PHY6426: CLASSICAL MECHANICS HOMEWORK …maslov/classmech/h10_07sol.pdf · PHY6426: CLASSICAL...
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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 [email protected]fl.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) Q = αp x P = βx 2 is this a canonical transformation? A criterion for canonicity is [P, Q] p,x = ∂P ∂p ∂Q ∂x - ∂P ∂x ∂Q ∂p =1 = -2βx α x = -2αβ =1 αβ = -1/2 b) Re-write the transformation as p = Qx α P = βx 2 . Then it is a transformation from (x, Q) to (p, P ) . Hence the generating function is a function of x and Q (type I) p = ∂F ∂x = Qx α P = - ∂F ∂Q = -βx 2 . Integrating the first equation, we get F = Qx 2 2α . Using the canonicity condition α = -1/2β, the equivalent form of F is F = -βQx 2 Then the result for P is obtained as well. c) Harmonic oscillator H = p 2 2m + 1 2 kx 2 . Hamilton’s equations ˙ x = ∂H ∂p = p m ˙ p = - ∂H ∂x = -kx
Transcript of PHY6426: CLASSICAL MECHANICS HOMEWORK …maslov/classmech/h10_07sol.pdf · PHY6426: CLASSICAL...
PHY6426: CLASSICAL MECHANICSHOMEWORK ASSIGNMENT #10: Solutions
Canonical Trasformations and Hamilton-Jacobi Equationdue by 9:35 a.m. Mon 11/05
Instructor: D. L. [email protected] 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)
Q =αp
x
P = βx2
is this a canonical transformation? A criterion for canonicity is
[P, Q]p,x
=∂P
∂p
∂Q
∂x−
∂P
∂x
∂Q
∂p= 1
= −2βx(α
x
)
= −2αβ = 1
αβ = −1/2
b) Re-write the transformation as
p =Qx
α
P = βx2.
Then it is a transformation from (x, Q) to (p, P ) . Hence the generating function is a function of x and Q (typeI)
p =∂F
∂x=
Qx
α
P = −
∂F
∂Q= −βx2.
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 =p2
2m+
1
2kx2.
Hamilton’s equations
x =∂H
∂p=
p
m
p = −
∂H
∂x= −kx
2
or
x =p
m= −
k
mx = −ω2x
Solution
x (t) = x0 cosωt +x0
ωsin ωt,
where x0 and x0 are the initial coordinate and velocity, correspondingly. Choose x0 = 0. Then
x (t) = x0 cosωt
p (t) = mx = −mωx0 sinωt
New coord. and momentum
Q (t) = αp
x= −αmω0 tan ωt
P (t) = βx2 = x2
0cos2 ωt.
2. Goldstein, Problem 9.32 (33 points)
Solution:
H = q1p1 − q2p2 − aq2
1+ bq2
2
F1 =p1 − aq1
q2
, F2 = q1q2.
Calculate the Poisson bracket for F1
[H, F1] =∂H
∂p1
∂F1
∂q1
+∂H
∂p2
∂F1
∂q2
−
∂H
∂q1
∂F1
∂p1
−
∂H
∂q2
∂F1
∂p2
.
Calculate the derivatives
∂H
∂p1
= q1
∂H
∂p2
= −q2
∂H
∂q1
= p1 − 2aq1
∂H
∂q2
= −p2 + 2bq2
∂F1
∂q1
= −
a
q2
∂F1
∂q2
= −
p1 − aq1
q2
2
∂F1
∂p1
=1
q2
∂F1
∂p2
= 0.
Substituting the derivatives into the Poisson brackets, we find
q1
(
−
a
q2
)
+ (−q2)
(
−
p1 − aq1
q2
2
)
− (p1 − 2aq1)1
q2
= 0
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 ofv and w is also a constant of motion. Check the Poisson bracket of F1 and F2
[F1, F2] =∂F1
∂p1
∂F2
∂q1
+∂F1
∂p2
∂F2
∂q2
−
∂F1
∂q1
∂F2
∂p1
−
∂F1
∂q2
∂F2
∂p2
=1
q2
q2 = 1
This bracket is another constant of motion but it is a trivial constant, which is not an algebraic function of q, p.
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