Solutions to Problems in Jackson,
Classical Electrodynamics, Third Edition
Homer Reid
January 13, 2003
Chapter 11: Problems 1-8
Problem 11.4
A possible clock is shown in the figure. It consists of a flashtube F and a photocell
P shielded so that each views only the mirror M, located a distance d away, and
mounted rigidly with respect to the flashtube-photocell assembly. The electronic
innards of the box are such that when the photocell responds to a light flash from
the mirror, the flashtube is triggered with a negligible delay and emits a short flash
toward the mirror. The clock thus “ticks” once every (2d/c) seconds when at rest.
(a) Suppose that the clock moves with a uniform velocity v, perpendicular to the
line from P F to M, relative to an observer. Using the second postulate
of relativity, show by explicit geometrical or algebraic construction that the
observer sees the relativistic time dilatation as the clock moves by.
(b) Suppose that the clock moves with a velocity v parallel to the line from P F
to M. Verify that here, too, the clock is observed to tick more slowly, by the
same time dilatation factor.
(a) Suppose that, relative to an observer, the clock is moving with speed v
perpendicular to the direction in which the light travels back and forth. Let dt
be the time measured by the observer for the light to travel from the flashtube
to the mirror. The vertical distance traveled by the light is d, but—as far as
the observer is concerned—during the time dt the mirror has also translated
a horizontal distance vdt. Hence the total distance the observer sees the light
1
Homer Reid’s Solutions to Jackson Problems: Chapter 11
2
p
travel (one-way) is d2 + (vdt)2 . But this distance must equal cdt, since the
light must have the universal speed c in any reference frame. Hence we have
p
cdt = d2 + (vdt)2
or
dt =
1
2
1 + vc2
!1/2
d
.
c
This is the time (as measured by the observer) in which the light makes the trip
from flashtube to mirror. The light takes the same amount of time to travel
back to the photocell, so the total period of the clock is just twice this, or
!1/2
2d
1
period =
v2
c
1 − c2
which is greater than the period of the clock at rest by the normal Lorentz
dilatation factor.
(b) Now we suppose that the clock is moving relative to an observer with speed
v parallel to the direction of motion of the light. We align the z axis with the
direction of motion. Then we may write down the space-time coordinates (in
the clock’s rest frame) of the three relevant events in the operation of the clock
(taking x4 = ct):
xa = (0, 0, 0, 0),
xb = (0, 0, d, d),
xc = (0, 0, 0, 2d),
(light leaves flashtube)
(light reaches mirror)
(light reaches photocell).
The transformation matrix from the clock’s reference frame to the observer’s
reference frame is


1 0 0
0
 0 1 0
0 

Γ=
 0 0 γ γβ 
0 0 γβ γβ
and using this we may write down the spacetime coordinates of the three events
above in the rest frame of the observer:
x0a = (0, 0, 0, 0)
x0b = 0, 0, γ(1 + β)d, γ(1 + β)d
x0c = 0, 0, 2γβd, 2γd
Evidently the difference in the time coordinates of events a and c (which is just
the observed period of the clock) is c∆t = 2γd, so again the observer observes
Homer Reid’s Solutions to Jackson Problems: Chapter 11
3
the clock to have a period of 2γd/c, longer by the factor γ than the period of
the clock in its rest frame.
Problem 11.5
A coordinate system K 0 moves with a velocity v relative to another system K. In K 0
a particle has a velocity u0 and an acceleration a0 . Find the Lorentz transformation
law for acceleration, and show that in the system K the components of acceleration
parallel and perpendicular to v are
1−
ak =
a⊥ =
1−
1+
v2
c2
v·u0
c2
1+
v2
c2
3/2
v·u0 3
c2
a0k
v
0
0
0
3 a⊥ + 2 × (a × u ) .
c
The initial components of the velocity in the moving frame (frame K 0 ) are
and u0⊥ . Using Jackson’s equations 11.31 to transform these to frame K, we
obtain
u0k
u0k + v
uk (0) =
1+
u⊥ (0) =
u0k v
c2
u0⊥
γv 1 +
(1)
uk v .
c2
(2)
After a time dt has passed in frame K, a time dt0 = dt/γ has passed in frame
K0 , and the velocity has increased by the amount a0 dt0 = a0 dt/γ. Then we can
write down the new components of the velocity in K 0 and again transform to
K:
uk (dt) =
u0k + a0k dt0 + v
(u0k +a0k dt0 )v
c2
u0⊥
(u0k +a0k dt0 )v
γv 1 +
2
c
(3)
1+
u⊥ (dt) =
Subtracting (1) from (3), we have
(4)
4
Homer Reid’s Solutions to Jackson Problems: Chapter 11
∆uk =
u0k + a0k dt0 + v
1+
(u0k +a0k dt0 )v
c2
=
1+
−
k
1+
2
u0k v
2
c
v
1 − c2
u0 v
+ 1 + ck2 a0k dt0
a0k dt0
u0 v 2
c2
u0k + v
Using the relation dt0 = dt/γ we can rewrite this as
3/2
2
a0k dt 1 − vc2
=
u0 v
u0 v 2
+ 1 + ck2
1+
1 + ck2
Dividing by dt and taking the limit as dt → 0, we obtain
v 2 1/2
c2
3/2
2
a0k 1 − vc2
∆uk
ak = lim
= .
u0 v 2
dt→0 dt
1 + ck2
a0k dt
So evidently I’m off by 1 in the exponent of the denominator. What am I doing
wrong?
Next, subtracting (2) from (4), we have


1  u0⊥ + a0⊥ dt0
u0⊥ 
∆u⊥ =
−
0
0
u0k v
γv 1 + uk v + ak dt0 v
1
+
2
2
c
c
c2

a0 dt0 v 
0 u
v
k
k
0
0
0
c2
1  a⊥ dt 1 + c2 − u⊥

=


u0k v 2
u0k v a0k dt0 v
γv
1 + c2
+ 1 + c2
2
c


v
0
0
0 0
0 0
0
a
dt
+
u
a
−
u
a
dt
2
⊥ k
k ⊥
c
1  ⊥

=


u0k v 2
u0k v a0k dt0 v
γv + 1 + c2
1 + c2
2
c
Again putting in dt = γdt0 , we have

v
0 0
0 0
0
u
a
−
u
a
a
dt
+
2
⊥
⊥
⊥
k
k dt 
c
1 
= 2 

2
0
0
u v a0k dtv
u v
γv
+ 1 + ck2
1 + ck2
γv c2

Homer Reid’s Solutions to Jackson Problems: Chapter 11
5
As before, when we divide by dt and take the limit as dt → 0 the second term
in the denominator becomes irrelevant and we obtain


h
i
1
v 0 0
∆u⊥

 0
0 0
a
+
a⊥ = lim
= u
a
−
u
a

⊥
⊥ k
k ⊥
u0 v 2
dt→0 dt
c2
γv2 1 + ck2


v2
h
i
v 0 0
 1 − c2  0
0 0
u
a
−
u
a
a
+
= 

⊥
⊥
⊥
k
k
u0 v 2
c2
1 + ck2
Jackson writes the second term in the brackets a little differently. To show that
his expression is equivalent to mine, we use the BAC-CAB rule:
v × (a0 × u0 ) = (v · u0 )a0 − (v · a0 )u0
= vu0k a0k + a0⊥ − va0k u0k + u0⊥
The parallel components cancel, and we are left with
= v(u0k a0⊥ − a0k u0⊥ )
which is the way I wrote it above.
But I’m still off by 1 in the exponent of the term in the denominator! What
am I missing?
Problem 11.6
Assume that a rocket ship leaves the earth in the year 2010. One of a set of twins
born in 2080 remains on earth; the other rides in the rocket. The rocket ship is
so constructed that it has an acceleration g in its own rest frame (this makes the
occupants feel at home). It accelerates in a straight-line path for 5 years (by its own
clocks), decelerates at the same rate for 5 more years, turns around, accelerates for
5 years, decelerates for 5 years, and lands on earth. The twin in the rocket is 40
years old.
(a) What year is it on earth?
(b) How far away from the earth did the rocket ship travel?
Let v(t) be the speed of the rocket, as observed on earth, at a time t as
measured on earth. Using the first result of problem 11.5, we see that the
acceleration of the rocket as measured from earth is
3/2
dv
v(t)2
a=
g,
=± 1− 2
dt
c
Homer Reid’s Solutions to Jackson Problems: Chapter 11
6
or
3/2
dβ
= ± 1 − β2
α
dt
where α = g/c and the plus or minus sign depends on whether we are accelerating or decelerating. Manipulating this a little, we obtain
dβ
= ±αdt.
(1 − β 2 )3/2
Integrating, we obtain
β 2
β0
(1 − β 02 )1/2 = ±α(t2 − t1 ).
(5)
β1
For the first leg of the rocket’s journey, we have t1 = β1 = 0 and we take the
plus sign in (5). Then we find
β
= αt
(1 − β 2 )1/2
or
β(t) =
αt
[1 + (αt)2 ]1/2
(6)
and
γ(t) = p
1
1 − β 2 (t)
=
p
1 + (αt)2
(7)
Next let’s work out the relation between time as measured on the rocket and
time as measured on earth. With primed (unprimed) quantities referring to the
rocket (to earth), the infinitesimal relation is
dt = γ(t)dt0
or
t02
−
t01
=
=
Z
Z
t2
t1
t2
t1
dt
γ(t)
dt
p
1 + (αt)2
du
1 αt2
√
α αt1
1 + u2
1
=
sinh−1 (αt2 ) − sinh−1 (αt1 )
α
=
Z
7
Homer Reid’s Solutions to Jackson Problems: Chapter 11
For the first leg of the journey this becomes
t2 =
1
sinh(αt02 ).
α
(8)
Now we know that the first leg of the journey lasts until t02 = 5 years, and we
have
9.8 m s−2
g
= 3.27 · 10−8 s−1 .
α= =
c
3 · 108 m s−1
Then the time on earth at the end of the first leg of the journey is, from (8),
1
3.153 · 107 s
−8 −1
t2 =
s
· sinh 3.27 · 10 s (5 yr)
3.27 · 10−8
1 yr
= 2.65 · 109 s
≈ 84 yr.
Finally, the distance the rocket travels during the first leg of its journey is
Z t
d=c
β(t)dt
0
Z t
αt
p
=c
dt
1 + (αt)2
0
Z
c αt
u
√
=
du
α 0
1 + u2
o
1/2
c n
1 + (αt)2
−1
=
α
= 3.0 · 1015 meters.
The behavior of the rocket on the subsequent three legs of the journey is
similar to that in the first leg. In particular, the total distance traveled away
from earth is twice that covered in the first leg, or 6.0 ·1015 meters, and the
total time elapsed on earth during the rocket’s journey is four times that elapsed
during the first leg, or 4·84=336 years. So it should be the year 2436 on earth
by the time the rocket returns home.
Homer Reid’s Solutions to Jackson Problems: Chapter 11
8
Problem 11.13
An infinitely long straight wire of negligible cross-sectional area is at rest and has
a uniform linear charge density q0 in the inertial frame K 0 . The frame K 0 (and the
wire) move with a velocity v parallel to the direction of the wire with respect to
the laboratory frame.
(a) Write down the electric and magnetic fields in cylindrical coordinates in the rest
frame of the wire. Using the Lorentz transformation properties of the fields,
find the components of the electric and magnetic fields in the laboratory.
(b) What are the charge and current densities associated with the wire in its rest
frame? In the laboratory?
(c) From the laboratory charge and current densities, calculate directly the electric
and magnetic fields in the laboratory. Compare with the results of part (a).
I don’t like q0 as a symbol for charge density, because it appears to have the
wrong units. I’ll use λ instead. We’ll take our z axis to coincide with the wire
and take v in the positive z direction.
(a) In the rest frame there is no current and the E field is static; hence B = 0.
The electric field is found by considering a Gaussian pillbox in the shape of
a right circular cylinder coaxial with the wire and of radius r and length dz.
There is no electric field normal to the upper and lower surfaces, and the field
normal to the radial bounding surface is uniform across the circumference. On
the other hand, the charge enclosed in the cylinder is λ dz. Then Gauss’ law is
I
E · dA = 4πQ
=⇒ 2πrdzEr = 4πλdz
2λ
=⇒ Er =
.
r
In terms of cartesian components we have
y0
x0
,
E
=
2λ
.
Ex = 2λ
y
x02 + y 02
x02 + y 02
(9)
9
Homer Reid’s Solutions to Jackson Problems: Chapter 11
The field-strength tensor in the laboratory frame is
F = ΛF 0 Λ̃

γ 0 0 βγ
0 −x0
 0 1 0 0   x0
1
0


= 2λ 02
0
x + y 02  0 0 1 0   y 0
βγ 0 0 γ
0
0


0
0
0 −x
−y
0
 x0
2γλ
0
0
βx0 

.
= 02
0
0
βy 0 
(x + y 02 )  y 0
0 −βx0 −βy 0
0

−y 0
0
0
0

0
γ

0 
 0
0  0
0
βγ
0
1
0
0

0 βγ
0 0 

1 0 
0 γ
Reading off from this the transformed fields, and dropping the primes on x and
y since the z boost leaves these coordinates unchanged, we have
2λ
(xi + yj)
x2 + y 2
2λ
= γ r̂
r
λ
B = βγ
(−yi + xj)
2
2(x + y 2 )
2λ
= βγ φ̂.
r
E=γ
(10)
(11)
(12)
(13)
(b) In the rest frame there is no current and the charge density is ρ0 = λδ(x)δ(y),
so
J 0µ = cλ(δ(x)δ(y), 0, 0, 0).
The transformed current density is
J µ = Λµν J 0ν

γ 0 0
 0 1 0

= cλ 
0 0 1
βγ 0 0

δ(x)δ(y)

0

= cγλ 
0
βδ(x)δ(y)


γβ
δ(x)δ(y)


0 
0





0
0
γ
0


.

(c) Computing the electric field in the laboratory frame is easy, since the charge
density is the same as we had before but with a factor of γ thrown in. Then
the E field is just (9) but with that factor of γ thrown in, i.e.
E=
2γλ
r̂
r
Homer Reid’s Solutions to Jackson Problems: Chapter 11
10
which agrees with (10).
For the magnetic field, we note that the current density in the lab frame is
J = cβγλδ(x)δ(y)ẑ. Then the current piercing a disc of radius r is I = cβγλ.
On the other hand, by symmetry the magnetic field in the azimuthal direction
around the circumference of this disc is constant, so we may use Ampere’s law,
∇ × B = (4π/c)J to write
2πrBφ =
4π
(cβγλ)
c
or
B=
2βγλ
φ̂
r
which agrees with (13).
Problem 11.15
In a certain reference frame a static, uniform, electric field E0 is parallel to the x
axis, and a static, uniform, magnetic induction B0 = 2E0 lies in the x − y plane,
making an angle θ with the axis. Determine the relative velocity of a reference
frame in which the electric and magnetic fields are parallel. What are the fields in
that frame for θ 1 and θ → (π/2)?
The untransformed fields are
E = E0 i,
B = 2E0 (cos θi + sin θj).
Let’s suppose we boost along the z axis. Then the fields transform according
to Jackson equation (11.149):
E0 = γE0 (1 − 2β sin θ)i + 2γβE0 cos θj
B0 = 2γE0 cos θi + γE0 (2 sin θ + β)j.
The angle between the fields is given by
cos θ =
=
E 0 · B0
|E0 ||B0 |
γ 2 E02 (1
2γ 2 E02 cos θ(1 + β 2 )
− 4β sin θ + 4β 2 )1/2 (4 + 4β sin θ + β 2 )1/2
Setting this equal to unity, we obtain
Homer Reid’s Solutions to Jackson Problems: Chapter 11
11
Problem 11.18
The electric and magnetic fields of a particle of charge q moving in a straight line
with speed v = βc, given by (11.152), become more and more concentrated as
β → 1, as indicated in Fig. 11.9. Choose axes so that the charge moves along
the z axis in the positive direction, passing the origin at t = 0. Let the spatial
coordinates of the observation point be (x, y, z) and define the transverse vector r ⊥ ,
with components x and y. Consider the fields and the source in the limit of β = 1.
(a) Show that the fields can be written as
E = 2q
r⊥
2 δ(ct − z);
r⊥
B = 2q
v̂ × r⊥
δ(ct − z).
2
r⊥
(b) Show by substitution into the Maxwell equations that these fields are consistent
with a 4-vector source density,
J α = qcv α δ (2) (r⊥ )δ(ct − z)
where the 4-vector v α = (1, v̂).
(c) Show that the fields of part a are derivable from either of the following 4-vector
potentials,
A0 = Az = −2qδ(ct − z) ln(λr⊥ );
A⊥ = 0
or
A0 = Az = 0;
A⊥ = −2qΘ(ct − z)∇⊥ ln(λr⊥ )
where λ is an irrelevant parameter setting the scale of the logarithm. Show
that the two potentials differ by a gauge transformation and find the gauge
function, χ.
(a) In the reference frame in which the particle is at rest at the origin, the fields
are
q
E0 = 02
(x0 i + y 0 j + z 0 k),
B0 = 0.
(x + y 02 + z 02 )3/2
Transforming back to the laboratory frame according to Jackson 11.148, the
electric field is
q
E = 02
(γx0 i + γy 0 j + z 0 k)
02
(x + y + z 02 )3/2
where in this expression the coordinates are still those of the observation point
in the moving frame. The transformation of these to the lab frame is x0 = x,
y = y 0 = y, z 0 = γ(z − βct). Then the correct expression for the transformed
field is
q
E= 2
(γxi + γyj + (z − ct)k).
2
[x + y + γ 2 (z − ct)2 ]3/2
Homer Reid’s Solutions to Jackson Problems: Chapter 11
12
In the limit β → 1, we have γ → ∞. For z 6= ct the γ 2 factor in the denominator
then ensures that all field components are zero. For z = ct, however, although
the z component of the field clearly vanishes, the behavior of the other components is not as immediately clear. To elucidate the behavior of, say, the x
component of the field at z = ct we integrate it from z = ct − to z = ct + :
Z ct+
Z ct+
dz
Ex dz = qγx
2
2
2 3/2
ct−
ct− [r⊥ + γ (z − ct) ]
Z γ/r⊥
qx
du
= 2
r⊥ −γ/r⊥ [1 + u2 ]3/2
γ
2qx
= 2 p 2
.
r⊥
r⊥ + γ 2 2
Taking the limit γ → ∞ for any finite , we find
Z ct+
2qx 2
.
lim
Ex dz =
γ→∞ ct−
r⊥
(14)
On the other hand, integrating between two points on the same side of z = ct,
say from z = ct + to z = ct + 2, we find
"
#
Z ct+2
2γ
γ
2qx 2
p
−p 2
Ex dz =
2 + 4γ 2 2
r⊥
r⊥
r⊥ + γ 2 2
ct+
which vanishes as γ → ∞.
Since Ex vanishes at any point z 6= ct but yields something nonzero when
integrated across that point, we conclude that it is just a δ function in (z − ct)
with coefficient given by (14):
Ex =
2qx
2 δ(z − ct).
r⊥
Ey =
2qy
2 δ(z − ct).
r⊥
and, similarly,
Combining these, we can write
E = 2q
r⊥
2 δ(ct − z).
r⊥
The B field is given by Jackson (11.150) with, in the ultrarelativistic limit,
β = k̂ :
v̂ × r⊥
δ(ct − z).
B = 2q
2
r⊥