# Solutions to Problems in Goldstein Class ```Solutions to Problems in Goldstein,
Classical Mechanics, Second Edition
Homer Reid
June 17, 2002
Chapter 8
Problem 8.4
The Lagrangian for a system can be written as
L = aẋ2 + b
p
ẏ
+ cẋẏ + f y 2 ẋż + g ẏ − k x2 + y 2 ,
x
where a, b, c, f, g, and k are constants. What is the Hamiltonian? What quantities
are conserved?
Problem 8.5
A dynamical system has the Lagrangian
L = q̇12 +
q̇22
+ k1 q12 + k2 q̇1 q̇2 ,
a + bq12
where a, b, k1 and k2 are constants. Find the equations of motion in the Hamiltonian
formulation.
Rewriting the Lagrangian in the form of Goldstein’s (8-16), we have
L = k1 q12 +
1
2
q̇1 q̇2
1
2
k2
k2
2
a+bq12
q̇1
q̇2
Homer Reid’s Solutions to Goldstein Problems: Chapter 8
2
From this we can immediately identify the T matrix and its inverse:
2
2
k2
−k2
a + bq12
2
−1
a+bq
1
T=
T =
2
k2 a+bq
2
4 − k22 (a + bq12 )
−k2
2
1
Then the Hamiltonian is
2
−k2
1
a + bq12
p1
2
a+bq
1
H=
p1 p2
− k1 q12
p2
2 4 − k22 (a + bq12 )
−k2
2
p21
a + bq12
2
=
− k2 p1 p2 + p2 − k1 q12 .
4 − k22 (a + bq12 )
a + bq12
Then the equations of motion are
∂H
2p1
a + bq12
q̇1 =
=
− k 2 p2
∂p1
4 − k22 (a + bq12 )
a + bq12
∂H
a + bq12
{−kp1 − 2p2 }
q̇2 =
=
∂p2
4 − k22 (a + bq12 )
∂H
= something ugly
ṗ1 = −
∂q1
∂H
ṗ2 = −
=0
∂q2
So in the Hamiltonian formulation there is one cylic variable, but I still think
this is much harder than the Lagrangian formulation for this problem.
Homer Reid’s Solutions to Goldstein Problems: Chapter 8
3
Problem 8.6
A Hamiltonian of one degree of freedom has the form
H=
ba
kq 2
p2
− bqpe−αt + q 2 e−αt (α + be−αt ) +
,
2a
2
2
where a, b, α, and k are constants. Note: I think there must be a misprint in the
book; the coefficient of p2 in the first term is printed there as 1/2α, which doesn’t
make sense dimensionally in light of the rest of the terms in the Hamiltonian. It
seems reasonable to assume that someone got their Greek and Roman letters mixed
up, as the units do work out correctly if we put 1/2a for the coefficient of that term.
(a) Find a Lagrangian corresponding to this Hamiltonian.
(b) Find an equivalent Lagrangian that is not explicitly dependent on time.
(c) What is the Hamiltonian corresponding to this second Lagrangian, and what
is the relationship between the two Hamiltonians?
(a) From the Hamilton equations of motion,
q̇ =
p
∂H
= − bqe−αt .
∂p
a
(1)
Then, using a reverse Legendre transformation,
L = pq̇ − H
2
p2
p
ba
kq 2
− bqpe−αt −
− bqpe−αt + q 2 e−αt (α + be−αt ) +
a
2a
2
2
2
2
ba
kq
p
− q 2 e−αt (α + be−αt ) −
.
=
2a
2
2
=
(2)
We would now like to eliminate p from this equation in favor of q̇. From (1) we
have
p = aq̇ + bqae−αt
p2 = a2 q̇ 2 + 2bq q̇a2 e−αt + b2 q 2 a2 e−2αt
so (2) becomes
aq̇ 2
1
baαq 2 −αt b2 aq 2 −2αt kq 2
+ bq q̇ae−αt + b2 q 2 ae−2αt −
e
−
e
−
2
2
2
2
2
aq̇ 2
kq 2
1
−αt
=
+ bqae
.
(3)
q̇ − αq −
2
2
2
L=
4
Homer Reid’s Solutions to Goldstein Problems: Chapter 8
(b) Since we can the total time derivative of any function f (q, q̇, t) to the Lagrangian without changing the resulting equations of motion, we consider
d ab 2 −αt
0
.
q e
L =L−
dt 2
The derivative term just cancels the second term in (3), leaving
L0 =
aq̇ 2
kq 2
−
2
2
(4)
which is just the Lagrangian of a one-dimensional harmonic oscillator.
(c) From (4), the new canonical momentum is
p=
∂L0
= aq̇
∂ q̇
Then the Legendre transformation defining the Hamiltonian reads
H = pq̇ − L =
=
kq 2
aq̇ 2
+
2
2
p2
kq 2
+
.
2a
2
Problem 8.9
The point of suspension of a simple pendulum of length l and mass m is constrained
to move on a parabola z = ax2 in the vertical plane. Derive a Hamiltonian governing
the motion of the pendulum and its point of suspension. Obtain the Hamilton’s
equations of motion.
We’ll denote the coordinates of the suspension point as (x, z) = (x, ax2 ).
Then, if θ is the angle the pendulum makes with the vertical (θ = 0 when the
mass point is precisely at 6:00, and grows in the positive direction as the mass
point moves counter-clockwise) then the coordinates of the mass point are
(xm , zm ) = (x + L sin θ, z − L cos θ)
= (x + L sin θ, ax2 − L cos θ).
The potential energy of the system is
L = mgz = mg(ax2 − L cos θ).
(5)
Homer Reid’s Solutions to Goldstein Problems: Chapter 8
5
The kinetic energy is
m 2
2
(ẋ + żm
)
2 m
m
(ẋ + Lθ̇ cos θ)2 + (2axẋ + Lθ̇ sin θ)2
=
2
o
mn
(1 + 4a2 x2 )ẋ2 + L2 θ̇2 + 2Lθ̇ẋ [cos θ + 2ax sin θ] .
=
2
T =
(6)
Then the Lagrangian for the system is, from (7) and (6),
L=T −L
o
mn
(1 + 4a2 x2 )ẋ2 + L2 θ̇2 + 2Lθ̇ẋ [cos θ + 2ax sin θ] − mgax2 + mgL cos θ.
=
2
For convenience in converting to the Hamiltonian, we may write this in the
language of Goldstein’s (8-16):
m
ẋ
(1 + 4a2 x2 )
L[cos θ + 2ax sin θ]
L = L0 (x, θ) +
ẋ θ̇
L[cos θ + 2ax sin θ]
L2
θ̇
2
(7)
where L0 (x, θ) = −mgax2 + mgL cos θ. Then from Goldstein’s (??) we can
write
1
1
&times;
H=
2m L2 (sin θ − 2ax cos θ)2
px
L2
−L[cos θ + 2ax sin θ]
− L0 (x, θ)
(px pθ )
pθ
−L[cos θ + 2ax sin θ]
(1 + 4a2 x2 )
1
1
=
&times;
2
2m L (sin θ − 2ax cos θ)2
2 2
2 2 2
L px − 2L[cos θ + 2ax sin θ]px pθ + (1 + 4a x )pθ − L0 (x, θ)
```