9
Homer Reid’s Solutions to Goldstein Problems: Chapter 9
which is of mixed F3 , F1 type. This is Legendre-transformed into a function of
the F1 type according to
F1 (q1 , Q1 , q2 , Q2 ) = F13 + p1 q1 .
The least action principle then says
p1 q̇1 + p2 q̇2 − H(qi , pi ) = P1 Q̇1 + P2 Q̇2 − K(Qi , Pi ) +
+
∂F13
∂F13
ṗ1 +
Q̇1
∂p1
∂Q1
∂F13
∂F13
q̇2 +
Q̇2 + p1 q̇1 + q1 ṗ1
∂q2
∂Q2
whence clearly
∂F13
= Q1
∂p1
∂F13
P1 = −
= −p1 − 2 Q2
∂Q1
= −p1 − 2 p2
∂F13
p2 =
= Q2
∂q2
∂F13
P2 = −
= −2 Q1 − q2
∂Q2
!
q1 = −
!
!
= −2 q1 − q2
!.
Problem 9.14
By any method you choose show that the following transformation is canonical:
1 !
( 2 P1 sin Q1 + P2 ),
α
1 !
y = ( 2 P1 cos Q1 + Q2 ),
α
x=
px
py
α !
( 2 P1 cos Q1 − Q2 )
2
α !
= − ( 2 P1 sin Q1 − P2 )
2
=
where α is some fixed parameter.
Apply this transformation to the problem of a particle of charge q moving in a plane
that is perpendicular to a constant magnetic field B. Express the Hamiltonian for
this problem in the (Qi , Pi ) coordinates, letting the parameter α take the form
α2 =
qB
.
c
From this Hamiltonian obtain the motion of the particle as a function of time.
We will prove that the transformation is canonical by finding a generating
function. Our first step to this end will be to express everything as a function
Homer Reid’s Solutions to Goldstein Problems: Chapter 9
10
of some set of four variables of which two are old variables and two are new.
After some hacking, I arrived at the set {x, Q1 , py , Q2 }. In terms of this set, the
remaining quantities are
"
#
1
1
1
y=
x − 2 py cot Q1 + Q2
(9)
2
α
α
" 2
#
α
1
α
px =
x − py cot Q1 − Q2
(1 0)
4
2
2
" 2 2
#
1
1
α x
P1 =
− xpy + 2 p2y csc2 Q1
(1 1 )
8
2
2α
α
1
P2 = x + py
(1 2 )
2
α
We now seek a generating function of the form F (x, Q1 , py , Q2 ). This is of mixed
type, but can be related to a generating function of pure F1 character according
to
F1 (x, Q1 , y, Q2 ) = F (x, Q1 , py , Q2 ) − ypy .
Then the principle of least action leads to the condition
px ẋ + py ẏ = P1 Q̇1 + P2 Q̇2 +
∂F
∂F
∂F
∂F
ẋ +
ṗy +
Q̇1 +
Q̇2 + y ṗy + py ẏ
∂x
∂py
∂Q1
∂Q2
from which we obtain
∂F
∂x
∂F
y=−
∂py
∂F
P1 = −
∂Q1
∂F
P2 = −
.
∂Q2
px =
(1 3)
(1 4)
(1 5)
(1 6)
Doing the easiest first, comparing (1 2 ) and (1 6) we see that F must have
the form
α
1
F (x, Q1 , py , Q2 ) = − xQ2 − py Q2 + g(x, Q1 , py ).
(1 7)
2
α
Plugging this in to (1 4) and comparing with (1 4) we find
"
#
1
1 2
g(x, Q1 , py ) = − xpy + 2 py cot Q1 + ψ(x, Q1 ).
(1 8)
2
2α
Plugging (1 7) and (1 8) into (1 3) and comparing with (1 0), we see that
∂ψ
α2
=
x cot Q1
∂x
4
Homer Reid’s Solutions to Goldstein Problems: Chapter 9
11
or
α2 x2
cot Q1 .
(1 9)
8
Finally, combining (1 9), (1 8), (1 7), and (1 5) and comparing with (1 1 ) we see
that we may simply take φ(Q1 ) ≡ 0. The final form of the generating function
is then
"
#
" 2 2
#
α
1
α x
1
1 2
F (x, Q1 , py , Q2 ) = −
x + py Q 2 +
− xpy + 2 py cot Q1
2
α
8
2
2α
ψ(x, Q1 ) =
and its existence proves the canonicality of the transformation.
Turning now to the solution of the problem, we take the B field in the z
direction, i.e. B = B0 k̂, and put
%
B0 $
− y î + x ĵ .
A=
2
Then the Hamiltonian is
1 $
q %2
p − A
2m&
c
"
#2 "
#2 '
qB0
qB0
1
=
px +
y + py −
x
2m
2c
2c
&"
#2 "
#2 '
1
α2
α2
=
px +
y + py −
x
2m
2
2
H(x, y, px , py ) =
where we put α2 = qB/c. In terms of the new variables, this is
(
%2 $ !
%2 )
1 $ !
α 2 P1 cos Q1 + α 2 P1 sin Q1
2m
α2
=
P1
m
= ωc P1
H(Q1 , Q2 , P1 , P2 ) =
where ωc = qB/mc is the cyclotron frequency. From the Hamiltonian equations
of motion applied to this Hamiltonian we see that Q2 , P1 , and P2 are all constant,
while the equation of motion for Q1 is
∂H
= ωc
−→
Q1 = ωc t + φ
∂P1
√
for some phase φ. Putting r = 2 P1 /α, x0 = P2 /α, y0 = Q2 /α we then have
Q̇1 =
x = r(sin ωc t + φ) + x0 ,
px
y = r(cos ωc t + φ) + y0 ,
py
mωc
[r cos(ωc t + φ) − y0 ]
2
mωc
=
[r sin(ωc t + φ) + x0 ]
2
=
in agreement with the standard solution to the problem.