Solutions to Problems in Goldstein, Classical Classical Me Mechanics chanics , Second Edition Homer Reid June 17, 2002
Chapter 8
Problem Problem 8.4
The Lagrangian for a system can be written as ˙y L = a = a x˙ 2 + b + c + cx˙ y˙ + f + f y 2 x˙ z˙ + g + g ˙ ˙y − k x
x + y , 2
2
where a,b where a,b,, c,f, g, and g, and k are constants. constants. What is the Hamiltonian Hamiltonian?? What quantitie quantitiess are conserved?
Problem Problem 8.5
A dynamical system has the Lagrangian q ˙22 L = q ˙ + + k1 q 12 + k + k2 ˙q 1 ˙q 2 , 2 a + bq + bq 1 2 1
where a where a,, b, k1 and k 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
1 L = k = k q + q ˙1 q ˙2 2 2 1 1
1
2 k2
k2 2 2 a+bq1
q ˙ 1
q ˙2
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Homer Reid’s Solutions to Goldstein Problems: Chapter 8
From this we can immediately identify the T matrix and its inverse: T =
2 k2
k2 2 2 a+bq1
−1
T
a + bq 12 = 4 − k22 (a + bq 12 )
2 a+bq12
−k2
−k2 2
Then the Hamiltonian is 1 a + bq 12 H = 2 4 − k22 (a + bq 12 )
2 1
a + bq = 4 − k22 (a + bq 12 )
2 a+bq12
p p p 1
2
2 1
2 1
a + bq
−k2
−k2 2
− k2 p1 p2 + p
2 2
2 1
1
p 1
p2
− k1 q 12
− k1 q 12 .
Then the equations of motion are ∂H q ˙1 = ∂p 1 ∂H q ˙2 = ∂p 2 ∂H p˙1 = − ∂q 1 ∂H p˙2 = − ∂q 2
a + bq 2 p − k p = 4 − k (a + bq ) a + bq a + bq 2 2
=
2 1
2 1
4 − k22 (a + bq 12 )
2 1
2 2
{−kp 1 − 2 p2 }
= something ugly =0
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.
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Homer Reid’s Solutions to Goldstein Problems: Chapter 8
Problem 8.6
A Hamiltonian of one degree of freedom has the form H =
p2 − bqpe 2a
αt
−
+
ba 2 q e 2
αt
−
(α + be
αt
−
)+
kq 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, q ˙ =
∂H p = − bqe ∂p a
αt
−
.
(1)
Then, using a reverse Legendre transformation, L = p q ˙ − H p2 p2 ba − bqpe αt − − bqpe αt + q 2 e = a 2a 2 2 2 p ba kq = − q 2 e αt (α + be αt ) − . 2a 2 2
−
−
−
αt
−
(α + be
αt
−
kq 2 )+ 2
−
(2)
We would now like to eliminate p from this equation in favor of q ˙ . From (1) we have p = a q ˙ + bqae
αt
−
p2 = a 2 q ˙2 + 2bq qa ˙ 2e
αt
−
+ b2 q 2 a2 e
αt
−2
so (2) becomes aq ˙2 1 baαq 2 αt + bq qae ˙ + b2 q 2 ae 2αt − e 2 2 2 aq ˙2 1 kq 2 = + bqae αt q ˙ − αq − . 2 2 2
L =
−
−
−
αt
−
−
b 2 aq 2 e 2
αt
−2
−
kq 2 2 (3)
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Homer Reid’s Solutions to Goldstein Problems: Chapter 8
(b) Since we can the total time derivative of any function f (q, ˙q, t) to the Lagrangian without changing the resulting equations of motion, we consider
d ab 2 L = L − q e dt 2
αt
−
.
The derivative term just cancels the second term in (3), leaving
L =
aq ˙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 ∂L p = = a q ˙ ∂ q ˙
Then the Legendre transformation defining the Hamiltonian reads H = p q ˙ − L = =
aq ˙2 kq 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 = ax 2 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)
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Homer Reid’s Solutions to Goldstein Problems: Chapter 8
The kinetic energy is m 2 2 ( x˙ + z˙ m ) 2 m m ˙θ sin θ)2 = (x + L ˙ θ˙ cos θ)2 + (2axx + L ˙ 2 m = (1 + 4a2 x2 )x˙ 2 + L2 ˙θ2 + 2Lθ˙x˙ [cos θ + 2ax sin θ] . 2
T =
(6)
Then the Lagrangian for the system is, from (7) and (6), L = T − L m = (1 + 4a2 x2 )x˙ 2 + L2 ˙θ2 + 2Lθ˙x˙ [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 L = L 0 (x, θ) + 2
˙ x˙ θ
(1 + 4a2 x2 ) L[cos θ + 2ax sin θ]
x˙
L[cos θ + 2ax sin θ] L2
θ˙
(7) where L0 (x, θ) = −mgax2 + mgL cos θ. Then from Goldstein’s (??) we can write
1 1 H = × 2 2m L (sin θ − 2ax cos θ)2 L2 ( px pθ ) −L[cos θ + 2ax sin θ]
−L[cos θ + 2ax sin θ] (1 + 4a2 x2 )
p x
pθ
1 1 × = 2m L2 (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)
− L0 (x, θ)
