p

q

q0

Rotation

p is periodic function of q configuration of the system is the same under q q + q0:p(t+ T ) = p(t); q(t+ T ) = q(t) + q0

eg rotating pendulum

6.3.1 Definition of the AngleAction variables

For either type of motion, define action variable

J =1

2pi

complete period

pdq J() , angular momentum= area in phase space swept out in one period/2pi .

[Note alternative convention without 2pi. Then in following.]Now use J to replace . So (q, p) (, J) via a generating function, F2(q, J) W (q, J) so that

p =W

q, =

W

J,

where is the coordinate conjugate to J the action variable.

p

q

J

77

Hamiltonian is

K(J) = H , ( cyclic) ,

and so equations of motion are

J = H

= 0 = J = constant

=H

J= (J) = = (J)t+ with (J) = H

J.

In one period, the change in is

=

d =

qdq

=

2W

qJdq

=

J

W

qdq as J is constant

=

J

pdq

= 2pi .

Hence if T (or ) is the period we have

2pi = = (t+ T ) (t) = [(t+ T ) + ] [t+ ] = T ,

so

=2pi

T,

thus can be identified as the (angular) frequency of the periodic motion.

So frequency may be obtained without finding a complete solution for the motion,once H is determined as a function of J .

Harmonic Oscillator

Frequency of the SHO using angleaction variables

H =p2

2m+

1

2m2q2 .

78

HJ equation is

1

2m

(W

q

)2+

1

2m2q2 = E ,

so

J =1

2pi

pdq

=1

2pi

W

qdq

=

2m

2pi

dq 12m2q2 q =

2

m2sin

=

.

So

H = = J = HJ

= ,

as required.

6.4 f dof Conservative System

Need separable HJ equation W =

i Wi(qi, ~) as canonical transformation is thenseparable

pi =Wiqi

= pi = pi(qi, ~) .

so if we assume that the orbit equation of the projection of the system point on the(qi, pi) plane is periodic (ie libration or rotation), then we can define action variablesas for the 1-dof case. Hence, as previously

i =2pi

Ti.

Note that this does not imply that the system motion is periodic unless the fre-quencies are commensurate. (ie ratio of any two frequencies is a rational number, cfclosed Lissajous figures and open Lissajous figures.)

Define action variables

Ji =1

2pi

pi dqi , (no sum) .

Formal variation, not necessarily actual variation as in 1-dimensional case

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If a qi is cyclic then pi = constant orbit is horizontal straight line motion of arbitrary period

Since qi always an angle, then

Ji = pi , cyclic variables qi .

In general

Ji =1

2pi

Wi(qi, ~)

qidqi Ji(~) ,

are independent (from independence of separate variable pairs (qi, pi)) and somay be taken to be the new constant momenta.

Thus we may write

W =

fi=1

Wi(qi, ~J) , H = H( ~J) K .

Conjugate angle variables are

i =W

Ji=Wi(qi, ~J)

Ji,

and satisfy equation of motion

i =H

Ji= i( ~J) , or i = it+ i .

So are the (angular) frequencies of the motion. We see this from how the qsdepend on the s.

The change in i due to small changes in qs are

i =j

iqj

qj , cf virtual work with no time Goldstein p459

=j

2W

Jiqjqj

=

Ji

j

Wjqj

qj

=

Ji

j

pj(qj , ~J)qj .

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So total change in i when each qj is taken through (eg) mj cycles (libration orrotation) is

i =

Ji

j

mj

pj(qj , ~J) dqj integration over mj cycles

= 2pi

Ji

j

mjJj

= 2pimi ,

or

~ = 2pi~m ,

(so again, as before, we may identify i as angular frequencies).

6.4.1 Libration

Under a change ~ = 2pi~m, qs and ps all return to their initial values, ie qj is a

periodic function of ~.

qj =

k1=k1=

kf =kf=

aj~k( ~J) ei

~k~ ~k

aj~k( ~J) ei

~k~ ,

or

qj(t) =~k

aj~k( ~J) ei

~k(~t+~) .

Note that qj is not a periodic function of time unless js are commensurate. For if

qi(t+ T ) = qi(t) then ei~k~T = 1 or need

~T = 2pi ~N .

This can be solved if the ~s are commensurate, then

j = Ni orij

=NiNj

T = 2pi.

6.4.2 Rotation

Under one complete cycle (t t+ 2pi/j of the pair (qj , pj)) qj is increased by itsperiod q0j , j is increased by 2pi = qj (j/2pi)q0j returns to its initial value andso is periodic in ~. So as before we have

qj(t) =q0j2pi

(jt+ j) +~k

aj~k( ~J) ei

~k(~t+~) .

81

6.5 The Kepler Problem using ActionAngle vari-

ables

V (r) = kr, (k > 0, bound state) .

1. k = GmM Gravitational (planets)

2. k = e1e24pi

Coulombe1 = e2 = e Hydrogen atom. With quantisation gives the Bohr atom.

In either case

~r = (r, 0, 0) , ~r = (r, r, r sin ) .

and

L = T V=

1

2m(r2 + r22 + r2 sin2 2) +

k

r.

Giving

pr =L

r= mr

p =L

= mr2

p =L

= mr2 sin = const. = Lz as is cyclic .

Thus

H = T + V

=1

2m

(p2r +

1

r2p2 +

1

r2 sin 2p2

) kr

= E = |E| < 0 as bound state .

The HJ equation is

1

2m

[(W

r

)2+

1

r2

(W

)2+

1

r2 sin2

(W

)2] kr= |E| .

Seek separated solution

W =Wr(r, ~) +W(, ~) +W(, ~) .

82

As is cyclic, we have p = W/ = const. = . This gives(Wrr

)2+2r2

= 2m

(|E|+ k

r

)(W

)2+

2sin2

= 2

W

= .

What does represent?

~L = (r, 0, 0) (mr,mr,mr sin )=(0, p

sin , p

),

giving

~L2 = p2 +p2

sin2 =

(W

)2+

2sin2

= 2 ,

so AM magnitude is constant.

6.5.1 Actionangle variables

Jr =1

2pi

prdr =

1

2pi

Wrr

dr =1

2pi

2m

(|E|+ k

r

)

2

r2dr

J =1

2pi

pd =

1

2pi

W

d =1

2pi

2

2sin2

d

J =1

2pi

pd =

1

2pi

W

d = .

Integrating

Jr = + k2

2m

|E|J = J = ,

Trick for Jr:[See eg Goldstein for proof using contour integrals.]

83

Integral easier to find if square root in denominator.

Jr|E| =

m

2pi2

r+r

rdr2m|E|r2 + 2mkr 2= 1

2pi

m

2|E| r+r

rdr(r r

)(r+ r)

,

(turning points on ellipse perihelion and apehelion are at r, r+ when pr vanishes).

Setting

r = 12(r+ + r) k

2|E| , =12(r+ r) .

Changing variables

r = r + cos ,

gives (r r)(r+ r) = 2 sin2 and r(pi) = r, r(0) = r+, so we have

Jr|E| =

1

pi

m

2|E| pi

0

(r + cos )d = k2

m

2

1

|E| 32 ,

and thus

Jr = const. +k

2

m

2|E| .

To determine the constant, consider a circular orbit, r = a, then pr = 0. So as

H =1

2m

(p2r +

2r2

),

then from Hamiltons equation

pr = Hr

= 2

mr3+

k

r2= 0 = a =

2

mk, and E = 1

2

mk2

2.

As in this case Jr = 0 (as pr = 0) then substituting E in the equation for Jr givesthe constant as .Trick for J:Use fact that motion is in a plane ~L = const. ([H, ~L] = 0 or d~L/dt = 0 or Lz = const.,z arbitrary.)

So in cylindrical coordinates (r, , z), z axis along ~L,

~L = (r, 0, 0) (mr,mr, 0) = mr2~ez ,

84