[Goldstein herbert] classical_mechanics_solution_m(book_see.org)

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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 the
inside of a paraboloid of revolution whose axis is vertical. Find the one-dimensional
problem equivalent to its motion. What is the condition on the particle’s initial
velocity to produce circular motion? Find the period of small oscillations about
this circular motion.
We’ll take the paraboloid to be deﬁned by the equation z = αr2
. The kinetic
and potential energies of the particle are
T =
m
2
( ˙r2
+ r2 ˙θ2
+ ˙z2
)
=
m
2
( ˙r2
+ r2 ˙θ2
+ 4α2
r2
˙r2
)
V = mgz = mgαr2
.
Hence the Lagrangian is
L =
m
2
(1 + 4α2
r2
) ˙r2
+ r2 ˙θ2
− mgαr2
.
This is cyclic in θ, so the angular momentum is conserved:
l = mr2 ˙θ = constant.
1

Homer Reid’s Solutions to Goldstein Problems: Chapter 3 2
For r we have the derivatives
∂L
∂r
= 4α2
mr ˙r2
+ mr ˙θ2
− 2mgαr
∂L
∂ ˙r
= m(1 + 4α2
r2
) ˙r
d
dt
∂L
∂ ˙r
= 8mα2
r ˙r2
+ m(1 + 4α2
r2
)¨r.
Hence the equation of motion for r is
8mα2
r ˙r2
+ m(1 + 4α2
r2
)¨r = 4α2
mr ˙r2
+ mr ˙θ2
− 2mgαr
or
m(1 + 4α2
r2
)¨r + 4mα2
r ˙r2
− mr ˙θ2
+ 2mgαr = 0.
In terms of the constant angular momentum, we may rewrite this as
m(1 + 4α2
r2
)¨r + 4mα2
r ˙r2
−
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 + 4α2
r2
)¨r + −
l2
mr3
+ 2mgαr = 0.
Evidently, ¨r will then vanish—and hence ˙r will remain 0, giving circular motion—
if
l2
mr3
= 2mgαr
or
˙θ = 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 on
˙θ for circular motion is independent of r.

Problem 3.2
A particle moves in a central force ﬁeld given by the potential
V = −k
e−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 and
E appropriate to each type of motion. When are circular orbits possible? Find the
period of small radial oscillations about the circular motion.
The Lagrangian is
L =
m
2
˙r2
+ r2 ˙θ2
+ k
e−ar
r
.
As usual the angular momentum is conserved:
l = mr2 ˙θ = constant.
We have
∂L
∂r
= mr ˙θ2
− k (1 + ar)
e−ar
r2
∂L
∂ ˙r
= m ˙r
so the equation of motion for r is
¨r = r ˙θ2
−
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
r
3/2
0
. (2)
What this means is that that if the particle’s initial θ velocity is equal to the
above function of the starting radius r0, then the second derivative of r will
remain zero for all time. (Note that, in contrast to the previous problem, in this
case 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
r2
0
−
k
m
(1 + a[r0 + x])
e−a[r0+x]
[r0 + x]2
(3)

Since x is small, we may write the second term approximately as
≈
k
m
e−ar0
r2
0
(1 + ar0 + ax)(1 − ax) 1 − 2
x
r0
≈
k
m
(1 + ar0)
e−ar0
r2
0
+
k
m
e−ar0
r2
0
a − a(1 + ar0) − 2
(1 + ar0)
r0
x
≈
k
m
(1 + ar0)
e−ar0
r2
0
−
k
m
e−ar0
r2
0
2a +
2
r0
+ a2
r0 x.
The first term here just cancels the first term in (??), so we are left with
¨x =
k
m
e−ar0
r2
0
2a +
2
r0
+ a2
r0 x
The problem is that the RHS here has the wrong sign—this equation is satisfied
by an x that grows (or decays) exponentially, rather than oscillates. Somehow
I 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 are
then released and allowed to fall into each other. Prove that they collide after a
time τ/4
√
2.
Since we are dealing with gravitational forces, the potential energy between
the particles is
U(r) = −
k
r
and, after reduction to the equivalent one-body problem, the Lagrangian is
L =
µ
2
[ ˙r2
+ r2 ˙θ2
] +
k
r
where µ is the reduced mass. The equation of motion for r is
µ¨r = µr ˙θ2
−
k
r2
. (4)
If the particles are to move in circular orbits with radius r0, (4) must vanish at
r = r0, which yields a relation between r0 and ˙θ:
r0 =
k
µ ˙θ2
1/3
=
kτ2
4π2µ
1/3
(5)

where we used the fact that the angular velocity in the circular orbit with period
τ is ˙θ = 2π/τ.
When the particles are stopped, the angular velocity goes to zero, and the
first term in (4) vanishes, leaving only the second term:
¨r = −
k
µr2
. (6)
This differential equation governs the evolution of the particles after they are
stopped. We now want to use this equation to find r as a function of t, which
we will then need to invert to find the time required for the particle separation
r to go from r0 to 0.
The first step is to multiply both sides of (6) by the integrating factor 2 ˙r:
2 ˙r¨r = −
2k
µr2
˙r
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 simply
that ˙r = 0 when r = r0, since initially the particles are not moving at all. With
the 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 in
fact we’re interested in solving for the time difference corresponding to given
boundary 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

We change variables to u = r/r0, du = dr/r0 :
=
µ
2k
1/2
r
3/2
0
0
1
u
1 − u
1/2
du
Next we change variables to u = sin2
x, du = 2 sin x cos x dx :
= 2
µ
2k
1/2
r
3/2
0
0
π/2
sin2
x dx
=
µ
2k
1/2
r
3/2
0
π
4
.
Now plugging in (5), we obtain
∆t =
µ
2k
1/2 kτ2
4π2µ
1/2
π
4
=
τ
4
√
2
as advertised.
Problem 3.6
(a) Show that if a particle describes a circular orbit under the influence of an
attractive central force directed at a point on the circle, then the force varies
as 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 all
three 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 orbit
is a circle of radius R centered at (x = R, y = 0) (so that the leftmost point
of the particle’s origin is the center of force). The equation describing such an
orbit is
r(θ) =
√
2R(1 + cos 2θ)1/2
so
u(θ) =
1
r(θ)
=
1
√
2R(1 + cos 2θ)1/2
. (9)

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 cos2θ + 2 cos2
2θ + 3 sin2
2θ . (10)
Adding (9) and (10),
d2
u
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
= 8R2
u3
. (11)
The diﬀerential equation for the orbit is
d2
u
dθ2
+ u = −
m
l2
d
du
V
1
u
(12)
Plugging in (11), we have
8R2
u3
= −
m
l2
d
du
V
1
u
so
V
1
u
= −
2l2
R2
m
u4
−→ V (r) = −
2l2
R2
mr4
(13)
so
f(r) = −
8l2
R2
mr5
(14)
which is the advertised r dependence of the force.
(b) The kinetic energy of the particle is
T =
m
2
[ ˙r2
+ r2 ˙θ2
]. (15)

We have
r =
√
2R(1 + cos 2θ)1/2
r2
= 2R2
(1 + cos 2θ)
˙r =
√
2R
sin 2θ
(1 + cos 2θ)1/2
˙θ
˙r2
= 2R2 sin2
2θ
1 + cos 2θ
˙θ2
Plugging into (15),
T = mR2 ˙θ2 sin2
2θ
1 + cos 2θ
+ 1 + cos 2θ
= mR2 ˙θ2 sin2
2θ + 1 + 2 cos2θ + cos2
2θ
1 + cos θ
= 2mR2 ˙θ2
In terms of l = mr2 ˙θ, this is just
=
2R2
l2
mr4
But this is just the negative of the potential energy, (13); hence the total particle
energy T + V is zero.
(c) Suppose the particle starts out at the furthest point from the center of force
on its orbit, i.e the point x = 2R, y = 0, and that it moves counter-clockwise
from this point to the origin. The time required to undergo this motion is half
the 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
dθ
dθ
= 2
π/2
0
dθ
˙θ
Using ˙θ = l/mr2
, we have
= 2
m
l
π/2
0
r2
(θ) dθ
=
4R2
m
l
π/2
0
(1 + 2 cos2θ + cos2
2θ) dθ
=
4R2
m
l
·
3π
4
=
3πR2
m
l
.

Problem 3.8
(a) For circular and parabolic orbits in an attractive 1/r potential having the same
angular momentum, show that the perihelion distance of the parabola is one
half the radius of the circle.
(b) Prove that in the same central force as in part (a) the speed of a particle at
any 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 case
r = 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)
= r ˙θ
sin θ
1 + cos θ
so
v2
= ˙r2
+ r2 ˙θ2
= r2 ˙θ2 sin2
θ
(1 + cos θ)2
+ 1
= r2 ˙θ2 sin2
θ + 1 + 2 cos θ + cos2
θ
(1 + cos θ)2
= 2r2 ˙θ2 1
1 + cos θ
=
2mkr3 ˙θ2
l2
=
2k
mr
(17)
in terms of the angular momentum l = mr2 ˙θ2
. On the other hand, for the circle
˙r = 0, so
v2
= r2 ˙θ2
=
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 the
same 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 another
elliptic orbit. Determine the new semimajor axis, eccentricity, and orientation of
major axis in terms of the old.
The orbit equation for elliptical motion is
r(θ) =
a(1 − 2
)
1 + cos(θ − θ0)
. (19)
For simplicity we’ll take θ0 = 0 for the initial motion of the particle. Then
perigee happens when θ = 0, which is to say the major axis of the orbit is on
the x axis.
Then at the point at which the impulse is delivered, the particle’s momentum
is entirely in the y direction: pi = pi
ˆj. After receiving the impulse S in the radial
(x) direction, the particle’s y momentum is unchanged, but its x momentum is
now px = S. Hence the final momentum of the particle is pf = Sî+pi
ˆj. Since the
particle is in the same location before and after the impulse, its potential energy
is 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 from
the origin, we have ∆L = r × ∆p = 0, i.e. the impulse does not change the
particle’s angular momentum:
Lf = Li ≡ L. (22)
With (20) and (22), we can compute the change in the particle’s eccentricity:
f = 1 +
2Ef L2
mk2
= 1 +
2EiL2
mk2
+
L2S2
m2k2
. (23)

What remains is to compute the constant θ0 in (19) for the particle’s orbit after
the collision. To do this we need merely observe that, since the location of the
particle 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 :
ai(1 − 2
i )
1 + i
=
af (1 − 2
f )
1 + f cos θ0
or
cos θ0 =
1
f
af (1 − 2
f )
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 gravitational
constant and the density of the dust, and r is the radius vector from the sun to the
planet (both considered as points). This additional force is very small compared to
the 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 ellipse
and ﬁnd the precession frequency. Is the precession the same or opposite
direction to the orbital angular velocity?
(a) The equation of motion for r is
m¨r =
l2
mr3
+ f(r)
=
l2
mr3
−
k
r2
− mCr. (24)
For a circular orbit at radius r0 this must vanish:
0 =
l2
mr3
0
−
k
r2
0
− mCr0 (25)

−→ l = mkr0 + m2Cr4
0
−→ ˙θ =
l
mr2
0
=
1
mr2
0
mkr0 + m2Cr4
0
=
k
mr3
0
1 +
mCr3
0
k
≈
k
mr3
0
1 +
mCr3
0
2k
Then the period is
τ =
2π
˙θ
≈ 2πr
3/2
0
m
k
1 −
mCr3
0
2k
= τ0 1 −
Cτ2
0
8π2
where τ0 = 2πr
3/2
0 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:
m¨x =
l2
m(r0 + x)3
−
k
(r0 + x)2
− mC(r0 + x)
≈
l2
mr3
0
1 − 3
x
r0
−
k
r2
0
1 − 2
x
r0
− mCr0 − mCx
Using (25), this reduces to
m¨x = −
3l2
mr4
0
+
2k
r3
0
− mC x
or
¨x + ω2
x = 0
with
ω =
3l2
m2r4
0
−
2k
mr3
0
− C
1/2
=
2l2
m2r4
0
−
k
mr3
0
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) = −
k
r
+
h
r2
is the same as that of the motion under the Kepler potential alone when expressed
in 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 small
compared 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 this
precession 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 orbit
is 0.206, and its period is 0.24 year).
The eﬀective one-dimensional equation of motion is
m¨r =
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
, and
write
m¨r =
[L + (mh/L)]2
mr3
+
k
r2
(26)
which is just the normal equation of motion for the Kepler problem, but with
the 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 ˙θ −→ L 1 +
mh
L2
= mr2 ˙θ 1 +
mh
L2
= mr2 ˙θ + mr2 ˙Ω
where
˙Ω =
mh
L2 ˙θ
=
2πmh
L2τ
is a precession frequency. If we were to go back and work the problem in the
reference frame in which everything is precessing with angular velocity ˙Ω, but
there is no term h/r2
in the potential, then the equations of motion would come
out the same as in the stationary case, but with a term ∆L = mr2 ˙Ω added to
the 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 =
2π
τ
m
L2
(h)
=
2π
τ
mka
L2
h
ka
=
2π
τ
1
1 − e2
h
ka
so
h
ka
= (1 − e2
)
τ ˙Ω
2π
= (1 − e2
)τfprec
where in going to the third-to-last line we used Goldstein’s equation (3-62), and
in 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
.

Problem 3.22
In hyperbolic motion in a 1/r potential the analogue of the eccentric anomaly is F
deﬁned by
r = a(e cosh F − 1),
where a(1 − e) is the distance of closest approach. Find the analogue to Kepler’s
equation 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
k
r − 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 cosh F − 1) −
l2
2m
= Ea2
e2
cosh2
F + ae(k − 2Ea) cosh F + Ea2
− ka −
l2
2m
It follows from the orbit equation that, if a(e − 1) is the distance of closest
approach, then a = k/2E. Thus
=
k2
e2
4E
cosh2
F −
k2
e2
4E
−
l2
2m
=
k2
4E
e2
cosh2
F − 1 +
2El2
mk2
=
k2
e2
4E
cosh2
F − 1 =
k2
e2
4E
sinh2
F = a2
e2
E sinh2
F.
Plugging into (27) and observing that dr = ae sinh F dF, we have
t =
m
2E
F
F0
a(e cosh F − 1) dF =
ma2
2E
[e(sinh F − sinh F0) − (F − F0)]
and I suppose this equation could be a jumping-oﬀ point for numerical or other
investigations of the time of travel in hyperbolic orbit problems.

Problem 3.26
Examine the scattering produced by a repulsive central force f = kr−3
. Show that
the 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 equation
for the orbit reads
d2
u
dθ2
+ u = −
m
l2
dU
du
= −
mk
l2
u
or
d2
u
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. The
first of these observations gives us a relation between A and B in the orbit
equation (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 the
condition
A cos γθs + B sin γθs = 0.
Using (30), this becomes
− cosγθs tan γπ + sin γθs = 0

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)1/2
s with s the impact parameter and E the particle
energy. 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.

Homer Reid
April 21, 2002
Chapter 7
Problem 7.2
Obtain the Lorentz transformation in which the velocity is at an inﬁnitesimal angle
dθ counterclockwise from the z axis, by means of a similarity transformation applied
to Eq. (7-18). Show directly that the resulting matrix is orthogonal and that the
inverse matrix is obtained by substituting −v for v.
We can obtain this transformation by ﬁrst applying a pure rotation to rotate
the z axis into the boost axis, then applying a pure boost along the (new) z
axis, and then applying the inverse of the original rotation to bring the z axis
back in line with where it was originally. Symbolically we have L = R−1
KR
where R is the rotation to achieve the new z axis, and K is the boost along the
z axis.
Goldstein tells us that the new z axis is to be rotated dθ counterclockise
from the original z axis, but he doesn’t tell us in which plane, i.e. we know θ
but not φ for the new z axis in the unrotated coordinates. We’ll assume the z
axis is rotated around the x axis, in a sense such that if you’re standing on the
positive x axis, looking toward the negative x axis, the rotation appears to be
counterclockwise, so that the positive z axis is rotated toward the negative y
1

axis. Then, using the real metric,
L =




1 0 0 0
0 cos dθ sin dθ 0
0 − sin dθ cos dθ 0
0 0 0 1








1 0 0 0
0 1 0 0
0 0 γ −βγ
0 0 −βγ γ








1 0 0 0
0 cos dθ − sin dθ 0
0 sin dθ cos dθ 0
0 0 0 1




=




1 0 0 0
0 cos dθ sin dθ 0
0 − sin dθ cos dθ 0
0 0 0 1








1 0 0 0
0 cos dθ − sin dθ 0
0 γ sin dθ γ cos dθ −βγ
0 −βγ sin dθ −βγ cos dθ γ




=




1 0 0 0
0 cos2
dθ + γ sin2
dθ (γ − 1) sin dθ cos dθ −βγ sin dθ
0 (γ − 1) sin dθ cos dθ sin2
dθ + γ cos2
dθ −βγ cos dθ
0 −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 z
axis of an inertial system. An observer at the origin observes the apparent length
of the rocket at any time by noting the z coordinates that can be seen for the head
and tail of the rocket. How does this apparent length vary as the rocket moves from
the extreme left of the observer to the extreme right?
Let’s imagine a coordinate system in which the rocket is at rest and centered
at 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 rest
frame of the observer is moving in the negative z direction with speed v = βc
relative to the rest frame of the rocket. Transforming the world lines of the
rocket’s top and bottom to the rest frame of the observer, we have
xt
µ = {0, 0, γ(L0/2 + vτ), γ(τ + βL0/2c)} (1)
xb
µ = {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 is
receiving light from two events, namely, the top and bottom of the rocket moving
past imaginary distance signposts that we pretend to exist up and down the z
axis. He sees the top of the rocket lined up with one distance signpost and the
bottom of the rocket lined up with another, and from the diﬀerence between the
two signposts he computes the length of the rocket. Of course, the light that
he 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 receives
simultaneously at time t were in fact emitted at diﬀerent proper times τ in the
rocket’s rest frame.
First consider the light received by the observer at time t0 coming from
the bottom of the rocket. Suppose in the observer’s rest frame this light were
emitted at time t0 − ∆t, i.e. ∆t seconds before it reaches the observer at the
origin; then the rocket bottom was passing through z = −c∆t when it emitted
this light. But then the event identiﬁed by (z, t) = (−c∆t, t0 − ∆t) must lie on
the world line of the rocket’s bottom, which from (2) determines both ∆t and
the 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 the
bottom of the rocket emits the light that arrives at the observer’s origin at the
observer’s time t0. At this proper time, from (2), the position of the bottom of
the 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 the
observer from his observations at time t0 in his reference frame:
L(t0) = γ(1 + β)L0
=
1 + β
1 − β
1/2
L0.

Problem 7.17
Two particles with rest masses m1 and m2 are observed to move along the observer’s
z axis toward each other with speeds v1 and v2, respectively. Upon collision they
are 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 it
be 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-momentum
of the system yields
γ1m1v1 − γ2m2v2 = γ3m3v3
γ1m1c + γ2m2c = γ3m3c
Solving the second for m3 yields
m3 =
γ1
γ3
m1 +
γ2
γ3
m2 (5)
and plugging this into the ﬁrst yields v3 in terms of the properties of particles
1 and 2:
v3 =
γ1m1v1 − γ2m2v2
γ1m1 + γ2m2
Then
β3 =
v3
c
=
γ1m1β1 − γ2m2β2
γ1m1 + γ2m2
1 − β2
3 =
γ2
1 m2
1 + 2γ1γ2m1m2 + γ2
2 m2
2 − [γ2
1m2
1β2
1 + γ2
2 m2
2β2
2 − 2γ1γ2m1m2β1β2]
(γ1m1 + γ2m2)2
=
γ2
1 m2
1(1 − β2
1 ) + γ2
2m2
2(1 − β2
2) + 2γ1γ2m1m2(1 − β1β2)
(γ1m1 + γ2m2)2
=
m2
1 + m2
2 + 2γ1γ2m1m2(1 − β1β2)
(γ1m1 + γ2m2)2
and hence
γ2
3 =
1
1 − β2
3
=
(γ1m1 + γ2m2)2
m2
1 + m2
2 + 2γ1γ2m1m2(1 − β1β2)
. (6)
Now, (5) shows that, for m3 to be zero when either m1 or m2 is zero, we must
have γ3 = ∞. That this condition cannot be met for nonzero m1, m2 is evident
from the denominator of (6), in which all terms are positive (since β1β2 1 if
m1 or m2 is nonzero).

Problem 7.19
A meson of mass π comes to rest and disintegrates into a meson of mass µ and a
neutrino 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
2π
c2
.
Working in the rest frame of the pion, the conservation relations are
πc2
= (µ2
c4
+ p2
µc2
)1/2
+ pνc (energy conservation)
0 = pµ + pν (momentum conservation).
From the second of these it follows that the muon and neutrino must have the
same momentum, whose magnitude we’ll call p. Then the energy conservation
relation becomes
πc2
= (µ2
c4
+ p2
c2
)1/2
+ pc
−→ (πc − p)2
= µ2
c2
+ p2
−→ p =
π2
− µ2
2π
c.
Then the total energy of the muon is
Eµ = (µ2
c4
+ p2
c2
)1/2
= c2
µ2
+
(π2
− µ2
)2
4π2
1/2
=
c2
2π
4π2
µ2
+ (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.

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 this
reaction in the laboratory system?
We’ll put c = 1 for this problem. The four-momenta of the pion and neutron
before the collision are
pµ,π = (pπ, γπmπ), pµ,n = (0, mn)
and the squared magnitude of the initial four-momentum is thus
pµ,T pµ
T = −|pπ|2
+ (γπmπ + mn)2
= −|pπ|2
+ γ2
πm2
π + m2
n + 2γπmπmn
= m2
π + m2
n + 2γπmπmn
= (mπ + mn)2
+ 2(γπ − 1)mπmn (7)
The threshold energy is the energy needed to produce the K and Λ particles
at 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 of
momentum, this must be equal to the magnitude of the four-momentum of the
initial 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) = 898
MeV, while its kinetic energy is K = T − m = 758 MeV.
The above appears to be the correct solution to this problem. On the other
hand, I first tried to do it a different way, as below. This way yields a different
and hence presumably incorrect answer, but I can’t figure out why. Can anyone
find the mistake?
The K and Λ particles must have, between them, the same total momentum
in the direction of the original pion’s momentum as the original pion had. Of
course, the K and Λ may also have momentum in directions transverse to the
original pion momentum (if so, their transverse momenta must be equal and
opposite). But any transverse momentum just increases the energy of the final
system, which increases the energy the initial system must have had to produce
the final system. Hence the minimum energy situation is that in which the K and
Λ both travel in the direction of the original pion’s motion. (This is equivalent
to Goldstein’s conclusion that, just at threshold, the produced particles are at

rest in the COM system). Then the momentum conservation relation becomes
simply
pπ = pK + pλ (8)
and the energy conservation relation is (with c = 1)
(m2
π + p2
π)1/2
+ mn = (m2
K + p2
K)1/2
+ (m2
Λ + p2
Λ)1/2
. (9)
The problem is to find the minimum value of pπ that satisfies (9) subject to the
constraint (8).
To solve this we must first resolve a subquestion: for a given pπ, what is the
relative allocation of momentum to pK and pΛ that minimizes (9) ? Minimizing
Ef = (m2
K + p2
K)1/2
+ (m2
Λ + p2
Λ)1/2
.
subject to pK + pΛ = pπ, we obtain the condition
pK
(m2
K + p2
K)1/2
=
pΛ
(m2
Λ + p2
Λ)1/2
=⇒ pK =
mK
mΛ
pΛ (10)
Combining this with (8) yields
pΛ =
mΛ
mK + mΛ
pπ pK =
mK
mK + mΛ
pπ. (11)
For a given total momentum pπ, the minimum possible energy the final system
can have is realized when pπ is partitioned between pK and pΛ according to
(11). Plugging into (8), the relation defining the threshold momentum is
(m2
π + p2
π)1/2
+ mn = m2
K +
mK
mK + mΛ
2
p2
π
1/2
+ m2
Λ +
mΛ
mK + mΛ
2
p2
π
1/2
Solving numerically yields pπ ≈ 655 MeV/c, for a total pion energy of about
670 MeV.

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 photon
collides with an electron of mass m at rest it will be scattered at some angle θ with
a new energy hν . Show that the change in energy is related to the scattering angle
by the formula
λ − λ = 2λc sin2 θ
2
,
where λc = h/mc, known as the Compton wavelength. Show also that the kinetic
energy of the recoil motion of the electron is
T = hν
2 λc
λ sin2 θ
2
1 + 2 λc
λ sin2
θ/2
.
Let’s assume the photon is initially travelling along the z axis. Then the sum
of the initial photon and electron four-momenta is
pµ,i = pµ,γ + pµ,e =




0
0
h/λ
h/λ



 +




0
0
0
mc



 =




0
0
h/λ
mc + h/λ



 . (12)
Without loss of generality we may assume that the photon and electron move
in the xz plane after the scatter. If the photon’s velocity makes an angle θ with
the z axis, while the electron’s velocity makes an angle φ, the four-momentum
after the collision is
pµ,f = pµ,γ + pµ,e =




(h/λ ) sin θ
0
(h/λ ) cos θ
h/λ



 +




pe sin φ
0
pe cos φ
m2c2 + p2
e



 =




(h/λ ) sin θ + pe sin φ
0
(h/λ ) cos θ + pe cos φ
(h/λ ) + m2c2 + p2
e



 .
(13)
Equating (12) and (13) yields three separate equations:
(h/λ ) sin θ + pe sin φ = 0 (14)
(h/λ ) cos θ + pe cos φ = h/λ (15)
h/λ + m2c2 + p2
e = mc + h/λ (16)
From the ﬁrst of these we ﬁnd
sin φ = −
h
λ pe
sin θ =⇒ cos φ = 1 +
h
λ pe
2
sin2
θ
1/2

and plugging this into (15) we find
p2
e =
h2
λ2
+
h2
λ 2
− 2
h2
λλ
cos θ. (17)
On the other hand, we can solve (16) to obtain
p2
e = h2 1
λ
−
1
λ
2
+ 2mch
1
λ
−
1
λ
.
Comparing these two determinations of pe yields
cosθ = 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. Prove
that the minimum value of E permitting formation of a pair of particles of mass m
is
Eth =
2m2
c4
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

spherical polar angles θ and φ = 0). Then the 4-momenta are
p1 = 0, 0,
E
c
,
E
c
p2 =
E
c
sin θ, 0,
E
c
cos θ,
E
c
pt = p1 + p2 =
E
c
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 portion
of 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 θ,
E + E
c
. (18)
At threshold energy, the two produced particles have the same four-momenta:
p3 = p4 = 0, 0, p, (m2
c2
+ p2
)1/2
(19)
and 4-momentum conservation requires that twice (19) add up to (18), which
yields two conditions:
2p = 1
c
√
E2 + E2 + 2EE cos θ −→ p2
c2
= 1
4 (E2
+ E2
+ 2EE cos θ)
2 m2c2 + p2 = E+E
c −→ m2
c4
+ p2
c2
= 1
4 (E2
+ E2
+ 2EE)
Subtracting the ﬁrst of these from the second, we obtain
m2
c4
=
EE
2
(1 − cos θ)
or
E =
2m2
c4
E(1 − cos θ)
as advertised.

Homer Reid
October 29, 2002
Chapter 9
Problem 9.1
One of the attempts at combining the two sets of Hamilton’s equations into one
tries to take q and p as forming a complex quantity. Show directly from Hamilton’s
equations 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 of
coordinates Q , P that are related to Q, P by a change of scale only, and that are
canonical?
Generalizing a little, we put
Q = µ(q + ip), P = ν(q − ip). (1)
The reverse transformation is
q =
1
2
1
µ
Q +
1
ν
P , p =
1
2i
1
µ
Q −
1
ν
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
µ = −
1
2iν
.
For µ = ν = 1, which is the case Goldstein gives, these conditions are clearly
not satisﬁed, so (1) is not canonical. But putting µ = 1, ν = − 1
2i we see that
equations (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
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 · r
n
− Ht.
(c) The transformation Q = λq, p = λP is obviously canonical. However, the same
transformation with t time dilatation, Q = λq, p = λP, t = λ2
t, is not. Show
that, 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 is
said 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
−
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)

Next,
pq,
1
q2
=
∂(pq)
∂q
∂ 1
q2
∂p
−
∂(pq)
∂p
∂ 1
q2
∂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 = (p2
1 + p2
2 + p2
3)n/2
− a(x2
1 + x2
2 + x2
3)−n/2
so
∂H
∂xi
= anxi(x2
1 + x2
2 + x2
3)−n/2−1
∂H
∂pi
= 2npi(p2
1 + p2
2 + p2
3)n/2−1
.
Then
{p · r, H} =
i
∂(p1x1 + p2x2 + p3x3)
∂xi
∂H
∂pi
−
∂(p1x1 + p2x2 + p3x3)
∂pi
∂H
∂xi
=
i
np2
i (p2
1 + p2
2 + p2
3)n/2−1
− anx2
i (x2
1 + x2
2 + x2
3)−n/2−1
= n(p2
1 + p2
2 + p2
3)n/2
− an(x2
1 + x2
2 + x2
3)−n/2
(6)
so if we deﬁne D = p · r/n − Ht, then
dD
dT
= {D, H} −
∂D
∂t
=
1
n
{p · r, H} −
∂D
∂t
Substituting in from (6),
= |p|n
− ar−n
− H
= 0.

(c) We put
Q(t ) = λq
t
λ2
, P(t ) =
1
λ
p
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
λ
˙q
t
λ2
=
1
λ
p
t
λ2
= P(t )
dP
dt
=
1
λ3
˙p
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
p
sin p , P = q cot p
is canonical.
The Jacobian of the transformation is
M =
∂Q
∂q
∂Q
∂p
∂P
∂q
∂P
∂p
=
−1
q cot p
cot p −q csc2
p
.

Hence
˜MJM =
−1
q cot p
cot p −q csc2
p
0 1
−1 0
−1
q cot p
cot p −q csc2
p
=
−1
q cot p
cot p −q csc2
p
cot p −q csc2
p
1
q − cot p
=
0 csc2
p − cot2
p
cot2
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 = arctan
αq
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


=


α
p
1
1+(αq
p )2 − αq
p2
1
1+(αq
p )2
αq p
α

 .
so
˜MJM =



α
p
1
1+(αq
p )
2 αq
− αq
p2
1
1+(αq
p )2
p
α





αq p
α
− α
p
1
1+(αq
p )2 + αq
p2
1
1+(αq
p )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 canonical
variables 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


=


1
2
q−1/2
cos p
1+q1/2 cos p
− q1/2
sin p
1+q1/2 cos p
q−1/2
sin p + 2 cos p sin p 2q1/2
cos p + 2q cos2
p − 2q sin2
p


=


1
2
q−1/2
cos p
1+q1/2 cos p
− q1/2
sin p
1+q1/2 cos p
q−1/2
sin p + sin 2p 2q1/2
cos p + 2q cos 2p

 .
Hence we have
˜MJM =


1
2
q−1/2
cos p
1+q1/2 cos p
q−1/2
sin p + sin 2p
− q1/2
sin p
1+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
− 1
2
q−1/2
cos p
1+q1/2 cos p
q1/2
sin p
1+q1/2 cos p


=


0 cos2
p+sin2
p+q1/2
cos p cos 2p+q1/2
sin p sin 2p
1+q1/2 cos p
−cos2
p+sin2
p+q1/2
cos p cos 2p+q1/2
sin p sin 2p
1+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 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 canonical
transformation, use the Legendre transformation to derive relations between
them.
(b) Find a generating function of the F4 type for the identity transformation and
of 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 plus
a 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

which is of mixed F3, F1 type. This is Legendre-transformed into a function of
the F1 type according to
F1(q1, Q1, q2, Q2) = F13 + p1q1.
The least action principle then says
p1 ˙q1 + p2 ˙q2 − H(qi, pi) = P1
˙Q1 + P2
˙Q2 − K(Qi, Pi) +
∂F13
∂p1
˙p1 +
∂F13
∂Q1
˙Q1
+
∂F13
∂q2
˙q2 +
∂F13
∂Q2
˙Q2 + p1 ˙q1 + q1 ˙p1
whence clearly
q1 = −
∂F13
∂p1
= Q1
P1 = −
∂F13
∂Q1
= −p1 − 2Q2
= −p1 − 2p2
p2 =
∂F13
∂q2
= Q2
P2 = −
∂F13
∂Q2
= −2Q1 − q2 = −2q1 − q2 .
Problem 9.14
By any method you choose show that the following transformation is canonical:
x =
1
α
( 2P1 sin Q1 + P2), px =
α
2
( 2P1 cos Q1 − Q2)
y =
1
α
( 2P1 cos Q1 + Q2), py = −
α
2
( 2P1 sin Q1 − P2)
where α is some fixed parameter.
Apply this transformation to the problem of a particle of charge q moving in a plane
that is perpendicular to a constant magnetic field B. Express the Hamiltonian for
this 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 generating
function. 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, the
remaining quantities are
y =
1
2
x −
1
α2
py cot Q1 +
1
α
Q2 (9)
px =
α2
4
x −
1
2
py cot Q1 −
α
2
Q2 (10)
P1 =
α2
x2
8
−
1
2
xpy +
1
2α2
p2
y csc2
Q1 (11)
P2 =
α
2
x +
1
α
py (12)
We now seek a generating function of the form F(x, Q1, py, Q2). This is of mixed
type, but can be related to a generating function of pure F1 character according
to
F1(x, Q1, y, Q2) = F(x, Q1, py, Q2) − ypy.
Then the principle of least action leads to the condition
px ˙x + py ˙y = P1
˙Q1 + P2
˙Q2 +
∂F
∂x
˙x +
∂F
∂py
˙py +
∂F
∂Q1
˙Q1 +
∂F
∂Q2
˙Q2 + y ˙py + py ˙y
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 have
the form
F(x, Q1, py, Q2) = −
α
2
xQ2 −
1
α
pyQ2 + g(x, Q1, py). (17)
Plugging this in to (14) and comparing with (14) we ﬁnd
g(x, Q1, py) = −
1
2
xpy +
1
2α2
p2
y cot Q1 + ψ(x, Q1). (18)
Plugging (17) and (18) into (13) and comparing with (10), we see that
∂ψ
∂x
=
α2
4
x cot Q1

or
ψ(x, Q1) =
α2
x2
8
cot Q1. (19)
Finally, combining (19), (18), (17), and (15) and comparing with (11) we see
that we may simply take φ(Q1) ≡ 0. The final form of the generating function
is then
F(x, Q1, py, Q2) = −
α
2
x +
1
α
py Q2 +
α2
x2
8
−
1
2
xpy +
1
2α2
p2
y cot Q1
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 = B0
ˆk, and put
A =
B0
2
− yî + xˆj .
Then the Hamiltonian is
H(x, y, px, py) =
1
2m
p −
q
c
A
2
=
1
2m
px +
qB0
2c
y
2
+ py −
qB0
2c
x
2
=
1
2m
px +
α2
2
y
2
+ py −
α2
2
x
2
where we put α2
= qB/c. In terms of the new variables, this is
H(Q1, Q2, P1, P2) =
1
2m
α 2P1 cos Q1
2
+ α 2P1 sin Q1
2
=
α2
m
P1
= ωcP1
where ωc = qB/mc is the cyclotron frequency. From the Hamiltonian equations
of motion applied to this Hamiltonian we see that Q2, 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(sin ωct + φ) + x0, px =
mωc
2
[r cos(ωct + φ) − y0]
y = r(cos ωct + φ) + y0, py =
mωc
2
[r sin(ωct + φ) + x0]
in agreement with the standard solution to the problem.

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