The Inverse Variational Problem in Classical Mechanics

i — 1,2,... n where
Principle,

Linear partial differential equations.

4.2

19

The method of characteristics. characteristics.

Linear partial differential equations. characteristics.

The method of

We shall sketch in short the method of solving the linear partial differential equations, which we shall need in the following, by means of characteristic curves. In subsection 3.2 we dealt with a system of differential equations (3.26), namely Xrr — = JT\E\ Jr\3>l, ■> ■ •■• •• % ^ nij *El) ^ 1 j *• ■* ■• 3-ns X * n s *) ' j >i

r = r1,.. = .n l , . .. . 7 l .

and first integrals or constants of motion (3.30), independent from each other &i — &i = Oj

(\X\ X i ,, . . . X n , X Xx, i , .■. .. .XXn n, ,Cjtj , ,

"(2) (Xi, . . . X X\,... Oj — Oj xn, t j , 0; = C j (X\ , . . . Xni n, Xi, . . ■ X n , t j ,

i = 1,..

I = l,..

.n,

(4.19) ( 4 W ) '

.n,

where a; and &i were constants chosen arbitrarily. Let us change our notation in the following way: zzM = (xi,Xi; (xi,ii; i = 1,... n),U n),p = 1,2,... 2n = in - 1 u = = (a,i,bi; (ai,bi;i i = l,...n) CM l,...n) fl =

(4.20) (4.20)

*M = (Q (1) ,Cf ) ; ? = l , . . . n ) . Then (3.26) can be written ^=^(z,t); at

i/=l,2,...2n = m - l ,

z = ( z 1 ; . . . zz2n 2n)) .(4.21)

Here for ^f = 1 , . . . n z (fv (fv =— ■Zjz+n u+n

and for v = n + 1 , . . . 2n V-V — Jv — n •

Moreover tfM(z,t)=CM,

/ i == l1,... , . . . 22n n = m -—l .1.

(4.22)

Here £ is the independent variable, while z are functions of t. As on the l.h.s. of (4.22) t and z appear on the same footing we could choose, as the

20

The Inverse Variational Problem in Classical

Mechanics.

independent variable, say Z\ as well. Before we do it, notice that we may write (4.21) in the form dt dt

_ dz\ dz^ __

1

fl tpi

dzm_i __ dzrn-i_

(4 231 (4.23)


Moreover, we may multiply the denominators in (4.23) by any function Xm(z, t) different from zero in a domain we are interested in where Xm ^ 0 dz\ dzi _ dz2 _ X\ X2

=_

dzm dzrn Xm

,4

^

24 (4.24)

where Xi - Xm(zi,...

)ipi{zi,... zm)ipi(zi,...

zm)

i-l,.,,m-l

and Zm — ^ -

The first integrals of this system of equations, integrals independent from each other, read, according to (4.22) 3Mz,2m) = C^I1 *^(z,z m) =

ft — 1l,...m-l. , . , . m — 1. fj,=

(4.25)

Using now z\ as independent variable the equations (4.21) become d,Z2 dz2 __ X^ X2 ~dz~[ dzi ~Xi ' Xx

dzm

_

Xm

" " dZ\ ~dz~x~ X\Xi

(4.26) { '

To keep the r.h.s. of (4.26) continuous one has to have mandatorily such initial conditions (z°,z^) that Xi(*°,«g,)^0. Xi(*°,«g,)^0.

(4.27)

A necessary condition for that the expressions (4.25) are really the first integrals of the system of equations (4.24) and (4.26) is that along the integral curves, which we shall call the characteristic curves, the expressions (4.25) should keep its value unchanged; in other words we should have —-^dzi —JLdz! + ... + + --^-^dz 5 - ^ dmz m == 00, , (M=l,...(m-1). H=l,...(m-1). (4.28) dzi dzm oz\ ozm But along the characteristic curves the differentials dzi are proportional to Xi, i — 1,... m; this follows from (4.24). Hence we should have

tx'lt'O1= 1

(4.29) (4-29)

'

Linear partial differential equations.

The method of characteristics. characteristics.

21

Since (4.24) holds true for any choice of CM, fi = 1 , . . . ( m - 1 ) , it follows that (4.29) is true for each characteristic curve. So we come to the conclusion that for each first integral the equation (4.29) is satisfied identically and also each * which satisfies (4.29) is the first integral, provided we put it equal to an arbitrary constant. Let us consider the equation m m

„ p.. j.

*[/] = E ^ = i=i

0 oZi

>

(4.30) (4-30)

where Xt, i = 1 , . . . m, are given functions of (z,zm) and / = f(z,zm) is the function to be evaluated. Taking into account all what was said before, the first integrals of (4.30) are equivalent to those of (4.24). The l.h.s. of (4.25) are the solutions of (4.30) and vice versa each solution of (4.30), put equal to an arbitrary constant, is the first integral of (4.24). To find the most general solution of the equation (4.30) it is in order to make the following remark. For an arbitrary function $* (( ** !! ,, .. .. ..

where ^

(4.31)

** mm -_ii)) ,,

are given by (4.25), we have

where "Q/^ are given by (4.25), we have t m —l

i ,-.

p

.

*"

T

* M=i

M

(4.32) (4.32)

^

The first integral (4.25) are also the partial solutions of (4.30). Therefore The firstX[#] integral (4.33) = 0.(4.25) are also the partial solutions of (4.30). Therefore X[9]=Q. (4.33) It can be shown that $ is the most general solution of (4.30). We are not going to prove it. The reader can find the proof in any text book on differential geometry.*) The geometrical interpretation is as follows. Each integral surface f(zx, ... zm), solution of the equation (4.30), is spanned by a bundle of characteristic curves, depending on (m — 1) parameters. Each characteristic curve arises as an intersection of *'s in (4.25) for fixed C"s. Each surface *' See e.g. R. Courant & D. Hilbert, Methoden der mathematischen Springer Verlag, 1937.

Physik, Berlin, J.

The Inverse Variational Problem in Classical

22

Mechanics.

/ ( z i , . . . zm), generated by such a bundle of curves is an integral surface. To construct such an integral surface it is enough to take a non-characteristic curve and let through each point of this curve pass a characteristic curve. 5.1

The Inverse Variational Problem.Helmholtz's Condi­ tions.

In the previous Section we showed how to get from the given Lagrange function *' L/\X\,

. . . Xrn 3>1 j • • •

%nyt)

the Euler-Lagrange Equations, the Equations of Motion. This Section is devoted to a problem, inverse - in some sense - to the former one: Given a system of Equations of Motion, what is the method to find out whether these equations correspond to some Lagrange function as its Euler-Lagrange Equations. This problem we shall call — the Inverse Variational Problem. In general, not for every system of Equations of Motion a Lagrange func­ tion exists, which yields these equations as its Euler-Lagrange Equations. It can, however, happen that there are modified Equations of Motion, which yield the same set of solutions as the original ones and for which a Lagrange function can be already found. We shall come back to these questions later, discuss them in detail and give some simple, illustrative examples. The necessary and sufficient conditions for that a given set of equations coincides with the Euler-Lagrange Equations of a certain Lagrange function was given by Helmholtz^' . These conditions form a system of identities. As we restrict ourselves to Lagrange functions of first order of derivatives of x with respect to t the equations under consideration have the general *' We shall use hereafter again the variables x instead of q as we shall consider them as independent from each other. They are functions of the independent time variable t. t) H. Helmholtz, Journ. Reine und Angewandte Mathematik 100 (1887) 137. As a matter of fact Helmholtz showed only that these identities are necessary for getting the Lagrange function; that they are also sufficient was proved by Mayer (A. Meyer, Berichte der Kaiserlichen Gesellschaft der Wissenshaften, Leipzig, Math.-Phys. Cl. (1896)). Hirsch and Boehm (A. Hirsch, Math. Annalen 50 (1898) 429, K. Boehm, Journ. Reine und Angewandte Mathematik 121 (1900) 124) generalized this system of identities for higher derivatives. The problem was elaborated once more in great details by Konigsberger (L. Konigsberger, Die Principien der Mechanik, Leipzig, 1901).

The Inverse Variational Problem.Helmholtz's

Conditions. Conditions.

23 23

form*' n

^2G +Gi(x,x,t) ^Gij(yL,±,t)x ij{x,ii,t)xj+Gi(x,iL,t)=0,

= 0,

ji = l1,, . . .. nn . .

(5.1)

t=i

Notice t h a t these equations are linear in x, otherwise they could not correspond to any Euler-Lagrange Equations of a Lagrange function L ( x , x , t ) , viz. dL _ , A d2L dxz ~^d^dij ~^diidijXjXj

d fdL\ dt \diiJ \dii)

.. +

A d2L _dL__ _d£_ ^ d^ ^d di i dj Xi jj ~X jd~x d~ x~~° ~ °

(( 55 2 ))

We have the following Theorem: The system of equations (5.1) is a system of Euler-Lagrange Equations for a certain Lagrange function iff G y and Gi, if j — 1 , . . . n, satisfy the following relations (Helmholtz's Conditions) Gij C jjj dG

oGij ij dxt dxk dGj dij

= —

Gji, J (jji

(5-3) (5.3)

=

dG

,5 ^

oGjk ik dii d±i

dGj dii

8Gi_dG£ dGj dGj dx dij d±i z

(5.4)

IV

fa

*

dt

)'

(v . _d_ d \ (dCh _ 8GA 2 Xk Xk + ^\^Y^ dxk dxk dtjXdxj dt) [dXj dxi) dxj

l

~

(5.5) (5.6)

[t>

*>

Proof;*) Let us assume t h a t there exists a Lagrange function L, for which the system of equations (5.1) coincides with its Euler-Lagrange Equations. Then Gij

Gi

d2L

(5.7)

dxi d±j yy ^, d2L . £-*>d±i L—I dii dxj dxi J jj

JJ

J

d2L _8L_ dL d±i d±idtdt dx; dxi' ' ■

(5.8)

■

From (5.7) follow immediately the conditions (5.3) and (5.4). *' see e.g. E. Engels, /( Nuovo Cimento 2 6 B (1975) 981 or P. Havas, Supplemento Vol. 5 Seria X del Nuovo Cimento (1957) 363.

al

t) The version of the proof of the theorem, as given here, was presented to me by Dr. J. Cislo.

24

The Inverse Variational Problem in Classical

Mechanics.

Let us differentiate (5.8) with respect to xk-. we get dG dG±t _=s-^dG^. YdGikx dx *-? dxkk ^r* dxj 9x, J

3

J J

ddk | Q2L d2L | dGik dt dxi dx dxkk dt dxi

2 d L Q2L dx dxtt '' dxkk dx

(5(5.9) 9)

where we made use of (5.7). If we interchange in (5.9) the indices i and k and so obtained relation add to (5.9) we get (5.5). If we subtract it from (5.9) we obtain dGi _ dGk / d L _ d L \ P-P^fJ^-J^) dx d±i \d±idx dx dxij 2

=2

dxkk dXi From (5.8) we obtain

\dxidxkk

(5,0) (5.10)

2

dxkkdxij

dd dGk _ . d + A d\ ( f & d2LL dG1_dG^ (Y±._
d2L )\ _ d2L dxidxk)

(5(5.11) n)

If we insert the l.h.s of (5.10) in the r.h.s of (5.11) we get the last Helmholtz's Condition, (5.6). In this way we showed that the relations (5.3)-(5.6) form the necessary conditions for the existence of a Lagrange function. Now we are going to show that these conditions are also sufficient. If dj and Gi i,j = 1 , . . . n satisfy (5.3)-(5.6) then there exists such a function L(x,x,t) that (5.7) and (5.8) are the Euler-Lagrange Equations for L. To start with notice that from (5.4) follows G g , °ij, = - | dx v

'

(5.12) (5-12)

due to the Converse of the Poincare Lemma. Condition (5.3) tells us that dKj _ dKj d±i dij or, or, using using again again the the Converse Converse of of the the Poincare Poincare Lemma, Lemma, that that

K ^-K(^t). Kj3 = = g-K(x,k,t).

(5.13) (5.13)

2 K dd2K d±i dij dxi dij

(5.14) ^' '

Hence Hence G.

lJ

=

The Inverse Variational Problem.Helmholtz's

Conditions. Conditions.

25 25

To proceed further we introduce some auxiliary functions Mi i = 1,2,3. defined as follows M M

d2K d2R

y ^ 8d*K 2 R , T" x ^—* d±i diidxk dxk ^r"

=C

dK_ — diidt dt dxi dxi dii

(5.15)

Let us insert (5.14) and (5.15) into (5.5); it follows that dMj dMi dMj l _ „ H = 0. dij dii

(5.16)

Hence d2Mj

d2Mj 9

dijdik

diidxk diidik

d2Mk diidij

d2Mi dikdij

consequently x = ]Y^ Mi = P ik(x,t)i Xik (x, t)i k+m{x,t). k + Mi(x, t).

(5.17)

k

Taking into account (5.16) we get (5.18)

\Kk k — — -A/ti. ~"ki ■

Let us insert (5.15) into (5.6). We have ^ / d3K *-" \diidxkdxj

dMj _ dMj_ dMj dxj dxi dii d3K 8

d3K

dii dxj dt

dij dx{ dxi dt

d3K \ . dijdxkdxi)

d2M d2M _ iW > > ) ±K i 2 ^f\dijdx diidxk) 2 k ^* \dijdxk diidxkj ^fd22Ml d22Mj\ d Mj\ {1(d Mz 22 \ dij dt dii dt) \ dij dt dii dt)

y

/

d3K

11

L— \dii dxk dxj t—

d3K dii dxj dt

d3K

(5.19)

\ .

dij dx dxkk dxi)

d3K dij dxi dt

or or dMj _ dMj_ _ 1 fy, . _d_ d_\ (dM _ dMj\ W « & = i2 ( E 4 a + » W M t _ « £ \ i \ ~ dxj dxi dx dtJ \dij dii dxj dxi \f-f dxkk dt I \dxj dxi ))

,5.20, (5.20)

26

The The Inverse Variational Problem in Classical

Mechanics.

Notice that we could obtain (5.20) without writing out (5.19) as t h e identity (5.6) holds for L as well as for K, so the terms containing K have to cancel. Let us insert (5.17) into (5.20), taking into account (5.18). We obtain sp y ^ fdXjk fd\jk £7? v dxj dxj £7?

dXjk \ . d\jk dxi ) dxi

djij_ _ _-s^. dfM _ df^i \ p . dXjj dxj dxi l£-* --J dxk dxj dxi dxk

dXjj dt dt

or

dXik

| ^ dxj oxj

dXkj

dX^ _

(5.21)

+ ^ + ^ = 0 dxi dxk axi axk

and QjH dm dxj

dfij _ dXjj dm dxi dt

(5.22)

as neither A^ nor fii depend on x Using the Converse of the Poincare Lemma we get from (5.21) and (5.18) t h a t there exist such functions \£i(x,£), i = 1,2,3, t h a t

cm_d%

V a a=f l * i _ f l * £

((5.23) 523)

lJ lJ

dxj dxi dxj dxi Then (5.22) can be written Then (5.22) can be written

_d_f

dXj y* ^ ("

_ 0*A

-

_d_ (

Mj at ) = ~ dxi IT) WiV("'

dVj\

)■ ~ dt-ft)

•

(5.24) (5 24)

-

Using once more the Converse of the Poincare Lemma Lemma we get from from (5.24)

(5.25) (5 25)

-

dt dxi ^-W-Wt' where $ = $ ( x , i ) . Finally from (5.17), (5.23) and (5.25) follows where $ =' "$"( xV, iV) .&Finally fromXj+(5.17), dt (5.23) dand (5.25) follows cj dxi) Xi

(5.26)

Let us define an auxiliary function Let us definei an auxiliary function where ipM and $ are functions + used before. We have then

* =d MY, .*^d M* > ^ 2

2

__ We have then where ^ and $ are functions used dM before. 4 2 dxj J 2 dt -~ dxi dxi dxi ^ dM . dM dM __ 3 ! -~4 5 i i 3a;,- J S i j d£ dxi '

(5.27)

( 5 - 27 ) (5.28)

J

3

J

\ ■ )

The Inverse Variational Problem.Helmholtz's Problem. Helmholtz's

Conditions. Conditions.

27 27

As M is linear in x we may write (5.14) in the form

G

dd22(K {K + + M) M)

=

, v (5 29) (5.29)

- = 4x^r

-

diidii

On the other hand from (5.15) and (5.28) follows _ yd y 2{K dHK + M)^ M) . *-* dii *—' d±i dxk dxk

dd22(K (K + + M) M) d±i dii dt

+

+

d(K d{K ++ M) M) dxi

(5.30)

From (5.29) and (5.30) we conclude that there exists a function L=K +M

(5.31)

which satisfies the equations (5.7) and (5.8). This accomplishes the proof of the theorem. It is worthwhile to emphasize the fact that the Conditions of Helmholtz were found in the second half of the 19-th century! In some papers a system of five Helmholtz's Conditions, instead of four is used. The 5-th condition reads 22 dGik1L_dGj _ dGj^ G% _ d22Gj \\ 1^l_ = i /t d Gj dxj dxi ~~2 2 \ \dik dik dij dij dik dik dii diiJJ

(5.32)

We are going to show that the identity (5.32) follows from the identities (5.3)-(5.5), thus it is superfluous as it should, taking into account the Theorem stated above. To prove this assertion*' notice that from (5.4) and (5.3) follows (5.14). Let us insert (5.14) into (5.5). We have dG% d±j dij

|

OGj - 2 ( y ± ^ dii ~ l\~j?

d x

dxk ®k

,

d

\ &K dt d±id±j ®t JI dii dij

((5.33) 533)

We look for the solution for Gi, i — 1,... n. The particular solution we get from (5.33) by taking i = j , 8Gj _ d8 fy dGj (y< . d2K \ _ d2K dij dii \~f d±i dxkdxij dxkdiij dxidii

d3K dx\di\dt dt

(5.34)

A solution of (5.33) with respect to Gi reads

r - VV "^

dt2R d2R dxk dxk dii dii

-— —

dxi dxi

d2R

(5.35)

dii dii dt

*) This result was known for some time. We follow in our derivations the reasoning due to Dr. P.C. Stichel.

28

The Inverse Variational Problem in Classical

Mechanics.

The general solution of (5.33) is G, = Gi — Gi Gi ++G° Gi ,,

(5.36)

8Gl + dGl = Q §2 + f? = 0. dij dxi

(5.37)

where (5.37) axi Let us insert (5.36) and (5.14) into (5.32), keeping in mind (5.35) and (5.36). We get OXJ

d3 K dxj diidik dxj diidik

d3 K dxidij dxidijdik dik

id d fy d2K dK | d2K | 2 dik dij \/-^1 dxidii diidt dxidxi dxi dxidt / 2 1i dd dd fy fy K dK | dd22K K dd2K dK | 2dikdii I " dxidij dxj dijdt 2dikdxi 3

3

dK _ d8 K dik dij dxi dx{ dik dii dxj

2

A J

%

N

| CA = | c0\ = JJ y

(5.38) (5-38)

y

2

1 / d8 G° 2 I dik dij

d_d^G]_\ G°j \ dik dii I

Since according to (5.37) ad2G° G° dik dij

d2G° dik dii

d2G°k dii dij

|

d922G° G»k dii dij

= Q

(5.39)

we get from (5.38) the desired result. Treating the Inverse Variational Problem for few point particles one cannot avoid to use the Helmholtz's Conditions, sometimes in a disguised form. They are useful in case one tries to remodel the original Equations of Motion, which do not admit a Lagrange function, in such a way that the newly obtained equations are already Euler-Lagrange Equations of a certain Lagrange function and, in addition, still have the same set of solutions as the original ones. Such approach was used e.g. by Havas*' and several other authors. In many mathematically more complicated cases, however, the identities of Helmholtz turn out to be of limited applicability in practice; for large number of equations the computations become tedious and complex. Everything depends, of course, very much on the model under consideration. *' P. Havas, Supplemento

al vol. V, Sena X del Nuovo Cimento (1957) 363.

The Inverse Variational Problem.Helmholtz's Problem. Helmholtz's

Conditions. Conditions.

29 29

If the matrix Gy in (5.1) is invertible we may write (5.1) in the normal form, i.e. Xi = fi(x,x,t), ^ 3 *) •&i — Ji\X~i

i ==l 1, ., . ... n. n.

i

(5.40)

If the 3-rd Newton's Law is observed, then (5.40) represents the Newton Equations and the r.h.s of (5.40) is the force divided by mass. The normal form (5.40) is useful in comparing systems of equations, say, Equations of Motions. Two systems of equations having the same normal form, have the same set of solutions. Such systems we shall call equivalent. We have the following statement:*' The system of equations (5.40) is equivalent to Euler-Lagrange Equations of certain Lagrange function iff there exist an invertible matrix G^ which satisfies the Helmholtz Conditions (5.3) and (5.4) as well as

£(^ + ^J + s|G« == - I E («.£+<*&)

^{fkdx-k+Xkdx-k)+dt\G-

and

(5.41)

+

^(?H ^(?H= i i + G G 3 V

*

l

I

y

'

(5.42) (5.42)

= i{^ ^ i)^{ ^- ^

To get formulae (5.41) and (5.42) from (5.5) and (5.6) one has to make use of (5.3) and (5.4) we well as from the formula (5.43)

Gi = = - £2_^ Gik
which one can obtain combining (5.1) and (5.40). The operator on the l.h.s. of (5.41) can be written

sp-ff JL fk

^{ d±k

+Xk

dxk

d +

9\ _ d dt) - dt

*' The statement as well as following it derivations are due to J. Cislo.

(5.44)

30

The Tfte Inverse Variational Problem in Classical

Mechanics.

under the proviso that it is taken along the trajectory (otherwise called substantial time derivative). For further use it seems to be advisable to put Helmholtz's Conditions in a more elegant form. Notice that by adding to both sides of (5.42) the expression

lE/^E^a^-^at)

(5.45)

we may write the r.h.s. of the new expression as

id Wr

dfl

dfl

r

\

(5.46)

We are going to show that the l.h.s. of the new expression becomes V (n

d k

f

dfk

r

(

7) "(5.47)

\

E H - s ^J j) 4A^ a '*a^ G

x

k

_G

J

At At the the start start the the expression expression under under consideration consideration looks looks as as follows, follows,

&r. {^Gikfk) ~ d^i y£Gjkfk)+ +IE^E(G«^-^H) = E

n

dfk v-^ n

K

rsr^ dGik

dfk

K

,1 a

/

L. ™

a/,

v-> dGjk ,

(5.48) ( 5 - 48 )

ft

a/A]

v ifc + /fc Gtk Gil G i I fk ^d^ dT d±l)\aiJ] *a^r a%~ k\\ i-

We are going to show that the expression in [...] parentheses in (5.48) vanishes. To this aim we use (5.41) as well as (5.3) and (5.4). Let us differentiate with respect to ±j. We get

y> fdhdGjj y /a/ f c aG t j | aii dik aij; ■^ \ ai;

2 ad*G cfcJkj aij d±i ai; d±i

|

a22GtJtJ \ || ag^ a o ^ {{ a^ d±i ai* Jy a^j dxk dxi

d2 c , 1 v ^ / f l g q A/it , dGjk dfk lJ + 2 + d± + tJ a7aiT ~^{'d^d^ ^irdi' dtd±i Y v dil d±j ' dii i , rr d*f a2/fck r d*f a 2k/ , \\ _

,

+

(j + « t ^T +(Jik TT.

WT~ Q.. Q Q. — UU .. WT~ + + (*jk djk Q . JI —

axj axj oxi ax;

oxi a^i oxi jJ

(5(5.49) 49)

-

The Inverse

Variational Problem.Helmholtz's

Conditions. Conditions.

31 31

If we interchange in (5.49) the indices i and I and so obtained expression subtract from (5.49) we obtain d2fk y dhdG^ y G/. &f k + y .. dh dG^++ dG^ dG^ || %I_Y( + dGj^dfk\ ^L^A) __ 2 2 *-f \ * dii dij ^r" dii dx dxi dx £-* dii dxkk dxi *-? \ * <9i( <9ij <9i;t dii) dii) 2 _ y d_h_d_G dhdG^ }L _dG _dGilll_l1y_y f d fk + dG dGjkihdh\ dfk\ = 2 2 *-? dii <9± dxfck dxii , £-* \ d±i dij ^-* <9±i d ®%i dxj dxi 9xidii diiJ J k \ = 1i fdhdGis f dfk ddj BfkdGiA dfkdGiA (| = 2

\di[ dik

dii dik )

i 1 T (c„ * ^hh 2

^-* V

l

dxi dij

&2fk Q2fk c cu. ) i dGij lk

dii dijJ

dGli dGlj

dxi

==0. 0.

dxi

or or

dG1L_dG &?„ ft?,,ll 9a;j dxi

dxi drr*

d y/dhcfdh ^ + | li_d_y 2

dij £-* ^-" \dii \ dii

_ dfk df\kc\==Q Q '

di Sii %

/)

(5.50)

which proves our conjections and simultaneously (5.47). Thus

E(^t-^t) = l|E(^t^-t)-(5-51) The remodeled system of Helmholtz's Conditions (5.3), (5.4), (5.41) and (5.51) can be written in an elegant matrix form. We introduce the following auxiliary matrices

* _ ! dxj £ y

^ and

*„.=
(5.52) (5.52)

Then we have*' T

GT = G, G, 9Gjj _ I9GH dG liz=dGu )

9iit 9±it

9 i; dij

(5.53) ( 5(5.54) 54)

TT |^-G G == - |- (|2(G G)) ,, vG$ $- -$$ G dt '

(5.55) (5.55)

G ** -- **TTG G == £i -j^( -G( G * $-- $$TTG G))..

(5.56)

•) see F . Pardo, J. Math. Phys. 3 9 (1989) 2054. *'

The Inverse Variational Problem in Classical

32

Mechanics.

To find the matrix G, which has | n ( n + 1) unknown entries in an n dimensional space, Douglas*) proposed the following procedure. Let us substitute the r.h.s. of (5.55) for ^ in (5.56). We obtain GM{1)

-Mil)TG

= 0,

(5.57)

where # » = J - i - + -$2 (5.58) * dt 4 Let us differentiate (5.57) with respect to t and in so obtained expression replace ^ again by the r.h.s. of (5.55). We obtain GM{2)

- M(2)T G = 0 ,

(5.59)

where M ( 2 )

^dM^ _ ^

( i )

+

^

( i ) $

^

6 Q )

We may continue this iterative procedure and get GM{3)

- M{3)TG

= 0,

(5.61)

where M(3)H^M^_|$M(2) +

|M(2)$)

G M ( 4 ) - M ( 4 ) T G = 0,

( 5 6 2 )

(5.63)

where M(4)

_ dM™_ _ i $ A f ( 3 )

+

^M(3)$ ^

(5 64)

and so on. All these equations are linear and homogeneous in G. The coefficient matrices M^1' i = 1,... n are known. We need \n(n + 1) equa­ tions to evaluate the matrix G. Douglas has shown that the existence of a solution of G for which det G ^ 0 implies the existence of the Lagrange function. *' J. Douglas, Trans. Am. Soc. 50 (1941) 71. I am grateful to Dr. J. Cislo for calling this to my attention.

Theorem of Henneaux.

5.2

33

Theorem of Henneaux.

Let us consider the case of a system of equations, which turn out to be the Euler-Lagrange Equations for a certain Lagrange function. Then we assert that there exists always infinitely many other Lagrange functions, yielding the same equations as their Euler-Lagrange Equations. This state of affairs is precisely defined by a Theorem of Henneaux. *' Before we formulate this theorem let us make four remarks and give some definitions. In these notes we shall deal with regular Lagrange functions only, such that the canonical momenta Pi = • 5 T - , dxi

i = 1,... n

(5.65)

can be defined and we may use either the variables (x, x) or (x, p); thus we assume that the system of variables is invertible i.e.

det

(S) *°

ae

(566)

-

The definitions we are going to use are as follows i) two systems of equations are called equivalent to each other, when they have the same set of solutions, ^ ii) two Lagrange functions are called s-equivalent, when their EulerLagrange Equations are equivalent, iii) two Lagrange functions Li(x,x,t) and L 2 (x, x,t) are called equiv­ alent if a) they are s-equivalent and 0) the Poisson brackets r .i

v~^ ( da

^b

da db \

.

.

*> M. Henneaux, Annales of Physics (NY) 140 (1982), 45. t) Notice the remark made in Section 5.1 that two systems of equations, having the same normal form, have the same set of solutions.

34

The Inverse Variational Problem in Classical

Mechanics.

of x and x satisfy the relations [Xi, Xj\fJ1

—

(X[Xi,Xj\L2

[xi,Xj}Ll

=

a[xi,Xj]L2,

/

0,

a £R

,

i,j =

(5.68) l,...n.

Notice that we assumed regularity of the Lagrange functions (see (5.65) and (5.66)). The Theorem of Henneaux tells us that: Two Lagrange functions Li(x,x,£) and L2(x,x,t), are equivalent iff dg{x,t) L2 = aLl +

. A dg = aLl +

^tT

dg

^dx-iXi+m>

(5 69)

"

where a £ i ? / 0 , x = ( i i , . . . xn). Proof: Let us introduce some auxiliary quantities zll = {xi,ui = xi,

i — 1, ...n),

a M (x,u,t) = ( — ,

fi = l,...2n

i = l,...n,0,...0)

(5.70) (5.71)

n

pr

_ daT ~ dzp

dap dzT

or /

d2L 82L dxTdus dxsdur r,s — 1 , . . . n

82L \ duTdus r, s = 1 , . . . n j

cr=

.

(5.72)

2

\

dL duTdus r,5 = l , . . . n

/

The (2n x 2n) matrix a is not singular because of (5.66). One can show by direct computation that -o-pT = {zp,zT}

(5.73)

35 35

Theorem of Henneaux. Henneaux.

where i,

b

dXs ,-, _ s *.i - v-» V^ ffdx

ddp P*s

dxsdxsdps\ dps\

(5.74)

^ ) =s E ^ ^ - ^ ^ 'J

(5-74)

v

are the s.c. Lagrange brackets, inverse to the Poisson brackets (5.67). *' *' In the book by E.T. Whittaker, A Treatise on the Analytical Dynamics etc., Cambridge Univ. Press, 1952, the Poisson brackets are indicated by ( , ) and the Lagrange brackets by [ , ]. We are going to use a different notation: the Poisson brackets are denoted by [ , ] and the Lagrange brackets by { , } For convenience of the reader we shall present here the connection between both brackets. We use the ad hoc notation (xi,X2, • • • X„,Xi,X X„,±i,X2,2 , .. .■. ■ X„) Xn= ) = (91,92,(91,92,• ••<72n) • <72n) • • We construct the matrix MM„ = = {
p.,u p.,v = = \,2,... 1,2,... In. 2n .

Then we have

k,>^] == ((M-% M-1)^ = [ ( M - 11)TT ] ^ \*»*>] tt = [(»*- ) ]IUf Since

we have we have

dL -_ dL Pi = -rrr r . axi axi

i = l,... l,...n, n,

{xi,Xj} {Xi,lj}

= -

{Xi,Xj}

=

g ? Q±. = Tij = Tj, = -{xi,Xj}

{ii,±j}

=

0,

g±.9x.dx.

p - gdijdxi-fv= Pij = -Pji i'', i . dx.

i,j = 1,... n

and

" = ( ^ ) = -

M T

This entails

(^^-^-(Alr-V-l) or

[xi,Xj]

= 0

[xi,Xj]

= - (

[ii,Xj] [xitij]

= =

T

~ %

=-[xi,Xj}

1 -(r^pr-(r^pr-1)..

= {Xj,Xi} ,

36

The Inverse Variational Problem in Classical

Mechanics.

For example for r, s = 1 , . . . n t

i1 _ r 1

1

''

s

' '

i] _ vV ^^ (dx'

d

Pi ®Pi

4 £-* \ dxr dxs t-

8d2L duTdxs

dxi dxi

d l

d r P \\ __ d?V Qgj P

dxs dxr JJ

dxs

d dp P*s

dxr

_

d2L dusdxT

We are going to show first that the condition of the theorem is sufficient. We assume (5.69) holds true. Then dLi _d_ dL2_d \ L(dL1fdL£\ dxi dt \ dii J)

=

d±i J J) \ dxi dt \ dii d d dg(x,t) d dq(x,t) dxi dt dt dii dt (dLx dd /(6Li\\ gM\ | = afdLi \ dxi dt \ dii di{ J / 22 2 dg d22g " || A || d2g . i *r? dxi dxk dxi dt ^— dxi dxj 3 fdln = a(dLi \ dxi

(5.75) ( 5 - 75 ) d22g dxi dt

SdLi\\ _ d (9LA\ dt \ dii J J

Thus L\ and L2 are s-equivalent. From (5.69) follows dL2 dii

dL\ di\

dg dxi

or using (5.71) aa,[L M[L22]J

=

V

=

dd Vv

rr r r i1 + — aa,[L aafl[Li] l] + — ,,

^ (5.76)

V(z,t)=g(x,t).

Since J)_dV_ _ d dV _ _d_dV_ dzT dzp dzp dzT we have from (5.72) apT[L = aa Op aapTpT[L [Li] 1] r\L22}\ =

(5.77)

Theorem of Henneaux. Henneaux.

37 37

and by virtue of (5.73) and (5.67) a[z «W,ZT]L p,zT]3L2

= [ZptZr]^. [zp,zT]Li .

(5.78)

This accomplishes the first part of the proof. Now we are going to show that the condition is also necessary. If L\ and L-2 are equivalent then (5.72) implies 0fir[L = aapTT\Lx}. [Li]. fffiT[L22]] =a<Xp

(5.79)

According to (5.72) and (5.79) d2L2 _ d2Lx - =a a rduSs a dur5~~ du dus ouraus ourous These relations entail

5(5.80) 80

-

3L2 8Lx fl = a^+g s{x,t).

ou ou$$ Again, according

d2L2 _ dxrdus dxrdus But (5.81) implies But (5.81) implies 822L2 d L2 dx dxTdu dus T

(5.81) (5.81)

ou ouss to (5.72) and (5.79), we have

s

8d2L2 =_ a (fJPLi_ d2L, _ J^n_\ 32LX \ dx du \dxrdus dxs durJ dxssdur r \dxrdus dxsdurJ d2L d2L22 dx dxssdu duTT

t d2L ( d2Lxx \dx \dxrTdu duss

d2L \ _ d2Lxx \ _ sduTJ dxdx sduT)

__ U9s_ _ d9r_

__ (Jajf dgs_ _ (Jdgr_ Jb $

(5g2) (5.82)

,(5.83) .„„, (o.ooj

Hence dgs _ dg^ dxs dxr and by the Converse of the Poincare Lemma we get and by the Converse of the Poincare Lemma we get dg(x,t) 9s = fc.Sfefi ~dxT

(5,4,

(5.84)

We We have have obviously obviously the the equations equations dL OL22 dxi

y d22L22 A

2-* dui dxj

u J

y^ 0d22LL22 „.. *— dui duj 3

_ 8a2L2 ^ = _Q dui dt

(5.85)

38

The Inverse Variational Problem in Classical

Mechanics.

From (5.81) and (5.84) we get d2L2 duidxj

= a

= ot

dd22L, U duidxj

dd22gg dxidxj diidxj

+

-+■

(5.86)

and 82L2 _ d2Lx dui dt dui dt

82g dxi dt

(5.87)

Let us substitute (5.86), (5.87) and (5.80) into (5.85). We have ^T-y d2Li £-* dxi £-? du\ dxj

dh2 dL dxi

A aA

4

J

2

A d2g £-* dxi &-? dxidx; dxj

3J

dd2U L, . *—' dui t-r" duidui dui : :

~

J

a

2 ad L, fl

dui duidtdt

33

2

dd2gg == QQ dxi dxi dtdt

J

37

or, taking into account the Euler-Lagrange Equations for L\ 8L2 dL 3iv dxi

A d2g 4 —^dxidxi ^r dxi dxi

u

/

33

ad22ffg dxidt

= a

9 dLj L! dx; dxi

(5.88)

J

3

Formula (5.88) can be also written dL d dg dL21__d_dl dxi dxi dt

dLi dxi

= adL1

(5 89)

(5.89)

From (5.89) follows

L2 = = aL aL11 + ^d-^A f(u,t). ^ ++ f(u,t). dt

(5.90)

Let us differentiate (5.90) with respect to us. We get dL2 _ dus

dLi dus

dg dxs

df dus

(5.91)

If we compare (5.91) with (5.81), taking into account (5.84) we conclude that

'= ' M - ^ Thus / can be incorporated into 4|, viz gS = = g(x,t) S p ( x , t+ ) + FF(t). (t).

(5.92)

The matrix a. Tr cra is conserved

quantity.

39

Hence (5.90) coincides with (5.69). This accomplishes the proof. The term dg(x,t) dt in (5.69) we shall call - the gauge term. We are not interested in Lagrange functions equivalent to the origi­ nal one. This set of Lagrange functions is a trivial generalization, void of physical significance. What we are interested in is the problem: are there such Lagrange functions, which yield equivalent Equations of Mo­ tion (s-equivalence) but are not equivalent (in the sense of the Theorem of Henneaux) to the original one as well as among themselves. It is worthwhile to notice that if we multiply a Lagrange function L by a number a ^ 0 and get the Lagrange function aL this affects the canonical momentum, the Hamilton function and the Poisson brackets; the canonical momentum as well as the Hamilton function have to be multiplied by a while the Poisson brackets have to be divided by a. The Equations of Motion (Euler-Lagrange Equations) do not change essentially, as we may divide them by a. What remains, however, unchanged under this procedure are the Newtonian Equations (Equations of Motion in their normal form).

5.3

T h e m a t r i x a. Tr aa is conserved quantity.

There remains the problem to be solved: what is the structure of the whole set of equations belonging to the same set of solutions. Definitely not all of them would be linear combinations of the other. But we have here to do with a special case of equations, Euler-Lagrange equations, which are linear in x. To make thing clear consider two s-equivalent regular Lagrange func­ tions L\ and L2. ^ The Euler Lagrange Equations read (i = 1 , . . . n) ^ d2Lx ., *-*'di~di'Xj

A d2L, . 4-diidXjXj

ud^LL_dL1_ +

diidt

loyjj

dx{ ~

and A ^

d2L2 .. dii d±iXj

+

A d2L2 . ^ dii dxiXj

+

^d^L2__8L1_ d^ dt dxi ~

*' See section 5.1 concerning this problem in context of Helmholtz conditions.

l

°'y4J

40

The Inverse Variational Problem, in Classical

Mechanics.

Since the Lagrange functions are assumed to be regular (Hessian different from zero a.e.), we may multiply both sides of (5.93) and (5.94) by the inverse matrix of the Hessian

J ki

L

-I ki

Then we get from (5.93) and (5.94) the equations in normal form, viz. d2L, 2-.di.dxjx3

.. _ A (1) (dig Z^Ww \ d x .

d2L,\ dxidtj-

.

v

Xk

(5.95)

■xt = v o{2) fe - T -^-± - *h.\ k

z_, vti

I

ai

.

l_,

= /i 2) (x,x,f).

d±. dx.

J

dx.

-

dt

(5.9fi)

Since both equations - by assumption - have to have the same set of solu­ tions the r.h.s. of (5.95) and (5.96) must be equal. In other words d2L2

E diidxj

3

_v

d2L2

dL2

diidt

dxi

fv

" Zf au [y

d2Ll

■

d2Ll

dLl

dx dx Xj + dx dt

i

J

dx

i

(59?)

\

i) '

where

™=s:(«2>r <®-±£fc(£T) k

k

\

■ <m / kl

Let us insert the r.h.s. of (5.97) into (5.94). We get

A

£

d2L2 ..

A

/A

Q—Q-^Y*

d2Lx

.

+

d2Lx

dLi\

\L. gjr^'l d±7d-t-^J

,

-

Let us multiply (5.93) on both sides by the matrix a and subtract it from (5.99). The result is d2L2

A

d2Lx

a-a- = L ^ a - a - .

. .

n

«,i,= i,..-n.

s

= 0 (599)

(5.100)

The matrix u. Trcr" is conserved quantity.

41

Hence by virtue of (5.97) and (5.100) A ^

d2L2 ., d±i 8x4Xj

+

y . d2L2 , ^ - 8±i 8x4Xj

3

d2L2 dii dt

+

dL2 _ dx, ~

3

(5101)

-pJt^+ttSt,^ |

d2Lx 8xk dt

8LA 8xk )

which proves the conjecture that the new equations are linear combinations of the former ones.*' We are going to show that Tr aa, where a is a natural number, is a Constant of Motion. The most simple case is that of n = l.t) Then (5.97) reads 82L2 . dx 8x

82L2 dx dt

8L2 _ 8x (5-102)

-*{x,xtt)(dLlx

I

\dxdx

d L l

dLl

) dx J

dxdt

Let us differentiate both sides of (5.102) with respect of x and use (5.100). We get 83L2 dx2 dx

83L2 dx2 dt . d ( 82LX\

X

a

2

+

d2L,\

d (

a

2

= dx-{ -di -) di{ -8i -) _da_.&U_ ~ dx~X dx2

z

. d Lx 8x28x

+aX

8a f. d2Lx x

~ dx [ dxdx

= 2

+

da d Lx dt dx2

8ZLX dx dt ~

+CJ

( d3Li

#%__^M +

dxdt

2

+a

2

x

dx ) \dx dx

.

d3L! \

+

di2dt)

or da (. 82LX x

dx \ 8xdx

PI*

dxdt

8a . 82LX

8LX \ +

x

2+

dx ) dx dx

da&Li

dt d±

*' I acknowledge the discussion of these topics with Dr. A. Borowiec. t> D.G. Currie & F.J. Saletan, J. Math Phys. 7 (1966) 967.

_ 2

.

(

u

lU6

- -^- >

42

The The Inverse Variational Problem in Classical

Mechanics.

Because of the Equations of Motion (5.93) equation (5.103) can be written da ox

da ox

da at

d dt

(N.B.:

x=

f(x,x,t)) (5.104)

valid on the trajectory. We divided (5.103) on both sides by qg^- ^ 0, a.e.. Equation (5.104) implies a ±a« = 0. ^-a = 0.

(5.105) dt at To prove our conjecture for n ^ 1 we shall use the Helmholtz Conditions in the elegant form (5.53)-(5.56). *) Notice that we have Ci

G 13

^

d2L d*L = d±i a ^d±j"

(5.106) (5 106)

-

and according to (5.43)

fk. fk ■ Gi = -Y^G - 2__,ikGik

(5.107)

k

We define G2 aG\,, G2 = = ad

a =
(5.108)

According to (5.55) J/-(

f

= -i(G* + *rG)

— = - | ( G # + #TG)

and and

(5.109) (5.109)

| ( < r t ! ) = * 0 + , f = -|(.G* + * ^ G ) .

(5.110) |(,0) = * * +„ * = -l(„G*+ *VG). (5.H0) If we multiply both sides of (5.109) from the left by cr and compose the result with (5.110) we get TT = \{a* | ( C TT*GT G - #3>
or

f- K ^»-•»,).

*' This way of proof was presented to me by Dr. J. Cislo. *'

(5.111)

The matrix
43 43

Hence we get _ da d_ T r — - = — Trcr =0 0. dt dt

5112 (5.112)

^t-Jt^^ -

(

)

aa l

If we multiply (5.111) by aa aa ~ ,*, a a natural number, we get a 1 aa°-l aa -

^ ^

dt

= =

± a* ~y

dt

= T _ = I^OQT ^( - ( T "^-^T " 1 ^ a) ) . V

2

(5.113)

'

This This entails

i^-o-

— Tro-a = 0. (5.114) at This accomplishes t h e proof. To wind u p t h e discussion of the matrix er we are going to show t h a t _ dp2,i &ij = ^ ,

(5.115)

where p is a canonical momentum where momentum defined by by dL Pi = ^ ,

i= = 1,2,3. 1,2,3.

OXi OXi

(5.116)

To show (5.115) let us use t h e Poisson as well as Lagrange brackets. We have d2 L . . a . = {Xi,Xkj = {Xk,Xi} = OXidx OXidxkk a

v-* fdxj / dxj dpj £-i \ dxi dxk

(5.117)

dpi dpj dxj \ _ dpi dxi dxk ) dxk '

Relations (5.117) a n d (5.100) entail {xi,xkk}}L2L2 {Xi,X

n n = ^2vu{xi,x y V t l { g f ,k}Xk}hi = ^2cr 22<Til{x ,xi}LlLl Ll it{xkk,Xl} /I Ii

.

(5.118)

Then by virtue of nn ^2[xi, r]{xi, / \Xj, x Xr\\Xj,

(5.119)

x Xss} j = = 5 0rs rs

ii=l =l

we have ^2[xk,Xj]Ll{Xi,Xk}L2 k

= Pij = CTji ■

(5.120) (5.120)

44

The Inverse Variational Problem in Classical

Mechanics.

Notice that . -, [x ,Xj}LlLl [xkk,Xj\ r

dxj

^ - i - == -= — dpuk

dx §±-k

(5.121)

dpij

Hence taking into account (5.117) and (5.121) we get from (5.112) • A dxk dp2,j _ ^ - dpij dxk _ A ^

/ dxk dp2,i \dpij dxk

dxk dp2)i\ \ _ dp2ti 2,j _ dpij dxk J) dpij

(5.122) %2 %3

This accomplishes the proof. 6.1

Instructive example of Cisto. Cislo.

In this section we shall explore a few examples using either the approach based on Helmholtz's Conditions or other methods, which nevertheless have in their background the idea of Helmholtz. To make things plain let us consider a simple model in (l+l)-dimensional space-time (due to J. Cislo) namely xx-ax — ax ==0, 0,

aa ee tRf1/ ^0 0. .

(6.1)

Without performing any computations one can state at once that (6.1) is not linked to any Lagrange function, as a term representing frictions appears there. Nevertheless, let us confirm this conjecture by a direct calculation.*' *' The same result follows from (5.5) of Helmholtz's conditions, namely

^ i + ^L = 2 (y dij

dii

\k

m

nik

®xk

+

™il) , 9t J

where Y^GijXj

+ G; +Gi=0. =0.

3

In our case G;J —> G = 1, G, —>
Hence (A) reduces to

(A)

Instructive

example of Cislo. Cislo.

The Euler-Lagrange Equation for the Lagrange function L(x,x,t), looking for, is d2L.. L.± dx22

d2L . dx dx dxdx

2 d^L_ d L _dL = ... . _ . dxdt dx dt dx dx ~

45 45

we are

(6.2)

Hence, comparing the coefficient at x in identity (6.2) we get d2L _ dx2

(6(6.3) 3)

w =' d2L . dx dx dxdx

-

&H^_dL_ _ a±. d2L _dL__ dx dxdtdt dx

(6.4)

From (6.3) follows L L== \x? ±x2 + f(x, f(x,t)xt)x + g(x, g(x,t).t).

(6.5) (6.5)

If we insert L of (6.5) into (6.4) we get set df

dg

dt

dx

(6.6)

The l.h.s. of (6.6) does not depend on x. If we differentiate both sides of (6.6) with respect to x we get a = 0. Thus there is no L corresponding to (6.1) provided a ^ 0. But let us consider instead of (6.1) the equation r(x,x,t)(x T(X, x, t)(x

— — ax) = 0 ,

(6-7) (6.7)

where r is a function to be found, which has to fulfill T ^ 0,oo

a.e... a.e...

(6.8) (6.8)

If (6.8) is not satisfied the set of solutions (6.7) and (6.1) does not need to coincide. Comparing (6.7) with the Euler-Lagrange variations of a Lagrange functions L we get d2L _ §dx2 - .

(6.9) (6-9)

d2L . d2L dL _ _ . X+ didi dx dx dxTt--dxdx dt dx = -TaX-

(6.10) (6 10) -

From (6.9) follows L—

\ T(x,u,t)(x

— u)du+ f{x,t)x f(x,t)x

g{x,t), + g(x,t),

(6.11) (6-11)

46

The Inverse Variational Problem in Classical

Mechanics.

where c is a constant chosen so that the integral does not diverge. Let us write

/ < M > * ^ and add to (6.11) dF{x,t)

8F(x,t) dF(x,t)

di

di

Then d F f(x, t)x + g(x, t) = ~ f ^^

++

g( ) g(x, Xj tt)

(6-12) (6 12) A , , , . W8F(x,t) g{x,t) =g(x,t) —— , . , . oF(x,t) g{x,t) =g(x,t) —— If If we we insert insert (6.12) (6.12) into into (6.11) (6.11) and and discard discard the the gauge gauge term term 4jr 4jr we we obtain obtain (we (we skip the „tilde" over g) skip the „tilde" over g) L=

r(x,u,t)(x

— u)du + g(x,t), +g(x,t),

(6.13)

Taking into account (6.13) we have

didi = Jc didx-

T{x UU t)dU Tx TxT{X '' ''t)du'

d2L

r d , TixTix UuAA dxdt = JLc MM >> 't)dU 't)du> dL

Tx dx-

= =

r

Jcc

8 ,

...

., dg ^T(X,U,t)(x-u)du+-. ^(x,u,t)(x-u)du+-.

Let us insert these relations into (6.10). We get

fr& fdr fdr

&T dr \\ ,, dg dg + uu du + du {-d-t Yx ) -Yx =

II {m

Tax -Tax -

8-x ) -£ = -

-

, (6.14)

^

We We differentiate differentiate both both sides sides of of (6.14) (6.14) with with respect respect to to x. x. We We obtain obtain *' *' 8T 8T . 8T . 8T + —x 8T . + -^rxa 8T . + ra = „ , —0. (6.15) dt + dxdx x + di dxxa + Ta = 0m ( 6 - 15 ) *> The same equation follows from the Helmholtz Condition (5.5). With G = r and g = —rax we get from dg _ dG . dx dx the relation (6.15).

dG dt

Instructive

example of Cislo.

47

To solve this equation for r we use the method of characteristics. We assume that there exists a function dV — ^Oa.e..

V(x,x,T(x,x,t))=0,

(6.16)

OT

From (6.16) follows dV dx

dV dr _ dr dx

which implies _

9x

-

8x

*K dr

In a similar way we may get derivatives of r with respect to x and t. The equation (6.15) takes now the form dV

dV .

dV .

dV

,„ „,

n

The equations for the characteristics are dt as

— = -1,

dx as

— =-x

dx ax

ax,

dr —=O.T as

or dx -dt = — - = x

dx dr , = — =ds ax ar

,„ „„. 6.18)

and the partial solutions read at — In |a;| = c,

(6.19)

ax — x = c2 ,

(6.20)

XT = c3.

(6.21)

Let us check (6.19) inserting the l.h.s. of it as V in (6.17). We have —a + ax— = 0. x Prom V = V{at-\n\x\,ax-x,xT) =0,

(6.22)

48

The Inverse Variational Problem in Classical

Mechanics.

taking into account (6.16) we find T = —a{at — In |±|, ax — x). x If we insert r, given by (6.23), into (6.14) we get r a(do_+ da_\

dbU)

Jc u\da

a = at — In |x|,

du + a a =

U

(6.23)

dg(x,t)

"'

dx

(6 24)

b = ax — x .

From this equation, choosing a specific a, we may evaluate g. Let us look at two simple examples: i)

a = 1, From (6.24) follows - = a

T= x

or

(6.25)

= ax + —

g

.

(6.26)

Then, from (6.13), discarding the gauge term, follows L=

(

1 ) du + ax =

Joo W

/

= i(ln \x\ — In |c|) — i + c +

(6-27) QI

.

The term —a; In \c\ — x + c = —(—a; In lei — x + ct) at is a gauge term and we may discard it too. Eventually we get L = x\n\x\ +ax.

(6.28) X

ii)

a = x - ax, From (6.24) follows r •

s

r = 1 - a-.

■

2

(6.29)

<9#

—a(a; — c) + ax — a x = — ax or ^

= - a 2 x + ac

(6.30)

Instructive

example of Cislo. Cislo.

49

and d F dF = —-ka -\a212x22 x 2 +acx+ — .. gg = + acx H—— dt Discarding the gauge terms we get from (6.13)

L=

/

(1 — a-)(x

.1 r^n Joo

uU

— u)du — acx == - \a\o2x2x22 ++acx

\x22

(6.31)

(6.32) (6.32)

\a22x22

= = \x — — axx axx In In \x\ \x\ — — \a x ,, where where the the gauge gauge terms terms 2 —xc + | c —xc + | c 2 + + axx axx In In \c\ \c\ + + axx axx = = d

— \c22tt + \ax22 In ^ax22)) — —(—xc —(—xc + + \c + ^ax In \c\ \c\ + + ^ax

(6.33) (6-33)

were omitted. Notice that (6.19) and (6.20) are Constants of Motion. Thus a is also a Constant of Motion. If we consider two equations corresponding to certain Lagrange functions, L\ and Li, we have n(x-ax)=0 TI(X — ax)

=0

(6.34)

T22(X T (x-ax)=0 - ax)

—0

(6.35)

and

resp.; then (6.35) can be rewritten (notice that from (6.23) follows ^- = f2-) — T I (x — ax) = 0, —T\(x 0,

(6.36) (6.36)

O'l

where — = a = Constant of Motion.

(6.37)

0"!

The Lagrange brackets for x and x are dp _dx<^>i dp &Pdp_ dp ix i\ - dx <^
(6.38)

(6.39)

50

The Inverse Variational Problem in Classical

Mechanics.

Hence {x,x} = r(x,x,t).

(6.40)

For two different Lagrange functions, L\ and L2, we have {x,x}Ll

= n

and

{x,x}L2

= a{x,x}Ll

{X,X}L2

= T2 = an

or (6.41)

Although a is a Constant of Motion, it so not, in general, a constant. This entails that for a non constant L\ and L2 belong to two different equivalence sets. As the equivalence sets are marked by a each equivalence set for a not constant consists only of one element.

6.2

Instructive example of Douglas.*'

We present now an interesting (2 + l)-dimensional space-time example of two equations for which and equivalent to them equations no Lagrange function whatsoever can be found. These equations read xi

-

x\ + x\,

(6.42)

x2

=

xi.

(6.43)

To prove our assertion we use Helmholtz's Conditions. The equations (6.42) and (6.43) are in the normal form fi=xl

+ xl,

f2= xi

(6.44)

Our aim is to find all Lagrange functions whose Euler-Lagrange Equa­ tions 2

Y,Glkxk+Gi

= 0,

i = l,2

(6.45)

it

are equivalent to the equations (6.42) and (6.43). To this end we are going to use equations (5.41) and (5.51). As /,, i = 1,2, do not depend on x we get ~6f

=°

*' J. Douglas, Trans. Am. Math. Soc. 50 (1971) 71.

(6-46)

Instructive

example of Pardo.

51

and

c

o

?«-"»&) ?( »5H*t£)="■-

<«7>

(6.47)

Equations (6.46) tells us that all entries of the matrix G are constants of motion. Equations (6.47), combined with equations (6.44), yield for i = 1 and j = 2 2Gnx2

-2G21 xxl 2\x

-G -G22 == 00 .. 22

(6.48)

Let us differentiate (6.48) twice with respect to t and take into account (6.46) and (6.44). We get Gnuxi22 - G- G ? 22ii£i i ==00,,

(6.49)

G - G =00. G unxi i i -G . 21(xl + xl) = 21{x\+x\)

(6.50)

Equations (6.48)-(6.50) form a system of homogeneous linear algebraic equation for G n , G\2 and G22. The determinant /-2x22 l-2x x2 \ xi

-2% -2xx x -±i -%i -x'\ -x\ - x\

- 1 \ —x + x\) +X!±i + x j i ! #^ 00.. 0 11 == -x 2{x\ +x\) 2{x\ 0 /

(6.51)

Hence Gik=G Gik —j=0, Gj — 0 ,

i,/c,i 1,2, i,k,j == 1,2,

(6.52)

where the equations (6.45) were also used. 6.3

Instructive example of Pardo.*)

We present below another (2 + 1)-dimensional space-time example, which also does not admit the Lagrange-Euler approach. The equations are Xi xi

=

xx2,2,

(6.53)

x2

-=

x%. x2.

(6.54)

*) F. Pardo, J. Math. Phys. 30 (1989) 2054. •)

52

The Inverse Variational Problem in Classical

Mechanics.

These equations are in the normal form. To prove our conjecture let us use the Helmholtz Conditions in the form given by (5.53)-(5.56). Here (see (5.52))

-CD-

(6.55)

• = ( ! ! )

Equations (5.55) and (5.56) read resp. G „ \\ ^ i 2\ N \ ___i i ^( 00 Gn 22 2Gi 2 y/ \VG11 Gn 2G12 G22 //


Gn n // G \Gi2 \Gi2

(6 56) (6.56)

and / 0 V-G12

if

Gxa\ 0 )

2

0

Gn\ 0 j

\-Gn

(6.57)

or Gn G n ==0 0, ,

G12 "G G12 — = — 2 ^i1i1 ,,

G22 = —G12, —— G12,

G12 — = 22*^11 ^ 1 1 ■•

(6.58) (6-58)

From (6.58) we conclude that Gn — =G =G Gn G12 G22 — 00.. 12— 22=

(6.59) (6.59)

hence

G

-(Z

\Cz21

- constant. £)=(» ?)■ P>——•

<J22 /

\U

PI

<66o) (6.60)

The matrix G is not invertible and the Lagrange function The matrix G is not invertible and the Lagrange function L — 2\0±l, PX2 i L=

/?#0

f3^0

(6.61) (6.61)

does not yield (6.53)-(6.54) as its Euler Lagrange Equations. does not yield (6.53)-(6.54) as its Euler Lagrange Equations. 7.1

Construction of an autonomous one-particle Lagrange function in (3+1) space-time dimensions yielding rotationally covariant Euler-Lagrange Equations coinciding with the Newton Equations.*'

In this Section we are going to present a much more representative example in (3+l)-dimensional space-time then those in former Section as far as *) J. Cisto, J. Lopuszariski, P.C. Stichel, Fortschr. Phys. 43 (1995) 733. *>

Construction

of an autonomous

one-particle Lagrange function

...

53

physics is concerned. In contradistinction to examples of Section 6 the Equations of Motion are not given; we require only that they represent Euler-Lagrange Equations for a certain, also not specified, autonomous Lagrange function. The Euler-Lagrange Equations should be rotationally forminvariant and should coincide with Newton's Equations (normal form of the equations) Xi = fi(x,x),

i = 1,2,3.

(7.1)

Our goal is to find most general form of the Lagrange function, conforming to the requirements imposed by us. In (7.1) x = x{t) = (xi,x2,x3), X _

dt

'

(7.2) { 6)

dt>

'-

and / are functions satisfying certain condition specified below, functions whose structure has to be found. It turns out that the class of variations of the form (7.1) is limited and that all Lagrange functions have the same form; namely L(x,x) = i x 2 - V ( x , x ) , they are singular and weekly rotationally symmetric.^

(7.4) If

d±i they are rotationally symmetric. We restrict ourselves to 3-dimensional Euclidean space, but some asser­ tions can be immediately extended to rotations in an n-dimensional Eu­ clidean space n = 2,4, To start with we give a description of the proper rotational group 50(3). In 3-dimensional Euclidean space this group is represented by (3 x 3 ) matrices R characterized by RRT

= 1, (7.5)

det R = 1. t) Weak invariance of the Lagrange function under some transformation of its parameter means that both Lagrange functions, the original one and that obtained as the result of the transformation, are equivalent to each other.

54

The Inverse Variational Problem in Classical

Mechanics.

Under the action of the element of SO (3) x, x and x transform as follows X{

r

ii

X j — 2-^/k ^ik%k

i

->■ x\ = Ylk Rik^k ,

Xi

r

Xj

— 2-^ik

(7-6)

^ik^k

As we assume that the Newton Equations (7.1) are rotationally covariant we have

^Rikik

= J2Rikfk(x,*).

= fi(Rx,Rx)

£

(7.7)

k

Hence / is a forminvariant vector with respect to the rotations.*) Let us assume that there exists a Lagrange function L°(x,x) such that (7.1) holds. If there exists another Lagrange function L, leading to the *' To make thing clearer; the relation (7.1) could have had e.g. the form Xi = ^^XjbjXi

,

b ^ O

(A)

i where b as a constant vector. Such equations are not rotationally symmetric as the direction given by TK is here distinguished among others. To see that let us apply the procedure used in (7.7); we shall immediately come to a contradiction. Starting with (A) we get

Y^Rik&k = ^2^2^2RjlxibjRikxk k

j

l

,

R^l

k

or ^2^2Rikxjbjxk k

=

j

^^^RjlXlbjRikXk j

I

k

and finally

^Zx3bi = YlJ2R3'XlbJ' j

j

i

which is obviously wrong. To make things correct we should write transformed (A) as ^Rikxk

= /,(Rx,fix)

k

where /i(x,x)#/,(x,x).

Construction

of an autonomous

one-particle Lagrange function

.... ..

55 55

same Equations of Motion we must have the identity x Xl

*

Xj + xx 9^. dii d±j Xj + 22 -- d± d±..dXj dXj i> -~dx~

f l

" - ~I .^

8 ( 7 -(7.8) )

As neither f nor L depend on x we must have d2L dxidij d±idij

(7-9)

l}

or*' L=|x L = | x 22 --J2^j{x)ij-B{x). J2 Aj(-x.)xj - B(x).

(7.10)

jj

*' *) For For convenience convenience of of the the reader reader we we derive derive here here formula formula (7.10) (7.10) from from (7.9). (7.9). We We start start with with

£f-1^ dx\

= 1

Formula (A) entails Formula (A) entails L = \x\ - Ai(x,X2,X 3 ,t)xi - Bi(x,X2,X3,t) . L = \x\ - y l i ( x , £ 2 , X 3 , t ) i i - Bi(x.,X2,±3,t) . Since by assumption Since by assumption

{A)

(B) (B)

^ = 0

at at

we have from (B) dAi . 0Bi n H xi -\ =0.

at at

at at

The l.h.s. is linear in x\ i\ therefore dAi _ dBj_ _ ~dt ~ ~dt ~

[C)

A i ==AAii((xx)).. Ai

(£>) (D)

From d922LL = dAj == QQ Mi 9X1 3X2 3xi 9X2 3X2 3X2 follows Ai = Ai =

Ai(x.,x Ai{x,x3s)

and from 2 d32LL Sxi 3x3 3X3 dii

we get

Mi _= dAi 3 il 3

Q

56

The Inverse Variational Problem in Classical

Mechanics.

Let us insert (7.10) into (7.8). To this end we compute 2 dL _ . dL_ dL l d±i — X-i1 Ai(X) ', dii d±i dxj dxj dL _ y~^ y , dAj_. 8Aj , _ 8B_ dB dii dxi ^-" dxi J ^r* dxi

dAj dxj dxj

(7.11) ^• 1 1 ^

J3

Hence

><-nm-%>>-£ ^fdAj *— V dxj

8AA . dxi J 3

J

j

dB dxi

(7-12)

From 92L _ dx\ dx?,

d92B! Bx __ dx2,

x

follows -Bi

= \x\ - A2(x,X3,t)X2 A2(x,X3,t)X2

- B22(x,±3,t) ( x , x 3 , t ) ..

(E) (E)

Because of (C) it follows from (E) 9A2 ^

_ 9B2_ = ^ = 0 .

(F) (F)

m ~ m ~

(

mm

As As 9L dx2 i?X2

dB! 9X2 aX2

.

. ,

'

. ,

wr- = - - r r - = a=2 - ,4 2 (x,x 2 ) we have d22LL 8x2 9x3 9x2

8A 6M22 _ 9x3 A2 = A2(x). = A2(x).

(G) (G)

Notice that from (B), (D), (E) and (G) follows L=\±\ i i 2| - .- 4 Ai(x)±i i ( x ) £ i - A 2(x)x 2 L = ±x2 + ±x From 92L _ dx\

92B2 dx\

_

follows eventually -B2

= i x ^ -- AA3(x)i ( x 3) x 3 -- BB((xx)) , = \x\

which inserted in (H) proves (7.10).

( x ,3). x3). -- BB2{x,x

(H) (A)

Construction

of an autonomous

one-particle Lagrange function

.... ..

57 57

Since f as well as x are vectors and the r.h.s. of (7.12) is linear in x, so . dAi dAi dAi ,s dAf -aji(x) = Q~ - Q— = M x )

(7.13) (7.13)

as well as bi = - — dxi

(7.14)

have to transform under rotations as a forminvariant tensor and vector resp.. Every antisymmetric tensor in 3-dimensional Euclidean space is equiv­ alent to a vector. In other words there exists always a vector v such that Ci;(x) s e#jfctffc(x), Oij(x) = y ^ y^e ijkVk(x-),

(7.15) (7-15)

£i23 £123 = — 1. 1 ■

k

Vector v(x) must be also forminvariant under rotations, hence we have aa x x H( )

ij( )

rr = |x| = (x% ( a f ++sx| | ++*i)a icf) »,,

= = Y €ijkXka(r), Y€iikXka(r}'

6i(x) = ii,jb6 ( rr)).. 6j(x)

(7.16) (7.16) (7.17)

From (7.13) and (7.16) we infer that

=Xia k YYY.e^CikJd^ Y, g^ = Xia ^-^ j

■

(7.18) (7-18)

J

k

The consistency requirement for (7.18) is

EEE^=w+^=«. EEE^=*)«r»i i

j j

k k

J J

(7.19) (7.9)

The solution of equation (7.19) (two last expressions) for a is a(r) = —- , (r) = ~^3 ' a

c - a constant. constant.

(7.20)

From (7.20) and (7.18) follows

\-^v-^k ®Ak _ _c ££ _c _®_ (\\ l^l^^ Qx ~ r 3 - dx. [rJ j

k

(7.21)

58

The Inverse Variational Problem in Classical

Mechanics.

Notice that as long as c / 0 A cannot transform as a rotationally forminvariant vector; a rotationally covariant vector namely has to have the form Ai(x) = XiA(r) = ^rA(r) = ~ , r oxi which would entail, taking into account the l.h.s. of (7.21), that

4-(-)=0.

(7.22)

(7-23)

oxi \rj Then c = 0. We conclude that for c ^ 0

i in (7.10) and consequently also L(x,x) itself can never be rotationally symmetric. The gauge term — ^ f i which we may always add to the Lagrange function, cannot restore the missing symmetry. Notice further that A cannot be regular everywhere for x 7^ 0 as long as c ^ 0. This can be shown as follows. The Green Theorem asserts that Y,Y.Y,

rneijk—QkdS = J2 I Qidxi

(7.24)

where S is any piece of a 2-dimensional surface in 3-dimensional Euclidean space, K is its boundary oriented in the positive sense and n = n(x),

n2 = 1

a normal unit vector field on the surface S. The theorem is true when Q(x) are ,,decent" functions. Let us replace Q by A and take for S a surface of a sphere of radius R^Q with the origin located at x = 0. Then according to (7.24), we should have

EEEjf/" | e ^ ^ 2 s i n ^ ^ = 0

(7.25)

Construction

of an autonomous

provided A is „decent" (7.21). We have

= -c [

one-particle Lagrange function

...

59

Let us insert on the l.h.s. of (7.25) the r.h.s of

[

^726^

sm6dOdifi = -4TTC,

Jo Jo (—sin8d8 = d(cos8),

cos7r = — 1,

cosO = l ) .

If c 7^ 0 (7.26) contradicts (7.25). This proves our conjecture. We are going now to solve (7.21) for c ^ 0. We have to find at least one particular solution of the inhomogeneous equation; we get the general solu­ tion by adding to the particular solution the solution of the homogeneous equation, which is

£*(*),

(7.27)

where $ is an arbitrary function of x. Function $ may also depend on other parameters characteristic for the model; it does not depend on x. Let us rewrite (7.21) in spherical coordinates, which turn out to be very convenient tool to find the special solutions. We have X\ = r costpsin8 , x2 = r sin ip sin 8 ,

(7.28)

£3 = r cos 8 For A the relations are *' A\ = sin 8 cos tpAT + cos 9 cos tpAg — sin (pAv , A2 = sin 6 sin
(7.29)

*' There is used also another notation adjusted more to geometrical considerations, namely: rsin^Ay, = A
60

The Inverse Variational Problem in Classical

Mechanics.

The reciprocal relations read Ar — sin 9 cos

A\ Ai — —sin sin^^3 9A3, , A^, = — — sin sin Ai Av — ■ ■

(7.30) (7.30)

In this variables our equations (7.21) have the form 11

dd

((

■ aa AS A 1t ■

s

r sm 9 \o9 rsm9 \d9 — ^

= =

ddAA s

A \ dip op>) )

ll

rlrl

sm9~(rA,): *4(r^),

(7.31) (7-31)

8AT 8 -W = d-r{rAe)The most simple partial solutions of (7.31) are A. A.^ — 0, U, Arr -— A v =

Ag^Q^ 0 A«

(7.32)

A Ar r ==i 4A*9 = 0, 0J

A 4 v,^0. ^0.

(7.33)

and

In case (7.32) we have

J_dAg_ _ 1 sm9 sin# dip

(7.34)

r

l(rAg) = 0. J:(r4*)=0.

(7.35)

These equations yield resp. Ag =c-sin, r Relations (7.36) and (7.37) entail

(7.36) fl).

(7.37)

ap sin 9 + rFg = Gg or Fg{r,9) -Hg{9) = ^{Gg{9,ip)-apsm9). -(Gg(9,ip) - c^sinfl). Fe(r,9) = \Hg{9) r r

(7.38)

Construction

of an autonomous

one-particle Lagrange function

...

61

Finally the partial solution reads Ag = c-sin6,

AT = A{p = 0,

(7.39)

where we dropped the solution of the homogeneous equation *) -He{9). T

In case (7.33) we have J^(sin0i4 v ) = - £ s i n 0 ,

(7.40)

j^(rAv)=0.

(7.41)

The solutions of (7.40) and (7.41) are 4 , = -cot<9, Ar=Ae=0, (7.42) r where again the homogeneous part was left out. " Let us return now to the Cartesian coordinates. In case (7.32) we have, according to (7.29), A\ = cos 6 cos ipA$ , A2 = cos 6 sin tpAg, Ax =

(7-43)

-sinOAe.

Taking into account that x2 arctan — =
sin ip —

x2

j- ,

cos

(Xl+Xl)2 X3

„

— = cos 6 , r

Xi

.

X2

— = sin 6 cos tp , r (x\ +x\)i = rsinO

*) If we replace in (7.34) and (7.35) Ag by ^He{8) c = 0.

.

xi

j- ,

{X\+Xl)2 - .

— = sin 6 sin w , r

these equations will be satisfied for

r) The case Av = Ag = 0, y4r ^ 0 leads to solutions of the homogeneous equations only and so we discard them.

62

The Inverse Variational Problem in Classical

Mechanics.

we conclude that (7.39) in Cartesian coordinates takes the shape (without the homogeneous part) a

!

A

a

(

4.

Z

2N\

X1Z3

A\ = c-u>cosu}sm(fcosO = c arctan— —— = r V X i J r3 = —c I arctan — I —— ( — ) \ %\) ax\ \T' or, generally containing the solution of the homogeneous equation,

* —(—S)S:(T) + ^

<7'">

The solution (7.39) of (7.44) is singular for r = 0 (not interesting) as well as then when ip jumps from the value 2ir to 0. Notice that this singular behaviour cannot be removed by the grad $ term. Solution (7.44) can be represented as depending on two fixed non-collinear vectors t / 1 ' and t/ 2 ) viz. Ai

=-

c arctan

/ ( b « xb<2))x\ d ( bWx \ ( |b(i)|(b(»)x) J ^ {WW\)

+

a$(x) , „ _ " f a T (7-45)

where b^ 1 ' = (0,0, u),

b ( 2 ' = (0, v, 0),

u,« - arbitrary numbers.

Let us evaluate B appearing in (7.10). According to (7.14) and (7.17) * * = -xbir) = dXl

^M dXi

(7.46)

dU

- —rb{r). ui \ —— =

dr thus B(x) = U(r)

(7.47)

where we have incorporated the constant into U(r). To summarize this case ((7.10), (7.44), (7.47)) we have L - J * "

+

c ( « - S ) | ( * ) - * < r >

+

^

.

(,48,

Construction

of an autonomous

one-particle Lagrange function

.... ..

63 63

Let us now discuss the other case, (7.33) and (7.42). Taking into account (7.42), (7.29) and the relations among x and w ip and 6 we get 2~,—2 ( — " + - # ¥ > ) > xi + + x\ xi \ \ r r 1 x\ )

M

= —

A2

— —~——2 (

1" Hf I '

xi + xi V r

/

^3 = 0 ,

(7.49)

H =H

* « (S) •

We deliberately kept here the solution —^~Z^f

r sin 0 rsmv of the homogeneous equation to reduce the singular behaviour of A. Let us us take take

H = ±c. H v=±c.

(7.50) (7.50)

v

Then Then Al

= -

C

- ^ 1 - .

rT £3 £3 = -FF rr

A2=°-^-. = °-^-.

rr X3 X3 =F =FTT '

(7.51) (7-51)

ASs = 0. A\ and A2 are singular for r = 0 as well as along the half-straight line £3 = r or X3 = —r. This solution depends on one fixed vector b . The most general form of (7.22) reads c

(x x b ) ' r |b|r Iblr ± bx '

A-

(7.52)

It is singular for r = 0 and for |b|r = =Fbx > 0. The singularity along half-straight line reminds on Dirac's monopole theory. According to (7.10), (7.47) and (7.52) the admissible Lagrange functions are L =

11 9 2

2

2X

L Lij = y j

+

c

bL . . .. . d$(x) d$(x) kL U{r) + r rr( r| bb| rr ++ bx) ^T' bx) ~ tit 1 '

y / k

c

CijkXjXk ■

(7.53) (7.53)

64

The Inverse Variational Problem in Classical

Mechanics.

Let us return to the solution (7.33) and (7.42) and discard the gauge term. We get c b x ( b x x)j X b 2 r 2 — (bx) 2

A Cr

(bx) 2 ==(b(bxxx)2 x)2 bb 2Vx 2 - (bx)2

^ r^- (bir

•.

-

(7(7.54) 54)

'

The solution (7.54) has a stronger singularity t h a n (7.52); it is singular along the whole straight line when b is parallel to x. The Lagrange function reads x 2

L

_ cc£x)bL ^x)bL_

dd ^

r (b x x ) 2

1

dt

(7.55)

The solution (7.54) was given and examined by Houard. *> The derivations of this Subsection are in many respects standard. Notice that the Euler-Lagrange Equations do not depend on b ' 1 ) or b ' 2 ) in (7.45) as well as on b of (7.53). Hence a different choice of these vectors leads to Lagrange functions, s-equivalent to the former ones and to L°. Let us return to the Equations of Motion (7.12)

=V (— - ^k) x- ^-' \ dxj dxi J 3

9U dxi

^

(7.56)

Since, according to (7.21), dAi dAj _ v—< Xk dAj _ dAj v-* £fc g^-^ dx~ dxi = -+C+cE^> 2^eijkr3 ' J

(7.57) (7-57)

k

we we insert insert (7.57) (7.57) into into (7.56) (7.56) and and get get Li _ •• - _ hi r

6

dU dU(r) ^ dxi

(7.58)

where L is the angular momentum, viz. (7.59) (7.59)

Li = ^2 X ! eHkXjXk ■ i4

kk

Notice the close relation of (7.56) and (7.58) to the expression of the Lorentz force in Classical Electrodynamics. In Electrodynamics one has the Lorentz Equation 3

mXi = Y^Fu,j",

i=l,2,3,

i/=0 i/=0

*) Mat/». PAys. *) C.S. C.S. Houard, Houard, J. J. Math. Phys. 18 18 (1977) (1977) 502. 502.

f^ = 00,,11,,22,,33,,

(7.60)

Construction

of an autonomous

one-particle Lagrange function

. .....

65

where m stands for the mass of the charged particle, exposed to exectromagnetic forces, and jo = j° -= PP JO

ji = = ~f ji ~jl

= = pii, PXi

= 1,2,3 i=

,

is the electric charge and electric current resp.. The (4 x 4) antisymmetric tensor F is F F o =i

dA (dAi _ ^l) dA0\_ dA0 _^ E-. E= l _- (fM* - ^ — - —l ) =- - — °,

\dx°

dx )

dxi

_ - y^dxi _ - dxJ _ Jj dx _ . _ - _ dx,.> ijkHk fa- - - ^e 2-e«*fl fc FlJ

(7.61)

(7.62)

where E and H are the electric end magnetic fields resp. and (Ao,A) is the the electromagnetic 4-vector potential. If we put the mass m and the electric charge p equal to one we get from (7.60)-(7.62) Fikik Fik±k

Xi = Fio-^2 Xi Fio-^2

k

= Ei { +

~®+X S3 X 5Z! eeHkXjHk Hk±iHk j

(7.63) (7.63)

k

or

*~ dxi+^\dxj

J

3 J

da)

3

'

(7.64)

To get (7.56) from (7.64) one has to put

dAp dA0

=

-T.— = dxi axi

dUQx\l dU(M) T. dxi axi

=

Ei

= Ei.

.

(7.65)

Taking into account (7.65) and defining the magnetic field by H ==- c-«£ ^f ^ k

(7.66)

we get from (7.63) and (7.59) the equation (7.58). The equations (7.66) and (7.58) show that our model has many features in common with classical model of Dirac's monopole; then (—c) would play the role of the magnetic charge (see also the relation (7.54)).

66

7.2

77»e The Inverse Variational Problem in Classical

Canonical variables. grange functions.

Mechanics.

Equivalence problem of the La-

Let us consider the classical Hamilton formalism. We have the canonical momenta (see (7.10)) _ dL _

<9$ - Ai Ai + +— axi oxi Then the Hamilton function reads

(7.67)

Pi — = TTTPi a- = — xXi i

H = J^P^i ^ p , * , -- LL == i* §x22 ++u[/(r) H= (r) ==

V

d$v 2+t/(r)

=K^-S)

The gauging of A K = A

Z

-

- -

(7(7.68) 68)

-

(7.69)

is physically irrelevant (as in Electrodynamics). The Hamilton function is conserved as dH dt

= Z.«(i< + lsr) =

=£*(*+£^)-.

(7.70)

where (7.53) was used; notice that

= e /] *»i< XiU = z2z2z2 } , / , } , HjkXjikXi yj vkXjxkxi —=0.o. i

i

j

(7.71)

k

Let us look to the Lagrange brackets of x and x. Taking into account (7.57) and (7.67) we have , -,, iXitXj) -

dAj _ dAj dAi _ ^ J

^-^ xnk \-^ c2_^eijk—,

(7.72)

k

i{**,%} - j T^EL-sx. i.x = 7=5y,

(7.73)

{«<,ij}

(7.74)

■^Xj j JJj j

=

—

0.

U-

These commutation relations are not standard.

Canonical

variables.

Equivalence

problem

of the Lagrange

67

functions. functions.

Let us compute Lagrange brackets for a rotated system. We have Z/ = i x

2

- £ ^ - £ / + !$',

(7.75)

i

where (see (7.6)) x\ = £ f e Rikxk, We have

x\ = Z ^kk Rikik, A' = A(x'), *' = $(x').

6 <"(7.76) >

^-SJ-i:(^-4 f|)^*-§£-£(4-4++*£)*■ Then, taking into account (7.13), we obtain consecutively dp dXkdp k {(xi x■}' y = YV ((^*M ' -_ ^ iMM^^ lI - dp'j'i Xi x Z^ \dx{ dxj dxj dxi J dxj Z^ y ^ i i 9XJ dx ; dxi J dxj

dA

(

' i ^

Wfl

dp dv

'i'i = = dxi dxi

= [7.77)

V ^m dx\dx>m) = 2 J 2 J a'mlRmiRlj = I

h

ml

\

TTl

i "= 2EJ E2 J( af'ml - ^RV ( * ^ ^ u - ay - ^ miRlj = I

TTl

1

1

^ E ^ -^ «*•*>' = ^? £) &

= ^ = tof ~ dx~'

= a^-Eafj^-E^^-^J

I

J

(7.78)

I

(7.79)

|{**, Xi, * Xj,* j} ' = — <), u ,

which coincides with (7.72)-(7.74). We conclude from that the Lagrange as well as the Poisson brackets stay invariant under rotations. According to (5.60) a = l(L1=L, L2 = V). The Euler-Lagrange Equations of the rotated Lagrange function L' read

^f

d2L> .

d2L' „ \

dV _

W / * * , 3+j + * * Xj^3 - g ^ O . ^£*\diidxi I a±i ax,- * dxtdxj dxr dij )) dxi dxt • j

K(7.80)

'

68

77ie The Inverse Inverse Variational Variational Problem Problem in inClassical Classical Mechanics. Mechanics.

From (7.75), taking into account (7.6), we have

*JL - T(i>-A>+^\^

=

= EN-4^U.. d2L' dii dx3

/ 0Ak &># \ ; V dx\+ dx'k dx[)

v

v

V

-

p Ul3Kki

'

(7.81)

8 L1

dx~d±~ = E ^ ^ = ^> dU d2$ . \ + Xk Hji dx dx\ 'i + dxdx) 'idxdx' '*k /) '■

d_V_ y ^ ^ Y 6A'k., I dx ^ V { ^ dx'j Xk 9a;,t ~ ~ ^, 4" * If we insert (7.81) into (7.80) we get 2

dA[ a $' \ „ „ EEE(-^ + a^J^^+^+ + E E E {--^ dx^J R^R^+ Xi+ v-v-v- f

dA v- v- v- dA 'i\ nry n n ■■ vv-- 9U 9U „ „ + E E E QJ-RvK-kiXj + Y, dZTRJi+ + Ei Ek EI Q^-^iK-kiXj +i J2 i^ r % + ° k X

'

* '

OX

5

"$, T

7RljRkiXj

=

&i't\

_

+ 1^2^1^ dx Jx i

k

I

ax

J

kaxl

V-V-V-/9A;

'

(7.82) (7.82)

= = EEE^-4J^^i+^ EEE^-4J^^i+^ + +E E ^ ^ = =o o-s^9U„

By By virtue virtue of of (7.13) (7.13) and and (7.14) (7.14) aa fci (Rx) = fci (Rx) = _ a 4 j _ 9 ^ _

v v

(dAn

dAm\

-~dx' dx'kk &:{ dx\~^ - 2 - Z . ««*»*» * » « «^- ^rma 9g^^JJ '. (Rx) =

^

- ^= - E ^ ^

(7 83) (7.83)

All Lagrange functions...

69

Thus (7.82) can be written, with help of (7.1),

E E E E E ***. ( ^ - ^ ) **,*,*,+ j

k

+

^

l

+

m

n

R k

\

R

E E ^ ~d^ ^

m

n /

=

(7.84)

which coincides with (7.12) (B(x) = £/(r)). Thus the second condition of the Theorem of Henneaux is satisfied (see Subsection 5.2 with a = 1) and we conclude that

ff-l+*a.

(7-85,

where 1/ is the rotated and L the original Lagrange function. In other words L' is equivalent to L. Notice that the Lagrange brackets (7.72)-(7.74) are independent of the constant vectors b ' 1 ' and b ' 2 ' or b. Therefore the Lagrange functions obtained by changing the values of b' 1 ), b^2) or b in the original Lagrange function are not only s-equivalent but they are just equivalent. In other words for V = L(u',v') and L = L(u,v), u' j£ u, v' ^ v, we have (7.85) too. Several authors devoted their papers to similar problems earlier then J. Cislo, P.C. Stichel and me.*' We did not here undertake the task to find all s-equivalent Lagrange functions to the one chosen at the start by us, having the structure (7.48) or (7.55). In the next Section we shall show how to accomplish this program but for a much simpler case, where the magnetic charge c = 0.

*) See e.g. E.L. Hill, Revs. Mod. Phys. 23 (1951) 253, J. - M. Levy-Leblond, Comm. Math. Phys. 12 (1969) 64, J. - M. Levy-Leblond, Am. J. Phys. 39/5 (1971) 502, M. Henneaux, Ann. Phys. N.Y. 140 (1982) 45.

70

8.1

The Inverse Variational Problem in Classical

Mechanics.

All Lagrange functions, s-equivalent to the Lagrange function L — | x 2 — C/(|x|), in (3+1) space-time dimen­ sions.

In this Section we are going to investigate a somewhat different problem from that in Section 7, although having many common features with it. To start with let us choose an autonomous Lagrange function whose rotationally invariant Euler-Lagrange Equations have a normal form (co­ incide with the Newton Equations). The most general Lagrange function of this type was given in (7.48), (7.53) or (7.55) and the corresponding Equations of Motion - in (7.63) and (7.64). In this section, however, we restrict ourselves to a special case when c = 0. Then (7.53) or (7.55) read, discarding the gauge terms, L = |x2 -

tf(|x|)

(8.1)

and the Equations of Motions (in the normal form) are Zi+Q*

=»<-/<(*) = 0 ,

i = l,2,3.

(8.2)

Notice that the Lagrange function (8.1) is now rotationally invariant. Our task*) will be to find all Lagrange functions whose Euler-Lagrange Equations are equivalent to (8.2) i.e. they give the same set of solutions as (8.2).t) We shall also investigate the problem of equivalence of the new Lagrange functions, found by us, among themselves and with respect to L, given by (8.1). The set of Equations of Motion, equivalent to (8.1) have the form

J2 ^u(x, x, t) Uj + g^'j

=0.

(8.3)

Here the matrix er has to be found. We require that dettr 7^ 0,oo

a.e.

(8.4)

for obvious reasons. *' J. Cislo,J. Lopuszariski, P.C. Stichel, Fortschr. Phys. 4 3 (1995) 733, M. Henneaux k. L.C. Shepley, J. Math Phys. 23 (1982) 2101, M. Ranada, J. Math. Phys. 32 (1991) 2764. t> N.B.: the Equations of Motion do not need to be any longer forminvariant with respect to rotations.

All Lagrange functions... J4I/ functions...

71 71

Our results in case of (3+l)-dimensional space-time for C / ( | x | )^# a xx22 ++/ 3P,, U(\x\)

a,/3 a, P - constants,

which we are going to prove below, can be summarized as follows. We show that &ij = g$ij 9$ij + G{L)LiLj = gS^ gSij + a°-, a°j,

(8.5) (8.5)

where g ^ 0 is a constant, L is defined by (7.53), viz. i-Ji — jT 7 Li = 3i

CijfcXjXk J7 j eijk XjXk

(8.6) (8.6)

k

and G is an arbitrary function, satisfying the relation G(AL) = A~ A- 33G(L),

(8.7) (8.7)

where A ^ 0 is a arbitrary constant. We have also 2 2 det<7 det
ff

2

(l+G|L|2).

(8.8) (8.8)

An important issue is the existence of a Lagrange function, corresponding to a prescribed a. So we prove that for any x )i , 4 = 4°(' jxi ,( x 4 (<xx' *, x) ) = ' )'

(8.9) (8.9)

satisfying the relations

5>°.±,= £ 4 ^ = 0 , 3

(8.10) (8.10)

3

(8.(8.11) n)

^4 = ^ i == ^ ^ >) S ijt dxk

dxj dxj

dxi dxi

dik Qifc

dij

d&{ dii

(8.12)

there exists a function L° for which d2L°

= CT0

(8.13)

The most general form of the Lagrange function L leading to (8.3) and satisfying PL 82L W7(|x|) dU{\x\) r^ *—• dii cte, dxj ^-> 9dxi i i <9i, ij

JJ

JJ

, gd22L £ . _ dl ^ =_ Q0 dij J 3 9a; dx ^*— r ' 9dii i j 9i,J

| ry

J3

J

(8.14)

72 72

The Inverse Variational Problem in Classical Mechanics.

as well as

d2lL

=

d22l

^

d&i dij dii

lJ

'

3d2L l

=_

dii dxj dij

(8.15)

dxi dxj dij

is, up to the gauge terms,

2 ^^(|x -r/(|xj))E(f-fUK + E(g-gLJ^, ^<w,)

i=9(

+

(8.16) (8-16)

where v is an arbitrary constant vector. To prove the results collected in formula (8.5)-(8.16) we start with the investigation of the Lagrange function L(x, x, t) such t h a t its Euler-Lagrange variation is of the form (8.3). In other words we demand t h a t

d dl dl v-^ , . J.. == aiAxlAX x t) X X3 dtdT -dV ^ ' ' i l dtdX--d^ Y {' 't){Xj

+ +

dU\ dx-) dx-J

is an identity. This implies the identities

d2l _ = >' 2 d% lL .

^y , 1

^—>diid±idxj dxj t— j

3 3

((8.17) 8J7)

d2L l

dl _y,y,

diid±idtdt

dxi dx{

d2l

dU

2 2

—'diidii dijdij dxj dxi —' o

(8.18)

Prom (8.17) and (8.4) follows From

det

(d^k)*°

ae

(8.19)

-

Consequently I is regular. We drop the ,,tilde" over L in the sequel for convenience of the reader. Equations (8.18) can be written

d dXi dii

\r^f_dILdU_ 2f1 V { dij dxjddxj ^r Xj

+

BL . \ Xj j ddxj Xj ) J

+

dL _ dL dt ~2dx~dx{ '

((8.20) 820)

Notice t h a t the ralation (8.20) remains true for every Newton's Equation Xi — JiyX.j Z) ,

All Lagrange functions. functions... ..

73

in our case

h Ji —

(8 21)

ggflxj)

= —tr

-

(8.21)

o OXi

Let us differentiate both sides of (8.20) with respect to x.Then the l.h.s. becomes symmetric with respect to the interchange of the indices k and i. This entails*' d2L

d82L

di dikk dxi

(8.22)

dxk dxi dii

From (8.22) follows e.g. d3L diidij

d3L dxk

d±k

diidxj

or, generally, d<J

dijjk jk ___ datj daij _ dxi dxk dxk

daki daki dxj

(8.23)

From (8.17) follows immediately d(7

ddjk doij _ daki_ daki ik _ doij_ d±k dii dij

(8.24)

By By virtue virtue of of (8.23) (8.23) equations equations (8.20) (8.20) can can be be also also written written dd

(^ (x-^

dL dL ,.

_ \r | _ vv -^

Xj

L

dxA^d±j

dU{\x\) dU(\x\)

a

J-z^ ^

dx.

2 L d82L

d±idt-

(8.25)

Let us differentiate both sides of (8.25) with respect to xk- The l.h.s. becomes symmetric under the interchange of the indices i and k, consequently, due to (8.22) and (8.23) we have dd22UU ^v-^ aij ^ °« dxTdx~

^

3 3

=

d2u vV-" ^ d2 U aGh3 j 4 - " dx~dx-

dxk dxj ~ ^ J J

3 3

dxi dx, J

t2^~dx7 - ~dx7 3j

(8.26) -

J

Let us once more differentiate (8.25) with respect to xk. account (8.22), we get Y ^ dakj Xj. _ v ^ ®a*3 ^ Xj

fona. (8 26)

daki

-^dx ~ 2? hdxf~dx~kd~x~i ~ot~W

Then taking into

( (8.27) Z{)

*-

J3 *) Notice that in case the force / does not depend on x relation (8.22) is always valid. This follows from (5.10), (5.43) differentiated with respect to x as well as (5.3) and (5.4).

74

The Inverse Variational Problem in Classical

Mechanics.

Notice*' that (8.17), (8.24) and (8.26) are Helmholtz's Conditions (5.3), (5.4) and (5.51). The missing piece is the equation (5.41)

££f E

&aij t

k

dx^k

, V ^ ®aV

, daV

■

n

(8.28) (8-28)

+ V5 dx: k ^ + ^9t = 0. k

expressing the fact that each entry of a is a Constant of Motion (not only the trace of a). We have only to replace the matrix er by the matrix G. The relation (8.23) follows from (8.28); we have to differentiate it with respect to xi, interchange the indices i and I and subtract so obtained expression from the former one (where i and I were not interchanged). So far every thing was true for every f (x, t), i.e. when we replace in the formulae above (—gradt/) by f. In the sequel, however, we are going to make use of the specific form of the second term in (8.2). We have «

OXJ

dxj

= ^ r r

, dr dr

/ J

W

, x

. '

3

x

(8.29) '

Hence d2U(\x\) oxk OXJ

ldf(r) r dr

(8.30)

In the next Subsection we shall investigate the case when £*0. dr The case

(8.31)

f = 0 dr leads to the equation for U(r), namely

(8.32)

d fld(/{r)\ ndU(r)\ dr \r

„ =Q

dr J

fliUir)

. \

{r^r = ~f{r))

(8.33)

or or 2 U(\x\) = ax2ax+ +f3, p,

where a and 0 a 3-dimensional a freely moving discuss the case

(8.34)

are constants. Relation (8.34) encompasses the case of harmonic oscillator" (notice that a can be < 0) and of particle in the 3-dimensional Euclidean space. We shall (8.34) in Subsection 8.3.

*' I am indebted to Dr. J. Cislo Oslo for a hint.

The case U(\x\) £/(|x|) ^ £ ax2 + fi, where a, 0 - constants. constants.

8.2

75 75

The case U(\x\) U(\x\) ^ a x 2 + 0, where a, 0 - constants.

According to (8.29) and (8.30) we have (8.35)

r x fi = f( f(.r)xj fj ) j >,

dfi 8fj dfk dfk 1 ldf(r) df(r) x x a— = OXj a— = ~~i— J + finhj = oxu r dr k j W+/W

'W"

^ =4 r *

J

(8.36) (8 36)

•

"

From (8.26) and and (8.36) follows (8.37) (8.37)

a j GjjXjXk /_j®kjXjXi . VijXjXk — = ^2 kjXjXi ■

^/ 3i

3j

us multiply both sides of (8.37) by Xk and and sum sum over over k. k. We We get Let us by Xk r22

x / ], ao~ijXj — 9gx%, X *ixi = i>

r 99 = =^2 ^2^2^jkXjXk. ^2 ajkXjXk . 33

3

(8.38) (8.38)

k k

We are going to show that g in (8.38) is a constant; *' this would mean that x is an eigenvector of a corresponding to the eigenvalue g. From (8.38) and (8.24) follows

ydaijx £— d±k

jx k]x =ydak J

= 3

^— d±i

3

8g_ 8g dii

3

or

(8.39) (8.39)

dg _ dg_ P-H = -J-xk .

dxk oxi oxk axi From (8.38) and (8.23) we obtain da daij tj

dg

E E -dx^ i

Xj+aik X +aik

or

=

dx-kXi+9Sik

3

dg a

_ dg_ X

i — Q

X

k •

& & » • dxk* = oxi Let us us differentiate (8.39) and and (8.40) with respect to xi and and xi x\ resp. 2 d g d*g „. x Xi + dxk dxi dxk oxi

dg d2g dg d'g li a~^° ~ dxi »■ axi a Xk dxk oxk oxi axi

+

dg dg ~57~°kl > axi axi

(8.40) (s 40) -

8 41 ((8.41) -

*' The proof of this conjecture was presented to me in this form by Dr. J. Cislo.

76

77ie The Inverse Variational Problem in Classical

d2g

Mechanics.

d2g

(8.42)

Relation (8.42) can be exploited in (8.41), viz. dg . _ jfo dg^_ . (/ d d22g9 _ d d22gg \\ ^ = d^^ \dxidx \dx {dxk

dxkdxij dxkdxij

d±i dxt

dxk

(8.43)

Let us multiply both sides of (8.43) by n and sum over I. We get ( &2g _ d2g \ x 2 = dg_ _ 8g_^ (_** *L.) x 2 = 0LXk - *LXi. \dxidx dx dii dx \dxidxkk dxkkdxi) dxi) dii dxkk Taking into account (8.39) we conclude from (8.44) that d2g9 dxi dxk

d2g dxk dii dxi

=

(8.44) (8.44)

(8.45)

This relation inserted in (8.43) yields d

d

9 x dxt d±6u

9 xx dx dx~kk6ti ■ If If nn > > 22 we we are are able able to to choose choose k k= — II ^ £ i. i. Then Then from from (8.46) (8.46) follows follows dg _ do ^d±i = 0, =

/Q Aa, (8.46) (8 46)

-

;8.47) (8.47)

i.e. g does not depend on x. Notice that Y ^d i

E dcnj k

\

J

k

d

(9Xi)

n

(8.48)

J

Due to (8.48) and (8.35) condition (8.28) reduces to

*-? dxk k

(8.49)

dt

k

Let us further consider the relation x"^ Y ^ d{xiOijXj)

_

•4-* ■A" dxk a xXi = EE J2J2 IiXia^ ~d^Ix3i ++ H 12^a*i HCTi ** i ~ i ^%i ++ E^ =

=

=

i*

j J

ii

jj

"

jj

x Y,YliXij-^ kEE ^xJ++229»-

i

(8.50) (8.50)

The case E/(|x|) C/(|x|) 7^ ^ a x 2 +fi, + / ? , where a, p - constants. constants.

77 77

On the other hand sr-yr^ sr-S^d(xiad[xiOijXj) JJxj)

2 d(gx 2 )

dg

22

2 . 2 . ^ ^ == ^- r^ ==^^x* + 2 ^^' 2-2tet 1

J

(8.51) 8 51

(- )

Combining (8.50) and (8.51) we get (8.52) 1

1

Let us multiply both sides of (8.49) by :Ej and Xj and sum over z and j . By virtue of (8.52) and (8.38) we get

< 8 ->

*(?&«. +£)-*?fr-30--

(8.53)

Since g
§H

OXk

(8.54)

Formulae (8.47) and (8.54) prove our assertion that g is a number, an eigenvalue of a for the eigenvector x. We are going now to show that g is also an eigenvalue of a for the eigenvector x. To this aim let us differentiate (8.38) with respect to Xk- We get

E xi jr~+ i

°lk

8 55 ^(8.55) ' ^

= gSik

'

i

We have also We have also Qg^ Q_ x 2-*<*«*'■> 2 , -dt ^x3~ = 2^ dt^3 3) 3

d(gXl)

= - ma r =U°--

(8 56) (8.56)

-

3

Let us multiply both sides of (8.49) by Xj, sum over j and use (8.56). We obtain (8.57) 33

k*

Let us further multiply (8.55) by ±k, sum over k and make use of (8.57). The result is ^"Y^OikXh V j f c Z f c -= k

g±i. fix,.

(8.58)

78

The Inverse Variational Problem in Classical

Mechanics.

This proves our conjecture. Moreover, the relation

E §■* ** = £ §^** = £ £(****) - 47^*° = ° (8'59) k

k

3

k

3

3

and (8.49) entail that ^ j T = 0-

(8-60)

In other words cr is a function of x and x only. Notice that the considerations of this Section are valid for any Euclidean space of dimension n > 2. Because of (8.38), (8.58), (8.47) and (8.54) it is convenient to introduce the notation *ij = gSij + °% •

(8.61)

Then by virtue of (8.17), (8.38) and (8.58) we have a% = ^ ,

(8.62)

E^=°-

(8-63)

E4^=°-

(8-64)

i Notice that because of (8.23) and (8.24) we have

E^»E^=0. 3
v ^ dan .

E ^=EftJ»i=°-

(8-65) (8-66)

Our next task will be to find the structure of
ij

=

G L

ii

The case fU(\x\) ( | x | ) jt ^ ao x 2 + /?, fi, w/tere where a, 0 - constants. constants.

79 79

where L is given by (7.53). Taking into account (8.62) we have G= = G(x, G ( x ,xx)).. G

a% ij = ~ GLiLj GLiLj,,

a

(8.67)

If we insert (8.67) into (8.65) and (8.66) we get dG

LkLiXj E d±~ d±~LkLiXj 3i

+GLi +GLi

e

x xx x kijXiXj kijXiXj ++GL GLkk^2 ^2 E E etij *#i i ji i ==0°

J2Ys J2Ys

3 3

i

li

i

li

or 01 v - 9G

(8.68) 8 6

E -Q-.xi = ° and and

3

v-

&G ■

v^dx- X3=0 9G . j

( - 8)

3

J

22 g^x> = °

resp..

(8.69)

res

( 8 - 69 )

P-

3 3 Let us apply the method of chracteristics to both equations, (8.68) and Let us The applyrelations the method of chracteristics to both equations, (8.68) and (8.69). (8.69). The relations —- = —^ and and -7^ = -r^-, i, 1,2,3 i,j3 ==1,2,3 (8.70) (!••/■■ ;

(JLJb I

Xi

Xj

entail the partial solutions XiXj X}Xj

XjXi — 6jj'jti-/Jt .

Hence (8.71)

G = G{L). G(L). We know already that, due to (8.67),

4 = G{L)LiLi. G{L)LiLi.

(8.72)

We recall that by virtue of (8.61) the relations (8.23) and (8.24) imply

0f& *£«,. ^ = && M = =^ = dxi

dxk axk

(8.73) (8.73)

dxj OXJ

da°i a8a°^k = da% 54 = ao| 1 Td±i F * = IT ^ dxk = d±j ftr* 9xfc dxj

, -

,

8(8.74) 74

80

The Inverse

Variational

Problem

in Classical

Mechanics.

Then from (8.72) and (8.73) follows /

/ /I

o r tlimXmLjLik 07-

+

GLj€kimxxmm + y7 ^ GLj€kim,

y ^ GCjimXfnLk

m 771

m 771

— /

/ /

nj

£lkmxm,LiLj mLiLj

—=

771 m

+ y ^ GCikm LrCikmxxm-L>j m-L>j

771

+ /

771

/

GLiCjkmxm

771

or 7/

J

/

/7

€lirn ^livaxm*rn^JJj^jlJJk

oj

1

->r *^ /7 ,. ^J^-kimxm

~r T /7 , GtjiTnXTnLi}z

771

T7l

771

m

™ =

— I^i I1 7/ ,.. / \

/Z

or

e

x

+ 7y

lkm mLj m^j

771

y

xx

— ((8.75) 8 75)

m

II

GCjkm GCjkm m m

777 771

/

To put (8.75) in a more perspicuous form let us choose kfc==2,2,

j; == l,i ,

ii == 33;;

then d8G G _ _ .

^ — I2LiXi ^ l ^ i ■^—L O JL2 OL2

SG _ ,_

.

_ . .

~ —M -L2Z2 + - ^-^—L + £L332Z: 33)H + 1{L-2X2 COL\ IJI

dG + -^-j—L f -^j-L 3Liii 3Li±i OL Oh33

x2 - 2Li±i + L3x3) - G{L2±2

= 0

or

L =0 ([12^: E | |Li^ ++ 3G3)G ^^ 1 , 1 = 0 as L x = 0. Relation (8.76) for L\i.\

(8.76) (8.76)

^7^ 0 implies

dG

(8.77)

i

This relation indicates t h a t G is a homogeneous expression with respect to L of (—3) degree, viz G(AL) = AA~33G(L) G(L)

(8.78)

for each constant A. To summarize: according to (8.61) (Jij = o"ij = g&ij gSn + G(L)LiLj, G(L)LiLj,

(8.79)

The case U{\x\) V{\x\) ^± a x 2 + /3, j3, where a, ft0 - constants. constants.

81

where g is a constant and G(L) is a homegeneous function of (-3) degree in L. What is the set of Lagrange functions, belonging to a given by (8.79)? We are going to show that for each a, which conforms to the constraits (8.61)-(8.64) and (8.73)-(8.74) there exists such a Lagrange function, satisfying equations (8.18), for which 8d22LL

_

, v (8.80)

The most general solution of (8.80) reads L =

2YIYI

cFijdiidxj + y^/9 ]P#fc(x,£)a; o-ijdiidxj k(x,t)xk fc + $(x,i) =

1

k > = |ffx \g± + + L° + ^Y. **(x, **(*, t)x t)ikk + *(x, $(x, t),

2

(8.81)

k

where L° is a certain particular solution of the equation

d'2L° l3

dii dij

(8.82)

'

where er° is given by (8.72) and (8.78). In (8.81) we made use of (8.79). We are going to show that the equation (8.82) has a solution. To this end let us introduce ad hoc three auxiliary functions L°, i = 1, 2,3, defined as follows by the equations | ^ = 4a ,

Si,

«'

3j == 1,2,3,

(8.83)

for t arbitrary but fixed. Due to (8.82) we have

|4-§&=o. ,,»_,,„ j,k =

OXk

l,2,Z.

OXj

Therefore, for a fixed i, there must exist such a function L°, which fulfills (8.83). Here we are exploiting the Converse of Poincare Lemma. Having the solution L° of (8.83) for each i we are looking for the solution of the equations fir0

^ - = L°. axi

(8.84) (8.84)

82

The Inverse Variational Problem in Classical

Mechanics.

Again it follows from (8.82) and (8.83) that dij dii and again using the converse of the Poincare Lemma we come to the conclusion that there must exist such a function L° for which (8.84) and, as a consequence of it, (8.82) are both satisfied. This accomplishes the proof of our conjecture. Next step is to examine the structure of L, given by (8.81), in more detail. It follows from (8.73) and (8.82) that 2 L° d2L° \\ = Q dd /f dW W (8.85) dii V dij dxk dxk dik dik dxj) dxj) \ dij Consequently the expression d2L° d2L° dij dxk dik dxj does not depend on x. Thus we have d2L° d2L° _ d2L° d2L° (8.86) dij dx dxk dxj dij dik dxj i = v dijdx dikdxj dijdxdx dikdxj k=v k k k k i=v where v is an arbitrary constant vector. Let us introduce the function L1 defined as follows ^

\dii

dii

(8.87)

i=vy

By virtue of (8.64) and (8.82) we have dV_ dij

_

= d22Lll

^y . d2L° . l t-r1 dij dii % dL° dl? ir~dij dxj d22L° d L° dij dij dii dii

8L° dL" ITdij dxj

dL° 8L° d±j dij

dL° 8L0 d±j dij

i=v

, % 8 88 (8.88) i=v 0

-

i=v

dL 0 lJ lJ ' dij dij dii dii ' i.e. L 1 satisfied relation (8.82). Moreover, L1 satisfies also the symmetry requirement (8.22), namely taking into account (8.88) and (8.86) we have d2Ll d2L° d82L° diidxj diidxj diidxj ■ _ (8.89) 2 (8 89) d L° d2L° dd2L2Ll l " dij dxi dij dxi . _ dij dxi

The case f/(|x|) ^ a x 2 + / 3 , where a, /3 - constants.

83

Notice that L° does not, in general, fulfill the relation (8.22) although it solves (8.82). As L 1 and L° are both particular solutions of (8.82), we may choose L1 instead of L° in (8.81). We have then L = \gy? + L 1 ( x , x ) + ^ * f e ( x , i ) i * + $ ( x , i ) .

(8.90)

Let us insert (8.90) into (8.18). We have then the identity

Z-* \ dxj

dxi)

2

^

a L°

v^

3

dt

dxi

2

d L°

= E te^-'i - E di-^

x i=v

_V-^-i.+V— +

^

a^i dfj

^ ^

V&E,

J

4 - 9 i j a±j

/a*i _ OVA . 5xi) Xj

-9t{r)xt + 2^yd

+

dxj

...

>

+5/(r)

^(891)

x+ J

i=v

9*, _ s$ _ dt dxi ~ J+

dt

U

dxi

'

This identity is linear in x therefore 89j _ d^j dxj dxi implies *< = ^ : * ( x , t )

(8.92)

and, because of (8.92),

^

= 5/(1x1)^ + ^

=^-^(1x1) +

^

implies $ = - 3 (7(|x|) + ^ + c

(8.93)

84

The Inverse Variational Problem in Classical

Mechanics.

where c is some constant. By virtue of (8.92) the term in (8.90) ^Yd x ~dx~k ~ k X kXk + ~dt ~di ~~~dt ~dt is a gauge term and can be discarded. Thus L = 59 ( | x 2 - t / ( | x | ) ) + L 11(x,x) ))

(8.94)

where we made use of (8.93). L1 is given by (8.87) where L° is an arbitrary solution of (8.82). According to (8.38) and (8.58) the constant g is an eigenvalue of a belonging at least to two eigenvectors, x and x. In view of (8.4) we must have g^O. 9^0-

(8.95)

The matrix er has three eigenvalues and eigenvectors. Two of them we know already. To find the third eigenfunction let us make the ansatz that it is proportional to L. With help of (8.79) we get 2 2 + GCL)^] GCL)^] )^. )^.

^aijL^ig

(8.96)

j3

Hence the eigenvalues of a are g and

g,

g + G(L)\L{ G(L)\L\2 (8.97)

2 2 deto- = <; <;2(
as indicated in (8.8). Let us derive from L (8.94) the Equations of Motion. We have dlL_ dj . dL° 8L° ^dii = gxt' + — d±i - — dii oxi

8d2L dii dxj

dxi

82L° dii dxj

dXi

(8.98) 8.98

k=v k=v

d2L° dx^ dxj .■ _ dxi

(8.99)

Q2 L

' a-3" = ^ *+ 4 o di~dx- = 9 ^ + a*i

dL_ _ _ dU{\x\) dUJM) dxi dxi

(8.100) yy^ , I d2L° ^— \dxjdxi £-* \dijdxi

d2L° dijdxi

.

\ . J J

(8.101) ' '

The case U(\x.\) The U(\x\) ^ a x 2 + fi, where a, fi - constants. constants.

85

Hence, taking into account (8.99)-(8.101), the Euler-Lagrange Equations read ^

d2L°

/

d2L°

\ .

„

at/(|x|) _ ^ / / d2L° L° 9dx{ ii

ad2L° L°

dij dxi ^£-? \\dxjdxi

d±j dijdxidxi

v-

o ■•

° \\ . .■_ /

3J

(8.102)

=^ + « + E4^« Since

y„.a*a ^ o gj ij r dr

,103) ((8.103)

v ^ o gt/(x|) JJ

jj

3j

we may add (8.103) to (8.102) and using (8.61) obtain eventually

8 1M (8.104)

12'* i > 4(*><++*£?)-<> ^H

<- >

J

j

i.e. equations (8.3), equivalent to (8.2). Let us check equations (8.28) telling us t h a t every entry of a is a Con­ stant of Motion. We have

^r = = EElf^r ^r JrL§^ ^ ++GG^^ as as

£—>

dt

dLkk dt —» dL

dt at

+ GL

+ GL^^ == °o

8 105 (8.105)

(- )

dt at

k

\~~^ cv~^ ^-^ ^

dLi

i

L-.

^ ^ v -^^ ^^. 4

h.

' v ^'v ^ 1 1IdLT dU kX x = 0 = -2^2^^ i K~dV j k j

oU vU

'

8 106 (8.106)

(-

)

k

Obviously we have also d m ^-Trcr —Tra dt

= = 0

(8.107)

as well as -~Tra T r < raa=0. =0. dt at

(8.108)

86

8.3

The Inverse Variational Problem in Classical

Mechanics.

T h e case U(\x\) = a x 2 + 0.

For the potential U(\x\) = ax 2 +P,

a,/3 - constants

(8.109)

the Equations of Motion are Xi + 2axi = 0,

i = 1,2,3.

(8.110)

To find the set of s-equivalent Lagrange functions L we have to make use of the identity 92L

^

..

-^

J

£-*• dii l d£jJ J

d2L

^— dii l dxj3 3

. 3

d2L

dL _

dii dt

dxi (8.1H)

= y^gjj(x,x,f)(ij- + 2axj) j

where deter ^ 0 , o o

a.e..

(8.112)

Relation (8.111) we may also write d2L di^i-= d

(

ai

>

= a

(8

»>

„ v-^ dL

^ dL . + X3+

8L\

2

dL

-U3)

,„

^[-^w,*' ^dx- m)= 0x--

<8-114>

We have, as before, d2L dxi dij

d2L dxj dii '

dajk _ doij _ duki axi oxk oxj da

Jk d±i

=daii

dxk

=

dak1 dij

' ,.„., '

v

*'

'

From (8.26), (8.27) and (8.109) follows 2a^2dtjSjk i

=2a^2^kjSji, i

(8.118)

The case U(\x\) C/(|x|) = = a x 2 + /3. 0.

4j - dxk

3

3

dt ~ U

^i dxiJ X} J

3

87

reSp

(8.119)

-

From (8.118) follows for a ^7^ 0 that (8.120)

ffijfc = 0~ik — ^
The same follows for arbitrary a from (8.113). From (8.119) by virtue of (8.117) and (8.120), we get „2 a v— ^» du v sr-^ ^ du . du / ~K^XJJ - / 7;—XJ: - —- = 0, 4 ^ dxjJ *— dii J dt 33

J

, (8.121)

J

3

where u stands for arbitrary entry of er. Equation (8.121) we may solve applying the method of characteristics. We have dx\ dxy _ dx2 2ax\ 2ax2 2ax2

dx\ dx22 _ _dx^ dx3 _ dxi _ _dx x,\ x2 ±3 X\

dx3 dx3 2ax3

,

,„ •.„„.. (8.122)

The partial solutions are the Constants of Motion \x2 + ax2 = Hj Hj, ,

ji = 1,2,3, (8.123)

x

XkiXk ~ XiX XiXk k == Lik Lik ~= ~L>ki -Lki — 22 e-ikiLi,, kixk ~ / i t-iklLl I1

%i%k 1 ZOiXiX AOtXiXk XiXk -+k

= Ai Aikk =

— -A-ki = A-ki.■

There are 9 of them but not all of them are independent from each other, e.g. we have*) AHiHj = 2*L% AHiHj 2aL% + A% or

(8.124) 4HX1H \H H22H H33 + A12A23A31

=E HXXA\ A\33 + +

H2A231

+

H3A A\22.

Nevertheless we may write u = u(Hi,H2,H ,L,L31, ,H3,Li2,L 3,Lu,L2323 3i,

A12, A), A12, A23, A31 3i),

(8.125)

where u stands for any 0{k . We are not going to give here a general solution of the stated problem, as the variety of solutions is very large. We shall inspect two special cases where (1) u , , £ 3L3li)) U ( 1 = u(Hi, U ( # i , tHf 22, , tHf 33, , £L12, l 2 , £L23 23

*) Relations (8.124) were given by Dr. J. Cislo.

(8.126) (8.126)

88

The Inverse Variational Problem in Classical

Mechanics.

and

(8.127)

u^=u{H H2,H ).,A23,A31). u^ = uu(H Hi, l2,AH 23,A 3l12 u 3,A 3,A

The functions (8.126) and (8.127) have to satisfy the conditions (8.116) and (8.117). Let us look at

dan _ <9<7i2 da\2 dx dx\ 8x22 dxi in case u ' 1 ' we get dau .. dan

dau dan

2a— r — Z 2 - » i - f dL l7 dH 2

dH2 =

dax2 i2 2a Xi ~xir a Hi dHi

12

dL12

, .

dan

^ 3 - 5dL 7— = 23

dL23 .. dai2 . da\ dai2 +X2 dLi2 ^?—^3^7—• dLzi <9Li2 dL3i

( 8 i 2 g )

(8.128)

In a similar way from

dan dx2

_

dai dai22 d±i

follows dan ■ dan X2 T - Xl -WTT + FdL dH2 0L12 x2 adai2 . = Xl ~aIT ~ ##! dHi

dan ~ AdL T - ^ 3 == oL 23 A AdaX2 dai2 x 7^7— * + -^7—^3 -Kj— 5L dL x3 ■■ dL dL ni 2

(8.129) (8-129)

3X 3X

It turns out t h a t the relations derived from relation (8.117), like (8.129), do not carry, in the case considered by us now, any new information as compared with the relation (8.116) and we shall ignore them.

The case £/(|x|) = a x 2 + 0.

89

R e l a t i o n s (8.116) yield i n a d d i t i o n t o (8.128) in c a s e « W t h e e q u a t i o n s „

dan ~aITXi oH3 „ dai2 Za ~xTrX2 alii „ dan dH3

. dan ~X2~XT dL23 . da\2 ~ x*~x? oL\i . do\i dL23

2a

da\3

.

da\3

9#2 9(713

.

2 a ^ - r — x 3 -x2 9.H3 _

9(722

.

dH3 „

, . dau dai3 . da\3 = 2a +X2 ~AT~ ^TTXl ~a? 0L31 dH\ oL\i , . <9cri2 „ da22 . daii ^ S ^ T — = 2 a — — X i + x2 — 0L23 oHi dLi2 , . <9ci2 „ da\3 . dan 0L31 dH2 dL\i 9(713

9(723

9Li2

.

0L23

oi/i

9(713 , .

9(713

2a

9(722

»

, .

9(722

9Z/23

9(733

„

— l-EioT— = 9L23 9L31

. 9(733

201—77-2:2 - x i — 9.H2 9Ll2

. dan ~a?— > dLi3 . da22 x3 —— , dL\3 . da\3 dL23

VXI

9(733

^7TXl 9i/i

, .

, .

+

X2

9(723

9L3i

9//2

. 9(733

9(723

.

X3

9(723

. 3(723

9Li2

dL3x

9(733

.

o7 9Li2

X3

9(723

, .

9Li2 9(723

l-X3^7— = 2 a — — x 3 -x2 91*23 9ii3

— 9Z/23

/o T on\

9(733

^7— > 9L3i 9(723

9Z-23 ,

9(723

t-^iF?91/31

N o t i c e t h a t for a 7^ 0 x a n d x c a n n o t b e e x p r e s s e d i n t e r m s of t h e p a r t i a l s o l u t i o n s ( 8 . 1 2 3 ) , i n p a r t i c u l a r in t e r m s of Hi a n d L ^ = —L^i, i, k — 1 , 2 , 3 ; if t h a t w o u l d b e t h e c a s e x a n d x s h o u l d solve t h e e q u a t i o n ( 8 . 1 2 1 ) , b u t this does not occur. T h e a n s a t z t o solve e q u t i o n s (8.128) a n d (8.130) is
1

2

3

(

8

m

)

F r o m (8.113) w e g e t t h e L a g r a n g e f u n c t i o n , c o r r e s p o n d i n g t o ( 8 . 1 3 1 ) , v i z

a

ii(Hi)\±i=Ui

'

-' k

(Xi -Ui)dlLi

+

' j^k

(8.132) j

w h e r e c is c h o s e n s o t h a t t h e i n t e g r a l in (8.132) m a k e s s e n s e . T h e f u n c t i o n s / a n d g h a v e t o b e e v a l u a t e d from t h e i d e n t i t i e s ( 8 . 1 1 4 ) . W e h a v e •sr^fdfi

9fj\

■

22 (3*7 " tel 3

Xj +

. d f i d g ^ r

"* " ax7 - 2 " I > ^ + J

, +2axj /

7777-^1 dui = 0 .

(8.133)

90

The 77se Inverse Variational Problem in Classical

Mechanics.

Since dHi we have

~n OXi

= X

i

fXi dan fXi "^— Ui du dUi = / U / ~E7T i i - I Je, J '« dHi dHi ii ii == UU ii Jcc.. I

— an <Tii| i ._ C i . = an — <7ii\ii=c^ .

dan —— dUi — du ~5 i ~ dui dui ii=Ui ii=u<

If we insert (8.134) into (8.133) we o b t a i n If we insert (8.134) into (8.133) we o b t a i n y(dU_dJl\i+d_h_ £—• \dxj \dxj dxij dxij J3 dt Z-" dt —2ay^ajjXj —2g } ^crjfXj - 2ax 2axia . ian\ =Ci ii\ i.=c ._,.

'

dg

- -— = 0.

(8.134) (8.134)

(8.135) (8.135)

OXi

Since the t h e l.h.s. of the t h e identity (8.135) is linear in x we we have

dfi_ = dfl dxj

dxi

or

A - M fdxi eS

(8.136) (8,36,

and dg d2$ dg n n v~— = -^— =—2a — 2a>> anXi anXi——2axian\laXiGiA- „„++——————75«

or or

9= =~ ~aa ^2 Y ^2PijXjXi ZJ ^ i * * -2a^, ~2aJ2 i

j^i

x

xi <Jii\±±i=Ci .=c. dxi (8.137) i <7ii\ dxi++— — .. (8.137)

i

In we insert (8.136) and (8.137) into (8.132) and discard the gauge terms we get IXi;

a

'

/ /

-

;

ii(Hi)\Xi=Ui

(*« _ Ui)dui + | J]]^Vj f c *•» ; ' ; * _ i Wi

a

~ YJ Yl a3kXjXk ~2aY.J *j£ /■Xi i

Jc

'

a

*i(Hi)\ii=c, dxi = (8.138) (8-138)

' i

+ |\ YJ 2_J ajkXjik UjkXjXk - a 2 J ajkXjXk (TjkXjXk++ const. const. +

The case C/(|x|) = a x 2 + p. p.

91

where Hll = ax} H ax} + + \x\ \x\ , (see (8.123)) and a is given by (8.131). We used also the formulae

^

= -«

and - — = 22axi. aii. OXi OXi

This is the most general form of the Lagrange function, under the proviso a conforms to (8.131), which function yields as its Euler-Lagrange Equations the Equations of Motion equivalent to (8.110). This Lagrange function satisfies the condition (8.115). The Hamilton function reads H = ]T ^ZuiHi) + Y, ^2^2a k B= Eii(tfi) + S jkff{±XjX i*(i*i**

+ M+axjX i**)k).•

8 139 ((8.139) - )

In case u' 2 ' we proceed in a similar way as in case of u^1)*' . We are not going to write out all the 8 equations for a. We again make an ansatz, namely C7ij—=aijctijAij (7jj Ji-ij ,

,

aija.ij==ajictji- -constants, constants,forfori i^ ^j ,j ,

(8.140) (8.140)

while no assumption is made as far as an, i — 1,2,3, are concerned. The relevant relations, obtained after performing some simple calculations, are dan dan ~dih da33 dHi « ,

_ ~ _ ~ =

9(7 2222 _ 9(7l 9(7122 _ <9<7 9 F 7 ~ O~AT2 ~ ai2 dan _ dai3 _ dH ~ Q* l »3 « 3, -~ «dA * l3» ■

5(722 _ 9(733 _

(8.141)

9(723 _

~dlh dH3 ~ dA^ ~~ ~~Q23 ° 23 Ms" ~ ~dih ggg _ „ dan _ dan _ dan 0A ~ 9 A " a A i 3 ~ '' 8~M2 dA23 12 2 3 ~ dAl3

. _ iz =i 1,2,3. - '/'"i-

*) The case u ' 2 ' , formula (8.127), is a slightly generalized version of an example given by S. Hojman and H. Harleston, J. Math. Phys. 22 (1981), 1414.

92

The Inverse Variational Problem in Classical

Mechanics.

Hence
+ f2{Hf2(H + 2), 2),

033 = aQ3i-#i + a ot3232H H22 3iHi

+

(8.142)

f3{H3).

The Lagrange function corresponding to a given by (8.140) and (8.142) can be computed in similar way as in case u^1'. The most general Lagrange function for u^2\ when fi(Hi) = 0, i = 1,2,3, is L = - jT >J &ij(tftf 3

i^j

L = - jT >J &ij(tftf

+ iaxjx? +

8aXi±iXjXj)-

+ iaxjx? +

8aXi±iXjXj)-

(8.143)

} ij*3

The case (3) (3) The ucase = u (L 1 2 , L23, L3U

A12,A23,A31)

U(3) = W ( 3 ) ( ^ 1 2 , ^ 2 3 , i 3 1 , A 1 2 , > l 2 3 , ^ 3 l )

(8.144) (8.144)

does not give any new results. It is a special case of (8.131) where an — const. % = 1,2,3. It has to be emphasized that ansatzes (8.131) and (8.140)-(8.142) do not exhaust all possible solutions. Notice, however, that our equations are linear, so the superposition of solutions yields again a solution*' 8.4

The Hamilton formalism for the model investigated in Subsections 8.1 — 8.3. Equivalence sets of Lagrange functions.

We shall first discuss the case considered in Subsection 8.2 where tC/ /((| |xx| |))^#aaxx22++ / 3 .

(8.145)

In this case the most general form of the s-equivalent Lagrange functions is given by (8.94) 2 - f/(|x|)) ++ Li L1 ++ ^ d*J^±, M , L = gi ^-U(W) 3 (|x

*' I profited from a discussion with Dr. J. Cislo on this problem.

. 146) (8 (8.146)

93

The Hamilton formalism for the model... model...

where (see (8.87))

^{dij

dxJkJX>

(8.147)

and the last term in (8.146) is the gauge term. The canonical momentum reads

dl 8L11 5$ 5$ Pi = «r= gxi + ~^7Pi = « r - = gxi + ~^7- + + ^— -77- = axi axi oxi = OXi OXi OXi .

dLo_dlS dLo _ d!P_ d±i dii dii i = v

9a$ $ dxi '

(8.148)

where we used (8.64). Thus p is fixed up to an arbitrary gradient of a function of x. By virtue of (8.82) we may, in principle, express x as a function of x and p . The form of this interrelation depends on the shape of L1 The Hamilton function is defined as usual (we skip in the following the gauge term) (8.149)

H = = 22 ^pi±i-L. H Pi^i ~ L ■ i

To compute (8.149) we first inspect the term

Ef-sy-i1.

(8.150) (8-150)

From (8.147) and (8.64) follows

£ | ^ - L l =1-00-Ear*'-* 33

(8(8.151) 151)

"

3 J

Thus the contribution coming from L1 just vanishes. Let us go back to (8.148). From (8.146) and (8.151) we conclude that 2 F = = g(^ + H + t/(|x|)). U(\x\)). ff(ix

(8.152)

Notice that in terms of x and x H looks like a Hamilton function of a particle in an external field; but in the variables p and x its form can become complex. The Hamilton function (8.152) is conserved as

— =9l2*i{xi+tei i

J=0.

(8.153)

94

The Inverse Variational Problem in Classical

Mechanics.

It is interesting to find out whether all Lagrange functions given by (8.146) are equivalent among themselves and equivalent to (8.1). To an­ swer this question we have to investigate the Lagrange brackets (or Poisson brackets). We have for the Lagrange brackets Y^ fdxr dpr *-— \dxi dxj

\Xi, Xj3 J-

d2L diidxj dii dxj

dxr dxj

dpr\ dxij

d2L dijdxi dij dx{

(8.154)

'

■j^Xj, Xj y \Xi) Xj j

— —

ij — ji — jJ C?ij — O'ji — l"*is\Xi)Xjj

\Xi) Xj j \Xi) Xj j

— —

U , U ,

where we used (8.15). Since according to (8.79) Oij = gS^ gSij + G(L)LiLj, G(L)LiLj,

(8.155)

the Lagrange functions belonging to various functions G are not equivalent to each other; they are only s-equivalent among themselves. The set, equiv­ alent to (8.1) consists of Lagrange functions for which G = 0. In general, the equivalence sets for G ^ 0 consist of only one single Lagrange function, characterized by fixed G. Let us now investigate the case treated in Subsection 8.3. We have U(\x\) = aaxx 22 ++ / ?3 ,,

(8.156)

where a and ft are constants. The most general form of the s-equivalent Lagrange function for the case u^ (see the formalae (8.126) and (8.138)) reads Xi

/

E H °ii(H du -Y Zu(Hi)+ °ii(Hii)\ )\ii=Ui ii=Uiduii-Y,/ n( i)

-'

+

i

+

x x z J z2 aVjkijXk - az2z2 a 2 J ^2 aaJk jkXjX + 22z2z2 ikxixk ~ j k k + const.

k k

jyik jyik

k k

jjtk jjtk

where where 2 Hi = \x —-— i-2 + ax]2 , dS(z) = a(z) , . and dz Hi = |x^ + aXi , — — = a(z) and
(8.157)

The Hamilton formalism for the model...

95

We may compute the canonical momenta Pi=^-=

/ "

OXi

0u{Hi)\ii=tt.

J

dm + Y^OijXj t&

(8.158)

Then the Hamilton function reads

H = Y, Xu(Hi) + Y, 5Z tr«t(k±*if + aXiX^ i

k

( 8 - 159 )

■

i^k

For Cij = Oij

relation (8.159) reduces to (N.B

dT,

"}tHi)

= 1 => XW = Hi)

H1 = | x 2 + a x 2 The Hamilton function in (8.159) is conserved viz. dll

\—*

-rr = 2_j

dl±i

Uii

~Af

+

V - ^ ^--V

1^,1^

i

i

- / .-

a

^Xi{xk

~

\

~

+ 2axk) = 0

k^i

as —— = Xi(Xi + 2axi). We are going now to show that Lagrange functions (8.157) belonging to different <x are not equivalent to each other. We have \Xi, Xj j — U ,

{xi,Xj} = aij ,

(8.160)

\Xi, Xj j — U .

The set of Lagrange functions marked by certain choice of er, where at least one au(Hi) is not a constant, consists only of one single Lagrange function and is inequivalent to other Lagrange function of this type which belong to different a. If all an, i = 1,2,3, are constants then all Lagrange functions belonging to a^ = c ,i,j — 1,2,3, c - arbitrary constant, form an equivalent set, not equivalent to the former.

96

Tfte The Inverse

Variational

Problem

in Classical

Mechanics.

For the case u' 2 ' the Lagrange function corresponding to cr given by (8.140) and (8.142) with /;(#*) = 0, i = 1,2,3, reads (see (8.143)) L = | 22_, J z Z aa
The canonical momenta are Pk Pfc = = iy~r- = 2^aifc(|a;itx "^aiki^ikif 2 + + aaxi k^x\x 2 + 2axA 2aa;itZtZi) :a;ia;i).•

(8.162)

i^k

Then the Hamilton function reads H = 2 j 2 j Q y ( | i ; 2 a ; 2 + \ax\x2

* ?vv

+aXi±iXji (8.163) ( 8 - 163 )

22

This Hamilton function is conserved. For a free particle (a = 0) we have for the cases it' 1 ' and u' 2 ' a 1 L(2)

ii{\u\){%i

--'

'

Y. aijilx2

= I £

- "i)dUi + 2 5 Z 5Z 3 «#i i **i resp.

°iiiiiJ (8.164)

and w H # ( 1 ) =Y1 = X!

CTi ii:i; , )uidui + /^2i(TijXiXj / aii{\u "sd™?)"**! + \'Y l5ZzC > J' 2

1 JciCi

H{2) H{2)

i

IJ2J2ai^2ii2j = IJlH^i^i i^j

i

res reg

pP-

The Equations of Motion are in both cases ^CTijXj

=

0.

j

For a*11' =

/ 0 0 1\ 0 1 0 \\ 11 00 00 //

Ji

&J m

(8.165)

Examples.

97

we have, discarding the gauge terms, L{1) = \x\ + xxxz

(8.166)

and the Equations of Motion are *3=0,

aJ2=0,

3-1=0.

(8.167)

1

Notice that Z/ ) in (8.166) is not equivalent to L = \-k2 .

(8.168)

Notice also that /0 1 0\ cr = 0 0 1 , \i 0 0/ corresponding to x2 = 0,

det«r = 1

x3=0,

(8.169)

ii=0

(8.170)

(in this order) is not admissible, as cr is here not a symmetric matrix.*' In this case there does not exists a Lagrange function yielding (8.170).

8.5

Examples.

According to (8.94) the most general Lagrange function for J7(|x|) / ax 2 +/3 reads

i-rt^-tfMHE^-sLK

(817I)

Here L° is a certain particular solution of the system of equations (8.82),

c%

(8172)

Hkr where 0% = G(L)LiLj,

(8.173)

G(XL) = A" 3 G(L)

(8.174)

are given by (8.72) and (8.78). *' see R. M. Santilli, Foundations York, 1978.

of Theoretical Mechanics.

I., Springer Verlag, New

98

8.5.1

The Inverse Variational Problem in Classical

Mechanics.

Example of Henneaux and Shepley.

Let us present here the nice example of Henneaux and Shepley*' , the only example preserving rotational symmetry of the Euler-Lagrange Equations. The model is as follows 0

d2L>

_

e

[L '_J e Li = y^ / , ijkXjik

3

]

(8.175) £ - a constant.

k

Then the equations of motion should be

^

+ H)+E^W,=°

(8-176)

which are evidently forminvariant under proper rotations. To find L1 let us make the ansatz for L° namely a = x2,

L° = L°(a,b,c),

b = xx,

c = x2

(8.177)

L° is then invariant with respect to rotations. Then d2L° diidij'

_ d2L° dV*XiXi+

o9 2

2

L° dbdc{XiXj+XiXj}

+

J

(8.177)

2

Ad

+4

L° . .

Qc2

5L°

XiXj +

^

d{j .

On the other hand ^

L i L j

=

e — / T ox 3, L (L2)2

2

"ijf

' XX^XjXj H- XiXj)

(8-178) X X{Xj

X X{Xj 1

If we compare (8.177) and (8.178) we get d2L° = db2 ~

ex 2 ILI3 '

*tf

£*1

flLP

|L|3 '

2

4

5c

2

a 2 L 0 _ exx 965c ~ ILI3 ' 9c

e_ |L| '

*) M. Henneaux and L.C. Shepley, J. Math. Phys. 2 3 (1982) 2101.

(8 179)

-

99 99

Examples. From (8.179) we conclude that that L = l\££~ L°0 =

fdz

_

^

+fiab) bh f % + / ( a , b) = = -(ac -b2)l*i{ac +h2)i+f f(a, b), aa JJ zi a Z2 a

where

2 L 2 = ac a c - 6b2 = zz,,

dc efc=-d,z. = -dz. aa From (8.180) and (8.179) follows d22L° £ b22 £ L° _ 1 E 2 2 <% ~ a ( a c - 6 ) i a (ac-62)f or

(8.180)

(8.181) 2 d 9 2 // _ <%22 ~ db ~

d2f

£C £C ( aacc--6622))§I (8.182) (8.182)

^db-2 = 0 . This implies

(8.183) (8.183)

f = r{a)b s{a). r(a)b + s{a). If we insert (8.183) into (8.180) we obtain 2 2 L°=-{ac-b L°= -(ac-b )5+r{a)b )i +r(a)b + s(a). sla). a According to (8.147)

(8.184)

(8.185) We do not need to perform this construction as for v = 0 1 L £ ( a c _2 6 2 ) i . eJH = L1 = =e^ z = -{ac-b )2

(8.186) x a As direct computations show function L\ given by (8.186) satisfies indeed the condition (8.22). The Lagrange function then reads L ==g(\x2-U{x))+e^. L 9(^-U{x))+e§.

(8.187)

Thus we conclude that (8.187) is the most general rotationally invariant form of the Lagrange function. The canonical momenta are dL

(

£ x2±i — (xx)a;j\ (xx)ij\

dxi dX

V

9

/ \ ° (. e(Lxx)\ ■ , e(Lxx)\

x

lLl

X | L |

/

'

(8.188) (8-188)

100

The Inverse Variationat Variational Problem in Classical

Mechanics.

To invert (8.188) with respect to x and p let us introduce an auxiliary vector (8.189)

Ji = i- ( x xXp p) « ) j.. Ji = 99

Let us multiply both sides of (8.188) by eijiXj. We get *' ee ((LL xx xx)) xx xx // e el1 \\ Ji = x x x i - - ^ j i f : = Li ( 1 + - - r r . Hence

L

x2 L

l l j t = ( x x x ) l - -5 ^r=u

(V i + - g\MJ mj.

j m

Hence L

or

(8.190) (8.191)

L

and and

(8.190)

U = Ji j ) T |

(8.191)

<-"-'("m)'-("+$ |J| = - ( £ <7

+ 55|L|)

(8.192) (8.192)

Taking into account (8.192) and (8.191) we have L

(8.193) (8-193)

i = jM-

From (8.188) we get, exploiting (8.193), 1 / e(L x x ) A; 1 / e(J x x ) A Pw fr^-Ua | , ) = ( aiTi J) •■ T ^ -9 ^ x2|L| J " g{ > x2|J| g V x 2 |L| / 5 V x 2 |J| J Formula (8.194) presents x expressed in terms of p and x. The Hamilton function in terms of p nad x is

H = |x2 2 ++C/(|x|) c/(|x|)= =i(H) § (H)-a g_l£ +g l+ilLi L +y(W) tf=ix +£,(,*,)

(8.194) 8.194

(8.195) (8.195)

aass xx

22 __ =

1 („2 f„2 3o((JJXx X (( JJXX XX )) V 1 x )) PP ££ V \\ 2 V X 22|J|| J | ++ ((XX22)2J2 5ff2 V X ) 2 J 2 )J ~i f/ 1

gg22 \\

2

2 ITI \ 2 g g jJ| | e 2x 2 2 2 2

x x x x yJ

If If ee = = 00 we we get get the the Hamilton Hamilton function function for for the the model model (8.1). (8.1). 2 *) *) (a (a xx b) b) xx cc = = (ac)b (ac)b - (bc)a; (bc)a; (L (L xx x) x) xx xx = = -x -x2h h

as as Lx Lx = = 00

(8.196) (8-196)

Examples.

8.5.2

Example

of

101

Stichel.

Let us present another interesting example given by P.C. Stichel *' . The model is 3 4a% == £ (r(L))" ( r ( L ) ) " 3LiLj,

T(L) = [ L 2 - ( b L ) 2 ] 2 = | L x b | ,

(8.197)

where b ^ 0 is an arbitrary fixed vector and e a constant. We are going to show that

i1 dd2Tr (bx) ( b x3 Fdiia ^dij-

L = r _ 3 L i LJ

*

-

((8.198) 8198)

The prove (8.198) it is sufficient to consider the particular case b = (0,0,1).

(8.199)

Then from (8.197) follows T(L) = (L 2 + L 2 )i = [x22i:i 2 + x22a;^2 - 2(xx)i3a;3]5 2(xx)i;3a;3]5 .

(8.200)

We have dT 1 2 ^r- = -^~ = ^[Si3(^ ^[^3(x2xl i3 - (xx)i 3 ) + Xix\ ±ix\ - XiX XiX3x3xz}. 3].

(8.201)

dT Xi°> 0 ^dx-Xi=

= '

(8-202)

5 2 —1 Xi = T,

(8.203)

OXi OX{

11

Hence

J2d~.

(8.202)

i

2-dii i

'

2

and

0 £aS^ ~ * <9£* 9a:-, = ' i

J

" i 9ij d iJ, %

J

*) Private *> Privatecommunication communicationfrom fromDr. Dr.P.C. P.C.Stichel. Stichel.

(8 204 (8.204)

- )

(8.205)

102

The Inverse Variational Problem in Classical

Consequently the shape of g?,g± d2T g-gdi^x-

= =

Mechanics.

must be (8.206) (8-206)

H^)^-f{X'iQ)L^-

Let us differentiate (8.201) taken for i = 1 with respect to X\. We get V *■ =

W

1 / •

X^

2

•

~T ~ f ? ( i i a ; 3 ~ x^^3?

\2

"^

= ^L\

r2

.

(8.207) (8.207)

If we compare (8.206) with (8.207) we conclude that

/(*.*) = f| or

„

,x

(bx), 2

/(x,x) = ^

(8.208)

which accomplishes the proof of relation (8.198) We have (8.209)

L° = ^ .2 (bx) ' Since T T is homogeneous of first degree in x we have

£ a^7^ = L■ ■

8 21 (8.210)

( - °)

^-dx~Xi=L

Choosing v\ ui = — uv22 = — 0 we get, making use of (8.185) rl _ rO L L _— e^ X (Xl±l — 3

-

+ X2X2)

- X3(XJ + X%)

c^p (z? + Z2) o

o—1

(8(8.211) 211)

-

5

Notice that the second term in (8.211) satisfies (8.22). It remaines to check

a2

_ ^ _ diidxj diidxj

2

/ £ r \ == _ d* _ (JL.\ \V(bx) ( b x ) 22 // dijdxi dijdxi

((J*L) £r \

\(bx)2/ ' V(bx)V

,(8.212) 8 2U) ° l >

{

To do that it suffices to show for b = (0,0,1) v

^

d2 ((L2 + L2)i\ _ d_ ((LJ+LiyA k diidxk \ x3 ) dxi y x3 )

(8.213)

Examples.

103

We have

^ifca^{^(x2^ _ h,3(-x2xj or

+ i2a;

3- 2 ( xi ) i 3^3)H =

+ 3(xx)± 3 x 3 - 2x2x3i) x33il

xkx\ - xkx3x3 xzT

E^P^-t^'

(8.214)

Let us differentiate both sides of (8.214) by ii*. j . We have

y ±i t—1 k

=

d2 d2

LirA ((Lj + LiyA

dxk dii \

\

x3

J

+ +

/

2 2d(i33(L 2d(x (Lj1+Ll)i) + L22)i)

({Ll + Lp'A _d_ ({L* Lpi\ dx% \

\

d = 2_d_ =2

x3

l{Lj f(LJ+Ll)i\ + Ll)i\

J /

= = /Q (j-i r\ (8.215)

which accomplishes the proof. So we may instead of (8.211) use £T 1 -L U = U -= p g j Ll Lo ~ ~ (bx^ The 1 he Lagrange Lagrange function function of or the the model model is is

L = 5 5( (I Ii i2 2- -Ct 77 ((| |xx| |) )) )++££II»»

(8.216) (8.216)

(8.217)

and the the Euler-Lagrange Euler-Lagrange Equations Equations are are and

99 (xi U ++^ ^J M p ) ) ++e Y. e j n\M)~ : r O3L LiL^ D - ^ i ;= =0. 0.

(8.218)

3

It is obvious that neither the Lagrange function nor the Equations of Motion, are rotationally invariant or covariant resp. We have

99 lx ++ g (bx) 2 dx t ~ V K ~=dxdxr [ * -9w^) Pi

J

(8.219) (8 219)

-

The r.h.s. of (8.219) is not linear in x. Nevertheless, relation (8.219) can be inverted i.e. x can be expressed in terms of x and p , but the procedure is tedious and the formula complicated.

104

The Inverse Variational Problem in Classical

8.5.3

Example

Mechanics.

of Ranada.

Next two examples for the case f/(|x|) ^ a x 2 + (3 are due to Ranada** . We choose first G = ( L{L\L i ^ L2Ls3) - 1

(8.220) (8.220)

According to (8.82)

L -P^ =L\L-rrr ^diidxj L

8 221 (8.221)

(- )

1

2 3

To solve this system of equations we start with d2L° _ £1 W dx\ = TT L2L3 '■ dx{ L 2L 3 We we are going to use the tne tormuia formula

(8.222) 8222

(

)

dZ dz _ J (b2 2 - ac) 2 - b - az 1 ft = *1 l n I {b - ac)i - b - az ) (8.223) n 2 2 22 2 ./ + 2bz 26z + + c ~ 2 ( 6 - a c ) 5 \(b - ax)i J az + \ {b ax)? + b +az + az )J

for ac-b2

< 0. In our case for

d (9L° _ J±--r dii 9±i 2

ac-b

= -\x\L\

dfei ff d±1

1

*J y L L22LL3 3 ' '

< 0 and according to (8.223) we get

\ „ dL° 1 / x2(x3xx1 -x3xxi) x) ln ■5^: T ) 3) = F " = -—7—: ? : = \ ) +gi{x,x + 5i(x,a: = 2,x23,a; Dxi Xi {X1xX2 - Xix XXX2) J (8.224) /onnAs da;i ^i \\ xX33(x x 2 2) J ( 3 l ,lrJ / ^~ l \ +gi{x,x , 2,x3^ = ). = ln<^ — ) +gi{x,x2,x3). Notice that it does not matter whether we choose +(x\L\)* +(x\L\)i or — (i (a:22L!,22)5 )2 in (8.223). From (8.224) follows L° = X\X xix33

L 2 (lnL 2 - 1)

+gi±i + +P1X1 +

— L33{\nL (lnL3 3 - 1) +

X\X xxx2 2

(8.225) (8.225)

hi(-x,x ). h1(-x.,x2,x 2,x33).

From From (8.224) (8.224) and and (8.221) (8.221) follows follows 2 2 L° d 1 = | ggi d L° 1 dgx= 11 9a ± <9±2 112:2 — :i i a : 2 + ^~ 9 i 2= T L3 ■ ' . i fl . =— ax\ ax2J. Math. i 1 i 2Phys. — xix L3 ♦' M.F. Ranada, 32 2(1991)ox 2764. 2 *' M.F. Ranada, J. Math. Phys. 32 (1991) 2764.

(8.226)

8.226 '

105

Examples.

From (8.226) we infer that 01 = = 0i(x,x3). a2r0

Examining Examining aa ?? ^^

in in aa similar similar way way we we find find that that

00il ==00ii((xx))..

(8.227)

Using (8.225) and (8.221) let us inspect d2L° _ 0X2 dx\

X! 1 x2x\x Xxx —x\x — 2 2 2 xxx2

d2h _ dx\ 0X2

L2 L\L LiL33

(8.228)

Hence dh-t . dki 1 . . . , . xi -- xx33xi)-—- = — ln(xix 2 - xXix (x33xi Xi)xx2)2 ) - (x ax xX22 0x22 jf

OXj2 d±2 J x\X xix33x\ x\ + + xx22({xx22xix i + + xx33xi)x xi)x x\iix3 2 2 + +x\iix

(8.229)

+/z (x,x = +h22(x,x 3) 3 ) = 1

1 L T == — In 33 x2

1l

L s 1In — jJ.+.

x2

LL\x

g 21((x, (x, xx■33))\ ==

= — I n L i + 3+g , x 33)) {x,x 2 (2x x2 or Li(lnLi -- 1) l) + + g32(x,x + h/i2(x,x (8.230) LiflnLr 3 )x 2(x,x3)x 2 2 + 2(-K,x3)3 ).■ x2x3 In a similar way we get from (8.225), where gi and h\ are replaced by (8.227) and (8.230) resp., and from (8.221) the relation hftix = =

d2L _ J_ dx2 dx3 Lx

dg2 _ 1 dx3 Lx Lx

or or (8.231) (8.231)

02 = 02 ( x ) .

g2 = g2(x). We have We have — L 2 (lnL 2 - 1) - — L° = X3X\ X3X1

-\

X2X3 X 2X3

XX\X i X 22

L3(\nL3

- 1)+

Li(lnLi - l) + 3i(x)£i c/i(x)ii +g +322{x)± (x)x2 2 + /i 2 (x,x 3 ).

(8.232)

106

The Inverse Variational Problem in Classical

Mechanics.

f\2 rr 0 p\2

From ~=r- in a similar way as before we find a x

3

(8.233) (8.233)

h2 =33(x) + h(x). h2 = g3(x) + h(x). Since Since XiJbi — U ,

/

y XiJbi ~ u, i i

we get eventually from (8.232) and (8.233) 11

L —

11

L\ InLi

X3X1

X2X3

1

L2lnL2

X\X2 X1X2

L3\nL3+

(8.234) (8.234)

+ ^3i(x)xi + 2__/gi{x)xi +/i(x). +h(x). i

We need, however, the function L1, given by (8.185). Prom (8.234) we obtain i i

r r L L = =

i±1i , L22 I I nnL- 3-

Xi Xi

x22 . L3z I I nnL\ --

x2 x2

L3

x33 x3 x3

L\

Ii Li In-—h In-—h L2

(8.235) 8 235

L2

/• T Li\ \ (x\ LT2 x■2 LT3 x3 +, I Xi — In — L2 + x—2 In — L3 + — x3 In —Li\ + \[— — + x— In Li — + x— In L—J x i In L \xi L33 x22 Li x33 L22J Let us check relation (8.22). We have e.g.

d2L} dxidx2

i ij 1 xi xi L3 Xi ii 1 xi L L33 X!

x\x xfx22 1 X\X ±1x22 xi L\ L\ V\ vi 1 x j I<3|x=v L3\ic=v %\

d2L} dx2dxi dx^~

x\x2 L\

x2 1 x 2L 3

( "

)

x=v x=v

1 vi x xix L 3L|3x\ii=v =v

and

-_

i

V2 _*2. x2 — 1 +. v2 x2 L 3 x2 x2 L3 x2

If (8.22) is satisfied we should have If (8.22) is satisfied we should have

( ** | i2\ V

xi

1

12/^3

-(v*

\x2

V xi x2J L3 \x2 Relation (8.237) is, in deed, true. Relation (8.237) is, in deed, true.

i

x\x\ 1 1 v2 x x22 L\ L% x2 L33\\k=v

1

1 L3\k=v L3\k=v V

A) I

a;i/l3|

xi)

x = v

I/ 3 |x= v

"

""

(8.236) (8.23b)

(8.237) (8237)

Examples.

107

The Lagrange function is 2 L = g(±x g(^2-U(\x\)) -U(\x\))

+ L\L\

(8.238)

and the Euler-Lagrange Equations are

9

(.. /..

3E/(|x|)\ 3*7(1x1) \

^v ^

1

+

{** - -Or) £

r, r ,

.

L

L^LZ ^=°

(8.239) 8

^ - 239 )

■

Both, the Lagrange function as well as the Equations of Motion are not rotationally forminvariant. The canonical momentum is dL /( . Pi = 7^ rF-- == f 5f [Ui ii Pi = dxi V dxi \ 8.5.4

1 11 /7 7I 22 ., L 7,22 In - = 4 - IInn -— gxi \ I n L L43 3 g xi V L3 L3

Second example

of

\\ ) /) ,, /

i = v / i=v/

etc.. (8.240) etc.. 8.240

Randda. Ranada.

The other example of Ranada*' Ranada*) .

°~k

(8.241) (8-241)

G = j , or

IS" = kL^ d2L°

JL_

diidij

L\

*

^

(8.242)

3

Let us start with d 2 L° _ 1 3if Li

(8.243)

This entails ix2 7° +Bi(x,x2,x L° = 1ly-2^- + AAi(x, i ( x , i±2,2:3 2 , i 3 ) i)xi ; i 4- J3i(x,ia,i • 3 )3).

(8.244)

■ki

To compute Ai A\ and 7?i i?i we proceed in a standard way. We compute e.g. d2L° _ iixg d±i dx2 L\

dAi __ L2 d±2 L\

or Ai = Ax =

x3xx fdM f dLi ^ xsxt _x xx3xil 1 / -JY+M+is)A2{-x.,x3) == 1 1 — x3 J L\ x3 Lx

*) M.F. Ranada, J. Math. Phys. 32 (1991) 2764.

+

+A2(x,x3),

(8.245)

108

The Inverse Variational Problem in Classical

Mechanics.

where we use dL\ — —Xzdx2 ■ After performing these not very exciting computations we end up with L° = | ^ ^

+ y]Fl(x)±i + F(x).

(8.246)

Xo -L/l

To get L1 we choose vi = v$ = 0 , vi ^ 0. We have V —■ ^ - " 9ij i

-i—M J

2

a;? L i °

and

Hence L ^ i ^ f

(8.247)

The Lagrange function reads L = 5(Ix2-[/(|x|)) + | ^ M

(g.248)

and the Euler-Lagrange Equations are (

9 Xi

dtf(|x|)\ +

v-, 1

r

r

LiL

[ ~ ~0T) ^

Lf ^ = ° ■

(8-249)

Both, the Lagrange function and the Equations of Motion, are no longer forminvariant with respect to rotations. The canonical momentum is Pi=xi

1 L2 + ——, x3 Li

etc..

(8.250) '

In both cases III and IV, the procedure to express x in terms of x and p is theoretically possible but of no practical use.

109 109

Examples.

8.5.5

Example

of Cislo.

We wind up with an example given by J. Cislo*) concerning the case of an oscillator in (2 + 1) dimensional space-time (a membrane). For simplicity reasons we put in (8.110) 2Q = 1. 2a =

Then the Euler-Lagrange Equations read Xi + Xi=0,

Xi ~\~ %i — U ,

i£ = = 1,2. 1,2.

(8.251)

The time independent Constants of Motion are

Hi =

\{x\+xi),

H22 =

\{x \(x22+x +x22), (8.252)

L = = X\X X\X22 — — xXxXxX 22 ,

A= = x\i,2 + X\x xxx2 2■. These Constants of Motion are not independent from each other e.g. we have L2 + A2 = 4H AHXXH H22 .■

(8.253)

The symmetric (2 x 2) matrix a consists of entries which all are Constants of Motion, consequently actij ti ^

(8.254)

=aij(Hi,H aij{H uH2,A). 2,A).

The conditions (8.116) and (8.117) read dann . da 2+ dH2 M? da22 . da22 Xl . or or

m mXl

+ +

dan . dai2 daX2 . x2 . Xl = Xl + oA oHi oA ~dA m ~dAX2' a fl a da . da\ . da X2 . da22 daX22X2. + daX2 . 22X2. Xl

~dA ~dAX2 = = dlf dif22X2

[dan fdffu

dax2\ da \

. .

I'da22 (da22

daX2\ daX2 X2\

. fda22 . +, (da22

+

fdan fda

8 255 ((8.255) - )

~dA -dAXl

dax2\ da \

. .

daX2\ daX2x\

. . = ()

= x2 X2 + n x2Xl X2 + {-jA-m) {d^-^A) UHT^A {-JA-m Xl = 0^'

{-WM-X2 ) \^-^2) 2

{dH;-^A) ' + \jmr^A-)Xl = *-

*' *' Private Private communication communication from from Dr. Dr. J. J. Cislo Cislo

(8.256)

(8.256)

110

The Inverse Variational Problem in Classical

Mechanics.

The relations (8.255) and (8.256) have the structure Y±i + Z±2=0,

(8.257)

where Y and Z are Constants of Motion. Relation (8.257) entails that for nonvanishing Y and Z ±2

should also be a Constant of Motion, which is not true. We conclude that d a n _ da12

.

-dir2-^A'

(8 258)

-

dan _ da12

.

.

(8 259)

-jA-m' gq

.

'

~22 _ da12

,

.

(8 260)

■dA-dlh'

-

da22 _ da12 -dih'^A

(8 261)

-

From (8.258) and (8.261) we get, using the Converse of the Poincare Lem­ ma, d\ * n = g]^ .

ff22

(8-262)

<9A = ^ ,

where A = \{HUH2,A).

(8.263) From (8.260) we infer that

d\ al2 = ^

(8.264)

Equation (8.259) does not supply any new information. But relation (8.258) yields still d2X _ d2X dHi 8H2 ~ 8A2 '

(8 265

"

)

Let us introduce the euxiliary quantities B = HX+H2,

C = H1-H2.

(8.266)

IIllll

Examples.

Then (8.265) can be written as a wave equation (membrane) dd22\x dd22\\ 5d22A \ _ 2 2 22 dB ~~dJP~dC dB dA ~-°' dC2 ~~ '

(8.267) (8 267) -

where B plays the role of the time variable and A and C stand for two independent space coordinates. Every X(A,B,C) satisfying (8.267) gives rise to a matrix cr, given by (8.262) (8.264). From relations (8.262) (8.264) one can construct the Lagrange function, Euler-Lagrange Equations as well as the Hamilton function. The variety is enormous. Let us inspect a very special and simple case \A = = a{\A a{\A22 + +H HiXHHa). 2).

(8.268) (8.268)

According to (8.262) - (8.264) we have a n = aH2 -= \a(x\ \a{x\ + x\), (8.269)

o-22 = aH cr aHix = - \a{x{ \a{x\ + x\), x\), <Ti2 — (T12 — aA aA == aa(xi±2 ( i i i 2 ++ x\X x\X2)2).■ Since

w = an

(8(8.270) 270)

d2L

-

-dxj=ai1 we get we get

L = -a{x\ +xl)±l L = -a{x\ +x\)x{

+ F(xi,x ,x )±i +F(xi,x22,x22)xi

+G(xi,x ,x ). +G(xi,x2,x2).2 2

(8.271) (8.271)

From (8.271) and (8.269) we obtain From (8.271) and (8.269) we obtain d32L dF OF ..,. , „. = aa; ax22xi a;i + + -sv-^- =CTi2 at2 == a{xix22 + x^x xxx22) n. ox\ ax2 ox2 or or (8.272) (8.272)

F = axix x2 + Fi(xi,x2). F = axtx22X2 + Fi(xi,x%). The The equation equation U

XJ

i

.9

O \J

l / . O

r

)\

112

77ie The Inverse Variational Problem in Classical

Mechanics.

yields + F2{xux,x x2%)x)x2 2

G = -ax\x\

(8.273) (8.273)

+ J(xi,x J{x1,x2).2 ) • +

If we insert (8.272) and (8.273) into (8.271) we get 2 1 1 L + ^^ Fiii Fiii + + J. J. (8.274) (8.274) L = = -a(± -a{x\ 2 + + 12)2:1 x\)x\ + + axix^xxj) axix2xix2 + + jax\x\ ~ax\x\ +

i=l

The Lagrange function L should satisfy

82L

A

y, J}

£-r" 9±i A* dii <9x, dij J

3

82L

3

dL _ Q

±

J -^-r* £—•3 idii i <9x,dx«

J

J

dxidxi

(8.275)

which entails Fi

=

—

dxi

(gauge (gauge term) term)

(8.276)

and J7 = x ? x2Xnf2.. — -ja ~CLX\ 4 l z

(8.277)

J

Hence

±J — r f l X i Xo 1 7 8 * 1 X 5 " I - i flX^Xi T

+ax\X2±\X +axiX X\X22 — \ax\x\

— \ax\x\

.

(8.278)

The Equations of Motion are 2 2

^2
3

= a(xix 2 + XiX2)(xi + Xi) 4- 2 a(x, + Xi)(x 2 + x 2 ) = 0.

These equations are2 not forminvariant under = a(xix + xix \a(x\rotations. + x2)(x2 The + x2Hamilton ) = 0. func2 )(xi + Xi) + tion Thesereads equations are not forminvariant under rotations. The Hamilton function reads H = \ax\x\ + \ax\x\ + jax 2 Xi + H = \ax\x\ + \ax\x\ + jax 2 Xi + (8.280) .2.2 . (8.280) ■\-ax\X2X\X2 + 3- a x2 1 xa 2 +1-ax\x\ +axix 2 xix 2 + -ax1x2 + -ax^x^ ■

The model of Subsections 8.1 - 8.4 for n ^ 3.

113

The canonical momenta are Pi = \ax\x2

\ax\x\2 + ^aiix

+ ax\x ax\x 2x22x2,

p2 = \ax\xi \ax\x2

+ \ax\x \ax\ 2x2 + ax\X2X\

(8.281) The r.h.s. are not linear in x. 9

The model of Subsections 8.1 - 8.4 for nn ^^ 3 .

In Subsections 8.1 - 8.4 we dealt mostly with the Euclidean space n = 3 and time as an independent variable. Except for some derivations (following e.g. the formula (8.66) down to (8.78)) our considerations apply to n > 2. In particular in case U ^£ a x 2 + P 0 for n = 2 all Lagrange functions are equivalent. This can be shown as follows. Relations (8.38) and (8.58) imply that x as well as x are eigenvectors of er, belonging to the same eigenvalue g. Therefore in case n = 2 av = gl. gl.

(9.1)

This makes the ansatz (8.3) trivial. A nontrivial issue is the case n = 1. The fundamental identity for an autonomous Lagrange function, which we are going to call £/, is , d2V dU(\x\) d2V dV _ , « . « . 2« , . dxdx dx dx dx ox ox ax ox'' ox If we differentiate (9.2) with respect to x and use the notation d2 V _

((9.2) 9.2)

(9.3)

we get .da _ dU(\x\) da _ ^ - « = 0 . dx dx ox ax dx ox The method of characteristics leads to the equation dx dx dx dx

-- + + dxm m= = °° dx

(9.4)

The Inverse Variational Problem in Classical

114

Mechanics.

valid for x / 0 and ^ / 0, which integrated yields \x2 + U(\x\) = H - conts..

(9.5)

a = a{H).

(9.6)

Thus

The general solution of (9.3) reads L' = [ cr{\u2 + U(\x\)){± - u)du + # ( x ) i + $ ( x ) ,

(9.7)

where $(x) and ty(x) are functions to be found. We may write

•« = £ =

then the term

is a gauge term and can be omitted. To find $(a;) we insert (9.7) into (9.2). We get r <9cr d£/ , d® / —-—udu— ./c 3ff i = „ dx dx or, taking into account that

dU —a = 0 dx

<m dx we get dU,

.

^(---l

. i = c

5*

dt/

) - ^ - ^ = o.

(9.8)

Relation (9.8) entails a$ _ dU dx ~ ~ ff'*=e dx~

where

^lx=c = ^(|c2 + c/(N)).

(9.9)

Hence ,V(«)


CT(V)

■/V(aso)

dF + const.,

(9.10)

All autonomous

s-equivalent one-particle Lagrange functions...

115

where V = \
rV(x)

a{\u2+

V =

U(\x\))(x-u)du-

Jc

a(V)dV.

(9.11)

JV{xo)

The Lagrange functions (9.11) for various a's are not equivalent; they are only s-equivalent. We have *

=

%

■

*

-

*

=

ox = jXa{\u2

+ U{\x\))udu + J%{\t?

+ U(\s\))^^-ds

= (9.12)

rH(x,x)

= I

a{H)dH,

J H(x0,±o)

where rccb &o a r e some constants. The derivations indicate that the one-dimensional case differs essentially from the model with n > 2.

10.1

All autonomous s-equivalent one-particle Lagrange func­ tions for (1+1) space-time dimensions.

In this Section we undertake the important task to find the largest set of s-equivalent Lagrange functions to a given one (which we do not, as a mat­ ter of fact, specify) in case of one point particle in a (l+l)-dimensional space-time. This set we are going to decompose into subsets of Lagrange functions, inequivalent to each other. The problem we intend here to tack­ le is important as the results obtained in this way allow us to apply the inequivalent classical Lagrange functions in the quantization procedure e.g. in Feynman's method of ,,integral over all paths". We have here an essential extension of our considerations of Section 9, concerned with the case n = 1. In Section 9 the form of the initial Lagrange function, say L°, was specified viz. L° = \x2 -

U(x).

116

The Inverse Variational Problem in Classical

Mechanics.

In this Section we do not specify the shape of the initial autonomous *' Lagrange function L. We do not need to bother whether it exists; it exists always*) If we have an equation in the normal form (10.1)

x = f(x,X,t) f(x,x,t), , X

this equation will be equivalent to the Euler-Lagrange Equation for a Lagrange function L if the identity d82L d2L . &L_d2_dL_ IL_dL cPL a^/o -e. 2 dx dxdx dxdt dx dx ' dx2 is fulfilled. Let us defferentiate (10.2) with respect to i . We get

(10.2)

dG . dG. &G dG . dG dG_dG_ dG dj_ Bf „ „ „, (10.3) dx dx dt dt dx In (10.3) the substantial time derivative of G was taken along the trajectory x = g(t,g(t,a,0), a, /?),

a, a,/3 0 - some constant parameters,

(10-4) (10.4)

which solves the equation (10.1). In (10.3) (10.5) G = 0 —. 0= (10.5) dx2 Equation (10.3) coincides with Helmholtz condition (5.41). Conditions (5.3) and (5.4) are trivially satisfied and (5.51) just vanishes. In contradistinction to n > 1 equation (10.3) has always a solution. From (10.3) we get the solution r(t,a,/3) = exp j - j ^ r (g(t,a,0), ^(t,a,0),t) r(t,a,0)=exp{-j^(g(t,a,0),^(t,a,0),tj\dt.

J dt.

(10.6)

Let us invert (10.4) as well as ±=—g(t,a,P) & = -jj.9(t,a,0)

(10.7)

with respect to a, (3 and x, x. We get the Constants of Motion a —

$(x,x,t),

0/3 == * ( !V(x,x,t). ,£,*)•

(10.8)

*' In Section 10.2 the case of Lagrange functions depending explicitly on t will be considered. t> I am obliged to Dr. J. Cislo for calling this to my attention. See e.g. F. Bolza, Lectures on the Calculus of Variations, New York, 1931.

All autonomous

s-equivalent one-particle Lagrange functions... functions...

117

Hence = a($(x,x,t),y(x,x,t))r($(x,±,t)y{x,x,t),t), a($(x,x,t),,i&(x,x,t))r($(x>±,t)iJ!{x,x,t),t),

G(x,x,t) G{x,x,t)

(10.9)

where a is an arbitrary function and $ and \1> are Constants of Motion. Integrating G in (10.9) twice with respect to x we get the desired result, i.e. the Lagrange function. We conclude that for n = 1 equation (10.1) is always equivalent to an Euler-Lagrange Equation for a certain Lagrange function. Let us return to the autonomous Lagrange function L and its EulerLagrange Equation d32L.. L ox21 dx

d2L . ox dx ox dx

d2L dx ox2z

BL dL _ „ ox dx

.n

„„ v (10.10)

a.e.

If there exist other Lagrange functions, say L', which give rise to EulerLagrange Equations equivalent to (10.10) (the same set of solutions) then we should have the identity d2V.. dx2

d2L' . dx dx dxdx

dL' _ dV dx

(d2L.. \\dx dx2 2

d2L . dxdx

dL\ dxdx) J

(10.11)

where a = a(x,x') ^ 0,oo a{x,x') T^O.OO

a.e.

(10.12)

or d2V _ dx2

d2L dx2

dx \ dx

(10.13)

J

dx \dx

J

(10.14)

Our goal is to find a as well as V Let us differentiate (10.13) with respect to x and (10.14) with respect to x. We get dad2L d3L 2 2+a2 +<J 2 d~x~dx dx dx dxd± dx dx~ da d 77— dxdx dx dx

—

d33L' V 2 2 " dxdx dx'dx'

(dL. V (ot d33L x. = 8d33L' T\\ xx. l—x-L *=— -r— x — L )+a—5 ) + a—2 2 —5 22 ■ \dx dx 8xdx dx dxdx \dx )) dx dx

{(10.15) }

(10.16); V

The Inverse Variational Problem in Classical

118

Mechanics.

If we subtract (10.16) from (10.15) the latter multiplied on both sides by x we obtain*' dadU _da_dH_ _ 8adH_dadH =0j dx dx

dx dx

(1017) (10.17)

H = H(x,x)

(10.18) = —x-L. -pr-x — L. dx ox The quantity (10.18) is the Hamilton function linked to L and is conserved by virtue of (10.10). The mothod of characteristics applied to (10.17) yields dH ., dH,. ..„_ „n -^—dx + -—- ax = dH an = 0 ox dx dx ox or a = a(H). a(H).

(10.19)

Thus, on the trajectory, a is a Constant of Motion. Let us denote d2L *—p = G(x,x).

(10.20)

Then from (10.13) we get L'V = = //

G(x,u){±-u)du+^(x), G(x,u)(x-u)du+%(x),

(10.21) (10.21)

where the constant*' c is so chosen that the integral on the r.h.s of (10.21) does not diverge. In (10.21) we discarded the gauge term. L' given by (10.21) should still satisfy (10.14); this allows us to fix the arbitrary function *$>{x). We have

^(S^ L ) = £(f G ( a : ' u ) t i d u - $ )Because of (10.18) and since dH _ d2L . 2 dx dx *' Notice that (10.17) is one of the Helmholtz conditions. t) We could assume that c = c(x), but we discard this possibility.

(10.22)

All autonomous

s-equivalent one-particle Lagrange functions.. functions....

119

we get from (10.22) taking into account (10.20) dH

8 ( /■* ,TT,dH

\

*& = *(/. ^ftrL*-*;

, (10.23) (io 23)

-

To make thing easier let us introduce ad hoc the function To make thing easier let us introduce ad hoc the function (10.24)

•o-(z) « = . -±1 ^ - .

(10 . 24)

Then (10.23) reads Then (10.23) reads

^S ->£= < U=Lt -. *- 1 |(= E << *f l>, --SS( fH ) f)U or or (10.25) (10.25)

**(«) (i) = const.. = -- S S (( fHf )) || ii ==cc + + const.. If If we we insert insert (10.25) (10.25) into into (10.21) (10.21) and and use use (10.20) (10.20) we we get get

L

'=f f _>-«>*-=
.

x u

~

(10.26)

f G(*,u)«ki-E(Jff). G(x,u)du-T,(H). = i /" This is the most general form of an autonomous Lagrange function, sequivalent to L. It is interesting that the Lagrange functions V are not equivalent to each other and to L. To see that we have to check the Lagrange brackets of x and x. We have dL g , =dx->

(10.27) (10-27)

, S P

p'== / G(x,u)du + xG{x,x) - —Tx — - i = p' xG{x,x)-—-— = / G(x,u)du (10.28) 7C off Ci-* "i-* yc and consequently <92L

(10.29)

{X,±}L = ^ ,

{x,x}LL'< = G{x,x) G(x,x) = a{H){x,x} a(H){x,x}LL {x,x}L[ — {x,x} L> = —{x,x} LI L> cr cr22

,

(10.30) (10.31)

120

The Inverse Variational Problem in Classical

Mechanics.

Since a is not, in general, a constant our conjecture is correct. For the example discussed in Section 9 we have {x,x}LLoo {x,±}

= =

1,

(10.32)

{x,x}LLI,

= =

a{H). . a{H)

(10.33)

Notice that the only quantity which chracterizes the various Lagrange functions (10.26) is cr(H). Thus each equivalence set consists of one single Lagrange function, provided a is not a constant. The Hamilton function is rH{x,x) rH{x,x)

H' = - p'x - V = = H(ff) T,(H) + const. = = /

JJH(xo,±o) H(xo,xo)

a(z)dz ,

(10.34)

where xo, in are some constants. Here we used (10.28) and (10.26). Our L' of (10.26) coincides with the Lagrange function obtained by the method of Engels*) using (10.13), (10.14), (10.18) and (10.19). The inverse problem for the case of dimension (1 + 1) was treated to large extent by many scientists, to mention only Darboux, Jacobi, Hirsch, Hamel or Kiirschak.t) 10.2

All s-equivalent one-particle Lagrange functions for (1+1) space-time dimensions.

As in Subsection 10.1 we do not specify the shape of the initial Lagrange function L, we drop only the restriction of autonomy. Thus we take for granted the existence of*' , L = L(x, L(x,x,t). i , t).

(10.35)

If there exist other Lagrange functions, say L', which give rise to EulerLagrange Equations equivalent to those of L, then we should have d2L' d2L „. -dx^^a-dx^^G{x^t)

. .,

d2L , ^ 2# T0 dx

a.e.

(10.36)

*) E. Engels, II Nuovo Cimento 2 6 B (1975) 481. *' " G. Darboux, Lecons sur la Thiorie Generate de Surface. Paris, 1894, C.G.J Jacobi, Zur Theorie der Variationsrechung und der Differentialgleichungen, Works vol 4, A. Hirsch. Math. Ann. 49 (1897) 49, G. Hamel, Math. Ann. 57 (1903) 231, J. Kiirschak, Math. Ann. 60 (1905) 157. t ' Of course L does not need in general to depend explicitly on t.

All s-eguivalent s-equivalent

one-particle Lagrange functions for (1+1) space-time dimensions.

d2V . dHJ__8L[_ 9HJ__dL[_ ft d2L d2 L x + dxdxX+x+ dxdt dx ~aa\dxdxX+dxdt

dxdx didt

dx ~ \d±dx

dxdt

8L\ dL\ dx J '

dx) '

121

(10.37) (10 37)

'

where where a(x,x,t) a{x,x,t)

/£ 00

a.e.. a.e..

(10.38) (10.38)

We showed in Subsection 5.3 that a is a Constant of Motion*' From (10.36) we get assuming that we already evaluated cr L' — u)du ++ ip(x,t)x ip(x,t)x ++ ip{x,t), ip(x,t), V = = // G(x,u,t)(x G(x,u,t)(x-u)du

(10.39) (10.39)

where c is a constant and tp and ip are here still arbitrary functions. Let us insert (10.39) into (10.37). Then I* (dG dG\ dip dip r fdG dG\ , (hp_dip__ u + Jc \dt +Udx) dt ds~ 22 22 d LL dL\ dL\ = ^ f dd LL . | d \dxdx dxdt dx J

[(10.40)

'

Since dp dip dt

dip dx

do not depend on x , we may put in (10.40) x = c. Then (10.40) reduces to dip dt

dtp dx

=

L (

d2 L \dxdx

|

d2L dxdt

dL\] dx J

i i=c =c

(10.41)

The most general solution to (10.41) reads dR(x,t) V^^x—' dRjx,t) dR(x,t) dt

frxx\ I" /f d22L Jc, [ \dxdx

2 ddH LL_^A,] dL\} ,, dxdt dx J *==

,

(10.42) (10 42) '

where is aa constant constant and and R R is is an an arbitrary arbitrary function. function. If If we we insert (10.42) where c' c is insert (10.42) into (10.39) and discard the gauge term we get into f 10.39) and discard the gauge term we get rX rX

V =

Jc Jc

p-x px

G{x,u,t){x-u)du+ G(x,u,t){x-u)du

+

Jc1

Gf\*=
*> D.G. Currie & E.J. Saleton, J. Math. Phys. 7 (1966) 967.

(10.43) (10.43)

122

The Inverse Variational Problem in Classical

Mechanics.

where x = f{x,x,t)

(10.44)

is the Newton Equation, obtained from the Euler-Lagrange Equations of L. The Hamilton function H reads H=

I G{x,u,t)uduf G({,c,t)f(£,c,t)d£. (10.45) Jc J c' Notice that in case L depends explicitly on t the Hamilton function H is no longer a Constant of Motion. So a is no longer a function of H, although it is still conserved. The commutation relations are similar to those in Subsection 10.1, so the Lagrange functions V are not equivalent to each other and to L. 11.1

Construction of the most general autonomous oneparticle Lagrange function in (3+1) space-time dimen­ sions giving rise to rotationally covariant Euler-Lagrange Equations.*^

In Subsections 7.1, 7.2 we investigated a similar problem: it was a special case, when the Euler-Lagrange Equations coincided with the Newton Equa­ tions. The study of the general case turns out to be much more complicated. It is simple for one particle in (1+1) dimensional space-time. Notice that even more complicated is the problem to find all Lagrange functions, s-equivalent to one chosen by us and having the structures of Lagrange function studied in this Section. The latter problem was solved in Sections 8.1 8,5, alas, for a very special case of a Lagrange function of a simple structure L = | x 2 - £/(|x|).

(11.1)

Let us formulate first more precisely our problem. We assume the ex­ istence of an autonomous Lagrange function*) for one point particle in a *) The considerations of Section 11 were presented to me by Dr. J. Cisto. Unfortunatelly, the main result, the formulae (11.8) and (11.9), are already contained in, J.C. Houard, J. Math. Phys. 18 (1977) 502 although without clear derivation. t> We could start with a Lagrange function depending explicitly on time; this would not cause much trouble. For clarity reasons, however, we refrain from doing this.

Construction

one-particle Lagrange function... function... 123

of the most general autonomous

(3 + l)-dimensional space-time. Its Euler-Lagrange Equations should be rotationally forminvariant. These equations read _ y^ ^ d2L ., ~2^ ~ 2-* dii dx, dxj XjXj

n Tl

+ +

J

3

d2 L , 8L dL Xj d±i dx* dx3 Xj~8x~' ~dx~'

^ ^

((11.2) 1L2)

J

3

where L(x, x) is the Lagrange function we are searching for. The forminvariance of (11.2) under proper rotations S € 50(3) means that ri(Sx,Sx,Sx) ri(Sx,Sx, Sx) = = ^ 5 i f c r^2S x,x). f c (ikxrk, (x,x,x).

(11-3) (11.3)

k

As x appears on the r.h.s of (11.2) linearly, condition (11.3) is equivalent to the following two conditions

Gi3(Sx,Sx) = J2J2hi7nbjn( mn\X->' -) J

m m

x

j

(11.4) (11.4)

nn

Gi(Sx, Sx) = ^2 J2 SimGm{x, (x,

xx), ),

(11.5)

m

where Ci 13 3

_ d2L ~ d±i dii dxj

(11.6) d2L

dL

y>

dxi

z—1 diidxj 33

, 3

(11.7)

JJ

Our procedure in solving the stated problem is as follows. We are going first to find all functions which satisfy condition (11-4), and consequently choose out of this set of functions those which conform to the condition (11.5). Let us summarize the results we were able to derive. The most general autononomous Lagrange function giving rise to rotationally forminvariant Euler-Lagrange Equations reads up to to gauge terms L= K i r ( x 2 , x x , x 2 ) + ^ i i ^ -[ [W W((xx22))PP((xx|| xx ,, b ) ] ,

(11.8)

i

where K and W are arbitrary functions, b ^ 0 is an arbitrary constant vector and 1 |x| EYlAb x)iXi , f W , b xxx) n , ,. ,> {Xi P (11.9) P ((xxx|, xb ), b=) —rarctan ,\x\R a r c t a n [E ^,i xXYx\sbbMj ,xXJ X) ; X.,4 jJ .■ (11.9)

124

The Inverse Variational Problem in Classical

Mechanics.

The Lagrange function (11.8) yields the following Euler-Lagrange Equations ^ d2K .. *-? d±i d±j J

^ d2K . _ dK ^—> dii dxj 3 dxi

-(2x2w - w) y A [ £ - ^ U + v

+

' *->dxj [ | x x x | 2 ] 2^W'

+

4(TX3X)J

W"

J

(11.10)

^ | | ,

where da The Lagrange function, given by (11.8), depends, besides x and x, also on one additional constant vector b. Of course, the Euler-Lagrange Equa­ tions do not depend on this additional parameter as this would break the rotational symmetry of these equations. We shall show that the Lagrange functions are weakly invariant with respect to a change of b. Thus all the Lagrange functions for different b's are equivalent. It is easy to see from (11.8) that the Lagrange functions are not, in general, rotationally, invariant. We show, however, that they are weakly invariant under rotational transformations. In other words they are all equivalent. 11.2

Evaluation of the function Gij.

First we are going to investigate the structure of G^. As we required con­ dition (11.4) to be fulfilled, it can be shown that he most general conjecture for Gij symmetrical in the indices i and j (due to (11.6)) is G^ = K^Sij

+ K{2)XiXj + K^iiXj

+ Kw(xiXj

+ xiXj)+ (11.11)

+K{-h){xiLj

+ XjLi) + K^\±iLj

+ XjLi),

where Li = ^ T ]T] UjkXjXk = (x x x)j i k

(11-12)

Evaluation of the function G ; J .

125

and the functions K^\ ... K^ depend only on the variables

p = '£iXiii=xx,

(11.13)

i.e. JifW = lf<*)(a,)9,7)

A = l,...6.

(11.14)

We are going to prove the conjecture (11.11). Let us notice first that the vectors x, x and L = x x x are linearly independent,*' unless x is parallel to x. They span a 3-dimensional space. Hence Lj

(11.15)

form a basis for the space of (3 x 3) tensors. If we take into account that ^LiLi=L2,

^XiL,=0,

i

y^ZjZ/j = 0,

i

i 2

^ii(xxL)i=0,

^]ii(xxl)i=L ,

i

i

^L,(Lxx)i=0,

^ii(Lxx),=0,

i

i

^i;(xxL)i=0,

(11.16)

i

^ii(Lxx)i=L2 i

we conclude that for the matrix Gji expanded in terms of (11.15) the co­ efficients in front of these expansions are invariant with respect to proper rotations, under the proviso that (11.4) is satisfied. For example the ex­ pansion - L ^ ^ G ^ x x L L f L x x ) , *

i

(11.17)

j

renders the coefficient in front of Xiij which is, in deed, invariant with respect to proper rotations.^ In other words, it is a function of x 2 , xx and *' Linear independence means that the relations Ax+jix + ^L = 0 implies A = fj. = v = 0, for x }f x. f)

1

—L - t f <4>[x(x x L)[x(L x x)] = KW , (L2)2

1126 26

The Inverse Variational Problem in Classical

Mechanics.

x 2 Moreover, it is uniquely determined by Gy. The same applies to other coefficients of the expansion. Notice, however, that in (11.11) we used a basis different from that in (11.15). But one can check that*' LiLj L/i±jj

2

=L + {xx)(xi±j — J J Sij 0{j -p (XX)(2^Xj + -f" XjXi) XjXiJ

2 - xX2±i±j X{Xj - iX XiXj X{Xj ..

(11.18)

So we may replace the entry LtLj in (11.15) by <%. This new basis was used in (11.11). This accomplishes the proof of our conjecture. The necessary and sufficient condition for the local existence of L, satisfying the conditions (11.6), is the symmetry of r)C

rG - dG* (11.19) (11.19) l]k ijk S f ^ - d±k with respect to permutations of the indices i, j , k. Hence we are looking for such K(x\ A = 1, 2 , . . . 6, for which this symmetry occurs. Let us introduce the auxiliary quantities L±(x,x) = l [ L ( x , x ) ± L ( - x , - x ) ] , (11.20) G*(x,x) = l[G | [ Gi3v(x,x) ( x , x ) ± Gy(-x, Gy(-X, --x)] x)]. Then, taking into account (11.6), 3 d& L± L± dik dii dij

_ dG% i ~ dik

G±± l k

(11.21)

is symmetrical with respect to interchange of the indices. According to (11.11) and (11.20) G± = = K^Sij K^dij G+-

+ K^XiXj

K^Xiij + K^x iXj

2

^ ^ ( tfx ( <x >(x(x — x L ) x[ xL)[x(L ( L x xx )x)] ] = 0

•) O

LiLj

}/ , . // , d/? , ,. /} i^iab^jcdXaX eiab^jcdXaXccXXbb±Xdd a b e d

=

z2zZz-,z2 a

=

e

d

zj2jzj2j a

=

b

b

2 2

e

5 i

S

*c "re

**J

Sac

6bj

Sbc

J'('

J a c 5 M

= S,d Old

xbbxxdd = *«* xaxc± 6bd

- ibcSad)XaX ad)xaXccXXbbXXbb

d

2

* i i i ) , (11.22) + ZjXi),

etc..

— =

S V °ij

K^{Xl±j + K^Hnij

2

ify 6ij(x (3C x 2 * 2 - ((xx) x x ) 2 ) + . .. .. = SijL SuL2

+ ...

... =

Evaluation

G% = K^ixiLj

of the function

Gij. Gij.

+ XjLi) XjU) + K^iiiLj

127 127

xjU). + XjL,).

(11.23)

Let us first concentrate upon Gf,k. We have the following Statement Statement 1 1 We We assert assert that that 22 „ ,+ , _ <9 AT(x22,xx,x ,xx,x2 ) a AT(x IJ dii dij

(11.24)

or, in accordance with (11.6) and (11.20), + x L+ = X ( x 2 , xxx, x , xx22) + Y X(x Y, ^F?(x)xi ( ) ^ + ^^++((*x)) ,,

(11.25)

i

where F + and J+ are arbitrary functions. Proof of the Statement 1: According to (11.14) dK^ dii

dK^ l dpXi+2

-~di-=-w

nBKW

(11.26) (1L26)

^rXidj-

Using (11.26) and (11.24) we get G

U =^

= ^TS^Xk + KWS^

+2 ——5ijik

{z) +K 5ikXj K^Sikij

+

+ K^Sjkxi+

K{3)Sjk±i+

PEW dKW_. . . H—^— l i i j i t + 2 — - — 2 ^ 2 ; * + "T^ T'&

no

XiXjXfc XiXjXfc -f-f-

Q _ XjXj'Xfc XiXjXf- "r -\-

oj

dp

r\st Z a/i X{XjX %i%j%k "V & k ~r

r\

(11.27) U 1 "*') _T

XiXjXk~T XiXjX^-j-

dp

X{XjXk -\- L

XiXjX XiXjXkk ■.

Let us make use of the symmetry of Gfj Gj,kk with respect to the interchange of the indices i, j , k and compare the coefficients in front of the tensors OijX OijXk,k ,

OijXk , OijXk,

XiXjXfc XiXjXk, ,

X%XjX XiXjXk k

e tc.. etc..

(11.28) (11.28)

These tensors are linearly independent (see the proof of conjecture (11.11)). As the consequence the following relations should be true dp '

(11.29)

128

The Inverse Variational Problem in Classical

Mechanics.

tf(3)=2^l,

(11.30)

2 ^ — = ^— 07 op

(11.31

Formulae (11.29)

(11.31) entail

-w=2jhr

(n 32)

-

It follows from (11.31) that there exists a function if (°) for which (locally) 2ffCfl = ^

,

JfW = * ? .

(11.33) (11.34)

07

From (11.29) and (11.34) we infer that there is a function if for which (locally) K{0) = 2 ^ ,

(11.35)

K W = 2~

(11.36)

07

Moreover, by virtue of relations (11.30), (11.33), (11.34), (11.35) and (11.36), we get eventually

^

=0 -

if(3)=4—,

^

( 4 )

=

2

|^-

(11-37) (11.38)

(1^)

Relations (11.21), (11.22) and (11.36)-(11.39) imply (11.24) and (11.25). This accomplishes the proof. Let us now investigate L~ and G~. We have the

129 129

Evaluation of the function Gij. Gij.

Statement 2

If G G~- defined denned by (11.23) satisfies the condition

dGj dG±L dGjjL =_ dG~ kj dxk d±i dii

(11.40)

G~ then G {{ has the form

^^kr'^M-^M

(11.41)

N.B.: Of course, if G~ has the form (11.41), it must satisfy (11.40). Proof of the Statement 2: To this aim we introduce two auxiliary quantities, namely A0

=

B0 Bo

2 i5) 2 ( 6 ){6) i f ( 5 ) xK + JxC+K x xxx, ,

(11.42)

2 2

= K^xx

(11.43)

+ K^x K^x

Then, due to (11.23),

J2 G-jXj = A0Li, j1

L Y, G&i = =B°BL°> YjGiJ^ i-■

(11.44) ( 1L44 )

j3

We are going to show that the functions B = BB0L0L2 2

A = A0L2 ,,

(11.45)

do not depend on x. The function AQ and Bo depend only a, (1 and 7. Taking this into account we have ^^

T i

'

dA 8Ao0 flB

^^

T

i

'

E4t = °i

i

fdA fdAo0

^odA dA00. .

\\ (11.46) (1146)

130

The Inverse Variational Problem in Classical

Mechanics.

Using (11.40), (11.44), (11.45) and (11.46) we get

- ^(EE<^.) = fdG-j i *->^

j

^

dLi\

Xk

OXk

\°

\dii

J

J

lJ

*-f \ oxi

axi

(11.47)

:

dxk) axk J

In a similar way we get £?-=0. dxk provided we exploit the relation

(11.48)

arj

«

(n 49

- >

J

Prom (11.47) and (11.48) as well as from (11.42), (11.43) and (11.45) we conclude that A = A(x.2), B = B(x 2 ). Taking once more into account (11.42), (11.43) and (11.45) we may express K^ and K^ in terms of A and B, viz. K^

=

-^[A(ah-B(a)p],

K^

=

-^[B(a)a-A(a)p}.

(11.50)

Thus the necessary form of K^ and K^ is given by (11.50). We are going now to show that Gy defined by (11.23) and (11.50) satisfies condition

Evaluation of the function Gij. Gij.

131 131

(11.40). We have G

ij H

= =

771^2 TT/T^ {Ma)[x {Mrfi*22(xi (xiLLjj + XjLi) - xx(xiLj + XjLi))+ L v > +B(a)[-xk(xiLj + XjLi) + x22{xiLj + ijLi)}} . +B(a)[-xk(xiLj + XjLi) + x {xiLj + ijLi)}} .

Notice that Notice that dd AA (f W\

dxi dxi U UV V ""

Hk tjikiiX tjikiiX11 Hk

(11.51) (11.51)

2Lj J^k J^k Hi Hi Lktkiiii Lktkiixi -- 2Lj L L2 2

" "

( () )

~'

Therefore

[A (h\ + JL(h\ A fh\ _i2i \JL(h\ 2 [dii V W o»i \ L2V . " [dii V W dxj U A = (T2\2 /L2^2 X 2_/) ^{LjLktku z2iLjLkekii =

+ LiLktkii)xi = =

(11.52)

= , _ 2 , 2 [Lj(xi± [L,(x*x2 2 - iixx) + Li(xjX. LiixjX.2 - XJXX.)] XjXk)] . A similar computation yields

i \JL(*±\ + -!L(h.X 2 2

2

2 1[d±i x\L )

[dii I W'

d±j \L2 )\ dij3 \\L /-i )_

"

(11.53)

2

2

= , 2 . 2 [Lj(x 4 x - ijxx) Xjxx) + Li{xjX. - XjX±)] Xjxx)] Notice that the replacement of x by x and x by x on the r.h.s. of (11.53) produces the r.h.s. of (11.52) with a minus sign. If we insert the l.h.s of (11.52) and of (11.53) into (11.51) we get

r-ij - -A\JL(i±\2 + -!L(k.\ _ " 2 [dxi \L J dxj \L2J u « " 2 [ai^W fej UvJ _B _d_ (LA 2 2 [d±i [dii \L )

_d_(LA dij \L2)

'

Since V^v^

we conclude that

JL(hb\ dij \L2J

d

(Lk\

v->. d

(Xiij - Xj±i\

. f(11.54)

132

The Inverse Variational Problem in Classical

Mechanics.

is symmetric under the interchange of j with k viz.,

A (M = A A (M (M 2 d±j U d±j \L2vJ

(1L55) (11.55)

d±kk \L*J U / d±

As the interchange of x with x in

m

(11.56)

causes only a change of sign of (11.56) we have also

A(M 2

(u(11.57) 57)

A(M

£(&)-£ @) dxj U /

dxk

-

W)

The formula (11.55) and (11.57) enables us to express G~- in (11.54) in a more compact form, namely

G" - G7l = -A— (Aj - B— (A j

(11.58)

This coincides with formula (11.41) of our assertion. Moreover, from relation (11.55) follows

A [A A [A (Ml (Ml == A A [A [A (Ml (M 2 dx [dxj v \ Lw / lJ a**k [a^-

dii L 22 /J / a±i [dxj ^ \U

«« m (11.59)

Due to the symmetry displayed in (11.55) and (11.57) as well as taking into account (11.59) we get by using (11.58) the relation (11.40). This accomplishes the proof. Our main task is still to find L~. We have the Statement 3

We assert that

IT = -\A{a)-

+ B(o)]P(x|x t b) + ^J ]pF-r(( x ))±i ii + - ( x ) , (11.60) + JJ~(x), i

where „r.r, , i. •, i, % 11 |x|(b l x l(b xx x)x x)x P , ' \ n; P (x x | x,b) X b ) = — a r Carctan t a n ' ' R \x\ ( x(xx bx )b)(x ( x xxx x) ) and F F~ and JJ~ are arbitrary functions.

(11.61) (1L61)

Evaluation of the function Gij. Gij.

133

Proof of the Statement 3: To show (11.60) and (11.61) we have to look for a function P for which dP _ Lj dij ~ L 2

(11.62) 1L62

W, = v

(

If P satisfies (11.62) then

_ ^ _

I

Txx

d_ (LA

- ^ [i^-H - £ (EO

(IL63 (11.63)

'

To see that (11.63) is really true one should carry out differentiations, indicated on both sides of (11.63), making use of (11.62) and L 2 = x 2 x 2 - (xx) 2 . The following expression should be then an identity XjLi + XiLj ^Lj + —j—g I ^Ci/yifcL ^2eikjXkL2 2 - 2LiX. 2LiX22±j Xj + 2Li(xx)xj x 2 (^ \ 2 2 = = -- 2LjXiX = - = -- jj ^eyfcifcL ^CijkXkli1 2LjXiX2 + + 2Lj{xx)±i 2Lj(xx)xi J1 ..

J = (11.64) (U 64) -

To check whether (11.64) ir really an identity it is enough to multiply both sides first by (5y and sum over i and j , then multiply both sides of (11.64), say, by Xi or ±i and sum over i ox j . The local existence of P follows from equations (11.55) by virtue of the Converse of the Poincare Lemma. It is obvious, due to (11.62), that we should have (11.65) j

J

3

J

and

Y.L*wrl-

(11.66) (1L66)

3

To proceed further we have to enter into some geometrical considerations. By virtue of (11.65) we have P(x,px P(x,x) P(x, px + crx) ax) = P(x, x)

(11.67)

>

134

The Inverse Variational Problem in Classical

Mechanics.

for L ^ 0 and for arbitrary numbers p and a, a 7^ 0. We infer from (11.67), that P is constant on the plane, spanned on the straight lines x and x. Thus P = P&),

(11.68)

where tp is the angle, measured between a fixed plane passing through the x-line and a plane, encompassing the x-line as well as the x-line. If we denote the rotation around the x-axis through the angle tp by Sv we have d ,„ .. x xx L ^(5,x),=0 = - ^ j - = R

/,,-„» (11.69)

Then due to (11.66), /dP\

x~,dPL]

1

_ (1I 70)

U L = ?^M=H

.

PM = ^

(11.71)

-

or

We are going now to express (p in terms of three independent vectors x, x and b. The x-line and the b-line mark the plane from which we are going to measure the angles. Vectors (x x b) and (x x x) are perpendicular to x; the angle between these two vectors is just ip. We have |x|(b x x)x S m ^|xxb||xxx|' (1L72) C

(x x b)(x x x) °S^=|xxb||xxx|-

(1L73)

|x|(b x x)x ^ (xxb)(xxx)'

(1L74)

t a

=

Prom (11.71) and (11-74) we conclude that n/ ^ n/ 1 • 1 \ 1 lxl(t> x x)x P(
Evaluation of the function Gij. Gij.

135

We shall need the formulae (a x b)c = (b x c)a = (c x a)b

(11.77)

(a x b) x c = (ac)b - (bc)a. (bc)a.

(11.78)

and

As a matter of fact, we used already (11.77) in getting (11.76). To compute the denominator on the r.h.s. of (11.76) we have to take into account that on one hand [(x x b) x L] 2 = (bL) 2 x 2 ,

(11.79)

on the other hand [(x x b) x L] 2 = (x x b) 2 L 2 - [(x x b)L] 22.

(11.80)

Hence, comparing (11.79) with (11.80), we conclude that the denominator [(x x b)L] 2 + x 2 (bL) 2 = (x x b ) 22L 22.

(11.81)

To compute the numerator notice first that according to (11.77) (x x x)(x x b) = [(x x b) x x]x and we have [(x x b) x x]x(x x b)i - [(x x b)x][(x x b) x x]j = [(x x b) x x]x(x x b)i - [(x x b)x][(x x b) x x]j = = [{[(x x b) x x] x (x x b)} x x] { = = [{[(x x b) x x] x (x x b)} x x]; = = (x x b) 22 (x x x)j = (x x = (x x b) (x x x)j = (x x

(11.82) (11.82)

b)22Li. b) Li.

Here we used twice the formula (11.78).*' Thus (11.82) yields the numerator of (11.76). Taking into account (11.81) and (11.82) we get

dP(x\x,b) _ U dii ~ L2

(11.83)

*' To get in (11.82) the second expression from the first one we have to identify in (11.78) x xxbb) ) xxx x, , aa ~~ ( (x

b ~ xx xx bb, ,

cc~~xx;;

to in the the second second expression expression to get get the the third third expression expression we we have have to to identify identity in a ~~ xx xxbb, ,

b~x,

cc~~ xx x bb .

136

The Inverse Variational Problem in Classical

Mechanics.

This accomplishes the proof. Combining our previous results of Statements 1 and 3, i.e. formulae (11.25) and (11.60), we finally get for the Lagrange functions satisfying condition (11.4) L = K(x2, xx,x 2 ) -

fe.4(x2)

LX

+ B(x 2 )l P(x|x, b)+ '

(11.84)

+ Y,Fi(x)xi + J(x), i

where F = F+ + F " ,

.7 = .7+ + J~ .

(11.85)

In the next Section we are going to choose from the variety of functions (11.84) those functions which, in addition, are compatible with condition (11.5). Before doing that, however,we shall write the Lagrange function L in a more shapely form. We write A as 2 ^ ( a ) - ^ =- M , a a

W

r(d,s^W. da

(11.86)

Then L = K + hW - ^pj xxP - BP + ]P Fi±i + J.

(11.87)

i

Let us prove an auxiliary identity, namely

E *,£o^-(nr-5)-r +

. r J § ^

(11.88,

The l.h.s of (11.88) reads

2W'xxP + wY,ii^i

(11-89)

Symmetry

of Gij and evaluation of the Lagrange function.

137

So we have to compute ^ p . dP _ x x ^-r' ldxi x2

p

[(b x x)x]{[(x x b)(x x x)]xx - x 2 [(x x b)(x x x)]} x 2 (x x b) 2 (x x x) 2 xx x2 _xx x2 xx ^ x2

[(b x x)x][b(x x x)]j[(xx)3:, - x 2 ^ ] x2(xxb)2(xxx)2

niQuI '

[

[(b x x)x][b x (x x x)]j[(x x x) x x], _ x 2 (x x b) 2 (x x x) 2 [(b x x)x](bx) x 2 (x x b) 2

Here we made use of formulae (11.77) and (11.78) as well as of the relation (a x b)(c x d) = (ac)(bd) - (ad)(be).

(11.91)

Relations (11.89) and (11.90) prove the identity (11.88). If we insert (11.88) into (11.87) we get

L = K + Y/ii-faT.(wp)

~ BP + J2iiNiW

+ J

'

( 1L92 )

where Ni^=Fi-W^)^^

(11.93)

Formula (11.92) does not yet present the final shape of the Lagrange func­ tion we are looking for. We postpone it to the next Subsection.

11.3

Symmetry of Gij and evaluation of the Lagrange func­ tion.

The condition (11.5) imposes some restrictions upon S, N and J in formula (11.92). We are going to prove the following. Theorem 1 We assert that the Lagrange function L, satisfying condi­ tions (11.4) and (H-5)> has the form (up to gauge terms). L = K(x2,xx,x2) + ^ i i ^ - [ P y ( x 2 ) P ( x | x , b ) ] . i

(11.94)

138

27ie The Inverse Variational Problem in Classical

Mechanics.

Proof of the Theorem: By virtue of (11.92) we have*) 2 y , d y, d2L L ,,

dL dL dxi

*—•d±i d±idxj dxj 33 *-? i ' 2 - dK if . _ dK V ^ dd2K J X *-? dii dxj = Saij dx--\dx-dx^ j

(11.95) (1L95)

ft

where where Qi = — Qi ^ - ( (J J + 5BPP) ) ++ i xx (V« x NN))..

(11.96)

A look at (11.95) and (11.96) makes it clear that only Q can spoil the forminvariance of G under rotations. Let us compute first Vx {BP). One *' In (11.92) the term

E x 1 £-(^).z is contained. The contribution coming from Z in (11.95) is dZ dxi

d2Z

^

>

^

d±i dxj

d a

v-v--

x, = - > J

> x,

*-" 4^ *-? *-£

J

dxj

AJ i tdxILM = _YEVI i ] dx * VLaw 9X

y v

■ •

d

xfc

d

,„ „„™ r™ (WP)

dxk d±iK

dx

J *v

/

by virtue of (11.62); notice that W does not depend on x. Since

dx

by virtue V of (11.62); « notice that W does not depend on x. Since the r.h.s. of (A) reduces to

-EE(*£<«')&

the r.h.s. of (A) reduces to

-EE (*£*£")&

'

=

{A) {A)

Symmetry

of Gi} and evaluation of the Lagrange function. function.

139 139

can show in a similar way as in (11.90) that*) d

x 9 p__ _pp3H i _ ((xx)(x x x ) ( x xxxx), )j dx{ dxi x2 x2(xxx)2

|

((xb)(x xb)(xx x bb)j ), x 2 (x x x) 2

(11.97)

Then -^-(BP) -£-(BP) axi

(-2x2B'-B)P^xz x b ) ( x x t ) ji ££ , ( x x ) ( x x x ) ii jBg ((xb)(xxfe) 2 2 22 22 x (x x x) x 2 (x x x) 2

=

(11.98)

Notice that x Q should not change under rotational transformations as required by (11.5). The change caused by rotation through an angle tp around x is - ]^^[[22xx22BB' '((xx22))--B B (( xx 22 ) ] \x\ \x\ Hence we infer that

(11.99)

(11.100)

2aB'(a)-B{a)=0 2aB'(a) - B{a) = 0 *' To get (11.97) we proceed as follows. We know the formula for ^

%

dx%

given by (11.90). We conclude from (11.75) that

and

P(x|x,b) = = - P- P( (xx| b| b, ,xx))

(A) (A)

P(Ax|x,b) = A A-"11PP((xx||xx,,bb))..

(B) (B)

From (A) and (11.90) we infer that

[{h

x)

„, i. , , „, ,- , * [(b x x)xl(xx) 5 >, Ap x|x,b)=-p xix,b)t+ :(V^ E ( ( -P(x|x,b) = -P(x|x,b)L J - ^ i d

*-r"

x b

x2 +

b ! dxi

2

x (x x

b ) r2

^ (C)

and from and from (B) (B) we we get get >Xi

4-

= - P

dxi

(D)

Assuming that x, x and b are not complanary vectors we make the ansatz (1) ) 3 3) — P ( x | x , b ) = uv^Xi ii + w i/ 2( 2>(x (x x X x)t x)i + W ?/ >(x ( x xx b)j b ) ; .. dxi dxi

(E) (E)

2) (2) To compute o' tA2) , uv^ and v3~> we we use use (D),(D), (11.90) (11.90) andand (C).(C). Eventually Eventually we we get get (11.97). (11.97).

140

The Inverse Variational Problem in Classical

Mechanics.

or B{x22) = C\x\, B(x C|x|,

C --a caoconstant. nstant.

(11.101)

We see that Q is forminvariant with respect to rotations if and only if axi

(11.102)

|x|(x x (x x by

and V t, x N = A(x22 )x.

(11.103)

If we take the rotation on both sides of (11.102) we get ( (xb)(x x b ) \ „ x c ( v , x ^ ^ ) 1= C ^ = J 0 . C \V |x|(x x b) = j C |x| = 0

( ^1XKW)

WF

-

Here we exploited (7.21) and (7.49). (11.104) implies Here we exploited (7.21) and (7.49). (11.104) implies C=0 C =0 and consequently (see (11.101)) and consequently (see (11.101)) B(x 2 ) = 0. B(x 2 ) = 0. Prom (11.102) follows then Prom (11.102) follows then 2 — == / / ( x2)X )^ dx~ ^ >

,

(11.104) (1L104)

(11.105) (11.105) (11.106)

or J(x) = J ( x 22 ) .

(11.107) 2

Thus we may include J(x ) into the term K in (11.92) . Taking all this into account we get from (11.92) L = K + YJ±i-fa(WP) i

+

YjiiN^-

(11.108) (11.108)

i

Now let us consider (11.103). The divergence of (11.103), taken on both sides, yields 0 = 2a\'(a) 2aA'(a)+2A(a) + 2\(a)

(11.109)

A

(11.110) (".HO)

or =j^3

Symmetry

properties of the Lagrange function. function.

141

Then the equation (11.103) reads V N= = c C ' ^ 33 v , xxN |x|

*

(11.111) (1L111)

'^F

From the first equality in (11.104) we guess the partial solution of (11.111), viz. , N N (( Pp)) = =c c

(, x( xbb))( (xxxx bb )) |x|(x x b) 2

(11.112)

The general solution consists of the r.h.s. of (11.112) and a gradient of any function of x, viz. c f rxb)(xxb) * ; ) ( x * g 2+ V * ( x ) . (11.113) |x|(x |x|(x xx b) b) 2 Let us recall (11.90) and compute with help of it and (11.112) N N ==

^ i ^ ( | x | p ) = x x | x r l p + | x | ^ i i l£ = i i

i i

b x b _ [( [(b x x)x](bx) _ ^Y ^ .■ (xb)(x ( x b ) ( x xx b)j )t _ 22 X1 l l | x | ( x x bb)) ^ | i | ( x x xx)) 22 |x|(x t-r \x\(x

(11.114) (11.114)

i

The relation (11.114) inserted in (11.108) yields i

* + + EY^/i^ [ (il(W ^ ++C''|x|)P] £^xi. L =K C'\x\)P} + Y,^^ii+ i

(11.115)(H-115

i

We can include C"|x| into W. Thus we get (11.94), where the gauge term has been omitted. 11.4

Symmetry S y m m e t r y properties of the Lagrange function.

In this Subsection we are going to prove the following Statement 4 The Lagrange function L is weakly invariant under transfor­ mations of rotation; this means that for any proper rotation S G S(Os) L(Sx,Sx)-L(x,x)=d/(x^;S) for some function / .

(11.116)

142

The Inverse Variational Problem in Classical Mechanics. P r o o f of t h e S t a t e m e n t 4: Let us consider Let us consider L(x, x) = L(Sx, 5 x ) ,

(11.117)

L(x,x) = Z ( 5 x , S x ) ,

(11.117)

where S € 5 0 ( 3 ) . With the notation

where 5 € 50(3). With the notation x' = 5 x ,

x' = 5 x ,

x' = 5 x ,

(11.118)

x' = 5 x ,

(11.118)

we have, according to (11.4) and (11.5),

we have, according to (11.4) and (11.5), g 2 L(x',x') _ S ^ x ^

- 2 ^ 2 ^ 2 ^ 2 ^ SkiSljSkmSln it

!

m

'

ramra

n

_ d 2 L(x,x)

-

(11.119)

"

and in in aa similar similar way and d2l

y.

,

^^— 9 i j <9±, <9i,

di_ _ ^ dL__Y* 3J

JJ

3 3

dxi eta;

da 2L£

.

9L dL

^—' ^ - ' dii S i ; dij d±, 3'

dxi dxi

JJ

33

(11.120) '

Thus we get from (11.119)

L-L

= Y/Dk(x;S)x Y/kD+^( k(x;S)x x-S),k+^(x-S),

(11.121)

kk

where D and $ are some functions. If we insert (11.121) into (11.120) we obtain

yfdDjjx-S) ^— V 9XJ

G>£,(x;5)\ dxi J

d*(x;S) dxi

j

_Q

(11.122)

As x appears linearly in (11.22) we have

or

dDi _ dDj_ dDj dlh 1 -7Tdxj = -Kdxi, OXj OXi

Di

_d$(x,5) ^dx~' OXi

=

$$ == const. const.

(11.123) (11.123)

v

$(5) == Cconst. HS) nSt °

-

'

((11.124) 1L124

)

Symmetry

properties of the Lagrange function.

143

Consequently we have, taking into account (11.121) L-L=^[¥(x;S)+t$(S)]

(11.125)

This accomplishes the proof It can be shown generally, applying the procedure above, for any quan­ tity which appears in the Lagrange function but does not appear in the Euler-Lagrange Equations that the Lagrange function is then weakly in­ variant under the change of this parameter. Let us, nevertheless, give another proof for the case of the vector b which method displays explicitly the form of the gauge term. To this end notice that P(x|x, v) + P(x|v, w) = P(x|x, w ) .

(11.126)

This follows as a result of adding together two angles. Let us denote by L the Lagrange function L = J Fs:(x 2 ,xx,x 2 ) + 5 3 i i ^ [ w ( x 2 ) P ( x | x , b ) ] .

(11.127)

i

From (11.94) and (11.127) we infer that L-L

= £ > A { W r ( x 2 ) [ p ( x | x , b ) ) - P(x|x,b))]} = i

= 5>^{W(x2)P(x|b,b)}=

(11.128)

= |[W(x2)P(x|b,b)]. We did not so far investigate other symmetries of the Equations of Motion given by (11.10), barring the rotational covariance. Now we shall examine the case when the Equations of Motion are invariant under time reflection x-»x,

x->--x,

x-»x.

(11.129)

The best way to discuss this stringent symmetry restriction is to inspect the behavior of the Lagrange function given by (11.94) under the time reflection. We get L = tf(x2, - x x , x 2 ) - 5 > ^ [ W ( x 2 ) P ( x | x , b ) ] . i

(11.130)

144

The Inverse Variational Problem in Classical

Mechanics.

If the second term in (11.130) does not vanish and we want to keep the absolute value of the Lagrange function unchanged we have to demand if(x2,-xx,x2) = -A-(x2,xx,x2);

(11.131)

the function K has to be an odd function with respect to xx. In this case the Equations of Motion remain unchanged. The consequence of (11.131) is that a K which only depends on x 2 and x 2 (e.g. K = | x 2 — V(|x|)) has to vanish. In case the second term in (11.94) equals zero (W = 0) the requirement of symmetry with respect to the time inversion reduces to K(x2,-xx,x2)

= ±K(x2,xx,x2).

(11.132)

Also in this case the Equations of Moton remain unchanged. We wind up our considerations by giving two special examples, which seem to be of some interest. If W = C\x\, where C is a constant then, according to (11.94) and (11.114)

r = *<*-.*) + 0Wj£$3.

(u.133)

Formula (11.133) coincides with (7.50) when K = | x 2 - U(\x\).

(11.134)

Thus (7.50) is a special case of (11.133). For W = x 2 we have the Lagrange function which depends on two constant vectors b and c / b, viz. L = K{x2,xx,x2)

12

+ (xx)P(x|x,b) + (xb)P(b|x,c),.

(11.135)

The largest set of Lagrange functions of one-particle sys­ tem in a (3+1) dimensional space-time, s-equivalent to a given Lagrange function yielding rotationally forminvariant Equations of Motion (formulation of the prob­ lem).

In this Section we are concerned with the following very important problem: to find the largest set of Lagrange functions of one particle in a (3 + 1)

145

The largest set of Lagrange functions. . .

dimensional space-time, s-equivalent to a given Lagrange function of the form presented by (11.94). A similar problem for one particle in a (1 + 1) dimensional space-time was solved in Section 10. The actual issue for (3 + 1) dimensions is much more complicated. The solution of it would be of paramount importance in Physics, as it could be used as a starting point for quantization procedure in, say, Feynman's approach to Quantum Mechanics. Unfortunately, we are not able to offer the reader a solution of this problem. It is not yet solved; it is, really, a difficult task; the reader will realize, why it is so, when he continues reading this Section. So we present here only a convenient formulation of the posed question. In Subsection 11.1 the most general autonomous Lagrange function, yielding rotationally symmetric Euler-Lagrange Equations was given (see formulae (11.8) and (11.9)), namely 2 L L= = K(x\xx,x K(x2,xx,x2))

+ J2^[W(x2)P(x\x,b)}, + Y/^[W(x2)P(x\x,b)},

(12.1)

i

where , , „, 1 (|x|[(b x x l x l l P x x, b = — arctan ' LV ' J .

|x|

(12.2)

[ (x x b)L J

The Equations of Motion (see (11.10) are ^ 82K .. 2— dxi dij j ditidxj j

y ^ d2K . *-? diidxj dii dxj j

J

dK dxi

3

-<**.-*>££(§)*,+

(12.3)

+ ( 2 x W "{2xW'+4{xx) + 4 ( x x ) 2 W2W")^, ")^, where „ NN W'(a)=

dW(a) dW(a)

L L= = xx xx xx ..

/

da

;

,

2

a = x2 , a = x2,

a

-

8= xk, P = xx,

-2 7 = x2 2 ,

7

= x ,

(12.4) (12.4)

146

The Inverse Variational Problem in Classical

Mechanics.

The equations (12.3) can be also written as an identity, in extenso, ^

d2K

„ . d2K 8K „. d2K n. 2 2x 2x + 28ii + 2:ClXi ^ + ***'1w8j + ^ + ^d^

/

?22 ([ft^ 3^ r + ^Wd~i 3

aT

+4x UiXjix^(x f^) J^)(x Jf^+x Jf^+ +L3Jf^) + JfV+i JfV .

+2

2

d d2K^

22 tI.I2 l289 KK + Xi|x|

W

++

+ A..

+4

.

2 d2K d K

™57^ s^ ™ ^ -2a,f-(2x2^-^)^.

2 2 , „. „. .„ .„ a a/s: ^ + 2a; x -

'

w

(12.5)

1 ] ( X^ e «« u * L 2 _ Li[2xjX2 - 2(xx)ii] 2(xx)i i ] 1 • .(si/W+i;./W+z./W)+ + (2x 2 ^' + 4 ( x x ) 2 W " ) ^ = 0 ,

where z< = /i(x,x) = Z i / W + i j W + LJ< S ), / « = /«(a>(9,7)

(12.6)

are the Newton Equations and we assumed that x | f i i.e. x, x and L are linearly independent and form a coordinate system in the 3-dimensional Euclidean space. The orthogonal set to x, x and L is

X x

— =

2X — p X £jx L I 7 r x' ,

,. j_i xX±

XX X xx x 2 .. - — X + — X , =_ -—x+—x,

L L

~" L *2

2

, (12.7) (12.7)

The largest set of Lagrange functions...

147

Let use rewrite (12.5) in a different manner, namely Y d2K

no

d2K

dK\

,

m

^w+2^)fi2)+{2awW'-w)if{3)+ n„

"/

d2K

d2K

d2K

2

AOd

dK]

K\

.,.,

LV

(12.8)

- ( 2 ^ -

W

) A

/

( 3 )

+ 4

^

+ 2 7

| | . ]

+

+Lj(2aH/'-W)^/(1»+2|^/(3) + + (27W" + 4/? 2 W")^ = 0 , where L 2 = a 7 - /3 2 .

(12.9)

If L 2 tends to 0 and all expressions remain regular during this limiting procedure, then (2aW - W)jf{3)

=0,

(12.10)

=0,

(12.11)

L 2 =0

{2aW - W)/3f{3) L 2 =0

and (2aW - W)f{1)

2_

+21W' + 4p2W" = 0.

(12.12)

We do not make any specific assumptions which would restrict the structure of the equations (e.g. of a sort of 2aW — W = 0 which implies W = cot? , c ^ O a constant). Then from (12.10) and (12.11) follows /<3)

=0 L 2 =0

(12.13)

148

T/ie The Inverse Variational Problem in Classical

Mechanics.

and from (12.12) ,(D| (1)|

13

IL2=O

+ *pw 2^W" 7W' + =_2iw 2aW'-W

((12.14)

'

X

'

We conclude from (12.13) and (12.14) that >( Q ,/3, 7 ), /< 3 >=L 2 $( 3 >(a,/?,7), j ^ + 2 / ^" ^T 2aW'-W S ^ ;

(1)

(12.15)

^

+ ^~WV*>P,1), ,,,,

(12.16) (me)

(1)

+

where $'3^ and i'1' have weaker singularities (poles) than 1 for L 2 = 0. Let us examine the coefficient of L, in (12.8). Using (12.15) and (12.16) we get (2aW - W)^ W)* ( 1 ) + + 2^L2$<3> = 0 or £(i) SW

= =

3 2 (1 — $< — $ ( 3>) = = LL2 $$( 1 > ) d-y 2aW -W

_22— _ —

(12171 (12.17)

Consequently 7

V

2aW"-W

(12.18)

;

and the force / is for 2aW / W tj fj

Xj

~

2 yW W" ■yW - 2[3 2/3'W" 2aW'-W 2aW> - W

(12.19)

-^W5Sp^-^)

rt,S, +

The Equations of Motion (12.8) can now be written L 9K 2 2(L L ((aa*Z *£ 2 ++2WB ^^ __ ++ 22M\^ dK ( L 2 )22)2

V a/3

d 7 ) d72aW'-W

^dpdj

^

W

|+

+ (2aW> + (2aW - W)j] Wft $< $<33>> ++ (/}!** ( 0 ^ 4+ + 2277 d2R ^ —)\ f(2)_ fW-

~22

' + 2/m ? 2" ^ / dc?22tftf , f^7 W"

- { 22aW>-W2 J [ W +

d922KK

8K\ 8K\

oa J [aaW ++ 22P/ W ++ 22 *7 J ++

OQ 2

a2/v: ^

+

a2x

^

-

2

air ax

^ = 0'

Wd~l

*7 J

('12 20') (12.20) ' ^

V

The largest set of Lagrange functions.. junctions... .

"_ / d2K _ (2 a 92R V dpdj

+

149

2 2 2 dK (L igd K\ dK ( L 2 )) 2 AB — \ P d~i 3 7 22 ) fry d 7 2aW - W

$<3> +

-(2aW'-W)p -(2aW'-W)p

+(^+<£^0)/ m + 2p2W"\

c,(lW'

(n

' H 2aW-W d2K da da 93 7

d2K 2a

(12.21)

AOd

+0

2

K\

){ Wih W)"

d2K dp dp dj dj

7

The The equations equations (12.20) (12.20) and and (12.21) (12.21) have have the the structure structure (3) A$ A$ ( 3 )

{2) + + Bf Bf{2)

(3) {2) D D $$ ( 3 ) + + Ef £/(2) = = F,F,

= = C, C,

(12.22) (12.22)

where where A- A

2R

(J 92K 22 (n V dp a/3

flP Ii 228

^d ' K dpdj

II 22 d ^K \ 1 d^ K d1 ) )

L 22 )) 22 22 (( L

d12aW'-W

II

+ (2aW -W)^, +(2aW'-W)j, „ „

2 nd d2K K n

„„

da 22Kx

„__nloW' ^ + 2p 2 / 2W W" (/ d2K K 7W' + 2aW"-WT r 3 / ? 2 °~ 2aW'-W 2

{

+

d2 K K fliTX dK\ + + d$d1 dpd fry) dp d 1 1J

a P 2+P

2

dd'K K dK dK frK_ frK 11 2 + + 2 ~ 8(3 dada'' ~ dadp ~ ~ dp 22 22 2 2 K ,d K\ (L if,»d ff dd K K\ dK dK (L2))2 a + 2 \^ Qdpdj d1 2aW 2aW -- W <9/3<9 7 + 5dj7 2 J) <9 w 7 -(2aW -W)p, -{2aW -W)p, P Pdadp

d2K K-2 — 28 4 — *3d%K dj dpdj dj22 dj <9/?<9 <9 7 7 2 fiW' <922if /7W + + 2P 2/32W"\ Ty"\ /

F

= 2 ( 2aW>-W ) {2adm _ 4p / ? - ^ - 2 7rr ^ dad~< dad~< dpdj dpdj

+

2 K\ jntad&K\

^W)

, (12.23) (12 23)

~

150

The Inverse Variational Problem in Classical

Mechanics.

For AE - DB ± 0 (3)

a.e.

CE-FB AE-BD'

(12.24) AF-DC AE-BD

(2) J

{

'

If we change the Lagrange function changing K and W the structure of the Lagrange function should remain the same; this is our requirement. The Newton Equations should also remain unchanged. Taking this into account we should have for K ->■ K' and W -> W w 31

C'E'-F'B1 A'E' - B'D' '

,

m

J

A'F'-D'C A'E1 - B'D' '

,„„„,., v

;

Where A',B', C", D', E' and F' correspond to expressions (12.23) where K and W were replaced by K' and W resp.. From (12.25) and (12.26) follows C'E' - F'B' = X(CE -

FB),

A'F' - D'C = \{AF - DC), A'E' - B'D' = \(AE -

(12.27)

BD),

where X = X(a,/3,j).

(12.28)

The relations (12.27) form the starting point for finding the answer to the question stated at the beginning of this Section. As an illustration of the method presented above let us inspect a model, investigated by us earlier, in Section 8, but from a slightly different point of view. Our starting point will be the Lagrange function (8.1) L=\1-U{a)=K{a,1)

(12.29)

and the Equations of Motion dU .. BU Xi + — = Xi + 2—Xi = 0.

(12.30)

Hence, according to (12.6), /(1) = - 2 ^ ,

/( 2 ) = /(3> = 0.

(12.31)

The largest set of Lagrange functions...

151

Our goal is to find the largest set of Lagrange functions V of the struc­ ture given by (12.1), s-equivalent to the Lagrange function (12.29). Notice that all those Lagrange functions have to have the same Newton's Equa­ tions (12.31) (normal form). To start with let us write (12.8) in the form of a system of three equations, taking into account (12.31), viz. „ / d2K

d2K

nn

, „„ d2K /

2a

d2K

-{ W^ -(2aW

, d2K 2

ind

+ 4p

K\dU

w)^

-W)^-

+ (fW

8K\

8U (12-32)

dK 2

nnd

+2

K

^

+1

d2K

=0

mh >

n

(12 33)

-

+ 2{32W") = 0,

(12.34) oa as |L| / 0. Here U as well as W are functions of a only. If we differentiate (12.34) with respect to 7 we get W = c = const.

(12.35)

Then (12.34) reduces to

C =

(12 36)

S °

-

which entails W =0

(12.37)

as

— ^ ±0

a.e.

oa by assumption. We conclude that the set of Lagrange functions, s-equivalent to (12.29) does not encompass Lagrange functions containing the term

appearing in (12.1). We infer also that the equations (12.32) and (12.33) remain unchanged and L = K(a,P,rr)-

(12-38)

152

The Inverse Variational Problem in Classical

Mechanics.

Let us differentiate (12.32) with respect to 7 and (12.33) with respect to j3 and subtract the latter from the former. We obtain d2K dp2

t

4

d2K = 0. da 97

(12.39)

If we introduce the new variables "f 7= =

a = z2-zi, - zi ,

z2 Z+2+Zi z\ ,

(12.40)

then the equation (12.39) becomes d2K 8d22KK n d22K K 2+ + ~dp dz\ dz2 =0 ~ w ^[-^[

,^nA% (12.41) (12 41)

-

a 2-dimensional wave equation (equation of the membrane) where P and z\ play the role of the space coordinates and z2 - of the time coordinate. Hence we conclude that every solution of equations (12.32) and (12.33) has to satisfy the wave eqaution (12.41) (or (12.39)). Notice, however, that not every solution of the wave equation satisfies the equations (12.32) and (12.33). One can easily check that K== - ( aa77- -/ 3P22))>i ,, K

(12.42)

a

discussed in the example 8.5 I., satisfies (12.41). Let us focus on the equation (12.33). If we denote (12.43) (12-43)

■g^ = Q, 07 equation (12.33) can be written in the form

(■>-*-m-<%+»%-°

(12.44)

or, using the theory of characteristics, d/3 dp

^ ^M 7 _- 22Q a 1 ' oa

_

dj

=

da =

"" ~Tp^M ~ ~7p™M Ww ""

(12.45) (12 45)

'

da oa

The partial solutions, obtained from (12.45), read

^ + U(a) == ci, %y+U(at) Cl,

aj - 2=c P2 2.= c2 . aj~P

(12.46)

The largest set of Lagrange functions... functions...

153 153

Hence the general solution of (12.44) is 2 ^=QC — = llQ(§7 ++ t / ( aU(a), ) , a 7 -ai/ -p 3 2 )),,

(12.47)

where Q is an arbitrary function. From (12.47) follows II Q{^u + U(a),au-/32)du

K=

+ F{a,P). F{a,P).

(12.48)

On the other hand let us insert (12.47) into (12.39). We get

=4 =4 +4

(1,(12.49) 2 49)

BQ A.dUdQ dUdQ ,. dQ d2K 8Q dQ = 4 = 4 1- 47 d(32 da da dc\ dc2

w £ ^ ^>

-

where notation of (12.46) was used. The solution of (12.49) reads fp Jb

(dUdQ fdUdQ

dQ\ dQ\

\dadcL

dc2J

+F1(a,j)l3

+ F2 (a, 7 ) .

(12.50) (12.50)

0=v

Let us compare (12.48) and (12.50). We have

f Q\1=udu + F(a,p) = fjdUdQ [\(dUdQ

8Q\ dQ\

(a

(12.51)

NJ

+F + F22 (a, +Fi(a,7)j8 (a,7). 1(a,j)(3 + Let us further differentiate both sides of (12.51) twice with respect to 0 and then once with respect to 7. We get 2dUd^

da dcf oc{

d2Q | 1 \dUa ' 0C2 dc2 [[da oa

1 d2Q \\ dci oci dc ac22

QdQ =

dc oc22

0.(12.52)

To get a solution of Q which depends only on c\ and c2 but not on a we conclude that d2Q dc\

=

d2Q _ Q dci dc2

or, after solving the equation (12.52), _ = a + —Z-T e +gI , Q Cl + r r - T T + Q= aci 2(c2)2 I9 1

(12.53) (12.53)

154

The Inverse Variational Problem in Classical

Mechanics.

where a, g and e are some constants. From (12.53) follows £ {

K = a(l+u),+

^-/

) k +

f7

+

F(a,fl.

(12.54)

If we insert (12.54) into (12.32) we realize that a =0

and

F(a,/3) = gU(a).

(12.55)

This leads us to the familiar result (see (8.187)) 6

K = g(^ + U(a)) +

-^.

(12.56)

As another example let us inspect the case when L = ^ii-^-[WP)

(12.57)

i.e. K = 0. Then the Equations of Motion reduce to >y(2aW' - W)f{3) = 0 , P(2aW - W)f{3) = 0 , (2aW - W)fw

(12.58)

+ (2jW + 4(32W") = 0 .

The only nontrivial case is when Xi =

2W'-f + W'P2 2aW'-W

Xi + f

m ±l

'

(12 59

- ^

2

where /( > is an arbitrary function of a, (3 and 7. The Equations of Motion read

} v

j 2

/

(12.60)

2

+(2k W - 4(xx) W")S- = 0. In other cases W is a constant or just zero and the equations (12.58) or (12.60) do not make sense.

Construction

of the most general two-particle Lagrange function..

.

155

Can we expect that equation (12.58) admits other Lagrange functions belonging to W ^ W ? To answer this question consider W(a) = W(a)\{a).

(12.61)

Then, according to (12.59), we should have W _ W'X + WX' 2aW< - W ~ 2a(W'X + WX')-WX' f\n\

/-in

W" 2aW -W~

W"X + 2W'\' + WX" 2a{W'X + WX1) - WX

'

If we rewrite these equations as differential equations for A we obtain X'W2 = 0 , (12.63) X"(W2 - 2aWW') + X'(2aWW"

- 4aW'2 + 2WW') = 0

or A = const.

(12.64)

The set of Lagrange equations, s-equaivalent to (12.57) is L' -XL,

A = const.

(12.65)

i.e. all are equivalent. 13.1

Construction of the most general two-particle Lagrange function in (1+1) space-time dimensions giving rise to Euler-Lagrange Equations covariant under Galilei transformation.*)

In the former Sections we investigated several aspects of the problem of Lagrange functions and Euler-Lagrange Equations for one particle in (1 +1) and (3 + l)-dimensional space. In the present Section we are going to consider the non-relativistic case of two point particles moving according to the rules of the Lagrange formalism in (1 + l)-dimensional space-time. The case of (3 -I- l)-dimensional space-time will be discussed in Section 14. We are not going to specify the form of these equations; the only restrictions imposed upon them are the following •) J. Lopuszariski & P.C. Stichel, Fortschr. Phys. 45 (1997) 1.

156

The Inverse Variational Problem in Classical

Mechanics.

i) these equations should be forminvariant with respect to the Galilei transformation, defined below, and ii) should be derivable from a Lagrange function as its Euler-Lagrange Equations. The unknown Lagrange function, we are starting with, (13.1)

L = L(x,x,R,R,t), L(x,x,R,R,t), depends on x H R M

xX

= = = =

x^-x™, x^-xM, (1)

TOia; m

1 == 1,2,

(2)

^+«*&2 , M ' mi +m mi^O, + m22^0,^ 0 , mtpQ, dx{t) Rk =_ dR(t) dt ~dT'

= -5-'

=

+ m z

x^ = x®(t), =x{i\t), x®

(13.2) (13.2) (13.3) (13.3)

m22 ^^ 00 ,,

^r

(13.4)

(13 5) (13.5)

-

The positions of the particles at the time t are given by x^^t) and x^(t). The real quantities mi and 7712 are arbitrary constants satisfying (13.4). The masses of the particles can, in general, not be evaluated unless the 3-rd Newton's Law is observed. Thus rrii i= 1,2 should not be necessarily identified with masses of the particles. The Galilei transformation for the variables x^\ x^ and t read (x^,x{2\t)

-> g(a,v,s)(xW,xW,t) ^(o.u.sJC^ 1 ),^ 2 ),*) =

(2) = (z (1) + a + vt, x{2) +a -)- a + vt,t vt, t + a). s).

(13.6)

In (13.6) a is the spatial translation, v - the velocity (special Galilei transformation, boost) and s - the time translation. The forminvariance of the Equations of Motion under the Galilei tranformation (13.6) means that the Euler-Lagrange variations of L should not depend on R, R and t. Thus we have --x d dL dL ,. == fix / (> : CX> a ;X> a ;R)j> R) dtdx--dxd t 5 ± - ^ ' ' ' '

(13.7) ^

d dL dL 9L „ , . .. »~. ;— —7- = r KF { (x,x,x,R). m R ) Jtd-R-dR dtdR OR = ' ' ' ' -

(13.8)

The question we are going to answer is: what is the most general form of L, when (13.7) and (13.8) are observed.

^

Construction

of the most general two-particle Lagrange function... function...

157

If follows from (13.1) and (13.7) that f(x, x, x, R) = (x, x)x + f2(x,x)R, (x, x)R, f(x,x,x,R) = ffo(x,x) fi(x,x)x 0{x, x) + /i

(13.9)

where /h! = £dx2> '

(13.10) (13-10)

h = - ^ dxdR dxdR

(13.11)

Since dh 8h ^ l= = dJl == QQ dR dx dR dx we conclude that we conclude that

(13.12)

hf2 ==Mx). h{x). In a similar way In a similar way F(x,x,x,R) F(x,x,x,R) where where

(13.13) (13.13) = F {x,x) + Fi(x,x)x = F00(x,i) + Fi(x,x)x

+ F {x,x)R, + Fa{x2ti)R,

(13.14) (13.14)

fF1i == ^^ = = ^/ 2 ( x- ) , dxdR 9a; dR 2 _ 9rP-T L F2 = ^-=-

(13 15) (13.15) -

^ = ^ = 0 , dx dR

(13.17)

(13.16) (13.16)

Since

we conclude that F2 = F2F(x). 2(x).

(13.18)

Now we are in position to integrate (13.10), (13.11) and (13.16), taking into account (13.13) and (13.18). Prom (13.16) and (13.18) follows L = g{x)R g(x)R22 + gi{x,x,R,t)R ffi(x, x, R, t)R + g2(x,x, g2(x,x,R,t), R, t),

(13.19)

2g = FF2.2.

(13.20)

where

158

The Inverse Variational Problem in Classical

Mechanics.

From (13.15) and (13.19) we get dgi(x,x,R,t) = dx

h{x)

or 9\ -9n{x)±

+ gio(x,R,t),

(13.21)

where gu(x) = f2(x).

(13.22)

Finally from (13.10) and (13.19), taking into account (13.21), follows

or 92 = g22(x,x) + g21(x,R,t)x

+ g20(x,R,t),

(13.23)

where g22{x,x)=

fi(x,u)(x-u)du

(13.24)

and the constant c is so chosen that the integral on the r.h.s. of (13.24) makes sense. If we insert (13.21) and (13.23) into (13.19) we obtain L = 922(x,x) + g2o{x,R,t) +

g2i(x,R,t)x+ (13.25) + g(x)R2 .

+g10(x,R, t)R + gn(x)xR

Let us once more use (13.7) and (13.9). We have d2L d2L • X+ R+ did-X dIdR didl-^

d2L

dL

r

,

=M

x

, ( 13 " 26 )

s

^ -

If we insert (13.25) in here we get, after some cancellations, d2g22{x,x) dxdx

. X

dg2i{x,R,t) OR

dg22{x,x)

dg20{x,R,t)

5

o

ox d

■

- -^R

2

=

ox fo(x,x).

dg2i{x,R,t) dt dg10(x,R,t) 3

ox

■

nio^

~

\\.a.£t)

H

Construction

of the most general two-particle Lagrange function...

159

The coefficients of R and R2 in the identity (13.27) have to vanish, which yields dg21{x,R,t) dR

dg10{x,R,t) dx

_n - °'

,.,„., (1328)

g(x) = g = const..

(13.29)

Taking into account (13.28) and (13.29) the identity (13.27) reduces to 82g22{x,x)

.

dg22{x,x) /o(l x) = Yx ' X ° ° (13.30) _ dg20(x,R,t) _ dg2i(x,R,t) dx dt The l.h.s. of (13.30) does not depend on R and t, the r.h.s., on the other hand, does not depent on x. Consequently x

didx aX X

d g22 . dg22 --^r-^-x+—— +f0 = g0{x) ox ox ox

, „ „ , (13.31)

and dg2o . dg2i + = 9o{x)

-^x-

^r

,,,,„! (13 32)

-

-

Due to (13.8) and (13.14) the identity X^x + ^RR + -5^- - | | = F0{x,i) (13.33) 8Rdx 8R dRdt dR is still left for inspection. With help of (13.25), taking into account (13.28) and (13.29) we get ■ 2dgu(x) ox

, dg10{x,R,t) at

dg20{x,R,t) dR

_

This can be also written as cw •-, -2dgii{x) F0(x,x) -x'—z = °X dg10(x,R,t) dg20(x,R,t) _ = dt dR— =

(13.34) h{x)

-

Let us differentiate (13.32) with respect to R and exploit (13.28). We get d (dgw(x,R,t)

dx-{—m

dg20(x,R,t)\ =0

m—J

-

,.„,, (13 35)

-

160

The Inverse Variational Problem in Classical

Mechanics.

This means that the differentiated function (13.35) depends at most on R and t. On the other hand we learn from (13.34) that it depends at most on x. Hence dgw — at ~W

dg dg20 20 —— = K — COnSt const.

K= an = - -3R

. . (13.36a)

'

To recapitulate: we have OR —M

dg2i(x,R,t)

_ dgio(x,R,t) dg10(x,R,t) dx dx

dg2i{x,R,t) dj^R^

_ dgdg 20{x,R,t) 20(xji,t)

—m

OR—

_

=

°'

= gQ{x)

=

9o{x)

/ t a a«u\

(13.36b) (13 36b)

-

(13 36c)

(13.36c)

and (13.36a). It is convenient to use the following notation gio(x, R, t) = gio(x, gio{x, R, t) - at, nt,

(13.37a)

g21 (x, (*, R, t) = g21 (x, R, t) - g0g{x)t. 0(x)t.

(13.37b)

Then (13.36a)-(13.36c) read dg20 _ dgw _ „

^dt = ~ 0,'

(13.38a)

dg i _ dg _ ff-^=0, dt dx

(13.38b) 13 8b

^OR2

20

dffio _ dg21 _

-dx--^R~°dx dR ~

( , , 13 38c (13.38c)

( -

)

The l.h.s.'s of (13.38a)-(13.38c) look as a 3-dimensional rotation of a vector g9 = (521,310,520) (921,910,920)

with respect to the variables Y =

(x,R,t).

Hence, by the Converse of the Poincare Lemma, 92i 921 ^

910 9io

= ~

d${x,R,t) d$(x,R,t) ^ dx

, '

(13.39a)

-=

d$(x,R,t) d${x,R,t) Q£ dR ',

(13.39b)

Construction

920 920

of the most general two-particle Lagrange function... function...

d$(x,R,t) = — gt

(13.39c) (13.39c)

'

at—

~

161

Let us once more recall the Lagrange function (13.25) and insert in there (13.37a)-(13.37b) and (13.39a)-(13.39c). If we discard the gauge term 4* we obtain L = g22{x,x) 922(x,x) + go(x)tx g0(x)tx + gn(x)xR

+ KtR gR22 ntR + gR

(13.40)

We may introduce _ dG(x) --90{X) * ( * ) == % * •■ -Hx-

(13.41) (13-4D

g0(x)tx + KtR = --~-[t{G{x) — [t(G{x) - KR)] + G(x) G{x) - KR

(13.42)

Then

and the modified Lagrange function (13.40) reads 2 + + gR gR2 , ,

L= L = U(x,x)-KR+^^-R U(x, X)-KR+ —f^R at at

(13.43)

where where (13.44) (13.44)

U(x, x) = g22(x, x) + G(x) U(x, x) = g22(x, x) + G(x) and and

dG(x) _ (13.45) ^ s f t =l ( flu(a:)« ) . —j— ax ax Thus (13.43) is the final form of the Lagrange function we were looking for. The Galilei forminvariant Equations of Motion are d22U . dxdx

d22U.. U dx2

^ _ ^dGf,_dU_ _ ^ L _ Q

dx

dx

(1346) (13.46)

d2G .o dG., n ?, d (dG . n ■ \ n 0—xi2 2 + + 2gR 2gR + = j—t (^x = 0 . (13.47) + ^—xx + + KK= l —x + + 2gR 2gR + + « t )Ktj=0. Let us write (13.46) and (13.47) in the form of linear algebraic equations with respect to x nad R, viz. d2U.. dx2

dG -~__ _ dU dx dx

d2U . dxdx

(13.48)

162

The Tfte Inverse /nuerse Variational Problem in Classical

dG.. ,t

Mechanics.

92G.2

-J

(1349 (13.49)

E * -0*'-'

»

We denote

*£-(£)'=»*<>-

(13.50)

a.e..

Taking into account (13.2)-(13.4) as well as (13.50) we get + + " P ff 5MM VV 35 ^^ dxdx ) M dx m2dGd2G.2 2 _dH]_ m2dG_d^G_. d2 U ( d2G . 22 \

+ +

21 M M 3^x ^a i ^

9G 9G 9a: dx

2 + 2 + " ^9i( ' V"s + a 9a;? a *! JJ +

2 U_., _ //_9a 2 t/ _ 9X/V dU\ \ dx dx dx JJ \dxdx dx

, „ „ , (13.51) (13 51)

"

m ■■(2) __ 11 \r, i fdU dd22 U rn^dG [Q "H fgg ^ -\ \ mi. 9(7 ff 9 7 7 + ~ " P -~v[ [ Mw{~dx~~dx~d± 1^ 9a ~ 9a T 9± a 7x)++W~dx~ 2 t f & KK+

1x

dGd2G ., m^dG_d^G_. mi 2

92£/ / G .,2\\ dHJ_( d92G

+ M dx dx2%1 + 2 X 9± 2 K \K++ adx^ Mto^ + M )J-~ +

2 _dG U . _dU\ _dUY _ 9 G /I d92E7 dx \dxdx dx J

(13.52) (13 52)

'

If the 3-rd Newton's Law is observed i.e. there exists such fi\ > 0 and fi2 > 0 that 1) {2) ^/ii£ i i ((1) +n +/J.2x2X^

== 00 ,,

(13.53)

then from (13.51)-(??) follows (\x //iim xm22 9 \ M + +

fj, M2mi\ /dU /9E7 2mi\ M ))\dx \dx

d2U \ dxdxX)

2 (firms Mgron gg dGd G .2\ /fiim2 H2fni\ /(dG dGd'G K + K+ 2X VV M M ~ M M / \ f t s )\dx~dx~~dx ~dx~~dx2X)~ j J.\d52U2 c / /( 9d22GG . 2,\\ + K+ X

-^ ^W\ M )-

__dG 9 G /t d922U [/ , _ 9 C E /A\ 1 _ 9a; \dxdx dx J

~

(13.54) (13 54)

-

Galilei forminvariance

163

of the Euler-Lagrange Equations... Equations...

If we choose mi mi = =m pi, ,

(13.55)

m22 = = H2, m H2 ,

then (13.54) reduces to d2UK+^. f ±2)_dG(dHl._dU\ d2G . 2 \ 8Go ( d2U . ^£( dx22 \ dx22 ) dx \dxdx \ dx dx

dU\ dx) dx)

(13.56)

Since we do not intend to restrict the freedom of choice of U(x,x) conclude from (13.56) that G = const, const. or just

K 0, « == 0,

G (2 = 0.

we

(13.57)

These relations will have some significance in the discussion of the next Subsection. 13.2

Galilei forminvariance of the Euler-Lagrange Equations for two particles in (1 + 1) space-time dimensions.

Let us start with equation (13.47) which entails that

is is If If

22 ^ ^ - x + 2gR + KKt=^-(r t= ^ ( r ++ I«i =S zUt ) ) = dx dt z ox at a Constant of Motion, where a Constant of Motion, where T = G{x) + 2gR. V = G{x) + 2gR. (13.57) is satisfed then G = 0 and (13.57) is satisfed then G = 0 and 1 2 r = g ( MMli * ( 1 ) + M 2 z ( 2 ) ) .

r = g( o;( ) + ^ ) ) .

The canonical momenta dx^ dxW

y

computed from (13.43) yield m P

,(2)p

8U dU dx dU dx

dG ■ dGdx

dTmi dTmi dt M '

dGdx

dTm22 dt M

(13.58)

(13.59) (13.59) (13.60)

The Inverse Variational Problem in Classical

164

Mechanics.

and W+pW ppW+pW

^ ==P. (13.61) (13.61) at In case the 3-rd Newton's Law is observed relations (13.60) and (13.61) imply = =

M = 2g. 2g.

(13.62)

Notice that p is not conserved as long as the 3-rd Noewton's Law does not hold. From (13.58) follows namely dp dt dt

^ == -—K. —K.

(13.63) ' The Lagrange function (13.43) is weakly invariant under the Galilei transformation 3,(0' xW

= xx(i) (i)

=

+t

+vvt

a , + a,

+

ii ==l , 1,2. 2.

(13.64) (13.64)

we have SL —Ka —— ——(—Kta), SLaa = = —no, (—Kta), at

(13.65a) (13.65a)

".-sEan* 1 " "!*">■

(13 65b) (13.65b)

5LV - -Kvt + —-V —v + 2gRv = — [(-±Kt2 + I > ] ,

(13.66a)

SL -r.{pvt). 5LVv = —(pvt). dt

(13.66b)

i.

i

-

The „Noether currents"*"> or generators of translation and the „boost" resp. are ja = pP + + lit Kt = = S, S , ja =

(13.65c)

*' When the Lagrange function L is weakly invariant under a symmetry transformation then „ _ dip{xM,xW,t) (A) oLi — dt On the other hand 6L = 4E" \dxM (4r-M%) + did) #TT^(')Nj))

Galilei forminvariance

jv = =pt+\nt pt+\nt22

165

of the Euler-Lagrange Equations... Equations...

(13.66c)

-v. -V.

Notice that due to (13.58)

■Jr =0 dt

(13.65d)

Jdjxv,

, 13.66d (13.66d)

and dp dT dT —— r ==-ft—t ++Pp+ + KtKt —-—-==-nt — K,t + p + Kt-p +p+Ht-p=0. = 0. at dt at dt at dt

The Hamilton function or the generator of the time translation reads Z^ L^,

gx(i) gx(i)

i = = ^-x ^-x dx ox

+ + 2^-xR 2^-xR dx ox

2 + + 2gR 2gR2 -- U U+ + KR KR -- ^-xR ^-xR dx ox

T—XR + gR2 = -=— x + -TT-XR

ox dx

ox dx

-U

+

We have 2 dH _ d2U , 2 do^U U ... dU.. dt dxdx dx2 dx dGdG..dGdG~ ■ ~ n + -^xR — x R + -^r-xR + 2gRR ox dx dx ox _ / d2U . \ dx dx +

d22U U± . dx2

KR.

d2G .2 ■ dx2 dU . dU.. ■ ~-x - -^—x -^rx + KR nR = ox dx dx ox

dG~_dU\. dGz_9U\ dx dx J

(d (d22GG « 9G.. 9G.. „ ==

\ -•

4£ -- dxto dxto

//

{!^x

+

2 -- gR gR2 = = (13.67a) (13.67a)

Hx-X + 29R + K ) R

(13.67b)

„n

= 0

>

or

di(') dxto dt dt v V

I

(B)

^£f V V dxto dam

dt dt did)) did))

++

dt dt [[^^

Bxto Bxto

J)

Taking and (B) (B) the the Taking into into account account the the Euler-Lagrange Euler-Lagrange Equations Equations we we get get from from (A) (A) and „Noether current"

i=Eii l i ( i ) -* w . i , J ! ' { )'

which is a conserved quantity as dj _ d I sr-^ 8L

\ _

Jt~ dt \^> dxM ~dt~ dW> ~~^J ^J ~~

166

The Inverse Variational Problem in Classical

Mechanics.

where we exploited (13.46) and (13.47). The Lie-Cartan relations for Galilei group expressed in terms of the Poisson brackets, using the formula*' r„

m

d A

dA

[*.*! + £ - £

(13.68) (13-68)

are d

r•

y . , f ilnl = - ^3a = - «

(13.69)

(see (13.65d)), [jv,H] - ^ == ~P~ -p-Kt Kt ==~i°-ja b'«, -H"] == ~~Qf

(13.70)

(see (13.66c)-(13.66d)) [ja,ja]=0, ba,ia] = 0 ,

(13.71)

[jvja] = =\ptT,p] = = --\T,p] = -2 -2g9 = = --A [7»,i«] b* - r,p] rr)P] = A

(13.72)

as \p,p] [p, p] = 0 and .„ , r _ „ _ , dG n mi dG m2 [T,p] = [r,p] = [G [G ++ 22gR,p) ^ p ] == -— + +2 2g-jr _ ++ 22 5^- | = 22g, 5 - -- -=5l ox M Ox M

(13 731 (13.73)

[jv,jv] == 0. [jv,jv] 0. If the 3-rd Newton's Law is valid and K — G = 0 we have instead of (13.43) the Lagrange function L = = U{x,x) [/(j;,i) + + \MR | M f2,l 2 ,

(13.74)

*' We derive here the formula (13.68). If we have a quantity A yl(x,p,t) "' ( x , p , t ) we have — A(x,p,t) v

dt

= >

—-j-x (l > H

^vai(')

TTTP 1 1 ' I H

a P (>r

;

a<

l(/i) J

Since 5p(0 '

ax(;)

expression (A), using (B), takes the form d dt dt

dA dt dt

which coincides with (13.68).

^p( sr-f dA 9H ^2f V 5 i ( 0 dpd) XaxM dpi*)

dA dH \ dpM dlw) dpWdxWj '

(*)

Construction

of the most general two-particle Lagrange function...

167

where we used also (13.62). Moreover (13.69), (13.70) and (13.72) are replaced by [ja,H] = \p,H] = 0, [jv,H] = \pt = nix{1) -n2x{2\H]

(13.2.69') = -p,

[jv,p] = -M,

(13.2.70') (13.2.72')

resp.. The quantities K and 1g in the algebra (13.69)-(13.73) are considered to be central generators, which commute with all other. In case the 3-rd Newton's Law is valid there is only one central generator M. 14.1

Construction of the most general two-particle Lagrange function in (3+1) space-time dimensions giving rise to Euler-Lagrange Equations covariant under Galilei transformations.*^

The assumptions made by us here are similar to those made in (1 + 1) dimensional case mutatis mutandis. We are looking for the most general form of a Lagrange function L = L(x,x,R,R,£) which we assume to exist and to give rise to Euler-Lagrange Equations, forminvariant with respect to the Galilei transformation (x(M^5(S,a,v,S))(x(M = (14.1a) = (5x ( i ) + a + vt, t + s)

i = l,2,

where 5 is an element of the 50(3) group, a 3 x 3 real orthogonal matrix, a is the vector of spatial translation and s - the time shift. In this Section we restrict ourselves mainly to the transformations (x(i),t)->g(l,a,v,S)(XW,t);

(14.1b)

at the very end we shall make use of the required rotational covariance. *) J. Lopuszariski & P.C. Stichel, Fortschr. Phys. 45 (1997) 1.

168

The Inverse Variational Problem in Classical

Mechanics.

We have the identities V d2£ . yV ■ V *-* dij dxk ^—" *-* dij dxk ~* + y . _ ^ L _ „ ^— dij dxj dik

d2L ■ R dij dRk 5±j 5 f t

J

k

d2L _ dL_ - — &&j dxj 9 i j dt 5t 5XJ

y, 92 L ^ *—' dii dij dRk k

(14.2)

J

x x ) i , + ^ / j2)2 ) (x,x xx ) f t = (x, x) + Y fffk = zj ff (x, i)+52 $ ((x, -*)** + E 4 ( ' )^

and and

fc

k

k

k

^ d2L . y, d2L X. + V 0 f t fix* Xk+ * Y 5 f t dxk z ^ dL + > — ^dRjdxk ^dRjdik

y > d2L ■ d2L y > d2L ■ + d2L V 5ft 5ft * 5 f t St ^ dRj 8Rk 2 * + dRj dt .. ^ 5 L xk + > Zz ^ft = r'd± ?d±]dRk

dL dL + 5ft + dRj (14.3)

x xx ) f t . = (x, x) *) + i f > (x, *)Su x)x* + + E £ F< i f$ >( (x, = ii ff > > (x, +£ £ FltHx, - )^ • k

k

From 5 ft dRi

5ifc dxk

we infer that 2, 2) 44 2 ) ==/< / ] , 2 ) ((*). x).

(14.4)

Prom From

djJl 5ff

= =

dRi 5 ft

djf_ dFJl^=Q dxk 5ijt

follows

*f = *f{x). if^^fCx).

(14.5) (14.5)

Let us integrate the relations d2L dijdxk ^ ^ dxjdRk

^tx'x>' = / ( 2 , (x)

J j k [ h

(14.6) (14.7)

Construction

of the most general two-particle Lagrange function... function...

d2L (2) : r— = F}, (x) . Jk dRjdRk The result is

169

(14.8)

L = 9gW W (x, x) + g<0101> (x, R, i) t) ++ £J2 9f 9?2)2) (x, R, t)±, tfy + 3

)1 0 )

+ 5^>9fj > ( x(,xR,1Rt ), ^^ 3++£^ ]^T< ^, S1 )1 () (x x) )i i; ^^ i + +

7

(M9) (14.9)

*

where ^ - = t f ( x , x ) ,

(14.9a)

si^W = /]"'«,

^M-Jff'W,

(14.9b)

2 f^f f( (xx)) = F^ ^( (xx ) ;

(14.9c)

the remaining terms are not fixed; they are at most linear in x and R. Let us preform a gauge transformation on (14.9) namely L-¥L L -> L = L L- - — ~ *$((xx ,, R R ,, i ) , at at where "3/ is defined as a certain solution of the equation 0 11 R ,I tt )) .. ^ff(° ))(x,R,t) ( x , R , t ) = -- ** (( xx ,I R

(14.10)

(14.11a)

Hence L = 5 ( 0 0 )(x,x)

+

^g(.02)(x,R,i)ii

+

i 1 ) s
+EE?fw^> where where

i (02) _

.7

fc

(14U) (14.11)

k (02) _

5 *

(14.11b)

170

The T/ie Inverse Variational Problem in Classical -(10) £(10)

=

yj

Mechanics.

(10)_ 5 5* * (10)

yj

(14.11c)i

.

d R

*

Keep in mind that the Equations of Motion are invariant with respect to the gauge transformations. Hence from (14.2) we get the identity d2L

^

82L

^ ++

d2 L +

■

dl

=/

, ( 0 )).. . „ , „ . _ , (x x)(14 12)

E^7^^ EaM^^ E^7^^ ^a-^^+a^-^ 5^-^ > '

"

or taking into account (14.11)

V *y» ay* ±±nk ++ y y ddHl y JlXk ± k++ y L- d±j dxkk *-*

k

2-J dx QXk ^ k

dd9

j£l^RRlXk+ tiKl

2_, dx gXkk *-?

k

y9jrRk+ °jr_*r_ydjrikik__ Rk+djryjT_ydjr ++yvjr £— 2— dR 8Rk

dt

K

l-

dxj

Q-I10)

k ft

/I

kk

Ii

J

L.

O (11)

K K

At

*— dxi dxj

J

(14.13)

3J

J J

Relation (14.13) is an identity. As the coefficients in front of Rj, Rjik RjRk do not depend either on x or R we have (2)

(14.14a)

9ki fffci == Cu Cki == Qk Cik = = const. const. oxk

and

(14.14b)

oxj

02) agj (x,R,t) ag<10>(x,R,t) =_Q0. gg^M,gT^M

.14c) (14 (14.14c)

Taking (14.14a)-(14.14c) into account, we get from (14.13)

ay»)(x,x) v ay°°)(x,x)

JOj(X,Xj - > > Joj(x,xj =

— —

22

9 ag(°°)(x,x)

XXfcfc HH

--

^^9ff(° y /a f f (° )(x,R,t) )(x,R,t) _ affff (° >(x,R,t)\ ^ .

V ~^M

= =

2

&i

dxk

02) ag((02) gg (x,R,t) at

^

dxj

)JXk

+

h

( w1 5 )

(14.15)

Construction

of the most general two-particle Lagrange function...

171

The l.h.s. of (14.15) does not depend on R and t, the r.h.s. is at most linear in x. Hence dg[02)(x,K,t) dt

,_, =gj W ,

c^

"

(14.16a)

dx~3

- l ^ ^ m

J

(x).

(14.16b)

i

Notice that from (14.14b) follows (by using the Converse of the Poincare Lemma)

•Tw - ^fer1

' I416c »

We still have to investigate the identity v ^ d2L > ■ ^dRjdxk

v^ d2L i ^dRjdRk

.

Xk + >

■

Hk H

d2L : dRjdt

dl dRj

=

(14.17)

= F o i (x,x) derived from (14.3). Taking again into account (14.11) we obtain 3$< 1 0 ) (x,R,t).

n )

¥ ^ ** ? ¥ ~^^*+ +

+^ ^ ^ ,

dg< 10) (x,R,*) ^3iJ t

^5ff(02)(x,R^). Xk \ dRj =

*+

( x ) .

3gj. 10) (x,R,t) 3t

^d$0)(x,R,t)h \ dRj

(14.18) _ ~

Rk

F0j(x,±)

or, because of (14.14c) /agj10)(x,R,Q _ c^10>(x,R,t)\ ^ ^- ^ aiifc dRj J k

+ E E

^

i A + fCl|M^ j(l[ , i ,

(1419>

172

The Inverse Variational Problem in Classical

Mechanics.

The r.h.s. of (14.19) does not depend on R therefore 9g( 1 0 )(x,R,i) dRk

agj 1 0 ) (x,R,t) dRj

which implies by using the Converse of the Poincare Lemma Pi ' ( x , R , t ) =

^

(14.20)

We have still

The l.h.s does not depend on R and t, the r.h.s — on x. Consequently

W^-Xll-tor*** =

at

=

(i4.2i)

= 5 ( 2 ) (x). ' Notice that (14.16a) entails ^

0 2

>(x,R,*)=0.

(14.22)

By virtue of (14.14c) and (14.22) d 2 g (° 2 >(x,R,t) dRk dt

2 { l0) g k (x,R,t)

=d

= Q

dxj dt

hence

agj10)(x,R,t) dt does not depend on x. Comparing this with (14.21) we conclude that -g{10) (x, R, t) = g{k2) (x) =

Kfc

= const.

(14.23)

We introduce a new notation which turns out to be handier in solving our problem. We define gf0) (x, R, t) = gf0) (x, R, t) -

Kjt,

(14.24)

Construction

of the most general two-particle Lagrange function... function...

~gf (x,>R R, gf2) (x, R, - g\0) (x)t. gSM2)J(x 1t)t)== jJ">(x > Rt) I *)-^(x)t.

173 173

(14.25) (14.25)

Instead of (14.14c) we have

agj02)(x,R,t) ^j 10) (x,R,Q
o

(14.26)

instead of (14.16a) 02 (02) S< / ( ° 2 )>(x,R,*)=S ( x , R , 0 = 5 ( O 2 ) ((x,R) x,R)

(14.27)

and instead of (14.23) ~9W=gW(x,R).

(14.28)

* r = s r (*.*)■ We may introduce

14 29 ((14.29) - )

(10) 10) G (10) (x, R) = G G((10) (x, (x,R) ( xR, , Rt), i-) -E^ K "J ^i ?i^* +
By virtue of (14.20), (14.24) and (14.28) the r.h.s of definition (14.29) can be justified as follows d(g(10> _ d(?(10> ~dR~~ dRj dRj

K] Kjt

~

(14.30) (14.30)

1 0 ) (x,R). = 3 J 11 00 )) ( x , R , t ) - ^ t = g(. 10) (x ) R).

Taking into account (14.27) and (14.30) the equation (14.26) can be written 02) 10 ffj 4 ° 2 ) (x,R) ( x , R ) = ^-G< >(x,R) + + ajQ(x). i(x).

(14.31) (14-31]

Let us collect (14.14a), (14.16c), (14.24), (14.25), (14.30) and insert it in (14.11). We have

L=/>°>(x,x)+Y, ^ L = fl<°°>(x,x) + £ v^ , , x + J2 aja (X)XJ

3

(1

2 i+E»rw«i+

^ 130

v^9G( 1 0 >(x,R)£ • R

+ J2 J ( )*i++J2 E j3

d&1 x,R)±

° ) j X ' R ) ^ - + £ e , f (x)tx,+

J3

i

^

•

K R t+ ^Q^. H* ~ i++EE *i^*+ i l J3

3

+E E ea \\. .( x( )x^)4^- + E E c«M c*M ■ +Y,E

174

The Inverse Variational Problem in Classical

Mechanics.

If we discard the gauge term rfG<10>(x,R) dt

we obtain L = (x,x) + £ 5 j.°>(x)**i + X > ( x ) ^ + I > ^ * __«#'>(*)

+E E

J ag<

.

'

+

( 14 - 32 )

. .

c R

*«** + E E ^ i ■

* i i 3 The Lagrange function (14.32) encompasses two terms which depend ex­ plicitly on t. To get rid of them let us examine the equation (14.16b), viz. gggfrjM ggj 0a) (x,R,0_ v m

^

^

-Li'Ml i

W-

According to (14.25) and (14.31) we have §j M ) = J - G ( 1 0 ' ( x , R ) + f l i ( x )

+ 3 (°»(x)i.

(14.33)

If we insert (14.33) into (14.16b) and differentiate both sides with respect to t we get

fri0)w..0

<w 5a;

fc

9a; j

or ffj0)W

= ^J»l0)(x).

(14.34)

Thus exploiting (14.34) we get for one of the terms of L in (14.32)

5 > f ( * ) t i i = | [ 3 ( 0 ) ( x ) t ] - 5 (°>(x).

(14.35)

j

Moreover

E ^J* = 1 E «>*** - E **^ •

(14-36)

Construction

of the most general two-particle Lagrange function.. function... .

175 175

Thus, using (14.35) and (14.36) and discarding the gauge terms,we arrive at the autonomous Lagrange function 8G BC

L KjRj+Y. Y -g-r^i+Y. L= = K(x,x)~Y^ A-(x,x)-]T KjRj+Y. Y -g^^+Y jj

ii

i 3

'

ii

Y CCijikR) ,(14-37) Y HRiRi >(14-37) ii

where a K(x,x) aj(x)x K(x, x) = g^(x,x) g(°°> (x, x)++yY, i ( xj ) ^ -- g^(x) 9{0) (*)

(14.38) (14-38)

.7 j

and Gf1}f > =SEGG G 3. 3.

(14.39)

Formula (14.37) is not yet the final form of the Lagrange function, we are looking for, as we did not yet take into account the consequences of the rotational part of the Galilei transformations. The Euler-Lagrange Equations derived from L of (14.37) read ^ d2K . £f dxs dxt '

d2K .. *-f dxs dxt f'

y , dCU- _ 8K_ ^ *-f *-f dx dxss ll dx dxss ''

T-,

+E -^-x +y+ 2C y y* -5—-^—Xiit+y 2c RA = -«*. E E s&** a r * S *-* *-?• oxi dx *-* ox ^r - —"■ t

tf

ir

t

tt.

t

1

st t

(14.40)

<14 41) (14.41)

-

tt

Notice that the equations (14.41) can be also written ^ Gj\G + 2j2C 2YJCjkjR kR ^L( (x)j{x) + = --««,,.. k) k\ =

(14.42) (14.42)

The equations (14.40) and (14.41) should be rotationally forminvariant under transformation x -> x x' — = Sx, xx --> > xx'' = = SS xx ,,

(14.43) (14.43)

x x— —► T x'x'==SSx x. .

This imposes some restriction on the construction of L. We are not going to enter into the lengthy calculations concerning the structure of K(x,x),

176

The Inverse Variational Problem in Classical

Mechanics.

we are going however, to straighten up other problem. It follows imme­ diately from the forminvariance of the equations (14.41) under rotation transformation that Kj = 0

(14.44)

and dj = —Sij ,

A - a constant.

(14.45)

Although G itself does not need to be a vector, | 2 i and af, ^ should be tensors with respect to th indices i, j , k. We are going, however, to show that*' Q = X2,

Gi(x) = G(a)zJ,

(14.46)

up to an inessential constant. The proof runs as follows. We start with the ansatz (see (11.11) and the considerations following it) U

^^5jkG^{a)+xjxkG^{a)+Y,t^iG{z){a).

Xk

(14.47)

i

By differentiating (14.47) with respect to Xi and making proper choice of indices i, j , k we get for i = 3, j = 1, Ar = 2 0 2 Gi ——— dx2ax3

„ d2G^ dG^ = 2xlx2x3—^r 2 + G{3> + 2x2s —— , da da

for i — 2, j = 1, k = 3

or QQ(3)

Gi3) + (x2+x2)-^-

= 0.

(14.48)

From (14.48) follows for x2 = x3 = 0 G(3)(Q)=0.

(14.49)

Thus we have -^-

= SjkG^(a)+xjXkG^(a).

*' the idea of the proof was given by Dr. J. Cislo.

(14.50)

Galilei forminvariance

177

of the Euler-Lagrange Equations... Equations...

Let us examine 8Q (9Gj\ Ad_ ffdG±\ ;A _. _d_ _d_ (dGj\

dxi \ dxk )

dxk \ dxi )

After performing the differentiation and comparing results with each other we get QQ(I)

2S 2SjkjkXi-^Xi-^--

dG^ + + 8ikXiSjkGxwlG{2).. ijXk-^+ SijXkGp) 8ijXkG(2) = = 25 iSijXk-Q— +

(14.51)

For j — k ^ i we get 2 3GW =

G(2)

(14.52)

oa da If we insert (14.52) into (14.50) we get dGj3 dG dx & *k

. MU „ .m . .

„ J

= -—(z,G oxk

(1)

dG^(a) 9a &*

(1453) (14.53)

( a ) + const).

If we put G'1) = G the proof of relation (14.46) is accomplished. Thus the final shape of the autonomous Lagrange function is L = K(x,x) + (xR)G(x 2 ) + v 8r( + 2 ( x x ) ( x R— ) ^U M +2(xx)(xR)

. ++ ^- ( R R ) ;

(14.54)

N.B. We did not discussed the structure of K(x, x ) . 14.2

Galilei forminvariance of the Euler-Lagrange Equations for two particles in (3 + 1) space-time dimensions.

The Euler-Lagrange Equations corresponding to the Lagrange function (14.54) read ^ d22K . ^ 832lK K .. l *-* ^ dx <9is s dx Sitt * L* ^ - dx dxss dx dxtt „< vft „ dG(a)^ +G(a)R. + Zxs—±-l-Y,xtRi

-

dtf = }£- >

(14.55) (1455)

178

The Inverse Variational Problem in Classical

^(xJG(a)

Mechanics.

(14.56)

+ XRXR J)=0. J)=0.

The latter equation implies that — (xjG{a) + +XRj) jt(xjG(a) XRj) = =pj, Pi ,

j3== 1,2,3 1,2,3

(14.57) (14.57)

are Constants of Motion. For an arbitrary function G(a) this conservation law differs from Newton's Third Law. In such a situation we are not able to evaluate the proper mass values of the particles under consideration. The constants mi and m^ (mi + va2 = M ^ 0) were chosen arbitrarily and do not need to correspond to real masses. Only in case when G = = /i n = const.

(14.58)

we may establish the true masses of the particles, viz. A

Hi mmii +M, +(i, Hi = -TT

(14.59a)

A ^2 V2 = = j^m j^rn22

(14.59b)

-- fi, H,

/Ui n% — A = total mass = M fit + fi%

(14.59c) (14.59c)

Then ■(1) Pi = m&fl +, itoif- (32 ) (14.60) Pj = p.\x) + V2Xj is the total momentum. The Lagrange function (14.54) is not Galilei invariant. It is, however, weakly invariant under the transformation (14.1). The ,,Noether currents" for translations and „boosts" can be computed as follows. We have on one hand

SL = J2 JttTT3J •■vivi ++°°( ( vv)), > SL^Yl

(14.61) (14-61)

j

Tj ^XJG Tj^XjG

+ XRJ + XRj

(14.62)

On the other hand

" - i E E j W - 1 (§)<*+%>•

(14.63)

Galilei forminvariance

of the Euler-Lagrange Equations. . ..

179 179

Hence the Noether currents" for translations are j(a)_lr._ ji a) = T =Pj i It >

((14.64) 14 64)

-

(see (14.57)), and for the boost — (14.65)

jf^Pit-Tj.

Both „currents" are Constants of Motion as a consequence of p in (14.57) being a Constant of Motion (see (14.56)). Both are also generators of the Galilei group. To see that we are going to use as auxiliary tool the relation . . _ . dA dA (14.66) where H is the time translation generator of the Galilei group and [ •, ■ ] denotes the classical Poisson brackets. The Lie-Cartan commutation relations for J* a ' = p, H and J^v> are as follows. In case of J^a\ p is conserved and does not depend explicitly on t; therefore \pk,H]=0. \Pk,H]=0.

(14.67)

In case of Ji"' this vector is also conserved but depends explicitly on t (see (14.65)). Taking into account that | £ = 0 (see (14.62)), we get v) d Vl{:\H] ,H} = = -p-kPk [J =- °^ ^ rr = ,,

(14.68)

{ V) == -\6 [{J4 i ,Pk} ,Pk] == -[ru -\Ti,Pk] -A<5jjfe, Pk} ik,

(14.69)

[j^^j^] [j^^^]

= -[t -[tpPii-r -rii,tp ,tpkk-T -Tkk]]

=

= -t{pi, = -t\pi, Trk] -- t[Ti, t\Ti,pPkk]] ++ [Ti, [Ti,rTkk] == t\6 t\sik -- t\S txsikik == 0, o, \pi,Pk]=0. \Pi,Pk] =0.

(14.70) (14.71)

To obtain the relation (14.69) one has to use the formula ^Pk]

=d^k+

X

W6ik

- d^k

+A

M

5ik = Xkk

(14.72)

180

The Inverse Variational Problem i n Classical Mechanics.

and to get (14.70) ane has to use

which is trivial as I7 depends on x(') and x(') only. Relation (14.71) follows also trivially. We do not present here the commutation relations of generators of the rotations S (see (14.1)) of the group SO(3) with themselves and other generators as they are standard. Relation (14.69) yields the extension of the group by the mass generator A. The Hamilton function is

This Hamilton function is a conserved quantity, due t o the Equations of Motion (14.54) and (14.55).

15.1

All two-particle Lagrange functions s-equivalent to a given autonomous Lagrange function yielding Galilei forminvariant Equations of Motion in (1+1) spacetime dimensions.

In Section 13 we presented the construction of the most general two-particle Lagrange functions in (1 1) dimensions, a Lagrange function giving rise to Euler-Lagrange Equations covariant under the Galilei transformations.*) The goal of this Section is to find all autonomous Lagrange functions, s-equivalent to the original Lagrange functi0n.t) Unfortunatelly, we are not able to acomplish our task and to construct the whole set of such Lagrange functions. Notice that under the s-equivalent Lagrange functions can be and are also such whose Euler-Lagrange Equations although equivalent t o the original ones are no longer Galilei covariant. In Subsection 15.2 we give a full account on the set of s-equivalent Lagrange

+

*)

J . Lopuszariski & P.C. Stichel, Fortschr. Phys. 45 (1997) 1.

t ) J . Cislo, J . Lopuszariski, P.C. Stichel, Fortschr. Phys. 46 (1998) 45

All Euler-Lagrange

Equations, forminvariant

under the Galilei transformations.

181

functions, all of which have Euler-Lagrange Equations forminvariant under the Galilei transformations. In Subsection 15.3 we give some examples which — we hope — will clarify our derivations. In Subsection 15.4 we give a natural extension of the set obtained in 15.2 to a partial subset of EulerLagrange Equations equivalent to the former which do not display Galilei covariance. Finally in Subsection 15.5 we touch upon another important problem where the equations in normal form (Newton's Equations) but not necessarily Euler-Lagrange Equations are forminvariant under the Galilei transformations. *) 15.2

All Euler-Lagrange Equations, forminvariant under the Galilei transformations.

We start with a Lagrange function given by (13.43), viz. L = U{x,x)-KR

+ G'{x)xR + gR2-

G'(x) = ^ ^ - , (15.1) ax where K and g are constants, U and G — arbitrary functions. Other quan­ tities are defined by (13.2)-(13.5). The Galilei forminvariant Equations of Motions are w t t .L:I s 0f +

var fl L:II

C 5 + ^ i - g = O1

= G'x + 2gR + G"x2 +K = 0.

(15.2) (15.3)

We assume that (13.50) holds true. Let us resolve the equations (15.2) and (15.3) with respect to x and R. We have

2

=M- 2fl (^*"^) +o,(tf, * ,+ ' B) ] s/ - (ir '* ) ' (is - 4) V =

dx2

\dxdx

dx)

(15.5)

IR{X,X).

Notice that the equations (15.4)-(15.5) remain true for each element of the set of s-equivalent Lagrange functions. *' An illuminating discussion with Dr. J. Cislo is acknowledged.

182

The Inverse Variational Problem in Classical

Mechanics.

We are going to show that the most general Lagrange functions, fulfilling the requirements presented above for g ^ 0 kept constant, are of the form

- S ( i ? i ) + i - ( 5 ' ( a ; ) ) 2 i 2 - ~G{x)+G'(x)xR+

(15 6)

'

+#.R2 - R.R, where cr is an arbitrary function and Hl(Xii)

= x9-^f±

- g (G'(z)) 2 - C7(M) - ^ G ( x ) ■ (15.7)

As far as the quantities k and G(x) are concerned the explanation is as follows. If the first Newton Equation*) (15.4) does not have the form f, = foix) + f2(x)x2

(15.8)

we have G

=

G,

(15.9)

h

=

K.

(15.10)

This is the most frequently encountered case. If, however, (15.4) has the from (15.8), i.e. we have a restriction imposed upon U in (15.1), there exists an additional set of solutions, namely a) for h=h{x)x2

(15.11)

we may have G' = G' + A e x p j - I f2{x)dx\

,

(15.12)

where A is an arbitrary constant; then it follows that & = «, *' First equation in normal form.

(15.13)

All Euler-Lagrange

Equations, forminvariant

under the Galilei transformations.

183

0) for fx=fo(x)+±2—\n\f0(x)\

(15.14)

we may have

G' = G'--P--

(15.15)

where 7 is an arbitrary constant, and K = K + J.

(15.16)

The latter set of solutions is not empty as we shall give examples where (15.8) accurs (Subsection 15.3). The Lagrange functions L, given by (15.6), give rise to Galilei covariant Euler-Lagrange Equations, which are linear combinations of the original equations (15.2) and (15.3). For g ^ 0 kept constant we have anI

+ a12II = ^ x ox1

+ G'R+P^x-^, ox ox

(15.17) ox

a2il + (T22II = G'x + 2gR + G"x2 + k ,

(15.18)

where a=(an

ai2

) =

_ U(Hl} 1 \

+

l°±poL

_ ^ l

2g(G'-G')

, 1

L

V

deta = a{Hl) ^ 0

G

+

| _ i ^ ^ : s (G'-G')G'

a.e.

V

(15.19)

j

' /

(15.20)

and V and Hi are given by (13.50) and (15.7) resp.. Since the trace of the cr-matrix should be a Constant of Motion (see Section 5.3) we infer from (7ll+(T22=a(ff1)

+ l+ §;

P = G'-G

(15.21)

that

P2 j - = Constant of Motion. (15.22) The case g J^ 0, g = 0 is easy to handle as any linear combination of elements of L with constant coefficients is again an element of L. It suffices

184

The Inverse Variational Problem in Classical

Mechanics.

to choose two proper elements of L belonging to the same g / 0 and subtract one from the other; then the g-term of the so obtained Lagrange function will vanish. The case g = g = 0 is rather simple to handle too. The most difficult case is that of g = 0 but g / 0; the model has to satisfy then certain restrictions: fr, given by (15.5), has to be of the form fR = iRfi(x) + IR,2{x)x2,

(15.23)

fRfl

=

~(G'fmfi+K),

(15.24)

fRfl

=

~(G'f,j

fx,o

=

- ~ ,

(15-26)

ha

=

~^G"

(15.27)

where

+ G"),

(15.25)

and the relation ^

= /»,o/fl,2

(15.28)

has to be satisfied. 15.2.1

Case

g^O.

According to (15.2)-(15.5) the functions fx and / # satisfy the identities d2U , d2U . -^Tfx+G'f R+-—xz ox ox ox

dU n — =Q, ox

G'h + ZgfR + G"x2 +K = 0.

,

15.29

(15.30)

Our goal is to find all possible Lagrange functions L = U{x, i)-kR

+ G'{x)xR + gR2 = 0

(15.31)

leading to Galilei covariant Euler-Lagrange Equations, having the same set of solutions as (15.2) and (15.3), and consequently to the some Newton's Equations (15.4) and (15.5). Taking this into account we get the following conditions to be fulfilled, viz. d2U, ~,„ d2U dU -^jfx+G'fR + ^—-x--—=Q, ox* ox ox ox

v(15.32)

'

All Euler-Lagrange Equations, forminvariant

under the Galilei transformations.

G'U + WR + G"x2 +k = 0.

185

(15.33)

Let us consider first the case when g ^ 0 and g ^ 0 too. Since g and g are constant quantities we may assume 9 = 97^0,

(15.34)

which does not restrict the generality of our investigation (scaling). Our first task is to solve (15.33). Let us subtract both sides of (15.30) from both sides of (15.33) resp.. We obtain P{x)fx + ^ ±

2

+ v = 0,

(15.35)

where P{x) = G'-G',

V = K-K.

(15.36)

If P = 0, (15.35) entails v = 0. Thus G' = G',

H = K.

(15.37)

When can we expect that /3 ^ 0 ? Should this be the case then we infer from the identity (15.35) that fx = fo(x) + f2(x)x2

,

(15.38)

where fo(x) = - ^ ,

(15.39)

hi*)^-\fx

(15-40)

We conclude that if fx, given by (15.4), has a different form from that given by (15.38)-(15.40) the quantity /3 and consequently v have to vanish. This is the most common case to be encountered. If, however, fx has the form given by (15.38)-(15.40) we may expect additional set of Lagrange functions corresponding to /3 ^ 0. Two separate cases have to be considered a) /o = 0: (15.39) entail v = 0 and (15.40) implies P = Aexp

{-Jf2{x)dx^,

where A / 0 is some constant,

(15.41)

186

The Inverse Variational Problem in Classical

Mechanics.

(3) /o ? 0: then 0 =~ , Jo

v^O

(15.42)

and /0/2 = ^

•

(15.43)

To summarize: we have usually the case (15.37); this can also happen when the Newtonian Equation (15.4) has the form (15.38), e.g. in case a) for A = 0 and in case fi) when v = 0. However, if the Newtonian Equation (15.4) has the shape (15.38) a new set of Lagrange functions is admissible, for which the Newtonian force has the shape a) fx = i2f2(x),

K = K,

G'=G

0) fx=fo(x)+x2-?-ln\fo(x)\, ox

+ \expt.-f

R = V + K,

f2(x)dx\

,(15.44)

G' = G - ■£ , (15.45) /o

where A nad v are arbitrary functions. We shall see scrutinizing some examples in Section 15.3 that the sets given in a) and fi) are not empty. So far we exploited only the equations (15.30) and (15.33). Now we are going to make use of (15.29) and (15.32) keeping in mind (15.34). We may write the Lagrange function (15.1) in the form L = L!+L2,

(15.46)

where Ia(x,x)

= U(x,x) - ^-(G')2x2

L2{x, x,R, R) = ^-(G')2x2 4ff

+ £-G,

- ^G 2g

+ G'xR + gR2 - KR.

(15.47) (15.48)

Notice that we may write L2 in the form L2 = gq2 - isq ,

(15.49)

q = R+—

(15.50)

G.

All Euler-Lagrange

Equations, forminvariant

under the Galilei transformations. transformations. 187

This simplifying substitution is, however, only valid for g ^ 0. We shall come back to that in Subsection 15.4. The Euler-Lagrange Equations for (15.46) are 2 var va.rflR L L :: G'x G'x + + 2gR 2gR + + G"x G"x2 + +KK = = 00, ,

(15.51) (15.51)

0| ( G ' i292 i0i 12 . i : | ( ^ ) - ^ » . ) 0=' 15 0.(15.52) '"•^slSrj-^+l' '^ ^ " » ^'' - 52 ) v a

t

Hence Hence according according to to d Ld_ ((dLA -^L\ dt\dx)

+

+

9

+G

i

+

(15.47) (15.47) and and (13.50) (13.50) dLl dLj -_ dx

-5-Ms=*-s

+ff

<

0V+

*>-» -

(15.53)

= 0.

If we compare (15.53) with (15.4) we see that L\ is a Lagrange function yielding (15.4). Conversely, let us assume that we have a Lagrange function L1=L Li =l(x,x), Li(x,x),

(15.54)

which gives rise to the Newton Equation (15.4) and a function G'{x) G'(x)

(15.55)

satisfying equations (15.33), keeping the restriction (15.34). Let us introduce the quantity L Z, = L Lix + LL22,

(15.56)

where x2 --^G -^G + + G'xR G'xR ++ gR gR22-kR.- h R . (15.57) \g \ ) 2g \g \ ) 2g Then L yields equations (15.4) and (15.5). To prove our conjecture let us use Euler-Lagrange Equations for L, viz. 2 2 L22 == ^(G'(X)) L — (G'(X)) X

G'x + 2gR + G"x2 + k = 00,,

5 i ++ 2 + 5< 5 1 ,i+! + s))= 0 !(§)-§ ++ | (<** ' **^ "" * =^-

(15.58)

<(15.59) 15 59)

-

188

The Inverse Variational Problem in Classical

Mechanics.

We conclude from (15.58) and (15.59) that d_ fdLi\ ggi !(§)-§-■ dt \ dx J dx = Q

(15.60)

According to our assumption L\ is a Lagrange function linked to (15.4). Consequently (15.60) yields (15.4) too. If we replace in (15.58) x by fx and confront so obtained relation with (15.33), we obtain (15.5), preserving condition (15.34). Formula (15.56) gives rise to all L of the form (15.31), leading to Equations of Motion, whose set of solutions coincides with the set of solutions of equations arising from (15.1), provided (15.34) is observed. In other words, all L and L lead to the same Newton's Equations (equations in the normal form). In Section 10 we learnt how to find the whole set of Lagrange functions V, s-equivalent to the given L. In the present case we have to replace L by Li and V by h\. Then using (10.26) and (15.47) we have Lx == x j"a{Ht(x,n))^*'"* f a{Ht(x,u))d2L^2'u) In

du - S(Jfft) £(#0 (15.61)

= £x- / r cr(ir 1 (x,«))D(ar,u)Ai-E(iri) J = 2— / aiHx (i, u))27(ar, u) du - E(if x ), where where

9 Jc

2 .dU - x— Ex _= a:.din ^ 7 - - TLi = x— (G 9a; ox 4g

2

- {/ - K— G , 2p 0

* - - 5 ~ ^ = -5-5""->'-w -

(15.62)

£&>-.<.,.

Finally by virtue of (15.61) and (15.57), we have X 2 L l = x* [[ a(tfi(*,u)) v(Hi{x,u)) ((~ 0 - i^-((G ' ) ) du - E(ffi)+ 3 7 \ (15.63) (15.63) 1 22 2 2 22 + - -(G'(z)) ^ GG + Gifl + gR RR +—(G'(
U(x,x) = x^ x f

l U U) U)
-^HlHl) ) -^

2 22 2 + j-g(G'(x)) ^(G'(x)) x -^ x -^G. gG.

9

duJ

(15.64) (15.64)

All Euler-Lagrange

Equations, forminvariant

under the Galilei transformations. transformations. 189

We are going to show how to get from (15.63) the a-matrix. start with the most frequent case 13 = 0, 0,

G = G,G,

K = k K = K

= g^0. 9=~9^0. g

Let us

(15.65)

T h e Euler-Lagrange Equations for (15.63) are II = G'x + 2gR + G"x2 +K + K = 0, 0, 11

v a rf fi i L: Z: var

~

[ ,„J&U

var.L:

[ ^ l ) ( ^ -

(G')2\

i

(15.66)

22 (G') (G') L 1].. -

2 r j + ^

j - +

G G + d + Gi+Gik

44g9 G' '"G"^+^-tdY-t9 + YgG=' + Glk = -%" + «*[{%-«£)» +,

=| " where

agi . fd22uU dHt _ .(d X \ dx2 ~ X

2

._&U_ ■ d U _ PGx^ ,GlG, Gl, dxdx 2g +

_dU_ K_G,dx 2g

2 {G') (G')2\ \

n

2g2 ) '

dm = .ffni_ Jk = ±^L

&GG„. __ 9L II±22 _ !E_du_ - ±G±G' ,

dx dxdx 2g dx dxdx 2g Consequently Consequently var x L = <j\il + <J\2lI = 0, var x L = o\\I + <J\2lI = 0 ,

where where

va,iR L = o\iI varj? L = a\il

+ (JiiII + o-nll

^(HO ff=M#i)

=

^)(i-|/r)=o,

-3x- = {^--2«7)> dx

™

(15.67)

dx 2g dx 2g

= 0, —0,

%{\-a{E ))\ ^ i(i-«m)y x

ddeta etu = = a(H cr(Hi). 1). It is obvious t h a t the trace of (15.71) is conserved.

, , ^

(15.68) (15 68)

'

(15.69) (15.69)

(15.70) (15.70)

(15.71) (1571) (15.72)

190

The Inverse Variational Problem in Classical

Mechanics.

In case a) (0 ^ 0), given by (15.44), we have fx=i2f2(x),

K = K,

r

(1573)

)

r

1

G = G' + Aexp I - / f2(x) dx \ = 0 + G'. The Euler-Lagrange Equations of L, given by (15.63), read v a r ^ Z : G'x + 2gR + G"x2 + K = 0

(15.74)

or, due to (15.73), varfi L: II + 0(x - fx) = 0.

(15.75)

By virtue of (15.4) we get from (15.75) var„i:

7/+ |

+

[ ( 2 ^ - (G'f)

2+

ff ov

*(ss*-f)- < H

=

i576

( »

We have further

var.L: ^ ) [^

- ^L j + i ^ - j

X+

= — ( G ' Z + 2ffjR + G"i 2 + K ) +

(15.77)

(G')2\

.^.r/^c/

+xpL_o_GI,i2_eu.G^Q ox ox

2g

dx

2g

Taking into account (15.74) and (15.76) we get from (15.77) ~

G' \2g0T

/

G'0\ ,

+a(Hl) I-^-Il] 2 9

„] (15.78)

=0.

All Euler-Lagrange Equations, forminvariant under the Galilei transformations. 191 191

Finally var.L:

(^.

+

,~ (G' G1

a(H1)\l+

~ ' G'G'0 G'G'p

(15.79) (15-79)

s

,,TT„,G'\ ,G'\

TT TT

n n

From (15.79) and (15.76) we infer that the
w

\ , G'0

„(II\G' „(ll\G'

,G' li x

\ M. V v \ p deta = a(H deter o-(Hi). 1).

G'G'0 \

_ 2L0 v v

/)

(15.80)

I'

(15.81)

The trace of (15.80)

02 1+ + ^^ )) + + ^ ^ is conserved (see Subsection 5.3). It is the right place to give an independent proof of that fe- is a conserved quantity. To show that notice that in cases a) and 0) there exists such a /? ^ 0 which fulfills equations (15.35). It is easy to check that L= L = j—x - i2 2—- ^— fpdx, 3dx , g^O, 9*0, 4ff 2gJ 4g 2g J is a Lagrange function leading to equation (15.35) and consequently to equation (15.4). From Section 10, using also the formula (15.47) for L\ as well as for L we get

d2L, _dU

(G1)2 _V _

. dL a{H)

-W=^-^lg--29-

W

_.p.P ~ a{H)Yg ■■

where a is a Constant of Motion as

dt

dt y ^ dx &c

J y

2g 25 V \

dx 3a;

J /

'

due to (15.35). Notice that from (15.82) follows immediately

(15.82) (15 82) '

192

The Inverse Variational Problem in Classical

Mechanics.

which accomplishes the proof. A similar consideration applied in case (3) (/3 / 0), given by (15.45), mutatis mutantis yields for the a-matrix the same expression, (15.80). To complete the case g / 0 let us inspect the case when g = 0. To this aim we use two Lagrange functions of the set L both with g ^ 0, say ZM = U(1){x,x)-h^R+{&l))'xR 2

+ gR2,

{2)

Z< > = U (x,x) - K&R + (G^)'xR

+ git ,

such that not all U^\ «W and ( G « ) ' are equal to U^\ resp.. If we subtract (15.83) from (15.84) we get 1(3) = fj{i)fa

±)

_ k(3)R

+

(15.83)

2

(15.84) h™ and (G<2))'

(G^yxR,

(15.85)

with the term „gR2" missing. A linear combination with constant coef­ ficients of two elements of L yields again an element of L and a linear combination of equations (15.32) and (15.33) as its Euler-Lagrange Equa­ tions. This means also that they yield the Newton Equations, (15.4) and (15.5). Hence H3' belongs to the set of L with g — 0 and is the desired Lagrange function with the missing term „gR2", unless I/' 3 ' = 3j|. 15.2.2

Case g = 0.

Let us first inspect the case 9 = 9 = 0.

(15.86)

In this case we have to have mandatorily G'?0,

G'?0

(15.87)

to conform to condition (13.50). Now the equations (15.30) and (15.33) have the structure of equation (15.35), viz. G'U + G"x2 + K = 0 ,

(15.88)

2

G'U + G"x + k = 0.

(15.89)

Then either a)

fx=i2f2(x),

« = 0,

G' = A e x p | - / f 2 d x \

,

(15.90)

K = 0,

G' = A e x p l - f f2dx\

,

(15.91)

All Buler-Lagrange

Equations, forminvariant

under the Galilei transformations. transformations. 193

or P)

=i2-^-ln\fo(x)\+fo(x), fx=x — \n\fo(x)\+f0(x), ox

K^O, n^O,

G' = = ~ , ~, (15.92) G' /o Jo

K ^ O , G' MO, G' ==-^-.- f . Jo

(15.93)

Hence G'=nG', G' = fiG' ,

(15.94)

K k == fiK, jUK ,

ft being a number which we may put equal to 1 without resctricting the gererality of our derivations (scaling). We have according to (15.1) and (15.31) L x) - KR++G'(X)±R, G'(x)xR, L = = U{%, U(X,X)-KR

(15.95)

L L= = U(x,x) U(X,X)-KR- KR + G'(X)XR. G'{X)XR.

(15.96)

The Euler-Lagrange Equations read , d2U . dxdx

8U .. dx2 9CJ„+G, dx2

|

d2U ± dx dx

_w__ dx dU dx

= Q

(15.97) (15.98)

and K= 0 G'x + G"x2 + K

(15.99)

as (15.88) and (15.89) coincide. For L(x, x) x)=L-L = L — L == U(x, U(x, x)x) -- U(x, U(x, x)x)

(15.100)

we get, by substracting (15.97) from (15.98), d2L.. dx2

d2L . dxdx

8L _ dx

(15.101)

To find L we use (15.101) and (15.99). We should have d2L „,, .,dG(x) ,8G(x) — =aM^L>.

(15.102) (15.102)

194

The Inverse Variational Problem in Classical

Mechanics.

From (15.102) we get

r

(15.103) l-G' f a(x,u)(x-u)du + h1{x)x + h2(x). L — G'\ a(x,u)(x — u)du + hi(x)x + h2(x). In the following we shall discard the term hi(x)x as a gauge term. If we insert L, given by (15.103), into (15.101) we get the identity rx

g

rx

G'—- I auduoudu-h'2 h2 -= bG"x aG"x22 + +&K, G" I/ audu + G'-—OK ,

(15.104)

where we made use of (15.99). Let us differentiate both sides of (15.104) with respect to x, We obtain -G"ax + G'^-x%(G"x2 + an) +&K)=0. = 0. (15.105) ox ox ox ox We are going to solve (15.105) by using the method of characteristics. If we introduce an auxiliary function $(x,x,a) $(x,x,a)=0,= 0,

(15.106)

relation (15.105) takes the shape - G'x^pG'x^- + (G"x2 + + &K)~
(15.107)

(15.108) (15108)

T G dx T=G

we we obtain obtain a = cG', cG',

(15.109)

where c can be a function of another partial solution of (15.108). For our purpose it is, however, enough to know only one single solution for &, as we learnt in Section 10 how to construct the whole set of s-equivalent Lagrange functions, starting from a given one in (1 + 1) dimensional space-time. So we restrict ourselves to c = 1. We are going to compute h2 using (15.104) and (15.109). We obtain ti2 = = -G'G'c -G'G'c22

-G'K

All Euler-Lagrange

Equations, forminvariant

under the Galilei transformations. transformations. 195

or h2 = -\{G') -l(G')22c22

-GK+

const.

(15.110)

Let us insert (15.109) and (15.110) into (15.103). We get I = \{G') L \{G'fx2x2 2

- {G'fcx (G'fcx - GK GK

(15.111)

or for c = 0 2 2 L == HG') \{G')2Xx2-GK. -GK.

(15.112)

The Lagrange functions (15.111) and (15.112) lead to (15.4). To get other solutions one has to use (10.13)

(T( = )(G )2 ^ ==.(H)g=^)( G')*,

S"

^S ^ ' '

(15U3)

(15.113)

where, according to (10.18) and (10.10), = zUG')2X + GK GK (15.114) ox ox is a Constant of Motion. To get the a-matrix we start with (10.26), (15.100) and (15.95). We get H=^X-L H = ^ X - L

L = x(G')2 I

a(H{x,u))du-Z(H)

+

U(x,x)-KR+G'xR(15.U5)

and varx L: L : 1I + a{H)G'II v a r TL L :: 11 77 = = 0. var^ 0.

= 0,

(15.116) (15.117)

Consequently (I

a(H)G'\

(15.118)

There is still pending the problem to be solved; what conditions have to be imposed upon the system when g = 0 but g / 0. If g = 0 we have to have obligatorily G' / 0. The equations (15.30) and (15.33) read 2 2 K = 0, G'fx+G"x + G"x +K 0,

(15.119)

196

The Inverse Variational Problem in Classical

G'fx + 2gfR + G"x2+K = 0.

Mechanics.

(15.120)

From (15.119) follows /* = /*,o(z) + /*,2(x)± 2 ,

(15.121)

f*fi = - § ,

(15-122)

ha = ~

(15-123)

where

Then from (15.120) we may derive / H , namely / * = /fi,o(z)+/*,2(*)± 2 ,

(15.124)

fRfi = -^(G'fx,o

(15.125)

where + ii),

fR,2 = -^z(G'fx,2 + G"). (15.126) 25 Keep in mind that g is supposed to be different from zero. Let us insert (15.122) and (15.123) into (15.125) and (15.126). We get G' 2gfR,o = ^ - K ,

(15.127)

2 ^ , = ~G>1 ( ^ ) .

(15.128)

If we differentiate (15.127) with respect to x and compare with (15.127) we get G ' ^ £ = -«/*,2

(15.129)

or, taking into account (15.122), ^ ~

= /x.o/ii.2 ■

(15.130)

This is the necessary condition for the existence of a solution involving g 7^ 0. We are going now to show that this is also a sufficient condition.

All Euler-Lagrange

Equations, forminvariant

under the Galilei transformations. transformations. 197 197

To start with, let us assume that (15.130) is, in deed, satisfied. Prom (15.127) and (15.128) we obtain G' «K = -2gf -2gfRRfi ,o + K— ,,

(15.131)

& = & = -2gG' -2gG'I I2& J^-dx. dx .

(15.132)

We have to check whether k is really a number and that (15.131) and (15.132) solve the equations (15.125) and (15.126). The equation (15.125) is trivially satisfied, taking into account (15.131) and (15.122). By virtue od (15.131), (15.132) and (15.130) we have dk

~dfR,o 0r.-dfufi

Tx = -29-dx~+

,

d (G'\

.dfRfi 00-<9//?,o

,2 „-/ H SR,2

G7 = -29~dx--2^-GT

s - * - s - « E ^ J = - ' r s ~ * « - s r ==( u(15.133) -ufl +R

Tx

= - t f ( ^ - / . ^ ) = 0 . Since, according to (15.131), k does not depend on x, it must be a constant. Let us now check on the equation (15.126). Using (15.132), (15.131) and (15.123) we get f & = G" = -2gG" -2gG" jI —"dx~^dx- 2gf 2~gf = ^G"G' ^G"G' - 2~gf 2gfR,2Ra Rft = R>2 G G (15.134) J (15.134) = —fxfiG —fxfiG - 2gf 2
which proves our assertion. For the case j = 0we have L leading to (15.119) or (15.121), viz. (15.135)

L = & X 22-KG. - K G . L=^Y-X

2

Consequently the most general solution for L is a(H(x,v))dv-■£(&)+ L = x{G')2 f[ a(H(x,v))dv - £(ff)+ Jo0 2 2

+ ^-(G') x 43 4c;

- ^G + G'xR + gR gk - kR, --^G 2g

where we used (15.56), (10.26), (15.57) and

g-«*>•*

22

(15.136)

198

The Inverse Variational Problem in Classical

Mechanics.

H is given by (15.114). To summarize: dealing with a Lagrange function L for which g = 0 we are able to construct a Lagrange function L involving g ^ 0, if and only if the function / R is of the form (15.123)-(15.125) and the condition (15.130) is satisfied. It is worthwhile to notice that the case of two particles in a (1 + 1) dimen­ sional space-time as well as the case of a time dependent point particle in a multidimensional Euclidean space was treated extensively by Douglas*' . His approach differs from ours. 15.3

Examples.

In this Subsection we are going to present some examples concering former Subsections and to convince the reader that the sets given in a) and /3) (formulae (15.11)-(15.16) are not empty. Example

of type a) (15.11)

(15.13).

L = \x2 + \R2

(15.137)

i.e. U=\x2,

K = 0,

G'=0,

g = \,

V= 1.(15.138)

The Equations of Motion are var^L : l = x = 0,

(15.139)

var f l L : II = R = 0.

(15.140)

Since

U = \x2 then Hi = \x2 is a conserved quantity. *' J. Douglas, Trans. Am. Math. Soc. 50 (1941) 71.

(15.141)

Examples.

199 199

Because of (15.139) we have the case a), /3 ^ 0 2 x = x2f2x(x), h{x),

/2(z) / 2 ( z ) = 0, 0,

K =00, K = , (15.142)

G' = /3 = A A ## 00 ,,

2

^ =A.

v

First we consider the case g = g = | . According to (15.63) f*

.

.

2 2 2 22i 2 + MR + gR22 L a{\v22))dv-Y,(\x dv - E(ix ) + + i\\A L= =x x I a{\v ) x + XxR + gR Jo

(15.143) (15.143)

The Euler-Lagrange Equations of (15.143) read

Hence

[a{\x22)) + \2]x + \R = 0, [a(±x

(15.144)

Xx + R = 0. 0.

(15.145)

.= (^ + A 2 J), J).

(15.146) (15.46,

deter let a = cr(|i; a(\x22))

(15.147)

-(^

2

A

) +A 2

in accordance with (15.80) and (15.81). To get the case where the term g = 0 we choose in (15.143) two functions i

~ = \x r odv-\Y,+ A—x2+\xR+\R2, L^ L(1) = \x / a d w - | E + — i 2 + A i ^ + | i ? 2 , Jo 2 7o 2

2 2 2 L< = -\x - § i [/" adv+^+^R adw+§£+|i? L{2)) = , , Jo Jo

(15.148) (15.149)

as a and A are arbitrary. We subtract (15.149) from (15.148) and we get L= =x x L

fx 2 a{\v )dv-Y,{\x22)) a{\v2)dv-Y,{\x Jo Jo

A2 22 + —x —x + + \xR. \xR. + 2 2

The Equations of Motion are [o{\x2)2) + \A22]x [cr(|a; ]a; + \R A/i = Q, 0,

(15.150)

Ax = 0.

(15.151)

200

The Inverse Variational Problem in Classical

Mechanics.

Consequently

a=(^2A)+A2 J). For special choice u{\x2)

(15.152)

= —A2 = —1, A = 1 we have

L = xR

(15.153)

and for 5=|,

ff=

(!

(15154)

i)

as well as for

5 = 0,

a=

(°i

< 15 - 155 )

I)

The latter means that the equations (15.139) and (15.140) are interchanged. Notice that (15.137) and (15.153) are not equivalent. Example

of type a)

L = ±ax2e-2ax

(15.11)-(15.13). - be-ax + cRxe-ax

+ \dk2 + —R,

ad - c2 = Ve2ax + 0.

(15.156) (15.157)

Hence U = \ax2e~2ax 9-

id, z

- be'ax ,

K=

G = --e~ax

aM c

, (15 158)

'

-

The Equations of Motion are varx L : I = a{x + ax2)e~ax vaiR L:II=

+cR-ab

-c(x + ax2)e~ax

= 0,

(15.159)

+d{R-—)=0.

(15.160)

The normal form of (15.159) and (15.160) is x = ax2 ,

(15.161)

Examples.

R=—

201 201

(15.162)

c

c resp.. Thus we have the case a) with 0^0. Hence resp.. Thus we have the case a) with 0 ^ 0. Hence __ abd 9 = 9. abd ~ c +a A t_t X„ x c , K=K= G= e , 9 = 3. c a P ~ adA-2 c2 ' £!

(15.163) , 15.163

2 ' We haveValso~ the ad-c Constant of Motion We have also the Constant of Motion

^i = ^

= |

5

(--^e-a-=flx.

We construct L according to (15.6) or (15.63), viz. 2 nrl - r 2

ff*x

L = ^±-e-2"* d

(Hl (x, / aa'Hx (x, v))(x v))(x -- v) v) dv+ dv+ Jo J0 (15.164) (15.164)

2 2 ax ax e- 2 ^ --b-{c + ^-Acc ++ A) \)2x2±e-2ax -{c++\)e\)e~+ + + i_( 9/7 r 2d c a a * * ++ ^R R . + (c ++ \)xRe\)xRe±R22 ++ ^^R.

For a = 1, c2 = a — 1, a > 1, d = 1, A = —c, a = 0 we have L=\x2 + \R2

(15.165)

as in example 1. The (T-matrix for equations (15.159) and (15.160) reads /

^-<*) l )

- < j ( # i ) $ e - Q I + £±Ae"ai

( c ++ A* )-e *—+A *<*

Ac(c+A;

" —

aZ> a

\

\

dXe- * ^ ^

\

V

/a(ffO

1 . +

ad-c2{

e

— 2 1

, Ace" " 1i + ^ ^ 1

[-a(i?x)^ + ^ ] e — \

"*"

)+

d(,ax

\

^

+

c

J ■

X)

/

(15.166) (15166)

202

The Inverse Variational Problem in Classical

Mechanics.

For A = 0 (15.166) reduces to a=('T{f)

$(l-*™«-~)

(15.167)

and for a{H\) = 1 we get an identity matrix as it should be. Example

of type /3)

(15.14)-(15.16).

x2 L = — + ax + \x2x2 + xxR + \R2 .

(15.168)

x2 U = — + ax + \x2x2 , K-0, 2 a » ■

,„ (15.169)

Hence

G=\x2,

g=\,

23 = 1.

The Equations of Motion are v a r x L : I = (1 + x2)x + xR + xx - a = 0 ,

(15.170)

va,TR L: II = xx + R + x2 = 0

(15.171)

or in normal form x = a,

(15.172)

R= -x2 -xa.

(15.173)

Because of (15.172) we have the case (5) with /3 =fi 0. We have x = /o + i2 — In |/o|, X

1 9

fo = a,

K

G=ia:2--i,

,

« = ./,

3 =3 = | ,

tf2

i/2

^7 = — ■

(15.174)

L1=U-±(G')2x2 + ^G=^+ax.

(15.175)

i2 Hi = —— ax

(15.176)

is conserved.

Examples.

203

Let us construct L using (15.63) rX

L=x Jo

n

a{Hl{x,v))dv-Y,{Hl{x,x)) 2

-v Ux

x2-

+ \(x--) K 2

- -x) + (x - -\ xR + \R

aJ

(15.177)

- vR.

The cx-matrix reads ( <*{Hi)-i{x-i)

-a(H1)x

+ x-^

+ ^x(x-^a)

a=\

\ .(15.1'

For 1/ = 0 and u(77i) = l w e get the identity matrix. The standard

model. 2

L = \x

- V{x) + \R2

(15.179)

or U = \x2 - V(x),

K

= G' = 0,

g = \.

(15.179)

The Equations of Motion are vaixL:

dV I = x+-— = 0 ,

(15.180)

v a r f l L : 77 = R = 0 .

(15.181)

Li = | x 2 - V(sc),

(15.182)

Hi = \x2 + V{x),

(15.183)

Since

then

which is conserved. Because of (15.180) we have the case (3 = 0,

G'-G'=0,

K = K = 0,

g = g=\.

(15.184)

L = x f a(Hi {x, v))dv - £(i?i (x, i)) + \R2 . 7o

(15.185)

Thus

204

The Inverse Variational Problem in Classical

Mechanics.

The <7-matrix is a

={

i)

o

For a(Hi) = Hi we have e.g. i=

Yi

+

L = J2^-

15.4

Yv

+

^2'^2-

arctan [ J ^-x

(15J86)

) - In (§x 2 + V) + | i ? 2

(15.187)

Generalization of the set of Lagrange functions L (ad­ mission of Euler-Lagrange Equations not covariant un­ der Galilei transformations).

In the previous Subsections we were interested in such Lagrange functions L, whose Euler-Lagrange Equations were Galilei covariant and led to the same set of solutions as the original equations, we started with. In this Subsection we are going to construct a set of Lagrange functions, say Lex, whose Euler-Lagrange Equations are no longer Galilei covariant, but still lead to the same set of solutions as those Lagrange functions be­ longing to the set L. To start with, let us turn back to formulae (15.46)-(15.50). For g ^ 0 we have L = Li + L 2 ,

(15.188)

where Li(x,x)

= U(x,x) - ~(G')2x2

L2(x,x, R, R) = j-{G')2x2

--^G

+ £-Q, + G'xR + gR2 - KR.

(15.189) (15.190)

If we define the new variables q = R+—G, "9

(15.191)

Generalization

of the set of Lagrange functions

L...

q = R+~G',

205

(15.192)

we may write L2 in a concise form L2{q,q) = gq2-Kq.

(15.193)

With this notation the Lagrange function L, (15.188), can be written as a sum of two Lagrange functions, Li(x,x) and L2(q,q), depending each on distinct variables, the interaction term missing. Thus we may apply to both terms of L the procedure established for one-partice case in Section 10. For L2 we have H2(q,q) =gq2 + nq = ,

(15.194)

= gR2 + G'xR + — {G')2x2 +

KR+-^G.

H2{q,q) is conserved as var, L2 : 2gq + K = 0.

(15.195)

The extended set Lex reads Lex = LX+L2,

(15.196)

where L1=x

rai{Hx(x,v)) Jo

f 5 r \dv2

L2 = 2gq [ a2(H2(q,v))dv Jo

{

^P-) 2g )

dv

~X2(H2(q,q)).

(15.197)

(15.198)

We may replace in L2, given by (15.190), the original quantities G and K by G and k resp., keeping g / 0 constant. Thus we may introduce new variables q and q instead of q and q, viz. q = R+^-G 2^

= q+^f0dx, 2g 7

^=i?+^G'=g+-^/?i, 2ff 2g

(15.199) (15.200)

206

The Inverse Variational Problem in Classical Mechanics.

H2=H2(q,q)=gqi+£q.

(15-201)

H2 is also conserved. The Lagrange function Lex = LX+L2

(15.202)

consists still of a sum of two Lagrange functions depending on different set of variables. While L\ does not change we have L 2 = 2gq~ f Jo

a2(H2(q,v)) dv - X2(H2(q,q~)).

(15.203)

To get the Euler-Lagrange Equations we may treat L\ and Lx separately. We have v a r x Z i : a1(H1)I-a1(H1)^-II

= 0,

(15.204)

where / and / / are given by (15.2) and (15.3) resp., va,rx L2 : a(H2(q, q))j-

[G'x + G"x2 +k + 2gRj = 0 .

Now compare relations (15.74)-(15.76) (similar considerations applied in case 0) yield the same result), viz. G'x + G"x2 + K + 2gR= ^-1

+ (l - ^ - \ 11 = 0.

(15.205)

Hence ~ vaxx L2 :

■ G18 CT2{H2(q,q))—I+ G> f

+a2(H2(q,$)-\l-^)lI

(15206)

G'B\

= 0.

As L\ does not depend on R and R it gives no contribution to the variation with respect to R. We just have va,iR L2 : cr2(H2{q, q)) (G'X + G"x2 +K + 2gR^j = 0

Generalization

of the set of Lagrange functions

L... L...

207

or, taking into account (15.74)-(15.76), v a r f l L 22:

aa2{H (q,q))~-I+ 2(H22{q,q))-^-I+

+a (q^)(^-^jn +
(15.207)

= 0.

Finally we get ~ G18 oi{Hi) + a2(H {H22)—— ) —

~ varxLex:

v

1+ JJ

\ L

C" -a\{H{)-—h -ax(Hi)-—h 29

L

5

(15.208) (15.208)

♦*<*>£ (i-^)]"-o. varflL Z e x : a2(H2)^I

+ a2(H2) (l - ^ - \ II = 0 ,

(15.209)

where H2=H2(q,^). The most general
-ax{H,)%+d2{H2)%

(l - ^ ) \

a= \^

2

a2(H2))if-f

a2{H2){i-^-) ){l-^)

deta = <jlcr {Hl(H (H 1)d12)d 2(H 2).2).

, (15.210) )) ' (15.211)

It is obvious that the cr-matrix (15.210) depends explicitly on R and R and so the Euler-Lagrange Equations, originating from Lex, break the Galilei covariance of the original equations. IfCT2= 1 we get the c-matrix (15.80). For /? = 0 a=\

f/ aMHi) i(ffi) V V

00

((ffff33{{H Haa))--ooii((H Hii))))gg ff

where H2 is given by (15.194).

(H2) 2(aff2»)

\\

, /) '

(15.212)

208

The Inverse Variational Problem in Classical

Mechanics.

It is easy to see that the matrices (15.210) and (15.212) are not the most general tr-matrices which would correspond to the Euler-Lagrange Equations yielding the same set of solutions as the original equations but not conforming to the requirement of Galilei covariance. To get complete set of such Lagrange functions one has to solve the Helmholtz Equations (Subsection 5.1). 15.5

Case of Galilei forminvariant Newton's Equations cor­ responding to Euler-Lagrange Equations which are not Galilei covariant.*) (formulation of the problem)

In Subsections 15.2 and 15.3 the starting point for searching for s-equivalent Lagrange functions was a certain Lagrange function which led us to EulerLagrange equations forminvariant under the Galilei transformation. The set of equivalent Euler-Lagrange Equations encompasses also equations which are not forminvariant under the Galilei transformation. It is obvious that the Galilei covariant Euler-Lagrange Equations put in the normal form (Newton's Equations) must be also forminvariant. No­ tice that these equations in their normal form remain unchanged for each element of the set of all Euler-Lagrange Equations of the s-equivalent La­ grange functions, we derived in Subsection 15.2 and 15.3. These Newton's Equations inserted into any system of Euler-Lagrange Equations yield a system of identities. This implies that in case we are given a system of Galilei covariant normal equations and this system, inserted into an arbi­ trarily chosen system of Euler-Lagrange Equations, discussed in Subsec­ tions 15.2 and 15.3 (see e.g (13.46) and (13.47) or (13.43)), does not satisfy these identities, then no one of Lagrange functions and no one system of Euler-Lagrange Equations, corresponding to these Lagrange functions, is sequivalent or equivalent to a Lagrange function of type (13.43) or Equations of Motion of type (13.46) and (13.47). This in turn implies that no one of these new set of Euler-Lagrange Equations can be forminvariant under the Galilei transformation. To make this plausible let us investigate the following example. Assume we are given the Newton's Equations x = (1 + x)2 = fx(x),

(15.213)

*) The idea comes from Dr. J. Cislo. 1

Case of Galilei forminvariant

Newton's Equations... Equations...

R= =x= = fR(x). fR(x).

209

(15.214)

Obviously these equations are forminvariant under the Galilei transforma­ tion. The most general Lagrange function leading to Euler-Lagrange Equa­ tions forminvariant under the Galilei transformation is given by formula (13.43), viz. L = U(x, U(x,x)±)-KR - KR + + G'XR G'XR + gR gR22

(15.215)

and the Equations of Motion are d2U d2U.. _,> _,——x+—jx ——x+ + G'R1—x 1 ox ox ox

dU nn — =0, =0, ox

, 15.216 (15.216)

G"x2 + + G'x + 2gR + K = 0.

(15.217)

If we insert (15.213) and (15.214) into (15.217) we get (G" + G')x2 + 2G'x + G' + 2qx 2gx + K = 0

(15.218)

ga == KK

(15.219) (15.219)

or = G' G' = = 0. 0. =

Let us insert (15.219) into (15.215). We get L = U(x,x). U(x,x).

(15.220)

Equation (15.216) reduces to d2U dx2

2

d2U dxdx

±

9V_^Q dx

(15.221)

which can be viewed as an equation for U. The solution of (15.221) inserted in (15.220) would reproduce (15.213) but not (15.214). Thus (15.215)(15.217) are incompatible with (15.213) and (15.214). We may, however, find a Lagrange function which will be in agreement with (15.213) and (15.214). This Lagrange function L will be not of the type (15.215). The same concerns (15.216) and (15.217). The Lagrange function we are talking about is for 1 + x / 0 L =

~^TTi

(1

+x

~ \iTi)

+R

+ R( (]nl n 1 1

15 222 ^ - I I + * ! - * ) ■ ( 15 - 222)

210

The Inverse Variational Problem in Classical

Mechanics.

We have var f l : —*— x - x - 1 = 0 , 1+ x 1

var^ : - —

^r

(15.223)

/• l +X j R + r - k T,

1 ^

\ .. , x+

1 + x) 2 I 1 + x 2(l + x)2J „ + 71 +^ x + 25{l7 7+^x)22 + 5 72l+x ^ - + * = °"

(15.224)

Equation (15.223) is Galilei covariant, (15.224), however, is not. (15.223) follows

From

X

(1 + x)2

= 1 ,

hence (15.224) reduces to - l - z + 4r+ii+77-^ + I =0 2 2 2 1+x l+x or

If we compare the structure of the Lagrange function (15.222) with (15.215) we see that the structure is essentially different. The same applies to (15.216)-(15.217) and (15.223)-(15.224). Another example similar to the former one is given by the Euler-Lagrange Equations x + (dx + C0)2 = 0,

(15.225)

R-^r-. -^--CIR\ Cix + Co

v

.„ , ^ + l l =0, \(Cia; + Co)2 J

(15.226) '

belonging to the Lagrange function L = — ln(Ci± + Co) - R{dx

+ Co),

(15.227)

where C\ and Co are some constants. Although equation (15.225) is Galilei covariant, equation (15.229) is not. Of course the Newton's Equations are both Galilei covariant. The structure of the Lagrange function (15.227) differs essentially form that of (15.215).

An Outlook. Application in the Feynman Approach to Quantum Mechanics. Mechanics.

211

We wind up this Section with a statement of J. Cislo.*) We quote his result without proof. Assume that we have a system of equations x = f(x,x), f{x,x),

(15.228)

R = — h{x,x) h(x,x)

(15.229)

in their normal form. They are obviously invariant under the Galilei transformations. Cislo asserts that this system of equations is equivalent to a certain system of Euler-Lagrange Equations of some Lagrange function iff det det

fa fa \ \\ \

0P

da da dt dt

d§_ __ i u dh , d0 dt _ i2 u 9dhi ~, dt 2 9i ~

\\ lf>d£ j lf>d£ 2^ dx I 2^ dx I

= = 00,,

(15.230)

where a

=

d±_ldLdj_ +

=dx--idtdi 2dtdx ~dx

13

dh 9h 1x d dh dx ~2 dtdx = ~dx~ 'Jtdi

\_(djy

4{di) 4\dx)

'

\df\dj_dh dh 4dx dx 4didi

+

((15.231) 15 231)

-

(15.232) (15 232) '

and

— = tJL.

-JL 1L

dt dx dx dt is the substantial time derivative. 16

(15.233)

An Outlook. Application in the Feynman Approach to Quantum Mechanics.

The approach to Quantum Mechanics proposed by Feynman is known as the ,,integral over all paths" method.t) It is based mainly on the following observation. Let us consider a classical system, the development in time of which is described by a Markovian process. The time dependent probability distribution function reads P(h,t P(b,t2\a,h). *' Private communication. t) R.P. Feynman, Revs. Mod. Phys. 20 (1948) 367.

(16.1)

212

The Inverse Variational Problem in Classical

Mechanics.

Formula (16.1) is the conditional probability to find the system under con­ sideration in the state b at the time t2 under the proviso that it was in state a at the time t\. This probability depends neither on the history of the system prior to t\ nor on the history of the time interval (t2 — *i). For such a probability distribution function the Smoluchowski Equation holds, viz. P{b, t2\a,h) = J2P(b>*2|c, t')P(c, t'\a,h),

h < t' < t2 .(16.2)

c

Let us now consider a quantum system described by the transition am­ plitude ip(b,t2\a,ti).

(16.3)

We have P(b,t2\a,t1)

= \
(16.4)

But instead of (16.2) we have 2

p

Qu

(M2|Mi)= £ > ( M 2 M > M ' M i )

,

(16.5)

c

or ip{b, t2\a, tx) = J2
(16.6)

c

The reason for the discrepancy between the classical and quantum descrip­ tion is that in classical picture of Nature the system must pass on his way from a at time t\ to b at time t2 through a certain classical state c at time t', {ti < t' < t2) if its space-time trajectory is continuous. This kind of reasoning can not be accepted in the quantum picture of Nature, as we did perform measurements of the state of the system at time ti and t2 but not at time t'. If we want (16.2) to hold true also in quantum picture of Nature we have to perform a measurement at t' too. For a particle moving in a three dimensional Euclidean space from (16.6) follows by iteration
(16.7)

An Outlook. Application

in the Feynman Approach to Quantum Mechanics.

213

Feynman postulates for infinitesimal At (we put h = 1) = A~lexpliL

ip{x',t + At\x,t)

(x, * ~ * j At\

(16.8)

So the contribution in (16.7) comes from xs close to xs+i, s = 1 , . . . n — 1, because of the strong oscillations of (16.8). Here L = L(x,x) is the classical autonomous Lagrange function. For a nonrelativistic particle (of mass=l) in the potential field V(x) we should have according to Feynman

(sy^M*^-™)*}

(169)

In the last formula we could also write V(x') as well as \{V{x!) + V(x)) instead of V(x). To justify Feynman's idea let us take the simplest model considered above for which H = | p 2 + V(x),

(16.10)

where p is still a classical momentum and start directly from the first principle of Quantum Mechanics, namely p(x',t2\x,t2)

= (x'\exp{-iH(t2-t1}\x.)

,

(16.11)

where (• | denotes the bra-state and | •) the ket-state of Dirac and H = H(x,p),

pi = - i - ^ ,

i = 1,2,3

(16.12)

is the Hamilton operator. We have (x'\exp{-iHt}

\x) = lim / d 3 x n _i . . . / d 3 x x • At ^°J J

(16.13)

■(x'te^^IXn-O-.-lxiXxile-^Mx). Let us use in (16.13) the formula e(a+b)t

_

Um

f e 4 e *n ] .

n—>oo \

/

( 1614 )

214

The Inverse Variational Problem in Classical

Mechanics.

Then, taking into account (16.10) we get (x'\exp{-iHt}\x)

=

/ d 3 x n _ i . . . / d3xi

= Jim Atn=l J

J

A

.{x'|e-^ *e-

iVAt

|xn_1)...|x1)(16.15)

•(x1|e-l^Aie-iVAt|x) = = lim

/ d3xn_i ... / d3xi •

n—►oo J

J

■ (x'| e~'^At

| x n _ ! ) . . . |xi) ( Xl | e - J ^ A t |x) ■ e~iVt

Let us concentrate on (x'|e-^At|x)= /d3p(x'|e-^At|p)(p|x) = ,

(16-16)

2

= yd3pe-l^At(x'|p)(p|x). As 3

(P|x)=(^)

2

(16.17)

we may write 2

(x'|exp{-iyAO|x) = =

{h)

/ d 3 p e x p{- ! ^p 2 - i p( x '- x )} =

= {h)'j *>-{-%{* + &<*■» , (x'-x)2 (Ai) 2

+

, i(x'-x)2\l 2 A< ;j

-(s)7"-{-"(-"^)>

(16-18)

An Outlook. Application in the Feynman Approach to Quantum Mechanics. Mechanics.

215

To evaluate the integral / ^ e x p | - ^ (

P

+

^ )

2

|

(16.19)

we use the saddle point method and get eventually f . A t .2- l , . ,„ . . v a ( i (x'-x)2l (( xx '' || ee xx pp || -- iz— ^ pp 2J| || xx )) = = (( 22^ ^A A 0 t e ^ ) - 2 e x p x| - vp | ^—^ '^ | j

(16.20)

Finally (x',i

-(dsy-K^-^M" ■(ski),-'K»fi^-vw)AI}=

(16 21 (16.21)

-»

= (ds)'-{«(-^)MThe formulae (16.9) and (16.21) coincide. The derivation of (16.21) from (16.11) holds for all Lagrange functions, equivalent to the Lagrange function L = | x 2 - V(x). If we namely multiply I, by a number a then

t

also gets an a-factor; this amounts, roughly speaking, to a change of the value of the Planck constant from h = 1 to h = a. Moreover, adding to L a term dt (we stick to the autonomous Lagrange functions) causes only a change in the phase of the transition amplitude.*' t) *' N.B.: it can happen that this change of the phase can have sometimes some physical quantum effects. t) I enjoyed a discussion with Dr. Z. Haba on these topics.

216

The Inverse Variational Problem in Classical

Mechanics.

Let us assume that the relation (16.8) postulated by Feynman holds true irrespectively of the structure of the classical Lagrange function. Then the following problem arises. The classical Equations of Motion have closer links to Physics than the Lagrange function, as the dynamical behavior of the model is determined by the Equations of Motion. But, as we learnt already, there are often many Lagrange functions yielding the same equivalent Equations of Motion. Which Lagrange function should one then choose? For the simplest (1+1) space-time dimensional case we have L = \x2 - V(x). V(x).

(16.22)

The Lagrange functions s-equivalent to (16.22) are 382L

* /

a(H)\i=u—

du-11(H)

(16.23) (16.23)

=

-C*&L*-™ where = a(z) S4-.W

H = \x2 + V(x), V(x),

^

(16.24)

and S is an arbitrary function. From (16.23) follows H

~Xdx

L

X

dH(H) dH .

JC

dH

±=u

, /"* dE(ff)

dU

+ X

,

dH , ,

or H' = EE((H #).

(16.25) 2

l

For Y,W(H) YS \H) = \H L (D =

i^±* ±

+

we have ±_ | i 2±y2(V{X) v{x? __ a ; ) _ ii_ v{xf

(16 26)

(16.26)

2

for E< Jp){H) >(if) = lntf we have for V(x) V(ar) > >0 Li2) L (2>

= = i ^ ^a ar rcctt aa n {j§^j ( ^ j )

- In (§± 2 + V(x)) V(a:)) .

(16.27)

An Outlook. Application

in the Feynman Approach to Quantum Mechanics. Mechanics.

217

The question arises: are the expressions

A

?"{'{i!lsf-rM)"}

and

A exp ^"■4 «» {' (s (aww1 1+ 5+ ^w s ^w1! v' w-"^
At\x,t)

is concerned? Let us have a closer look to the case (16.26). We have for the canonical momentum w ppW

- = = \x3 +\±^iV{x). = ^ — xV{x).

(16.28)

Equation (16.28) is an algebraic equation of 3-rd degree with respect to x. For V(x) > 0 we may introduce new variables _ gW 1 ± = v^IvT*. r'=§rjgft-vm. " \v\JWV

(16.29) <#>*>

Then (16.28) reads t3+t-r + t - r = 00. .

(16.30)

We are only interested in t real as a function of r. We have dt_ _ _ 1

?dr = T-K^ > 0 1 + Zt2 >

dr l + 3t2 for t real. For r = 0, t (1) = 0, *<2'3) = ±i and dtW dt' 1 )

(16.31) (16-31) '

v

= 1

So t^(r) is a monotonously growing function crossing the point t = 0, r = 0 at angle 45° and for large r behaving like i(1» = (r)3 t(D (r)i

(16.32a)

218

The Inverse Variational Problem in Classical

Mechanics.

or 1) x = (6p( )')s x= (6p (1)

For large p^ H^

(16.32b)

the Hamilton function (V ~ 0) is = \H2~\tfp^ ~ \tf p{l)*

f ~1.36p^ .

For V(x) = 0 the dependence of x on p ' 1 ' is given by (16.32a)-(16.32b). For V(x) < 0 we have = 00, , t33 -- t -- rr =

(16.33)

where t and r are defined as in (16.29) . In this case A _ 1 dr ~ ~ 1 - U2

(16.34)

Forr = Owe 0 w e hhave a v eti(( 11 ) = 0 ;;ii(<2 )2 = ) =l ,li,(i3()3 )== - l1, , *™ &£■ dr

|i . In In the the interval interval

= = -1, ^ r=0

dr

= r=0

2 / T < r << 22 // T 7 3 3V3 V3 the behaviour of t(r) is singular in the sense that to each r belong 3 real values of t. For the domain r > | v / | and r < — \\J\, there is a one to one correspondence between iP-' real and r. Two roots of (16.33) t^ and t (3 ' are in this domain complex conjugate. The shape of i' 1 ' is similar to an stretched capital letter "S". The asymptotic behaviour for large r is given by (16.32a)-(16.32b). If V(x) is bounded from below and V(x) < 0 we may add to V(x) a constant A > \V(x)min\. Then we have the first case considered by us, V(x) = V(x) + A > 0, and the relation between real p^1' and x is one to one. Another example which can be pursued in more detail than the former example is (3+1) dimensional model. We have 2 L L= = §x2 - U(\x\) U(\x\)

(16.35a)

7 ( 2|-U(\x\))+s^. L' = g{\x2 _ J g(^ x|))+eU.

(16.35b)

and

An Outlook. Application

in the Feynman Approach to Quantum Mechanics.

219

Here

F =

' K?) 2 -^ + ^ + f / ( | x | ) -

(i6 36)

-

where Ji = - ( x x p ) i ,

» = 1,2,3

(16.37)

and e is a parameter. To get H' one has to replace p by p and J by Ji = | - [ ( x x p); + (p x x)i]. Both expressions (16.35a) and (16.35b) inserted into Feynman's formula should yield the same result, which is not at all obvious. We leave these questions open for further investigation. If it should happen that for two inequivalent but s-equivalent Lagrange functions the expressions for the transition amplitude should be essentially different i.e. the quantization procedure would be different in both cases, that would indicate that the Equations of Motion do not determine the quantization scheme; the decisive factor would be then the Hamilton functions and the connected with it canonical formalism. A possible way out, which would make sense in Physics, would be to prove that two Feynman transition amplitudes, belonging to two inequiva­ lent Lagrange functions, give in the limit of time tending towards (+) or (-) infinity resp. the same results (up to an inessential phase factor). In other words the use of two different Lagrange functions would not affect the Smatrix. To check whether this can really happen needs a lot of work. The hope that this is feasible comes from Relativistic Local Quantum Field The­ ory; there Wess and Zumino made some calculations based on perturbation theory and the theory of Borchers classes (based on Axiomatic Quantum Field Theory) and got reasonable, promissing partial results.*' I cannot say anything more ... ,,So steh' ich hier, ich kann nicht anders, helfe mir Gott. Amen," 7 '

*' I am thankfull to Dr. J. Wess for a discussion. t) Martin Luter in Worms (1521).

Index

Euler, Theorem of — 2-3 Euler-Lagrange Equations (Equations of Motion) 13, 22-23

Acceleration 4,9 Boost 4 Boost, Infinitesimal — 6 Bracket, Lagrange — 35-36, 43, 67 Bracket, Poisson — 33-36, 43

Feynman's Method of ,,Integral over All Paths" 115, 145, 211-215 Force 8-9, 12-13 Forminvariant Vector 54,57 Forminvariant Tensor 57

Cartesian Coordinate System 1-2, 9, 14 Central Charges 7, 167, 182? Characterises, Method of — 19-22, 47-48 Constants of Motion 14, 191-192 Constraints, Holonomic —, Anholonomic 10

Galilei Group 4, 156, 164, 167, 179-180 Galilei Group, Infinitesimal — 5-7 Gauge Term 39 Hamilton Formalism 66 Helmholtz's Conditions 23-29, 30-32, 208 Helmholtz's Conditions, Douglas Method of Solving — 30-32 Henneaux, Theorem of — 33-39

Degrees of Freedom 10 Dirac Monopole 63, 65 Douglas' Method of Solving Helmholtz's Conditions 32 Dynamics, Analytical — 7 Energy, Kinetic — 12 Energy, Potential — 13 Equations, Equivalent System of — 29, 33 Equations, Linear Partial Differential — 19-22 Equations of Motion 13, 22-23 Euclidean Space 1,4

Kinematics 1-7 Lagrange Lagrange Lagrange 14 Lagrange 33, 39 220

Equations 10-13 Function 13-14 Function, Autonomous — Functions, Equivalent —

Index

Lagrange Functions, s-Equivalent — 33, 39 Lagrange Function, Weakly Symmetrical — 53 Lorentz Force 64-65 Mass of a Particle 9 Matrix, a — (and Its Trace) 40-43 Newton's Laws (Equations) 8-10, 15 ,,Noether Current" 164, 178 Particle, Material Point 8 Poincare's Lemma 16 Poincare's Lemma, Converse of — 16-18 Rotations (SO(3), Group of Proper —) 1-5, 53 Rotations, Infinitesimal — 4-6 Translations, Group of— 1, 4 Translations, Infinitesimal — 6 Translations, Time — 6 Variational Problem, Inverse — 22 Velocity 4, 8-9 Work 12

221