PHY6426: CLASSICAL MECHANICS
HOMEWORK ASSIGNMENT #10: Solutions
Canonical Trasformations and Hamilton-Jacobi Equation
due by 9:35 a.m. Mon 11/05
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Please help your instructor by doing your work neatly.
1. Goldstein, Problem 9.10 (note that x is used for q) (33 points)
Solution
a)
αp
x
P = βx2
Q =
is this a canonical transformation? A criterion for canonicity is
∂P ∂Q ∂P ∂Q
−
=1
∂p ∂x
∂x ∂p
α
= −2αβ = 1
= −2βx
x
αβ = −1/2
[P, Q]p,x =
b) Re-write the transformation as
Qx
α
P = βx2 .
p =
Then it is a transformation from (x, Q) to (p, P ) . Hence the generating function is a function of x and Q (type
I)
Qx
∂F
=
∂x
α
∂F
P = −
= −βx2 .
∂Q
p =
Integrating the first equation, we get
F =
Qx2
.
2α
Using the canonicity condition α = −1/2β, the equivalent form of F is
F = −βQx2
Then the result for P is obtained as well.
c) Harmonic oscillator
H=
1
p2
+ kx2 .
2m 2
Hamilton’s equations
p
∂H
=
∂p
m
∂H
ṗ = −
= −kx
∂x
ẋ =
2
or
ẍ =
ṗ
k
= − x = −ω 2 x
m
m
Solution
x (t) = x0 cos ωt +
ẋ0
sin ωt,
ω
where x0 and ẋ0 are the initial coordinate and velocity, correspondingly. Choose ẋ0 = 0. Then
x (t) = x0 cos ωt
p (t) = mẋ = −mωx0 sin ωt
New coord. and momentum
p
= −αmω0 tan ωt
x
P (t) = βx2 = x20 cos2 ωt.
Q (t) = α
2. Goldstein, Problem 9.32 (33 points)
Solution:
H = q1 p1 − q2 p2 − aq12 + bq22
p1 − aq1
F1 =
, F2 = q 1 q 2 .
q2
Calculate the Poisson bracket for F1
[H, F1 ] =
∂H ∂F1
∂H ∂F1
∂H ∂F1
∂H ∂F1
+
−
−
.
∂p1 ∂q1
∂p2 ∂q2
∂q1 ∂p1
∂q2 ∂p2
Calculate the derivatives
∂H
∂p1
∂H
∂p2
∂H
∂q1
∂H
∂q2
∂F1
∂q1
∂F1
∂q2
∂F1
∂p1
∂F1
∂p2
= q1
= −q2
= p1 − 2aq1
= −p2 + 2bq2
a
q2
p1 − aq1
= −
q22
1
=
q2
= −
= 0.
Substituting the derivatives into the Poisson brackets, we find
p1 − aq1
1
a
+ (−q2 ) −
− (p1 − 2aq1 )
=0
q1 −
2
q2
q2
q2
Jacobi identity (Goldstein, eq. 9.75e) states
[u, [v, w]] + [v, [w, u]] + [w, [u, v]] = 0.
3
If u = H and v and w are the two constants of motion, so that [H, v] = [H, w] = 0 then the Poisson bracket of
v and w is also a constant of motion. Check the Poisson bracket of F1 and F2
∂F1 ∂F2
∂F1 ∂F2
∂F1 ∂F2
∂F1 ∂F2
+
−
−
∂p1 ∂q1
∂p2 ∂q2
∂q1 ∂p1
∂q2 ∂p2
1
=
q2 = 1
q2
[F1 , F2 ] =
This bracket is another constant of motion but it is a trivial constant, which is not an algebraic function of q, p.