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Classical Mechanics Solutions 2nd Edition Goldstein

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Page 1: Classical Mechanics Solutions 2nd Edition Goldstein

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Page 19: Classical Mechanics Solutions 2nd Edition Goldstein

Solutions to Problems in Goldstein,

Classical Mechanics, Second Edition

Homer Reid

December 1, 2001

Chapter 3

Problem 3.1

A particle of mass m is constrained to move under gravity without friction on theinside of a paraboloid of revolution whose axis is vertical. Find the one-dimensionalproblem equivalent to its motion. What is the condition on the particle’s initialvelocity to produce circular motion? Find the period of small oscillations aboutthis circular motion.

We’ll take the paraboloid to be defined by the equation z = αr2. The kineticand potential energies of the particle are

T =m

2(r2 + r2θ2 + z2)

=m

2(r2 + r2θ2 + 4α2r2r2)

V = mgz = mgαr2.

Hence the Lagrangian is

L =m

2

[

(1 + 4α2r2)r2 + r2 θ2]

− mgαr2.

This is cyclic in θ, so the angular momentum is conserved:

l = mr2θ = constant.

1

Page 20: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 3 2

For r we have the derivatives

∂L

∂r= 4α2mrr2 + mrθ2 − 2mgαr

∂L

∂r= m(1 + 4α2r2)r

d

dt

∂L

∂r= 8mα2rr2 + m(1 + 4α2r2)r.

Hence the equation of motion for r is

8mα2rr2 + m(1 + 4α2r2)r = 4α2mrr2 + mrθ2 − 2mgαr

or

m(1 + 4α2r2)r + 4mα2rr2 − mrθ2 + 2mgαr = 0.

In terms of the constant angular momentum, we may rewrite this as

m(1 + 4α2r2)r + 4mα2rr2 − l2

mr3+ 2mgαr = 0.

So this is the differential equation that determines the time evolution of r.If initially r = 0, then we have

m(1 + 4α2r2)r + − l2

mr3+ 2mgαr = 0.

Evidently, r will then vanish—and hence r will remain 0, giving circular motion—if

l2

mr3= 2mgαr

orθ =

2gα.

So if this condition is satisfied, the particle will execute circular motion (assum-ing its initial r velocity was zero). It’s interesting to note that the condition onθ for circular motion is independent of r.

Page 21: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 3 3

Problem 3.2

A particle moves in a central force field given by the potential

V = −ke−ar

r,

where k and a are positive constants. Using the method of the equivalent one-dimensional potential discuss the nature of the motion, stating the ranges of l andE appropriate to each type of motion. When are circular orbits possible? Find theperiod of small radial oscillations about the circular motion.

The Lagrangian is

L =m

2

[

r2 + r2θ2]

+ ke−ar

r.

As usual the angular momentum is conserved:

l = mr2θ = constant.

We have

∂L

∂r= mrθ2 − k (1 + ar)

e−ar

r2

∂L

∂r= mr

so the equation of motion for r is

r = rθ2 − k

m(1 + ar)

e−ar

r2

=l2

m2r3− k

m(1 + ar)

e−ar

r2. (1)

The condition for circular motion is that this vanish, which yields

θ =

k

m(1 + ar0)

e−ar0/2

r3/20

. (2)

What this means is that that if the particle’s initial θ velocity is equal to theabove function of the starting radius r0, then the second derivative of r willremain zero for all time. (Note that, in contrast to the previous problem, in thiscase the condition for circular motion does depend on the starting radius.)

To find the frequency of small oscillations, let’s suppose the particle is exe-cuting a circular orbit with radius r0 (in which case the θ velocity is given by(2)), and suppose we nudge it slightly so that its radius becomes r = r0 + x,where x is small. Then (1) becomes

x =k

m

(

1 + ar0

)e−ar0

r20

− k

m(1 + a[r0 + x])

e−a[r0+x]

[r0 + x]2(3)

Page 22: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 3 4

Since x is small, we may write the second term approximately as

≈ k

m

e−ar0

r20

(1 + ar0 + ax)(1 − ax)

(

1 − 2x

r0

)

≈ k

m(1 + ar0)

e−ar0

r20

+k

m

e−ar0

r20

(

a − a(1 + ar0) − 2(1 + ar0)

r0

)

x

≈ k

m(1 + ar0)

e−ar0

r20

− k

m

e−ar0

r20

(

2a +2

r0+ a2r0

)

x.

The first term here just cancels the first term in (??), so we are left with

x =k

m

e−ar0

r20

(

2a +2

r0+ a2r0

)

x

The problem is that the RHS here has the wrong sign—this equation is satisfiedby an x that grows (or decays) exponentially, rather than oscillates. SomehowI messed up the sign of the RHS, but I can’t find where–can anybody help?

Problem 3.3

Two particles move about each other in circular orbits under the influence of grav-itational forces, with a period τ . Their motion is suddenly stopped, and they arethen released and allowed to fall into each other. Prove that they collide after atime τ/4

√2.

Since we are dealing with gravitational forces, the potential energy betweenthe particles is

U(r) = −k

r

and, after reduction to the equivalent one-body problem, the Lagrangian is

L =µ

2[r2 + r2θ2] +

k

r

where µ is the reduced mass. The equation of motion for r is

µr = µrθ2 − k

r2. (4)

If the particles are to move in circular orbits with radius r0, (4) must vanish atr = r0, which yields a relation between r0 and θ:

r0 =

(

k

µθ2

)1/3

=

(

kτ2

4π2µ

)1/3

(5)

Page 23: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 3 5

where we used the fact that the angular velocity in the circular orbit with periodτ is θ = 2π/τ .

When the particles are stopped, the angular velocity goes to zero, and thefirst term in (4) vanishes, leaving only the second term:

r = − k

µr2. (6)

This differential equation governs the evolution of the particles after they arestopped. We now want to use this equation to find r as a function of t, whichwe will then need to invert to find the time required for the particle separationr to go from r0 to 0.

The first step is to multiply both sides of (6) by the integrating factor 2r:

2rr = − 2k

µr2r

or

d

dt

(

r2)

= +d

dt

(

2k

µr

)

from which we conclude

r2 =2k

µr+ C. (7)

The constant C is determined from the boundary condition on r. This is simplythat r = 0 when r = r0, since initially the particles are not moving at all. Withthe appropriate choice of C in (7), we have

r =d r

d t=

(

2k

µ

)1/2 √

1

r− 1

r0

=

(

2k

µ

)1/2 √

r0 − r

rr0. (8)

We could now proceed to solve this differential equation for r(t), but since infact we’re interested in solving for the time difference corresponding to givenboundary values of r, it’s easier to invert (8) and solve for t(r):

∆t =

∫ 0

r0

(

dt

dr

)

dr

=

∫ 0

r0

(

dr

dt

)

−1

dr

=( µ

2k

)1/2∫ 0

r0

(

rr0

r0 − r

)1/2

dr

Page 24: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 3 6

We change variables to u = r/r0, du = dr/r0 :

=( µ

2k

)1/2

r3/20

∫ 0

1

(

u

1− u

)1/2

du

Next we change variables to u = sin2 x, du = 2 sinx cos x dx :

= 2( µ

2k

)1/2

r3/20

∫ 0

π/2

sin2 x dx

=( µ

2k

)1/2

r3/20

π

4.

Now plugging in (5), we obtain

∆t =( µ

2k

)1/2(

kτ2

4π2µ

)1/2(π

4

)

4√

2

as advertised.

Problem 3.6

(a) Show that if a particle describes a circular orbit under the influence of anattractive central force directed at a point on the circle, then the force variesas the inverse fifth power of the distance.

(b) Show that for the orbit described the total energy of the particle is zero.

(c) Find the period of the motion.

(d) Find x, y, and v as a function of angle around the circle and show that allthree quantities are infinite as the particle goes through the center of force.

Let’s suppose the center of force is at the origin, and that the particle’s orbitis a circle of radius R centered at (x = R, y = 0) (so that the leftmost pointof the particle’s origin is the center of force). The equation describing such anorbit is

r(θ) =√

2R(1 + cos 2θ)1/2

so

u(θ) =1

r(θ)=

1√2R(1 + cos 2θ)1/2

. (9)

Page 25: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 3 7

Differentiating,

du

dθ=

sin 2θ√2R(1 + cos 2θ)3/2

du

dθ=

1√2R

[

2 cos 2θ

(1 + cos 2θ)3/2+ 3

sin2 2θ

(1 + cos 2θ)5/2

]

=1

2√

2R

1

(1 + cos 2θ)5/2

[

2 cos 2θ + 2 cos2 2θ + 3 sin2 2θ]

. (10)

Adding (9) and (10),

d2u

dθ2+ u =

1√2R(1 + cos 2θ)5/2

[

(1 + cos 2θ)2 + 2 cos2θ + 2 cos2 2θ + 3 sin2 2θ]

=1√

2R(1 + cos 2θ)5/2[4 + 4 cos 2θ]

=4√

2R(1 + cos 2θ)3/2

= 8R2u3. (11)

The differential equation for the orbit is

d2u

dθ2+ u = −m

l2d

duV

(

1

u

)

(12)

Plugging in (11), we have

8R2u3 = −m

l2d

duV

(

1

u

)

so

V

(

1

u

)

= −2l2R2

mu4 −→ V (r) = −2l2R2

mr4(13)

so

f(r) = −8l2R2

mr5(14)

which is the advertised r dependence of the force.

(b) The kinetic energy of the particle is

T =m

2[r2 + r2θ2]. (15)

Page 26: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 3 8

We have

r =√

2R(1 + cos 2θ)1/2

r2 = 2R2(1 + cos 2θ)

r =√

2Rsin 2θ

(1 + cos 2θ)1/2θ

r2 = 2R2 sin2 2θ

1 + cos 2θθ2

Plugging into (15),

T = mR2θ2

[

sin2 2θ

1 + cos 2θ+ 1 + cos 2θ

]

= mR2θ2

[

sin2 2θ + 1 + 2 cos 2θ + cos2 2θ

1 + cos θ

]

= 2mR2θ2

In terms of l = mr2θ, this is just

=2R2l2

mr4

But this is just the negative of the potential energy, (13); hence the total particleenergy T + V is zero.

(c) Suppose the particle starts out at the furthest point from the center of forceon its orbit, i.e the point x = 2R, y = 0, and that it moves counter-clockwisefrom this point to the origin. The time required to undergo this motion is halfthe period of the orbit, and the particle’s angle changes from θ = 0 to θ = π/2.Hence we can calculate the period as

τ = 2

∫ π/2

0

dt

dθdθ

= 2

∫ π/2

0

θ

Using θ = l/mr2, we have

= 2m

l

∫ π/2

0

r2(θ) dθ

=4R2m

l

∫ π/2

0

(1 + 2 cos 2θ + cos2 2θ) dθ

=4R2m

l· 3π

4

=3πR2m

l.

Page 27: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 3 9

Problem 3.8

(a) For circular and parabolic orbits in an attractive 1/r potential having the sameangular momentum, show that the perihelion distance of the parabola is onehalf the radius of the circle.

(b) Prove that in the same central force as in part (a) the speed of a particle atany point in a parabolic orbit is

√2 times the speed in a circular orbit passing

through the same point.

(a) The equations describing the orbits are

r =

l2

mk(circle)

l2

mk

(

1

1 + cos θ

)

(parabola.)

Evidently, the perihelion of the parabola occurs when θ = 0, in which caser = l2/2mk, or one-half the radius of the circle.

(b) For the parabola, we have

r =l2

mk

(

sin θ

(1 + cos θ)2

)

θ (16)

= rθsin θ

1 + cos θ

so

v2 = r2 + r2θ2

= r2θ2

[

sin2 θ

(1 + cos θ)2+ 1

]

= r2θ2

[

sin2 θ + 1 + 2 cos θ + cos2 θ

(1 + cos θ)2

]

= 2r2θ2

[

1

1 + cos θ

]

=2mkr3θ2

l2

=2k

mr(17)

in terms of the angular momentum l = mr2θ2. On the other hand, for the circler = 0, so

v2 = r2θ2 =l2

m2r2=

k

mr(18)

Page 28: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 3 10

where we used that fact that, since this is a circular orbit, the condition k/r =l2/mr2 is satisfied. Evidently (17) is twice (18) for the same particle at thesame point, so the unsquared speed in the parabolic orbit is

√2 times that in

the circular orbit at the same point.

Problem 3.12

At perigee of an elliptic gravitational orbit a particle experiences an impulse S (cf.Exercise 9, Chapter 2) in the radial direction, sending the particle into anotherelliptic orbit. Determine the new semimajor axis, eccentricity, and orientation ofmajor axis in terms of the old.

The orbit equation for elliptical motion is

r(θ) =a(1 − ε2)

1 + ε cos(θ − θ0). (19)

For simplicity we’ll take θ0 = 0 for the initial motion of the particle. Thenperigee happens when θ = 0, which is to say the major axis of the orbit is onthe x axis.

Then at the point at which the impulse is delivered, the particle’s momentumis entirely in the y direction: pi = pij. After receiving the impulse S in the radial(x) direction, the particle’s y momentum is unchanged, but its x momentum isnow px = S. Hence the final momentum of the particle is pf = S i+pij. Since theparticle is in the same location before and after the impulse, its potential energyis unchanged, but its kinetic energy is increased due to the added momentum:

Ef = Ei +S2

2m. (20)

Hence the semimajor axis length shrinks accordingly:

af = − k

2Ef= − k

2Ei + S2/m=

ai

1 + S2/(2mEi). (21)

Next, since the impulse is in the same direction as the particle’s distance fromthe origin, we have ∆L = r × ∆p = 0, i.e. the impulse does not change theparticle’s angular momentum:

Lf = Li ≡ L. (22)

With (20) and (22), we can compute the change in the particle’s eccentricity:

εf =

1 +2EfL2

mk2

=

1 +2EiL2

mk2+

L2S2

m2k2. (23)

Page 29: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 3 11

What remains is to compute the constant θ0 in (19) for the particle’s orbit afterthe collision. To do this we need merely observe that, since the location of theparticle is unchanged immediately after the impulse is delivered, expression (19)must evaluate to the same radius at θ = 0 with both the “before” and “after”values of a and ε:

ai(1 − ε2i )

1 + εi=

af (1 − ε2f )

1 + εf cos θ0

or

cos θ0 =1

εf

af (1 − ε2f )

ai(1 − εi)− 1

.

Problem 3.13

A uniform distribution of dust in the solar system adds to the gravitational attrac-tion of the sun on a planet an additional force

F = −mCr

where m is the mass of the planet, C is a constant proportional to the gravitationalconstant and the density of the dust, and r is the radius vector from the sun to theplanet (both considered as points). This additional force is very small compared tothe direct sun-planet gravitational force.

(a) Calculate the period for a circular orbit of radius r0 of the planet in this com-bined field.

(b) Calculate the period of radial oscillations for slight disturbances from this cir-cular orbit.

(c) Show that nearly circular orbits can be approximated by a precessing ellipseand find the precession frequency. Is the precession the same or oppositedirection to the orbital angular velocity?

(a) The equation of motion for r is

mr =l2

mr3+ f(r)

=l2

mr3− k

r2− mCr. (24)

For a circular orbit at radius r0 this must vanish:

0 =l2

mr30

− k

r20

− mCr0 (25)

Page 30: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 3 12

−→ l =√

mkr0 + m2Cr40

−→ θ =l

mr20

=1

mr20

mkr0 + m2Cr40

=

k

mr30

1 +mCr3

0

k

≈√

k

mr30

[

1 +mCr3

0

2k

]

Then the period is

τ =2π

θ≈ 2πr

3/20

m

k

[

1 − mCr30

2k

]

= τ0

[

1 − Cτ20

8π2

]

where τ0 = 2πr3/20

m/k is the period of circular motion in the absence of theperturbing potential.

(b) We return to (24) and put r = r0 + x with x r0:

mx =l2

m(r0 + x)3− k

(r0 + x)2− mC(r0 + x)

≈ l2

mr30

(

1 − 3x

r0

)

− k

r20

(

1 − 2x

r0

)

− mCr0 − mCx

Using (25), this reduces to

mx =

[

− 3l2

mr40

+2k

r30

− mC

]

x

orx + ω2x = 0

with

ω =

[

3l2

m2r40

− 2k

mr30

− C

]1/2

=

[

2l2

m2r40

− k

mr30

]1/2

where in going to the last line we used (25) again.

Page 31: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 3 13

Problem 3.14

Show that the motion of a particle in the potential field

V (r) = −k

r+

h

r2

is the same as that of the motion under the Kepler potential alone when expressedin terms of a coordinate system rotating or precessing around the center of force.For negative total energy show that if the additional potential term is very smallcompared to the Kepler potential, then the angular speed of precession of the ellip-tical orbit is

Ω =2πmh

l2τ.

The perihelion of Mercury is observed to precess (after corrections for known plan-etary perturbations) at the rate of about 40′′ of arc per century. Show that thisprecession could be accounted for classically if the dimensionless quantity

η =k

ka

(which is a measure of the perturbing inverse square potential relative to the grav-itational potential) were as small as 7× 10−8. (The eccentricity of Mercury’s orbitis 0.206, and its period is 0.24 year).

The effective one-dimensional equation of motion is

mr =L2

mr3− k

r2+

2h

r3

=L2 + 2mh

mr3+

k

r2

=L2 + 2mh + (mh/L)2 − (mh/L)2

mr3+

k

r2

=[L + (mh/L)]2 − (mh/L)2

mr3+

k

r2

If mh L2, then we can neglect the term (mh/L)2 in comparison with L2, andwrite

mr =[L + (mh/L)]2

mr3+

k

r2(26)

which is just the normal equation of motion for the Kepler problem, but withthe angular momentum L augmented by the additive term ∆L = mh/L.

Such an augmentation of the angular momentum may be accounted for by

Page 32: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 3 14

augmenting the angular velocity:

L = mr2θ −→ L

(

1 +mh

L2

)

= mr2θ

(

1 +mh

L2

)

= mr2θ + mr2Ω

where

Ω =mh

L2θ=

2πmh

L2τ

is a precession frequency. If we were to go back and work the problem in thereference frame in which everything is precessing with angular velocity Ω, butthere is no term h/r2 in the potential, then the equations of motion would comeout the same as in the stationary case, but with a term ∆L = mr2Ω added tothe effective angular momentum that shows up in the equation of motion for r,just as we found in (26).

To put in the numbers, we observe that

Ω =

(

τ

)

( m

L2

)

(h)

=

(

τ

) (

mka

L2

) (

h

ka

)

=

(

τ

) (

1

1 − e2

) (

h

ka

)

so

h

ka= (1 − e2)

τ Ω

= (1 − e2)τfprec

where in going to the third-to-last line we used Goldstein’s equation (3-62), andin the last line I put fprec = Ω/2π. Putting in the numbers, we find

h

ka= (1 − .2062) ·

(

0.24 yr)

· 40′′(

1

3600′′

) (

1 revolution

360

) (

1 century−1

100 yr−1

)

yr−1

= 7.1 · 10−8.

Page 33: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 3 15

Problem 3.22

In hyperbolic motion in a 1/r potential the analogue of the eccentric anomaly is Fdefined by

r = a(e coshF − 1),

where a(1 − e) is the distance of closest approach. Find the analogue to Kepler’sequation giving t from the time of closest approach as a function of F .

We start with Goldstein’s equation (3.65):

t =

m

2

∫ r

r0

dr√

kr − l2

2mr2 + E

=

m

2

∫ r

r0

r dr√

Er2 + kr − l2

2m

. (27)

With the suggested substitution, the thing under the radical in the denom-inator of the integrand is

Er2 + kr − l2

2m= Ea2(e2 cosh2 F − 2e coshF + 1) + ka(e coshF − 1) − l2

2m

= Ea2e2 cosh2 F + ae(k − 2Ea) coshF +

(

Ea2 − ka − l2

2m

)

It follows from the orbit equation that, if a(e − 1) is the distance of closestapproach, then a = k/2E. Thus

=k2e2

4Ecosh2 F − k2e2

4E− l2

2m

=k2

4E

e2 cosh2 F −(

1 +2El2

mk2

)

=k2e2

4E

[

cosh2 F − 1]

=k2e2

4Esinh2 F = a2e2E sinh2 F.

Plugging into (27) and observing that dr = ae sinh F dF , we have

t =

m

2E

∫ F

F0

a(e coshF − 1) dF =

ma2

2E[e(sinhF − sinh F0) − (F − F0)]

and I suppose this equation could be a jumping-off point for numerical or otherinvestigations of the time of travel in hyperbolic orbit problems.

Page 34: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 3 16

Problem 3.26

Examine the scattering produced by a repulsive central force f = kr−3. Show thatthe differential cross section is given by

σ(Θ)dΘ =k

2E

(1 − x)dx

x2(2 − x)2 sin πx

where x is the ratio Θ/π and E is the energy.

The potential energy is U = k/2r2 = ku2/2, and the differential equationfor the orbit reads

d2u

dθ2+ u = −m

l2dU

du= −mk

l2u

ord2u

dθ2+

(

1 +mk

l2

)

u = 0

with solution

u = A cos γθ + B sin γθ (28)

where

γ =

1 +mk

l2. (29)

We’ll set up our coordinates in the way traditional for scattering experiments:initially the particle is at angle θ = π and a great distance from the force center,and ultimately the particle proceeds off to r = ∞ at some new angle θs. Thefirst of these observations gives us a relation between A and B in the orbitequation (28):

u(θ = π) = 0 −→ A cos γπ + B sin γπ = 0

−→ A = −B tan γπ. (30)

The condition that the particle head off to r = ∞ at angle θ = θs yields thecondition

A cos γθs + B sin γθs = 0.

Using (30), this becomes

− cosγθs tan γπ + sin γθs = 0

Page 35: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 3 17

or

− cos γθs sin γπ + sin γθs cos γπ = 0

−→ sin γ(θs − π) = 0

−→ γ(θs − π) = π

or, in terms of Goldstein’s variable x = θ/π,

γ =1

x − 1. (31)

Plugging in (29) and squaring both sides, we have

1 +mk

l2=

1

(x − 1)2.

Now l = mv0s = (2mE)1/2s with s the impact parameter and E the particleenergy. Thus the previous equation is

1 +k

2Es2=

1

(x − 1)2

or

s2 = − k

2E

[

(x − 1)2

x(x − 2)

]

.

Taking the differential of both sides,

2s ds = − k

2E

[

2(x − 1)

x(x − 2)− (x − 1)2

x2(x − 2)− (x − 1)2

x(x − 2)2

]

dx

= − k

2E

[

2x(x − 1)(x − 2) − (x − 1)2(x − 2) − x(x − 1)2

x2(x − 2)2

]

= − k

2E

[

2(1 − x)

x2(x − 2)2

]

. (32)

The differential cross section is given by

σ(θ)dΩ =| s ds |sin θ

.

Plugging in (32), we have

σ(θ)dΩ =k

2E

[

(1 − x)

x2(x − 2)2 sin θ

]

dx

as advertised.

Page 36: Classical Mechanics Solutions 2nd Edition Goldstein

Solutions to Problems in Goldstein,

Classical Mechanics, Second Edition

Homer Reid

April 21, 2002

Chapter 7

Problem 7.2

Obtain the Lorentz transformation in which the velocity is at an infinitesimal angledθ counterclockwise from the z axis, by means of a similarity transformation appliedto Eq. (7-18). Show directly that the resulting matrix is orthogonal and that theinverse matrix is obtained by substituting −v for v.

We can obtain this transformation by first applying a pure rotation to rotatethe z axis into the boost axis, then applying a pure boost along the (new) zaxis, and then applying the inverse of the original rotation to bring the z axisback in line with where it was originally. Symbolically we have L = R−1KRwhere R is the rotation to achieve the new z axis, and K is the boost along thez axis.

Goldstein tells us that the new z axis is to be rotated dθ counterclockisefrom the original z axis, but he doesn’t tell us in which plane, i.e. we know θbut not φ for the new z axis in the unrotated coordinates. We’ll assume the zaxis is rotated around the x axis, in a sense such that if you’re standing on thepositive x axis, looking toward the negative x axis, the rotation appears to becounterclockwise, so that the positive z axis is rotated toward the negative y

1

Page 37: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 7 2

axis. Then, using the real metric,

L =

1 0 0 00 cos dθ sin dθ 00 − sin dθ cos dθ 00 0 0 1

1 0 0 00 1 0 00 0 γ −βγ0 0 −βγ γ

1 0 0 00 cos dθ − sin dθ 00 sin dθ cos dθ 00 0 0 1

=

1 0 0 00 cos dθ sin dθ 00 − sin dθ cos dθ 00 0 0 1

1 0 0 00 cos dθ − sin dθ 00 γ sin dθ γ cos dθ −βγ0 −βγ sin dθ −βγ cos dθ γ

=

1 0 0 00 cos2 dθ + γ sin2 dθ (γ − 1) sin dθ cos dθ −βγ sin dθ0 (γ − 1) sin dθ cos dθ sin2 dθ + γ cos2 dθ −βγ cos dθ0 −βγ sin dθ −βγ cos dθ γ

.

Problem 7.4

A rocket of length l0 in its rest system is moving with constant speed along the zaxis of an inertial system. An observer at the origin observes the apparent lengthof the rocket at any time by noting the z coordinates that can be seen for the headand tail of the rocket. How does this apparent length vary as the rocket moves fromthe extreme left of the observer to the extreme right?

Let’s imagine a coordinate system in which the rocket is at rest and centeredat the origin. Then the world lines of the rocket’s top and bottom are

xt′µ = 0, 0, +L0/2, τ xb′

µ = 0, 0,−L0/2, τ .

where we are parameterizing the world lines by the proper time τ . Now, the restframe of the observer is moving in the negative z direction with speed v = βcrelative to the rest frame of the rocket. Transforming the world lines of therocket’s top and bottom to the rest frame of the observer, we have

xtµ = 0, 0, γ(L0/2 + vτ), γ(τ + βL0/2c) (1)

xbµ = 0, 0, γ(−L0/2 + vτ), γ(τ − βL0/2c) . (2)

Now consider the observer. At any time t in his own reference frame, he isreceiving light from two events, namely, the top and bottom of the rocket movingpast imaginary distance signposts that we pretend to exist up and down the zaxis. He sees the top of the rocket lined up with one distance signpost and thebottom of the rocket lined up with another, and from the difference between thetwo signposts he computes the length of the rocket. Of course, the light thathe sees was emitted by the rocket some time in the past, and, moreover, the

Page 38: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 7 3

light signals from the top and bottom of the rocket that the observer receivessimultaneously at time t were in fact emitted at different proper times τ in therocket’s rest frame.

First consider the light received by the observer at time t0 coming fromthe bottom of the rocket. Suppose in the observer’s rest frame this light wereemitted at time t0 − ∆t, i.e. ∆t seconds before it reaches the observer at theorigin; then the rocket bottom was passing through z = −c∆t when it emittedthis light. But then the event identified by (z, t) = (−c∆t, t0 − ∆t) must lie onthe world line of the rocket’s bottom, which from (2) determines both ∆t andthe proper time τ at which the light was emitted:

−c∆t = γ(−L0/2 + vτ)t0 − ∆t = γ(τ + βL0/2c)

=⇒ τ =

(

1 + β

1 − β

)1/2

t0−L0

2c≡ τb(t0).

We use the notation τb(t0) to indicate that this is the proper time at which thebottom of the rocket emits the light that arrives at the observer’s origin at theobserver’s time t0. At this proper time, from (2), the position of the bottom ofthe rocket in the observer’s reference frame was

zb(τb(t0)) = −γL0/2 + vγτb(t0)

= −γL0/2 + vγ

(

1 + β

1 − β

)1/2

t0 −L0

2c

(3)

Similarly, for the top of the rocket we have

τt(t0) =

(

1 + β

1 − β

)1/2

t0 +L0

2c

and

zt(τt(t0)) = γL0/2 + vγ

(

1 + β

1 − β

)1/2

t0 +L0

2c

(4)

Subtracting (3) from (4), we have the length for the rocket computed by theobserver from his observations at time t0 in his reference frame:

L(t0) = γ(1 + β)L0

=

(

1 + β

1 − β

)1/2

L0.

Page 39: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 7 4

Problem 7.17

Two particles with rest masses m1 and m2 are observed to move along the observer’sz axis toward each other with speeds v1 and v2, respectively. Upon collision theyare observed to coalesce into one particle of rest mass m3 moving with speed v3

relative to the observer. Find m3 and v3 in terms of m1, m2, v1, and v2. Would itbe possible for the resultant particle to be a photon, that is m3 = 0, if neither m1

nor m2 are zero?

Equating the 3rd and 4th components of the initial and final 4-momentumof the system yields

γ1m1v1 − γ2m2v2 = γ3m3v3

γ1m1c + γ2m2c = γ3m3c

Solving the second for m3 yields

m3 =γ1

γ3m1 +

γ2

γ3m2 (5)

and plugging this into the first yields v3 in terms of the properties of particles1 and 2:

v3 =γ1m1v1 − γ2m2v2

γ1m1 + γ2m2

Then

β3 =v3

c=

γ1m1β1 − γ2m2β2

γ1m1 + γ2m2

1 − β23 =

γ21m2

1 + 2γ1γ2m1m2 + γ22m2

2 − [γ21m2

1β21 + γ2

2m22β

22 − 2γ1γ2m1m2β1β2]

(γ1m1 + γ2m2)2

=γ21m2

1(1 − β21) + γ2

2m22(1 − β2

2) + 2γ1γ2m1m2(1 − β1β2)

(γ1m1 + γ2m2)2

=m2

1 + m22 + 2γ1γ2m1m2(1 − β1β2)

(γ1m1 + γ2m2)2

and hence

γ23 =

1

1 − β23

=(γ1m1 + γ2m2)

2

m21 + m2

2 + 2γ1γ2m1m2(1 − β1β2). (6)

Now, (5) shows that, for m3 to be zero when either m1 or m2 is zero, we musthave γ3 = ∞. That this condition cannot be met for nonzero m1, m2 is evidentfrom the denominator of (6), in which all terms are positive (since β1β2 < 1 ifm1 or m2 is nonzero).

Page 40: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 7 5

Problem 7.19

A meson of mass π comes to rest and disintegrates into a meson of mass µ and aneutrino of zero mass. Show that the kinetic energy of motion of the µ meson (i.e.without the rest mass energy) is

(π − µ)2

2πc2.

Working in the rest frame of the pion, the conservation relations are

πc2 = (µ2c4 + p2µc2)1/2 + pνc (energy conservation)

0 = pµ + pν (momentum conservation).

From the second of these it follows that the muon and neutrino must have thesame momentum, whose magnitude we’ll call p. Then the energy conservationrelation becomes

πc2 = (µ2c4 + p2c2)1/2 + pc

−→ (πc − p)2 = µ2c2 + p2

−→ p =π2 − µ2

2πc.

Then the total energy of the muon is

Eµ = (µ2c4 + p2c2)1/2

= c2

(

µ2 +(π2 − µ2)2

4π2

)1/2

=c2

(

4π2µ2 + (π2 − µ2)2)1/2

=c2

2π(π2 + µ2)

Then subtracting out the rest energy to get the kinetic energy, we obtain

K = Eµ − µc2 =c2

2π(π2 + µ2) − µc2

=c2

2π(π2 + µ2 − 2πµ)

=c2

2π(π − µ)2

as advertised.

Page 41: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 7 6

Problem 7.20

A π+ meson of rest mass 139.6 MeV collides with a neutron (rest mass 939.6 MeV)stationary in the laboratory system to produce a K+ meson (rest mass 494 MeV)and a Λ hyperon (rest mass 1115 MeV). What is the threshold energy for thisreaction in the laboratory system?

We’ll put c = 1 for this problem. The four-momenta of the pion and neutronbefore the collision are

pµ,π = (pπ, γπmπ), pµ,n = (0, mn)

and the squared magnitude of the initial four-momentum is thus

pµ,T pµT = −|pπ|2 + (γπmπ + mn)2

= −|pπ|2 + γ2πm2

π + m2n + 2γπmπmn

= m2π + m2

n + 2γπmπmn

= (mπ + mn)2 + 2(γπ − 1)mπmn (7)

The threshold energy is the energy needed to produce the K and Λ particlesat rest in the COM system. In this case the squared magnitude of the four-momentum of the final system is just (mK + mΛ)2, and, by conservation ofmomentum, this must be equal to the magnitude of the four-momentum of theinitial system (7):

(mK + mΛ)2 = (mπ + mn)2 + 2(γπ − 1)mπmn

=⇒ γπ = 1 +(mK + mΛ)2 − (mπ + mn)2

2mπmn= 6.43

Then the total energy of the pion is T = γπmπ = (6.43 · 139.6 MeV) = 898MeV, while its kinetic energy is K = T − m = 758 MeV.

The above appears to be the correct solution to this problem. On the otherhand, I first tried to do it a different way, as below. This way yields a differentand hence presumably incorrect answer, but I can’t figure out why. Can anyonefind the mistake?

The K and Λ particles must have, between them, the same total momentumin the direction of the original pion’s momentum as the original pion had. Ofcourse, the K and Λ may also have momentum in directions transverse to theoriginal pion momentum (if so, their transverse momenta must be equal andopposite). But any transverse momentum just increases the energy of the finalsystem, which increases the energy the initial system must have had to producethe final system. Hence the minimum energy situation is that in which the K andΛ both travel in the direction of the original pion’s motion. (This is equivalentto Goldstein’s conclusion that, just at threshold, the produced particles are at

Page 42: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 7 7

rest in the COM system). Then the momentum conservation relation becomessimply

pπ = pK + pλ (8)

and the energy conservation relation is (with c = 1)

(m2π + p2

π)1/2 + mn = (m2K + p2

K)1/2 + (m2Λ + p2

Λ)1/2. (9)

The problem is to find the minimum value of pπ that satisfies (9) subject to theconstraint (8).

To solve this we must first resolve a subquestion: for a given pπ, what is therelative allocation of momentum to pK and pΛ that minimizes (9) ? Minimizing

Ef = (m2K + p2

K)1/2 + (m2Λ + p2

Λ)1/2.

subject to pK + pΛ = pπ, we obtain the condition

pK

(m2K + p2

K)1/2=

(m2Λ + p2

Λ)1/2=⇒ pK =

mK

mΛpΛ (10)

Combining this with (8) yields

pΛ =mΛ

mK + mΛpπ pK =

mK

mK + mΛpπ. (11)

For a given total momentum pπ, the minimum possible energy the final systemcan have is realized when pπ is partitioned between pK and pΛ according to(11). Plugging into (8), the relation defining the threshold momentum is

(m2π + p2

π)1/2 + mn =

(

m2K +

(

mK

mK + mΛ

)2

p2π

)1/2

+

(

m2Λ +

(

mK + mΛ

)2

p2π

)1/2

Solving numerically yields pπ ≈ 655 MeV/c, for a total pion energy of about670 MeV.

Page 43: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 7 8

Problem 7.21

A photon may be described classically as a particle of zero mass possessing never-theless a momentum h/λ = hν/c, and therefore a kinetic energy hν. If the photoncollides with an electron of mass m at rest it will be scattered at some angle θ witha new energy hν′. Show that the change in energy is related to the scattering angleby the formula

λ′ − λ = 2λc sin2 θ

2,

where λc = h/mc, known as the Compton wavelength. Show also that the kineticenergy of the recoil motion of the electron is

T = hν2(

λc

λ

)

sin2 θ2

1 + 2(

λc

λ

)

sin2 θ/2.

Let’s assume the photon is initially travelling along the z axis. Then the sumof the initial photon and electron four-momenta is

pµ,i = pµ,γ + pµ,e =

00

h/λh/λ

+

000

mc

=

00

h/λmc + h/λ

. (12)

Without loss of generality we may assume that the photon and electron movein the xz plane after the scatter. If the photon’s velocity makes an angle θ withthe z axis, while the electron’s velocity makes an angle φ, the four-momentumafter the collision is

pµ,f = pµ,γ + pµ,e =

(h/λ′) sin θ0

(h/λ′) cos θh/λ′

+

pe sinφ0

pe cosφ√

m2c2 + p2e

=

(h/λ′) sin θ + pe sin φ0

(h/λ′) cos θ + pe cosφ

(h/λ′) +√

m2c2 + p2e

.

(13)

Equating (12) and (13) yields three separate equations:

(h/λ′) sin θ + pe sinφ = 0 (14)

(h/λ′) cos θ + pe cosφ = h/λ (15)

h/λ′ +√

m2c2 + p2e = mc + h/λ (16)

From the first of these we find

sinφ = − h

λ′pesin θ =⇒ cosφ =

[

1 +

(

h

λ′pe

)2

sin2 θ

]1/2

Page 44: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 7 9

and plugging this into (15) we find

p2e =

h2

λ2+

h2

λ′2− 2

h2

λλ′cos θ. (17)

On the other hand, we can solve (16) to obtain

p2e = h2

(

1

λ− 1

λ′

)2

+ 2mch

(

1

λ− 1

λ′

)

.

Comparing these two determinations of pe yields

cos θ = 1 − mc

h(λ′ − λ)

or

sin2 θ

2=

mc

2h(λ′ − λ) =

1

2λc(λ′ − λ)

so this is advertised result number 1.Next, to find the kinetic energy of the electron after the collision, we can

write the conservation of energy equation in a slightly different form:

mc +h

λ= γmc +

h

λ′

=⇒ (γ − 1)mc = K = h

(

1

λ− 1

λ′

)

= h

(

λ′ − λ

λλ′

)

= h

(

2λc sin2(θ/2)

λ[λ + 2λc sin2(θ/2)]

)

=h

λ

(

2χ sin2(θ/2)

1 + 2χ sin2(θ/2)

)

where we put χ = λc/λ.

Problem 7.22

A photon of energy E collides at angle θ with another photon of energy E. Provethat the minimum value of E permitting formation of a pair of particles of mass mis

Eth =2m2c4

E(1 − cos θ).

We’ll suppose the photon of energy E is traveling along the positive z axis,while that with energy E is traveling in the xz plane (i.e., its velocity has

Page 45: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 7 10

spherical polar angles θ and φ = 0). Then the 4-momenta are

p1 =

(

0, 0,E

c,E

c

)

p2 =

(Ec

sin θ, 0,Ec

cos θ,Ec

)

pt = p1 + p2 =

(Ec

sin θ, 0,E + E cos θ

c,E + E

c

)

It’s convenient to rotate our reference frame to one in which the space portionof the composite four-momentum of the two photons is all along the z direction.In this frame the total four-momentum is

p′t =

(

0, 0,1

c

E2 + E2 + 2EE cos θ,E + E

c

)

. (18)

At threshold energy, the two produced particles have the same four-momenta:

p3 = p4 =(

0, 0, p, (m2c2 + p2)1/2)

(19)

and 4-momentum conservation requires that twice (19) add up to (18), whichyields two conditions:

2p = 1c

√E2 + E2 + 2EE cos θ −→ p2c2 = 1

4 (E2 + E2 + 2EE cos θ)

2√

m2c2 + p2 = E+E

c −→ m2c4 + p2c2 = 14 (E2 + E2 + 2EE)

Subtracting the first of these from the second, we obtain

m2c4 =EE2

(1 − cos θ)

or

E =2m2c4

E(1 − cos θ)

as advertised.

Page 46: Classical Mechanics Solutions 2nd Edition Goldstein

Solutions to Problems in Goldstein,

Classical Mechanics, Second Edition

Homer Reid

October 29, 2002

Chapter 9

Problem 9.1

One of the attempts at combining the two sets of Hamilton’s equations into onetries to take q and p as forming a complex quantity. Show directly from Hamilton’sequations of motion that for a system of one degree of freedom the transformation

Q = q + ip, P = Q∗

is not canonical if the Hamiltonian is left unaltered. Can you find another set ofcoordinates Q′, P ′ that are related to Q,P by a change of scale only, and that arecanonical?

Generalizing a little, we put

Q = µ(q + ip), P = ν(q − ip). (1)

The reverse transformation is

q =1

2

(

1

µQ+

1

νP

)

, p =1

2i

(

1

µQ− 1

νP

)

.

The direct conditions for canonicality, valid in cases (like this one) in which the

1

Page 47: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 9 2

transformation equations do not depend on the time explicitly, are

∂Q

∂q=

∂p

∂P

∂Q

∂p= − ∂q

∂P

∂P

∂q= − ∂p

∂Q

∂P

∂p=

∂q

∂Q.

(2)

When applied to the case at hand, all four of these yield the same condition,namely

µ = − 1

2iν.

For µ = ν = 1, which is the case Goldstein gives, these conditions are clearlynot satisfied, so (1) is not canonical. But putting µ = 1, ν = − 1

2i we see thatequations (1) are canonical.

Page 48: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 9 3

Problem 9.2

(a) For a one-dimensional system with the Hamiltonian

H =p2

2− 1

2q2,

show that there is a constant of the motion

D =pq

2−Ht.

(b) As a generalization of part (a), for motion in a plane with the Hamiltonian

H = |p|n − ar−n,

where p is the vector of the momenta conjugate to the Cartesian coordinates,show that there is a constant of the motion

D =p · rn

−Ht.

(c) The transformation Q = λq, p = λP is obviously canonical. However, the sametransformation with t time dilatation, Q = λq, p = λP, t′ = λ2t, is not. Showthat, however, the equations of motion for q and p for the Hamiltonian in part(a) are invariant under the transformation. The constant of the motion D issaid to be associated with this invariance.

(a) The equation of motion for the quantity D is

dD

dT= D,H +

∂D

∂t

The Poisson bracket of the second term in D clearly vanishes, so we have

=1

2pq,H −H

=1

4

pq, p2

− 1

4

pq,1

q2

−H. (3)

The first Poisson bracket is

pq, p2

=∂(pq)

∂q

∂(p2)

∂p− ∂(pq)

∂p

∂(p2)

∂q

= (p)(2p) − 0

= 2p2 (4)

Page 49: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 9 4

Next,

pq,1

q2

=∂(pq)

∂q

∂(

1q2

)

∂p− ∂(pq)

∂p

∂(

1q2

)

∂q

= 0 −(

− 2

q3

)

q

=2

q2(5)

Plugging (4) and (5) into (3), we obtain

dD

dt=p2

2− 1

2q2−H

= 0.

(b) We have

H = (p21 + p2

2 + p23)

n/2 − a(x21 + x2

2 + x23)

−n/2

so

∂H

∂xi= anxi(x

21 + x2

2 + x23)

−n/2−1

∂H

∂pi= 2npi(p

21 + p2

2 + p23)

n/2−1.

Then

p · r, H =∑

i

∂(p1x1 + p2x2 + p3x3)

∂xi

∂H

∂pi− ∂(p1x1 + p2x2 + p3x3)

∂pi

∂H

∂xi

=∑

i

np2i (p

21 + p2

2 + p23)

n/2−1 − anx2i (x

21 + x2

2 + x23)

−n/2−1

= n(p21 + p2

2 + p23)

n/2 − an(x21 + x2

2 + x23)

−n/2 (6)

so if we define D = p · r/n−Ht, then

dD

dT= D,H − ∂D

∂t

=1

np · r, H − ∂D

∂t

Substituting in from (6),

= |p|n − ar−n −H

= 0.

Page 50: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 9 5

(c) We put

Q(t′) = λq

(

t′

λ2

)

, P (t′) =1

λp

(

t′

λ2

)

. (7)

Since q and p are the original canonical coordinates, they satisfy

q =∂H

∂p= p

p = −∂H∂q

=1

q3.

(8)

On the other hand, differentiating (7), we have

dQ

dt′=

1

λq

(

t′

λ2

)

=1

λp

(

t′

λ2

)

= P (t′)

dP

dt′=

1

λ3p

(

t′

λ2

)

=1

λ3

1

q(

t′

λ2

)

=1

Q3(t′)

which are the same equations of motion as (8).

Problem 9.4

Show directly that the transformation

Q = log

(

1

psin p

)

, P = q cot p

is canonical.

The Jacobian of the transformation is

M =

(

∂Q∂q

∂Q∂p

∂P∂q

∂P∂p

)

=

( − 1q cot p

cot p −q csc2 p

)

.

Page 51: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 9 6

Hence

MJM =

( − 1q cot p

cot p −q csc2 p

)(

0 1−1 0

)( − 1q cot p

cot p −q csc2 p

)

=

( − 1q cot p

cot p −q csc2 p

)(

cot p −q csc2 p1q − cot p

)

=

(

0 csc2 p− cot2 pcot2 p− csc2 p 0

)

=

(

0 1−1 0

)

= J

so the symplectic condition is satisfied.

Problem 9.5

Show directly for a system of one degree of freedom that the transformation

Q = arctanαq

p, P =

αq2

2

(

1 +p2

α2q2

)

is canonical, where α is an arbitrary constant of suitable dimensions.

The Jacobian of the transformation is

M =

∂Q∂q

∂Q∂p

∂P∂q

∂P∂p

=

(

αp

)

1

1+( αq

p)2 −

(

αqp2

)

1

1+(αq

p)2

αq pα

.

so

MJM =

(

αp

)

1

1+( αq

p)2 αq

−(

αqp2

)

1

1+(αq

p)2

αq pα

−(

αp

)

1

1+(αq

p)2 +

(

αqp2

)

1

1+( αq

p)2

=

0 1

−1 0

= J

so the symplectic condition is satisfied.

Page 52: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 9 7

Problem 9.6

The transformation equations between two sets of coordinates are

Q = log(1 + q1/2 cos p)

P = 2(1 + q1/2 cos p)q1/2 sin p

(a) Show directly from these transformation equations that Q,P are canonicalvariables if q and p are.

(b) Show that the function that generates this transformation is

F3 = −(eQ − 1)2 tan p.

(a) The Jacobian of the transformation is

M =

∂Q∂q

∂Q∂p

∂P∂q

∂P∂p

=

(

12

)

q−1/2 cos p1+q1/2 cos p

− q1/2 sin p1+q1/2 cos p

q−1/2 sin p+ 2 cos p sin p 2q1/2 cos p+ 2q cos2 p− 2q sin2 p

=

(

12

)

q−1/2 cos p1+q1/2 cos p

− q1/2 sin p1+q1/2 cos p

q−1/2 sin p+ sin 2p 2q1/2 cos p+ 2q cos 2p

.

Hence we have

MJM =

(

12

)

q−1/2 cos p1+q1/2 cos p

q−1/2 sin p+ sin 2p

− q1/2 sin p1+q1/2 cos p

2q1/2 cos p+ 2q cos 2p

×

q−1/2 sin p+ sin 2p 2q1/2 cos p+ 2q cos 2p

−(

12

)

q−1/2 cos p1+q1/2 cos p

q1/2 sin p1+q1/2 cos p

=

0 cos2 p+sin2 p+q1/2 cos p cos 2p+q1/2 sin p sin 2p1+q1/2 cos p

− cos2 p+sin2 p+q1/2 cos p cos 2p+q1/2 sin p sin 2p1+q1/2 cos p

0

=

0 1

−1 0

= J

so the symplectic condition is satisfied.

Page 53: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 9 8

(b) For an F3 function the relevant relations are q = −∂F/∂p, P = −∂F/∂Q.We have

F3(p,Q) = −(eQ − 1)2 tan p

so

P = −∂F3

∂Q= 2eQ(eQ − 1) tan p

q = −∂F3

∂p= (eQ − 1)2 sec2 p.

The second of these may be solved to yield Q in terms of q and p:

Q = log(1 + q1/2 cos p)

and then we may plug this back into the equation for P to obtain

P = 2q1/2 sin p+ q sin 2p

as advertised.

Problem 9.7

(a) If each of the four types of generating functions exist for a given canonicaltransformation, use the Legendre transformation to derive relations betweenthem.

(b) Find a generating function of the F4 type for the identity transformation andof the F3 type for the exchange transformation.

(c) For an orthogonal point transformation of q in a system of n degrees of freedom,show that the new momenta are likewise given by the orthogonal transforma-tion of an n−dimensional vector whose components are the old momenta plusa gradient in configuration space.

Problem 9.8

Prove directly that the transformation

Q1 = q1, P1 = p1 − 2p2

Q2 = p2, P2 = −2q1 − q2

is canonical and find a generating function.

After a little hacking I came up with the generating function

F13(p1, Q1, q2, Q2) = −(p1 − 2Q2)Q1 + q2Q2

Page 54: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 9 9

which is of mixed F3, F1 type. This is Legendre-transformed into a function ofthe F1 type according to

F1(q1, Q1, q2, Q2) = F13 + p1q1.

The least action principle then says

p1q1 + p2q2 −H(qi, pi) = P1Q1 + P2Q2 −K(Qi, Pi) +∂F13

∂p1p1 +

∂F13

∂Q1Q1

+∂F13

∂q2q2 +

∂F13

∂Q2Q2 + p1q1 + q1p1

whence clearly

q1 = −∂F13

∂p1= Q1 X

P1 = −∂F13

∂Q1= −p1 − 2Q2

= −p1 − 2p2 X

p2 =∂F13

∂q2= Q2 X

P2 = −∂F13

∂Q2= −2Q1 − q2 = −2q1 − q2 X.

Problem 9.14

By any method you choose show that the following transformation is canonical:

x =1

α(√

2P1 sinQ1 + P2), px =α

2(√

2P1 cosQ1 −Q2)

y =1

α(√

2P1 cosQ1 +Q2), py = −α2

(√

2P1 sinQ1 − P2)

where α is some fixed parameter.Apply this transformation to the problem of a particle of charge q moving in a planethat is perpendicular to a constant magnetic field B. Express the Hamiltonian forthis problem in the (Qi, Pi) coordinates, letting the parameter α take the form

α2 =qB

c.

From this Hamiltonian obtain the motion of the particle as a function of time.

We will prove that the transformation is canonical by finding a generatingfunction. Our first step to this end will be to express everything as a function

Page 55: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 9 10

of some set of four variables of which two are old variables and two are new.After some hacking, I arrived at the set x,Q1, py, Q2. In terms of this set, theremaining quantities are

y =

(

1

2x− 1

α2py

)

cotQ1 +1

αQ2 (9)

px =

(

α2

4x− 1

2py

)

cotQ1 −α

2Q2 (10)

P1 =

(

α2x2

8− 1

2xpy +

1

2α2p2

y

)

csc2Q1 (11)

P2 =α

2x+

1

αpy (12)

We now seek a generating function of the form F (x,Q1, py, Q2). This is of mixedtype, but can be related to a generating function of pure F1 character accordingto

F1(x,Q1, y,Q2) = F (x,Q1, py, Q2) − ypy.

Then the principle of least action leads to the condition

pxx+ pyy = P1Q1 + P2Q2 +∂F

∂xx+

∂F

∂pypy +

∂F

∂Q1Q1 +

∂F

∂Q2Q2 + ypy + pyy

from which we obtain

px =∂F

∂x(13)

y = − ∂F

∂py(14)

P1 = − ∂F

∂Q1(15)

P2 = − ∂F

∂Q2. (16)

Doing the easiest first, comparing (12) and (16) we see that F must havethe form

F (x,Q1, py, Q2) = −α2xQ2 −

1

αpyQ2 + g(x,Q1, py). (17)

Plugging this in to (14) and comparing with (14) we find

g(x,Q1, py) =

(

−1

2xpy +

1

2α2p2

y

)

cotQ1 + ψ(x,Q1). (18)

Plugging (17) and (18) into (13) and comparing with (10), we see that

∂ψ

∂x=α2

4x cotQ1

Page 56: Classical Mechanics Solutions 2nd Edition Goldstein

Homer Reid’s Solutions to Goldstein Problems: Chapter 9 11

or

ψ(x,Q1) =α2x2

8cotQ1. (19)

Finally, combining (19), (18), (17), and (15) and comparing with (11) we seethat we may simply take φ(Q1) ≡ 0. The final form of the generating functionis then

F (x,Q1, py, Q2) = −(

α

2x+

1

αpy

)

Q2 +

(

α2x2

8− 1

2xpy +

1

2α2p2

y

)

cotQ1

and its existence proves the canonicality of the transformation.Turning now to the solution of the problem, we take the B field in the z

direction, i.e. B = B0k, and put

A =B0

2

(

− y i + x j)

.

Then the Hamiltonian is

H(x, y, px, py) =1

2m

(

p − q

cA)2

=1

2m

[

(

px +qB0

2cy

)2

+

(

py − qB0

2cx

)2]

=1

2m

[

(

px +α2

2y

)2

+

(

py − α2

2x

)2]

where we put α2 = qB/c. In terms of the new variables, this is

H(Q1, Q2, P1, P2) =1

2m

[

(

α√

2P1 cosQ1

)2

+(

α√

2P1 sinQ1

)2]

=α2

mP1

= ωcP1

where ωc = qB/mc is the cyclotron frequency. From the Hamiltonian equationsof motion applied to this Hamiltonian we see thatQ2, P1, and P2 are all constant,while the equation of motion for Q1 is

Q1 =∂H

∂P1= ωc −→ Q1 = ωct+ φ

for some phase φ. Putting r =√

2P1/α, x0 = P2/α, y0 = Q2/α we then have

x = r(sinωct+ φ) + x0, px =mωc

2[r cos(ωct+ φ) − y0]

y = r(cosωct+ φ) + y0, py =mωc

2[r sin(ωct+ φ) + x0]

in agreement with the standard solution to the problem.

Page 57: Classical Mechanics Solutions 2nd Edition Goldstein

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