitrary dyadic C. The equation X ·11= B can be solved in corresponding manner. Let us now study the linear equation A~X

= B,

(2.81)

where A is a complete dyadic. Multiplying this by (A~ A) ~ and applying (2.47) we easily have for the solution = =-lTx= A: B= X = A xB - ----:=A.

2detA

For example, if A = I, the expression (2.54) is obtained.

(2.82)

2.5. SOLVING DYADIC EQUATIONS

2.5.2

33

Quadratic equations

Let us study the quadratic equation (2.83) which is quite easy to solve if A is a complete dyadic. Double-cross multiplying both sides with itself and applying (2.66) gives us the two solutions

=

x=±

=x= AxA

~,

(2.84)

V8detA which fails for planar dyadics satisfying detA = o. If A is s~ictly planar, :4(2) t= 0, there is no solution at all, as was seen earlier. If A is strictly linear, there exist infinitely many strictly planar solutions X. In fact, in this case A defines a class of strictly planar dyadics satisfying X . = X = o. For = 0, any linear dyadic is a solution.

AT AT .

A

Square roots of dyadics The quadratic equation (2.85) is more difficult to solve than (2.83). Its solutions, square roots of:4, may be infinEe or zero in number, or something in be.!ween, depending on the dyadic A. For exa~ple, all dyadics of the form ±(1- 2uu) are square roots of the unit dyadic 1 for all unit vectors u. A solution procedure can be based on the solution of (2.83) through ~e following identity, which can be derived by applying the identity for A~(B~C):

(2.86) The procedure is first to solve for the unknown scalar a = spmX and then solve (2.86). An equation for a can be derived as follows. Denoting

D(o) = a1 + X 2

= a1 + A, we can write (X~I)(2) = D(a), [(X~I)(2)](2)

(2.87)

= [det(X~I)]X~I = D(2),

det[(X~I)(2)] = [det(X~I)]2

= detD.

(2.88) (2.89)

CHAPTER 2. DYADICS

34

For D(a) ~mplete, from (2.88) and (2.89) we can express the unknown dyadic x~I in terms of the unknown dyadic D(a): _

_

D(2)

X~I==

__ det(X ~ I)

==±

D(2)

_,

(2.90)

J detD

whence by taking the spm() operation of this and applying the identity (trX)2 = 2spmX + tr(X 2) we obtain the following quartic equation for the unknown scalar a: a4

-

2a 2spmA - 8a detA + (spmA)2 - 4trA detA ==

o.

(2.91)

Values of a substituted in (2.90) solved for X

X =±

1 = ((spmD)I _ 2D(2)T)

(2.92)

2VdetD give us possible square root solutions corresponding to those cases for which the dyadic D(a) is complete. Written explicitly, the solution is

X

= A1 / 2 = ±

~

_ ((spm(A + 0:1))1 -

2Vdet(A + 0:1)

2(A + o:I)(2)T) .

(2.93) If the completeness condition is not valid, there are either no or infinitely many solutions corresponding to that particular solution of (2.83). As an example of a dyadic with no square roots we may write

A=uv+ww,

(2.94)

is

an orthon~rmal set of unit vectors. In fact, this corwhere u, v, w responds to detA = 0, spmA = 0, whence (2.91) ~nly has the solution a == o. Thus, 15 equals the strictly planar dyadic A, whence (2.88) does not possess a solution. A dyadic of this kind is called a shearer.

Square roots of the unit dyadic As an example, le~us c..?nsider the problem of finding all the square roots of the unit dyadic A == I. In this case, the quartic (2.91) can be written as

(a-3)(a+l)3=O, or the roots are a

== 3 and

(2.95)

a = -1. The equation (2.86) now reads

(2.96)

2.5. SOLVING DYADIC EQUATIONS

35

whose right side equals 41 and zero for the respective two a values. The solution formula (2.93) can n<:.w be applied only for the case a = 3 and it results in the obvious roots ±1. The other roots are obtained as solutions of (2.97) which is valid if X~I is any linear dyadic. The solutions X satisfying both this and X 2 == are of the general form

I

..:..ab ) X==± (1-2a·b with a . b

,

(2.98)

=I o.

Square roots of the null dyadic

Solutions of (2.99) must be linear dyadics, because, as is known from solutions of (2.72), at least one.-?f two dyadics must be linear if their product is the null dyadic. Writing X = ab in (2.99), we see that any trace-free linear dyadic, i.e. a dyadic of the form aa x c, is a solution of (2.99). 2.5.3

Shearers

Solutions of the cubic equation (2.100) i.e. cubic roots of the null dyadic, are important because of their physical significance. If X is a linear dyadic satisfying (2.100), obviously it also satisfies (2.99), or it is trace free. Thus, the other solutions must be strictly planar and can easily be shown to ~tisfy the three scalar equations trX = 0, spm X == 0 and detX = o. If X is written as a dyadic binomial, it is easily concluded that the most general solution of (2.100) is of the form

x

== a x (bb + ca), or X == (ab + cc) x a.

(2.101)

(2.101) also includes the solutions of (2.99) for c = 0 or b = o. Dyadics of the form (2.101) in general, including the trace-free linear dyadics, are called trace-free shearers, which are special cases of shearers, defined by the two equations

detA == 0,

spmA

== o.

(2.102)

CHAPTER 2. DYADICS

36

Note that this defini!ion d~ffers from that adopted by Gibbs, who calls dyadics of the form A + Ctl shearers. From the Cayley-Hamilton identity (2.65) we can write another condition equivalent to (2.102), for the definition of the shearer: (2.103) It is easily seen that, in the trace-free case, (2.103) corresponds to (2.100). The most general shearer can be written in the forms

A = a x (be + da), or A = (ad + eb) x a.

(2.104)

It is not difficult to check that (2.102) and (2.103) are satisfied for any vectors a, b, c, d. As was seen just before, a strictly planar shearer does not have a square root.

2.6

The eigenvalue problem

Right and left eigenvalue problems with eigenvalues and eigenvectors cq, a, and, respectively, {3i, hi, are of the form (2.105) (2.106) Because the dyadic A-,1 is planar when, equals satisfy the equation

- det(A -

,1) = ,3 - ,2 trA +, spmA -

ai or (3i, the eigenvalues detA =

o.

(2.107)

Because both right and left eigenvalues satisfy the same problem, they have the same values which are denoted by~i. There are either one, two or three different values for Cti. Because b i · A· aj = (aihi)· aj = hi· (ajaj), we see that if Cti f; Ctj, the left and right eigenvectors are orthogonal, i.e. they satisfy b, . aj = o. Eigenvectors hi and ai corresponding to a solution Cti of (2.107) can be constructed using dyadic methods. The construction depends on the mul~tude ~f the particular eigenvalue, which depends on whether the dyadic A - ail is strictly planar, strictly linear or null. Let us consider these cases separately. For this we need the following identities:

a x (A~B) = B x (a· A) + A x (a· B), (A~B) x a

= (:4. a)

xB

+ (B· a) x :4,

(2.108) (2.109)

37

2.6. THE EIGENVALUE PROBLEM

with the special cases a x (A~A) = 2A x (a A),

(2.110)

(A~A) x a =

(2.111)

2(A· a) x A.

These can be derived with the general method described in Section 2.3, for creating dyadic identities.

=JA -

• Strictly planar A - aJ Defining B, aJ)~(A - aJ) i= 0, from (2.110), (2.111) we see that hi x B; = B i x a, = 0, whence there exists a scalar i= 0 ~uch that B; = eibiai. Thus, from the knowledge of Oi the dyadic B i is known and the eigenvectors can be written in the form a, = C • B, and hi = B i . c with suitable vector c. In this case Oi is a single root of (2.107).

ei

-'!- oil. There exist non-null vectors c, d such that A is of the form 0.1+ cd. This is a special type of dyadic, which is called uniaxial. (2.107) leaves us with the equation (;'-0)2(,-0-c·d) = 0, which shows us that the eigenvalue Qi = Q is a double root of (2.107). Assuming c . d -# 0, the third root is a simple one, Qj = Q + c . d, for which the eigenvectors can be obtained through the expression above. The left and right eigenvectors corresponding to the double root are any vectors satisfying the conditions b . c = 0 and d . a = 0, respectively. For c . d = 0, all three eigenvalues are the same, but there only exist two linearly independent eigenvectors, those just mentioned.

• Strictly linear

• Null dyadic A - Oil. In this case A = aI, there is a triple eigenvalue and any vector i~ an eigenvector. This happens if A is a multiple of the unit dyadic 1.

°

The previous classification was made in terms of the dyadic A - ail, or multitude of a particular eigenvalue ai. Let us now consider the number of different eigenvalues of a dyadic iI, which can b~ 1, 2 or 3. Because detA = a10203, spmA = {}102+02Q3+a301 and trA = a1 +a2+a3, the Cayley-Hamilton equation (2.65) can be written as (2.112)

The order of terms is immaterial here.

38

CHAPTER 2. DYAD/CS

-=-

• One eigenvalue 01 = 02 == 03. Because (A 01/)3_ = 0, from the solution of equation (2.100) we conclude th~t 11 - ~II is a trace-free shearer, whence the most general form for A is 01I.+ (ab + cc) x a. Taking the trace operation it is seen that 01 = trA/3. The number of eigenvectors is obviously that of the shearer term. Here we can separate the cases. - Strictly planar trace-free shearer with a . b x c :f:. o. It is easy to show that there exists only one eigenvector i!eft a~ ri~t), which can be obtained from the expression (11 - 011) ~ (A -

oIl) = 2(a . b x c)(a x c)a. Thus, the left eigenvector is a x c and the right eigenvector a. Only this type of dyadic has just one eigenvector. - Linear shearer, w..Eich c~n be written with c = 0 in the above expression. Now 11 - all satisfies (2.99). There exist two eigenvectors, which are orthogonal to vectors a from the left and b x a from the right.

• Two eigenvalues a1 =F a2 = a3. The dyadic A satisfies an equation of the form (2.103): B2(B - trB = 0 with B = 02/, as is

I)

A-

~asily

verified. Thus..!. the dyadic B must be a general shearer and is of the form 021 + a x (be + da). The eigenvalue 01 equals a2 + a x b . c, whence the shearer here cannot be trace free in order that the two eigenvalues do not coincide. In this case there exist two linearly independent eigenvectors.

A

• Three eigenvalues 01 :f:. 02 :f:. a3. In this case there exist three linearly independent eigenvectors. In fact, assuming the eigenvector a3 to be a linear combination of a1 and a2, which are linearly independent, (2.105) will lead to the contradictory conditions a3 aI, a3 = a2.

=

As a summary, the following table presents the different cases of dyadics with different numbers of eigenvalues (N in horizontal lines) and eigenvectors (M in vertical columns).

1 2 3

2

3

none

none

1

MIN 01

+ (ab + cc) x a 0/ + aa x b 0/

0/

+ (ab + cd) 0/

+ab

xa

none

ab + cd + ef

2.7. HERMITIAN AND POSITIVE DEFINITE DYADICS

39

Because the left or right eigenvectors of a dyadic with three linearly independent eigenvectors form two bases {hi}, {ail, the dyadic can be written in either base as (2.113) From this we conclude that the left and right eigenvectors are in fact reciprocals of each other: ai = hi, hi = ai. If two dyadics commute, they have the same eigenvectors.

2.7

Hermitian and positive definite dyadics

Hermitian and positive definite dyadics are often encountered in electromagnetics. In fact, lossless medium parameter dyadics are hermitian or antihermitian depending on the definition. Also, from power considerations in a medium, positive definiteness of dyadics often follows. 2.7.1

Hermitian dyadics

By definition, the hermitian dyadic satisfies

fIT=A*,

(2.114)

whereas the antihermitian dyadic is defined by AT =

-A*.

(2.115)

Any dyadic can be written as a sum of a hermitian and an antihermitian dyadic in the form (2.116) Any hermitian dyadic can be written in the form E ±cc* and antihermitian, in the form L: ±jcc*. Conversely, these kinds of dyadics are always hermitian and antihermitian, respectively. From (2.114), (2.115) it follows that the symmetric part of a hermitian dyadic is real and the antisymmetric part imaginary, whence the most general hermitian dyadic H can be written in the form (2.117) with real and symmetric S and real h. Any antihermitian dyadic can be written as jH, where H is a hermitian dyadic.

40

CHAPTER 2. DYADICS

Hermitian dyadics form a subspace in the linear space of dyadics. The dot product of two hermitian dyadics is not necessarily hermitian, but the double-cross product is, as is seen from (2.118) where A and B are hermitian. Also, the double-dot product of two hermitian dyadics is a re~ num~r, as J.s easy to s~ from the sum expression. Thus, the scalars trA, spmA, detA are real if A is hermitian. This implies that the inverse of a hermitian dyadic is hermitian. The following theorem is very useful when deriving identities for hermitian dyadics: = A : aa * = 0 for all a, =} = A = o. (2.119) This can be pr~ed by setting first a = b + e and then a = b + je, whence the condition if : be = 0 for all ~ectors b, c will result fro~ (2.119), making the matr~ components of if vanish. For comparison, A : aa = 0 for all a implies A antisymmetric, as is easy to prove. From (2.119) we can show that if aa" is real for all vectors a, is hermitian. In fact, this implies A: aa" - A* : a*a = 0 or (A - A*T) : aa" = 0, whence A is hermitian. A hermitian dyadic always has three eigenvectors no matter how many eigenvalues it has, as can be shown. The right and left eigenvectors corresponding to the same eigenvalues are complex conjugates of each other, because from A· a == oa we have A* .a" = a" . it == a*a*. But eigenvalues are real and eigenvectors conjugate orthogonal, because (ai -aj)ai .aj = 0, whence ai -a1 = 0 and a, .a; = 0 for ai =f. aj. Thus, the general hermitian dyadic can be written in terms of its eigenvalues and eigenvectors as

A;

A

(2.120)

2.7.2

Positive definite dyadics

By definition, a dyadic D is positive definite (PD), if it satisfies

D : aa"

> 0,

for all a f:.

o.

(2.121 )

A PD dyadic is always hermitian, as is evident. Other properties, whose

proofs are partly omitted, follow.

41

2.8. SPECIAL DYADICS

• PD dyadics are complete. If D were planar, there would exist a vector a such that Ir-e. = 0, in.contradiction with (2.121). Thus, the inverse of a PD dyadic always exists. • PD dyadics possess positive eigenvalues. This is seen by dot multiplying the expansion (2.120) by ajaj/aj . aj, and the result is aj, which must be greater than 0 because of (2.121). • if A is PD, its symmetric part is PD.

• A is

PD exactly when its invariants positive.

trA,

spmA,

detA

are real and

PD implies A~B PD.

• A and B

• A dyadic of the form

A.A*T is positive semidefinite and PD if A is

complete.

2.8

Special dyadics

In this section we consider some special classes of dyadics appearing in practical electromagnetic problems. Rotation and reflection dyadics emerge in symmetries of various structures whereas uniaxial and gyrotropic dyadics are encountered when electromagnetic fields in special materials are analysed. Parameters of some media like the sea ice can be approximated in terms of a uniaxial dyadic, while others like magnetized ferrite or magnetoplasma may exhibit properties which can be analysed using gyrotropic dyadics. 2.8.1

Rotation dyadics

In real vector space, the rotation of a vector by an angle () in the right-hand direction around the axis defined by the unit vector u can be written in terms of the following dyadic:

R( u, 0) = uu + sin O( u x I)

+ cos (}(I -

uu),

(2.122)

It can also be written in the form eu xIB, as is seen if the dyadic exponential function is written as a Taylor series. The rotation dyadic obeys the properties R(-u,O)

= R(u, -B) = RT(u, 0) = :R- 1 (u , 0),

(2.123)

CHAPTER 2. DYADICS

42

R( U, ( 1) . R( u, ( 2 ) == R( u, 81

+ (2 ) ,

(2.124)

R( U, 8) ~ R(u, (J) == 2R(u, 8), -

detR( u,

(J) ==

1-

(2.125)

-

"3 R( u, 8) : R( u, (J) == 1.

(2.126)

It is not difficult to show that the propertie~detR == 1 and R T == R,-l uniquely define the form (2.122) of the dyad~ R(u, (J), so that they could be given as the definition. Also, (2.125) with R 1= 0 would do. In general, (J and/or u may be complex, which means that the geometrical interpretation is lost. The resulting dyadic is also called the rotation dyadic in the complex case. The dot product of two rotation dyadics with arbitrary axes and angles is another rotation dyadic. This is seen from -

-

-

-

1-

-

-

-

(R 1 • R2)~(Rl . R2) = 2(Rl~Rd· (R2~R2)

--

= 2(R 1 • R 2),

(2.127)

where use has been made of the identity (2.67). It is not very easy to find the axis and angle of the resulting rotation dyadic. This can be done perhaps most easily by using a representation in terms of a special gyrotropic dyadic. The gyro tropic dyadic was defined in (2.35) and it is the most general non-symmetric dyadic which commutes with its own transpose. The special gyrotropic dyadic of interest here is of the type

G(q) =

-

-

I + q x I,

(2.128)

and the rotation dyadic can be written as (2.129) In fact, because -

-

(1- q x 1)-1

1

-

= -l+q - 2 (1 + qq + q

-

x 1),

(2.130)

(2.129) can be seen to be of the form (2.122) if we write q == u tan(B/2), or u == q/.;q:q and (J = 2 tan-l.;q:q. Thus, q is not a CP vector. For real q, its length q determines the angle of rotation. q = 0 corresponds to B == 0, q == 00 to (J == 1r and q == 1 to () == 1r /2. It is straightforward, although a bit tedious, to prov~ th!. following identity between the dot product of two rotation dyadics HI, R2 and the corresponding q vectors:

q (R 1· R) 2 ==

ql

+ q2 + q2

X ql . 1 - qt . q2

(2.131 )

43

2.8. SPECIAL DYADICS This~quation

shows us that two rotations do not commute in general, since R 1 . R 2 and R 2 . R 1 lead to the same q vector only if q1 and q2 are parallel vectors, i.e. the two rotation dyadics have the same axes. 2.8.2

Reflection dyadics

The symmetric dyadic of the form

I - 2uu

C{u) =

(2.132)

with a unit vector u is called the reflectio~dyadic because, when u is real, mapping the position vector r through C . r = r - 2u(u . r) obviously performs a reflection in the plane u . r == O. The reflection dyadic can be also presented as a negative rotation by an angle 1r around u as the axis

C(u) = -R(u,1r),

(2.133)

as is seen from the definition of the rotation dyadic (2.122). In fact, the unit dyadic 1 and the negative of the reflection dyadic are the only rotation dyadics that are symmetric. The reflection dyadic satisfies C 2(u)

= I,

or C

-1

= C,

(2.134)

C(u)~C(u) == -2C(u), (2.135) trC = 1, spmC = -1, detC = -1, (2.136) The most general square roo!. of the unit dyadic is not the reflection dyadic, but a dyadic of the form ±(1 - 2ab) with either a · b = 1 or ab = o. It is easy to see that both rotation and reflection dyadics preserve the inner product of two vectors. In fact, because they both satisfy AT .A == I, we have (A . a) . (A. b) = a- (AT. A) . b = a· b, (2.137)

for any vectors a, b. The cross product is transformed differently through rotation -

-

(R. a) x (R· b) =

1--(R~R) . (a x b) 2

-

== R· (a x b),

(2.138)

than through reflection

(C· a) x (C· b)

= -C· (a

x b).

(2.139)

This equation shows us that a reflection transformed electromagnetic field is not an electromagnetic field, because if the electric and magnetic fields are transformed through reflection, the Poynting vector is not. The converse is, however, true for the rotation transformation.

CHAPTER 2. DYADICS

44

2.8.3

Uniaxial dyadics

By definition, a uniaxial dyadic is of the general form

-

D

-

= aI + abo

(2.140)

A condition for !} to be uniaxial is obviously that there exist a scalar a such that D - al is a linear dyadic, or (D - al)~(D - aI) = o. (2.141) It is easi~ to define conditions not for D itself but for its trace-free part

C =

r:- trpi.

D is uniaxial exactly when C is uniaxial, or of the form C = {31 + ab, with (J = -a . b/3, whence it satisfies

(C - (3I)~(C - (3I) =

o.

(2.142)

Taking the trace operation leaves us an equation for {3:

=

=

-T

(32 _ spmC _ C : C ---3---6-·

(2.143)

Either root of (2.143) inserted in (2.142) would result in a dyadic equation which is satisfied for any trace-free uniaxial dyadic C. Applying (2.50) and (2.51) gives us the two equations C 2 ±J.Jspmc -3- C

spmC

+ 2 - 3 - "l --

0.

(2.144)

Conversely, if a trace-free dyadic C satisfies either of (2.144), or what is

J

J

(C =F j SP~lC = 0, it is uniaxial, because equivalent, (C ± 2j SP~l(j either of the dyadic factors must be a linear dyadic. The left-hand axis vector a is the right-hand eigenvector of a uniaxial dyadic and conversely. The corresponding eigenvalue is a + a . b. If the vectors a and b are fixed, all uniaxial dyadics of the form

I) .

I)

D(alf3) = al + {jab

(2.145)

form a set of co-uniaxial dyadics defined by the vector pair (a, b). These dyadics form a linear space, in which all sums and dot products of couniaxial dyadics are co-uniaxial. In fact, we can easily show that (2.146)

2.8. SPECIAL DYADICS

45

D(a11,81) . D(0:21,82) = D(a10:210:1IJ2 + 0:2(31

+ (31,82 a· b).

(2.147)

From the symmetry of (2.147) we see that the dot product of two couniaxial dyadics commutes. Further we have

D(20: 1a2 + (0:1(32

+ 0:2,81)a · bl -

0:1,820:2,81).

(2.148)

Thus, the inverse of a uniaxial dyadic is co-uniaxial: (2.149) and it exists if the determinant (2.150)

is non-zero. Thus, any power of a uniaxial dyadic is co-uniaxial:

(2.151) For n = 0, the resulting unit dyadic is obviously co-uniaxial. For n < 0, (2.151) is valid only if (2.150) is non-zero. (2.151) is also valid for noninteger values of n, whence it defines co-uniaxial roots of uniaxial dyadics. Of course, there exist other roots as well. For a · b side of (2.151) approaches D(o:nln,8o:n-1) 2.8.4

~

0, the right-hand

Gyrotropic dyadics

The gyrotropic dyadic G was defined in S~tion 2.3.1 as any dyadic commuting with an antisymmetric dyadic g x 1: G x g = g x G.

(2.152)

The vector g is called the axis of G and all gyrotropic dyadics with the same axis vector are called coaxially gyrotropic (CG). In the following g is assumed NCP and normalized so that g · g = 1. The most general gyrotropic dyadic has the form (2.35) and is written in the coaxial form

(2.153) Coaxially gyrotropic dyadics form a linear space with the properties (2.154)

46

CHAPTER 2. DYADICS

G(oll,8111'1) . G(021,8211'2) G(0I Q21'11'21,81,82

=

+ 1'11'2 + 01,82 + ,81 021011'2 + 1'1 02)'

(2.155)

Because of the symmetry in the indices in (2.155), two CG dyadics always commute in the dot product. The square is a special case of (2.155): (2.156) The CG square-root of a gyrotropic dyadic can be obtained from (2.156) if the parameters 0, ,8, l' are solved in terms of p, 0', 7 from the equations

(2.157) a

= (32 +,2 + 20,8,

(2.158)

= 20,.

(2.159)

7

The solutions for 0 can be written (2.160) and the other parameters are obtained from (2.157)-(2.159). The double-cross product of two CG dyadics is the CG dyadic G(011,811'1)~G(021,821'2)

= G((OI + (81)02 + (02 + (82)011

2'1,2 - 01,82 - ,81 021(01 + ,81),2 +'1 (02 + (82)).

(2.161)

This has the special case

(2.162) The inverse dyadic of a gyrotropic dyadic is CG and of the form

= G-

1(al,B!'Y)

= 0 ,2 - of) -, = G(a2 + '121 (a + ,B)(a2 + '12 ) Ia 2 + '12 ) '

(2.163)

This obviously exists for a non-zero determinant of the gyrotropic dyadic:

(2.164) The eigenvalues of the gyrotropic dyadic G(ol,8I,) of (2.153) are thus o + {3 and Q ± i,. The former corresponds to the left and right eigenvector g. The left and right eigenvectors bj,, a± corresponding to the eigenvalues

47

2.9. TWO-DIMENSIONAL DYADICS

a ± j, can be constructed with the aid of conclusions following (2.111), provided, I- 0 and (3 =F i, I- 0:

=f

2i,«(3 =F i,)(I -

gg ± ig x I).

(2.165)

The dyadic I - gg ± j g x I is really a linear dyadic, because it is of the form G(ll - 11 ± i). which inserted in (2.161) gives zero. Explicit expressions for the eigenvectors can be written in terms of any unit vector v satisfying v . v == 1 and v . g == 0, because if the dyadic vv~ gg is evaluated through (2.51), we can write

I-

gg±ig x

I

== vv + gg~vv ± j(vv x g - v x gv) == (v =fiv x g)(v ±jv x g).

(2.166)

Thus, the eigenvectors corresponding to the eigenvalues a ± j, are, respectively, a = v ± jv x g and b = v =f jv x g, which are easily seen to be circularly polarized. If the gyrotropic dyadic is hermitian, the vector g must be real and the scalars a, f3 real and , imaginary, which makes the eigenvalues real. This is all easily seen if we write the general gyrotropic dyadic (2.153) in the form

aI + f3gg +,g x I = (a + f3)gg+ a2 + j-, ( v-l jv . x g )( v-Jvxg . ) +a--2i, . )( v-l-jv . x g).) (2.167) - (v-Jvxg Expanding the right-hand side, it is seen that the unit vector v really has no effect if it satisfies v . g o.

=

2.9

Two-dimensional dyadics

By definition, a planar dyadic a =1= 0 satisfying

A is two dimensional, if there is a vector

A·a= a·A = o. We restrict a to be NCP, whence there exists the unit vector u

(2.168)

= a/ va-:-a..

CHAPTER 2. DYADIOS

48

2.9.1

Eigendyadics

The dyadic eigenvalue problem uu~A ==

AA

(2.169)

is essential in the classification of two-dimensional dyadics. The solutions can be found by operating (2.169) by uu~ and applying (2.47) to be A == 1 and A == -1. The eigendyadics are obviously two dimensional and they span the whole two-dimensional subspace, because any of its members can be expressed as -

1-

-

I--

A = 2(A + A~uu) + 2(A - A~uu).

(2.170)

It can easily be seen that the first term corresponds t~ the eigenvalue A == 1 and the second one to A == -1. Applying (2.51) for A~uu, we can further write (2.170) in the form -

1

--

-

-

1-

-

--

A = 2[(trA)I t + A - AT] + 2[A + AT - (trA)ItJ,

(2.171)

with the two-dimensional unit dyadic

It

==

1- uu.

(2.172)

Denoting the dyadic corresponding to 90° rotation around u by ] == u x I,

(2.173)

it is easy to show that the most general antisymmetric two-dimensional AT must be a multiple of J. From the form of the first term in dyadic (2.171) it is seen that the most general eigendyadic of (2.169) corresponding to A = 1 is a multiple of the two-dimensional (complex) rotation dyadic

A-

R( u, 0) == cos OIt + sin oJ.

(2.174)

The eigendyadics corresponding to A = -1 are symme~ic and trace free, as can be seen from the second term of (2.171) and tr1t == 2. They can be interpreted as multiples of (~mplex) reflection dyadics. Denoting basic reflection dyadics by K and L, the most general two-dimensional dyadic can thus be written as (2.175)

2.9. TWO-DIMENSIONAL DYADICS

49

There is no obvious way to define the dyadics ~ and L. Any multiple of a reflection is necessarily of the form ,K + 8£ and there can be shown to exist a two-dimensional vector p such that this dyadic is of the form pp - uu~pp, or another two-dimensional vector q such that the same dyadic is of the form qq x u + u x qq. If the vector denoted by u in the reflection dyadic (2.132) is replaced by pi Vff7P, its two-dimensional part is most easily seen to be a multiple of the above form.

Base dyadics

2.9.2

Taking an orthonormal base of vectors u, v, w satisfying u x v = w, with the two-dimensional sub-base v, w, the dyadics can be defined as

It = vv + WW,

(2.176)

] = wv -

VW,

(2.177)

K=vv-ww,

(2.178)

L=vw+wv.

(2.179)

It must be remembered that and J are independen~and K, L dependent on the chosen base. Denoting the base dyadics by B i, they can be shown to satisfy the following orthogonality conditions:

It

B · ' B· - 28.. 1,), 1,'

)

whence we may write for any two-dimensional dyadic -

A

1 -

- -

-

--

= "2 [(A : It)It + (A: ])] +

(2.180)

-

if the expansion

-

---(A: K)K + (A: L)L].

(2.181)

Because for any two two-dimensional dyadics we have

A~B = (A~B : uu)uu

= (A~uu) : Buu,

(2.182)

and with the eigenproblem (2.169) solutions

UU~It = It, uu~J

= J,

uu~K

= -K, uu~L = -L,

(2.183)

the double-cross product of different base dyadics. can be obtained from the double-dot product expression (2.180). Thus, B, ~ B j = 0 if i ::I j and

It~It = 2uu, J~J = 2uu, K~K = -2uu, L~L = -2uu.

(2.184)

The dot product Bi·B j of different base dyadics B, (in the vertical columns) and B j (in the horizontal rows) can be obtained from the following multiplication table:

50

CHAPTER 2. DYADICS

]

K

It

It It

]

K

J

]

.t,

K

K

-L

L It

I

L

K

]

i , j

2.9.3

I I -K -]

It

The inverse dyadic

Further, from (2.51) we can write for any two-dimensional dyadic

(2.185)

which is the two-dimensional counterpart of the Cayley-Hamilton identity (2.65): (2.186) The funct~n spmA serves as a two-dimensional determinant function, because detA == 0 for ~anar dyadics such as these. The two-dimensional inverse of t~ d~dic A~~ be ~rectly~~tte~ from (2.1~)_by writing it in the form A . [A - (trA)ltl == [A - (trA)Itl . A = -(spmA)lt, as

A- 1 ==

==

=

spmA This

is

=T

(trA)It~A == A ~uu.

(2.187)

spmA

a generalization of the planar inverse (2.74), because :4(2)

=

(spmA)uu. The two-dimensional inverse can be written in terms of the expansion (2.175) as

A -1 = aIt - r/J - "(Ka2 2.9.4

+ {32 -

6£.

(2.188)

,2 - 82

Dyadic square roots

oi

Finally, let us find the two-dimensional square root a two-dimensional dyadic, i.e. the solution to the equation X . X == A. The problem is much easier than the general three-dimensional case. From (2.67) we have (X~X)2 == 2X2~X2 and, hence, spmA == (spmX)2. From (2.186) we have

A = X 2 = (trX)X -

(spmX)I t = (trX)X =f

V

spm:4

It.

(2.189)

2.9. TWO-DIMENSIONAL DYADICS

51

Taking the trace operation allows us to solve for trX with the result (2.190) This actually stands for two pairs of values with two different signs attached to the first square root independent of the ± sign. Thus, (2.189) can be solved to give two pairs of solutions without showing the first double sign:

(2.191)

Here, the two ± signs correspond to each other. The solution fails if the denominator vanishes. This corresponds to special solutions satisfying trX = ~, wh~ch from (2.189) are seen to be valid only for dyadics of the form 11 = alt. Square roots in this case are obtained through the following expressions: X~X

= (X~X : uu)uu = 2(spmX)uu =

[(trX)2 - tr(X 2)]uu (X ± valt)~ (X ± valt)

= -2auu,

= -2auu ± 2JA(trX)uu + 2o:uu

(2.192) = O. (2.193)

vc:Jt

in this case must be linear and we can write Thus, the dyadic X ± the corresponding solutions in terms of any two two-dimensional vectors a, b satisfying a . b # 0

= X =

=-

±VQ(It

ab

2a· b).

(2.194)

Obviously, this solution satisfies both X 2 = aft and trX = o. It also includes as ~ecial cases multiples of the reflection dyadics X = =fVCiK and = ":fVCiI if we choose a = b = wand a = b = v - w, respectively, where v, ware orthonormal two-dimensional vectors.

References (1947). Vector and tensor analysis. Wiley, New York. T.B. (1961). Handbook of vector and polyadic analysis. Reinhold, New York. BRAND, L. DREW,

52

CHAPTER 2. DYADICS

GIBBS, J.W. (1881,1884). Elements of vector analysis. Privately printed in two parts, 1881 and 1884, New Haven. Reprint in The scientific papers of J. Willard Gibbs, vol. 2, pp. 84-90, Dover, New York, 1961. GIBBS, J.W. (1886). On multiple algebra. Proceedings of the American Association for the Advancement of Science, 35, 37-66. Reprint in The scientific papers of J. Willard Gibbs, pp. 91-117. Dover, New York, 1961. GIBBS, J.W. (1891). On the role of quaternions in the algebra of vectors. Nature (London), 43, 511-3. Reprint in The scientific papers of J. Willard Gibbs, pp. 155-60. Dover, New York, 1961. GIBBS J.W. and WILSON, E.B. (1909). Vector Analysis, (2nd edn). Scribner, New York. Reprint, Dover, New York, 1960. LAGALLY, M. and FRANZ, W. (1956). Vorlesungen tiber Vektorrechnung, (5th edn). Akademische Verlagsgesellschaft, Leipzig. LEHTI, R. (1962). On the introduction of dyadics into elementary vector algebra. Societas Scientiarum Fennica, Communicationes PhysicoMathematicae, 36, (2), 1-41. LINDELL, I.V. (1968). A collection of dyadic identities in threedimensional vector calculus. Helsinki University of Technoloqu, Radio Laboratory Report 824. LINDELL, I.V. (1973a). On the square roots of dyadics. Helsinki University of Technology, Radio Laboratory Report S57. LINDELL, I.V. (1973b). Coordinate independent dyadic formulation of wave normal and and ray surfaces of general anisotropic media. Journal of Mathematical Physics, 14, (1), 65-67. LINDELL, I.V. (1981). Elements of dyadic algebra and its application in electromagnetics. Helsinki University of Technology, Radio Laboratory Report 8126. LOHR, E. (1939). Vektor und Dyadenrechnung fur Physiker und Techniker. de Gruyter, Berlin. PHILLIPS, H.B. (1933). Vector analysis. Wiley, New York. TAl, C.T. (1987). Some essential formulas in dyadic analysis and their applications. Radio Science, 22, (7), 1283-8. WILLS, A.P. (1931). Vector Analysis with an introduction to tensor analysis. Prentice-Hall, 1931.- Reprint Dover, New York, 1958.

Chapter 3

Field equations The essence of electromagnetics lies in the Maxwell equations, which were formed by James Clark Maxwell in 1864 and are still far from being solved. (Some people insist on calling them Maxwell's equations, but, following JACKSON (1975), we adopt similar usage as in 'the Bessel functions.) The Maxwell equations form a highly symmetric set of partial differential equations, which can be cast in many forms using different mathematical formalisms, such as scalars, vectors, tensors, quaternions, differential forms, and Clifford algebras. Maxwell originally applied the quaternionic form of thinking and scalar form of writing and the present vector notation applied by electromagnetists is mainly due to HEAVISIDE from the 1880s. It is typical to all formalisms that, while a higher degree of abstraction allows simplicity in writing and interpreting equations, it is counterbalanced by the inconvenience of learning the use of new operations. This is why new formalisms have had a hard time penetrating into the electromagnetics literature. A hundred years ago this was true of the vector notation and it was mostly only overcome by the stubbornness of Oliver Heaviside. The same seems now to be the case in notations applying differential forms or Clifford algebras, which have certain advantages over the vector notation. This has provoked Georges A. Deschamps, the great promoter of differential forms, to exclame 'Too bad vector calculus was ever invented!' The present exposition, however, lies firmly on the basis of vector analysis. Some logical connections, hard to see in plain vector notation but simple in differential form notation, will be emphasized using matrix operators.

3.1

The Maxwell equations

The Maxwell equations written in terms of real vector quantities, the electric and magnetic field vectors E, H and the electric and magnetic flux densities D, B, as functions of the position vector r and time t, are \7

a

x E(r, t) == - at B(r, t),

(3.1)

CI-IAP'TER 3. F1ELD EQUATIONS

54

\7 x H(r, t)

==

8

at D(r, t) + J(r, t),

(3.2)

\7·B(r,t) == 0,

(3.3)

\7 . D(r, t) == {}(r, t).

(3.4)

Here, the current vector J and the charge density {} are the postulated sources of electrornagnetic fields and they satisfy the continuity condition 8 \7 . J(r, t) == - at {}(r, t),

(3.5)

which is implicit in the equations (3.2) and (3.4) and can be obtained by divergence and a/at operations. The electromagnetic force could be calculated without any reference to electromagnetic fields, but the introduction of vectors E, B, which are calculated first, facilitates the problem. The force density is then obtained from the Lorentz expression

F(r, t) == {}(r, t)E(r, t)

+ J(r, t)

x B(r, t).

(3.6)

The electromagnetic fields E, B have thus the physical significance in transmitting the force between currents and charges. Because they are not easily calculated from the sources, the problem is alleviated by introducing a second pair of auxiliary field vectors H, D. Thus, we arrive at the Maxwell equations (3.1)-(3.4), which are a set of two vector and two scalar equations for four vector or twelve scalar unknowns. This is not enough for a unique solution of the fields. In fact, there should exist an addi tional set of two vector equations between the field vectors E, B, H, D. These equations are dependent on the medium and are called the constitutive equations or medium equations. For example, for the most general linear, local and non-dispersive media the additional equations can be written as follows:

D ==E·E+~·H,

(3.7)

+ j1. H.

(3.8)

B == ( . E

Here, the dyadic parameters f, ~, (, Ii depend on properties of the mediurn. All physical phenomena within the medium are hidden behind these four dyadics. In fact, the macroscopic electromagnetic fields do not distinguish between two media if they have the sarne dyadic parameters, even if the physical processes behind those parameters are totally different. The problems considered here are concerned with effects of rnedia on electrornagnetic fields rather than the converse case.

3.1. THE MAXWELL EQUATIONS

3.1.1

55

Operator equations

The logic leading to the Maxwell equations can be seen more easily using a mathematical formalism higher than the vector algebra. For example, using differential forms the equations take a very simple form. This notation can be simulated using mixed matrix operators as follows. Starting from the electromagnetic source defined as a current-charge four- vector i: (3.9)

the equation of charge conservation (3.5) can be written as D1 . i

= (V ~). at ( J(}

)

= V .J + ~ at e = o.

(3.10)

with a four-vector operator 0 1 defined by (3.11) The physical fields E, B satisfy the combination of two Maxwell equations (3.1), (3.3). Writing the vector pair as a single six vector e: (3.12) and defining a second operator D2 by (3.13)

we can write

v v t I%t o

\7

(3.14)

Here, 0 on the right-hand side means a combination of the null vector and the scalar zero. It is easy to check that the following operator product results in the null operator: (3.15) where we denote (3.16)

CHAPTER 3. FIELD EQUATIONS

56

Comparing (3.10) and (3.15) shows us that the former is satisfied identically if the source four-vector i is written in terms of another six-vector h in the form (3.17) i==KD 2K·h. Writing h == (H, D)T as a matrix of two vectors, this equation equals (3.18) This together with (3.14) defines the Maxwell equations and can, thus, be interpreted as being just a definition of the source in terms of the fields

H, D. Finally, the medium equation can be written in an operator form e == M(h),

(3.19)

where M contains the information of the medium. For a linear medium, the M operator is a matrix of four dyadics.

3.1.2

Medium equations

The most general linear medium can be described in terms of four dyadic parameters defined in (3.7), (3.8). These equations can also be written as a relation between the vector pairs e == (E, B)T and h == (H, D)T:

(3.20)

H == M ·E+Q·B,

(3.21) The relation between the two sets of medium dyadics can easily be derived by substitution: _

_

_

=-1

?==P-L·Q _

=-1

(== -Q

or

_

·M, _

·M,

== - = --1 '(, P==f-~'ll = --1M = -Ji . (,

_

_

==-1

~=L'Q

(3.22)

--1

~==Q

(3.23)

==

=

==

--1

--1

L==~'ll

Q==Ji

,

(3.24)

(3.25)

The general linear medium is also called magnetoelectric or bianisotropic. In t~e ~ase of no special directions in the_medium all medium dyadics €, p" ~, '( are multiples of the unit dyadic I and the medium is

3~1.

THE MAXWELL EQUATIONS

57

called bi-isotropic:- Spec~al ca~es of bi-isotropic media are the i~tr0l!.ic ch!.ral1!!:.edium ~ith '( = -~ = (I and the Tellegen medium with '( = ~ = (I. For ~ = 0, '( = 0, the medium is anisotropic and if the dyadics €, p. are multiples of /, it is simply isotropic. The constitutive equations of the bi-isotropic medium in the frequency domain can be also written in the form

D

= EE + (X -

jK)VEoJ-Lo H,

(3.26) (3.27)

The dimensionless coefficients X and K are called the Tellegen and the chirality parameters, respectively, and for a lossless medium they turn out to be real numbers. A chiral medium can be produced by inserting in a suitable base material particles with specific handedness, i.e, particles whose mirror image cannot be brought into coincidence with the original particle, like particles of helical form. The Tellegen medium can be produced by combining permanent electric and magnetic dipoles in similar parallel pairs and making a mixture with such particles. Such a medium was first suggested by TELLEGEN in ~948. When the medium parameters are actually operators containing time differentiation a/at, the medium is called dispersive, or, more exactly, time dispersive. If they contain the space differentiation \7, the medium is called space dispersive or non-local. If the linear relations do not hold, the medium is, of course, non-linear. Every physical medium can be understood to present time-dispersive properties due to the inevitable inertia of its molecules. In particular, at frequencies high enough every physical medium should act as if unpolarizable, like the vacuum. However, for some frequency ranges medium parameters may depend very little on frequency and they may be considered non-dispersive as a first approximation.

3.1.3

Wave equations

Eliminating all but one of the field vectors from the Maxwell equations of first order, wave equations of the second order are produced. Let us ~o:r.:sider the general bianisotropic medium with the parameter dyadics f, ~, (, M. Substituting (3.7), (3.8) in the Maxwell equations (3.1), (3.2), we obtain the following equations provided the medium is time independent:

( = =8) at V'xI+(-

- at8 '

·E=-Jl·-H

(3.28)

58

CHAPTER 3. FIELD EQUATIONS

=8) ·H=="E·-E+J. - at8 :t

= ( \7xI-~at

Differentiating the second of these equations by

(3.29) and substituting

-it H from the first one gives us the wave equation for the electric field

(3.30) Eliminating the E vector gives us the corresponding equation for the magnetic field:

The wave operators W e(\7, %t) and W m(\7, === W (\7 -8) e

'8t

== -

(= Vx

I -

= 8) C~ 8t

•

--1 -Jl

it) are defined by

. ( \7

x =I + =i -8 ) ~ at

2

-

_ 8l

at 2 '

(3.32)

As special cases we have for the homogeneous anisotropic medium (3.34)

=

a

W m (V, 8t) ==

=-lx f

= 82

x VV - Jl 8t 2 '

(3.35)

and for the homogeneous isotropic medium (3.36) == 8 1= = 82 W m(\7, at) == -: I~\7\7- J-LI 8t 2 ·

(3.37)

The other two Maxwell equations (3.3), (3.4) must be added as initial conditions to these wave equations and for anisotropic media they are

V . (€. E)

+ V . (~ . H) == {},

(3.38)

V . (=P . H)

+ V . (( . E)

(3.39)

==

o.

3.2. FOfJRIER TRANSFOR!v1ATIONS

59

References BLADEL,

J.

(1964). Electromagnetic fields, McGraw-Hill, New York. J. D. (1975). Classical electrodynamics, (2nd edn). Wiley,

VAN

JACKSON,

New York.

J .A. (1986). Theory of electromagnetic waves, Wiley, New York. B.D.H. (1948). The Gyrator, a new electric network element. Philips Research Reports, 3, 81-101. KONG,

TELLEGEN,

3.2

Fourier transformations

To simplify the solution of problems it is often useful not to consider fields in time and space but to make Fourier transformations from time and/or space to the corresponding Fourier variable domain. Thus, differential equations can be transformed to algebraic equations which are easier to solve. Difficulties are normally encountered when transformation back to physical time and/or space is performed. This step could in many cases be made numerically. However, Fourier transformed solutions give information in terms of sinusoidal signals or plane waves and are useful as such. 3.2.1

Fourier transformation in time

The transformation pair in time can be written as

f(w) =

i:

F(t) == -1

F(t)e-iwtdt,

JOO

21r

f(w)eiwtdw.

(3.40)

(3.41)

-00

Since the fields in time domain are real, this sets a condition for the complex field dependence on frequency. In fact, the condition

f( -w) == j*(w)

(3.42)

follows directly from the expressions defining the transform. This means that the complex fields in the frequency domain must in fact be real functions of the variable jw, a fact that can also be seen directly from (3.40). Thus, odd functions of w must be pure imaginary and even functions of w real and, conversely, the imaginary part of a field is an odd function of w and the real part an even function of w. The Maxwell equations in the frequency domain are \7 x E

== -jwB,

(3.43)

CHAPTER 3. FIELD EQUATIONS

60

v

x H == jwD

+ J,

(3.44)

V . B == 0,

(3.45)

e.

(3.46)

V . D ==

(3.45) and (3.46) are in fact unnecessary, because they follow from (3.43), (3.44) and the equation of continuity: \7. J

= -jW{}.

(3.47)

(3.43) and (3.44) are the time-harmonic Maxwell equations which can be applied directly to sources with sinusoidal time dependence.

Operator equations To express the Maxwell equations (3.43), (3.44) concisely, the following operator formalism is of great use for time-harmonic fields. Similar operators could also have been introduced for time-dependent fields. To obtain more symmetry into the equations, the magnetic current density vector J m is introduced in (3.43): V x E == -jwB - J m . (3.48) The rnagnetic source, first introduced by OLIVER HEAVISIDE in 1885, can be understood as being an equivalent source adopted for convenience, without making any speculations about the existence of magnetic monopoles, a subject of great controversy to physicists. The magnetic charge is automatically introduced through an equation of continuity similar to (3.47): (3.49) It is possible to write the equation pair (3.48), (3.44) as an operator equation in many different ways, for example:

L1 . f - jwL 2 . f == g,

(3.50)

where we define for the bianisotropic medium:

II =

( -v

£

= x I - jw(

-)

- jw~

V7 x I-

,

(3.51)

0

(3.52) For an anisotropic medium (3.50) can be written concisely as

(

-jWf.

-v x I

V'

XI) . (E) (J) H == J

-jw=p.

1n

(3.53) .

3.2. F'OURIER TRANSFORMATIONS

61

Helmholtz equations The wave equations (3.30), (3.31) are reduced to Helmholtz equations after Fourier transformation in time when the dyadic wave operators are transformed to the dyadic Helmholtz operators through replacing a/at by

jw:

H e(\l)

= W .rv, jw),

= W m(\7, jw).

H rn(\7)

(3.54)

They can also be obtained from the operator form (3.50) through substitution, based on the fact that the dyadic matrix operator product L1 . L2"l . L1 is diagonal. In fact, from (3.50) we can write L 1 . L21 . L 1 . f - j w L 1 . f

= L1 . L21 . g,

(3.55)

and substituting L1 . f from (3.50), the equation

L1 . L21 . L1 . f + w2L2 . f

= jwg + L1 . L21 . g.

(3.56)

The first matrix can be written in diagonal form after some algebra: (3.57) with the Helmholtz dyadic operators defined as

= -(\1 x I -

H eC\7)

H m (\7) =

I + jw() + W €, -(\7 x I + jw() . €-l . (\7 x I - jW~) + W2M. jW€) . t:' . (\7 x

2

(3.58)

(3.59)

1

Because of the diagonal property of L1 . L2 . L1 and L2 matrices, (3.56) is split into two Helmholtz equations for the two field vectors E and H: -

-

-

1

H e (\1) · E = jwJ + (\7 x I - jw~)· (Ji- . J m ) ,

Hm(\l)· H = -(\7 x

I + jw(). (€-l

. J) + jwJ m.

(3.60)

(3.61)

For the homogeneous anisotropic medium the Helmholtz operators are reduced to (3.62) (3.63) Fina~y,

for the homogen~o~s hi-isotropic medi~m with the parameter dyadics ~ = (X - jK.)JJ-tofo1, '( = (X + jK)JJ-tof o I, the operators can be written as = -1= (3.64) H e(\7) = J-L H(\7),

CHAPTER 3. FIELD EQUATIONS

62

with

(3.65)

For the isotropic medium k± == k and we have simply

He\') == /~\7\7

+ k2 /

== -[(\7

x /)2 -

k 2 / ] == -(\7

x/ - kI). (\7 x/ + k/). (3.67)

It is worth noting that for the hi-isotropic medium, the Helmholtz operators can be factorized, i.e. written as a product of two operators of first order. This property is helpful in finding solutions for the Helmholtz equation, a subject to be discussed in Chapter 5.

3.2.2

Fourier transformation in space

When solving electrornagnetic problems, Fourier transformation with respect to the space variable r often helps in finding the solution. This corresponds to expanding the fields and sources as a continuous sum of plane wave quantities in the Fourier variable k space. Denoting by Vr and Vk the respective physical and transformation spaces, the transformation to both directions is written as

i..

f(k) =

F(r)ei k r dVn

F(r) == -1)3 ( { f(k)e-ikrdVk . 27r

JvJ.~

(3.68)

(3.69)

Thus, the Maxwell equations in the k, w domain are obtained by replacing \7 through -jk: (3.70) k x E == wB - jJ m ,

k x H == -wD

+ jJ,

(3.71) (3.72)

k·D==je·

(3.73)

We make no attempt to introduce different notation for original and transformed quantities, because the argument space is always obvious from the context. The last two equations are unnecessary because they follow from the first two and

k . J ==

w{}

(3.74)

3.3. ELECTROMAGNETIC POTENTIALS

63

It is obvious that since F(r) is a real function of jw, the doubly transformed quantity f(k) must in fact be a real function of two variables, jk and jw. Fourier transforming the Helmholtz equations for the bianisotropic medium results in

[(kxI +w().JL-1.(kxI-w()+w 2 €].E = jwJ-j(kxI +w~).Ji-l·Jm, (3.75)

Here, the linear medium equations (3.7), (3.8) have also been Fourier transformed. For non-dispersive media, medium parameters are the same constants for Fourier-transformed fields as for the original fields. For time-dispersive media, the parameters become functions of the frequency, whereas for space-dispersive media, they would be functions of the Fourier variable k. The equations in k space correspond to a representation of electromagnetic fields as a combination of plane waves, sometimes also called the plane-wave spectrum. These equations can be applied directly if the problem only deals with plane waves, for example, if only media with plane interfaces and boundaries are present and the excitation is a plane wave. Otherwise, the equations can be applied to solve a problem in Fourier space, after which the solution can be transformed back to the physical space. This, however, is hampered by computational difficulties because the inverse transformation can very seldom be made in analytic form and a numerical one is ill behaved. Fourier transformations in two space variables will be applied to planar structures in Chapter 7.

References HEAVISIDE,

3.3

O. (1925). Electromagnetic theory. Ernest Benn, London.

Electromagnetic potentials

The electromagnetic problem can often be simplified by introducing potential functions in terms of which the electromagnetic fields can be expressed. The potential problem is simpler than the original electromagnetic field problem, if, for example, the original vector problem is reduced to a scalar problem, or if the dyadic operator of the original field problem is reduced to a scalar operator of the potential problem.

64

3.3.1

CHAPTER 3. FIELD EQUATIONS

Vector and scalar potentials

Applying the operator formalism introduced in Section 3.1.1 it can easily be shown that, defining a differential operator 0 3 as follows,

D3 =

-V'

-1=8

(

V x8 ]

0

),

(3.77)

the identity

O2.0 3 =

( V0x J

-1.Q. ).( v

=8

I at

8L

V'xI

-oV)

=0

(3.78)

is valid. Here we have denoted the null matrix by '0'. Because from the Maxwell equations we have O2 . e = 0, we are tempted to express the electromagnetic fields in terms of a four-vector a == (A, ¢)T in the form e = 0 3 . a,

(3.79)

or, equivalently,

B == V' x A,

(3.80)

which satisfies O2 . e = 0 automatically. The potential a is not unique, because of the identity (3.81 ) which allows us to add to the potential a a term of the form KD1'tjJ where 'tjJ is any scalar function, without changing the resulting field e = 0 3 . a = 0 3 . (a + KD1 1/J). Thus, the potential function can be transformed into a form subject to a suitably chosen additional scalar condition to make the potential problem simpler to solve. This is called the gauge transformation of the potential and the additional condition the gauge condition. The governing equation for the potential is obtained by combining the equation KD2K . h == i with the medium equation, which can be expressed as h = N- 1 . e and the equation for the potential reads

(3.82) Although the definition of the potential quantities is independent of the medium, it turns out that the equations governing the potentials are not

3.3.

ELECTROMAGNETIC POTENTIALS

65

essentially simpler than the wave equations for the electromagnetic fields unless the medium is bi-isotropic, in which case we have N- 1

.!. (

_

-

1)

-(

J.L€ - ~(

J.L

~

(3.83)

'

in which case the equation (3.82) becomes

8

\7 x (\7 x A) - (~- () at \7 x A

+ (Jl€ -

a

(J-Lf - ~()(at V· A

82

~()(at2A

+ V 2 ¢) =

a

+ at \74»

== J.LJ, (3.84) (3.85)

-J.Lf}.

These equations are decoupled if, following choose the gauge condition

CHAMBERS

(1956), we

(3.86)

in which case we can write

-

[V 21 + (~-

8

-

() at v x 1- (J-Lf -

[V 2

-

(J-Lf -

82

-

~() at 21 ] . A = -J.LJ,

~()~]¢ = 8t 2

-

J.L

J.Lf-~(

(3.87) (3.88)

f}.

It is seen that the operator in the vector potential equation is dyadic unless we have ~ == (, which is the condition of the Tellegen medium. Writing ( == € == XVJ.Lo€o, we have for the gauge condition in this case (3.89)

and the potential equations become (3.90)

[ \7

2

2

8

- (J.L€ - X l-£ of o) -8

2

t

2]4> ==

-

J-l

2

I-£f - X I-£o€o

g.

(3.91)

For isotropic, homogeneous media with parameters e, J-l, the gauge condition (3.86) is simplified to

8

\7 . A ==
at~'

(3.92)

CHAPTER 3. FIELD EQUA1JONS

66

called the Lorenz gauge condition because it was first suggested by Ludwig Lorenz. The potential equations in this case become 2 82 e [\7 - EJ-L-J¢ == --. 8t 2 f.

(3.93)

Because of the scalar operators in contrast to dyadic operators, vector and scalar potentials are most useful in problems involving homogeneous biisotropic media. Problems with inhomogeneities like scattering bodies can also be handled if the inhomogeneities are replaced by equivalent sources.

3.3.2

The Hertz vector

In the previous presentation the field was studied in terms of one vector and one scalar function totalling four scalar functions. Because of the gauge condition, the functions are not independent and there is some redundancy in their solution. In fact, from the knowledge of the vector potential, the scalar potential for hi-isotropic media can be solved by integrating the gauge condition (3.86), provided the value of the scalar potential is known at some time instant t == to: t

¢(r, t) - ¢(r, to) = -

~(

JlE :

J

\7 . A(r, t')dt'.

(3.94)

to

A similar idea is to express the vector and scalar potentials in terms of another potential, the Hertz vector n. Writing the potential pair A, ¢ in the form (3.95) the gauge condition (3.86) is satisfied identically. The Hertz vector is seen to obey the wave equation of the form

82

2

[\7 - (JlE - ~() 8tz1IT

=-

J.L

JlE _ ~(p,

(3.96)

where p denotes the temporal integral of the current function t

per, t) =

J

(3.97)

J(r, t)dt.

to

The electromagnetic fields are obtained by substitution:

E

= \7\7. II -

82 (w - ~() 8t 2II

= \7 x

J.L

(\7 x II) - Jlf _ ~(p.

(3.98)

3.3.

ELECTROMAGNETIC POTENTIALS

67

(3.99) All the previous potential expressions can easily be written for timeharmonic fields by substituting a/at -+ jw.

3.3.3

Scalar Hertz potentials

The electromagnetic potential problem can be further reduced to solving two scalar functions. These functions can be defined in terms of components of the electric and magnetic fields along a certain direction in space, denoted by the unit vector u. Let us only consider the special case of a hi-isotropic medium and time-harmonic fields arising from both electric and magnetic current sources. Writing the Maxwell equations in the form

\7 x (

~)

(~ ~

1

= jw

)

(~ ~) ( ~ )

+(

o 1

-1 0

) (J~ ) (3.100)

or, in shorthand, with respective symbols defined by

V' x f == jwJNf

+ Jg,

(3.101)

we can express the transverse fields f t in terms of the longitudinal field components f u . In fact, taking the transverse component of (3.101) gives us (3.102) Operating this by -u x [(u· 'V)I t equation

[(u· 'V)2 - w2(JN)2]ft = [(u· 'V)It

-

-

jwJNu x I]. leaves us with the

jwJNu x I]. ['Vtf u

-

Ju x gt], (3.103)

which states that if the longitudinal components f u of the field are known, the transverse components f t can be solved from a one-dimensional transmission-line type of an equation. This prompts one to look for the solution, outside the sources, in terms of two scalar Hertz potentials

IT -- ( II IT e nt

)

(3.104)

in the form

fu

= [(u. 'V)2 -

f t = [(u· V')It

-

w 2 (J N)2]rr,

(3.105)

jwJNu x I] . 'VII,

(3.106)

CHAPTER 3. FIELD EQUATIONS

68

because these satisfy (3.103) when g = 0 and

(3.107) or W2Jl(~ - () \72 + W2(JlE _ ~2)

(3.108)

The total field can thus be expressed in the form

f = [\7 x 11

+ jwJN] . \7 x (uIT),

(3.109)

or, more explicitly,

( E) == (

V' x

H

~ -Jwd

JWEI

Scalar Hertz potential expansions have also been developed for anisotropic media in the Ii terature. They lead to fourth-order differential equations to be satisfied by the potentials, whereas for hi-isotropic media the equations are of the second order. This point is clearly seen from considerations of the Green dyadic in Chapter 5.

References

J.

(1964). Electromagnetic fields, McGraw-Hill, New York. L.G. (1956). Propagation in a gyrational medium. Quarterly Journal of Mechanics and Applied Mathematics, 9, (3), 360-70. JONES, D.S. (1964). The theory of electromagnetism, Pergamon, Oxford, 17-9. NISBET, A (1957). Electromagnetic potentials in a heterogeneous nonconducting medium. Proceedings of the Royal Society of London, 240, (4), 375-81. PRZEDZIEDZKI, S. and LAPRUS, W. (1982). On the representation of BLADEL,

VAN

CHAMBERS,

electromagnetic fields in gyrotropic media in terms of scalar Hertz potentials. Journal of Mathematical Physics, 23, (9), 1708-12. WEIGLHOFER, W. and PAPOUSEK, W. (1985). Skalare Hertzsche Potentiale fur anisotrope Medien. Archiv der Elektronik und Ubertraqunqstechnik, 39, (6), 343-6.

3.4.

BOUNDARY, INTERFACE AND SHEET CONDITIONS

3.4

69

Boundary, interface and sheet conditions

To define fields uniquely, in addition to the previous differential equations, more information on the solution is needed. Differential equations alone have an infinity of solutions, whence only by requiring suitable conditions to be satisfied, can the uniqueness of the solution be ensured. The additional conditions must not be too stringent, however, otherwise there might not exist a solution at all. The question of uniqueness will be considered in more detail in the subsequent section. 3.4.1

Discontinuities in fields, sources and media

The Maxwell equations in differential equation form are not valid everywhere in the classical sense if the medium parameters are discontinuous, because some of the field components may become discontinuous and their derivatives infinite. However, adopting generalized functions such as the delta function, differential operators in the Maxwell equations can be understood in a more general way, allowing us to include all discontinuities in the same equations. Differential operators operating on functions with a step discontinuity at a surface S may result in discontinuous functions involving the 'surface delta function' Ds(r) defined by

J

f(r)8 s(r)dV

v

=

J

f(r)dS,

(3.111)

s

for functions I(r) continuous at S. In fact, for the gradient operation we can write, following VAN BLADEL (1964), the result in two parts, the continuous part and the discontinuous part (3.112) where the 'surface gradient' \7 s is defined as! (3.113)

Here It and 12 denote values of the field I on each side 1 and 2 of the interface with unit normal vectors nj , n2 = -nt. {} denotes the operation outside the surface S of discontinuity. The expression (3.113) can be readily checked by integrating (3.112) over S. If I is continuous at S, we have /1 = 12, whence \7 s 1 = 0, 1 The operator V' s must not be mistaken as the gradient along the surface S, sometimes also denoted by V's.

70

CI-IAP1'ER 3. FIELD EQUATIONS

otherwise the step discontinuity is recovered in integration. Defining in the same way the surface divergence and curl operators: (3.114) (3.115)

the Maxwell equations in the frequency domain can be written explicitly as {\7 x E} + (\7s x E)8 s + jwB == -{J m } - Jmsb s' (3.116)

{\7 x H} + (\7s x H)8 s

-

jwD == {J}

+ J s8s '

(3.117)

{\7 . D} + (\7s . D)8 s == {p} + Ps 8s ,

(3.118)

{\7 . B} + (\78 • B)8 s == {Pm} + Pmsbs.

(3.119)

Here we have assumed that there are no delta discontinuities in the fields themselves, only in their derivatives. Obviously, the delta discontinuous parts of the equations above must be equal on both sides. This gives us the following equations relating the discontinuous fields to the discontinuous sources: \7 s x E ==

III X

E1

+ ll2

X

E 2 == -J ms'

(3.120) (3.121 )

(3.122) (3.123) These equations also express the relations of delta-discontinuous fields to the step discontinuities of the media, as is seen by inserting medium equations of both sides of the discontinuity. Vector citcui ts The expressions (3.120), (3.121) are easily remembered when interpreted as generalized circuit equations. The quantity II x H can be interpreted as a vector current corning to the surface S from the direction of n, i.e. flowing in the direction of -no In this way, (3.121) tells us that the currents III x Hj and n2 x H 2 both flow towards the surface S and add up to the current J s : On the other hand, the tangential electric field E t can be interpreted as a vector voltage. If (3.120) is written otherwise: (3.124)

71

3.4. BOUNDARY, INTERFACE AND SHEET CONDITIONS

it can be interpreted as the difference of vector voltages on each side of the surface 5, equating the voltage over the surface, which from side 1 to side 2 equals D1 X J m s .

nxJ

n

1

~

nxH 1

1

~ J

E t1

s

nxH

n

~ E t2

Fig. 3.1 Vector circuit corresponding to a discontinuity in an electromagnetic field.

We could, alternatively, have chosen the magnetic circuit interpretation in terms of magnetic currents and magnetic voltages. 3.4.2

Boundary conditions

A closed surface S containing the region V is called a boundary for electromagnetic fields if the fields are zero outside V (behind S) for any sources

in V. From Huygens's principle to be discussed in Chapter 6, it follows that there must be secondary or induced surface sources on S which outside V exactly cancel the fields produced by the original sources in V. The required surface sources can also be obtained from the surface conditions on S by requiring the fields E 2 and H 2 to vanish behind S. Suppressing the indices ()1 in (3.120), (3.121), we can write

= n x H, J m s = -n x E Js

(3.125) (3.126)

on the boundary. The boundary terminates the fields at the surface S just like a terminating impedance at the end of a transmission line and it can be represented in terms of surface impedance. Henceforth, n points into the region V where the fields are non-zero. A boundary with a surface impedance independent of the fields is a mathematical idealization like many other concepts in electrornagnetic theory. Sometimes an interface between media 1 and 2 can be approximately replaced by such a boundary when only fields in medium 1 are of interest. In doing so, the fields in medium 2 are set equal to zero and a suitable boundary impedance on the interface surface is introduced. While this method

72

CHAPTER 3. FIELD EQUATIONS

is only approximate, because a field-independent surface impedance does not correspond to the true interface conditions, it usually leads to a great simplification of the problem, and thus is useful if the error made is not significant. In practice this is done when the refraction factor of medium 2 is much smaller than that of medium 1. In exact form, the field dependence of the surface impedance must be taken care of through an impedance operator. Applying the vector circuit concept to the boundary condition, we write a linear relation between the surface current J s and the tangential electric field E t in the form (3.127) !!ere,

Zs

Zs

is a two-dimensional surface impedance dyadic satisfying n Zs =: . n = 0 and Y s is its two-dimensional inverse. Similarly, we can write

Ht

=: Zms .

Jm s'

Jm s

=:

= Y ms . H

=-1

=: Zms .

H,

(3.128)

where Zms and Y ms are the corresponding two-dimensional magnetic impedance and admittance dyadics. In terms of fields, the boundary conditions take the following equivalent forms: (3.129) nxH=Ys·E, Et=:Zs·(nxH),

n

X

E = -Y ms . H,

H,

=:

-Zms . (n x E),

(3.130)

implying the equivalence of electric and magnetic dyadics:

= =Zmsxx n n , Y s =:

-

-

Zs == y ms~nn.

Applying the definition of the inverse of a two-dimensional dyadic -

spmA =:

1- - -A~A : I, 2

(3.131)

A: (3.132)

we can write relations between impedance and admittance dyadics

Y = s

Z-l s

=TX =:

Z s x nn , spm Z,

-

y ms

-

==ZT x

-Z-lms-X n n ms -

(3.133)

spmZms

Special boundary conditions The i~tropic '!!Jundary is defined to have the impedance dyadic of the form Zs == whence Zms == Y s and Vms == Zs. A perfect electric

z.t.,

3.4. BOUNDARY, INTERFACE AND SHEET CONDITIONS

73

conductor (PEe) surfa~ is characterized by Zs = 0 and a perfect magnetic conductor (PMC) by Zms = 0 or, what is equivalent, by y s == O. A 'good conductor' corresponds to a'surface impedance which is small with respect to the impedance of the medium "1 = Il/ e. Similarly, a 'good magnetic conductor' surface corresponds to a surface admittance dyadic whose components are small with respect to 1/'T}. The radiation condition, also known as the Silver-Muller condition, is an asymptotic relation between the radiation fields far from the source. It can be written as an isotropic impedance boundary condition on the sphere at infinity with n = u, and the impedance dyadic

J

n; ==

J Ila/ fa·

(3.134)

The isotropic boundary is a special case of the more general bi-isotropic boundary, neither of which has any preferred direction in the tangent plane normal to n. Because the 0.E-Iy tw~dimensi~nal dyadics with this property are linear combinations of It and] = n x I, the most general bi-isotropic boundary impedance dyadic must be of the form (3.135) with two impedance parameters ~a and Z.£.= From Section 2.9 we know that this kind of dyadic satisfies Zs~nn == Zs, whence for the bi-isotropic surface we have the simple rules (3.136) The most general surface impedance dyadic contains four scalar parameters. If the impedance depends on the direction of the fields, it is called anisotropic. A reciprocal anisotropic surface has a symmetric dyadic (3.137) with orthogonal two-dimensional unit vectors u, v. This means that the impedance is Zu for the tangential electric field polarized along u and, correspondingly, Z; for the field along v. As an example we may consider a corrugated surface, which is composed of shorted thin slots parallel to a unit vector u. For a tangential field E t parallel to u it behaves very like a PEe surface, whereas for E t parallel to the orthogonal direction v, the impedance is some scalar Zv depending on th~ nature of the slots. Thus, the surface impedance dyadic has the form Zs = uuO + vvZv. Being a linear dyadic it does not have a planar

74

CHAPTER 3. FIELD EQUATIONS

invers!. in a strict mathematical sense, although we may write one in the form Y s = uu 00 + vv/ Zv. Still worse from the mathematical point of view is a balanced, or tuned, corrugated surface, with Zv = 00.

Scalar potential fields For scalar potential fields, the impedance boundary condition is defined as a linear relation between values of the potential and its normal derivative at the boundary: o¢ + {3n . \7¢ ==

o.

(3.138)

This condition can be interpreted physically in terms of acoustic pressure ¢ and velocity, which is proportional to \74>. For 0: = 0 one has n- \74> = 0, which is called the hard boundary condition, because the velocity of air is zero in the normal direction at a rigid wall. For f3 = 0 one has 4> = 0, which is called the soft boundary condition. The condition for radiated potentials at infinity is called the Sommerfeld radiation condition with n = U r : u; . V¢

3.4.3

+ jk¢ = o.

(3.139)

Interface conditions

Let us consider discontinuities of the fields across an interface S between two media 1 and 2. If there are no surface sources, from (3.120)-(3.123) we have the following conditions at S: (3.140) (3.141) (3.142) (3.143)

These equations present the dependence between fields on one and the other side of the interface. This dependence can be written in dyadic form depending on the parameters of the two media. The counterpart of an interface in transmission-line theory is a junction between two transmission lines.

3.4·. BOUNDARY, INTERFACE AND SI-IEET CONDITIONS

75

Isotropic media For an interface between two isotropic media, the fields in medium 2 can be expressed in terms of fields in medium 1. Since the normal vector appears twice, n in the following equations stands for either 01 or n2: (3.144) (3.145) Here, D( a) denotes a symmetric uniaxial dyadic, which can also be defined in the notation of Section 2.8 as (3.146) = I + (a - l)nn = D(lla - 1). D(l) == I and also that this kind of dyadic satisfies

D(a)

It is easy to see that the simple condition

D(a)· D(f3) == D(f3)· D(a) == D(a{j),

(3.147)

whence the inverse dyadic is (3.148) The isotropic interface relations for electric and magnetic fields can be written symmetrically in terms of relative permittivities and permeabilities (3.149)

D(MIr)· HI == D(Jl2r) . H 2 ,

(3.150)

from which it is also easy to see that the inverse dyadic of (3.144) is obtained by an interchange of indices 1 and 2.

Anisotropic media For the case of two anisotropic media, the interface condition is just a little more involved. We define the more general uniaxial dyadic of vector variable D(a) = 1 +na, (3.151 ) and n is defined as either ni or n2. The special case a = a~ can be expressed in terms of the previous symmetric dyadic (3.146) as D(an) = D(a + 1). Other properties are D(a) . D(b) == D(a + (1 + n· a)b),

(3.152)

CHAPTER 3. FIELD EQUATIONS

76

D- 1 (a)

== D(

-a

1 +n·a

).

(3.153)

D- 1(a).D(b)=D(- a-b ). 1 + n· a

(3.154)

The two uniaxial dyadics D(a), D(b) do not commute in general. They do so only when a is a multiple of b. In terms of the defined uniaxial dyadic, the interface conditions with relative medium parameters can be expressed in symmetric form (3.155) -

D(n·

Ji2r -

-

n)· H 2 == D(n·

Ji l r

-

n)· HI,

(3.156)

and, hence, in the non-symmetric form (3.157)

(3.158) It is easy to show that, again, the inverse of the dyadics in (3.157), (3.158) is of the same form with indices 1 and~ interchanged. If n is an eigenvector of all medium dyadics, the uniaxial D dyadics become symmetric. This condition is satisfied when there is an optical axis in the direction of n in both media. The dyadics ~n (3.157).2. (3.158) obviously reduce to those of isotropic media for

Ei

==

Ei

l , Iii

==

/-lil.

Bianisotropic media The previous relations can finally be generalized to bianisotropic media. In fact, instead of (3.157), (3.158) we have for the interface condition of two bianisotropic media (3.159) with (3.160)

(3.161)

77

3.4. BOUNDARY, INTERFACE AND SHEET CONDITIONS

1

=

-

=

=

--

C21 == -D n· [(£2 : nn)((1 - (2) - ((2 : nn)(£1 - £2)], 22 1 = = = d 21 == -D n· [(E2 : nn)(171 - 172) - ((2 : nn)((1 - €2)], 22 D 22 == (€2 : nn)(JL2 : nn) - ((2 : nn)(~2 : nn).

(3.162) (3.163) (3.164)

All dyadics in (3.159) are symmetric only if n is an eigenvector of all difference dyadics. Also, the anisotropic conditions (3.157), (3.158) are special cases of the bianisotropic condition (3.159) when ~i == 0 and (i == For bi-isotropic media, the condition (3.159) is simplified to

o. (3.165)

with (3.166)

1'21 ==

f2(1 €2tL2 -

(2 f1 1"'

':,2~2

~ u21

==

f2J-l1 -

(2~1

f2tL2 -

(2~2

,

(3.167)

which, again, has the isotropic conditions (3.144), (3.145) as the limiting case (i -~ (i - t o. 3.4.4

Sheet conditions

Thin rnaterial slabs or shells can be approximately handled through sheet conditions. This amounts to approximating shells or slabs of finite thickness and finite medium parameters by infinitely thin sheets with infinite medium parameters. Relations between fields on each side of the sheet are specified in terms of sheet conditions. Some special structures like dense metallic grid or mesh surfaces can also be approximated by suitable sheet conditions. The sheet conditions give linear relations between the tangential electric and magnetic fields across the sheet. A counterpart of an impedance sheet in circuit theory is a shunt or series impedance, or, more generally, a two port. The relation with circuit quantities is obtained again by considering E t as the vector voltage and n x H as the vector current flowing in the -n direction.

Impedance sheet The impedance sheet condition is similar to the terminating boundary impedance condition but with non-zero fields on both sides of the surface.

78

Cf/APTER 3. FIELD EQUATIONS

Assuming the magnetic surface current to be zero on the sheet we have (3.168) The corresponding vector circuit is a shunt impedance since the vector voltage is continuous at the sheet. Writing from the Maxwell equations (3.169)

we have by denoting the surface admittance of the sheet by Y s (3.170) Thus, the impedance sheet condition in this case can be expressed in terms of a dyadic shunt admittance Y sAs an example we may consider a thin slab of dielectric material with parameters frE o, flo in air. The polarization current in the slab is (3.171) and if the thickness t of the slab approaches zero so that (E r finite, we have as the limit the surface current

-

l)t remains (3.172)

Thus, the shunt admittance of the dielectric sheet is (3.173) In the lossless case with f r real this is a capacitive reactance. If the dielectric has conductive loss, e == frf o - [a /w, we can write for the shunt admittance (3.174) which can be interpreted as being due to two parallel sheets: a resistive sheet with the admittance at and a pure reactive sheet.

Magnetic impedance sheet The dual case of the previous example is a sheet with rnagnetic current and with no electric current: J., == O. In this case the tangential magnetic field is continuous and the tangential electric field discontinuous. The condition (3.175)

3.4. BOUNDARY, INTERFACE AND SHEET CONDITIONS

79

means that the vector currents at both sides of the sheet are the same, whence the equivalent circuit must be a series impedance. The sheet condition can be derived from (3.120) in the form

This corresponds to the series surface impedance dyadic s = y ms ~ nl nj , As an example let us consider a thin slab of magnetic material, fo, J-LrJ-Lo with magnetic polarization current approximated through the magnetic surface current

Z

(3.177)

Writing (3.178)

we see that the vector voltage over the series impedance is the series impedance equals

nl

x J m s and

(3.179)

Combined sheet A generalization of the electric and magnetic impedance sheet is a sheet, where both electric and magnetic surface currents can flow. Because both tangential fields are now discontinuous across the sheet, the right-hand sides of (3.127) and (3.128) are taken as average values of the fields at the two sides. Thus, for the sheet conditions we can write the following pair of equations: (3.180) (3.181)

For the general sheet, Ysand Y ms are independent dyadics. The above equations can be expressed in a form relating vector voltages and currents to each other: (3.182)

80

CHAPTER 3. FIELD EQUATIONS

where we have assumed ZII = Z22 due to symmetry and ZI2 = Z2I due to reciprocity of the isotropic slab. From a simple comparison, we can write expressions for the dyadic impedance parameters (3.183) -

-

Z12 = Zs -

1

-

4nlnl~Y rns :

(3.184)

The slab can be simply described by an equivalent cir~it of.! type by noting that two series impedance dyadics are each 2(ZlI - ZI2) =

nj n, ~yrn s s which is the same as the series impedance of a magnetic impedance slab. The shunt impedance in this case equals

ZI2

Zs -

=

xy in1n1 x ms· 1

The two ports of the T circuit are disconnected in the case which corresponds to the condition

ZI2

== 0,

(3.185)

This actually means that the two ports are terminated by a boundary surface impedance dyadic == ~nIni ~y rn s : which corresponds to a PEe plane positioned in the symmetry plane of the sheet short circuiting the surface admittance Y s and halving the magnetic surface admittance.

ZII - ZI2

E

t

v sh

E

t

Fig. 3.2 T circuit representation of a sheet of dielectric and magnetic material with series impedances and a shunt admittance.

As an example of a combined sheet we may take a slab with both relative permittivity and permeability dJ.fferen0rom unity. Comp~ing expressions we can write Y s == jw(€ - €o)tl and Y ms == jW(J-L - J-Lo)tl, whence the T circuit series impedances are Zse == j'f/o(J-Lr -1 )kot and the shunt admittance Ysh == j4kot ( f.r - 1)/1]0, Fig. 3.2. The disconnection condition becomes (f r - 1)(J-Lr - 1)(kot )2 + 4 == 0, which requires either J-Lr < 1 or f r < 1.

3.4. BOUNDARY, INTERFACE AND SHEET CONDITIONS

81

Anisotropic sheets are sometimes used when different transmission and reflection properties are needed for different field polarizations. A simple example is a planar sheet of equidistant parallel metallic wires, which reflects well the polarization parallel to the wires and transmits the orthogonal polarization. The analysis is, however, not touched here. 3.4.5

Boundary and sheet impedance operators

The surface impedance concept is an approximation when replacing the physical interface of two media, because it assumes a constant ratio of tangential fields on the other side of the interface for all incident fields. At best this is approximately valid when the refraction factor of the second medium is much lower than that of the first medium, as demonstrated by USLENGHI (1991), because then the transmitted field is almost transverse electromagnetic (TEM) to the normal direction and the ratio of the tangential fields equals the wave impedance of the second medium. In more exact form the surface impedance can be written as an operator taking the field dependence into account. Likewise, the sheet impedance parameters can also be generalized to operators. Let us consider the electromagnetic field vectors in a homogeneous isotropic medium. In particular, we are interested in a relation between the field components tangential to a plane with normal unit vector n. This relation is carried over to another medium across a planar interface because of continuity. Starting from the Maxwell equations, we obtain the following second-order equations through elimination of the normal field components

n·E, n·H: 2=

(V't V't + kIt) . (n x H) == -jw€(n· V')E t,

(3.186)

(nn~V'\7 + k 2 It ) . E t == -jwJ.L(n· V')n x H.

(3.187)

These do not yet present conditions for the tangential field vectors at the plane because of the normal differentiation operator n . \7 on the righthand sides. However, the normal differentiations of source-free fields can be obtained from the scalar Helmholtz equations (V'2

+ k 2)E t == 0 =>

(\72 + k2)(n x H ) = 0

=}

+ k 2)E t ,

(3.188)

= -(\7;+k 2)(nxH).

(3.189)

(n . V')2E t. == -(\7;

(n·\7)2(nxH)

Accepting a pseudo-differential operator involving a square root of differential operators, we may further write (3.190)

82

CHAPTER 3. J?IELD EQUATIONS

(n · \1)(n

X

H)

= ±h/\1; + k 2 (n X H). the case \7t = 0 we obtain

(3.191)

The sign is chosen so that for the boundary impedance expressions, which is denoted by taking ± - t +. Inserting (3.190), (3.191) in (3.186), (3.187) results in relations similar to those of the surface impedance conditions (3.129):

E,

= Zs(V't) . (n x

n x H

H),

= Y s (\7t ) . E t ,

(3.192) (3.193)

with the definition of the following impedance and admittance operators, _

2=

Z

(\1 ) s

t

= !.!. \1 t \1 t + kIt k V\!; + k 2 '

Y.(\1t) =

nn~\1\1 +

c:

k17VV'; + k 2

(3.194)

(3.195 )

These two dyadics can be easily shown to be two-dimensional inverses of each other so that (3.192) and (3.193) actually represent the sarne equation. In fact, after showing that spmZs == ",2, we can write _

x==T

s -Z-l(\7 ) _ nnxZ s t _ spm Z,

_

-

y s (" ) Vt .

(3.196)

It is also immediately seen ~at if ~he fields do_not vary tangentially, i.e, for \7 t == 0, we simply have Zs == flIt and Y s == I t / ", . If the variance is slight, we may treat the operator \7 t like a srnall quantity and use Taylor expansions for the square roots to arrive at the following two-term series expansions:

(3.197)

(3.198)

The truncated operators Y s (\7) and Z s (\7) are not exact inverses of each other, but the error is a fourth order operator, only. These impedance and adrnittance operator expressions can be applied for problems with fields varying sufficiently slowly in the transverse direction.

3.4. BOUNDARY, INTERFACE AND SHEET CONDITIONS

83

Wire grid A useful approximate impedance condition for a grid or mesh of thin metallic wires, with cells much smaller than the wavelength, was formulated by KONTOROVICH (1963). In practice, this condition has been observed to give good results with grid sizes as large as a quarter of a wavelength, Only the simple theory for a planar grid with square cells and bonded junctions is discussed here. The grid plane is approximated by a continuous surface with the sheet current condition (3.169). In the Kontorovich average boundary condition theory, the tangential electric field is given by (3.199) This is of a form similar to the operator surface impedance condition above with the operator (3.200) The geometry of the grid is taken into account through the parameter '"Y

== :: In A

(_a),

(3.201 )

21TT o

where a is the spacing between the wires, To is the wire radius and A the wavelength. For ~ --t 0 the grid b~comes a conducting plane with infinitesirnal cells and, in fact, we have Zs -+ a corresponding to a PEe plane. The approximation assumes the inequalities To « a and a < 0.25,;\ for reasonable accuracy. Because there are no magnetic currents on the grid, we have n x (E 1 E 2 ) == 0, and the circuit equivalent is a shunt impedance operator. The grid theory will be applied in Chapter 7.

References BLADEL, J. VAN (1991). Singular electromagnetic fields and sources, Clarendon Press, Oxford. CASEY, K.F. (1988). Electromagnetic shielding behavior of wire-mesh screens. IEEE Transactions on Electromagnetic Compatibility, 30, (3), 298-306. IDEMEN, M. (1973). The Maxwell's equations in the sense of distributions. IEEE Transactions on Antennas and Propagation, 21 (5), 736-8.

CHAPTER 3. FIELD EQUArpIONS

84

KONTOROVICH, M.I. (1963). Averaged boundary conditions on the surface of a grid with quadratic cells (in Russian). Radiotekhnika i Elektronika, 7, (9), 239-49. English translation in Radio Engineering and Electronic Physics, 8, 1242-5, 1964. LEONTOVICH, M.A. (1944). Approximate boundary conditions for the electromagnetic field on the surface of a good conductor. Bulletin of the Academy of Sciences of the USSR, Physical Series, 9, 16. OKSANEN, M.I., TRETYAKOV, S.A. and LINDELL, I.V. (1990). Vector circuit theory for isotropic and chiral slabs. Journal of Electromagnetic Waves and Applications, 4, (7), 613-43. SENIOR, T.B.A. (1981). Approximate boundary conditions. IEEE Transactions on Antennas and Propagation, 29, (5), 826-9. SENIOR, T.B.A. (1985). Combined resistive and conductive sheets. IEEE Transactions on Antennas and Propagation, 33, (5), 577-9. USLENGHI, P.L.E. (1991). A detailed analysis of the Leontovich boundary condition for coated bodies. Progress in electromagnetics research symposium, p. 794. Cambridge MA. WAIT, J.R. (1978). Theories of scattering from wire grid and mesh structures. In Electromagnetic scattering (ed. P.L.E. Uslenghi), 253-87. Academic Press, New York. WAIT, J.R. (1982). Ceo-electromagnetism. Academic Press, New York.

3.5

Uniqueness

In solving an electromagnetic problem numerically, it is essential to make sure that the problem has a unique solution. Non-uniqueness may lead to ill-behaved procedures, like singular or almost singular matrices, which either means a halt in running the computer program or a high error level. The uniqueness of a problem defined through a differential equation is dependent on the type of associated boundary conditions. If a linear deterministic differential equation of the general form L f = g possesses two solutions /1 and /2 which are different 11 - 12 = 10 1= 0, there exists a non-null solution to the corresponding homogeneous equation L[; = o. Thus, the question whether or not the solution is unique equals the question whether or not the homogeneous equation possesses a non-null solution l.: The homogeneous equation can be understood as a generalized eigenvalue problem when one of the parameters of the problem, say Pn is defined as the eigenvalue parameter A: Pn

== A.

(3.202)

3.5.

UNIQUENESS

85

By a 'parameter' we mean any geometrical or physical parameter like a measure, material parameter, operating frequency or propagation factor. Normally the problem does not have a non-null solution 10 for arbitrary value of A, but only if the value of A is among the set of eigenvalues Ai, in which case the corresponding I; can be understood as an eigenfunction of the problem. The eigenvalue problem is not necessarily of the standard form £11 == >"£21, because the parameter A may, for example, appear in an exponent or under a square-root sign. If the assigned value of the parameter Pn does not coincide with any of the eigenvalues Ai of the homogeneous problem, the solution of the deteministic problem, if it exists, is unique. On the other hand, if the parameter value coincides with one of the eigenvalues, then we may add any multiple of the corresponding eigensolution to a solution of the deteministic problem and the result is again a solution, which thus is non-unique. For example, for a problem of a source inside a metallic enclosure, the field is unique if the operating frequency does not equal one of the resonance frequencies of the enclosure; otherwise we may add to the solution the field of the resonance mode with any amplitude. The idea of uniqueness is easily understood but to check any specific problem for uniqueness is a difficult task in general, because it requires solving the homogeneous (sourceless) problem for eigenvalues. For some cases the uniqueness can, however, be simply established through a suitable integral theorem which ascertains that for some range of parameter values the homogeneous problem only has the null solution. Although these theorems do not give all possible values of parameters for uniqueness, they usually cover a range of practical values. A few such theorems are considered in the following. 3.5.1

Electrostatic problem

Let us consider the sourceless electrostatic problem, the scalar potential obeying the Laplace equation \124>0 = O. For a closed problem with a surface S bounding the volume V of interest, the following integral expression can be written:

J

1'\7¢012dV

v

=

J

J

f

v

v

s

'\7 . (¢o'\7¢o)dV -

¢o'\72¢odV =

¢o(n· '\7¢o)dS.

(3.203) If the surface integral vanishes because of suitable boundary conditions, we have TV 4>0 == 0 in V, which makes the original problem unique since the corresponding sourceless electric field Eo = - \1 ¢o vanishes in V. We can list a number of boundary conditions giving unique solutions.

CHAPTER 3. FIELD EQUATIONS

86

• 0 = 0 on S.

condition n· '\7
This corresponds to the Neumann boundary on S.

= F(r)

• For the combined condition which is Dirichlet on part of S and Neumann on the rest of S the uniqueness is still valid, because, again, the right-hand side of (3.203) vanishes. • Finally also a mixed condition of the form a + (3n· \7 = F(r) may lead to a unique solution. To see this, the integral expression (3.203) is written in the following form:

f s

2

la<po + {3n· '\7
f s

2 la<pol dS 2a{3

f l{3n· s

2

'\7
2a{3

ss.

(3.204)

Uniqueness is obtained because the first surface integral vanishes owing to the boundary conditions and provided the condition a{3 > 0 is satisfied. In fact, in this case the right-hand side is non-positive while the left-hand side is non-negative, whence both sides must vanish. Considering a an eigenvalue parameter, we might ask' about its values giving non-null solutions 0' The previous reasoning shows us only that they are not of the form A/{3 where A is any positive real number. The previous is relevant to interior problems. If the volume V is bounded by a surface S and the sphere at infinity, we have the exterior problem. For the surface integral in infinity to vanish it is sufficient to have the far-field condition

Scalar electromagnetic problem

The electromagnetic time-harmonic scalar potential

0 = o. This is a typical eigenvalue problem L<po = A<po and possesses a non-null

87

3.5. UNIQUENESS

solution if the quantity k satisfies the condition k 2 = - Ai, where Ai is one of the eigenvalues. The spectrum of eigenvalues depends on the boundary values of the problem. In some cases we know certainly that the solution is unique, for example when k 2 = w 2 J.-l€ is real and the eigenvalues are all non-real, or conversely. These two cases can be considered separately. k 2 not real: From the integral identity

(k 2 - k*2) j v

l4>ol2dS = f(4) on . '14>: - 4>:n. V4>o)dS.

(3.206)

s we see that the surface integral vanishes for the Dirichlet and Neumann boundary conditions, 1> = F(r) and n . V1> = F(r). The volume integral vanishes for sure only if k 2 = w2 tu. is non-real, or if the medium is lossy and the frequency is real. Also, for impedance conditions a1> + Bn- V1> = F(r) on S the surface integral can be shown to vanish provided a/ {3 is real. k 2 real: In this case we apply the identity

a

f

l4>ol2dS =

f 4>:

(a4>o+(3n,V4>o)dS-(3 j(IV4>oI2_ k214>oI2)dV. (3.207)

s s v Assuming ~(a/{3) 1= 0 (and hence a 1= 0), we see that for the lossy impedance boundary condition a1> + (3n . V1> = F( r) on S, the second surface integral vanishes for the homogeneous solution 1>0. Because the quantities on each side of the equation have different phase angles, the integrals must be zero. Thus,
(3.208)

r~oo

are in the form of impedance conditions with, a = jk and f3 = 1. The reason for the uniqueness is that the eigenvalues in this case are complex and cannot coincide with the real k 2 values. The eigenvalues can also be real and the problem still unique if the frequency does not coincide with any of the resonant frequencies. Because these depend not only on the boundary conditions but also on the shape of the boundary surface, it does not seem possible to find a condition without actually solving the eigenvalue problem.

CHAPTER 3. FIELD EQUATIONS

88

3.5.3

Vector electromagnetic problem

For the electromagnetic problem enclosed by a surface S, the solution Eo, H o , again, is zero except when w is one of the resonance frequencies of the cavity resonator defined by the surface S. This can be seen from the identity

f

n . Eo x H:dS

s

= jW€*

J

2

IEol dV - jwlJ.

v

J

IHoI2 dV.

(3.209)

v

For the PEC or PMC boundary conditions the surface integral vanishes. If f. -arg(€), or the equivalent, ~{k2} =1= 0, the medium is lossy and both volume integrals must vanish separately, whence Eo = and H, = o. For real k 2 this conclusion does not result from the identity. In fact, at resonance, the volume integrals are proportional to magnetic and electric energies which are known to be equal. In this case, the uniqueness cannot be deduced simply because it depends on the shape of the resonator. For lossy impedance boundary conditions the surface integral does not vanish, but the integrand is H which has a positive real part at every point of the boundary surface. Assuming now a lossless medium, (3.209) has a real part which is non-zero only on the left-hand side. This leads to the vanishing of H o, and hence Eo, on the boundary and, from the Huygens' theorem (to be discussed in Chapter 6), at every point in V. Thus, lossy boundary, lossless medium and real frequency imply uniqueness for electromagnetic problems. In other words, a cavity resonator with a lossy boundary has non-real eigenfrequencies, and a real frequency never coincides with them. If the boundary is a sphere receding to infinity we have basically a similar problem even if the discrete spectrum of eigenva~es becomes continuous at the limit. Taking the lossy surface impedance as Zs = TJ o1t , the impedance condition becomes the Silver-Muller radiation condition and thus we have obtained uniqueness for the radiation field problem. This, of course, relies more on physical intuition than mathematics, those wishing a more saticfactory proof can consult JONES (1979) or COLTON and KRESS (1983).

arg(J-t)

°

H: ·Zs ·

o,

References D. and KRESS, R. (1983). Integral equation methods in scattering theory. Wiley, New York. JONES, D.S. (1979). Methods in electromagnetic wave propagation, pp.474-7, 513-4, 532-4. Clarendon Press, Oxford. COLTON,

CONDITIONS FOR MEDIUM PARAMETERS

3.6.

89

LEIS, R. (1986). Initial boundary value problems in mathematical physics. Wiley, Chichester. MULLER, C. (1969). Foundations of the mathematical theory of electromagnetic waves, pp.267-85. Springer, Berlin. STRATTON, J.A. (1941). Electromagnetic Theory, pp.486-8. McGrawHill, New York.

Conditions for medium parameters

3.6

In macroscopic electromagnetics all physics of media are hidden behind the medium parameters. The four dyadics of the linear medium have algebraic properties reflecting their different physical properties like passivity, reciprocity or losslessness, which are studied in this Section. The discussion is restricted to time-harmonic fields only. 3.6.1

Energy conditions

The real part of the complex Poynting vector

S =!E x H*

(3.210)

2

is classically accepted to represent the directed average power flow density of the electromagnetic field. The origin of this flow is obtained from the divergence of the Poynting vector. From the Maxwell equations we can write the divergence in terms of the electromagnetic fields in the form

yo . S ==

1 -iCE. J * + H * . J m ) + 2J.w I( E· D * 4

H * . B),

(3.211)

whose real part has the following termwise interpretation, each at the point of the argument r:

•

~{V'.

5} rate of electromagnetic energy generated at a point;

• ~~{E· J* + H* . J m } amount of power needed to sustain the electric and magnetic currents at that point;

• We == ~~{E. D*} electric energy density; • W m == ~~{H* . B} magnetic energy density.

The interpretations for We and W m above are, however, only valid for non-dispersive media. In macroscopic form we can write by integration

- !s.ndS= s

~j(E'J*+H*'Jm)dV-2jWj(We-Wm)dV. v

v

(3.212)

90

CHAPTER 3. FIELD EQUATIONS

The real part of this equation expresses the power balance in the volume V. The left-hand side gives the power flow through the surface S into the volume V (direction -n), the real part of the second term corresponds to the power needed to sustain the currents in V and the real part of the third term gives the power dissipation in the material medium, zero if the medium is lossless.

Active and passive media Electromagnetic media can be classified as being active or passive corresponding to their properties of increasing or reducing the energy of an electromagnetic field. The property is inherent in the medium parameters. Although there may exist non-isotropic media active for some fields and passive for some other fields, let us only consider media which are either active or passive for all fields. This property can be seen directly from the sign of the divergence of the real part of the Poynting vector. The following characterizations are valid if the energy conditions are satisfied for all possible electromagnetic fields:

• •

> 0 active medium (medium gives energy to the field);

~{V'.

5}

~{V·

S} $ 0 passive medium (medium does not give energy to the

field);

< 0 lossy medium (field gives energy to the medium);

•

~{V'.

•

~{V'. S} = 0 lossless medium (energy is not transferred between the medium and the field).

S}

These energy conditions are transferred to conditions for the medium parameters of a linear medium if we write from (3.211) for a point r without sources J, J m :

* -H jw ~{V·S } =~ { -(E·n 2

JW (E. 4

jw (E H). ( 4

* ·B) } =

n* - E* . D + H· B* - H* . B) =

!:-~T r -(T ) . ( 7 ~

-

"""iT ~

=* =:;T J-l - J-l

E ) *, H

(3.213)

and require that this quantity be positive, negative or zero, for all fields E, H, corresponding to the different cases above.

3.6. CONDITIONS FOR MEDIUM PARAMETERS

91

Lossless media For lossless media, ~{\7 . S} = 0 must be valid for all vectors E, H. For example, when H = 0, this should be valid for all vectors E. Invoking the theorem A : aa* = 0 for all a => A = 0 we see that the dyadic j(f* _ 'fT) must vanish. Thus, € = f.T* is hermitian, because its antihermitian part vanishes. In a similar way, other conditions are obtained and the conditions for lossless media can be written as (3.214)

For bi-isotropic media, the conditions simplify to

e* = (, J.L*

€* = €,

= J.L.

(3.215)

e

Writing = (X + jK,)JP,o€o and ( = (X - j~)JJ.Lof.o, we see that, for a lossless bi-isotropic medium all four parameters €, u, X and Ii, must be real.

Lossy media For lossy bianisotropic media, the antihermitian Eart of the 6 x 6 matrix Ji, null in the lossless defined by the medium parameter dyadics f, case, must be positive definite. For example, for a conductive anisotropic medium, the conduction dyadic ~ = jw(i - ~T*) must be positive definite for lossy media. For a lossy bi-isotropic medium we can write an inequality limiting the magnitudes of X and ~ parameters. In fact, because the matrix

e, (,

-j(€ - €*) ( -j«( - e*)

-j(e - (*) ) -j(J.L - J.L*)

(3.216)

must be negative definite, its eigenvalues must be negative numbers. Writing the parameters in terms of real and imaginary parts, €=€re+j€im,

e=ere+jeim,

(=(re+j(im,

J.L=J.Lre+jJ.Lim, (3.217)

we actually arrive at the three conditions €im

< 0,

J.Lim

< 0,

(eim

+ (im)2 + (ere -

(re)2

< 4€imJ-Lim,

(3.218)

of which the last one can be written 2

€" + Kim 2 < rlm rm ---. 11.'

(3.219) J.Lo€o Thus there is a physical limit for the magnitudes of the imaginary parts of the parameters X and ~ due to the passivity of the medium. The medium cannot be lossy through the X and ~ parameters alone, because J-Lim = 0 and €im = 0 are seen to imply Xim = 0 and ~im = O. Xim

CHAPTER 3. FIELD EQUATIONS

92

Active media For active media, all the conditions given for lossy media are valid with inequality signs turned around. In particular, the conduction dyadic (j = jw(f. - f.T*) must be negative definite for active media, which corresponds to negative resistance in circuit theory.

Positive energy density function Yet another condition for the medium parameters is obtained by requiring that the energy function be positive. For a dispersive and lossless bi-isotropic medium, the following is an extension of the energy density expression in a dispersive anisotropic medium given by YEH and LIU (1972), and CHEN (1983):

W=!(EH).~( 8

dw

w(€+€*) w«(+e*)) (E)* = w(€+(*) w(J-L+J-L*) H

!(E H) . ( . E (X + j2K)VJ.Lo€o ) ( E ) * . 4 (X-J2K,)y'J-L of o J-L H

(3.220)

In the last expression we have assumed that the real parameters f, J.l and X are independent and the imaginary parameter j K, is linearly dependent on w, which appears a fair assumption for low enough frequencies. To have a positive energy function, the matrix should be positive definite, which leads to the following three conditions for the real parameters, (3.221) which in particular gives a limiting condition for the magnitudes of the parameters X and 1'\,. It must be emphasized, however, that these are valid for low frequencies only and the w dependence of the parameters according to any particular model of the medium affects the inequalities at higher frequencies.

Boundary conditions The power absorbed by a boundary surface with the impedance dyadic can be obtained from

~{n· S} = ~n. (E x H* + E* x H) = ~H. (Zs * + "is T) . H*· 4

4

Zs

(3.222)

Here the unit vector n points towards the impedance surface. If (3.222) is positive for all fields H, the power is flowing from the fields into the

3.6. CONDITIONS FOR MEDIUM PARAMETERS

93

absorbing boundary and not the other way_ Thus, the condition for a lossy boundary is positive definiteness of the hermitian part of the surface impedance dyadic. This means a positive real symmetric part and a negative imaginary antisymmetric part. For a lossless boundary the hermitian part of the surface impedance dyadic is zero. In other words, the impedance satisfies

(3.223) or Zs must be an antihermitian dyadic. For a lossless bi-isotropic = + Zbn x we thus see that Za must be imagiimpedance, nary and Zb real. In the isotropic case, the boundary impedance must be imaginary to be lossless, as is well understood from circuit theory.

z.t,

Zs

3.6.2

I,

Reciprocity conditions

Besides energy quantites, reaction is another quadratic function of electromagnetic fields and sources. It was introduced by RUMSEY in 1954 to account for the interaction of fields and antennas. Let us consider two sources ga and gb and their fields fa, fb. The reaction of the field a on the source b is a complex number defined by

< fa, gb >=

J

(E a · .r,

- n, . Jmb)dV,

(3.224)

Vi,

where the volume Vb contains all of the sources b. (The original notation < a, b > does not distinguish between the source and the field.) The reaction is a measurable quantity. For example, if the source b is a current element J = u1bL6(r - rs ), the reaction of the unknown field Ea(r) is (3.225) which is Ib times the voltage induced by the field E a in the dipole b. In general, setting a test source b in the field due to a source a and measuring the reaction gives a method to determine the field a. Reaction is symmetric under conditions called reciprocal. The difference of two reactions can be expanded from the Maxwell equations

J

(E, · J b

V

-

H, . J mb - E, . J a + H, . J ma)dV

=

94

CHAPTER 3. FIELD EQUATIONS

J J(s, ·

n· (E a X n, - Eb X Ha)dS+

s

jw

(IT -~) · s,

- s, · ((7' + e) ·n, +

V

n, · (P' + () · s, - n, · (IV - If) · n,) dV,

(3.226)

where S is the surface of the volume V containing all sources and n is the outward normal unit vector of S. Reciprocity requires (3.226) to vanish for all fields. Being a property of the medium, we can see the conditions for the medium to be reciprocal: =

=!T €=€,

-

(=-~,

:;;T

]i='ji,

(3.227)

because the volume integral in (3.226) vanishes. For the bi-isotropic medium, this reduces to ~ = -(, or X = o. Thus, the parameter X can be called the non-reciprocity parameter. Because a medium with X # 0 was introduced by TELLEGEN in 1948, it is also called the Tellegen parameter. Requiring the surface integral of (3.226) to vanish, we can obtain conditions for a reciprocal boundary. Inserting the impedance boundary condition (3.228) E t = Zs · (n x H), the boundary term in (3.226) can be written as

J

n- (E a X

s

J

n, - s, X Ha)dS = n. · iz, -

ZS T) · HadS,

(3.229)

s

which is seen to vanish for any fields H a , H, with the condition

= =r Zs = Zs ·

(3.230)

Thus, a symmetric impedance dladic is reciprocal. The bi-isotropic impedance dyadic Zs = z.t, + Zbn x I is reciprocal only for Zb = 0, or when the impedance is actually isotropic. References BOOKER, H.G. (1982). Energy in electromagnetism. Peter Peregrinus, London. CHEN, H.C. (1983). Theory of electromagnetic waves, pp. 65-74. McGraw-Hill, New York.

3.6. CONDITIONS FOR MEDIUM PARAMETERS

95

HAMMOND, P. (1981). Energy methods in electromagnetism. Clarendon Press, Oxford. HARRINGTON, R.F. (1961). Time-harmonic electromagnetic fields, pp. 116-20. McGraw-Hill, New York. JACKSON, J.D. (1975). Classical electrodynamics, (2nd edn), pp. 23644. Wiley, New York. KONG, J .A. (1986). Electromagnetic wave theory, pp. 396-415. Wiley, New York. LINDELL, I.V. (1972). Some properties of lossless bianisotropic media. Proceedings of the IEEE, 60, (4), 463-4. MONTEATH, G.D. (1973). Applications of the electromagnetic reciprocity principle. Pergamon, Oxford. RUMSEY, V.H. (1954). Reaction concept in electromagnetic theory. Physical Review, 94, (6), 1483-91. TELLEGEN, B.D.H. (1948). The Gyrator, a new electric network element. Philips Research Reports, 3, 81-101. YEH, K.C. and LID, C.H. (1972). Theory of ionospheric waves, pp.4956. Academic Press, New York.

Chapter 4

Field transformations There exist various transformations which can be applied to electromagnetic field problems to transform them into other electromagnetic problems. Thus, from known solutions of certain problems solutions of certain other problems can be obtained without going through the solving procedure. For example, the duality transformation changes an electromagnetic problem to a magnetoelectric problem and the affine transformation changes the metric of the space so that an isotropic space becomes anisotropic. Of course, only transformations which do not change the Maxwell equations are of interest, because otherwise the transformed fields would not be electromagnetic fields. A few such transformations are considered in this chapter.

4.1

Reversal transformations

The simplest examples of field transformations are changes of sign in polarity, time, space and frequency, which transform sources, fields and medium parameters so that the Maxwell equations remain invariant. Each of them will be discussed briefly below. 4.1.1

Polarity reversal

In this transformation all electromagnetic source and field quantities q are changed as q ~ qc, starting from the electric charge, which reverses its sign: flo

= -e·

qc

= -q.

(4.1) Assuming that the medium and boundary parameters do not change, from the linearity of the Maxwell equations it is easily seen that they do not change if all source and field quantities q change sign in this transformation:

(4.2)

The sign of the electric charge is actually a convention which could be changed, the present choice being due to Benjamin Franklin from the eighteenth century. If the medium and boundary quantities are the same,

98

CHAPTER 4. FIELD TRANSFORMATIONS

the fields for sources with change in sign are also changed in sign. Products of these quantities do not change because the two minus signs cancel. Thus power, energy and impedance remain unchanged in this transformation. For example, the force between two sources equals that of two similar sources of opposite sign.

4.1.2

Time reversal

In this transformation, the sign of time is reversed: tT = -to Let us assume that the electric charge does not change sign in this transformation: eT(t) = e(-t). It turns out that all other transformed quantities can be determined from the Maxwell equations. For example, we have from \l . JT = 8eT/8tT the rule JT(t) = -J( -t) and from V . DT = or the rule DT(t) = D( -t). Assuming €T = € and "=fiT = we s~ from ~ediu~ equations that the other parameters must obey eT = and (T = (. However, for dispersive media, the relation is not so simple. Also, if assuming magnetic charges, they must obey the law gmT = -Um, JmT = J m . Thus, all quantites fall into two groups: those which are invariant in the T transformat!on, e,}m, E, D, € and~, and those which are anti-invariant, em, J, H, B, ~ and (. Combining time reversal with charge reversal gives us a similar transformation in which the magnetic charge is invariant, (}mCT = Um, and the electric charge anti-invariant, ocr = -g. Because reversal of time corresponds to changing the direction of motion of electric charges, it reverses the currents and magnetic fields but does not change the electric fields. The change of sign in magnetic sources can be understood if magnetic monopoles are omitted as non-physical and magnetic dipoles are replaced by electric current loops, which change sign in time reversal. In the frequency domain, the T transformation corresponds to change of sign of the frequency as is obvious from the Fourier integral representation

F,

J

-e

00

qT(W) =

qT(t)e-jwtdt = q(-w) = q*(w).

(4.3)

-00

Because the T transformation also implies conjugation, it is seen that the real parts of complex functions are even and the imaginary parts are odd functions of frequency.

4.1.

REVERSAL TRANSFORMATIONS

4.1.3

99

Space inversion

In this transformation, space points r are reversed in the origin to points r p = -r. Assuming again that the sign of the electric charge is not changed in the reversal, (}p(r) = {}(-r), we can find from the Maxwell equations, again, that the quantities H2 ~, J m , €, Ii do not change sign, wheras the quantities J, E, D, em, ~, "( transform like Jp(r) = -J(-r). Space inversion is a special case of the affine transformation discussed in more detail in a subsequent section. 4.1.4

Transformations of power and impedance

To see how power flow changes as a result of these transformations, let us study the Poynting vector S(t) = E x H. It is easy to see that we have

Sc

= S,

(4.4)

or the power flow is hot dependent on the sign of the charges nor the direction of the current. On the other hand we have

ST(t) = -S( -t),

(4.5)

which means that the sign of the energy flow is reversed in the time reversal, as is most obvious. Finally, we can write

Sp(r)

= -S( -r).

(4.6)

The frequency sign change has an effect on the complex Poynting vector:

ST(W)

= -S?fc(-w).

(4.7)

Thus, the real power flow changes sign in this transformation, whereas the reactive imaginary part does not. Similarly, all impedance quantities change like ZT(W) = -Z*(-w). (4.8) Again, the real part changes sign in time reversal. This can be understood so that power absorbing passive medium becomes active in time reversal, feeding power to the outside world, which is just like watching a movie backwards. References JACKSON, J.D. (1975). Classical electrodynamics, (2nd edn), pp. 245-51. Wiley, New York. MITTER, H. (1990). Elektrodynamik, (2nd edn), pp. 209--15. BI Wissenschaftsverlag, Mannheim.

CHAPTER 4. FIELD TRANSFORMATIONS

100

4.2

Duality transformations

The duality transformation makes use of the symmetry between the electric and magnetic quantities in the Maxwell equations. In its classical form, the electric quantities of an electromagnetic problem are changed to magnetic quantities and vice versa. In changing the fields and sources of the problem the parameters of media and boundaries are also changed. In many cases it is necessary that the medium remains unchanged, which makes the transformation dependent on the medium in question. However, in the form considered here, duality transformations can only be found for bi-isotropic media. 4.2.1

Simple duality

In the literature, a simple form of duality is often applied to transform an equation corresponding to a certain electromagnetic problem. The outcome is another equation corresponding to a dual problem. If the solution of the former is known, there is no need to solve the latter. Instead, by performing the duality transformation to the known solution, the unknown dual solution can be readily obtained. In the simple version, one just makes the substitution E ---+ H and conversely H ---+ E, comp~eted ~ the additional substitutions B ~ -D, J ~ -J m and € ~ -ii. ~ ~ -(. However, in this form the duality transformation changes the nature of the quantities so that their dimensions are changed. For example, the electric field is transformed into a magnetic field. Thus, it is not possible to add the original and transformed field quantities. Let us consider duality transformations which transform quantities so that their nature does not change. Thus, the electric field is transformed into another electric field. Let us also limit the theory to the frequency domain.

4.2.2

Duality transformations for isotropic media

The original idea behind duality is to interchange electric and magnetic quantities so that one of the Maxwell equations is transformed to the other one and conversely. Because the equations can be multiplied by non-null constants without changing their validity, let us require that the corresponding quantities in the two Maxwell equations are transformed into each other through multiplying by constant factors a and {3:

\7

x H == jwD

+J

---+

\7 X Ed == -jwBd - J md.

(4.10)

4.2. DUALITY TRANSFORMATIONS

101

The equation pair is invariant if we define Ed

= oH, B d = -aD,

Jmd

= -aJ,

(4.11 ) (4.12)

The divergence equations transform to (4.13) (4.14) if we define (4.15) Further, from the condition that the medium equations transform to one another the parameters of the dual problem can be identified:

(~ ~) (--~ d

=

--;E)'

(4.16)

Finally, from the impedance boundary conditions Et =

Zs . (n

x H)

(4.17)

with the unit normal vector n pointing out from the boundary, we have for the dual dyadics

(4.18) 4.2.3

Left-hand and right-hand transformations

The definitions of the dual quantities above are valid for any values of the constants Q, (3. The simple duality transformation, applicable for the derivation of formulas, is obtained by taking a {3 = 1. The coefficients can also be determined by imposing certain conditions for the transforrnation. It appears natural to require the following properties.

=

1. The duality transformation is an involution, i.e., the dual of a dual quantity equals the original quantity itself. This implies a{3 = 1, as is seen by checking through all the equations.

2. A chosen isotropic medium is defined to be self-dual: its parameters €s, ue are not changed in the duality transfo~ation. T~e ~edium chosen is usually the free space with fS = €a!, P-s = /-lol, = 0,

es

CHAPTER 4. FIELD TRANSFORMATIONS

102

O. The conditions fSd = Eol, JlSd = J-Lol, ~Sd = 0, (Sd = 0 imply the condition a//3 = -J-Lo/E o = -TJ~. More generally, we could actually take any isotropic chiral medium with the parameters fS, J-Ls, ~s = -(s = -j"'SJJ-Lofo as the self-dual medium, because, as can be seen from (4.16), the dual of the chirality parameter coincides with that of the original medium: Kd = KS.

(S =

From the two conditions a(3 = 1 and a/ (3 for the parameters: I Q

.

= 11]s,

at = -j"7S,

= -1]~ we have

two solutions

f3' = _1_ . ,

(4.19)

f3T = -~.

(4.20)

J"7s

J1]s

Two possible solutions means that, corresponding to a chosen self-dual medium, there in fact exist two duality transformations, neither of which is a more 'natural' one than the other. These two transformations will be denoted as follows: let the duality tranformation corresponding to the superscript l be called the left-hand duality transformation, and that corresponding to r the right-hand duality transformation. These labels refer to polarizations of two self-dual propagating fields, as will be explained su bsequently. From the previous knowledge we can write the following double duality transformation table for the fields and sources f --t t' and f --t IT:

It j"7sH -jTJsD -jTJsJ -j'TJse

I

IT

E

-j1]sH jTJsD jTJsJ j'TJse

B

Jm em

It E/ins -B/jTJs -J m / j "7s em/j'TJS

I

IT

H D

-E/j1]S B/j"7s Jm/jflS em/j'TJS

J

e

and for the medium and impedance parameters, which transform in the same way in both transformations

I

It = t: Jl/TJ.~ -(

f

~ Ys

I Jl (

-2Zsxx n n Zs

fls

t' = IT fI~! -~

2= x

fls Y s x nn

It is seen clearly from the above table that the transformation equals its inverse whence only one half of the table is actually needed. The two duality transformations possess the following properties.

4.2. DUALITY TRANSFORMATIONS

103

• For a lossless medium (real1Js) the dual of Ex H* is E* x H, whence the real part of the Poynting vector is self dual. Thus, the field pattern representing the energy flow in a lossless medium is unchanged in both duality transformations. • The PEC and PMC boundaries are dual to each other. • The dual of a dielectric object in air with f.s = f.o, tts = tto, with the relative dielectric factor f. r = A, is a magnetic object with the relative permeability ttr = A, since the dual of €/ f. o in this case equals =PI tto and conversely. • The wave number of a plane wave, k = w#' in the dual of an isotropic medium equals the original propagation factor kd = k. • The wave impedance "l of a propagating plane wave in the dual of an isotropic medium equals "ld = "l~/"l. • The dual of an electric dipole J = uIL6(r) is the magnetic dipole J m = -juT}sIL6(r). Application of the duality transformations

4.2.4

Either of the two duality transformations can be applied for an electromagnetic problem in two ways because of the involutory property of the transformation: 1. attach the label d (representing either l or r superscript) to each quantity of the problem and then replace these quantities from the above definitions,

2. replace the original quantities in terms of transformed quantities from the definitions, substitute these in the problem and discard the labels

d. Both ways lead to the same result. It is, however, quite easy to get confused with the two possibilities and end up with a wrong result. This means that one has to be careful and pursue only one path. Of course, only one of the two possible duality transformations, either the left-hand or the right-hand transformation, must be systematically applied. Dipole fields

Let us consider as an example of the method 1 above the problem of finding the far field expression of a magnetic dipole from that of an electric dipole.

CHAPTER 4. FIELD TRANSFORMATIONS

104

The electric field of an electric dipole J (r) == uIL8( r) in the far field region in air is e -jkr E == uojwJ-to--ILsinO. (4.21) 41fT

Since the self-dual medium is air, we have tts = 'fJo. Making the lefthand duality transformation (i.e. adding the superscript I and replacing transformed quantities from their definitions), for every quantity of this equation we obtain

(4.22) or

-jkr

H = uojW€o_e-ImLsinO,

(4.23)

41fT

which is the expression for the magnetic field arising from a magnetic dipole in the far field region. The same result is obtained for the right-hand transformation and also for the simple duality transformation.

Scattering problem As a second example, let us consider the duality transformation of a problem of electromagnetic plane wave scattering from a dielectric obstacle in air with e; = A, J-tr = 1. Let the self-dual medium be free space and the incident wave

E t·( r ) -- E oe-jk.r ,

H i () r

= H oe-jk.r .

(4.24)

Both duality transformations change the obstacle to a magnetic scatterer with J.1r = A and e; = 1. Moreover, they also change the incident plane wave to

Ei(r)

= ±j7/oHoe-jkr,

Hi(r) = ±

~o e- jk r.

(4.25)

)"'0

Also, the scattered field Esc, H sc of the first problem can be transformed to give the scattered field of the second problem. It must be noted, that in this way we obtain a solution to the dual problem involving the dual scatterer and the dual incoming field, which in general is different from the solution for the original incoming field, unless the incoming field is self dual. This can be applied as follows. • If the incoming field is left-hand self dual, i.e. it coincides with its own left-hand transformation, the left-hand duality transformation will not change the incoming field and the dual scattering problem has the original incoming field.

4.2. DUALITY TRANSFORMATIONS

105

• The same is valid if the incoming field is right-hand self dual, in which case we can apply the right-hand duality transformation. • If the incoming field is none of these, we can write it as a sum of left-hand and right-hand self-dual fields, treat these separately as above and add the results. Thus, the scattered field for the dual scatterer and original incoming field can be obtained through two duality transformations, one for each self-dual parts of the problem.

4.2.5

Self-dual problems

Quantities that are invariant in either of the two duality transformations are called self-dual quantities. The two self-dual concepts will be distinguished by the subscript + corresponding to the rand - corresponding to the l transformation. Medium parameters and impedances are not affected by the handedness of the transformation. Obviously the sum of any quantity and its proper dual is a self-dual quantity and their difference is an antiself-dual quantity. It is easy to see that a quantity self dual with respect to the l transformation is antiself dual with respect to the r transformation and conversely. Any quantity can be written as a sum of a self-dual part and an antiself-dual part, or, what is the same thing, as a sum of two self-dual parts (4.26) Self-dual fields

The self-dual electromagnetic fields can be written as

H±

1

1

= -2(H =f -.-. E),

(4.27)

J'TJs

and they are seen to satisfy (4.28) Further, we can write

B± =

~(B ± j"1sD),

D±

1

1

1

2

J'TJs

J'TJs

= -(D ± -. -B) = ±-.-B±.

(4.29)

As an example of self-dual fields (also called wave fields) let us consider a self-dual plane wave propagating in an isotropic chiral medium in the direction of the unit vector u. Because the fields of a plane wave satisfy (4.30)

CHAPTER 4. FIELD TRANSFORMATIONS

106

with

1

H, == - u x Eo, 71s

Eo == -f/sU x H o,

(4.31 )

~(E 1= i'7s H ) = ~(E 1= iu x E). 2 2

(4.32)

the self-dual fields are of the form

E± =

The two self-dual fields corresponding to the two duality transforms are obviously circularly polarized, because they satisfy E± . E± == O. We can form the polarization vectors for the two fields and obtain

p(E±) == ±u.

(4.33)

The upper sign corresponds to a right-hand polarized and the lower sign a left-hand polarized CP wave when looking in the direction of propagation U, thus justifying the labels 'left-hand duality transformation' and 'righthand duality transformation' with respect to which these two respective polarizations are self dual. Self-dual sources and media Self-dual sources can be written similarly as 111 J m± = 2(J m ± i'7sJ) = ±j'7sJ±,

J± = 2(J ± i'7S J m ) ,

(4.34)

111 (4.35) f2± = -2 (f2 ± -.-em), f2m± = -2 (em ± jT/se). == ±jT/Sf2±. JT/s A single electric or magnetic current cannot be a self-dual source, instead, there must be a suitable combination of both of them. The self-dual source combination is obtained from any electric current source J, by adding the magnetic current source J m± == ±jf/sJ. For example, a dipole current J == uIL8(r), requires the magnetic dipole current J m == ±uj71sIL6(r) for a self-dual source. The magnetic dipole must be parallel to the electric dipole and have ±90° phase shift and amplitude 71s1. For fields outside the sources, the magnetic dipole can be replaced by an equivalent electric current loop perpendicular to the electric dipole. The isotropic chiral medium assumed self dual with respect to the two duality transformations was denoted by the parameters f.s, J-ls, ~s == -(s == -jKJJlo€o. Since only the impedance ne == VJ.ls/€S is needed in defining the two duality transformations, all isotropic chiral media with the same impedance appear self dual with respect to these transformations.

4.2. DUALITY TRANSFORMATIONS

107

Self-dual boundaries

The self-dual boundary impedance dyadic --

Zsd

2-x = 7Js(Ys xnn)

Zs

must satisfy the equation

= Z T-= Zs. spm Z,

2 s = 7Js---=:-

(4.36)

Taking the spm operation of this, we have (spmZ s )2 = 7J~, which has two solutions spmZ s = ±1J~' Substituting in (4.36), we have the condition = whence the impedance dyadic must be either symmetric or antisymmetric to be self dual. Let us study these cases separately. T~ antisyrnmetric self-dual two-dimensional dyadic is obviously of the form Zs = Zsll X 7, where n is the unit vector normal to the surface. Equating spmZs with -7J~ gives us finally two antisymmetric self-dual surface impedance solutions

Z;

±Zs,

». = ±j'T/sn x I.

(4.37)

Studying the most general symmetric boundary impedance dyadic (4.38) where Ul and U2 are the two orthonormal eigenvectors of Zs perpendicular to n, from spmZ. 11~ we have Z lZ2 11~ for the sel!:.duality_condition. As one limiting case we have the isotropic impedance Zs = 'T/s7t, like the Silver-Muller radiation condition at the spherical boundary in infinity. As another limit we have the anisotropic surface with Zl ~ 0, Z2 ~ 00, which approximates a dense tuned corrugated surface, i.e. a conducting surface with grooves parallel to U1 and a quarter wavelength of depth.

=

4.2.6

=

Self-dual field decomposition

The decomposition of electromagnetic fields in self-dual constituents E+ and E_ in (4.27), also called wave fields, leads to a useful way of treating electromagnetic problems in isotropic chiral media, instead of the conventional analysis based on electric and magnetic fields. In fact, substituting fields and sources in terms of self-dual components, for homogeneous isotropic chiral media, the Maxwell equations are seen to split into two non-coupled sets, each self-dual system of fields and sources acting in an isotropic non-chiral medium of its own. The governing equations can be written together with double subscripts as follows: (4.39)

CHAPTER 4. FIELD TRANSFORMATIONS

108

(4.40)

where the effective non-chiral parameters of the two media are denoted by K.

f± == f(1 ± -),

n

K,

It± == 1t(1 ± -), n

n

==

Jfrltr.

(4.41 )

It must be emphasized, however, that this is valid only for homogeneous media, in inhomogeneous media the two self-dual fields couple to each other. It is also seen that in an isotropic chiral medium the wave impedances of the two self-dual fields are the same and independent of the chirality parameter K "l± ==

.: J¥ - == f±

(4.42)

- =="l, e

whereas the wave numbers of plane waves in the two media are different: (4.43)

The total electric and magnetic fields can be obtained any time from the wave fields as follows: (4.44)

Power relations Consider the real part of the Poynting vector in terms of the self-dual wave fields for a homogeneous lossless chiral medium: (4.45)

It is seen that the power is propagated independently in the two self-dual fields because there are no cross terms present. Applying the real polarization vector concept from Chapter 1, of a complex vector 3, p(a) == -j(a x a*)/(a· a"), we can write (4.46)

Here, the quantities W±

f±E ±. E*± = 2 It±H±. H*±' =2

v±

1

= ---

JJ-L±€±

(4.47)

109

4.2. DUALITY TRANSFORMATIONS

can be identified as energy densities and phase velocities of the two wave fields in their respective isotropic nonchiral media. The two real polarization vectors p(E±) point to the positive (right-hand) directions of the two wave-field vectors and their lengths, which give the ratio of energy velocity to the phase velocity, vary from 0 for LP to 1 for CP fields. Thus, we can state that the power in each of the self-dual fields propagates in a direction which is perpendicular to the plane of polarization and in the right-hand direction of the '+' field and the left-hand direction of the '-' field. This further justifies the basis for labeling the two duality transformations as 'right hand' and 'left hand'. It is seen that self-dual LP fields are standing waves because the power does not propagate at all. The energy velocities of the two wave fields are different from the phase velocities except for CP waves. Circularly polarized self-dual fields

We can make further conclusions by splitting each of the wave fields into two CP components with labels R, L to be defined shortly. Let us write (4.48)

with two pairs of CP unit vectors u~, u~. The vectors are pairwise coplanar, u~, u~ with E+ and u~, u~ with E_ and they satisfy (4.49) 2

R . u± L -_ 1u± R 1 -u±

I

2

L1 u±

= 1.

(4.50)

The polarization vectors of the CP vectors are unit vectors and parallel or antiparallel to those of the corresponding E± vectors. They can be seen to obey the relations

L R) = -p (u±L) = -Ju± . R x u±, P (u±

(4.51 ) (4.52)

Thus, the total power flow expression can be written in terms of CP components of the two self-dual fields as 1

*

2!R{E x H }

IE~12

R

IE~12

L

= w,», IE+1 2 p(u+) + w,», IE+1 2 p(u+)

IE~12 R IE~ 12 L - W_v-IE_12P(u-) - W_v-IE_12P(u-).

(4.53)

110

CHAPTER 4. FIELD TRANSFORMATIONS

This shows us that the field can be written in terms of four components, two self-dual fields each with two coplanar CP components, which do not couple energy between one another in a homogeneous medium. From the signs of the different terms it is seen that the + waves are right handed and the - waves left handed with respect to their individual directions of propagation. However, since u~ point oppositely to u~, the polarizations of the total + and - fields depend on the amplitude ratios of the partial waves. Defining the labels so that IE~ I ~ IE~ I, the direction of the energy flow of the + wave points along p(u~) wave whereas for IE~I ~ IE~I the energy of the - wave flows in the direction of p(u~). (This makes sense only if the self-dual fields are not LP fields, which do not propagate at all.) Thus, it is seen that the power in the electromagnetic field in a homogeneous isotropic chiral medium propagates in two self-dual fields each with two uncoupled CP components so that the + wave contains two oppositely propagating right-hand waves and the - wave two oppositely propagating left-hand waves. The velocities of the two + waves are different from those of the two - waves if the chirality parameter is non-zero. If both self-dual fields carry energy in the same direction, it is a consequence that the components E~ and E~ are two forward waves and the components E~ and E~ two backward waves with different velocities of propagation.

Fig. 4.1 Electromagnetic field in an isotropic chiral medium can be decomposed into its self-dual parts each with two oppositely polarized CP components.

4.2.7

Duality transformations for bi-isotropic media

A non-reciprocal bi-isotropic medium cannot serve as the self-dual medium in the previous duality transformation. However, a more general form for the duality transformation can be obtained by writing (LINDELL and

4.2. DUALITY TRANSFORMATIONS VIITANEN,

111

1992)

(:)d=(~ ~)(:),

(4.54)

(~)d=(~ ~)(~),

(4.55)

(

J~ )

d

= .:

i) ( J~ ) ·

(4.56)

Inserting these in the Maxwell equations for dual quantities

we should obtain the original Maxwell equations, which gives rise to equations for the coefficients A, ...,L:

H=L=A, G=K=-B, F=J=-C, E=I=D.

(4.58)

For the transformation to be an involution: ()dd = (), we must have

A 2 + BC

= D2 + BC =

1,

C(A + D)

= B(A + D) =

O.

(4.59)

One possible solution is G = B = 0, which is not, however, very interesting because it reproduces the original problem. The second solution satisfies A = -D

= ±Vl - BG.

(4.60)

The coefficients Band C can be determined if we choose one particular hi-isotropic medium with the parameters fS, J.Ls, = (xs - jItS)JJ.Lof o and (s = (Xs + jlts}JJLo€o as the self-dual medium. Solving Band C we arrive, again, at two duality transformations

es

(:)d

=

c~O (~~/~S -~~o) (:) ·

sIne

JJ.Lof o = Xs-=

I

(4.61)

XS -.

(4.62) tisee ns These expressions generalize the previous duality transformations in the special case of a reciprocal chiral medium, as is seen if XS = 0, implying e = 0, is substituted in (4.61): (

~ ) d = (±J7JS =Fjt7JS ) ( : ) ·

(4.63)

112

CHAPTER 4. FIELD TRANSFORMATIONS

Again, the upper sign can be called the left-hand transformation and the lower sign the right-hand transformation in analogy with the previous convention. The self-dual fields are thus 1

'0

E± == --(e=FJ E =f j"lsH), 2 cos ()

(4.64)

which are seen to be generalizations of the self-dual fields for reciprocal bi-isotropic media. References HARRINGTON, R.F. (1961). Time-Harmonic electromagnetic fields, pp. 98-100. McGraw-Hill, New York. JACKSON, J.D. (1975). Classical Electrodynamics, (2nd edn), p. 252. Wiley, New York. JAGGARD, D.L., SUN, X. and ENGHETA, N. (1989). Canonical sources and duality in chiral media. IEEE Transactions on Antennas and Propagation, 36, (7), 1007-13. KONG, J.A. (1986). Electromagnetic wave theory, pp. 367-76. Wiley, New York. LINDELL, I.V. (1981). Asymptotic high-frequency modes of homogeneous waveguide structures with impedance boundaries. IEEE Transactions on Microwave Theory and Techniques, 29, (10), 1087-93. LINDELL, I.V. and SIHVOLA, A.H. (1991). Generalized WKB approximation for stratified isotropic. chiral structures. Journal of Electromagnetic Waves and Applications, 5, (8), pp. 857-72. LINDELL, I.V. and VIITANEN, A.J. (1992). Duality transformations for general bi-isotropic (nonreciprocal chiral) media. IEEE Transactions on Antennas and Propagation, 40, (1).

4.3

Affine transformations

An affine transformation changes the metric of the space. An example of such a transformation is one that squeezes the space so that all distances in one direction are changed in the same proportion. A sphere is then transformed to an ellipsoid and a cube to a parallelepiped. Also, making the space the mirror image of itself in a plane is an affine transformation. !.he affine transformations considered here are defined by a constant dyadic A, which moves every space point r to another point according to the law r-+ra=A·r.

(4.65)

4.3. AFFINE TRANSFORMATIONS

113

The question remains how to transform sources, fields, media and boundary conditions so that the Maxwell equations remain valid in the transformed space. Knowing this, it is possible to obtain the solution for a transformed problem from that of the original problem by just transforming the solution.

Transformation of fields and sources

4.3.1

Let us assume that in conjuction with the affine transformation of the space, the fields are transformed as

E(r) ~ Ea(r) = B . E(r a) = B . E(A . r),

(4.66)

= C . H(A· r).

(4.67)

H(r) ~ Ha(r) = C· H(r a)

The dyadics Band C should be chosen to depend on A so that the transformed fields satisfy the Maxwell equations. In fact, after the affine transformation the equations in the frequency domain should read as \7

x Ea(r) = -jwBa(r) - Jma(r), \7 x Ha(r)

= jwDa(r) + Ja(r).

(4.68) (4.69)

To be sure of this, we apply the knowledge that the original fields satisfy the Maxwell equations in the fa space. The expression of the gradient in the transformed space, \7 a, can be obtained from the definition (4.70)

Replacing

fa

by

A· r = r· AT, we have (4.71)

whence

=-IT

\7 a = A

. \7,

or

=T

\7 = A

. \7 a.

(4.72)

The curls of a field vector in the original and transformed spaces are related through the dyadic identity 1---(K~K) . (a x b) = K(2) . (a x b) = (K . a) x (K . b), 2

which allows us to write

(4.73)

CHAPTER 4. FIELD TRANSFORMATIONS

114

= (detA)A- I . [Va

X

(A-IT. B· E(r a ) ) ] .

(4.74)

Thus, the equations (4.68), (4.69) can be written as Va

= - IT X [A

= - IT

X [A

=

] = ----=. jwA Ba(r) -

. B· E(r a )

A ---=. Jma(r),

detA

(4.75)

detA

= ] . C . H(r a)

jwA A = ---= .Da(r) + ---= .Ja(r).

Va

= -jwB(r a) -

(4.76) detA When these equations are compared with the Maxwell equations in the transformed space, \7 a

X

Va

E(r a ) X

detA

H(r a )

Jm(r a ) ,

= jwD(r a ) + J(ra ) ,

(4.77) (4.78)

by identifying the corresponding terms, t~ following relation between the dyadics Band C and the original dyadic A are obtained: _

B

=T

_

= aA,

C

=T

= {3A

,

(4.79)

where a and (3 may be any scalar constants. Thus, we arrive at the following affine transformation for the fields and sources: Ea(r) = aAT . E(A. r), (4.80) =T

=

Ha(r) = (3A . H(A . r),

(4.81)

= a(detA)A- 1 . B(A . r),

(4.82)

Da(r) = (3(detA)A- 1 • D(A. r),

(4.83)

Jma(r) = 0:'(detA)A-1 . Jm(A . r),

(4.84)

Ba(r)

Ja(r)

= (3(detA)A- 1 . J(A. r),

(4.85)

which depends on the dyadic A and two scalars 0:', {3. The coefficients 0:', {3 control the amplitudes of the transformed fields. Let us restrict their values by requiring that the affine transformation defined by the inverse dyadic A-I giv~the inverse transformation, i.e. return the quantities transformed through A back to the original ones. It is readily seen that this property requires the condition for the parameters (4.86)

4.3. AFFINE TRANSFORMATIONS

115

to be valid. Several important transformations fall under the general case of the affine transformation. For example, rotation around an axis by an angle, space reversion at a point and reflection at a plane are special cases of the affine transformation. 4.3.2

Transformation of media

The affine transformation also changes the medium parameters and their formulas are obtained by comparison of terms. If the constitutive equations for the general bianisotropic medium are written after the affine transformation as (4.87) D, = fa . E a + ~a . H a,

B, = (a . E,

+ Ji a . H,;

(4.88)

and the transformed fields are substituted from the above equations, we are able to identify the transformed medium dyadics in terms of the original medium dyadics in the form (4.89)

ea = (detA)A- e· A-IT, 1

(4.90)

.

(a = (detA)A- 1 . ( . A-IT, ~a = ')'(detA)A- 1 . J7. A-IT, where the parameter "y = af3 has either the value of 1 or -1.

(4.91) (4.92)

The above formulas (4.89) - (4.92) show that an isotropic medium with parameters e, J.L is transformed to an anisotropic medium in the general affine transformation: = r€(detA)(AT . A)-I,

(4.93)

~a = T'J.L(detA)(AT . A)-I.

(4.94)

fa

From these expressions it is seen that the transformed fa and Ila dyadics are multiples of the same symmetric dyadic and they satisfy the relation -

--1

--1

-

€

=

€a·Jia =lla ·€a=-I, J.L

or

fa

lla

e

J.L

(4.95)

It is alluring to search for an effine transformation that would transform a given anisotropic medium into an isotropic medium. However, from the previous equations we see that this is only possible for a limited class of

116

CHAPTER 4. FIELD TRANSFORMATIONS

anisotro£ic media. In fact, solving for €, Jl from (4.89), (4.92) with fa = €aI,

Jia = J.Lal shows us that the original medium dyadics f and Ji must satisfy an equation of the form

= =-1 E·J-l

€a = == -1, J-la

(4.96) -T

or € and Jl must be multiples of the same dyadic. Also, because A .A is a symmetric dyadic.?-both € and Ii must be multiples of the same symmetr~ complete dyadic S. Only in this case it is possible to find a dyadic if defining the affine transformation needed-..!o make the anisotropic medium isotropic. It is easy to see that the dyadic A can then be taken as symmetric and a multiple of the dyadic (1/2. In fact, we can write (4.97) Thus, only a very special type of anisotropic medium can actually be transformed into an isotropic medium and the corresponding class can be called affinely isotropic. This can be generalized to a relation for bianisotropic media which can be transformed to bi-isotropic media by adding a similar requirement to the skew parameter dyadics, i.e, that they be multiples of the same symmetric dyadic as € and Ii. Such a medium might be called affinely hi-isotropic. Finally, we might be interested to study what kind of anisotropic medium can be transformed to a symmetric uniaxial anisotropi~ medium with medium dyadics of the form fa == €lI + €2 U U and Jia == J.L1I + J-l2 U U. From the above expressions it can be shown, after some algebra, that the original medium dyadics must be of the general symmetric form (4.98) where S is a symmetric complete dyadic and ~ is a certain unit vector. In this case, the required transformation dyadic A can be taken as a multiple =1/2

=

of S and the unit vector u will be a multiple of S1/2 . v. A medium of this kind of can be called affinely uniaxial. 4.3.3

Involutory affine transformations

Finally, let us consider affine transformations that are involutions like the duality transformation, which means that the original electromagnetic field is obtained through two consecutive transformations, or, the equivalent,

4.4.

REFLECTION TRANSFORMATIONS

117

the affine transformation equals its inverse. To find the condition, let us consider the double transformation (4.99) from which we have the condition -2

-

11 =/.

(4.100)

The same condition is obtained from all transformation formulas. Of course, the condition (X2 = {32 = 1 must also be valid. Thus, for an involutory affine transformati~ the coefficients (X and {3 can only have values +1 or =1 and the dyadic A must be ~ squa~e root of the unit dyadic, whence detA = ±1. The trivial solutions A = ±1 are two special cases of the most general square root of the unit dyadic, which is of the uniaxial form

A = ±(/ 4.4

ab)

with

a ·b

= 2.

(4.101)

Reflection transformations

As an important special case of the involutory affine transformation we consider the most general symmetric square root of the unit dyadic, which can be written in the form

c = 1- 2uu,

(4.102)

where u is a real unit vector. This actually defines a reflection transformation, which maps any vector r to the vector r - 2u(u . r), i.e. its mirror image with respect to the plane with r . u = O. The uniaxial reflection dyadic C satisfies the basic conditions =2

=

C =1,

4.4.1

CT

= C- 1 = C ,

detC

= -1.

(4.103)

Invariance of media

The reflection transformation is often applied in conjunction with an image principle, in which original and transformed fields are combined. This makes sense only if the medium does not change in the reflection transformation, a property which is not valid for arbitrary media. Let us~tudy what kind of media are invariant in reflection. Requiring ~a = ~ for A = C in (4.89), we obtain the condition (4.104)

CHAPTER 4. FIELD TRANSFORMATIONS

118

Taking the determinant function of both sides and assuming that det€ "I 0 (otherwise the medium could not be polarized in a certain direction), we arrive at 'Y == -1. Thus, for an involutory transformation we have two possibilities: either a == -1 and (3 = 1 or a = 1 and (3 = -1. Requiring the invariance of medium parameters in reflection transformation gives us four dyadic equations for the bianisotropic parameters: (4.105) (4.106) can be any dyadics which commute, and ~ and "( any dyadics

Thus, f and Ji that anticommute, with the reflection dyadic C. Writing each dyadic in terms of components parallel and transverse to U we can see that the medium parameter dyadics must be of the form (4.107) "( == UCt

+ d.u,

Ii = J-Luu u u + lit,

(4.108)

where the dyadics Et, lit and the vectors at, ... .d, are transverse to u. From the above the following can be seen. • A bi-isotropic medium is invariant in the reflection transformation if and only if it is isotropic, i.e. if == ( == O. In fact, a chiral medium changes handedness in reflection and thus cannot be invariant.

e

• A bianisotropic medium is invariant only if u is one of the eigenvectors of the medium dyadics €, ii from both left and right and the other eigenvectors are transverse to u. This is the case, for example, for a uniaxial symmetric medium with an_ o~tical axis normal to the reflection plane. Moreover, the dyadics ~, "( must be planar and of a special form to be invariant in reflection. 4.4.2

Electric and magnetic reflections

It was seen that there exist two reflection transformations of electromagnetic fields and sources. Let us call the transformation defined by a == 1 and (3 == -1 the electric reflection, because the electric fields and sources are transformed to their mirror images, whereas the magnetic fields and sources are transformed to their negative mirror images:

Ec(r) == C· E(C· r),

Hc(r) == -C· H(C· r),

(4.109)

4.4.

REFLECTION TRANSFORMATIONS

Be(r) == -C· B(C· r),

(4.110)

Jme(r) == -C· Jm(C . r).

(4.111)

De(r) == C· D(C· r), Je(r) == C· J(C· r),

119

The second possible reflection transformation with a == -1, f3 == 1 is called the magnetic reflection and defined by

Ee(r) == -C· E(C· r),

He(r) == C· H(C· r),

(4.112)

De(r) == -C . D(C . r), Je(r) == -C· J(C· r),

Be(r) == C· B(C· r), Jme(r) == C . Jm(C . r).

(4.113) (4.114)

It is important to note that the naive reflection transformation where all fields and sources are transformed either through C or -C alone, does not satisfy the Maxwell equations in a medium which is invariant in the reflection transformation. Without different signs in electric and magnetic field vectors, the Poynting vector S == (1/2)E x H* would not transform to its mirror image but to the negative of its mirror image. 4.4.3

The mirror image principle

Any electromagnetic field and the corresponding source quantity q can be written as a sum of two self-reflecting parts, each invariant in one of the two reflection transformations: (4.115)

Here, the superscript e denotes the quantity invariant in the electric reflection (electrically symmetric) and m, in the magnetic reflection (magnetically symmetric). For example, the electric and magnetic fields can be written as (4.116) with 1

--

1

--

He(r) == 2[H(r)-C.H(C.r)],

p

1

--

1

--

Em(r) == 2[E(r)-C.E(C.r)], (4.117)

Ee(r) == "2[E(r)+C.E(C.r)],

Hm(r) == 2[H(r)+C.H(C.r)]. (4.118)

Denoting r == u( u . r) + p, on the plane u r == 0 we have C . r == C . p == == r, whence the arguments of the field vectors and their reflection images

are the same. Thus, we can write

Ee(p)

1-

-

-

= "2[1 + CJ· E(p) = It . E(p),

(4.119)

CHAPTER 4. FIELD TRANSFORMATIONS

120

1-

-

1-

-

Em(p) = 2[1- C] · E(p) = uu E(p), He(p)

= 2[1- C] · E(p) = uu· H(p),

Hm(p)

= 2[1 + C]· H(p) = It·

1-

-

-

H(p).

(4.120) (4.121) (4.122)

From the conditions u· E e == 0,

u x He == 0,

(4.123)

we see that electrically symmetric fields satisfy the PMC boundary conditions on the plane of symmetry. Correspondingly, from

u x Em == 0,

u· H'" == 0,

(4.124)

we see that magnetically symmetric fields satisfy the PEC boundary conditions. Electromagnetic fields can be decomposed into electrically and magnetically symmetric parts by decomposing their sources into symmetric parts 1

--

Je(r) == 2[J(r)

+ C· J(C· r)],

1

--

J~(r)

== 2[J m(r) - C· Jm(C· r)], 1

--

Jm(r) == 2[J(r) - C· J(C· r)], 1

--

J:(r) == 2[J m(r) + C· Jm(C· r)].

(4.125) (4.126) (4.127) (4.128)

Because the symmetric fields satisfy either PMC or PEC boundary conditions at the reflection plane u · r == 0, the problem 'original source and plane boundary' can be replaced by the problem 'original and reflectiontransformed sources with no plane boundary'. This is called the mirror image principle because the transformed source is either the mirror image or the negative of the mirror image of the original source. The image sources for the PEC plane are obtained through magnetic reflection: (4.129) and for the PMC plane through electric reflection (4.130)

4.4. REFLECTION TRANSFORMATIONS

121

The advantage of the image principle is in replacing a boundary value problem by a source problem, which in most cases is easier to handle. The present image principle is, however, only associated with PEC or PMC planes and cannot be simply extended to impedance surfaces or interface problems. A method for solving problems of this kind in terms of a more complicated image principle will be discussed in Chapter 7. 4.4.4

Images in parallel planes

The reflection transformation above was defined with respect to a plane passing through the origin. A more general reflection transformation can be written in the form r

--+

rc

= C . r + 2uu· r o ,

(4.131)

which defines reflection with respect to a plane with the normal unit vector u and passing through the point roo The image sources corresponding to a PMC plane are now (4.132) Jmi(r)

= -C . Jm(C . r + 2uu· r o ) ,

(4.133)

and corresponding to a PEe plane, Ji(r) = -C . J(C . r Jmi(r)

+ 2uu . r o ) ,

= C· Jm(C· r + 2uu· r

(4.134) (4.135)

o) .

These expressions make it possible to extend the image theory to a region bounded by two parallel PEC or PMC planes. As a simple example take the case of two PEe planes 1 and 2 going through the points rl = -UTl and r2 = ur2 with the distance d = TI + T2. If the source J(r) lies between the planes, we can substitute the plane 1 by an unknown image source J 1 (r) and the plane 2 by another unknown image source J 2 (r ). The original plus the image sources must be magnetically symmetric with respect to each of the two planes: JI(r) = -C· J(C· r

+ 2rl) - C· J 2(C· r + 2rl),

(4.136)

J 2(r) = -C· J(G· r

+ 2r2) - C· JI(G· r + 2r2)'

(4.137)

From these equations we can obtain, by elimination, equations for a single unknown source, either J 1 or J 2: Jl(r) - J 1(r - 2ud) = -C· J(C· r - 2ud)

+ J(r -

2ud),

(4.138)

CHAPTER 4. FIELD TRANSFORMATIONS

122

J 2(r ) - J 2(r

+ 2nd) = -C· J(C· r + 2nd) + J(r + 2nd).

(4.139)

The difference equations for the unknown image currents (4.138), (4.139) can be solved to give

J1(r) = -C·

J 2(r) = -C·

00

00

n=O

n=1

00

00

n=O

n=1

L J(C· r - 2nud + 2rl) + L J(r - 2nnd), L J(C· r + 2nud + 2r2) + L J(r + 2nud).

(4.140)

(4.141)

Each of these sources consists of two sets of periodic sources extending to infinity. The theory also works backwards: an infinite periodic structure can be replaced by a finite structure bounded bY,PEC or PMC planes. In addition to sources, boundaries and obstacles also transform by reflection with respect to a PEC or PMC plane. To see this, we can temporarily replace the obstacle by its equivalent source, for example, a dielectric obstacle by its polarization current J p = jw( f - f o )E. Since the polarization current transforms like the electric field, the transformed electric field induces exactly the transformed polarization current at the mirror image position of the dielectric sphere. 4.4.5

Babinet's principle

A disadvantage of the duality transformation is that a PEC boundary is always transformed to a PMC boundary, which is unphysical. However, if the boundary is planar, the duality transformation can be applied in conjunction with the image principle, so that PMC boundaries can be avoided. This combination leads to a method called Babinet's principle, which relates two problems involving complementary PEC boundary planes. Two planes containing PEC and open regions (holes) are called complementary if the holes in one of the planes correspond to PEC regions in the other and conversely. It is possible to obtain a solution for the diffraction field due to a metallic planar structure, a disk for example, if the solution is known for the corresponding complementary structure, such as a hole in the metallic plane. However, the sources of the two problems must be dual to each other. If self-dual, they are the same sources in both problems. Let the original source be J (r) = f (r) in the half space u- r > 0 bounded by the plane 8 defined by u . r = o. Let 8 = 8 1 + 8 2 consist of a PEC (metal) part 8 1 and open part 8 2 . If the original source is written as a sum of two combined sources JA(r), In(r) defined by

4.4. REFLECTION TRANSFORMATIONS

1

--

JA(r)

= 2[f(r) + C· f(C· r)],

JB(r)

= "2 [f(r)

1

-- C· f(C· r)],

123

(4.142) (4.143)

the problem can be split into two problems. The source J B is magnetically symmetric and the part 8 2 of the boundary plane S can be covered with PEC because the fields satisfy the correct boundary conditions. Thus, B corresponds to a problem of reflection from an intact metallic plane 8. In the A problem, the 8 2 part of the interface can be covered by PMC, while 8 1 is still a PEC surface, or it is a problem with an inhomogeneous plane boundary. The original problem of whole space and an interface plane has thus been transformed into two problems with plane boundaries. The complementary boundary problem with 8 1 empty and PEe on 8 2 can be approached through a similar method backwards after making the duality transform to problem A with the inhomogeneous boundary. Let us start from a problem with the complementary boundary and the dual source: (4.144) Likewise, this source can be split into two parts

Jmc(r)

jrJ = = = -"2 [f(r) + C· f(C· r)],

(4.145)

JmD(r)

jrJ = = = -"2[f(r) - C· f(C· r)],

(4.146)

of which the first one (C) corresponds to PEe boundary conditions on the whole plane, whence the plane can be either completed or removed without changing the fields. The second D source gives rise to PMC conditions on the 8 1 part of the surface. This problem is thus dual to the A problem with both dual sources and dual boundary conditions. Hence, the fields are also dual: (4.147) On the other hand, the fields Band C are easily calculated because of the planar PEe conditions. Thus, the original problem 1 and the problem 2 with complementary boundary and dual source are related through

(4.148)

124

CHAPTER 4. FIELD TRANSFORMATIONS

This means that there is a simple relation between the diffraction patterns of the two complementary problems, because the problems Band D involve no diffraction. However, the sources must be dual in the complementary problems, which is a limitation. For example, the diffraction from a hole in a metallic plane for a horizontally polarized plane wave has the same pattern as the diffraction from a complementary metallic disk for a vertically polarized plane wave. If the original source is a self-dual or antiself-dual source, this is no problem, because the dual source is then either the original source or the negative of it. The sign of the self-dual source depends of course on the chosen duality transformation. For an incoming plane wave the self-dual field is circularly polarized.

References BOOKER, H.G. (1946). Slot aerials and their relation to complementary wire areals (Babinet's principle). lEE Proceedings, 93, (3A), 620-6. JONES, D.S. (1964). The theory of electromagnetism, pp. 569-72. Pergamon, Oxford. KONG, J.A. (1986). Electromagnetic wave theory, pp. 367-76. Wiley, New York. SENIOR, T .B.A. (1977). Some extensions of Babinet's principle in electromagnetic theory. IEEE Transactions on Antennas and Propagation, 25, (3), 417-20.

Chapter 5

Electromagnetic field solutions In this chapter we consider solutions of the time-harmonic Maxwell equations for simple sources like the point source, line source and plane source in various media. Solutions for more complicated sources can be constructed from solutions corresponding to these simple sources.

5.1

The Green function

The field due to a source of unit amplitude plays a basic role in electromagnetics, because fields for arbitrary sources in linear media can be obtained by integrating such a field function. If the source is a unit point source, line source or plane source, the corresponding fields are called, respectively, the three-dimensional, two-dimensional or one-dimensional Green functions. The equation for the three-dimensional scalar Green function in homogeneous medium is written as L(\7)G(r - r')

= -<5(r - r').

(5.1)

We shall only consider such linear operators L(\1) which are polynomial functions of the operator V'. The minus sign in front of the delta function is a convention which saves us from a minus sign of the Green function. In two dimensions, the vector variable is p and in one dimension, the scalar coordinate z, instead of the three-dimensional position vector r. The twoand one-dimensional Green functions will be treated as special cases of the three-dimensional problem. Knowing the Green function, the field of any source in a homogeneous medium can be written as an integral:

f(r)

=-

J

G(r - r')g(r')dV',

(5.2)

v because operating this equation by L(\1) from the left and applying properties of the delta function, the equation L(\1)f(r) = g(r)

(5.3)

126

CHAPTER 5. ELECTROMAGNETIC FIELD SOLUTIONS

results. Here, the order of the differential operator L and the integration was simply interchanged, which is valid if fields are calculated outside the source region. Field computation inside the source region requires a knowledge of the Green function singularities, to be discussed in a subsequent section. Since the Green function performs a linear mapping from the source function g(r) to the field function f(r), for a scalar field problem, the Green function is a scalar quantity, whereas for a vector problem a dyadic Green function is obviously required. The Green function is essentially dependent on the medium. It is very difficult to derive expressions for other than homogeneous media. Also, closed-form expressions have only been obtained for homogeneous bi-isotropic and uniaxially anisotropic media and not for more general media.

5.1.1

Green dyadics of polynomial operators

Let us consider, in general terms, the solution of dyadic equations of the type L(\7) . G(r) = -8(r)I. (5.4) The dyadic operator £(\7) is assumed to be a polynomial of \7, L(\7)

= £2 + £3 . \7 + £4

: V'V'

+ ...,

(5.5)

where £2 is a constant dyadic, £3 a constant triadic, £4 a constant tetradic etc. The order of the operator is the highest power of \7 in the polynomial. In the following we consider some methods of reducing the dyadic equation (5.4) to a scalar equation. Let us define the dyadic operator L(~) to be complete if there ex~ts a vector p such that the algebraic dyadic L(p) is complete. Otherwise £(\7) is called a planar operator. For example, the operator \7\7 +k 2 is complete for k i- 0, because det(p~ + k 2 ] ) = k 4 (p . P + k 2 ) is_not identically zero. On the other hand, \7\7~I is planar because det(pp~I) = 0 for all p. The rule that the dot product of two dyadics is complete only if the dyadics are complete, valid for algebraic dyadics, does not hold for operator dyadics. For example, the ~ot product of the planar operator \7\7 and the linear dyadic rr equals 4I,_which is complete. In the following, we only consider complete operators L(\7). All dyadic identities are valid for dyadic operators if the result can be understood as an operator. The following identity is essential in solving Green dyadic equations:

1

£(2)T (\7)

. £(\7)

= £(\7) . £(2)T (\7) = [det£(V)] I,

(5.6)

5.1. THE GREEN FUNCTION

127

because the dyadic equation can be reduced to solving a scalar Green function. In fact, starting from a scalar problem of the form detL(V') G(r)

= -8(r),

(5.7)

multiplying both sides by the unit dyadic and inserting (5.6) results in the equation [detL(V)]G(r)I = L(2)T(V) · (L(V)G(r)] =

L(V). [L(2)T(V)G(r)] = -6(r)I.

(5.8)

Comparing this to (5.4) shows us that the Green dyadic corresponding to the operator I(V) can be written in terms of G(r), the solution to the scalar equation (5.7) in the form

G(r) = L(2)T(V) G(r).

(5.9)

This conclusion presumes uniqueness, which is obtained by imposing Sommerfeld and Silver-Muller radiation conditions at infinity. If, instead of L(V), the operator in (5.4) is of the form L(2)T(V), the corresponding Green dyadic F(r), i.e, the solution of

(5.10) can be written as

F(r) = I(V) G(r),

(5.11)

where G(r) satisfies (5.7). The two dyadic Green functions G and F corresponding to the operators Land L(2)T have the simple relation (5.12) as can be quite easily seen. Knowing F, the other Green dyadic G can be directly written down. Thus, instead of solving a dyadic equation (5.1) for nine scalar functions it is sufficient to solve_a single scalar function from the equation (5.7), pro~ded the operator I(V) is complete so that the polynomial operator detI(V) is non-zero. However, the price is paid in the order of the operator. In general, if the order of L is!!" the o~der of L(2) is 2n and that ~ detL, 3n. For the Helmholtz operator L(\7), of special interest here, detL(\i') turns out to be at most of the fourth order. If the fourth-order equation can

CHAPTER 5. ELECTROMAGNETIC FIELD SOLUTIONS

128

be solved, the dyadic Green function is straightforwardly obtained through application of the dyadic operator. The order may be reduced if the original oper~or dyadic can be ~ritten in terms of another polynomial dyadic operator A(V') in the form £(\7) = A(2)T (V').Jf this can be done, the equation is of the form (5.~), with A(V') replacing 1(\7), and the operator~n (5.7) is in this case detA(\7), which is generally of lower order than detI(\7).

5.1.2

Examples of operators

Let us consider some operator triples £(\7), £(2)T(\7) and det£(\7), appearing in electromagnetic problems. In the first example, the order of all three operators happens to be the same: (5.13) £(\7) = \7\7 + k 2

I,

[;(2)(\7)

= k2(\7\7~I + k2I),

(5.14)

detL(\7) = k4(\7 2 + k2). The second example involves a constant dyadic of the first order: 1(\7) = \7 x I + a,

(5.15)

a in a dyadic operator (5.16)

= \7\7 + (\7 ·0) x I - \7(o~I) + 0(2), det£(V') = \7\7 : 0 + 0(2) : (\7 x I) + deto.

£(2) (\7)

(5.17) (5.18)

If 0 is symmetric, one term can be seen to drop off from the expressions I)2~ and detL. If ~ is antisymmetric, the operator L(V') is planar and detI(\7) = o. The third example involves a constant dyadic 0 in a dyadic operator of the second order: (5.19) L(\7) = \7\7 + k2

a,

[)2)(\7) = k2(\7\7~a + k2a(2»), 4(\7\7

detL(\7) = k

: 0(2)

+k

2

deta ).

All operators are seen to be of the second order. The fourth example involves two constant dyadics

(5.21)

a and 73:

a,

(5.22)

+ k2(\7\7~{J)~a + k4a(2),

(5.23)

£(\7) = \7\7~~ + k2 £(2)(\7) = (\7\7 : {J(2»)\7\7

(5.20)

5.2. GREEN FUNCTIONS FOR HOMOGENEOUS MEDIA

detL(V') = k2 [(V'V' : ~(2»)(V'V' :~) + k 2V'V' : (~2)~~)

129

+ k4det~]

.

(5.24) is linear or (i antisym-

It is seen to lead to fourth-order operators unless 7J metric. As the last example we consider a composite operator: L(V)

= A(V') . p.B(V') + 0,

I/ 2 )(V') = A(2) . ~(2) • B(2) + (A. ~. B)~a + 0(2),

(5.25) (5.26)

detL(V') = (detA)(detp)(detB) + (A(2) . p(2) · B(2)) : Zi+

(:4 . =g . B) : a(2) + deta.

(5.27)

!.he orders ~f these ~rived operators depend on the original operators A(V') and B(V'). If LiV) is theyelmholtz operator of a bianisotropic medium, the operators A(V') and B(V') are of the form (5.16), which leads to fourth-order operators [)2)(V) and detL(V). References LINDELL, I. V. (1973). On the theory of Green's dyadics of polynomial operators. Helsinki University of Technology, Radio Laboratory Report 858.

5.2 5.2.1

Green functions for homogeneous media Isotropic medium

The dyadic Green function for the isotropic homogeneous space with parameters e, J1. is one of the most useful mathematical concepts in formulating electromagnetic problems. In the isotropic case, the Green function, originally a function of a vector variable, G(r - r') is a function of a scalar variable only, G(D), where D is the distance between the two points rand r', in real space a positive number D = [r - r'[. This property is due to the fact that there is no preferred direction in the space, i.e, the Green function is invariant to all rotations of the space. The scalar Green function

Consider the basic scalar Helmholtz operator problem in isotropic space: H(V)G(D) = (V 2

+ k2)G(D)

= -h(r - r"),

(5.28)

130

CHAPTER 5. ELECTROMAGNETIC FIELD SOLUTIONS

The solution satisfying the Sommerfeld radiation condition is of the form

e- j k D G(D)==-D'

(5.29)

47r

where the distance function D is denoted by

D(r, r") == v(r - r") . (r - r').

(5.30)

We may check the result (5.29) by writing \7. \7G(r - r') 1 [ D2

and for D

=1=

= \7. [-Uk + ~)G\7DJ =

1 J \7 D . \7 D - Uk + D 1 )G\72D, + Uk + D)2

(5.31)

0,

\7D

= r ; r' ,

\72D =

~.

(5.32)

These substituted in (5.31) gives us V . VG(r - r") == -k 2G for either branch of D and any r', which may also be a complex vector. At the point with D == 0 the Green function is singular, and so also is the term V' 2 G(D ). This latter singularity is obviously more pronounced and is thus responsible for producing the delta function on the right-hand side of (5.28). The nature of the singularity can be checked by multiplying (V'2+k 2 )G(r - r /) with an arbitrary function f(r), continuous at r == r', and integrating over the real space. Considering first only real vectors r', we can put r' == 0 for simplicity and without losing anything from generality. Thus, D == 0 at the point r == 0 and the integrand vanishes everywhere except at the origin. Surrounding this by a small sphere with radius To, we obtain, applying partial integration in the small volume Vo with the surface So,

f(O)

fUr. \7 ~dS + \7 f(O) · JU; dV = - f(O) f d; = -47rf(0). T

So

T

\1;,

(5.33)

To

So

The last volume integral disappears because of radial symmetry. Thus, from the definition of the delta function, the solution (5.29) for (5.28) has been established for r' == 0 and, through analytical continuation, for any

r'.

5.2. GREEN FUNCTIONS FOR HOMOGENEOUS MEDIA

131

The Green dyadic The Green dyadic in an isotropic and homogeneous medium satisfies the vector Helmholtz equation

H(~) . G(D) == -6(r - r')I,

(5.34)

where the Helmholtz operator is defined as =

x=

H(~)

== ~\!xI

+ k 2=I.

(5.35)

We may now apply the previous method for the solution of (5.34). In fact, identifying H(\!) with k-2L(2)(~) of (5.14), the solution can be written as G == k 2F with F defined in (5.11) with the operator £(\7) defined in (5.13) and the scalar equation in (5.15). The scalar Green function is given by (5.29) divided by k 4 , whence the dyadic Green function is

G(D)

= (1 +

:2 'V'V)G(D).

(5.36)

(5.36) is a very useful result and it forms the basis for integral equation formulation of electromagnetic problems, as first done by LEVINE and SCHWINGER (1950). The Green dyadic is usually given in this form without performing the differentiations, because in field computation the gradients can often be simplified through partial integrations. The Green dyadic appears highly singular owing to the double differentiation of the 1/D dependent scalar Green function. It is important to know how fields can be calculated inside a source region where the distance function becomes zero. The singularity will be studied in a subsequent section. The electromagnetic fields due to and outside a current function J(r) can be written as integrals of the Green dyadic:

E(r) = -jwl1

H(r) =

J

J

G(r - r') . J(r')dV',

(5.37)

v

'VG(r - r') x J(r')dV'.

(5.38)

v From duality, the fields from a magnetic current Jm(r) are

J

H(r) = -jWf

G(r - r') . Jm(r')dV'.

(5.39)

'VG(r - r') x Jm(r')dV'.

(5.40)

v

E(r)

=-

J v

132

5.2.2

CHAPTER 5. ELECTROMAGNETIC FIELD SOLUTIONS

Bi-isotropic medium

A slightly different method can be applied to finding the Green dyadic for the bi-isotropic medium with the parameters €, u, K, and X. The Helmholtz operator applicable in this case can be written in product form

in terms of two first-order dyadic operators

(5.42)

a

which are both of the form (5.16) with = =fk±I. The solution is obtained by first solving the scalar Green function G(D) satisfying

det(H)G(D) = -det(L+)det(L_)G(D) = -8(r - r').

(5.43)

Because the operator is factorized in terms of two determinant operators

(5.44) we first try to find the solution for G in terms of a linear combination

(5.45) of solutions for two second-order equations (5.46) By substitution we have a condition for the coefficients A+ and A_: 2

2

2

1

(A+ + A_)V 6 + (A+k_ + A_k+)6 = k+k- 6. Requiring that the coefficients of both equations which can be solved to give

\728

(5.47)

and 8 vanish, we have two

(5.48) This inserted in (5.45) gives us the scalar Green function: G(D)

1

= - k+k_

1 k~ _ k~ [G+(D) - G_(D)].

(5.49)

5.2. GREEN FUNCTIONS FOR HOMOGENEOUS MEDIA

133

Finally, we can write for the Green dyadic corresponding to the biisotropic medium the expression

G(D) =

H(2)T Ge(D)

=

L+ (2)T . L_ (2)T G(D),

L±(2)T = V'V' ± k±V' x I + k~I.

(5.50) (5.51)

Substituting the operators, we have for the final expression = 1 [ G+ G(D) = k+ + L V'V'(k;

V'(G+ - G_) x I

G_

+ L)+

+ (k+G+ + k_G_)I].

(5.52)

The Green dyadic involves two scalar Green functions (5.46), each with a different exponential factor. Together they represent a combination of two radially propagating electromagnetic wave components with two different phase velocities. It is easy to check that for the non-chiral limiting case k ; -.... k : ~ k = koVn2 - X2 , the bi-isotropic Green dyadic reduces to one similar to the isotropic space Green dyadic (5.36). The same result could be obtained by splitting the unit source into two self-dual parts and computing the corresponding self-dual fields using the corresponding isotropic Green dyadics, whence adding up the results for the total field, the Green dyadic (5.52) can be identified. 5.2.3

Anisotropic medium

The Green dyadic problem for any bianisotropic media can also be reduced to a scalar Green function problem, but the equation will be of the fourth order in general, whence it is not solvable in terms of special functions. In some special cases, however, the problem can be reduced to solving secondorder equations. In the non-isotropic case, the space is not invariant to all rotations, whence we cannot write the Green functions as functions of a single scalar variable D. Instead, they must be written as functions of the vector variable r - !'. ~et us restrict, for simplicity, the problem to anisotropic media with = '( = o.

e

General anisotropic medium

The Green dyadic problem for the general anisotropic medium corresponding to the electric Helmholtz operator is =

= , = -8(r - r ,= )1,

He(V') . Ge(r - r)

134

CHAPTER 5. ELECTROMAGNETIC FIELD SOLUTIONS

and the electric field corresponding to a current source J (r) can be written as

E(r) = jw

J

Ge(r - r') . J(r')dV'.

(5.54)

v The method of reducing (5.53) to a scalar equation presented in the previous section can be applied in a straightforward manner. The operator H e(V) is of the form (5.22), whence the Green dyadic can be expressed in the form (5.9) with (5.23) inserted:

= G e (r

- r /)

=

[(\7\7 : iL( -2)\7\7 + w2(\7\7~iL-I) ~f + W4f(2)

r

Ge(r - r').

(5.55)

The scalar Green function satisfies

detHe(V) Ge(r - r/) = -6(r - r"), detH e (\7) = w2(\7\7:

]l =) (\7\7 : f)

detJl

(5.56)

+ w4\7\7 : (f(2) ~ II-I) + w6detf.

(5.57) Unfortunately, in general, this is a partial differential equation of fourth order and there are no special functions to express its general solution. For some special dyadics €, Jl the fourth-order operator detH e (Vi') can be factorized to a product of two second-order operators. Let us study this possibility closer. Because of missing first- and third-order terms, the scalar operator (5.57) can be written in the following form after some algebraic manipulations:

==

w4

detHe(V) = w2 [€ : VV' + a][JL(-2) : V'V + -detf]a

(5.58) This expression equals (5.57) for any finite and non-zero value of the parameter Q. Obviously, in order for the operator detH e(V) to be of factorized form, the last term in (5.58) should be zero. Because the symmetric part of the dyadic which double-dot multiplies the VV operator must the~v~ish, the dyadic itself must be antisymmetric. But since it is of the form A~ A, it cannot be strictly planar like the antisymmetric dyadic, whence we conclude that the dyadic itself must vanish. Thus, we have found the condition for

135

5.2. GREEN FUNCTIONS FOR HOMOGENEOUS MEDIA

factorizing the determinant operator by requiring that there exist a scalar a such that the following condition between the f and p, dyadics be valid: (5.59) This implies that the dyadic inside the brackets must be a linear dyadic, or there exist vectors a, b and a second scalar (3 such that we can write

f- 1 • ~ = (31 + abo

(5.60)

This condition resembles that for a medium which was earlier called affinely uniaxial. In fact, if either ~ or fi is a symmetric dyadic, the medium parameter dyadics can be obtained from those of a uniaxial medium through an affine transformation. Affinely isotropic medium

Affinely isotropic media with symmetric parameter dyadics ~ = €o~r, ~ = J.Lo~r satisfying the condition =tir = 'Yfr also satisfy the condition (5.59) with

(5.61)

and, thus, the corresponding scalar Helmholtz operator can be factorized. To avoid solving fourth-order differential equations at all, we can write the original dyadic Helmholtz ~perator He (V) in the form A (2) (V) with a suitable polynomial operator A(V). In fact, comparing (5.53) and (5.20), we have = _ 1 2 = (2) (2) (5.62) H e(V') - k 2 2d = (\7\7 + k o 'Y€r ) • oJ.Lo'Y et€r Thus, the Green dyadic must be of the form AT(V')Ge(r - r'), or (5.63) and the scalar Green function Ge(r - r') satisfies the equation =

_

1

2

= (2)

detA(V)G e - k 3 3( d - )3/2 det [VV + ko'Y€r o'Y

Jjo et€r

o ( VV': = krr:« €r J-Lo ydet€r

3/2

with ka = koV'Ydet€r.

+ ka2)Ge =

-6 ( r-' r),

JGe

_ -

(5.64)

CHAPTER 5. ELECTROMAGNETIC FIELD SOLUTIONS

136

The scalar function Ge(r - r') can be obtained through an affine transformation of the isotropic Green dyadic by writing \7 a == Er 1/2 . \7 and r., == Er -1/2 . r, which transform (5.64) into 3/2

( \72a + k a2)G e == _J.-Lo

~

ydet€r ~( _ r ') == k o r o

3/2

_~~( k o ra

_ ')

ra ·

(5.65)

0

The solution for the scalar Green function is, obviously, (5.66)

Da

= J(r a - r~) · (ra - r~) =

Jf

r -1 :

(r - r)(r - r').

(5.67)

Thus, the Green dyadic for the affine isotropic medium is obtained through substitution in (5.63): (5.68) As a check, setting ii; == I, f r == I, we have k a == k o and D a == D, whence the free-space Green dyadic can be seen to arise as a special case of (5.68).

Reciprocal uniaxial medium It is also possible to find the Green dyadic of a reciprocal uniaxial medium in terms of special functions. Let us consider a uniaxial dielectric medium with a symmetric permittivity dyadic with an axis defined by the unit vector v, defining the electric Helmholtz operator: (5.69) Taking a == k~f.of.tf.v in (5.58), the determinant of the Helmholtz operator can be written in factorized form as (5.70)

The corresponding scalar Green function can be written in terms of the scalar Green functions of the two operators in integral form. It does not, however, seem to be possible to obtain any closed-form expression for the scalar Green function.

137

5.2. GREEN FUNCTIONS FOR HOMOGENEOUS MEDIA

Nevertheless, the dyadic Green function can be and has been solved after considerable algebraic effort (LINDELL 1974; CHEN 1983; WEIGLHOFER 1990), details of which must be omitted here = 1 VV Ge(r - r , ) = {La ( kl

= + EvEr De

-1)

G(D e )

= JEv'lr -1

:

+ {La V

X

itv« 47l"k '

(5.71)

t

(r - r')(r - r'),

(5.72) (5.73)

q =

V X

Vln(kop)

vxr

= (vxr )2'

(5.74)

The solution appears singular on the axis p = (1- vv) . r = 0 because the function q is proportional to 1/p. However, the function T can be shown to vanish like p2 so that the result is actually continuous on the axis. The Green _dyadic corresponding t~ the more general co-uniaxial medium with J-LI replaced by ~ = J-Lo(J-Lt1t + J-Lvvv) can be written in the same form as (5.71) with just a small change in the definition of T (WEIGLHOFER 1990): Dm

=

J

{Lv/ir -1 :

(r - r')(r - r').

(5.75)

The resulting expression (5.71) can be checked for the special case J-Lt = 1ft, /-Lv = --rEv, which corresponds to an affine isotropic medium. In this case, we have De = D m, whence T = 0 and the curl term vanishes in (5.71). The remaining expression can be seen to coincide exactly with that of (5.68), if we note that kaD a == ktD e and k~ = k;E v' References BASSIRI, S., ENGHETA, N. and PAPAS, C.H. (1986). Dyadic Green's function and dipole radiation in chiral media. Alta Frequenza, 55, (2), 83-8. BLADEL, J. VAN (1964). Electromagnetic fields. McGraw-Hill, New York. CHEN, H. (1983). Theory of electromagnetic waves, pp. 370-88. McGraw-Hill, New York. FELSEN, L.B. and MARCUVITZ, N. (1973). Radiation and scattering of waves. Prectice Hall, Englewood Cliffs, NJ.

138

CHAPTER 5. ELECTROMAGNETIC FIELD SOLUTIONS

GVOZDEV, V.V. and SERDYUKOV, A.N. (1980). Green's function and the field of moving charges in a gyrotropic medium. Optics and Spectroscopy (USSR), 47, (3), 301-4. LEVINE, H. and SCHWINGER, J. (1950). On the theory of electromagnetic wave diffraction by an aperture in an infinite plane conducting screen. Communications on Pure and Applied Mathematics, 3, 355-91. LINDELL, I.V. (1974). Coordinate-independent Green dyadic for general symmetric uniaxial media. Helsinki University of Technology, Radio Laboratory Report 861. TAl, C.T. (1973). Dyadic Green's functions in electromagnetic theory; Intext Educational Publishers, Scranton, PA. WEIGLHOFER, W. (1988). Green's tensor in anisotropic media, American Journal of Physics, 56, (12), 1095-7. WEIGLHOFER, W. (1990). Dyadic Green's functions for general uniaxial media. lEE Proceedings, 137H, (1), 5-10.

5.3

Special Green functions

From the Green function corresponding to the homogeneous threedimensional space we can derive Green functions corresponding to some special cases by integrating the three-dimensional Green function. In particular, we consider the two- and one-dimensional Green functions as well as the half-space Green function limited by a PEC or PMC plane boundary. 5.3.1

Two-dimensional Green function

When sources, boundaries and media are independent of one Cartesian space coordinate, the fields also have the same independence and the problem is two dimensional. The basic source is a constant line source of unit amplitude and the corresponding field is called the two-dimensional Green function. Its functional form can be obtained either by solving the two-dimensional Helmholtz equation or through integration of the threedimensional Green function. The scalar Green function for an isotropic medium is the solution for (5.28) in two dimensions (the x-y plane):

(5.76) and it can be written as

(5.77)

5.3.

139

SPECIAL GREEN FUNCTIONS

where Hi 2 ) is the Hankel function of second kind and zero order. The distance in two dimensions is

D = V(p - pi) . (p - pi).

(5.78)

The function (5.77) can be obtained from (5.29) by integrating along the z coordinate from -00 to +00:

_J oo

G 2 (p) -

41r -ex:>

z2

Vp2 + z2

J oo

~

e-jkv p2+

-

dz - -

1

e-

72 -

21r 1

and applying the integral identity (3.388.4) from RYZHIK (1980),

j(X 2 - 1t- 1 e- iJ""dx = -j

v: (~)

j k pr

~dr,

1'-1/2

1

(5.79)

GRADSHTEYN

r(1J)Hi~~_v(1L),

and

(5.80)

1

with t/ = 1/2 and Jl = kp, r{I/2) = Vi. The two-dimensional Green function satisfies the radiation condition in two dimensions for p --+ 00. This condition requires that the function behave like the Hankel function of second kind for large argument values: H(2)(kp) -+ n

J

2 e- j kPej(-If/Hmr/2) ,

1rkp

(5.81)

i.e. vanish like 1/ v'P and propagate outwards like a cylindrical wave with the propagation factor k = wViif.. For a vector problem in two dimensions the Green dyadic can be formed in much the same way as in three dimensions. In fact, the same dyadic formalism can be applied, but in this case the scalar Green function is given by (5.77): -

-

1

-

1

G 2(D) = (1+ k 2 V'V')G 2(D) = u zu z G 2 (D )+ (1 t + k 2 V'tV't)G 2(D). (5.82)

The gradients are in two dimensions because they operate on a function depending on two space dimensions only. (5.82) shows us that the problem is split into two parts, because u, u z G 2 gives a z directed electric field for a z directed current, whereas the last term gives a transverse field for a transverse current. Thus, the field problem has two components that do not couple to one another. Because the magnetic field is orthogonal to a two-dimensional electric field if it is z directed or transversal (E . V't x E = 0 for E = uzEz or E = E t ) , we can speak of splitting the problem into transverse electric (TE) and transverse magnetic (TM) field problems.

140

CHAPTER 5. ELECTROMAGNETIC FIELD SOLUTIONS

5.3.2

One-dimensional Green function

The one-dimensional problem involves plane parallel boundaries and sources that are constant on parallel planes. Thus, the fields are also constant on parallel planes and depend only on the normal (z) coordinate. The one-dimensional scalar Green function satisfies the equation

(:2 +

k2)G 1(z - z')

= -6(z -

z').

(5.83)

The solution which represents a plane wave propagating outwards from the plane z = z' is G (z - z') = _l_e-jklz-z'l (5.84)

2jk

1

'

which is also directly obtained by integrating the three-dimensional Green function in the x - y plane

ff 211"

G 1 (z) =

o

00

f

00

G( J p2 + z2)pdpd¢ =

0

~

e-ik.jp2+ z2 d( J p2

+ Z2).

(5.85)

0

The corresponding Green dyadic in one space dimension is , = , = Id2 G1(z - z) = (1 + UzUz k 2 dz2)G 1(z - z)

= ItG

1 (z

-

,

Z ) -

=

1 , u, u, k 2 <5(z - z ).

(5.86)

It is easy to solve the electric field due to and outside of a constant transverse surface current source J (r) = J s<5(z) in the form

E(r) = -jWJL

J

G1(z - z')· Js(z')dz' =

-~Jse-iklzi.

(5.87)

The electric field is continuous across the surface current layer and outside the layer it is always tangential to the plane. A normal current source J(z) = uzJ(z) does not produce any field outside the current layer, but, instead, a field is generated inside the layer. The magnetic field experiences a step discontinuity across the surface, as is observed by taking the curl of the electric field expression above: (5.88)

This means that a z directed current component in the sheet does not produce a magnetic field.

5.3. SPECIAL GREEN FUNCTIONS

5.3.3

141

Half-space Green function

Green functions can also be expressed for problems involving inhomogeneous media, provided the field can be solved for the general source. A number of such problems has been treated in a book by TAl (1971). As a simple example, let us consider the problem of half space bounded by a PEC or PMC plane at u . r = o. We can apply the two reflection transformations to construct the corresponding Green dyadics in terms of the free-space Green dyadic. Because the field E(r) due to the source J(r) can be made to satisfy the PMC and PEC boundary conditions by adding the respective transformed fields ±C· E(C . r), the total field can be written as

E±(r)

= E(r) ± C . E(C . r).

(5.89)

From this, it is easy to form the corresponding Green dyadics. In fact, we can write =

I

=

I

= ==

I

G±(r, r) = G(r - r ) ± C· G(C· r - r ),

(5.90)

where the upper signs refer to the PMC and the lower signs to the PEC boundary. To check the electric field boundary conditions for r = C . r = p, we can write

(5.91) whence u, ·G+ = 0, which is the PMC condition and u, x G_ = 0, which is the PEC condition. It is also seen that the Green dyadic is not a function of the difference r - r', as for homogeneous media. Applying suitable reflection transformations, Green dyadics for problems involving more than one conducting or magnetically conducting plane can be constructed. Green dyadics corresponding to plane parallel material interfaces are discussed in Chapter 7.

References I.S. and RYZHIK, I.M. (1980). Table of integrals, series and products, Academic Press, New York. TAl, C. T. (1971). Dyadic Green's functions in electromagnetic theory, Intext Pub1ishers, Scranton, PA.

GRADSHTEYN,

CHAPTER 5. ELECTROMAGNETIC FIELD SOLUTIONS

142

5.4

Singularity of the Green dyadic

To be able to apply the Green function concept to integral equations, its singularity must be known exactly. Basically the question is how to calculate fields inside the source region where the source and field points coincide. Let us study two simple cases for which the fields within the source can be easily determined. 5.4.1

Constant volume current

For a constant electric current source J (r) = J 0 filling the whole space, the resulting electromagnetic field is obviously also constant and the curl terms in the Maxwell equations vanish. Thus, the fields in an isotropic medium are 1 (5.92) Eo(r) == --A-J o, Ho(r) == O. JWf. For a constant magnetic current J m o we can write the fields from duality: 1

Ho(r) == --A-J m o . JWIl

(5.93)

These expressions can be generalized for the anisotropic medium

E o (r) ==

1 =-1 . J 0' JW

(5.94)

- -. f .

and even for the bianisotropic medium (5.95)

Ho(r)

= ~Cji - '( . ~-1 • ~)-1 JW

.

(-J m o

+'(. ~-1 . J o).

(5.96)

When studying the singularity of the isotropic Green dyadic it is of interest to find the electric field E e5 at a point r within a small element of constant current J o , zero outside a small volume ~ bounded by a surface Se5. The field can be obtained by subtracting from Eo in (5.92) the field of the complementary problem: constant current J o filling the whole space except the volume Ve5. The field in the cavity can be determined with the isotropic Green dyadic without any knowledge of its singularities since the field point is now outside the source region. Let us find the field at the origin r == 0 inside the cavity, for simplicity of notation. Taking the limit of a small volume ~ with its largest measure

5.4. SINGULARITY OF THE GREEN DYADIC

satisfying k6

-+

143

0, we obtain after evaluating an integral for the cavity field

J( =

1 k

, ') e

-jkr'

,

I+-V'V' - - d V ·J o 2

E o - E6 == -jW{l

47rr'

= L .J -+-.-, (5.97

J0

-+ - - . JW€

0

)

JW€

V-VII

L=

J

=-

V'V'G(r')dV'

V - V6

f

n'V'G(r')dS',

(5.98)

So

with n denoting the unit normal vector pointing outside of V6 • The surface integral at infinity vanishes if we assume losses however small, because the exponential decay of the Green function is faster than the algebraic growth of the integrand. Since the term -Jo/jWf. equals Eo, the last term in (5.97) must be -E6 and we can write for the field inside the current element the expression

E6 ==

-~L.

.r, ==

JW€

f

D'.' J JW€

oV , _ 1_ d5 ,. 41rr'

(5.99)

Btl

This can be interpreted as the electric field due to a static surface charge !Js(r/) distributed on the small surface 56: !Js (r

/)

n' . J o

== -.-.

(5.100)

JW

The field E6 depends on the charge distribution, which again depends on the current vector J o and tE.e form of the surface 56. The result is best expressed through the dyadic I, called the depolarization dyadic (STRATTON 1941). If the volume v:s is not small, or, the !.quivalent, k is not small, the depolarization dyadic should be written as L(k), an analytic function of k, which can be expanded in a Taylor series as

= L(k) == L

=:

=:

+ kL l + k

2= L2

+ ...

(5.101)

with (5.102)

Because £2 and the following terms contain powers of kr' i~the integrand, the integrals are orders of magnitude smaller than that of I.

144

CHAPTER 5. ELECTROMAGNETIC FIELD SOLUTIONS

The dyadic L is symmetric, because it is the volume integral of a symmetric dyadic. Further, it satisfies the condition

(5.103)

Because the solid angle of a surface element d.S' seen from the origin is dO' = n' . u~dS' /r,2, the integral in this equation gives the solid angle of the sur~ce S6 as seen from the field point r = 0 and it equals 471". The dyadic I in general depends on the position of the origin within the volume V6 except for ellipsoidal ~olumes for which it is actually independent of that position. Beyond that, is only dependent on the form of the s~face S6 and not on its size, a~hough it must be small so that the dyadic L(k) can

r

be approximated by £. It is intuitively clear that in the case of a non-constant current distribution J(r), the field due to and within a small volume element Vb of this current is given predominantly by the value of the current density at the volume and not by its derivatives. This is because the largest correction term obtained from the Taylor expansion for the current function J(r) is r6 . V'J(O), where r6 is the largest distance within the volume Vb, and it is negligible for small Ir61, whereas the basic term is independent of Ir61. Of course, for strongly changing current functions V6 must be small enough. The field at a zero of the current function cannot be obtained in this manner.

5.4.2

Constant planar current sheet

Another canonical source is the constant surface current J so on the plane = O. The electric field is everywhere tangential and continuous across the surface, as is seen from the expression (5.87): u . r = 0 with u . J so

(5.104) This can be interpreted in terms of a_current source across a vector transmission line of dyadic impedance 1]/t so that the loading impedance is (".,/2)lt, whence (5.104) describes the resulting vector voltage. On the other hand, according to (5.88) the magnetic field is discontinuous at the surface satisfying (5.105)

5.4. SINGULARITY OF THE GREEN DYADIC

145

which actually is the interface condition for the magnetic field valid at any interface. Because of antisymmetry, we must have H 2 = -Hi and, hence, (5.106)

Also, in the middle of the current sheet the magnetic field must be zero. Thus there are two limiting values for the tangential magnetic field when the field point approaches the plane from each side and their average value is zero. This can be assumed to be valid also on curved surfaces if the tangent to the surface is continuous. From duality we have corresponding properties for fields around a magnetic surface current. Let us now consider the field inside the constant planar current element J so on a small disk A6 on the x-y plane A as due to two parts: (1) the constant surface current on the whole A minus (2) the field inside a hole A6 in the plane surface current. As in the volume current case, we consider the expression for the field Eo - E 6 due to the surface current in a decreasing hole A6 with the rim curve C6:

Eo -

s, = -jWJ.L

J [1 + :2

V'V']G(lp - p'l)dS' . J so =

A-AcS

(5.107)

where n denotes the unit normal vector at the rim point p pointing outwards from the disk A6. The term -(TJ/2)J so = Eo can be identified as the field due to the total surface current, whence the last term equals - E6. When the maximum measure of the disk satisfies k6 --+ 0, we can apply static approximation and write (5.108)

This, again, can be interpreted as the field due to a static line charge along the rim curve C6:

ec (p')

n' . J so = --.-. JW

(5.109)

It is seen that the integral for E6 diverges as the area A6 grows smaller, because the integrand grows like l/lp - p'l.

146

CHAPTER 5. ELECTROMAGNETIC FIELD SOLUTIONS

5.4.3

Singularity for a volume source

When trying to determine fields inside a continuous volume source J(r), the singularity of the Green function can be expressed in terms of previous analysis. In fact, enclosing the field point within an infinitesimal volume Va, the field can be expressed in two parts: (1) from the volume current surrounding V6 and (2) from the local current in V6. The latter was seen to be affected only by the shape of Va and the density of the current and not the size of the element or the total current integral. It was also seen to be dependent on the location of the field point within V6, in general. This division in two parts can be transferred into the Green dyadic itself. The cavity field (1) can be computed in terms of the non-singular part of the Green dyadic, which is called the principal value term and denoted by PYa. The field (2) within the current element can be given as (5.99). Thus, the Green dyadic can be written in two terms corresponding to these two fields 1(5.110) G(D) = PV{jG(D) - k 2 L6(r - r'). The dyadic L depends on the geometrical form of the infinit~imal volume and the position of the field point within the volume. L may be called the normalized dyadic solid angle of the volume V6 . In the following list, expressions of I dyadics for some simple volume geometries, are given according to YAGHJIAN (1980): • for any point within the sphere or for the centre point within any Platonic polyhedron, L = iI; • for any point within an ellipsoid, L = A{Lxuxux+Lyuyuy+Lzuzuz), where the L; are the depolarizing factors (STRATTON 1941) and the axes of the ellipsoid are the coordinate axes, A = l/{L x + L y + L z ) ; • for a parallelepiped L = (flxuxu x +OyUyU y +Ozu z u z )/ 41r, where ni is the solid angle subtended at the centre point by sides perpendicular to the axis i, Ox + ny + Oz = 41r; • for a pillbox of arbitrary cross section, thin in the z direction, L = UzU z;

• !9r acircular cylinder of any cross section, long in the z direction,

L = I t / 2; • for a finite circular cylinder, axis in the z direction, L = (I cos O)uz u, + cos O(J - u , u z ) , where 0 is half of the angle subtended by the bottom of the cylinder at the centre point.

!

5.4.

SINGULARITY OF THE GREEN DYADIC

147

Volume integral equations The singularity of the Green dyadic gives a kind of internal boundary condition which can be applied for the construction of volume integral equations for electromagnetic fields. In fact, consider a dielectric object as an example with permittivity € == €o€r. The polarization in the object can be represented by the polarization current J p = jWf o ( €r -1 )E. The field inside the object comes from two parts: from the incident field E i from outside the object and the field from the polarization current of the object, which can be written as an integral

E(r) = Ei(r) - jW/Lo Ei(r) + k~PV6

!(€r -

!

G(r - r') · Jp(r/)dV' =

v

1)G(r - r /). E(r/)dV' - (€r - 1)£. E(r). (5.111)

v This defines an integral equation for the unknown fiel~E(rl inside the obstacle volume V. For a spherical volume VeS we have L == 1/3 and the volume integral equation can be written as

Ei(r) = €r; 2E(r) -

k~PV6 !(€r -

1)G(r - r /). E(r/)dV'.

(5.112)

v

5.4.4

Singularity for a surface source

Because the field inside a small surface current element grows infinite as the element size decreases, the previous principal value decomposition does not work. Actually, this can be seen from the fact that since in a surface current the volume current densi~ becomes infinite, in the volume field integral, the term containing the I dyadic becomes infinite. Since the infinity is cancelled by another infinity in the principal value integral, there should be a better means to take out the singularity from the integral. Such a method is called the finite part method (DAVIES and DAVIES, 1989). Mathematically it amounts to subtracting from the current function Js(r') another function, Jso(r'), causing the same singularity, but whose contribution can be computed separately. As such a function we might take the constant planar surface current flowing on the plane A tangential to the surface S at the field point r, with the current of the field point Jso(r') == Js(r), constant in r'. Thus, we can write (5.113)

148

FP

CHAPTER 5. ELECTROMAGNETIC FIELD SOLUTIONS

J

G(r - r/) . Js(r/)dS' =

J

G(r - r/)' [Js(r/) - Js(r)]dS'. (5.114)

S+A

S

The last integral means an integration on the surface S of the source Js(r') plus another integration on the tangent plane A at r' = r of the constant source -Js(r). With this as an interpretation we can define the Green dyadic in the form

= , G(r - r ) =

=

1

=

FPG(r - r") + 2jk 18(r - r").

(5.115)

Of course, it is numerically inconvenient to integrate along two surfaces. This is why the Green dyadic is normally discarded in surface integral equations where the unknown is the surface current function. Integral equations for surface sources will be discussed in more detail in Chapter 6.

References J. VAN (1961). Some remarks on Green's dyadic for infinite space. IRE Transactions on Antennas and Propagation, 9, (6), 563-6. BLADEL, J. VAN (1991). Singular electromagnetic fields and sources, pp. 59-112. Clarendon Press, Oxford. DAVIES, K.T.R. and DAVIES, R.W. (1989). Evaluation of a class of integrals occurring in mathematical physics via a higher order generalization of the principal value. Canadian Journal of Physics, 67, 759-65. GRADSHTEYN, I.S. and RYZHIK, I.M. (1980). Tables of integrals, series and products. Academic Press,. New York. LEE, S.W., BOERSMA, J., LAW, C.L. and DESCHAMPS, G.A. (1980). Singularity in Green's function and its numerical evaluation. IEEE Transactions on Antennas and Propagation, 28, (3), 311-7. MULLER, C. (1969). Foundations of the mathematical theory of electromagnetic waves. Springer, Berlin. STRATTON, J.A. (1941). Electromagnetic theory, p. 213. McGraw-Hill, New York. YAGHJIAN, A.D. (1980). Electric dyadic Green's functions in the source region. Proceedings of the IEEE, 68, (2), 248-63. YAGHJIAN, A.D. (1985). Maxwellian and cavity electromagnetic fields within continuous sources. American Journal of Physics, 53, (9), 859-63. BLADEL,

5.5

Complex source point Green function

Complex space source points have been adopted to electromagnetic theory because, as first shown by DESCHAMPS in 1971, Gaussian beams can be

COMPLEX SOURCE POINT GREEN FUNCTION

5.5.

149

simply approximated through a field arising from a point source in complex space. In fact, if the source point vector r' of the distance function D in (5.30) is made complex, the Green function (5.29) continues to satisfy the Helmholtz equation (5.28), provided the delta function of complex argument is correctly interpreted. Let us study the condition D = 0 in some detail. 5.5.1

Complex distance function

The condition D == v(r - r') . (r - r') = 0 is not only satisfied at the complex point r = r', but there are also points r in real space where the distance function vanishes. Denoting r' = a + jb they can be found from the equation D 2 = (r - a) . (r - a) - b . b - 2j(r - a) . b = o. In fact, equating real and imaginary parts we have the two conditions

[r - a] = [b],

(r - a)· b =

o.

(5.116)

The first of these represents a sphere centred at a with radius [b], and the second one, a plane orthogonal to the vector b going through the point r = a. Thus, the points r satisfying D = 0 lie on a circle with radius [b], centred at a and perpendicular to b. If b --+ 0, the circle obviously shrinks to the point r = a. So we see that, instead of a point of singularity, the complex-source-point Green function actually has a circle of singularity in real space. It can also be called the branch circle of the distance function D, be-

cause, if the value of the function is traced along a curve which passes through the circle of singularity, it does not return to the original value but differs from that by the sign. This is clearly seen if we consider points close to the branch circle by taking b = uzb and r == a

If q

« Ibl,

+ u , b + q,

q == q(U x cos 1/J

+ u, sin 1/J).

(5.117)

we have the approximation (5.118)

Thus, when "p is changed by 21r, the distance function D does not return to its original value but changes its sign. The branch circle corresponds to the branch point of a function of one variable in the complex plane. This function can be made unique by introducing branch cut Iines and Riemann surfaces. Similarly, the distance function D and any functions of D can be made unique by a branch cut surface, which can be any surface with a rim along the branch circle. The

150

CHAPTER 5. ELECTROMAGNETIC FIELD SOLUTIONS

branch cut surface can also extend to infinity, In analogy with Riemann sheets we can also think of having two real spaces glued together along the branch circle, each with a unique value for the distance function D. Moving the point through the branch cut surface moves r between the two spaces. The idea of two real spaces to interpret two branches of a function is not new, actually it was applied in a classical diffraction paper by SOMMERFELD (1896). Following this idea, the point source in real space can be interpreted as actually being an infinitesimally small circular source which separates two real spaces, one with positive and the other one with negative distances D.

5.5.2

Point source in complex space

A point source in complex space may also be a source of confusion, because the delta function for complex arguments is not a very familiar concept. Intuitionally, we could imagine a function which is zero at all other points, real or complex, except 'the point' of the delta function. However, with this definition, we lose the Cauchy integral theorem that any path of integration ran be changed without changing the value of the integral provided the function is analytic. Of course, the delta function is not analytic but can be obtained as a limit of a sequence of analytic functions. The Cauchy property is very important to maintain, because we often want to change the integration path to obtain the best convergence. Therefore it is important to define the delta function as a limit of an analytic function. It is known that, for a real argument, the delta function can be defined as a limit of an analytic function in infinitely many ways. Perhaps the most popular analytic function is the Gaussian pulse function (5.119)

for which the limit can be written as 8T ( x ) --+ 8(x), as 7 --+ 00. For complex x = X r e + jXinl we can write (5.119) in the form (5.120) and in three dimensions as a product of three similar functions, or more generally as I

bT(r - r ) =

/2 e (7)3 -;

-TD

2

,

D

= J (r -

r') . (r - r').

(5.121 )

Studying (5.119) we can see what kind of function the delta sequence approaches as T ~ 00. The two lines X r e = ±XiIn define four sectors in

5.5. COMPLEX SOURCE POINT GREEN FUNCTION

151

which the function (5.120) has a different behaviour. In regions Ixrel > IXilnl, the function approaches zero except at the boundary lines. On the other hand, in the other two sectors IXreI < IXinll the amplitude and rate of oscillation of the function (5.120) grow without limit. For example, along the line Xinl = a the real part of the delta sequence function is

~{Or(X)} =

.[fe-rx?"era2 cos(2rax r e ) .

(5.122)

The amplitude is bounded by the Gaussian function e-rx~(' and the rate of oscillation grows without limit as r a --+ 00. In Figure 5.1 we see an example of the delta function 8r (z - ja) for argument values on the real aXIS Z = x. Xloa

8

If \\ f\ f\

A

i

i

~ i

I I

I

i \ I

J

! \ "A-

i

\V

I: i\\/ Iii '\

-2

,

r

-4

-6

~2

A\

-1.5

-I

-O.S

O.S

I.S

Fig. 5.1 Approximation for the complex delta function at

a = 2 for 7' = 5. The envelope gives the absolute value and dashed line the real part of 0.,. (x - ja).

The analytic extension of the delta function is essential in extending integration of functions to the complex plane. Thus the integral

J X2

f(x)o(x - xo)dx = f(x o),

(5.123)

Xl

valid for real-valued parameters Xl < X o < X2, can be extended to integration from Xl to X2 along a path on the complex X plane. Also, if X o is complex, the delta function for real x has a similar property. Finally, a point source in complex space at r = a + jb can be pictured as a spherical source in real space, centred at r = a and with the radius [b], In fact, the limit of exp( -r(r 2 - b2 ) ) exp(2jrb . r) is zero for r > [b] and of infinite amplitude and rate of oscillation for r < [b], 5.5.3

Green function

Because the distance function D for complex source point r' = a + jb vanishes on the branch circle, the Green function G(r - r') (5.29) has a

152

CHAPTER 5. ELECTROMAGNETIC FIELD SOLUTIONS

singularity at the branch circle in real space. Also, the function is two valued unless a branch cut surface is defined. This can be made with respect to the condition at infinity. Dipole mode branch cut Requiring the Sommerfeld radiation condition Ur ·

\lG(r - r')

+ jkG(r - r') = o(l/r)

(5.124)

to be satisfied, the branch is defined by the condition ~{D} ~ 0, corresponding to an outgoing wave solution radiating towards infinity. Conversely, choosing the condition ~{D} ~ 0, we obtain the incoming wave solution radiating from infinity. The equality sign in both definitions corresponds to the branch cut surface separating these two solutions: ~{D} = 0, or the equivalent, b· (r - a)

= 0,

Ir-al ~ Ibl·

(5.125)

These conditions define the circular disk inside the branch circle. We can actually picture the energy coming through the branch cut disk to the space with the outgoing field branch. Where does it come from? From the other space with the incoming field branch! Thus, the energy from the complex space point source circulates between these two branch spaces by passing through the branch cut disk and reflecting from the infinity. This is also valid for a point source in real space, where the energy obviously also passes from one (incoming field) space to the other (outgoing field) space through the source point. This choice of branch cut (5.125) defines the dipole mode of the complex space point source. Beam mode branch cut

Another possibility is the beam mode in which the field comes from the left of the branch circle, passes through the circle and radiates to the right. Now the radiation condition above is only valid on the right-hand side, whereas on the left-hand side we must have the incoming wave condition. The corresponding condition for the distance function branch can be shown to be ~{D} ~ 0, and the limiting case ~{D} = defines the branch cut surface: (5.126) b· (r - a) = 0, [r - a] ~ [b],

°

which is the complement of the disk in the plane of the branch circle, Le. the plane with a circular hole.

5.5.

COMPLEX SOURCE POINT GREEN FUNCTION

153

For large values of [b], the field defined by ~{D} ~ 0 can be shown to satisfy the Gaussian beam condition, because assuming b = uzb we can write for paraxial points z » Ipi the approximation

D

= V(z -

jb)2

+ p2 ~ Z -

2

jb

+ 2(Z:+ b2) (z + jb).

(5.127)

The Green function can be written in this approximation as (5.128)

In the last exponential factor it is seen that the function decreases as the Gaussian function of p only if b < o. Also, it is seen that the surface of constant phase is not a plane because of the term quadratic in p. The surface of constant phase can be approximated by a spheroid sufficiently far from the origin and close to the z axis, in which case D can be approximated by z in the denominator of the Green function. In fact, requiring that the real part of the distance function D = p2 + (z - jb)2 be a constant c, we arrive at the equation of a spheroidal phase surface:

V

Z2

p2

2+~b2 =1. C c +

(5.129)

This can be approximated by a sphere whose radius is a function of z by fitting the two surfaces at z = c. The radius R of the sphere can be written as 2 c2 + b2 z2 + b R';::1,---=--(5.130) C

Z

The surface is a plane both at the origin and at infinity and has the minimum radius at z = ±b.

References DESCHAMPS, G.A. (1971). The Gaussian beam as a bundle of complex rays. Electronics Letters, 7, (23), 684-5. FELSEN, L. (1976). Complex source point solutions of the field equations and their relation to the propagation and scattering of Gaussian beams. In Symposia Mathematica, Istituto Nazionale Alta Matematica, XVIII, Academic Press, London. LINDELL, I.V. and ALANEN, E. (1984). Exact image theory for the Sommerfeld half-space problem, Part I: Vertical magnetic dipole. IEEE Transactions on Antennas and Propagation, 32, (2), 126-33.

154

CHAPTER 5. ELECTROMAGNETIC FIELD SOLUTIONS

LINDELL, I.V. and NIKOSKINEN, K.I. (1990). Time-domain Green function corresponding to a time-harmonic point source in complex space. Electromagnetics, 10, 313-25. SOMMERFELD, A. (1896). Mathematische Theorie der Diffraction. Mathematische Annalen, 47, 317-74.

5.6

Plane waves

Plane waves are simple exponential function solutions of the Maxwell equations, which also serve as basic building bocks for Fourier-transformed problems. The plane wave field is known as soon as one of the four field vector functions is known, for example, the electric field vector E(r), which depends on two constant vectors Eo, k:

E( r ) -- E oe-jk.r .

(5.131)

Substituting this in the Maxwell equations corresponding to a source-free region and in the medium relations shows us that the other three field vectors must have a similar exponential dependence on r:

H( r ) -- H oe -jk.r ,

(5.132)

D( r ) -- D oe -jk.r ,

(5.133)

B( r ) -- B oe-jk.r .

(5.134)

All constant vectors are connected through the field and medium equations.

5.6.1

Dispersion equations

Let us consider conditions for the constant vectors corresponding to the general bianisotropic medium. From the medium equations we have (5.135) (5.136) Substituting these in the Maxwell equations gives us two vector relations for the three constant vectors k, Eo and H o: (5.137) (5.138)

5.6.

PLANE WAVES

155

Eliminating Eo or H, results in the following relations:

H;

= = x I - () . Eo,

k = =-1 J-l • (w

(5.139)

(5.140) Thus, from the knowledge of the vectors k and Eo of a plane wave field (5.131), all the other field vectors can be determined. Substituting (5.139) and (5.140) in (5.137) and (5.138), respectively, we have the following two equations for the two vectors k and Eo or k and

Ho:

= D

k

e( - ) ·

w

= k Dm ( -

Eo == 0,

w

) ·

H, = 0,

(5.141)

o.; w~) = (~w x I + ~) . /1-1 . (~w x I - () + I,

(5.142)

Dm(~) = (~ x I - () · €-1 · (~ x I +~) + /1. w w w

(5.143)

The dyadics De, D m are called dispersion dyadics and (5.141) can be recognized as eigenvalue equations, although there is no obvious eigenvalue parameter. The dispersion dyadics are related to-.!.he Helmholtz operators H e(\7), H m(\7), defined in (3.58) and (3.59), by De(k/w) = H e(-jk)/w 2 ,

= =. 2 Dm{k/w) = Hm{-Jk)/w .

For the solution of (5.141) the dispersion dyadics must be be planar. This results in two conditions for the k vector called dispersion equations:

=

k detD e ( - ) = 0,

(5.144)

= k detD m ( - )

(5.145)

w

w

= 0.

It can be shown that (5.144) and (5.145) are in fact the same equation. This is seen from the expansion of the following determinant function:

det(A . B

+ I)

= det

A detB + (A . B)(2) : I + (A . B) : I + 1 ==

detA detB

+ :4(2)

: B(2)T

+ A : B T + 1,

which is obviously invariant to the interchange of the two dyadics B. Choosing

= - 1 k B = Ji- . (w

= = x I - (),

(5.146)

A and

(5.147)

CHAPTER 5. ELECTROMAGNETIC FIELD SOLUTIONS

156

we see that the determinant functions in the equati~ns (5.144), (5.145) can be written as multiples of the form (5.146) with A and B interchanged, whence (5.144) and (5.145) are the same equation. € and Ii have been tacitly assumed to be complete dyadics. The dispersion equations (5.144) and (5.145) are only fourth-order algebraic equations .!.or t~e wave vector k, because hi~her order terms are of the form. (k x I) : S with a symmetric dyadic S and, consequently, vanish. The dispersion equation can be visualized in k space by a set of wave vector surfaces. For the general bianisotropic medium there are four such surfaces, which may have complex values. In special cases, some of the surfaces may coincide. In fact, for anisotropic and bi-isotropic media the number of wave vector surfaces is two and for isotropic media just one. The dispersion equation in the general hianisotropic case is not very simple in appearance if the coefficients are expanded. In special cases, it can he essentially simplified.

5.6.2

Isotropic medium

Let us consider the simplest special case. The dispersion dyadics for an isotropic medium can be written as D

-

1

-

== I - --kk~I, 2 W

/L f

(5.148)

and the dispersion equation simply reads

k· k ==

W

2

(5.149)

Jlf .

From the form of the dispersion dyadics (5.148) with (5.149) substituted

= = 1 = kk D == I - (k . kI - kk) == , 2J1€ W2~€

(5.150)

W

it is seen that any vector Eo satisfying the condition k . Eo == 0 will satisfy (5.141) and be a possible electric field vector of a plane wave. In this case, the wave vector surfaces reduce to one single sphere.

5.6.3

Bi-isotropic medium

The hi-isotropic medium case is just slightly more complicated to handle but more general because it has two dispersion surfaces. The dispersion dyadic in (5.141) can be given the factorized form

De(~) = fl + .!:.[~I + ~ x I]. [(I - ~ x I] = _l_ D + . D_, 2 W

/L

W

W

W

/L

(5.151)

5.6. PLANE WAVES

157

with (5.152)

Thus, the dispersion equation is split into two equations (5.153) and the wave-vector surfaces consist of spheres with radii k = k.; and k = k.: The electric field polarizations corresponding to the two eigenwaves are obtained from the dispersion dyadic (5.151) by substituting k = uk+ or k = ue., with the unit vector u denoting the direction of propagation. Because the dyadics D_ and D+ commute, the two eigenpolarizations satisfy D+ E o + = 0 and D _ . E o - = 0, or the equivalent,

u x E o - = -jE o _

u x E o + = jE o+,

.

(5.154)

The fields are circularly polarized in the plane perpendicular to u, right hand for E o + and left hand for E o - , because u . E o ± = 0 and we can write for their polarization vectors

p(E o±)

= ~o±

x E:± JE o ± . E o ±

= E o± ~

(=fiu x* E o±)* JE o ± . E o ±

= ±u.

(5.155)

Thus, plane waves propagating in a bi-isotropic medium are either right-hand circularly polarized with the wave vector k., = uk+ = uk o ( Jn 2 - X2 + K) or left-hand circularly polarized with the wave vector k., = nk: = uk o ( Jn 2 - X2 - K). A linearly polarized field can be split into these two components and because their phase velocities are different, their sum is a linearly polarized vector whose direction is turned much like that of a wave propagating in an anisotropic medium. It is easily seen that the angle of rotation is proportional to k.; - k : = 2k o "" and the Tellegen parameter has no effect.

5.6.4

Anisotropic medium

For the anisotropic medium, the dispersion equation reads = kkx=-l = k detD e ( - ) = d e t ( f - 2 x J.L )=0, w w

(5.156)

which is biquadratic in k/w:

-

kk -

kk

w

w

(Ii : 2

)(l : 2) -

-(2)Tx-(2)

Ii

kk

xl: 2

w

+ detf-detli =

O.

(5.157)

CHAPTER 5. ELECTROMAGNETIC FIELD SOLUTIONS

158

Writing k = uk with u a u\lit vector, the biquadratic equation can be solved for k == k(u) in an analytic form:

(~r = ~jI-1~f-1T : uu± ~V(f1-1~f-1T : uu)2

-

(f1-1~f1-1 : UU)(f-1~f-1

: uu).

(5.158)

There are two pairs of solutions corresponding to two waves propagating with the same velocity in opposite directions in each pair. This is seen from changing the sign of the direction of propagation, u, which does not change the solutions. Thus, the wave vectors of an anisotropic medium form two surfaces symmetric with respect to the origin. Unlike the general bianisotropic medium, the anisotropic medium is reciprocal for plane waves in that they propagate with the same wave number in opposite directions. As a special ~ase we consider the dielectrically anisotropic medium with f == €ofr, Ii = J.LoI, for which the solution (5.158) can be written by defining n as the refraction factor for the plane wave, n == klk o , with It = I - uu,

n2 =

~etfr

2f r

:

uu

(fr -1 : It =f V(f;l : It)2 - 4(fr : uu/detEr)).

(5.159)

The polarization of the field vector Eo can be determined through (5.141) when the k vector is known. If the dyadic De(klw) is strictly planar, there is a unique polarization for Eo. Applying the dyadic identity (A~A) x a = 2(A. a) x A, we see that D e (2) x Eo = 0, whence there exists a vector a such that De (2) = aE o , from which the polarization of the eigenvector Eo can be determined by dot multiplication from the left. This breaks down, however, if the dyadic De is linear, because then we have a = o. In this case, there are two linearly independent polarizations corresponding to the same wave vector k and any linear combination of these will do. For this direction of k the wave vector surfaces coincide and the direction is called an optical axis of the medium. Such an axis may also be in a complex direction corresponding to a complex unit vector u. Wave vectors k with complex direction vectors u correspond to inhomogeneous plane waves, whose planes of constant amplitude and phase are non- parallel.

Uniaxial medium Let us consider plane wave propagation in a uniaxial anisotropic medium with the symmetric uniaxial medium dyadics (5.160)

159

5.6. PLANE WAVES

Here, v is a real unit vector and defines the axis of the medium. It will be seen that it also defines the optical axis, as can be expected since there are no other special directions in the medium. The refraction index n can be solved as a function of the propagation direction u from the expression (5.159). In fact, writing uu det€r =

€r :

=

-1

€r

:

=1

=

t

ft(U

x v)2

+ fv(U · v)2

€~€v

2fv - (f v - ft)(U X

v)2

€t€v

,

(5.161)

we have from (5.159) for the general solution

n=

(5.162)

From this we obtain the two solutions nl

n2

=

= V€i = Jlr : WW,

(5.163)

f.tf. v

€t(u x v)2 + €v(u· v)2

€r :

uu

(5.164)

In the last expressions we have introduced the unit vector w in Uu~vv

ww = ( uxv )2'

(5.165)

which is meaningful except for U = v, in which case the solutions nl and n2 coincide. The solution 1 is independent of the wave direction U and can thus be called the ordinary wave. Its wavenumber surface is a sphere. The other solution 2 depends on the wave direction and can be called the extraordinary wave. Its wavenumber surface is a spheroid if f.t and f.v are positive real numbers, as can be easily demonstrated. The optical axis of the medium is obtained for n1 = n2, which occurs for U = v, in the direction of the axis of the uniaxial medium, as expected. The field polarizations corresponding to the two possible wave vectors k, = uniko, i = 1,2 are obtained by forming the dyadics De(2) corresponding to the two solutions ni. From the equation (5.141) applied to the uniaxial anisotropic medium we have for the two dispersion dyadics

Dei = De(~)

= f o ( -n~I~uu + .i. + fUVV) ,

(5.166)

CHAPTER 5. ELECTROMAGNETIC FIELD SOLUTIONS

160

from which we can write (5.167) implying that E 1 is parallel to w, or u x v, and

= (2) €4€t(€t + €v) 2 D e2 = = (u X v) WW~(€r' U)(€r' u), 0

(5.168)

e : uu

which means that E 2 is parallel to the vector w x (€r . u). The two waves are seen to be orthogonally polarized, because El . E 2 = 0, except for propagation along the optical axis, U = v, in which case the previous expressions fail and any vector orthogonal to u can represent the polarization just like in isotropic media. For propagation transverse to the axis, U . v = 0, the polarization of the wave 2 is seen to lie parallel to the v axis. Gyrotropic medium The expression (5.158) can also be applied to a gyrotropic medium, which is a generalization of the symmetric uniaxial medium. Assuming only dielectric gyrotropy such as in magnetoplasma, we consider the medium dyadics

~ = €o[€t(I - vv) + €vVV + ,V x I],

=p,

= Mol.

(5.169)

,v

The term x I is responsible for the gyrotropic effect. Applying dyadic identities we can evaluate =-1 x=-l €

=-1 x=-IT J..L x € :

x€

:

uu =

2€t(U X v)2 + 2€v(u . v)2 2 (2 2) ,

1 [€t€v(l + (u . v)2) uu = - ( 2 2) J..Lo€o€v €t + "Y

€o€v €t

+

(

U

xv

+, )2]

·

(5.170)

(5.171)

These substituted in (5.158) give us, after some algebra, the following expression for the two possible refraction indices:

(5.172) Here we have denoted S2=(uxv)2,

c=uov,

(5.173)

5.6. PLANE WAVES

161

The famous Appleton-Hartree-Lassen formula for wave propagation in magnetoplasma is straightforwardly obtained if the parameters €t, €v and , are given the expressions corresponding to the physics of the magnetoplasma. As a check, the previous result for the uniaxial medium is readily recovered by setting, = O. The expression (5.172) for n can be applied for any direction of propagation u, although the relation is not very simple. Let us consider two special cases.

Longitudinal propagation 8

For the wave propagating along the axis (u = v), we have, setting = 0, C = 1 in (5.172), for the refraction indices of the two plane waves:

n1 = €t ± it

(5.174)

In comparison with (2.164), it is seen that these are two of the three eigenvalues of the gyrotropic dyadic f r itself. The corresponding polarizations are easily seen from the dispersion dyadic by comparison with (2.165):

D± (~) w

= -€on~vv~I + ~ = €o[:py(I =f jv x I) + (€v ± j-y)vv] = €o[=t=j,(w 1= jw x v)(w ± jw x v)

+ €v vv],

(5.175)

or, better still, by forming the dyadics

D± (2)

= =Fj€~€v,(w ± jv x w)(w ± jv x w).

(5.176)

This time, the vector w is any unit vector satisfying w . v = O. Thus, the electric polarization vectors E± are multiples of the two vectors (5.177)

which are CP vectors orthogonal to the direction of propagation u = v. By forming the corresponding polarization vectors

E± x El p(E±) =.E E* = ±v, J ±.

±

(5.178)

we see that of the two plane waves in longitudinal propagation, E+ has the right-hand and E_, the left-hand polarization.

Transversal propagation

Considering wave propagation transverse to the v axis with u · v = 0, we have, by setting c = 0, 8 2 = 1 in (5.172), for the refraction indices of the two possible plane waves: (5.179)

162

CHAPTER 5. ELECTROMAGNETIC FIELD SOLUTIONS

In this case, the wave 1 is not affected by the gyrotropy of the medium since the parameter , is absent in ni' This wave is called the ordinary wave and the other one (2), the extraordinary wave. The polarizations E 1 , E 2 are obtained from the dispersion dyadics

Eo[(EtU - ,U X

v)u + (,u +

(Et - Ev)U X

v)(u

X

v)],

= Eo [ 2 ( 2 2 x D 2 (k) = - E t U U - , +Et(Et-€v))vv-, UUxVV+Et"Yv Et

W

Eo [(Et U Et

-,u

X V)(Et U

+,u x v) + (,2 + Et(Et

- Ev))VV].

(5.180) x = I]

=

(5.181)

For the ordinary wave 1, we have :;;:(2) D1

2

= Eo ( ,

2

+ Et(Et - Ev))VV,

(5.182)

whence the corresponding polarization is in the direction of (u x v) x u = v. This also explains the notion 'ordinary wave': the wave behaves like in an isotropic medium since the electric field is only affected by the dielectric coefficient along the axis v, which equals Ev Eo. For the extraordinary wave 2 we have from the dispersion dyadic after some algebra (5.183)

from which the polarization of E 2 is seen to be in the direction of Et U x ,U. For, --+ 0, the medium becomes uniaxial and the polarization is along u x v, which coincides with that of the ordinary wave in the uniaxial medium. Optical axes One may finally ask about the optical axes of the gyrotropic medium. To have only one solution for the refraction index, the square-root term must be equal to zero in (5.172), which gives us the equation V -

(5.184) Writing

s2

= 1 - c2 leads us to the four solutions

c = u. v =

± f.v-Y ± J(f.~ + -y2)«f.t ,2 + Et (Et -

fv

)2 + -y2),

€v)

where the two double signs are independent of each other.

(5.185)

5.6.

PLANE WAVES

163

Real axis directions are obtained only if the expression above gives real values for e and satisfies lei ~ 1. For a lossless gyrotropic medium with hermitian l, this equation does not have a real solution, because Et, Ev are real and f is an imaginary number and, thus, c becomes complex. However, if Et and €v are complex, there may exist a real optical axis. This happens, for example, in lossy magnetoplasma like the ionosphere, where the ordinary and extraordinary waves may have the same refraction factor for a certain frequency and direction of propagation. References CHEN,

H.C. (1983). Theory of electromagnetic waves. McGraw-Hill, New

York. KONG, J.A.

(1986). Electromagnetic wave theory, pp. 60-79. Wiley,

New York. LINDELL, I.V. (1973). Coordinate independent dyadic formulation of wave normal and ray surfaces of general anisotropic media. Journal of Mathematical Physics, 14, (1), 65-7.

Chapter 6

Source equivalence From the uniqueness property of electromagnetic fields it is clear that two different fields always require two different sources. However, the reverse is not true, i.e. there may well be different sources which give rise to the same fields within a certain region, V. Such sources J 1 , J 2 will be called equivalent with respect to V and denoted by J 1 J 2. Equivalence of sources with respect to a region of interest is obviously an important property, because if an original complicated source can be shown to be equivalent with some simpler source, the problem can be simplified. The original source can also be a secondary source like a polarization source arising in a polarizable medium. Source equivalence is often applied without recognition. For example, multipole expansions and surface sources based on Huygens' principle are equivalent to certain physical sources, because they give the same field in the space outside of these sources. Let us restrict ourselves here to timeharmonic sources and fields. f"V

6.1

Non-radiating sources

If two sources are equivalent with respect to a region V, their difference obviously produces the null field in V. A non-null source like this is called non-radiating (NR) with respect to V. Fields in regular regions, whose medium parameters are analytic functions of position and whose interfaces and boundary surfaces have continuous tangents, can be shown to be analytic functions of position (MORREY and NIRENBERG 1957). This means that they can be analytically continued within the regular regions and, hence, the null-field actually extends to all of the regular region containing the NR region. 6.1.1

Electric sources in isotropic medium

Let us consider sources and their fields in a homogeneous isotropic infinite medium. Bounded electric NR sources produce the null field outside their support. Any two equivalent electric sources differ from one another

166

CHAPTER 6. SOURCE EQUIVALENCE

through an NR electric source. The most general form of a time-harmonic NR electric source can be shown to be (DEVANEY and WOLF 1973)

(6.1) where H(V') is the Helmholtz operator

(6.2) and f is a vector function whose support defines the complement of the region of null field. J N R defined by (6.1) obviously has the same support as f and thus produces the null field outside this support. Let us study the necessity and sufficiency of this statement. First, the necessity is seen from the form of the Helmholtz equation for the electric field,

(6.3) In the NR case, the electric field vanishes outside a certain region, which is the support of the vector E. Thus, dividing (6.3) by jW/-L, we see that the NR source function must necessarily be of the form (6.1) with f = E/jw/-L. To show the sufficiency, let us substitute the expression (6.1) in the Helmholtz equation (6.3) and write it in the form

H('V) . (E - jWJ-Lf)

= O.

(6.4)

Because f vanishes at infinity, E - jWJ-Lf satisfies the radiation condition of E implying that the solution is unique and since there are no source terms on the right-hand side, we arrive at the null solution

E - jWJ.Lf

= O.

(6.5)

Thus, the electric field, and consequently the magnetic field, vanish outside the support of f, i.e. the region where f i= 0, whence every source of the form (6.1) is an NR source.

Special sources Although the form (6.1) is the most general one for the NR source, it is often helpful to recognize special forms of NR sources. For example, the gradient of any scalar function ¢ of bounded support happens to be an NR source, because we can write (6.6)

6.1. NON-RADIATING SOURCES

167

Thus, irrotational sources do not excite fields beyond the source itself. Actually the function 4> need not be of bounded support, it suffices that its gradient be of bounded support. As an example of such a source we may consider radiation from an exploding atomic bomb in a homogeneous atmosphere or outside atmospheres in space. (This example is actually not one with time-harmonic sources, but serves as a demonstration of the same principle with a more general time dependence.) Such an electron flow corresponds to a radially symmetric current function J(r) = urJ(r) and it can be written as a gradient of another function J(r) = \7 J(r )dr, which is of bounded support if the function J(r) is. Thus the radially symmetric source is obviously non-radiating. The EMP (electromagnetic pulse) arising from a nuclear detonation in the air is due to the inhomogeneity of the atmosphere, which breaks the radial symmetry. Thus, non-radiating sources are not just mathematical oddities but can really exist in the physical world (to a certain accuracy, at least). Another special form of NR source is obtained from (6.1) by expanding the double curl operation and deleting the non-radiating gradient term:

J;)

(6.7) As a consequence we may note that the two sources (6.8)

are equivalent: J 1 J 2, because they radiate the same fields and their difference does not radiate beyond the sources. The possibility of NR sources poses an awkward problem to remote sensing of electromagnetic sources: it is theoretically impossible to determine the source by measuring the radiation in such a part of the space which does not include all the sources. Of course, knowing the field E(r) in all space, its source can be obtained uniquely by substituting the field in the left-hand side of the Helmholtz equation, whence the right-hand side will give the source function. Because any NR source with the same support can be added to the original source without changing the measurable field outside the sources, the problem of source determination becomes inherently non-unique. Thus, additional information about the source is necessary for its identification. Alternatively, we may look for the simplest possible source, for example, the one which has the minimum source energy. To find this, we have to require that a certain energy norm functional of the source function be rninimized. f'..J

168

6.1.2

CHAPTER 6. SOURCE EQUIVALENCE

Sources in bianisotropic media

Combined electric and magnetic sources Above we considered NR electric sources J N R in isotropic media. A similar expression for the most general magnetic NR source J ~ R can be wri tten from duality. More generally, a combination of electric and magnetic sources (J,Jm)NR can be non-radiating. The form of the most general electromagnetic NR source for the general bianisotropic medium can be obtained by writing the Maxwell equations in the form

-jW€ ( -\7 x jW(

(6.9)

I-

whence in analogy to the electric NR source we can write

(

J~

)

NR

= L(V) . (

:: )

=

-jW€.~+(v=xI-jW~Lfm),

(6.10)

( -(\7 x I + jw() . f, - jwJi· fm

where f e and f m are two vector functions of bounded support V. The proof goes as in the electric case: the necessity is obtained from (6.9) because if E and H are null outside V, the right side gives sources which thus must be of the form (6.10) with fe, f m vanishing outside V. For sufficiency, (6.10) substituted in (6.9) results in (6.11 ) whence from uniqueness we have E 1'.

= f,

and H

= fm, which vanish outside

Electric and magnetic sources The most general expression for the electric NR source in a bianisotropic medium is obtained from the previous (6.10) by setting the magnetic source equal to zero:

(

JNR) =£(v). ( f e

a

fm

)

= (

-jW€.~+(V=xI-jw~~.fm

-(\7 x I

+ jw() . f,

),

- jwJi· f m

(6.12)

6.1. NON-RADIATING SOURCES

169

from which, solving for the function f m in the form

fm =

1 --1 = ---;-Ji . (\7 x = I + jw() . fe, JW

(6.13)

and substituting this in one of the equations of (6.12), the resulting expression for the most general electric NR source in a bianisotropic medium is obtained: JNR = [-jw'£ -

~(V x I - jwe) · p,-1 . (V x I + jw()] . f e •

JW

(6.14)

The corresponding expression for the most general magnetic NR source is obtained correspondingly by starting from J = 0, J;;'R = [jwp, + ~(V x JW

I + jw() · ,£-1 . (V x I -

Setting € = EI, =P = J,tI and sources in isotropic spaces,

jw{)] . fm'

(6.15)

e

= ( = 0 we immediately obtain the NR

(6.16) J;;'R

= -jwllfm

-

~V

JWE

x (V x fm),

(6.17)

of which (6.16) can be recognized as equivalent to the previous expression of DEVANEY and WOLF (1973), with the substitution f e = jwJ,tf. References BLEISTEIN, N. and COHEN, J.K. (1977). Nonuniqueness in the inverse source problem in acoustics and electromagnetics. Journal of Mathematical Physics, 18, (2), 194-201. DEVANEY, A.J. and WOLF, E. (1973). Radiating and nonradiating classical current distributions and the fields they generate. Physical Review D, 8, (4), 1044-7. LINDELL, I.V. (1988). TE/TM decomposition of electromagnetic sources. IEEE Transactions on Antennas and Propagation, 36, (10), 1382-

8. MORREY, C.P. JR. and NIRENBERG, L. (1957). On the analyticity of the solutions of linear elliptic systems of partial differential equations. Communications on Pure and Applied Mathematics, 10, (2), 271-90. PORTER, R.P. and DEVANEY, A.J. (1982). Holography and the inverse source problem. Journal of the Optical Society of America, 72, (3), 327-30.

170

6.2

CHAPTER 6. SOURCE EQUIVALENCE

Equivalent electric and magnetic sources

In the previous section we obtained the most general expression for the NR combination of electric and magnetic sources, (J,Jm)NR. Thus, we may state that since the field due to J is cancelled by that of J m in the NR pair, outside their supports, the sources J and -J m corresponding to the same NR pair are equivalent: (J,O) rv (O,-J m ) . This shows us that there exists a magnetic source equivalent to a given electric source and conversely. Actually there exists an infinity of equivalent sources because we can always add an NR source to a given equivalent source.

6.2.1

Bianisotropic medium

Starting from the electric source J and setting f m = 0 in (6.10), we may solve the magnetic source whose negative is equivalent to the electric source:

= . f, J::! == (V' x =I + jw()

1 = = --1 . J. = -~(V' x I + jw() . f

(6.18)

JW

Correspondingly, starting from a magnetic source J m and setting f e = 0 in (6.10), we may solve for the equivalent electric source: Jeq

=

=

== -(V' x I -

jw~)

1 = ~(V' x I JW

. f m ==

= . --1 Jl . J

jw~)

(6.19)

m.

To see the equivalence, let us denote the fields due to J f:. 0, J m == 0 by

E e, He:

(

- jW€ - \7

X

I - jW(

V' x I-

-

jw~

- jw"ji

)

s,

J

. ( He )

=(

(6.20)

0 ) .

Denoting the fields due to J = 0, J m == J::!, f:. 0 by Em, H m and substituting the magnetic source J~ from (6.18) in (6.9), we can write this in the form

. ( - \7 x-t:w~ I - JW(

=

~ -!W~

\7 x - JWJ.l

) .( Em -

j~~-l • J

Hm

) == ( J ) , 0

(6.21) similar to those for the electric source J, whence from uniqueness we have the following relation between the two sets of fields:

Em -

1

--1

-;-f

JW

.J

== E e ,

(6.22)

Thus, the magnetic fields of the original and equivalent sources are equal not only outside the sources but everywhere, whereas the electric fields

6.2. EQUIVALENT ELECTRIC AND MAGNETIC SOURCES

171

coincide outside the sources o~ly. The converse is true if we start from a magnetic source and replace it by the equivalent electric source (6.19). It is easy to see that there cannot exist complete equivalence of electric and magnetic sources so that E, = Em and He = H m would be satisfied everywhere. Of course, a magnetic source equivalent to an electric NR source obtained in the foregoing manner must be NR. To verify this, we substitute the expression of an NR source (6.14) in (6.18) and write the result as

Jeq = -~(V' x m JW

I + jW() . (-1 . JNR =

1 = = --1 1 = = --1 = = - -:-(\7 x I + jw()' f . [-jWf - -:-(V' x I - jwe)· JL . (V' x I + jw()]· fe = JW JW -

=

1

[jwli + -:-(V' x I JW with fm

= --1 = + jw() . f . (\7 x I -

=

jwe)] . fm,

(6.23)

1 --1 = = = -:-ll . (\7 x I + jw() . fee

(6.24) JW Obviously, (6.23) is of the form (6.15), which means that the equivalent source of an electric NR source is really a magnetic NR source.

6.2.2

Isotropic medium

For the isotropic medium we can write the equivalent sources (6.18), (6.19) simply as Je q = V'.x J m , Jeq = _ V' x J . (6.25) JWJL m jWE. There are many interesting and important consequences from these expressions. Let us itemize a few of them. • If V' x J = 0, the equivalent magnetic source is J'::!, = 0 and, hence, the electromagnetic field vanishes outside the original source. Thus, irrotational sources do not radiate. Such a source can be expressed as the gradient of a scalar funct.ion, which is an example of a class of NR sources, as was seen in the previous section. • If J is a constant vector function within a volume bounded by a surface S, the corresponding J';2 is non-zero only on S. For example, if J is a small axially directed cylindrical source, the equivalent J m is a magnetic surface current flowing around the cylinder perpendicular to the axis. This leads to equivalence of a dipole and a loop, which is considered in more detail below.

172

CHAPTER 6. SOURCE EQUIVALENCE

• Applying the equivalence condition twice we can see that the source J is equivalent to another electric source Jeq = 'l x 'l x J / k2 • For a cylindrical dipole this is a double surface current source. The source J - Jeq is an example of an NR source, because its expression is of the general NR form (6.1). • Because the source \l(\l · J) is irrotational and does not radiate, we conclude that the electric source Jeq = - \72J / k2 is also equivalent with the source J. Dipole and loop equivalence

As an example let us consider the electric current equivalent to the magnetic dipole current function (6.26)

The equivalent electric current is, from (6.25),

Jeq(r)

ImL = -. -\78(r) JWJ.L

x u.

(6.27)

An idea of this function is obtained if the magnetic dipole is considered to be the limit of a cylindrical constant magnetic volume current function parallel to the z coordinate:

(6.28) Here, U(x) is the Heaviside step function and a, 2h = L are the radius and length of the cylinder, respectively, with Jm = I m / 1ra 2 • The corresponding equivalent electric current can be written as

(6.29) which is easily pictured as a surface current flowing on the cylindrical surface p = a perpendicular to the z axis. The total current on the surface is Ieq = JmL = ImL ..!.., (6.30) jWJ.L

A

jWJ.L A

= 1ra being the cross-sectional area of the cylindrical dipole. 2

This expression is also valid for other than circular cross sections of the dipole. If fields are considered at distances large enough and the dipole is small with respect to the wavelength in all dimensions, the cylindrical

6.2. EQUIVALENT ELECTRIC AND MAGNETIC SOURCES

173

surface current can be concentrated to a loop with the current I. Thus, magnetic dipole and electric current loop are equivalent sources. From duality, we can also write the magnetic loop current equivalent to an electric dipole current I: /eq _ _ ~ m

-

[cae A'

(6.31)

The equivalent source of the constant cylindrical magnetic source can be applied to get an idea how an equivalent source of a general current source can be formed. In fact, any magnetic current function Jm(r) can be split into dipoles small enough that the current density of each of them can be considered constant, whence the equivalent sources for each dipole can be pictured as forming a distribution of surface currents in space which obviously result in a continuous equivalent magnetic volume current source. For example, a line magnetic current source can be replaced by a current tube with surface current flowing transverse to the original magnetic line current. Also, a magnetic surface current can be replaced by a double electric surface current, consisting of two surface currents flowing in opposite directions on each side of the original magnetic surface current in the transverse direction. 6.2.3

Chiral medium

As a final special case let us apply the previous expressions for the chiral medium with the parameter dyadics

(6.32) Defining the self-dual sources as

(6.33) we can write from (6.10) for the most general NR sources the expressions (6.34) with

(6.35) From (6.34) we see that, in the chiral medium, the NR sources J~R are uncoupled. Thus, a combined NR source always consists of two self-dual sources which are individually NR sources.

174

CHAPTER 6. SOURCE EQUIVALENCE

As another implication we may state that a plus source and a minus source cannot be equivalent if they both radiate, because their difference is an NR source only if the plus and minus components are individually NR sources. This means that the sources J and J m are equivalent in a chiral medium only if both their plus and minus components are equivalent. The equivalent source expressions (6.18), (6.19) reduce to the following form for equivalent sources in chiral medium: 1

Jeq = --(\7 m ·

X

JWE

1

-

I-

-

",k0 I) . ' J

-

Jeq = -.-(\7 x I - "'koI) . J m . JWJ1-

(6.36)

(6.37)

These expressions mean that, unlike for the isotropic medium, an irrotational source is not necessarily NR.

References LINDELL, I. V. (1990). Nonradiating and equivalent sources in bianisotropic and chiral media. Helsinki University of Technology, Electromagnetics Laboratory Report 77. MAYES, P.E. (1958). The equivalence of electric and magnetic sources. IRE Transactions on Antennas and Propagation, 6, (4),295-6.

6.3

Multipole sources

The multipole expansion of a bounded electromagnetic source is one example of an equivalent source producing the same field as the original source in the region outside the original source and the multipole. Since the expansion involves an infinite sum, even though it represents the true field analytically, the series may diverge and thus give numerical trouble for points close to the source or the multipole. The greatest advantage of the expansion is obtained for such sources which can effectively be approximated through just a couple of terms in the expansion or for field points sufficiently far from the source. The multipole expansion consists of a series of point sources, which are located at the same point, usually determined by symmetry or some other obvious choice. The location of the multipole can, however, be selected on the basis of obtaining the simplest approximation for the expansion. Through this process, the best location will in general be found to be in complex space.

175

6.3. MULTIPOLE SOURCES

6.3.1

Delta expansions

It is known that slowly varying functions can be approximated effectively through the Taylor series, in terms of power functions. On the other hand, it is possible to express properties of highly localized functions in terms of delta function expansions of the form

f(x) ~ foo(x)

+ fl 0'(X) + f20"(X) + ... + fno(n) + ...

(6.38)

Because this expansion does not mean that values for the function on the left-hand side can be obtained from the series on the right-hand side, the equivalence sign rv has been applied. The significance of the delta expansion lies in the fact that the integral of the product of f(x) and an analytic function g(x) can be expressed as

J 00

f(x)g(x)dx = fog(O) - hg'(O) + hg"(O) + ... + (-It fng(n) (0) + ...,

-00

(6.39) following from the properties of the delta function and its derivatives:

J 00

g(x)t5(n) (x)dx

= (-ltg
n

= 0,1,2, ..,

(6.40)

-00

For more exact definitions in terms of distributions the reader is advised to consult a recent book by VAN BLADEL (1991). The expansion (6.39) resembles the one obtained when the function g(x) is expanded in a Taylor series within the integral:

g(x)

1 1 2 = g(O) + g'(O)x + -g"(0)x + ... + ,g(n)(O)xn + ..., 2 n.

(6.41)

resulting in the following integral expression:

J 00

f(x)g(x)dx = g(O)

-00

... + g(n)(O)

J J~xn

f(x)dx + g'(O)

n.

J

f(x)dx + ..,

xf(x)dx+

(6.42)

Identifying terms in (6.39) and (6.42) gives us coefficients for the delta expansion of f(x), whence (6.38) can be written as (6.43)

CHAPTER 6. SOURCE EQUIVALENCE

176

It can be noted that the two function sets, {8(m)(x)} and {x n } are biorthogonal, because they satisfy

J 00

x n8(m)(x)dx = (-1)mm!8mn ·

(6.44)

-00

Thus, the nth coefficient in the delta series expansion (6.38) can be simply obtained through multiplying by z ", integrating and applying the orthogonality (6.44). The orthogonality in one dimension can be generalized into three dimensions. In fact, the n-adic rr...r and m-adic V'V'... V8(r) can be shown to be orthogonal, for m 1= n, when integrated over the whole threedimensional s~ce. This property is easily understood because, for example, VVr = VI = 0 everywhere and V(rr) = 0 at the origin. Similarly, all expressions with di~erent numbers of V and r vanish at the origin. On the other hand, \7r = I and also all other expressions with the same number of V and r give nonzero results. This property can be applied when expanding a function of r as a delta series. To write the result in coordinate-independent form we adopt the following n-dot product @ between two n-ads (polyads of rank n):

(6.45) Special cases are the dot product of vectors (n = 1) and double-dot product of dyadics (n = 2). With this, we can write

f(r)

= F o8(r ) + F 1 . V8(r) + ... + F n ® '---v--' VV'..\7 8(r) + ...,

(6.46)

n

where the coefficients F n are n-ads. Their expressions can be found from orthogonality, through multiplying the expression by the m-ad rr...r and integrating over the whole space. 6.3.2

Multipole expansion

A multipole expansion of a given source is just a delta series representation of the source function. As can be well understood, the expansion is practical only if the source is localized enough. For calculating the far field, the higher order derivatives of the delta function give rise to more rapidly decreasing fields, whence the series converges more rapidly than for the near field. Because there also exist NR point sources, the multipole series

177

6.3. MULTIPOLE SOURCES

is non-unique. NR multipoles can always be added to make the series look different. Of course, it is of interest to obtain the simplest possible form. Let us write the delta expansion for a concentrated current source function J(r) in the form

J(r)

= 8(r)Po(0) -

L 00

=

1 2.

\78(r)· P 1 (0) + ,\7\78(r) : P2 (0) - ... (-1)n -,-\7\7...\78(r)@Pn (O),

(6.47)

n.

n=O

with n operators \7 in the nth term. The quantities P n are n + l-adics, i.e. polynomials of n + I-ads: Po is a vector, PI a dyadic etc. They represent the moments of the current function:

Pn(O)

=

j r'r'...r'J(r')dV',

(6.48)

v with n vector r' multiplicants. The multipole considered so far is located at the origin. The expansion is changed if the point of the multipole is moved from the origin. Taking r = a as the point of the multipole, we can write another expansion:

L 00

J(r) =

(-l)n -,-\7\7...\78(r - a)@Pn(a),

n=O

n.

(6.49)

with current moments

Pn(a) =

j(r' - a)(r' - a)...(r' - a)J(r')dV'.

(6.50)

v The location of the expansion can be chosen at will. Usually the choice is made in some obvious way, by taking a point of symmetry or the origin of the coordinate system. However, by a proper choice of the multipole location the convergence of the expansion can be improved. In general, such a point a will be in complex space. The dipole term in the current function is 8(r-a)Po(a), with the current moment

Po(a)

=j

J(r')dV'

= jWPe'

(6.51)

v Because the dipole moment vector Pe does not depend on a, the vector moment Po(a) is actually independent of the location of the multipole. For

CHAPTER 6. SOURCE EQUIVALENCE

178

a current function small in volume and not varying too rapidly, the dipole term is the most dominant. The dyadic moment of the multipole series can be written as

P I(a)

=

j(r' - a)J(r')dV' =

PI (0) - aPo(O) =

v JW=

TQe - Pm

X

=. I - Jwape,

(6.52)

where the vector Pm denotes the magnetic dipole moment and the symmetric dyadic Q e the electric quadrupole moment in common notation. It is seen that the expression (6.52) is changed if the vector a is changed. However, if the electric dipole moment is zero, Plea) is independent of a. More generally, it can be stated that the lowest moment in the multipole series does not depend on the position. 6.3.3

Dipole approximation of a multipole source

Let us consider 'simple' sources, which can be well approximated by a couple of first terms in the multipole series. If the electric dipole term is non-zero, we can optimize the location a of the multipole by defining it so that the dyadic moment Plea) is minimized. This can be made more exact by defining a norm in the space of dyadics, for example,

II All =

VA:A\

(6.53)

which is called the Erhard Schmidt norm in mathematics. The position vector a can now be obtained by minimizing the nonnegative real number

v(a) == [PI(0) - aPo(O)] : [PI(0) - aPo(O)]* == PI (0) : pi(o) - a- pi(o), PocO) - a" . PI (0)· pri(O) + (a· a*)Po(O) . P;(O). (6.54) This can be done by writing a = a, + jai, differentiating yea) with respect to a; and a, and equating the results to zero. The same result is obtained if the vector a * is considered independent of the vector a and differentiating v with respect to a" by keeping a constant. As a result, the following expression for a is obtained:

PI(O) . Pri(O) a == PoCO) . Pri(D) ==

J r'J(r')dV' . J J*(r')dV' J J(r')dV' . J J*(r')dV' .

(6.55)

6.3.

MULTIPOLE SOURCES

179

If the phase of the current function is constant, a is a real vector, otherwise it is complex, in general. To see that the choice of the vector a in (6.55) really decreases the norm of the dyadic moment, we can write after some algebra

= Po(O)Pri(O) ) IIPI (a )11 = PI (0) . ( I - Po(o). Prien)

I

P l (O) . Pri(O) 12 1 - II PI (0)1I 2 IPo(0)12 , I

II = (6.56)

showing us that the norm decreases unless PI (0) . Pri (0) = 0, in which case the square root equals unity and the norm remains the same. If the original current function has constant polarization J(r) = uJ(r) with u . u" = 1, the vector Po(O) is parallel to u and PI (0) is of the form quo Substituting in the above expression we see that both the square root and PI (a) vanish. In this case transfer of the multipole from the origin to the point a is really worth while. Cubic current source As an example, let us consider radiation from a current wave of constant polarization, J(r) = uJe- j k x , non-zero inside and zero outside the cube defined by 0 < x, y, z < L. The polarization of the current is given by the cornplex unit vector u satisfying u- u* = 1. Substituting J(r) in (6.48) we have, after integration, for the first two moments

Po(O)

sin r

= uJL 3 -

,

(6.57)

T

with T = kL/2. The position of the multipole can be obtained from the expression (6.55): (6.58) From the expression (6.52) it is easily seen that PI (a) = 0, or moving the dipole moment (6.57) to the point r = a makes the second multipole term vanish. The optimum point of the multipole is seen to be at an imaginary distance from the orgin. When the side of the cube is small, kL
(6.59)

CHAPTER 6. SOURCE EQUIVALENCE

180

The shift is small when compared with the wavelength. Hence, for small current sources the optimum point of the approximating dipole is almost at the centre of the cube. The advantage for choosing the complex point (6.58) for the multipole follows when the size of the cube is not very small. This is seen by comparing the radiation patterns. Assuming that the current is polarized in the direction of the z axis: u = u z , the exact pattern function can be obtained by integration: F((), ¢) = sin (J (Sin[T(l -.sin lJ cos 1) (sin[T cos 0]) I. r( 1 - SIn () COS ¢) r SIn () SIn ¢ T COS 0 I (6.60) The approximating by a dipole at the origin gives us simply F ((J, ¢)

~

Isin () I,

(6.61)

whereas by putting the dipole at r = a of (6.58) leads to

F(lJ,
I.

(6.62)

0.9 0.8 0.7

0.6 0.5 0.4

::~ 0.1

00

20

40

60

80

Fig. 6.1 Radiation pattern for the cubic current wave in three approximations: (a) exact, (b) dipole at a complex point and (c) dipole at the origin. The parameter T equals kL/2, where L is the side of the cube.

Expanding the pattern expressions (6.60), (6.61) and (6.62) in Taylor series in powers of r = kL/2, it is seen that (6.60) and (6.62) coincide up to terms of the second order,

F( lJ,
~ Isin lJ ( 1 - ~2 (1 -

sin lJ cos ))

I'

(6.63)

whereas (6.61) gives only the first-order term. This difference is clearly seen from the graphical representation of (6.60), (6.61) and (6.62). The

6.3. MULTIPOLE SOURCES

181

increase in directivity of a radiator when shifted in a complex direction can be explained in the same way as the Gaussian beam emerging from a point source in complex space.

6.3.4

Electric and magnetic dipole approximation

Writing the dyadic moment (6.52) in the equivalent form (6.64) we see how moving the multipole changes the magnetic dipole and the electric quadrupole terms. If the source is not well approximated by a single electric dipole, a combination of electric and magnetic dipoles may be tried. This leads to minimization of the norm of the electric quadrupole term, i.e. the symmetric part of PI in a manner similar to the previous one. The equation for a now becomes = * =0, (Qe-ape-Pea)'Pe

(6.65)

and its solution can be written as (6.66) The resulting quadrupole dyadic takes the form (6.67)

and its norm is obviously less than or equal to that of Qe. The magnetic dipole moment is changed to jw

Pm(a) = Pm

-""2 a

jw

X

Pe = Pm

+ 21Pel2 P e X

= Qe Pe 0

o

(6.68)

Thus, it is seen that when moving the multipole from the origin to the point (6.66), the electric dipole is unchanged but the magnetic dipole changes from Pm to Pm(a), while the electric quadrupole is minimized. From duality, we could write similar formulas also for the magnetic current J m , in which case the magnetic dipole is the dominating term and the electric dipole changes when the position of the multipole is moved.

182

CHAPTER 6. SOURCE EQUIVALENCE

References

(1964). Electromagnetic fields, McGraw-Hill, New York. J. VAN (1977). The multipole expansion revisited. Archiv der Elektronik und Ubertragungstechnik, 31, 407-11. BLADEL, J. VAN (1991). Singular electromagnetic fields and sources. Clarendon Press, Oxford. HAFNER, C. (1990). The generalized multipole technique for computational electromagnetics. Artech House, Boston, MA. LINDELL, I.V. (1987). Complex space multipole expansion theory with application to scattering from dielectric bodies. IEEE Transactions on Antennas and Propagation, 35, (6), 683-9. LINDELL, I.V. and NIKOSKINEN, K.I. (1987). Complex space multipole theory for scattering and diffraction problems, Radio Science, 22, (6), 9637. PAPAS, C.H. (1965). Theory of electromagnetic wave propagation. McGraw-Hill, New York. BLADEL,

J.

VAN

BLADEL,

6.4

Huygens' sources

Huygens' sources are surface sources equivalent to physical sources, whether primary or secondary, existing behind the surface. To apply the equivalence, the region behind the surface with Huygens' sources can be replaced by any medium, because there will be no electromagnetic interaction across that surface. Thus, a part of electromagnetic problem can be separated by terminating the fields in the Huygens source. In practice, Huygens' sources are often not known because they depend on the fields at the surface, which are not known in general. However, applying surface integral equations, Huygens' sources can be computed. Also, if sufficient knowledge of the problem exists, they can be approximated and radiation fields computed with reasonable accuracy. For example, radiation from aperture antennas like reflectors, lenses or slots in a metallic plane is usually calculated through a straightforward approximation of the Huygens source. 6.4.1

The truncated problem

Let us consider an electromagnetic differential equation problem in the form (6.69) Lf = {Lv + Lo)f = g, where the linear operator L and the field f and source g quantities can be associated with any version of the electromagnetic problem. For, example,

6.4. HUYGENS'SOURCES

183

the Maxwell equations in bianisotropic media lead to the definitions

Lv

=(

0 = -V' x I

V xI

),

0

E(r) )

f = ( H(r)

,

Lo

. = -JW

(€(=p~) ,

J(r) )

g = ( Jm(r)

.

(6.70)

(6.71)

Let us now truncate the problem by introducing a pulse function Pv(r), where V refers to some region in space of sufficient regularity . Pv(r) = 1, rEV,

Pv(r) = 0, r ~ V.

(6.72)

Multiplying (6.69) by the pulse function Pv, we have (Lf)Pv = gPV, which means that the source has been truncated and those parts of the source which lie outside V have been removed. Because" Lv is a first-order differential operator, we can write L(fPv)

= (Lf)Pv + (LvPv)f = gPv + (LvPv)f.

(6.73)

This is actually a theorem equivalent to Huygens' principle and the last term equals Huygens' source. 6.4.2

Huygens' principle

The equation (6.73) can be interpreted as follows: the source gPv is the truncated source and f Pv the truncated field, equal to the original field inside V and zero outside V. Hence, the term (LvPv)f represents an equivalent source, which radiates inside V the same field as the removed source g - gPv. Outside V it is obviously equivalent to the source -gPv , because it cancels the field due to the source gPv. The source (LvPv)f appears only at the boundary surface S of the volume V. Let us be more specific with the operator. Taking the definitions (6.70), (6.71), we can write the Huygens source term in bianisotropic medium from (6.73) as

=(

0 = VPv x - \l Pv x I 0

I ) .(

=(

n x Hbs ), -n x Ebs (6.74) where n is the unit normal on S pointing into the volume V. Here we denote by bs(r) = n . V'Pv(r) the surface delta function. Writing (6.73) in explicit form in this case results in the equation (LvPv)f

(

-jWE - V x

I - jw(

-)

- jw~ V' x 1- jwfi

E ) H

EPv . ( HPv ) =

184

CHAPTER 6. SOURCE EQUIVALENCE

( J~;v ) + ( ~nXxH;;s ).

(6.75)

The solution can be written in terms of Green dyadics, functions of rand

r', corresponding to the homogeneous bianisotropic medium, even if we do

not know their analytic expressions. V + denotes a volume containing V and 5,

Gem ). ( JPv

G mm

JmPV

+ -

<: n

X

H6s ) dV '. (6.76) E8 s

In the isotropic medium we~now the expressions of the scalar and dyadic Green functions G(r - r"), G(r - r"), whence we can write

EPv ) ( HPv

J(

v+

=

xl) ( JmPv JPv+ n ' x H8 s ) dV'. - n Eb s

-jWJLG _"'J'G x I

"'J'G -jWf.G

' x

.

(6.77)

Let us introduce, for convenience, the following shorthand notation:

£[JIJml v = -jWfJ.

J

J(r')· G(r - r')dV'-

v

J

Jm(r') · (V' x G(r - r'))dV',

(6.78)

v

where V is any volume. The function £[JIJmJv gives a vector function of r, which equals the electric field produced by those parts of the electric current function J(r) and the magnetic current function Jm(r) which lie inside the volume V. For surface sources, the corresponding notation is

£[JsIJmsls = -jWJL

J

I

=

I

I

Js(r)· G(r - r )dS-

s

J

Jms(r') . (V' x G(r - r'))dS',

(6.79)

s and the expression gives us the field from those parts of the surface source functions which lie on the surface S.

6.4.

HUYGENS'SOURCES

185

With this notation, Huygens' source problem can be conveniently written as (6.80) E(r)Pv(r) = e[JIJml v + e[n x HI- n x El s . The interpretation is dependent on the location of the field point r. In fact we have three cases. 1. r is inside the volume V. The left-hand side of (6.80) gives us E(r), which arises from those parts of the sources J, J m that are contained in V plus the surface sources J s = n x H, J m s = -n x E on S, the boundary of V. Thus, for points r inside V, the surface sources are equivalent to the parts of volume sources outside V. 2. r is outside the volume V. The left-hand side of (6.80) is now zero.

With respect to the space outside V, the sources inside V together with the surface sources are NR. 3. r is at the surface S. This case will be considered later through limit processes. For a smooth surface at r, the left-hand side of (6.80) gives us E(r)/2. We can picture the field point being in the middle of the surface source, whence only half of it is cancelling the field from the interior sources, thus resulting in half of the original field. This excludes points on S where the tangent is discontinuous (wedge, apex), because the normal vector n is not uniquely defined at those points. Corresponding formulas can be written for the magnetic field from duality: H(r)Pv(r) = H[JIJmlv + 11[n x HI- n x E]s, (6.81) with similar definitions

J l =J

1t[JIJ m lv =

J(r')·V' xG(r-r')dV' -jW€

v

1t[JslJ m s s

Js(r') . V' x G(r - r')dS' -

J

Jm(r')·G(r-r')dV', (6.82)

v jW€

J

s

Jms(r')· G(r - r')dS'.

s (6.83)

6.4.3

Consequences of Huygens' principle

Let us consider some implications of the previous principle.

186

CHAPTER 6. SOURCE EQUIVALENCE

Analyticity of field functions The representations (6.80), (6.81) give field functions E(r), H(r), which are analytic at points r outside sources and the boundary S. To see this, we may note that the Green dyadic is an analytic function when r is not in the domain of integration. Thus, the field functions are continuous and bounded functions together with their derivatives. The field functions cannot take infinite values at such points. They might be infinite, however, at the boundary S, or inside the source region if the source function has some discontinuity. A consequence of the analyticity is that the field functions can be continued analytically along a path whose points are in the region of analyticity. Also, all analytic functions of fields are analytic. For example, if the electric field or one of its components vanishes in some small part of the region of analyticity, it vanishes in all of it. Thus, the field is transverse electric in the whole region of analyticity, not just in a part of it.

Equivalence of sources From (6.80) and (6.81) we see that a Huygens source on the surface S containing the volume V is actually equivalent to the volume sources outside V, with respect to V. If all volume sources are in V, the Huygens source on S is equivalent to no sources and, thus, is NR in V. This means that the electric surface current is then equivalent to the negative of the magnetic surface current with respect to V. With respect to the outside V, however, Huygens' sources are equivalent to the negative of the volume sources in V. Changing the direction of n in Huygens' sources is tantamount to changing the reference volume from V to its outside and the signs of the surface sources. In this case we see that the principle that volume sources are equivalent to Huygens' surface sources is also valid when the volume is the outside a closed surface. If V is bounded by two closed surfaces 8 1 and 8 2 , Huygens' priciple applies when the surface sources are taken on both surfaces 8 1 and 8 2 •

Use of Huygens' principle Huygens' principle is usually applied in one of two ways: to simplify complicated problems through suitable approximations or to derive integral equations for the problem. If we have a complicated problem with boundaries and material interfaces and we are interested in fields radiated outside this system, we may enclose it within a closed surface S and use the approximation method. The fields radiated outside S can be calculated from Huygens' sources depending on the tangential electric and magnetic fields

6.4. HUYGENS' SOURCES

187

on S. Usually these are not known exactly, but in certain cases they can be approximated. For example, when S is at a PEe surface with an aperture of small extent, the tangential electric field is zero everywhere on S except at the aperture. Also, if the surface contains the aperture of a large reflector antenna, we can approximate the surface fields by calculating them through geometrical optics procedures. 6.4.4

Surface integral equations

Huygens' principle also allows one to define integral equations for the unknown surface sources by letting the field point approach the surface S. Because of the singularity of the Green dyadic the approach must be made with care. Let us consider the limiting cases when the field point approaches the surface S containing the electric and magnetic surface currents J 8, J rris , from the two normal directions D1, n2. From the interface conditions (6.84)

(6.85) we see that the tangential magnetic and electric fields are not continuous through the surface unless the corresponding surface source is zero. To build an equation for the surface sources it is not useful to consider limiting values of fields but fields exactly at the surface. It was seen that the Green dyadic is too singular to let the fields be calculated inside a surface source as we wish to do here. Even the principal value method does not work, because it does not exist. The finite part method does the job, but it involves additional integration. To find converging field integrals we can modify definitions of the surface source functions (6.79), (6.83) by breaking the double gradient through partial integration. In fact, we may write for an open surface S, when the field point r is not on S

j G(r - r'). Js(r')dS' s j GJsdS' s

=j

GJsdS' +

:2 jC'V\1G)' s.es

s

s

:2 j (\1'G)(\1' · Js)dS' + :2 j s

=

\1' . tr, \1'G)dS',

(6.86)

s

and the last term can be reduced to a line integral around the boundary curve C of S. For a closed surface, this curve reduces to a point and produces zero if the field point is outside the surface. For a point on the surface the integral is responsible for the singularity. Discarding this term

CHAPTER 6. SOURCE EQUIVALENCE

188

we may give new definitions for the functions (6.79), (6.83), valid for closed surfaces

s:

J +J J J

f[JslJmsl s = -jWf-L

~

JW€

s

J

\1'G(r - r')\1' . Js(r')dS'

S

~

\1'G(r - r') x Jms(r')dS',

(6.87)

S

1i[J sIJmsl s

JWJ-L

G(r - r')Js(r')dS'-

= -jWf

G(r - r')Jms(r')dS'-

s

J

\1'G(r - r')V" . Jms(r')dS' -

s

V"G(r - r') x Js(r')dS'.

(6.88)

s

These field functions are still singular when the field point r is on the surface S, but the singularity is of lower order and can be removed by removing a circular disk of area A6 whose measure 6 is small. The limit b --+ 0 is again called the principal value of the integral. It is not difficult to check that there is the following relation between the PV and limiting values of the field function, lim £[JsIJmsl s r-:s

= PV£[JsIJmsls + ~\7. .I, 2JWE

lim 'H[JsIJmsl s

= PV'H[JslJmsls + 2~ \1. J ms + m 2 JWJ-L

r~s

- m x J ms, 2

x

.r.,

(6.89) (6.90)

when the field point r approaches S in the direction of the unit normal vector m, which may be either n = V'Pv(r) or its opposite. To form the surface integral equations, we take Huygens' expressions of the fields and let the field point approach the surface S. Substituting the limit in terms of the PV integral above, we have 1

2E(r) = £[JIJmJv 1

+ PV£[n x HI- n x E]s,

(6.91)

+ PV?-l[n x Hln x E]s.

(6.92)

2H(r) = 'H[JIJmlv

These equations presume that the tangent surface is continuous at the field point r, so that n = \7 Pv is continuous. Because tangential fields on a closed surface S are enough to determine the fields inside S through Huygens' principle, it is sufficient to make integral equations for the tangential field components. Operating (6.91) and (6.92) by nx we have 1

"2n x E(r)

= n x £[JIJmJv + n x PV£[n x HI- n x E]s,

(6.93)

6.4. HUYGENS'SOURCES

'12 n

x H(r) = n x 1i[JIJ m]v

189

+n

x PV1i[n x HI- n x E]s,

(6.94)

either of which represents a relation between the tangential components of the electric and magnetic field on any closed surface whose inside is a homogeneous isotropic space. It is customary to call (6.93) EFIE, the electric field integral equation, and (6.94), MFIE or the magnetic field integral equation. It can be easily deduced that, although (6.93) and (6.94) are two vector equations for two vector unknowns, the tangential field vectors on S cannot be solved from them uniquely. In fact, these equations depend only on those volume sources that are inside 5, whereas the true solutions certainly must also depend on sources which are outside S. Thus, (6.93) and (6.94) must actually contain the same information, like the equations 2x - y = 0 and 2y - 4x = O. To solve the fields on the surface, something more must be known. For example, if one of the tangential fields is known on on S, either of (6.93), (6.94) can be used to solve the other tangential field and then, applying Huygens' principle (6.80), (6.81) the fields can be calculated at any point in V though integration. For example, if n x E is known on 5, the procedure would be • n x E is known on 5, • n x H can be computed from either EFIE or MFIE on S,

• E and H can be determined in V from (6.80), (6.81). The integral equation is a substitute for formulating the problem in terms of the Helmholtz equation plus boundary conditions. The advantage of the integral equation formalism lies in the more compact domain of the equation: surface instead of volume, which in radiation problems extends to infinity. 6.4.5

Integral equations for bodies with impedance surface

Considering scattering of electromagnetic waves from a PEC obstacle, the surface 5 can be taken to coincide with the surface of the obstacle. To apply the previous procedure, the PEe boundary condition has n x E = 0 on 5, whence the tangential magnetic field can be solved from either the EFIE or the MFIE equation. Assuming the incident fields £[JIJml v = Ei, 1l[JIJ m lv = Hi known, the respective EFIE and MFIE equations read

n x E i = n x [jWJLPV

J

G(n' x H)dS'+

s

CHAPTER 6. SOURCE EQUIVALENCE

190

~pvj(V'G)V'. (n'

JWf

x H)dS'],

(6.95)

(n' x H) x V' GdS'.

(6.96)

s

n x HZ. = '12 n x H - n x PV

j s

The normal unit vector n points to the region where the fields are calculated, i.e. out of the conducting body. The latter (MFIE) equation is also called the Maue integral equation, although, as pointed out by YAGHJIAN and WOODWORTH (1989), MADE (1949) was not the first to derive this equation. The equations can also be written in terms of the surface current J s induced on the scatterer, by substituting n x H = J s :

n x E i = n x [jwj.tPV

j GJsdS' + ~PV j(VG)V . JsdS'],

(6.97)

JWf

s

S

1 s n x HZ. = '2J

-

n x PV

j

J s x V' GdS'.

(6.98)

s

The latter (MFIE) equation is often preferred because there is no need to differentiate the surface current function on S. However, if the obstacle is a thin body, there are numerical difficulties because the cross product within the integral tends to be a small quantity, which might give numerical error. Thus, the EFIE equation is more often applied for conducting wires and sheets. Integral equations for the PMC scatterer can be written through the ~uality transformation. For an ~pedance surface with surface impedance Zs obeying the condition E t = Zs . J s' neither n x E nor n x H vanish on S. Instead, we may consider two-dimensional vector functions

(6.99) whence the boundary condition is B = 0 on S. The functions A and B can be taken as new unknowns, 1=

J s = -n x H = "2Ys . (A - B),

(6.100)

which inserted in the EFIE and MFIE equations together with setting B = 0, give rise to two equivalent equations, of which either one can be applied for solving for the unknown A and, hence, the tangential fields.

6.4. HUYGENS'SOURCES

6.4.6

191

Integral equations for material bodies

For bodies with non-zero inner field, there is no boundary condition on S, but instead, interface conditions giving relations between the inner and outer fields. In fact, because of the continuity of the transverse fields through the interface, we may express Huygens' principle for both the inside and outside the body and obtain a pair of surface integral equations for the two field functions. This, however, can only be made for homogeneous bodies. For inhomogeneous bodies, the volume integral equation method given in Chapter 5 must be applied. Let us denote the medium of the object by 1 and the outside medium by 2, and the Green functions corresponding to these two media are G 1 and G 2 . If all volume sources are in the medium 2, the surface integral equations can be written from (6.93), (6.94) as

where ll2 = -nl points into the medium 2. The notation £2 corresponds to the medium 2 in the Green function. For the medium 1, we may write similar formulas except that there are no volume sources: (6.103)

(6.104) All these expressions assume a continuous tangent plane at the field point on the surface S. From the two pairs of integral equations (6.101)-(6.104) we may choose two, one on each side of S and the resulting two integral equations constitute a true equation pair for the two unknown surface fields. So, the problem of scattering from dielectric bodies is more complicated than that from conducting bodies in this formulation. Extensive work has been done on the question of obtaining uniqueness for the solutions of these surface equations so that NR surface sources interfering with the solution would be controlled. This subject will, however, not be touched here. A number of references is given in the report by YAGHJIAN and WOODWORTH (1989) on the conducting scatterer problem.

CHAPTER 6. SOURCE EQUIVALENCE

192

References

D. and KRESS, R. (1983). Integral equation methods in scattering theory. Wiley, New York. MAUE, A.W. (1949). Zur Formulierung eines Allgemeinen Beugungsproblems durch eine Integralgleichung. Zeitschrift fur Physik, 126, 601-18. POGGIO, A.J. and MILLER, E.K. (1973). Integral equation solutions of three-dimensional scattering problems. In Computer techniques for electromagnetics, (ed. R. Mittra), pp. 159-264. Pergamon, New York. YAGHJIAN, A.D. and WOODWORTH, M. (1989). Derivation and application of dual-surface integral equations for three-dimensional multiwavelength perfect conductors. Rome Air Development Center Report RADC-TR-89-142. COLTON,

6.5

TE/TM decomposition of sources

Electromagnetic problems are inherently vector problems and in this sense more complicated than, for example, the scalar problems encountered in acoustics. In some geometries there are eigenpolarizations which do not couple to each other, which makes it possible to reduce the vector problem into scalar problems. This presumes that, given an incoming field with one of the eigenpolarizations, the resulting field has the same polarization everywhere. Thus, the polarization of the field is known and only the amplitude of the field, a scalar quantity, remains to be solved. As examples of problems for which such eigenpolarizations exist we may mention horizontally layered media, in which case the fields can be split into TE and TM fields with respect to the normal direction n. Also problems with two-dimensional (cylindrical) structures, invariant in the axial direction u, have TE and TM eigenpolarizations with respect to u. Further, problems involving reciprocal uniaxial anisotropic media share the same property with respect to their axes. Thus, it would be of advantage to be able to split the source of the original problem into two parts, one of which radiates only a TE field and the other, only a TM field, with respect to a given direction u in space. Such a splitting is called TE/TM decomposition and will be considered in this section. It is easily seen that an electric current of constant linear polarization J(r) = uJ(r) gives rise to a TM field, as is evident from the field integral

H(r)

=

J

VC(r - r') x uJ(r')dV'.

v

(6.105)

6.5.

TE/TM DECOMPOSITION OF SOURCES

193

From duality we also know that a magnetic current of the same polarization Jm(r) = uJm(r) gives rise to a field which has TE polarization outside of the sources. Replacing this by an equivalent electric current from (6.25), we see that the current of the form u x V g(r) gives rise to a TE field. Thus, if we can find functions f (r ), g(r) and J N R (r) such that our original source can be written in the form

J(r) = uf(r)

+ u x Vg(r) + JNR(r),

(6.106)

where J N R( r) is of the form (6.1) and does not radiate outside of the defined volume, the original source has been given a TE/TM decomposition of J(r). 6.5.1

Decomposition identity

The decomposition for a given source J can be formulated by applying the following identity: 1 1 1 J = (J - k 2 \7 x \7 x J) + k 2 uu . (\7 x \7 x J) - k 2 U x \7(u· \7 x J)+

V[k12 (u . \7)(u . J )] - (u.\7)2 -k- J,

(6.107)

where the argument of J is r. We see that, with the exception of the last term, the terms on the righthand side are either of the form uj, u x Vg or NR type. The operator (u . \7/ k)2 does not destroy this property. To proceed, we have basically three alternatives.

1. Apply (6.107) recursively by substituting the right-hand side for J in the last term over and over again. This leads to a series representation of J, which for a point source corresponds to a point (multipole) decomposition. 2. Move the last term of (6.107) to the left-hand side. This results in an ordinary differential equation for the current, which can be solved. For a point source this corresponds to a decomposition in terms of line sources.

3. Add the term -(V~ + k 2)J/k2 to both sides of (6.107). When combined with the last term, the result is an NR term and can be omitted. Thus, a two-dimensional partial differential equation for J is obtained, which can be solved. For a point source, this corresponds to a decomposition in terms of planar sources.

CHAPTER 6. SOURCE EQUIVALENCE

194

Thus, the identity (6.107) can be split into three equations for the TE, TM and NR components of the original current. The equations can be written in three different forms as given above and these will be considered separately. 6.5.2

Line source decomposition

To decompose a current J (r) in terms of line currents, let us identify from the identity (6.107) the TM, TE and NR components on the right- hand side and write equations for the corresponding current components:

+ k 2]JT M = uu· V' x

[(u· V)2 [(u· V')2

+ k 2]J TE =

-(u

X

V' x J,

V')(u x V) . J.

(6.108) (6.109)

The one corresponding to the NR currents is composed of the rest of the right-hand side in (6.107). The solutions for (6.108) and (6.109) can be written in terms of the one-dimensional Green function satisfying the transmission-line equation (6.110) G (z - z') 1

= _l_e-iklz-z'l. 2jk

(6.111)

Taking u = u,; we can write the solutions

J 00

JTM(r)

rv

-uu·

G 1(z - z')(V' X V' x J)dz',

(6.112)

-00

J 00

JTE(r)

rv

uu~V'V'·

G1(z - z')Jdz'.

(6.113)

-00

Here, under the integral sign, J depends on x, y, z'; and V's in (6.112) also operate on these variables. Obviously, the current components are of the right form (6.106) to be able to produce TM and TE fields. A better form is obtained if the following identities are applied: uu·(V'xV'xJ) = [(u.V)2+k 2]uu·J-(V2+k 2)uu.J+uu·V'V't· J, (6.114) - (uu~VV')· J

= [(u· V)2 + k 2 ]J t -

(\7 2

+ k 2 )J t + V'V't· J

- uu \7V't· J. (6.115)

6.5.

TEITM DECOMPOSlrrION OF SOURCES

195

In these expressions, the subscript t denotes the vector component transversal to u. Ignoring the NR terms in these expressions, the solutions (6.112), (6.113) can be written as

f f 00

J ™ (r)

rvuu·J-u\7

t ·

dJ 1 ( z-z ' ) ' dz,G dz,

(6.116)

-00

00

J TE( r )

Jt

rv

+ u V' t .

dJ G t ( z - z ' ) ' dz' dz.

(6.117)

-ex>

It is seen that the sum of these expressions equals J, whence the NR currents discarded in the course of the analysis have in fact cancelled each other.

Decomposition of a dipole Let us apply the decomposition of a dipole current J(r) = vIL6(r). Because the u directed component produces a TM field, we may only concentrate on a transversal dipole with v . u = O. From the integral expressions (6.112), (6.113) we immediately have the decomposition:

JTM(r)

rv

-uIL[v· V'6(p)]u. V'Gt(z),

JTE(r)

rv

(6.118)

vIL6(r) - JTM(r).

(6.119)

These expressions can be interpreted through a transmission-line analogy. In fact, a two-conductor transmission line with line separation L, parallel to u and in the plane of v, can be characterized in space by the current density function

J(r) = -uLI(z)v· \76(p).

I(z)

(6.120)

I(z) IL

)U

Fig. 6.2 TE component of dipole decomposition in line sources consists of the original dipole plus a transmission-line source.

196

6.5.3

CHAPTER 6. SOURCE EQUIVALENCE

Plane source decomposition

Another possibility for decomposing the current J(r) is in terms of planar currents. This can be done by again identifying from the identity (6.107) the TM, TE and NR components on the right-hand side and writing the following equations for the corresponding current components:

\7;J™

= -uu· \7 x (\7 x J) = \7;u(u· J) - uu \7(\7 t . J),

V';J T E == -(uu~V'V')· J = \7;J - V'(V't . J)

+ uu- \7(V't

. J).

(6.121) (6.122)

The equation for the NR current is not of interest. The solutions for (6.121) and (6.122) can be written in terms of the twodimensional Green function satisfying the stationary plane current equation

V';G 2(p - p') == -8(p - p'), G2(p - p') =

-21n (k lp 21r

(6.123)

p'l).

(6.124)

The solutions are JTM (r) '" uu . / (''\7 x '\7 x J)G 2 (p - p')dS'

=

s

uu· J

+ uu· '\7 /('\7t' J)G 2 (p -

p')dS',

(6.125)

S

JTE(r) '" -u x / '\7(u x '\7 . J)G 2 (p - p')dS'

=

s Jt

- uu

'\7 /('\7t' J)G 2 (p - p')dS',

(6.126)

s Here, under the integral sign, J depends on the variables p', Z, and \7 also operates on the same variables. Again, the current components are of the right form (6.106) to produce TM and TE fields as seen in the expressions (6.121), (6.122). Also, the sum of the final expressions equals J, whence the NR current terms discarded earlier have actually cancelled each other.

Decomposition of a dipole Taking the dipole source J = vIL8(r) with v . u = 0, without losing the generality, we can write from (6.125), (6.126), neglecting the NR terms v·p

JTM(r) == uIL8' (z )(v . VG 2(p)) = uIL8' (z )- 2 21rp

r-..I

6.5. TE/TM DECOMPOSITION OF SOURCES

=

v

IL8(z)[I - 2u pup ]

· -2

21rp

197

1

+ -vIL8(r),

(6.127)

2

JTE(r) = vIL8(r) - JTM(r).

(6.128)

The last expression of the surface current term in (6.127) can be identified as a d.c. surface current flowing on a resistive sheet as excited by a dipole current source at the origin. The flow lines on the plane can be shown to be circles, each starting and ending at the origin. Thus, the TE component of the dipole consists of one-half of the original dipole plus the stationary surface current generated by the original dipole. The TM part consists of the other half of the original dipole plus the negative of the previous surface current. This decomposition in terms of planar currents was given in Fourier components in the classic paper by CLEMMOW in 1963. Point source decornposit ion

6.5.4

The third possibility for decomposing a current J(r) can be made by applying (6.107) recursively, substituting J in the last term over and over again. The resulting series expansions can be written in the following forms: J

TM

J + kU'\?t .L •

= uu· J

(X)

(

u· ")2n+l --:-;;J, v

J

j

TE,

== J t

-

k"u\7t

.

(6.129)

J

n=O

~

(u.

n=O

J

\7)2n+l

c: --:-;;-

(6.130)

J.

These decompositions are limited in the region of the original source J(r), because only differential operations are involved.

Decomposition of a dipole If these expressions are applied to a transversal dipole J == v I L8( r), the following multipole expansions are obtained:

JT M (r) J

TE

(r)

=

U

IL[ . \1~( )][6'(z) _ 6(3)(z) k v U P k k3

IL

6' (z)

+

6(5)(Z) _ k5

6(3) (z)

] ..• ,

6(5)(z)

= vIL8(r)-uT[v.\76(p)][-k--k3+k5- ...].

(6.131 )

(6.132)

It is not evident that the point decomposition will converge in all cases. In fact, comparing with the delta expansion (6.43), we see that the amplitude of the moments of the multipole increases as n!/k n + 1 . However, the method can be useful, as was demonstrated by LINDELL (1988).

198

CHAPTER 6. SOURCE EQUIVALENCE

References CLEMMOW, P.C. (1963). The resolution of a dipole field into transverse electric and magnetic waves. lEE Proceedings, 110, (1), 107-11. HARRINGTON, R.F. (1961). Time-Harmonic electromagnetic fields, pp. 129-32. McGraw-Hill, New York. LINDELL, I.V. (1988). TE/TM decomposition of electromagnetic sources. IEEE Transactions on Antennas and Propagation, 36, (10), 13828. WILTON, D.R., (1980). A TE-TM decomposition of the electromagnetic field due to arbitrary sources radiating in unbounded regions. IEEE Transactions on Antennas and Propagation, 28, (1), 111-4.

Chapter 7

Exact image theory Image sources can be defined as equivalent sources replacing physical structures such as regions of different media or boundaries with impedance conditions. As a classical example we have the mirror image due to a perfectly conducting plane surface. In this case, the image source located at its mirror image position replaces the conducting plane. It was seen in Chapter 4 that by applying the reflection transformation, the correct boundary conditions on the plane are ensured through the introduction of this kind of image source. The image principle has also been applied in electrostatics to problems with charges in front of a dielectric half space. In this case, the image charge is in the mirror image position and its amplitude depends on the dielectric constants of the two media. In this case, the image source is determined through the property that the correct interface conditions for the electric field are satisfied. A similar principle is also valid in magnetostatics. When trying to extend these to time dependent electromagnetic problems, trouble arises because a simple image source at the mirror image location cannot satisfy the correct interface conditions for both the electric and the magnetic field. Thus, the image concept must be generalized in one way or another. In contrast to different existing approximate theories, the present exact image theory (EIT) was constructed in the 1980s.

7.1

General formulation for layered media

In the present section an introduction to the EIT theory will be given in a general form applicable to layered media with plane parallel interfaces. In the following sections, specific geometries are presented as special cases of the general formulation.

7.1.1

Fourier transformations

Let us consider the problem of piecewise homogeneous media, with all interfaces and boundaries parallel to the xy plane, in terms of Fourier transformation in the two dimensions of the plane. The transformation

CHAPTER 7. EXACT IMAGE THEORY

200

does not seem to be necessary, however, and it would be interesting to find a method of image construction without applying the Fourier transformation at all. As indicated in a paper by KELLER in 1981, some problems of this kind can be handled directly without the Fourier transformation, but the shortcut requires a certain theorem. In waiting for such an approach for the general problem, the Fourier method will be applied here to produce the results with some labour. z

z·············?' 0/

- p

Fig. 7.1 Geometry of a multilayered medium with homogeneous layers of isotropic material.

Vector transmission-line equations The fields in the layer n with parameters satisfy the Maxwell equations

En, J-Ln

and sources J m n , I

n

(7.1) \7 x H; =

jWEnE n

+I n.

(7.2)

The two-dimensional Fourier transformation is defined as

fer)

--+

F(z,K) =

J

f(r)eiKPdS p ,

(7.3)

SfJ

with the inverse transformation

F(z,K)

--+

fer) =

(2~)2

J

F(z,K)e-iKPdSK ,

(7.4)

SK

where the integration domains Sp, SK are the p and K planes, respectively. Applying the transformation to (7.1), (7.2), we obtain the ordinary differential equations

(7.5)

7.1. GENERAL FORMULATION FOR LAYERED MEDIA

u, x H~ - jK x H n

-

201

(7.6)

jW€nEn = I n ,

where the prime denotes differentiation with respect to the coordinate z. To help in finding solutions, these equations are written as transmissionline equations for the vector field components transverse to the normal unit vector u;, denoted bye, h, by eliminating the normal components u, . E, U Z • H. The resulting equations are ,

en

+ J.(3n=Zn' ( -U z X

h n) = u ,

X

· Jmn

. + 1Jn k K Jen,

(7.7)

n

- u,

X

h in

· + J'(3n Y n' en = -Jen

1 uz -kTJn

X

K·Jmn,

(7.8)

n

where the electric source has been denoted by I n = jen + uzjen, and the magnetic source by J m n = jmn + uzjmn in components transverse and parallel to U z . Also, we denote 'T/n=

g, V4:

(7.9)

and the dyadics in (7.7), (7.8) have the form

can be identified as the propagation factor !!! the Fourier plane wave component in u, direction. The planar dyadics Zn, Y n can easily be seen to satisfy the relation

fin

=Zn' =Y n

_ -

1 4= 2 2 x k 2 {32 (knIt - knKK - knuzuzxKK)

= =It,

(7.11)

n n

because of the identity (7.12)

Y n is the two-dimensional inverse ofZn and conversely. Thus, we can write = 1 = --1 Yn=2Uzuz~Zn=Zn, TJ n

spm Z; =

=

2

=

=-1

Zn=""nuzuz~Yn=Yn, 1

_

spmr";

= ".,~,

(7.13) (7.14)

in accord with the expression for the inverse of two-dimensional dyadics as discussed in Chapter 2.

CHAPTER 7. EXACT IMAGE THEORY

202

The equations (7.7), (7.8) are obviously vector counterparts of transmission-line equations with the propagation factor f3n in the z direction. A comparison can be made with the scalar transmission-line equations

+ j{3ZcI(z)

= u(z),

(7.15)

I'(z) + j{3YcU(z)

= i(z),

(7.16)

U'(z)

with characteristic impedance Zc = l/Ye and propagation factor {3, as well as distributed generator voltage u and current i quantities. This shows us that in the present case, en can be interpreted as a vector voltage and -U z x h., as a vector current. The ch~acteristic impedance of the vector transmission line is a dyadic quantity Zn. Eliminating the vector u, x h n , an equation for en can be derived from (7.7), (7.8): (7.17) with the distributed source function 8n

2 'f/n K ., . , . (UzX K) Jmn· . (7 . 18) = J.n« (k n =] e>: KK)· ·Jen+-k Jen+uzxJmn+J

kn

n

Likewise, we may derive (7.19) with the distributed magnetic source function 8m n

=

-U z

. x J.,en - J.( u, x K) Len

j (k n2=I t + -kn"1n

-

KK)· 1 K"J · . Jmn + -kmn n"1n

(7.20) Alternatively, these complicated-looking source expressions can be derived directly from the Helmholtz equation expressions, in homogeneous space outside the sources. For the electric field we can write (7.21) In fact, Fourier transforming the transverse component of this equation gives us exactly the vector transmission-line equation (7.17) with the expression (7.18) for the source above. In the same manner, the magnetic source is obtained from the magnetic field expression. The sources 8 and 8 m can be seen to correspond to each other through the duality transformation: Sd

= j'f/Sm.

7.1. GENERAL FORMULATION FOR LAYERED MEDIA

203

Reflection and transmission Solutions of these equations in each region n can be written in terms of two waves propagating in opposite directions: "(7.22) (7.23) For a wave propagating outside the sources in medium n there is a linear relation between the electric and magnetic field amplitudes. In fact, we can write from (7.7), (7.8) for sourceless regions (7.24) (7.25) with Y n, Zn defined in (7.10). The amplitude vectors depend on the sources and the boundary conditions. At the interfaces, there is the continuity condition for the tangential fields. For example, at the boundary between media 1 and 2 at z = d we can write (7.26) (7.27) At the interface, the boundary condition gives rise to a linear relation for the four amplitude vectors at, aI, at, a2" of the form (7.28) The reflection and transmission dyadics are easily obtained for d = 0 from the above relations in the form

(7.29)

= = = = = = -1 T21=It+Rll=2Z2·(Z2+Z1) = = = = = =1 T = It + R = 2Z (Z2 + Zl)12

22

1·

=

=

=-1 Y2 ) ,

=

=

=1

= 2Y l · (Y 1 +

= 2Y 2 · (Y 1 + Y 2)- .

(7.30) (7.31 )

If, instead of z = 0, the interface is at z = d, there are additional phase factors in the coefficient dyadics: (7.32)

CHAPTER 7. EXACT IMAGE THEORY

204

R22 ( d) == R22(O)e-j{J22d,

T12 (d) = T12 (0)e- j (,Bl - ,B2)d , T2 1 (d) = T2 1 (O)e-j(,Bl-,B2)d.

(7.33)

(7.34) (7.35)

The reflection dyadic expression (7.29) can be generalized to the case where, instead of the medium 2, in the half space z < 0 there is a stratified structure of some kind and in the half space z > 0, theJ.1omogeneous medium 1, if only we know the 'loading' admittance dyadic Y L due to the stratified structure, which is defined through the tangential fields at z == 0 by -uzxh=-YL·e, or e==-ZL·(-uzxh), (7.36) ~ith

--1

Z L == Y L

R· aI'

.

In fact, from reasoning similar to that above and

at ==

we can write

from which

R can be solved in the form (7.38)

This expression is the same as (7.29) above if we identify Y L w~h Y 2, ~ut the last expression is only valid if the impedance dyadics Zl and Z L commute. In the more general case (for example, anisotropic loading impedance) the impedance expression reads (7.39)

By substituting the free-space admittance or impedance dyadics (7.10) for Y n,

Zn we have y

_. ~ KK n -

a K2 Pn"ln

~ uzuz~KK

+ k n"ln

K2'

zn -- f3n"ln KK kn"ln uzuz~KK k K2 + f3n K2 .

(7.40)

(7.41)

n

Thus, the eigenvectors of these two-dimensional dyadics are K, and u; x K, which correspond to TM and TE fields, respectively. (To check

this, we may note that, outside of sources, from the divergence condition \7 . E = 0, for a TE field in Fourier space we have K . e == 0, implying that

7.1. GENERAL FORMULATION FOR LAYERED MEDIA

205

e must be parallel to the vector u, x K. Analogously, for a TM field we have u; . h = 0 or U z x K . e = 0, whence e must be parallel to K.) The eigenvectors K and u, x K are also shared by the twodimensional reflection and transmission dyadics, provided the admittance and impedance dyadics of all media and impedance boundaries have the same eigenvectors. In this case, these symmetric dyadics can be written in terms of their eigenvalues as (7.42)

T=TTM K K +TTEUzUz~KK K2 K2'

(7.43)

For example, substituting (7.41) in (7.38), the reflection dyadic associated with an interface of two media takes on the form _

R 11 -

{32112 _ {3111 1 KK k2 k1 __ {32!l2 lb.!l1. K2 k2 k1

+

+

k 2 Tl2

{32 k 2!l2 {32

ssu. {31 + ssss. -

xKK K2

UzUzx

(7.44)

{31

The problem can actually be handled in scalar form if we make the TE/TM decomposition to the original source, since the two polarizations do not couple to each other except if there are anisotropic media or surface impedances which possess other eigenvectors.

7.1.2

Image functions for reflection fields

To find the fields in physical space one must perform the inverse Fourier transformation. The result, often called the Sommerfeld integral, cannot be written in closed form because the reflection and transmission coefficients are too complicated functions of the Fourier parameter K. There have been attempts to approximate the reflection coefficients by simpler functions for which the inverse transformation can be found, for example by PARHAMlet ale (1980). However, the accuracy of the result depends on the approximations and cannot be easily predicted. The Sommerfeld integrals, despite their long history, are still a challenge to computers, since the integrands in most cases are highly oscillating. The idea behind EIT is to represent the reflection coefficient functions first as certain integral transformations of other functions which allow the inverse Fourier transformation from K space to p space to be performed exactly and what remains is the new transformation integral. This may just seem to be a substitution of one integral by another, but because the result can be interpreted as integration of an image source, it gains quasi-physical insight, which is helpful in setting up equations. Also, there is some numerical

CHAPTER 7. EXACT IMAGE THEORY

206

advantage in this approach. Thus, it becomes an interesting problem to find image sources corresponding to different geometries. This method was possibly first sketched, although not labelled in terms of images, for the classical half-space 'problem by BOOKER and CLEMMOW in 1950. In fact, their result can be interpreted in terms of an exponentially diverging image. More recently, KUESTER and CHANG (1979) and MOHSEN (1982) applied a similar idea by making approximations for the reflection coefficient functions. The converging exact form of the image principle was introduced in 1983 by LINDELL and ALANEN. The EIT method can be described for any layered structures, although it was first applied to the Sommerfeld problem of two homogeneous halfspaces. Its success depends on whether the reflection coefficient functions can be represented analytically in terms of suitable integral transformations. The basic image functions can then be obtained by comparing the resulting field expression to the free-space field expression. Electric source problem

The EIT image functions are obtained by comparing actual field expressions with those arising from a line current source along the z axis. The freespace transverse field arising from a dipole source J(r) = vIL8(r - uzh), where v is a unit vector not necessarily equal to u ,, in physical space, obeys the law

e(r)

= -jkTJ ( =It + k12 \7 t \7 )

. G(r - uzh)vIL.

(7.45)

Making the Fourier transformation leaves us with the expression =

e(K, z)

= -jkTJOe "

e-jt3lz-hl

"/3

2)

vIL,

(7.46)

where the (electric) operator 0 e is defined as

=

Oe

= KK .Ku, d = It - k2 -)/il dz.

The corresponding field from a line current I(z) z = -Zo to -00, can be written as the integral

= vI(z),

(7.47)

extending from

(7.48)

7.1. GENERAL FORMULATION FOR LAYERED MEDIA

and in Four-ier space in the region =

z

>

with no sources:

-Zo

J

00.

e(K, z) = -jk.,.,Oe .

207

,

e- J /3(z+z )

2jf3

I(z')dz'.

(7.49)

Let us now try to compare the free-space field expression (7.49) with the expression of the reflected field in the homogeneous half-space z > 0, due to a dipole source at height h: J(r) = vIL6(p)6(z - h), when the stratified structure exists for z < O. The reflected field denotes the difference of the total field and the field from the dipole in free space (the incident field). The effect of the stratified half space is characterized, in Fourier space, through the reflection dyadic R at z = O. Another analysis for the transmission field for~he region :...< 0 will be subsequently made with the transmission dyadic if replacing R. The field incident from the dipole to the stratified structure, in the region h > z > 0, is

e·(K z) t,

=

e- j /3 h

.

.

J /3 z = a-eJ {3z = -Jok'YlO e · --vILe 2j{3 , "/

(7.50)

giving rise to the reflected field .

=.

==

= e- j /3(z+h)

er(K,z) = a+e- J ,8z = R·a-e- J ,8z = -jk.,.,R·Oe·C·

2jf3

vIL. (7.51)

The mirror reflection dyadic (7.52) arises because the differentiation d/dz in the operator O; now operates on exp( -j{3z) and not on exp(j{3z). To obtain an image source giving the exact reflected field (7.51) as radiating from a line source in free space, in the form (7.49), the operator Oe should appear in front of the expression. Applying the eigenvector expansion of the reflection dyadic R, with eigenvectors K and U z x K corresponding to the respective TM and TE polarizations, we may write

k2 (1 _ KK t

_ oKUz~) J k 2 dz

. (RTEUZUz~KK R™(KK )) K2 + K2 + u, u; , (7.53)

208

CHAPTER 7. EXACT IMAGE THEORY

or

o; = o; . Re ,

R·

(7.54)

with the modified (electric) reflection dyadic defined as

Re = RTEUzUz~KK ) K2 + R™(KK K2 + u, u, .

(7.55)

This in fact implies that the operator Oe commutes with the modified reflection dyadic R e • Now we are able to write (7.51) in a form resembling that of (7.49): =

er(K,z)=-jkr,oe'

e- j /3( z + h ) = = 2j{3 Re·C·vIL,

(7.56)

allowing a representation for the reflected field as arising from an image current source.

Dyadic image function The EIT method is based on an integral representation of the modified reflection dyadic, defined by a function H( (): 00

It(K) = flAK, ()e- j /3 H«() d( .

(7.57)

o Here, ( is an integration variable and H(() is~ function of ( but not of the Fourier parameter K. The dyadic function leeK, () is called the electric dyadic image function and its functional form depends on the stratified structure in question. Note that for the moment we consider K and {3 as ~dependent variables so that K only ~pears in the vector form. Thus, R; is a function of both fJ and K, and depends on K but not on (3. If a representation of the above kind cannot be found, it is sufficient to define a more general form

Ie

00

Re(K) =

L n=l

00

jlen(K, ()e- j /3 H" «() d( , o

(7.58)

which gives us the possibility of expressing the structure in terms of a series of image sources. Inserting (7.57) in (7.56), its form becomes similar to that of (7.49): (7.59)

7.1. GENERAL FORMULATION FOR LAYERED MEDIA

209

Defining the following relation between the two integration variables z' and (:

z'

= h+H«(),

(7.60)

we can finally identify the image current function in the Fourier space in the form I 1 = = (7.61) I(z) H,«(/AK,(). C· vIL.

=

The inverse Fourier transformation to the transverse component of the reflection field can now be performed exactly, from which the total reflected electric field can be obtained in the form 00

Er(r)

= -jk1]

(1 + k~ vv) ·/ / G(r - r' + v

uzH«())Ji(r',()dV'd(.

0

'\1.'CI1'J Defining the mirror reflection dyadic C by (7.52), the image source corresponding to the original dipole can obviously now be written in physical space as (7.63) Thus, thefunctional form of the image source depends on the dyadic image function fe' which again depends on the geometry of the problem. It is obvious from linearity that for an original volume source J(r) the image source has the general form (7.64) with the mirror image of the original source defined as

Jc(r)

= C · J(C . r).

(7.65)

The dyadic image function is actually an operator operating on the mirror image of the primary source. The field from the image source must be calculated as a fourfold integral over the three space coordinates and the parameter (. For a point source, the image is a line source, which starts from the mirror image point at z = -h and proceeds along a line parallel to the z axis: (7.66) z = -h- H«(). In many cases, H«() may take complex values so that the image line lies in complex space. This does not, however, essentially complicate the field calculation process.

CHAPTER 7. EXACT IMAGE THEORY

210

Scalar image functions Since the reflection dyadic can be written in terms of its eigenvalues, the dyadic image function can also be written in terms of certain scalar functions. In fact, let us write the modified reflection dyadic (7.55) as

R

x n; k12 uzuzxKK,

TM=

I -

(7.67)

with the third reflection coefficient (7.68)

For each scalar reflection coefficient function, corresponding scalar image functions ITE ( (), ITM (() and f 0 ( ( ) can be defined through

Jf (X)

RTE(IJ) =

TE«( )e- i (3 H «) d ( ,

(7.69)

fTM «()e-i (3 H ((} d (,

(7.70)

o

J (X)

R™ (/3) =

o

J (X)

Ro(lJ) =

fo«()e- i (3 H (() d ( .

(7.71 )

o In terms of these scalar functions, the dyadic image operator can be represented as (7.72)

Thus, the image source can be explicitly written as (7.73)

This form is not unique, since the equivalent source is not unique. Other possible forms for the image source will be given subsequently.

211

7.1. GENERAL FORMULATION FOR LAYERED MEDIA

Duality transformations The image functions have certain relations through the simple duality transformation (Section 4.2.1), following from corresponding relations between the reflection coefficients. The duality transformation of the reflection dyadic can be obtained from a consideration of incident and reflected fields, related through e+=R·e-. (7.74) The dual of this relation can be written

h+:::.:: Rd· h ", Inserting

-U z

x h±

- u, x h+ = (uzuz~Rd)· (-u z x h-).

or

= ±Y . e±

(7.75)

leaves us with the relation

UzUz~Rd =

-

-

--1

-y. R· y

(7.76)

This takes a simple form provided the eigenvectors of Y and R are the same, which is the case for example in isotropic problems, because the dyadics then commute. Thus, in this case, the relation reads (7.77)

This corresponds to the following duality relations between the scalar reflection coefficients: R™ d

--

-

RT E ,

k2

Rod = K2(RI

M

-

RI

E

)

= Ro. (7.78)

These, finally, lead to duality relations between the image functions:

!od(()

= !o(().

(7.79)

Note that the dual of a dual always gives the original quantity. Thus, in dealing with the general isotropic medium, we need only derive, for example, the functions ITE, 10 since ITM can be obtained through the duality transformation. However, if the analysis is limited to, for example, dielectric media, this property cannot be applied.

Magnetic source problem The duality transformation analysis can be applied for the construction of the image of a magnetic current source Jm(r). In fact, making the duality transformation for the image expression of the electric source (7.64) gives us (7.80)

212

CHAPTER 7. EXACT IMAGE THEORY

with

=

=

f m(j\7 t , () == f ed(jV t , () == - f

TE

=

1

(()! + fo(() k 2 u, u, ~VV.

(7.81 )

The total reflection field due to the magnetic image source can written similarly to that in (7.62):

JJ 00

Er(r) = -V' x

G(r - r' + uzH«())Jmi(r', ()dV'd(.

v

(7.82)

0

Again, the image line corresponding to a magnetic point source lies on a line parallel to the z axis: z == -h - H((),

(7.83)

extending from the mirror image point z == -h to infinity, probably on the complex z plane.

Further properties of the image functions The three image functions fTE((), fTM(() and fo(() depend on the reflection coefficient functions and, thus, on the geometry of the problem. This being the case, they must be determined for each problem separately. However, they share certain general properties in all problems. Since there exists a relation between the three reflection coefficients R T E , R™ and R o , we might expect a relation between the corresponding image functions, aside from that due to the duality properties. To find this, we write K 2 == k 2 - {32, and apply the relation (7.84) which leads to

J

J

o

0

00

00

k2!o«( )e- i f3 H« )d( -

J

!o«(){32e- i f3 H«) d( =

00

k 2[JTM«)

-

jTE«)]e- j ,8H ({ ) d ( .

(7.85)

o Since all the image functions are zero together with their derivatives at ( ~ 0, we can easily make partial integrations in the second integral term and discard the terms at the endpoint ( == o. Also, we may anticipate

7.1. GENERAL FORMULATION FOR LAYERED MEDIA

213

functions vanishing at infinity, whence the corresponding endpoint terms are also discarded. Thus, equating the functions in the transformation integrands we finally have the resulting relation for the image function fo(() in the form of a differential equation, the prime denoting differentiation with respect to (: (7.86) For the special case H(() = ( this becomes simply (7.87) with the boundary conditions 10(0) = 10(00) = o. Another form of this condition can be simply written after inverse Fourier transformation as the operator equation, which is valid inside a field integral, where ~ t operates on the current function: (7.88) This together with the following identity

u, u, ~ ~~ + ~~ t

+ (~2 + k2 )u z u,

= ~;I + u, U z . (~~

+ k 2 I),

(7.89)

which is easy to verify, giYes us the possibility of transforming the dyadic image operator function le given in (7.72) in the form (7.90) in a different form

= . le(J'7 t , ( )

= 1

TE

=

=

1

(()1 + fo(()uzu z· (I + k 2 '7'7).

(7.91)

To arrive at this expression, one must note that the terms with '7'7 t and '7 2 + k 2 in the image source function result in NR image components and can be omitted, as explained in Chapter 6. Further, we can easily develop other possible forms, for example: (7.92) For the magnetic dyadic image operator we can correspondingly write from duality (7.93)

214

CHAPTER 7. EXACT IMAGE THEORY

=

.

!m(JV t , ( ) = - !

f m(rVt, () =

TM

=

=

1

(()I+!o(()uzuz·(I+ k2VV),

fTM «()It -

U z u,

(7.94)

. (fTE«()I + k12 fo«()V'V't) .

(7.95)

With any of the above expressions for the dyadic image operators, the image sources can be compactly expressed as Ji(r, () = fe(jV t , () . Jc(r), Jmi(r, ()

(7.96)

= 7m(jV t , () . Jmc(r).

(7.97)

The difference of the electric and magnetic dyadic image operators is seen from previous expressions to be (7.98)

Ie

which shows that the anti-self-dual part of the operator is just a function of ( times the unit dyadic. Correspondingly, we have

== fe(jV t , (

)

1 += !m(jV t , ( ) = k2fo(()[2uzuz~VV -

:2fo«()UzUz~V'V' + (fTM(() for twice the self-dual part of 7.1.3

fTE(()]I,

2=

VtI]

= (7.99)

7e '

Image functions for transmission fields

The EIT method can also be applied to problems of fields traversing a stratified layer. Let us assume that the transverse electric field (Fourier component) penetrating into medium 2 through a layer of any kind from rnedium 1 is eT(K, z) = a"2 ej 132z = a 1ej 132Z • (7.100)

T.

For simplicity in notation, the thickness of the layer in between the media 1 and 2 is taken as zero. This can be understood as a convention for the coordinates in the two media: medium 1 is defined for z > 0 and medium 2 for z < O. All information on effects du~ to the layer on transmitted waves is included in the transmission dyadic T. Let us follow a similar line of reasoning as in the derivation of the reflection image expressions. The transmitted field can be written in Fourier space as (7.101)

7.1. GENERAL FORMULATION FOR LAYERED MEDIA

215

in analogy with (7.50), (7.51). The indice~1 and 2 denote parameters corresponding to media 1 and 2. The dyadic Del in (7.47), when operating on the incident field a 1ei,Bl z, can be written as (7.102) Since the transmission image source must be in homogeous space filled with material 2, we should be able to write (7.101) in the form resembling (7.56) (7.103)

It is seen that the exponential term containing the 'wrong' f31 coefficient must be treated otherwise than in the reflection problem, unless the media 1 and 2 are the same. The modified transmission dyadic turns out to have quite a complicated-looking appearance in the general case:

(7.104) with

Tu-- €2 T ™

(7.105)

,

tl

T[

= p,d32 T T E ,

(7.106)

J.L2/31

To == (€2{31 T ™ _ J-Ll(32 T T E) €1(32

J-L2(31

ki .

K2

(7.107)

It can be argued that, in To, K2 in the denominator is cancelled by the preceding bracketed factor for any layer. In fact, if the term in brackets is written as a Taylor expansion according to powers of K, it is clear that the series only contains even powers, because the terms are independent of the sign of K (actually they are independent of the direction of the vector K). Thus, the series starts by A + BK 2 + .... Because for K = 0 (normal plane-wave incidence), the TE and TM polarizations are both TEM, the transmission coefficients are actually the same, and from (31 = k1 and {32 =: k 2 the bracketed term can easily be shown to vanish. Thus, A = 0 and K 2 cancels out.

216

CHAPTER 7. EXACT IMAGE THEORY

Transmission into a similar medium We can now define the transmission dyadic image function he analogously with (7.57). For the special case when the media 1 and 2 are the same, whence k I = k2 = k, {3I = {32 = {3 and T; = TTM, T[ = TTE, we have 00

T e = fhe(K, ()e- i{3H«)d(,

(7.108)

o

with (7.109)

!

00

h T E(()e- i{3H((}d(,

(7.110)

TTM ({J) = f h™ (()e- i{3H((}d(,

(7.111)

TTE({J) =

o 00

o

To({J) =

~: (TTM -

!

00

TTE) =

ho(()e-i13H(Od(.

(7.112)

o

Transmission into a different medium For the general case, the previous expressions do not work and, instead, we must define a dyadic image function based not on the transmission dyadic Te but on Te e- i {31h , which means that the image functions also become functions of the distance parameter h: 00

Tee-il31h = !h.e(K, h, ()e- i 132H(h,Od(.

(7.113)

o Applying the expression (7.104) in the form (7.114) the dyadic image function can be expressed in terms of three scalar image functions, in the corresponding form

7.1. GENERAL FORMULATION FOR LAYERED MEDIA

217

uZ~ + Uhu z + K)ho(h, () fJ~?'

(7.115)

ho(h, ()

The scalar transmission image functions are defined through

J ex>

Tu,e- j {3 1h

= €2 T ™

e- j {31h

=

€1

hu,(h, ()e- j {32 H (h '( )d (

,

(7.116)

o

J ex>

Tle- j (31h = J.1.1fJ2 TTE e- j (31h

=

J-L2f31

hI(h, ()e- j {32 H (h '( )d ( ,

(7.117)

o

J ex>

Toe- j (31h = fJ2 To e- j (31h =

ho(h, ()e- j (32 H (h'() d( .

(7.118)

o

Other forms for the transmission image dyadic are also possible. The last term in (7.115) will be omitted, because a term starting with -jK- jf32uz, or in physical space with V', corresponds to an NR source. Thus, we can neglect the corresponding term in the transmission dyadic as well, and apply the expression (7.119)

Also, note the difference between the coefficients To and To = f32To. Application of this expansion leads to a transmission image theory in which the form of the image function corresponding to a point source is dependent on the position of the source, which was not the case for the reflection image. In fact, writing the transverse component of the transmitted electric field in the Fourier space

J ex>

- jk2112

•

eJf32z-

Oe2 · 2jfJ2 heCK, h, ()e- i 132H (h,() · vILd(

(7.120)

o

we have for the total transmitted field in physical space corresponding to the original dipole J(r') = vIL6(r' - uzh), ex>

ET(r) = -jk2 112

JJ

C(D«()) . Ji(r', ()dV'd(,

o v

(7.121)

CHAPTER 7. EXACT IMAGE THEORY

218

with

= Vp2 + [z -

D(()

Ji(r', () =

H(z', ()]2,

he (jV'~, Z', () . V I L8(r', (),

(7.122) (7.123)

which is dependent on the position z' = h of the dipole. Note that this Green dyadic depends on the parameters of medium 2. More generally, for a three-dimensional source J(r') we obtain the four-dimensional image (7.124)

7.1.4

Green functions

Instead of applying image functions each time when fields are being computed, one can define a dyadic Green function which takes into account the reflection or change in transmission due to the stratified geometry in question. This gives us an opportunity to apply solution routines designed for free-space problems by adding to the free-space dyadic Green function a new dyadic Green function.

Reflection problem A new Green dyadic K(r) corresponding to the reflection from the structure can be obtained by integrating out the parameter ( in the field expressions. It can be seen that, to construct the dyadic image Green function, two scalar Green functions are required instead of only one as in the homogeneous-space problem. The reflection image Green dyadic is obtained from the reflection field expression: oo

Er(r)

= -jkT]

JJ

G(r - r'

v -jkT] [

(i

+ uzH(()) . Ji(r', ()dV'd( =

0

G(r - r'

+ uzH(()) ·

- jkT]

fe(j'\7~, ()d() . Jc(r')dV' =

J

K(r - r') · Jc(r')dV',

(7.125)

+ uzH(()) . !e(j'\7~, ()d(.

(7.126)

v oo

K(r) =

J

G(r

o

7.1. GENERAL FORMULATION FOR LAYERED MEDIA

219

The expression of the image current Ji contains functions of the ( variable and the V operator as separate factors, which makes it possible to integrate the ( variable out and define the scalar Green functions

! ! ! 00

KTE(r) =

G(r + uzH((n!TE(()d(,

(7.127)

G(r + uzH((n!TM (()d(,

(7.128)

G(r + uzH((n!o(()d(.

(7.129)

o

00

K™ (r) =

o

00

Ko(r) =

o The \7' operators can be moved in front of the Green function through the following manipulations in the field integrals, recalling that the field point r never coincides with the image source point r':

!

Ko(r -

r')(uzuz~V'V'). Jc(r')dV' =

v u, x

! v

u, x V

V' [Ko(u z

X

V')· Jc]dV' - u, x !(V'Ko)(U z' V' x Jc)dV'

v

J v

Ko(u z X V' . Jc)dV' =

(uzuz~VV) ·

=

Jv

Ko(r - r')Jc(r')dV'.

(7.130) Thus, the dyadic image Green function can be defined in terms of these scalar image Green functions as

=K(r)_

(=

1 V V ) K TM (r)+k2(uZuZxVV)Ko(r), 1 x I+ k2

or, taking another representation for the dyadic operator

(7.131 )

7e , (7.132)

The reflected field due to the image source can be expressed simply as

Er(r)

= -jk.,.,

J

K(r - r') · Jc(r')dV'.

v

(7.133)

CHAPTER 7. EXACT IMAGE THEORY

220

Note that the source here is the mirror image of the original source and not the original source. We could have included the mirror image operator c in the definition of the image Green dyadic. It is seen that only two image scalar Green functions are needed in the computation of the reflected fields, either of the functions K™ and K TE together with the function K o. From duality, there exist obvious relations between the image Green functions: KIM (r) = _KTE(r), (7.134)

KJE(r) = _K™ (r),

(7.135)

Kod(r) = Ko(r).

(7.136)

Thus, values for the Green functions K T E and K™ can be computed applying the same procedure with dual parameter values.

Asymptotic expressions for the reflection problem Asymptotic expressions of the Green functions are of interest when computing fields far from the source region. Also, extracting the asymptotic term from the Green function leaves a function which is more easily approximated because it decays more quickly than the original Green function. If the argument r becomes large, so that we may assume IH(()I « [r] for those ( values for which the image function is not negligible, we may write the two-term Taylor expansion for the distance function as

(7.137) which inserted in the Green function gives us the basic approximation (7.138) Thus, we have for the first-term approximation of any of the Green functions (7.127)-(7.129)

J

J ~

00

G(r + uzH(())f(()d(

o

~ G(r)

e-jkUr·U.H(()

f(()d(.

(7.139)

0

The last integral expression can be interpreted in terms of the reflection coefficient function, applying the definitions (7.69)-(7.71). In fact, we can write 00

J

e-jkUT.·UzH«)

o

f(()d(

= R({3)I l3=k cos ()

= R(kcosO),

(7.140)

7.1. GENERAL f'ORMULATION FOR LAYERED MEDIA

221

denoting the angle of the vector r and the z axis by 0, whence we have the asymptotic expressions for the image Green functions

KTE(r) ~ RTE(kcosO)G(r),

(7.141 )

K™ (r) ~ R™ (k cos O)G(r),

(7.142)

Ko(r)

~

Ro(k cos O)G(r).

(7.143)

The corresponding approximation for the reflection dyadic image Green function

K(r)

~ R™(kcosO)

R o ( k cos 0)

(1 + :2 V'V') G(r)+

:2(

u, u, ~ V'V')G( r).

(7.144)

can be written, if we make the far field approximation \7 --t -jku r and apply (7.68) (note that the Fourier parameter is now K = k sin 0), in the form (7.145) This can be called the reflection-coefficient method, or ReM approximation, in which the reflection far field is obtained by replacing the exact image by the mirror image of the original source, and multiplying it by the dyadic R T E u., u., + R T M uo.uo. When computing the reflection field, the argument r of the Green function is replaced by the difference r - r', where r' denotes a point of the mirror image source. The angles and distances are then measured from the point r' and not the origin.

Transmission problem The transmission field can also be written in terms of a new transmission Green dyadic taking care of the geometry behind the transmission plane. In fact, writing the transmission field (7.121) in the form

- jk2 1/2

J

KT(r - r') · J(r')dV',

v

(7.146)

222

CHAPTER 7. EXACT IMAGE THEORY

we can define the transmission Green dyadic by

J 00

KT(r - r')

=

G(D«(» . he(j'V~, z', ()d(,

(7.147)

o

D(() = J[r - p' - UZH(Z', ()] . [r - p' - UZH(ZI, ()].

(7.148)

The asymptotic far field expression for the new Green dyadic can be written similarly with the reflection problem: D(() ~ r - u, . [p' + uzH(z', ()],

(7.149)

00

KT(r - r')

~ G(r)eik2U,·P' . Jhe(j'V~, z', ()eik2U.OU.H(z"()d(.

(7.150)

o

The last integral can be interpreted in terms of the definition of the transmission dyadic, (7.113): (7.151 )

with the parameters in the transmission dyadic and exponent expressions given as (7.152) (7.153)

(7.154)

z.. z·

Fig. 7.2 The asymptotic far field approximation for the transmission image expression can be interpreted in terms of refracted rays starting from a source point at an apparent height z".

7.1. GENERAL FORMULATION FOR LAYERED MEDIA

The angles we can write

()1

and

()2

223

are defined above. To obtain an interpretation,

= ,== Kr(r - r) ~ G(D) . T e ,

(7.155)

D ~ J(r - pi - uzz") . (r - pi - uzz"),

(7.156)

"

{31

I

z = -z = {32

k1 cos ()1 I z k2 cos ()2

tan ()2 =-z . tan ()1 I

(7.157)

This expression can be interpreted geometrically as representing the point on the z axis from which a ray in the direction u, is emanating, when the layers between the planes in the medium 1 and 2 are squeezed to a plane.

References

P. (1986). Applications of complex image theory. Radio Science, 21, (4), 605-16. BOOKER, H.G. and CLEMMOW, P.C. (1950). A relation between the Sommerfeld theory of radio propagation over a flat earth and the theory of diffraction at a straight edge, lEE Proceedings, 97, pt.III, 18-27. KELLER, J.B. (1981). Oblique derivative boundary conditions and the image method. SIAM Journal of Applied Mathematics, 41, (2), 294-300. KUESTER, E.F. and CHANG, D.C. (1979). Evaluation of Sommerfeld integrals associated with dipole sources above the earth. University of Colorado Electromagnetics Laboratory, Scientific Report 43. LINDELL, I.V. and ALANEN, E. (1983). Exact image theory for the Sommerfeld half-space problem with a vertical magnetic dipole, Proceedings of the 13th European Microwave Conference, Nuremberg, pp. 727-32. Microwave exhibitions and publishers, Tunbridge Wells. LINDELL, I.V., NIKOSKINEN, K.I., ALANEN, E. and HUJANEN, A.T. (1989). Scalar Green function rnethod for microstrip antenna analysis based on the exact image theory, Annales des Telecommunications, 44, BANNISTER,

(9-10), 533-42. MOHSEN, A. (1982). On the evaluation of Sommerfeld integrals. lEE Proceedings, 129H, (4), 177-82. PARHAMI, P., RAHMAT-SAMII, Y. and MITTRA, R. (1980). An efficient approach for evaluating Sommerfeld integrals encountered in the problem of a current element radiating over lossy ground. IEEE Transactions on Antennas and Propagation, 28, (1), 100-4. WAIT, J.R. (1985). Electromagnetic wave theory. Harper and Row, New York.

CHAPTER 7. EXACT IMAGE THEORY

224

7.2

Surface problems

As an application of the general theory, let us consider problems involving planar impedance surfaces and sheets. If the surface is impenetrable, it can be characterized by an impedance surface condition, as discussed in Chapter 3. For penetrable surfaces, the corresponding impedance sheet characterization is valid. As a last example in this section, the thin metallic grid problem is considered. In all these cases, the image functions can be quite easily written in explicit form involving only simple exponential functions. 7.2.1

Impedance surface

The planar impedance surface serves as the simplest example of a problem for which the image method can be applied. The idea has also been applied in the book by FELSEN and MARCUVITZ (1973, pp. 557-559). Reflection coefficients

In the case of an isotropic planar impedance su.:!ace at !. = 0 with surf~e impedance Zs, defining the impedance dyadic Zs = Zslt, the dyadics Zs and Z obviously commute and the reflection dyadic can be written from (7.38)

R

= (Z s + Z)-l. (Z s - Z) = RTEUzUz~KK R™KK K2 + K2 '

(7.158)

with the eigenvalues RTE

=Z

_!l!! S

Z,

{3

2!l!! = 1- __Z_tc_ (3 +

+W

r'

-

RTM _

Z - f1!J. S

Zs

k

-

+~ -

-1

+

2~ 17

(3 +

¥.

(7.159) These coefficients obviously satisfy the duality condition (7.78) because the dual of Zs/TJ is TJ/Zs. As another check, both coefficients can be seen to tend to -1 as Z, ~ 0 and to +1 as Z, ~ 00, corresponding respectively to perfect electric and magnetic conductor surfaces. The third reflection coefficient takes the following forms:

(7.160)

7.2. SURFACE PROBLEMS

225

and it is seen to vanish in both the perfect electric and magnetic conductor limiting cases. Image functions The image functions corresponding to the reflection coefficients can be found in a straightforward manner if the transformation is defined through the integral

J 00

R(f3) =

f«()e-j(j(d(.

(7.161)

o

This means that the function H(t) of (7.57) in this case is simply defined as H«() = (. The image functions can then be written as (7.162)

fTM «()

= -6+«() + 2j Zsk e-jZ,k(/.,U+«(),

(7.163)

rJ

fo«() =

~k~~;

(e-jZ,k(/., - e-j.,k(/Z.) U+«().

(7.164)

The unit step function U+«() and the corresponding delta function 6+«() are defined at the limits U«( - ~) and 6«( - ~) as Ll --+ O. Thus, all image functions vanish for ( ~ o. They can also be checked to satisfy the condition (7.87). In terms of these functions, the reflected field can be expressed as arising from an image ~ource, by applying one of the many forms for the dyadic image function fe. For example, with (7.92) we have the following explicit expression for the image source corresponding to a primary source J(r): .

Ji(r, () (fTM«()UzUz

= fe(j~t' () . Je(r) =

+ fTE«()I t + u, f~~()Vt) · Jc(r),

(7.165)

in which the image functions must be substituted from (7.162)-(7.164).

Convergence of image functions To obtain a meaningful field integral, the integrand should converge. Because the Green function can be made converging through a proper choice of the branch of the distance function, it suffices that the image functions

226

CHAPTER 7. EXACT IMAGE THEORY

are non-divergent. It is seen immediately that for real parameter ( values, for a lossless medium (real and positive k and T/), the exponential functions are all non-diverging when Z; is real. On the other hand, if Z, has an imaginary part, some of the exponential functions turn out to be diverging. In particular, for a lossless medium with real k and "7, the condition for the TE image (7.162) to be non-divergent is obviously ~{(/Zs} == ~{ (ys} ~ 0, or the surface impedance should be inductive or pure resistive if ( is real and positive. Correspondingly, for the TM image, from (7.163) we have the condition ~{(Zs} ~ 0, or the surface impedance should be capacitive or pure resistive. Thus, we see that unless the surface is purely resistive, one of the functions fT E((), fT M(() diverges for real and positive ( and the fo( () function diverges. The image source diverging for real and positive ( can be interpreted as giving rise to a surface-wave field. The existence of a surface wave is associated with the poles of the reflection coefficient. Writing

A R({J) = {J - (Jp'

(7.166)

a surface wave with z dependence as exp( -jf3p z ) exists if the condition ~{,Bp} < 0 is valid. Because the pole corresponds to either ,BJE == -k".,/Zs or f3;M = -kZs/T/, there exists a TE surface wave for ~{Ys} > 0, which defines a capacitive surface impedance. On the other hand, there exists a TM surface wave for ~{Zs} > 0, which defines an inductive surface impedance. These conditions are the same as those for diverging image functions for real and positive t. Exponentially diverging image functions corresponding to surface waves along the structure will also be encountered in other problems. To obtain non-diverging image functions we may choose an integration path in complex ( plane. It is sufficient to choose a path along the imaginary axis ( = j(', although this does not make the image function converge. However, since the distance function D is now complex, its branch can be chosen so that the Green function becomes exponentially decaying, whence the field integrand converges. However, for imaginary (, there is a singularity in the Green function when the source and field points are at the impedance plane, because the distance function D becomes zero for p = ('. This singularity is clearly demonstrated by the following example. Vertical electric dipole above an impedance plane Let us consider a horizontal impedance plane and the field from a vertical electric dipole (VED) J(r) = u zIL6(z - h). According to the EIT method,

227

7.2. SURFACE PROBLEMS

we can replace the impedance surface with the image source Ji(r, () = fe(j\lt' () . [-u zIL8(p)8(z

+ h)] =

- jTM«()u zIL8(p)8(z + It).

(7.167)

---+--I----t!-·--~-~

0.9 I-----f---+----+----+--

I--+--I,--+-!:

Integrand Ifor the field due! to YED

0.8t--------of----t---·r~"tlecteif·trom·· an···'.pedancc····sunaeel--:--.....-----1 fOf krho :k 1, kh *= 0, Zs/eta = 1 :

0.71----HJ----+----+---+----+--=---+---=----+---+----+------f

0.6t-------H4---of-----+----of---+-------f----r----f----+----t

0.5 I-----H+---+-----+---+-----+---+------+----+---+---~ kz=O.pl 0.4 t----+++---+----+---+----+---+------+----+------.-+-----..1

I 0.31---+-++--+----+---+----+---+-----+---+----+-----..1

/;

0.2 0.1

kz=OJl

~ \ kz~0.3 .....

~

kz=l

"-~I

4

3

5

6

8

7

9

10

Fig. 7.3 Normalized integrand of the reflected field integral without the delta term, as a function of k( for different values of kz, at h = 0 and p = o. A peak arises when z + h ~ 0, i.e, the source and field points are both close to the surface.

The image can be expressed in terms of the function ITM alone, because the VED source produces a TM field. The reflected field can be written as

JJ 00

Er{r)

= -jk'fJ

G{r - r' + u z () · Ji(r',()dV'd( =

v

0

. (= + k1)

JkrJ I

2

\7\7

. uzIL K TM (r

+ uzh),

when the Green function is defined as in (7.128) with H«() = (:

J 00

K™(r+ u h) = z

o

-jkD«)

e fTM(()d( 41rD«()

=

(7.168)

228

CHAPTER 7. EXACT IMAGE THEORY

J » 00

Z _ G(D(O)) + 2jk--!..

e

o with the distance function defined by D(() =

-jkD«)

47rD(()

e- j Z t< k
J p2 + (z + h + ()2.

(7.169)

(7.170)

To ensure the convergence of the integral in (7.169), the branch for the distance function is defined by ~{D} ::; o. It is obvious that for a capacitive surface impedance, the fT M function makes the integrand converge exponentially. For an inductive surface impedance we write ( == -j(' and rewrite (7.169) as K™ (r + uzh) = -G(D(O))

+ 2kZ. TJ

J 00

o

-jkD( _j(')

e . 47fD(-J(')

e-Z.~k(' /l1d('.

(7.171)

Again, the integrand converges exponentially because of the complex distance function D, even if Zs is imaginary. The singularity is clearly integrable, because at z + h == 0 it is of the type 1/ p2 - ('2. Actually, integration along a short distance ~ around the singularity point (' == p leaves us with the expression

J

p+l::t./2

J

2kZ. e- j kD(-j(') e-Z,k(' IT/d(' :::::;: (1 _ j) TJ 47fD(-j(')

~ Z,kp/2T/ ~ ep 41r '

(7.172)

p-A/2

which shows us that its contribution varies roughly as ponentially decaying function.

7.2.2

1/vp times

an ex-

Impedance sheet

Reflection images The previous isotropic impedance surface geometry can be transformed to impedance sheet geometry by adding a homogeneous half-space behind the surface. If the admittance in both homogeneous spaces is denoted by Y, and the surface admittance by Y s, the reflection dyadic at the interface is, accordingly, of the form

R = [Y -

(Y s + y)]. [Y

+ (Y s + y)]-l = -Ys· (2Y + Ys)-l.

For isotropic surface admittance Y s

(7.173)

= Yslt = Z;l It, we simply have (7.174)

7.2. SURFACE PROBLEMS

229

When the admittance dyadic Y is substituted in terms of its TE and TM eigenvalues (7.40), the reflection dyadic for the isotropic sheet can be written as Ys KK R = _ Ys uzuz~KK (7.175) y: + 2k K2 . y: + yt K2 1311

s

k11

s

Thus, the reflection coefficients are

RT E

RT M

_ _

-

=_

E:L 2Z.'C

fJ {3 + 2kZ" ~

R

k

o

2

2kTJ Z s 4Z;

= ({3 + 2kZI! )({3 + E:L) = ".,2 ~

(7.176)

{3+E:L' 2Z...

2Z"

-1

+

2kZt!

{3 +

(1+ {3

11 2kZ~ , ~

2kZI! ~

(7.177)

1) ( )

+ {3 + E:L .

7.178

2Z ll

Note that the duality check does not work here because the sheet is not the most general one (magnetic surface impedance is not included). The corresponding image functions are, again, combinations of exponential functions:

;s

k fTE() = -j 2 e- jT/k(,/2ZI! U+(),

(7.179)

fTM () = -b+() + j2Zs k e- j2kZI!(,/T/ U+(),

(7.180)

TJ

fo(() =

".,t~~~;

(e- j2Zl!k(,/T/ + e- jT/k(,/2ZI!) U+().

(7.181 )

To check the results, let us consider the case Zs ~ 0, which is readily seen to lead to fTM = -8+(() and vanishing of fo(()' The function fTE is in this case of the form [a exp jn:( with a ~ 00, which inside the field integral can be seen to act like the function -8+():

f

~

f

~

jo:ejoc(,P()d( = j

f

~

ejTP(Tjo:)dT - jP(O)

ejT dT = -P(O).

0 0 0

(7.182) This property is also seen from the corresponding reflection coeffient RT E , which tends to the value 1 when Z, ~ O. Thus, the limit equals the perfectly conducting sheet case as expected. On the other hand, for Zs = 00, all image functions vanish, which again is in accord with the vanishing of the impedance sheet.

230

CHAPTER 7. EXACT IMAGE THEORY

Transmission images Transmission through the sheet is defined by the transmission dyadic (7.183) with the eigenvalues

(7.184)

2kZ,

TT M

= 1 + RT M =

(3 +

'7

2kZ, ,

(7.185)

'7

and the third transmission coefficient is simply

(7.186) The image functions can be defined through the transformations (7.110)(7.112) with H«() = (, or

I 00

T({3)

=

h(()e-i{J(d(,

(7.187)

o

whence we have

Let us again check the limiting cases. For Zs = 0 we have the perfect conductor sheet, whence h™ and fo functions are seen to vanish immediately and h T E after a consideration similar to that above. For Za = 00 both h functions reduce to 6+(t) in accord with the vanishing of the sheet.

7.2. SURFACE PROBLEMS

231

Convergence Since the reflection image functions for the impedance sheet are similar to those of the impedance surface case, the same convergence considerations apply. In fact, a surface wave is again connected with the divergence of the corresponding function for real and positive (. In particular, because the pole expressions only differ by a factor of 2, the same conclusions concerning the appearance of a surface-wave for a certain complex Z, apply as for the impedance surface.

Dielectric slab Modelling a thin dielectric slab by an impedance sheet, we may write for the sheet impedance when the thickness ~ of the slab becomes small, according to (3.173): .

Z.

-j1J

= lX. = kLl(f r _

(7.191)

1)"

The TE image function (7.179) can be written in this case as

d (k2(~ f TE (() = de exp -2-(f r

-

)

1) .

(7.192)

This function diverges unless e; < 1 or ( is imaginary. It is well known that the basic surface wave (actually a waveguide mode) can exist in a dielectric slab for any thickness of the slab. 7.2.3

Wire grid

The problem of a planar wire grid with square cells and small mesh size can be treated by the theory of KONTOROVICH (1963) much in the same way as the impedance s~eet, although the surface impedance dyadic is no longer a multiple of It. However, it turns out to have the previous eigenvectors, which means that the grid does not couple TE and TM waves, which makes the theory simple. On the other hand, if the cells were of more general anisotropic form, such coupling would occur and the theory should be modified accordingly. The condition at the grid can be written for the transverse fields e, h on each side as a double equation, which in Fourier space reads (7.193) Defining the grid impedance dyadic

Z 9 through the sheet equation (7.194)

232

CHAPTER 7. EXACT IMAGE THEORY

we can write its expression from (3.200) after replacing the operator "'J t by -jK:

Z _. 9 -

JTJ~

(1 _KK) _ZTEUzUz~KK + ZTM K K t

2k2

-

K2

9

9

(7.195)

K2'

with

= (ygTE)-l, Z'{M = j i~ (k 2 + (32) = (y;M)-l. ZJE == jTJ~

(7.196) (7.197)

Y g, the two-dimensional inverse of Z g, can now be directly substituted for Y 8 in the impedance sheet expression (7.173) to produce the reflection dyadic. The reflection coefficients are found after some algebraic manipulations to be of the form i! RTE _ _2_",_ (7.198) - {3- i!!' 2",

R T M == R o -

1

VI + 4~2

~(RTE _ RTM) = K2

j! __ 8 _._ ( {3 +

-jkK

1 + 4~2

1-

·k)

+

J s

,

(7.199)

{3 - j ks

(_2_ _ 1 __1_) if ' {3 - ~

{3 - jks

{3 +

(7.200) with the definition

S=~+J1+ 1 2~ 4K

2

•

(7.201)

If the TM reflection coefficient is compared with that obtained for the vertical electric dipole by LINDELLet ale (1986), a difference in sign is noted. This is due to the difference in the definition of the reflection coefficient, which in the reference was defined for the Hertzian potential and not for the electric field as in the present analysis. As a check we see that for K -+ 0 or s -+ 00, corresponding to a very dense grid, we have R -+ -It, which is the reflection dyadic of the perfectly conducting plane. Also, for the limiting case K --+ 00 (vanishing grid), we have s --+ 1 and all the reflection coefficients are seen to vanish. Again, the transmission coefficients are in a simple relationship to the reflection coefficients: TE TTE == 1 + R (7.202) , TTM

= 1 + R™,

(7.203)

which makes it possible to express the transmission image functions in terms of the reflection image functions as in the case of the impedance sheet discussed previously.

233

7.2. SURFACE PROBLEMS

Image functions The reflection image functions for the wire grid can be written down without problem, since the reflection coefficients are of such a simple form. In fact, applying the same integral transform "as for the impedance sheet, we can write

fTE«)

FM « ) = fo«()

VI:

= I :~1I:2

= _~e-kf,/2K U+«), 2K

411: 2

(2e-

~ (ek(,/s -

e-Skf,)

kf,/2K - e-skf, - ek

U+«),

r,/8) U+«().

(7.204) (7.205) (7.206)

The transmission image functions have the corresponding form: hT E (( ) = b+(()

+ fTE((),

(7.207)

h T Nf (() == 8+(()

+ fTM (().

(7.208)

To check these different image expressions, let us again set K ~ 00, or s ~ 1 for the vanishing grid, whence all reflection image functions are seen to vanish. The case K -+ 0 or s = 1/ K -+ 00 for an infinitely dense grid or conducting plane, leads to fTE(() = fTM (() = -b+((), because the function ae- Q ( is a delta sequence for a -+ 00. The function fo is seen to vanish in both of these limiting cases. Thus, the two cases give no image or negative mirror image corresponding to total transmission and total reflection, respectively, which checks with the physical pictures of vanishing grid and conducting plane. When studying the convergence of the image functions, it is seen that, as for the impedance surface, for real k one of the exponential functions is bound to diverge unless ( is taken to be imaginary. Obviously, this again corresponds to a TM surface wave (waveguide mode) attached to the grid.

Vertical electric dipole above a metallic grid As an example of application of the image concept let us consider a horizontal metallic grid with a vertical electric dipole J(r) = uzI L6(r - uzh). The grid gives rise to the reflection image current, analogously to (7.167),

Ji(r, () = -uzfTM (()ILb(p)6(z + h) == (7.209)

CHAPTER 7. E..X ACT IIv/AGE TJIEORY

234

while the transmission image current can be written as J~{r,()

= -u zh™(()IL6{p)6(z -

h)

=

- u zIL6(p)8(z - h)6+() - u zILjTM()8(p)6(z - h).

(7.210)

As noted above, the TM image function diverges unless the integration is performed along the imaginary direction on the ( plane. Let us write

( = -j('.

(7.211)

The reflected electric field expressed in terms of the reflection image, with the complex distance function defined as

D

= V(p -

p')2 + (z - z' - j(')2,

reads

~{D} ~ 0,

(7.212)

JJ 00

Er(r) = -kTJ

G(D)· Ji(r', -j(')dV'd(' =

o v

JV 00

kTJILuz . ·

G(

p2 + (z

+h -

j(')2)JTM (-j(')d('.

(7.213)

o The field can easily be computed numerically, because the integrand converges exponentially owing to the exponential decay of the Green function. Let us look at some asymptotic cases of the field. When the source and/or the field point are/is far from the surface, i.e. for k(z + h) » 1, the field can be approximated according to the ReM method given in Section 7.1.4. The reflection field can be approximately expressed as

(7.214) and interpreted as the field arising from a dipole at the mirror image point with the original amplitude multiplied by the TM reflection coefficient associated with the geometrical optics ray going from the original dipole to the field point with a reflection at the grid. The field close to the surface with p » z + h can be evaluated, applying the approximation (7.215) in the form

J ex:>

G(D)JTM(_j(')d('

o

~

7.3. THE SOMMERFELD HALF-SPACE PROBLEM

4jVl

k

+ 4K 2

Vik

21r P

(seW~erfc(wd + ~eW~erfC(W2)) ,

235

(7.216)

S

with erfc denoting the complementary error function and

Wl

=

V2ipk(z+h-iPS),

W2

=

V2ipk(Z+h+i~).

(7.217)

This can be identified as a sum of two Sommerfeld surface-wave solutions with imaginary numerical distances - j k p S 2 /2 and j kp/28 2 , as discussed in FELSEN and MARCUVITZ (1973, pp. 509-510). In the region 2~2 «: kp «: 2/~2 with k(z + h) «: 1 we have IWII ~ 1 and IW21 « 1, whence (7.216) can be shown to reduce to

(7.218)

which is obviously of the form of a radial surface wave, decaying in the z direction and propagating in the radial direction along the grid.

References

N. (1973). Radiation and scattering of waves. Prentice-Hall, Englewood Cliffs, N J . KONTOROVICH, N.I. (1963). Averaged boundary conditions at the surface of a grating with square mesh. Radio Engineering and Electronic Physics, 8, (9), 1446-54. LINDELL, I.V., AKIMOV, V.P. and ALANEN, E. (1986). Image theory for a dipole excitation of fields above and below a wire grid with square cells, IEEE Transactions on Electromagnetic Compatibility, 28, (2), 10710. WAIT, J.R. (1978). Theories of scattering from wire grid and mesh structures. In Electromagnetic scattering (ed. P. Uslenghi), pp.253-87. Academic Press, New York. FELSEN, L.B. and MARCUVITZ,

7.3

The Sommerfeld half-space problem

The simplest electromagnetic field problem involving sources and material media, beyond that of the homogeneous space, is the problem with a source in front of a planar interface of two homogeneous half spaces. This problem

CHAPTER 7. EXACT IMAGE THEORY

236

was first studied seriously by SOMMERFELD in 1909, which is why it is often called the Sommerfeld half-space problem. Let us find the image functions associated with two half-spaces 1 (z > 0) and 2 (z < 0) with respective medium parameters denoted by €I€o, J.LIJ.Lo and €2€, J.L2J.L. The primary source is assumed to lie in the half-space 1. Let us denote the parameter ratios by € = €2/€1, J.L = J.L2/J.Ll' For static fields, the image principle is well documented in elementary textbooks. For example, if a point charge Q is located at the point r = uzh in the air (€o,J.Lo), the field reflected from the dielectric ground medium (€€o, J.Lo) can be shown to equal that arising from the reflection image Qr located at the mirror image point ri, when the medium is all air: Qr

€-1

= - € + 1 Q,

ri

= C . uzh =

-uzh.

(7.219)

Also, the field transmitted into the ground in the static case can be found from the transmission image charge QT, 2€ QT =

f

+ 1Q,

ri

=

uzh,

(7.220)

which appears at the location of the original charge when the medium is all ground. It turns out that the correct interface condition, continuity of the transversal total electric field, is exactly satisfied with these two image sources. For the time-dependent problem the image solution is not that obvious. If the original static charge is set in harmonic motion, the idea of setting the static image in the corresponding harmonic motion does not work, since, at the interface, only the condition for the electric field and not for the magnetic field is guaranteed. Thus, possible image sources in this case must necessarily be more complicated, which gives a motivation for the EIT method.

7.3.1

Reflection coefficients

As was seen in the previous section, construction of the image sources requires three reflection coefficient functions and their integral transforms to be found. In the case of an interface between two homogeneous and isotropic media with the parameters €l, ftl' €2, ft2, the reflection coefficient functions were already given in (7.44) based on the wave impedances of the TE and TM polarized fields: ZTE

= kTJ , f3

ZTM =

{3TJ I k

(7.221)

7.3.

THE SOMMERFELD HALF-SPACE PROBLEM

237

(7.222)

(7.223)

Here we denote

JL2

(7.224)

J-L=-, J-Ll

and q

= ~,

B = jk~ - k? = k1 Jw- - 1,

(7.225)

where k, = WVf.iJ-Li. The reflection function R( a, q) applied above is defined by

H+1

- aqR( a, q) 1'::271' aq

+ Vq2 + 1

(7.226)

The reflection coefficients obviously satisfy the duality condition (7.78). For the third reflection coefficient we can evaluate after some effort

with

€-1 E=f.+ i '

JL-I i

M--- J-L+

(7.228)

The image functions can be found in a convenient series form if we expand the reflection coefficient functions in terms of a reflection parameter r defined by r -- R(I ,q) --

q-

H+1 .

q+H+1

(7.229)

In fact, the following expansion can be easily verified: A+1' 1 + Ar'

Rio: q) = - ,

a-I

A=--. a+l

(7.230)

CHAPTER 7. EXACT IMAGE THEORY

238

Applying the Taylor expansion, we can write 00

R(a,q) = (A

+ r) "(-Art = ALJ n=O

+ 4

00

"(-Ar)n.

a -1L.t

Obviously, the series expansion converges best for small 7.3.2

(7.231)

n=l

IArl

values.

Reflection image functions

The reflection image sources are obtained in terms of a function of two variables, f( a, p), which can be identified through the following integral identity, found in any collection of Laplace transforms, for example, ABRAMOWITZ and STEGUN (1964).

J J2~(P) 00

2n

e-qpdp =

[R+!- q]2n = (-rt,

(7.232)

o

where J 2n (p) is the Bessel function of order 2n. In fact, if we write

J 00

R(a,q)

=

[f(a,p)

+:

~~ 8+(p)] e-pqdp,

(7.233)

o

the image function f(a,p) can be identified from the integral identity and the series expansion of the reflection coefficient in the form (7.234) The special case a = 1 gives us (7.235)

Values for the image function f(a,p) can be easily calculated from the above series, because Bessel functions with the same argument but different indices can be quickly computed with the aid of simple recursive formulas, as explained in ABRAMOWITZ and STEGUN (1964, pp.385-386). There exists a differential equation for the image function f( a, p), from which it is easy to derive the Taylor expansion for the same function. In

7.3. THE SOMMERFELD HALF-SPACE PROBLEM

239

fact, it can be easily checked that the reflection coefficient function satisfies the algebraic equation 0-1 0+1

R(o,q)---=a

(q -

#-+1)2 -

(2 )2 o -1q -1

~+~

r + ~+~ )2,(7.236) 0 -lq-l

=-a(2

which inserted in the integral identity (7.233) gives rise to the following differential equation for the image function:

d2 (o? - l)-d2!(0,P) - !(o,p) p

a-1

b+(P)]. = -a[!(l,p) + --1 a+

(7.237)

To define the solution uniquely, the following boundary conditions are needed: (7.238) !(0,0) = !(a,oo) = O. It is seen from the differential equation that the function f( a, p) is continuous at p = 0+ but its first derivative (with respect to p) experiences a step discontinuity (7.239) Taylor series expansion

Applying the differential equation (7.237), a Taylor series expansion can be written for the image function. The function /(1, p) has the known expansion

f( l p) ,

= _2 J2(P)U P

+

(p)

=_~

L...J n=O

(-l)n(p/2)2n+l U ( ) '( + 2)'. +P , n.n

(7.240)

which can be substituted in the right-hand side of the differential equation (7.237). Obviously, like !(1,p), !(o,p) is also an odd function ofp and can be written as a power series with unknown coefficients: (7.241) The coefficients can be solved by equating the factors of equal powers of p on each side of the differential equation (7.237), which gives us a recursive formula. The second derivative gives rise to a delta discontinuity: (7.242)

240

CHAPTER 7. EXACT IMAGE THEORY

Equating the coefficients of the delta functions gives us first

(7.243)

Further coefficients are obtained from the recursive formula

A

_ n+l -

a

An

+

a2 -1

+ I)! + 2)!n!'

(-1)n(2n

a 2 -1 22n+l(n

(7.244)

whose solutions start with

A __ a(a2 + 4a + 5)

A _ a(a+3) 1 -

4(a+ 1)3'

A _ a(50 3 3 -

8(a

2 -

+ 1)4

+ 250 2 + 470 + 35) 64(0 + 1)5 .

'

(7.245)

Asymptotic expansion To write an expansion for the image function f( 0, p) valid for large p values, we must first take out the term corresponding to the pole of the reflection coefficient. In fact, writing (7.226) in the form

_ _ 1_(f.q- #+1)2 R(a, q) - 2 1 f.

-1

q2 - f2-1

(7.246)

shows us that there are poles at 1

qp ±Vf2=1' ---

(7.247)

which correspond to exponential image functions. With the convention ~{Vf2=1} > 0 the pole qp = 1/ Vf2=11eads to a converging image, while

7.3. THE SOMMERFELD HALF-SPACE PROBLEM

241

the other one can be shown to be cancelled by a zero of the numerator.

0.3r------r------r----~-----r----~------.,

----+----+-----+-._---+----~

0.2t - - - + - f - - - \ - - t - - - - - - t - - - - - - + - -

0.1 t--ff-----t-~---__t_~--_t----+_---_f_---~

0.05H----~----+-~---t--·~-=::::--t-------+----~

10

5

15

20

25

30

Fig. 7.4 Examples of the image function !(a,p) (absolute value) for different complex values of the parameter

Q.

Extracting the pole leaves us with an expression for the reflection coefficient function that has branch points at qb = ±j. The image function can be obtained by applying the Mellin inverse and Watson's lemma as was done by LINDELL and ALANEN (1984, part II) to produce an expression valid for large p values:

-p/~ 4 . ( 1r') f( ) ~ - (a 2 2a2 a,p _1)3/2 e + /'iiap3 / 2 sm P+"4 -

3 3(3a - 8) ( 1f) - . . . . . . . ....--cos P + 3

2/'iia p5 / 2

105(91a6

4

-

15{23a4

472a4 + 896a 2 . IiC 256y 21fo? p9/2 -

-

144a2

+ 128) SIn . ( 1f) p+- +

32V2ia5 p7 / 2

-

512)

( 1f) cos P + - ... 4

4

(7.248)

CHAPTER 7. EXACT IMAGE THEORY

242

Fig. 7.5 Approximate values of the image function f(a:,p) for a: = 10, as calculated from the 3 term Taylor expansion and 3 term asymptotic expansion, compared with the 15 term Bessel expansion, which may be considered as exact for p < 30. In Fig. 7.4, values for the function 1(0'., p) are seen for different values of the parameter a. For real a the function is real, for complex a, only absolute values are given. It is seen that for large 10'.1, the oscillations in the function are reduced and the exponential term in (7.248) is dominant.

TE and TM image functions The relation between the image function f(a,p) and the functions fTE((), fTM (() defined in the Section 7.1 can be found if the integral expression (7.233) is compared with (7.69), (7.70). In this case we have

H(() == (,

(7.249)

and the relation between the integration parameters p and ( is pq ==

{31P B

==

. a /:

JfJl~,

(7.250)

or (7.251)

243

7.3. THE SOMMERFELD HALF-SPACE PROBLEM

O.02r-----y----r---:=---y----y-----r---------, O...------+-----+-+----~-+__+~--f"'o...-~~'7"_i_-~~----"7'I

-O.02H-----t------f--+--O.04H----t----+---+-----+--~~---+------+--------4

-0.06 . . - - + - - - - + - - 1 - - - - + _ _ - + - 7 ' - - -..- - - - + _ _ - - - - + _ _ - - _ _ _ _ 4 -O.08~---+-+---~'f-----+----+----+---

-0.1 t - - + - - - - + - - # - + - - - , + _ _ - - - + _ - - - + - - - - - + - - - - - - - 1

p

15

10

5

20

25

30

Fig. 7.6 Approximate values of the image function f(a,p) as calculated from the Bessel expansion with different numbers of terms. The accuracy of the Bessel series breaks down at approximately p = 2N, where N is the number of terms in the series.

Thus, we can write for the two image functions fTE and expressions fTE() = jBf(JL,.iB() + M6+() =

_~ fn (J.L-l)n + J..l2 - 1

n=l

fTM ( ) =

~~n t2

-

1 L..J

n=l

1

J..l

+

(

p.-lfy ((), JL + 1 +

-jBf(€,jB() - E6+()

(€ - l)n J t

J 2n (j B( )

+1

2n (j B ( )

C

_

€ -

t

1

the

(7.252)

=

fy (().

+1 +

fTM

(7.253)

The third image function fo() can be obtained directly through the definition of R o {(31): (7.254)

(7.255)

244

CHAPTER 7. EXACT IMAGE THEORY

It is readily seen that the fo«() function is invariant in the duality transformations, while f( €, p) and f(J.L, p) transform to one another. It can also be verified, with some effort, that the differential equation (7.87) for foe(), valid for any layered structure, is really satisfied in the present Sommerfeld half-space problem case. To prove this, we must invoke the differential equation (7.237):..The dyadic image operator fe' needed in the construction of the image source, can be written in any of the forms (7.90)-(7.92), for example, according to (7.92) as

=fe(J\lt, . ()

= f

TM

«()uzu z + I

TE=

It

+ lo«() k12 u. u, . vv..

(7.256)

Making a partial integration in the field integral, it is possible to transfer z differentiation u, . \7 to ( differentiation, whence it is possible to write instead of (7.256) (7.257) which appears to be a very useful form, because it involves only one differentiation of the current source J. The derivative of the function fo«() can be found analytically.

Convergence Since the Bessel function In(p) is known to converge for Ipl -+ 00 only for real values of p, to obtain a converging image, the integration variable ( should be chosen so that the parameter p in (=L.= P jB jk1VJ.Lf-l

(7.258)

is real. Thus it is wise to keep p as the integration parameter since p = 0... 00 defines the image line. When B is not imaginary, ( becomes complex, which means that the image line is defined on the complex ( plane by

z

= -h -

p

(= -h - jB'

(7.259)

The image source and the reflection field The dyadic image operator fe(j\lt, () is now known and the expression for the image source in the Sommerfeld half-space problem can be expressed in terms of the image functions and the original source J (r) in the form (7.260)

7.3. THE SOMMERFELD HALF-SPACE PROBLEM

245

whence the reflection field can be written as 00

Er(r)

= -jk l 171

(1 + :? vV').J JG(r-r' +Uz()Ji(r', v

()dV'd(. (7.261)

0

This is the general solution for the reflected field and includes all special cases of sources, medium parameters and field points. For example, a VED source (vertical electric dipole above a horizontal interface) J (r) = uzIL8(r - uzh) gives rise to a vertical image source: (7.262) Because the field is everywhere transverse magnetic, only the function fT M appears. On the other hand, a HED source (horizontal dipole) J(r) = vIL8(r - uzh) with v· u, = 0 gives rise to both horizontal and vertical image currents:

IL

Ji(r, () = v fTE (()I tsi» )o(z + h) - uzf~(() k2 lv- V'o(p)]o(z+ h). (7.263) 1

The vertical component is a double line current, like a transmission line, because of the term v· V'6(p).

Limiting cases of the theory Let us consider some limiting cases to test these results. For e ~ 1 and J.L ~ 1 we obviously have E ~ 0, M.~ 0 and B ~ 0, whence fTM«() ~ 0, fTE«) ~ 0 and fo«) ~ O. Thus, the image source vanishes, which is in accord with the vanishing interface. Letting € ~ 00 and J.L = 1, we have fTE«() = jBf(l,jB(), which for oo B ~ k 2 ~ 00 acts like 6+ «) f(l, p)dp = -8+ «) in the field integral. Also, we have fTM «) ~ -b+«) because the continuous term in (7.253) vanishes. Further, we have fo«() --+ 0 for the same re~on, whence th.:

fo

dyadic operator from (7.256) can be simply written as le(jV't' () ~

-1

and the image source becomes Ji(r, () ~ -Jc(r) = -C . J(C . r). This is obviously the image in a PEe plane. For large but finite €, we can take the first term of the asymptotic expansion (7.248) to approximate fTM = -8+(() - jBj(€,jB()

_ o+(()

~ -o+(() + (€22jB€2 e- j B(!V,2- 1 ~ _ 1)3/2 + 2jB e- j B( / , . e

(7.264)

CHAPTER 7. EXACT IMAGE THEORY

246

This is similar to the impedance-surface function (7.163), if we identify Zs = VJl2/€2, which is a small quantity. In the same way, fTE and I; functions can be checked for the same result. This shows us clearly, that at the limit of large e the interface can be approximated by an inpedance surface. Finally, the reflection-coefficient method (ReM) can be shown to arise as the far field limiting case of the EIT formalism. This was already done in general form for the image Green dyadic in Section 7.1 and is not repeated here.

Green functions The different Green functions defined in Section 7.1.4 can be expressed in terms of just one scalar Green function K (a, o, z) for the Sommerfeld half-space problem. In fact, defining

J 00

K(a,p,z)

=

G(D)f(a,p)dp,

(7.265)

o with (7.266) we can write

K™(p,Z)

= -K(€,p,z) -

KTE(p, z) = K(J-t, p, z)

EG(r),

(7.267)

+ MG(r),

(7.268)

(7.269) The Green function can be computed for different argument p, z values and stored for computer memory instead of working with the image functions. For large arguments we can approximate as in Section 7.1.4, which results in the limiting case of the Green function

K(a,p,z)

--t

(

z

a-I) .

G(r) R(a, JJU=l) + - r

JL€ - 1

a

+1

(7.270)

7.3. THE SOMMERFELD HALF-SPACE PROBLEM

247

0.1 /" ........ ",

0.05 , I

~

I I

o

~j "'"\?~~ V "<, ....

..

It I / ,

...... _-,.-"

.........-..--

..--.

~-,

~F"""

'",,""-,,"

~---

-0.05

kz=.1

-0.1 -0.15

4

2

8

6

10

12

14

16

18

20

Fig. 7.7 Real (solid) and imaginary (dashed) parts of the Green function K(f.,p,z) for

f.

= 2 and kz = 0.1, together with the asymptotic approximations (curves with larger amplitudes).

7.3.3

Antenna above the ground

As an example of applying the reflection image theory, let us consider a thin horizontal ),,/2 dipole antenna above the ground with the purpose of finding the effect of the ground on its impedance. The general problem is quite involved and could be treated through an integral equation for the surface current of the antenna, in which the effect of the ground is taken into account in terms of a second Green dyadic. Let us, however, simplify the problem by assuming a known sinusoidal current distribution for the dipole, which is a fair assumption if the dipole is thin enough and not very close to the ground. Thus, the original source is (7.271) and zero elsewhere. Knowing the total field E(r) at the dipole, the impedance can be computed from the stationary functional

Z

= - J;

J

E· JdV

V

>"/4

=-

J J E(x)· o

->"/4

u; cos(k1x)dx.

(7.272)

248

CHAPTER 7. EXACT IMAGE THEORY

Writing the total field in terms of the incident and reflected fields E = E, + E r , the impedance functional can be written as Z = Zi + Zr, where Zi is the impedance of the dipole in free space and Zr represents the effect of the ground. To compute Zr, E can be replaced by the expression (7.261) for E r in (7.272) in terms of the image source applying (7.257)

= fe(j\l t, () · Jc(r) =

Ji(r, () [fTE«()u:c1o COS(klX)

+ f~«()uz ~: sin(k 1 x)]8(y)8(z),

(7.273)

valid for [z] < >../4. (7.272) can thus be written as A/4

A/4

J JJ

z: = jk1'f/l

00

G(D) [fTE«() cos(k1x')cos(k1x)+

-A/4 -A/4 0

fi:() J JJ

sin(k1x') COS(k1X)] d(dxdx'

>-/4 A/4

jk1'f/l

00

G(D)

-A/4

[!€ fTM «() +

=

[f TE«( ) cos(k1x) cos(k1x')+

->-/4 0

f €

1 8+«()] sin(k1x)sin(k1X')] d(dxdx'.

(7.274)

For the last expression we have made partial integrations and applied (7.87) for the expansion of f~' (() to obtain

:2 f:«() = - fo«() + fTM «() -

fTE«() =

1

_! fTM «() _ f

f f

1 8+«( ) _ fTE«().

(7.275)

At this point it is wise to change the integration variables from x, x' to and r = x - x', The distance D does not depend on the parameter ~ at all: D = J(x - x')2 + (2h + ()2 = Jr 2 + (2h + ()2. (7.276)

e = x + x'

Writing 2 cos(k1x) cos(k1x') = COS(kl~)+cos(kl r) and 2 sin(k1x) sin(k1x') = - cos(k1e) + COS(k1T), the parameter e can be integrated out in (7.274). Denoting e = k1r we have 001r

Zr=jZ~JJ G(D) [fTE(SinO + (1r-O)cosO)+ o

0

249

7.3. THE SOMMERFELD HALF-SPACE PROBLEM

1

[- fTM f

e- 1 + -6+]( -

sinO + (1r - 0) cosO) ] dOd(.

f

(7.277)

Values for this integral are quite easily computed for different values of complex € and height h. The ( integration can also be performed separately and the final result expressed in terms of the image Green function K( f, 0) as defined in (7.265) and in LINDELLet al. (1985):

z. = j2k~1

J 1r

+ (11" -

[K(l,O)(sinO

0) cos 0)+

o 1 [-K(f,O) €

€ - 1 + -G(Do(O))](-sinO + (1r f

0) cosO) ] dO,

J

(7.278)

00

K(f,O) =

G(D)j(f,p)dp,

(7.279)

o

klD = J02 klD o = k1luz(x - x')

+ (2k l h + k1( )2,

+ u z 2hl =

J02

+ (2k 1 h )2.

(7.280)

For large k1h values, (7.278) can be shown to give the limiting value -j'f/l

V€ -

1

Z -+----e r 21rk1hV€+1

-j2k 1 h

(7.281)

'

which can also be directly obtained by applying the reflection-coefficient method. 7.3.4

Transmission coefficients

Considering the problem of wave transmission through a planar interface between two isotropic media 1 and 2, the transmission coefficients can be written in the form (7.282)

TTE

= 1 + RT E =

2p,{J1 J-t/31

T. = o

kr (TTM _ TTE) = K2 kr (R™

K2

(7.283)

+ /32 '

_ RT E ) = R

o·

(7.284)

CHAPTER 7. EXACT IMAGE THEORY

250

The transmission image s~urce can be written in terms of the dyadic transmission image function he (7.115), which can be found if the transmission dyadic is first written in the form

(7.285) and then in not exactly the same but the equivalent form in the sense that it produces the same transmitted field:

(7.286) where, denoting q

= {31/B

T«

and applying the definition (7.226), we have

= fT TM = €f32€{32 f3 = f[l 1 + 2

Ri«, q)],

(7.287) (7.288)

(7.289) Note, again, the difference between To and To. Only the latter is needed here. The expressions (7.287)-(7.289) will be expanded in terms of powers of the reflection parameter r as given above, to find the transmission image functions needed for the construction of the transmission image source.

7.3.5

Transmission image functions

The transmission image source corresponding to a volume source J(r /) was shown in Section 7.1 to be of the form

(7.290) giving the total transmission field in the form

JJ 00

ET(r)

= -jk2 172

G(D) · Ji(r', ()dV'd(,

(7.291)

o v

D

= V(p -

p')2 + (z - H(z', ())2.

(7.292)

7.3. THE SOMMERFELD HALF-SPACE PROBLEM

251

The dyadic transmission image function he(K, h, () and the function H(z, () are defined through the integral expression

T

j f31h e e-

00

Jh (K e

"h

()e- j f32 H ( h "

) d ~, l"

(7.293)

o

with the definitions (7.115)-(7.118), written again for convenience,

(7.294)

J J J 00

T"e- i 13t h

=

hu(h, ()e- i 132H{h,Od(,

(7.295)

h[(h, ()e- i 132H{h'()d(,

(7.296)

ho(h, ()e- i{32H{h,Od(.

(7.297)

o

00

T[e- i 13t h =

o

00

Toe-il3th =

o Because of the exponential factor exp(-j{31h), an integral identity of a more general type than that applied for the reflection problem must be found. As such we adapt an identity which can be found, for example, in the reference GRADSHTEYN and RYZHIK (1980, equation 6.646), and through a change of variables can be written in the form

(7.298) where ( is the integration parameter, r = ({31 - (32)/«(31 parameter and H denotes the function H(h, ():

,

d

H = d(H

()

n.;

+ ..82) the reflection (

= H(h,()

(7.299)

Let us now define a transmission image function F( a, h, () through the integral expression

(7.300)

CHAPTER 7. EXACT IMAGE THEORY

252

From the expansion (7.231) applied in (7.298) and (7.300), the transmission image function can be written explicitly in the form 20'

F(O', h, () == -Jo(B()0'+1

0

24: 1

~ ( - : : ~) (~: ~) n

n

J2n (B() .

(7.301 )

By making partial integration in (7.300), we can write

J(0~16+(()+F'(0,h,(») 00

e-

j 2H {3 d(

.

(7.302)

o

In these expressions, the prime denotes differentiation with respect to the parameter (. (7.302) can be applied directly to find the image functions corresponding to the transmission coefficients T u and T] defined by (7.287), (7.288):

hu(h,()

= ~6+(() + F'(f,h,(), €+1

h/(h,() == _2_ 6+(() + ~F'(/-L,h,(). J.L + 1 J.L

(7.303) (7.304)

The third coefficient To defined by (7.289) corresponds to the image function ho which can be written with the aid of (7.300) as follows:

ho(h,()

= jH'(h, ()~[F(/-L,h, () j-t-f

F(f, h, ()].

(7.305)

Thus, substituting in (7.294), the transmission image source for the general current source J (r) is finally obtained in the form

2 ( --b+() J.L + 1 I

J.l€ - 1

1 (J.L, z, () + -F I

J.L

)=ItJ + t

H (z, ()--[F(J.L, z, () - F(€, z, ()]u z(\7 t . J). J.l-€

(7.306)

7.3. THE SOMMERFELD HALF-SPACE PROBLEM

253

The transmission field can now be computed from the expression

JJ oo

ET(r)

= -jk2 Tf2

G(D) Ji(r',()dV'd(, 0

(7.307)

o v

D= V(p-p')2+(z-H(z',())2.

(7.308)

It is noted that, as for the reflection case, a VED source gives rise to a vertical (u, directed) image source while a horizontal source brings about both horizontal and vertical image components, unless the horizontal source is solenoidal (V' . J = 0), in which case the vertical component vanishes. Corresponding expressions for a magnetic source can be directly written applying the duality transformation. Asymptotic expression

Similarly to the reflection field case, we can write a simple asymptotic far field expression for the transmission field starting from the approximation

D r

~r

+ u , . [p' + uzH(z', ()] =

+ r' sin fJ sin fJ' cos(

+ H(Z', () cos fJ.

(7.309)

Here, we have applied spherical coordinates to the source and field points.

z

h

Fig. 7.8 Geometry of far field radiation through an interface. Inserting (7.309) into (7.307), with the far field approximation for the Green dyadic, leads us to

JJ oo

ET(r)

~ -jk2 Tf2 G (r ) ·

eik2UrO[P'+UzH(z',O]Ji(r',()dV'd(,

o v

(7.310)

CHAPTER 7. EXACT IMAGE THEORY

254

in which (7.290) can be substituted. The parameters of the Green dyadic are those of the medium 2. Comparison with (7.293) gives us the field in terms of the transmission dyadic:

ET(r)

~ -jk2 1J2 G(r ) .

f

eik2U"'P'Te(K)e-i{3tz' . J(r')dV',

(7.311)

v with

(7.312) (7.313) (7.314) (7.315) The expression (7.311) is the field arising from an image source whose amplitude is the original one multiplied by the transmission dyadic, which depends on the angle ()2 of the field point. An interpretation is obtained by considering the transmission from a point source J(r) = v I L6(p )6(z - h), for which we can write the expression (7.311) in the form (7.316) Writing (7.317)

D

= v(r -

uzh') . (r - uzh'),

h' = {31 h = tan O2 h, /31 tan ()1

(7.318)

we see that, actually, the position of the apparent image source can be interpreted geometrically as the point on the z axis from where the ray appears to be emanating. It has been demonstrated by this author (LINDELL, 1988) that the original source may also be in complex space and yet the image theory is valid. In fact, representing a Gaussian beam by a point source in complex space, the well-known Goos-Hanchen shift at the interface reflection can be seen to arise together with the correct transmission beam.

255

7..3. THE SOMMERFELD HALF-SPACE PROBLEM

7.3.6

Radiation from a loop antenna into the ground

As an example of the transmission image theory, let us consider a horizontal loop antenna in the air above non-magnetic ground with J.t = 1. Assuming the loop perimeter 21rb to be equal to one wavelength in the air, or k1b = 1, and the wire thin enough, the current can be approximated by the sinusoidal function

J(r)

= utplo cos
b)6(z - h).

(7.319)

The exact transmission image current from (7.306) is then

Ji(r, () = (6+«() + F'(l, z, ()]J(r)H' (z, ()[F(l, z, () - F( e, z, ()]u z [V' . J(r)] = I o6(z - h)6(p - b)

[[6+(() + p' (1, z,

H' [F(l, z, () b

()]u
cos ep-

- F( e, z, ()]u z sin cp ] .

(7.320)

When considering radiation into the ground, the far field approximation can be obtained from (7.311) with the same symbols

ET(r)

~ -jk2 T/2 G (r ) .

Jejk2U~.pITe(K)e-j{31h

. J(p')dS'.

(7.321)

s This gives rise to the following radiation field expression:

(7.322) with

JejKbcos(
A

= (J -

uru r) .

T

e •

cos ep'dep',

(7.323)

o

K

= upK,

K

= k2 sin(J2.

(7.324)

The integral can be evaluated after some algebra in the form

JejKbcos(
cos ep'dep' =

u
epJ~ (Kb) + u p21r sin ep Jl~b)

o

(7.325)

CHAPTER 7. EXACT IMAGE THEORY

256

by invoking the integral identity

J 21r

ejz cos
= 2j n 1f'J n(z).

(7.326)

o

Thus, we can evaluate, writing the transmission dyadic T e from (7.285), 2 with {32 = k2 cos ()2, {31 = k1 € sin ()2:

VI -

(7.327)

O.9t-----'~-+----+----+---+------+----+-·---+---+---~

eps,lon=81

0.8 ~-+---+----t'-~r-----+---+---___+--

0.7 t - - - . . - - + - - + - + - - - - + - - + - - - - - + _ _ - - + - - - - + - - - _ + _ - - + _ - - - - 4 0.6 t-----ttt-+---t--,,-.~--1H----+__--+----_t_

0.5...-----+lrI-+--+_#_. 0.4 ~~H++-_++___+_--o\-+-+--_+___t'---___+-~_+_--_+_--+__-~ 0.3 t---Htt----+,f---r---+-~-~-__++_--_+_ 0.2 t----+-'A'--II--I--+--~-+--H-__f+_7'_--+--~-+_~-_+_--+_-____i

10

20

30

40

so

60

70

90

80

Fig. 7.9 Radiation pattern for a one-wavelength loop in air above the dielectric ground with e = 81, at different values of the height h.

The radiation pattern is obtained by normalizing the transmission field by the field amplitude at (}2 = 0: (7.328)

7.3. THE SOMMERFELD HALF-SPACE PROBLEM

257

whence we have the radiation pattern defined by the vector

F( O2 , cp) = e- j({31-k1)h (y'f +

Vf

1) (U

f32

cos
2€filfi2 J1(}(b)) Uo k 2( ff31 + f32) Kb .

(7.329)

The radiation pattern in the xz plane, with cp = 0 is shown in Figs. 7.9 and 7.10 for some values of k1h and €. It is seen that the larger the permittivity e, the narrower the transmitted radiation. The nulls in the radiation pattern correspond to nulls of the function J~ (y'fO). 1

0.9

l/\ \ \\,

\'0~r-------

0.8

'\ \ \~ H \ 1 \ \

0.7

\

0.6

\

0.5 0.4 0.3

0.2 0.1

~o

\

sr

\ I~O

\ \

\ \

kh kl :

i\

\

1\

\

\ \ \ \ '\

-,

'~

\

<,

I'---..

"r-,

I\~~r--,,- ...... """'--'"'~ .. ~e-

10

20

30

40

50

60

--UleTa

70

80

90

Fig. 7.10 Same as in Fig. 7.9, but the height h is kept constant, k1h = 1, and e is varied from 2 to 80.

7.3.7

Scattering from an object in front of an interface

To demonstrate the application of the EIT in formulating a problem let us consider a scatterer in air in front of a half-space of isotropic medium. The incident field E, (r) at the scatterer is the field from the original source plus its image due to the half-space without the effect of the scatterer and it is assumed to be known from the previous analysis. For a distant source, the incident field is approximately a plane wave plus its reflection from the interface. For simplicity, let us assume that the scatterer is a dielectric object located at position r = uzh and sufficiently small so that its scattered field

CI-IAPTER 7. EXACT IMAGE THEORY

258

can be approximated by the field arising from a dipole with the dipole moment (7.330) where a, a dyadic, is the tensor dielectric polarizability. Explicit expressions for Q are available for isotropic and anisotropic spheres and ellipsoids in the literature. E t is the total field at the location of the scattering object: the sum of the incident field E; and the reflected field due to the polarized dielectric object itself. The last component is unknown and EIT can be applied to its computation. The goal is to solve for the unknown dipole moment vector p. The corresponding current density vector can be written as

J(r)

= jwp8(r -

uzh).

(7.331)

The image of this dipole due to the interface can be written as (7.332) The mirror image of the current dipole is

(7.333) The total electric field at the object, needed for the determination of the dipole moment, can be written as

(7.334) where the field E s from the image of the dipole is

Es(r) = -jwJ-to

JJ

= G(r - r I

+ u z( ) . Ji(r " ,z)dV d(,

(7.335)

v c

where V denotes the integration volume of the mirror image and C is a line in the complex ( plane ranging from the origin to infinity in such a way that the image function integrals converge, i.e. along a path such that arg{(}

= -1r/2 -

arg{ ~}.

(7.336)

Applying the reflection Green dyadic:

=K(r) =

(=I + k V'V' 1

2

) K TM (r)

+ k12 (uzuzxx V'V')Ko(r),

(7.337)

7.3. THE SOMMERFELD HALF-SPACE PROBLEM

259

the field from the image source can now be expressed more concisely as

Es(r)

= -jWlto

J

K(r - r') · Jc(r')dV'.

(7.338)

v Inserting the proper mirror image current function from (7.333) we have (7.339) which contains the unknown dipole moment vector p, for which a final equation can now be formed. In fact, combining (7.330), (7.334) and (7.338) for the point r = uzh, we arrive at an algebraic equation for the moment vector: (7.340) This has the solution (7.341) Since all the terms on the right-hand side are known, it is possible to compute the moment vector for any exciting field E i . The only limitation for the applicability of this result is the dipole approximation of the scatterer, which may be poor if the object is located too close to the interface even if it is valid in free space. However, for small objects the range of validity probably extends to distances from the interface comparable with the object size. Computations have been made by LINDELLet ale (1991). References

M. and STEGUN, I.A. (1964). Handbook of mathematical formulas. Dover, New York. GRADSHTEYN, I.S. and RYZHIK, I.M. (1980). Tables of integrals, series and products. Academic Press, New York. LINDELL, I.V. (1988). On the integration of image sources in exact image method of field analysis. Journal for Electromagnetic Waves and Applications, 2, (7),607-19. LINDELL, I.V. and ALANEN, E. (1984). Exact image theory for the Sommerfeld half-space problem, Part I: Vertical magnetic dipole. IEEE Transactions on Antennas and Propagation, 32, (2), 126-33. Part II: Vertical electric dipole. IEEE Transactions on Antennas and Propagation, 32, (8),841-7. Part III: General formulation. IEEE Transactions on Antennas and Propagation, 32, (10), 1027-32. ABRAMOVITZ,

CHAPTER 7. EXACT IMAGE THEORY

260

LINDELL, I.V., ALANEN, E. and BAGH, H. VON (1986). Exact image theory for the calculation of fields transmitted through a planar interface of two media. IEEE Transactions on Antennas and Propagation, 34, (2), 129-37. LINDELL, I.V., ALANEN, E. and MANNERSALO, K. (1985). Exact image method for impedance computation of antennas above the ground. IEEE Transactions on Antennas and Propagation, 33, (9), 937-45. LINDELL, I.V., SIHVOLA, A.H., MUINONEN, K.D. and BARBER, P.W. (1991). Scattering by a small object close to an interface. I: Exact image theory formulation. Journal of the Optical Society of America A, 8, (3), 472-6. SOMMERFELD, A. (1909). tiber die Ausbreitung der Wellen in der drahtlosen Telegraphie. Annalen der Physik, 28, 665-736.

7.4

Microstrip geometry

As another geometry let us consider the substrate of the microstrip consisting of a dielectric slab backed by a PEe plane, forming the basis for microstrip circuits and antennas. The microstrip was introduced as a lightweight, inexpensive and easily fabricable circuit and antenna structure in the early 1950s and its significance has never decreased. Despite its simplicity, the structure has been a challenge to scientists in electrical engineering for decades, because a simple method exact enough for the computation has not emerged. The microstrip can be analysed with the EIT method after finding the proper image functions. To find the fields in the air above the structure, the dielectric slab with the ground plane is replaced by the image of the original source. Thus, for example, integral equations arising in microstrip problems can be formulated with these image sources. Defining Green functions for the structure gives an added advantage of reducing the number of integrations in field calculations. However, the Green functions must be stored in the computer in numerical form.

7.4.1

Reflection coefficients and image functions

Consider the typical microstrip geometry: planar interface at z = 0 between a homogeneous half space (air) z > 0 with the permittivity f o and a dielectric layer of thickness d and permittivity fE o above a perfectly conducting plane at z = -d. Permeability is assumed to be JLo in the whole

7.4. MICROSTRIP GEOMETRY

261

space.

Fig. 7.11 The microstrip geometry: dielectric slab with ground plane. Reflection coefficients

The reflection coefficients at the dielectric interface z = 0 of the microstrip structure can be derived from the transmission-line analogy, because the impedance dyadic at the interface, in Fourier space, is obviously (7.342) with Z2 denoting the characteristic impedance dyadic of the dielectric medium as given in (7.10). (7.38) gives us the reflection dyadic with the TE and TM eigenvalues TE

R R™

j /31 tan /32d - /32 r T E - e- 2j /32d = j(31 tan(32 d + (32 = 1 - rTEe-2j(:J2d' = r™ _e- 2j /32d €{31 1 - r T M e- 2j /32d ·

= j/32 t an/32d-f.(31 j (32 tan {32d

+

(7.343) (7.344)

Here we denote the partial reflection coefficients corresponding to the interface reflection by r T E ., r™ and define TE

r

(31 - (32

= (31 + (32 = r,

(7.345)

Note that r T E equals the reflection parameter r of Section 7.3. The third reflection coefficient R; can also be easily derived:

(7.346)

CHAPTER 7. EXACT IMAGE THEORY

262

with

k 21 r o = K 2 (r™

€

+1

- r T E ) = -f-(r™

+ E).

(7.347)

Here, again we denote E = (e - 1)/(€ + 1). r o given here equals what was denoted by R; in the Sommerfeld problem (7.57) with J-l = 1. The expressions given here can also be quite straightforwardly generalized for the case J-l f:. 1. Let us only consider the problem where the original source lies at the interface like a current in a microstrip circuit. To obtain the corresponding image functions, we should find the integral representations for the reflection coefficient functions of {31 in the form 00

R(fJd =

f !(()e-

j

(7.348)

(31( d (,

o

i.e., with H(()

= ( in

(7.57).

Transverse electric image function To find the function fTE((), the integral identity (7.298) must be written in another form:

f

00

;~ e- (32 = (~ ~ ~) m 12m(BV(2 j

h

h2)U+(( - h)e- j (31( d ( . (7.349)

o

Here, In(x) denotes the modified Bessel function. In forming the image function expressions the function Fn(h, () to be defined below appears helpful. It has no connection with the function F( a, h, () of the previous section. The definition is

Fo(h,()

=

Bh

V(2 - h 2

Fm(h, () = -

h(BV(2_h2)U+((-h),

2V(~ _ h2 [(( -

((+h)I2m_l(BV(2_ h2)]

(7.350)

h)I2m+l(BV(2 - h2)-

(~~~)m U+((-h),

m>O.

(7.351)

The identity (7.349) can be written in terms of this function as

f

00

r me- j (32 h

=

o

[Fm(h, ()

+ 8mo8+(( -

h)]e- j {31( d ( ,

(7.352)

7.4. MICROSTRIP GEOMETRY

263

where Dmo is the Kronecker symbol. This can be applied in finding the image function fTE((). First, we expand the total TE reflection coefficient as a Taylor series 00

R TE

= r + L [rn+ 1 - rn-I] e- i /32(2n d ) ,

(7.353)

n=l

which can be interpreted as a sum of partial waves reflecting between the interfaces in the microstrip. Applying now the integral identity (7.352), we may write the image function as an expression involving the newly defined function Fn(h, (): 00

fTE(() = -0+(( - 2d)

+ F1(O, () + L[Fn +1(2nd, () - Fn - 1(2nd, ()]. n=l

(7.354) The result can also be expressed as a Taylor series by expanding the modified Bessel functions as 1 (x)n+2k L k!(n + k)!"2 ' k=O 00

In(x) =

whence the function Fn(h, () can be written for n

B Fn(h, () =

-"2

(7.355)

> 0 as

x k [y2n+k+1 y2n+k-l] (2n + k + 1)! - (2n + k _ 1)! U+(( - h),

t; kT 00

(7.356) with

B(( + h)

x=---

2

'

B(( - h) y= --2--·

(7.357)

The representation of the image function fTE(() as a series of modified Bessel functions In' individually divergent, may seem non-physical, but the sum function turns out to converge. However, such an expression may pose problems to the computer, because numbers of increasingly large values are being added and subtracted. It is wise to start summing from the large

264

CHAPTER 7. EXACT IMAGE THEORY

values of indices, i.e. from the smallest values of the terms.

lID

o

-e:=3.o

co

··..········£=2.2 ------e:=1.5

o

.0

0.5

d

= o.ix,

'.0

J.3

("II

ci I

.,. ci I

co

ci I

co

ci I

Fig. 7.12 Normalized TE image function for the microstrip for different relative permittivities

f

of the substrate.

Transverse magnetic image function When finding the TM image function the duality transform for the TE function does not work since we assumed the special case J.-L = 1. Thus, fTM must be evaluated from the start. Writing the reflection coefficient (7.344) as a Taylor series

= r™ + L 00

R™

[r™(n+1) _ r™(n-l)] e- j t32(2nd) ,

(7.358)

n=O

we may apply the identity (7.352) if the reflection coefficient r T M is expressed in terms of r T E = r: T

TM

E+r = ----.

(7.359)

1 +Er

Now we need the expansions n

(1

(1 - x )

+

n x)

=

,

n. i LJ ( _ ')'.'x , i=O n z u:

(7.360)

"'"

_ n _ ~ (n - 1 + k ) ! k - L...- ( _ )Ik l x, k=O n 1 ..

n

> 0,

(7.361)

7.4. MICROSTRIP GEOMETRY

which both converge for

265

Ixl < 1.

Thus, we may write

(r™)n

= L Cr(E)r k,

00

(7.362)

k=O where the coefficient functions are polynomials defined by

C n( ) = k

X

~ (-I)i n (n + k L..J i=O

i-I)! (_ )n+k-2i x , n > 0,

(7.363)

(n_ ~.)'.~..'(k _ ~.)'.

c2(x) = bOk.

(7.364)

The symbol nlk here denotes 'smaller of the numbers n, k'. The coefficient function C;:(x) can also be defined through the expansion 00

= L Cr(tanh 0) tanh k 4>,

tanhn(O - 4»

(7.365)

k=O and it obeys the following rules

= nC~(x),

kC;:(x) C~(x)

= (_x)k+l

n, k

- (_x)k-l,

> 0,

(7.366)

k > 0,

(7.367)

C[;(x) = (_x)n,

(7.368)

Cf(x) = n[(_x)n+l - (_x)n-l].

(7.369)

With this function we can write the reflection coefficient function in the following form 00

R™

=

00

00

L C~(E)rk + L L [C;+l(E) - C~-l(E)] r ke- {32(2nd). j

k=O

n=lk=O

(7.370) Now applying the integral identity (7.352) we finally have an expression for the TM image function: 00

fTM «()

= C~(E)6+«() + L

[C:+1(E) - C:- 1 (E )] 6+«( - 2nd)+

n=O

00

00

00

L Cl(E)Fk(O, o + L L k=l

n=lk=l

[Ck+1(E) - Ck-1(E)] Fk(2nd, ().

(7.371)

CHAPTER 7. EXACT IMAGE THEORY

266

The image function fo The image function fo«() needed in the EIT method can be constructed much in the same way but with somewhat more complicated expressions. Writing again for the reflection coefficient R; a Taylor series expression

L 00

L 00

(rTEe-2iI32d)m

m=O

f:

(rTMe-2iI32d)n =

n=O

1

f

[e-2im,82d _ e- 2i(m+2),82 d]

x

m=O

m

L {rTE)m-i [E{r™)i + (r™)i+l] ,

(7.372)

i=O

we may again apply the identity if the reflection coefficient r™ is expressed in terms of the parameter r. The resulting image function is

fo(()

= f+1 fffD~(E)X f.

m=O k=l i=O

[Fm +k- i (2m d, () - Fm +k- i(2(m

+ 2)d, ()],

(7.373)

D~(E)

(7.374)

with

Di(E)

= Ect(E) + ct+ 1(E),

k

> 0,

= o.

These formulas are not too difficult to use in practice although the modified Bessel functions In(x) are diverging, because the sum functions can be seen to converge, except for simple exponentially diverging parts, which can be extracted and handled otherwise. In Fig. 13, examples of the

7.4. MICROSTRIP GEOMETRY

267

image functions are given for some parameter values.

-£=3.0 ··..········£=2.2

N

o

d

= o.ix,

_·_··-£=1.5

......

_._ ..-..~':::':"': __ 1.5

2.0

2f············3.0

".:::::::::•..•.:.::_

let'.

._""""""'"''''

4.0

3.5

1

. d I

CD

oI

Fig. 7.13 Normalized derivative of the image function 10 for the microstrip for different relative permittivities

£

of the substrate.

There is the connection (7.87) between the three image functions: (7.375)

which will save us from numerical differentiation of the fo function. The fTM(() and fo(() functions do not, however, converge as such. The lack of convergence is due to certain poles of the reflection coefficient, which correspond to propagating waveguide modes of the grounded slab waveguide, a situation similar to that of the impedance surface. Their contribution can, however, be extracted and made to converge by reorientating their location on the complex plane.

7.4.2

Fields at the interface

The image dyadic operator can be written in the many ways discussed in Section 7.1, for example, as =

.

le(JV't, ()

= I TE (()/= + fo(()uzu z · (/= + k12 VV),

(7.376)

1

which is perhaps the most preferable because it contains the simple function ITE. The reflected fields in the half-space z > 0 due to the image sources can be written as

JJ 00

Er(r)

= -jwllo

G(D)· Ji(r', ()dV'd(,

v

0

(7.377)

CHAPTER 7. EXACT IN/AGE THEORY

268

with

D == D(r, r', () == J(p - p')2 + (z -

Z, -

()2,

~{D} ::;

o.

(7.378)

Surface image sources Restricting the original current source on the planar interface z == 0, J (r) == J s (p)8(z), the transversal component of the electric field due to the image source represents the reaction of the grounded slab to the original field and is of importance for microstrip circuit analysis. This field added to the field due to the original source in free space gives the total transversal field, which is required for formulation of the integral equation of the original source problem. The formulation is made simpler by explicitly expressing the surface charge density

es

\7. J s

== - - . JW

(7.379)

in the image source and field expressions (7.64), (7.377)

Ji(r, ()

= !TE(()J s(p)t5(z) -

u, {~ !o(()l!s(p)t5'(z),

(7.380)

1

l!i(r, () = !TE(()l!s(p)t5(z) + :2!O(()l!s(p)t5I1(z),

(7.381 )

1

!!

00

Er(r) = -jwlLo

v

G(D)Ji(r',()dV'd(+

0

00

f~! !l\7'G(D))l!i(r', ()dV'd(.

(7.382)

v 0

Here, the integration volume V is any volume containing the original source surface S. Note that the double differentiation within the Green dyadic of (7.377) is reduced through partial integration in (7.382) so that only one differentiation of G(D) is present. This is necessary in reducing the singularity for the practical computation of the integrals. When taking the integration volume V large enough so that the original sources are zero on its boundary, we may perform further partial integrations with respect to z and ( variables so that the differentiations can be transferred from the delta functions to the functions of (. In this case, the equivalence 8/8z == a/a( can be applied when operating on the source function of both ( and r inside the field integral (7.382),

7.4. MICROSTRIP GEOMETRY

269

allowing us to replace (7.380) and (7.381) by the following equivalent surface image sources

Jsi(p, () = fTE(()Js(p) - u, {~ f~(()es(p),

(7.383)

1

(7.384)

The transverse electric field When calculating only the transverse component e of the electric field E, the u, directed terms in the expressions of J si can be omitted. The z integration in (7.382) can now be easily performed because of the delta functions and the transverse reflection field at the interface z = 0 reads 00

er(p)

= -jwJlo j S

j G(D)Jsi(p',()dS'd(+ 0

00

€~ j s

j['\1'G(D)]ei(p',()dS'd(,

(7.385)

0

with (7.386) For the double differentiation of the fo(() function needed in the above expressions, an analytic form is desirable, to save us from numerical differentiation in the actual calculation

with a new function defined as

Qn(x, ()

= F n +1 (x, () + 2Fn (x, () + F n - 1 (x, ().

(7.388)

CHAPTER 7. EXACT IMAGE THEORY

270

Denoting Do = Ip - p'l, the transverse total field in the same plane as the surface currents can also be written by defining two-dimensional Green functions:

e(p) = -jWJ.Lo j[G(D o) + KTE(Do)]Js(p')dS'+ s

~ f. o

j V'[G(Do) + L(Do)]es(p')dS'.

(7.389)

s

It is seen that the field can be expressed in terms of the free-space Green function G for the direct field and two new Green functions KTE and L for the reflected field. These Green functions can be calculated in terms of the image functions as follows: 00

KTE(D o) = j G(D)fTE«()d(

= -G(Dt}+

o

(7.390)

J 00

L(Do) =

G(D) [JTE«() +

:rf~'«()] d( = KTE(Do) -

EG(Do)+

o

00

j G(D)[QmH-i(2md, () - Qm+k-i(2(m + 2)d, ()]d(,

(7.391)

o

with (7.392)

Note that the definition of the Green function L can be expressed as K T M K; in earlier notation. In solving integral equations for the surface current problem it is not necessary to deal with the image sources at all if we directly compute the Green functions defined above and store them in the computer memory for further use.

7.4. MICROSTRIP GEOMETRY

7.4.3

271

The guided modes

As shown by LINDELLet ale (1987), the image function fo«() does not converge as ( approaches infinity owing to certain poles of the reflection coefficient. To obtain convergence, it is necessary to extract a finite number of terms, responsible for the loss of convergence, and treat them separately. Each of these terms corresponds to a propagating mode guided by the grounded slab. For a sufficiently small thickness d and/or if e is close to 1, there is only one such mode. This case is of particular interest to microstrip antenna design, because the guided mode increases coupling between the antenna elements, which is not generally what is wanted. If only one guided mode propagates, we can write the image function fo«() in two parts (7.393) with foe converging and fod exponentially diverging

(7.394) Here, a is the basic solution of the modal equation: (7.395) The amplitude coefficient A can be obtained from the residue of the reflection coefficient R o :

A_

j2€k~ad2[(€

- (t - 1)[(k1d)2

- 1)(k1d)2 - (ad?]

+ (ad)2][(t + ad)(k1d)2 + (€ + 1)(ad)3]'

(7.396)

Also, the corresponding Green function L(D o ) can be calculated in two parts: (7.397) with

:r J 00

Ld(D o ) =

J 00

G(D)f:d(()d( = ja::

o

G(D)eOl(d(.

(7.398)

0

Because this integral does not converge on the negative z axis, we must change the integration path in an imaginary direction by writing ( = -j(':

a:: J 00

Ld(D o ) =

G(D')e- jOl('

o

«.

(7.399)

CHAPTER 7. EXACT IMAGE THEORY

272

(7.400) Although the exponential term now oscillates and does not converge, the integral converges because the argument of the Green function exponent becomes imaginary when (' passes the value Ip - p/l. Thus, the determination of the Green function L(Dd) is defined in two converging integrations.

7.4.4

Properties of the Green functions

Singularities

Finally, the nature of the singularity of the new Green functions in the transversal field integrals can be questioned. Applying the Green dyadic expression, the double-nabla term is seen to be too singular to give the field within the surface current source itself, even with the definition of principal value integral, which is applied in volume current integrals by YAGHJIAN (1980). When one of the nablas is transferred by partial integration in front of the current function, the principal value can be defined. In fact, we may write for the limit of the field point r approaching the surface S from the normal direction n according to VAN BLADEL (1964)

lim j[VIG(D)]!(r/)dS I

r~s

s

= PV j[V"G(D)lf(r')dS' + !nf(r). (7.401) 2 s

Thus, the normal component of the field is discontinuous through the surface current. Considering only the transverse field component, we see that the principal value integral equals the limit when the field point approaches the surface. Thus, we may write for a point p on the surface S in the transversal field expression (7.385)

JJ ex>

e(p) = -jWJ-LoPV

G(D)Jsi(p',()dS'd(+

s :oPV

! 5

0

ex>

!l\7I G(D)]ei(pl,()dS'd(.

(7.402)

0

For the Sommerfeld problem with a planar interface between two homogeneous half spaces and the present microstrip problem, the same singularity properties can be seen to apply. Thus, the free-space Green function G in (7.402) must be replaced by the correct Green functions, taking into account the properties of the half space z < o.

7.4. MICROSTRIP GEOMETRY

273

Computation of Green functions For numerical computation of the Green functions, it is wise first to extract their asymptotic far field limit expressions, leaving simpler difference functions for raw computation. The asymptotic forms of KTE(p), L(p) are obtained from in the limit p --+ 00 by approximating the distance function (7.403) whence we may write

J 00

KTE(p) =

J 00

G(D)fTE(()d(

~ G(p)

o

fTE(()d(.

(7.404)

0

From the definition of the reflection coefficients (7.348) we have

J 00

fTE(()d(

= R TE({31) I

= -1.

(7.405)

~l=O

o

Since G(p) is singular at p = 0 whereas KTE(p) is not, we can write (7.406) because (7.407) approaches the same limit p as D. The function ~K vanishes quickly and thus takes little storage space. Likewise, we can write for the second Green function L(p) 00

L(p)

~ KTE(p) + :~G(p) J f~«()d(.

(7.408)

o Applying (7.375), implying f~'

= kr(fTM -

fTE - fo),

(7.409)

gives us 00

L(p)

~ G(p) JU™(() -

fo«()]d( = G(p) [R™

- RoL,=o

= -G(p).

o

(7.410)

CHAPTER 7. EXACT IMAGE THEORY

274

E; = 1.5 real part. d=O.1A

N

ci I

piA

•

cil-t-----,~---r----y---r-----,..---r---,....----., 0.0 1.0 2.0 ....0 J.O

Fig. 7.14 Microstrip Green function KTE(p) for two values of the slab thickness d normalized by the free-space wave length.

The second Green function can now be written

L(p) == -G(D 1 )

+ ~L(p),

(7.411 )

with rapidly vanishing term ~L(p). The waveguide mode term must be taken care of separately. For large p values we can write the limiting expression

(7.412)

In Fig. 7.14, examples of computed Green functions are given for some parameter e and d values. It is seen that the Green functions are unsensitive to the thickness d of the substrate at large distances p except for the waveguide term.

References ALANEN, E., LINDELL, I.V. and HUJANEN, A.T. (1986). Exact image method for field calculation in horizontally layered medium above a conducting ground plane. lEE Proceedings, 133H, (4),297-304. BLADEL, J. VAN (1964). Electromagnetic fields. McGraw-Hill, New York.

7.5. ANISOTROPIC HALF SPACE

275

LINDELL, I.V., ALANEN, E. and HUJANEN, A.T. (1987). Exact image theory for the analysis of microstrip structures. Journal for Electromagnetic Waves and Applications, 1, (2), 95-108. LINDELL, I.V .. NIKOSKINEN, K.I., ALANEN, E. and HUJANEN A.T. (1989). Microstrip antenna analysis through scalar Green functions. Annales des Telecommunications, 44, (9-10), 533-42. YAGHJIAN, A.D. (1980). Electric dyadic Green's functions in the source region. Proceedings of the IEEE, 68, (2), 248-63.

7.5

Anisotropic half space

To extend the image theory to problems with anisotropic media, let us consider the simple case of two homogeneous half-spaces of which one (1) is isotropic and the other one (2) uniaxially anisotropic with the axis normal to the planar interface. The dielectric dyadic of the medium 2 can thus be written as (7.413) In this special anisotropic case, the TE and TM fields do not couple to each other, which simplifies the analysis. As a practical problem involving this type of medium the sea ice may be mentioned, since it contains almost vertically elongated pockets of brine, which makes the medium approximately macroscopically anisotropic. The theory given in previous sections must now be modified since the parameter €2 is not a scalar. Let us very briefly consider the transmissionline analogy in the present uniaxially anisotropic case. From the Maxwell equations, the following Fourier-transformed vector equations for the transverse field components in medium 2 can be straightforwardly derived:

/

=

e2(K, z) = A 2 . (-u z x h2)

- u, x

h~(K, z) = B 2 · e2 -

1 + U z X j2m + --Kj2e,

(7.414)

W€2z

j2e -

_l_(u z

X

K)j2m.

(7.415)

WJ.-L2

Here, the dyadics A2 and B 2 are defined by (7.416) B 2 = -jW€2t

+ _J-uzuz~KK = -j=g2 . Y2.

(7.417)

WJ.L2 All these dyadics obviously have the same eigenvectors K, U Z x K as before, whence they commute in the dot product. Defining Y 2 to be the

CHAPTER 7. EXACT IMAGE THEORY

276

two-dimensional inverse of Z2, we can derive from the transmission-line equations the second-order equation for the transverse electric field:

with the propagation dyadic ~2 satisfying

({3fM)2 ~~ + ({3fE)2 U U12

KK

Z

.

(7.419)

The eigenvalues correspond to the two different propagation factors of the two eigenpolarizations TM and TE. They have the explicit expressions

{3fE = /k~t

- K2, (7.420)

with the two wave numbers defined as

k2 z

= WJJl2€2z,

(7.421)

The TE case appears simpler, because it corresponds to an isotropic medium with the parameters E2t, {to For the isotropic limiting case €2t ---+ E2z ---+ E2, these expressions obviously reduce to the earlier isotropic results. Finally, the impedance dyadic can be solved from

(

Z T M )2 KK

2

K2

+

~KK K2

(ZTE)2 UzU z 2

(7.422)

in terms of its eigenvalues ZTE _ 2

-

WJ.l2

f3'[E'

Z2T M

TM {32_ __ -

•

(7.423)

WE2t

The reflection image Let us now consider reflection from the planar interface of an isotropic medium with scalar parameters El, {t ~nd a uniaxial anisotropic medium with the dyadic parameters E2 = EEl, {tI. The relative permittivity dyadic

7.5. ANISOTROPIC HALF SPACE

277

is denoted by € = f.2/ fl = f z u, u, -h. ftlt. If the source is in medium 1, the reflection dyadic can be written in terms of its TE and TM eigenvalues as

RT E R™

Z 2T E

a

ZTE

-

1

aTE

fJ 1 -

fJ2

()

= Zr E + Zr E = fJ1 + fJrE '

_ Z 2T M

ZTM

-

1

_

{3TM 2.

7.424

a

- ftfJl

- Zr M + Zr M - fJrM + ftfJ1 .

(7.425)

The reflection image functions can now be identified by comparison with the isotropic problem of Section 7.3. Defining the integral transforms by

J 00

R(fJd =

f(()e- if3t ( d( ,

(7.426)

o we are able to find the image functions corresponding to the reflection coefficients. The TE case is straightforward since the anisotropic medium acts just like an isotropic medium with €2 = €2t, which gives us the possibility of writing the image function in terms of the !(a,p) function defined in (7.234):

ts, = Jk~t

- kr·

The path of integration goes in the complex ( plane from 0 to the parameter Pt defined by Pt

= jB t (

(7.427) 00

so that (7.428)

is real and positive for fTE(() to converge. Also, writing the TM reflection coefficient in the form

R T M = _ y'f;..flifJ1 y'f;..fli{31 -

J fJ~ + B'1 , J f3r + B;

(7.429)

the corresponding image function can again be written from analogy with isotropic media as fTM (()

=- ~ ~ ~ o+(() €z€t

jBzf( .fi;€i,jBz().

(7.430)

The image corresponds to that of an isotropic medium with effective parameters f e = J€z€tfl' J.le = J€z/€tJ.ll' The path of integration must now follow another line such that the parameter (7.431)

CHAPTER 7. EXACT IMAGE THEORY

278

is real and positive. Thus, there is a difference compared with the isotropic medium case in that the images of TE and TM sources may take different paths in complex space. However, for lossless media both lines coincide. To find the image for the general source, the third reflection coefficient

R; should be expanded in a form from which the image function can be identified:

R

o

=

2k2 {3T E «r M {32 k 12 T M TEl 2 fJ2 - €t 1 K2 (R - R ) = K2 (,BrM + ft,Bt}(,Br E + ,Bt).

(7.432)

After a few hours of algebra, the following form for R; can be found:

(7.433)

This expression requires some checks to be credible. For the limiting case f.t --+- f. z = f. the medium becomes an isotropic medium. In this case the expression (7.433) can be shown to reduce to

(?434)

which coincides with the corresponding expression of the isotropic Sommerfeld problem (7.255) given in Section 7.3. As another check we may take the limit (31 --+- 0, whence we can show that both (7.432) and (7.433) tend to the same value R; --+ 2. Finally, for K --+ 00, both (7.432) and (7.433) can be seen to give zero. The limit zero means that in the corresponding image function 10«() there

7.5. ANISOTROPIC HALF SPACE

279

are no delta singularities. 0.4 ----.------.---~--,..--___.,...---_.___-__r_--_r_-__r_-___. 0.351----+-~~,--~l;J--~-..:-+---1----+---+---+----+----f

0.25 ~--I-+++--+-+-+-----\-+___--.;lt-t----Po~

---1------4---4----+----1

0.21---l-l-../-I..-J~---+l~-+\--~\. -~I--..lor--+--__+--+---+----l

0.15 I---l~-I+-----I--'\ --+--\---+---+-I---~--+--+_-__+_-___f 0.1 ~I-/---+----+-_\__+-~_+_----\-If__-_+_~__+--+_.-_+_-__I

0 0

1

2

3

4

5

6

7

8

9

=

/Bzl for f2t 2fl, The argument is p = jk 1 ( . The delta function part is not shown.

Fig. 7.15 Examples of the normalized image function li™ J..L2

10

= J..Ll and different values of

fz

=

f2z/fl.

o

The image function fo{ associated with the reflection coefficient R; can now be written without problem, because the terms in (7.433) are of the same form as (7.429): fo«() = _8_. ft - 1

{f: (v'V€i€t +- l)n n=l

1

nJ2n (j B t (

)

+

(

References CLEMMOW, P.C. (1963). The theory of electromagnetic waves in a simple anisotropic medium. lEE Proceedings, 110, (1), 101-6. LINDELL, I.V., SIHVOLA, A.H. and VIITANEN, A.J. (1990). Exact image theory for uniaxially anisotropic dielectric half-space. Journal of Electromagnetic Waves and Applications, 4, (2), 129-43.

Appendix A Problems The following set of problems covers a selection of topics discussed in Chapters 1 - 6. The symbols used in the problems are defined in the corresponding Chapters.

Complex vectors 1.1 Show through the corresponding time-harmonic vector that the com-

plex vector a = al + ja2 with real vectors aI, a2 satisfying al ·a2 = 0, is in axial form, Le., the vectors al and a2 are on the two symmetry axes of the ellipse. 1.2 Show that the transformation a ---. eitPa, with real ¢, does not change the ellipse of the complex vector but moves the phase vector A(O) on the ellipse. 1.3 Show that the length of the polarization vector p(a) of a complex

vector a has the following geometrical properties (a) Ip(a)1 = sin '1/;, where 1/J is any angle of the equilateral quadrangle whose diagonals are the axes of the ellipse of a, (b) Ip(a)1

= 2A/1rlaI 2 , where A is the

area of the ellipse.

1.4 Study the polarization of the following complex vectors a in terms of

the polarization vector p(a): (a) a =

Ux

cos a

+ ju y sin o,

(a a real number)

(b) a = b + ju x b, (b a complex vector, u a real unit vector satisfying u · b = 0) (c) a = b x b*, (b a complex vector) 1.5 Show that p(a x p(a)) = p(a) when a is not a linearly polarized vector.

APPENDIX A PROBLEMS

282

1.6 Show that any complex vector a can be written as the projection of a circularly polarized vector b on the plane of a. Find the possible expressions for b.

1.7 Show that any complex vector a can be written as a = b and C are circularly polarized and [b] = [c].

+ c, where b

1.8 Find the most general complex vector b satisfying p(b) = p(a) when a is a given complex vector.

1.9 Show that if a· b = 0 and a- b" = 0, one of the vectors a, b must be linearly polarized or zero. 1.10 Show that the reciprocal basis of the reciprocal basis equals the original basis of complex vectors aj , a2, a3.

1.11 Determine the reciprocal basis of at = a, a2 = a", a3 a is circularly polarized.

= a x a"

when

1.12 Show by expanding in the base a, a", ax a" that the solutions vectors b± to the equations a x b± = ±jb± are of the form

when a is not a circularly polarized vector. The coefficients Q:± may be arbitrary. 1.13 Study the relation between the real and imaginary parts of the complex vector k when it satisfies k . k = k~ with real ko .

Dyadics 2.1 Prove the following identity:

axI=Ixa 2.2 Prove the following identity: (a x

I) : (b x I) =

2a . b

2.3 Prove the following identity: (a x A) : (B x b) = a (A~B) . b

APPENDIX A PROBLEMS

283

2.4 Prove the following identity:

(a x

I) ~ (b x I) = ab + ba

2.5 Prove the following identity:

(A · a) x

(:4. b) =

fI(2) .

(a x b)

2.6 Show that det(A~A) = 8(detA)2. This implies that A~A is complete

only when A is complete. 2.7 Expand the inverse of the dyadic

A

Check that A · A-I = 2.8 Study the solutions

Q,

I

= aI +a X I.

is really satisfied.

A of the following dyadic equation: A~A = aA,

when A is restricted to be a symmetric dyadic. 2.9 Solve the following dyadic equation for the dyadic X:

(aI + a x I)~X =

a x

I.

2.10 Defining the uniaxial dyadic as

D

= aft + (3uu,

write its Cayley-Hamilton equation and find the eigenvalues and eigenvectors. 2.11 Defining the gyrotropic dyadic as

G({3, R, 0) = (3Ull + Re Jo,

J = II X I,

where the dyadic exponential function is understood as

= = -It + -J() + ,11J 2 ()2 + ,J3()3 + ... 2. 3.

eJ 9

= It cos() + J sinO, It = I derive its eigenvalues and eigenvectors.

uu,

284

APPENDIX A PROBLEMS

2.12 Show that the two conditions ]i-I

= RT ,

detR = 1

for a real dyadic R are sufficient to guarantee that if b = R· a, we have Ibl2 = lal 2 and [b x b*I 2 = [a x a*1 2 . These mean that in the transformation a --+ b = R · a the magnitude and polarization of the complex vector a do not change. Thus, the transformation only moves the ellipse to another position and can be interpreted as a rotation operation. 2.13 The gyrotropic dyadic can be defined by

G«(3, R, 0) =

(3Ull

+ ReJo.

Determine its square root dyadic satisfying the condition = = R, 0)]1/2 = =G(f3, R, 0). [0(,6, R, 0)]1/2 . [G«(3, 2.14 Show the following properties of the dyadic

A:

(a)

11: aa =

(b)

A: (ab - ba) = 0 for all vectors a, b implies A symmetric. A: aa" = 0 for all vectors a implies A = O.

(c)

0 for all vectors a implies A antisymmetric.

Field equations 3.1 Derive the Helmholtz dyadic operator H eC'J) for the bi-isotropic medium and show that it can be factorized in the form

What are the operators Hi (\7) and H2(V)? 3.2 An electromagnetic shield is comprised of three layers of media: two

dielectric layers of permittivity f1 and thickness t 1 and, in the middle, a third magnetic layer with permeability 1-£2 and thickness t2. Determine the relation between these parameters so that there would be no reflection of a normally incident plane wave from the shield. The thicknesses are assumed to be very small and flt1, 1-£2t2 finite. Hint: consider the input impedance of the equivalent network.

APPENDIX A PROBLEMS

285

3.3 In some frequency regions the bi-anisotropic medium can be approxi-

mated by a lossless and nondispersive medium whose rnedium parameter dyadics are independent of frequency. Consider the expression of the energy density

w = "41 (E

E* ) , H) · M· ( H*

M=

(€(=p,e) .

and require that the energy density be positive (W > 0) for all possible fields E, H. Show that this leads to the condition that the following four dyadics connected to the material six-dyadic M must be positive definite:

3.4 What is the condition for the parameters corresponding to that above if we require a sharper condition W ~ W o , where

is the energy density in vacuum? 3.5 Consider the special case of a bi-isotropic medium of the previous problems 3.3 and 3.4. Derive the conditions of losslessness for the scalar medium parameters €, j.L, K, X without applying the result of the problem 3.3. 3.6 Find the conditions of losslessness for the impedance parameters Zl

and Z2 of a hi-isotropic impedance surface with the impedance dyadic

when the boundary consition is

nxE=

-Zs ·H.

Field transformations 4.1 Study the special duality transformation

APPENDIX A PROBLEMS

286

defined by the transformation matrix

T(o:)

1 = VI -sin2a

(

2

f!

1 cos a

y'2sina'f]o ) 1 .

(a) Show that detT(a) = 1. (b) Show that T- 1(a) = T(o + 1r).

(c) Derive the transformation_rul~sfor the medium parameters and show that the parameter transforms to itself for all a.

e.-(

(d) Show that a reciprocal isotropic medium with parameters E and J-L is in general transformed to a nonreciprocal bi-isotropic medium with ~d, (d =1= O. (e) Which a transforms =t1 to itself? What are the other transformed parameters? (f) Study whether a given nonreciprocal bi-isotropic medium can always be transformed to a reciprocal isotropic medium with Xd = (~d + (d)/2y!J.Lo€o = O. Find the dependence of the angle a on the parameters u, € and X in this case. 4.2 Applying a suitable affine transformation, solve the basic electrostatic

problem in an anisotropic dielectric: point charge Q at the origin Er is assumed symmetric, real and positive definite. r == 0 in a medium with the permittivity dyadic frEo. The dyadic

In particular, solve the scalar potential ¢( r) satisfying the Poisson equation

\7.

~r • \7¢(r)] =

-

Q 8(r)

Eo

together with the electric field E(r) = -\l¢(r) and the flux density vector D(r) = €. E(r). The following identity may be of some help: 8(A · r)

=

8(rl, IdetAI

where A is a real dyadic. 4.3 Find the image charge for the previous problem when the original

charge lies at the point r = uzh, h > 0, and the previous anisotropic medium is bounded by a perfectly conducting plane at z == O.

287

APPENDIX A PROBLEMS

Electromagnetic field solutions 5.1 Find the two-dimensional Green dyadic for the hi-isotropic medium.

In particular, find the solution for H(\1) . G(p) = -8(p)I,

with the Helmholtz operator defined by H(\1) = -(\1 x

I - jweI) · (\7 x I + jw(I) + k 2I.

Apply the symbols

and k± = k(cos'l9 ± ~r),

where

X

= nsin'l9,

~

= nn«,

n = JJ.Lr€r.

Note that the dyadics are not two-dimensional, only the Green dyadic does not depend on z. 5.2 Find the delta singularity of the Green dyadic in the bi-isotropic medium by comparing its expression to that of the isotropic medium. Note that delta singularities arise from the scalar Green functions obeying 1/r law in the double differentiations. 5.3 Find the field in real space from a dipole in complex space. Assume a

dipole with the current density

J(r) = u yI L8(x )8(y)8(z - ja) where a » A is real. Find the field (amplitude and polarization) close to the z axis for large Izi values. 5.4 Show that any plane wave propagating in the general uniaxial anisotropic medium with the parameter dyadics

is either TE or TM to v, Le., either satisfies v . E = 0 or v . H = 0, unless the material satisfies the condition /Lt€v = €tj.,£v. Explain the special behavior occurring at this special material condition.

288

APPENDIX A PROBLEMS

5.5 Study the plane-wave propagation in a bi-anisotropic medium with the

parameter dyadics defined as

with This kind of a medium can be fabricated by taking similar righthanded and left-handed helices and mixing them in a base medium so that N left-handed helices are parallel to the z axis and 2N right handed helices are isotropically orthogonal to the z axis. In particular, find the wave-number surfaces of the two plane waves. Study the optical axis directions in which the wave numbers are the same. What happens when the parameter K, approaches the value n = J tLr€r? (n is the refraction factor of a plane wave in isotropic medium with Kr == 0.) Also determine the eigenpolarizations for propagation along the z axis. Because of axial symmetry, write k == ukN(8), with k = wJ/i€, u is a unit vector which makes the angle () with the z direction. N(O) is the refraction factor of the plane wave propagating at the angle () in the present chiral medium.

Source equivalence 6.1 Find the equivalent electric source corresponding to a magnetic dipole,

J m == u z l mL6(r), in an isotropic chiral medium with parameters J.-t and K, == K,rk/ko.

€,

6.2 Show that a radial source of the form J(r) == urf(r) does not radiate outside the support of the function f(r).

6.3 Find the equivalent magnetic source of an electric surface-current source

6.4 Find the equivalent magnetic volume current of the coaxial current

Assume that the equivalent magnetic current is in the volume between the surface currents and that it is of the form Jm(r) == utpJm(p).

APPENDIX A PROBLEMS

289

6.5 Show that the approximation of a current sourceJ(r) by a dipole of moment P at a position r = a can also be done by minimizing the error function R(r) = J(r) -P6(r-a) in the following sense: require JR(r)dV = 0 and minimize the norm of JrR(r)dV. 6.6 Formulate the Huygens principle to electrostatic problems with charges and magnetic currents as sources. The space is assumed homogeneous and isotropic. Start from the equations

"V x E

= Jm ,

\l · D = g,

D

= fE.

Write the expression for the electric field in a volume V surrounded by the surface S with sources truncated in V and Huygens sources on S. Study the possibility to replace the magnetic Huygens current by equivalent electric charge on the surface S.

Appendix B Solutions The following set of solutions covers most of the problems given in Appendix A. The solutions are given in compact form to save space.

Complex vectors 1.1 The extremal values of the square of the corresponding time-harmonic vector are found from

~ [A(t) · A(t)] = ~[a~ cos2 wt + a~ sin 2 wt] = 0, which reduces to (a~ - a~) sin wt cos wt = o. For a circularly polarized vector (a~ = a~) this is satisfied for all t, which means that it has discrete axes. Otherwise we have solutions t = nat12w. For n even the axis occurs at ±al, for n odd at ±a2. 1.2 The time-harmonic vector of aejc/J equals ~{aejw(t+c/J/w)}, which shifts

the time origin by /w but otherwise does not change the ellipse of the vector a. 1.3 Assume first a is not circularly polarized. Write b = aei 8 in axial representation, with 8 such that b· b = lal2 • Because p(a) = p(b), the axial vectors b re and b im satisfy

(a) From the geometry we have tan('l/J/2) = Ibimi/ibrel and 2 tan ('l/J/2)

Ip(a)1 = 1 + tan 2 (t/J/ 2)

.

= sin t/J.

APPENDIX B SOLUTIONS

292

(b) The area of the ellipse can be written as

which equals the required relation. These results are also valid in the limit when a is circularly polarized. 1.4 The polarization vector p(a) can be expanded in each of the cases as

(a) p(a) = -uz2sino:coso: = -u zsin20:. The axial ratio of the ellipse is e = tan 0:. It is seen that all polarizations are obtained because Ip(a)1 goes through all values between 0 and 1 when 0: is changed. Linear polarization (e = 0 or ±oo) corresponds to Q == n1r/2 and circular (e == ±1 to Q == n1r/4), n odd.

(b) Assuming b not linearly polarized, p(b) is parallel to u and we can write b x b*

+ (u x b)

x (u x b*) - jb x (u x b*) - jb* x (u x b) 2j(lbl2 + ju · b x b*)

p(b) - u u. p(b)

= 1-

= -u = p(a).

Thus, a is circularly polarized, which is easily checked: a-a = O. This is valid in the limit also when b is linearly polarized. (c) Because (b x b*) x (b x b*)* = 0 and, hence, p(a) == 0, a is linearly polarized (or zero). 1.5 a· p(a) == a" · p(a) = 0 in all cases. Expanding

[axp(a)] x [axp(a)]* = p(a)[axa* ·p(a)] == jp(a)(a.a*)[p(a)·p(a)], [a x p(a)] · [a x p(a)]* == (a· a*)[p(a) · p(a)]

we have

[a x p(a)] x [a x p(a)]* j[a x p(a)] . [a x p(a)]*

=. JP

(a) (a. a*)[p(a) · p(a)] j(a· a*)[p(a) . p(a)]

= p(a).

1.6 If a is not linearly polarized, p(a) ::I 0 and we can define a unit vector n = p(a)/lp(a)1 normal totheplaneofa. Ifb = a+no:, from b-b == 0 we can solve for 0: and have two possibilities: b == a±jnva:a. This

is also valid for the linearly polarized case when n is any unit vector in the plane orthogonal to a.

APPENDIX B SOLUTIONS

293

1.7 For a circularly polarized we can write b = e = a/2. Applying the

result of the previous problem, we can directly write b

= ~(a + jn..;a:i ),

because [b] =

lei, as can

c

= ~(a -

jn..;a:i ),

be readily checked.

1.9 Because a x (b x b") = bfa b") - b*(a. b) = 0, either b x b* = 0, whence b is linearly polarized, or a is a multiple of b x b", which is known to be a linearly polarized vector. 1.11 For a circularly polarized, p(a) = u is a real unit vector. Thus, a3 = jlaI 2p (a ) = jlal 2u and u x a = (a x a*) x a/jlal 2 = ja. The

reciprocal basis is

1.13 Writing k re + jk im, we have by separating the real and imaginary parts of k . k = k~,

Thus, the vectors k re and kim are orthogonal and Ikrel > Ikiml. The equation Ikre l2 -lkiml2 = k~ is that of a hyperbola. In parametrized form, the most general solution for k is k

= ko(ucosh 8 + jv sinh 8),

where u and v are any two real orthogonal unit vectors.

Dyadics 2.1 Consider mapping of a vector b:

[a x

I - I x a] . b =

ax

(I. b) - I· (a x b) = a x b -

a xb

=0

Since the result is zero for any vector b, the dyadic in square brackets must be zero, which gives the identity.

294

APPENDIX B SOLUTIONS

2.2 Using the result of the previous problem, we can write

(a x 1) :(1 x b)

= L L[(a x Ui)Ui] : [Uj(Uj i

==

L L[(a x Ui) . Uj][Ui · (u, x b)] j

i

== a- LL(Ui x Uj)(Ui X Uj)' b == a- (I~I). b == i

x b)]

j

a (21). b == 2a· b.

j

2.3 It follows directly from

(a x cd) : (ef x b) == a- (c x e)(d x f) . b

==

a·

(cd~ef) . b

2.4 Start from the expression

(a x uu)~(b x vv)

= [(a x u)

= [ba- (u x v)

+ (a

x (b x v)](u x v)

x b) . uv](u x v)

= ba- (uu~vv) - vv x uu· (a x b)

Replace uu and vv by (a x == 2ba

I:

I) ~ (b x I) == ba · (I~ I) - I x I . (a x b) -1 x (a x b)

= 2ba - ba + ab == ab

+ ba

2.5 Start from the expression (cd~ef)

· (a x b) = (c x e)[(d x f) . (a x b)]

== (c x e)[(d· a)(f· b) - (d· b)(f· a)]

= (cd. a) x (ef. b) Replace cd and ef by

(cd- b) x (ef· a)

A:

(A~A) · (a x b) == 2(A. a) x (A. b). 2.8 One can immediately see that a = 0 and all linear dyadics A make one set of solutions. For a =1= 0, denoting B = 2A/a we have a new problem B(2) = B. Its solutions B =1= 0 must be complete dyadics. In fact, assuming B planar, B(2) must be linear, whence B is also

APPENDIX B SOLUTIONS

295

linear implying B == B(2) == 0, a contradiction. Since detB f.: 0 for a complete dyadic, taking the determinant gives us detB(2) = det 2 B == detB, or detB = 1. Hence, B(2) == B-1 T == B. Because B is assumed symIEetric, it must be a square root of the unit dyadic, i.e., of the form I - 2(abja· b) with a- b -# o. 2.10 The invariants are

whence the Cayley-Hamilton equation can be written as

Thus, the eigenvalues are Al == (3 and A2,3 == a. The eigenvector == u and the other two a2,3 are any vectors orthogonal to u.

a1

2.11 Because the dyadic G consists of a transverse part and a multiple of uu, it is obvious that one of the eigenvectors is u and the rest are orthogonal to u. The first eigenvalue is Al == (3 and a1 == u. Thus, we are left with the two-dimensional eigenvalue problem

=

eJ O • a

-

-

A

= (1t cos 0 + J sin 0) · a = R a.

Because we have

(eJO)~(eJO) = 2uu,

spm(eJo) = 2,

the Cayley-Hamilton equation reads

and the solutions A f.: 0 are

A2,3

== Re±j(J. The eigenvectors satisfy

This equals u x a2,3 == ±ja2,3, whence the eigenvalues are circularly polarized (a2,3·a2,3 = 0), with a2 right-hand and a3 left-hand rotation with respect to the u direction.

296

APPENDIX B SOLUTIONS

2.12 We have first =T= b b * =a·R ·R·a* =a·a*, v

whence the magnitude of the vector is not changed. Secondly, because of -

--1

R

R(2)T

-(2)T

=----=:-=R

detR

-T

=R,

we have

b x b"

= (R· a) x (R· a") = R(2) . (a x a") = R· (a x a"),

and thus

(bxb*).(bxb*)*

=

(axa*).RT.R.(axa*)*

= (axa*).(axa*)*,

which tells us that the form of the ellipse has not changed. 2.13 Since the dyadic consists of orthogonal axial and transverse parts, both of these can be treated separately. The square-root of the axial part is obviously ±.Jljuu sin~ its square is {Juu. The transverse part

gives correspondingly- ±J[ieJ 9 / 2 , whence the following four dyadics (double signs are not related)

are square roots of the gyrotropic dyadic. 2.14 For each case we apply the fact that if a· A· b

= 0 for all a,

b, then

A=O. (a) A: (a+b)(a+b) = 0 for all vectors a, b implies A : (ab-l-ba) = a· (A + AT) . b = 0, whence A = -AT and A is antisymmetric. (b)

A: (ab -

(c)

= 0 and A : (a+ jb)(a+ jb)* = -jA : (ab" - ba") = 0 imply A : ab" = 0 for all vectors a, b *,whence =A = O.

ba) = asymmetric.

A: (a+ b)(a+ b)*

(A -

AT) · b

= 0, whence A =

= A: (ab" + ba")

AT and A is

APPENDIX B SOLUTIONS

297

Field equations 3.1 The Helmholtz operator can be written as

-JlH e = (\7 x I - jweI) · (\7 x I = (\7 x 1)2 + jw( - €)\7 x 1 -

+ jw(I) W

2

W

2

Jl€I

(jl€ - €()1

This is a polynom of \7 x 1 and can be written in the form

-JlH 1(\7) · H 2(\7) = (\7 x

I + af) . (\7 x I + (3I).

= (\7 x 1)2 + (a + {j)\7 x I + a{jI The coefficients a and {3 can be solved by comparing the expressions and the result is

In the reciprocal chiral (Pasteur) medium with ( = the coefficients reduce to

-e =

j"'JJ.to€o

3.2 From the 7r network with shunt admittances Y1 and series impedance Z2 with loading impedance 1Jo we have the input admittance

When this is equated with l/'TJo, we can solve for Z2 in the form

2'TJ;Y1 Z2 = 1 _ 2y;2· 'TJo 1 Substituting Z2 = jkot(jlr - l)'TJo, Y1 = jkot(€r - 1)/"10, we have k ot2(JLr - 1)

2kotl(fr - 1)

= 1 + k~tH€r -

1)2'

3.3 Because of the lossless character, the medium six-dyadic M is hermitian, which implies that all its eigenvalues are real. The positive energy function requires that M is positive definite (PD), whence its

298

APPENDIX B SOLUTIONS

eigenvalues are positive. Also, its diagonal elements € and =p, must be PD. From matrix algebra we know that the inverse M-1 must be PD, whence its diagonal dyadics are PD. Solving for the inverse dyadics by solving two vector equations, we arrive at the required condition that

€, ~, €-

e·~-1 .(,

Ii - ( .€-1 .

e

must all be positive definite dyadics. 3.4 Same as above when € and Ii are replaced by € - €o! and Ii - /-Lo!, respectively. However, because the sign > was replaced by ~, PD (positive definite) must be replaced by PSD (positive semidefinite). In fact, for the vacuum, the equality sign must apply. 3.5 Consider the energy expression of the problem 3.3. Setting H = 0 and E = 0 we conclude that € > 0 and J-L > 0, respectively, corresponding to f ~ f o and /-L ~ /-Lo of the problem 3.4. Denoting a

= v'€E,

b = JjLH,

we can write the inequality W > 0 as

lal 2 + Ibl 2 ± 2IXrl~{a· b"} ± 21"'rl~{a· b*} > 0, which should be valid for all vectors a, b. Here we have denoted (Xr - j"'r)y'ii€ and ( = (Xr + j"'r)y'ii€. Choosing a = ejab, the inequality becomes

e=

IXrl cos a + I"'r\ sin a < 1, which is valid for all real angles a if 1, Xr2 +2 «; <

In the sharper case of the problem 3.4, the inequality is replaced by

These conditions can be readily obtained as special cases of the results of the two previous problems. In fact, if f =p.-l .( is PD and (f-€oI) -~. (Ji- J-L 0 1)-1.( is PSD, in the bi-isotropic case these reduce > 0 and (€- €o) (J-L - J-Lo) ~ 0, to the respective conditions €J-L which coincide with the results derived with a change of symbols.

e.

e(

e(

APPENDIX B SOLUTIONS

299

3.6 The condition of losslessness is n- ~{E x H*} = 0 for all possible fields on the surface. Evaluating -n· E x H*

= H* . Zs' H = ZllHI 2 + Z2n· H x H*

= (Zl

+ jZ2n· p(H))IHI 2 = o.

The real part of the bracketed term must be null for any field H. Writing Zl = R I + jXI and Z2 = R2 + jX2, we have

Because the real vector p(H) can have any direction, we must have separately R 1 = 0 and X 2 = O. The lossless surface impedance dyadic is of the form Zs = jX1I + R2 n x I, where Xl and R 2 have real values.

Field transformations 4.1

(a) detT(a)

=

l-sfn2a(l- 2sinacoso:)

= 1.

(b) Inverting the T(a) matrix changes the signs of sin a and cos a in T(o}, which is tantamount to replacing a by a + it . (c) Denoting T =

e+ (, the transformation formulas are

~ 2 ~¥ (1 - sin2a)€d = € + 22' cos a - 2v2- cos o,

~o

(1 - sin 2O:)~d =

(1 - sin 20:)1'd =(1

~o

Ji + 2€1J~ sin 2 0: -

+ sin 20:)1' -

2J2T~0 sin o,

VZ€1Jo sin 0:-

J2 7l

cos 0:

'fJo

(d) From the previous equations it is seen that if T = 0, in general we have Td =1= 0, which means that a reciprocal medium is transformed to a nonreciprocal medium. (e) =tid = =ti when sin 0: las read

= 0, in which case the transformation formu-

300

APPENDIX B SOLUTIONS

lid

=

Ji,

ed - (d =

e-(

The double sign corresponds to cos a == ±1. (f) Requiring in the bi-isotropic case Td == TdI = 0 and denoting T = XJP,of o, 11 = Vp,/f, we have the equation

-.£ = yi217- 1 sin a + 17 cos Q; /Ji€ 1 + sin 2a for the angle a. The right-hand side can be seen to obtain continuously all values from -00 at a = -1r / 4 to +00 at a = 31r/ 4. Thus, there is a unique solution a in the interval -1r/ 4 < a < 31r /4 corresponding to any value of X and a transformation from any nonreciprocal bi-isotropic medium to a reciprocal biisotropic medium exists. 4.2 Writing \I . [fr . \I¢(r)] =

(11 . \I) . (11 . \I)¢(r),

where A == €r 1/2 is a real, symmetric and positive definite dyadic, we denote A · V = V', whence r' = A -1 . r. Writing also

't/J(r') = (r) == (A· r"), the Poisson equation becomes

V,2'l/J{r')

= - Q<5{r) = €o

Q _<5{r'). €odetA

This is an equation for a point charge in an isotropic medium with the permittivity (detA)f o . The solution for the potential is known:

'l/J{r') r'

=

Q _, 41rf oT'detA

= .;rt:ri = V(A-1 · r) · (A-1 · r) = Vir -1

:

rr

Thus, the potential in the original medium can be written as

APPENDIX B SOLUTIONS

301

Surfaces of constant potential are ellipsoids f r -1 electric field is orthogonal to these surfaces:

:

rr = const. The

but the flux density vector D is not parallel to E:

D(r) = fr€o' E(r)

Q= 41rD3 (detfr)r.

Actually, the flux density vector is radially directed as in an isotropic medium. Let us check the solution. Noting that 'V · r

= 3, when r =F 0, we have

€ (2) . r Q 3 \7. D(r) = - [ - - 3 r 41r D3 D5

.

r] = 0

.

Thus, the source must be at r = O. The singularity of the source is found when integrating over any small volume V containing the origin, with unit vector n normal to its surface B:

J

V · D(r)dV

=

f

n· D(r)dB

s

V

= ~detEr

f (Ir(2~:'

:r)3/2 dB

s

Now we can show through the same transformation that = deter

f (I (2)n :.rr)3/2 r f n'· r' , f ' dB = (r' . r')3/2 dB = dO. = 411", 8

r

8'

because d.S' = r,2dO,' In' . u~. This leads to

J

V · D(r)dV

= Q,

v whence the source is the point charge, 'V . D nonradiating point source can be added.

= Qc5(r) to which any

Electromagnetic field solutions 5.1 Apply the formal procedure G(p) = -H- 1(\7)8(p) =

H(2~(V) 8(p). detH('V)

APPENDIX B SOLUTIONS

302

The Helmholtz operator can be written as

The Green dyadic can be written as

G(p)

= H(2)T (\7)G(p)

= L+(2)T (\7) ·

L_ (2)T (D)G(p),

L±(2)T (\I) = \IV ± k±\I x I + k~I. Because detL±(\7) = =fk±(\72+kl), the scalar Green function obeys

G(p) =

_

1

_

6(p)

detL+(\7)detL_ (\7)

Writing

whose solutions are known to be

the Green dyadic can finally he written in the form

5.2 The Green dyadic expression for the hi-isotropic medium is

In analogy to the isotropic medium we can write

APPENDIX B SOLUTIONS

303

whence the singularity of the Green dyadic can be expressed as -

G(r)

-

= PVG(r) -

1-

k+k_ L8(r).

5.3 The vector potential due to a dipole source whether in real or complex

space is

A(r) = uyj.tILG(D),

D = v(r - juza) . (r - juza)

and the electric field

E(r) = -jw[A(r) + In the far field (kD

»

:2 VV · A(r)].

1) we have

VVG(D) ~ _k 2 (r - jUza1~ - juza) G(D), so that u y· \I\IG(D) is of the order (yj D)G(D) z axis. Thus, we can approximate

E(r)

~

-jwA(r)

« G(D)

close to the

= -uyjwp,ILG(D).

Inserting the expression (5.128) for G(D) from the book, the electric field is seen to be a Gaussian beam with polarization u y close to the z axis. 5.4 The fields of any plane wave satisfy the conditions

Eliminating the transverse fields we have

Thus, unless the bracketed term vanishes, we must have either E; = 0 or H; = o. The bracketed term vanishes when € and Ii are multiples of one another, i.e., when the uniaxial medium is affine-isotropic. In this case the two spheroidal wave-number surfaces coincide and any combination of TE and TM waves is a possible plane wave.

APPENDIX B SOLUTIONS

304

5.5 The eigenvalue equation for the plane wave can be written as D(k)· E = 0,

= (1_ I + jkKr)2 + k 21 = D+(k) . D_(k), D±(k) = k x I + jk(K r =F I), k = w#-.

D(k)

The two dyadics D±(k) obviously commute. In this case the problem can be split into two parts:

The determinants are expanded as detD±(k±) = -jk k±k± : (~r =f I) + jk3det(~r =f I), which leads to the wave-number equations k±k±: (I=fKr) = k2det(I=fKr) = k2(1- ",;)(1 =F "'r). To find the rotationally symmetric eigensurfaces, denoting k± kN± u with the unit vector satisfying u . U z = ·cos(), we can write for the refraction factors

(1 - ~~)(1 =f ~r) 1 ± "'r cos 28 ' which defines two spheroids. The two spheroids have common points at 8 = 1r/2, because N+{1r/2) = N_(1r/2) = Jl - It~. This means that, instead of one or two optical axes, there exists an 'optical plane'. The axial refraction-factor values are N±(O) = 1 =f "'r. When 'ltrl ~ 1, the refraction factors tend to zero and the wavelengths become infinite. The eigenvectors satisfy

When u = u ,, the eigenvectors have no z component, as is easy to see by multiplying the equation by U z · . Thus, in this case we have

lV±(O)

E± = . (

=fJ 1 =f Itr

. ) u; x E± = ±Juz x E±,

and the eigenwaves are circularly polarized (E± · E± = 0) with '+' corresponding to the right-hand polarization and '-' to the left-hand polarization. Eigenpolarizations for waves in other directions are elliptical and have more complicated expressions.

APPENDIX B SOLUTIONS

305

Source equivalence 6.1 The equivalent electric current in a chiral medium is

ImL () = -.-"\16 r x u, JW~

. + JU z

K,r

ry

I m L6( r ) .

The curl term corresponds to a loop current just like in a nonchiral medium and the second term corresponds to a dipole with

I = jK,r1m/ry. 6.2 A radial function satisfies "\1 x J = 0, which means that the equivalent magnetic current is zero. Thus, the radiation field outside the source is zero. 6.3 Writing

\1 x (utpJm(p)) = u)[pJm(p)]' = jwp.J(r), P we have the solution in integral form as

=- J p

Jm(p)

jw~I

1

p

-2-[8(p-a)-8(p-b)]dp

o

1r

jw~I = -2 -[U(p-a)-U(p-b)]. itp

Thus, the equivalent magnetic volume current has the amplitude jWJ.LI/21rp in the region a ~ p ~ b. 6.4 From JR(r)dV = 0 we have the condition P = J J(r)dV. Minimization of the norm of rR(r)dV = rJ(r)dV - aP is the same operation as minimizing the norm of the moment 'PI (a), whence the resulting expression for a coincides with (6.55).

J

J

6.5 Truncating the region by the pulse function Pv(r) gives us

\7

X

Er

= (\7 x E)Pv + (VPv) x E = Jmr + JmH,

\7. Dr = (\I. D)Pv

+ (\7Pv)· D

=

{}T

+ {}H,

with the Huygens sources JmH

= (\lPv)xE = nxE8 s ,

{}H

= (VPv)·D = n· D 6s

= n·€E8s ·

APPENDIX B SOLUTIONS

306

The electric field can be written as a sum of the field from charges and magnetic currents. The component from the charge is familiar:

n, = -v» = -

J

\lG(r - r,)e(;') dV'

The electric field from the magnetic current is dual to the magnetic field from an electric current:

Em =

J

\lG(r - r')

Jm(r')dV'.

X

Thus, the total field in the volume V is

E

=-

J

J

v

v

\lG(r - r') e(;') dV' +

\lG(r - r') x Jm(r')dV'

-f \lG(r - r')n/ · E(r/)dS' + f \lG(r - r') x (n' x E(r'))dS' s

s

Appendix C Vector formulas General formulas \7(af(r)) = a\lf(r) \7[f(r)g(r)]

= g(r)V f(r) + f(r)V' g(r)

V . [af(r)]

= a'\1· f(r)

\I · If(r)g(r)] = [\I f(r)] . g(r)

+ f(r) [V' . g(r)]

\l x [af(r)] = aV' x f(r) '\1 x [f(r)g(r)]

= ['\1 f(r)] x g(r) + f(r) ['\1 x g(r)]

\7 . [f(r) x g(r)] = [V' x f(r)] . g(r) - f(r) . [\7 x g(r)] V x [f x g) = f['\1 . g] - g[\7 . f] \l x \l f(r)

+ [g . '\1]f - [f · '\1]g

=0

\l . [V' x f(r)] = 0

\7 x (V' x f) = V'(\7 . f) - V' . \7f = V'(V' · f) - \72f

J'\/ · J'\/

fdV =

v

f f· dS

x f· dS =

s

(Gauss)

s

fr c

de

(Stokes)

308

APPENDIX C VECTOR FORMULAS

Cartesian coordinates

X,

y,

Z

a a {} 'VI = U x oxl + u y 8y1 + u, ozl {}

{}

{}

'V · f = ox l x + OylY + azlz

Cylindrical coordinates p, ip, z

o

1

o

{}

a

a

'1!=up - !+ucp--a !+uz - f p ip az ap 1

a

1

= Pap(pip) + pov/cp + ozlz

'V · f

1

up

pUcp

u;

'Vxf=p

p

;cp

fp

pfcp

t,

t

f)

tz

'12 _ ~ ~ ( {} +~0 I +0 I p2 Ocp2 OZ2 I - p op pop \lp = \I(r - Uzz) =

I-

UzU z

=

2

It,

p = Ipl = [r - u. z], 1

=

'1'Y p = '1u p = -(1 p

2

\I p

'Yep

1 P

= -UCP'

UzU z

2

\I. P = 2,

\7 x p = 0

'Y p = up

1 x 1 - upu p) = -uzuzxupup = -ucpucp p P

1 = \I . up = -, p

\I x

up

=0

\7. Ucp = 0,

'Y

1

X Ucp = -U z

P

APPENDIX.C VECTOR FORMULAS

309

'\l2g2(p) = -8(p) uzUz~V'V'g2(P)

1

= -PV 21l" p2(upu p -

V'G2(p) = -up ~Hi2)(kIPI)

G2(p) = :jH£2)(kl pl), V'V'G2(p)

1=

u'Pu'P) - 2 I tc5(p)

k

(2)

= -upupk2G2(klpl)+PV(upup-u'Pu'P) 4jpH1

1: '\l'7G2(p) = -k 2G2(p) -

'\l2G2(p) =

Spherical coordinates '\lI = '7. f =

-8r I

18

1

a

a

1

+ Uo -r af) I + u., .-f) -a I r SIn sp 1 a. 1 a

2 2-a (r Ir) + - .-0 ae (sin elo) + - .-e-a f

ruO r sin Ou
Ur

a ar fr

\7xf=_lr 2 sin

V'2 f =

8(p)

cp

T, (),

a

Ur

1-

(klpl)-21tc5(p)

e

a f

80 r 8

a

8

f
.!.-2 ~ (r 2 af) + _1_ ~ (Sin (} af) + 2 1 2 fj2 f r or Br r 2 sin (} ao oe r sin eacp2 \7r

= I,

= 3,

'\l . r

1 = '\lu r = -(1 - uru r ) , r

'7(a x r) =

'7xr=O 2

'l. u; = -, r

-1 x a =

"V · (a x r) = 0,

'\l x u; = 0

-a x 1

\J x (a x r) = 2a

e- j k r G(r) = 41fr' '7G(r) = -

V'V'G(r)

= -urur k 2G(r) '7 2G(r) =

U

r

r

r = [r]

(1 + jkr)G(r)

1

=

1=

PV r 2 (1 + jkr)(I - 3urur)G(r) - 3Ic5(r)

I : 'l'7G(r) =

-k 2G(r) - 8(r)

Appendix D Dyadic identities Definitions

A = A·A,

A- 2

2

=(2)

A

l=x=

= (A- 1 ) 2 = (A 2 ) - 1

A(-2) =

= 2"A xA,

(11- 1 ) (2) =

(A(2»)-1

trA = A: I -

-

1- -

-

spmA = trA(2) = 2A~A : 1 -

detA

1- -

-

= 6A~A: A

detA f= 0 ~ A complete detA = 0

~

A planar

=0

~

}1 linear.

}1(2)

Identities

A~B = B~A = [(A: I)(B: I) - A: BT ] 1(A : I)J3T

-

(B : I)AT

A~1 =

+ [A . B + B . A]T

(A : 1)1 - AT

A~ (a x I) = a(A~1) + I x (a· A) 1~1 =

21

312

APPENDIX D DYADIC IDENTITIES

(a x I)~I

=a x I

8~I =

-8 (8 symmetric, trace free) (a x I) ~ (b x I) = ab + ba 8~ (a x I) = (8 . a) x I (8 symmetric) (;1 x a)~(B x a) =

(A~B). aa

(a x A)~(a x B) = aa- (A~B)

(a x

I) ~ (a x I) =

2aa

A~(B~C) = (A: C)B + (A: B)C - B· AT. C - C· AT. B

I~(A~B)

= (Ii: I)B + (B : I)A - (A· B + B· A) I~(I~A) = A+ (A: 1)1 I~(J~1) =

41

(A~A)~(A~A) = 8(A(2))(2) = 8detAA

(A(2)) (2)

=

AdetA

det(A~ A) = 8det2A

(A~B)· (C~D)

=

(A. C)~(B·

D)

+ (A. D)~(B. C)

(A~A)· (B~B) = 2(A. B)~(A. B) (A~B)2

= (A2)~ (B 2 ) + (A. B)~ (ft. (A~A)2

(A~I)2

=

=

B)

2(A2)~ (A 2)

(A2)~I + A~A

.. A:)I ( A~ A)T ·A = A· (A~A)T = !(A~A 3 1I(2)T .

A = II . A(2)T = detA 1

APPENDIX D DYADIC IDENTITIES

-A- I =

313

3(A~A)T A~A: A

A(2)T =

detA

=X=(2)* T = I (AxA ) A- =

(A complete)

(planar inverse)

A(2) : A(2)*

=

=

A-I · A

=

jf(2)*. A(2)T

=I -

(A planar)

A(2) : A(2)*

= ATxuu A-I = x

(two - dimensional inverse)

spmA

A-I · A =

It = I - uu (A two -

dimensional)

A : B = B : 11 = (If · [iT) : I (A~B)

: C = A : (B~C) (with all permutations)

spmA = trA(2) = spm(A. B) detA

1-

~[(A: 1)2 - A : "AT] 2

= tr(A(2) . B(2») = -

= "3A3 : 7 -

1 -

-

jf(2) : B(2)T

-

1 -

-

"2(A2 : I)(A : 7) + 6(A : 7)3

det(A~A) = 8det(A(2)) = 8(detA)2 det(A · B) = detA detB

det(A · B + oJ) = det(B · A+ oJ)

-

a x (A~B)

= B x (a A) + 11 x (a. B)

(A~B) x a

= (A· a) x B + (B· a) x 1-1

-.

1--

(A· a) x (A· b) = -(A~A) · (a x b) 2

-

= jf(2) • (a x b)

1 -(a· A) x (b· A) = -(a x b) . (A~A) = (a x b) .1-1(2) 2

Subject index active medium 90, 92 admittance dyadic 72 affine transformations 112-7 affinely isotropic medium 116, 135 affinely uniaxial medium 116 analyticity of fields 186 anisotropic half space problem 275-9 anisotropic medium Green dyadic 133 dispersion equation 157 antenna above ground 247 antihermitian dyadics 39 antisymmetric dyadic 19, 21, 27 Appleton-Hartree formula 161 axial decomposition 9 Babinet's principle 122-4 base dyadics 49 beam mode radiation 152 bi-isotropic boundary 73 bi-isotropic medium 57 bianisotropic medium 56 boundary conditions 71-4, 92-3 boundary operator 81-2 branch circle 149 cavity field 143 Cayley-Hamilton equation 29, 37 characteristic impedance 202 circular polarization 3 co-polarized vectors 7 co-uniaxial dyadics 44 coaxially gyrotropic dyadics 45 complete dyadic 21, 31 complex delta function 150 distance function 149 vector bases 14 vectors 1-17 continuity condition 54

corrugated surface 73 CP vector 3 cross-polarized vectors 7 cross-product square 28 cross-square of dyadics 281 delta expansion 175 depolarization dyadic 143, 146 determinant of dyadic 28, 281 dipole approximation of multipole 178 dipole mode radiation 152 discontinuities of fields 69-70 dispersion equation 154 dispersive medium 57 dot product 22 double-cross product 25, 281 double-dot product 24 duality transformation 100-12 dyad 17 dyadic 18 characteristic impedance 202, 204 eigenvalues 36-9 equations 30-6 identities 26, 281-3 image functions 208 products 18 square root 33 EFIE 189 eigenvalues of dyadics 36-9, 48 EIT 199-279 electric reflection 118 elliptic polarization 3 ellipticity of vector 11 EMP 167 energy conditions 89-93 equivalent sources 165-97 exact image theory 199-279 extraordinary wave 159, 162

318 finite part 147 Fourier transformations 59-63 gauge 64 Gaussian pulse 150 Gibbs' identity 14 Green dyadic anisotropic medium 133 hi-isotropic medium 132 one dimensional 140 singularity 144-8 two dimensional 138 Green function dyadic 126, 131 scalar 125, 129 isotropic interface 246 microstrip 272 gyrotropic dyadic 24, 42, 45 gyrotropic medium 160 half-space Green function 141 half-space image problem 235-59 handedness of vector 3 Helmholtz equations 61-2 hermitian dyadics 39 Hertz potentials 67 Hertz vector 66 Huygens' principle 183-7 Huygens' source 182-3 image principle, mirror 119 impedance dyadic 72 wire grid 231 uniaxial half space 276 impedance sheet 77-8 impedance surface 224 interface conditions 74-77 invariants of dyadics 28 inverse of dyadic 29, 50, 283 involutory affine transformations 117 isotropic boundary 72 layered media 199 line source decomposition 194 linear dyadic 22, 31

SUBJECT INDEX linear dyadic equations 31-2 linear mapping 20 linear polarization 2 longi tudinal propagation 161 loop antenna radiation 255 Lorentz force 54 Lorenz gauge 66 lossless medium 91 LP vector 2 magnetic impedance sheet 78-9 magnetic reflection 118 Maue integral equation 190 Maxwell equations 53-4 medium conditions 89-94 medium equations 54, 56-7 MFIE 189 microstrip guided modes 271 microstrip image problem 260-74 mirror image principle 119 multilinear identity 4 multipole expansion 176 multipole sources 174-81 NCP vector 9 NLP vector 3 non- radiating source 165 norm of dyadic 25 NR source 165 object in front of interface 257 optical axis 158, 162 ordinary wave 159, 162 p vector 10 parallel vectors 5 passive medium 90 PD dyadics 40 perpendicular vectors 5 planar current sheet 144 planar dyadic 22, 31 planar inverse of dyadic 32, 281 planar unit dyadic 32 plane source decomposition 196 plane waves 154-63

SUBJECT INDEX point source decomposition 197 polarity reversal 97 polarization match factor 7 polarization vectors 10 polyad 17 polynomial operator 126 posi tive definite dyadics 40 potentials 63-68 power orthogonality 7 Poynting vector 89 principal value 146 propagation factor 202 q vector 12 quadratic dyadic equations 33-5 radiation condition 73 reciprocal base .19, 22 reciprocal surface 73 reciprocity 93-4 reflection dyadic 43, 203 impedance sheet 228 impedance surface 224 interface 205 isotropic interface 237 wire grid 232 microstrip 261 quad uniaxial half space 277 reflection Green functions 219 reflection image function 210 asymptotic expansion 240 quad isotropic interface 238 Taylor expansion 239 wire grid 233 uniaxial half space 277 reflection parameter 237 reflection transformations 117-24 rotation dyadic 41 scalar potential 64 scattering from object 257 self duality 101-10 self-dual boundaries 107

319 decomposition 107-8 fields 105-6 medium 101, 106 sources 106 self-reflecting fields 119 shearer 35 sheet conditions 77-81 Silver-Miiller condition 73, 88 singularity of surface source 147 singularity of volume source 146 Sommerfeld integrals 205 Sommerfeld surface wave 235 source equivalence 165-97 space inversion 99 spm of dyadic 28, 281 square root of dyadic 33, 51 strictly planar dyadic 22, 31, 37 surface integral equations 187-91 symmetric dyadic 19, 27 TE/TM decomposition of sources 1927

Tellegen medium 57 Tellegen parameter 94 time reversal 98 time-harmonic vector 1 trace of dyadic 24,281 trace-free shearer 35 transmission dyadic 203 impedance sheet 230 isotropic interface 250 wire grid 232 transmission Green dyadic 222 impedance sheet 229 impedance surface 225 transmission image functions 217 isotropic interface 250 impedance sheet 230 wire grid 233 transmission-line equations 202 transversal propagation 161 two-dimensional dyadics 47 two-dimensional inverse 281 uniaxial

320 anisotropic half space 275 dyadic 44 medium Green dyadic 136 medium dispersion equation 158 uniqueness 84-8 unit dyadic 21 unit vector representation 13 vector circuits 70 vector potential 64 vector transmission line 200 vector voltage 202 volume integral equation 147 wave equations 57-8 wave vector surfaces 156 wire grid 83 image theory 231

SUBJECT INDEX

Author index Abramovitz, M. 259 Akimov, V. P. 235 Alanen, E. 153,223, 235, 259, 260, 274, 275

Hammond, P. 95 Harrington, R. F. 95, 112, 198 Heaviside, O. 63 Hujanen, A. T. 223, 274, 275

Bagh, H. von 260 Bannister, P. 223 Barber, P. W. 260 Bassiri, S. 137 Bladel, J. van 59,68,83, 137, 148, 182, 274 Bleistein, N. 169 Boersma, J. 148 Booker, H. G. 94, 124, 223 Brand, L. 51

Idemen, M. 83

Casey, K.F. 83 Chambers, L.G. 68 Chang, D. C. 223 Chen, H. C. 94, 137, 163 Clemmow, P.C. 198, 223, 279 Cohen, J. K. 169 Colton, D. 88, 192 Davies, K. T. R. 148 Davies, R. W. 148 Deschamps, G.A. 15, 16, 148, 153 Devaney, A. J. 169 Drew, T.B. 51

Jackson, J.D. 59, 95, 99, 112 Jaggard, D. L. 112 Jones, D.S. 68, 88, 124 Keller, J. B. 223 Kong, J.A. 59,95, 112, 124, 163 Kontorovich, M.l. 83, 235 Kress, R. 88, 192 Kuester, E. F. 223 Lagally, M. 52 Laprus, W. 68 Law, C. L. 148 Lee, S. W. 148 Lehti, R. 52 Leis, R. 89 Leontovich, M.A. 83 Levine, H. 138 Lindell, LV. 16, 52, 83, 95, 112, 129, 138, 153, 154, 163, 169, 174, 182, 198, 223, 235, 259, 260, 274, 275, 279, Liu, C. H. 95 Lohr, E. 52

Engheta, N. 112, 137 Felsen, L. B. 137, 153, 235 Franz, W. 52 Gibbs, J.W. 16,52 Gradshteyn, 1. S. 141, 148, 259 Gvozdev, V. V. 138 Hafner, C. 182

Marcuvitz, N. 137, 235 Maue, A. W. 192 Mayes, P. E. 174 Miller, E. K. 192 Mitter, H. 99 Mittra, R. 223 Mohsen, A. 223 Monteath, G. D. 95 Morrey, C. P. 169

AUTHOR INDEX

316 Muinonen, K. O. 260 Muller, C. 16, 89, 148 Nikoskinen, K. 1. 154, 182, 223, 275 Nirenberg, L. 169 Nisbet, A. 68 Oksanen, M. 1. 83 Papas, C. H. 137, 182 Papousek, W. 68 Parhami, P. 223 Phillips, H.B. 52 Poggio, A. J. 192 Porter, R. P. 169 Przedziedzki, S. 68 Rahmat-Samii, Y. 223 Rumsey, V. H. 95 Ryzhik, 1. M. 141, 148, 259 Schwinger, J. 138 Senior, T. B. A. 83, 124 Serdyukov, A. N. 138 Sihvola, A. H. 112, 260, 279 Sommerfeld, A. 154, 260 Stegun, 1. A. 259 Stratton, J. A. 89, 148 Sun, X. 112

Tai, C.T. 52, 138, 141 Tellegen, B.H.D. 59, 95 Tretyakov, S. A. 83 Uslenghi, P. L. E. 83 Viitanen, A. J. 112, 279 Wait, J. R. 83, 223,235 Weiglhofer, W. 68, 138 Wills, A.P. 52 Wilson, E.B. 52 Wilton, D. R. 198 Wolf, E. 169

Woodworth, M. 192 Yaghjian, A. D. 148, 192, 275 Yeh, K. C. 95