PHY6426/Fall 2007: CLASSICAL MECHANICS
Homework set #11: solutions
Due: Monday, Nov. 19.
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Please help your instructor by doing your work neatly.
1. Consider a particle of mass m which is constrained to move on the surface of a sphere of radius R. There are
no external forces of any kind on the particle.
a) Derive the Hamiltonian of the particle. Is it conserved? [15 ponts]
b) Using the Hamiltonian equations of motion, prove that the motion of the particle is along a great circle of
the sphere. [15 points]
NB: A great circle on a sphere is a circle on the sphere’s surface whose center is the same as the center of the
sphere
Solution:
x = R sin θ cos φ
y = R sin θ sin φ
z = R cos
θ
ẋ = R θ̇ cos θ cos φ − sin θ φ̇ sin φ
ẏ = R θ̇ cos θ sin φ + sin θ φ̇ cos φ
ż = −Rθ̇ sin θ
The Larangian
L=T =
Momenta
mR2 2
m 2
ẋ + ẏ 2 + ż 2 =
θ̇ + φ̇2 sin2 θ .
2
2
∂L
= mR2 θ̇
∂ θ̇
∂L
=
= mR2 φ̇ sin2 θ
∂ φ̇
pθ =
pφ
The Hamiltonian
H = θ̇pθ + φ̇pφ − L =
p2φ
p2θ
+
2mR2 mR2 sin2 θ
H does not depend on φ → pφ is conserved
ṗφ = −
∂H
= 0.
∂φ
By a suitable choice of the initial conditions, pφ can always be made equal to zero. e.g., starting the motion at
θ = 0. Velocity
φ̇ =
pφ
∂H
=
=0
∂pφ
mR2 sin2 θ
mR2 φ̇ sin2 θ = 0.
As θ cannot be equal to zero for any instant of time, we conclude that φ̇ = 0 or φ = const. This equation defines
a great circle.
2
2. Goldstein, Problem 10.5. [30 points]
mω 2
q + α2 cot ωt − mωqα csc ωt
2
cos ωt
1
mω 2
− mωqα
q + α2
=
2
sin ωt
sin ωt
S =
Hamilton-Jacobi equation
∂S
H p=
,q
∂q
1
=
2m
"
∂S
∂q
2
2
2 2
+m ω q
#
=−
∂S
∂t
mω 2
∂S
=
q + α2 csc2 ωt − mωqα cot ωt csc ωt
∂t
2
∂S
= mωq cot ωt − mωα csc ωt
∂q
2
∂S
2
2
= (mωq cot ωt) + (mωα csc ωt) − 2m2 ω 2 qα cot ωt csc ωt
∂q
" #
2
i
∂S
1 h 2 2 2
1
2
2 2 2
+m ω q
m ω q cot2 ωt + 1 + (mωα csc ωt) − 2m2 ω 2 qα csc ωt
=
2m
∂q
2m
1 2 2 2
=
m ω q + α2 csc2 ωt − 2m2 ω 2 qα csc ωt
2m
−
Comparing the first and last lines, we see that they are the same.
Show that S generates a correct solution to the equations of motion:
∂S
= mω (q cot ωt − α csc ωt)
∂q
∂S
β =
= mω (α cot ωt − q csc ωt)
∂α
p =
Solve the last equation for q = q (α, β, t)
β
sin ωt
mω
= q (t = 0) = α
β
= q̇(t = 0) = −
m
q = α cos ωt −
q0
q̇0
β
p = mω
α cos ωt −
sin ωt cot ωt − α csc ωt
mω
β
cos ωt = mq̇
= −mω α sin ωt +
mω
2
3. A tennis ball of mass m is bouncing off the floor. The total energy of the ball is E. The ball is moving strictly
along the vertical. The collision between the ball and the floor is perfectly elastic.
a. Derive the action-angle variables for the tennis ball and determine the period of motion. [30 points]
p2
+ mgz
2m
p
p = ± 2m (E − mgz)
E =
3
The action variable
I
Z 0
Z z0 p
p
J =
pdz =
(−) 2m (E − mgz)dz +
2m (E − mgz)dz
0
z
{z
}
| 0
{z
} |
bottom→top
top→bottom
= 2
Z
0
where
z0
p
2m (E − mgz) = 2
p
2m2 g
Z
z0
0
(z0 − z)1/2 dz =
√
3/2
4 2 √
E
≡ CE 3/2 ,
m g
3
mg
√
4 2 1
√
.
z0 ≡ E/mg; C ≡
3
mg
Expressing the Hamiltonian in terms of J
H =E=
J 2/3
C 2/3
Frequency
ν=
∂H
2
=
∂J
3C 2/3 J 1/3
Period
√
3C 2/3 J 1/3
3 2/3 1/3 1/2
3
34 2 1
1
1/2
√
=
= C C E
= CE
=
E 1/2
T =
ν
2
2
2
2 3
mg
√ E 1/2
= 2 2√
.
mg
b. Suppose that the tennis ball is a quantum-mechanical object. Using the results of part [a], surmise the
dependence of the quantized energy levels on the level number. [10 points]
Assuming the naive quantization rule J = h̄n, we obtain for the energy of quantized levels
J 2/3
En = 2/3 =
C
√ !2/3
0
3 2
h̄2/3 n2 3 m1/3 g 2/3 .
8
The exact quantum-mechanical formula (see, e.g., S. Flugge, Practical Quantum Mechanics I, Problem 40)
gives in the limit of large n
En =
√ !2/3
0
3π 2
h̄2/3 n2 3 m1/3 g 2/3 .
8
By comparing the two results, we see that the correct quantization rule must have been J = πh̄n. The
wave-function of the problem is known as the Airy function.