Field Theory exam II – Solutions

Field Theory exam II – Solutions
Problem 1
(a) Consider 2 point charges, one with charge q located at ⃗x 1 = (1, 0, 1), and
the other one with charge −q at ⃗x 2 = (1, 0, −1). Compute the multipole
moments q lm in terms of the associated Legendre polynomials P lm . Show
that q lm = 0 if l + m is even.
(b) Let q lm be the multipole moments of an arbitrary charge distribution, and
¯q lm the ones for the same distribution rotated by an angle β around the
z-axis. Express ¯q lm in terms of q lm and β.
(c) Now consider a charge distribution consisting of n times the configuration
from (a), but each rotated around the z-axis by an angle β k = 2πk/n,
where k = 0 . . . n − 1. Compute the multipole moments q lm by either
combining the previous results or by explicit calculation. Compute the
dipole moment ⃗p as well.
(d) Show that q lm = 0 if m/n is not an integer number.
Required knowledge
• Definition of the multipole moments
∫
q lm = Ylm(θ, ∗ φ)r l ρ(r, θ, φ)d 3 x (1)
• The spherical harmonic functions
Y lm (θ, φ) = N lm P lm (cos(θ))e imφ ,
N lm ≡
√
2l + 1 (l − m)!
4π (l + m)!
(2)
• The relations between multipole moments q lm and dipole moment ⃗p
• The identities
Solution
q ∗ 11 = −N 11 (p x − ip y ) q ∗ 10 = N 10 p z (3)
P lm (−x) = (−1) l+m P lm (x) (4)
n−1
∑
r k = 1 − rn
1 − r
k=0
(a) The given charge distribution can be written as
if r ≠ 1 (5)
ρ(⃗x) = qδ(⃗x − ⃗x 1 ) − qδ(⃗x − ⃗x 2 ) (6)
The charge positions in spherical coordinates read
r 1 = r 2 = r = √ 2 φ 1 = φ 2 = 0 θ 1 = 1 4 π θ 2 = 3 4 π (7)
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From (1) we obtain
∫
q lm =
∫
=
With (2) and e −0i = 1 we get
Y ∗
lm(θ, φ)r l ρ(r, θ, φ)d 3 x (8)
Y ∗
lm(θ, φ)r l q (δ(⃗x − ⃗x 1 ) − δ(⃗x − ⃗x 2 )) d 3 x (9)
= qYlm(θ ∗
1 , φ 1 )r1 l − qYlm(θ ∗
2 , φ 2 )r2 l (10)
= qr l (Ylm(θ ∗
1 , 0) − Ylm(θ ∗
2 , 0)) (11)
q lm = qr l N lm (P lm (cos(θ 1 )) − P lm (cos(θ 2 ))) (12)
We have
( ) 1
cos(θ 1 ) = cos
4 π = √ 1
(13)
2
Using (4) we finally arrive at
cos(θ 2 ) = cos(π − θ 1 ) = − cos(θ 1 ) (14)
q lm = qr l N lm P lm (cos(θ 1 )) ( 1 − (−1) l+m) (15)
Obviously, q lm = 0 if l + m is an even number. Otherwise,
q lm = 2qr l N lm P lm (cos(θ 1 )) (16)
(b) From (2) we get the transformation of the spherical harmonic functions
under rotation
Y lm (θ, φ) = N lm P lm (cos(θ))e imφ (17)
= N lm P lm (cos(θ))e im(φ−β) e imβ (18)
= Y lm (θ, φ − β)e imβ (19)
⇒ Y ∗
lm(θ, φ) = Y ∗
lm(θ, φ − β)e −imβ (20)
If ρ is an arbitrary charge distribution and ¯ρ the same distribution rotated
by an angle β around the z-axis, we have ¯ρ(r, θ, φ) = ρ(r, θ, φ − β). For
the multipole moments, we compute from (1)
∫
¯q lm = Ylm(θ, ∗ φ)r l ¯ρ(r, θ, φ)d 3 x (21)
∫
= Ylm(θ, ∗ φ − β)e −imβ r l ρ(r, θ, φ − β)d 3 x (22)
= e −imβ ∫
= e −imβ ∫
Y ∗
lm(θ, φ − β)r l ρ(r, θ, φ − β)d 3 x (23)
Y ∗
lm(θ, φ)r l ρ(r, θ, φ)d 3 x (24)
= e −imβ q lm (25)
where (23) and (24) are equal because the integral runs over all angles φ.
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(c) If we rotate the charge configuration from (a) by the angle β k = 2πk/n
around the z-axis, we can get the corresponding multipole moment using
(25)
q k lm = e −imβ k
q lm = e −im2π k n qlm (26)
where q lm is the multipole moment of the distribution from (a). As one can
easily see from (1), the combined multipole moment ˜q lm of the n rotated
configurations is just the sum of the single multipole moments, i.e.
n−1
∑
n−1
∑
n−1
∑
˜q lm = qlm k = e −im2π k n qlm = q lm e −im2π k n (27)
k=0
k=0
k=0
n−1
∑
= q lm a k , a ≡ e −2iπ m n (28)
k=0
If m/n is not an integer, we have a ≠ 1 and, by using (5),
1 − a n
˜q lm = q lm
1 − a = q 1 − e −2iπm
lm
= 0 (29)
1 − a
If m/n is an integer on the other hand, we have a = 1 and obtain
n−1
∑
˜q lm = q lm 1 k = nq lm (30)
k=0
From part (a), we know that q 11 = 0, since l + m = 1 + 1 = 2 is even.
Further, since 0/n is integer, ˜q 10 = nq 10 . Using this in (3), we get the
dipole moment
˜p x = ˜p y = 0 (31)
˜p z = N10 −1 10 = N10 −1 10 = nN10 −1 N 10 P 10 (cos(θ 1 )) (32)
= 2nqrP 10 (cos(θ 1 )) = 2nqr cos(θ 1 ) = 2nqz 1 (33)
= 2nq (34)
(d) This was already shown in (c), see (29).
Problem 2
Imagine two conducting hollow spheres around the origin, with radius R 0 and
R 1 (where R 1 − R 0 > 0), and a thickness negligible in comparison to the radius.
Assume that the spheres have a conductivity σ > 0, and that the system is in
a stationary state without any currents.
(a) Use Ohm’s law to show that the electrostatic potential Φ is constant on
each of the spheres, that the electric field vanishes inside the conducting
material, and that it is orthogonal to the surface outside the material.
(b) Use Gauss’ law to show that the inner surface of the inner sphere is uncharged,
and that the electric field inside the inner sphere is zero.
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(c) Assume that the inner surface of the outer sphere carries a total charge Q 1 .
Compute the electric field ⃗ E(r) between the two spheres, the total charge
Q 0 on the outer surface of the inner sphere, and the charge densities on
the surfaces.
(d) Compute the voltage U between the two spheres in terms of Q 1 , and the
capacitance C = |Q 1 /U| of the capacitor formed by the concentric spheres.
(e) From the electric field, compute the total energy W of the electric field
between the two spheres in terms of U and C.
Required knowledge
• Ohm’s law: in a conductor with conductivity σ, electric field and current
are related by ⃗ J = σ ⃗ E
• Gauss law: given a vector field ⃗w and a volume V with surface ∂V ,
∫ ∫
⃗∇ · ⃗w = ⃗w · ⃗nda (35)
• Coulomb’s law (in Gauss units)
V
∂V
⃗∇ · ⃗E = 4πρ (36)
• Definition of electrostatic potential ⃗ E = − ⃗ ∇Φ
• Energy density of the electric field (in Gauss units)
Solution
u = 1
8π ⃗ E 2 (37)
(a) Because we assume ⃗ J = 0, it follows that ⃗ E = 0 inside the conducting material,
otherwise there would be a current ⃗ J = σ ⃗ E due to Ohm’s law. From
⃗E = − ⃗ ∇Φ, it follows that Φ is constant inside the conducting material.
Since the tangential component of ⃗ E is continuous across surfaces (due to
Faraday’s law), it is also zero near the surface outside the conductor, and
therefore orthogonal to the surface.
(b) First we note that since the system is spherically symmetric, all quantities
can only depend on r. This implies that the electric field has only radial
components E r (r), and that the surface charge densities are constant along
each of the sphere’s surfaces. For any sphere V r with radius r and surface
∂V r , we can combine Gauss and Coulomb’s laws to compute
∫
∫
4πr 2 E r (r) = E ⃗ · ⃗n da = ∇ ⃗ · E ⃗ d
∫∂V 3 x ′ = 4π ρ d 3 x ′ (38)
r V r V r
= 4πQ(r) (39)
where Q(r) is the total charge inside a radius r. Inside the conducting
material, we have ⃗ E = 0, and, due to (36), also ρ = 0. Thus the only
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charges sit on the surfaces of the hollow spheres. In the following, we
denote the arbitrary small thickness of the hollow sphere shells by 2ɛ and
assume that their outer and inner surfaces are located at r = R 0 ± ɛ and
r = R 1 ± ɛ. We have already shown in (a) that E r (R 0 ) = 0. Evaluating
(39) at r = R 0 , we thus find that Q(R 0 ) = 0. On the other hand, Q(R 0 )
is nothing but the total charge on the inner surface of the inner hollow
sphere. Since the corresponding surface charge density is constant, it has
to vanish as well. Evaluating (39) at r < R 0 − ɛ, we find that E r (r) = 0,
since the total charge Q(r) inside the radius r < R 0 − ɛ is zero. Since
the nonradial components of the electric field are zero as well due to the
spherical symmetry, we have E ⃗ = 0 inside the inner sphere.
(c) Using (39) again, we find that
Therefore,
Q 0 + Q 1 = Q(R 1 ) = R 2 1E r (R 1 ) = 0 (40)
Q 0 = −Q 1 σ 0 = − Q 1
4πR 2 0
σ 1 = Q 1
4πR 2 1
(41)
Between the inner and outer spheres, i.e. for R 0 +ɛ < r < R 1 −ɛ, equation
(39) yields
E r (r) = 1 r 2 Q(r) = 1 r 2 Q 0 = − 1 r 2 Q 1 (42)
(d) The voltage U between the spheres is the difference
∫ R1
U = Φ(R 1 ) − Φ(R 0 ) = −
R 0
E r (r)dr =
[
= Q 1 − 1 ] R1
( 1
= Q 1 − 1 )
r
R 0
R 0 R 1
The capacitance follows as
(e) The total energy in the field is
W =
Problem 3
∫ R1
= Q2 1
2
∫ R1
R 0
1
r 2 Q 1dr (43)
(44)
( C =
Q 1
1
∣ U ∣ = − 1 ) −1
(45)
R 0 R 1
∫ R1
u(r)4πr 2 dr = 1 r 2 E ⃗ 2 (r)dr = Q2 1
R 0
2 R 0
2
[
− 1 ] R1
( 1
− 1 )
r
R 0 R 1
= Q2 1
R 0
2
∫ R1
1
dr (46)
R 0
r2 = Q2 1
2C = C 2 U 2 (47)
A particle with charge q moves with constant relativistic velocity v = βc along
the x-axis.
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(a) By applying a Lorentz-boost from the rest frame of the particle to the
laboratory frame, prove that the electromagnetic fields are given by
⃗E = γq (⃗x − t⃗v)
r
′3
⃗B = γq ⃗v × ⃗x
cr
′3
r ′2 = γ 2 (x 1 − vt) 2 + x 2 2 + x 2 3
(b) Compute the corresponding Poynting vector ⃗ S in terms of ⃗v, ⃗x, t, and r ′ .
Also compute | ⃗ S|.
Required knowledge
• If K and K ′ are two inertial coordinate frames, and K ′ moves with velocity
βc in in x-direction with respect to system K, and the coordinate axes are
not rotated against each other, then the relativistic transformation is the
Lorentz-boost given by
x ′ 0 = γ(x 0 − βx 1 )
x ′ 1 = γ(x 1 − βx 0 ) (48)
x ′ 2 = x 2 x ′ 3 = x 3
• The electromagnetic fields transform under the above Lorentz-boost like
E 1 = E 1 ′ B 1 = B 1
′
E 2 = γ (E 2 ′ + βB 3) ′ B 2 = γ (B 2 ′ − βE 3) ′ (49)
E 3 = γ (E ′ 3 − βB ′ 2) B 3 = γ (B ′ 3 + βE ′ 2)
• The electric field of a charge q at rest at the origin is given by
⃗E =
q ⃗x (50)
|⃗x|
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• The Poynting vector in vacuum (in Gauss units) is given by
Solution
⃗S = c
4π ⃗ E × ⃗ B (51)
(a) In the following, the quantities with a prime are defined with respect to the
rest frame of the particle, unprimed ones with respect to the laboratory
frame. Since the particle moves with constant velocity, it’s rest frame is
an inertial frame. In the rest frame of the particle, there is no magnetic
field and the electric field is given by
⃗E ′ = q
r ′3 ⃗x′ r ′ ≡ |⃗x ′ | (52)
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Using the transformation rule (49) and the Lorentz-boost (48), we get the
electric field in the laboratory frame
⎛ ⎞ ⎛ ⎞ ⎛
⎞
E 1
′
⃗E = ⎝γE 2
′ ⎠ = q x ′ 1
⎝γx ′ ⎠
r ′3 2 = q γ (x 1 − βx 0 )
⎝ γx
r ′3 2
⎠ (53)
γx 3
γE ′ 3
= qγ
r ′3 (
⃗x − x 0
⃗ β
)
γx ′ 3
= qγ (⃗x − t⃗v) (54)
r
′3
For the magnetic field, we get
⎛ ⎞
0
⃗B = ⎝−βγE 3
′ ⎠ = β ⃗ × E ⃗ = qγ
βγE 2
′ r ⃗ )
′3 β ×
(⃗x − x 0β ⃗ (55)
= qγ
r ⃗ ′3 β × ⃗x = qγ ⃗v × ⃗x
cr
′3
(56)
Using (48), we finally find
r ′2 = x ′2
1 + x ′2
2 + x ′2
3 = γ 2 (x 1 − vt) 2 + x 2 2 + x 2 3 (57)
(b) Using (51), (55), and (54) we obtain
⃗S = c E
4π ⃗ × B ⃗ = 1
= 1
4π
E
4π ⃗ (
×
⃗v × E ⃗ )
= 1 ( ( ) )
⃗E 2 ⃗v − ⃗E · ⃗v ⃗E
4π
(58)
( qγ
r ′3 ) 2 (⃗y 2 ⃗v − (⃗y · ⃗v) ⃗y ) (59)
where ⃗y ≡ ⃗x − t⃗v. Further,
Problem 4
| S| ⃗ = 1 ( qγ
4π
= 1
4π
) 2
√
r ′3 ⃗y 4 ⃗v 2 − 2⃗y 2 (⃗y · ⃗v) 2 + ⃗y 2 (⃗y · ⃗v) 2 (60)
( qγ
) 2
√
|⃗y|
r ′3 ⃗y 2 ⃗v 2 − (⃗y · ⃗v) 2 (61)
A particle with charge q, mass m, and velocity ⃗v crosses a region with a homogeneous
magnetic field ⃗ B. Assume that the field is weak such that the particle
moves approximately on a straight line with constant velocity.
(a) Show that the instantaneous power radiated by the particle in terms of
the force F ⃗ acting on the particle is given by
P = 2q2 γ 2 ( ( ) ) 2
⃗F 2
3m 2 c 3 − ⃗F · β ⃗
(b) From this, compute the instantaneous power radiated by the particle while
inside the field region.
(c) Compute the total energy W radiated by the particle in terms of the proper
time τ F the particle spent inside the field region.
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Required knowledge
• The instantaneously radiated power of an accelerated charge q is
P = − 2 q 2 ( ) ( )
dpµ dp
µ
3 m 2 c 3 dτ dτ
(62)
where p µ
particle.
= (p 0 , ⃗p) is the 4-momentum and τ the proper time of the
• Proper time in terms of laboratory time
dt
dτ = γ ≡ ( 1 − β 2) − 1 2
(63)
• The relation between force and change of 4-momentum
d
dt ⃗p = ⃗ F
d
dt p 0 = ⃗ β · ⃗F (64)
• The Lorentz force acting on a charge inside a magnetic field
⃗F = q ⃗ β × ⃗ B (65)
Solution
(a) From (62), we get
P = − 2 q 2 ( ) ( )
dpµ dp
µ
3 m 2 c 3 = − 2 q 2 γ 2 ( ) ( )
dpµ dp
µ
dτ dτ 3 m 2 c 3 dt dt
( (d⃗p
= 2 q 2 γ 2 ) 2 ( ) dp
0 2
)
3 m 2 c 3 −
dt dt
With (64), this becomes
P = 2 q 2 γ 2 ( ( ) ) 2
⃗F 2
3 m 2 c 3 − ⃗β · F ⃗
(66)
(67)
(68)
(b) Inside the magnetic field region, the force acting on the particle is the
Lorentz force (65). We compute
⃗F · ⃗β
( )
= q ⃗β × B ⃗ · ⃗β = 0 (69)
⃗F 2 = q 2 ( ⃗ β × ⃗ B
) 2
= q
2⃗ β
2 ⃗ B 2 sin 2 (α) (70)
where α is the angle between β ⃗ and B. ⃗ Inserting in (68) yields the result
P = 2 q 4 γ 2
3 m 2 c ⃗ 3 β 2 B ⃗ 2 sin 2 (α) (71)
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(c) Assuming that the particle moves with approximately constant velocity
in a homogeneous magnetic field, the radiated power P and the Lorentz
factor γ are also constant. Thus we only have to compute the laboratory
time t F spent in the magnetic field from the corresponding proper time
τ F . Since γ is constant, (63) gives us t F = γτ F . The total radiated energy
is thus
2 q 4 γ 3
W = t F P = γτ F P = τ F
3 m 2 c ⃗ 3 β 2 B ⃗ 2 sin 2 (α) (72)
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