T'RAN9FOLMATION

396
a rcsult
explicitly in P, 1s Jacoben other two forIns
cxpresscd of
But neither q nor D cAn be However, as the
lbe written.
generating function may not are invertible, and transfortnatiz,
/ = l 0 , the transfornation
2(e- 1)(e+ 1)
4(e- 1)2(e+ 1)2 + P? VP'+ 4(e-1) (e?+ 1)2|
4(e+ 1)?
the transformation Q= V2q e cos p; P = sinp
Fxample 6.6.3. Show that function.
canonical. Find generating
twhcre a is a constant, is :
transformation will be canonical if the expression Pdo- d
Solution. The given
an exact differential. Now,

PdQ- pdg = V2qe sin p|(24)-1/e cospd - 2qe sin pdp- pd

sinp cosp dq 24sin pdp - pdg
=
o l6q sin2p - pajdg + l2qsin 2p padp =dF.
transformation is canonical. From the given
Since it is an exact differential, the given
transformation equations, we have

Q?
COS p ’ sinp = 2qe2a
V2gea

Thus the generating function is given by

1 Q2 2qe20 Q?
F(g, 2) = G9 sin2p- pq = qtan
2ge2a Q

6.6.4. Let us consider the transformation (g, p) ’ (Q, P) as Q = logsinp

Lq
qcotp. Show thut it is cunoical. Obtain the four major types of generating functions
ussvciated with this transfornation.
Solution. First let us check that the transformation is canonical. We have
pöq - P 8Q = poq -q cotp cotp öp - 6q
(p+ cotp)6q - qcotp &p
o{q(p+ cotp)} = u perfect different ial
 MECHANICS 397
A T'EXT BoOK OF

Ths the given transformation is canonical aud the generating function is

F= qp +qcotp = qc o P. t +P.
Fromthe transformation cquations, we have

’ sinp = qe
cosp = V1 - (e20 ’ cotp = /1 - gPe22/(ge).
Therefore, the generating function Fi(g, Q) is given by
Fi(g,) qp +q cotp = qsin(ge)+ e-20 - g1/2.
ASubstitution of Fi(g, P) into Eq. (6.21), we have

p=
g
cos
(V1-geo) P=
=Ve-20- ².
Next, let s find F2(g, P). From Eq.(6.22), we have
Fz(g, P) = Fi(g, Q)+PQ= qp+ q cotp +P Q.
Evaluating each of these terms, using (q, P) as variables, we see that
p= gtan-1 9
p' QP=-Plog va? +P2, qcotp= P
whích leads to

Falg, P) =gtan+P1-log Vg +P.

As a check, we see for Eq.(6.23) that

p= = tann-14 =-log .q²+ P2.

ASinilar analysis using Eq.(G.24) yiclds the third generating function, namely

Fs(Q,p) = F(g,Q)- Pq = qcotp = e"cosp.

We find that, in accordance with Eq. (6.25),

P= =e-sinp.
398
"TRANSFORMATiON THEsc
Finally, from Eq.(6.26), we obtnin
Fa(p, P) = Blg, P) - py = qp +q cotp +PQ- I
QP+q cotp =Plog()+P.
Acheck of Eq. (6.27) shows that
OFA
Q= aF, =log(): -Ptanp.
q=aP
In all cases, the generating function is not an explicit function of time. Therefore.
see fromn Eq.(6.23), for example, that
K= H+ SF2(g, P, t) ’ K(Q,P,t) = H(9,p, t),
9t
regardless of which particular dynamical system is under cons ration.
Example 6.6.7. Under what condition will the transformation P= ap-+bq,Q .
would be canonical, where a, b,c, d are constants?
Solution. It would be canonical if [Q,P) = 1, when it is given that (g, P) = 1. To
this, we calculate (Q,P|. Now, see
IR,P] =(p+ d¡; ap +b] = cblp,gl + dalg, p) = ad - bc.
Bor ,P to be unity we must have ad- bc = 1.
Example 6.6.8. Show that the transformation q = V2P sin Q;p= V2P cosQ i:
canonical by using Poisson brackets.
Solution. The inverse of the given transformation is given by P = (p + ) and
tan Q= Y. Now, using the definition of Poisson bracket, we have [Q,Ql =
[P.Pl=0
and

[Q, P) = cos?
(cos² Q)a
(1+5)cosQ=(1+tan Q) cos Q=1.
Hence the given transformation is canonical.
Example 6.6.9. Using the fundamental Poisson brackets, find the values of m and
n. so that thetransfornmation Q = q" cos np and P = g sin np may be a canoncai
transformation. Obtain the generating function.
Solution. From the given transformation equations, we get
= mg" cos Np,
n
-ng" sin np;
mam- sin np, =nq" cos np.
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TEXT BOOK OF MECHANICS
A

(2, P] =1. Therefore,

Thegiven transformation is chnonical if (Q. Q] (P, P=0and
[Q.P= 1

cos np) (ng" cos np) -- ng" sin np)(mym sin np) =l
mnq?m=cos' np +sin² np = mnqm -l = 1.
which
coefficient of on both sides
dimensionless. Equating the = 2.
The right hand side is 1/2, and then mng = 1 gives n
if we take 2m -l=0, i.e., m =
is possible only transformation equations become
With these values of m and n, the '/2 sin 2p.
Q= g'/ cos 2p and P= function must be of the third type. n
and Q, the generating
Since q is a function of p using the second equation of Eq.(6.27), we
get
function,
order to find generating
,Q) =-g=-Q' sec?2p
’ Fsp. Q) = -/Qsec?2p dp = -; tan 2p.
2p, is
transformation Q = Vg cos 2p; P = Vq sin this
Example 6.6.10. Show that
the
of generating functions associated with
major types
canonical. Obtain the four
transformation.
Poisson bracket, we have Q, Q = [P, Pl =0
definition of
Solution. Now, using the
and 1
sin 2p
cos 2p- 2Vq cos 2p + 2/q sin 2p
1

2/a 2/4
Q,P) = ag ôp
cos 2p + sin 2p =1.
function, using
transformation is canonical. Inorder to find generating
Hence the given
(6.21), we get
the secondequation of Eq.
SF sin 2p = Vg-Q2
2 (9,2) = -P=-v-
=cos()-V-2,
F4,Q) =- /Va-Q2 dQ
Eq.(6.22), we have
Next, let us find Flg, P). From
-vq-Q +PQ.
Fzlg, P)= Fi(4,4) +PQ=2 cos
402
TRANSFORMATION
that
using (g, P) as vaariables, we sce
EvaBuating cach of thesc terms,
sec2p
P=Qtan 2), g = yQ? +P?
which lecads to
Vãcen.
cos-(cos 2p)- ;vacos 2pv4-qcos? 2p +P
1
Fz(g,P) = 2

4
sin 4p + P/q cos 2p.

A similar analysis using Eg.(6.24) yields the third generating

function, namely

- pq.
Fs(Q.p) = F(g,)- pq =
Evaluating each of these terms, using (g, P) as variables, we see that
Q= Pcot 2p, q = VQ?+ P2 = P cosec2p
which leads to

Fs(Q.p) = sec 2p cos

sec 2p. sec 2p - Q2- pQ sec 2p.
Finally, from Eq.(6.26), we obtain
FAlp, P) = Fz(g,P) - pq = -4 sin 4p + P¡ cos 2p
P
-;+PVP
2 cosec 2p cos 2p.
In all cases, the generating function is not an explicit
function of time. Therefore, e
see from Eq.(6.23), for example, that

K= H +OF2(4, P, t) ’ K(Q, P, t) = H(g,p, t),

regardless of which particular dynamicalsystem is under consideration.
6.6.3 Lagrange Bracket Form
Here we shall give anotlher form to the conditions that the
(0. P) nay be contact transfornation. Let the transformation (g.P)
P(g,p, t); i = 1,2,... n be canonical. Then transformation Q; = O:la,p, );K*

Pióq1 - P;6Qi = a perfect differential.

Example 0.0.12. Show hat the transformation Q=/2g cosp; P== V2qk sinp is
canonical.
Solution. From the given transformation equations, we have,
k
+ y2gk cos pdp; 8Q = V2qk sin póq - |2k4 sin pöp
1
dP
V2q-sin pdq
k 1 2q
SP = sin põq + V2gk cos pöp; dQ = V2qk cospdp + V sin pdp
V2q
1
dP5Q = -sin pdq + V2kcos pdp 2gk sin póq - /sin póp)
1
sin p cos pdðg - sinpdgÑp + cos´ pdpog - 2q sinp cos pdpbq.
24
-sin pÑg +2qk cos pópV2ak cos pdp+ 24
1
6PdQ =
2q in pßy
1
-sinp cos pdpb -sinpdpQq + cos pdpQp - 2q sinp cos pdpåp.
2
To pruve that the given transformation is canonical, we are to show the Eq.(6.30). Now

aPSQ 6PAQ = -sin" pdgóp - õs dp +cos papôg - bydg

/6.8 Equation of Motion
Theorem 6.3. (Poisson's first theorem ):The total time rate of evaluation of ar;
dynamical variable ulg,p, t) is
du Ju
dt
= (u, H]+ (6.37

which is the equation of motion of u in terms of the Poisson brackets.

Proof: Using the Hamiltons's.
IMIM: canonical equations of motion (6.21), the total ti
derivative of the dynamical variable u= ulg.p, t) is given by
du Ôu Ou u
dt t
k=l

Ou oH Ou aH] Ou Qu
+
Ot
= (u, H]+ at
This is the generalised cquation of motion for the arbitrary function u in the PoisOr
bracket form, where H is the Hamiltonian of the system. In particular,
 Using Charpit's method, we get.
mdq dt du
dp
-mwlg 1 0
mdq = dt

md dt: u = C2
VG- m²ug?
idq
= w|dt; u= C2; c=

log +y/1-S]=iwt
iut ++C3;c; U F C2

imwg + v - m²wg? | = wt + c3; u=2

log C1
loglp + iwg] - iwt= = u(g,p, t).
/DEDUCTION 6.8.1. Canonical equations of motion in Poisson bracket for
The Poisson brackets of the dynamical variables qk, Pk
given by
With the Hamiltonian H. an

j=l
OH OH
-öjk k; from Eq. (6.21)

j=1
aH OH
=Ps; from Eq. (6.21).
Thus, if, in Eq. (6.37), u is replaced by qk and Pk,
variables qk and Pk do not depend explicitly on time Eq. (6.37) gives when thedynamta
t
dq;
dt = (gj, H); |= dpj
dt
(6.39)
Equations (6.39) can thus be referred to as the
bracket form. Also, when the Poisson canonical eguation of motion in ro
lincar nomentum is conserved, which bracket Pj, H) vanishes, then P; =constant, ie.
implies that corresponding co-ordinate by
 418 TRANSFORMATION
Solution. The Hamiltonian of the systenm is given by H(q,p) =p"/2-1/2. Tlherekae
Hamiltonian is
using E. (G.37), the total tinoderivative of the
+ [H, H| =Dt
dt
motion. Als0, using E. (6.3T), tor the given dynuni.
1hus, H is n constant of the
varinbl F,thctotal timc detivative is given by
OF
OF
dt
+ (F, H| Dt
2 0.
-H+5- 2g
Thus, F is a constant of the motion.
system the lHamillonian is given by II =
XAxamplc 6.8.4. For a dynamical corTesponding gencnlid
generalised co-ordinatcs and
´a), uherc q's and p's are constant. Show that IË =
is n
2P3 13P2 und E
momenta respcctively and u
P1 sin(put), are constants of molion. is
uqË cOs(/t) -
total time derivative of the given ilnmiltonian
Solution. Using Eq. (6.37), the
dH 9H = 0
+ [1, H]=
motion. Using Eq. (6.37), for the given dynanical variat.i
Thus, I is a constant of the given by
is
F,the total tine derivative
far, H
dFi
dt at +|Fi, I)=LDDp: Op, Oq.
= -(0:p1 -0'a)+ (p3P2 + 43 q2)
+(-2"P3 - 42us) = 0.
for the given dynanical
Thus, F, is a constant of the notion. AlsO, using Eq. (6.37),
given by
variable F2, the total time derivative is
dFi OF2 [OF; OH
dt
+|Fz, H] = +

-'n sin(ut) - p: cos(ut) + cos(1ut) -

+fo-p-0-p'a] +(o
Thus, P2 is a constant of the notion.