Chapter 6

Harmonic Expansion ofElectromagnetic Fields

6.1 Introduction

For a given current source J(r, t), the vector potential can in principle be found by solving the

inhomogeneous vector wave equation,(∇2 − 1

c2∂2

∂t2

)A (r, t) = −µ0J (r, t) ,

provided the Lorenz gauge is chosen. In source free region, the electromagnetic fields E and H can

then be deduced from the vector potential through

B = ∇×A,

ε0µ0∂E

∂t= ∇×B, (in source-free region J = 0).

In this Chapter, we develop spherical harmonic expansion of the electromagnetic fields. In exper-

iments, the electromagnetic fields, not the potentials, are measured. Radiation electromagnetic

fields can be decomposed into two basic vector components, Transverse Magnetic (TM) and Trans-

verse Electric (TE), where “transverse” is with respect to the direction of wave propagation r.These fundamental modes are normal to each other and provide convenient base vectors in analysis

of radiation fields.

6.2 Wave Equations and Green’s Function

The set of Maxwell’s equations,

∇ ·E =ρ

ε0, (6.1)

1

∇×E = −∂B∂t, (6.2)

∇ ·B = 0, (6.3)

∇×B = µ0

(J+ ε0

∂E

∂t

), (6.4)

can be reduced to two inhomogeneous wave equations for the two potentials Φ and A as follows.

Substitution of

E = −∇Φ− ∂A

∂t, (6.5)

into Eq. (6.1) yields

∇2Φ +∂

∂t∇ ·A = − ρ

ε0. (6.6)

Also, substitution of B = ∇×A and Eq. (6.5) into Eq. (6.4) yields

∇2A− 1

c2∂2A

∂t2−∇

(∇ ·A+

1

c2∂Φ

∂t

)= −µ0J. (6.7)

If the Lorenz gauge

∇ ·A+1

c2∂Φ

∂t= 0, (6.8)

is chosen for ∇ ·A (i.e., the longitudinal component of the vector potential), then Eqs. (6.6) and

(6.7) are completely decoupled, and reduce, respectively, to(∇2 − 1

c2∂2

∂t2

)Φ = − ρ

ε0, (6.9)

(∇2 − 1

c2∂2

∂t2

)A = −µ0J. (6.10)

If the Coulomb gauge

∇ ·A = 0,

is chosen instead, that is, if the longitudinal component of the vector potential is chosen to be zero,

such decoupling cannot be achieved,

∇2ΦC = − ρ

ε0, (6.11)

∇2A− 1

c2∂2A

∂t2− 1

c2∂

∂t∇Φ = −µ0J. (6.12)

Since the longitudinal current Jl satisfies the charge conservation law,

∂ρ

∂t+∇ · Jl = 0, (6.13)

it follows that1

c2∂

∂t∇2ΦC = µ0∇ · Jl, (6.14)

2

and the wave equation for the vector potential in Coulomb gauge involves only the transverse

component of the vector potential,

∇2At −1

c2∂2At

∂t2= −µ0Jt. (6.15)

As briefly pointed out in Chapter 3, the scalar potential Φ in Eq. (6.11) is not subject to retardation

due to finite propagation speed of electromagnetic disturbance. The part of the electric field

generated from the gradient of the scalar potential

E = −∇ΦC −∂A

∂t,

is also nonretarded because of instantaneous propagation. Physical (retarded) electric field is con-

tained in the vector potential. In the Lorenz gauge, both potentials are appropriately retarded and

we will therefore employ Lorenz gauge. (As we will see in the following section, the instantaneous

Coulomb field is in fact cancelled by a term in the retarded transverse vector potential.)

The scalar potential Φ and three cartesian components of the vector potential Ai all satisfy the

wave equation in the form (∇2 − 1

c2∂2

∂t2

)Ai = −fi(r, t), (6.16)

where fi(r, t) is a source function, either the charge density or the current density. The Green’s

function for the scalar wave equation G(r, t) can be found as a solution for the following singular

wave equation, (∇2 − 1

c2∂2

∂t2

)G(r− r′, t− t′) = −δ(r− r′)δ(t− t′), (6.17)

with the boundary condition that

G = 0 at r =∞ and t = ±∞.

Once the Green’s function is found, the solution to the original wave equation can be written down

in the form of convolution,

Ai(r, t) =

∫G(r− r′, t− t′)fi(r′, t)dV ′dt′. (6.18)

We seek a solution for the Green’s function by the method of Fourier transform. Let the Fourier

transform of G(r, t) be g(k, ω),

g(k, ω) =

∫dV

∫dτe−i(k·R−ωτ)G(R, τ), (6.19)

G(R,τ) =1

(2π)4

∫d3k

∫dωg(k, ω)ei(k·R−ωτ), (6.20)

3

where R = r− r′, τ = t− t′. Then, Eq. (6.17) in the Fourier-Laplace space (k, ω) becomes(−k2 +

ω2

c2

)g(k, ω) = −1,

or

g(k, ω) =1

k2 − (ω/c)2. (6.21)

Substitution of g(k, ω) into Eq. (6.20) yields

G(R, τ) =1

(2π)4

∫dk

∫dω

ei(k·R−ωτ)

k2 − (ω/c)2.

Integration over ω can be done easily,∫ ∞−∞

e−iωτ

k2 − (ω/c)2dω = 2π

c

ksin(kcτ). (6.22)

(Note that with a new variable s = −iω, the integral reduces to

−ic2∫ i∞

−i∞

esτ

s2 + (ck)2ds =

2π

ck

1

2πi

∫ i∞

−i∞

ckesτ

s2 + (ck)2ds =

2π

cksin(kcτ), (6.23)

which is the standard inverse Laplace transform. Causality is appropriately handled in inverse

Laplace transform.) Then,

G(R, τ) =c

(2π)3

∫ ∞0

dk

∫ π

0dθ

∫ 2π

0dφk sin(ckτ)eikR cos θ sin θ, (6.24)

where θ is the polar angle in the k space measured from the direction ofR = r− r′, and R = |r− r′| .The integrations can be carried out as follows:

G(R, τ) =c

(2π)2

∫ ∞0

kdk sin(ckτ)

∫ π

0eikR cos θ sin θdθ

=c

4π2R

∫ ∞0

2 sin(ckτ) sin(kR)dk

=c

4πR[δ(cτ −R)− δ(cτ +R)] , (6.25)

where in the final step, use is made of∫ ∞0

cos(ak)dk = πδ(a).

The functionc

Rδ(cτ −R) =

1

Rδ

(τ − R

c

), (6.26)

4

describes an impulse propagating radially outward at a speed c, and the function

c

Rδ(cτ + t) =

1

Rδ

(τ +

R

c

), (6.27)

describes an impulse propagating radially inward. For radiation fields due to a source spatially

limited, only the outgoing solution is physically meaningful and we adopt it as the Green’s function,

G(R, τ) =1

4πRδ

(τ − R

c

)=

1

4π|r− r′|δ(t− t′ − |r− r

′|c

), (m−1s−1). (6.28)

It is straightforward to check that G(R, t) satisfies the inhomogeneous wave equation in Eq. (6.17)

if the following identity is recalled,

∇2 1

|r− r′| = −4πδ(r− r′

).

Having found the Green’s function for the wave equation, the general solution to the inhomo-

geneous wave equation, (∇2 − 1

c2∂2

∂t2

)Ai = −fi(r, t), (6.29)

can be written down as

Ai(r, t) =1

4π

∫dV ′

∫dt′fi(r

′, t′)

|r− r′| δ(t− t′ − |r− r

′|c

). (6.30)

If the source function f(r, t) is separable in the form,

fi(r, t) = fi(r)e−iωt, (6.31)

Eq. (6.30) reduces to

Ai(r, t) =1

4πe−iωt

∫eik|r−r

′|

|r− r′| fi(r′)dV ′, (6.32)

where k = ω/c. The spatial part of this solution,

Ai(r) =1

4π

∫eik|r−r

′|

|r− r′| fi(r′)dV ′, (6.33)

satisfies the Helmholtz equation, (∇2 + k2

)Ai = −fi(r). (6.34)

Corresponding Green’s function is

G(r− r′) =eik|r−r

′|

4π |r− r′| , (6.35)

5

which satisfies the Helmholtz equation,

(∇2 + k2

)G(r− r′

)= −δ

(r− r′

). (6.36)

The formulation developed here agrees with that in the preceding Chapter which has been derived

through a qualitative argument based on the retarded nature of electromagnetic disturbance. In

static cases ω = 0, k = 0, we trivially recover the static potentials,

Φ(r) =1

4πε0

∫ρ(r′)

|r− r′|dV′,

A(r) =µ04π

∫J(r′)

|r− r′|dV′.

In the Green’s function

G(r− r′, t− t′) =1

4π |r− r′|δ(t− t′ − |r− r

′|c

), (6.37)

the retarded nature of electromagnetic disturbance can be clearly seen. For a source located at

r′ and observer at r, electromagnetic phenomena observed at time t is due to source disturbance

created at the time |r− r′| /c seconds earlier than t.

6.3 Coulomb and Lorenz Gauges

The divergence of the vector potential ∇ ·A can be assigned an arbitrary scalar function without

affecting the electromagnetic fields E and B. The Maxwell’s equations do not specify ∇·A. In fact,potential transformation involving a scalar function λ,

A′ = A+∇λ, Φ′ = Φ− ∂λ

∂t, (6.38)

does not affect the electromagnetic fields,

E′ = −∇Φ′ − ∂A′

∂t= −∇Φ− ∂A

∂t= E,

B′ = ∇×A′ = ∇×A = B.

In Lorenz gauge characterized by

∇ ·A+1

c2∂Φ

∂t= 0, (6.39)

the function λ must satisfy the homogeneous wave equation,(∇2 − 1

c2∂2

∂t2

)λL = 0, (6.40)

6

and in the Coulomb gauge choice ∇ ·A = 0, λ must satisfy Laplace equation

∇2λC = 0. (6.41)

In classical electrodynamics, the scalar function λ can be chosen to be zero because a reasonable

boundary condition λL,C → 0 at r →∞ for both equations(∇2 − 1

c2∂2

∂t2

)λL = 0, ∇2λC = 0,

allows only λL,C = 0.

In macroscopic electromagnetic analysis, Lorenz gauge defined by

∇ ·A+1

c2∂Φ

∂t= 0,

is more convenient because all potentials and electromagnetic fields are explicitly retarded. In

contrast, the scalar potential in the Coulomb gauge is not retarded and consequent Coulomb electric

field is also nonretarded. Such instantaneous fields are clearly unphysical, and should be cancelled

somehow by a counterpart. In this section, it is shown that the electromagnetic fields formulated in

Lorenz gauge, which are all retarded appropriately, are indeed consistent with those in the Coulomb

gauge. The instantaneous Coulomb electric field emerging in the Coulomb gauge is cancelled.

In Coulomb gauge, the scalar potential satisfies time dependent Poisson equation,

∇2Φ(r, t) = −ρ(r, t)

ε0. (6.42)

Its solution is nonretarded,

Φ(r, t) =1

4πε0

∫ρ(r′, t)

|r− r′|dV′, (6.43)

and resultant Coulomb electric field is also nonretarded,

E(r, t) = −∇Φ(r, t)

=1

4πε0

∫(r− r′)ρ(r′, t)

|r− r′|3dV ′. (6.44)

As we have seen, the vector potential in Coulomb gauge (∇ ·A = 0) is transverse and obeys the

wave equation, (∇2 − 1

c2∂2

∂t2

)At = −µ0Jt = −µ0(J− Jl), (6.45)

where Jt is the transverse component of the current density. Note that the longitudinal current Jland the charge density ρ are related through the charge conservation law,

∂ρ

∂t+∇ · Jl = 0. (6.46)

7

Solution for the transverse vector potential is retarded,

At(r, t) =µ04π

∫1

|r− r′|Jt(r′, t− |r− r

′|c

)dV ′. (6.47)

The electric field in the Coulomb gauge is thus give by

EC(r, t) = −∇Φ− ∂At

∂t

=1

4πε0

∫(r− r′)ρ(r′, t)

|r− r′|3dV ′ − µ0

4π

∂

∂t

∫1

|r− r′|Jt(r′, t− |r− r

′|c

)dV ′. (6.48)

The instantaneous Coulomb field (the first term in the RHS) must be somehow cancelled by a term

stemming from the time derivative of the transverse vector potential for the electric field to be

consistent with that from Lorenz gauge which guarantees that all fields are retarded.

It is convenient to work with spatial Fourier transforms of the potentials and fields. Since the

inverse Laplace transform of the Fourier-Laplace Green’s function

g(k, ω) =1

k2 − (ω/c)2, (6.49)

is

− c2

2π

∫ ∞−∞

e−iτω

ω2 − (ck)2dω = c2

sin(ckτ)

ck, (6.50)

the particular solution for an inhomogeneous wave equation(1

c2d2

dt2+ k2

)A(k, t) = f(k, t),

can be given in the form of convolution,

A(k, t) =c

k

∫ ∞0

sin(ckτ)f(k, t− τ)dτ. (6.51)

Applying this to the transverse vector potential, we find

At(k, t) = µ0c

k

∫ ∞0

sin(ckτ) [J(k, t− τ)− Jl(k, t− τ)] dτ. (6.52)

Since in the Coulomb gauge, the scalar potential is given by

Φ(k, t) =ρ(k, t)

ε0k2, (6.53)

8

the electric field becomes

EC(k, t) = − ik

ε0k2ρ(k, t)− ∂

∂tAt(k, t)

= − ik

ε0k2ρ(k, t)− µ0

c

k

∂

∂t

∫ ∞0

sin(ckτ) [J(k, t− τ)− Jl(k, t− τ)] dτ. (6.54)

Noting

∂

∂t

∫ ∞0

sin(ckτ)Jl(k, t− τ)dτ

= −∫ ∞0

sin(ckτ)∂

∂τJl(k, t− τ)dτ

= ck

∫ ∞0

cos(ckτ)Jl(k, t− τ)dτ, (6.55)

and

Jl(k, t) =ik

k2∂

∂tρ(k, t), (6.56)

we find

EC(k, t) = − ik

ε0k2ρ(k, t)− µ0

c

k

∂

∂t

∫ ∞0

sin(ckτ)J(k, t− τ)dτ

+ik

ε0k2ρ(k, t)− ik

ε0

c

k

∫ ∞0

sin(ckτ)ρ(k, t− τ)dτ

= − ikε0

c

k

∫ ∞0

sin(ckτ)ρ(k, t− τ)dτ − µ0c

k

∂

∂t

∫ ∞0

sin(ckτ)J(k, t− τ)dτ. (6.57)

Note that the instantaneous fields (the first and third terms in the RHS) exactly cancel each other

and the electric field is appropriately retarded and consistent with the result in the Lorenz gauge,

EL(r, t) = − 1

4πε0∇∫ρ(r′, τ)

|r− r′|dV′ − µ0

4π

∂

∂t

∫J(r′, τ)

|r− r′| dV′, (6.58)

where

τ = t− |r− r′|

c.

Fourier transform of Eq. (6.58) indeed recovers

EL(k, t) = − ikε0

c

k

∫ ∞0

sin(ckτ)ρ(k, t− τ)dτ − µ0c

k

∂

∂t

∫ ∞0

sin(ckτ)J(k, t− τ)dτ, (6.59)

which is identical to the field formulated in the Coulomb gauge.

The magnetic field is generated by the transverse vector potential,

B = ∇×A = ∇×At, (6.60)

(note that ∇×Al = 0 by definition) and is thus explicitly retarded irrespective of the choice of the

9

gauge.

6.4 Elementary Spherical Waves

For a current source oscillating at a frequency ω, the vector potential can be calculated from Eq.

(6.33),

A(r,t) =µ04πe−iωt

∫eik|r−r

′|

|r− r′| J(r′)dV ′.

Here, we wish to expand the spatial Green’s function,

G(r− r′) =eik|r−r

′|

4π |r− r′| , (6.61)

in terms of spherical harmonics as done in static cases,

Static : G(r− r′) =1

4π |r− r′|

=∑l,m

1

2l + 1

(r′)l

rl+1Ylm(θ, φ)Y ∗lm(θ′, φ′), r > r′. (6.62)

Evidently, the objective of the harmonic expansion is to identify electric and magnetic multipole

components in the radiation electromagnetic fields. The Green’s function

G(r− r′) =eik|r−r

′|

4π |r− r′| ,

satisfies the singular Helmholtz equation,

(∇2 + k2

)G = −δ(r− r′). (6.63)

We wish to construct a series expansion of the Green’s function in terms of solutions to the homo-

geneous Helmholtz equation, (∇2 + k2

)f(r) = 0. (6.64)

Let the function f(r) be separated in the form

f(r) = R(r)F (θ, φ).

Substitution into Eq. (6.64) yields(d2

dr2+

2

r

d

dr+ k2 − l(l + 1)

r2

)R(r) = 0, (6.65)

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2 θ

∂2

∂φ2+ l(l + 1)

]F (θ, φ) = 0, (6.66)

10

where l(l + 1) is a separation constant. The angular function F (θ, φ) is unchanged from the static

case,

F (θ, φ) = Ylm(θ, φ). (6.67)

To find solutions for the radial function R(r), let us assume

R(r) =u(r)√r. (6.68)

Substitution into Eq. (6.65) yields the following equation for u(r),

d2

dr2+

1

r

d

dr+ k2 −

(l + 12)2

r2u(r) = 0, (6.69)

whose solutions are Bessel functions of half-integer order,

u(r) = Jl+12(kr), N

l+12(kr). (6.70)

It is customary to introduce spherical Bessel functions defined by

jl(kr) =

√π

2

Jl+12(kr)

√kr

, nl(kr) =

√π

2

Nl+12(kr)

√kr

, (6.71)

where the numerical factor√π/2 is to let both functions approach the form of spherical wave,

|jl(kr)| , |nl(kr)| →1

krfor kr 1. (6.72)

Noting that the asymptotic forms of the ordinary Bessel functions are

Jn(x)→√

2

πxcos

(x− 2n+ 1

4π

), (6.73)

Nn(x)→√

2

πxsin

(x− 2n+ 1

4π

), (6.74)

we observe that jl(x) and nl(x) asymptotically approach

jl(x)→ 1

xcos

(x− l + 1

2π

), x 1 (6.75)

nl(x)→ 1

xsin

(x− l + 1

2π

), x 1. (6.76)

In radiation analyses, it is convenient to introduce spherical Hankel functions defined by

First kind: h(1)l (x) = jl(x) + inl(x), (6.77)

11

Second kind: h(2)l (x) = jl(x)− inl(x). (6.78)

The asymptotic forms of these functions are:

h(1)l (x)→ (−i)l+1 e

ix

x, x 1, (6.79)

h(2)l (x)→ il+1

e−ix

x, x 1, (6.80)

which describe outgoing and incoming spherical waves, respectively.

Having found elementary spherical harmonic solutions for the homogeneous Helmholtz equation,

we are now ready to construct the Green’s function in terms of those elementary solutions. Since

any linear combinations of the four radial functions jl(kr), nl(kr), h(1)l (kr) and h(2)l (kr) satisfy the

Helmholtz equation, and the Green’s function should be bounded everywhere, we may assume

G(r, r′) =∞∑l=0

l∑m=−l

Almjl(kr′)h

(1)l (kr)Ylm(θ, φ)Y ∗lm(θ′, φ′), for r > r′, (6.81)

and

G(r, r′) =

∞∑l=0

l∑m=−l

Almjl(kr)h(1)l (kr′)Ylm(θ, φ)Y ∗lm(θ′, φ′), for r < r′, (6.82)

where Alm is an expansion coeffi cient of the (l,m) harmonic and continuity on the surface r = r′

is imposed. Note that for a small argument, only jl(x) remains bounded,

jl(x) ' xl

(2l + 1)!!, x 1. (6.83)

The spherical Bessel function of the second kind nl(x) diverges at small x as

nl(x) ' −(2l − 1)!!

xl+1, x 1. (6.84)

The radially converging (incoming) solution h(2)l (kr) is discarded because we are interested in

radiation of electromagnetic field from localized sources.

The coeffi cient Alm can be determined through the familiar procedure exploiting the disconti-

nuity in the radial derivative. Substituting Eqs. (6.81) and (6.82) into the original equation

(∇2 + k2)G = −δ(r− r′) = −δ(r − r′)

rr′ sin θδ(θ − θ′)δ(φ− φ′),

multiplying both sides by Y ∗l′m′(θ, φ), and integrating the result over the entire solid angle by noting

the orthogonality of the harmonic functions,∫Ylm(θ, φ)Y ∗l′m′(θ, φ)dΩ = δll′δmm′ , (6.85)

12

we obtain

Alm

(d2

dr2+

2

r

d

dr+ k2 − l(l + 1)

r2

)gl(r, r

′) = −δ(r − r′)

rr′, (6.86)

where

gl(r, r′) =

jl(kr

′)h(1)l (kr), r > r′,

jl(kr)h(1)l (kr′), r < r′

(6.87)

The derivative of the radial function gl(r, r′) is discontinuous at r = r′ and thus second order

derivative yields the singularity compatible with the RHS,

d2

dr2gl(r, r

′) = ik1

(kr)2δ(r − r′), (6.88)

where use is made of the Wronskian of the spherical Bessel functions,

jl(x)[h(1)l (x)]′ − j′l(x)h

(1)l (x) = i

[jl(x)n′l(x)− j′l(x)nl(x)

]=

i

x2. (6.89)

We thus find a rather simple result,

Alm = ik, (6.90)

and the desired spherical harmonic expansion of the Green’s function is

eik|r−r′|

4π |r− r′| =

ik∞∑l=0

l∑m=−l

jl(kr)h(1)l (kr′)Ylm(θ, φ)Y ∗lm(θ′, φ′), r < r′,

ik∞∑l=0

l∑m=−l

jl (kr′)h

(1)l (kr)Ylm(θ, φ)Y ∗lm(θ′, φ′), r > r′.

(6.91)

In the limit of static fields k → 0 ( that is, ω → 0), we indeed recover

1

4π |r− r′| =

∞∑l=0

l∑m=−l

1

2l + 1

(r′)l

rl+1Ylm(θ, φ)Y ∗lm(θ′, φ′), r > r′

∞∑l=0

l∑m=−l

1

2l + 1

rl

(r′)l+1Ylm(θ, φ)Y ∗lm(θ′, φ′), r < r′.

(6.92)

The expansion in Eq. (6.91) allows us to write the vector potential in the form

A(r) =µ04π

∫eik|r−r

′|

|r− r′| J(r′)dV ′

= ikµ0

∞∑l=0

l∑m=−l

h(1)l (kr)Ylm(θ, φ)

∫J(r′)jl(kr

′)Y ∗lm(θ′, φ′)dV ′. (6.93)

13

For cartesian components Ai(r), (i = x, y, z), Eq. (6.93) gives

Ai(r) = ikµ0

∞∑l=0

l∑m=−l

h(kr)Ylm(θ, φ)

∫Ji(r

′)jl(kr′)Y ∗lm(θ′, φ′)dV ′.

However, for non-cartesian components (e.g., components in the spherical coordinates), it is nec-

essary to single out the component of the current density J(r) in the same direction as the vector

potential A at the observing position

A(r) = ikµ0

∞∑l=0

l∑m=−l

h(1)l (kr)Ylm(θ, φ)eA

∫eA · J(r′)jl(kr

′)Y ∗lm(θ′, φ′)dV ′, (6.94)

where eA is the unit vector in the direction of A, eA = A/A. The problem with this representation

is that the direction of the vector potential eA is not known a priori. It is more convenient if we can

find basic electromagnetic field vectors which are normal to each other everywhere. The Transverse

Electric (TE) and Transverse Magnetic (TM) eigenvectors will serve exactly for this purpose and

form a convenient dyadic for radiation electromagnetic fields.

6.5 TE (Transverse Electric) and TM (Transverse Magnetic) BaseVectors

In electromagnetic wave problems, boundary conditions are usually imposed on the electric and

magnetic fields E and B, rather than the potentials Φ and A. (Note that the electromagnetic fields

are gauge invariant, while the potentials are not.) Also, in experiments, what is usually measured

is the electric field. For these reasons, it is desirable to formulate spherical harmonic expansion of

the electric field.

In Chapter 3 on magnetostatics, we saw that ∇ ·A = 0 allows us to write the vector potential

in the form

A = ∇× F,

where F is a vector field. In the Lorenz gauge, the divergence of A does not vanish,

∇ ·A = − 1

c2∂Φ

∂t,

which suggests

A = ∇f +∇× F,

where f is a scalar field. However, the field f does not contribute to the magnetic field B = ∇×Asince ∇×∇f ≡ 0 and we may continue to assume A = ∇× F even for time varying fields. (Thescalar field f does affect the formulation of the electric field.) The vector field F may further be

14

decomposed into radial and transverse components as follows,

F = rφ+ r×∇ψ, (6.95)

where φ and ψ are scalar functions. If the vector potential satisfies the Helmholtz equation, so do

φ and ψ,

(∇2 + k2)φ = 0, (∇2 + k2)ψ = 0, (6.96)

and thus can be expanded in spherical harmonics. (Note that Laplacian ∇2 and operator r × ∇commute.) The magnetic field in terms of the scalar functions is given by

B = ∇×A = ∇× (r×∇φ)− r×∇∇2ψ= ∇× (r×∇φ) + k2r×∇ψ, (6.97)

where use is made of

∇ · (r×∇ψ) = 0. (6.98)

The second term in the RHS is evidently transverse to r, that is, the scalar field ψ is a generating

function for transverse magnetic (TM) modes.

The scalar field φ generates transverse electric (TE) modes, since in the current free region, the

electric field is given by

− iωc2E = ∇×B

= −r×∇∇2φ−∇× (r×∇∇2ψ)

= k2 [r×∇φ+∇× (r×∇ψ)] . (6.99)

The TE and TMmodes are the desired base vectors because vector functions r×∇φ and∇×(r×∇ψ)

are normal to each other. To prove this, let us assume

φ(r) =∑

Almgl(r)Ylm(θ, φ), ψ(r) =∑

Blmgl(r)Ylm(θ, φ). (6.100)

Expanding

∇× [gl(r)r×∇Ylm] = ∇gl(r)× [r×∇Ylm(θ, φ)] + gl(r)∇× (r×∇Ylm)

= ∇gl(r)× [r×∇Ylm(θ, φ)]− gl(r)l(l + 1)

rYlmer − gl(r)∇Ylm

= −gl(r)l(l + 1)

rYlmer −

d

dr[rgl (r)]∇Ylm, (6.101)

we see that the two vectors r × ∇Ylm and ∇ × [gl (r) r×∇Ylm] are indeed normal to each other,

since (r×∇Ylm) · r = 0 and (r×∇Ylm) · ∇Ylm = 0. (The product

r (r×∇Ylm) · ∇Y ∗lm = −2imd

dθ|Ylm|2 ,

15

evidently does not vanish. Interpret this.)

The operator

r×∇Ylm(θ, φ) = iLYlm(θ, φ), L =− ir×∇, (6.102)

is the angular momentum operator and its explicit form is

r×∇Ylm(θ, φ) = −eθ1

sin θ

∂Ylm∂φ

+ eφ∂Ylm∂θ

. (6.103)

Noting∂eφ∂φ

= − sin θer − cos θeθ,

we can readily show that

(r×∇) · (r×∇Ylm) = (r×∇)2Ylm =1

sin θ

∂

∂θ

(sin θ

∂Ylm∂θ

)+

1

sin2 θ

∂2Ylm

∂φ2= −l(l+ 1)Ylm, (6.104)

or

L2Ylm = l(l + 1)Ylm. (6.105)

Other useful properties are:

∇ = er∂

∂r− i

r2r× L, (6.106)∫

(r×∇Ylm) · (r×∇Y ∗l′m′) dΩ = l(l + 1)δll′δmm′ , (6.107)

∇ · (r×∇Ylm) = 0, (6.108)

L2LYlm = LL2Ylm, ∇2LYlm = L∇2Ylm, L× LYlm = iLYlm, (6.109)

Lx = −i(y∂

∂z− z ∂

∂y

), Ly = −i

(z∂

∂x− x ∂

∂z

), Lz = −i

(x∂

∂y− y ∂

∂x

)= −i ∂

∂φ, (6.110)

L± ≡ Lx ± iLy = e±iφ(± ∂

∂θ+ i cot θ

∂

∂φ

), (6.111)

L+Ylm (θ, φ) =√l (l + 1)−m (m+ 1)Yl,m+1 (θ, φ) , (6.112)

L−Ylm (θ, φ) =√l (l + 1)−m (m− 1)Yl,m−1 (θ, φ) , (6.113)

LzYlm (θ, φ) = mYlm (θ, φ) . (6.114)

We are now ready to expand the raw form of the vector potential

A(r) =µ04π

∫J(r′)eik|r−r

′|

|r− r′| dV ′

= µ0ik∞∑l=0

l∑m=−l

h(1)l (kr)Ylm(θ, φ)

∫jl(kr

′)Y ∗lm(θ′, φ′)J(r′)dV ′, (6.115)

16

in terms of the TE and TM base vectors. For TE modes, the vector potential should be in the form

ATE = r×∇φ. (6.116)

Therefore, it is necessary to identify TE component of the source current J(r′), which can be

effected by

JTElm (r) = jl(kr)(r×∇Y ∗lm) · J(r). (6.117)

The TE component of the vector potential can thus be assumed in the form

ATE(r) = µ0ik

∞∑l=1

l∑m=−l

almh(1)l (kr)r×∇Ylm(θ, φ)

∫jl(kr

′)[r′ ×∇′Y ∗lm(θ′, φ′)] · J(r′)dV ′, (6.118)

where alm is expansion coeffi cient. Note that in the summation over l, the monopole term l = 0

vanishes because ∇Y00 = 0. It is convenient to choose alm so that

alm

∫(r×∇Ylm) · (r×∇Y ∗lm)dΩ = 1, (6.119)

or

alm =1

l(l + 1). (6.120)

Corresponding TE base vector is

uTElm (r) =1√

l(l + 1)r×∇Ylm(θ, φ), (6.121)

which is normalized as ∮uTElm (r) ·

[uTElm (r)

]∗dΩ = 1. (6.122)

The TE component of the vector potential is therefore given by

ATE(r) = µ0ik∞∑l=1

l∑m=−l

1

l(l + 1)h(1)l (kr)r×∇Ylm(θ, φ)

∫jl(kr

′)[r′ ×∇′Y ∗lm(θ′, φ′)] · J(r′)dV ′

= −µ0ik∞∑l=1

l∑m=−l

1

l(l + 1)h(1)l (kr)r×∇Ylm(θ, φ)

∫jl(kr

′)(r′ × J(r′)

)· ∇′Y ∗lm(θ′, φ′)dV ′,(6.123)

The current density J(r′) may be separated into the conduction current Jc and magnetization

current Jm = ∇×M (r),

J(r) = Jc (r) +∇×M (r) , (6.124)

where M (r) is the magnetic dipole moment density.

For the TM modes, it is convenient to let the TM base vector have the same dimensions as TE

17

base vector. For this purpose, we rewrite the decomposition of the vector potential in the form

A = r×∇φ+1

k∇× (r×∇ψ). (6.125)

Following the same procedure developed for the TE modes, we find

ATM(r) = iµ01

k

∞∑l=1

l∑m=−l

1

l(l + 1)∇×[h

(1)l (kr)r×∇Ylm(θ, φ)]

∫∇′×[jl(kr

′)r′×∇′Y ∗lm(θ′, φ′)]·J(r′)dV ′.

(6.126)

Corresponding TM base vector in the radiation zone is

uTMlm (r) =1√

l(l + 1)er × [r×∇Ylm(θ, φ)], (6.127)

which is also normalized as ∮uTMlm (r) ·

[uTMlm (r)

]∗dΩ = 1. (6.128)

Note that in Eq. (6.126),

∇′ × [jl(kr′)r′ ×∇′Y ∗lm(θ′, φ′)] · J(r′) (6.129)

is the TM component of the current density.

For a small source kr′ 1, the spherical Bessel function jl(kr′) can be approximated by

jl(kr′) ' (kr′)l

(2l + 1)!!. (6.130)

Then, the moment for the TE modes becomes (the primes are omitted for brevity)∫[jl(kr)r×∇Ylm(θ, φ)] · Jc(r)dV

= − kl

(2l + 1)!!

∫rl(r× Jc) · ∇Y ∗lmdV

= − kl

(2l + 1)!!

∫∇(rlY ∗lm) · (r× Jc)dV. (6.131)

This contains the current circulation r× Jc which produces magnetic multipole moments. Themagnetic multipole moment is in general defined by

mlm =1

l + 1

∫∇(rlY ∗lm) · (r× Jc)dV. (6.132)

In terms of the magnetic moment mlm, the integral may be rewritten as∫[jl(kr)r×∇Ylm(θ, φ)] · Jc(r)dV = − (l + 1)kl

(2l + 1)!!mlm. (6.133)

18

The contribution from the magnetization current ∇×M can be calculated in a similar manner,

m′lm = −∫rlY ∗lm∇ ·MdV,

and the (l,m) component of the TE vector potential is given by

ATElm (r) = −µ0ikl+1

l(2l + 1)!!(mlm +m′lm)h

(1)l (kr)r×∇Ylm(θ, φ). (6.134)

The radiation power associated with the vector potential can readily be calculated as follows:

PTElm = r2∫Z∣∣HTE

lm

∣∣2 dΩ, (6.135)

where the magnetic field HTElm (r) in the radiation region kr 1 is approximately given by

HTElm (r) ' 1

µ0ik×ATElm

=kl+1(−i)l(mlm +m′lm)

l(2l + 1)!!

1

rei(kr−ωt)n× [r×∇Ylm(θ, φ)]. (6.136)

Then ∣∣HTElm

∣∣2 =k2(l+1)

[l(2l + 1)!!]2∣∣mlm +m′lm

∣∣2 |r×∇Ylm(θ, φ)|2 . (6.137)

Noting ∫|r×∇Ylm(θ, φ)|2 dΩ = l(l + 1), (6.138)

we find the radiation power in the form

PTElm =

√µ0ε0

l + 1

l

k2(l+1)

[(2l + 1)!!]2|mlm +m′lm|2

=1

4πε0

4π

c

l + 1

l

k2(l+1)

[(2l + 1)!!]2|mlm +m′lm|2 (6.139)

=1

4πε04πc

l + 1

l

k2(l+1)

[(2l + 1)!!]2|mG

lm +m′Glm|2, (6.140)

where the latter expression facilitates comparison with the formulation in the Gaussian unit system

in which the magnetic multipole moment is defined by

mGlm =

1

c

1

l + 1

∫∇(rlY ∗lm) · (r× J)dV, (6.141)

Electric multipoles radiate TM modes. This may be seen from the TM component of the current

19

density, ∫∇× [jl(kr)r×∇Y ∗lm(θ, φ)] · J(r)dV

=

∫jl(kr)[r×∇Y ∗lm(θ, φ)] · ∇ × JdV

= −∫∇× [jl(kr)rY

∗lm] · ∇ × JdV

= −∫jl(kr)Y

∗lmr · [∇× (∇× J)]dV

= −∫jl(kr)Y

∗lmr ·

[∇(∇ · J)−∇2J

]dV

=

∫Y ∗lm

d

dr[rjl(kr)]∇ · JdV − k2

∫jl(kr)Y

∗lmr · JdV, (6.142)

where it is noted that the function jl(kr)Y ∗lm(θ, φ) satisfies the Helmholtz equation,

(∇2 + k2

)jl(kr)Y

∗lm(θ, φ) = 0.

Again, if the current is separated into the condution current and magnetization currents, J =

Jc +∇×M, Eq. (6.142) can be rewritten as∫Y ∗lm

d

dr[rjl(kr)]∇ · JdV − k2

∫jl(kr)Y

∗lmr · JdV

=

∫Y ∗lm

d

dr[rjl(kr)]∇ · JcdV − k2

∫jl(kr)Y

∗lmr · JcdV + k2

∫jl(kr)Y

∗lm∇· (r×M) dV.

(6.143)

For a small source kr′ 1, the first term in the RHS containing electric multipole moment is

dominant. Recalling∂ρ

∂t+∇ · Jc = 0,

and

jl(kr) '(kr)l

(2l + 1)!!for kr 1,

we may approximate Eq. (6.143) by∫Y ∗lm

d

dr[rjl(kr)]∇ · JcdV − k2

∫jl(kr)Y

∗lmr · JcdV + k2

∫jl(kr)Y

∗lm∇· (r×M) dV

' ic(l + 1)kl+1

(2l + 1)!!

(qlm + q′lm

), (6.144)

where the second term is of order ka 1 (a being the source size) and thus ignorable, qlm is the

20

electric multipole moment defined earlier,

qlm =

∫rlY ∗lm(θ, φ)ρ(r)dV, (6.145)

and q′lm is an effective electric quadrupole moment created by the magnetic dipole moment density,

q′lm = − ik

(l + 1) c

∫rlY ∗lm(θ, φ)∇· (r×M) dV. (6.146)

The radiation vector potential of TM mode is thus given by

ATMlm (r) = −µ0ckl

l(2l + 1)!!qlm∇×

[h(1)l (kr)r×∇Ylm(θ, φ)

], (6.147)

and corresponding radiation power is

PTMlm = µ0c3 k2(l+1)

[(2l + 1)!!]2l + 1

l|qlm + q′lm|2

=1

4πε04πc

k2(l+1)

[(2l + 1)!!]2l + 1

l

∣∣qlm + q′lm∣∣2 . (6.148)

6.6 Radiation of Angular Momentum

In chapter 5, radiation of angular momentum from a circulating charge was discussed. In general,

if a charge undergoes circular motion, it radiates (loses) angular momentum as well as energy, for

the two quantities are intimately related. In this section, a general formulation for radiation of

angular momentum is given. Since the momentum flux density associated with electromagnetic

fields is given by1

cE×H∗, (6.149)

The angular momentum flux density associated with electromagnetic fields is

1

cr× (E×H∗). (6.150)

In the radiation zone, this must be proportional to 1/r2 if radiation of angular momentum accom-

panies radiation of energy. Therefore, terms proportional to 1/r3 in the Poynting flux, which have

been ignored in the calculation of radiation power, should be retained.

As a concrete example, let us consider TMmodes due to electric multipoles. The vector potential

of TM mode is

ATMlm (r) = −µ0ckl

l(2l + 1)!!qlm∇×

[h(1)l (kr)r×∇Ylm(θ, φ)

]. (6.151)

21

Corresponding magnetic and electric fields are

HTMlm (r) =

1

µ0∇×ATMlm

= −c kl+2

l(2l + 1)!!qlmh

(1)l (kr)r×∇Ylm(θ, φ), (6.152)

and

ETMlm (r) = − i

ωε0∇×HTM

lm

= − ic

ωε0

kl+2

l(2l + 1)!!qlm∇× [h

(1)l (kr)r×∇Ylm(θ, φ)], (6.153)

where use is made of the identity

∇ ·[h(1)l (kr)r×∇Ylm(θ, φ)

]= 0.

The expressions for the electromagnetic fields are exact. The radiation flux of angular momentum

associated with an (l,m) mode is

1

cr× (Elm×H

∗lm)TM

=ic

ωε0

k2(l+2)

[l(2l + 1)!!]2|qlm|2r×

[[∇× (h

(1)l (kr)r×∇Ylm(θ, φ))]× [h

(1)∗l (kr)r×∇Y ∗lm(θ, φ)]

]= − ic

ωε0

k2(l+2)

[l(2l + 1)!!]2|qlm|2

(r · ∇ × [h

(1)l (kr)r×∇Ylm(θ, φ)]

)h(1)∗l (kr)r×∇Y ∗lm(θ, φ)

=ic

ωε0

k2(l+2)

[l(2l + 1)!!]2|qlm|2l(l + 1)

∣∣∣h(1)l (kr)∣∣∣2 Ylm(θ, φ)r×∇Y ∗lm(θ, φ). (6.154)

In radiation zone kr 1, this reduces to

ic

ωε0

k2(l+1)

[(2l + 1)!!]2l + 1

l|qlm|2Ylm(θ, φ)r×∇Y ∗lm(θ, φ)

1

r2, (6.155)

and the integration over the solid angle yields the rate of angular momentum radiation,

−dLlmdt

=1

4πε04πc

k2(l+1)

[(2l + 1)!!]2l + 1

l|qlm|2

m

ωez, (6.156)

where use is made of ∫Ylm(θ, φ)r×∇Y ∗lm(θ, φ)dΩ

= −im∫Ylm(θ, φ)Y ∗lm(θ, φ)dΩez

= −imez, (6.157)

22

Comparing with the radiation power given in Eq. (6.148), we see that they are related through a

simple ratio,

− LzP

=m

ω. (6.158)

Evidently, this is not consistent with the familiar quantum mechanical result,

− LzP

=~√l(l + 1)

~ω=

√l(l + 1)

ω. (6.159)

The discrepancy is not surprising because the quantum mechanical formula is for a single pho-

ton while the result obtained in classical electrodynamics pertains to (infinitely) many photons.

Quantum mechanical formula for N photons is

− LzP

=

√N2m2 +Nl(l + 1)−m2

Nω, (6.160)

and the classical result can be recovered in the limit of N 1.

6.7 Some Examples

Example 1 Spherical Dipole Antenna

Figure 6-1: Spherical antenna.

Two ideally conducting hemispheres of radius a insulated from each other at the equator with

a small gap are connected to an oscillating voltage source V0e−iωt. In this case, the surface current

flows in the θ direction and creates an azimuthal magnetic fields Bφ. The resultant radiation fields

23

are thus TM (Transverse Magnetic) and we may assume the magnetic field in the form

B = r×∇ψ, (6.161)

where ψ is a scalar function. Since the system is axially symmetric without dependence on the

azimuthal angle φ, the scalar function ψ can be expanded in spherical harmonics as

ψ(r, θ) =∑l≥1

alh(1)l (kr)Pl(cos θ), r > a, (6.162)

which yields the magnetic field

B = r×∇ψ= −

∑l

alh(1)l (kr)P 1l (cos θ)eφ, r > a. (6.163)

Note thatd

dθPl(cos θ) = − sin θ

d

d(cos θ)Pl(cos θ) = −P 1l (cos θ). (6.164)

The electric field associated with the magnetic field can be found from the Maxwell’s equation,

ε0µ0∂E

∂t= ∇×B, (6.165)

which yields

E(r) =ic2

ω

[1

r sin θ

∂

∂θ(sin θBφ)er −

1

r

∂

∂r(rBφ)eθ

]. (6.166)

Substituting Bφ(r) in Eq.(6.163) yields the transverse component of the electric field,

Eθ(r, θ) =ic2

ω

∑l

al

[l

rh(1)l (kr)− kh(1)l−1(kr)

]P 1l (cos θ), (6.167)

where the following recurrence formula for the spherical Bessel functions has been used,

d

dxh(1)l (x) =

l

xh(1)l (x)− h(1)l+1(x)

= h(1)l−1(x)− l + 1

xh(1)l (x). (6.168)

Since the sphere is ideally conducting, the θ component of the electric field should vanish except

at the gap. Therefore, at the sphere surface r = a, the electric field may be approximated by

Eθ(r = a, θ) =V0aδ(θ − π

2

), (6.169)

24

which satisfies the obvious requirement that

a

∫ π

0Eθ(a, θ)dθ = V0. (6.170)

Then, for arbitrary angle θ, the following relation must hold

V0aδ(θ − π

2

)=ic2

ω

∑l

al

[l

rh(1)l (ka)− kh(1)l−1(ka)

]P 1l (cos θ). (6.171)

Multiplying both sides by P 1l (cos θ) sin θ and integrating over θ, we readily find the expansion

coeffi cient al,

al =ω

ic22l + 1

2l(l + 1)

P 1l (0)

lh(1)l (ka)− kah(1)l−1(ka)

V0, (6.172)

where

P 1l (0) =

(−1)

l−12

l!!

(l − 1)!!, l odd

0, l even

(6.173)

and use has been made of the integral∫ 1

−1P 1l (x)P 1l′ (x)dx =

2

2l + 1l(l + 1)δll′ . (6.174)

The disappearance of even harmonics is expected because of the up-down anti-symmetry of the

problem. (Note that P 1l (x) is an odd function of x if l is odd.)

The surface current density Js on the sphere surface can be found from the boundary condition

for the magnetic field,

Js = n×Hφ(r = a, θ)

= −Hφ(a, θ)eθ

= − 1

µ0

∑l≥1

alh(1)l (ka)P 1l (cos θ)eθ. (6.175)

The current at the gap θ = π/2 where the voltage V0 appears is

I = 2πaJs(θ = π/2)

= −2πa

µ0

∑l≥1

alh(1)l (ka)P 1l (0)

= i2πaω

µ0c2

∑l≥1

2l + 1

2l(l + 1)

h(1)l (ka)

lh(1)l (ka)− kah(1)l−1(ka)

[P 1l (0)

]2V0. (6.176)

25

This defines the radiation admittance of the spherical dipole antenna,

Y =I

V0

= iπ

√ε0µ0

∑l≥1

2l + 1

l(l + 1)

kah(1)l (ka)

lh(1)l (ka)− kah(1)l−1(ka)

[P 1l (0)

]2. (6.177)

A spherical antenna has limited practical applications. However, the procedure developed in

this example is applicable to more practical problems such as half wavelength dipole antennas. A

thin rod antenna can be analyzed rigorously using the prolate spheroidal coordinates.

Example 2 Circular Loop Antenna

Figure 6-2: Circular loop antenna. ka is arbitrary.

Let us consider a circular oscillating current I0e−iωt carried by a thin conductor ring of radius

a. The current density may be written as

J = I0e−iωt δ(r − a)

aδ(θ − π

2

)eφ. (6.178)

As shown in Chapter 5, if the ring radius is much smaller than the wavelength ka =ω

ca 1, the

problem reduces to radiation by a magnetic dipole with a dipole moment

mz(t) = πa2I0e−iωt.

For an arbitrary value of ωa/c, higher order multipole fields must be retained. Since the loop

current radiates TE modes, the TE vector potential in Eq. (6.118) can be directly applied with a

result

ATE(r) = µ0ik∑l

1

l(l + 1)h(1)l (kr)r×∇Yl0(θ, φ)Ml, (6.179)

26

where

Ml =

∫jl(kr

′)[r′ ×∇Y ∗l0(θ′)] · J(r′)dV ′, (6.180)

is the TE component of the current density. Note that because of the axial symmetry, only m = 0

components are nonvanishing. Since

r′ ×∇Y ∗l0(θ′) =dYl0dθ′

eφ′

= −√

2l + 1

4πP 1l (cos θ′)eφ′ ,

we find

Ml = −2πaI0

√2l + 1

4πjl(ka)P 1l (0). (6.181)

The vector potential is therefore given by

ATE(r) = −iµ0I0ka∑l

2l + 1

2l(l + 1)jl(ka)h

(1)l (kr)P 1l (0)P 1l (cos θ)eφ, (6.182)

and the magnetic field in the radiation zone kr 1 is

H(r) ' 1

µ0ik×ATE

= −akI0eikr

r

∑l≥1

(−i)l+1 2l + 1

2l(l + 1)jl(ka)P 1l (0)P 1l (cos θ)eθ. (6.183)

The radiation power associated with the l-th harmonic mode is

P l = r2Z0

∫|Hl|2 dΩ

= Z0(akI0)2

(2l + 1

2l(l + 1)

)2 [jl(ka)P 1l (0)

]22π

∫ π

0[P 1l (cos θ)]2 sin θdθ

= πZ0(akI0)2[jl(ka)P 1l (0)

]2 2l + 1

l(l + 1), l = 1, 3, 5, · · · (6.184)

In the long wavelength limit ka 1, the dipole term (l = 1) is dominant, and we recover

P = πZ0(akI0)2 (ka)2

6

=1

4πε0

2ω4m2

3c5, (6.185)

where m = πa2I0 is the magnetic dipole moment of the ring current.

27

6.8 Spherical Harmonic Expansion of a Plane Wave

A plane wave of planar polarization is characterized by constant amplitudes of electromagnetic

fields and unidirectional propagation assumed here in the z-direction,

E0eikzex, H0e

ikzey. (6.186)

The purpose of this section is to decompose the plane wave into spherical harmonics. Once achieved,

such a representation will be very useful in analyzing scattering of electromagnetic wave by an object

placed in the plane wave. Scattered waves are evidently no longer plane waves but they consist

of many (often infinitely many) spherical harmonic waves. A key observation to be made is that

there is one-to-one correspondence between spherical harmonic component in the incident plane

wave and the one in the scattered wave. This is because a scattered harmonic mode characterized

by mode numbers (l,m) can only be produced by a mode in the incident wave having exactly the

same angular dependence. For example, TM (l,m) harmonic component in the scattered wave is

generated by the same TM harmonic component contained in the incident wave.

Figure 6-3: Scattered electromagnetic waves consist of spherical waves. The incident plane wavecan be decomposed into spherical waves to facilitate analysis.

The first step is to expand the propagator function eikz = eikr cos θ in terms of spherical har-

monics. This can be effected by considering the limiting case of the scalar Green’s function,

G(r, r′) =eik|r−r

′|

4π |r− r′| = ik∞∑l=0

l∑m=l

jl(kr)h(1)l (kr′)Y ∗lm(θ, φ)Ylm(θ′, φ′), r′ > r. (6.187)

In the limit of r′ →∞, that is, when the source is far away, h(1)l (kr′) approaches

h(1)l (kr′)→ (−i)l+1 e

ikr′

kr′. (6.188)

28

Noting also

k∣∣r′−r∣∣→ kr′ − k · r, r′ r,

we find

e−ik·r = 4πi∞∑l=0

l∑m=−l

(−i)l+1jl(kr)Y ∗lm(θ, φ)Ylm(θ′, φ′), (6.189)

where the angles (θ′, φ′) are those of the wavevector k = (k, θ′, φ′). Since we have assumed a plane

wave propagating in the z-direction, it follows that θ′ = 0 and φ′ becomes irrelevant. Noting

Pml (1) = δm0,

and taking the complex conjugate of Eq. (6.189) yields the following identity,

eikz = eikr cos θ =

∞∑l=0

il(2l + 1)jl(kr)Pl(cos θ). (6.190)

The electric field assumed to be in the x-direction can be converted into components in the

spherical coordinates as

E0ex = E0∇(r sin θ cosφ)

= E0(sin θ cosφer + cos θ cosφeθ − sinφeφ). (6.191)

This field can be represented as a sum of TE and TM modes. To effect such a representation, we

assume the following expansion,

E0eikr cos θex =

∑lm

almjl(kr)r×∇Ylm(θ, φ) +1

k

∑lm

blm∇× [jl(kr)r×∇Ylm(θ, φ)] ,

or substituting the expansion in Eq. (6.190) for eikr cos θ in the LHS,

E0

∞∑l=0

il(2l + 1)jl(kr)Pl(cos θ)(sin θ cosφer + cos θ cosφeθ − sinφeφ)

=

∞∑l=1

l∑m=l

almjl(kr)r×∇Ylm(θ, φ) +1

k

∑lm

blm∇× [jl(kr)r×∇Ylm(θ, φ)] , (6.192)

where alm and blm are expansion coeffi cients for TE and TMmodes, respectively. The TE coeffi cient

alm can be determined by multiplying both sides by r×∇Y ∗lm(θ, φ) and integrating the result over

the solid angle. Exploiting the following orthogonality relationships,∫[r×∇Ylm(θ, φ)] · [r×∇Y ∗l′m′(θ, φ)]dΩ = l(l + 1)δll′δmm′ ,

[r×∇Y ∗lm(θ, φ)] · ∇ × [jl(kr)r×∇Ylm(θ, φ) = 0,

29

we find

alm = E0il(2l + 1)

l(l + 1)

∫Pl(cos θ)(cos θ cosφeθ − sinφeφ) · (r×∇Y ∗lm)dΩ. (6.193)

The presence of cosφ and sinφ functions in the integral makes only m = ±1 components nonvan-

ishing. For m = +1, we find

al,1 = E0il(2l + 1)

l(l + 1)

∫Pl(cos θ)(cos θ cosφeθ − sinφeφ) · (r×∇Y ∗l,1)dΩ, (6.194)

where

Y ∗l,1(θ, φ) = −Yl,−1(θ, φ)

= −

√2l + 1

4πl(l + 1)P 1l e

−iφ. (6.195)

Since ∫Pl(cos θ)(cos θ cosφeθ − sinφeφ) · (r×∇Y ∗l,1)dΩ

= −iπ

√2l + 1

4πl(l + 1)

∫ π

0Pl(cos θ)

(cos θ

sin θP 1l (cos θ) +

dP 1l (cos θ)

dθ

)sin θdθ, (6.196)

and ∫ π

0

dP 1l (cos θ)

dθPl(cos θ) sin θdθ

= −∫ π

0P 1l (cos θ)

d

dθ[Pl(cos θ) sin θ]dθ

= −∫ π

0P 1l (cos θ)Pl(cos θ) cos θdθ +

∫ π

0[P 1l (cos θ)]2 sin θdθ, (6.197)

the integral reduces to ∫ π

0Pl(cos θ)

(cos θ

sin θP 1l (cos θ) +

dP 1l (cos θ)

dθ

)sin θdθ

=

∫ π

0[P 1l (cos θ)]2 sin θdθ

=2

2l + 1l(l + 1). (6.198)

Then, finally,

al,1 = −il+1√π(2l + 1)

l(l + 1)E0. (6.199)

30

The coeffi cient al,−1 can be found in a similar manner,

al,−1 = al,1 = −il+1√π(2l + 1)

l(l + 1)E0. (6.200)

The appearance of m = ±1 components in the spherical harmonic expansion of a plane wave is

understandable, for a plane wave can be decomposed into two circularly polarized waves of opposite

helicity.

The coeffi cient blm of the TM mode can be found in a similar manner. It is more convenient to

work with the magnetic field,

B =1

iω∇×E

=1

iω

∞∑l=1

al,±1∇× [jl(kr)r×∇Yl,±1] +k

iω

∞∑l=1

bl,±1jl(kr)r×∇Yl,±1, (6.201)

where use is made of the identity,

∇×∇× [jl(kr)r×∇Yl,±1] = ∇∇ · [jl(kr)r×∇Yl,±1]−∇2[jl(kr)r×∇Yl,±1]= 0 + k2jl(kr)r×∇Yl,±1. (6.202)

The magnetic field associated with the plane wave is

B0eikr cos θey =

E0c

∑l

il(2l + 1)jl(kr)Pl(cos θ) (sin θ sinφer + cos θ sinφeθ + cosφeφ) .

Then,

E0c

∞∑l=1

il(2l + 1)jl(kr)Pl(cos θ) (sin θ sinφer + cos θ sinφeθ + cosφeφ)

=1

iω

∞∑l=1

al,±1∇× [jl(kr)r×∇Yl,±1] +k

iω

∑l,±1

bl,±1jl(kr)r×∇Yl,±1. (6.203)

Multiplying both sides by r×∇Y ∗l,±1 and integrating the result over the solid angle, we find

bl,±1 = E0il+1(2l + 1)

l(l + 1)

∫Pl(cos θ)

(−cos θ

sin θsinφ

∂Y ∗l,±1∂φ

+ cosφ∂Y ∗l,±1∂θ

)dΩ

= ∓il+1√π(2l + 1)

l(l + 1)E0

= ±al,±1. (6.204)

(Calculation steps are left for exercise.) Then the desired spherical harmonic expansion of a plane

31

wave is

E0eikzex = E0

∑l,±1

il

2i

√4π(2l + 1)

l(l + 1)

[jl(kr)r×∇Yl,±1(θ, φ)± 1

k∇× [jl(kr)r×∇Yl,±1(θ, φ)]

].

(6.205)

The accompanying magnetic field is expanded as

B0eikzey = −E0

c

∑l,±1

il

√π(2l + 1)

l(l + 1)

[1

k∇× [jl(kr)r×∇Yl,±1(θ, φ)]± jl(kr)r×∇Yl,±1(θ, φ)

].

(6.206)

These expansions of the plane wave greatly facilitate analysis of scattering of plane electromagnetic

waves by an object. An important observation to be made is that the electromagnetic waves

re-radiated (scattered) by an object also consist of TE and TM modes in the form

TE mode: h(1)l (kr)r×∇Yl,±1(θ, φ), (6.207)

TM mode:1

k∇× [h

(1)l (kr)r×∇Yl,±1(θ, φ)], (6.208)

since the boundary conditions for electric and magnetic fields at the object require that scattered

wave should have exactly the same angular dependence as the incident plane wave. In examples to

follow, we analyze scattering of a plane wave by ideally conducting sphere and by dielectric sphere.

6.9 Scattering by an Ideally Conducting Sphere

Let us consider a highly conducting sphere of radius a placed in a plane electromagnetic wave as

shown in Fig.6-4. In two extreme cases, ka 1 (low frequency limit) and ka 1 (high frequency

limit), the problem can readily be solved without detailed analysis based on the spherical harmonic

expansion. If ka 1, the dipole approximation can be used. As we have seen, the electric dipole

moment of a conducting sphere placed in a static electric field E0 is

p = 4πε0a3E0, (6.209)

and the magnetic dipole moment of a superconducting sphere placed in a static magnetic field H0

is

m = −2πa3H0. (6.210)

These expressions remain valid in oscillating fields as long as the oscillation frequency is suffi ciently

small so that ka = ωa/c 1. Each dipole radiates linearly independent modes. The radiation

powers of the modes are thus additive. The power radiated by the electric dipole is

PE =1

4πε0

2ω4p2

3c3= 4πε0

2ω4a6

3c3E20 , (6.211)

32

and that due to the magnetic dipole is

PM =1

4πε0

2ω4m2

3c5=

1

44πε0

2ω4a6

3c3E20 , (6.212)

where H0 = E0/Z, Z =√µ0/ε0 is substituted in the magnetic dipole moment. The total re-

radiated (scattered) power is

P = 4πε02ω4a6

3c3E20 ×

(1 +

1

4

), (6.213)

and corresponding scattering cross-section is given by

σ =P

Si=

P

cε0E20=

(8π

3+

2π

3

)a2(ak)4. (6.214)

The dependence σ ∝ k4, which is common to all dipole radiation by objects much smaller than the

wavelength of incident wave, is known as Rayleigh’s law.

The reader may wonder why the magnetic dipole radiates a power comparable with that by

electric dipole in this case. As we have seen, the radiation fields due to a magnetic dipole is of

higher order by a factor (ka)2 compared with those due to electric dipole. The reason is as follows.

The low order vector potential

A(r) ' µ04πr

eikr∫J(r′)(1− ik · r′)dV ′, (6.215)

must be applied separately for electric and magnetic dipoles because the surface currents on the

conducting sphere induced by the incident electric and magnetic fields are entirely different. The

current due to the electric field flows between the poles while the current induced by the magnetic

field is azimuthal. The surface current in the polar direction is of order

Jsθ ' aωε0E0 = akH0, (6.216)

while the azimuthal current is of order

Jsφ ' H0. (6.217)

The polar current is smaller than the azimuthal current by a factor ak. (This is not surprising

because current flows in the azimuthal direction without any hindrance while in the polar direction,

current flow results in charge accumulation at the poles.) Therefore, the radiation fields from the

electric and magnetic dipoles are comparable.

In the opposite limit of ka 1 (high frequency or short wavelength limit), geometric and

physical optics approximation can be applied. The problem reduces to reflection by the illuminated

surface and diffraction by the other half surface in the shadow. Each surface contributes equally

to the scattering cross-section,

σ = 2πa2, (6.218)

33

although the angular distribution of scattered Poynting fluxes are entirely different. We will revisit

this problem in chapter 7.

Figure 6-4: Geometry assumed in analysis of scattering by a conducting sphere.

For arbitrary value of ka, the spherical harmonic expansion of the plane wave can be exploited as

follows. The scattered wave can be decomposed into TE and TM modes having the same angular

dependence as those contained in the incident plane wave. We therefore assume the following

expansion for the electric and magnetic fields of the scattered wave,

Esc(r) = −E0∑l,±1

il+1

√π(2l + 1)

l(l + 1)(AlMl,±1 ±BlNl,±1) , r > a (6.219)

Bsc(r) = −B0∑l,±1

il

√π(2l + 1)

l(l + 1)(AlNl,±1 ±BlMl,±1) , r > a

where Ml,±1 and Nl,±1 are the TE and TM base vectors,

Ml,±1(r) = h(1)l (kr)r×∇Yl,±1, (6.220)

Nl,±1(r) =1

k∇× [h

(1)l (kr)r×∇Yl,±1], (6.221)

and Al and Bl are expansion coeffi cients to be determined. If the sphere is ideally conducting, the

boundary conditions at the sphere surface are that the tangential component of the electric field

and normal component of the magnetic field both vanish. Since the total field at the surface is the

sum of incident and scattered waves, the explicit forms of the boundary conditions are:

(Einc+Esc)t = 0, at r = a, (6.222)

and

(Binc+Bsc)n = 0, at r = a. (6.223)

34

In fact, the coeffi cients Al and Bl can be determined from (Einc+Esc)t = 0 alone,

Al = − jl(ka)

h(1)l (ka)

, (6.224)

Bl = −

d

dr[rjl(kr)]

∣∣∣∣r=a

d

dr[rh

(1)l (kr)]

∣∣∣∣r=a

= − [kajl(ka)]′

[kah(1)l (ka)]′

, (6.225)

where

[xjl(x)]′ =d

dx[xjl(x)], [xh

(1)l (x)]′ =

d

dx[xh

(1)l (x)]. (6.226)

Note that for gl(kr) = jl(kr) or h(1)l (kr),

∇× [gl(kr)r×∇Yl,±1]

= −gl(kr)l(l + 1)

rYl,±1er −

d

dr[rgl(kr)]∇Yl,±1, (6.227)

and the coeffi cient Bl has been determined from the tangential component of the electric field of

the TM mode. The boundary condition for the magnetic field is automatically satisfied. (Verify

this statement.)

x 1086420

1

0.8

0.6

0.4

0.2

0

Figure 6-5: Normalized scattering cross-section of an ideally conducting sphere. σ/(2πa2) as afunction of ka.

The radiation power of the l-th harmonic mode can be readily found as

Pl = 2cε0E20

π(2l + 1)

l(l + 1)r2(|Al|2

∫|Ml1|2 dΩ + |Bl|2

∫|Nl1|2 dΩ

)= cε0E

20

2π(2l + 1)

k2

(|Al|2 + |Bl|2

), (6.228)

35

where the factor 2 accounts for an equal contribution from the (l,m = ±1) modes. Then the

scattering cross-section is given by

σ(ka) =∑l

Plcε0E20

=2π

k2

∞∑l=1

(2l + 1)(|Al|2 + |Bl|2

)

=2π

k2

∞∑l=1

(2l + 1)

∣∣∣∣∣ jl(ka)

h(1)l (ka)

∣∣∣∣∣2

+

∣∣∣∣∣ [kajl(ka)]′

[kah(1)l (ka)]′

∣∣∣∣∣2 , (m2). (6.229)

σ(ka) normalized by 2πa2 (this is the scattering cross-section in the geometrical optics limit ka 1)

is plotted in Fig.?? in the range 0 < x(= ka) < 10. In the long wavelength limit, ka 1, the

spherical Bessel functions approach

jl(x)→ xl

(2l + 1)!!, [xjl(x)]′ → l + 1

(2l + 1)!!xl for x 1, (6.230)

h(1)l (x)→ −i(2l − 1)!!

xl+1, [xh

(1)l (x)]′ → i

l(2l − 1)!!

xl+1for x 1. (6.231)

The lowest order terms of l = 1 remain finite in this limit,

A1 = −i13

(ka)3, B1 = i2

3(ka)3, (6.232)

which yields a scattering cross-section

σ ' 10π

3k4a6, ka 1. (6.233)

In short wavelength limit ka 1, the figure indicates that σ indeed approaches 2πa2.

In radar engineering, the Poynting flux scattered back toward a radar is of main interest. In

the geometry assumed in Fig. 6-4, the direction of back scattering corresponds to θ = π. Since

Yl,1(θ, φ) + Yl,−1(θ, φ) = −2i

√2l + 1

4πl(l + 1)P 1l (cos θ) sinφ, (6.234)

dP 1ldθ

+1

sin θP 1l = 0, and

dP 1ldθ

= (−1)ll(l + 1)

2at θ = π, (6.235)

the scattered field at θ = π becomes

Esc(r) = −E0∑l,±1

il+1

√π(2l + 1)

l(l + 1)(AlMl,±1 ±BlNl,±1)

= −iE0eikr

kr

∑l

2l + 1

2(−1)l(Al −Bl)ex. (6.236)

At θ = π, the scattered field is plane polarized in the same direction as the incident electric field.

36

The Poynting flux at θ = π is

Sr(θ = π) =cε0E

20

4r2k2

∣∣∣∣∣∑l

(−1)l(2l + 1)(Al −Bl)∣∣∣∣∣2

.

The radar scattering cross section is defined by

σradar =4π

cε0E20r2Sr(θ = π) =

π

k2

∣∣∣∣∣∑l

(2l + 1)(−1)l(Al −Bl)∣∣∣∣∣2

, (6.237)

where 4π is the total solid angle.

Example 3 Scattering by a Dielectric Sphere

Scattering by a sphere of dielectric and magnetic properties can be analyzed in a similar manner.

We assume a sphere of radius a having relative permittivity εr and relative permeability µr placed

in a plane wave. In general, εr and µr are both complex to account for dissipation (absorption) of

electromagnetic energy. The incident wave is described by the fields

E0eikzex = −E0

∑l,±1

il+1

√π(2l + 1)

l(l + 1)

[jl(kr)r×∇Yl,±1(θ, φ)± 1

k∇× [jl(kr)r×∇Yl,±1(θ, φ)]

],

(6.238)

and

H0eikzey = −E0

Z0

∑l,±1

il

√π(2l + 1)

l(l + 1)

[1

k∇× [jl(kr)r×∇Yl,±1(θ, φ)]± jl(kr)r×∇Yl,±1(θ, φ)

].

(6.239)

where Z0 =√µ0/ε0. The scattered wave continues to be assumed in the form

Ee(r) = E0∑l,±1

il−1

√π(2l + 1)

l(l + 1)(AlMl,±1 ±BlNl,±1) , r > a (6.240)

He(r) = −E0Z0

∑l,±1

il

√π(2l + 1)

l(l + 1)(AlNl,±1 ±BlMl,±1) , r > a (6.241)

where Z0 =√µ0/ε0 is the impedance in the free space. The fields inside the sphere should be

bounded at r = 0 and thus may be assumed as

Ei(r) = E0∑l,±1

il−1

√π(2l + 1)

l(l + 1)

(Cljl(k

′r)r×∇Yl,±1 ±Dl1

k′∇× [jl(k

′r)r×∇Yl,±1]), (6.242)

Hi(r) = −E0Zs

∑l,±1

il

√π(2l + 1)

l(l + 1)

(Cl

1

k′∇× [jl(k

′r)r×∇Yl,±1]±Dljl(k′r)r×∇Yl,±1

), (6.243)

37

where

Zs =

√µrεrZ0, k

′ =√εrµrk, (6.244)

to account for the impedance and index of refraction of the sphere. Continuity of the tangential

components of the electric field E and magnetic field H yields the following conditions:

jl(ka) +Alh(1)l (ka) = Cljl(k

′a), (6.245)

[kajl(ka)]′ +Bl[kah(1)l (ka)]′ =

1√εrµr

Dl[k′ajl(k

′a)]′, (6.246)

jl(ka) +Blh(1)l (ka) =

√εrµrDljl(k

′a), (6.247)

[kajl(ka)]′ +Al[kah(1)l (ka)]′ =

1

µrCl[k

′ajl(k′a)]′. (6.248)

where

[xjl(x)]′ =d

dx[xjl(x)], [xh

(1)l (x)]′ =

d

dx[xh

(1)l (x)]. (6.249)

Solving for the coeffi cients Al and Bl, we find

Al =µrjl(k

′a)[kajl(ka)]′ − jl(ka)[k′ajl(k′a)]′

h(1)l (ka)[k′ajl(k′a)]′ − µrjl(k′a)[kah

(1)l (ka)]′

, (6.250)

Bl =εrjl(k

′a)[kajl(ka)]′ − jl(ka)[k′ajl(k′a)]′

h(1)l (ka)[k′ajl(k′a)]′ − εrjl(k′a)[kah

(1)l (ka)]′

. (6.251)

The case of ideally conducting sphere can be recovered in the limit εr → ∞ and µr → 0. The

scattering cross-section can readily be found,

σ =2π

k2

∞∑l=1

(2l + 1)(|Al|2 + |Bl|2

). (6.252)

In the long wavelength limit, ka, k′a 1, the leading order terms are the dipoles, l = 1,

A1 = i2

3

µr − 1

µr + 2(ka)3, B1 = i

2

3

εr − 1

εr + 2(ka)3. (6.253)

For a dielectric sphere with µr = 1, the scattering cross-section in the low frequency regime is thus

given by

σ ' 8π

3

∣∣∣∣εr − 1

εr + 2

∣∣∣∣2 k4a6, ka 1. (6.254)

In the limit of εr 1, we recover the electric dipole portion of the cross-section of a conducting

sphere,

σ =8π

3k4a6. (6.255)

Note that the case of conducting sphere can be recovered, mathematically, in the limit µr = 0 for

38

A1 and εr = ∞ for B1. The case of conducting sphere should be analyzed by incorporating the

impedance

Z =

√−iωµ

−iωε+ σc,

where σc is the conductivity. Ideal conductor is characterized by Z = 0 which can be realized by

letting σc →∞, or µ→∞, or ε→∞.

6.10 Scattering by a Cylinder

In this section, we consider scattering of electromagnetic waves by a cylindrical object having a

length l suffi ciently longer than the wavelength, kl 1. If the incident wave propagates perpen-

dicular to the cylinder axis, the problem becomes two dimensional. For general incident angle, the

incident wave can be decomposed into normal and axial components. The axial component suffers

little scattering and it is suffi cient to consider only the case of normal incidence.

Figure 6-6: Two possible polarizations of the incident field relative to the cylinder axis, Ei parallelto the axis (top) and perpendicular (bottom).

Scattering by a conducting cylinder can be analyzed in a manner similar to the case of conducting

sphere. If the incident wavelength is much shorter than the radius of the cylinder, ka 1, the

geometrical optics approximation applies. Regardless of the polarization of the incident wave

relative to the cylinder axis, the scattering cross-section per unit length of the cylinder is given by

σ

l= 2a+ 2a = 4a, ka 1, (6.256)

where 2a is the contribution form the illuminated surface and another 2a is the contribution from

the shadow surface due to forward diffraction. Analysis for this case, which is quite parallel to the

39

case of a conducting sphere, is left for an exercise..

In long wavelength limit ka 1, the polarization direction of the incident wave becomes

important. This is understandable because the current induced on the cylinder surface sensitively

depends on the orientation of the incident electric field. As intuitively expected, the surface current

flows much more easily along the axis than in azimuthal direction. Therefore, the scattering cross-

section when Ei ‖ ez should be much larger than the case Ei ⊥ ez.The axial components of the electric and magnetic fields satisfy the following 2-dimensional

scalar Helmholtz equation, (∂2

∂ρ2+

1

ρ

∂

∂ρ+

1

ρ2∂2

∂φ2+ k2

)(EzHz

)= 0. (6.257)

General solutions can be found by assuming a form R(ρ)eimφ, where the radial function satisfies

the Bessel equation (d2

dρ2+

1

ρ

d

dρ− m2

ρ2+ k2

)R(ρ) = 0, (6.258)

R(ρ) = Jm(kρ), Nm(kρ).

For wave analysis, it is convenient to define the Hankel functions of the first and second kinds by

H(1)m (kρ) = Jm(kρ) + iNm(kρ), (6.259)

H(2)m (kρ) = Jm(kρ)− iNm(kρ). (6.260)

The asymptotic form of the first kind is

H(1)m (kρ)→

√2

πkρexp

(ikρ− i2m+ 1

4π

), (6.261)

which has an amplitude dependence 1/√ρ appropriate for cylindrical waves. H(1)

m (kρ) describes an

outgoing wave and H(2)m (kρ) an incoming wave.

We first analyze the case of incident electric field along the cylinder axis Ei ‖ ez. The incidentwave is assumed to be propagating in the negative x-direction,

Ei(r) = E0e−ikρ cosφez. (6.262)

The scattered electric field is also in the z-direction and we assume

Escz =∑m

amH(1)m (kρ)eimφ. (6.263)

If the cylinder is ideally conducting, the boundary condition is that the tangential component of

40

the electric field vanish at the cylinder surface,

E0e−ika cosφ +

∑m

amH(1)m (ka)eimφ = 0. (6.264)

Multiplying by e−im′φ and integrating over φ, we find the expansion coeffi cient am,

am = − E0

2πH(1)m (ka)

∫ 2π

0e−i(ka cosφ+mφ)dφ

= −(−i)m Jm(ka)

H(1)m (ka)

E0, (6.265)

and the scattered electric field is

Escz = −E0∑m

(−i)m Jm(ka)

H(1)m (ka)

H(1)m (kρ)eimφ. (6.266)

The Poynting flux in the radiation zone kρ 1 is

Sρ =E20Z0

2

πkρ

∑m

∣∣∣∣∣ Jm(ka)

H(1)m (ka)

∣∣∣∣∣2

, (6.267)

and the total scattered power is given by

P = ρ

∫ 2π

0Sρdφ

=E20Z0

4

k

∑m

∣∣∣∣∣ Jm(ka)

H(1)m (ka)

∣∣∣∣∣2

. (6.268)

The scattering cross-section is

σ

l=

4

k

∑m

∣∣∣∣∣ Jm(ka)

H(1)m (ka)

∣∣∣∣∣2

.

The function

f(x) =1

x

∑m

∣∣∣∣∣ Jm(x)

H(1)m (x)

∣∣∣∣∣2

, (6.269)

is plotted below. In the short wavelength regime it approaches unity and the scattering cross section

is σ/l = 4a as expected from the geometrical optics approximation. In the long wavelength regime

ka 1, m = 0 mode is dominant and the cross-section diverges in a manner

σ

l' π2

k

1[ln

(ka

2

)+ γE

]2 ,

41

where γE = 0.5772 · ·· is the Euler’s constant. Note that in the limit x→ 0, J0(x)→ 1 and

H(1)0 (x)→ i

2

π

[ln(x

2

)+ γE

], x→ 0.

Physically, the symmetric mode (m = 0) corresponds to radiation by a long cylindrical antenna

having a length much larger than the wavelength. Fig.6-7 shows σ/l normalized by 4a as a function

of x = ka.

1

x

10∑m=−10

∣∣∣∣ Jm(x)

Jm(x) + jYm(x)

∣∣∣∣2

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

ka

Figure 6-7: Scattering cross-section per unit length of a long conducting cylinder of radius a. σ/4alas a fucntion of x = ka. The electric field is polarized along the cylinder.

If the incident electric field is perpendicular to the cylinder axis, it is more convenient to use

the magnetic field which is axial. The incident magnet field is

B0e−ikρ cosφez, (6.270)

and we assume the scattered magnetic field in the form

Bscz (r) =

∑m

bmH(1)m (kρ)eimφ. (6.271)

Corresponding electric field can be found from the Maxwell’s equation,

1

c2∂Esc

∂t= ∇×Bsc. (6.272)

42

The φ component of the scattered electric field is thus

Escφ (ρ, φ) = −ic∑m

bmk

∂

∂ρ[H(1)

m (kρ)]eimφ, (6.273)

while the φ component of the incident electric field is

Eiφ(ρ, φ) = cB0 cosφe−ikρ cosφ. (6.274)

The boundary condition of vanishing tangential component of the electric field at the cylinder

surface yields

B0 cosφe−ika cosφ − i∑m

bmk

∂

∂a[H(1)

m (ka)]eimφ = 0. (6.275)

Multiplying by eim′φ and integrating over φ, we obtain

bm = −B0(−i)m[Jm(ka)]′

[H(1)m (ka)]′

, (6.276)

where the prime means differentiation with respect to the argument ka and the following transfor-

mation is used, ∫ 2π

0cosφe−i(ka cosφ+mφ)dφ = i(−i)m2π[Jm(ka)]′.

A resultant scattering cross-section is

σ

l=

4

k

∑m

∣∣∣∣∣ [Jm(ka)]′

[H(1)m (ka)]′

∣∣∣∣∣2

. (6.277)

Fig.6-8 shows σ/l normalized by 4a. In contrast to the preceding case Ei = E0ez, the cross-section

in the long wavelength regime ka 1 is bounded and given

σ

l' 3π2

4k3a4, (6.278)

since if ka 1, the dominant harmonics are

b0 =iπB0

4(ka)2, b±1 = ±πB0

4(ka)2. (6.279)

In short wavelength regime ka 1, the cross-section approaches 4a as in the case of axial polar-

ization.

43

x 108642

0.8

0.6

0.4

0.2

0

Figure 6-8: (σ/l)/4a of a long conducting cylinder when the electric field is normal to the cylinder.

Problems

6.1 A conducting sphere is placed in a low frequency plane electromagnetic wave such that ka 1

where a is the sphere radius. Finding the effective electric and magnetic dipole moments, show

that the scattering cross section of the sphere is given by

σ =(8 + 2)π

3k4a6,

where 8 parts is due to electric polarization and 2 parts due to magnetic dipole. According

to the radiation magnetic field due to a small source in Eq. (5.51),

H(r) ' 1

4πc

eikr

r

(p× n+

1

cn× (n× m)− 1

2cn× (n ·

...Q)

),

the radiation power due to magnetic dipole is supposed to be of higher order than that due to

electric dipole by a factor (ka)2. In scattering by a conducting sphere, they are comparable.

Resolve this apparent paradox.

6.2 A conducting sphere of radius a has a complex index of refraction√ε/ε0 = nr+ini. Determine

the scattering cross-section as a function of ka.

6.3 Show that the transverse component of the current density J is given by

Jt =1

4π∇×∇×

∫J(r′)

|r− r′|dV′,

44

and consequently the longitudinal component by

Jl = − 1

4π∇∇ ·

∫J(r′)

|r− r′|dV′ = − 1

4π∇∫ ∇′ · J(r′)

|r− r′| dV′.

Hint:

∇2 1

|r− r′| = −4πδ(r− r′).

6.4 The scattering cross section of small objects (ka 1 with a being the size) obeys the

Rayleigh’s law σ ∝ k4 which is often used to explain blueness of sky and redness of sunset.

Scattering of sunlight by air molecules requires fluctuation in the air density. Explain why.

You may treat the atmosphere as a fluid.

6.5 A conducting sphere of radius a is coated with a material having a relative permittivity εr and

permeability µr (both may be complex). The thickness of coating is δ. Analyze low frequency

scattering in the limit ka 1 and show that the radiation fields are characterized by the

dipole terms

A1 = −i(kb)3

3

2 + ρ− 2µr(1− ρ)

2 + ρ+ µr(1− ρ),

B1 = −i(kb)3

3

2(1− ρ)− 2εr(1 + 2ρ)

2(1− ρ) + εr(1 + 2ρ),

where b = a + δ, ρ = a3/(a + δ)3. Discuss possible effects of coating on the scattering cross

section.

6.6 For a tangential electric field on the surface of a sphere of radius a,

Es (θ, φ) = Eθ(θ, φ)eθ + Eφ(θ, φ)eφ,

find a general expression for the radiation electric field.

6.7 A conducting sphere of radius a has a narrow planar gap cut at polar angle θ0. Show that

the radiation admittance of the sphere is

Y =i

Z02πka sin2 θ0

∞∑l=1

2l + 1

2l(l + 1)[P 1l (cos θ0)]

2 h(2)l (ka)

dda [ah

(2)l (ka)]

, (Siemens)

where Z0 =√µ0/ε0.

6.8 In the Lorenz gauge, the scalar and vector potentials are

Φ (r) =1

4πε0

∫eik|r−r

′|

|r− r′| ρ(r′)dV ′,

A (r) =µ04π

∫eik|r−r

′|

|r− r′| J(r′)dV ′,

45

where the charge density and current density are related through

−iωρ+∇ · J = 0.

Show that the electric field is given by

E (r) = −∇Φ− ∂A

∂t

=iωµ04π

(1+

1

k2∇∇

)·∫eik|r−r

′|

|r− r′| J(r′)dV ′,

where 1 is the unit tensor. Furthermore, show that the concrete form of the tensor operator

in the spherical coordinates is

(1+

1

k2∇∇

)eikR

4πR=

2

(kR)2− 2i

kR0 0

0 1 +i

kR− 1

(kR)20

0 0 1 +i

kR− 1

(kR)2

eikR

4πR,

where R = |r− r′| . Note that the radial (longitudinal) component is proportional to

1

R

(2

(kR)2− 2i

kR

).

Hint:∂er∂θ

= eθ,∂er∂φ

= sin θeφ.

6.9 The Lorenz gauge is characterized by

∇ ·A+1

c2∂Φ

∂t= 0.

(a) Show that∂

∂t

(∇ ·A+

1

c2∂Φ

∂t

)= 0,

yields the Coulomb’s law,

∇ ·E =ρ

ε0.

(b) What does

∇(∇ ·A+

1

c2∂Φ

∂t

)= 0,

yield?

(c) Show that

∇ ·A+1

c2∂Φ

∂t= 0,

46

is consistent with the charge conservation law,

∂ρ

∂t+∇ · J = 0.

(d) Repeat a and b for the Coulomb gauge, namely, interpret

∂

∂t∇ ·A =0, ∇∇ ·A = 0.

47