Goldstein Chapter 8

5/19/2018 Goldstein Chapter 8

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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= ax2 +by

x+ cxy+fy2xz+gyk

x2 +y2,

wherea,b, c,f, g,and k are constants. What is the Hamiltonian? What quantitiesare conserved?

Problem 8.5

A dynamical system has the Lagrangian

L= q21 + q22a+bq21

+k1q21 +k2q1q2,

wherea, b, k1andk2are constants. Find the equations of motion in the Hamiltonianformulation.

Rewriting the Lagrangian in the form of Goldsteins (8-16), we have

L= k1q2

1+

1

2q1 q2

2 k2k2 2a+bq2

1

q1q2

1


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Homer Reids Solutions to Goldstein Problems: Chapter 8 2

From this we can immediately identify the T matrix and its inverse:

T=

2 k2k2

2

a+bq21

T1 =

a+bq21

4k22(a+bq21)

2

a+bq21

k2

k2 2

Then the Hamiltonian is

H=1

2

a+bq21

4k22(a+bq21)

p1p2

2

a+bq21

k2

k2 2

p1p2

k1q

21

=

a+bq21

4k22(a+bq21)

p21a+bq21

k2p1p2+p22

k1q

21 .

Then the equations of motion are

q1 = H

p1=

a+bq21

4k22(a+bq21)

2p1a+bq21

k2p2

q2 = H

p2=

a+bq21

4k22(a+bq21)

{kp12p2}

p1 = H

q1 = something ugly

p2 = H

q2= 0

So in the Hamiltonian formulation there is one cylic variable, but I still thinkthis is much harder than the Lagrangian formulation for this problem.


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Problem 8.6

A Hamiltonian of one degree of freedom has the form

H= p2

2abqpet +

ba

2q2et(+bet) +

kq2

2 ,

where a,b,, and k are constants. Note: I think there must be a misprint in thebook; the coefficient ofp2 in the first term is printed there as 1/2, which doesntmake 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 put1/2afor 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 whatis the relationship between the two Hamiltonians?

(a)From the Hamilton equations of motion,

q=H

p =

p

abqet. (1)

Then, using a reverse Legendre transformation,

L= pqH

=p2

a bqpet

p2

2a bqpet +

ba

2q2et(+bet) +

kq2

2

= p2

2a

ba

2q2et(+bet)

kq2

2 . (2)

We would now like to eliminate p from this equation in favor of q. From (1) wehave

p= aq+bqaet

p2 =a2q2 + 2bqqa2et +b2q2a2e2t

so (2) becomes

L=aq2

2 +bqqaet +

1

2b2q2ae2t

baq2

2 et

b2aq2

2 e2t

kq2

2

=aq2

2

+bqaetq1

2

qkq2

2

. (3)


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(b) Since we can the total time derivative of any function f(q, q, t) to the La-grangian without changing the resulting equations of motion, we consider

L =L d

dt

ab

2q2et

.

The derivative term just cancels the second term in (3), leaving

L =aq2

2

kq2

2 (4)

which is just the Lagrangian of a one-dimensional harmonic oscillator.

(c)From (4), the new canonical momentum is

p=L

q =aq

Then the Legendre transformation defining the Hamiltonian reads

H= pqL= aq2

2 + kq

2

2

= p2

2a+

kq2

2 .

Problem 8.9

The point of suspension of a simple pendulum of length l and massmis constrainedto move on a parabolaz = ax2 in the vertical plane. Derive a Hamiltonian governingthe motion of the pendulum and its point of suspension. Obtain the Hamiltonsequations of motion.

Well denote the coordinates of the suspension point as (x, z) = (x,ax2).Then, if is the angle the pendulum makes with the vertical ( = 0 when themass point is precisely at 6:00, and grows in the positive direction as the masspoint moves counter-clockwise) then the coordinates of the mass point are

(xm, zm) = (x+L sin , zL cos )

= (x+L sin ,ax2 L cos).

The potential energy of the system is

L= mgz = mg(ax2 L cos ). (5)


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The kinetic energy is

T =m

2(x2m+ z

2m)

=m

2

(x+L cos )2 + (2axx+L sin )2

= m2

(1 + 4a2x2)x2 +L22 + 2Lx [cos + 2ax sin ]

. (6)

Then the Lagrangian for the system is, from (7) and (6),

L= T L

=m

2

(1 + 4a2x2)x2 +L22 + 2Lx [cos + 2ax sin ]

mgax2 +mgL cos.

For convenience in converting to the Hamiltonian, we may write this in thelanguage of Goldsteins (8-16):

L= L0(x, ) +m

2

x

(1 + 4a2x2) L[cos+ 2ax sin]L[cos + 2ax sin ] L2

x

(7)

where L0(x, ) = mgax2 +mgL cos . Then from Goldsteins (??) we can

write

H= 1

2m

1

L2(sin 2ax cos)2

(px p)

L2 L[cos+ 2ax sin ]

L[cos+ 2ax sin ] (1 + 4a2x2)

pxp

L0(x, )

= 1

2m

1

L2(sin 2ax cos)2

L2p2x

2L[cos+ 2ax sin ]pxp+ (1 + 4a2x2)p2

L0(x, )

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