Physics 214 Solution Set 3 Winter 2017

1. [Jackson, problem 9.2] A radiating quadrupole consists of a square of side a with charges±q at alternate corners. The square rotates with angular velocity ω about an axis normalto the plane of the square and through its center. Calculate the quadrupole moments, theradiation fields, the angular distribution of radiation, and the total radiated power, all in thelong-wavelength approximation. What is the frequency of the radiation?

ω

+q+q

−q

−q

a

Figure 1: A radiating quadrupole consisting of a square of side a with charges±q at alternate corners.

The charge distribution consists of point charges at the four corners of a square in the x–yplane, as depicted in Figure 1. The charges are located at the following positions in Cartesiancoordinates,

+q :a√2(cosωt , sinωt , 0) , + q : − a√

2(cosωt , sinωt , 0) , (1)

−q :a√2(sinωt , − cosωt , 0) , −q :

a√2(− sinωt , cosωt , 0) . (2)

The quadrupole moment Cartesian tensor is given by1

Qij =∑

k

qk

[3x

(k)i x

(k)j − (r(k))2δij

], (3)

where k labels each charge and i and j label the components of the position vector ~x. Notethat the charges all lie in the x–y plane (corresponding to z = 0). Moreover, r(k) is the distanceof the kth charge from the origin, located at the center of the square. Hence,

(r(k))2 = (x(k)1 )2 + (x

(k)2 )2 + (x

(k)3 )2 = 1

2a2 , for all k .

1If one uses eq. (4.9) of Jackson, then one should express the charge distribution ρ(~x, t) as a sum of deltafunctions, whose arguments vanish at the locations of the four charges. Integrating over all space then yieldseq. (3). Note that I am using Qij for the real physical quadrupole moment Cartesian tensor. I have introducedthe tilde ( ˜) to avoid confusion with the complex quadrupole moment Cartesian tensor to be defined in eq. (4).

1

This means that ∑

k

qk(r(k))2 = 1

2a2

∑

k

qk = 0 ,

since there are an equal number of positive and negative charges. Plugging in the location ofthe four charges in eq. (3), we obtain:

Q13 = Q23 = Q33 = 0 , Q11 =32· 2 a2q

[cos2 ωt− sin2 ωt

]= 3a2q cos 2ωt ,

Q22 = −3a2q cos 2ωt , Q12 =32· 4 a2q sinωt cosωt = 3a2q sin 2ωt ,

after employing some well known trigonometric identities. Thus, the electric quadrupole tensoris given by

Qij(t) = 3a2q

cos 2ωt sin 2ωt 0sin 2ωt − cos 2ωt 0

0 0 0

.

In a Cartesian basis, all the elements of the physical multipole tensors are real. We can introducethe complex time-dependent multipole tensor Qij(t) [without the tilde] by defining

Qij(t) = 3a2q e−2iωt

1 i 0i −1 00 0 0

. (4)

One can check that the physical quadrupole Cartesian tensor is given by2

Qij(t) = Re [Qij(t)] . (5)

To make contact with the convention for harmonic sources employed by eq. (9.1) of Jackson,we note that the complex electric quadrupole tensor is defined to be

Qij(t) =

∫(3xixj − r2δij)ρ(~x, t) d

3x ,

where ρ(~x, t) = ρ(~x) e−iωt. However, this would not yield the correct time dependence exhibitedin eq. (4). However, the solution is simple—we write:

ρ(~x, t) = ρ(~x) e−2iωt .

That is, one must replace ω with 2ω in the formulae that appear in Chapter 9 of Jackson. Notethat this also implies that

k =2ω

c. (6)

Thus, it follows thatQij(t) = Qij e

−2iωt , (7)

2The simple relation, given in eq. (5), between the complex time-dependent quadrupole tensor Qik(t) and

the corresponding physical tensor Qij(t) provides the motivation for using Cartesian tensors in this problem.However, one can also solve this problem by employing the quadrupole tensor in the spherical basis, as shownat the end of this solution.

2

where

Qij =

∫(3xixj − r2δij)ρ(~x) d

3x = 3a2q

1 i 0i −1 00 0 0

. (8)

Using eq. (8), the matrix ~Q whose components are defined by

Qi =3∑

j=1

Qijnj ,

are easily evaluated. Using n = (sin θ cosφ , sin θ sinφ , cos θ), we find

Q1 = 3a2q sin θ eiφ , Q2 = 3a2qi sin θ eiφ , Q3 = 0 .

That is,~Q = 3a2q sin θ eiφ(x+ iy) . (9)

We can now employ eqs. (6) and (9) in eqs. (9.44), (9.45) and (9.49) of Jackson. First weevaluate

n× ~Q = 3a2q sin θ eiφ det

x y z

sin θ cosφ sin θ sin φ cos θ1 i 0

= −3a2qi sin θ eiφ[cos θ(x+ iy)− sin θ eiφz

].

Therefore, eq. (9.44) of Jackson [in SI units] yields

~H = −ck3

8π

eikr

ra2q sin θ eiφ

[cos θ(x+ iy)− sin θ eiφz

].

The physical magnetic fields are then given by

Re( ~H e−2iωt) = −ck3a2q

8πrsin θ

x cos θ cos(kr − 2ωt+ φ)− y cos θ sin(kr − 2ωt+ φ)

−z sin θ cos(kr − 2ωt+ 2φ)

. (10)

The electric fields are obtained by using eq. (9.39) of Jackson. In SI units,

~E = Z0~H × n , (11)

where Z0 =√

µ0/ǫ0 is the impedance of free space.3 Thus,

~E = −Z0k3

8π

eikr

ra2q sin θ eiφ det

x y z

cos θ i cos θ − sin θ eiφ

sin θ cosφ sin θ sin φ cos θ

= −Z0k3

8π

eikr

ra2q sin θ eiφ

x(sin2 θ sinφ eiφ + i cos2 θ)

−y(sin2 θ cos φ eiφ + cos2 θ)− iz sin θ cos θ eiφ.

3If this is not enough motivation a switch to gaussian units, then I don’t know what more I can say!

3

The physical electric fields are then given by

Re(~E e−2iωt) = −Z0k3a2q

8πrsin θ

x[sin2 θ sinφ cos(kr − 2ωt+ 2φ)− cos2 θ sin(kr − 2ωt+ φ)

]

−y[sin2 θ cos φ cos(kr − 2ωt+ 2φ) + cos2 θ cos(kr − 2ωt+ φ)

]

+z sin θ cos θ sin(kr − 2ωt+ 2φ)

. (12)

As a check, it is easy to verify that eq. (11) is also satisfied by the physical fields given ineqs. (10) and (12).

Next, we compute the time-averaged power radiated per unit solid angle. Using eqs. (9.45)and (9.46) of Jackson,

dP

dΩ=

c2Z0

1152π2k6

[~Q

∗· ~Q− |n · ~Q|2

], (13)

Using eq. (9), we compute:

~Q∗· ~Q = 18a4q2 sin2 θ , |n · ~Q|2 = |3a2q sin2 θ e2iφ|2 = 9a4q2 sin2 θ .

Hence,~Q

∗· ~Q− |n · ~Q|2 = 9a4q2 sin2 θ(2− sin2 θ) = 9a4q2 sin2 θ(1 + cos2 θ) .

Eq. (13) then yieldsdP

dΩ=

c2Z0a4q2k6

128π2sin2 θ(1 + cos2 θ) .

Using k = 2ω/c [cf. eq. (6)], we obtain

dP

dΩ=

Z0a4q2ω6

2π2c4sin2 θ(1 + cos2 θ) . (14)

The total radiated power is obtained by integrating over solid angles. Using∫

dΩ sin2 θ(1 + cos2 θ) = 2π

∫ 1

−1

(1− cos4 θ) d cos θ =16π

5,

we end up with

P =8Z0a

4q2ω6

5πc4. (15)

As a check, we can use eq. (9.49) of Jackson,

P =c2Z0k

6

1440π

∑

i,j

|Qij|2 . (16)

Using eq. (8) ∑

i,j

|Qij |2 = 36a4q2 .

Inserting this back into eq. (16) along with k = 2ω/c, we recover eq. (15) as expected.

4

A second derivation that employs the quadrupole tensor in the spherical basis

One may be tempted to compute the quadrupole tensor in the spherical basis using eq. (4.6)of Jackson. Starting from eqs. (1) and (2), one immediately obtains,

q2,±2(t) =qa2

2

√15

2πe∓2iωt , q2,±1(t) = q20(t) = 0 . (17)

These are the physical time-dependent multipole moments in the spherical basis. Note that itis a complex quantity that satisfies

q∗ℓm(t) = (−1)mqℓ,−m(t) , (18)

which follows from the corresponding property of the spherical harmonics,

Y ∗ℓm(θ, φ) = (−1)mYℓ,−m(θ, φ) . (19)

However, in order to use the results of Section 9.10 of Jackson, we must employ the complexmultipole moments defined by,

Qℓm(t) =

∫Y ∗ℓm(θ, φ) r

ℓ ρ(~x, t) d3x = Qℓm e−2iωt , (20)

where the complex charge density is ρ(~x, t) = ρ(~x) e−2iωt and the physical charge distributionis Re [ρ(~x) e−2iωt]. Eq. (20) defines the time-independent quantity Qℓm,

Qℓm ≡∫

Y ∗ℓm(θ, φ) r

ℓ ρ(~x) d3x . (21)

However, the physical time-dependent multipole moment, qℓm(t), is not equal to Re [Qℓm e−2iωt].Indeed, qℓm(t) is complex when m 6= 0, whereas Re [Qℓm e−2iωt] is real.

In order to determine the Qℓm, the simplest approach starts with the time-independentquantities, Qij [cf. eq. (8)], which are the analogs of the Qℓm in the Cartesian basis. The Qij

are related to the Qℓm by the same formulae given in eq. (4.6) of Jackson,

Q22 =1

12

√15

2π

(Qxx − 2iQxy −Qyy

), (22)

Q21 = −1

3

√15

8π

(Qxz − iQyz

), (23)

Q20 =12

√5

4πQzz , (24)

Q2,−1 =1

3

√15

8π

(Qxz + iQyz

), (25)

Q2,−2 =1

12

√15

2π

(Qxx + 2iQxy −Qyy

). (26)

However, unlike in the case of eq. (4.6) of Jackson, the Qij that appear on the right hand sideof eqs. (22)–(26) may be complex quantities.

5

Inserting the matrix elements of eq. (8) into eqs. (22)–(26), we immediately obtain

Q2m = qa2√

15

2πδm2 .

That is, Q22 is the only nonvanishing component. We employ this result in eq. (9.169) ofJackson [keeping in mind that Q′

ℓm = 0 as there is no intrinsic magnetization in this problem]and obtain

aE(2, m) =ck4

5i√6Q2m = −1

2icqa2k4

√1

5πδ2m .

Inserting the above result into eq. (9.151) of Jackson yields

dP

dΩ=

Z0c2q2a4k6

40π| ~X22|2 ,

where according to Table 9.1 on p. 437 of Jackson,

| ~X22|2 =5

16π(1− cos4 θ) =

5

16πsin2 θ(1 + cos2 θ) .

Integrating over all angles using eq. (9.120) of Jackson yields

P =Z0c

2q2a4k6

40π.

Finally, using k = 2ω/c, we see that we have reproduced eqs. (14) and (15).

2. [Jackson, problem 9.8]

(a) Show that a classical oscillating electric dipole ~p with fields given by eq. (9.18) of Jacksonradiates electromagnetic angular momentum to infinity at the rate

d~L

dt=

k3

12πǫ0Im [~p ∗

× ~p] .

In class, we derived the following result for the radiated angular momentum per unit time ingaussian units (which was denoted by ~τ ):

~τ = − r3

8πRe

∫ [(n× ~E∗)(n · ~E) + (n× ~B)(n · ~B∗)

]dΩ , (27)

where ~E and ~B are the complex electric and magnetic field vectors (after removing the harmonic

e−iωt factor). To rewrite this in SI units, we must replace ~E →√4πǫ0 ~E and ~B → √

4πµ0~H ,

where c = 1/√ǫ0µ0. Note that we must also replace ρ → ρ/

√4πǫ0 and ~J → ~J/

√4πǫ0, which

means that ~p → ~p/√4πǫ0 [cf. Table 3 on p. 782 of Jackson]. Thus, in SI units, we have

~τ = −12r3Re

∫ [ǫ0(n× ~E∗)(n · ~E) + µ0(n× ~H)(n · ~H∗)

]dΩ , (28)

6

We now make use of the electric dipole fields given by eq. (9.18) of Jackson,

~H =ck2

4π(n× ~p )

eikr

r

(1− 1

ikr

),

~E =1

4πǫ0

k2(n× ~p )× n

eikr

r+ [3n(n · ~p )− ~p ]

(1

r3− ik

r2

)eikr

.

Note that n · ~H = 0 and

n · ~E =1

4πǫ0

2n · ~p

(1

r3− ik

r2

)eikr

= − ik

2πǫ0r2n · ~p eikr +O

(1

r3

).

Thus, we only need to keep the O(1/r) terms in n× ~E∗. Using the vector identity,

n×(n× ~p )× n

= n× ~p ,

for a unit vector n, it follows that,

n× ~E∗ =k2

4πǫ0n× ~p ∗

e−ikr

r+O

(1

r

).

Hence, it follows that

~τ = Reik3

16π2ǫ0

∫dΩ n · ~p (n× ~p ∗) ,

where we have dropped terms that vanish in the limit of r → ∞. In component form,

τi = Reik3

16π2ǫ0ǫijkpℓp

∗k

∫dΩnjnℓ , (29)

where there is an implicit sum over the repeated indices j, k, and ℓ. Using eq. (9.47) of Jackson,

∫dΩnjnℓ =

4π

3δjℓ .

Inserting this result into eq. (29) yields

τi = Reik3

12πǫ0(~p× ~p ∗)i .

Finally, noting that Re(iz) = − Im z for any complex number z, and ~p × ~p ∗ = −~p ∗× ~p, weend up with4

~τ =d~L

dt=

k3

12πǫ0Im(~p ∗× ~p) . (30)

4In class, I wrote ~τ = −d~L/dt, where the right hand side denotes the rate of angular momentum lost by

the radiating sources. Jackson denotes d~L/dt as the rate of angular momentum transported to infinity, whichis the negative of the rate of angular momentum lost by the radiating sources.

7

(b) What is the ratio of angular momentum radiated to energy radiated? Interpret.

Eq. (9.24) of Jackson states that the total power radiated is given by

P =c2Z0k

4

12π|~p|2 , (31)

where Z0 =√

µ0/ǫ0 is the impedance of free space. In particular, note that c ǫ0Z0 = 1. Thus,using ω = kc it follows that

~τ

P=

Im(~p ∗× ~p)

ω|~p|2 . (32)

To interpret eq. (32), consider the case where the electric dipole moment possesses a definitevalue of m in the spherical basis. Recall that

q1,±1 = ∓√

3

8π(px ∓ ipy) , q10 =

√3

4πpz .

Consider three cases:

1. If q11 6= 0 and q10 = q1,−1 = 0, then

~p =p√2(−1 , −i , 0) =⇒ ~p ∗× ~p = i|~p|2z ;

2. If q10 6= 0 and q11 = q1,−1 = 0, then

~p = p(0 , 0 , 1) =⇒ ~p ∗× ~p = 0 ;

3. If q1,−1 6= 0 and q11 = q10 = 0, then

~p =p√2(1 , −i , 0) =⇒ ~p ∗× ~p = −i|~p|2z ,

where p ≡ |~p|. That is,

~p ∗× ~p = im|~p|2z , for m = −1, 0,+1 .

Inserting this result into eq. (32) yields

τzP

=dLz/dt

dU/dt=

m

ω.

In the quantum mechanics of electromagnetic radiation, photons possess an energy U = ~ωand a spin angular momentum Sz = m~, so that Sz/U = m/ω. The analogy is quite striking!

(c) For a charge e rotating in the x–y plane at radius a and angular speed ω, show that thereis only a z component of radiated angular momentum with magnitude dLz/dt = e2k3a2/(6πǫ0).What about a charge oscillating along the z axis?

8

For a charge e rotating in the x–y plane at radius a and angular speed ω, the components ofthe electric dipole vector are given by, ~p = ea(cosωt , sinωt , 0). This result can be rewrittenas

~p = Re

ea e−iωt (1 , i , 0)

.

Thus, we may define a complex electric dipole vector,

~p(t) = ~p e−iωt , where ~p = ea (1 , i , 0) .

It then follows that

~p ∗× ~p = det

x y z

ea −iea 0ea iea 0

= 2ie2a2 z .

Hence, Im(~p ∗× ~p ) = 2e2a2 z. Inserting this result into eq. (30) yields

τz =dLz

dt=

e2k3a2

6πǫ0.

For a charge oscillating along the z-axis, the real physical charge density is

ρ(~x, t) = qδ(x)δ(y)δ(z − z0 cosωt) .

Hence, ~p = z qz0 cosωt = z qz0Re e−iωt. Thus, we identify the corresponding complex electric

dipole moment vector (with the harmonic factor stripped off) as

~p = z qz0 .

Note that this is in fact a real vector, in which case ~p ∗× ~p = ~p× ~p = 0. Hence, for this case,~τ = 0 and no angular momentum is radiated.

The above two cases correspond to m = 1 and m = 0, respectively, which were treatedexplicitly at the end of part (b).

(d) What are the results corresponding to parts (a) and (b) for magnetic dipole radiation?

For magnetic dipole radiation, we use eqs. (9.35) and (9.36) of Jackson,

~E = −Z0k2

4π(n× ~m )

eikr

r

(1− 1

ikr

),

~H =1

4π

k2(n× ~m )× n

eikr

r+ [3n(n · ~m )− ~m ]

(1

r3− ik

r2

)eikr

.

As noted by Jackson at the top of p. 414, one can obtain results for magnetic dipole radiationfrom that of electric dipole radiation by the following set of interchanges,

~E → Z0~H , Z0

~H → −~E , ~p → ~m/c .

9

Applying these interchanges on the results obtained in eqs. (30) and (31) yields

~τ =µ0k

3

12πIm(~m ∗ × ~m) , (33)

P =µ0ck

4

12π|~m|2 . (34)

The corresponding ratio of these two quantities is:

~τ

P=

Im(~m ∗× ~m)

ω|~m|2 . (35)

The analysis of part (b) is nearly identical, with the magnetic dipole moment in the sphericalbasis replacing the electric dipole moment. Thus, again, we conclude that

~m ∗× ~m = im|~m|2z , for m = −1, 0,+1 .

which again yieldsτzP

=dLz/dt

dU/dt=

m

ω.

ADDED NOTE:

In the literature, you will sometimes see another expression for the rate of angular momen-tum transport by radiation, in place of eq. (28). To derive this expression, we first note that in

the radiation zone, the Jefimenko equations imply that n · ~E = n · ~H = 0 at O(1/r). Thus,

n · ~E = O(

1

r2

), n · ~H = O

(1

r2

). (36)

It follows that we only need to keep expressions for n× ~E and n× ~H at O(1/r) in eq. (28).For harmonic fields, eq. (9.5) of Jackson is

~E =iZ0

k~∇× ~H = −Z0 n× ~H +O

(1

r2

). (37)

where we have used the leading behavior of ~H ∝ (1/r)eikr−iωt. Likewise,

n× ~E = −Z0 n×

(n× ~H

)= Z0

~H +O(

1

r2

), (38)

after expanding out the triple product and using eq. (36).Inserting eqs. (37) and (38) into eq. (28), and using ǫ0Z0 = µ0/Z0 =

√ǫ0µ0 = 1/c, it follows

that

~τ = −r3

2cRe

∫ [~H∗(n · ~E)− ~E(n · ~H∗)

]dΩ .

10

Writing n = ~x/r and noting the vector identity,

~H∗(n · ~E)− ~E(n · ~H∗) = −1

r~x× (~E × ~H∗) ,

we end up with5

d~τ

dΩ=

r2

2cRe ~x× (~E × ~H∗) . (39)

Eq. (39) provides an alternative expression for the rate of angular momentum transport and isequivalent to eq. (28) in the limit of r → ∞.6

The infinitesimal area element is da = r2 dΩ, so eq. (39) can be rewritten as

d~τ

da=

1

2cRe ~x× (~E × ~H∗) . (40)

Since ~τ = d~L/dt, we interpret d~τ/da as the angular momentum flux that is transported fromthe sources out to the observer located a long distance away. Eq. (40) should be compared withthe expression for the angular momentum density [cf. problem 7.27 of Jackson],

1

µ0c2~x× (~E × ~B) =

1

c2~x× (~E × ~H) .

The above result is applicable to the real fields. The corresponding result for the time-averagedangular momentum density of a distribution of harmonic electromagnetic fields is given by

~L =1

2c2Re ~x× (~E × ~H∗) .

We conclude thatd~τ

da=

1

2cRe ~x× (~E × ~H∗) = c ~L +O

(1

r3

).

That is, the angular momentum flux in the radiation zone is equal to c times the angularmomentum density, although this identification is correct only at the lowest nontrivial order inthe inverse distance expansion.7

5For the record, the corresponding result in gaussian units is obtained by replacing ~H with ~B and multiplyingEq. (39) by c/(4π).

6In this limit, ~τ approaches a constant value which is equal to the rate of angular momentum transportedto the surface at infinity by the radiation.

7This added note was inspired by a treatment in Emil Jan Konopinski, Electromagnetic Fields and Relativistic

Particles (McGraw Hill Inc., New York, 1981). In particular, see the discussion on p. 226, including the veryenlightening footnote at the bottom of that page.

11

3. [Jackson, problem 9.12] An almost spherical surface is defined by

R(θ) = R0 [1 + β P2(cos θ)] (41)

has inside of it a uniform volume distribution of charge totaling Q. The small parameter βvaries harmonically in time at frequency ω. This corresponds to surface waves on the sphere.Keeping only lowest order terms in β and making the long-wavelength approximation, calculatethe nonvanishing multipole moments, the angular distribution of radiation, and the total powerradiated.

First, we need to evaluate the charge density ρ(~x, t). It is a constant ρ0 for r ≤ R(θ) and zerootherwise. Since the total charge Q is conserved (and hence time independent),

Q =

∫d3x ρ(~x, t) = ρ0

∫r2 dr d cos θ dφΘ(R(θ)− r) ,

where the step function is defined as,

Θ(x) =

1 , for x > 0 ,

0 , for x < 0 .

Thus,

Q = 2πρ0

∫ 1

−1

d cos θ

∫ R(θ)

0

r2 dr =2πR3

0ρ03

∫ 1

−1

d cos θ [1 + β P2(cos θ)]3 .

Assuming that |β| ≪ 1 and dropping terms of O(β2), it follows that

Q =2πR3

0ρ03

∫ 1

−1

d cos θ[1 + 3β P2(cos θ) +O(β2)

]=

4πR30ρ0

3

[1 +O(β2)

],

after using the orthogonality relation,∫ 1

−1

d cos θ Pℓ(cos θ)Pℓ′(cos θ) =2

2ℓ+ 1δℓℓ′ .

The parameter β varies harmonically with time. Using complex notation,

β = β0e−iωt

Hence, including all terms up to and including O(β0),

ρ(~x, t) =3Q

4πR30

Θ(R0 +R0β0P2(cos θ)e−iωt − r) . (42)

Next, we compute the elements of the multipole tensor in the spherical basis,

Qℓm(t) =

∫d3x rℓ Y ∗

ℓm(θ, φ) ρ(~x, t)

=3Q

4πR30

∫dΩY ∗

ℓm(θ, φ)

∫ R(θ)

0

rℓ+2 dr

=3QRℓ

0

4π(ℓ+ 3)

∫dΩY ∗

ℓm(θ, φ)[1 + β0Pℓ(cos θ)e

−iωt]ℓ+3

12

Working to first order in β0, we can approximate[1 + β0Pℓ(cos θ)e

−iωt]ℓ+3

= 1 + (ℓ+ 3)β0Pℓ(cos θ)e−iωt +O(β2

0) .

Writing

Pℓ(cos θ) =

√4π

2ℓ+ 1Yℓ0(θ, φ) ,

it follows that

Qℓm(t) =3QRℓ

0

4π(ℓ+ 3)

∫dΩY ∗

ℓm(θ, φ)

[1 + (ℓ+ 3)β0 e

−iωt

√4π

5Y20(θ, φ)

]

=3QRℓ

0

(ℓ+ 3)√4π

[δℓ0 δm0 +

(ℓ+ 3)β0 e−iωt

√5

δℓ2 δm0

], (43)

after using√4π Y00(θ, φ) = 1 and employing the orthogonality relation of the spherical har-

monics [cf. eq. (3.55) of Jackson],∫

Y ∗ℓm(θ, φ) Yℓ′m′(θ, φ) dΩ = δℓℓ′ δmm′ .

For the radiation problem, we are only interested in the harmonic piece of Qℓm(t). We thereforedenote the coefficient of e−iωt by Qℓm. Hence, we identify

Qℓm =3Qβ0R

20√

20πδℓ,2 δm,0 .

That is, the only non-zero electric multipole moment is

Q20 =3Qβ0R

20√

20π. (44)

An alternative derivation of eq. (44)

Since we are working to first order in β0, it is convenient to expand the Θ-function thatappears in eq. (42) using the fact that δ(x) = dΘ(x)/dx. Thus, to O(β0),

ρ(~x, t) =3Q

4πR30

[Θ(R0 − r) +R0β0P2(cos θ) e

−iωt δ(R0 − r)]. (45)

Using eq. (9.170) of Jackson, we can evaluate the multipole moments,

Qℓm(t) =

∫d3x rℓ Y ∗

ℓm(θ, φ) ρ(~x, t)

=3Q

4πR30

∫d3x rℓ Y ∗

ℓm(θ, φ)[Θ(R0 − r) +R0β0P2(cos θ) e

−iωt δ(R0 − r)]

=3QR0

4πR30

[√4π

Rℓ+30

ℓ+ 3δℓ0 δm0 +

√4π

5Rℓ+3

0 β0 e−iωt δℓ2 δm0

], (46)

which reproduces eq. (43).

13

As for the other possible multipole moments, we first note that there is no magnetizationin this problem so that Q′

ℓm = M ′ℓm = 0. [cf. eqs.(9.170) and (9.172) of Jackson]. However,

there is a non-zero harmonic current density due to motion of electric charges. The azimuthalsymmetry of the problem implies that ~J(~x, t), when written in spherical coordinates, has no φ

component and is independent of φ.8 That is,

~J(~x, t) =[Jr(r, θ) n+ Jθ(r, θ) θ

]e−iωt ,

where n ≡ ~x/r is the unit vector in the radial direction. Using eq. (9.172) of Jackson (in SI

units), with ~J(~x, t) = ~J(~x) e−iωt,

Mℓm = − 1

ℓ+ 1

∫d3x rℓ Y ∗

ℓm(θ, φ) ~∇ · (~x× ~J(~x)) . (47)

Using,~x× ~J(~x) = r n× ~J(~x) = r Jθ(r, θ) φ ,

we conclude that~∇ · (~x× ~J(~x)) =

1

sin θ

∂Jθ

dφ= 0 .

Hence, it follows from eq. (47) thatMℓm = 0 .

The angular distribution of the radiated power can be obtained from eqs. (9.151) and (9.169)of Jackson,

dP

dΩ= 1

2Z0c

2k2ℓ+2 ℓ+ 1

ℓ[(2ℓ+ 1)!!]2|Qℓm|2 | ~Xℓm|2 , (48)

where~Xℓm =

1√ℓ(ℓ+ 1)

~LYℓm(θ, φ) ,

is a vector spherical harmonic. Integrating over solid angles is trivial since the ~Xℓm are nor-malized to unity. Thus,

P = 12Z0c

2k2ℓ+2 ℓ+ 1

ℓ[(2ℓ+ 1)!!]2|Qℓm|2 . (49)

Inserting the value for Q20 obtained in eq. (44) into the above formulae, and noting that

| ~X20|2 =15

8πsin2 θ cos2 θ ,

according to Table 9.1 on p. 437 of Jackson, it follows that

dP

dΩ=

3Z0c2Q2β2

0R40k

6

2000π

15

8πsin2 θ cos2 θ ,

and

P =3Z0c

2Q2β20R

40k

6

2000π.

8An explicit expression for ~J(~x, t) will be given in an added note following this solution.

14

ADDED NOTE:

In this added note, we shall obtain an explicit form for ~J(~x, t) which is valid to first order

in β. As noted previously, the azimuthal symmetry of the problem implies that ~J(~x, t) has noφ component and is independent of the azimuthal angle φ. That is,

~J(~x, t) =[Jr(r, θ)n + Jθ(r, θ)θ

]e−iωt ,

where n ≡ ~x/r is the unit vector in the radial direction. Using the continuity equation,

~∇ · ~J +dρ

dt= 0 ,

we can compute ~∇ · ~J using the result for ρ obtained in eq. (45). Hence,

~∇ · ~J = −dρ

dt=

3iωQβ0

4πR20

P2(cos θ) e−iωt δ(R0 − r) . (50)

In spherical coordinates, we have

~∇ · ~J =1

r2 sin θ

[sin θ

∂

∂r

(r2Jr

)+ r

∂

∂θ(sin θ Jθ) + r

∂Jφ

∂φ

].

Since ~J arises due to charges in motion, it must be proportional to β. Thus, to leading order inβ, it must also be true that ~J is proportional to Θ(R0−r) since we can drop any β-dependencein the argument of the Θ-function (as the dropped terms will only contribute at higher orderin β). Hence, as Jφ = 0, we can write:

~J(~x, t) = β0 e−iωtΘ(R0 − r)

[J1n+ J2θ

]. (51)

Using eq. (51), we compute

~∇ · ~J = β0 e−iωt

1

r2∂

∂r

(r2Θ(R0 − r)J1

)+

1

r sin θΘ(R0 − r)

∂

∂θ(sin θ J2)

= β0 e−iωt

−J1 δ(R0 − r) + Θ(R0 − r)

[1

r2∂

∂r

(r2J1

)+

1

r sin θ

∂

∂θ(sin θ J2)

]. (52)

Comparing this result with eq. (50), we can immediately equate the coefficients of the deltafunction, which yields

J1 = −3iωQβ0

4πR20

P2(cos θ) . (53)

Inserting this result back into eq. (52), and comparing again with eq. (50), we conclude thatthe overall coefficient of the step function must vanish. That is,

−3iωQ

4πR20

P2(cos θ)2

r+

1

r sin θ

∂

∂θ(sin θ J2) = 0 .

15

Using P2(cos θ) =12(3 cos2 θ − 1), we obtain

−3iωQ

4πR20

sin θ(3 cos2 θ − 1

)+

∂

∂θ(sin θ J2) = 0 .

Hence,

J2 sin θ = −3iωQ

4πR20

∫ (3 cos2 θ − 1

)d cos θ =

3iωQ

4πR20

cos θ(1− cos2 θ) ,

or equivalently,9

J2 =3iωQ

8πR20

sin 2θ . (54)

Inserting eqs. (53) and (54) back into eq. (51) yields the final result, which is valid at first orderin β0,

~J(~x, t) = −3iωQ

8πR20

e−iωt Θ(R0 − r)[(3 cos2 θ − 1)n− sin 2θ θ

].

4. [Jackson, problem 9.16] A thin linear antenna of length d is excited in such a way that asinusoidal current makes a full wavelength of oscillation as shown in the figure below.

Figure 2: A thin linear antenna with a sinusoidal current that makes a full wavelength of oscillation.

(a) Calculate exactly the power radiated per unit solid angle and plot the angular distribu-tion of radiation.

Choose the z-axis to lie along the antenna, and let z = 0 correspond to the center of theantenna. Then, ~J(~x, t) = ~J(~x) e−iωt, where

~J(~x, t) = I sin

(2πz

d

)δ(x) δ(y) z , for |z| ≤ 1

2d , (55)

where d is the length of the antenna. In class, we derived the following results for the complexmagnetic and electric fields (assumed to be harmonic) in gaussian units,

~B(~x, t) =iω

c2rei(kr−ωt) n×

∫d3x′ ~J(~x ′) e−ik~x ′

·n +O(

1

r2

),

~E(~x, t) = B(~x, t)× n+O(

1

r2

),

9In evaluation of the indefinite integral, the constant of integration must be set to zero, since J2 must benon-singular at θ = 0 and at θ = π.

16

where n ≡ ~x/r and r ≡ |~x|. In SI units,10 the above results take the following form,

~H(~x, t) =iω

4πc rei(kr−ωt) n×

∫d3x′ ~J(~x ′) e−ik~x ′

·n +O(

1

r2

), (56)

~E(~x, t) = Z0~H(~x, t)× n+O

(1

r2

). (57)

Using eq. (9.21) of Jackson, the time-averaged power radiated per unit solid angle is givenby

dP

dΩ= 1

2Re

[r2 n · ~E × ~H ∗

].

In light of eq. (57), we compute

( ~H × n)× ~H ∗ = − ~H ∗× ( ~H × n) = n( ~H · ~H ∗)− ~H(n · ~H ∗) = n| ~H|2 ,

where at the last step we n · ~H ∗ = 0, which is a consequence of eq. (56). Hence,

dP

dΩ= 1

2Z0r

2| ~H|2 . (58)

Thus, our task is to compute the integral,∫

d3x′ ~J(~x ′) e−ik~x ′·n .

By assumption, the sinusoidal current makes a full wavelength, which implies that

k =2π

d. (59)

Inserting eq. (55) into the integral above and employing rectangular coordinates,∫

d3x′ ~J(~x ′) e−ik~x ′·n = z I

∫ d/2

−d/2

sin kz e−ikz cos θ dz ,

where θ is the angle between ~n and the positive z-axis (which corresponds to the usual polarangle of spherical coordinates). The following indefinite integral appears in many integrationtables, ∫

eaz sin kz dz =eaz(a sin kz − k cos kz)

a2 + k2.

Using eq. (59), the limits of integration are |z| ≤ π/k,∫ π/k

−π/k

sin kz e−ikz cos θ dz =e−ikz cos θ(−ik cos θ sin kz − k cos kz)

(−ik cos θ)2 + k2

∣∣∣∣π/k

−π/k

=e−iπ cos θ − eiπ cos θ

k sin2 θ

= −2i sin(π cos θ)

k sin2 θ.

10To convert from gaussian to SI units, we must replace the fields ~E →√4πǫ0 ~E, and ~B → √

4πµ0~H, where

c = 1/√ǫ0µ0 and the current ~J → ~J/

√4πǫ0 [cf. Table 3 on p. 782 of Jackson].

17

Using ω = kc and n× z = − sin θ φ, it follows that

~H(~x, t) =Iω

2πkc rei(kr−ωt) sin(π cos θ)

sin2 θn× z = − I

2πrei(kr−ωt) sin(π cos θ)

sin θφ .

Plugging this result into eq. (58), we end up with

dP

dΩ=

Z0I2

8π2

[sin(π cos θ)

sin θ

]2. (60)

A plot of the angular distribution is shown in Figure 3.

-0.5 0.5

-1.0

-0.5

0.5

1.0

Figure 3: A polar plot of the antenna pattern of a thin linear antenna with a sinusoidal currentthat makes a full wavelength of oscillation. The angular distribution of the radiated power is givenby eq. (60) and is proportional to sin2(π cos θ)/ sin2 θ. Normalization has been chosen such thatZ0I

2 = 8π2. This plot was created with Mathematica 11 software.

(b) Determine the total power radiated and find a numerical value for the radiation resis-tance.

The total power is

P =

∫dP

dΩdΩ = 2π

∫ 1

−1

dP

dΩd cos θ ,

18

since the angular distribution obtained in eq. (60) is independent of the azimuthal angle φ.Defining x ≡ cos θ, and employing sin2 θ = 1− cos2 θ in the denominator of eq. (60),

P =Z0I

2

4π

∫ 1

−1

sin2(πx)

1− x2dx =

Z0I2

8π

∫ 1

−1

1− cos(2πx)

1− x2dx ,

after employing a well-known trigonometric identity. We now apply the method of partialfractions to write

1

1− x2=

1

2

(1

1− x+

1

1 + x

).

The resulting two integrals are equal after making a variable change x → −x in the first integral.Thus,

P =Z0I

2

8π

∫ 1

−1

1− cos(2πx)

1 + xdx .

Next, we make a change of variables, t = 2π(1 + x), which converts the above integral into thefollowing form,

P =Z0I

2

8π

∫ 4π

0

1− cos t

tdt .

This integral can be evaluated in terms of the cosine integral, which is defined as

Ci(x) = −∫ ∞

x

cos t

tdt .

It then follows that:11 ∫ x

0

1− cos t

tdt = γ + ln x− Ci(x) , (61)

where γ ≃ 0.5772 is the Euler constant. Thus,

P =Z0I

2

8π[γ + ln(4π)− Ci(4π)] .

Using the following numerical value, Ci(4π) = −0.006,12 we obtain

P =Z0I

2

8π(3.114) .

The corresponding radiative resistance (in ohms) is equal to the coefficient of 12I2 [cf. the text

below eq. (9.29) of Jackson]. Thus, using Z0 = 376.7 ohms [given below eq. (7.11)′ of Jackson],

Rrad = (3.114)Z0

4π= 93.3 ohms . (62)

11See e.g. formula 8.230 no. 2 on p. 895 of I.S. Gradshteyn and I.M. Ryzhik, Table of Integrals, Series, and

Products (8th edition), edited by Daniel Zwillinger and Victor Moll (Academic Press, Elsevier, Inc., Waltham,MA, 2015). Eq. (61) can also be found on p. 41 [cf. problem 3 on this page] of N.N. Lebedev, Special Functionsand Their Applications (Dover Publications, Inc., Mineola, NY, 1972).

12For example, one can consult Milton Abramowitz and Irene A. Stegun, Handbook of Mathematical Functions

(Dover Publications, Inc., Mineola, NY, 1965), which provides numerical tables of the cosine integral. Alter-natively, one can use a mathematical program such as Mathematica or Maple to evaluate the cosine integraldirectly.

19

5. [Jackson, problem 9.17] Treat the linear antenna of the previous problem by the multipoleexpansion method.

(a) Calculate the multipole moments (electric dipole, magnetic dipole, and electric quadrupole)exactly and in the long-wavelengths approximation.

The current density is given by eq. (55). It is convenient to rewrite this in spherical coordinates.Note that z = n for cos θ = 1 (i.e., θ = 0) and z = −n for cos θ = −1 (i.e., θ = π), where n isa unit vector pointing in the radial direction, and r ≡ |~x| is the radial coordinate. Hence, wemay write13

~J(~x) =I

2πr2sin

(2πr

d

)[δ(cos θ − 1) + δ(cos θ + 1)

]Θ(1

2d− r)n , (63)

where I have inserted the Heavyside step function since the current I(z, t) = 0 for |z| > 12d. In

obtaining eq. (63), I used the fact that

sin

(2πz

d

)= sin

(2πrε(z)

d

)= ε(z) sin

(2πr

d

),

where the sign function ε(z) is defined as

ε(z) =

+1 , for z > 0,

−1 , for z < 0.

Finally, we note that n = ǫ(z)z along the z-axis.We shall make use of eqs. (9.167) and (9.168) of Jackson for the electric and magnetic

multipole coefficients. In the absence of magnetization, in MKS units,

aE(ℓ,m) =k2

i√

ℓ(ℓ+ 1)

∫Y ∗ℓm(θ, φ)

cρ(~x)

∂

∂r

[rjℓ(kr)

]+ ik ~x· ~J(~x)jℓ(kr)

d3x , (64)

aB(ℓ,m) =k2

i√

ℓ(ℓ+ 1)

∫Y ∗ℓm(θ, φ)~∇·

(~x× ~J(~x)

)jℓ(kr)d

3x . (65)

It is convenient to integrate by parts in evaluating the first term of the integrand in eq. (64).The surface term can be dropped, since the charge density is localized. Since d3x = r2 dr dΩ,after integrating by parts, one obtains

− ∂

∂r

[r2ρ(~x)

]dr = −

(∂ρ(~x)

∂r+

2

rρ(~x)

)r2 dr .

It then follows that

aE(ℓ,m) =k2

i√

ℓ(ℓ+ 1)

∫Y ∗ℓm(θ, φ)jℓ(kr)

−c

(2 + r

∂

∂r

)ρ(~x) + ik ~x· ~J(~x)

d3x , (66)

13Note that this differs from eq. (9.179) of Jackson by a relative sign. This difference is due to the fact thatfor the antenna showed in Figure 9.6 of Jackson, we have I(−z) = I(z). In contrast, in this problem, eq. (55)yields I(−z) = −I(z).

20

which is the version we obtained in class.14

First consider the computation of aB(ℓ,m). Using the vector identity,

~∇·(~x× ~J = ~J ·(~∇× ~x)− ~x·(~∇× ~J) = −~x·(~∇× ~J) ,

after using ~∇× ~x = 0. However, the current density given in eq. (63) is purely radial, which

implies that ~∇ × ~J = 0. Hence, we conclude that ~∇·(~x × ~J) = 0, which implies that

aB(ℓ,m) = 0. That is, all the magnetic multipole coefficients vanish.To evaluate the electric multipole coefficients, aE(ℓ,m), we can either use eq. (64) or eq. (66).

We shall first employ eq. (64), and then in an addendum we will provide details of the calculationthat makes use of eq. (66).

For harmonic sources [cf. eq. (9.15) of Jackson], ~∇· ~J = iωρ . Using eq. (63), we see that ~J

is purely radial, ~J = Jrn, and

~∇· ~J =1

r2∂

∂r

(r2Jr

)=

I

r2[δ(cos θ − 1) + δ(cos θ + 1)

]

×1

dcos

(2πr

d

)Θ(1

2d− r)− 1

2πsin

(2πr

d

)δ(1

2d− r)

.

Noting that sin(2πr/d)δ(12d − r) = sin π δ(1

2d − r) = 0, we can drop the delta function in the

previous equation. We conclude that

ρ(~x) =1

ikc~∇· ~J(~x) =

I

ikcr2dcos

(2πr

d

)[δ(cos θ − 1) + δ(cos θ + 1)

]Θ(1

2d− r) , (67)

after making use of ω = kc. We also note that eq. (63) yields

~x· ~J(~x) =I

2πrsin

(2πr

d

)[δ(cos θ − 1) + δ(cos θ + 1)

]Θ(1

2d− r) . (68)

after using ~x·n = r.Plugging eqs. (67) and (68) into eq. (64), and evaluating the integral using spherical coor-

dinates, d3x = r2 dr dΩ,

aE(ℓ,m) =Ik

d√ℓ(ℓ+ 1)

∫Y ∗ℓm(θ, φ)

[δ(cos θ − 1) + δ(cos θ + 1)

]dΩ

×∫ d/2

0

r dr

− cos

(2πr

d

)1

r

∂

∂r

[rjℓ(kr)

]+

k2

2πsin

(2πr

d

)jℓ(kr)

.(69)

We first evaluate the angular integral above. Writing dΩ = d cos θ dφ, consider the integral

1

2π

∫Y ∗ℓm(θ, φ)

[δ(cos θ − 1) + δ(cos θ + 1)

]d cos θ dφ .

14One small advantage of using eq. (64) instead of eq. (66) is that no delta functions arise in the computation[cf. eq. (78) below]. By employing eq. (64), Jackson can simply set the limits of the radial integration to0 ≤ r ≤ 1

2d, and otherwise ignore the implicit Heavyside step function in his analysis of the linear, centerfed

antenna on pp. 445–446.

21

Since Y ∗ℓm(θ, φ) ∝ e−imφ, the φ integral yields

∫ 2π

0

e−imφdφ = δm0 .

Thus, in light of eq. (3.57) of Jackson and the properties of the Legendre polynomials,

1

2π

∫Y ∗ℓm(θ, φ)

[δ(cos θ − 1) + δ(cos θ + 1)

]d cos θ dφ = δm0

[Y ∗ℓ0(0, φ) + Y ∗

ℓ0,φ(π)]

= δm0

(2ℓ+ 1

4π

)1/2 [1 + (−1)ℓ

]. (70)

Plugging eq. (70) back into eq. (69),

aE(ℓ,m) =Ik

d

√π(2ℓ+ 1)

ℓ(ℓ+ 1)δm0

[1+(−1)ℓ

] ∫ d/2

0

− cos

(2πr

d

)∂

∂r

[rjℓ(kr)

]+k2rd

2πsin

(2πr

d

)jℓ(kr)

dr .

The first term of the integrand above can be rewritten using an integration by parts,

∫ d/2

0

cos

(2πr

d

)∂

∂r

[rjℓ(kr)

]= cos

(2πr

d

)rjℓ(kr)

∣∣∣∣d/2

0

+2π

d

∫ d/2

0

r sin

(2πr

d

)jℓ(kr)dr

= −12d jℓ

(12kd

)+

2π

d

∫ d/2

0

r sin

(2πr

d

)jℓ(kr)dr .

We then end up with

aE(ℓ,m) =Ik

2

√2ℓ+ 1

πℓ(ℓ+ 1)δm0

[1+(−1)ℓ

]πjℓ

(12kd

)+

[k2 −

(2π

d

)2]∫ d/2

0

sin

(2πr

d

)jℓ(kr) r dr

.

(71)Finally, setting kd = 2π [cf. eq. (59)], we arrive at the final result,

aE(ℓ,m) =Ik

2

√π(2ℓ+ 1)

ℓ(ℓ+ 1)δm0

[1 + (−1)ℓ

]jℓ(π) . (72)

We now consider the long wavelength approximation, kd ≪ 1. We will o the computationin two ways. First we will start with eq. (71) and use the small argument approximation forthe spherical Bessel function,

jℓ(kr) ≃(kr)ℓ

(2ℓ+ 1)!!.

Changing variables to x ≡ 2r/d,

aE(ℓ,m) ≃ Ik

2(2ℓ+ 1)!!

√π(2ℓ+ 1)

ℓ(ℓ+ 1)δm0

[1 + (−1)ℓ

](kd

2

)ℓ1− π

∫ 1

0

xℓ+1 sin(πx)dx

,

22

after dropping the O(k2) term in the factor that multiplies the integral in eq. (71). Integratingby parts yields

π

∫ 1

0

xℓ+1 sin(πx)dx = 1 + (ℓ+ 1)

∫ 1

0

xℓ cos(πx)dx .

Hence, we obtain a slightly simpler result,

aE(ℓ,m) ≃ − Ik

2(2ℓ+ 1)!!

√π(ℓ+ 1)(2ℓ+ 1)

ℓδm0

[1 + (−1)ℓ

](kd

2

)ℓ ∫ 1

0

xℓ cos(πx)dx . (73)

As a check of eq. (73), we can perform thee computation using eqs. (9.169)–(9.170) ofJackson (after setting the magnetization to zero),

aE(ℓ,m) ≃ ckℓ+2

i(2ℓ+ 1)!!

(ℓ+ 1

ℓ

)1/2

Qℓm , (74)

where

Qℓm =

∫rℓY ∗

ℓm(θ, φ)ρ(~x) d3x . (75)

Inserting eq. (67) into eq. (75), and making use of eq. (70),

Qℓm =I√π(2ℓ+ 1)

ikcdδm0

[1 + (−1)ℓ

] ∫ d/2

0

rℓ cos

(2πr

d

)dr ,

after using ω = kc. Changing variables to x = 2r/d,

Qℓm =I√

π(2ℓ+ 1)

2ikcδm0

(d

2

)ℓ [1 + (−1)ℓ

] ∫ 1

0

xℓ cos(πx)dx .

Plugging this result into eq. (74) yields

aE(ℓ,m) ≃ − Ik

2(2ℓ+ 1)!!

√π(ℓ+ 1)(2ℓ+ 1)

ℓδm0

[1 + (−1)ℓ

](kd

2

)ℓ ∫ 1

0

xℓ cos(πx)dx ,

in agreement with eq. (73)We now evaluate these results explicitly for the electric dipole (ℓ = 1) and the electric

quadrupole (ℓ = 2). Due to the factor of 1 + (−1)ℓ, we immediately see that only even ℓmultipoles survive. Hence, the electric dipole coefficient vanishes. Thus, we henceforth focuson the electric quadrupole coefficient. First, we use the exact result given in eq. (72). Using

j2(x) =

(3

x3− 1

x

)sin x− 3

x2cosx ,

it follows that j2(π) = 3/π2. Hence, for kd = 2π and ℓ = 2, eq. (72) yields

aE(2, 0) = Ik

√15

2π3. (76)

23

Let us compare this result with eq. (73), which was obtained in the long wavelength approxi-mation.

aE(2, 0) ≃ −Ik

√π

30

(kd

2

)2 ∫ 1

0

x2 cos(πx)dx .

Performing the integral,∫ 1

0

x2 cos(πx)dx =1

π3

[2πx cos(πx) + (pi2x2 − 2) sin(πx)

]∣∣∣∣1

0

= − 2

π2,

we end up with

aE(2, 0) ≃ Ik

√2

15π3

(kd

2

)2

.

This result should only be valid for kd ≪ 1. Nevertheless, to compare with eq. (76), we bravelyput kd = 2π to obtain

aE(2, 0) ≃ Ik

√2π

15, (77)

which is larger than the exact result given in eq. (76) by a factor of 2π2/15 ≃ 1.316. Not toobad!

ADDENDUM:

As promised, we exhibit the necessary calculations to obtain aE(ℓ,m) starting from eq. (66).In this method, one needs to keep track of the Heavyside step function, since it will generate adelta function when computing ∂ρ/∂r that cannot be ignored, as noted in footnote 14.

In this method, we use eq. (67) to compute

−(2 + r

∂

∂r

)ρ(~x) =

[δ(cos θ−1)+δ(cos θ+1)

] I

ickrd

2π

dsin

(2πr

d

)Θ(1

2d−r)+cos

(2πr

d

)δ(r−1

2d)

.

The delta function piece can be simplified by using cos(2πr/d)δ(r − 12d) = cosπ δ(r − 1

2d) =

−δ(r − 12d). Hence,

−(2 + r

∂

∂r

)ρ(~x) =

[δ(cos θ−1)+ δ(cos θ+1)

] I

ickrd

2π

dsin

(2πr

d

)Θ(1

2d− r)− δ(r− 1

2d)

.

(78)Using eq. (68), we end up with

−c

(2 + r

∂

∂r

)ρ+ ik~x· ~J =

2πI

ikrd2

[1−

(kd

2π

)2]sin

(2πr

d

)Θ(1

2d− r)− d

2πδ(r − 1

2d)

×[δ(cos θ − 1) + δ(cos θ + 1)

]. (79)

We now insert eq. (79) into eq. (66). Using eq. (70), and performing some algebraic simplifica-tions, it follows that

aE(ℓ,m) =Ik

2

√2ℓ+ 1

πℓ(ℓ+ 1)δm0

[1+(−1)ℓ

]πjℓ

(12kd

)+

[k2 −

(2π

d

)2]∫ d/2

0

sin

(2πr

d

)jℓ(kr) r dr

,

which reproduces eq. (71).

24

(b) Compare the shape of the angular distribution of the radiated power for the lowestnonvanishing multipole with the exact distribution obtained in the previous problem.

Using eq. (9.151) of Jackson, the angular distribution of power for a pure electric multipole oforder (ℓ,m) is given by,

dP (ℓ,m)

dΩ=

Z0

2k2|aE(ℓ,m)|2| ~Xℓm|2 .

We shall apply this result to the exact form of the pure electric multipole of order (ℓ,m) = (2, 0)obtained in eq. (76), which we rewrite again here,

aE(2, 0) = Ik

√15

2π3.

Using Table 9.1 on p. 437 of Jackson,

| ~Xℓm|2 =15

8πsin2 θ cos2 θ ,

Hence,dP (2, 0)

dΩ=

225Z0I2

32π4sin2 θ cos2 θ . (80)

This should be compared with the exact result obtained in eq. (60).

-1.0 -0.5 0.5 1.0

-1.0

-0.5

0.5

1.0

Figure 4: A polar plot of the antenna pattern of a thin linear antenna with a sinusoidal current thatmakes a full wavelength of oscillation. Normalization has been chosen such that Z0I

2 = 8π2. Theangular distribution of the radiated power, shown in red, is given by eq. (60). This is compared withthe corresponding angular distribution of the electric quadrupole component, shown in blue, whichis given by eq. (80). This plot was created with Mathematica 11 software.

25

(c) Determine the total power radiated for the lowest multipole and the correspondingradiation resistance using both multipole moments from part (a). Compare with part (b) ofthe previous problem. Is there a paradox here?

The total power radiated by a pure electric multipole of order (ℓ,m) is given by eq. (9.154) ofJackson,

P (ℓ,m) =Z0

2k2|a(ℓ,m)|2 .

In part (b) we obtained two expressions for aE(2, 0). The first expression was exact for kd = 2π[cf. eq. (76)],

aE(2, 0) = Ik

√15

2π3. (81)

The second was computed in the long-wavelength limit, but with kd = 2π [cf. eq. (77)],

aE(2, 0) ≃ Ik

√2π

15. (82)

If we use the exact electric quadrupole result [eq. (81)], then we obtain

P (2, 0) =15Z0I

2

4π3.

The corresponding radiative resistance (in ohms) is equal to the coefficient of 12I2 [cf. the text

below eq. (9.29) of Jackson]. Thus, using Z0 = 376.7 ohms [given below eq. (7.11)′ of Jackson],

Rrad =15Z0

2π3= 91.1 ohms , (83)

which is remarkably close to the exact result, Rrad = 93.3 ohms, given in eq. (62). In contrast,had we used eq. (82), we would have obtained Rrad = 2πZ0/15 = 157.8 ohms, which is a terribleapproximation, as one might have expected.

There is no paradox here. The discussion in Jackson on pp. 446–448 makes clear thatkeeping the lowest nonvanishing multipole but computing it exactly (i.e., without assumingthat kd ≪ 1) yields an accurate result to the exact antenna problem even for values of kd aslarge as 2π. Presumably, if one computes the next non-trivial multipole (in this problem, thatwold be ℓ = 4) its numerical contribution, the result would be a rather small correction to thepower even when kd = 2π.

Perhaps the paradox that Jackson is alluding to is based on the expectation that,

P (2, 0) < Pexact ,

since according to eq. (9.155) of Jackson, the total power is equal to an incoherent sum ofcontributions from all the multipoles. Indeed in our computations above, we did confirm thatP (2, 0) < Pexact, or equivalently the radiation resistance of the electric quadrupole contributiongiven in eq. (83) is less than the exact result obtained in eq. (62). In contrast, the opposite(incorrect) conclusion would have been drawn had we used the expression for P (2, 0) based onsetting kd = 2π in the long wavelength limit [e.g., eq. (82)]. Of course, this latter result is anartifact of a poor approximation.

26