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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 particles initialvelocity to produce circular motion? Find the period of small oscillations aboutthis circular motion.

Well take the paraboloid to be defined by the equation z = r2. The kineticand potential energies of the particle are

T =m

2(r2 + r22 + z2)

=m

2(r2 + r22 + 42r2r2)

V = mgz = mgr2.

Hence the Lagrangian is

L =m

2

[(1 + 42r2)r2 + r2 2

]mgr2.This is cyclic in , so the angular momentum is conserved:

l = mr2 = constant.

1

Homer Reids Solutions to Goldstein Problems: Chapter 3 2

For r we have the derivatives

L

r= 42mrr2 + mr2 2mgr

L

r= m(1 + 42r2)r

d

dt

L

r= 8m2rr2 + m(1 + 42r2)r.

Hence the equation of motion for r is

8m2rr2 + m(1 + 42r2)r = 42mrr2 + mr2 2mgr

or

m(1 + 42r2)r + 4m2rr2 mr2 + 2mgr = 0.

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

m(1 + 42r2)r + 4m2rr2 l2

mr3+ 2mgr = 0.

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

m(1 + 42r2)r + l2

mr3+ 2mgr = 0.

Evidently, r will then vanishand hence r will remain 0, giving circular motionif

l2

mr3= 2mgr

or =

2g.

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


Problem 3.2

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

V = k ear

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 + r22

]+ k

ear

r.

As usual the angular momentum is conserved:

l = mr2 = constant.

We have

L

r= mr2 k (1 + ar) e

ar

r2

L

r= mr

so the equation of motion for r is

r = r2 km

(1 + ar)ear

r2

=l2

m2r3 k

m(1 + ar)

ear

r2. (1)

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

=

k

m(1 + ar0)

ear0/2

r3/20

. (2)

What this means is that that if the particles initial 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, lets suppose the particle is exe-cuting a circular orbit with radius r0 (in which case the 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

)ear0r20

km

(1 + a[r0 + x])ea[r0+x]

[r0 + x]2(3)


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

km

ear0

r20(1 + ar0 + ax)(1 ax)

(1 2 x

r0

)

km

(1 + ar0)ear0

r20+

k

m

ear0

r20

(a a(1 + ar0) 2(1 + ar0)

r0

)x

km

(1 + ar0)ear0

r20 k

m

ear0

r20

(2a +

2

r0+ a2r0

)x.

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

x =k

m

ear0

r20

(2a +

2

r0+ a2r0

)x

The problem is that the RHS here has the wrong signthis equation is satisfiedby an x that grows (or decays) exponentially, rather than oscillates. SomehowI messed up the sign of the RHS, but I cant find wherecan anybody help?

Problem 3.3

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

2.

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

U(r) = kr

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

L = 2

[r2 + r22] +k

r

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

r = r2 kr2

. (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 :

r0 =

(k

2

)1/3

=

(k2

4pi2

)1/3(5)


where we used the fact that the angular velocity in the circular orbit with period is = 2pi/ .

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

r = kr2

. (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 = 2kr2

r

or

d

dt

(r2

)= +

d

dt

(2k

r

)

from which we conclude

r2 =2k

r+ 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

)1/2 1

r 1

r0

=

(2k

)1/2 r0 rrr0

. (8)

We could now proceed to solve this differential equation for r(t), but since infact were interested in solving for the time difference corresponding to givenboundary values of r, its easier to invert (8) and solve for t(r):

t =

0r0

(dt

dr

)dr

=

0r0

(dr

dt

)1

dr

=(

2k

)1/2 0r0

(rr0

r0 r)1/2

dr


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

=(

2k

)1/2r3/20

01

(u

1 u)1/2

du

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

= 2(

2k

)1/2r3/20

0pi/2

sin2 x dx

=(

2k

)1/2r3/20

pi

4.

Now plugging in (5), we obtain

t =(

2k

)1/2 ( k24pi2

)1/2 (pi4

)=

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.

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

r() =

2R(1 + cos 2)1/2

so

u() =1

r()=

12R(1 + cos 2)1/2

. (9)


Differentiating,

du

d=

sin 22R(1 + cos 2)3/2

du

d=

12R

[2 cos 2

(1 + cos 2)3/2+ 3

sin2 2

(1 + cos 2)5/2

]

=1

2

2R

1

(1 + cos 2)5/2[2 cos 2 + 2 cos2 2 + 3 sin2 2

]. (10)

Adding (9) and (10),

d2u

d2+ u =

12R(1 + cos 2)5/2

[(1 + cos 2)2 + 2 cos2 + 2 cos2 2 + 3 sin2 2

]=

12R(1 + cos 2)5/2

[4 + 4 cos 2]

=4

2R(1 + cos 2)3/2

= 8R2u3. (11)

The differential equation for the orbit is

d2u

d2+ u = m

l2d

duV

(1

u

)(12)

Plugging in (11), we have

8R2u3 = ml2

d

duV

(1

u

)

so

V

(1

u

)= 2l

2R2

mu4 V (r) = 2l

2R2

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 + r22]. (15)


We have

r =

2R(1 + cos 2)1/2

r2 = 2R2(1 + cos 2)

r =

2Rsin 2

(1 + cos 2)1/2

r2 = 2R2sin2 2

1 + cos 22

Plugging into (15),

T = mR22[

sin2 2

1 + cos 2+ 1 + cos 2

]

= mR22[sin2 2 + 1 + 2 cos 2 + cos2 2

1 + cos

]= 2mR22

In terms of l = mr2, 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 particles angle changes from = 0 to = pi/2.Hence we can calculate the period as

= 2

pi/20

dt

dd

= 2

pi/20

d

Using = l/mr2, we have

= 2m

l

pi/20

r2() d

=4R2m

l

pi/20

(1 + 2 cos 2 + cos2 2) d

=4R2m

l 3pi

4

=3piR2m

l.


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

)(parabola.)

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

(b) For the parabola, we have

r =l2

mk

(sin

(1 + cos )2

) (16)

= rsin

1 + cos

so

v2 = r2 + r22

= r22[

sin2

(1 + cos )2+ 1

]

= r22[sin2 + 1 + 2 cos + cos2

(1 + cos )2

]

= 2r22[

1

1 + cos

]

=2mkr32

l2

=2k

mr(17)

in terms of the angular momentum l = mr22. On the other hand, for the circler = 0, so

v2 = r22 =l2

m2r2=

k

mr(18)


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() =a(1 2)

1 + cos( 0) . (19)

For simplicity well take 0 = 0 for the initial motion of the particle. Thenperigee happens when = 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 particles momentumis entirely in the y direction: pi = pij. After receiving the impulse S in the radial(x) direction, the particles 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 = k2Ef

= k2Ei + S2/m

=ai

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

Next, since the impulse is in the same direction as the particles distance fromthe origin, we have L = r p = 0, i.e. the impulse does not change theparticles angular momentum:

Lf = Li L. (22)With (20) and (22), we can compute the change in the particles eccentricity:

f =

1 +

2EfL2

mk2

=

1 +

2EiL2

mk2+

L2S2

m2k2. (23)


What remains is to compute the constant 0 in (19) for the particles 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 = 0 with both the before and aftervalues of a and :

ai(1 2i )1 + i

=af (1 2f )

1 + f cos 0or

cos 0 =1

f

{af (1 2f )ai(1 i) 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

r2mCr. (24)

For a circular orbit at radius r0 this must vanish:

0 =l2

mr30 k

r20mCr0 (25)


l =

mkr0 + m2Cr40

= lmr20

=1

mr20

mkr0 + m2Cr40

=

k

mr30

1 +

mCr30k

k

mr30

[1 +

mCr302k

]

Then the period is

=2pi

2pir3/20

m

k

[1 mCr

30

2k

]

= 0

[1 C

20

8pi2

]

where 0 = 2pir3/20

m/k is the period of circular motion in the absence of the

perturbing potential.

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

mx =l2

m(r0 + x)3 k

(r0 + x)2mC(r0 + x)

l2

mr30

(1 3 x

r0

) k

r20

(1 2 x

r0

)mCr0 mCx

Using (25), this reduces to

mx =

[ 3l

2

mr40+

2k

r30mC

]x

orx + 2x = 0

with

=

[3l2

m2r40 2k

mr30 C

]1/2

=

[2l2

m2r40 k

mr30

]1/2

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


Problem 3.14

Show that the motion of a particle in the potential field

V (r) = kr

+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

=2pimh

l2.

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

=k

ka

(which is a measure of the perturbing inverse square potential relative to the grav-itational potential) were as small as 7 108. (The eccentricity of Mercurys 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


augmenting the angular velocity:

L = mr2 L(

1 +mh

L2

)= mr2

(1 +

mh

L2

)= mr2 + mr2

where

=mh

L2=

2pimh

L2

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 , 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 = mr2 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

=

(2pi

) ( mL2

)(h)

=

(2pi

) (mka

L2

) (h

ka

)

=

(2pi

) (1

1 e2) (

h

ka

)

so

h

ka= (1 e2)

2pi

= (1 e2)fprecwhere in going to the third-to-last line we used Goldsteins equation (3-62), andin the last line I put fprec = /2pi. Putting in the numbers, we find

h

ka= (1 .2062) (0.24 yr) 40 ( 1

3600

) (1 revolution

360

) (1 century1

100 yr1

)yr1

= 7.1 108.


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 Keplersequation giving t from the time of closest approach as a function of F .

We start with Goldsteins equation (3.65):

t =

m

2

rr0

drkr l

2

2mr2 + E

=

m

2

rr0

r drEr2 + kr l22m

. (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) l

2

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 k

2e2

4E l

2

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

FF0

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.


Problem 3.26

Examine the scattering produced by a repulsive central force f = kr3. Show thatthe differential cross section is given by

()d =k

2E

(1 x)dxx2(2 x)2 sin pix

where x is the ratio /pi and E is the energy.

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

d2u

d2+ u = m

l2dU

du= mk

l2u

ord2u

d2+

(1 +

mk

l2

)u = 0

with solution

u = A cos + B sin (28)

where

=

1 +

mk

l2. (29)

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

u( = pi) = 0 A cos pi + B sin pi = 0

A = B tan pi. (30)

The condition that the particle head off to r = at angle = s yields thecondition

A cos s + B sin s = 0.

Using (30), this becomes

coss tan pi + sin s = 0


or

cos s sin pi + sin s cos pi = 0 sin (s pi) = 0 (s pi) = pi

or, in terms of Goldsteins variable x = /pi,

=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 particle

energy. Thus the previous equation is

1 +k

2Es2=

1

(x 1)2or

s2 = k2E

[(x 1)2x(x 2)

].

Taking the differential of both sides,

2s ds = k2E

[2(x 1)x(x 2)

(x 1)2x2(x 2)

(x 1)2x(x 2)2

]dx

= k2E

[2x(x 1)(x 2) (x 1)2(x 2) x(x 1)2

x2(x 2)2]

= k2E

[2(1 x)

x2(x 2)2]

. (32)

The differential cross section is given by

()d =| s ds |sin

.

Plugging in (32), we have

()d =k

2E

[(1 x)

x2(x 2)2 sin ]

dx

as advertised.



Homer Reid

April 21, 2002

Chapter 7

Problem 7.2

Obtain the Lorentz transformation in which the velocity is at an infinitesimal angled 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 = R1KRwhere 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 d counterclockisefrom the original z axis, but he doesnt tell us in which plane, i.e. we know but not for the new z axis in the unrotated coordinates. Well assume the zaxis is rotated around the x axis, in a sense such that if youre 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


axis. Then, using the real metric,

L =

1 0 0 00 cos d sin d 00 sin d cos d 00 0 0 1

1 0 0 00 1 0 00 0 0 0


=


1 0 0 00 cos d sin d 00 sin d cos d 0 sin d cos d

=

1 0 0 00 cos2 d + sin2 d ( 1) sin d cos d sin d0 ( 1) sin d cos d sin2 d + cos2 d cos d0 sin d cos d

.

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?

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

xt = {0, 0, +L0/2, } xb = {0, 0,L0/2, } .

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

xt = {0, 0, (L0/2 + v), ( + L0/2c)} (1)xb = {0, 0, (L0/2 + v), ( L0/2c)} . (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


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 in therockets rest frame.

First consider the light received by the observer at time t0 coming fromthe bottom of the rocket. Suppose in the observers 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 = ct when it emittedthis light. But then the event identified by (z, t) = (ct, t0 t) must lie onthe world line of the rockets bottom, which from (2) determines both t andthe proper time at which the light was emitted:

ct = (L0/2 + v)t0 t = ( + L0/2c) = =

(1 +

1 )1/2

t0L02c

b(t0).

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

zb(b(t0)) = L0/2 + vb(t0)

= L0/2 + v{(

1 +

1 )1/2

t0 L02c

}(3)

Similarly, for the top of the rocket we have

t(t0) =

(1 +

1 )1/2

t0 +L02c

and

zt(t(t0)) = L0/2 + v

{(1 +

1 )1/2

t0 +L02c

}(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) = (1 + )L0

=

(1 +

1 )1/2

L0.


Problem 7.17

Two particles with rest masses m1 and m2 are observed to move along the observersz 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 v3relative 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 m1nor m2 are zero?

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

1m1v1 2m2v2 = 3m3v31m1c + 2m2c = 3m3c

Solving the second for m3 yields

m3 =13

m1 +23

m2 (5)

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

v3 =1m1v1 2m2v2

1m1 + 2m2

Then

3 =v3c

=1m11 2m22

1m1 + 2m2

1 23 =21m

21 + 212m1m2 +

22m

22 [21m2121 + 22m2222 212m1m212](1m1 + 2m2)2

=21m

21(1 21) + 22m22(1 22) + 212m1m2(1 12)

(1m1 + 2m2)2

=m21 + m

22 + 212m1m2(1 12)(1m1 + 2m2)2

and hence

23 =1

1 23=

(1m1 + 2m2)2

m21 + m22 + 212m1m2(1 12)

. (6)

Now, (5) shows that, for m3 to be zero when either m1 or m2 is zero, we musthave 3 = . That this condition cannot be met for nonzero m1, m2 is evidentfrom the denominator of (6), in which all terms are positive (since 12 < 1 ifm1 or m2 is nonzero).


Problem 7.19

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

(pi )22pi

c2.

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

pic2 = (2c4 + p2c2)1/2 + pc (energy conservation)

0 = p + p (momentum conservation).

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

pic2 = (2c4 + p2c2)1/2 + pc

(pic p)2 = 2c2 + p2

p = pi2 22pi

c.

Then the total energy of the muon is

E = (2c4 + p2c2)1/2

= c2(

2 +(pi2 2)2

4pi2

)1/2

=c2

2pi

(4pi22 + (pi2 2)2)1/2

=c2

2pi(pi2 + 2)

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

K = E c2 = c2

2pi(pi2 + 2) c2

=c2

2pi(pi2 + 2 2pi)

=c2

2pi(pi )2

as advertised.


Problem 7.20

A pi+ 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 hyperon (rest mass 1115 MeV). What is the threshold energy for thisreaction in the laboratory system?

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

p,pi = (ppi, pimpi), p,n = (0, mn)

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

p,T pT = |ppi|2 + (pimpi + mn)2

= |ppi|2 + 2pim2pi + m2n + 2pimpimn= m2pi + m

2n + 2pimpimn

= (mpi + mn)2 + 2(pi 1)mpimn (7)

The threshold energy is the energy needed to produce the K and particlesat rest in the COM system. In this case the squared magnitude of the four-momentum of the final system is just (mK + m)

2, and, by conservation ofmomentum, this must be equal to the magnitude of the four-momentum of theinitial system (7):

(mK + m)2 = (mpi + mn)

2 + 2(pi 1)mpimn

= pi = 1 + (mK + m)2 (mpi + mn)2

2mpimn= 6.43

Then the total energy of the pion is T = pimpi = (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 cant figure out why. Can anyonefind the mistake?

The K and particles must have, between them, the same total momentumin the direction of the original pions momentum as the original pion had. Ofcourse, the K and 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 and both travel in the direction of the original pions motion. (This is equivalentto Goldsteins conclusion that, just at threshold, the produced particles are at


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

ppi = pK + p (8)

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

(m2pi + p2pi)

1/2 + mn = (m2K + p

2K)

1/2 + (m2 + p2)

1/2. (9)

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

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

Ef = (m2K + p

2K)

1/2 + (m2 + p2)

1/2.

subject to pK + p = ppi, we obtain the condition

pK(m2K + p

2K)

1/2=

p(m2 + p

2)

1/2= pK = mK

mp (10)

Combining this with (8) yields

p =m

mK + mppi pK =

mKmK + m

ppi. (11)

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

(m2pi + p2pi)

1/2 + mn =

(m2K +

(mK

mK + m

)2p2pi

)1/2+

(m2 +

(m

mK + m

)2p2pi

)1/2

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


Problem 7.21

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

= 2c sin2 2,

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

T = h2(

c

)sin2 2

1 + 2(

c

)sin2 /2

.

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

p,i = p, + p,e =

00

h/h/

+

000

mc

=

00

h/mc + h/

. (12)

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

p,f = p, + p,e =

(h/) sin 0

(h/) cos h/

+

pe sin0

pe cosm2c2 + p2e

=

(h/) sin + pe sin 0

(h/) cos + pe cos

(h/) +

m2c2 + p2e

.

(13)

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

(h/) sin + pe sin = 0 (14)

(h/) cos + pe cos = h/ (15)

h/ +

m2c2 + p2e = mc + h/ (16)

From the first of these we find

sin = hpe

sin = cos =[1 +

(h

pe

)2sin2

]1/2


and plugging this into (15) we find

p2e =h2

2+

h2

2 2 h

2

cos . (17)

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

p2e = h2

(1

1

)2+ 2mch

(1

1

).

Comparing these two determinations of pe yields

cos = 1 mch

( )

or

sin2

2=

mc

2h( ) = 1

2c( )

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

= mc +

h

= ( 1)mc = K = h(

1

1

)

= h

(

)

= h

(2c sin

2(/2)

[ + 2c sin2(/2)]

)

=h

(2 sin2(/2)

1 + 2 sin2(/2)

)

where we put = c/.

Problem 7.22

A photon of energy E collides at angle 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 ) .

Well 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


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

p1 =

(0, 0,

E

c,E

c

)

p2 =

(Ec

sin , 0,Ec

cos ,Ec

)

pt = p1 + p2 =

(Ec

sin , 0,E + E cos

c,E + E

c

)

Its 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

pt =

(0, 0,

1

c

E2 + E2 + 2EE cos , 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 = 1cE2 + E2 + 2EE cos p2c2 = 14 (E2 + E2 + 2EE cos )

2

m2c2 + p2 = E+Ec m2c4 + p2c2 = 14 (E2 + E2 + 2EE)

Subtracting the first of these from the second, we obtain

m2c4 =EE2

(1 cos )

or

E = 2m2c4

E(1 cos )as advertised.



Homer Reid

October 29, 2002

Chapter 9

Problem 9.1

One of the attempts at combining the two sets of Hamiltons equations into onetries to take q and p as forming a complex quantity. Show directly from Hamiltonsequations 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 = (q + ip), P = (q ip). (1)

The reverse transformation is

q =1

2

(1

Q+

1

P

), p =

1

2i

(1

Q 1

P

).

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

1


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

= 12i

.

For = = 1, which is the case Goldstein gives, these conditions are clearlynot satisfied, so (1) is not canonical. But putting = 1, = 12i we see thatequations (1) are canonical.


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

2Ht.

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

H = |p|n arn,

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 = q, p = P is obviously canonical. However, the sametransformation with t time dilatation, Q = q, p = P, t = 2t, 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

2{pq,H} H

=1

4

{pq, p2

} 14

{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)


Next,

{pq,

1

q2

}=(pq)

q

(

1q2

)p

(pq)p

(

1q2

)q

= 0( 2q3

)q

=2

q2(5)

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

dD

dt=p2

2 1

2q2H

= 0.

(b) We have

H = (p21 + p22 + p

23)

n/2 a(x21 + x22 + x23)n/2

so

H

xi= anxi(x

21 + x

22 + x

23)

n/21

H

pi= 2npi(p

21 + p

22 + p

23)

n/21.

Then

{p r, H} =

i

{(p1x1 + p2x2 + p3x3)

xi

H

pi (p1x1 + p2x2 + p3x3)

pi

H

xi

}

=

i

{np2i (p

21 + p

22 + p

23)

n/21 anx2i (x21 + x22 + x23)n/21}

= n(p21 + p22 + p

23)

n/2 an(x21 + x22 + x23)n/2 (6)

so if we define D = p r/nHt, thendD

dT= {D,H} D

t

=1

n{p r, H} D

t

Substituting in from (6),

= |p|n arn H= 0.


(c) We put

Q(t) = q

(t

2

), P (t) =

1

p

(t

2

). (7)

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

q =H

p= p

p = Hq

=1

q3.

(8)

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

dQ

dt=

1

q

(t

2

)

=1

p

(t

2

)= P (t)

dP

dt=

1

3p

(t

2

)

=1

31

q(

t

2

)=

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 =

(Qq

Qp

Pq

Pp

)

=

( 1q cot pcot p q csc2 p

).


Hence

MJM =


)(0 11 0

)( 1q cot pcot p q csc2 p

)

=


)(cot p q csc2 p

1q cot p

)

=

(0 csc2 p cot2 p

cot2 p csc2 p 0)

=

(0 11 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 = arctanq

p, P =

q2

2

(1 +

p2

2q2

)

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

The Jacobian of the transformation is

M =

Qq Qp

Pq

Pp

=

(

p

)1

1+( qp

)2(

qp2

)1

1+(qp

)2

q p

.

so

MJM =

(p

)1

1+( qp

)2 q

(

qp2

)1

1+(qp

)2p

q p

(

p

)1

1+(qp

)2+(

qp2

)1

1+( qp

)2

=

0 1

1 0

= J



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 =

Qq Qp

Pq

Pp

=

( 12) q1/2 cos p1+q1/2 cos p q1/2 sin p1+q1/2 cos p

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

=

( 12) q1/2 cos p1+q1/2 cos p q1/2 sin p1+q1/2 cos p

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

.

Hence we have

MJM =

( 12) q1/2 cos p1+q1/2 cos p q1/2 sin p+ sin 2p

q1/2 sin p1+q1/2 cos p

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

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

(12) q1/2 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



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

F3(p,Q) = (eQ 1)2 tan pso

P = F3Q

= 2eQ(eQ 1) tan p

q = F3p

= (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 ndimensional vector whose components are the old momenta plusa gradient in configuration space.

Problem 9.8

Prove directly that the transformation

Q1 = q1, P1 = p1 2p2Q2 = 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


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) + F13p1

p1 +F13Q1

Q1

+F13q2

q2 +F13Q2

Q2 + p1q1 + q1p1

whence clearly

q1 = F13p1

= Q1 X

P1 = F13Q1

= p1 2Q2= p1 2p2 X

p2 =F13q2

= Q2 X

P2 = F13Q2

= 2Q1 q2 = 2q1 q2 X.

Problem 9.14

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

x =1

(

2P1 sinQ1 + P2), px =

2(

2P1 cosQ1 Q2)

y =1

(

2P1 cosQ1 +Q2), py = 2

(

2P1 sinQ1 P2)

where 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 take the form

2 =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


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

2py

)cotQ1 +

1

Q2 (9)

px =

(2

4x 1

2py

)cotQ1

2Q2 (10)

P1 =

(2x2

8 1

2xpy +

1

22p2y

)csc2Q1 (11)

P2 =

2x+

1

py (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 = Fpy

(14)

P1 = FQ1

(15)

P2 = FQ2

. (16)

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

F (x,Q1, py, Q2) = 2xQ2 1

pyQ2 + g(x,Q1, py). (17)

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

g(x,Q1, py) =

(1

2xpy +

1

22p2y

)cotQ1 + (x,Q1). (18)

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

x=2

4x cotQ1


or

(x,Q1) =2x2

8cotQ1. (19)

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

F (x,Q1, py, Q2) = (

2x+

1

py

)Q2 +

(2x2

8 1

2xpy +

1

22p2y

)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 =B02

( y i + x j

).

Then the Hamiltonian is

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

2m

(p q

cA)2

=1

2m

[(px +

qB02c

y

)2+

(py qB0

2cx

)2]

=1

2m

[(px +

2

2y

)2+

(py

2

2x

)2]

where we put 2 = qB/c. In terms of the new variables, this is

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

2m

[(

2P1 cosQ1

)2+(

2P1 sinQ1

)2]

=2

mP1

= cP1

where c = 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= c Q1 = ct+

for some phase . Putting r =

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

x = r(sinct+ ) + x0, px =mc

2[r cos(ct+ ) y0]

y = r(cosct+ ) + y0, py =mc

2[r sin(ct+ ) + x0]

in agreement with the standard solution to the problem.

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