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Physics 210A PS 7 Solutions
Anthony Karmis
March 9, 2009
Jackson 6.
Part (a)
Defining the Fourier transform by:
A(~x, t) =
(2π)
3
A(
k, t)e
−~k·~x
d
3
k
Therefore, the diffusion equation becomes:
t
A = −
k
2
μσ
A
This is solved by
A(
k, t) =
A(
k, 0)e
−k
2 t/μσ .
The inverse transform gives us:
A(~x, t) =
(2π)
3
A(
k, 0)e
−k
2 t/μσ e
i~k·~x d
3 k
where
A(
k, 0) =
A(~x
′ , 0)e
−i
~ k·~x
′
d
3 x
′
The result is:
A(~x, 0) =
(2π)
3
A(~x
′ , 0)e
−k
2 t/μσ e
i
~ k·(~x−~x
′ ) d
3 kd
3 x
′
G(~x − ~x
′ , t)
A(~x
′ , 0)d
3 x
′
With the Green function:
G(~x − ~x
′ , t) =
(2π)
3
e
−k
2 t/μσ e
i
~ k·(~x−~x
′ ) d
3 k
Part (b)
Introducing the Fourier transform
G(~x, t) =
(2π)
3
G(
k, ω)e
i(
~ k·~x−ωt) d
3 kdω
The inhomogeneous equation in Jackson becomes:
(−iω)
2 − |ik|
2 /μσ
G = e
−i
~ k·~x
′
Which gives the Green function:
G(
k, ω) =
e
−i
~ k~x
′
k
2 /μσ − iω
We may invert the transform by performing the ω integral. We have:
G(
k, t) =
2 π
∞
−∞
G(
k, ω)e
−iωt dω =
ie
−i
~ k·~x
′
2 π
e
−iωt
ω + ik
2 /μσ
dω
We can compute this integral by contour integration. For t > 0, Jordan’s
lemma tells us to close the contour in the lower-half plane. As a result, we pick
up the residue at ω = −ik
2 /μω. The result is:
G(
k, t) = (− 2 πi)
ie
−i
~ k·~x
′
2 π
e
−k
2 t/μσ = e
−k
2 t/μσ e
−i
~ k·~x
′
However, for t < 0, we close the contour in the upper-half plane. Since there
are no enclosed poles, we end up with G = 0. Therefore, we have:
G(~x − ~x
′ , t) =
Θ(t)
(2π)
3
e
−k
2 t/μσ e
i
~ k·(~x−~x
′ ) d
3 k
Part (c)
Taking μ and σ to be uniform in all space, we may perform the above integral
by completing the square:
G(~x − ~x
′ , t) =
Θ(t)
(2π)
3
e
−μσ|~x−~x
′ |
2 / 4 t
e
−t|
~ k−iμσ(~x−~x
′ )/ 2 t|
2 /μσ d
3 k
Θ(t)
(2π)
3
πμσ
t
3 / 2
e
−μσ|~x−~x
′ |
2 / 4 t
= Θ(t)
μσ
4 πt
3 / 2
e
−μσ|~x−~x
′ |
2 / 4 t
Jackson 6.
Part (a)
We assume the sphere is magnetized and spinning along the ˆz-axis. Since the
magnetic moment is m~ =
M V , where V =
4
3
πr
3 is the volume of the sphere, we
see that the magnetization is simply
M = M zˆ. As seen in Jackson, the sphere
has a uniform magnetic induction
B =
2
3
μ 0
M in its interior. In terms of m, we
have:
B =
μ 0 m
2 πR
3
ˆz
Since there will be no current in the frame where the sphere is at rest, Ohm’s
law in the reference frame tells us that σ
E
′ = σ(
E+~v×
B) = 0, where the primed
variables are in the rotating reference frame, and the nonprimed variables are
in the at-rest lab frame. Therefore, we have:
E = −~v ×
B = −(~ω × ~r) ×
B = −
μ 0
mω
2 πR
3
(~r − zˆ(ˆz · ~r))
In cylindrical coordinates, this is:
E = −
μ 0 mωρ
2 πR
3
ρˆ
Using ρ = ǫ 0
E, we have:
ρ = ǫ 0
E = ǫ 0
ρ
∂ρ (ρEρ) = −
mω
πc
2 R
3
Part (b)
First, we note that the interior electric field can be integrated to obtain the
interior electrostatic potential:
Φ(ρ) = −
E · d
l = −
Erhodρ = Φ 0 +
μ 0
mωρ
2
4 πR
3
Converting back to spherical coordinates, we have:
Φ(r, θ) = Φ 0
μ 0
mω
4 πR
3
r
2 sin
2 θ
μ 0
mω
6 πR
3
r
2 [P 0 (cosθ) − P 2 (cosθ)]
where we have converted sin
2 θ into Legendre polynomials. This can be
written explicitly as a Legendre expansion:
Φ(r, θ) =
0
μ 0 mω
6 πR
3
r
2
P
0
(cosθ) −
μ 0 mω
6 πR
3
r
2 P 2
(cosθ)
Since charge neutrality requires that the monopole moment be zero, this sets
0
for us, and we have:
Φ(r, θ) = −
μ 0 mωr
2
6 πR
3
P 2 (cosθ)
This potential clearly only has a quadrupole term:
4 π
μ 0 mωR
2
6 π
Y 2 , 0 (θ, φ)
r
3
Comparing this with the multipole expansion,
4 πǫ 0
l,m
4 π
2 l + 1
ql,m
Y
l,m
(θ, φ)
r
l+
We see we have:
q 2 , 0
= − 4 πǫ 0
4 π
μ 0
mωR
2
6 π
4 π
2 mωR
2
3 c
2
Converting this to cartesian tensors yields:
Q 33 = 2
4 π
q 2 , 0 = −
4 mωR
2
3 c
2
and
Q
11
= Q
22
Q
33
Part (c)
The surface charge may be computed by first obtaining the jump in the normal
component of the electric field at the surface of the sphere. Working in spherical
coordinates, and taking the gradient of the potential above, we have:
E
out
r
μ 0 mωR
2
2 πr
4
P 2 (cosθ)
E
in
r
μ 0 mωr
3 πR
3
[P 0 (cosθ) − P 2 (cosθ)]
The surface charge is then computed as:
σ = ǫ 0
E
out
r
− E
in
r
|r=R
μ 0 ǫ 0 mω
3 πR
2
[
P 2 (cosθ) − (P 0 (cosθ) − P 2 (cosθ))
]
mω
3 πc
2 R
2
[
P 0 (cosθ) −
P 2 (cosθ)
]
Part (d)
Although the sphere is rotating, both the electric and the magnetic field are
static. Hence, the line integral of the electric field gives simply the electrostatic
potential. In this case, we have:
ξ =
pole
equator
E · d
l = Φequator − Φpole = Φ(θ =
π
) − Φ(θ = 0)
Using our results from Part (b), this becomes:
ξ = −
μ 0 mω
6 πR
[P
2
(0) − P
2
(1)] =
μ 0 mω
4 πR
Part (c)
The surface contribution is:
c
2
∇ × (Φ
H) = −
c
2
∇Φ) ×
H +
∇ ×
H
d
3
x
μ 0
c
2
E ×
Bd
3
x +
c
2
E · ~x)
Jd
3
x
The first integral is given by Jackson 5.62, and we saw the second in Part
(b). Therefore, we have:
c
2
∇ × (Φ
H) = −
3 c
2
E × m~
Therefore, we see that:
P =
3 c
2
E(0) × m~
Jackson 6.
Part (a)
The momentum conservation equation reads:
d
dt
P
f ields
P
mech.
j
T
ij
n j
da = −
T ) · ~nda
This tells us that the pressure due to the fields incident on the plane is Tij nj.
If we choose the wave to be propagating in the z-direction and the plane normal
to it, then the pressure is T 33
. Jackson 6.120 yields:
P = T
33
ǫ 0
E
2
2 B
2
which is the energy density.
Part (b)
The flux is just the rate at which energy is carried through space. Hence, the
flux is Φ = uc, where u is the energy density. Let the mass density of the sail
be δ. Then, the acceleration is:
a =
F
m
P
δ
cδ
(1400 W/m
3 )
(3 × 10
8 m/s) (0. 001 kg/m
2 )
≈ 5 × 10
− 3 m/s
2
