PHY4222/Spring 08: CLASSICAL MECHANICS II
HOMEWORK ASSIGNMENT #3:
due by 12:50 p.m. Wed 01/30/08
Instructor: D. L. Maslov
maslov@phys.ufl.edu 392-0513 Rm. 2114
Please help your instructor by doing your work neatly.
Every (algebraic) final result must be supplemented by a check of units. Without such a
check, no more than 75% of the credit will be given even for an otherwise correct solution.
Equations of motion also have units!
1. 3. A two-dimensional (2D) harmonic oscillator is described by the Hamiltonian
H=
1
1
p21 + m2 ω 2 x21 +
p22 + m2 ω 2 x22 ,
2m
2m
where x1,2 are the Cartesian coordinates of a particle and p1,2 are the conjugated momenta. Property a) in
Problem TM7.30 (p. 284) states that for any function g of the coordinates, momenta, and time
∂g
dg
=
+ [g, H],
dt
∂t
where [g, H] is the Poisson bracket. If g does not depend on time explicitly, i.e., ∂g/∂t = 0, this property reduces
to
dg
= [g, H].
dt
This equation says that g is conserved, if
[g, H] = 0.
Using property (1), show that
a) the z−component of the angular momentum,
Lz = x 1 p 2 − x 2 p 1
of a 2D harmonic oscillator is conserved [15 points];
[Lz , H] =
2
X
∂Lz ∂H
∂Lz ∂H
−
∂x
∂p
∂pi ∂xi
i
i
i=1
∂Lz
∂Lz
∂Lz
∂Lz
= p2 ;
= −x2 ;
= −p1 ;
= x1
∂x1
∂p1
∂x2
∂p2
∂H
∂H
∂H
∂H
= mω 2 x1 ;
= p1 /m;
= mω 2 x2 ;
= p2 /m.
∂x1
∂p1
∂x2
∂p2
[Lz , H] = p2 (p1 /m) + x2 mω 2 x1 − p1 (p2 /m)
−x1 mω 2 x2 = 0.
b) all components of a 2×2 matrix
Aij =
1
pi pj + m2 ω 2 xi xj , i, j = 1, 2
2m
(1)
2
are conserved [19 points].
Diagonal elements:
A11 =
1
p21 + m2 ω 2 x21
2m
Notice that H = A11 + A22 . Hence
[A11 , H] = [A11, A11 + A22 ] = [A11 , A11 ] + [A11, A22 ].
A11 and A22 depend on different sets of coordinates and momenta. Therefore, [A11, A22 ] = 0. A Poisson bracket
of any function with itself is zero. Therefore, [A11 , A11 ] = 0 and [A11 , H]. Same for [A22 , H].
Off-diagonal elements:
A12 =
1
p 1 p 2 + m 2 ω 2 x1 x2 .
2m
2
X
∂A12 ∂H
∂A12 ∂H
[A12 , H] =
−
∂x
∂p
∂pi ∂xi
i
i
i=1
1
∂A12
1
∂A12
∂A12
∂A12
= mω 2 x2 ,
= mω 2 x1 ,
= p2 /2m;
= p1 /2m.
∂x1
2
∂x2
2
∂p1
∂p2
Using expressions for the derivatives of H, calculated earlier, we have
[A12 , H] =
1
mω 2 x2 (p1 /m) − (p2 /2m) mω 2 x1
2
1
+ mω 2 x1 (p2 /m) − (p1 /2m) mω 2 x2 = 0.
2
Same for A21 .
4. Bonus: 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 points]
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 =
mR2 2
m 2
ẋ + ẏ 2 + ż 2 =
θ̇ + φ̇2 sin2 θ .
2
2
3
Momenta
∂L
= mR2 θ̇
∂ θ̇
∂L
= mR2 φ̇ sin2 θ
=
∂ φ̇
pθ =
pφ
The Hamiltonian
2
H = θ̇pθ + φ̇pφ − L =
pφ
p2θ
+
2
2mR
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 =
∂H
pφ
=0
=
2
∂pφ
mR 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.
7-17.
y
C
A
h
θ
qθ
mg
B
Using q and θ (= ω t since θ (0) = 0), the x,y coordinates of the particle are expressed as
x = h cos θ + q sin θ = h cos ω t + q ( t ) sin ω t ⎤
⎥
y = h sin θ − q cos θ = h sin ω t − q ( t ) cos ω t ⎦⎥
(1)
x& = − hω sin ω t + qω cos ω t + q& sin ω t ⎤
⎥
y& = hω cos ω t + qω sin ω t − q& cos ω t ⎥⎦
(2)
from which
Therefore, the kinetic energy of the particle is
T=
1
m x& 2 + y& 2
2
(
)
1
m h 2ω 2 + q2ω 2 + q& 2 − mhω q&
2
(3)
U = mgy = mg ( h sin ω t − q cos ω t )
(4)
=
(
)
The potential energy is
Then, the Lagrangian for the particle is
L=
1
1
1
mh 2ω 2 + mq2ω 2 + mq& 2 − mgh sin ω t + mgq cos ω t − mhω q&
2
2
2
(5)
Lagrange’s equation for the coordinate is
&&
q − ω 2 q = g cos ω t
(6)
The complementary solution and the particular solution for (6) are written as
qc ( t ) = A cos ( iω t + δ ) ⎤
⎥
⎥
g
qP ( t ) = − 2 cos ω t ⎥
2ω
⎦
(7)
so that the general solution is
q ( t ) = A cos ( iω t + δ ) −
Using the initial conditions, we have
g
2ω 2
cos ω t
(8)
g
⎤
=0 ⎥
⎥
⎥
q& ( 0 ) = − iω A sin δ = 0
⎦
q ( 0 ) = A cos δ −
2ω 2
(9)
Therefore,
δ = 0, A =
g
(10)
2ω 2
and
g
q (t) =
2ω 2
( cos iωt − cos ωt)
(11)
( cosh ωt − cos ωt )
(12)
or,
q (t) =
g
2ω 2
q(t)
g
2ω 2
t
In order to compute the Hamiltonian, we first find the canonical momentum of q. This is
obtained by
p=
∂L
= mq − mω h
∂q&
(13)
Therefore, the Hamiltonian becomes
H = pq& − L
= mq& 2 − mω hq& −
1
1
1
&
mω 2 h 2 − mω 2 q2 − mq& 2 + mgh sin ω t − mgq cos ω t + mω qh
2
2
2
so that
H=
1 2 1
1
mq& − mω 2 h 2 − mω 2 q2 + mgh sin ω t − mgq cos ω t
2
2
2
(14)
Solving (13) for q& and substituting gives
H=
p2
1
+ ω hp − mω 2 q2 + mgh sin ω t − mgq cos ω t
2m
2
The Hamiltonian is therefore different from the total energy, T + U. The energy is not
conserved in this problem since the Hamiltonian contains time explicitly. (The particle
gains energy from the gravitational field.)
(15)
7-25.
z
r
z
m
y
θ
x
In cylindrical coordinates the kinetic energy and the potential energy of the spiraling
particle are expressed by
1
⎤
m ⎡ r& 2 + r 2θ& 2 + z& 2 ⎤⎦ ⎥
2 ⎣
⎥
⎥
U = mgz
⎦
(1)
i.e., z& = kθ& ⎤
⎥
⎥
r = const.
⎦
(2)
T=
Therefore, if we use the relations,
z = kθ
the Lagrangian becomes
L=
⎤
1 ⎡ r2 2
m ⎢ 2 z& + z& 2 ⎥ − mgz
2 ⎣k
⎦
(3)
⎡ r2
⎤
∂L
= m ⎢ 2 + 1⎥ z&
∂z&
⎣k
⎦
(4)
Then the canonical momentum is
pz =
or,
z& =
pz
(5)
⎡ r2
⎤
m ⎢ 2 + 1⎥
k
⎣
⎦
The Hamiltonian is
H = pz z& − L = pz
pz
⎡ r2
⎤
m ⎢ 2 + 1⎥
⎣k
⎦
−
pz2
⎡ r2
⎤
2m ⎢ 2 + 1⎥
⎣k
⎦
+ mgz
(6)
or,
H=
1
2
Now, Hamilton’s equations of motion are
pz2
⎡ r2
⎤
m ⎢ 2 + 1⎥
⎣k
⎦
+ mgz
(7)
−
∂H
= p& z ;
∂z
∂H
= z&
∂p z
(8)
so that
−
∂H
= − mg = p& z
∂z
∂H
=
∂p z
pz
⎡ r2
⎤
m ⎢ 2 + 1⎥
⎣k
⎦
= z&
(9)
(10)
Taking the time derivative of (10) and substituting (9) into that equation, we find the
equation of motion of the particle:
&&
z=
g
⎡r
⎤
⎢ k 2 + 1⎥
⎣
⎦
2
It can be easily shown that Lagrange’s equation, computed from (3), gives the same result
as (11).
(11)