Chapter 1

Electrostatics I. Potential due toPrescribed Charge Distribution,Dielectric Properties, Electric Energyand Force

1.1 Introduction

In electrostatics, charges are assumed to be stationary. Electric charges exert force on other charges

through Coulombs law which is the generic law in electrostatics. For a given charge distribution,

the electric eld and scalar potential can be calculated by applying the principle of superposition.

Dielectric properties of matter can be analyzed as a collection of electric dipoles. Atoms having

no permanent dipole moment can be polarized if placed in an electric eld. Most molecules have

permanent dipole moments. In the absence of electric eld, dipole moments are oriented randomly

through thermal agitation. In an electric eld, permanent dipoles tend to align themselves in the

direction of the applied eld and weaken the eld. Some crystals exhibit anisotropic polarization and

the permittivity becomes a tensor. The well known double di¤raction phenomenon occurs through

deviation of the group velocity from the phase velocity in direction as well as in magnitude.

1.2 Coulombs Law

Normally matter is charge-neutral macroscopically. However, charge neutrality can be broken

relatively easily by such means as mechanical friction, bombardment of cosmic rays, heat (e.g.,

candle ame is weakly ionized), etc. The rst systematic study of electric force among charged

bodies was made by Cavendish and Coulomb in the 18th century. (Cavendishs work preceded

Coulombs. However, Coulombs work was published earlier. Lesson: it is important to publish

your work as early as possible.) The force to act between two charges, q1 at r1 and q2 at r2; follows

1

the well known Coulombs law,

Figure 1-1: Repelling Coulomb force between like charges.

F = const.q1q2r2

= const.q1q2

jr1 r2j2; (N) (1.1)

where r = jr1 r2j is the relative distance between the charges. In CGS-ESU (cm-gram-second

electrostatic unit) system, the constant is chosen to be unity. Namely, if two equal charges q1 =

q2 = q separated by r = 1 cm exert a force of 1 dyne = 105 N on each other, the charge is dened

as 1 ESU ' 13 10

9 C. The electronic charge in ESU is e = 4:8 1010 ESU. In MKS-Ampere(or SI) unit system, 1 Coulomb of electric charge is dened from

1 Coulomb = 1 Ampere 1 sec,

where 1 Ampere of electric current is dened as follows. If two innitely long parallel currents of

equal amount separated by 1 meter exert a force per unit length of 2 107 N/m on each other

(attracting if the currents are parallel and repelling if antiparallel), the current is dened to be 1

Ampere. Since the force per unit length to act on two parallel currents I1 and I2 separated by a

distance d is given byF

l= B1I2 =

02

I1I2d; (N/m) (1.2)

the magnetic permeability 0 = 4107 H/m in MKS-Ampere unit system is an assigned constantintroduced to dene 1 Ampere current. (The permittivity "0 is a measured constant that should

be determined experimentally.)

In MKS-Ampere unit system, the measured proportional constant in Coulombs law is approx-

imately 9:0 109 N m2 C2: It is customary to write the constant in the form

const. =1

4"0;

2

Figure 1-2: In MKS unit system, I = 1 Ampere current is dened if the force per unit length betweeninnite parallel currents 1 m apart is 2 107 N/m. The magnetic permeability 0 = 4 107H/m is an assigned constant to dene 1 Ampere current.

where

"0 = 8:85 1012C2

N m2

=Faradm

;

is the vacuum permittivity. Since

c2 =1

"00; m2s2;

the permittivity "0 can be deduced from the speed of light in vacuum that can be measured

experimentally as well.

1.3 Coulomb Electric Field and Scalar Potential

The Coulombs law may be interpreted as a force to act on a charge placed in an electric eld

produced by other charges, since the Coulomb force can be rewritten as

F =1

4"0

q1(r2 r1)jr1 r2j3

q2

= q2Eq1 ; (N) (1.3)

where

Eq1 =1

4"0

q1(r2 r1)jr1 r2j3

;

NC=Vm

(1.4)

is the electric eld produced by the charge q1 at a distance r1 r2: The eld is associated with thecharge q1 regardless of the presence or absence of the second charge q2: (The factor 4 is the solid

3

angle pertinent to spherical coordinates. If it is ignored as in the CGS-ESU system, it pops up in

planar coordinates. Corresponding Maxwells equation is

r E =

"0; in MKS-Ampere,

or

r E = 4; in CGS-ESU.

Which unit system to choose is a matter of conveneince and it is nonsense to argue one is superior

to other.)

Since two charges q1 and q2 exert the Coulomb force on each other, assembling a two-charge

system requires work. A di¤erential work needed to move the charge q2 against the Coulomb force

is

dW = F dr2 = 1

4"0

q1q2(r2 r1)jr1 r2j3

dr2:

Noting

rr21

jr1 r2j= +

r1 r2jr1 r2j3

;

integration from r2 =1 to r2 can be readily carried out,

W =1

4"0

q1q2jr1 r2j

; (J) (1.5)

This is the potential energy of two-charge system. Note that it can be either positive or negative. If

W is positive, the charge system has stored a potential energy that can be released if the system is

disassembled. Energy released through nuclear ssion process is essentially of electrostatic nature

associated with a system of protons closely packed in a small volume. On the other hand, if W

is negative, an energy jW j must be given to disassemble the system. For example, to ionize ahydrogen atom at the ground state, an energy of 13:6 eV is required. In a hydrogen atom, the total

system energy is given by

W =1

2mv2 e2

4"0rB= e2

8"0rB< 0;

where rB = 5:3 1011 m is the Bohr radius and

1

2mv2 =

e2

8"0rB= 13:6 eV,

is the electron kinetic energy. The electric potential energy is

e2

4"0rB= 27:2 eV.

4

The potential energy of two-charge system may be written as

W =1

4"0

q1jr1 r2j

q2; (J)

which denes the potential associated with a point charge q1 at a distance r;

=1

4"0

q

r; (J C1 = V); (1.6)

and the electric eld is related to the potential through

E = r; (V m1): (1.7)

Both potential and electric eld E are created by a charge q regardless of the presence or absence

of a second charge.

1.4 Maxwells Equations in Electrostatics

In electrostatics, electric charges are stationary being held by forces other than of electric origin

such as molecular binding force. Since charges are stationary, no electric currents and thus no

magnetic elds are present, B = 0:

For a distributed charge density (r), the electric eld can be calculated from

E(r) =1

4"0

ZV

(r0)(r r0)jr r0j3

dV 0 = 1

4"0rZ

(r0)

jr r0jdV0; (1.8)

since the di¤erential electric eld due to a point charge dq = (r0)dV 0 located at r0 is

dE =1

4"0

(r0)dV 0

jr r0j3(r r0):

Note that

r 1

jr r0j = r0 1

jr r0j = r r0

jr r0j3;

where r0 is the gradient with respect to r0. Eq. (1.8) allows one to calculate the electric eld for agiven charge density distribution (r):

In order to nd a di¤erential equation to be satised by the electric led, we rst note that the

surface integral of the electric eld IE dS;

can be converted to a volume integral of the divergence of the eld (Gaussmathematical theorem,

5

not to be confused with Gausslaw),ISE dS =

ZVr EdV

= 1

4"0

ZVdVr2

ZV 0

(r0)

jr r0jdV0:

However, the function 1= jr r0j satises a singular Poissons equation,

r2 1

jr r0j = 4(r r0): (1.9)

Therefore, ISE dS = 1

"0

ZVdV

ZV 0(r r0)(r0)dV 0 = 1

"0

Z(r)dV: (1.10)

This is known as Gauss law for the electric eld. Note that Gauss law is a consequence of the

Coulombs inverse square law. FromZr EdV = 1

"0

Z(r)dV; (1.11)

it also follows that

r E =

"0: (1.12)

This is the di¤erential form of the Gausslaw for the longitudinal electric eld and constitutes one

of Maxwells equations.

For the electric eld due to a static charge distribution,

E(r) =1

4"0

ZV

(r0)(r r0)jr r0j3

dV 0 = 1

4"0rZV

(r0)

jr r0jdV0;

its curl identically vanishes because

rE = 1

4"0rr

Z(r0)

jr r0jdV0 0:

(Note that for any scalar function F; r rF 0:) Therefore, the second Maxwells equation inelectrostatics is

rE = 0: (1.13)

This is a special case of the more general Maxwells equation

rE = @B@t; (1.14)

which determines the transverse component of the electric eld. Evidently, if all charges and elds

are stationary, there can be no magnetic eld. It should be remarked that a vector eld can be

uniquely determined only if both its divergence and curl are specied. (This is known as Helmholtzs

6

theorem.) In electrodynamics, two vector elds, the electric eld E and magnetic eld B, are to befound for given charge and current distributions (r; t) and J(r; t): Therefore it is not accidental

that four Maxwells equations emerge specifying the four functions r E; rE; r B; and rBwhich determine the longitudinal and transverse components of E and B.Digression: A vector A can be decomposed into the longitudinal and transverse components Aland At which are, by denition, characterized by

rAl = 0; r At = 0:

The longitudinal component can be calculated from

Al (r) = 1

4rZ r0 A (r0)

jr r0j dV0;

since

r2 1

jr r0j = 4r r0

:

The transverse component is given by

At (r) = A (r)Al (r)

=1

4rr

ZA (r0)

jr r0jdV0;

where the identity

rrA =rr Ar2A;

is exploited. It is known that for a vector to be uniquely dened, its divergence and curl have to

be specied. Since r A =r Al; divergence of a vector spcies the divergence of the longitudinalcomponent. Likewise, rA =rAt specifes the curl of the tarnverse component. As a concreteexample, let us consider the electric eld. Its divergence is

r E =

"0;

and the solution to this di¤erential equation is

El (r) = 1

4"0rZV

(r0)

jr r0jdV0:

The tranverse component is specied by the magnetic eld as

rE = @B@t:

The vanishing curl of the static electric eld allows us to write the electric eld in terms of a

7

gradient of a scalar potential ,

E = r; (1.15)

for the curl of a gradient of a scalar function identically vanishes,

rr 0:

For a given charge distribution, the potential has already been formulated in Eq. (1.8),

(r) =1

4"0

ZV

(r0)

jr r0jdV0: (1.16)

In terms of the scalar function (r); Eq. (1.12) can be rewritten as

r2 = "0; (1.17)

which is known as Poissons equation. In general, solving the scalar di¤erential equation for is

easier than solving vector di¤erential equations for E.

The physical meaning of the scalar potential is the amount of work required to move adiabati-

cally a unit charge from one position to another. Consider a charge q placed in an electric eld E.The force to act on the charge is F = qE and if the charge moves over a distance dl; the amount

of energy gained (or lost, depending on the sign of qE dl) by the charge is

qE dl = qr dl:

Therefore, the work to be done by an external agent against the electric force to move the charge

from position r1 to r2 is

W = qZ r2

r1

E dl = qZ r2

r1

r dl = q [(r2) (r1)] ; (1.18)

where (r2) (r1) is the potential di¤erence between the positions r1 and r2: The work isindependent of the choice of the path of integration. The potential is a relative scalar quantity and

a constant potential can be added or subtracted without a¤ecting the electric eld.

In a conductor, an electric eld drives a current ow according to the Ohms law,

J = E; (A m2) (1.19)

where (S m1) is the electrical conductivity. In static conditions (no time variation and no ow

of charges), the electric eld in a conductor must therefore vanish. A steady current ow and

constant electric eld can exist in a conductor if the conductor is a part of closed electric circuit.

However, such a circuit is not static because of the presence ow of charge. For the same reason,

a charge given to an isolated conductor must entirely reside on the outer surface of the conductor.

Since E (static) = 0 in a conductor, the volume charge density must also vanish according to

8

= "0r E = 0: A charge given to a conductor can only appear as a surface charge (C m2)

on the outer surface. The time scale for a conductor to establish such electrostatic state may be

estimated from the charge conservation equation

@

@t+r J = 0: (1.20)

The current density can be estimated from the equation of motion for electrons

m@v

@t= eEmev; (1.21)

where c (s1) is the electron collision frequency with the lattice ions. Noting J = nev with nthe density of conduction electrons, we obtain

@

@t+ c

J =

ne2

mE: (1.22)

Therefore, the equation for the excess charge density in a conductor becomes

@

@t

@

@t+ c

+ !2p = 0; !2p =

ne2

"0m; (1.23)

which describes a damped plasma oscillation 0e ti!pt with an exponential damping factor =

c=2: The typical electron collision frequency in metals is of order c ' 1013= sec and the time

constant to establish electrostatic state is indeed very short. (If the low frequency Ohms law is

used, an unrealistically short transient time emerges,

@

@t+

0 = 0; = 0e

t= ;

where = 0= ' 1019 sec.)The principle that a charge given to a conductor must reside entirely on its outer surface is also

a consequence of Coulombs 1=r2 law. For a conductor of an arbitrary shape, the surface charge

distribution is so arranged that the electric eld inside the conductor vanishes everywhere. In Fig.

1-3, one conducting spherical shell surrounds a smaller one. The spheres are connected through

a thin wire. A charge originally given to the inner sphere will end up at the outer surface of the

larger sphere and the electric eld inside should vanish if Coulombs law is correct. Experiments

have been conducted to measure residual electric elds inside and it has been established that the

power in the Coulombs law F _ 1=r is indeed very close to 2 within one part in 1016: is

probably exactly equal to 2.0. If not, there would be a grave consequence that a photon should

have a nite mass. This is because if a photon has a mass mp; corresponding Compton wavenumber

kC = mpc=~ will modify the wave equation for all potentials, including the Coulomb potential, inthe form

r2 k2C 1

c2@2

@t2

=

"0: (1.24)

9

Figure 1-3: Charge initially given to the inner conductor will all end up at the outer surface of theouter conductor. The absence of the electric eld inside the sphere is a consequence of Coulombslaw.

In static case @=@t = 0; this reduces to

r2 k2C

=

"0; (1.25)

and for a point charge (r) = q(r); yields a Debye or Yukawa type potential,

(r) =1

4"0

q

rexp (kCr) ; (1.26)

and electric eld

Er =q

4"0

1

r2+kCr

ekCr: (1.27)

The power in the Coulombs law Er _ q=ra experimentally established is 2 = O(1016);and this corresponds to an upper limit of photon mass of mp < 1051 kg. The lower limit of

photon Compton wavelength is C > 4 105 km at which distance a signicant deviation from the

Coulombs law, if any, is expected. (Incidentally, in a plasma, the static scalar potential does have

a Debye form,

(r) =1

4"0

q

rexp (kDr) ; (1.28)

where

kD =

sne2

"0T; (1.29)

is the inverse Debye shielding distance, n is the plasma density and T (in Joules) is the plasma

10

temperature. Likewise, in a superconductor, static magnetic eld obeys

r2 k2L

B = 0; (1.30)

where kL = !p=c is the London skin depth with !p =pne2="0me the electron plasma frequency.)

The potential energy of a two-charge system is

W =1

4"0

q1q2jr1 r2j

= 2 12

1

4"0

q1q2jr1 r2j

; (1.31)

where1

2

1

4"0

q1q2jr1 r2j

; (1.32)

is the potential energy associated with each charge. For many charge system, this can be generalized

in the form

W =1

2

Xj

jqj =1

2

1

4"0

Xi6=j

qiqjjri rj j

; (1.33)

where

j =1

4"0

Xi6=j

qijri rj j

; (1.34)

is the potential at the location of charge qj due to all other charges. For distributed charge with

a local charge density (r) (C/m3); the potential energy of a di¤erential charge dq = dV , which

can be regarded as a point charge if dV is su¢ ciently small, is

dW =1

2dq =

1

2dV:

Since = "0r E;dW =

1

2"0r EdV: (1.35)

Integrating over the entire volume, we nd

W =1

2"0

ZVr EdV

=1

2"0

ZV[r (E) (r )E] dV

=1

2"0

ISEdS+

ZV

1

2"0E

2dV: (1.36)

where use is made of Gausstheorem,ZVr FdV =

ISFdS; (1.37)

for an arbitrary well dened vector eld F. The surface integral vanishes because at innity, bothpotential and electric eld vanish. Therefore, the potential energy associated with a distributed

11

charge system is

W =

Z1

2"0E

2dV; (1.38)

which is positive denite. In the expression for the potential energy of discrete charge system,

the potential energy due to a point charge itself is excluded while in the integral form, spatial

distribution of charge is assumed even for point charges and the so-called self energy is included.

The quantity

ue =1

2"0E

2; (J/m3) (1.39)

is the electric energy density associated with an electric eld. This expression for the energy density

holds regardless of the origin of the electric eld, and can be used for elds induced by time varying

magnetic eld as well.

The self-energy of an ideal point charge evidently diverges because the electric eld proportional

to 1=r2 does in the limit of r ! 0. However, if the electron is assumed to have a nite radius re;

the self-energy turns out to be of the order of

We '1

4"0

e2

re: (1.40)

This should not exceed the rest energy of the electron mec2: Equating these two, the following

estimate for the electron radius emerges,

re '1

4"0

e2

mc2= 2:85 1015 m. (1.41)

Although this result should not be taken seriously, the scattering cross section of free electron

placed in an electromagnetic wave (the process known as Thomson scattering) does turn out to be

=8

3r2e ;

and the concept of electron radius is not totally absurd. For proton whose radius is also of order

1015 m, the analogy obviously breaks down.

The Poissons equation in Eq. (1.17) is the Eulers equation to make the energy-like integral

stationary

U =

Z 1

20E

2 dV =

Z 1

20(r)2

dV: (1.42)

Indeed, the variation of this integral

U =

Z(0r r ) dV

= Z

0r2+ dV;

12

becomes stationary (U = 0) when the Poissons equation holds,

r2+

"0= 0:

In other words, electrostatic elds are realized in such a way that the total electric energy becomes

minimum. In general, electric force act so as to reduce the energy in a closed (isolated) system.

Macroscopically, electric force tends to increase the capacitance as we will see in Chapter 2.

If the charge density distribution is known, the potential (r) can be readily found as a solution

of the Poissons equation. However, if the charge density is unknown a priori, as in most potential

boundary value problems, the potential and electric eld must be found rst. Then the charge

density is to be found from = "0r E; or in the case of surface charge on a conductor surface,

= "0En; (C m2);

where En is the electric eld normal to the conductor surface.

1.5 Formal Solution to the Poissons Equation

As shown in the preceding section, if the spatial distribution of the charge density (r) is given,

the solution for the Poissons equation

r2(r) = (r)0; (1.43)

can be written down as

(r) =1

40

ZV

(r0)

jr r0jdV0: (1.44)

This is understandable because the di¤erential potential due to a point charge dq = dV is

d =1

4"0

(r0)dV 0

jr r0j ;

and the solution in Eq. (1.44) is a result of superposition.

It is noted that the Poissons equation for the scalar potential still holds even for time varying

charge density,

r2C(r; t) = (r; t)

"0; (1.45)

if one employs the Coulomb gauge characterized by the absence of the longitudinal vector potential

r A = 0: (1.46)

13

Figure 1-4: Di¤erential potential d due to a point chargedq = dV 0.

In the Coulomb gauge, the transverse vector potential satises the wave equation,r2 1

c2@2

@t2

At = 0Jt; (1.47)

where Jt is the transverse current. The solution for the scalar potential in the Coulomb gauge is

non-retarded,

C(r; t) =1

40

ZV

(r0; t)

jr r0jdV0; (1.48)

and so is the resultant longitudinal electric eld,

El(r; t) = rC =1

4"0

ZV

r r0

jr r0j3(r0; t)dV 0: (1.49)

Such non-retarded (instantaneous) propagation of electromagnetic disturbance is clearly unphysical

and should not exist. In fact, the non-retarded Coulomb electric eld is exactly cancelled by a term

contained in the retarded transverse electric eld

Et(r; t) = @At@t

= 04

@

@t

ZJt(t )jr r0j dV

0; =jr r0jc

;

as will be shown in Chapter 6.

The potential in Eq. (1.44) is in the form of convolution between the function

G(r; r0) = G(r r0) = 1

4jr r0j ; (1.50)

14

and the source function (r0)=0: The function G(r; r0) is called the Greens function and introduced

as a solution to the following singular Poissons equation

r2G = (r r0); (1.51)

subject to the boundary condition that G = 0 at r = 1: (G subject to this boundary condition

is called the Greens function in free space. It is a particular solution to the singular Poissons

equation. Later, we will generalize the Greens function so that it vanishes on a given closed

surface by adding general solutions satisfying r2G = 0:) It is evident that the Greens function

is essentially the potential due to a point charge, for the charge density of an ideal point charge

(without spatial extension) q located at r0 can be written as

c = q(r r0): (1.52)

Here (r r0) is an abbreviation for the three dimensional delta function. For example, in thecartesian coordinates,

(r r0) = (x x0)(y y0)(z z0); (1.53)

in the spherical coordinates (r; ; ),

(r r0) = (r r0)[r( 0)][r sin ( 0)] (1.54)

=(r r0)rr0 sin

( 0)( 0) (1.55)

=(r r0)rr0

(cos cos 0)( 0); (1.56)

and in the cylindrical coordinates (; ; z),

(r r0) = ( 0)[( 0)](z z0) (1.57)

=( 0)

( 0)(z z0): (1.58)

The singular Poissons equation

r2G = (r r0); (1.59)

can be solved formally as follows. Let the Fourier transform of G(r r0) be g(k); namely,

g(k) =

ZG(r r0)eik(rr

0)dr: (1.60)

Its inverse transform is

G(r r0) = 1

(2)3

Zg(k)eik(rr

0)d3k: (1.61)

Noting that the operator r is Fourier transformed as ik and (r r0) as unity, we readily nd from

15

Eq. (1.59),

g(k) =1

k2: (1.62)

Substitution into Eq. (1.61) yields

G(r r0) =1

(2)3

Z1

k2eik(rr

0)d3k

=1

(2)3

Z 1

0dk

Z

0eikjrr

0j cos sin d

Z 2

0d

=1

(2)2

Z

0(jr r0j cos ) sin d

=1

4jr r0j ; (1.63)

where the polar angle in the k-space is measured from the direction of the relative position vector

r r0; andd3k = k2dk sin dd; (1.64)Z 1

1eikxdk = 2(x); (1.65)

(ax) =1

jaj(x); (1.66)

are noted. (The same technique will be used in nding a Greens function for the less trivial wave

equation, r2 1

c2@2

@t2

G(r; r0; t; t0) = (r r0)(t t0); (1.67)

in Chapter 6.)

If the charge density distribution is spatially conned within a small volume such that r r0;

the inverse distance function 1=jr r0j may be Taylor expanded as follows:

1

jr r0j =1

rr

1

r

r0 + 1

2!rr

1

r

: r0r0

=1

r+r r0r3

+1

2r5

Xi;j

(3rirj r2ij)r0ir0j + (1.68)

and so is the potential

(r) =1

40

ZV

(r0)

jr r0jdV0 =

1

40

1

r

ZV(r0)dV 0 +

r

r3ZVr0(r0)dV 0 +

3rr r212r5

: Q+ ;

(1.69)

where 1 is the unit dyadic. The quantity q =R(r0)dV 0 is the total charge contained in the (small)

volume V and the corresponding lowest order monopole potential is

monopole =1

40

q

r: (1.70)

16

The vector quantity

p =

Zr(r)dV; (1.71)

is the dipole moment and the corresponding dipole potential is

dipole =1

40

p rr3: (1.72)

The tensor

Q = Qij =

Zrirj(r)dV; (1.73)

denes the quadrupole moment and the corresponding quadrupole potential is

quadrupole =1

40

3rr r212r5

: Q =1

40

1

2r5

Xi;j

(3rirj r2ij)Qij : (1.74)

Example 1 Electric Field Lines of a Dipole

The potential due to a dipole moment directed in z direction p = pez at the origin is given by

(r; ) =1

40

p

r2cos :

The equipotential surfaces are described by

cos

r2= const.

The electric eld can be found from

E = r = 1

40

2p cos

r3er +

p sin

r3e

: (1.75)

The equation to describe an electric eld line is, by denition,

dr

Er=rd

E; (1.76)

which givesdr

r= 2 cot d;

and thus

r = const. sin2 : (1.77)

Evidently, the electric eld lines are normal to the equipotential surfaces. In Fig.1-5, one equipo-

tential surface and three electric eld lines are shown.

Example 2 Linear quadrupole

17

10.50­0.5­1

1

0.5

0

­0.5

­1

Figure 1-5: Equipotential surface and electric eld lines of an electric dipole pz: The z-axis isvertical.

A linear quadrupole consists of charges q at z = a and 2q at z = 0 as shown in Fig.1-6. Thecharge density is

(r) = q[(z a) 2(z) + (z + a)](x)(y):

Therefore, only Qzz is nonvanishing,

Qzz =

Zz2(r)dV = 2a2q;

and the quadrupole potential at r a is given by

(r) =1

40

1

2r5(3z2 r2)Qzz =

1

40

2qa2

r33 cos2 1

2=

1

40

2qa2

r3P2(cos );

where

P2(cos ) =3 cos2 1

2;

is the Legendre function of order l = 2.

Alternatively, the potential can be found from the direct superposition of the potentials pro-

duced by each charge,

(r; ) =q

40

1p

r2 + a2 2ar cos 2r+

1pr2 + a2 + 2ar cos

:

For r > a; the functions 1=pr2 + a2 2ar cos can be expanded in terms of the Legendre functions

18

Figure 1-6: Linear quadrupole.

Pl(cos ) as1p

r2 + a2 2ar cos =1

r

1Xl=0

ar

l(1)lPl(cos ); r > a:

Retaining terms up to l = 2 (quadrupole), we recover

(r; ) ' 1

40

2qa2

r3P2(cos ); r a:

At r a; the leading order potential is of quadrupole. The total charge is zero, thus no monopole

potential at r a. Also, the charge system consists of two dipoles of equal magnitude oriented in

opposite directions. Therefore, the dipole potential vanishes at r a as well. Equipotential prole

is shown in Fig. ??.

1.6 Potential due to a Ring Charge: Several Methods

A given potential problem can be solved in di¤erent coordinates systems. Of course, the solution

is unique, and answers found by di¤erent methods should all agree. For the purpose of becoming

familiar with several coordinates systems and some useful mathematical techniques, here we nd

the potential due to a ring charge having radius a and total charge q uniformly distributed using

several independent methods.

Method 1: Direct integrationThe potential is symmetric about the z-axis because of the uniform charge distribution. There-

fore, we may evaluate the potential at arbitrary azimuthal angle and here we choose = =2,

that is, observing point in the y z plane. The di¤erential potential due to a point charge

19

­1

­0.5

0

0.5

1

­1 ­0.5 0.5 1

Figure 1-7: Equipotential surface of a linear quadrupole. The z axis ( = 0) is in the verticaldirection. The thick line is at +1 unit potential, and the thin line is at 1:

dq = qd0=2 located at angle 0 is

d =1

40

dq

jr r0j =1

40

q

2pr2 + a2 2ar cos

d0; (1.78)

where is the angle between the vectors r = (r; ; = =2) and r0 = (a; 0 = =2; 0): cos reduces

to

cos = cos cos 0 + sin sin 0 cos( 0) = sin sin0: (1.79)

Integrating over 0 from 0 to 2; we nd

(r; ) =1

40

q

2

Z 2

0

d0pr2 + a2 2ar sin sin0

: (1.80)

Changing the variable from 0 to through

2 = 0 +

2;

the integral can be reduced to

4pr2 + a2 + 2ar sin

Z =2

0

1p1 k2 sin2

d =4p

r2 + a2 + 2ar sin K(k2); (1.81)

20

Figure 1-8: Uniform ring charge.

where

k2 =4ar sin

r2 + a2 + 2ar sin ; (1.82)

21

is the argument of the complete elliptic integral of the rst kind dened by

K(k2) Z =2

0

1p1 k2 sin2

d: (1.83)

The nal form of the potential is

(r; ) =q

220

1pr2 + a2 + 2ar sin

K(k2): (1.84)

f (x) =

Z =2

0

1p1 x sin2

d

2

2.5

3

3.5

4

4.5

0 0.2 0.4 0.6 0.8 1

Figure 1-9: K(x) the complete elliptic integral of the rst kind. It diverges logarithmically at x . 1;K(x) ' ln

4=p1 x

:

The function K(k2) is shown in 1-9. It diverges as x = k2 approaches unity, that is, near the

ring itself, as expected. However, the divergence is only logarithmic,

limk2!1

= ln

4p

:

Example 3 Capacitance of a Thin Conductor Ring

Let us assume a thin conducting ring with ring radius a and wire radius ( a) shown in

Fig.1-10. The potential on the ring surface can be found by letting r = a; = =2: In this limit,

22

the argument k2 approaches unity,

k2 =4a(a )

(a )2 + a2 + 2a(a ) ' 1 2a

2;

and the ring potential becomes

ring 'q

420aln

8a

: (1.85)

Then the self-capacitance of the ring can be found from

C =q

ring=

420a

ln

8a

; a : (1.86)

This formula is fairly accurate even for a not-so-thin ring because of the mere logarithmic depen-

dence on the aspect ratio, a=: For example, when a= = 5 (which is a fat torus rather than a thin

ring); Eq. (1.86) still gives a capacitance within 2 % of the correct value. An exact formula for the

capacitance of a conducting torus will be worked out later in terms of the toroidal coordinates.

Figure 1-10: A thin conductor ring with major radius a and minor radius ; a :

Method 2: Multipole Expansion in the Spherical CoordinatesAs the second method, we directly solve the Poissons equation in the spherical coordinates,

r2 = (r)0

= 10

q

2a2(r a)(cos ): (1.87)

Except at the ring itself, the charge density vanishes. Therefore, in most part of space, the potential

satises Laplace equation,

r2 = 0; r 6= a; 6=

2;

and we seek a potential in terms of elementary solutions to the Laplace equation. Since @=@ = 0;

23

the Laplace equation reduces to

@2

@r2+2

r

@

@r+1

r21

sin

@

@

sin

@

@

= 0: (1.88)

Assuming that the potential is separable in the form (r; ) = R(r)F (), we nd

r2

R

d2R

dr2+2

r

dR

dr

+1

F

1

sin

d

d

sin

dF

d

= 0: (1.89)

Since the rst term is a function of r only and the second term is a function of only, each term

must be constant cancelling each other. Introducing a separation constant l(l + 1); we obtain two

ordinary di¤erential equations for R and F;

d2R

dr2+2

r

dR

dr l(l + 1)

r2R = 0; (1.90)

1

sin

d

d

sin

dF

d

+ l(l + 1)F = 0; (1.91)

which can be written asd

d

(1 2)dF

d

+ l(l + 1)F = 0; (1.92)

where = cos : Eq. (1.92) is known as the Legendres di¤erential equation. The solutions for

R(r) are

R(r) = rl and1

rl+1; (1.93)

and the solutions for F () are the familiar Legendre functions,

F () = Pl(cos ) and Ql(cos ); (1.94)

where Ql(cos ) is the Legendre function of the second kind. Some low order Legendre functions

for real x (jxj 1) are listed below.

P0(x) = 1; P1(x) = x; P2(x) =3x2 12

; P3(x) =5x3 3x

2; (1.95)

Ql(x) =1

2Pl(x) ln

1 + x

1 x Wl1(x); W1(x) = 0; W0(x) = 1; W1(x) =3

2x; (1.96)

For a general complex variable z; the denition for Ql(z) is modied as follows:

Ql(z) =1

2Pl(z) ln

z + 1

z 1 Wl1(z); W1(z) = 0; W0(z) = 1; W1(z) =3

2z; (1.97)

In the present problem, the potential should remain nite everywhere except at the ring. Since

Ql(cos ) diverges at = 0 and ; it should be discarded and we assume the following series solutions

24

for the potential separately for interior (r < a) and exterior (r > a);

(r; ) =

8>>><>>>:PlAl

ra

lPl(cos ); r < a

PlBl

ar

l+1Pl(cos ); r > a

(1.98)

where Al and Bl are constants to be determined. The potential should be continuous at r = a

because there are no double layers to create a potential jump. Therefore, Al = Bl: To determine

Al; we substitute the assumed potential into the Poissons equation,

@2

@r2+2

r

@

@r+1

r21

sin

@

@

sin

@

@

= 1

0

q

2a2(r a)(cos );

to obtain Xl

d2

dr2+2

r

d

dr l(l + 1)

r2

AlRl(r)Pl(cos ) =

1

0

q

2a2(r a)(cos ); (1.99)

where

Rl(r) =

8>>><>>>:ra

l; r < a

ar

l+1; r > a

(1.100)

If we multiply Eq. (1.99) by Pl0(cos ) sin and integrate over from = 0 to ; the summation

over l disappears because of the orthogonality of the Legendre functions,Z

0Pl(cos )Pl0(cos ) sin d =

Z 1

1Pl()Pl0()d =

2

2l + 1ll0 : (1.101)

We thus obtain

Al

d2

dr2+2

r

d

dr l(l + 1)

r2

Rl(r) =

1

0

q

2a2(r a)2l + 1

2Pl(0): (1.102)

The LHS should contain a delta function (r a) to be compatible with that in the RHS. This canbe seen as follows. The radial function Rl(r) has a discontinuity in its derivative at r = a;

dRldr

r=a+0

= l + 1a;dRldr

r=a0

= +l

a: (1.103)

Therefore, the second order derivative d2Rl=dr2 yields a delta function,

d2Rldr2

= 2l + 1a

(r a): (1.104)

25

We thus nally nd the expansion coe¢ cient Al;

Al =q

40aPl(0) =

q

40a

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

; (l = 0; 2; 4; 6; ); (1.105)

and the potential,

(r; ) =q

40a

Xl even

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

Rl(r)Pl(cos ): (1.106)

Figure 1-11: The derivative of the radial function R(r) is discontinuous at r = r0 and thus itssecond order derivative yields a delta function.

The disappearance of the odd harmonics is understandable because of the up-down symmetry

of the problem. At r a; the leading order term is the monopole potential,

l=0(r) =q

40a

a

r;

followed by the quadrupole potential,

l=2(r; ) = q

40a

1

2

ar

3P2(cos );

and so on. The dipole potential (l = 1) vanishes because of the symmetric charge distribution,

(r) = (r);p =

Zr(r)dV = 0:

For problems with axial symmetry (@=@ = 0) as in this example, knowing the potential along

the axis (z) is su¢ cient to nd the potential at arbitrary point (r; ) in the spherical coordinates.

This is because for all of the Legendre functions, Pl(1) = 1; Pl(1) = (1)l: In the case of the ring

26

charge, the axial potential can be readily found,

(z) =q

40

1pz2 + a2

: (1.107)

For jzj > a; this can be expanded as

(z) =q

40

1

jzj

1 1

2

az

2+3

8

az

4

: (1.108)

Therefore, at arbitrary point (r > a; ); the potential is

(r; ) =q

40

1

r

1 1

2

ar

2P2(cos ) +

3

8

ar

4P4(cos )

; r > a: (1.109)

For interior region (r < a); the potential becomes

(r; ) =q

40

1

a

1 1

2

ra

2P2(cos ) +

3

8

ra

4P4(cos )

; r < a: (1.110)

They agree with the potential given in Eq. (1.106).

27

Method 3: Cylindrical CoordinatesIn the cylindrical coordinates (; ; z); the Poissons equation for the potential due to a ring

charge becomes @2

@2+1

@

@+@2

@z2

(; z) = q

20

( a)a

(z): (1.111)

Since the z-coordinate extends from 1 to1; we seek a solution in the form of Fourier transform,

(; z) =1

2

Z 1

1(; k)eikzdk; (1.112)

where (; k) is the one-dimensional Fourier transform of the potential having dimensions of Vm.Since

@

@z(; z);

is Fourier transformed as

ik(; k);

and (z) as unity, the equation for (; k) becomes an ordinary di¤erential equation,d2

d2+1

d

d k2

(; k) = q

20

( a)a

: (1.113)

Elementary solutions of the di¤erential equation at 6= a;d2

d2+1

d

d k2

(; k) = 0; (1.114)

are the zero-th order modied Bessel functions,

(; k) = I0(k); K0(k); (1.115)

shown in Fig.1-12.

The Fourier potential (; k) which is continuous at = a and remains bounded everywhere

may be constructed in the following form,

(; k) =

8><>:AI0(k)K0(ka); < a

AI0(ka)K0(k); > a

(1.116)

The coe¢ cient A can be determined readily from the discontinuity in the derivative of (; k) at

= a;d

d

=a+0

= AkI0(ka)K00(ka);

d

d

=a0

= AkI 00(ka)K0(ka); (1.117)

28

x 21.510.50

5

4

3

2

1

Figure 1-12: Modied Bessel functions I0(x) (starting at 1) and K0(x) (diverging at x = 0):

from which it follows that

d2

d2

=a

= AkI0(ka)K

00(ka) I 00(ka)K0(ka)

( a) = A

a( a); (1.118)

where the Wronskian of the modied Bessel functions

I0(ka)K00(ka) I 00(ka)K0(ka) =

1

ak; (1.119)

has been exploited. From

d2

d2

=a

= Aa( a) = q

20

( a)a

; (1.120)

we nd

A =q

20; (1.121)

and the nal form of the potential is

(; z) =q

420

Z 1

1

8><>:I0(k)K0(ka)

I0(ka)K0(k)

9>=>; eikzdk; < a

> a

(1.122)

Noting that the modied Bessel functions are even with respect to the argument, this may be

further rewritten as

(; z) =q

220

Z 1

0

8><>:I0(k)K0(ka)

I0(ka)K0(k)

9>=>; cos(kz)dk; < a

> a

(1.123)

29

The convergence of the integral is rather poor since the function I0K0 decreases with k only

algebraically. An alternative solution for the potential can be found in terms of Laplace transform

rather than Fourier transform,

(; z) =

Z 1

0(; k)ekjzjdk: (1.124)

The Laplaces equation thus reduces tod2

d2+1

d

d+ k2

(; k) = 0; (1.125)

whose solution is the ordinary Bessel function J0(k): We thus assume for (; k)

(; k) = A(k)J0(k); (1.126)

where A(k) may still be a function of k: The Poissons equation becomes@2

@2+1

@

@+@2

@z2

Z 1

0A(k)J0(k)e

kjzjdk = q

20

( a)a

(z): (1.127)

Notingd2

dz2ekjzj = k2ekjzj 2k(z) (1.128)

we thus nd Z 1

0kA(k)J0(k)dk =

q

40

( a)a

: (1.129)

However, the Bessel function forms an orthogonal set according toZ 1

0kJ0(ka)J0(k)dk =

( a)a

; (1.130)

which uniquely determines the function A(k);

A(k) =q

40J0(ka): (1.131)

Therefore, the nal form of the potential is

(; z) =q

40

Z 1

0J0(ka)J0(k)e

kjzjdk: (1.132)

The convergence of this solution is much faster than the solution in Eq. (1.123) and thus more

suitable for numerical evaluation.

30

Method 4: Toroidal CoordinatesThe toroidal coordinates (; ; ) are related to the cartesian coordinates (x; y; z) through the

following transformation, 8>>>>>>>>>><>>>>>>>>>>:

x =R sinh cos

cosh cos ;

y =R sinh sin

cosh cos ;

z =R sin

cosh cos :

(1.133)

This coordinate system is one example in which the Laplace equation is not separable, that is, the

potential cannot be written as a product of independent functions, (; ; ) 6= F1()F2()F3():However, the potential is partially separable and can be sought in the form

(; ; ) =pcosh cos F1()F2()F3(): (1.134)

Figure 1-13: Cross-section of the toroidal coordinates (; ; ): ! 0 corresponds to a thin ring ofradius R:

In the toroidal coordinates, the = constant surfaces are toroids having a major radius R coth

and minor radius R= sinh : !1 degenerates to a thin ring of radius R: In the other limit, ! 0

describes a thin rod on the z-axis. The = constant surfaces are spherical bowls as illustrated in

Fig.1-13.

Assuming the partial separation for the potential in Eq. (1.134) leads to the following ordinary

31

equations for the functions F1(); F2() and F3();

1

sinh

d

d

sinh

dF1d

+

1

4 l2 m2

sinh2

F1 = 0; (1.135)

d2

d2+ l2

F2 = 0;

d2

d2+m2

F3 = 0:

Comparing Eq. (1.135) with the standard form of the di¤erential equation for the associated

Legendre function Pml (cos ); Qml (cos );

1

sin

d

d

sin

d

d

+ l(l + 1) m2

sin2

(Pml ; Q

ml ) = 0; (1.136)

we see that solutions for F1() are

F1() = Pml 1

2

(cosh ); Qml 1

2

(cosh ): (1.137)

The functions F2 and F3 are elementary,

F2() = eil; F3() = e

im; (1.138)

and the general solution for the potential may be written as

(; ; ) =pcosh cos

Xl;m

AlmPl 1

2(cosh ) +BlmQl 1

2(cosh )

eil+im: (1.139)

Of course, the potential is a real function. The solution above is an abbreviated form of a more

cumbersome expression,

(; ; ) =pcosh cos

1Xl=0

AlPl 1

2(cosh ) +BlQl 1

2(cosh )

(Cl cos l +Dl sin l)

1Xm=0

(El cosm+ Fl sinm) :

We now consider a conducting toroid with a major radius a and minor radius b: Its surface is

described by 0 = const. where

a = R coth 0; b =R

sinh 0: (1.140)

Because of axial symmetry, only the m = 0 term is present. If the toroid is at a potential V; the

potential o¤ the toroid can be written as

(; ) =pcosh cos

1Xl=0

AlPl 12(cosh ) cos l; (1.141)

32

where the Legendre function of the second kind Ql 12(cosh ) has been discarded because it diverges

at ! 0 which corresponds to the zaxis. (The potential along the zaxis should be bounded.)The expansion coe¢ cients Al can be determined from the boundary condition, = V at = 0;

V =pcosh 0 cos

1Xl=0

AlPl 12(cosh 0) cos l; (1.142)

or 1Xl=0

AlPl 12(cosh 0) cos l =

Vpcosh 0 cos

:

Multiplying both sides by cos l0 and integrating over from 0 to ; we obtain

Al = VQl 1

2(cosh 0)

Pl 12(cosh 0)

p2

l; (1.143)

where

0 = 1; l = 2 (l 1);

and the following integral representation has been exploited,Z

0

cos lpx cos

d =p2Ql 1

2(x): (1.144)

Then, the nal form of the potential is

(; ) =

p2V

pcosh cos

1Xl=0

Ql 12(cosh 0)

Pl 12(cosh 0)

Pl 12(cosh ) cos(l)l: (1.145)

The capacitance of a conducting torus can be found from the behavior of the potential at a

large distance from the torus which should be in the form of monopole potential,

(r !1) = q

40r: (1.146)

At a large distance from the torus r R; both and approach zero,

r2 = x2 + y2 + z2 =R2(sinh2 + sin2 )

(cosh cos )2 ! 4R2

2 + 2; (1.147)

pcosh cos '

p2 + 2p2

'p2R

r; (1.148)

Pl 12(cosh ) ' 1: (1.149)

33

Therefore, the asymptotic potential is

(r R) =2V R

r

1Xl=0

Ql 12(cosh 0)

Pl 12(cosh 0)

=q

40r; (1.150)

from which the capacitance of the torus can be found,

C q

V= 80R

1Xl=0

Ql 12(cosh 0)

Pl 12(cosh 0)

l = 80a tanh 0

1Xl=0

Ql 12(cosh 0)

Pl 12(cosh 0)

l; (1.151)

where a is the major radius of the torus. The function Ql 12(cosh 0) can be evaluated from Eq.

(1.144) and Pl 12(cosh 0) from

Pl 12(cosh 0) =

1

Z

0

d

(cosh 0 + sinh 0 cos )l+ 1

2

: (1.152)

Example 4 Capacitance of a Fat Torus

We wish to nd the capacitance of a conducting torus having a major radius of a = 50 cm and

minor radius of b = 10 cm. The torus is dened by cosh 0 = 5:0 and R = a tanh 0 = 49 cm. For

cosh 0 = 5:0; the Legendre functions numerically evaluated are:

l Ql 12(5) Pl 1

2(5)

Ql 12(5)

Pl 12(5)

0 1:00108 0:74575 1:34234

1 0:05063 2:03557 0:02487

2 0:00384 13:32184 0:00029

The ratio Ql 12(5)=Pl 1

2(5) rapidly converges as l increases and it su¢ ces to truncate the series at

l = 2: The capacitance is

C = 8"0R(1:3423 + 2 0:0249 + 2 0:0003 + ) ' 80R 1:393 = 80a 1:36:

The approximate formula worked out earlier gives

C ' 420a

ln

8a

b

= 80a 1:34;which is in reasonable agreement with the numerical result.

34

1.7 Spherical Multipole Expansion of the Scalar Potential

The potential due to a prescribed charge distribution,

(r) =1

4"0

Z(r0)

jr r0jdV0;

can be expanded in terms of the general spherical harmonics as follows. For this purpose, it is

su¢ cient to expand the Greens function,

G(r; r0) =1

4

1

jr r0j ; (1.153)

which satises the singular Poissons equation,

r2G = (r r0): (1.154)

The Greens function G satises Laplace equation except at r = r0;

r2G = 0; r 6= r0; (1.155)

and we rst seek elementary solutions for Laplace equation. Assuming that the function G(r; ; )

is separable in the form

G(r; ; ) = R(r)F1()F2(); (1.156)

and substituting this into Laplace equation@2

@r2+1

r

@

@r+1

r21

sin

@

@

sin

@

@

+

1

r2 sin2

@2

@2

G(r; ; ) = 0; (1.157)

we obtain three ordinary equations,d2

dr2+1

r

d

dr l(l + 1)

r2

R(r) = 0; (1.158)

1

sin

d

d

sin

d

d

+ l(l + 1) m2

sin2

F1() = 0; (1.159)

d2

d2+m2

F2() = 0; (1.160)

where l(l+1) and m2 are separation constants. Solutions of the radial function R(r) are as before,

R(r) = rl;1

rl+1:

In free space, the Greens function must be periodic with respect to the azimuthal angle : Therefore,

F2() = eim with m being an integer.

35

Equation (1.159) is known as the modied Legendre equation and its solutions are

F1() = Pml (cos ); Q

ml (cos ); (1.161)

where

Pml (x) = (1 x2)m=2dm

dxmPl(x); (1.162)

Qml (x) = (1 x2)m=2dm

dxmQl(x): (1.163)

That Pml (cos ) given in Eq. (1.162) satises the modied Legendre equation can be seen as follows.

The ordinary Legendre function Pl(cos ) = Pl() satises

d

d

(1 2)dPl()

d

+ l(l + 1)Pl() = 0:

Di¤erentiating m times yields

(1 2)dm+2Pldm+2

2(m+ 1)dm+1Pldm+1

+ [l(l + 1)m(m+ 1)] dmPldm

= 0:

Let us assume

F1() = (1 2)m=2f();

and substitute this into Eq. (1.159). f() satises the same equation as dmPldm ;

(1 2)d2f

d2 2(m+ 1) df

d+ [l(l + 1)m(m+ 1)] f = 0:

Therefore, f() = dmPldm :

Since the Legendre function Pl(x) is a polynomial of order l; the azimuthal mode number m

is limited in the range 0 m l: For later use, we combine the functions F1() and F2() and

introduce the spherical harmonic function dened by

Ylm(; ) = const. Pml (cos )e

im; (1.164)

where the constant is chosen from the following normalization,ZYlm(; )Y

lm(; )d = 1: (1.165)

Noting Z

0[Pml (cos )]

2 sin d =2

2l + 1

(l +m)!

(l m)! ; (1.166)

we nd

const. =

s2l + 1

4

(l m)!(l +m)!

;

36

Ylm =

s2l + 1

4

(l m)!(l +m)!

Pml (cos )eim: (1.167)

For historical reasons, it is customary to write Ylm(; ) in the form

Ylm(; ) = (1)ms2l + 1

4

(l m)!(l +m)!

Pml (cos )eim; for 0 m l;

Yl;m(; ) = (1)mY lm(; ) ; for l m 0;

or for arbitrary m; positive or negative,

Ylm(; ) = (1)(m+jmj)=2s2l + 1

4

(l jmj)!(l + jmj)!P

jmjl (cos )eim: (1.168)

Some low order forms of Ylm(; ) are:

l = 0 m = 0 Y00 =1p4

l = 1 m = 0 Y10 =

r3

4cos

m = 1 Y1;1 = r3

8sin ei

l = 2 m = 0 Y20 =

r5

4

3 cos2 12

m = 1 Y2;1 = r15

8sin cos ei

m = 2 Y2;2 =1

4

r15

2sin2 e2i

Having found the general solutions of the Greens function, we now assume the following ex-

pansion for G;

G(r; r0) =Xlm

Almgl(r; r0)Ylm(; );

where the radial function gl(r; r0) is

gl(r; r0) =

8>>>><>>>>:rl

r0l+1; r < r0;

r0l

rl+1; r > r0:

37

In this form, the function gl(r; r0) remains bounded everywhere. The expansion coe¢ cient Alm can

be determined by multiplying the both sides of

r2G = (r r0) = (r r0)

rr0(cos cos 0)( 0); (1.169)

by Y l0m0(; ) and integrating the result over the entire solid angle,

Almd2

dr2gl(r; r

0) = (r r0)

rr0Y lm(

0; 0);

where orthogonality of Ylm(; );ZYlm(; )Y

l0m0(; )d = ll0mm0 ; (1.170)

is exploited to remove summation over l andm: Since the radial function gl(r; r0) has a discontinuity

in its derivative at r = r0; it follows that

d2

dr2gl(r; r

0) = (2l + 1)(r r0)

r2: (1.171)

Therefore,

Alm =1

2l + 1Y lm(

0; 0); (1.172)

and the desired spherical harmonic expansion of the Greens function is given by

G(r; r0) =1

4 jr r0j =Xlm

1

2l + 1gl(r; r

0)Ylm(; )Ylm(

0; 0): (1.173)

For a given charge distribution (r); this allows evaluation of the potential in the form of spherical

harmonics,

(r) =1

4"0

Z(r0)

jr r0jdV0

=1

4"0

Xlm

4

2l + 1

1

rl+1Ylm(; )

Zr0lY lm(

0; 0)(r0; 0; 0)dV 0

=1

4"0

Xlm

4

2l + 1

1

rl+1Ylm(; )qlm; in the region r > r

0; (1.174)

where

qlm Zr0lY lm(

0; 0)(r0; 0; 0)dV 0; (1.175)

is the electric multipole moment of order (l; m):

Example 5 Planar Quadrupole

A planar quadrupole consists of charges +q and q alternatively placed at each corner of a

38

square of side a as shown in Fig.1-14. The charge density may be written as

(r) = q

(r) (r a)

a2(cos ) () +

(r p2a)

2a2(cos )

4

(r a)

a2(cos )

2

!:

The potential in the far eld region r a is of quadrupole nature and given by

(r; ; ) =1

"0

1

5r3(Y2;2q2;2 + Y2;2q2;2) ;

where

q2;2 =

Zr2Y 2;2(; )(r)dV =

1

2

r15

2a2qei=2:

The potential reduces to

(r; ; ) ' 3

8"0

a2

r3sin2 sin(2); r a:

The reader should check that this is consistent with the direct sum of four potentials in the limit

r a:

Figure 1-14: Planar quadrupole in the x y plane.

1.8 Collection of Dipoles, Dielectric Properties

Dielectric properties of material media originate from dipole moments carried by atoms and mole-

cules. Some molecules carry permanent dipole moments. For example, the water molecule has a

dipole moment of about 6 1030 Cm due to deviation of the center of electron cloud from the

center of proton charges. When water is placed in an external electric eld, the dipoles tend to be

39

aligned in the direction of the electric eld and a resultant electric eld in water becomes smaller

than the unperturbed external eld by a factor "="0; where " is the permittivity of water. Even a

material composed of molecules having no permanent dipole moment exhibits dielectric property.

For example, a dipole moment is induced in a hydrogen atom placed in an electric eld through

perturbation in the electron cloud distribution which is spatially symmetric in the absence of ex-

ternal electric eld. In this section, the potential and electric eld due to a collection of dipole

moments will be analyzed.

The potential due to a single dipole p located at r0 is

(r) =1

40

(r r0) pjr r0j3 : (1.176)

Consider a continuous distribution of many dipoles. It is convenient to introduce a dipole moment

density P =np (Cm/m3 = C/m2) where n is the number density of dipoles. An incremental

potential due to an incremental pointdipole dp = P(r0)dV 0 is

d =1

40

(r r0) P(r0)jr r0j3 dV 0: (1.177)

Integration of this yields

(r) =1

40

Z(r r0) P(r0)jr r0j3 dV 0: (1.178)

Sincer r0jr r0j3 = r

0

1

jr r0j

; (1.179)

where r0 means di¤erentiation with respect to r0; the integral in Eq. (1.178) can be rewritten asZ(r r0) P(r0)jr r0j3 dV 0 =

Zr0

P(r0)

jr r0j

dV 0

Z r0 P (r0)jr r0j dV

0:

The rst term in the right can be converted into a surface integral through the Gauss theorem,ZVr0

P(r0)

jr r0j

dV 0 =

IS

P(r0)

jr r0j dS0;

which vanishes on a closed surface with innite extent on which all sources should be absent.

Therefore, the potential due to distributed dipole moments is

dipole(r) = 1

40

Z r0 P(r0)jr r0j dV

0: (1.180)

Comparing with the standard form of the potential,

(r) =1

4"0

Z(r0)

jr r0jdV0;

40

we see that

e¤ = r P;

can be regarded as an e¤ective charge density. To distinguish it from the free charge density,

e¤ dened above is called bound charge density. In general, e¤ cannot be controlled by external

means. In a dielectric body, the bound charge appears as a surface charge density. Adding the

monopole potential due to a free charge density free(r); we nd the total potential

(r) =1

40

Zfree(r

0)

jr r0j dV0 1

40

Z r0 P(r0)jr r0j dV

0: (1.181)

Noting

r2 1

jr r0j = 4(r r0);

we nd

r E = r2 = free0

r P0

: (1.182)

In this form too, it is evident that r P can be regarded as an e¤ective charge density. Eq.

(1.181) can be rearranged as

r (0E+P) = free: (1.183)

Introducing a new vector D (displacement vector) by

D = 0E+P; (1.184)

we write Eq. (1.183) as

r D = free; (1.185)

which is equivalent to the original Maxwells equation

r E =all0; (1.186)

provided that the total charge density all consists of free charges and dipole charges and that

higher order charge distributions (quadrupole, octupole, etc.) are ignorable. The vector D was

named the displacement vector by Maxwell. Its fundamental importance in electrodynamics will be

appreciated later in time varying elds because it was with this displacement vector that Maxwell

was able to predict propagation of electromagnetic waves in vacuum. As briey discussed in Sec-

tion 2, Maxwells equations would not be consistent with the charge conservation principle if the

displacement current,@("E)

@t=@D

@t;

were absent.

In usual linear insulators, both D and P are proportional to the local electric eld E which

41

allows us to introduce an e¤ective permittivity ";

D = "0E+P ="E: (1.187)

In some solid and liquid crystals, the permittivity takes a tensor form,

D = E or Di = ijEj ; (1.188)

because of anisotropic polarizability. Double refraction phenomenon in optics already known in

the 17th century is due to the tensorial nature of permittivity in some crystals as we will study in

Chapter 5.

Figure 1-15: In an electric eld, the hydrogen atom exhibits a dipole moment due to the displace-ment of the center of electron cloud from the proton.

The origin of eld induced dipole moment may be seen qualitatively for the case of hydrogen

atom as follows. In an electric eld, the distribution of the electron cloud around the proton

becomes asymmetric because of the electric force acting on the electron. The center of electron

cloud is displaced from the proton by x given by

m!20x = eE;

where !0 is the frequency of bound harmonic motion of the electron and m is the electron mass.

(~!0 is of the order of the ionization potential energy.) Then, the dipole moment induced by theelectric eld is

p = ex =e2

m!20E:

If the atomic number density is n; the dipole moment density P is

P = np =ne2

m!20E; (1.189)

42

and the displacement vector is given by

D = "0E+P

= "0

1 +

ne2

"0m!20

E = "E: (1.190)

The permittivity can therefore be dened by

" = "0

1 +

!2p!20

!; (1.191)

where

!p =

sne2

"0m; (1.192)

is an e¤ective plasma frequency. (The plasma frequency pertains to free electrons in a plasma. !pintroduced here is so called only for dimensional convenience.)

In the incremental contribution to the permittivity normalized by "0,

!2p!20= n

e2

m"0!20;

the quantitye2

4"0m!20;

is of the order of r30 where r0 is the Bohr radius. This can be seen from the force balance for the

electron,

mr0!20 =

e2

4"0r20;

which indeed givese2

4"0m!20= r30:

The quantity

= 4"0r30 =

e2

m!20; (1.193)

is called the atomic polarizability. A more rigorous quantum mechanical calculation yields for this

quantity

=9

2 4"0r30; (1.194)

indicating a substantial correction (by a factor 4.5) to the classical picture based on the assumption

of rigid shift of the electron cloud. For molecualr hydrogen H2; the polarizability at T = 273 K is

approximately given by

' 5:4 4"0r30:

43

Satisfactory quantum mechanical calculation was performed by Kolos and Wolniewicz relatively

recently (1967). This is in a fair agreement with the permittivity experimentally measured in the

standard conidition 20 C, 1 atmospheric pressure,

" '1 + 2:7 104

"0:

The atomic polarizability and the macroscopic relative permittivity "r are related through a

simple relationship known as the Clausius-Mossoti equation,

n

"0= 3

"r 1"r + 2

; (1.195)

where n is the number density of atoms. This may qualitatively be seen as follows. A dipole

moment induced by an atom,

p = ex =e2

m!20E = Eext; (1.196)

produces an electric eld

E0 = p

4"0r3; (1.197)

at a distance r: If the number density of atoms is n; the average inter-atom distance R can be found

from4

3R3 =

1

n: (1.198)

Then,

E0 =

4"0R3Eext; (1.199)

and the total electric eld is

E =

1

4"0R3

Eext (1.200)

=

1 n

3"0

Eext: (1.201)

Therefore, using this total eld in calculation of the polarization eld P; we nd

P =n

1 n

3"0

E =n="0

1 n

3"0

"0E; (1.202)

which denes the relative permittivity "r as

"r 1 =n="0

1 n

3"0

: (1.203)

Solving for n="0; we obtain Eq.(1.195). Note that the relative permittivity is an easily measurable

quantity and it reects, somewhat surprisingly, microscopic atomic polarizability through a simple

44

relationship.

In an oscillating electric eld, the displacement x is to be found from the equation of motion,

m

@2

@t2+ !20

x = eE0ei!t; (1.204)

which yields

x(t) =eE0e

i!t

!2 !20: (1.205)

A resultant permittivity is

"(!) = "0

1

!2p!2 !20

!: (1.206)

This exhibits a resonance at the frequency !0: The resonance plays an important role in calculation

of energy loss of charged particles moving in a dielectric medium as will be shown in Chapter 8.

Most molecules have permanent electric dipole moments. In the absence of electric elds,

dipoles are randomly oriented due to thermal agitation. When an electric eld is present, each

dipole acquires a potential energy,

U = p E = pE cos ; (1.207)

where is the angle between the dipole moment p and electric eld E: Dipoles are randomly

oriented but should obey the Boltzmann distribution in thermal equilibrium,

n(cos ) = n0 exp

U

kBT

= n0 exp

pE cos

kBT

; (1.208)

where n(cos )=n0 is the probability of dipole orientation in direction, n0 is the number density

of the dipoles and kB = 1:38 1023 J/K is the Boltzmann constant. The average dipole numberdensity oriented in the direction of the electric eld can be calculated from

n =

Rn(cos ) cos dRexp

pE cos kBT

d: (1.209)

In practice, the potential energy is much smaller than the thermal energy pE kBT: Therefore,

the exponential function can be approximated by

exp

pE cos

kBT

' 1 + pE cos

kBT; (1.210)

and we obtain

n ' n0p

3kBTE: (1.211)

45

Resultant dipole moment density is

P = np =n0p

2

3kBTE: (1.212)

This denes a permittivity in a medium consisting of molecules having a permanent dipole moment

p;

" = "0

1 +

n0p2

3"0kBT

: (1.213)

Note that the correction term

" =n0p

2

3kBT; (1.214)

is inversely proportional to the temperature. This property can be exploited in separating con-

tributions from atomic and molecular polarizability in a given medium. In liquids and solids, the

additional permittivity given above can be comparable with "0 primarily because of the large num-

ber density n0: For example, a water molecule has a dipole moment of p = 6 1030 C m. At

room temperature, the additional permittivity due to molecular polarizability of water is approxi-

mately " ' 11"0: The measured permittivity of water at room temperature is 81"0: The atomic

polarizability is thus dominant.

1.9 Boundary Conditions for E and D

In electrostatics, the electric eld E obeys the Maxwells equations,

r E =

"0;

and

rE = 0:

In problems involving dielectrics, it is more convenient to introduce the displacement vector D,

r D = free; (1.215)

where free is the free charge density that can be controlled by external means.

Let us consider a boundary of two dielectrics having permittivities "1 and "2; respectively.

Integration of r D = free over the volume of a pancake dV = dS dn (dn is the thickness of thepancake in the direction perpendicular to the surface dS) on the boundary yieldsZ

r DdV =ID dS =

ZfreedV: (1.216)

In the limit of innitesimally thin pancake, this reduces to

(Dn1 Dn2)dS = freedndS = freedS; (1.217)

46

Figure 1-16: Gausslaw applied to a thin pancake at a boundary between two dielectrics.

where free = freedn is the free surface charge density residing on the boundary. Note that the

volume charge density in the presence of surface charge can be written as

= (n n0): (1.218)

General boundary condition for the normal component of the displacement vector is thus given by

D1n D2n = free: (1.219)

In the absence of free surface charge, the normal component of the displacement vector should be

continuous,

Dn1 = Dn2; no free charge. (1.220)

Figure 1-17:HE dl = 0 applied to a rectangle at the boundary of two dielectrics. The tangential

component of the electric eld Et is continuous. This holds in general for time varying elds aswell.

47

The equation rE = 0 demands that the tangential component of electric eld be continuousacross a boundary of dielectrics. This can be seen by integrating rE = 0 over a small rectangulararea on the boundary, Z

rE dS =IE dl = 0; (1.221)

which yields

(E1 E2) dl = 0; E1t = E2t: (1.222)

In fact, the continuity of the tangential component of the electric eld holds for general time varying

case,

rE = @B@t; (1.223)

for the area integral ZSB dS;

vanishes in the limit of innitesimally thin rectangle, S ! 0: In contrast to charge and current

densities, a singular magnetic eld involving a delta function is not physically realizable because

then the magnetic energy simply diverges. Note that the square of a delta function is not integrable,Z2(x)dx = (0) =1:

In a conductor, there can be no static electric elds and E = 0 must hold inside conductors:

Therefore, at a conductor surface, the tangential component of the electric eld should vanish and

the electric eld lines fall normal to the surface. The normal component of the electric eld and

the surface charge density are related through

En =

"0; at conductor surface. (1.224)

The potential of a conducting body is constant. The surface electric eld depends on the amount

of charge carried by the body and also curvature of the surface. A trivial case is the electric eld

at the surface of charged conducting sphere of radius a;

E =1

4"0

q

a2:

In general, the eld increases as the curvature radius decreases. The electric eld at the tip of

needle can be strong enough to allow emission of electrons as in tunneling electron microscopes. In

power engineering, conductor surfaces should be made smooth and round as much as possible to

avoid breakdown.

Example 6 Dielectric Sphere in an Electric Field

Let us consider an uncharged dielectric sphere having a uniform permittivity " placed in an

uniform external electric eld E0 = E0ez: The sphere perturbs the external eld because a bound

48

surface charge = r P will be induced on the sphere. The interior and exterior potentials maybe expanded in the spherical harmonics,

(r; ) = 0 +Xl

Al

ra

lPl(cos ); r < a (interior); (1.225)

(r; ) = 0 +Xl

Bl

ar

l+1Pl(cos ); r > a (exterior); (1.226)

where

0(r; ) = E0z = E0r cos ; (1.227)

is the potential associated with the external electric eld E0 = E0ez. Since there is no double layer

to cause potential jump at the sphere surface, the potential is continuous at the surface r = a;

from which it follows Al = Bl: This also follows from the continuity of the tangential component

( component) of the electric eld,

E = 1

r

@

@: (1.228)

The continuity of the normal (radial) component of the displacement vector requires

"

E0 cos +

Xl

All

aPl(cos )

!= "0

E0 cos +

Xl

All + 1

aPl(cos )

!: (1.229)

Since cos = P1(cos ); only the l = 1 terms are non-vanishing and we readily nd

A1 =" "0"+ 2"0

aE0: (1.230)

Then the solutions for the potentials are

(r; ) =

8>>>><>>>>:0(r; ) +

" "0"+ 2"0

E0r cos ; r < a

0(r; ) +" "0"+ 2"0

E0a3

r2cos ; r > a

(1.231)

The electric eld in the dielectric sphere is uniform and given by

Eiz =3"0

"+ 2"0E0; r < a: (1.232)

This is smaller than the external eld E0 since " > "0: The interior displacement vector is

Diz = "Eiz =3"

"+ 2"0"0E0; (1.233)

49

which is larger than D0 = "0E0: The perturbed exterior potential,

" "0"+ 2"0

a3

r2E0 cos ; (1.234)

is of dipole form with an e¤ective dipole moment

pz = 4"0" "0"+ 2"0

a3E0; (1.235)

located at the center of the sphere.

In the limit of " "0; the potential reduces to

(r; ) =

8>><>>:0; r < a

0(r; ) +a3

r2E0 cos ; r > a

(1.236)

which describes the potential when a conducting sphere is placed in an external electric eld. A

sphere of innite permittivity is mathematically identical to a conducting sphere.

The polarization vector P can be found from

Dz = "0Ez + Pz; (1.237)

Pz =3(" "0)"+ 2"0

"0E0; r < a: (1.238)

Since P = 0 outside the sphere, the divergence of the polarization vector yields the bound charge

density induced on the sphere surface,

e¤ = r P

=2(" "0)"+ 2"0

"0E0 cos (r a): (1.239)

The total dipole moment carried by the sphere is

pz =4a3

3Pz

=4(" "0)a3"+ 2"0

"0E0;

which agrees with that deduced from the exterior potential.

50

1.10 Electric Force

The energy density associated with an electric eld

1

2"0E

2; (J/m3) (1.240)

manifests itself as either pressure or tensile stress depending on the direction of the force relative

to the electric eld. A charge density placed in an electric eld E experiences a force per unit

volume,

f = E

= "0(r E)E= r ("0EE) "0E rE: (1.241)

In electrostatics, r E = 0; and thus the curvature E rE and gradient ErE =1

2rE2 of the

electric eld are identical,

r(E E) = 2ErE+ 2E rErE2 = 2E rE: (1.242)

Therefore,

f = r "0EE

1

2"0E

21

; (1.243)

where 1 is the unit sensor. The tensor

Tij = "0EiEj 1

2"0E

2ij ; (1.244)

is called the Maxwells stress tensor associated with the electric eld. For a linear dielectric, "0may be replaced with its permittivity ": The force vector is

f = r T; or fi =@

@xjTij : (1.245)

Let us consider a simple case: an electric eld in the z-direction, E = Ezez: The tensor Tij is

diagonal with the following components,

T =

0BBBB@12"0E

2z 0 0

0 12"0E

2z 0

0 0 +1

2"0E

2z

1CCCCA : (1.246)

51

The force in the directions perpendicular to the eld are

fx = @

@x

1

2"0E

2z

; (1.247)

fy = @

@y

1

2"0E

2z

; (1.248)

which appear as a pressure acting from a higher energy density region to lower energy density

region. The force in the direction of the electric eld is

fz = +@

@z

1

2"0E

2z

; (1.249)

which appears as tension acting from a lower energy density region to higher energy density region.

The force to act on a volume V can be found from the integral

F =

ZVfdV =

IST dS

=

IS

"0E(E n)

1

2"0E

2n

dS; (1.250)

where n is the unit normal vector on the closed surface S; dS = ndS: For a charged conducting

sphere, the electric eld is radially outward everywhere. At the surface, the force per unit area is

1

2"0E

2r ; (N/m

2);

acting radially outward. Since E = 0 inside a conducting sphere, the force acts from lower energy

density region to higher energy density region as expected of tensile stress. Of course, no net force

acts on the sphere.

Example 7 Dielectric Hemispheres in an External Electric Field

As a less trivial example, we consider a dielectric sphere consisting of identical hemispheres with

a narrow gap at an equatorial plane. It is placed in an electric eld with the gap plane perpendicular

to the eld. We wish to nd a force to act between the hemisphere. The force, if any, should be

axial in the direction perpendicular to the gap. The z-component of the integral,

F =

I "E(E n) 1

2"E2n

dS

is

Fz =

I "EzEn

1

2"E2nz

dS; (1.251)

where on the spherical surface of one of the hemispheres,

Ez = E03

"+ 2"0

" cos2 + "0 sin

2 ; (1.252)

52

Er = E03"

"+ 2"0cos ; (1.253)

E = E03"0

"+ 2"0sin ; (1.254)

E2 = E2r + E2

= E209

("+ 2"0)2"2 cos2 + "20 sin

2 : (1.255)

In the gap, the eld is uniform,

Figure 1-18: A dielectric sphere with a negligible gap at an equator placed in an electric eld normalto the gap surface. The hemispheres attract each other. The D-led lines (not E eld lines) shownare relevant to the preceding Example as well.

Egapz ="

"0Eiz =

3"

"+ 2"0E0; (1.256)

where

Eiz =3"0

"+ 2"0E0; (1.257)

is the internal electric eld in the dielectric sphere given in Eq. (1.232). The integral over the

hemispherical surface is

Fz1 = 2a2Z =2

0

"0EzEr

1

2"0E

2 cos

sin d

= 2a29"0E

20

("+ 2"0)2

Z =2

0

"2 cos2 + ""0 sin

2 12("2 cos2 + "0 sin

2 )

cos sin d

= a29"0E

20

("+ 2"0)21

4("2 + 2""0 "20): (1.258)

53

This force is repelling. The contribution from the at surface in the gap is

Fz2 = a29"0E

20

("+ 2"0)2"2

2; (1.259)

and the net force is

Fz = a29(" "0)24("+ 2"0)2

"0E20 : (1.260)

The minus sign indicates an attractive force between the hemispheres. In the limit " "0; the

force becomes

Fz = 9

4a2"0E

20 :

This corresponds to the case of solid or closed conducting hemispheres.

1.11 Force in Capacitor

For two electrode systems, the potential di¤erence between the electrodes V and the charges Qresiding on each electrode are related through the capacitance C;

Q = CV: (1.261)

An incremental energy required to increase the charge by dQ is

dU = V dQ =1

CQdQ = CV dV: (1.262)

Therefore the amount of energy stored in a capacitor is

U =1

2

Q2

C=1

2CV 2: (1.263)

This is applicable for a single electrode as well if the second electrode is at innity (self-capacitance).

For example, a conducting sphere of radius a has a self-capacitance of

C = 4"0a: (1.264)

If it carries a charge Q; the amount of potential energy associated with it is

U =Q2

8"0a: (1.265)

Of course, the energy is stored in the space surrounding the sphere, and the energy can alternatively

be calculated from the integral,

U =

Z 1

a

1

2"0E

2r4r

2dr =Q2

8"0a; (1.266)

54

where

Er =Q

4"0r2; r > a (1.267)

is the electric eld.

If a capacitance is described as a function of geometrical factor ; such as electrode separation

distance and electrode size, the electric force tends to act in such a way to increase the capacitance,

F =1

2V 2@C()

@; if the voltage is held constant, (1.268)

F = 1

2Q2

@

@

1

C()

; if the charge is held constant. (1.269)

For example, the capacitance of parallel plates capacitor consisting of circular disks is

C(d; a) = "0a2

d; (1.270)

where a is the disk radius and d is the separation distance. The disks evidently attract each other

so as to reduce the distance d or increase the capacitance with a force

Fd = 1

2"0a2

d2V 2 = 1

2

Q2

"0a2: (1.271)

The radial force is

Fa = "0a

dV 2 =

d

"0a3Q2; (1.272)

which acts so as to increase the radius of the disks. Note that when the charge is xed, the force

acts to reduce the energy stored in the capacitor, while when the voltage is xed, the force acts

to increase the energy. A power supply is a large reservoir of energy and a capacitor connected to

a power supply (the case of xed voltage) is not a closed system. The earlier statement based on

the variational principle that an electrostatic equilibrium is the minimum energy state of course

pertains to closed systems.

55

Problems

1.1 Verify that for each of the expressions of the three dimensional delta functions,

(r r0) = (x x0)(y y0)(z z0);

(r r0) = ( 0)0

( 0)(z z0);

(r r0) = (r r0)rr0

(cos cos 0)( 0);

the volume integral is unity, Z(r r0)dV = 1:

1.2 The electron cloud in a hydrogen atom is described by the charge density distribution

(r) = e

a3exp

2ra

;

where a = 5:3 1011 m is the Bohr radius. Show that the potential energy of a hydrogen

atom is

U = 1

4"0

e2

a:

1.3 A planar quadrupole in the x y plane consists of four charges, +q and q alternativelyplaced at the corners of a square of side a: What is the potential energy of the quadrupole?

1.4 Four charges e; e; e and e are placed at the corners of a tetrahedron having side a:Findthe electric dipole and quadrupole moments.

1.5 Show that for l = 1;1X

m=1rY1;m(; )q1;m =

3

4r p;

56

and a resultant dipole potential is consistent with the direct expansion of the potential,

dipole =1

4"0

r pr3:

1.6 An octupole consists of eight charges (four +q and four q) alternatively placed at the cornersof a cube of side a: Determine the far eld potential (r; ; ) at r a and also the potential

energy of the octupole.

1.7 Equal octants on a spherical surface of radius a are maintained alternatively at potentials V

and V: Determine the lowest order far eld potential (r; ; ) at r a: Useful expansion

is

(r) =Xlm

ar

l+1Ylm(; )

Is(

0; 0)Y lm(0; 0)d0; r > a;

where s (; ) is the surface potental and d0 = sin 0d0d0:

1.8 The potential due to a long line charge (C/m) is

() =

2"0ln + constant,

where is the distance from the line charge. Verify this from the basic formula,

(r) =1

4"0

Z(r0)

jr r0jdV0

=1

4"0

Z 1

1

p2 + z02

dz0:

Then, show that the capacitance per unit length of a parallel-wire transmission line with wire

separation distance d and common wire radii a ( d) is approximately given by

C

l' "0

ln

d

a

; (F/m).57

What is the inductance per unit length of the transmission line? (Hint: The product

C

l

L

l;

is constant and equal to "00: For a coaxial cable lled with insulating material having

permittivity "; the capacitance isC

l=

2"

ln(a=b);

and the product ClLl is equal to "0:)

1.9 An insulating circular disk of radius a carries a total charge q uniformly distributed over

its area a2: By rst nding the potential on the axis of the disk, (z); generalize it to a

potential at arbitrary position (r; ) in terms of the spherical harmonic functions Pl(cos ):

(The case of charged conducting disk will be analyzed in Chapter 2.) Check whether the

Poissons equation,

r2 = "0;

is satised by your solution. (It should be.)

1.10 Solve the preceding problem using the cylindrical coordinates (; z).

1.11 The Legendre polynomial Pl(x) can be generated from

Pl(x) =1

2ll!

dl

dxl(x2 1)l; (Rodriguesformula).

Using this repeatedly, verify the orthogonality of Legendre functions,Z 1

1Pl(x)Pl0(x)dx =

2

2l + 1ll0 :

1.12 Evaluate numerically the capacitance of a conducting torus having a major radius of 10 cm

and minor radius of 3 cm.

1.13 Two concentric circular rings of radii a and b in the same plane carry charges q and q;respectively. Show that the far eld potential is of quadrupole nature. What is the potential

energy of the system?

1.14 Two coaxial conductor rings of radii a and b and axial separation distance c carry charges q

and q: What is the dominant far eld potential? Find the mutual capacitance.

1.15 The permittivity of a molecular hydrogen (H2) gas under the standard condition, 0 C, one

atmospheric pressure, is

" = "0 +" =1 + 2:7 104

"0:

58

When " is written in the form

" = const:!2p!20"0 = const:

ne2

me!20;

where n is the molecule density, me = 9:1 1031 kg is the electron mass, and !0 is thefrequency of bound harmonic motion of electron ~!0 = 27:2 eV, what should the constant

be? For atomic hydrogen, the constant is 4.5 as predicted by quantum mechanics.

1.16 Two dipole moments p1 and p2 are a distance r apart. Show that the potential energy of the

two dipole system is given by

U =1

4"0

1

r3

p1 p2 3

(p1 r)(p2 r)r2

:

1.17 Two parallel charge sheets of opposite polarity (C/m2) separated by a small distance form a double layer. (a) Show that the potential jump across the double layer is

=

"0=

";

where = (C m1) is called the moment of double layer. (b) A circular double layer of

radius a and moment is on the x y plane. Find the potential (r; ):

1.18 Concentric spherical capacitor of radii a (inner) and b (outer) is lled with a nonuniform

dielectric material whose permittivity depends on the radius r; "(r): Determine "(r) if the

radial electric eld between the electrodes is to be constant. Assume that the permittivity at

r = b is "0:

1.19 The problem of ring charge can alternatively be solved by the method of Laplace transform

in the cylindrical coordinates as shown in Chapter 1,

(; z) =q

4"0

Z 1

0J0(ka)J0(k)e

kjzjdk:

Show that the solution above satises the Poissons equation,@2

@2+1

@

@r+@2

@z2

(; z) = q

2"0a( a)(z):

1.20 A conducting sphere of radius a is placed in a uniform electric eld E0 = E0ez whose potential

is 0 = E0z: The dipole moment induced on the sphere is

pz = 4"0a3E0;

59

which produces a potential

0 =1

4"0

pzr2cos :

(a) The exterior electric eld is

E(r; ) = r0 r0

= E0ez +pz

4"0r3(2 cos er + sin e) ; r > a:

Find a change in the total electric energy and interpret your result. Note that E = 0 in

the sphere.

(b) A dipole p in an electric eld E has a potential energy

U = p E = 4"0a3E20 :

Recover this result from the direct integral,

U =1

2

I0dS;

over the sphere surface. Note that at a conductor surface, = "0Er(r = a):

1.21 A charge q1 is placed at a distance z = a from a dielectric plate having a permittivity ":

Another charge q2 is at the mirror position z = a: Find the force on each charge. As youwill nd, the forces are not equal in magnitude. Explain.

Hint: If a charge q is placed at a distance z = a from the surface of a dielectric plate, the

potential in the air region is

air =1

4"0

q

jr aj " "0"+ "0

q

jr+ aj

; z > 0

60

and that in the dielectric is

dielectric =1

4"

2"

"+ "0

q

jr aj ; z < 0;

where a = aez:

1.22 The permittivity of an unmagnetized plasma is

" (!) = "0

1

!2pe!2

!;

where !pe =pne2=me"0 is the plasma frequency. Show that a spherical plasma exhibits a

dipole (l = 1) ocsillation at a frequency

! =!pep3:

What are the frequencies of higher multipole modes?

61

Chapter 2

Electrostatics II. Potential Boundary

Value Problems

2.1 Introduction

In Chapter 1, a general formulation was developed to find the scalar potential Φ(r) and consequent

electric field E = −∇Φ for a given static charge distribution ρ(r). In a system involving conductor

electrodes, often the potential Φ is specified on electrode surfaces and one is asked to find the po-

tential in the space off the electrodes. Such problems are called potential boundary value problems.

In this case, the surface charge distribution on the electrodes is unknown and can only be found

after the potential and electric field have been found in the vicinity of the electrode surfaces from

σs = ε0En, (C/m2)

where

En = −∂Φ

∂n,

is the electric field component normal to the conducting electrode surface with n the normal coor-

dinate.

If the potential is specified on a closed surface, the potential off the surface is uniquely deter-

mined in terms of the surface potential. This is known as Dirichlet’s boundary value problem and

most problems we will consider belong to this category. Solving Dirichlet’s problems is greatly

facilitated by finding a suitable Green’s function for a given boundary shape. However, except

for simple geometries (e.g., plane, sphere, cylinder, etc.), finding Green’s functions analytically is

not an easy task. For complicated electrode shapes, potential problems often have to be solved

numerically.

1

Specifying the normal derivative ∂Φ/∂n on a closed surface also uniquely determines the poten-

tial elsewhere. This category of boundary value problems is called Neumann problem. Physically,

specifying the normal derivative of the potential on a closed surface corresponds to specifying the

surface charge distribution on the surface through

σ = −ε0∂Φ

∂n.

Then, the problem is reduced to finding the potential due to a prescribed charge distribution as

worked out in Chapter 1.

Specifying both the potential itself and its normal derivative everywhere on a closed surface is

in general overdetermining. However, in some problems, the potential is known in one part of a

closed surface and its normal derivative in the remaining part. This constitutes the so-called mixed

boundary value problem.

By introducing suitable coordinates transformation, some potential problems can be reduced

to one dimensional, that is, the potential becomes a total function of a single coordinate vari-

able. This happens if the Laplace equation and potential are completely separable, Φ(u1, u2, u3) =

F1(u1)F2(u2)F3(u3). There are some 30 known rectilinear coordinate systems developed in the past

for specific purposes. As one example, we will study the oblate spheroidal coordinates because of

its wide variety of applications in electrostatics and magnetostatics.

2.2 Dirichlet Problems and Green’s Functions

If a charge is given to a conductor, the potential of the conductor becomes constant everywhere after

a short transient time as shown in Chapter 1. Electrostatic state is thus quickly established. Since

the volume charge density ρ should vanish in a conductor, all of the charge given to a conductor

must reside entirely on the conductor surface in the form of singular surface charge density σ

(C/m2). The corresponding volume charge density involves a delta function

ρ = σδ(n−ns),

where n is the coordinate normal to the surface and ns indicates the location of the surface.

After static condition is established, the volume charge density and the electric field in a con-

ductor both vanish. The potential of a conductor thus becomes constant Φ = Φc = const. If a

charge q is given to an isolated conductor, the potential of the conductor relative to zero potential

at infinity is uniquely determined and the proportional constant C defines the self-capacitance of

the conductor,

C =q

Φc, (F). (2.1)

2

Let us consider a trivial case, a conducting sphere of radius a carrying a charge q. The potential

outside the sphere is given by

Φ(r) =q

4πε0

1

r, r ≥ a. (2.2)

The sphere potential is

Φs =q

4πε0a, (2.3)

which determines the self-capacitance of the sphere,

C = 4πε0a, (F). (2.4)

The outer potential Φ(r) can be written in the form

Φ(r) =a

rΦs, r > a, (2.5)

which indicates that the potential is uniquely determined if the sphere potential Φs is known. In

general, if the potential is specified everywhere on a closed surface, the potential elsewhere off

the surface is uniquely determined in terms of the surface potential Φs(rs) where rs denotes the

coordinates on the closed surface. This is known as Dirichlet’s theorem and finding a potential for

given boundary potential distribution on a closed surface is called Dirichlet’s problem.

The same problem can also be solved in terms of the electric field on the sphere surface,

Er =1

4πε0

q

a2, (2.6)

which can be replaced with a surface charge,

σ = ε0Er =q

4πa2, (C/m2). (2.7)

The potential due to the uniform surface charge is

Φ(r) =1

4πε0

∮σ

|r− r′|dS

=σ

4πε02πa2

∫ π

0

1√r2 + a2 − 2ar cos θ

sin θdθ

=q

4πε0r, (r > a) (2.8)

where θ is measured from the direction of r. (This is allowed because of symmetry. For r < a, the

integral yields

Φ(r) =q

4πε0a= const., (r < a, interior)

3

which is also an expected result.) Since

Er = −∂Φ

∂r=∂Φ

∂n, (2.9)

where n is the normal coordinate on the surface directed away from the volume of interest, the

potential can be rewritten as

Φ(r) =1

4π

∮S

1

|r− r′|∂Φ

∂ndS′. (2.10)

As this simple example indicates, potential boundary value problems can be solved in terms of

either the surface potential Φs or its normal derivative, ∂Φ/∂n. The latter method may be regarded

as a boundary value problem for the electric field.

Let us revisit the potential due to a prescribed charge distribution,

Φ(r) =1

4πε0

∫ρ(r′)

|r− r′|dV′. (2.11)

The potential can be understood as a convolution between the charge density distribution ρ(r) and

the function

G(r, r′) =1

4π

1

|r− r′| , (2.12)

which is the particular solution to the singular Poisson’s equation

∇2G = −δ(r− r′), (2.13)

subject to the boundary condition that G vanish at infinity. The function G is called Green’s

function. Physically, the Green’s function defined as a solution to the singular Poisson’s equation

is nothing but the potential due to a point charge placed at r = r′. In potential boundary value

problems, the charge density ρ(r) is unknown and one has to devise an alternative formulation

in terms of boundary potential Φs(r). It is noted that the Green’s function in Eq. (2.12) is the

particular solution to the singular Poisson’s equation and we still have freedom to add general

solutions satisfying Laplace equation,

G = Gp +Gg, (2.14)

where Gp is the particular solution and Gg is a collection of general solutions satisfying

∇2Gg = 0. (2.15)

This freedom will play an important role in constructing a Green’s function suitable for a given

boundary shape as we will see shortly. In doing so, we exploit the following theorem:

4

Theorem 1 Green’s Theorem: For arbitrary scalar functions φ and ψ, the following identity holds,∫V

(φ∇2ψ − ψ∇2φ

)dV =

∮S

(φ∇ψ − ψ∇φ) · dS. (2.16)

Proof of this theorem goes as follows. Gauss’theorem applied to the function φ∇ψ gives∫V∇ · (φ∇ψ)dV =

∮S

(φ∇ψ) · dS. (2.17)

The LHS may be expanded as∫V∇ · (φ∇ψ)dV =

∫V

(∇φ · ∇ψ + φ∇2ψ

)dV. (2.18)

Therefore, ∫V

(∇φ · ∇ψ + φ∇2ψ

)dV =

∮S

(φ∇ψ) · dS. (2.19)

Exchanging φ and ψ, ∫V

(∇ψ · ∇φ+ ψ∇2φ

)dV =

∮S

(ψ∇φ) · dS. (2.20)

Subtracting Eq. (2.20) from Eq. (2.19) yields∫V

(φ∇2ψ − ψ∇2φ

)dV =

∮S

(φ∇ψ − ψ∇φ) · dS, (2.21)

which is the desired identity.

Figure 2-1: Φs(r′) is the potential specified on a closed surafce S, n is the coordinate normal to the

surface directed away from the volume wherein the potential Φ(r) is to be evaluated.

We now apply the formula to electrostatic potential problems. Let ψ be the Green’s function

5

ψ = G, satisfying

∇2G = −δ(r− r′), (2.22)

and φ = Φ be the scalar potential satisfying Poisson’s equation

∇2Φ = − ρ

ε0. (2.23)

Then, the terms in the LHS of Eq. (2.21) become∫V

Φ∇2r′GdV ′ = −∫V

Φ(r′)δ(r− r′)dV ′

= −Φ(r),

provided the coordinates r resides in the volume V where we wish to find the potential, and∫G∇2ΦdV ′ = − 1

ε0

∫Gρ(r′)dV ′. (2.24)

The RHS of Eq. (2.21) reduces to ∮S

(Φ∂G

∂n−G∂Φ

∂n

)dS, (2.25)

where Φ is the potential on the closed surface and n is the coordinate normal to the surface directed

away from the volume of interest as indicated in Fig.2-1. Therefore, the solution for the potential

Φ(r) is given by

Φ(r) =1

ε0

∫VGρ(r′)dV ′ −

∮S

(Φs∂G

∂n−G∂Φs

∂n

)dS. (2.26)

At this stage, the Green’s function is still arbitrary except it should satisfy the singular Poisson’s

equation in Eq. (2.22). The first term in the RHS allows evaluation of the potential for a given

charge distribution as we saw earlier. The surface integral involves the potential on the closed

surface Φs and its normal derivative, namely, the normal component of the electric field at the

surface.

In usual boundary value problems, the potential on a closed surface is specified as a function

of the surface coordinates. In this case, it is convenient to choose the Green’s function so that it

vanishes on the surface,

G = 0 on S.

Then the last term in Eq. (2.26) vanishes, and the solution for the potential becomes

Φ(r) =1

4πε0

∫V

ρ(r′)

|r− r′|dV′ −∮S

Φs∂G

∂ndS′, G = 0 on S. (2.27)

6

In particular, if there are no charges in the region of concern ρ = 0, the potential is uniquely

determined in terms of the surface potential alone,

Φ(r) = −∮S

Φs∂G

∂ndS′, ρ = 0 in V, G = 0 on S. (2.28)

We have a freedom to make such a choice for the Green’s function that it vanish on the closed

surface S through adding general solutions to the particular solution of the singular Poisson’s

equation. Therefore, solving a potential boundary value problems for a given closed surface S boils

down to finding a Green’s function satisfying

∇2G = −δ(r− r′), G = 0 on S. (2.29)

Once such an appropriate Green’s function is found for a given surface shape S, the potential at

arbitrary point can be found from Eq. (2.28) for a specified potential distribution Φs(r) on the

surface.

In the following, Green’s functions for some simple surface shapes will be found. It is noted

that three dimensional Green’s functions have dimensions of 1/length, two dimensional Green’s

functions are dimensionless, and one dimensional Greens functions have dimensions of length.

2.3 Examples of Green’s Functions

2.3.1 Plane

Suppose that the potential is specified everywhere on an infinite (x, y) plane, Φs(x, y). The plane

is closed at infinity and the method of Green’s function is applicable. The Green’s function is to

be found as a solution to the equation

∇2G = −δ(x− x′)δ(y − y′)δ(z − z′), (2.30)

with the boundary condition G = 0, z = 0. Mathematically, the Green’s function is equivalent to

the potential due to a point charge placed near a grounded conducting plate that can be readily

worked out using the method of image as shown in Fig.2-2,

G(r, r′) =1

4π

(1√

(x− x′)2 + (y − y′)2 + (z − z′)2− 1√

(x− x′)2 + (y − y′)2 + (z + z′)2

), (2.31)

where the second term in the RHS is the contribution from the image charge at the mirror point.

Note that the Green’s function is reciprocal and remains unchanged against the coordinates inter-

7

change,

G(r, r′) = G(r′, r).

This is expected from the fact that the delta function in the original singular Poisson’s equation is

even,

δ(r− r′) = δ(r′−r).

In the upper region z > 0,

∂G

∂n= − ∂G

∂z′

∣∣∣∣z′=0

= − 1

2π

z

[(x− x′)2 + (y − y′)2 + z2]3/2. (2.32)

Therefore, for a surface potential Φs(x′, y′) specified as a function of (x′, y′), the potential in the

region z > 0 is given by

Φ(r) =z

2π

∫ ∞−∞

dx′∫ ∞−∞

dy′Φs(x

′, y′)

[(x− x′)2 + (y − y′)2 + z2]3/2. (2.33)

Figure 2-2: Image charge −q for a large, grounded conducting plate. The potential due to q and−q vanishes at the plate.

Let us apply this formula to the boundary condition on the (x, y) plane,

Φs(ρ) =

V, ρ < a

0, ρ > a(2.34)

where ρ =√x2 + y2 is the radial distance on the plane as shown in Fig. 2-3. Physically, the

boundary condition describes a large conducting plate which is grounded except for a circular

region of radius a whose potential is maintained at V. The potential on the z-axis can be found

8

Figure 2-3: A large conducting plate is grounded except for a circular region which is at a potentialV.

easily,

Φ(z) =zV

2π

∫ a

0

2πρ′dρ′

(ρ′2 + z2)3/2

= V

(1− z√

z2 + a2

), z > 0. (2.35)

The axial potential in the lower region z < 0 can be found by observing the up-down symmetry

and for both regions,

Φ(z) = V

(1− |z|√

z2 + a2

). (2.36)

Then, the potential at arbitrary point (r, θ) is

Φ(r, θ) =

V

[1− r

a|P1(cos θ)|+ 1

2

(ra

)3|P3(cos θ)| − 3

8

(ra

)5|P5(cos θ)|+ · · ·

], r < a

V

[12

(ar

)2|P1(cos θ)| − 3

8

(ar

)4|P3(cos θ)|+ · · ·

], r > a

(2.37)

Note that at r a, the potential is of dipole type,

Φ(r a) ∝ 1

r2|cos θ| . (2.38)

This problem should not be confused with the potential due to an isolated charged conducting disk

which will be discussed later. The potential and electric field in the upper half region are identical

to those realized by an ideally thin circular capacitor whose top plate is at a potential V and the

lower plate at −V. The appearance of the dipole potential is thus an expected result. (For a thincapacitor with plate separation distance δ, the electric field between the plates diverges but the

product Eδ = 2V remains constant. Such a structure is called a double layer.)

9

2.3.2 Sphere

Figure 2-4: The image of charge q with respect to a grounded conducting sphere is q′ = −qa/r′located at a2r′/r′2 where r′ is the location of the charge q.

In finding a Green’s function for a given surface shape, the method of images is most conveniently

exploited. In the case of a sphere having a radius a, the Green’s function can be found as a solution

for the potential due to a charge q placed at a distance r′ from the center of a grounded conducting

sphere. In the case of a sphere having radius a, an image charge

q′ = − ar′q, (2.39)

placed at

r′′ =a2

r′2r′, (2.40)

together with the charge q, makes the surface potential vanish. This is illustrated in Fig.2-4. The

potential due to charge q placed near a grounded conducting sphere is thus equivalent to that due

to two charges, q and its image charge q′, and is given by

Φ(r) =q

4πε0

1

|r− r′| −a

r′

|r− r′′|

=

q

4πε0

1√r2 + r′2 − 2rr′ cos γ

− 1√(rr′/a)2 + a2 − 2rr′ cos γ

, (2.41)

10

which readily yields the Green’s function for a sphere,

G(r, r′) =1

4π

1√r2 + r′2 − 2rr′ cos γ

− 1√(rr′/a)2 + a2 − 2rr′ cos γ

. (2.42)

Here γ is the angle between the two position vectors r = (r, θ, φ) and r′ = (r′, θ′, φ′). Its cosine

value is

cos γ = cos θ cos θ′ + sin θ sin θ′ cos(φ− φ′). (2.43)

The Green’s function indeed vanishes on the sphere surface r = a or r′ = a. Again, the Green’s

function is invariant against coordinates exchange, r↔ r′, that is, Green’s functions are reciprocal.

For exterior (r > a) potential problems, the normal gradient ∂G/∂n is

∂G

∂n= − ∂G

∂r′

∣∣∣∣r′=a

=1

4π

a− r2

a

(r2 + a2 − 2ar cos γ)3/2, r > a (2.44)

and for interior (r < a) problems,

∂G

∂n= +

∂G

∂r′

∣∣∣∣r′=a

=1

4π

r2

a− a

(r2 + a2 − 2ar cos γ)3/2, r < a. (2.45)

Note that the normal coordinate n is directed away from the volume of interest. If the surface

potential is specified as a function of θ′ and φ′, Φs(θ′, φ′), and there are no charges, the exterior

potential at an arbitrary point r = (r, θ, φ) can be found from

Φ(r) = −∮

Φs∂G

∂ndS

=1

4π

∮a(r2 − a2)

(r2 + a2 − 2ar cos γ)3/2Φs(θ

′, φ′)dΩ′

=a(r2 − a2)

4π

∫ π

0sin θ′dθ′

∫ 2π

0dφ′

Φs(θ′, φ′)

(r2 + a2 − 2ar cos γ)3/2. (2.46)

Recalling the expansion of the function 1/ |r− r′| in terms of the spherical harmonic functions

1

|r− r′| =∑l,m

4π

2l + 1

r′l

rl+1Ylm(θ, φ)Y ∗lm(θ′, φ′), r > r′ (2.47)

11

the exterior potential can be decomposed into multipole potentials,

Φ(r, θ, φ) =∑l,m

(ar

)l+1Ylm(θ, φ)

∮Φs(θ

′, φ′)Y ∗lm(θ′, φ′)dΩ′, r > a. (2.48)

The interior potential can be found using the expansion

1

|r− r′| =∑l,m

4π

2l + 1

rl

r′l+1Ylm(θ, φ)Y ∗lm(θ′, φ′), r < r′, (2.49)

Φ(r, θ, φ) =∑l,m

(ra

)lYlm(θ, φ)

∮Φs(θ

′, φ′)Y ∗lm(θ′, φ′)dΩ′, r < a. (2.50)

Example 2 Charge near a Floating Conducting Sphere

To become familiar with the Green’s function method, let us consider a somewhat trivial problem

of finding the potential when a charge q is placed at a distance d from the center of a floating

conducting sphere of radius a. The charge q and its image q′ = −adq at (a/d)2d make the sphere

potential 0 as we have just seen. However, since the floating sphere should carry no net charge, a

charge −q′ = adq must be placed at the center of the sphere which raises the sphere potential to

Φs =−q′

4πε0a=

q

4πε0d, d > a.

Therefore, the exterior potential can be found by summing contributions from q, its image q′ and

the charge −q′ at the center,

Φ(r) =1

4πε0

(q

|r− d| −qa/d

|r− (a/d)2d| +qa/d

r

).

In this expression, the function

1

4π

(1

|r− d| −a

d

1

|r− (a/d)2d|

),

is the Green’s function which vanishes on the sphere surface, r = a. The last term is in the form

Φsa

r,

where

Φs =q

4πε0d, (independent of a)

12

is the surface potential. Indeed,

−∮

Φs∂G

∂ndS =

a(r2 − a2)4π

∫ π

0sin θ′dθ′

∫ 2π

0dφ′

Φs(θ′, φ′)

(r2 + a2 − 2ar cos γ)3/2

= Φsa(r2 − a2)

2

∫ π

0

1(r2 + a2 − 2ar cos θ′

)3/2 sin θ′dθ′

= Φsa

r,

where θ′ is measured from the direction of the vector d. (This is allowed because of the symmetry.)

Example 3 Specified Potential on a Sphere Surface

Figure 2-5: Φs = +V for 0 < θ < π/2, −V for π/2 < θ < π.

Let us find the potential outside a spherical shell of radius a whose top half is maintained at

potential +V and lower half at −V,

Φs(θ′) =

+V, 0 ≤ θ′ ≤ π

2,

−V, π

2≤ θ′ ≤ π,

(2.51)

as shown in Fig.2-5. Because of axial symmetry, only m = 0 terms survive the integration over

the azimuthal angle φ′. Also, because of up-down antisymmetry, only odd l terms survive the

integration over the polar angle θ′. Noting∮Φs(θ

′, φ′)Y ∗l0(θ′, φ′)dΩ′

= 2π × 2V

√2l + 1

4π

∫ 1

0Pl(µ)dµ, l = 1, 3, 5, · · ·,

we readily find the exterior potential,

Φ(r, θ) = V

[3

2

(ar

)2P1(cos θ)− 7

8

(ar

)4P3(cos θ) + · · ·

], r ≥ a. (2.52)

13

The interior potential is

Φ(r, θ) = V

[3

2

r

aP1(cos θ)− 7

8

(ra

)3P3(cos θ) + · · ·

], r ≤ a. (2.53)

The surface charge density on the sphere can be found from the normal component of the

electric field,

σ = −ε0∂Φ

∂r

∣∣∣∣r=a+0

=ε0V

a

[3P1(cos θ)− 7

2P3(cos θ) + · · ·

].

The total surface charge on the upper hemisphere

q = 2πa2∫ π/2

0σ(θ) sin θdθ,

simply diverges (albeit only logarithmically) and it is not possible to define the capacitance of the

hemispheres. This is because of the assumption of ideally small gap separating the two hemispheres.

If a small gap δ a is assumed, a finite capacitance containing a factor ln(a/δ) emerges.

2.3.3 Interior of Cylinder of Finite Length

Figure 2-6: Cylinder of a finite length.

14

The Green’s function for the interior of a cylinder of radius a and length l shown in Fig.2-6 can

be found as a solution for the following singular Poisson’s equation

∇2G =

(∂2

∂ρ2+

1

ρ

∂

∂ρ+

1

ρ2∂2

∂φ2+

∂2

∂z2

)G = −δ(ρ− ρ

′)

ρδ(φ− φ′)δ(z − z′), (2.54)

with the boundary condition

G = 0, ρ = a, z = 0 and l. (2.55)

Since the Green’s function should be periodic with respect to φ and should also be invariant with

respect to exchange of φ and φ′, the angular dependence can be assumed to be cos[m(φ−φ′)] wherem is an integer. Assuming the following separation of variables,

G(r, r′) =∑m

Rm(ρ)Zm(z) cosm(φ− φ′), (2.56)

we see that the radial function Rm(ρ) and the axial function Zm(z) satisfy, respectively,(d2

dρ2+

1

ρ

d

dρ− m2

ρ2+ k2

)Rm(ρ) = 0, (2.57)

(d2

dz2− k2

)Zm(z) = 0, (2.58)

where k2 is a separation constant which can be either positive or negative.

Let us first consider the case k2 > 0. Solutions for Rm(ρ) which satisfies the boundary condition

Rm(ρ = a) = 0 is the m-th order Bessel function,

Rmn(ρ, ρ′) = Jm

(xmnρa

)Jm

(xmnρ

′

a

), (2.59)

where xmn is the n-th root of Jm(x) = 0. (The Bessel function of the second kind Nm(x) is discarded

because it diverges on the axis, ρ = 0.)

Solutions for the axial function Zm(z) are e±kz or sinh(kz) and cosh(kz). The boundary con-

dition for Zm(z) is it vanish at z = 0 and l. Therefore, we can construct the axial function as

follows,

Zm(z, z′) =

sinh(kmnz) sinh[kmn(l − z′)], 0 < z < z′ < l,

sinh[kmn(l − z)] sinh(kmnz′), 0 < z′ < z < l,

(2.60)

where kmn = xmn/a. A more fancy way to write Zm(z, z′) is

Zm(z, z′) = sinh[kmn min(z, z′)] sinhkmn[l −max(z, z′)]. (2.61)

15

The Green’s function may thus be assumed in the form

G(r, r′) =∑m,n

AmnRmn(ρ, ρ′)Zmn(z, z′) cos[m(φ− φ′)]. (2.62)

The expansion coeffi cient Amn can be determined from the discontinuity in the derivative of the

axial function Zmn(z, z′) at z′,

d

dzZmn

∣∣∣∣z=z′+0

= −kmn cosh[kmn(l − z′)] sinh(kmnz′),

d

dzZmn

∣∣∣∣z=z′−0

= +kmn cosh(kmnz′) sinh[(kmn(l − z′)].

Then, a singularity appears in the second order derivative,

d2

dz2Zmn = −kmn sinh(kmnl)δ(z − z′), (2.63)

which is compatible with the delta function in the RHS of the original singular Poisson’s equation

in Eq. (2.54). Eq. (2.54) now reduces to

∑mn

Amnkmn sinh(kmnl)Jm(kmnρ)Jm(kmnρ′) cosm(φ− φ′) =

δ(ρ− ρ′)ρ

δ(φ− φ′). (2.64)

Multiplying both sides by ρ′Jm(kmnρ′) cosmφ′ and integrating over ρ′ and φ′, we find

A0n =1

πa2kmn

1

J2m+1(kmna) sinh(kmnl), m = 0, (2.65)

Amn =2

πa2kmn

1

J2m+1(kmna) sinh(kmnl), m ≥ 1, (2.66)

where use has been made of the following integral,∫ a

0ρJ2m(kmnρ)dρ =

a2

2J2m+1(kmna). (2.67)

The final form of the desired Green’s function is

G(r, r′) =1

πa

∞∑m=0

∞∑n=1

Jm

(xmnρa

)Jm

(xmnρ

′

a

)Zmn(z, z′)

xmnJ2m+1(xmn) sinh(kmnl)cos[m(φ− φ′)]εm, (2.68)

16

where

εm =

1, m = 0

2, m ≥ 1

If one does not like the appearance of εm, the summation over m can be changed to from −∞ to

∞,

G(r, r′) =1

πa

∞∑m=−∞

∞∑n=1

Jm

(xmnρa

)Jm

(xmnρ

′

a

)Zmn(z, z′)

xmnJ2m+1(xmn) sinh(kmnl)cos[m(φ− φ′)]. (2.69)

If it is assumed that k2 = −κ2 < 0, appropriate general solutions to(d2

dρ2+

1

ρ

d

dρ− m2

ρ2− κ2

)Rm(ρ) = 0, (2.70)

(d2

dz2+ κ2

)Zm(z) = 0, (2.71)

are

Rm(ρ, ρ′

)=

[Km (κma) Im

(κmρ

′)− Im (κma)Km

(κmρ

′)] Im (kmρ) , ρ < ρ′ < a, (2.72)

Rm(ρ, ρ′

)= [Km (κma) Im (κmρ)− Im (κma)Km (κmρ)] Im

(kmρ

′) , ρ′ < ρ < a, (2.73)

with κm = mπ/l and

Zm (z) = sin (κmz) sin(κmz

′) , (2.74)

from which the Green’s function can be constructed. Remaining calculation is left for exercise. The

reader should appreciate how a delta function δ (ρ− ρ′) appears from the term

d2Rmdρ2

. (2.75)

One may wonder about Green’s function for the exterior region of a cylinder of finite length.

This problem appears to be a diffi cult one and analytical expressions are not available to the

author’s knowledge. It may be the case the problem can only be solved numerically.

2.3.4 Long Cylinder (3-Dimensional)

Three dimensional Green’s function for a long cylinder satisfies

∇2G =

(∂2

∂ρ2+

1

ρ

∂

∂ρ+

1

ρ2∂2

∂φ2+

∂2

∂z2

)G = −δ(ρ− ρ

′)

ρδ(φ− φ′)δ(z − z′), (2.76)

17

which is to be solved for the boundary conditions

G(ρ = a) = 0, G(z = ±∞) = 0. (2.77)

Following the same procedure as in the preceding example, the interior solution for interior ρ, ρ′ < a

may be assumed as

G(r, r′) =∑m,n

AmnJm(kmnρ)Jm(kmnρ′) cos[m(φ− φ′)] exp[−kmn|z − z′|], (2.78)

where

kmn =xmna, (2.79)

and xmn is the n-th root of Jm(x) = 0. Since

d2

dz2e−kmn|z−z

′| = k2mne−kmn|z−z′| − 2kmnδ(z − z′), (2.80)

∞∑m=−∞

cos[m(φ− φ′)] = 2πδ(φ− φ′), (2.81)

we readily find the interior Green’s function (ρ, ρ′ < a)

G(r, r′) =1

π

∞∑m=−∞

∞∑n=1

Jm(kmnρ)Jm(kmnρ′)

kmnJ2m+1(kmna)cos[m(φ− φ′)]e−kmn|z−z′|. (2.82)

For exterior of a long cylinder, solutions to the equation(∂2

∂ρ2+

1

ρ

∂

∂ρ+

1

ρ2∂2

∂φ2+

∂2

∂z2

)G = −δ(ρ− ρ

′)

ρδ(φ− φ′)δ(z − z′),

can be found in terms of Fourier transform with respect to the z-coordinate. Let G(r, r′) be

G(r, r′) =∑m

eim(φ−φ′)∫Rm(ρ, ρ′; k)eik(z−z

′)dk. (2.83)

The radial function Rm(ρ, ρ′; k) satisfies(d2

dρ2+

1

ρ

d

dρ− m2

ρ2− k2

)Rm(ρ, ρ′; k) = −δ(ρ− ρ

′)

2πρ. (2.84)

Elementary solutions are the modified Bessel functions Im(kρ) and Km(kρ) and we can construct

18

following solutions which remain bounded in the region a < ρ <∞,

Rm(ρ, ρ′; k) =

A(k)Im(kρ) +B(k)Km(kρ), a < ρ < ρ′ <∞,

C(k)Km(kρ), a < ρ′ < ρ <∞,(2.85)

The boundary conditions are Rm(ρ = a) = 0 and Rm(ρ) be continuous at ρ = ρ′,

A(k)Im(ka) +B(k)Km(ka) = 0, (2.86)

A(k)Im(kρ′) +B(k)Km(kρ′) = C(k)Km(kρ′). (2.87)

Then,

Rm(ρ, ρ′; k) =

A(k)

(Im(kρ)− Im(ka)

Km(ka)Km(kρ)

), a < ρ < ρ′ <∞,

A(k)1

Km(kρ′)

(Im(kρ′)− Im(ka)

Km(ka)Km(kρ′)

)Km(kρ), a < ρ′ < ρ <∞.

(2.88)

The unknown function A(k) can be found from the discontinuity in the derivative at ρ = ρ′,

d2

dρ2Rm(ρ, ρ′; k)

∣∣∣∣ρ=ρ′

= kA(k)K ′m(kρ′)Im(kρ′)−Km(kρ′)I ′m(kρ′)

Km(kρ′)δ(ρ− ρ′) (2.89)

= − A(k)

Km(kρ′)

δ(ρ− ρ′)ρ′

, (2.90)

where again use has been made of the Wronskian of the modified Bessel functions,

I ′m(x)Km(x)− Im(x)K ′m(x) =1

x. (2.91)

We thus find

A(k) =Km(kρ′)

2π, (2.92)

and Rm(ρ, ρ′; k) reduces to

Rm(ρ, ρ′; k) =

1

2πKm(kρ′)

(Im(kρ)− Im(ka)

Km(ka)Km(kρ)

), a < ρ < ρ′ <∞,

1

2π

(Im(kρ′)− Im(ka)

Km(ka)Km(kρ′)

)Km(kρ), a < ρ′ < ρ <∞.

(2.93)

19

The exterior Green’s function of a long cylinder is given by

G(r, r′) =1

2π

∑m

eim(φ−φ′)∫Rm(ρ, ρ′; k)eik(z−z

′)dk. (2.94)

2.3.5 Long Cylinder (2-Dimensional)

Cross-section of a long cylinder. ±λ are the line charge and its image, respectively, that togethermake the cylinder surface an equipotential surface,

Φ(ρ = a) = λ/(2πε0) ln(a/ρ′).

For boundary value problems in which z-dependence is suppressed, it is convenient to formulate

a two dimensional Green’s function. Two dimensional Green’s function for a long cylinder is to be

found from

∇2G(r, r′) = −δ2(r− r′), (2.95)

where δ2(r− r′) is the two-dimensional delta function. In the cylindrical geometry, it is given by

δ2(r− r′) =δ(ρ− ρ′)

ρδ(φ− φ′), (2.96)

and the Green’s function satisfies(∂2

∂ρ2+

1

ρ

∂

∂ρ+

1

ρ2∂2

∂φ2

)G = −δ(ρ− ρ

′)

ρδ(φ− φ′). (2.97)

In this case, the method of image can be exploited very conveniently. Let us consider a long line

charge λ (C/m) placed at (ρ′, φ′) parallel to a long, grounded conducting cylinder of radius a. A

negative line charge −λ placed at (ρ′′, φ′) where

ρ′′ =a2

ρ′, (2.98)

20

makes the cylinder surface an equipotential surface at a potential

Φs =λ

2πε0ln

(a

ρ′

). (2.99)

Since we are seeking a potential that vanishes on the cylinder surface ρ = a, the constant potential

Φs can be subtracted from the potential due to two line charges λ and −λ,

Φ(r, r′) = − λ

2πε0

[ln∣∣r− r′∣∣− ln

∣∣r− r′′∣∣+ ln

(a

ρ′

)], (2.100)

where ∣∣r− r′∣∣ =√ρ2 + ρ′2 − 2ρρ′ cos(φ− φ′), (2.101)

∣∣r− r′′∣∣ =√ρ2 + ρ′′2 − 2ρρ′′ cos(φ− φ′)

=

√ρ2 +

(a2

ρ′

)2− 2

a2ρ

ρ′cos(φ− φ′). (2.102)

The desired Green’s function is

G(r, r′) = − 1

2πln

( √ρ2 + ρ′2 − 2ρρ′ cos(φ− φ′)√

(ρρ′/a)2 + a2 − 2ρρ′ cos(φ− φ′)

). (2.103)

For exterior Dirichlet problems, the normal derivative at the cylinder surface is

∂G

∂n= − ∂G

∂ρ′

∣∣∣∣ρ′=a+0

=1

2π

a− ρ2

aρ2 + a2 − 2aρ cos(φ− φ′) , ρ > a, (2.104)

and for interior,

∂G

∂n=∂G

∂ρ′

∣∣∣∣ρ′=a−0

=1

2π

ρ2

a− a

ρ2 + a2 − 2aρ cos(φ− φ′) , ρ < a. (2.105)

If the potential on a long cylindrical surface is specified as a function of the angle φ, Φs(φ), the

potential off the surface can be calculated from

Φ(ρ, φ) = −a∮

Φs(φ′)∂G

∂ndφ′. (2.106)

21

For the interior (ρ < a) , the potential is given by

Φ(ρ, φ) =1

2π

∫ 2π

0Φs

(φ′) a2 − ρ2a2 + ρ2 − 2aρ cos(φ− φ′)dφ

′

=1

2π

∫ 2π

0Φs

(φ′)(

1 + 2∞∑m=1

(ρa

)mcos[m

(φ− φ′

)]

)dφ′, (2.107)

where use is made of the following expansion,

a2 − ρ2a2 + ρ2 − 2aρ cos(φ− φ′) = 1 + 2

∞∑m=1

(ρa

)mcos[m

(φ− φ′

)].

Example 4

Figure 2-7: Φ = V for 0 < φ < π, Φ = −V for −π < φ < 0 on the surface of a long cylinder.(Example of 2-D Green’s function.)

As an example, let us consider a long conducting cylinder consisting of two equal troughs. The

upper half in the region 0 < φ < π is at a potential V and the lower half −π < φ < 0 is at a

potential −V as shown Fig.2-7. The exterior potential ρ > a is given by

Φ(ρ, φ) = Vρ2 − a2

2π

(∫ π

0

1

ρ2 + a2 − 2aρ cos(φ− φ′)dφ′ −∫ 0

−π

1

ρ2 + a2 − 2aρ cos(φ− φ′)dφ′).

(2.108)

22

The first integral can be effected by changing the variable from φ′ to θ through φ′ − π/2 = θ,

∫ π/2

−π/2

1

ρ2 + a2 + 2aρ sin(θ − φ)dθ

=2

ρ2 − a2

tan−1

(ρ2 + a2) tan

(θ − φ

2

)+ 2aρ

ρ2 − a2

π/2

−π/2

=2

ρ2 − a2

[tan−1

((ρ2 + a2)(cotφ− tanφ) + 2aρ

ρ2 − a2

)+ tan−1

((ρ2 + a2)(cotφ+ tanφ) + 2aρ

ρ2 − a2

)]=

2

ρ2 − a2

[tan−1

(2aρ sinφ

ρ2 − a2

)− π

2

], (2.109)

where use has been made of the identities,

tan(π

4± x

2

)= cotx± tanx,

tan−1 x+ tan−1 y = tan−1(x+ y

1− xy

),

tan−1 x =π

2− tan−1

(1

x

).

Similarly, the second integral yields

2

ρ2 − a2

[tan−1

(2aρ sinφ

ρ2 − a2

)+π

2

], (2.110)

and the potential becomes

Φ(ρ, φ) =2V

πtan−1

(2aρ sinφ

ρ2 − a2

), ρ > a. (2.111)

The interior potential is

Φ(ρ, φ) =2V

πtan−1

(2aρ sinφ

a2 − ρ2

), ρ < a. (2.112)

2.3.6 Wedge

A wedge is formed by two large plates intersecting at an angle γ as illustrated in Fig.2-8.

The potential due to a point charge q at (ρ′, φ′, z′) with the boundary conditions Φ = 0 at the

plates φ = 0 and φ = γ, and ρ = ∞, |z| = ∞ essentially gives the Green’s function. We thus seek

23

Figure 2-8: A wedge formed by two large conducting plates intersecting at an angle γ.

a solution to the Poisson’s equation(∂2

∂ρ2+

1

ρ

∂

∂ρ+

1

ρ2∂2

∂φ2+

∂2

∂z2

)G = −δ(ρ− ρ

′)

ρδ(φ− φ′)δ(z − z′), (2.113)

subject to the those boundary conditions. As in the case of 3-dimensional Green’s function for a

long cylinder, we Fourier transform the Green’s function,

G(r, φ, z) =1

2π

∫ ∞−∞

g(ρ, φ, k)eik(z−z′)dk. (2.114)

The angular dependence of the Green’s function can be assumed to be

sin

(mπ

γφ

)sin

(mπ

γφ′), (2.115)

which indeed vanishes at φ = 0 and φ = γ. We thus assume

g(ρ, φ, k) =∑m

AmRm(ρ) sin

(mπ

γφ

)sin

(mπ

γφ′), (2.116)

to obtain

∑m

Am

[d2

dρ2+

1

ρ

d

dρ− 1

ρ2

(mπ

γ

)2− k2

]Rm(ρ) sin

(mπ

γφ

)(sin

mπ

γφ′)

= −δ(ρ− ρ′)

ρδ(φ− φ′).

(2.117)

24

The radial function can be composed of the modified Bessel functions,

Rm(ρ) =

Imπ/γ(kρ)Kmπ/γ(kρ′), ρ < ρ′,

Imπ/γ(kρ′)Kmπ/γ(kρ), ρ′ < ρ.

(2.118)

The derivative of the radial function Rm(ρ) has discontinuity at ρ = ρ′, and the second order

derivative yieldsd2Rm(ρ)

dρ2= − 1

ρ′δ(ρ− ρ′), (2.119)

where the Wronskian of the modified Bessel functions,

Iν(x)K ′ν(x)− I ′ν(x)Kν(x) = −1

x,

has been substituted. The expansion coeffi cient Am is thus determined as

Am =1

πγ, (2.120)

and the desired Green’s function is

G(r, r′) =2

πγ

∑m

∫ ∞0

Rm(ρ, ρ′) cos[k(z − z′)]dk sin(νφ) sin(νφ′), (2.121)

where

ν =mπ

γ. (2.122)

We will encounter an application of wedge potential in the section of inversion method later in this

Chapter.

Example 5 Line Charge parallel to a Long Dielectric Cylinder

Consider a long line charge with charge density λ (C m−1) placed parallel to a long dielectric

cylinder of radius a. The distance between the line charge and the cylinder axis is b. The potential

outside the cylinder can be sought by assuming an image line charge λ′ at the position a2/b and

another image −λ′ at the center,

Φρ>a (ρ, φ) = − λ

4πε0ln(ρ2 + b2 − 2ρb cosφ

)− λ′

4πε0ln

(ρ2 +

(a2

b

)2− 2ρ

a2

bcosφ

)+

λ′

4πε0ln ρ2,

where φ is measured relative to the location of line charge λ. The interior potential is free from

25

singularity and may be assumed to be due to an image λ′′ at the location of the line charge λ,

Φρ<a (ρ, φ) = − λ′′

4πε0ln(ρ2 + b2 − 2ρb cosφ

)+

λ′

4πε0ln b2.

Note that the constant potential (the last term)

λ′

4πε0ln b2,

is needed to match same outer potential at r = a,

Φρ>a (r = a) = −λ+ λ′

4πε0ln(a2 + b2 − 2ab cosφ

)+

λ′

4πε0ln b2.

The pertinent boundary conditions are Eφ and Dr be continuous at r = a which yield

λ+ λ′ = λ′′,

and

ε0(λ− λ′

)= ελ′′.

Then

λ′ =ε0 − εε0 + ε

λ, and λ′′ =2ε0ε0 + ε

λ.

The attracting force to act on the unit length of the line charge is given by

F

l=

λ

2πε0

(λ′

b− (a2/b)− λ′′

b

)=

λ2

2πε0

a2

b (b2 − a2)ε0 − εε0 + ε

, (N m−1).

A magnetically dual problem is the case of long current I parallel to a magnetic cylinder with

a permeability µ. The image currents are

I ′ =µ− µ0µ+ µ0

I at ρ =a2

b,

and

I ′′ =2µ

µ+ µ0I at the axis.

2.4 Other Useful Rectilinear Coordinates

The familiar three coordinate systems, cartesian, spherical, and cylindrical, are frequently used in

analyzing potential problems. However, there are some 30 known coordinate systems developed for

specific problems. For simple electrode shapes, potential problems can be rendered one dimensional

26

by a suitable choice of coordinates. However, in some coordinates, solutions to Laplace equations

are not always completely separable. We have encountered one such example in Chapter 1, the

toroidal coordinates, in analyzing the potential due to a ring charge. In this section, some coordinate

systems useful for potential problems will be introduced.

2.4.1 Oblate Spheroidal Coordinates (η, θ, φ)

Figure 2-9: Oblate spheroidal coordinates (η, θ, φ). η → 0 degenerates to a thin disk of radius a.θ = cons. describes the surface of a hyperboloid.

The oblate spherical coordinates (η, θ, φ) are related to the cartesian coordinates through the

following transformation, x = a cosh η sin θ cosφ

y = a cosh η sin θ sinφ

z = a sinh η cos θ

(2.123)

A surface of constant η is the surface of an oblate spheroid described by

x2 + y2

(a cosh η)2+

z2

(a sinh η)2= 1, (2.124)

as shown in Fig.2-9. In the limit of η → 0, the surface degenerates to a thin disk of radius a

with negligible thickness, and in the opposite limit η 1, the surface approaches a sphere with

27

a radius r = a cosh η ' a sinh η. This coordinate system is convenient if electrode shapes are an

oblate sphere or disk. A surface of constant θ is a hyperboloid described by

x2 + y2

(a sin θ)2− z2

(a sin θ)2= 1. (2.125)

The metric coeffi cients are

hη =

√(∂x

∂η

)2+

(∂y

∂η

)2+

(∂z

∂η

)2= a

√cosh2 η − sin2 θ, (2.126)

hθ =

√(∂x

∂θ

)2+

(∂y

∂θ

)2+

(∂z

∂θ

)2= hη, (2.127)

hφ =

√(∂x

∂φ

)2+

(∂y

∂φ

)2+

(∂z

∂φ

)2= a cosh η sin θ. (2.128)

The Laplace equation in the oblate spherical coordinates can thus be written down as

1

hηhθhφ

∂

∂η

(hθhφhη

∂Φ

∂η

)+

∂

∂θ

(hηhφhθ

∂Φ

∂θ

)+

∂

∂φ

(hηhθhφ

∂Φ

∂φ

)= 0, (2.129)

which reduces to

1

cosh2 η − sin2 θ

(∂2

∂η2+ tanh η

∂

∂η+

∂2

∂θ2+ cot θ

∂

∂θ

)Φ +

1

cosh2 η sin2 θ

∂2Φ

∂φ2= 0. (2.130)

Assuming a separated solution Φ(η, θ, φ) = F1(η)F2(θ)eimφ, (m = integer), we obtain(

d2

dη2+ tanh η

d

dη− l(l + 1) +

m2

cosh2 η

)F1(η) = 0, (2.131)

(d2

dθ2+ cot θ

d

dθ+ l(l + 1)− m2

sin2 θ

)F2(θ) = 0, (2.132)

where l(l + 1) is a separation constant. Eq. (2.132) is the standard form of the Legendre equation

and solutions for F2(θ) are

F2(θ) = Pml (cos θ), Qml (cos θ). (2.133)

Eq. (2.131) can be rewritten as(d2

dη2+

sinh η

cosh η

d

dη− l(l + 1) +

m2

1 + sinh2 η

)F1(η) = 0, (2.134)

28

which is also the Legendre equation with a variable i sinh η. Therefore, solutions for F1(η) are

F1(η) = Pml (i sinh η), Qml (i sinh η), (2.135)

and general solution to Laplace equation can be constructed from these elementary solutions.

If a point charge q is placed at(η′, θ′, φ′

), the potential in terms of the oblate spheroidal

coordinates can be found as

Φ(r) =1

4πε0

q

|r− r′|

=q

ε0a

∞∑l=0

l∑m=−l

(l − |m|)!(l + |m|)!

Pml (i sinh η)Qml (i sinh η′)

Pml (i sinh η′)Qml (i sinh η)

Ylm (θ, φ)Y ∗lm

(θ′, φ′

),

η < η′

η > η′

.(2.136)

Derivation of this expression is left for exercise. The Wronskian of the Legendre functions,

Pml (x)d

dxQml (x)−Qml (x)

d

dxPml (x) =

1

x2 − 1

(l + |m|)!(l − |m|)! , (2.137)

should be useful. Furthermore, the Green’s function for an oblate spheroidal surface described by

η = η0 can readily be worked out to be:

for η0 < η < η′,

G(r, r′

)=

1

a

∞∑l=0

l∑m=−l

(l − |m|)!(l + |m|)!P

ml (i sinh η)Qml

(i sinh η′

)Ylm (θ, φ)Y ∗lm

(θ′, φ′

)−1

a

∞∑l=0

l∑m=−l

(l − |m|)!(l + |m|)!

Pml (i sinh η0)

Qml (i sinh η0)Qml (i sinh η)Qml

(i sinh η′

)Ylm (θ, φ)Y ∗lm

(θ′, φ′

),(2.138)

and for η0 < η′ < η,

G(r, r′

)=

1

a

∞∑l=0

l∑m=−l

(l − |m|)!(l + |m|)!P

ml

(i sinh η′

)Qml (i sinh η)Ylm (θ, φ)Y ∗lm

(θ′, φ′

)−1

a

∞∑l=0

l∑m=−l

(l − |m|)!(l + |m|)!

Pml (i sinh η0)

Qml (i sinh η0)Qml (i sinh η)Qml

(i sinh η′

)Ylm (θ, φ)Y ∗lm

(θ′, φ′

).(2.139)

Example 6 Charged Conducting Disk

A thin disk of radius a is described by η = 0 in the oblate spherical coordinates. If a constant

η surface is an equipotential surface, the potential off the surface is a function of η only, that is,

the potential problem becomes one dimensional. This is the most advantageous merit of using a

coordinate system most suitable for particular potential problems. The relevant solution which

29

Figure 2-10: A charged conducting disk of radius a. A disk is described by η = 0, 0 ≤ θ ≤ π.

vanishes at η =∞ is the lowest order Legendre function of the second kind,

Φ(η) = AQ0(i sinh η) +B, (2.140)

where A and B are constants. Since

Q0(i sinh η) = i[tan−1(sinh η)− π

2

]= −i cot−1(sinh η), (2.141)

and the boundary condition is

Φ(η = 0) = V (disk potential),

we readily find the potential at an arbitrary η,

Φ(η) =2V

πcot−1(sinh η). (2.142)

Note that cot−1(0) = π/2. The far field potential at η 1 or r a can be found from the

asymptotic form of the function cot−1 x,

cot−1 x ' 1

x− 1

3x3+ · · ·, x 1. (2.143)

The leading far field potential is monopole as expected,

Φ(η 1) ' 2V

π

1

sinh η' 2V

π

a

r. (2.144)

30

Comparing with the standard monopole potential

Φ(r) =1

4πε0

q

r, (2.145)

we readily find the total charge carried by the disk,

q = 8ε0aV,

and the self-capacitance of the disk,

C = 8ε0a, (F). (2.146)

This expression was first found by Cavendish.

The surface charge distribution on the disk is quite nonuniform because like charges repel each

other. Charge is distributed in such a manner that the tangential electric field on the disk surface

vanishes. The surface charge density can be found from the normal component of the electric field,

σ = ε0En = ε0Eη, (2.147)

where

Eη = − 1

hη

∂Φ

∂η

∣∣∣∣η=0

=2V

πa

1

| cos θ| . (2.148)

Note thatd

dxcot−1 x = − 1

1 + x2. (2.149)

The surface charge density diverges at the edge of the disk where θ = π/2. The charge residing on

the disk surface can be found from the following surface integral,

q = ε0

∫ π

0dθ

∫ 2π

0dφσhθhφ

∣∣∣∣η=0

=2ε0aV

π

∫ π

0sin θdθ

∫ 2π

0dφ

= 8ε0aV.

This is consistent with the charge found earlier using the monopole potential.

If one uses a coordinate system other than the oblate spheroidal system, solutions will be much

more involved. Let us employ the cylindrical coordinates (ρ, φ, z). Because of axial symmetry, φ

31

dependence can be suppressed and we seek a solution in the form of Laplace transform,

Φ(ρ, z) =

∫ ∞0Φ(ρ, k)e−k|z|dk. (2.150)

The Laplace equation without φ dependence(∂2

∂ρ2+

1

ρ

∂

∂ρ+

∂2

∂z2

)Φ(ρ, z) = 0, (2.151)

becomes (d2

dρ2+

1

ρ

d

dρ+ k2

)Φ(ρ, k) = 0, (2.152)

which suggests that

Φ(ρ, k) = A(k)J0(kρ). (2.153)

The boundary conditions are:

Φ(ρ ≤ a, z = ±0) = V (constant).

The following integral has a peculiar property,

∫ ∞0

sin ax

xJ0(bx)dx =

π

2, if a > b,

sin−1(a/b), if a < b.

(2.154)

Exploiting this property, we can construct the following solution for the potential,

Φ(ρ, z) =2V

π

∫ ∞0

sin ka

kJ0(kρ)e−k|z|dk. (2.155)

The potential in the disk plane (z = 0) is

Φ(ρ, z = 0) =

V, if ρ < a,

2V

πsin−1(a/ρ), if ρ > a.

(2.156)

Example 7 Dipole Moment of a Conducting Disk in an External Electric Field

If a thin conductor disk is placed perpendicular to an external field, the dipole moment is zero

because of negligible thickness of the disk even though charge separation does take place in such

a manner that disk surfaces are oppositely charged. The external electric field is little disturbed

by the disk in this case. The maximum disturbance occurs when the disk surface is parallel to the

32

Figure 2-11: Conducting disk in an external electric field parallel to the disk surface.

field.

We assume a uniform external electric field in the x−direction and a thin conducting disk placedin the x− y plane with its axis in the z−direction as shown in Fig.2-11. The potential associatedwith the external uniform electric field is

Φ0 = −E0x

= −E0a cosh η sin θ cosφ. (2.157)

The “radial” function cosh η is actually P 11 (i sinh η) and the presence of the disk should yield

a perturbation proportional to the Legendre function of the second kind Q11(i sinh η) since the

perturbed potential should have the same angular dependence as Φ0(η, θ, φ) to satisfy the boundary

condition at the disk. Thus we assume

Φ(η, θ, φ) = −E0a cosh η sin θ cosφ+AQ11(i sinh η) sin θ cosφ, (2.158)

where Q11(i sinh η) is actually a real function,

Q11(i sinh η) = cosh η

(cot−1(sinh η)− sinh η

cosh2 η

). (2.159)

The constant A can be determined from the boundary condition that the disk potential be zero,

that is, Φ(η = 0) = 0. We thus find

A =2

πaE0,

33

and the potential becomes

Φ(η, θ, φ) = −aE0(

cosh η − 2

πQ11(i sinh η)

)sin θ cosφ. (2.160)

Far away from the disk at r a or η 1, the potential approaches

limη1

Φ(η, θ, φ)→ −aE0 cosh η sin θ cosφ+4E0a

3

3π

sin θ cosφ

r2, (2.161)

where the asymptotic form of Q11(i sinh η),

Q11(i sinh η) ' 2

3

1

sinh2 η=

2

3

(ar

)2, (2.162)

has been substituted. Comparing the dipole term in Eq. (2.161) with the standard dipole potential

Φdipole =1

4πε0

p · rr3

, (2.163)

we can readily identify the dipole moment induced by the disk,

p = 4πε04a3

3πE0‖, (2.164)

where E0‖ is the component of the external electric field tangential to the disk surface. Note that the

dipole moment is proportional to a3. The moment is equally applicable for low frequency oscillating

electric field as long as the wavelength associated with the oscillating filed is much longer than the

disk radius, ka = 2πλ a 1. A resultant scattering cross-section of a conducting disk (sphere too)

placed in a low frequency electromagnetic wave is proportional to a6.

Example 8 Leakage of Electric Field through a Small Hole in a Conducting Plate

Consider a parallel plate capacitor whose grounded, lower plate has a small circular hole of

radius a as shown in Fig.2-12. We wish to find how the hole perturbs the potential. This problem

has important applications in analyzing leakage of microwaves through a small hole in waveguide

walls.

The unperturbed electric field E0 between the plates is assumed downward with a corresponding

potential

Φ0(z) =

E0z, z > 0

0, z < 0

(2.165)

34

Figure 2-12: The lower plate of a parallel plate capacitor has a small hole of radius a. The electricfield leaks throught the hole.

where z = a sinh η cos θ. We note

sinh η = −iP1(i sinh η). (2.166)

Therefore, the perturbed potential can be sought in term of the Legendre function of the second

kind Q1(i sinh η) which is equivalent to

Q1(i sinh η) = sinh η cot−1(sinh η)− 1. (2.167)

We thus assume the following form for the potential in both regions,

Φ(η, θ) =

aE0 sinh η cos θ +A

[sinh η cot−1(sinh η)− 1

]cos θ, 0 < θ <

π

2

−A[sinh η cot−1(sinh η)− 1

]cos θ,

π

2< θ < π,

which ensures continuity of the potential at the hole (η = 0). Continuity of the normal component

of the electric field at the hole requires

∂Φ

∂η

∣∣∣∣ η=0z=+0

=∂Φ

∂η

∣∣∣∣ η=0z=−0

,

from which we readily find the constant A,

A = −aE0π.

35

In the region below the lower plate (z < 0), the potential is

Φ(η, θ) =aE0π

[sinh η cot−1(sinh η)− 1

]cos θ,

π

2< θ < π. (2.168)

Its asymptotic form is of dipole nature,

Φ(r a)→ −E0a3

3π

1

r2cos θ > 0, (2.169)

(note that cos θ < 0 in the region below the plate) and the effective dipole moment of the hole is

p =4πε0a

3

3πE0, (2.170)

which is downward. The far-field potential in the upper region (z > 0) is

Φ ' Φ0 +E0a

3

3π

1

r2cos θ, (2.171)

in which the dipole term is due to an effective dipole moment upward. The potential at the center

of the hole is

Φ(η = 0, θ = 0 or π) =aE0π. (2.172)

The results of this example, together with those of Example 14 in Chapter 3 (leakage of magnetic

field through a hole in a superconducting plate), will have important implications on diffraction of

electromagnetic waves by an aperture in a conducting plate. Since the effective dipoles are opposite

to each other in the two regions z > 0 and z < 0, it follows in general that

Ez(−z) = −Ez(z),

that is, the electric field normal to the plate is an odd function of z. This means that the surface

charges σ = ε0n ·E (C/m2) induced on both sides of the plate at z = +0 and z = −0 are identical,

where n is the unit normal vector at the plate surface. (Note that n changes its sign from one side

to other.) The component tangential to the plate,

Et = n×E,

is an even function of z,

Et(−z) = Et(z).

Of course, on the surface of the conducting plate, Et vanishes but it does not in the hole. For

36

magnetic fields resulting from a hole in an ideally conducting plate, we will see that the normal

component should vanish at the plate surface

Hz = 0, at z = ±0,

and off the plate, it is even with respect to z,

Hz(−z) = Hz(z),

while the tangential component Ht = n×H is an odd function of z,

Ht(−z) = −Ht(z).

It follows that the surface currents

Js = n×H, (A/m)

on both surfaces of the plate are identical.

2.5 Method of Inversion

The method of inversion is useful when an electrode has a spherical shape, either complete spheres

(e.g., two spheres touching) or incomplete sphere (e.g., spherical bowl, solid hemisphere, etc.). For

a given sphere of radius a which we call inverting sphere, the inverted position of a point at r is

defined by

ri =a2

r2r. (2.173)

(See Fig.2-13.) A sphere is inverted into another sphere. If the center of the inverting sphere is

chosen on the surface of a sphere to be inverted, the inverted surface becomes a plane as shown

in Fig.2-14. This is where the merit of method of inversion is found because potential problems of

planar electrodes are often simpler than those involving spheres.

Consider a charge q placed at r′ = (r′, θ′, φ′). The potential at position r = (r, θ, φ) is

Φ =q

4πε0

1√r2 + r′2 − 2rr′ cos γ

, (2.174)

where

cos γ = cos θ cos θ′ + sin θ sin θ′ cos(φ− φ′).

37

Figure 2-13: Point P at (r, θ, φ) is inverted with respect to the sphere of radius a to Q at (a2/r, θ, φ),i.e., at the image position.

Figure 2-14: If an inverting sphere is centered on a surface of a sphere to be inverted, the sphere isinverted to an infinite plane.

In the inverted space with respect to a sphere of radius a, a charge q′ will appear at( ar′

)2r′, (2.175)

and the position r is inverted to (ar

)2r. (2.176)

The potential at the inverted position is

Φi =q′

4πε0

1√a4

r2+a4

r′2− 2

a4

rr′cos γ

=q′

4πε0

rr′

a21√

r2 + r′2 − 2rr′ cos γ. (2.177)

38

In general, if a function Φ(r, θ, φ) satisfies the Laplace equation, the potential function

a

rΦ

(a2

r, θ, φ

), (2.178)

also satisfies the Laplace equation.

It should be noted that an equipotential spherical surface is in general not inverted to an

equipotential sphere. However, a spherical surface at zero potential is inverted to a zero potential

spherical surface. Since the reference potential can be chosen arbitrarily without affecting the

electric field, one can always choose the potential of an equipotential spherical surface at zero

potential. For example, the potential of a charged conducting sphere of radius a is

Φs =1

4πε0

q

a, (2.179)

relative to zero potential at infinity. However, we can subtract Φs from the potential everywhere

and choose the sphere potential at zero and the potential at infinity as

Φ∞ = − 1

4πε0

q

a.

The electric field remains unchanged through uniform shift of the potential. If an inverting sphere

is chosen in such a way that it has a radius 2a centered at the surface of the conducting sphere of

radius a, the conducting sphere is inverted to an infinite plane touching the both spheres as shown

in Fig. 2-15. Since the sphere potential is chosen at zero, the potential of the plane is also zero.

The potential at infinity is inverted to

− 1

4πε0

q

a× 2a

r= − 1

4πε0

2q

r, (2.180)

where r is the radial distance from the center of the inverting sphere with radius 2a. This is a

potential due to a point charge −2q. Therefore, a charge

−2q = −8πε0Φsa, (2.181)

appears at the center of the inverting sphere.

Example 9 Capacitance of Touching Spheres

Using the method of inversion, we can find the capacitance of two conducting sphere touching

each other as shown in Fig.2-15. The potential of the touching spheres is denoted by Φs. If the

inverting sphere has radius 2a and its center at the touching point, the two spheres become two

39

Figure 2-15: Touching spheres are inverted to parallel plates by a sphere of radius 2a centered atthe touching point. Images appear in the inverted space.

parallel planes separated by a distance 4a. A charge

−q = −8πε0aΦs, (2.182)

appears at the midpoint between the plates after inversion which can be analyzed easily using the

method of multiple images. The following image charges appear: q at |z| = 4a, −q at |z| = 8a, q

at |z| = 12a, · · ·. The amount of total charge on the surface of the original spheres can be found byre-inverting the image charges,

Q = 2q

(2a

4a− 2a

8a+

2a

12a− · · ·

)= q ln 2

= 8πε0aΦs ln 2. (2.183)

Therefore, the self-capacitance of the touching spheres is

C =Q

Φs= 8πε0a ln 2. (2.184)

The potential Φi(ρ, z) in the inverted space shown in Fig.2-16can be found in the form of Fourier

transform,

Φi(ρ, z) =

∫ ∞0

A(k) sinh[k(2a− |z|)]J0(kρ)dk, (2.185)

where A(k) is a weighting function to be determined. It is noted that the elementary solution to

the Laplace equation is

J0(kρ)e±kz, (2.186)

40

Figure 2-16: Geometry in the inverted space.

and the assumed form of the potential certainly satisfies the Laplace equation as well. The weighting

function A(k) can be determined by noting

d2

dz2sinh[k(2a− |z|)] = −2k cosh(2ak)δ(z), (2.187)

and ∫ ∞0

kJ0(kρ)dk =1

ρδ(ρ). (2.188)

The charge density of the point charge q at the origin is

ρc =q

2πρδ(ρ)δ(z). (2.189)

Then, A(k) can be determined from the Poisson’s equation

∇2Φi = −ρcε0, (2.190)

as

A(k) = − q

4πε0

1

cosh(2ak), (2.191)

and the potential in the inverted space is

Φi(ρ, z) = − q

4πε0

∫ ∞0

sinh[k(2a− |z|)]cosh(2ak)

J0(kρ)dk. (2.192)

The potential in the original configuration can be found by reinverting Φi through the transforma-

tion

z →(

2a

r

)2z, ρ→

(2a

r

)2ρ,

where

r2 = ρ2 + z2,

41

is the distance from the center. The result is

Φ(ρ, z) = − q

4πε0

2a

r

∫ ∞0

sinh

[k

(2a−

(2a

r

)2|z|)]

cosh(2ak)J0

[k

(2a

r

)2ρ

]dk, (2.193)

with r2 = ρ2+ z2. Recalling that we have subtracted Φs = q/4πε0a (the sphere potential) from the

potential everywhere to make the sphere potential vanish, we finally obtain

Φ(ρ, z) = Φs

1− (2a)2

r

∫ ∞0

sinh

[k

(2a−

(2a

r

)2|z|)]

cosh(2ak)J0

[k

(2a

r

)2ρ

]dk

. (2.194)

Figure 2-17: Geometry in the original space.

Example 10 Capacitance of Spherical Bowl

As a second example, we consider a hollow spherical bowl of radius a with an angle 2θ subtended

at the center shown in Fig.2-18. As inverting sphere, one can choose a sphere having a radius 2a sin θ

centered at the edge of the bowl. After inversion, the bowl becomes a semi-infinite plane as shown

and a charge q = −8πε0a sin θΦs will appear at the center of the inverting sphere. Potential

problems involving a semi-infinite conducting plate can be analyzed as a limiting case of a wedge.

For a charge q placed at (ρ′, φ′, z′) near a wedge intersecting at an angle α, the potential is given

42

Figure 2-18: A bowl (radius a, center angle 2θ) is inverted to a semi-infinite plane by a sphere ofradius 2a sin θ centered at the edge of the bowl.

by

Φ(ρ, φ, z) =2q

πε0α

∑m

∫∞0 Iν(kρ)Kν(kρ′) cos[k(z − z′)]dk sin(νφ) sin(νφ′), ρ < ρ′

∑m

∫∞0 Iν(kρ′)Kν(kρ) cos[k(z − z′)]dk sin(νφ) sin(νφ′), ρ > ρ′

(2.195)

where ν = mπ/α. Noting∫ ∞0

Iν(kρ)Kν(kρ′) cos[k(z − z′)]dk =1

2√

2ρρ′

∫ ∞η

e−νζ√cosh ζ − cosh η

dζ, (2.196)

where

cosh η =ρ2 + ρ′2 + (z − z′)2

2ρρ′, (2.197)

and the sum formula ∞∑m=1

pm cos(mx) =1

2

(1− p2

1− 2p cosx+ p2− 1

), (2.198)

43

we see that the potential reduces to

Φ(r) =q

4πε0

1

α√

2ρρ′

∫ ∞η

sinh(παζ)

×[

1

cosh(πζ/α)− cos[π(φ− φ′)/α]− 1

cosh(πζ/α)− cos[π(φ+ φ′)/α]

]× 1√

cosh ζ − cosh ηdζ. (2.199)

For a plate α = 2π, and this becomes

Φ(r) =q

4π2ε0

[1

Rcos−1

(−cos[(φ− φ′)/2]

cosh(η/2)

)− 1

R′cos−1

(−cos[(φ+ φ′)/2]

cosh(η/2)

)], (2.200)

where

R =√ρ2 + ρ′2 − 2ρρ′ cos(φ− φ′),

R′ =√ρ2 + ρ′2 − 2ρρ′ cos(φ+ φ′).

The potential in the vicinity of the charge q can be found by letting ρ′ = ρ− r, φ = φ′ = π− θ, η =

r/2a sin θ 1,

Φ(r) =q

4πε0

1

r− q

4πε0

1

4πa sin θ

(1 +

θ

sin θ

), (2.201)

where r is the distance from the charge. The correction due to the presence of the conducting plate

is therefore

∆Φ = − q

4πε0

1

4πa sin θ

(1 +

θ

sin θ

), (2.202)

and in the physical space, the far field potential due to a charged conducting bowl is in the form

Φ(r a) = − q

4πε0

1

2πr

(1 +

θ

sin θ

)=

1

π

a

r(sin θ + θ) Φs. (2.203)

Comparing with the standard monopole potential

Φ =Q

4πε0r,

we finally find the capacitance of the bowl,

C = 4ε0a(θ + sin θ). (2.204)

44

For a sphere, θ = π, we recover C = 4πε0a. For a disk of radius R, θ ' sin θ ' R/a 1, and we

also recover

C = 8ε0R.

The capacitance of a solid (or closed) hemisphere can be found in a similar manner and given

by

C = 8πε0a

(1− 1√

3

). (2.205)

This is left for an exercise.

2.6 Numerical Methods

Analytic solutions in potential problems can only be found for a limited number of applications, and

in practice, it is often necessary to resort to numerical analysis. In this section, we will estimate the

capacitance of a square conductor plate of side a.Mathematically speaking, this problem constitutes

an integral equation for the potential Φ, which is constant at the conductor,

Φ =1

4πε0

∮σ(r′)

|r− r′|dS′ = − 1

4π

∮1

|r− r′|∂Φ

∂n′dS′ = V = constant, (2.206)

where

σ = ε0En = −ε0∂Φ

∂n,

is the unknown surface charge density. The capacitance can be found from

C =1

Φ

∫σdS. (2.207)

As a very rough estimate, we recall that the capacitance of a circular disk of radius a is given

by

C = 8ε0a, (2.208)

and approximate the capacitance of the square plate by

C = 8ε0reff = 8ε0 × 0.564a, (2.209)

where reff is the radius of a circular disk having the same area as the plate,

πr2eff = a2, reff = 0.564a.

The finite element numerical method given below yields C ' 8ε0 × 0.547a.

45

Figure 2-19: A square conducting plate of side a is divided into 25 sub-areas. Because of symmetry,the number of unknown potentials is reduced to 6.

The capacitance is the ratio between the total charge Q and the plate potential V , C = Q/V.

We divide the plate into n× n sub-areas of equal size each with side a/n. Each sub-plate is at anequal potential V but charges on the sub-plates differ. To illustrate the procedure, we choose n = 5

(25 sub-areas) as shown in Fig. 2-19. Because of symmetry, there are 6 unknown charges to be

found. The potential on each sub-plate can be calculated by summing contributions from charges

on all sub-plates including the charge on itself. The self-potential of one unit can be estimated as

follows. Consider a square plate of side δ carrying a uniform surface charge density σ (C/m2). The

potential at the center of the plate can be found from

Φ =σ

4πε0

∫ δ/2

−δ/2dx

∫ δ/2

−δ/2dy

1√x2 + y2

=σ

4πε04

∫ δ/2

0

[ln(√

x2 + δ2 + δ)− lnx

]dx

=σ

4πε04δ ln

(1 +√

2)

=q

4πε0δ× 4 ln

(1 +√

2)

=q

4πε0δ× 3.5255, (2.210)

where q = σδ2 is the charge carried by the sub-plate. With this preparation, we can write down

the potential of sub-plate A as follows:

4πε0ΦAδ =

(3.5255 +

2

4+

1

4√

2

)qA +

(2 +

2

3+

2√17

+2

5

)qB

+

(1 +

2√20

)qC +

(1√2

+2√10

+1

3√

2

)qD +

(2√5

+2√13

)qE +

1

2√

2qF

= 4.2023qA + 3.5517qB + 1.4472qC + 1.5753qD + 1.4491qE + 0.3536qF . (2.211)

46

Other potentials ΦB ∼ ΦF , which are all equal, can be written down in a similar way and we obtain

6 simultaneous equations for qA ∼ qF which can be solved easily. A resultant total charge is

Q = 1.743× 4πε0Φδ,

and the capacitance is

C ' 0.3486× 4πε0a

= 0.547× 8ε0a. (2.212)

Accuracy will improve if a larger number of sub-areas are used.

The method can be applied to estimate the capacitance of a conducting cube as well. With 150

sub-areas (25 sub-areas on each side), the following capacitance emerges,

C ' 0.65× 4πε0a, (2.213)

where a is the side of the cube. An estimate based on a sphere having the same surface area gives

C = 4πε0reff = 0.69× 4πε0a, (2.214)

where

reff =

√6

4πa = 0.69a.

A well known finite element method of solving the Laplace equation is based on the fact that the

potential at the center of a cube may be approximated by the average of 6 surrounding potentials

on each face of the cube,

Φ0 =1

6

6∑i=1

Φi.

This follows from the Taylor expansion of the potential,

Φ(x± δ, y, z) = Φ(x, y, z)± δ ∂Φ

∂x+

1

2

∂2Φ

∂x2± · · ·,

Φ(x, y ± δ, z) = Φ(x, y, z)± δ ∂Φ

∂y+

1

2

∂2Φ

∂y2± · · ·,

Φ(x, y, z ± δ) = Φ(x, y, z)± δ ∂Φ

∂z+

1

2

∂2Φ

∂z2± · · ·,

Adding these 6 equations, we find

Φ(x± δ, y, z) + Φ(x, y ± δ, z) + Φ(x, y, z ± δ) = 6Φ(x, y, z) +∇2Φ +O(δ4). (2.215)

47

Therefore, if Φ satisfies the Laplace equation, ∇2Φ = 0,

Φ center '1

6

6∑i=1

Φi, (2.216)

valid to order δ3.

For 2-dimensional problems in which z-dependence is suppressed, we have

Φcenter '1

4

4∑i=1

Φi. (2.217)

The equation can be applied to each sub-unit having a volume δ3 (3-D) or area δ2 (2-D). Resultant

simultaneous equations can be solved numerically.

Example 11 Potential in a Long Cylinder

Consider a conducting cylinder having a cross-section as shown in Fig. ??. The periodic upper

electrode is at a potential V and the flat lower electrode is grounded. In order to apply the finite

element method, we divide the cross section into sub-sections and allocate 10 nodes points as shown.

Applying Eq. (2.217) to the potentials Φi(i = 1− 10), we obtain

4Φ1 = 100 + Φ2 + 2Φ3,

4Φ2 = Φ1 + 2Φ4,

4Φ3 = 100 + Φ1 + Φ4 + Φ5,

4Φ4 = Φ2 + Φ3 + Φ6,

4Φ5 = 100 + Φ4 + Φ6 + Φ9,

4Φ6 = Φ4 + Φ5 + Φ10,

4Φ7 = 300 + Φ8,

4Φ8 = 200 + Φ7 + Φ9,

4Φ9 = 2Φ5 + Φ8 + Φ10,

4Φ10 = 2Φ6 + Φ9.

Solutions are: Φ1 = 63.7, Φ2 = 31.0, Φ3 = 61.8, Φ4 = 30.2, Φ5 = 53.5, Φ6 = 27.9, Φ7 = 97.1,

Φ8 = 88.2, Φ9 = 55.7, Φ10 = 27.9 all in Volts. A larger number of node points will improve

accuracy.

48

Cross-section of long cylinder with periodic anode structure.

49

Problems

2.1 A ring charge of total charge q and radius b is coaxial with a long grounded conducting

cylinder of radius a (< b). Determine the potential everywhere.

2.2 A ring charge of total charge q and radius b is coaxial with a long uncharged dielectric cylinder

of permittivity ε and radius a. Determine the potential everywhere.

2.3 A charge q is placed at an axial distance b from a conducting disk of radius a. Determine

the potential everywhere. Consider two cases, (a) the disk is grounded, and (b) the disk is

floating.

2.4 Show that a charge q at a distance d from the center of a floating conducting spherical shell

of radius a raises the sphere potential to

Φs =q

4πε0d, if d > a (q outside the sphere),

or

Φs =q

4πε0a, if d < a (q inside).

2.5 A large grounded conducting plate has a hemispherical bob of radius a. A charge q is placed

at an axial distance d from the center of the bob. Find the force on the charge.

2.6 Show that the capacitance per unit length of a parallel wire transmission line with a common

wire radius a and separation distance d is

C

l=

πε0

ln

(d+√d2 − 4a2

2a

) .

2.7 A coaxial cable having inner and outer radii a and b is bent to form a thin toroidal capacitor

with a major radius R ( a, b). Find the capacitance.

2.8 Show that the mutual capacitance between conducting spheres of radii a and b separated by

a large distance d a, b is approximately given by

Cab ' 4πε0ab

d,

and that the capacitance of the sphere of radius a is affected by the sphere of radius b as

Caa ' 4πε0a

(1 +

ab

d2

).

50

2.9 Rigorous analysis of potential problems involving two conducting spheres can be made by

using the bispherical coordinates defined by

x =a sin θ cosφ

cosh η − cos θ,

y =a sin θ sinφ

cosh η − cos θ,

z =a sinh η

cosh η − cos θ.

η = constant surface is a sphere described by

x2 + y2 + (z − acotanhη)2 =

(a

sinh η

)2,

and θ = constant surface is

(ρ− acotanθ)2 + z2 =( a

sin θ

)2,

which is spindle-like shape.

(a) Finding the metric coeffi cients hη, hθ, and hφ, show that the Laplace equation in the

bispherical coordinates is

∂

∂η

(1

cosh η − cos θ

∂Φ

∂η

)+

1

sin θ

∂

∂θ

(sin θ

cosh η − cos θ

∂Φ

∂θ

)+

1

sin2 θ(cosh η − cos θ)

∂2Φ

∂φ2= 0.

(b) Show that the general solution to the Laplace equation is in the form

Φ(η, θ, φ) =√

cosh η − cos θ∑l,m

(Alme(l+ 1

2)η +Blme

−(l+ 12)η)Pml (cosh η)eimφ.

As in the oblate spheroidal coordinates, in this coordinate system too, the Laplace

equation is not separable.

2.10 Using the inversion method, show that the capacitance of a solid or closed conducting hemi-

sphere of radius a is given by

C = 8πε0a

(1− 1√

3

).

2.11 Find a 2D Green’s function for the interior of long cylinder having a semicircular cross-section

of radius a.

51

Hint: Assume

G(r, r′) =

∑mAm(ρ/ρ′)m sin(mφ) sin(mφ′), ρ < ρ′ < a,∑m [Bm(ρ/ρ′)m + Cm(ρ′/ρ)m] sin(mφ) sin(mφ′), ρ′ < ρ < a,

where r = (ρ, φ), r′ = (ρ′, φ′).

2.12 A conducting disk of radius a is placed parallel to an external electric field E0. The dominant

perturbation to the potential is dipole as shown in Example 6. What is the leading higher

order correction?

2.13 Cylindrical capacitors have cross-sections as shown. Estimate graphically the capacitance per

unit length for each. For (a), analytic expression for the capacitance is

C

l=

2πε0

cosh−1(ρ21 + ρ22 − d2

2ρ1ρ2

) ,where ρ1 = 4a, ρ2 = 2a, and d = a. For (b), one has to resort to numerical analysis for an

exact value.

2.14 Find numerically the capacitance of a conducting cube of side a. What do you estimate for

the lower and upper bounds of the capacitance?

2.15 Derive Eq. (??), the expression for the potential due to a point charge in the oblate spheroidal

coordinates (η, θ, φ) . The charge is at(η′, θ′, φ′

).

2.16 Derive Eqs. (2.138) and (2.139), the Green’s function for an oblate spheroid described by

η = η0 (const.)

2.17 The prolate spheroidal coordinates (η, θ, φ) is convenient to solve potential problems involving

prolate spheroids (sphere elongated along the z axis). The coordinate transformation is

52

defined by

x = a sinh η sin θ cosφ,

y = a sinh η sin θ sinφ,

z = a cosh η cos θ.

In the limit of η → 0, η = const. surface describes a thin rod having a length 2a, and in the

limit η → ∞, it approaches a sphere with radius a cosh η ' a sinh η. Show that the metric

coeffi cients are:

hη = a

√sinh2 η + cos2 θ = hθ,

hφ = a sinh η cos θ.

Then, show that general solution of Laplace’s equation ∇2Φ = 0 in the prolate spheroidal

coordinates is in the form

Φ (η, θ, φ) =∑l,m

[AlmPml (cosh θ) +BlmQ

ml (cosh η)] [ClmP

ml (cos θ) +DlmQ

ml (cos θ)] eimφ.

In the lowest order l = 0, possible one dimensional solutions are

Φ (η) = Q0 (cosh η) = ln coth(η

2

),

Φ (θ) = Q0 (cos θ) = ln cot

(θ

2

).

53

Chapter 3

Magnetostatics

3.1 Introduction

Magnetostatics is a branch of electromagnetic studies involving magnetic elds produced by steady

non-time varying currents. Evidently currents are produced by moving charges undergoing trans-

lational motion. An e¤ective current (called magnetization current) is also produced if magnetic

dipoles are nonuniformly distributed. To realize steady currents, a large number of charges must

be involved so that collection of charges can be regarded as a continuous uid. In non-time varying

cases (@=@t = 0), the charge conservation principle requires that

@

@t+r J = r J =0; in steady state.

This implies that a steady current is transverse. A static magnetic eld can also be produced by a

nonuniform magnetization (magnetic dipole density in the case of a collection of magnetic dipoles)

M (A/m) which produces a magnetization current

JM = rM; (A m2):

The magnetization current is transverse (divergence-free), since r JM = r (rM) 0:The magnetic eld B exerts a force perpendicular to the velocity of charged particles. The force

to act on a charge q is

F = q (E+ v B) :

The magnetic eld does not do any work on charged particles since the rate of work v F is

independent of B;

v F = qv E:

(Note that v (v B) =0:) In contrast to the electric eld which is a true (or polar) vector, the

1

magnetic eld is a pseudo (or axial) vector, or more precisely, constitutes a pseudotensor,

Bij =

264 0 Bz By

Bz 0 BxBy Bx 0

375 :The magnetic eld due to a prescribed current distribution can be calculated using the Biot-

Savarts law. This law is generic because all other laws in magnetostatics, such as the Amperes law

and vanishing divergence of the magnetic eld rB = 0; follow from it. Calculation of the magneticeld is facilitated by introducing a magnetic vector potential A; which is related to the magnetic

eld through B = r A: This expression is consistent with r B = 0; since r (rA) 0

identically. It also follows that r A = 0 in static cases.

One important objective of magnetostatics is to derive formulae for the self-inductance and

mutual inductance for given current congurations. Use of the vector potential greatly facilitates

calculation of the magnetic ux,

=

ZB dS =

ZrA dS =

IA dl;

and inductance,

L =

I:

3.2 Biot-Savarts Law and Vector Potential

Figure 3-1: Geometry to illustrate Biot-Savarts law.

2

For a given, non-time-varying current density distribution J(r); the magnetic eld B(r) can be

calculated from the Biot-Savarts law,

B(r) =04

ZJ(r0) (r r0)

jr r0j3dV 0; (T) (3.1)

where

0 = 4 107 (T m A1)

= 4 107 (H m1); (3.2)

is the vacuum magnetic permeability. As explained in Chapter 1, 0 is an assigned constant

introduced to dene 1 Ampere current. Noting

r r0

jr r0j3= r

1

jr r0j

; (3.3)

we may rewrite the law in the form

B(r) =04r

ZJ(r0)

jr r0jdV: (3.4)

Then, one of the Maxwells equations

r B = 0; (3.5)

immediately follows because of the identity for an arbitrary vector eld A;

r (rA) 0:

Physical meaning of r B =0 is that there be no magnetic charges in contrast to the electric eldr E = ="0; and that the magnetic eld is transverse without longitudinal component. Eq. (3.4)

implies that the magnetic eld is a curl of a vector eld dened by

A(r) =04

ZJ(r0)

jr r0jdV0: (3.6)

This vector quantity is called the magnetic vector potential. In terms of the vector potential, the

magnetic eld is thus given by

B(r) =rA: (3.7)

The magnetic eld given in Eq. (3.1) is a pseudo vector because it is invariant against coordi-

nates inversion, r! r: In contrast, the Coulomb electric eld,

E (r) =1

4"0

Z(r r0) (r0)jr r0j3

dV 0;

3

is a true (or polar) vector, because it changes sign on coordinate inversion.

As will be shown in Chapter 4, for time varying current density, the vector potential is to be

modied by taking into account the retardation due to nite propagation time of electromagnetic

disturbances,

A(r;t) =04

ZJ(r0; t )jr r0j dV 0;

where

=jr r0jc

:

B(r; t) =rA and r B = 0 continue to hold in general.The curl of the magnetic eld is

rB = rrA = r (r A)r2A: (3.8)

The divergence of the vector potential vanishes for static elds, since

r A =04

Zr

1

jr r0j

J(r0)dV 0

= 04

Zr0

1

jr r0j

J(r0)dV 0

= 04

Zr0

J(r0)

jr r0j

dV 0 +

04

Z1

jr r0jr0 J(r0)dV 0

= 0;

where it is noted that Zr0

J(r0)

jr r0j

dV 0 =

IJ(r0)

jr r0j dS0 = 0;

and that for steady current ow, r J = 0:Therefore, for static magnetic elds,

rB = r2A: (3.9)

However,

r2A =04

Zr2

1

jr r0j

J(r0)dV 0

= 04

Z4(r r0)J(r0)dV 0

= 0J(r): (3.10)

This means that the vector potential for a prescribed current distribution J(r)

A(r) =04

ZJ(r0)

jr r0jdV0;

4

is the particular solution for the vector Poissons equation

r2A = 0J(r); (3.11)

which is identical to the familiar Amperes law

rB = 0J: (3.12)

In summary, the Biot-Savarts law is generic in the sense other well known laws in magnetostatics

follow from it. In particular, we have derived the following vectorial relationships from the Biot-

Savarts law:

r B = 0; absence of magnetic monopoles (always)

rB = 0J; Amperes law for static elds

B = rA; always

r A = 0; for static elds; always in Coulomb gauge (by denition)

r2A = 0J static Amperes law in terms of the vector potential A

The Amperes law in Eq. (3.12) is valid only for static elds. For more general time-varying elds,

it is generalized as

rB = 0

J+ "0

@E

@t

; (3.13)

where

D = "0@E

@t; (A/m2) (3.14)

is the displacement current density originally conceived by Maxwell. The divergence of Eq. (3.13)

yields (note that "0r E = )

r (rB) = 0 = 0

r J+ @

@t

;

that is, the Maxwells equation in Eq. (3.13) is consistent with the charge conservation law,

@

@t+r J = 0: (3.15)

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

a¤ecting the electric and magnetic elds. However, for static elds, the divergence of the vector

potential identically vanishes. For time varying elds, the following choice is often made,

r A = 1c2@

@t; Lorentz gauge (3.16)

where is the scalar potential. As we will see, this choice for r A allows complete decoupling

5

between and A in the sense that the potentials satisfy similar inhomogeneous wave equations,r2 1

c2@2

@t2

= 1

"0(r; t); (3.17)

r2 1

c2@2

@t2

A = 0J(r; t): (3.18)

If the Coulomb gauge, characterized by r A = 0 (absence of longitudinal vector potential), is

used instead, such decoupling cannot be achieved. In Coulomb gauge, the scalar potential C and

transverse vector potential A? satisfy the following,

r2C = 1

"0(r; t); (3.19)

r2 1

c2@2

@t2

A? = 0J?(r; t); (3.20)

where J? is the transverse current. Solution of C (r; t) is non-retarded,

(r; t) =1

4"0

Z(r0; t)

jr r0jdV0; (3.21)

and so is the longitudinal electric eld,

EC (r; t) = rC =1

4"0

Z(r r0) (r0; t)jr r0j3

dV 0: (3.22)

As we will see in Chapter 4, this non-retarded electric eld is cancelled and observable electric eld

is in fact always retarded. Observable electromagnetic elds, E and B; should be independent of

the choice of gauge.

In the vector Poissons equation for the vector potential,

r2A = 0J; (3.23)

if the current density J is unidirectional, so is the vector potential and A k J holds. For example,if only Jz exists, the resultant vector potential is also in the z-direction, Az: If the current density

is azimuthal J; so is the vector potential, and only A exists. It is noted that in noncartesian

coordinates, r2A

i6= r2Ai; (3.24)

6

in general. For example, in the spherical coordinates,

r2A r(r A)r (rA)

=

r2Ar

2

r2Ar

2

r2@A@

2 cot r2

A 2

r2 sin

@A@

er

+

r2A

1

r2 sin2 A +

2

r2@Ar@

2 cos

r2 sin2

@A@

e

+

r2A

1

r2 sin2 A +

2

r2 sin

@Ar@

+2 cos

r2 sin2

@A@

e: (3.25)

If the current is purely azimuthal, J = Je; the azimuthal component of the vector potential

satises

r2A 1

r2 sin2 A = 0J: (3.26)

In axially symmetric cases with @=@ = 0; elementary solutions for

r2A 1

r2 sin2 A = 0; (3.27)

are

A (r; ) = rlP 1l (cos ) ;1

rl+1P 1l (cos ) : (3.28)

Then the lowest order harmonic is l = 1 (dipole) which is consistent with the absence of magnetic

monopoles.

3.3 Boundary Conditions

The vanishing divergence of the magnetic eld r B = 0 requires that the normal component of

the eld B be continuous at any boundary,

B1n = B2n: (3.29)

The static Amperes law,

rB = 0J; (3.30)

indicates that the tangential component can be discontinuous in the presence of a surface current,

n (B1 B2) = 0Js; (3.31)

where n is a unit vector normal to the surface and Js (A/m) is the surface current density. Thesurface current density Js is the total current density regardless of its origin, including translationalmotion of charge (conduction current), collection of magnetic dipoles (magnetization current), etc.

As will be shown later, for practical applications, it is convenient to separate the conduction current

Jcond = v ((r) the charge density and v = dr=dt the velocity of charge density) and rewrite the

7

Amperes law in terms of a vector H,

rH = Jcond; (3.32)

where

H =B

0M; (A m1) (3.33)

withM the magnetization vector or the magnetic dipole moment density. The boundary condition

for the tangential component of the vector H is

n (H1 H2) = Jcs; (3.34)

where Jcs is the conduction surface current density having dimensions of A m1.

3.4 Some Examples

In this section, vector potentials will be worked out for some simple current congurations.

Example 1 Ring Current

Figure 3-2: Ring current. Because of axial symmetry, nding the vector potential on the y zplane will su¢ ce.

Consider a thin circular ring current I with radius a: The current is unidirectional in the

direction and the current density can be written as

J = I(r a)

a (cos ) : (3.35)

8

Therefore, the vector potential is also unidirectional and only the azimuthal component A is

nonvanishing. When Ar = A = 0; in Eq. (3.25), the component of the vector Poissons equation

r2A = 0J(r); (3.36)

in the spherical coordinates isr2 1

r2 sin2

A = 0J = 0I

(r a)a

(cos ) ; (3.37)

where

r2 = @2

@r2+2

r

@

@r+

1

r2 sin

@

@

sin

@

@

; (3.38)

is the scalar Laplacian. Note that @=@ = 0 because of axial symmetry of the problem. Since

bounded general solutions to the di¤erential equation@2

@r2+2

r

@

@r+

1

r2 sin

@

@

sin

@

@

1

r2 sin2

A(r; ) = 0; (3.39)

are

rlP 1l (cos );1

rl+1P 1l (cos ); (3.40)

the solution for A(r; ) may be assumed in the form

A(r; ) =

8>>>>><>>>>>:

Xl

Al

ra

lP 1l (cos ); r < a

Xl

Al

ar

l+1P 1l (cos ); r > a

(3.41)

where the continuity of A at r = a is required from the continuity in the radial component of the

magnetic eld,

Br(r; ) =1

r sin

@

@(sin A): (3.42)

The appearance of P 1l (cos ); even though the problem is axisymmetric, is due to the nature of the

vector Laplace equation. Eq. (3.39) is formally identical to scalar Laplace equation with m = 1:The coe¢ cient Al can be determined by multiplying Eq. (3.39) by P 1l0 (cos ) sin and integrating

the result over = 0 to ;

Xl

Al

d2

dr2+2

r

d

dr l(l + 1)

r2

gl(r)

Z 1

1P 1l ()P

1l0d = 0I

(r a)a

P 1l0 (0); (3.43)

9

where = cos and

gl(r) =

8>>><>>>:ra

l; r < a

ar

l+1; r > a

(3.44)

Noting Z 1

1P 1l ()P

1l0 ()d =

2

2l + 1

(l + 1)!

(l 1)!ll0 =

2

2l + 1l(l + 1)ll0 ; (3.45)

P 1l (0) =

8>>><>>>:(1) l12 l!!

(l 1)!! ; l odd

0; l even

(3.46)

and the singularity at r = a;d2

dr2gl(r) = (2l + 1)

(r a)a

; (3.47)

we nally obtain

Al = 0IP 1l (0)

2l(l + 1)= 0I(1)

l12(l 2)!!2(l + 1)!!

; l = 1; 3; 5; (3.48)

and

A(r; ) = 0I1Xl=1

gl(r)P 1l (0)

2l(l + 1)P 1l (cos ): (3.49)

Note the absence of the monopole potential, l = 0:

Far away from the ring r a; the vector potential is dominated by the dipole term (l = 1) as

expected,

A(r; ) '0mz

4r2sin ; (3.50)

or in vector form,

A =04

m rr3

; (3.51)

where

m = a2I ez; (3.52)

is the magnetic dipole moment of the ring current,

m =1

2

Zr JdV

=1

2

Z(e)rJr2 sin drdd

=I

2

Z(sin ez cos e)

(r a)(cos )a

r3 sin drdd

= a2I ez:

10

The vector potential of the ring current can alternatively be found from the following direct

integration,

A(r) =04

ZJ(r0)

jr r0jdV0; (3.53)

where for a lamentary ring current, the volume integral can be replaced with line integral,

J(r0)dV 0 = I dl = Ia d0e0 :

Because of the axial symmetry, the vector potential can be evaluated at an arbitrary location and

we choose = =2; a point on the yz plane. Then only the xcomponent of Iad0e0 contributesto the integral,

A(r; ) =04Ia

Z 2

0

sin0pr2 + a2 2ar cos

d0;

where

cos = cos cos2

+ sin sin

2

cos2 0

= sin sin0:

By changing the variable from 0 to through

2 = 0 +

2;

the integral can be manipulated into the following form,

A(r; ) =0Ia

1pr2 + a2 + 2ar sin

1

k2(2 k2)K(k2) 2E(k2)

; (3.54)

where

k2 =4ar sin

r2 + a2 + 2ar sin ; (3.55)

and K(k2) and E(k2) are the complete elliptic integrals of the rst and second kind, respectively,

dened by

K(k2) =

Z =2

0

1p1 k2 sin2

d; E(k2) =

Z =2

0

p1 k2 sin2 d: (3.56)

Using this result, the inductance of a thin conductor ring of loop radius a and wire radius b

with a b can be estimated. The magnetic ux enclosed by the ring current itself is

=

ZSB dS =

ICA dl: (3.57)

11

Figure 3-3: A thin conductor ring. The magnetic ux enclosed by the ring can be found from =

HA dl where A = A is the vector potential on the surface of the ring.

On the inner surface of the wire, r = a b; and = =2: The argument of the elliptic functions k2

approaches unity and we may approximate the elliptic functions by

limk2!1

K(k2) = ln

4p1 k2

= ln

8a

b

; lim

k2!1E(k2) = 1; (3.58)

where

1 k2 = 1 4a(a b)(a b)2 + a2 + 2a(a b) '

b2

4a2: (3.59)

Then, the vector potential on the wire surface is

A '0I

2

ln

8a

b

2; (3.60)

and the magnetic ux enclosed by the ring is

= 2(a b)A

' 0aI

ln

8a

b

2: (3.61)

The external self inductance of the ring is therefore given by

Lext =

I= 0a

ln

8a

b

2: (3.62)

External inductance pertains to the magnetic ux outside the conductor. Magnetic eld also

exists in the conductor. For a straight conducting wire, the internal inductance per unit length is

Lil=08; (H/m). (3.63)

12

The internal inductance Li is dened by

1

2LiI

2 = magnetic energy in the conductor. (3.64)

For a straight wire of radius a carrying a current I uniformly distributed across the cross-section,

the magnetic eld in the conductor can be found by using Amperes law,

2B() = 02

a2I;

B() = 0I

2a2: (3.65)

As will be shown, the magnetic energy density is

1

20B2; (J m3): (3.66)

Then, the magnetic energy stored per unit length of the wire isZ a

0

1

20B2()2d =

1

2

08I2; (J m1), (3.67)

which is independent of the wire radius a:This denes the internal inductance per unit length,

Lil=08; (H m1). (3.68)

If the current density is not uniform as in strong skin e¤ect, the internal inductance must be

evaluated according to Eq. (3.64).

In the cylindrical coordinates (; ; z); the vector potential A(; z) satises@2

@2+1

@

@+

@2

@z2 1

2

A(; z) = 0I( a)(z): (3.69)

The solution for this equation can readily be found in terms of Fourier transform involving the

modied Bessel functions I1(k) and K1(k) as

A(; z) =0Ia

8><>:R10 I1(k)K1(ka) cos(kz)dk; < a;

R10 I1(ka)K1(k) cos(kz)dk; > a:

(3.70)

This expression or its alternative based on Laplace transform,

A(; z) =0Ia

2

Z 1

0J1(ka)J1(k)e

kjzjdk; (3.71)

is useful in calculating the self-inductance of a uniformly wound solenoid.

13

Example 2 Rotating Charged Conducting Sphere

If a conducting sphere of radius a carrying a surface charge density rotates about its axis at

an angular frequency !, the current density becomes singular and is given by

J = !a(r a) sin : (3.72)

Solutions for the vector potential satisfying

Figure 3-4: Rotating charged conductor sphere.

@2

@r2+2

r

@

@r+

1

r2 sin

@

@

sin

@

@

1

r2 sin2

A(r; ) = 0!a(r a) sin ; (3.73)

can be sought in the form

A(r; ) =

8>>><>>>:PlAl

ra

lP 1l (cos ); r < a

PlAl

ar

l+1P 1l (cos ); r > a

(3.74)

where Al is a constant. The radial function

gl(r) =

8>>><>>>:ra

l; r > a

ar

l+1; r < a

(3.75)

14

has discontinuity in its derivative at the surface r = a and thus

d2

dr2gl(r) = (2l + 1)

(r a)a

: (3.76)

The sin dependence of the singular current density requires that only the l = 1 harmonic be

present,

P 11 (cos ) = sin :

Therefore, the solution for the vector potential is

A(r; ) =0!a

2

3

8>>><>>>:r

asin ; r < a

ar

2sin ; r > a

The outer eld is of pure dipole with an e¤ective dipole moment

mz =4!a4

3=!qa2

3; q = 4a2:

The exterior magnetic eld is given by

B = rA=

0mz

4r3(2 cos er + sin e) :

The interior magnetic eld is uniform,

B =20!a

3ez

=20!a

3(cos er sin e); r < a: (3.77)

The polar component of the magnetic eld B is discontinuous at the surface with a jump

B(r = a+ 0)B(r = a 0) = 0!a sin = 0Js; (3.78)

where

Js = !a sin ; (A/m)

is the surface current density.

Example 3 Rotating Charged Sphere

15

If a uniformly charged sphere of radius a and uniform charge density (C m3) rotates about

its axis at an angular frequency ! (rad s1); the current density J(r; ) is

J(r; ) =

8><>:!r sin ; r < a;

0; r > a:

(3.79)

The vector potential satises

@2

@r2+2

r

@

@r+

1

r2 sin

@

@

sin

@

@

1

r2 sin2

A(r; ) =

8><>:0!r sin ; r < a;

0; r > a:

(3.80)

In the region outside the sphere r > a where there is no current, the solution may be assumed to

be

A(r; ) =Xl

Al

ar

l+1P 1l (cos ); r > a: (3.81)

Interior solution may be sought in the form of a sum of general and particular solutions. The

particular solution can be readily found by assuming

A(r; ) = Br3 sin : (3.82)

Substituting this into@2

@r2+2

r

@

@r+

1

r2 sin

@

@

sin

@

@

1

r2 sin2

A(r; ) = 0!r sin ; (3.83)

we nd the particular solution,

AP (r; ) = 1

100!r

3 sin ; (3.84)

and the interior solution may be assumed to be

A(r; ) =Xl

Bl

ra

lP 1l (cos )

1

100!r

3 sin ; r < a; (3.85)

where the rst term in the RHS is the general solutions. Continuity of the magnetic eld at the

sphere surface requires that both A and @A=@r be continuous. Then, only the l = 1 component

is nonvanishing and we easily nd

A1 =1

150!a

3; B1 =1

60!a

3: (3.86)

16

As in the preceding example, the eld outside the sphere is pure dipole. This is due to the particular

angular dependence of the current density J _ sin : Comparing the outer vector potential

A(r; ) =1

150!a

5 sin

r2; (3.87)

with the standard dipole eld, we nd the dipole moment of the rotating charged sphere

mz =4

15!a5: (3.88)

The reader should check this result agrees with the dipole moment directly calculated from

m =1

2

ZVr J dV: (3.89)

Example 4 Long Straight Currents

Let us assume a straight current I directed along the z axis. The cartesian component of the

vector potential Az satises scalar Poissons equation

r2Az = 0Jz; (3.90)

where

Jz = I(x)(y): (3.91)

The Poissons equation is mathematically identical to that for the scalar potential due to a long

line charge,

r2 =

"0(x)(y);

which has a solution

() =

2"0ln ;

where =px2 + y2 is the radial distance in the cylindrical coordinates. By analogy, we nd

Az() = 0I

2ln : (3.92)

A resultant magnetic eld is the familiar one,

B = rAz =0I

2e: (3.93)

17

For an innite array of identical z directed currents I located in the plane x = 0 at y = 0; a;2a; , the potential is given by

Az(x; y) = 0I4

ln(x2 + y2)[x2 + (y a)2][x2 + (y + a)2]

= 0I

4ln

1 +

x2

y2

1 +

x2

(y a)2

1 +

x2

(y + a)2

y2(y a)2(y + a)2

= 0I

4ln

cosh

2x

a

cos

2y

a

+ constant, (3.94)

where the following closed forms of the innite products are used,

1 +

x2

y2

1 +

x2

(y a)2

1 +

x2

(y + a)2

=

cosh2xa

cos

2ya

1 cos

2ya

; (3.95)

y2(y a)2(y + a)2 =1 cos

2y

a

constant. (3.96)

Calculation of the magnetic eld B(x; y) is left for exercise.

3.5 Inductance

The magnetic ux enclosed by a closed current circuit is in general proportional to the current

I: The external inductance is dened by the ratio,

Le =

I: (3.97)

In linear systems without ferromagnetic materials, the inductance is a constant determined solely

by geometrical factors. The magnetic ux can be calculated in terms of either the magnetic eld

B or the vector potential A since

=

ZSBdS =

ZS(rA)dS =

ICA dl: (3.98)

As an example, consider a parallel wire transmission line with a common wire radius a and

center to center separation distance d: In the absence of skin e¤ect (uniform current density), the

vector potential outside the wires is given by

Az =0I

2ln

21

; 1; 2 > a: (3.99)

Then, the magnetic ux enclosed by the unit length of the transmission line is

l= 2 1

2

IAz(1 = a; 2)d; (3.100)

18

Figure 3-5: Parallel wire transmission line. The current in the conductors is assumed to be uniform(no skin e¤ect).

where is the azimuthal angle about the center of the current I; and

2 =pd2 + a2 2ad cos: (3.101)

The integral needed is

1

2

Z 2

0ln

"d

a

2+ 1 2d

acos

#d = 2 ln

d

a

: (3.102)

Then the ux per unit length is

l=0I

ln

d

a

; (Wb m1) (3.103)

and the external inductance per unit length is

Lel=0ln

d

a

; (H m1). (3.104)

This holds for an arbitrary ratio of d=a (> 1) as long as the current ows uniformly in both wires

(no skin e¤ect). Each wire has an internal inductance of

Lil=08; (H m1). (3.105)

Therefore, the total inductance per unit length of the transmission line is

L

l=0

ln

d

a

+1

4

; (H m1), (uniform current). (3.106)

19

If the current is concentrated at the wire surface as in the case of strong skin e¤ect, the internal

inductance vanishes and the external inductance is modied as

Lel=0ln

d+

pd2 a22a

!; (H m1), (strong skin e¤ect), (3.107)

which is dual of the capacitance of parallel wire transmission line,

C

l=

"0

ln

d+

pd2 a22a

! ; (F m1). (3.108)

The similarity between the capacitance and inductance is due to the fact that the scalar potential

and the cartesian component of the vector potential Az both satisfy the scalar Laplace equation,

r2 = 0; r2Az = 0; (3.109)

in the case of parallel wire transmission line.

Example 5 Inductance of a Solenoid

The inductance of closely wound solenoid of circular cross-section as shown in Fig.3-6 can be

found using the vector potential of a single current loop of radius a,

A(; z) =0Ia

8>>>><>>>>:

Z 1

0I1(k)K1(ka) cos(kz)dk; < a;

Z 1

0I1(ka)K1(k) cos(kz)dk; > a:

(3.110)

Consider a solenoid of radius a and length l with a total number of windings of N: The vector

potential at arbitrary axial location z on the solenoid surface = a is

A(a; z) =0NIa

l

Z l

0dz0Z 1

0I1(ka)K1(ka) cos[k(z z0)]dk: (3.111)

A di¤erential segment of length dz has a number of windings Ndz=l and the magnetic ux linked

to the segment is

d = 2aAN

ldz

= 0I

N

l

22a2dz

Z l

0dz0Z 1

0I1(ka)K1(ka) cos[k(z z0)]dk: (3.112)

The total ux linked to the solenoid is therefore

= 0I

N

l

22a2

Z l

0dz

Z l

0dz0Z 1

0I1(ka)K1(ka) cos[k(z z0)]dk; (3.113)

20

and the inductance is given by

L =

I(3.114)

= 0

N

l

22a2

Z l

0dz

Z l

0dz0Z 1

0I1(ka)K1(ka) cos[k(z z0)]dk; (H). (3.115)

This is exact as long as windings are dense enough so that only the azimuthal component of the

vector potential is present.

Figure 3-6: Solenoid with uniform winding density.

For a long solenoid l a; we note

limx!0

I1(x)K1(x) =1

2; (3.116)

and Z 1

0cos[k(z z0)]dk = (z z0): (3.117)

Then, the inductance reduces to the familiar form of the inductance of a long solenoid,

L ' 0N2a2

l= 0

a2

lN2; (H) ; (long solenoid l a). (3.118)

An expression valid to order (a=l)2 can be worked out using the approximation that the magnetic

eld is axial everywhere inside the solenoid. The magnetic eld on the axis of a circular loop current

is

Bz(z) =0I

2

a2

(z2 + a2)3=2: (3.119)

21

For a solenoid of winding number density n, length l and radius a; the axial magnetic eld is

Bz(z) =0nIa

2

2

Z l

0

dz0

[(z0 z)2 + a2]3=2

=0nI

2

l zp

(l z)2 + a2+

zpz2 + a2

!: (3.120)

For a long solenoid, the magnetic eld inside the solenoid can be approximated by the axial eld.

Therefore, the ux linkage is

' 0n2I

2a2

Z l

0

l zp

(l z)2 + a2+

zpz2 + a2

!dz

= 0n2a2I

pl2 + a2 a

; (3.121)

and the self-inductance is given by

L ' 0a2

N

l

2 pl2 + a2 a

; l a; (3.122)

' 0a2

N

l

2l a+ 1

2

a2

l

: (3.123)

The exact inductance in Eq. (3.114) may be written in the form

L = 0(aN)

l

2

f(a=l);

where f(a=l) is the correction factor given by

fal

=

2

l

Z l

0dz

Z l

0dz0Z 1

0I1(ka)K1(ka) cos[k(z z0)]dk

=8

Z 1

0

sin2(x=2)

x2I1

alxK1

alxdk:

This had been tabulated by Nagaoka. The function f(a=l) is shown in Fig.3-7.

3.6 Neumanns Formula for Mutual Inductance

The mutual inductance among electric circuits can be conveniently calculated using Neumanns

formula. Consider two closed current loops C1 and C2: To nd the mutual inductance of the two

loops, we let the loop 1 carry a current I1 and calculate the magnetic ux linked to the loop 2. The

vector potential created by the current I1 at a point on the loop 2 is

A =0I14

IC

dl1r12

; (3.124)

22

x 21.510.50

1

0.8

0.6

0.4

0.2

0

Figure 3-7: Nagaoka factor f(a=l) for a solenoid of radius a and length l: In the gure, x = a=l:

where r12 is the distance between the segment dl1 and the position on loop 2 as illustrated in

Fig.3-8. Then, the magnetic ux enclosed by loop 2 is given by

12 =

IA dl2

=0I14

I Idl1 dl2r12

: (3.125)

The mutual inductance between the two loops is dened by

Figure 3-8: Geometry for two circuits mutual inductance (Neumanns formula).

23

M12 =12I1

=04

I Idl1 dl2r12

: (3.126)

It is evident that the mutual inductance is reciprocal,

M12 =M21; (3.127)

and depend only on the geometrical shapes and their mutual orientation.

Example 6 Coaxial Parallel Square Loops

Figure 3-9: Coaxial parallel square loops. The mutual inductance can be found from the sum ofcontributions from all parallel and anti-parallel pairs.

The mutual inductance between two identical square loops, which are coaxial and whose sides

are parallel, can be calculated as a sum of mutual inductances of parallel wires. If two parallel

wires of length a are a distance d apart, the mutual inductance is

M =04

Z a

0dl1

Z a

0dl2

1p(l1 l2)2 + d2

=02

Z a

0ln

x+

px2 + d2

d

!dx

=02

"a ln

a+

pa2 + d2

d

!pa2 + d2 + d

#; (H). (3.128)

Applying this results to all parallel pairs of the square loops, we nd

M12 =20

"a ln

a+

pa2 + d2

d

!pa2 + d2 + d

a ln

a+

p2a2 + d2pa2 + d2

!+p2a2 + d2

pa2 + d2

#; (H). (3.129)

Example 7 Mutual Inductance of Coaxial Circular Loops

24

Figure 3-10: Coaxial circular loops.

If two circular loops of radii a and b are coaxial and their planes are a distance c apart, the

mutual inductance can readily be found from the vector potential at the loop B due to a current I

in the loop A,

A =0Ia

1p(a+ b)2 + c2

1

k2(2 k2)K(k2) 2E(k2)

; (3.130)

where

k2 =4ab

(a+ b)2 + c2: (3.131)

The magnetic ux enclosed by the loop B is thus

= 2bA

=20abIp(a+ b)2 + c2

1

k2(2 k2)K(k2) 2E(k2)

; (3.132)

and the mutual inductance is

Mab =1

20p(a+ b)2 + c2

(2 k2)K(k2) 2E(k2)

; (H). (3.133)

An alternative expression for the mutual inductance can be found in terms of spherical harmonic

expansion. The vector potential at the loop B is, if a <pb2 + c2;

A(r =pb2 + c2; 0) = 0I

Xl odd

ap

b2 + c2

l+1 P 1l (0)

2l(l + 1)P 1l (cos 0); (3.134)

where

cos 0 =bp

b2 + c2: (3.135)

25

If a >pb2 + c2;

A(r =pb2 + c2; 0) = 0I

Xl odd

pb2 + c2

a

!lP 1l (0)

2l(l + 1)P 1l (cos 0): (3.136)

The ux linked to the loop B is therefore given by

= 2bA(r =pb2 + c2; 0); (3.137)

and the mutual inductance is

Mab = b0Xl odd

ap

b2 + c2

l+1 P 1l (0)

l(l + 1)P 1l (cos 0); if a <

pb2 + c2; (3.138)

and

Mab = b0Xl odd

pb2 + c2

a

!lP 1l (0)

l(l + 1)P 1l (cos 0); if a >

pb2 + c2: (3.139)

Figure 3-11: Geometry for calculating the mutual inductance of two circular loops whose axesintersect at an angle :

If two circular loops are not coaxial but their axes intersect as shown in Fig.3-11, the mutual

inductance can be calculated as follows. First, we note the vector potential due to a circular current

of radius a sin residing on a sphere of radius a is given by (see Eq. (3.49))

A(r; ) = 0IXl

ar

l+1 sin

2l(l + 1)P 1l (cos)P

1l (cos ): (3.140)

26

The radial magnetic eld is

Br =1

r sin

@

@(sin A)

=0I sin

2r

Xl

ar

l+1P 1l (cos)Pl(cos ); (3.141)

where use is made of the identity,

1

sin

d

d[sin P 1l (cos )] = l(l + 1)Pl(cos ): (3.142)

The magnetic ux linked to the loop B can be found by integrating the radial magnetic eld over

the incomplete spherical surface at radius b: Denoting the angular location on the sphere by (0; 0)

where 0 is measured from the axis of the loop B and 0 is the angle about it, we note that the

argument of the Legendre function Pl(cos ) becomes

cos = cos 0 cos + sin 0 sin cos0:

Then the Legendre function can be expanded as,

Pl(cos 0 cos +sin 0 sin cos) = Pl(cos

0)Pl(cos )+21Xm=1

Pml (cos 0)Pml (cos ) cosm; (3.143)

and the magnetic ux through the loop B is

= b2Z

0d0Z 2

0d0 sin 0Br(r = b)

= 0Ib sin sinXl

1

l(l + 1)

ab

l+1P 1l (cos)P

1l (cos)Pl(cos ); (3.144)

where the integration over reduces toZ

0sin 0Pl(cos

0)d0 =sin

l(l + 1)P 1l (cos): (3.145)

The desired mutual inductance is

Mab = 0b sin sinXl

1

l(l + 1)

ab

l+1P 1l (cos)P

1l (cos)Pl(cos ); (H). (3.146)

3.7 Multipole Expansion of Static Vector Potential

The vectorial nature of the vector potential prohibits us from expanding the vector potential directly

in terms of the spherical harmonics in contrast to the case of scalar potential. However, the

vector potential can be indirectly expanded in spherical harmonic functions. In magnetostatics,

27

the Coulomb gauge r A = 0 allows us to write A in terms of another vector function F as

A = r F; (3.147)

since r (r F) 0 holds identically. The vector F introduced may be regarded as a generatingfunction for the vector potential and it may not have any physical meanings. The vector eld F

can be decomposed into radialor longitudinalcomponent, and transversecomponent in the

form

F = r + rr; (3.148)

where and are scalar functions. If the vector potential satises vector Laplace equation r2A =

0; then and both satisfy scalar Laplace equation,

r2 = 0; r2 = 0; (3.149)

and thus can be expanded in spherical harmonics,

; =Xl;m

Almr

l +Blmrl+1

Ylm(; ): (3.150)

The transverse function does not contribute to the magnetic eld B: (Why? Note the identity

r (rr) = 0.) Then, can de discarded and

A = rr : (3.151)

The di¤erential operator

rr = e1

sin

@

@+ e

@

@; (3.152)

is the well known angular momentum operator and operates only on the harmonic function Ylm(; ):

Evidently, Eq. (3.151) generates components A and A only, but the remaining radial component

Ar can be calculated from r A = 0: Note that the radial component Ar; which exists in

r (rr);

does not contribute to the magnetic eld.

Let us revisit the vector potential due to a ring current. Because of the axial symmetry, the

scalar function may be expanded in the form

(r; ) =

8>>>><>>>>:Pl

al

ra

lPl(cos ); r < a

Pl

al

ar

l+1Pl(cos ); r > a

(3.153)

28

and the vector potential can be calculated as

A = rr = e@

@

= eXl

algl(r)P1l (cos ); (3.154)

where

gl(r) =

8>>><>>>:ra

l; r < a

ar

l+1; r > a

(3.155)

and use is made ofd

dPl(cos ) = P 1l (cos ): (3.156)

The vector potential is therefore consistent with that worked out in Example 1.

If the current density J(r) is conned in a limited spatial region, the vector potential at a

su¢ ciently large distance can be Taylor expanded as

A(r) =04

ZJ(r0)

jr r0jdV0

=04

1

r

ZJ(r0)dV 0 +

04

1

r3

Z(r r0)J(r0)dV 0 + (3.157)

where r r0 is assumed. The rst term in RHS vanishes for static current, sinceZJidV

0 =

Zrr0i JdV 0

=

Zr (r0iJ)dV 0

Zr0ir JdV 0

= 0;

Note that for static currents, rJ = 0 holds. The absence of monopole vector potential is consistentwith the absence of magnetic charges (magnetic monopoles), rB = 0: Therefore, the lowest orderfar eld vector potential is the dipole eld,

A(r) ' 04

1

r3

Z(r r0)J(r0)dV 0: (3.158)

Using

r (r0J) = r0(r J) (r r0)J;

29

and Zr0i(r J)dV 0 =

Zr0irjJjdV

0 = rj

Zr0iJjdV

0

= rj

Zr0i(rr0j J)dV 0

= rj

Zr (r0ir0jJ)dV 0 rj

Zr0j(rr0i J)dV 0

= rjZr0jJid

0 = Z(r r0)JdV 0;

the integral can be reduced toZ(r r0)J(r0)dV 0 = r 1

2

ZJ(r0) r0dV 0 (3.159)

and the dipole vector potential can be written in the form

A(r) =04

m rr3

; (3.160)

where m is the magnetic dipole moment,

m =1

2

ZVr JdV; (A m2): (3.161)

For a lamentary loop current, the volume integral reduces to line integral,

JdV = Idl;

and the dipole moment can be found from the contour integral,

m =1

2I

ICr dl = IS; (3.162)

where

S =1

2

ICr dl; (3.163)

is the area enclosed by the loop.

3.8 Magnetic Materials

If magnetic dipoles are continuously distributed, it is convenient to introduce a magnetic dipole

moment density,

M(r) = n(r)m(r); (A m1); (3.164)

where n(r) is the number density of dipoles and m(r) is the dipole moment. Although individual

dipoles do not carry macroscopic current, a collection of many dipoles can e¤ectively produce

30

a macroscopic current. However, such an e¤ective current is divergence-free and thus does not

contribute to charge transfer. The vector potential in the presence of distributed magnetic dipoles

can be approximated by

A(r) =04

ZJc(r

0)

jr r0jdV0 +

04

ZM(r0) (r r0)

jr r0j3dV 0;

where Jc(r) is the conduction current due to translational motion of charges. The second term in

RHS is the contribution from the distributed magnetic dipoles. Singling out the conduction current

Jc in magnetostatics is similar to separating the electric charges into free and bound charges in

dielectrics,

(r) =1

4"0

Zf (r

0)

jr r0jdV0 +

1

4"0

ZP(r0) (r r0)jr r0j3

dV 0:

Notingr r0

jr r0j3= r0 1

jr r0j ;

ZM(r0) (r r0)

jr r0j3dV 0 =

ZM(r0)r0 1

jr r0jdV0

= Z

r0 M(r0)

jr r0j r0 M(r0)jr r0j

dV 0

=

IS

M(r0)

jr r0j dS+Z r0 M(r0)

jr r0j dV 0

=

Z r0 M(r0)jr r0j dV 0; (3.165)

we nd that the curl of the magnetic eld becomes

rB = r2A = 0Jc + 0rM: (3.166)

where use is made of the identity

r2 1

jr r0j = 4(r r0):

Rearranging Eq. (3.166) yields

rB

0M

= Jc: (3.167)

It is customary to introduce a vector H dened by

H =B

0M; (3.168)

31

and an alternative form of Amperes law is

rH = Jc: (3.169)

In vacuum and nonmagnetic media, M = 0; and

B = 0H;

holds. The permeability of linear magnetic materials is dened by

B = H:

Permanent magnets are highly nonlinear. In permanent magnets, there are no conduction current,

Jc = 0; and thus

rH = 0; (3.170)

must holds. This means that the vector H can be assigned a scalar potential,

H = rm: (3.171)

In permanent magnets, the two vectors B and H are in general oriented in opposite directions as

shown in the following simple example.

Example 8 Magnetized Sphere

If an iron sphere of radius a is uniformly magnetized with an axial uniform magnetization

M = M0ez;the elds B and H can be found as follows. The scalar potential m satises the

Laplace equation,

r2m = 0:

In the sphere r < a the uniform magnetization produces uniform magnetic elds B and H;

Bz = 0(Hz +M0); r < a; (3.172)

where Bz and Hz are constants. Then the solution for the scalar potential may be assumed as

m(r; ) =

8>><>>:Hzr cos ; r < a

Hza3

r2cos ; r > a

(3.173)

where continuity of m; that is, continuity of the tangential component of H; H; at the surface

r = a is taken into account. The continuity of the radial component of the magnetic eld Br yields

0(M0 +Hz) = 20Hz; (3.174)

32

or

Hz = 1

3M0: (3.175)

This is the intensity of the vector H inside the sphere. The interior magnetic eld is

Figure 3-12: Spherical permanent magnet. The B prole is shown qualitatively in the left gureand H prole in the right gure. Note the discontinuity in the H eld lines at the surface.

Bz =2

30M0: (3.176)

The eld outside the sphere is of dipole type,

B = 0H =0M0a

3

3r3(2 cos er + sin e) : (3.177)

In the absence of conduction current, rH = 0 requiresIH dl = 0; (3.178)

along an arbitrary closed path. If the path intersects with the magnetized sphere, the vector H

must change its sign at the sphere surface in order to satisfyHHdl = 0: The magnetic eld B

found in this example is mathematically identical to that in Example 2 worked out for a rotating

charged conducting sphere. This is not surprising because an e¤ective current density associated

with a uniform magnetization indeed yields

Jeff = rM= M0(r a) sin e; (3.179)

which is identical to that due to a rotating surface charge on a conducting sphere.

33

In the absence of conduction current, Jc = 0; the following integral over the entire volume

should vanish, Zall space

B HdV = 0: (3.180)

This is because the LHS can be written in terms of the scalar potential m as

ZB rmdV =

Zr (Bm)dV +

Zmr BdV = 0: (3.181)

The magnetic energy associated with a permanent magnet can thus be calculated from either

Em =Zall space

1

20H

2dV; (3.182)

or

Em = Zmagnet

1

20M HdV: (3.183)

In the case of spherical permanent magnet, the total magnetic energy is

Em =1

20H

2z

4

3a3 +

1

20

Zr>a

H2r +H

2

dV

=2

270M

20a3 +

4

270M

20a3

=2

90M

20a3; (3.184)

where the elds in the exterior region r > a

Hr =2M0a

3

3r3cos ; H =

M0a3

3r3sin ;

have been substituted. The energy agrees with that from

Zmagnet r<a

1

20M HdV = 1

60M

20

4

3a3 =

2

90M

20a3:

Example 9 Image Currents

If a current I is placed parallel to a at surface of highly permeable bloc, an image current

in the same direction as the current itself appears. This is because of the boundary condition

that the tangential component of the magnetic eld should vanish at the surface. The image of a

current segment perpendicular to the surface is opposite to the current also because of the boundary

condition.

34

Image currents I 0 in iron (left) and superconductor (right).

If the permeability of the bloc is nite, the vector potential in air region can be calculated by

summing contributions from the current I itself and an image current in the iron,

I 0 = 0+ 0

I;

and the vector potential in the bloc by assuming a current

I 00 =2

+ 0I;

at the location of the current.

For a superconducting surface, the normal component of the magnetic eld should vanish.

Therefore, the image of a current parallel to superconducting surface is opposite, while for a vertical

current, the image is in the same direction. Fig. ?? illustrates the two cases.

3.9 Magnetic Force and Stress Tensor

The volume force to act on a current density J is

f = JB; (N m3): (3.185)

Substituting J = rB=0; we may rewrite f as

f =1

0(rB)B

=1

0

(B r)B 1

2rB2

: (3.186)

However,

r (BB) = (r B)B+ (B r)B= (B r)B: (3.187)

35

Therefore, the force density can be written in the form

f =1

0r BB B2

21

; (3.188)

where 1 is the unit tensor. The magnetic stress tensor is dened by

Tm =1

0

BB B2

21

: (3.189)

This is reminiscent to the electric stress tensor we have encountered in Chapter 1,

Te = "0

EE E2

21

: (3.190)

The magnetic energy density is

um =B2

20; (J/m3 = N/m2) (3.191)

and the total magnetic energy is

Um =

ZB2

20dV

=1

20

ZB (rA)dV

=1

20

Zr (AB)dV + 1

20

ZA (rB)dV

=1

2

Z(A J)dV; (J). (3.192)

For a closed current loop, this may be written as

Um =1

2I

ICA dl;

where the vector potential can be divided into that part due to the current I itself and another

part due to currents in other circuits. Recalling the denition of self and mutual inductances, the

magnetic energy associated with a single current loop can be written as

Um1 =1

2LI2 +

1

2IXi

MiIi; (3.193)

where Mi is the mutual inductance between the circuit under consideration and other circuits. For

two loops, the total magnet energy stored is given by

Um =1

2L1I

21 +

1

2L2I

22 +M12I1I2; (3.194)

36

because of reciprocity M12 = M21: The mutual interaction energy M12I1I2 can be either negative

or positive depending on the directions of current ows I1 and I2: However, the total magnetic

energy Um is evidently positive denite.

As electric forces act so as to increase self and mutual capacitances, magnetic forces tend to

increase inductances. The force in the direction of geometrical metric factor i contained in the

inductances can be calculated from

Fi =@

@i

1

2L(i)I

2

; (3.195)

for the case of self-inductance, and mutual interaction force from

Fi =@

@i(M12(i)I1I2) : (3.196)

For example, the self-inductance of a circular current loop of loop radius a and wire radius is

L = 0a

ln

8a

2 + li

2

; (3.197)

where li is the factor of internal inductance due to the magnetic energy stored in the wire. For

uniform current distribution (no skin e¤ect), li = 1=2: If the loop carries a current I; it tends to

expand so as to increase a with a force

Fa =1

2I2

@

@aL(a)

=1

20I

2

ln

8a

1 + li

2

: (3.198)

Example 10 Toroidal Magnet

Consider a toroidal magnet with a major radius R; minor radius a; and number of windings N:

By the Amperes law, the magnetic eld in the magnet can be readily found,

B = 0NI

2; R a < < R+ a:

The magnetic ux is

=0NI

Z R+a

Ra

qa2 (a )2

d

= 0NIR

pR2 a2

: (3.199)

Therefore, the inductance of the magnet is

L = 0N2R

pR2 a2

: (3.200)

37

The magnet tends to shrink in the major radius direction with a force

FR =1

20(NI)

2 @

@R

R

pR2 a2

=

1

20(NI)

2

1 Rp

R2 a2

< 0: (3.201)

In the minor radius direction, the magnet tends to expand with a force

Fa =1

20(NI)

2 @

@a

R

pR2 a2

=

1

20(NI)

2 apR2 a2

> 0: (3.202)

In large fusion devices, mechanical support structures must be able to handle these magnetic forces

which are enormous.

3.10 Boundary Value Problems

As discussed briey, the Maxwells equation for the magnetic eld B;

r B = 0; (3.203)

requires that the normal components of B be continuous across a boundary of two magnetic media,

B1n = B2n: (3.204)

For the eld H;

rH = Jc; (3.205)

the tangential components may be a provided there exists a surface conduction current Js (A/m)

owing on the boundary surface,

n (H1 H2) = Js; (3.206)

In the absence of conduction surface current, the tangential components ofH should be continuous.

On the surface of highly permeable medium such as iron, the magnetic eld lines falls almost

normal to the surface provided iron body and current circuits are not linked topologically. Linked

and unlinked examples are shown in Fig. 3-13. In unlinked cases, the eld H in iron should vanish

if iron 0: Since

B1 cos 1 = B2 cos 2; or 1H1 cos 1 = 2H2 cos 2; (3.207)

and

H1 sin 1 = H2 sin 2; (3.208)

38

Figure 3-13: Iron rings topologically unlinked (left) and linked (right) to a current loop I: Forunlinked ring, H = 0; while for linked ring, H 6= 0 in iron even if 0: The magnetic eld linesfall normal to iron surface if a current is unlinked.

we nd1

tan 1=

2tan 2

: (3.209)

Therefore, if 2 1; the angle 1 must approach 0. The eld H2 in iron should vanish also.

In linked cases, the eld H remains nite even in iron and should satisfyIH dl = I; (3.210)

along a path in iron. The magnetic eld lines do not necessarily fall normal to the iron surface.

For example, if a current is placed at the center of hollow iron cylinder, the magnetic eld line at

the iron surface is tangential everywhere.

Example 11 Ring Current around a Long Iron Core

We consider a circular ring current I of radius b which is coaxial with a long, highly permeable

iron cylinder of radius a as shown in Fig.3-14. If the iron cylinder has no return legs, it is not

topologically linked to the current ring. Therefore, the boundary condition for the magnetic eld

at the iron surface is

Bz( = a; z) = 0: (3.211)

Because of axial symmetry, the vector potential A(; z) can be found from the following equation,@2

@2+1

@

@+

@2

@z2 1

2

A(; z) = 0I( b)(z): (3.212)

Fourier transformation through

A(; z) =1

2

Z 1

1A(; k)e

ikzdk; (3.213)

39

Figure 3-14: A ring current coaxial with an iron cylinder.

reduces the problem to one dimensional,d2

d2+1

d

d k2 1

2

A(; k) = 0I( b): (3.214)

Bounded solutions in each region may be assumed as

A(; k) =

8><>:AI1(k) +BK1(k); a < < b;

CK1(k); > b;

(3.215)

where the coe¢ cients A;B; and C may be functions of the Fourier variable k: Continuity of A(; k)

at = b yields

AI1(kb) +BK1(kb) = CK1(kb): (3.216)

The axial magnetic eld is

Bz =1

@

@(A): (3.217)

Since1

d

d[I1(k)] = kI0(k);

1

d

d[K1(k)] = kK0(k); (3.218)

the boundary condition Bz = 0 at the cylinder surface gives

B =I0(ka)

K0(ka)A; (3.219)

40

and

C =

I1(kb)

K1(kb)+

I0(ka)

K0(ka)

A: (3.220)

The coe¢ cient A can be found from the discontinuity of the radial derivatives of the vector potential,

d2

d2A(; k)

=b

= Ak

I1(kb)

K1(kb)K 01(kb) I 01(kb)

( b)

= A

bK1(kb)( b); (3.221)

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

I 0m(x)Km(x) Im(x)K 0m(x) =

1

x:

Comparing with the RHS of the original equation in Eq. (3.214), we nd

A = 0IbK1(kb); (3.222)

and the solution of the physical vector potential is

A(; z) =0Ib

8>>>><>>>>:

Z 1

0K1(kb)

I1(k) +

I0(ka)

K0(ka)K1(k)

cos(kz)dk; a < < b;

Z 1

0K1(k)

I1(kb) +

I0(ka)

K0(ka)K1(kb)

cos(kz)dk; > b:

(3.223)

The rst terms indicate the vector potential due to the current ring alone while the second terms are

corrections due to the presence of the iron cylinder. The iron cylinder increases the ring inductance

by an amount

L = 20b2

1Z0

I0(ka)

K0(ka)K21 (kb)dk: (3.224)

The magnetic eld can be calculated from

B = rA:

The radial component is, for a < < b;

B(; z) = @A@z

=0Ib

Z 1

0K1(kb)

I1(k) +

I0(ka)

K0(ka)K1(k)

k sin(kz)dk; (3.225)

41

and for > b;

B(; z) =0Ib

Z 1

0K1(k)

I1(kb) +

I0(ka)

K0(ka)K1(kb)

k sin(kz)dk: (3.226)

The axial component in the region a < < b is

Bz(; z) =1

@

@(A)

=0Ib

Z 1

0K1(kb)

I0(k)

I0(ka)

K0(ka)K0(k)

k cos(kz)dk; (3.227)

and for > b;

Bz(; z) = 0Ib

Z 1

0K0(k)

I1(kb) +

I0(ka)

K0(ka)K1(kb)

k cos(kz)dk: (3.228)

If the permeability of the iron cylinder is not innite but nite , the vector potential in the

region a < < b is modied as

A(; z) =0Ib

1Z0

(K1(kb)I1(k) + f(k)K1(kb)K1(k)) cos(kz)dk; (3.229)

where

f(k) =( 0)kaI0(ka)I1(ka)

( 0)kaI1(ka)K0(ka) + 0:

Derivation of this modication is left for an exercise.

Example 12 Magnetic Shielding

Figure 3-15: Cross-section of a cylindrical iron shell placed in an external magnetic eld. The eldinside the shell is greatly reduced.

42

An iron cylinder having inner and outer tic eld radii a; b and permeability is placed in

an external magnetic eld B0 with its axis perpendicular to the eld. The magnetic eld inside

the cylinder is greatly reduced if 0 even if the thickness of the cylinder is small. Since no

conduction currents are present, the magnetic eld H can be generated from a scalar potential mwhich satises the Laplace equation,

r2m = 0; (3.230)

or @2

@2+1

@

@+1

2@2

@2

m(; ) = 0: (3.231)

The external magnetic eld can be generated from

0 = H0 cos: (3.232)

Since the magnetic eld due to the presence of the iron cylinder should have the same angular

dependence, we may assume the following solutions in each region,

m(; ) =

8>>>><>>>>:c1 cos; 0 < < a;c2+

c3

cos ; a < < b;

c4cosH0 cos; b < :

(3.233)

The boundary conditions are both m and @m=@ be continuous at = a and b: Then,

c1a = c2a+c3a; (3.234)

c2b+c3b=c4bH0b; (3.235)

0c1 = c2

c3a2

; (3.236)

c2

c3b2

= 0

c4b2+H0

: (3.237)

These simultaneous equations can be readily solved. For the interior magnetic eld, only c1 is

needed. Its solution is

c1 = 4H0b2

0

1

b21 +

0

2 a2

1

0

2 : (3.238)

If = 0; we recover c1 = H0: If 0; we nd

c1 = 4H00

b2

b2 a2 ; (3.239)

43

and thus the interior magnetic eld

Hi = 4H00

b2

b2 a2 H0: (3.240)

Example 13 Superconducting Disk in an External Magnetic Field

In this example, we analyze how a superconducting disk disturbs a uniform external magnetic

eld. The boundary condition for the magnetic eld at a surface of superconducting body is that

the normal component vanish because magnetic eld cannot penetrate into superconductors. The

same boundary condition should also apply for ordinary conductors in oscillating magnetic eld if

the skin depth given by

=1p

f0; (m) (3.241)

is small compared with conductor dimensions. In both cases, electric currents ow at the conductor

surface in a manner to cancel the external magnetic eld.

Figure 3-16: Superconducting disk in an external magnetic eld with its face normal to the eld.

If the disk is ideally thin, the greatest disturbance to the magnetic eld occurs when the disk

axis is parallel to the magnetic eld because in this case, the magnetic dipole moment induced

on the disk surface becomes maximum. The potential of the external uniform magnetic eld is as

before,

m0 = H0z = H0r cos : (3.242)

It is convenient to use the oblate spheroidal coordinates (; ; ): A thin disk is described by = 0:

Because of axial symmetry, @=@ = 0; general solutions to the Laplaces equation may be assumed

as

m(; ) =Xl

[AlPl(i sinh ) +BlQl(i sinh )]Pl(cos ): (3.243)

Since the external eld has cos = P1(cos ) dependence, only the l = 1 harmonic is relevant, and

44

the potential reduces to

m(; ) = [AP1(i sinh ) +BQ1(i sinh )] cos : (3.244)

Noting z = a sinh cos in the oblate spheroidal coordinates, and P1(i sinh ) = i sinh ; we nd

A = iaH0: (3.245)

The coe¢ cient B can be found from the boundary condition B = 0 at the disk surface, namely,

@m@

= 0 at = 0: (3.246)

Since

Q1(i sinh ) = sinh arccot(sinh ) 1; (3.247)

we readily nd

B =2

aH0; (3.248)

where arccot(0) = =2 is noted. The solution for m(; ) is thus given by

m(; ) = aH0

sinh + 2

(sinh arccot(sinh ) 1)

cos : (3.249)

The behavior of the potential at r a can be found using the asymptotic form of Q1(i sinh );

Q1(i sinh )! 1

3 sinh2 ; 1; (3.250)

with result

m( 1; ) ' H0r cos 2a3H03r2

cos : (3.251)

The dipole moment of the disk placed perpendicular to an external magnetic eld is thus given by

m = 42a3

3H0: (3.252)

Let us check if this is consistent with the dipole moment expected from

m =1

2

Zr JdV = 1

2

Zr JsdS;

where

Js = nH= 2

H0

pa2 2

e; (3.253)

45

which exists on both sides of the disk. Then,

mz = 2H0

Z a

0

2pa2 2

2d

= 83H0a

3;

which agrees with that identied from the far eld potential. Note thatZ 1

0

x3p1 x2

dx =2

3:

Example 14 Leakage through a Hole in a Superconducting Plate

A uniform magnetic eld parallel to a superconducting plate exists in one of the regions sepa-

rated by the plate. The plate has a hole of radius a:We wish to nd the magnetic eld in the other

region. This problem is similar to Example 7 in Chapter 2. The leakage led is expected to be of

dipole nature. Since the unperturbed eld can be described by the potential

0 = H0x = H0a cosh sin cos; (3.254)

in the oblate spheroidal coordinates (; ; ); we assume m

m(; ; ) = H0a cosh sin cos+AQ11(i sinh ) sin cos; (3.255)

where

Q11(i sinh ) = cosh

arccot(sinh ) sinh

cosh2

: (3.256)

It is convenient to dene the function arccotx without discontinuity in its derivative. The asymp-

totic forms are:

arccotx ' +1

x 1

3x3+ ; x! 1; (3.257)

arccotx ' 1

x 1

3x3+ ; x! +1: (3.258)

? 0 corresponds to z ? 0; respectively. In the region z ! 1; the uniform eld should vanish

and the coe¢ cient A can thus be readily found,

A =aH0

: (3.259)

In the upper region, the far eld potential becomes

m(; ; ) = H0x+2H0a

3

3

r xr3

; r a; z > 0: (3.260)

46

while in the lower region

m(r) = 2H0a

3

3

r xr3

; r a; z < 0: (3.261)

The e¤ective magnetic dipole moments for the region below the hole is

m = 8a3

3H0; z < 0: (3.262)

In the region z > 0; the e¤ective magnetic dipole moment is

m =8a3

3H0; z > 0: (3.263)

The formulae derived here may be conveniently used in analyzing leakage radiation from a small

hole drilled in the wall of waveguides.

Figure 3-17: Leakage of magnetic eld through a hole in a superconducting plate.

47

Problems

3.1 Find the magnetic eld at the center of (a) an equilateral triangular loop and (b) square loop

both having side a and carrying current I:

3.2 Find the magnetic eld at point P:

3.3 A current I ows along a at spiral coil described by

() =R

2N;

where N is the number of turns and R is the outermost radius of the coil. Show that the

magnetic eld along the axis is given by

Bz(z) =0NI

R

"ln

R+

pR2 + z2

z

! Rp

R2 + z2

#:

3.4 An elliptic loop of axes a and b (a b) carries a current I: Show that the magnetic eld at

the center of the loop is given by

Bz = 0Il

4S;

where

l = 4aE

1 b2

a2

;

is the circumference of the loop and S = ab is its area. Ek2is the complete elliptic

integral of the second kind dened by

Ek2=

Z =2

0

p1 k2 sin2 d:

Discuss the limiting cases of a = b and a b:

3.5 Find the mutual interaction force between a circular loop current I1 of radius a and a coplanar

straight long current I2 when the center of the loop is at a distance b from the line current.

48

3.6 Show that an alternative expression for the vector potential due to a circular loop current

(radius a; current I) is

A(; z) =0Ia

2

Z 1

0J1(ka)J1(k)e

kjzjdk:

3.7 The loop in the preceding problem is placed with its plane parallel to the surface of a large

magnetic bloc of permeability at a distance d: Using the method of image, show that the

inductance of the loop increases by

L = 0a 0+ 0

2

k kK(k2) 2

kE(k2)

;

where

k =ap

a2 + d2:

3.8 A circular current loop of radius a is concentric with an iron sphere of innite permeability

of radius b. Show that the presence of the iron sphere increases the loop inductance by

L = 0b1Xl=0

(2l 1)!!(2l)!!

2 ba

4l+2:

3.9 Show that the mutual inductance between two circular loops of radii a and b with parallel

49

axes a distance c apart and loop planes a distance d apart is given by

M = 0ab

Z 1

0J1(ka)J1(kb)J0(kc)e

kddk:

3.10 Evaluate the Nagaoka factor numerically for a=l = 0:1; 0:5 and 1.

3.11 A current ows along a circular tunnel in a highly permeable body. Do the magnetic eld lines

fall normal to the tunnel surface? What about the case in which a return current coexists?

Draw qualitatively the magnetic eld lines for both cases.

3.12 A conductor having a rectangular cross-section a b carries a uniform axial current density

J0 = J0ez: Find the magnetic eld:

50

3.13 A current sheet I with a width 2a is placed on one of iron walls as shown. Show that the

vector potential in the gap is given by

Az = 0I

4a

Z a

alnhcosh

d(x x0)

cos

dyidx0:

3.14 A current I is placed on the midplane of air gap formed by iron as shown. Find the vector

potential and magnetic eld in the gap.

3.15 A thin superconducting ring of major radius a and minor (wire) radius b ( a) is placed in

an external magnetic eld with its plane perpendicular to the eld. Determine the vector

potential and the magnetic dipole moment induced by the ring.

3.16 If the magnetic eld and current density are parallel to each other, the Lorentz force JBvanishes and Amperes law reduces to

rB = B;

where (m1) is a constant. Such arrangement is pertinent to designing superconducting

wires which are fragile and also to low pressure plasma equilibria such as Reversed Field

Pinch (RFP) and Spheromak. Solve the equation in (a) the cylindrical coordinates assuming

B = 0; @=@z = @=@ = 0; and (b) in spherical coordinates with the boundary condition Br= 0 at r = a:

51

Chapter 4

Time Varying Fields, Simple Waves

4.1 Introduction

In this Chapter, the charge density and current density J are generalized to be varying with time.

The charge conservation law@

@t+r J = 0; (4.1)

imposes a constraint between the two quantities (charge density in C m3) and J (the current

density in A m2) and various electrodynamic laws must be consistent with this basic law. A time

varying magnetic ux induces an electric eld through Faradays law,ICE dl = d

dt

ZSB dS: (4.2)

Likewise, a time varying electric eld induces a magnetic eld through the displacement current,ICB dl =0"0

d

dt

ZSE dS; (4.3)

even in the absence of the current J = 0. As is well known, the displacement current played a

crucial role in Maxwells prediction that electromagnetic waves can propagate through vacuum.

The generalized set of Maxwells equations,

r E =

"0; (4.4)

rE = @B@t; (4.5)

r B = 0; (4.6)

rB = 0J+ "0

@E

@t

; (4.7)

1

is consistent with the charge conservation law, since the divergence of the LHS of the last equation

indeed vanishes (recall that r (rB) =0 identically) which requires that

r J+ "0@

@tr E = r J+ @

@t= 0: (4.8)

4.2 Faradays Law

In 1831, Faraday discovered that an electric current was induced along a conductor loop when a

magnetic ux enclosed by the loop changed with time. This important discovery gave an answer

to the old question prior to Faradays time whether a magnetic eld could induce an electric eld

because it had been known that an electric eld, via an electric current, could induce magnetic

eld. What Faraday found was that an electric eld (or electromotive force, emf) was induced by

a time varying magnetic ux. The integral form of the Faradays law,

emf =IE dl = d

dt

ZSB dS; (4.9)

was later put into a di¤erential form by Maxwell,

rE = @B@t: (4.10)

The negative sign is due to Lenz and indicates that the emf is so induced as to oppose the change in

the magnetic ux. It should be noted that many of the experiments originally done by Faraday were

actually due to motional emf in which motion of conductors across a magnetic eld was responsible

for generation of emf without apparent time variation of the magnetic eld itself. An object moving

in a magnetic eld experiences an e¤ective electric eld given by

Eeff = v B: (4.11)

This may be seen by noting that change in the magnetic ux enclosed by a loop consists of two

parts, one due to time variation of the magnetic eld,

d1 =

Z@B

@t dSdt;

and the other due to the change in the shape of the loop,

d2 =

ZB @S

@tdt

=

IB (v dl)dt

= I(v B) dldt:

2

Figure 4-1: Induction of emf in time varying magnetic eld (upper gures) and motional emf instatic magnetic eld (lower gure).

Then the total ux change is

d

dt=

Z@B

@t dS

I(v B) dl;

and the emf induced along the loop is given by

emf =IE dl =

Z@B

@t dS+

I(v B) dl: (4.12)

An emf can be generated by letting a conductor move across a stationary magnetic eld as done in

most electric generators.

4.3 Displacement Current and Wave Equation

As shown in Introduction, the displacement current density

"0@E

@t;

plays a crucial role for the Maxwells equations to be consistent with the charge conservation law,

@

@t+r J = 0:

3

The magnetic eld induced by the displacement current may be best visualized in a capacitor being

slowly charged as shown in Fig.4-2. The current ows on the surface of electrodes. The radially

Figure 4-2: The surface currents on the capacitor electrode are consistent with the magnetic eldsboth outside and inside.

outward surface current on the outer surface of the upper plate,

Js1 =I

2; (4.13)

is consistent with the boundary condition for the magnetic eld,

n1H = Js1; (4.14)

where

H = I

2e; (4.15)

is the magnetic eld expected from the Amperes law. In the space between the electrodes, no

conduction current exists but there exists an azimuthal magnetic eld,

B() = 1

20"0

@Ez@t; (4.16)

as if there were a uniform conduction current equal to

Jz = "0@Ez@t: (4.17)

The magnetic eld is required to exist to satisfy the boundary condition at the inner electrode

surface,

n2 H = Js2; (4.18)

4

where

Js2 = I

2a2; (4.19)

is the radially inward surface conduction current on the inner surface of the upper plate. Since

I =dq

dt; (4.20)

and

Ez = q

"0a2; (4.21)

we nd

H = "0@Ez@t

1

2: (4.22)

This is consistent with the Maxwells equation,

1

@

@(H) = "0

@Ez@t; (4.23)

or its integral form,

2H = 2"0@Ez@t: (4.24)

As is well known, the displacement current was instrumental for Maxwell to predict that elec-

tromagnetic elds obey a wave equation. To see what wave equations the electromagnetic elds

should satisfy, let us take a curl of the Faradays law,

rE = @B@t;

r (rE) = @@trB: (4.25)

The LHS can be expanded as

r (rE) = r(r E)r2E:

Since

r E =

"0; rB = 0

J+ "0

@E

@t

;

Eq. (4.25) reduces to the following inhomogeneous wave equation,r2 1

c2@2

@t2

E =

1

"0r+ 0

@J

@t: (4.26)

where c is the speed of light in vacuum,

c2 =1

"00: (4.27)

5

Likewise, the magnetic eld obeysr2 1

c2@2

@t2

B = 0r J: (4.28)

If the displacement current were absent (which, incidentally, is equivalent to the assumption that

c!1), both elds would merely satisfy vector Poissons equation,

r2E = 1

"0r+ 0

@J

@t;

r2B = 0r J;

which do not exhibit any propagation nature with a nite speed.

Equation (4.26) can be solved symbolically as

E =1

r2 1

c2@2

@t2

1

"0r+ 0

@J

@t

; (4.29)

where1

r2 1

c2@2

@t2

; (4.30)

is the propagator integral operator which yields a retarded solution for the electric eld,

E(r; t) = 1

4"0

Z r0(r0; t )jr r0j dV 0 0

4

Z1

jr r0j@

@(t )J(r0; t )dV 0

= 1

4"0rZ(r0; t )jr r0j dV 0 0

4

@

@t

ZJ(r0; t )jr r0j dV 0; (4.31)

where

=jr r0jc

; (4.32)

is the time required for electromagnetic disturbances to propagate over a distance jr r0j betweenthe source at r0 and the observer at r: Note thatZ r0(r0; t )

jr r0j dV 0

=

Z r0(r0; t )jr r0j

r r0

jr r0j3(r0; t )

dV 0

= rZ(r0; t )jr r0j dV 0:

In Eq. (4.31),

(r; t) =1

4"0

Z(r0; t )jr r0j dV 0; A (r; t) =

04

ZJ(r0; t )jr r0j dV 0;

6

are the retarded scalar and vector potential, respectively. For a single charged particle moving at

a velocity vp (t) ; the scalar potential becomes

(r; t) =e

4"0

Zr0 rp

t 1

c jr r0j

jr r0j dV 0

=e

4"0

1

(1 n ) jr rpj

ret; (4.33)

where jret means every time dependent quantity is to be evaluated at the retarded time t0 to bedetermined implicitly from

t t0 1c

r rp t0 = 0:Likewise, the vector potential due to a moving charge is

A (r; t) =e04

v

(1 n ) jr rpj

ret: (4.34)

These potentials were rst formulated by Lienard and Wiechert. We will use the potentials in

Chapter 8 in formulating the radiation electromagnetic elds due to moving charges.

Equation (4.26) can also be written as

E =1

r2 1

c2@2

@t2

1

"0r+ 0

@J

@t

=1

r2 1

c2@2

@t2

1

"0r+ 0

@Jl@t

+ 0@Jt@t

; (4.35)

where Jl is the longitudinal component of the current density satisfying

@

@t+r Jl = 0; (4.36)

and Jt = J Jl is the transverse component which does not a¤ect the charge density because ofthe identity (or denition)

r Jt = 0: (4.37)

Since r Jl = 0; it followsr(r Jl) = r2Jl;

and a symbolic solution for the longitudinal current density is

Jl = 1

r2r@@t: (4.38)

7

Substituting into Eq. (4.35), we obtain

E =1

"0r2r+ 1

r2 1

c2@2

@t2

0@Jt@t: (4.39)

This formulation is consistent with the choice of Coulomb gauge for the scalar and vector potentials

as we will see in more detail in Chapter 6. Solution for the electric eld is

E(r; t) = 1

4"0rZ(r0; t)

jr r0jdV0 0

4

@

@t

ZJt(r

0; t )jr r0j dV 0: (4.40)

Note that in this formulation, the Coulomb electric eld (the rst term in the RHS) is non-retarded

and instantaneous. This unphysical result is a consequence of the Coulomb gauge which singles

out the transverse current Jt. In fact, the non-retarded longitudinal electric led is cancelled by

a term contained in the last term and all physically observable electromagnetic elds are retarded

because of the nite propagation speed c:

4.4 Fields and Potentials

By now, it is clear that there exists two kinds of electric eld, one dened in terms of the scalar

potential,

E1 = r; (4.41)

and another originating from time varying magnetic eld,

rE2 = @B

@t= @

@trA: (4.42)

The eld E1 is longitudinal because r E1 = 0: In Eq. (4.42), the longitudinal components of

E and A vanish. In fact Maxwells equations do not impose any conditions on the longitudinal

component of A; and choice of r A is arbitrary. Therefore, we are allowed to assume

E = r @

@tA

= r @

@tAl

@

@tAt;

where Al and At are the longitudinal and transverse component of the vector potential. Substitu-

tion in r E = ="0 yieldsr E =r2 @

@tr Al =

"0: (4.43)

In Coulomb gauge, Al is so chosen as to satisfy

Coulomb gauge: r Al = 0: (4.44)

8

(Note that Al may not be zero identically.) Eq. (4.43) becomes

r2C (r; t) = (r; t)

"0; (4.45)

where C is the scalar potential in the Coulomb gauge. It should be emphasized that both Cand are time dependent and in Coulomb gauge the scalar potential responds to a change in the

charge density instantaneously. In Lorenz gauge, r Al is assigned as

Lorenz gauge: r Al +1

c2@L@t

= 0: (4.46)

In this case, the scalar potential obeys a wave equation,r2 1

c2@2

@t2

L =

"0: (4.47)

This yields a retarded solution,

L (r; t) =1

4"0

Z1

jr r0jr0; t 1

c

r r0 dV 0: (4.48)

Proof is straightforward if the following are noted:

r2 1

jr r0j = 4r r0 ;

r2r0; t 1

c

r r0 = 1

c2@2

@t2

r0; t 1

c

r r0 :For the vector potential, we rewrite the Ampere-Maxwells law

rB = 0J+ "0

@E

@t

;

in terms of the potentials as

r2A 1

c2@2A

@t2r

(r A) + 1

c2@

@t

= 0J: (4.49)

In Coulomb gauge with r A =0; this reduces to

r2A 1

c2@2A

@t2= 0J+

1

c2@

@tr

r2A 1

c2@2A

@t2= 0Jt; (4.50)

since the longitudinal component of the current Jl vanishes through the continuity equation

@

@t+r Jl = 0:

9

Note that@

@t= "0

@

@tr E = "0

@

@tr2C ;

and thus

Jl "0@

@trC = 0:

In Coulomb gauge, the vector potential is transverse.

In Lorenz gauge, Eq. (4.49) becomes a wave equation for A ;

r2A 1

c2@2A

@t2= 0J; (4.51)

where J = Jl + Jt; A = Al +At: In Lorenz gauge, the potentials are symmetric in the sense that

both and A satisfy the same wave equation,r2 1

c2@2

@t2

=c

A

!= 0

c

J

!; (4.52)

and appropriately form a covariant four vector (=c;A) : In contrast, the formulation of electro-

magnetic elds in terms of Coulomb gauge is not invariant under the Lorentz transformation. All

potentials and elds are retarded in Lorenz gauge while in the Coulomb gauge, the scalar potential

is non-retarded. The appearance of non-retarded scalar potential is due to the choice r A = 0; or

the assumption that the vector potential is purely transverse.

4.5 Poynting Vector: Energy and Momentum Conservation

The complete set of Maxwells equations is

r E =

"0;

rE = @B@t;

r B = 0;

rB = 0J+ "0

@E

@t

:

It is noted that the charge density contains all kinds of charges, free charges, bound charges, etc.

Likewise, the current density J contains all kinds of current, conduction currents, magnetization

currents, etc. If free charges are singled out, the rst equation can be written in terms of the

displacement vector D;

r D = free; (4.53)

where

D = "0E+P = "E; (4.54)

10

with P being the polarization vector, or the electric dipole moment density. The current density

is likewise decomposed into the part due to the motion of free charges and the other due to time

variation of the polarization vector,

@@tr P+r Jp = 0;

from which

Jp =@P

@t: (4.55)

Therefore, the magnetic induction equation can be rewritten as

rB = 0J+ "0

@E

@t+@P

@t

= 0

J+

@D

@t

; (4.56)

where the current density J consists of conduction and magnetization currents,

J = Jc+Jm= Jc+r M:

If the conduction current is singled out, Eq. (4.56) can be rewritten in terms of the vector H;

rH = Jc +@D

@t; (4.57)

where

H =B

0M:

In macroscopic applications, free charges and conduction currents are the quantities that can be

controlled by external means and Eqs. (4.53) and (4.57) are often more convenient than the original

forms.

The Maxwells equations are also consistent with energy conservation law. To show this, we

dene the Poynting vector by

EH; (W m2): (4.58)

As its dimensions imply, the Poynting vector indicates the ow of electromagnetic energy per unit

area per unit time, that is, power density. The divergence of the Poynting vector is

r (EH) = (rE) H (rH) E

= @B@tH

Jc +

@D

@t

E

= @@t

1

2H2 +

1

2"E2

E J;

provided that the permittivity " and permeability are independent of the frequency !: If not, the

11

energy densities should be modied as

1

2

@ [!" (!)]

@!E2;

1

2

@ [! (!)]

@!H2; (4.59)

respectively. We will return to this problem shortly in the following Section. Integrating over a

volume, we ndZVr (EH) dV =

IS(EH) dS = d

dt(Um + Ue)

ZE JdV; (4.60)

where

Ue =1

2

Z"E2dV; Um =

1

2

ZH2dV; (4.61)

are the total electric and magnetic energies stored in the volume andZE JdV; (J s1 =W) (4.62)

is the rate of electromagnetic energy conversion into other forms of energy, e.g., creation of heat

through Joule dissipation and acceleration of charged particles. Therefore,

IS(EH) dS; (4.63)

can be interpreted as the electromagnetic power ow into the volume.

For a system consisting of charged particles, the momentum conservation can be shown in a

similar way. The mechanical momentum Pm follows the equation of motion,

dPmdt

=

Z(E+ JB) dV; (N). (4.64)

Substituting

= "0r E

and

J =1

0rB "0

@E

@t;

we nd

dPmdt

=

Z "0(r E)E+

1

0rB "0

@E

@t

B

dV

=

Z r !T 1

c2@

@tEH

dV; (4.65)

where !T = Tij = "0EiEj

1

2"0E

2ij +1

0BiBj

1

20B2ij ; (N m

2) (4.66)

12

is the Maxwells stress tensor and use is made of the following vector identities,

rE2 = r(E E) = 2E (rE) + 2 (E r)E;

r (EE) = (r E)E+ (E r)E

rB2 = r(B B) = 2B (rB) + 2 (B r)B

r (BB) = (r B)B+ (B r)B = (B r)B:

Eq. (4.65) suggests that the vector

1

c2EH; (N s m3) (4.67)

can be regarded as the electromagnetic momentum density and

1

cEH; (N m2) (4.68)

as the electromagnetic momentum ux density. For a system of charged particles, momentum

conservation thus requires inclusion of the electromagnetic momentum as well as mechanical mo-

mentum.

From the momentum density in Eq. (4.67), the angular momentum density associated with

electromagnetic elds is naturally dened by

1

c2r (EH); (J s m3) (4.69)

and the total angular momentum associated with electromagnetic elds by

L =1

c2

Zr (EH)dV ; (J s). (4.70)

Its ux density is

R = 1

cr (EH); (J m2): (4.71)

Both the momentum and angular momentum densities are proportional to the Poynting vector,

namely, the energy ow. If for example a system is losing energy through radiation of electromag-

netic energy, the system is necessarily losing momentum and angular momentum as well. Consider

a charged particle undergoing circular motion, e.g., electron in a magnetic eld. Since the elec-

tron is continuously accelerated by the centripetal force, it radiates electromagnetic energy. At

the same time it loses its angular momentum to radiation. Therefore, it is natural to expect that

electromagnetic elds (with proper polarization) carry an angular momentum with them.

13

4.6 Plane Electromagnetic Waves

One special mode of electromagnetic waves in free space is a plane wave in which the amplitude

of electric and magnetic eld remains constant. Without loss of generality, we may assume wave

propagation in the z direction and an electric eld in the x direction,

E(z; t) = E0ei(kz!t)ex:

In free space, the electric eld satises the wave equation,@2

@z2 1

c2@2

@t2

E(z; t) = 0; (4.72)

provided!

k= c: (4.73)

From

rE = @B@t;

we nd

kE = !B; (4.74)

where k = kez: The magnetic eld is thus in the y direction,

B =k

!E0e

i(kz!t)ey; (4.75)

and its amplitude is

B0 =E0c; or H0 =

E0c0

=E0Z; (4.76)

where

Z =

r0"0= 376:8; () (4.77)

is the impedance of free space.

The electric and magnetic energy densities associated with a plane wave are the same, for

1

2"0E

2 =1

2"0Z

2H2 =1

20H

2: (4.78)

This equipartition of wave energy is similar to that in mechanical waves in which kinetic and

potential energies are equal. The total wave energy density is therefore given by

u = 2 12"0E

2 = "0E2; (J m3) (4.79)

14

and the Poynting ux may be written in terms of either the electric or magnetic eld as

Sz = ExHy = c"0E

2x =

E2xZ; (W m2) (4.80)

or

Sz = c0H2y = ZH

2y : (4.81)

For a harmonic wave with an amplitude E0, the average (rms) wave energy density is given by

uave =1

2"0E

20 ; (W m2) (4.82)

and corresponding rms Poynting ux is

Szave =1

2c"0E

20 =

1

2

E20Z; (W m2): (4.83)

Electromagnetic waves radiated by a localized source approach plane waves at a su¢ ciently

large distance but they can never be pure plane waves. Plane waves can be constructed from two

circularly polarized waves with opposite helicities, one rotating with positive helicity and another

with negative helicity. Helicity of electromagnetic waves is related to the angular momentum

associated with the waves. Evidently, a plane polarized wave carry zero angular momentum. A

more general theory of electromagnetic radiation will be developed in Chapter 5.

In a dielectric medium, the Maxwells equations are modied as

rE = @B@t;

rB = 0@("E)

@t; (4.84)

where " is the permittivity which in general depends on the wave frequency and spatial position

and also the electric eld. The origin of the permittivity is in the current induced by the electric

eld in a material medium. In the magnetic induction equation

rB = 0J+ "0

@E

@t

;

if the current density is proportional to the electric eld through a conductivity ; J =E; we have

rB = 0E+ "0

@E

@t

= 0"

@E

@t; (4.85)

where the permittivity is given by

" = "0

1 + i

!"0

: (4.86)

It should be noted that the current density is due to deviation of electron orbit from bound harmonic

15

motion in molecules. The equation of motion for an electron placed in an oscillating electric eld is

m

d2

dt2+ !20

x = eE0ei!t; (4.87)

where m is the electron mass and !0 is the frequency of bound harmonic motion. The current

density is

J = nedxdt=

i!

!2 !20eE

m;

and the conductivity becomes

=i!

!2 !20ne2

m; (4.88)

where n is the number density of electrons. Then the permittivity is given by

"(!) = "0

1

!2p!2 !20

!; (4.89)

where !p is the plasmafrequency dened by

!2p =ne2

"0m: (4.90)

The phase velocity of electromagnetic waves in a dielectric is

!

k=

1p"(!)0

; (4.91)

and the group velocity isd!

dk=

1

1 +1

2

!

"(!)

d"

d!

!

k: (4.92)

(At frequencies remotely separated from the resonance frequency !0; the group velocity coincides

with the energy propagation velocity. Near the resonance, however, the group velocity exceeds c and

it loses the meaning of energy propagation velocity. The concept of signal velocity was introduced

by Brillouin.) The impedance is accordingly modied as

Z(!) =E

H=

r0"(!)

: (4.93)

By denition, the group velocity is equal to

d!

dk=

Poynting uxEnergy density

;

16

where the Poynting ux is

S =E2

Z=

s"(!)

0E2; (W m2):

Therefore the wave energy density in a dielectric medium is

u =

1 +

1

2

!

"(!)

d"

d!

"(!)E2

=1

2

d

d![!"(!)]E2 +

1

20H

2

=1

2"0E

2 +1

2

!2p(!2 + !20)

(!2 !0)2"0E

2 +1

20H

2; (J m3)

where the relationship1

2"(!)E2 =

1

20H

2; (4.94)

is used. This result is valid only if the group velocity can be regarded as energy propagation velocity

which may not hold near the resonance ! ' !0 if the dielectric is dissipative.The origin of the additional factor

1

2!d"

d!E2;

in the electric energy density is due to electron kinetic and potential energies in an oscillating

electric eld. From the equation of motion of a bound electron,@2

@t2+ !20

x = e

mE;

we readily nd the electron kinetic energy density,

1

2nmv2 =

1

2

ne2

m

!2

(!2 !0)2E2 =

1

2

!2!2p(!2 !0)2

"0E2; (4.95)

and potential energy density1

2nm!20x

2 =1

2

!20!2p

(!2 !0)2"0E

2: (4.96)

Therefore, the total energy density associated with the electric eld is

1

2"0E

2 +1

2

!2!2p(!2 !0)2

"0E2 +

1

2

!20!2p

(!2 !0)2"0E

2; (J m3) (4.97)

which is consistent with that conveniently calculated from

1

2

d

d![!"(!)]E2: (4.98)

Since the wave under consideration is strongly dispersive, there is no simple energy equipartition

as in the case of nondispersive waves. Note that the electron kinetic and potential energies are the

17

result of forced oscillations.

The fact that

E @D@t

= E @ ("E)@t

; (4.99)

is not always equal to1

2"@E2

@t; (4.100)

can be seen if we realize that the permittivity " is in general frequency dependent. Waves become

dispersive and wave amplitude is bound to decrease although slowly. In this case,

@ ("E)

@t= i!" (!)E0ei!t

' (i!0 + )" (!0) + i

@"

@!0

E0e

i!t

= i!0" (!0)E0ei!t +" (!0) + !0

@"

@!0

dE0dtei!t; (4.101)

where (< 0) is the damping rate of the amplitude,

dE0dt

= E0: (4.102)

Therefore, time variation of electric energy density should be calculated from

E@ ("E)@t

=1

2

" (!0) + !0

@"

@!0

dE2

dt; (4.103)

and the electric energy density becomes

ue =1

2

" (!0) + !0

@"

@!0

E2: (4.104)

Likewise, the magnetic energy density should be generalized as

um =1

2

(!0) + !0

@

@!0

H2: (4.105)

These expressions were originally formulated by von Laue.

4.7 Wave Reection and Transmission - Normal Incidence

Reection and transmission of plane electromagnetic waves at a boundary of two dielectrics can be

conveniently analyzed in terms of impedance mismatch. Let a plane wave of amplitude Ei in air be

incident normally on a at dielectric surface of impedance Z2 as shown in Fig.4-3: The impedance

of air is very close to the vacuum impedance, Z1 = 377 : The incident Poynting ux is split into

18

those of reected (Er) and transmitted (Et) waves,

E2iZ1

=E2rZ1

+E2tZ2; (W m2): (4.106)

At the boundary, continuity of the electric elds, which are all tangential to the surface, yields

Figure 4-3: A plane wave incident normal to a dielectric boundary.

Ei + Er = Et: (4.107)

Solving these equations for Er and Et; we nd

Er =Z2 Z1Z2 + Z1

Ei; (4.108)

Et =2Z2

Z2 + Z1Ei: (4.109)

In terms of the magnetic elds,

Z1H2i = Z1H

2r + Z2H

2t ; (4.110)

Hi +Hr = Ht; (4.111)

yield

Hr =Z1 Z2Z1 + Z2

Hi; Ht =2Z1

Z1 + Z2Hi: (4.112)

Note that the polarity of either electric or magnetic eld of the reected wave is reversed. This

is understandable because the incident and reected elds must satisfy the following vectorial

relationships,

kEi = !0Hi;

19

kEr = !0Hr:

Either Er or Hr must change sign on reection for the reected Poynting ux to be in opposite

direction relative to the incident ux.

It should be pointed out that the analysis presented above is solely based on energy conser-

vation and no consideration was given to momentum conservation. In fact, if only the momenta

associated with the three waves (incident, reected, and transmitted) are considered, momentum

is not conserved. Since the momentum ux densities associated with each wave are

Incident wave :1

c1

E2iZ1

= "1E2i ;

Reected wave : "1E2r ; (4.113)

Transmitted : "2E2t ;

the following momentum unbalance emerges,

"1(E2i + E

2r ) "2E2t =

2(Z21 Z22 )(Z1 + Z2)2

"1E2i =

2(Z1 Z2)Z1 + Z2

"1E2i : (4.114)

This is actually taken up by the dielectric body as mechanical momentum. Recall that an innitely

massive body can absorb momentum without absorbing energy. If Z2 > Z1; the dielectric is pushed

to the left. In fact, at the boundary, the electric energy densities are discontinuous, and the

di¤erence is

1

2"2E

2t

1

2"1E

2t = ("2 "1)

2Z22(Z1 + Z2)2

E2i

=2(Z21 Z22 )(Z1 + Z2)2

"1E2i

=2(Z1 Z2)Z1 + Z2

"1E2i :

This appears as a force per unit area on the boundary surface acting from the higher energy density

side to the lower because electric force in the direction perpendicular to the eld appears as pressure.

Reection at a conductor surface can be analyzed in a similar manner by modifying the im-

pedance appropriately. For a medium having a conductivity ; the impedance is given by

Z =

ri!0 i!"0

: (4.115)

This can be seen from the Maxwells equation in the presence of conduction current,

rH = J+ "0@E

@t= E i!"0E;

20

and the e¤ective permittivity in a conductor is dened by

"e¤ = "0

i!:

Therefore, the impedance in a conductor is

Z =

r0"e¤

=

ri!0 i!"0

:

For ordinary conductors, the conduction current far dominates over the displacement current even

in microwave frequency range. Then,

Z 'ri!0

= (1 i)r!02

: (4.116)

A complex impedance indicates strong dissipation of electromagnetic energy. The magnitude of

the impedance

jZj =r!0; (4.117)

is much smaller than the free space impedance Z0 =p0="0 ' 377 and electromagnetic waves

incident on a conductor surface su¤ers strong reection. However, reection can never be complete.

Bath room mirrors coated with aluminum has power reection coe¢ cient of about 90% at optical

frequency ! ' 1014 rad/sec.

Example 1 Impedance Matching

A dielectric lm a quarter wavelength thick coated on a surface of another dielectric (say, optical

glass) can eliminate reection of electromagnetic waves normally incident if the impedance of the

lm is chosen to be a geometric mean,

Zf =pZ0Zg: (4.118)

This condition follows from cancellation between two reected waves, one at the air-lm boundary

and another at the lm-glass boundary,

Zf Z0Zf + Z0

+ eiZg ZfZg + Zf

= 0;

where the phase factor ei = 1 is due to additional propagation distance f42 = f

2of the wave

reected at the lm-glass boundary. Here, f is the wavelength in the dielectric lm. Solving

Zf Z0Zf + Z0

=Zg ZfZg + Zf

;

21

for Zf ; we nd

Zf =pZ0Zg:

Likewise, reection from a conductor plate can be avoided by placing a thin conducting plate

of thickness d at a quarter wavelength in front of the conductor surface if the conductance of the

plate is chosen to be

=1

Z0d: (4.119)

4.8 Reection and Transmission at Arbitrary Incident Angle

For an arbitrary incident angle 1; reection and transmission at a dielectric boundary can be

analyzed by exploiting the boundary conditions for the electric and magnetic elds as follows. An

incident wave continues to be assumed plane polarized. The refracted angle 2 is related to the

incident angle 1 through the well known Snells law,

sin 1sin 2

=n2n1; (4.120)

where n1 =p"1="0 and n2 =

p"2="0 are the indices of refraction of respective media. The Snells

law follows from the conservation of the wavenumber parallel to the boundary,

k1 sin 1 = k2 sin 2; (4.121)

and the change in the wave propagation velocity,

!

k1=

1p"10

;!

k2=

1p"20

; (4.122)

ork1n1=k2n2: (4.123)

The normal wavenumber in the medium 2 is

kz = k2 cos 2 = k2

s1

n1n2

2sin2 1:

This becomes pure imaginary when

sin 1 >n2n1; n2 < n1;

This is the condition for total reection which occurs when a wave is incident on a medium with a

smaller index of refraction (e.g., from glass to air), n2 < n1: An imaginary wavenumber indicates

exponential damping in the region n2 from the surface,

E0ekzz;

22

where kz is the damping factor in the axial (z) direction,

kz = k2

sn1n2

2sin2 1 1:

The wavenumber component along the surface is

kk = k2n1n2sin :

If the second medium is vacuum (or air) n2 = 1; k2 = !=c = k0; and

kk = k0n1 sin :

In total reection, electromagnetic elds in the region of smaller index of refraction exponentially

decay from the surface. Such waves are called evanescent.Important applications of evanescent

waves are being found in high resolution microscopy even with visible light.

In analysis of wave reection and transmission at a boundary between di¤erent media, use of

the continuity of tangential electric and magnetic elds, Et;Ht; is su¢ cient because continuity of

normal components Dn;Bn is redundant. From Maxwells equation,

rH =@D

@t; or kH = !D;

it is evident that continuity of Dn demands continuity of n (kH) = k (nH). However, thetangential component of H; nH; is continuous and the wavenumber parallel to the boundary isalso continuous. (Note that in k (nH) = (kn) H; only the parallel component of k appears.)The latter (continuity of k n) follows from the conservation of wave momentum parallel to the

boundary which manifests itself in the form of well known Snells law. Likewise, from

rE = @B@t; or kE = !B;

we see that continuity of Bn automatically follows because of continuity of tangential component

of the electric eld.

4.8.1 H in the Incident Plane (E tangential to the boundary)

We rst consider the case in which the magnetic eld of the incident wave is in the incident plane,

or the electric eld is parallel to the boundary surface. The reected and refracted electric elds

are also parallel to the surface and the continuity of the tangential component of the electric eld

gives

Ei + Er = Et: (4.124)

23

Figure 4-4: A plane wave with incident angle 1 when the magnetic eld is in the incidence planeand the electric eld is parallel to the boundary surface.

Since for each wave, the following vectorial relationship holds,

kE = 0!H; (4.125)

the continuity of tangential component of the H eld yields

k1 cos 1(Ei Er) = k2 cos 2Et: (4.126)

Recalling the Snells law, we thus nd

Er =sin(2 1)sin(1 + 2)

Ei; (4.127)

Et =2 cos 1 sin 2sin(1 + 2)

Ei: (4.128)

In the limit of normal incidence (small angles 1; 2), we recover

Er =sin(2 1)sin(1 + 2)

Ei 'sin 2 sin 1sin 1 + sin 2

Ei =n1 n2n2 + n1

Ei =Z2 Z1Z2 + Z1

Ei; (4.129)

where

Zi =

r0"i; (4.130)

is the impedance of respective media.

When the conditions for total reection are met, the impedance dened by the ratio between

24

the tangential component of the electric and magnetic eld,

ZTE2 =E2

H2 cos 2= i

r0"2

1sn1n2

2sin2 1 1

; (4.131)

is also pure imaginary which indeed ensures total reection of electromagnetic waves, iZTE2 Z1Z1 + i

ZTE2 = 1:

In total reection, the phase di¤erence between the incident and reected waves is

TE = 2 tan1

Z1ZTE2

!= 2 tan1

psin2 1 (n2=n1)2

cos 1

!: (4.132)

4.8.2 E in the Incident Plane (H tangential to the boundary)

Figure 4-5: The electric eld is in the incidence plane and the magnetioc eld is paralllel to theboundary.

In this case, the magnetic eld is parallel to the boundary plane and the continuity of tangential

component of the magnetic eld is simply

Hi +Hr = Ht: (4.133)

For each wave, the electric eld is related to the magnetic eld through

E = 1

!"kH: (4.134)

25

Therefore, continuity of the tangential component of the electric eld yields

1

n1cos 1(Hi Hr) =

1

n2cos 2Ht =

1

n2cos 2(Hi +Hr): (4.135)

Solving for Hr; we nd

Hr =sin 1 cos 1 sin 2 cos 2sin 1 cos 1 + sin 2 cos 2

Hi (4.136)

=tan(1 2)tan (1 + 2)

Hi: (4.137)

The transmitted magnetic eld is

Ht =

1 +

tan(1 2)tan (1 + 2)

Hi =

2 sin 1 cos 1sin(1 + 2) cos(1 2)

Hi:

The reected electric eld is

Er =tan(2 1)tan (1 + 2)

Ei; (4.138)

and transmitted electric eld is

Et =2 sin 2 cos 1

sin(1 + 2) cos(1 2)Ei: (4.139)

If

1 + 2 =

2; (4.140)

the reected wave vanishes completely. Under this condition, Snells law becomes

n2n1=sin 1sin 2

=sin 1cos 1

= tan 1; (4.141)

and this particular angle

B = tan1n2n1

; (4.142)

is called the Brewsters angle. If incident wave is plane polarized with the magnetic eld oriented

parallel to, say, a surface of glass, reection can be avoided at the Brewsters angle. This principle

is often exploited in designing reecting mirrors in lasers so that output laser beam has a high

degree of planar polarization.

Because the reection coe¢ cient depends on wave polarization, unpolarized wave with random

polarization becomes partially polarized on reection and transmission at a dielectric boundary.

The magnetic reection coe¢ cient derived in Eq. (4.136) will play an important role in analyzing

the transition radiation discussed in Chapter 8.

When the conditions for total reection are met, the impedance dened by the ratio between

26

the tangential component of the electric and magnetic eld is

ZTM2 =E2 cos 2H2

= ir0"2

sn1n2

2sin2 1 1: (4.143)

The phase di¤erence between the incident and reected waves is

TM = 2 tan1

Z1ZTM2

!= 2 tan1

psin2 1 (n2=n1)2(n2=n1)2 cos 1

!: (4.144)

4.9 Circularly and Elliptically Polarized Plane Waves

Planar polarization discussed in the preceding section is a highly idealized mode of propagation of

electromagnetic waves. A plane polarized wave can be decomposed into two circularly polarized

plane waves of opposite helicity, one with positive helicity and another with negative helicity.

Helicity of an electromagnetic wave is closely related with the angular momentum carried by the

wave.

Circularly polarized waves propagating in the z-direction can be described by the electric eld

vectors,

E(z; t) = E0(ex iey)ei(kz!t); (4.145)

where the positive sign is for positive helicity and minus sign is for negative helicity. The sum of

these two waves of opposite helicity trivially yields a plane polarized wave with the electric eld in

the x-direction. Corresponding magnetic elds are

B =1

!kE

=k

!E0(ey iex)ei(kz!t); (4.146)

and Poynting ux is

S = EH = 2E20Zez; (4.147)

where the factor 2 accounts for the two independent modes of equal amplitude.

A general form of mixed helicity may be written as

E(z; t) = (E1ex + E2iey)ei(kz!t); (4.148)

where E1 and E2 are complex amplitude. Corresponding magnetic eld,

B(z; t) =k

!(E1ey E2iex)ei(kz!t); (4.149)

is of course normal to the electric eld,

E B = 0: (4.150)

27

However, the scalar product E B in general does not vanish.

Figure 4-6: Trace of the head of the rotating electric eld vector associated with a circularlypolarized wave with positive helicity. In the case of negative helicity, the direction of rotationrelative to the wavevector is reversed.

Example 2 Reection and Transmission of Circularly Polarized Wave

Consider a circularly polarized wave incident at an angle i to a at surface of a dielectric. The

electric eld of the incident wave may be decomposed into two components, one in the incident

plane and another perpendicular to the incident plane,

Ei = Ek0 + iE?0:

The reected wave of the parallel component is

Ekr =tan(2 1)tan(1 + 2)

Ek0;

while the reected perpendicular component is

E?r =sin(2 1)sin(2 + 1)

E?0:

The transmitted (refracted) components are

Ekt =2 sin 1 cos 2

sin(1 + 2) cos(1 2)Ek0;

E?t =2 cos 1 sin 2sin(1 + 2)

E?0:

Both reected and transmitted waves are elliptically polarized. In particular, if the incident angle is

at the Brewsters angle, 1 = =22; Ekr vanishes and the reected wave becomes plane polarized.

28

On total reection of a circularly polarized electromagnetic wave, the reected wave becomes

elliptically polarized because the phases of TE and TM components di¤er as evident from Eqs.

(4.132) and (4.144).

4.10 StokesParameters

Consider an electromagnetic wave propagating in the z direction with electric eld components

E(r; t) = (Exex + Eyey:)ei(kz!t): (4.151)

The amplitudes Ex and Ey may be complex allowing for nite phase di¤erence,

Ex = jExj eix ; Ey = jEyj eiy : (4.152)

If = xy = 0; the led is a simple superposition of two linearly polarized waves. If is nite,the wave is in general elliptically polarized. In optics, direct measurement of the phase di¤erence

is not easy. What is normally measured is the intensity or the quadratic quantities of the electric

eld,

jExj2 ; jEyj2; jExj jEyj cos(x y); and jExj jEyj sin(x y): (4.153)

Stokesparameters are dened, with I = jExj2 + jEyj2 the total intensity, by

s0 =1I

jExj2 + jEyj2

= 1;

s1 =1I

jExj2 jEyj2

;

s2 =2I jExj jEyj cos(x y);

s3 =2I jExj jEyj sin(x y);

(4.154)

and satisfy

s21 + s22 + s

23 = 1; (4.155)

if the waves are purely coherent. If not, s2 and s3 are to be modied as

s2 =2

IReExEy ; s3 =

2

IImExEy ; (4.156)

where the bar indicates time average. For incoherent waves,

s21 + s22 + s

23 < 1: (4.157)

Natural light is characterized by a collection of many waves with random phases and complete

depolarization, s1 = s2 = s3 = 0 even though each wave may be highly monochromatic. For a

plane wave polarized in the x direction,

s1 = 1; s2 = s3 = 0:

29

For a plane wave polarized in the direction = =4 (along the plane x = y),

s1 = s3 = 0; s2 = 1:

For a circularly polarized wave with positive (negative) helicity,

s1 = s2 = 0; s3 = 1:

In experiments, Stokes parameters can be determined by rotating a polarizer plate. jExj2

and jEyj2 can be readily found by aligning the polarization direction along x and y direction,respectively. If the polarizer is at angle from the x axis, the intensity measured at that angle is

I() = (jExjeix cos + jEyjeiy sin )jExjeix cos + jEyjeiy sin

= jExj2 cos2 + jEyj2 sin2 + 2 jExj jEyj cos sin cos(x y):

By choosing = =4; for example, we have

I =

4

=1

2

jExj2 + jEyj2 + 2jExjjEyj cos(x y)

: (4.158)

Finally, exploiting a uniaxial (or biaxial) crystal, a so-called quarter wavelength plate can be fab-

ricated to induce =2 relative phase delay between Ex and Ey due to the di¤erent propagation

velocities of ordinary and extraordinary modes. The intensity in this case is

I 0() = (jExjeix cos + ijEyjeiy sin )jExjeix cos ijEyjeiy sin

= jExj2 cos2 + jEyj2 sin2 2 jExj jEyj cos sin sin(x y)

If = =4; this reduces to

I 0 =

4

=1

2

jExj2 + jEyj2 2jExjjEyj sin(x y)

: (4.159)

Therefore, jExjeix and jEyjeiy and corresponding Stokesparameters can be determined by mea-suring the four intensities,

I ( = 0) ; I =

2

; I =

4

and I 0

=

4

:

The tensor dened by

Iij = EiEj ; (i; j = x; y) (4.160)

is called polarization tensor. Unless the eld is purely coherent, the time averaged intensity EiEjmay still vary slowly with time. Iik is Hermitian and thus can be diagonalized through the eigen-

30

values 1 and 2 which are the roots of

det (Iij ij) = 0: (4.161)

Corresponding two polarization eigenvectors n(1) and n(2) can be determined from

Iijn(1)j = 1n

(1)i ; Iijn

(2)j = 1n

(2)i : (4.162)

In terms of the eigenvectors, the polarization tensor can be written in the form

!I = 1n

(1)n(1) + 2n(2)n(2): (4.163)

As an example, let us consider superposition of two plane polarized waves, one in the x direction

and another in the direction (cos ; sin ) in the x y plane. The relative phase between the twowaves is assumed to be random and the intensities of the waves are I1 and I2: The electric eld is

E = (E1 + E2ei cos ; E2e

i sin ); (4.164)

where I1 = E21 ; I2 = E22 ; and is random phase. Then

Iij =

I1 + I2 cos

2 I2 cos sin

I2 cos sin I2 sin2

!: (4.165)

Eigenvalues are

1;2 =1

2

I1 + I2

q(I1 + I2)2 4I1I2 sin2

: (4.166)

The ratio

= min=max =I1 + I2

p(I1 + I2)2 4I1I2 sin2

I1 + I2 +p(I1 + I2)2 4I1I2 sin2

; (4.167)

may be called the degree of depolarization. = 1 corresponds to completely unpolarized state,

while = 0 corresponds to a plane polarized wave. When I1 = I2 = I; the polarization tensor

reduces to

Iij = I

1 + cos2 cos sin

cos sin sin2

!: (4.168)

Eigenvalues are

1;2 = 1 cos ; (4.169)

and the degree of depolarization is

=1 cos 1 + cos

: (4.170)

The eigenvectors are

n(1) =

cos

2; sin

2

;n(2) =

sin

2; cos

2

: (4.171)

31

Derivation of the eigenvectors is left for exercise.

The Stokesparameters si and the polarization tensor are related through

Iij =1

2I

ij +

3Xm=1

sm(m)ij

!=1

2I

1 + s1 s2 is3s2 + is3 1 s1

!; (4.172)

where

(1)ij =

1 0

0 1

!; (2)ij =

0 1

1 0

!; (3)ij =

0 ii 0

!; (4.173)

are Paulis spin matrices. When s1 = 1; s2 = s3 = 0; Iij reduces to

Iij = I

1 0

0 0

!;

which describes a plane polarized wave in the x direction. When s2 = 1; s1 = s3 = 0;

Iij =1

2I

1 1

1 1

!:

This describes the case of plane polarization in the direction = =4: When s3 = 1; s1 = s2 = 0;

Iij =1

2I

1 ii 1

!;

which describes circular polarization with positive helicity. Finally, s3 = 1; s1 = s2 = 0 and

corresponding

Iij =1

2I

1 i

i 1

!;

describes circular polarization with negative helicity.

4.11 Propagation along a Conductor Rod

In this section, we analyze propagation of electromagnetic waves along a conductor rod of radius

a and conductivity : A simple transverse Magnetic (TM) mode is considered with the following

eld components, E; Ez; and H: The axial electric eld Ez outside the conductor rod satises the

scalar wave equation, @2

@2+1

@

@+@2

@z2

Ez(; z; t) =

1

c2@2

@t2Ez(; z; t): (4.174)

32

Figure 4-7: Field proles of electromagentic wave propagating along a conductor rod.

For a harmonic wave with time dependence ei!t; this reduces to@2

@2+1

@

@+@2

@z2+!2

c2

Ez(; z) = 0: (4.175)

Furthermore, since the wave is propagating along the rod, we may single out the z dependence in

the form eikz and reduce the wave equation tod2

d2+1

d

d+ 2

Ez() = 0; (4.176)

where

2 =!c

2 k2: (4.177)

Elementary solutions to Eq. (4.176) are the Bessel functions J0(); N0() and their linear com-

binations,

H(1)0 () = J0() + iN0(); (4.178)

H(2)0 () = J0() iN0(); (4.179)

known as Hankel functions of the rst and second kind, respectively. Their asymptotic behavior at

large argument is

H(1)0 (x)!

r2

xexp

hix

4

i;

H(2)0 (x)!

r2

xexp

hix

4

i;

indicating radially outward and inward propagation, respectively. In the present case, both k

and are complex because of dissipation in the rod. It is reasonable to assume radially outward

33

propagation (radiation from the rod) and we choose

Ez() = E0H(1)0 (): (4.180)

From r E = 0; we nd the radial component of the electric eld,

E() = ik

E0H

(1)1 (); (4.181)

and from rE = i!0H; the azimuthal component of the magnetic eld,

H() = i!"0E0H(1)1 (); (4.182)

where use is made ofd

dx

xH

(1)1 (x)

= xH

(1)0 (x);

d

dxH(1)0 (x) = H(1)

1 (x):

The boundary condition at the rod surface is

Ez( = a)

H( = a)= Z =

ri!0

(surface impedance); (4.183)

orH

(1)0 (a)

H(1)1 (a)

= i!"0Z: (4.184)

For a given wave frequency !; this equation determines and thus k; the axial wavenumber and

in this respect, it is a dispersion relation.

For a small impedance Z; approaches 0, and the axial wavenumber k becomes

k =!

c;

as expected. Noting

limx!0

H(1)0 (x)! 1 + i

2

(ln(x=2) + E) ; lim

x!0H(1)1 (x)! 2

1

x; (4.185)

we see that the axial electric eld becomes negligible and the transverse elds approach those of

TEM mode,

E() = E0a

; H() =

E()

Z0; (4.186)

where Z0 =p0="0 is the impedance of free space. The elds in coaxial cables used for transmission

of electromagnetic waves can be approximated by those given above,

E =V

ln(b=a)

1

; H =

EZ=

I

2; (4.187)

34

where V is the potential di¤erence between the inner and outer conductor with radii a and

b;respectively, I is the current and Z =p0=" with " the permittivity of the insulating mate-

rial lling the cable. The characteristic impedance of the cable is

Zcable =V

I=

p0="

2ln

b

a

: (4.188)

Similarly, the impedance of a parallel wire transmission line with conductor radius a and separation

distance D is approximately given by

Z 'p0="0

ln

D

a

; D a: (4.189)

4.12 Skin E¤ects in Conductors

In a conductor, the conduction current dominates over the displacement current and a simple Ohms

law,

J = E; (4.190)

may be assumed where (S/m) is the conductivity. Eliminating the electric eld between the

Maxwells equations

rE = @B@t; rB ' 0E;

yields the following di¤usion equation for the magnetic eld,

r2B = 0@B

@t:

If the eld is oscillating at a frequency !; we obtain the following ordinary di¤erential equation,

r2B+ i!0B = 0:

Penetration of the magnetic eld into a conductor slab can be described by

d2Bydz2

+ i!0By = 0: (4.191)

This has a bounded solution

B(z) = B0eikz;

where

k =pi!0 =

1 + ip2

p!0: (4.192)

The magnetic eld decays exponentially from the conductor surface and the quantity

=

s2

!0; (m) (4.193)

35

is called the skin depth. Damping of the electromagnetic elds is evidently due to Ohmic dissipation

in the conductor.

For a cylindrical conductor rod with radius a( c=!) in which an axial current Jz()ei!t is

excited, Jz obeys d2

d2+1

d

d+ i!0

Jz() = 0; (4.194)

This has a bounded solution

Jz() = E0J0(k)

J0(ka); (4.195)

where E0 is the electric eld at the rod surface. The impedance per unit length of the rod can thus

be dened byZ

l=E0I=

k

2a

J0(ka)

J1(ka); (Ohms/m): (4.196)

where use is made of

I = 2

Z a

0Jz()d = 2

a

kE0

J1(ka)

J0(ka): (4.197)

In the low frequency limit jkj a 1; series expansion of the Bessel functions can be exploited,

J0(x) ' 1x2

4; J1(x) '

x

2

1 x

2

8

:

The impedance reduces toZ

l=

1

a2 i! 0

8; (4.198)

whereLil=08

(H/m) (4.199)

is the well known internal inductance of a cylindrical rod carrying a uniform current (no skin e¤ect).

In the high frequency limit (strong skin e¤ect), the ratio J0(ka)=J1(ka) approaches i; and theimpedance is

Z

l=

1

2a(1 i): (4.200)

This is also reasonable since with strong skin e¤ect, the current ow is limited in a thin layer with

an area 2a: Note that the inductive reactance is identical to the resistance in this limit.

If a conductor cylinder is placed in an oscillating axial magnetic eld Hz0ei!t as in inductive

rf heating, the current ows in the azimuthal direction. J() satisesd2

d2+1

d

d 1

2+ i!0

J() = 0; < a:

An appropriate solution is

J() = kJ1(k)

J0(ka)Hz0; < a; (4.201)

36

where k is still given by

k =pi!0:

4.13 Skin E¤ect in a Plasma

A charge neutral plasma contains equal amount of positive and negative charge density. When

placed in an electric eld oscillating at a high frequency, a plasma current due to electron motion

is induced. The conductivity can be found by letting !0 = 0 in Eq. (4.88) (because in a plasma,

electrons are free),

=i

!

ne2

m; (4.202)

and corresponding permittivity

" = "0

1

!2p!2

!: (4.203)

In a collisional plasma with electron collision frequency c; this is modied as

" = "0

1

!2p! (! + ic)

!; (4.204)

as can be easily worked out by introducing a nite collision frequency in the equation of motion.

For a frequency much smaller than the plasma frequency, the permittivity becomes negative,

" ' "0!2p!2; (4.205)

and wave propagation is forbidden. The wavenumber is complex in this case

k = i!pc; (4.206)

and the wave amplitude decays in a manner

E0ei(kz!t) = E0 exp

!pczei!t; (4.207)

where z is the distance in the plasma from the vacuum-plasma boundary. This means that an

electromagnetic wave incident on a plasma cannot penetrate into plasma except for a distance of

the order of the skin depth dened by

=c

!p: (4.208)

The wave is completely reected if the plasma is collisionless. Reection of low frequency radio

waves by the ionoshperic plasma and is a well known example.

37

In a collisional plasma with ! !p; c; the skin depth is modied as

' c

!p

r!

2c: (4.209)

Derivation of this formula is left for exercise.

Strictly speaking, the collisionless skin depth = c=!p is valid for a cold plasma with negligible

electron temperature. To implement the e¤ects of nite electron temperature, it is necessary to

employ a kinetic theory to nd a conductivity : The electron velocity distribution function f

(r;v; t) obeys the kinetic equation,

@f

@t+ v rf e

m(E+ v B) @f

@v= 0: (4.210)

As it is, it is a nonlinear equation because the electromagnetic elds E and B associated with a

wave a¤ect the distribution function f: Let us assume a wave propagating along the z direction

with electric eld polarized in the x direction Ex (z; t) = E0ei(kz!t): The distribution function may

be linearized as f = f1 + FM ; where f1 is the perturbation and FM is unperturbed Maxwellian

distribution. Noting@FM (v

2)

@v= mv

TeFM (v

2); (4.211)

and thus

(v B)@FM (v2)

@v= 0;

we nd

f1 =

e

TeEx

i (kvz !)vxFM : (4.212)

The current density can be found from the rst order moment,

Je = neZvf1dv: (4.213)

Only Jx is nonvanishing, and given by

Jx = ine2

TeEx

Zv2x

kvz !FMdv

= i ne2

mkvTeEx

1p

Z 1

1

et2

t dt

= i ne2

mkvTeExZ () ; (4.214)

where = !=kvTe with vTe =p2Te=m being the thermal velocity of electrons and Z () is known

38

as the plasma dispersion function. The permittivity is therefore given by

" = "0

1 +

!2p!kvTe

Z ()

!: (4.215)

If the electron temperature is negligible 1; the function Z () approaches

Z () ' 1;

and we recover the case of cold plasma,

" = "0

1

!2p!2

!:

In the opposite limit 1;

Z () ' ip;

and

" ' ip"0

!2p!kvTe

:

The damping factor (inverse skin depth) in this limit is given by

Im k =1

'

1=6

2

!pc

2=3 !

vTe

1=3: (4.216)

This is often called anomalous skin e¤ect. (This is probably misnomer because there is nothing

anomalous in the derivation.)

4.14 Waves in Anisotropic Dielectrics

Some crystals exhibit anisotropy in polarizability. The permittivity in such media becomes a tensor,

and the displacement vector D and electric eld E are related through a dielectric tensor ";

D = " E; or Di ="ijEj ; (4.217)

where " is a diagonal tensor consisting of three permittivities in each axial direction, x; y; and z;

" =

0B@ "x 0 0

0 "y 0

0 0 "z

1CA : (4.218)

In uniaxial crystals, polarization occurs preferentially along one axis, say, "x = "y 6= "z: In

anisotropic media, the phase and group velocities are in general oriented in di¤erent directions

and the well known double refraction phenomenon (already known in the 17th century) can be

39

explained in terms of the dielectric anisotropy.

The Maxwells equations to describe wave propagation in an anisotropic medium in which there

are no sources (free = 0; J = 0) are

rE = @B@t; rB = 0

@D

@t= 0"

@E

@t: (4.219)

Then

rrE = 0" @2E

@t2; (4.220)

r (r E)r2E+0" @2E

@t2= 0: (4.221)

Note that in an anisotropic medium, the divergence of the electric eld, r E; does not necessarilyvanish, although

r D = free = 0; (4.222)

must hold. After Fourier decomposition, Eq. (4.221) reduces to

k(k E) k2E+ !20" E = 0; (4.223)

or

(k2ij kikj !20"ij)Ej = 0: (4.224)

The dispersion relation of electromagnetic waves in a uniaxial crystal is therefore given by the

determinant,

det(k2ij kikj !20"ij) = 0: (4.225)

Introducing "ij = "ij="0 and the index of refraction,

n =c

!k; (4.226)

we rewrite the dispersion relation as

det(n2ij ninj "ij) = 0: (4.227)

In uniaxial crystals, polarizability along one axis (say, z-axis) di¤ers from those along other

axes. We assume

"x = "y "?; "z "k: (4.228)

Since the crystal is symmetric about the z-axis, the wave vector k can be assumed to be in the

x z plane without loss of generality,

kx = k?; kz = kk; (4.229)

40

or

n? =ck?!; nk =

ckk!: (4.230)

The dispersion relation then becomesn2k "? 0 n?nk0 n2 "? 0

n?nk 0 n2? "k

= 0: (4.231)

Expanding the determinant, we nd

n2 = "?; (4.232)

and

"kn2k + "?n

2? "k"? = 0: (4.233)

The rst mode of propagation is independent of the propagation angle as if in an isotropic medium.

It is called the ordinary mode because it maintains the properties of electromagnetic waves in

isotropic media. The electric eld is in the y-direction and the dispersion relation in Eq. (4.232) is

equivalent to the following di¤erential equation,@2

@x2+@2

@z2 0"?!2

Ey(x; y) = 0: (4.234)

In this mode, the phase and group velocities are in the same direction (in the direction of k)

although in magnitude they di¤er because of the frequency dependence of the permittivity "?(!):

The second mode has a peculiar property and for this reason is called extraordinary mode. Let

the angle between k and the z-axis be : The dispersion relation in Eq. (4.233) becomes

"k cos

2 + "? sin2 ck

!

2 "k"? = 0: (4.235)

The phase velocity given by

!

k= c

scos2

"?+sin2

"kek; (4.236)

is directed along k: The group velocity is

d!

dk=

@!

@kek +

1

k

@!

@e

=!

kek +

c2

!=k

1

"k 1

"?

sin cos e: (4.237)

This is evidently not parallel to the phase velocity unless "k = "? (isotropic dielectric). The

magnitude of the group velocity is always larger than that of the phase velocity.

Electromagnetic waves in a plasma conned by a magnetic eld is another example of anisotropic

medium which accommodates variety of waves. A simple case of cold plasma will be considered in

41

Problem 4.11.

Figure 4-8: Double refraction of randomly polarized incident wave by a uniaxial crystal. Propaga-tion of the O-mode (ordinary mode) is una¤ected but that of the X-mode (extraordinary mode) is.The group velocity of the X-mode in the crystal deviates from the incident direction.

An important consequence of the presence of the ordinary and extraordinary modes is the well

known double refraction caused by some crystals. Consider a light beam with random polarizations

incident normal to a surface of a uniaxial crystal as shown in Fig. 4-8. Unless the optical axis (z-

axis in the geometry assumed) coincides with the direction of the incident beam, the beam is split

into two beams at the surface. In the crystal, one beam propagates along the direction of the

incident beam and another at an angle. The phase velocities of the ordinary and extraordinary

modes are in the same direction as the incident beam but the group velocity of the extraordinary

mode deviates from the incident direction. Wave energy propagates at the group velocity and thus

beam splitting occurs.

Some isotropic dielectrics can become uniaxial media if placed in an electric eld. This is

because the permittivity is in general nonlinear and the component in the direction of the eld

becomes eld-dependent,

" = "(0) + En; (4.238)

where is a constant, n = 1 is for Pockels e¤ect and n = 2 is for Kerr e¤ect. Pockels e¤ect in

some liquids is widely used for laser switching and optical modulation.

42

Problems

4.1 A sphere of radius a carries a charge q which is decreasing due to emission of charge in

every radial direction. Show that there is no magnetic eld even though there exists a radial

conduction current density.

4.2 A cylindrical permanent magnet with radius a and a uniform axial magnetization Mz is

rotating at an angular frequency ! about its axis. What is the emf induced in the closed

circuit shown?

Problem 4.2. Unipolar generator.

4.3 The permittivity of an unmagnetized plasma is given by

"(!) = "0

1

!2pe!2

!;

where !pe is the electron plasma frequency,

!pe =

sne2

"0me:

Show that the energy density of a plane electromagnetic wave in a plasma is

u =1

2"0E

2 +1

2"0!2pe!2E2 +

1

20H

2 = "0E2;

and interpret the term1

2"0!2pe!2E2:

43

4.4 An isotropic dielectric has a permittivity in the form

"(!) = "0

1

!2p!2 !20

!:

Show that the group velocity does not exceed c: What if a nite dissipation is allowed,

"(!) = "0

1

!2p!2 + 2i ! !20

!;

where is a damping constant? For simplicity, assume !p = !0 and plot Re (d!=dk) as a

function of ! for various =!0:

4.5 Show that if a dielectric medium is loss free, its dielectric tensor should be Hermitian,

"ij = "ji:

Note: In the absence of external magnetic eld, the tensor is symmetric "ij = "ji: In this

case, the loss-less condition is Im ("ij) = 0:

4.6 Find the permittivity and thickness of a dielectric lm to be coated on a glass surface to

eliminate reection of light wave of wavelength = 550 nm. Assume that the glass has an

index of refraction of n = 1:5:

4.7 A light beam linearly polarized is incident on a glass (nglass = 1:5) surface at an angle i = 45:

Find the amplitude and polarization of the reected and refracted waves. Consider both cases

of polarization, electric eld in the incident plane and magnetic eld in the incident plane.

4.8 A light beam of circular polarization in glass is incident on a at glass-air boundary at an

angle i = 60: Find the polarization of totally reected beam.

4.9 A uniaxial crystal has "xx = "yy = 1:3"0 and "zz = 1:5"0: The optical axis is at angle 70 to

a at surface. A light beam of random polarization is incident normal to the surface. Find

the propagation direction of the extraordinary mode in the crystal.

4.10 A uniaxial crystal has the following permittivity tensor,

!" =

0B@ 3"0 0 0

0 3"0 0

0 0 2"0

1CA :A light beam in air is incident on a at surface(x y plane)of the crystal at an angle ifrom the normal which coincides with the optical axis z: Show that the refraction law for the

44

extraordinary mode is

tan r =

p"? sin iq

"k"k sin2 i

;where "? = 3; "k = 2: What is the refraction law for the ordinary mode?

4.11 Find the impedance per unit length of a copper wire of radius 3 mm at f = 60, 103 and 106

Hz. Copper conductivity is = 5:8 107 S/m at room temperature.

4.12 Using the equation of motion for electrons in a cold magnetized plasma,

me@v

@t= e (E+ v B0) ;

where B0 (external magnetic eld) is in z direction, show that the dielectric tensor is given

by

!" = "0

266666641

!2p!2 2 i

!2p!(!2 2) 0

i!2p

!(!2 2) 1!2ps

!2 2 0

0 0 1!2p!2

37777775where !p =

pne2=me"0 is the plasma frequency and e = eB0=me is the cyclotron frequency.

The wavevector k may be assumed to be k = k?ex+kkez without loss of generality because of

axial symmetry. Note that the tensor is Hermitian. (In this analysis, the electron temperature

is ignored.)

4.13 A laser beam passes through a glass window of refractive index 1.50 into water with refractive

index 1.33. The beam is E-polarized (electric eld in the incident plane). Design the glass

window to avoid reection at both surfaces.

45

Chapter 5

Radiation from Macroscopic Sources(Antennas and Apertures)

5.1 Introduction

Strictly speaking, radiation of electromagnetic waves from antennas and apertures should be an-

alyzed as a boundary value problem incorporating the boundary conditions for the electric and

magnetic elds. This is because the current distribution in antennas and electromagnetic elds in

apertures are likely to be modied by the elds created by antennas and apertures, and incorporat-

ing the boundary conditions should be able to solve radiation problems self-consistently. However,

such a procedure is usually very much involved even in simple geometry and often requires solving

integral equations. In this Chapter, we will assume that feedback from the elds on the prescribed

current and eld distribution is negligibly small. This simplication makes numerous radiation

problems analytically tractable.

5.2 Radiation Vector Potential

In Chapter 4, we have seen that the vector potential obeys a wave equationr2 1

c2@2

@t2

A = 0J (r; t) ; (5.1)

provided the Lorenz gauge is chosen,

r A+ 1c2@

@t= 0: (5.2)

Eq. (1) has the following retarded solution,

A(r; t) =04

ZJ(r0; t )jr r0j dV 0; (5.3)

1

where

=1

c

r r0 ; (5.4)

is the time required of electromagnetic disturbance to propagate from the source position r0 to the

observing position r: It is noted that J and A consist of longitudinal and transverse components.

Figure 5-1: The vector potential dA at a distance jr r0j from the current segment JdV 0 locatedat r0 is created by the current = jr r0j =c seconds earlier because of nite propagation speed ofelectromagnetic disturbance.

If the current is oscillating at a frequency !; and if the current density is separable in the form

J(r; t) = J(r)ei!t; (A m2) (5.5)

the vector potential becomes

A(r; t) =04

ZJ(r0)

jr r0j expi!

t jr r

0jc

dV 0; (5.6)

For a localized, small radiation source, the observation point is at r r0; and the vector potential

may be approximated by

A(r; t) ' 04

ei(kr!t)

r

ZJ(r0)eikr

0dV 0; (5.7)

where k = !=c; k = ker; and the following approximation is used,

kr r0 ' kr k r

r r0 = kr k r0 for r r0: (5.8)

2

The magnetic eld associated with the vector potential can be calculated from

B = rA; (5.9)

and the transverse radiation electric eld from

1

c2@E

@t= rB: (5.10)

If a localized source is radiating electromagnetic energy at a constant power P; the surface

integral of the Poynting ux on an arbitrary sphere with radius r should yield the power,

P =1

Z0

IjEj2 r2d = Z0

IjHj2 r2d = constant (in W); (5.11)

where

d = sin dd; (5.12)

is the di¤erential solid angle. Therefore, as far as radiation of energy is concerned, retaining both

electric and magnetic elds proportional to 1=r will su¢ ce: Since the vector potential already has

1=r dependence, the di¤erential operator r in calculating the electric and magnetic elds can be

approximated by

r1

rei(kr!t)

= r

r3ei(kr!t) + ik

1

rei(kr!t) ' ik1

rei(kr!t);

provided kr 1, that is, in the region at least several wavelengths away from the source region.

The region kr 1 is called radiation or far-eld zone. This indicates that as far as radiation power

is concerned, the magnetic eld may be approximated by

B = rA' ikA; kr 1: (5.13)

In the far eld region kr 1 e¤ects of radiation source become ignorable and electric and mag-

netic elds are essentially detached from the source. The radiation power calculated through the

approximation is pure real indicating a net power radiated by the source.

In order to nd a reactive power due to electromagnetic energy stored in the vicinity of a

source, one has to integrate the Poynting ux on a closed surface near the source. The radiation

zone approximation fails completely in this case. Also, even in radiation zone, higher order terms

are often required in evaluating radiation of angular momentum L,

dLdt=1

c

Ir (EH)r2d; (J). (5.14)

Since

r (EH) = E(r H)H(r E);

3

it can be seen that radial components, either Er or Hr; proportional to 1=r2; yield a constant rate

of radiation of angular momentum,

r (EH) _ 1

r2:

Example 1 Center-Fed Half Wavelength Dipole Antenna

Figure 5-2: Half wavelength center-fed antenna.

Radiation by a center-fed half wavelength dipole antenna is of practical importance because

of its common use in communication. It is formed by bending inner and outer conductors at a

distance =4 from the open end of a coaxial cable. The standing current wave along the antenna

may be approximated by

I(z; t) = I0 cos(kz)ei!t; k =

2

; (5.15)

if feedback from the radiated eld is ignorable. Then, the vector potential at kr 1 is

Az(r; t) =0I04r

ei(kr!t)Z =4

=4cos(kz0)eikz

0 cos dz

=0I04r

ei(kr!t)2

k sin2 cos2cos

; (5.16)

and the radiation magnetic eld in the region is

B ' ikA= i0I0

4rei(kr!t)

2

sin cos2cos

e: (5.17)

4

Corresponding Poynting ux is

Sr = Z jHj2

=ZI2042r2

1

sin2 cos2

2cos

; (5.18)

and angular distribution of the radiation power is given by

dP

d= r2Sr

=ZI2042

1

sin2 cos2

2cos

: (5.19)

Integration over the entire solid angle yields the total radiation power,

P =

ZdP

dd

=

Z

0sin d

Z 2

0dZI2042

1

sin2 cos2

2cos

=

ZI202

Z

0

1

sin cos2

2cos

d: (5.20)

The integral can be done numerically and amounts to 1.219. Then

P = 73:1 ()I20 = RradI2; (W) (5.21)

where

Rrad =

p0="02

1:219 = 73:1 () (5.22)

is the well known radiation resistance of the center-fed half wavelength dipole antenna. The angular

dependence of the Poynting ux

f() =1

sin2 cos2

2cos

; (5.23)

is shown below.

The calculation presented above entirely ignores the reactive power which may exist due to stor-

age of electric and/or magnetic energy in the vicinity of the antenna. To account for such reactive

power, we must deviate from the far-eld analysis and integrate the Poynting ux directly on a

surface close to the antenna surface. Instead of calculating the Poynting ux on the antenna surface

of nite radius a, we calculate the Poynting ux on a cylindrical surface of radius a surrounding an

ideally thin antenna of length =2 as shown in Fig. 5-4. In this approximation, the magnetic eld

on the antenna surface may be replaced by the static form without retardation,

H(z) =I(z)

2a=

I02a

cos(kz); 4< z <

4: (5.24)

5

­1.0 ­0.8 ­0.6 ­0.4 ­0.2 0.2 0.4 0.6 0.8 1.0

­0.3­0.2­0.1

0.10.20.3

x

y

Figure 5-3: Angular () dependence of the Poynting ux radiated by a half wavelength dipoleantenna.

The electric eld on the antenna surface is zero within our assumption of ideally conducting antenna

except at the gap at the midpoint. However, the electric eld due to the current lament assumed

at the axis is nite at the surface a distance a away. It can be calculated from

1

c2@

@tE = rB = r(r A)r2A; (5.25)

where A is the vector potential on the surface. It can be found from the integral,

Az(z) =04

Z =4

=4

I0 cos kz0

R(z; z0)eikR(z;z

0)dz0; (5.26)

where

R(z; z0) =p(z z0)2 + a2; (5.27)

is the distance between a point on the surface ( = a; z) and a current segment I(z0)dz0 at z0:With

this approximation, Eq. (5.25) reduces to

i !c2Ez(z) =

@2Az@z2

+ k2Az; (5.28)

since in current-free region the vector potential satises the Helmholtz equation

r2 + k2

Az = 0: (5.29)

The integral in

Ez(z) = ic20I04!

Z =4

=4cos(kz0)

@2

@z2+ k2

eikR(z;z

0)

R(z; z0)dz0; (5.30)

6

Figure 5-4: The Poynting ux on the surface of =2 antenna of nite radius a is approximated bythat on the cylindrical surface of radius a surrounding a thin antenna.

can be performed by noting@

@z

eikR(z;z0)

R(z; z0)= @

@z0eikR(z;z

0)

R(z; z0); (5.31)

and by integrating by parts twice with the result

Ez(z) = iZ0I04

eikR1

R1+eikR2

R2

; (5.32)

where

R1 =

sz

4

2+ a2; R2 =

sz +

4

2+ a2: (5.33)

The radial outward Poynting ux on the antenna surface is therefore given by

S = EzH

= iZ0I20

82a

eikR1

R1+eikR2

R2

cos(kz); (5.34)

and the power leaving through the antenna surface is

P = 2a

Z =4

=4Sdz

= iZ0I20

4

Z =4

=4

eikR1

R1+eikR2

R2

cos(kz)dz: (5.35)

7

x 0.10.080.060.040.020

2

1.5

1

0.5

0

Figure 5-5: Real (lower curve) and imaginary (upper curve) parts of the integral.

The integral has to be performed numerically. Introducing

x =4z

; A =

4a

;

we rewrite the integral in the form

f(A) =

Z 1

1

0BB@expi

2

p(1 x)2 +A2

p(1 x)2 +A2

+

exp

i

2

p(1 + x)2 +A2

p(1 + x)2 +A2

1CCA cos2x dx; (5.36)

which is shown in Fig. 5-5 as a function of the normalized antenna radius A = 4a=: For A < 0:01;

the radiation impedance is constant and approximately equal to

Zrad ' 73:1 i42:0 ()

= 73:1 + j42:0 () (in engineering notation)

The real part comfortably agrees with the radiation resistance calculated earlier in the far eld zone.

The reactive part of the impedance is inductive due to dominant magnetic energy compared with

the electric capacitive energy. However, the reactance is a very sensitive function of the antenna

length. It vanishes if the antenna length is chosen at l ' 0:49 and further decrease in the lengthmakes the reactance capacitive. Radiation from a center-fed antenna can be analyzed by assuming

a standing wave form,

I(z) = I0 sin[k(l jzj)];

and is left for exercise.The axial electric eld Ez(z) in Eq. (5.32) can be alternatively (perhaps

more conveniently) found from

Ez = @

@z @

@tAz; (5.37)

8

where is the retarded scalar potential given by

(z) =1

4"0

Zeijrr

0j

jr r0jr0dV 0

=1

4"0

ZeiR

Rlz0dz0; (5.38)

with

R =

q(z z0)2 + a2;

and l the linear charge density that can be found as

@l@t

+@I (z)

@z= 0;

l (z) =i

cI0 sin kz; (C m

1): (5.39)

Then,

Ez = @

@z @

@tAz

=iI04"0c

Z =4

=4

@

@z0

eiR

R

sinkz0dz0 +

i!0I04

Z =4

=4

cos kz0

ReikRdz0

=iI04"0c

eikR1

R1+eikR2

R2

=iZ0I04

eikR1

R1+eikR2

R2

; (5.40)

which agrees with Eq. (5.32). In radiation zone, the scalar potential is irrelevant but in near eld

region, it should be considered together with the vector potential in a self consistent manner.

5.3 Radiation from Sources Smaller than Wavelength: MultipoleRadiation

If the size of a radiating source is small compared with the wavelength kr0 1, the far eld vector

potential at r r0

A(r; t) ' 04r

ei(kr!t)ZJ(r0)ei

0krdV 0;

can be expanded in a series of k r0;

A(r; t) ' 04r

ei(kr!t)ZJ(r0)

1 ik r0 +

dV 0:

In magnetostatics, the rst integral vanishes,ZJ(r0)dV 0 = 0;

9

as shown in Chapter 3. However, for time varying current density, it does not, sinceZJidV =

Zrxi JdV =

Zxir JdV

=

Zxi@

@tdV =

d

dtpi;

where

pi =

ZxidV; (5.41)

is the i-th component of the electric dipole moment,

p =

ZrdV; (C m).

In general, for time varying sources, ZJ(r;t)dV =

d

dtp: (5.42)

The lowest radiation eld from a small source is therefore of electric dipole nature.

The integral in the next order, Z(k r)JdV;

is a combination of electric quadrupole and magnetic dipole as seen from the following manipulation,

kj

ZrjJidV = kj

Zrjrri JdV

= kj

Zr (rirjJ)dV kj

Zrirrj JdV kj

Zrirjr JdV

= kjZriJjdV + kj

Zrirj

@

@tdV; (5.43)

or going back to the vector formZ(k r)JdV =

Zr(k J)dV + d

dt

Z(k r)rdV: (5.44)

However,

k (r J) = r(k J) J(k r): (5.45)

Therefore, Z(k r)JdV = 1

2kZ(r J)dV + 1

2k ddt

ZrrdV

= km+1

2k ddtQ; (5.46)

10

where

m =1

2

Zr JdV; (A m2 = J T1) (5.47)

is the magnetic dipole moment and

Q=Zrr (r) dV; (C m2)

is the electric quadrupole moment tensor. Then, the radiation vector potential to order k r0 is

A(r) ' 04r

_p+ikm1

2ik _Q

eikr; kr0 1 (small source) (5.48)

where the dot indicates time derivative. The radiation magnetic eld can be found from

0H = B ' ikA; (5.49)

H(r) ' 1

4c

n p+ 1

cn(n m) 1

2cnn ...Q eikr

r; (5.50)

where n = r=r is the radial unit vector and the vector k =kn has been replaced by a time derivative

operator wherever appropriate,

k =!

cn = i

n

c

@

@t: (5.51)

The radiation power can be calculated from

P = Z0

ZjHj2 r2d; (W). (5.52)

For example, the power radiated by an electric dipole oscillating at a frequency ! is given by

P =Z0!

4

(4c)2

Z pn2 d=

1

4"0

1

4c3

Z pn2 d; (5.53)

wherep=

d2

dt2p:

The solid angle integration can be performed by exploitingZninjd =

4

3ik; i; j = x; y; z: (5.54)

The result is

electric dipole: Pp =1

4"0

2!4p2

3c3=

1

4"0

2p2

3c3; (W) (5.55)

11

where p = jpj is the magnitude of the dipole moment, e.g.,

p2 = p2x + p2y + p

2z: (5.56)

Similarly, the power radiated by a magnetic dipole is

magnetic dipole: Pm =1

4"0

2!4m2

3c5=

1

4"0

2 m2

3c5; (W) (5.57)

which is of higher order by a factor (v=c)2 ' (ka)2 compared with the radiation power due to

electric dipoles where v is the typical velocity of charges associated with the current J = nev and

a is the typical size of the radiation source.

The radiation power due to an electric quadrupole is given by

P =

r0"0

1

1621

4c4

Z n n ...Q2 d: (5.58)

The xcomponent of the vectorq = n (n Q) ; (5.59)

is

qx = ny(nxQxz + nyQyz + nzQzz) nz(nxQxy + nyQyy + nzQzy): (5.60)

Using Zninjnknld =

4

15(ijkl + ikjl + iljk); (5.61)

we nd Zq2xd =

4

15

Q2xy +Q

2xz + 4Q

2yz + (Qyy Qzz)2

: (5.62)

Similarly, Zq2yd =

4

15

Q2xy +Q

2yz + 4Q

2xz + (Qxx Qzz)2

; (5.63)

and Zq2zd =

4

15

Q2xz +Q

2yz + 4Q

2xy + (Qxx Qyy)2

; (5.64)

where the symmetry of the quadrupole moment Qij = Qji is noted. Adding the three components

yieldsZq2d =

4

15

6(Q2xy +Q

2xz +Q

2yz) + (Qxx Qyy)2 + (Qyy Qzz)2 + (Qzz Qxx)2

=

4

15

243Xij

Q2ij (Xi

Qii)2

35 ; (5.65)

12

and the radiation power is

electric quadrupole: PQ =1

4"0

1

60c5

243Xij

Q2

ij X

i

Qii

!235 ; (W). (5.66)

Example 2 Charge Undergoing Nonrelativistic Circular Motion

Figure 5-6: Charge e undergoing nonrelativistic circular motion. The dominant radiation processis of electric dipole nature.

This example of radiation from a nonrelativistic single charge is discussed here because radiation

elds are of multipole nature. In the lowest order, dipole radiation is expected followed by electric

quadrupole radiation and higher. The magnetic dipole radiation is absent because the magnetic

dipole associated with the circulating charge is constant. The current due to a moving single charge

is

J = ev (t) (r rp) ;

where e is the charge, rp (t) is the instantaneous location of the charge, and v (t) = drp=dt is the

velocity. If a charge e is undergoing circular motion with nonrelativistic velocity, v = !a c; the

dipole moment is

p (t) = erp (t) = ea(cos!tex + sin!tey)

= ea [sin cos( !t)er + cos cos( !t)e sin( !t)e] ; (5.67)

where the conversions

ex = rx = r(r sin cos)= sin coser + cos cose sine;

13

ey = ry = r(r sin sin)= sin siner + cos sine + cose;

are substituted. It is convenient to introduce complex notation,

p =1p2ea (sin er + cos e + ie) e

i(!t):

The factor 1=p2 is required for the amplitude jpj to remain ea: The vector potential is

A(r; t) =04r

_peikr =04r

(i!)pei(kr!t+); (5.68)

and the radiation magnetic eld is

H(r; t) ' 1

0ikA

= iea!k4r

1p2(e + i cos e) : (5.69)

Note that the eld rotates with the charge as explicitly indicated by the factor ei(kr!t+): The

magnetic eld is transverse (TM mode). In general, electric multipoles radiate TM modes. The

radiation eld is elliptically polarized,HH

= i cos : (5.70)

In particular, at = 0; the eld is circularly polarized with positive helicity,

HH

= i; (5.71)

At = =2; the component vanishes and the led is plane polarized. At = ; the eld is again

circularly polarized but with negative helicity,

HH

= i: (5.72)

The radiation power is constant because the magnitude of the dipole moment is constant, and is

given by

P =1

4"0

2!4(ea)2

3c3: (5.73)

However, a!2 = j _vj is the centripetal acceleration experienced by the charge and the power can berewritten as

P =1

4"0

2(e _v)2

3c3: (5.74)

This is the well known Larmors formula for radiation power emitted by a nonrelativistic charge

under acceleration _v regardless of its direction relative to the velocity v.

14

Let us consider higher order modes. The magnetic dipole moment of the circulating charge is

constant,

m =1

2

Zr JdV

=1

2e!r0

Zr e[r rp(t)]dV

=1

2e!r20ez (constant).

Therefore, there is no magnetic dipole radiation. The non-vanishing components of the quadrupole

moment tensor of the circulating charge are

Qxx = ea2 cos2 !t; Qyy = ea

2 sin2 !t; Qxy = Qyx = ea2 cos!t sin!t: (5.75)

Corresponding radiation power is

P =1

4"0

1

60c5

243Xij

...Q2ij

Xi

...Qii

!235=

1

4"0

8

5

(ea2)2!6

c5: (5.76)

It should be noted that the quadrupole radiation elds all oscillate at 2! (twice the circulation

frequency) and corresponding radial wavenumber is 2k = 2!=c: The quadrupole radiation power is

smaller than the dipole radiation power by an order (ka)2 as expected, for ka is the small expansion

parameter employed in the course of derivation. The appearance of the quadrupole radiation elds

even for pure circular motion of the charge is strictly due to retardation of radiation elds which

causes frequency modulation in the electromagnetic elds. In relativistic cases, this e¤ect is much

more pronounced and radiation intensity of higher harmonics becomes signicant. Radiation from

relativistic charges will be analyzed in Chapter 8.

Example 3 Radiation of Angular Momentum in the Preceding Example

The charge loses its kinetic energy to radiation elds at the rate

ddt

1

2Mv2

= P =

1

4"0

2!4(ea)2

3c3: (5.77)

Since the angular momentum of the charge is

Lz =Mvr0; (5.78)

radiation elds carry away angular momentum at the rate

dLzdt

= Mr0dv

dt=P

!: (5.79)

15

In general, if radiation elds have eim dependence, the ratio becomes

1P

dLzdt

=m

!: (5.80)

Here, we wish to verify Eq. (5.79) through direct calculation of the rate of angular momentum

radiation,

dLdt=1

cr2Zr (EH)d: (5.81)

Since an electric dipole radiates TM mode, it follows that

r (EH) = H(E r); (5.82)

and the radial component of the electric eld Er; that has been ignored in calculation of energy

radiation, must be retained. For the ux of angular momentum r (EH)=c to be proportional

to r2; the radial electric eld must be proportional to 1=r2; since the dominant radiation magnetic

eld is proportional to 1=r:

Since the radial electric eld is of higher order than the radiation eld, we may continue to use

the radiation magnetic eld,

H =1

0rA

' i

4reikrk _p = 1

4creikrn p: (5.83)

The electric eld is to be found from

"0@

@tE = rH =

1

0rrA =

1

0

rr A+ k2A

: (5.84)

Noting

r A =04r eikr

r_p

=04

ik

r 1

r2

eikr _pr;

we nd

Er =eikr

2"0cr2_pr =

eikr

2"0cr2(n _p):

Therefore, the radiation ux of the angular momentum is

1

cr (EH)

= 1c(r E)H

=1

4"0

1

2c3r2(n p)(n _p); (5.85)

16

and the rate of angular momentum radiation is given by

dLdt=

1

4"0

1

2c3

Z(n p)(n _p)d: (5.86)

In the case of circulating charge,

p(t) =1p2p0(sin er + cos e + ie)e

i(!t); p0 = ea; (5.87)

we have

_p = ip2!p0(sin er + cos e + ie)e

i(!t); (5.88)

and

n p = 1p2!2p0(cos e + ie)e

i(!t): (5.89)

Then, the rate of angular momentum loss and radiation power are related through

dLdt

=1

4"0

!3p204c3

Zsin2 dez

=1

4"0

2!3(ea)2

3c3ez

=P

!ez; (5.90)

as expected.

Example 4 Circular Loop Antenna

Figure 5-7: Small circular loop antenna with ka 1:

If a circular loop antenna of radius a carries a slowly oscillating current I0ei!t where !a c

or ka 1; the lowest order radiation eld is of magnetic dipole type. The magnetic dipole moment

17

is

mz = a2I0e

i!t; (5.91)

and radiation magnetic eld can be found from

H(r) =1

4c2reikrn (n ::

m)

=eikr

4c2r!2m0 sin e; (5.92)

where

m0 = a2I0:

The peak radiation power is

P =1

4"0

2!4m20

3c5; (5.93)

and the time averaged power is

Pave =1

4"0

!4m20

3c5: (5.94)

As the frequency increases, the dipole approximation breaks down. We will return to this problem

in Chapter 7.

5.4 Radiation from Apertures

Radiation from an aperture such as microwave horn antennas should be analyzed as a vector

boundary value problem. A rigorous formulation along this line will be made in Chapter 7. Here, a

nearly equivalent (but less rigorous) formulation based on a ctitious magnetic currentdeveloped

by Schelkuno¤ in 1936 will be given. The absence of magnetic charge (magnetic monopole) and its

ow (magnetic current) is well established. However, as the tangential component of the magnetic

eld on a given boundary surface can be replaced by a surface electric current, the tangential

component of electric eld can be replaced by a ctitious magnetic current.

Figure 5-8: The tangential magnetic eld on an aperture can be replaced by a surface electriccurrent, while an electric eld by a ctitious magnetic current.

18

Let us consider an area S on which tangential components of electric and magnetic elds Etand Ht are specied as illustrated in Fig. 5-8. In the method developed by Schelkuno¤, the normal

components of the electromagnetic eld are ignored, and the formula does not fully agree with that

based on the Greens dyadic in Chapter 7. The magnetic eld can be replaced by a surface electric

current

Js = nHt;

owing on a superconducting surface. Here n is the unit normal on the surface. The radiation

magnetic vector potential due to the surface current is given by

A(r) =04

ZJs

jr r0jeikjrr0jdS

=04

ZnHs

jr r0j eikjrr0jdS: (5.95)

For the tangential surface electric eld, Es; Schelkuno¤ introduced an electric vector potential,

F(r) =1

4

ZEs njr r0je

ikjrr0jdS; (5.96)

and called the quantity Es n as a surface magnetic current. The electric eld in terms of bothvector potentials is given by

E(r) =ic2

!rrAr F

=i

4"0!rr

ZnHs

jr r0j eikjrr0jdS +

1

4r

ZnEsjr r0je

ikjrr0jdS: (5.97)

In the radiation zone r r0; kr 1; this reduces to

E(r) ' i

4reikr

Zfk (nEs) Zk [er(nHs)]g e

ikr0dS; (5.98)

where

Z =p0="0 = 377 : (5.99)

Note that the electric eld derived above is explicitly transverse to the radial direction. In Chapter

7, a similar formula based on the Greens dyadic for the wave equation will be derived,

E(r) =i

4reikr

Z[k (nEs) + !nBs k(n Es)] e

ikr0dS: (5.100)

This is not explicitly transverse but is in fact identical to Eq. (5.98) as will be shown in Chapter 7.

For a small hole in a conducting plate, the electric and magnetic dipole moments derived earlier

can be conveniently used to analyze leakage of electromagnetic energy through the hole. The electric

19

dipole moment of a small hole in a conducting plate is

p = 4"0a3

3E?; (5.101)

where E? is the normal component of the unperturbed electric eld behind the plate. The magnetic

dipole moment of the hole is

m = 42a3

3Hk =

8a3

3Hk; (5.102)

where Hk is the component of the magnetic eld parallel to the plate. If the elds are oscillating,

so are the dipole moments and they radiate electromagnetic waves. The oscillation frequency !

must be low enough for the dipole approximation to hold, !a c; or ka 1; that is, the hole size

be much smaller than the wavelength :

Example 5 Radiation from an Open End of Coaxial Cable

Figure 5-9: Radiation from an open end of coaxial cable. (The lowest order TEM mode in thecable is assumed.)

The lowest order electromagnetic eld propagating in an coaxial cable is azimuthally symmetric

and the eld components are given by

E(; t) =V0

ln(b=a)

ei(kz!t)

; (5.103)

H(; t) =E(; t)

Z0; (5.104)

where a and b are the inner and outer radii and Z0 =p"=0 is the impedance of the insulating

material lling the cable. If the cable is terminated at an open end, voltage and current standing

waves are formed. The magnetic eld vanishes at an open end and the electric eld is doubled. The

electric eld radiated by the open end can be calculated from Eq. (??) with ignoring the magnetic

20

eld,

E(r) ' i

4reikrk

Z(nEs)e

ikr0dS

=i

4reikr

kV0ln(b=a)

er ez

Z b

ad

Z 2

0d0elk sin cos(

0)e0

' e

ikr

8r

k2V0ln(b=a)

(b2 a2) sin e; (5.105)

where kb 1 has been assumed. The radiation power is given by

P = r2Z jEj2

Z0d

=1

64

1

Z0

k2V0ln(b=a)

(b2 a2)22

Z

0sin3 d

=

24

1

Z0

k2

ln(b=a)(b2 a2)

2V 20 : (5.106)

This denes the radiation resistance

Rrad =24

ln(b=a)

k2(b2 a2)

2Z0: (5.107)

Rrad is large because of the assumption kb 1 and indicates that radiation from an open end

of a coaxial cable is extremely ine¢ cient. Note that in the integration over the solid angle, the

entire space is covered, 0 < < ; 0 < < 2: If a coaxial cable is connected to a hole in a large

conducting plate, the integration is limited to 0 < < =2: However, the radiation electric eld

should be doubled because of image e¤ect. The Poynting ux is quadrupled. However, since

integration covers the half space, the radiation power is only doubled.

Example 6 Leakage of Microwave through a Small Hole

Branching microwave power to another waveguide is often done with a device called directional

coupler shown in Fig.5-11. An even number of small holes are drilled a quarter wavelength apart

through a wall separating two waveguides. It is evident that microwave propagates toward the

region A because of constructive interference in that direction, while in region B, waves su¤er

destructive interference. In this example, we analyze radiation from a single small hole drilled at

the center of a wider wall of a rectangular waveguide.

If the hole radius is much smaller than the wavelength, the hole acts as oscillating dipoles

with electric and magnetic dipole moments,

p = 4"03

3E?; m = 8

3

3Hk: (5.108)

21

Figure 5-10: For radiation from an open end of coaxial cable in a large conducting plate, theradiation electric eld should be doubled compared with the case of free end. This is due to animage current formed by the conducting plate.

The fundamental mode in a rectangular waveguide is the TE10 mode whose elds are

Ey(x) = E0 sinax; Hx =

E0Z10

sinax; Hz _ cos

ax; (5.109)

where

Z10 =Z0p

1 (!c=!)2; (5.110)

is the impedance of the TE10 mode with !c = c=a the cuto¤ frequency. Since Hz vanishes at the

hole, the dipoles of the hole become

p = 4"03

3E0ey; m = 8

3

3

E0Z10

ex; (5.111)

Substituting this into the radiation magnetic eld due to dipoles in Eq. (??), we obtain

H(r) ' 1

4c

::pn+ 1

c(::mn) n

eikr

r

= k23E0e

ikr

3Z0r

(1 2

p1 (!c=!)2) cose

+(2p1 (!c=!)2 cos ) sine

; (5.112)

where Z0 =p0="0: Calculation of radiation power, which is proportional to

6; is left for exercise.

Example 7 Resonant Slot Antenna

22

Figure 5-11: Top: Directional coupler. g is the axial wavelength in the waveguide. Propagationtoward region B is prohibited because of destructive interference. Bottom: A small hole at themidpoint of the wider wall radiates as electric and magnetic dipoles.

Slot antennas are widely used in microwave radiation. In this example, a simple resonant half

wavelength slot antenna is considered. As explained in Example 5, radiation electric eld from an

aperture in a large conducting plate is twice the eld due to an open aperture because of image

e¤ect. In Fig. 5.4, the electric eld across the slot is assumed to be

Ex = E0 cos(kz): (5.113)

Then the radiated electric eld is found from

E(r) ' i

4reikrk

Z(nEs)e

ikr0dS

= i2aE04r

eikrk ezZ =4

=4cos kz0eikz

0 cos dz0; (5.114)

where the antenna is assumed to be in the z-direction. The integral is identical to that encountered

in the case of half wavelength antenna, and the radiation electric eld reduces to

E(r; ) =ieikr

aE0r

cos2cos

sin

: (5.115)

23

Figure 5.12: =2 resonant slot antenna.

24

Problems

5.1 A quarter wavelength antenna is erected vertically from the ground which may be regarded

as a perfect conductor in the lowest order approximation. Show that the radiation resistance

is 73=2 = 36:5 :

5.2 Assuming the following current distribution in a center-fed dipole antenna of length 2l and

radius a,

I(z) = I0 sin[k(l jzj)]; l z l;

nd, through numerical analysis, the length at which the antenna impedance becomes pure

real. Consider the cases of a= = 0:001 and 0.01.

5.3 Show that if the permittivity and permeability tensors are symmetric, "ij = "ji; ij = ji;

r (Ea Hb Eb Ha) = 0;

where Ea(b) is the electric eld due to antenna a(b) and Ha(b) is the magnetic eld due to

antenna a(b): This is known as the reciprocity theorem.

5.4 An antenna array is formed by six dipole antennas half wavelength apart. The phase of

feeding currents can be controlled Im = I0ei(m1); where m = 1 6: Determine the angularand phase dependence of the far-eld Poynting ux f(; ) and observe that the radiation

pattern can be rotated by changing the phase :

Six-antenna array (top view).

5.5 An antenna of length l carries a travelling wave I(z; t) = I0ei(kz!t): Show that the radiation

power is given by

P =Z04I20

Z

0

sin2[(kl=2)(1 cos )] sin3 (1 cos )2 d; (W),

25

where Z0 =p0="0:

5.6 Two half wavelength antennas are a distance d apart as shown. One is driven by a power

supply and the other is passively driven by the rst antenna through a mutual impedance

approximately given by

Z12 = iZ04

Z =4

=4

eikr1

r1+eikr2

r2

cos(kz2)dz2;

where

r1 =pd2 + (z2 =4)2; r2 =

pd2 + (z2 + =4)2:

Determine the angular dependence of the radiation pattern. (This problem shows the principle

of Yagi-Uda antenna. The driven antenna is called a parasitic antenna, or reector/director

depending on its e¤ect on the radiation pattern. In practice, several parasitic antennas are

installed to increase directivity as in TV antennas.)

Hint: To nd I2 in terms of I1; use

V2 = 0 = Z12I1 + Z22I2;

where Z22 = Z11 = 73:1 i42:0 :

Yagi-Uda antenna. One is active and the other is passive (driven by the rst antenna).

5.7 A rectangular aperture of area a b in a conducting plate is illuminated uniformly by a shortwavelength incident wave. Find the di¤racted electric eld.

5.8 Calculate the power radiated by a hole in the wall of a rectangular waveguide using the eld

given in Eq. (5.112).

5.9 Two identical charges undergo circular motion at an orbit radius a and rotation frequency !0staying on a diameter. Find (a) the radiation power and (b) the rate of angular momentum

radiation. Assume a!0 c (nonrelativistic).

5.10 Generalize the preceding problem for the case of N identical charges distributed along a circle

at equal spacing. In the limit N !1; radiation is expected to disappear. Does it?

26

5.11 An electric dipole moment is suddenly created at r = 0; t = 0; p (t) = pzU (t) where U (t) is

the Heaviside step function. (Consider two charges q and q originally overlapped that aresuddenly pulled apart at t = 0.) Find exact expressions for the electric and magnetic elds.

5.12 A magnetic dipole m precesses at a frequency ! and angle 0 from precession axis. Find the

radiation power. (Radiation from pulsersis believed to be due to this mechanism.)

27

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

Chapter 7

Di¤raction: Boundary ValueProblems of Electromagnetic Waves

7.1 Introduction

In electrostatics, a prescribed potential distribution on a closed surface uniquely determines the potential

elsewhere according to the Dirichlets formulation,

(r) = IS

@G

@ns(r

0)dS0; G = 0 on S;

provided G(r; r0), the Greens function, is so chosen that it vanishes on the closed surface S: The normal

derivative of the Greens function @G=@n can be regarded as a projection or mapping operator to convert a

given distribution of the surface potential s(r0) into the potential at an arbitrary observing position r: For

scalar potentials, the projection operator is a scalar function, or more precisely, 1 1 matrix function.For any vector elds, too, prescribing a vector on a closed surface uniquely determines the vector led

elsewhere. However, a vector eld changes its direction as well as its magnitude over a closed surface. A

vector eld prescribed on a closed surface is to be converted into a vector eld with denitive direction and

magnitude at the observing position. A projection operator that transforms one vector eld on a closed

surface into another elsewhere must be a tensor (or dyadic) composed of appropriate eigenvectors for the

vector eld of concern. For electromagnetic elds, the TE and TM eigenvectors identied in Chapter 5 can

be conveniently used for this purpose.

In this Chapter, a basic formulation will be developed for vector boundary value problems of electromag-

netic elds, E and B: It will then be applied to some di¤raction problems in which boundary electromagnetic

elds are known. However, in most practical problems, it is di¢ cult to know precisely boundary elds be-

cause of the feedback from the di¤racted elds themselves. A rigorous analysis of di¤raction problems usually

requires solving integral equations based on the boundary conditions for electromagnetic waves.

7.2 Vector Greens Theorem

For arbitrary vector elds E and F, by exploiting the expansion

r (Er F) = rE r FE r r F; (7.1)

1

r (FrE) = r F r E F r rE; (7.2)

the following identity can easily be proven,ZV

(E r r F F r rE)dV =IS

(FrEEr F) dS: (7.3)

Since

E r r F = E [r(r F)r2F] (7.4)

= r [E(r F)] (r E)(r F)E r2F; (7.5)

Eq. (7.3) can be modied asZV

(F r2EE r2F)dV =IS

[(Fr EEr F) n (nrE) F nE r F] dS; (7.6)

where n is the unit normal vector on the closed surface directed away from the volume V; dS = ndS: (See

Fig. 7.1.) This is a mathematical identity and holds for arbitrary vector elds E and F.

Figure 7-1: For Es; the electric eld specied on the closed surface S; the electric eld o¤ the surface E isuniquely determined. n is the normal unit vector directed away from the region of interest.

We now let the vector E be the electric eld and F be one component of Greens dyadic. In source freeregion, the electric eld E satises the Helmholtz equation,

(r2 + k2)E = 0; (7.7)

and the Greens dyadic satises the singular inhomogeneous Helmholtz equation

(r2 + k2)G = (r r0)1; (7.8)

where 1 is the unit dyadic dened in terms of appropriate eigenvectors for the electromagnetic elds. Then,

2

Eq. (7.4) yields an expression for the electric eld E in terms of the electric eld specied on the closed

surface S;

E(r) = IS

[(n E)r G+ (nrE) G+ (nE) (rG)] dS; (7.9)

where r E = 0 (no charges on S) has been assumed. Note that r G is a vector and r G is a dyadic.

For a diagonal Greens dyadic

G = G1; (7.10)

it follows that

r G = rG; (7.11)

(nrE) G = (nrE)G; (7.12)

(nE) (rG) = (nE)rG: (7.13)

Therefore, Eq. (7.9) reduces to

E(r) = IS

[(n E)rG+ (nrE)G+ (nE)rG] dS: (7.14)

The scalar Greens function G satises

(r2 + k2)G = (r r0); (7.15)

and its explicit expression in terms of spherical harmonic expansion is

G(r; r0) =1

4 jr r0jeikjrr0j

= ik1Xl=0

lXm=l

(jl (kr

0)h(1)l (kr)

jl (kr)h(1)l (kr0)

)Ylm (; )Y

lm

0; 0

;

(r > r0

r < r0

): (7.16)

As shown in Chapter 6, the solution for the vector potential that satises the Helmholtz equationr2 + k2

A = 0J; (7.17)

can be decomposed into TE and TM modes. In the outer region r > r0; we have the following decomposition,

A(r) = 0ik1Xl=1

lXm=l

1

l(l + 1)h(1)l (kr)rrYlm(; )

Zjl(kr

0)r0 r0Ylm(0; 0) J(r0)dV 0

+0i

k

1Xl=1

lXm=l

1

l(l + 1)r [h(1)l (kr)rrYlm(; )]

Zr0 [jl(kr0)r0 r0Ylm(0; 0)] J(r0)dV 0:(7.18)

This can be written in a more compact form using the Greens dyadic G;

A(r) =04

Zeikjrr

0jjr r0j J(r

0)dV 0 = 0

ZGJ(r0)dV 0; (7.19)

3

where the Greens dyadic is

G(r; r0) =eikjrr

0j

4 jr r0j1

= ikXlm

1

l(l + 1)h(1)l (kr)jl(kr

0)rrYlm(; )r0 r0Ylm(0; 0) (7.20)

+i

k

Xlm

1

l(l + 1)r [h(1)l (kr)rrYlm(; )]r0 [jl(kr0)r0 r0Ylm(0; 0)]: (7.21)

Note that the TE eigenvector

h(1)l (kr)rrYlm(; ); (7.22)

and TM eigenvector

r [h(1)l (kr)rrYlm(; )]; (7.23)

are orthogonal to each other for given mode numbers l and m and the Greens dyadic is evidently diagonal.

For di¤raction by an aperture in a large, at conducting screen, Eq. (7.14) simplies to

E(r) = 2Zaperture

(nE)rGdS: (7.24)

This formula was originally found by Smythe based on an equivalent layer of magnetic dipole (equivalent to

a sheet of magnetic current),

dm =2

i!0nEdS; (7.25)

which produces a vector potential

A (r) =2

i!

ZrG (nE)dS: (7.26)

(Recall that an oscillating magnetic dipole creates a vector potential,

A (r) = 0rGm '0eikr

4rikm;

provided kr 1:) Corresponding electric eld is

E(r) = @@tA =2

Zaperture

rG (nE)dS

= 2Zaperture

(nE)rGdS:

(Explain why we can use

E(r) = @@tA (r) ;

here instead of going through the usual procedure

1

c2@E

@t= rB = r (rA);

in terms of longitudinal and transverse components of A:) The tangential component of the di¤racted

magnetic eld in the aperture vanishes (that is, in the aperture, the tangential magnetic eld is equal to that

4

of the incident wave which is una¤ected by the aperture) and only the tangential component of the electric

eld contributes to di¤raction. Likewise, the normal component of the di¤racted electric eld vanishes in

the aperture. These follow from the symmetry property of the di¤racted electric and magnetic eld in front

of and behind the conducting plate. The di¤racted electric eld normal to the plate is odd with respect to

the normal coordinate z,

Ez(z) = Ez(z); (7.27)

and so is the tangential magnetic eld,

ez B (z) = ez B (z) ; (7.28)

but the tangential electric eld and normal magnetic eld are even,

ez E (z) = ez E (z) ; Bz(z) = Bz(z): (7.29)

Therefore in the aperture, the tangential component of the di¤racted magnetic eld should vanish. On the

surface of the conducting plate away from the aperture, tangential component of di¤racted magnetic eld

can exist with a corresponding surface current

nH = Js; (A m1):

Since the normal unit vector n changes sign from one side (z < 0) to other (z > 0) ; Js = nH is even

with respect to z; Js (z = 0) = Js (z = +0) : The surface charges induced on both sides of the conductingplate are also equal since = "0n E is even with respect to z: We have encountered such symmetry

properties in Chapters 2 and 3 when we analyzed leakage of electric and magnetic elds through a hole in a

superconducting plate. In both cases, e¤ective dipoles for elds in regions z < 0 and z > 0 are opposite to

each other and perturbed electric and magnetic elds due to the hole have such symmetry properties which

hold in general even for radiation elds as long as the plate can be regarded as ideally conducting.

The Smythes formula, Eq. (7.24), may be derived purely mathematically as follows. For a at boundary,

the normal unit vector n is constant, say n = ez. First, we noteI(nE)rGdS =

Ir (GnE) dS +

IGr (nE) dS =

IGr (nE) dS; (7.30)

since for a closed surface, Ir [G (nE)]dS = 0: (7.31)

Recalling r E =0 and exploiting the fact that for a unidirectional vector n;

r (nE) = (Er)n (n r)E+ n (r E)E (r n)= (n r)E; (7.32)

r (n E) = (Er)n+(n r)E+ n (rE) +E (r n)= (n r)E+ n (rE) ; (7.33)

5

we nd for a constant vector n;

n (rE) = r (nE) +r (n E) : (7.34)

Then IGn (rE) dS =

IGr (nE) dS +

IGr(n E)dS

=

I(nE)rGdS

I(n E)rGdS;

and Eq. (7.14) reduces to

E(r) = IS

[(n E)rG+ (nrE)G+ (nE)rG] dS

= 2Zwhole at surface

(nE)rGdS;

or

E(r) = 2Zwhole at surface

[(n E)rG+ (nrE)G] dS:

In the case of di¤raction by an aperture in a conducting plate, nE =0 except in the aperture. Therefore,

E(r) = 2Zaperture

(nE)rGdS: (7.35)

This should be equal to

E(r) = 2Zconducting plate

[(n E)rG+ i!(nB)G] dS; (7.36)

since in the aperture where no eld discontinuity exists, both n E and nB vanish. (They are odd functionsof z:)

Rigorous analysis of di¤raction by an aperture in a conducting screen is extremely di¢ cult mainly because

in Eqs. (7.14) and (7.24), the elds on the surface and in aperture are the di¤racted elds which are to be

found! However, in some limiting cases, sensible analytic solutions can be found. In short wavelength limit

ka 1 where a is the size of aperture, geometric optics approximation may be used. In this case, the eld

in the aperture may be approximated by the incident eld. In long wavelength limit ka 1; the eld in the

aperture is totally unknown. However, di¤racted power can be calculated accurately using e¤ective dipoles.

Then the eld intensity in the aperture can be estimated to be kaEi Ei where Ei is the incident electric

eld.

7.3 Some Examples of Aperture Di¤raction

7.3.1 Long Slit

Let a plane wave be incident on a at opaque screen having a long slit with an opening width a: The problem

is essentially two dimensional and the di¤racted wavevector k may be assumed to lie in the x y plane.Since the slit extends from z0 = 1 to 1; we are not allowed to assume r r0 and the di¤racted electric

6

Figure 7-2: Di¤raction by a long slit of width a in a conducting screen. In short wavelength limit ka 1;the electric eld in the slit opening may be approximated by that of the incident wave. Smythes formula inEq. (7.24) is applicable.

eld has to be calculated from

E(r) = 2Z(nE)r0Gdy0; (7.37)

where G is the two-dimensional Greens function satisfying

(r2 + k2)G2 = 2(r r0) = (x x0)(y y0): (7.38)

Solution for G2 is

G2 =i

4H(1)0

hkpx2 + (y y0)2

i; (7.39)

where H(1)0 (x) is the Hankel function of the rst kind. Its asymptotic form is

H(1)0 (x) '

r2

kxeikxi

4 ; x 1: (7.40)

Using n0= n (unit normal directed into the di¤racted region) and noting

r0G2 = rG2;

we rewrite Eq. (7.37) as

E(r) = 2rZG(n0E)dy0: (7.41)

Substituting

H(1)0

hkpx2 + (y y0)2

i'

r2

keiki

4iky

0

=

r2

keiki

4iky

0 sin; (7.42)

7

where

=px2 + y2; (7.43)

and is the angle between k and x-axis, we obtain

E(r) ' a2k (n0E0)

r2

keiki

4sin

; (7.44)

with dened by

=ka

2sin: (7.45)

The function sin= is the familiar Fraunhofer di¤raction form factor to describe the angular () dependence

of the eld intensity. The radiation (di¤racted) power per unit length of the slit is

P

l=

Z =2

=2c"0 jEj2 d: (7.46)

In short wavelength limit ka 1; the integration limits may be approximated by from 1 to 1; and weobtain an expected result

P

l= c"0 jE0j2 a; ka 1: (7.47)

In the opposite limit ka 1; the electric eld in the slit opening cannot be replaced by the incident eld

because the current induced in the conducting plate greatly disturbs the incident eld. The problem is dual

of scattering of electromagnetic waves by a thin, long conducting plate which will be analyzed as a separate

problem.

7.3.2 Circular Aperture

Figure 7-3: Geometry of circular aperture of radius a in an opaque screen.

Radiation from a circular aperture on a large conducting plate is a classical problem and of practical im-

portance. In short wavelength limit ka 1 (a the hole radius), the eld in the aperture may be approximated

8

by that of the incident wave E0: The di¤racted electric eld is given by

E(r) =i

2rk (n0E0)eikr

Zeikr

0dS; (7.48)

where in the geometry shown,

k r0 = kr0 sin sin0:

The integral can be performed as follows:Z a

0

r0dr0Z 2

0

eikr0 sin sin0d0

= 2

Z a

0

J0(kr0 sin )r0dr0

=2a

k sin J1(ka sin ): (7.49)

The di¤racted electric eld thus reduces to

E(r) =ia2

rk (n0E0)eikr

J1(ka sin )

ka sin : (7.50)

Since the amplitude of the Bessel function J1(x) decreases with x in a manner J1(x) _ 1=px; the electric

eld is appreciable only at small : In order of magnitude, di¤racted Poynting ux has an angular spread

about the z axis,

sin ' ' 1

ka 1:

The rst intensity minimum occurs at the rst root of J1(x) = 0; ka sin ' 3:83; or

' 3:83

ka= 1:22

D; D = 2a (diameter of the hole). (7.51)

This is the familiar di¤raction limited resolving power of circular apertures such as mirrors in telescopes and

parabolic microwave antennas. The radiation power associated with the di¤racted eld is

P = c"0r2

ZjEj2 d

= c"0 jE0j2 2a2Z =2

0

J21 (ka sin )

sin2 sin d: (7.52)

Since ka 1; the upper limit of the integral may be extended to 1 and sin may be approximated by :

Noting Z 1

0

J21 (ka)

d =

1

2; (independent of ka);

we nd

P = c"0 jE0j2 a2: (7.53)

This is an expected result in the short wavelength limit (ka 1) and simply corresponds to the incident

power going through the circular aperture.

In long wavelength limit ka 1; the above analysis breaks down completely, for the electromagnetic elds

in the aperture are entirely di¤erent from those associated with the incident wave. For normal incidence on a

conducting plate, the electric eld vanishes at the plate but the magnetic eld is doubled because of complete

9

reection. This does not mean there exists a tangential magnetic eld of H =2H0 in the aperture because

the aperture should not a¤ect the incident magnetic eld and the tangential component of the magnetic

eld in the aperture is H = H0:What we can do is to nd an e¤ective magnetic dipole moment of the hole

using 2H0 as the unperturbed eld as we did in analyzing the leakage of magnetic eld through a hole in a

superconducting plate. The dipole moment of the hole is

m = 8a3

3 2H0 =

16a3

3H0; (7.54)

which radiates at a power

P =1

2 1

4"0

2!4m2

3c5

=64

27(ka)4a2Z0 jH0j2 ; (7.55)

where the factor 12 is due to radiation into half space (behind the screen). The transmission cross-section is

t 'P

Z0 jH0j2=64

27(ka)4a2 _ k4a6: (7.56)

The dependence _ k4 (or !4) is the common feature of Rayleigh scattering of electromagnetic waves bysmall objects (in this case an aperture). _ a6 indicates extremely sensitive dependence of the cross sectionon the hole radius. To order k6; the cross section is

' 64

27(ka)4a2

1 +

22

25(ka)2 +

; ka 1: (7.57)

Since the radiation power is proportional toZ (nEa)eikr0dS

2 ' E2aa4;where Ea is the electric eld in the aperture, it is evident that the eld is of the order of

Ea ' kaE0 E0;

being much smaller than the incident eld E0: This is expected since the incident wave is essentially short-

circuited at the plate and only a small eld can leak through the aperture. The electric eld in the aperture

has been worked out by Bouwkamp,

Ex (x; y) = i4k

3

xypa2 x2 y2

E0; (7.58)

Ey (x; y) = i4k

3

x2 + 2y2 2a2pa2 x2 y2

E0 = i4k

3

2 + 2 sin2 2a2pa2 2

E0: (7.59)

Ex is an odd function of x and y and does not contribute (integrates to 0). The eld diverges at the rim

= a but is integrable. Integration over the aperture yieldsZ a

0

d

Z 2

0

dEy(x; y) = i8ka3

3E0: (7.60)

10

The di¤racted electric eld can then be found using the Smythes formula which is equivalent to the eld

due to an e¤ective magnetic dipole moment in Eq. (7.54).

Di¤raction in the regime ka ' O(1) is di¢ cult to analyze and requires numerical analysis. The trans-mission cross-section peaks at ka ' 1:56 and its value is max ' 1:8 a2. This behavior is similar to thecase of scattering by a conducting sphere.

7.3.3 Radiation from an Open End of a Rectangular Waveguide

Figure 7-4: Radiation from an open end of a rectangular waveguide. The electric eld in the opening isEy(x) = E0 sin(x=a) and the tangential component of the magnetic eld is Hx(x) = E(x)=Z10 where Z10is the impedance of the TE10 mode.

The electric eld of the TE10 mode in a rectangular waveguide is given by

E(x; z; t) = E0 sinaxei(kzz!t)ey; (7.61)

where a is the width of the waveguide and

kz =!

c

p1 (!c=!)2; (7.62)

is the axial wavenumber with

!c =c

a; (7.63)

being the cuto¤ frequency of the TE10 mode. The accompanying magnetic eld transverse to the axis is

H(x; z; t) = E0Z10

sinaxei(kzz!t)ex; (7.64)

where

Z10 =

r0"0

p1 (!c=!)2; (7.65)

is the impedance of the TE10 mode. (The axial magnetic eld Hz associated with the TE10 mode does not

11

contribute to radiation from the aperture.) If a waveguide is truncated, the open radiates electromagnetic

waves. The Smythes formula is not applicable because it only pertains to radiation from apertures in a

large conducting plate. In the general di¤raction formula

E(r) = IS

(n E)r0G+G(nr0 E) + (nE)r0G

dS; (7.66)

where

G(r; r0) =eikjrr

0j4 jr r0j ;

we assume r r0. Then,

G(r; r0) ' 1

4reikrikr

0;

and

r0G = rG ' ik 1

4reikrikr

0:

Also,

rE = @B@t

= i!B:

Then,

E(r) ' ieikr

4r

IS

k (n0E) + !n0B k(n0E)

eikr

0dS; (7.67)

where n0= n is the normal unit directed toward the region where the electric eld is to be evaluated. It isnot obvious that the contribution from the second and third terms in the integrand,I

S

!n0B k(n0E)

eikr

0dS; (7.68)

is explicitly transverse (perpendicular to k) which should be the case since we are calculating radiation eld.

To prove this, we rst show that

r IS

(nB)G (r; r0) dS;

can be expressed in terms of the normal component of the electric eld as follows.

r IS

G (nB) dS = IS

r0G (nB) dS

=

IS

r0G (B n) dS

=

IS

r0GB

ndS

=

IS

r0 (GB)

ndS

IS

Gr0 B

ndS

= 0 +i!

c2

IS

G (E n) dS;

where use is made of the identity IS

r0 A

ndS = 0:

12

Therefore,

rrIS

G (nB) dS = rr IS

G (nB) dS + k2IS

G (nB) dS

=i!

c2rIS

G (E n) dS + k2IS

G (nB) dS

= i!c2

IS

(E n)r0GdS + k2IS

G (nB) dS;

and the di¤racted eelctric eld can alternatively be calculated from

E(r) = IS

(n E)r0G+G(nr0 E) + (nE)r0G

dS

= IS

rG (nE) + i

!"0rr (GnH)

dS:

Then Eq. (7.67) may be rewritten as

E(r) =ieikr

4rkIS

n0 E 1

!"0k (n0 H)

eikr

0dS; (7.69)

which is explicitly transverse to k: As mentioned in Chapter 5, this formula was rst derived by Schelkuno¤ in

term of a ctitious magnetic current to replace the tangential component of the electric eld on a boundary.

The price to be paid for introducing a magnetic current is that the Maxwells equation, r B = 0; has to beviolated. In the derivation presented here, no magnetic current is assumed.

Denoting the surface integral over the open end of the waveguide by

I(; ) =

Z a

0

dx

Z b

0

dy sinaxeik(x sin cos+y sin sin)

=i

k sin sin

=a

(=a)2 k2 sin2 cos2 (1 + eika sin cos)(1 eikb sin sin); (7.70)

we nally nd the electric eld radiated from the open end of a rectangular waveguide,

E(r) = ikeikr

4rI(; )

" cos p

1 (!c=!)2+ 1

!sine +

1p

1 (!c=!)2+ cos

!cose

#: (7.71)

Calculation of the radiation power is left for an exercise.

7.3.4 Scattering by a Conducting Sphere Revisited

This problem has been solved rigorously in the preceding Chapter in terms of spherical harmonic expansion.

However, the result is not very illuminating physically particularly in the short wavelength limit ka 1

wherein summation of a large number of functions involving spherical Bessel functions and their derivatives

must be performed. In this limit, geometrical optics approximation should be able to yield the scattered

electric eld provided the boundary conditions for the electromagnetic elds are appropriately incorporated.

The incident plane wave seesthe cross-section of the sphere a2 which corresponds to the scattering cross-

section due to scattering by the illuminated hemispherical surface facing the incident wave. Scattering by

the illuminated surface is nothing but reection by a spherical convex mirror. The Poynting ux associated

with reection is uniform (isotropic) and independent of the angles and . The total correct cross-section

13

Figure 7-5: Scattering by a conducting sphere in short wavelength limit ka 1:

as revealed from the rigorous analysis was 2a2: The additional a2 is due to shadow scatteringfrom the

hemispherical surface in the shadow of the incident wave where the eld vanish. This vanishing eld can be

interpreted as cancellation between the eld of the incident wave and that of the scattered wave. Shadow

scattering is dual of di¤raction by a circular aperture in a conducting screen which radiates a power

Psh = c"0 jE0j2 a2; ka 1: (7.72)

Although the scattered power through shadow scattering is identical to that by the illuminated surface, the

Poynting ux is sharply peaked behind the sphere as in the case of di¤raction with an angular spread of

order ' =a 1:

If the sphere is ideally conducting, the boundary conditions for the electric and magnetic elds are

Et = 0; Bn = 0: (7.73)

Let the electric eld of the incident wave be Ei and that of scattered eld be Es: The vanishing tangential

component of the total eld requires that

n (Ei+Es) = 0: (7.74)

The vanishing normal component of the magnetic eld requires that

n (Bi+Bs) = 0: (7.75)

In the spirit of geometrical optics valid in the short wavelength regime,

Ei =c

kkiBi; Es =

c

kksBs;

14

where ki is the wavevector of the incident wave and ks is that of the scattered wave. From

n (Ei+Es) = n (kiBi+ksBs) = 0; (7.76)

and

n ki= n ks; n ki= n ks; (7.77)

it follows that

n (BiBs) = 0: (7.78)

Likewise, from

n (kiEiksEs) = 0;

we nd

n (EiEs) = 0:

With these preparations, we now apply Eq. (7.67) to the illuminated and shadow surfaces separately.

The contribution from the illuminated surface is

Eill(r) = ieikr

4r

Z (k k0) (n0Ei) + (k k0)(n0Ei)

eikr

0dS;

where the incident magnetic eld has been eliminated through

!Bi = k0Ei:

In the geometry shown in Fig. 7-5, the incident wave propagates in the negative z-direction,

Ei(r) = E0eikz0 = E0e

ikr0 cos 0 : (7.79)

Therefore, the electric eld di¤racted by the illuminated surface is

Eill(r) = ieikr

4r

Z (k k0) (n0E0) + (k k0)(n0E0)

eikr

0 cos 0ikr0dS: (7.80)

Similarly, the contribution from the shadow surface is

Esh(r) = ieikr

4r

Z (k+ k0) (n0E0) (k k0)(n0E0)

eikr

0 cos 0ikr0dS: (7.81)

The total di¤racted eld is given by the sum of Eill and Esh : However, the angular dependence of the eld

intensities is entirely di¤erent as explained in the introduction, namely, jEillj is insensitive to and ; whilejEsh j sharply peaked in the direction = (forward scattering). Therefore, the scattered power can be

calculated separately as if the elds were incoherent.

Noting r0 = a on the sphere surface, we nd the phase function reduces to

kr0 cos + k r0 = ka(1 + cos ) cos 0 + sin sin 0 cos( 0)

= ka f(; 0;; 0): (7.82)

15

By assumption ka 1: Therefore, the exponential function

eikaf ; (7.83)

rapidly oscillates as 0 and 0 are varied. The slow angular dependence in the amplitude function

(k k0) (n0E0) + (k k0)(n0E0)

can be ignored in the integration over 0 and 0 and can be taken out of the integral,

Eill(r) ' ieikr

4r

(k k0) (n

0Ei) + (k k0)(n0Ei)

00;

00a2Zeikaf(

0;0) sin 0d0d0;

where 00 and 00 indicate the angular location on the sphere surface which makes the dominant contribution

to the phase integral Zeikaf(

0;0) sin 0d0d0: (7.84)

Since ka 1; major contribution to the integral comes from the angular location where the function f(0; 0)

becomes stationary,@f

@0=@f

@0= 0: (7.85)

This determines

0 =

2; 0 = :

This is a trivial result well expected from optical reection. At an observing angular location (; ); only

the wave reected at (0 = =2; 0 = ) can be detected.

Let us Taylor expand the phase function f(0; 0) about the stationary phase point 0 = 0 = =2 and

0 = 0 = ;

f(; 0;; 0) ' cos

2

2

0

2

2 sin2

2

(0 )2 +

!: (7.86)

Integration over 0 yields

Z =2

0

exp

"ika cos

2

0

2

2#sin 0d0

' sin

2

Z 1

1exp

"ika cos

2

0

2

2#d0

= sin

2

pp

ika cos(=2): (7.87)

Note that the integration limits can be extended to 1 because only the region 0 ' =2 contributes to the

16

integral. Similarly, Z 2

0

exp

ika cos

2

sin2

2

(0 )2

d0

'Z 1

1exp

ika cos

2

sin2

2

(0 )2

d0 (7.88)

'pp

ika cos(=2) sin(=2): (7.89)

Therefore, Zeikaf(

0;0)dS

= a2i

ka cos(=2)exp (2ika cos(=2)) : (7.90)

At the particular angular location (=2; ); the magnitude of the vector jk k0j is

2k cos

2

; (7.91)

and the normal vector n0 is in the direction of k k0;

n0 =

1;

2;

: (7.92)

Then,

(k k0) (n0E0) + (k k0)(n0E0)

= 2k cos

2

[n0(n0E0) + n0(n0E0)]

= 2k cos

2

[2n0(n0E0)E0] : (7.93)

The vector

2n0(n0E0)E0; (7.94)

has a magnitude of E0; that is,

j2n0(n0E0)E0j = E0;

and is perpendicular to the scattered wavevector k. (Prove this statement.) Therefore, the electric eld

scattered (or reected) by the illuminated surface is

Eill(r) =a

2reikr2ika cos(=2) [2n0(n0E0)E0] : (7.95)

The amplitude jEill(r)j is independent of the angles and as expected from isotropic reection and scatteredpower can readily be found as

Pill = c"0E20r2 a2r

2 4

= c"0E20 a2: (7.96)

17

The scattering cross-section due to scattering (essentially reection) by the illuminated surface is therefore

ill = a2: (7.97)

Scattering by the shadow surface is identical to di¤raction by a circular aperture on a conducting screen

analyzed earlier, for the boundary conditions

Ei +Es = 0; Bi +Bs = 0; (7.98)

indicate that the elds of the incident wave can be used on the surface of the shadow hemisphere. (The

sign inversion does not a¤ect the di¤racted power.) The shape of the shadow surface is irrelevant as long as

its area projected normal to the incident wave is circular with an area a2: To see this, let us consider the

electric eld scattered by the shadow surface in Eq. (7.81),

Esh(r) = ieikr

4r

Z (k+ k0) (n0E0) (k k0)(n0E0)

ei(k0k)r

0dS0: (7.99)

Since there is no stationary phase point in the shadow (it only occurs in the illuminated surface) and the

function ei(k0k)r0rapidly oscillates because of the assumption ka 1; the integral will be nonvanishing

only if k0k ' 0 which means predominantly forward scattering. The change in the wavevector k = kk0is thus perpendicular to k and k0;

k = k sin e

= k sin (cosex + siney); (7.100)

where is polar angle of scattered wavevector k now measured form the negative z axis, e is the radial

unit vector in the cylindrical coordinates and is the azimuthal angle about the z axis. The phase function

becomes

(k0 k) r0 = ak sin sin 0 cos0

; (7.101)

and the eld amplitude may be approximated by

(k+ k0) (n0E0) (k k0)(n0E0) ' 2k cos 0E0; (7.102)

which is independent of and 0. ThenZ (k+ k0) (n0E0) (k k0)(n0E0)

ei(k0k)r

0dS0

' 2kE0a2Z

0

sin 0d0Z 2

0

d0eiak sin sin 0 cos(0) cos 0

= 2kE0Z 2

0

d0Z a

0

deik sin cos(0)

= 2k (ez E0)Z a

0

d

Zd0eik sin cos(

0); (7.103)

where = a sin 0; n0 is the unit vector normal to the circular at surface, and k E0 ' k0 E0 = 0 is

18

noted. The electric eld scattered by the spherical surface in the shadow is thus given by

Esh(r) = ieikr

2rk (ez E0)

Z 2

0

d0Z a

0

eik sin cos(0)d: (7.104)

This is the negative of the eld di¤racted by a circular aperture given in Eq. (7.48). Integrations over 0

and 0 yield

Esh(r) = ia

rk (ez E0)

J1(ka sin )

k sin ' ia

rk (ez E0)

J1(ka)

k; 1: (7.105)

The radiation power associated with shadow di¤raction is identical to that in the case of circular aperture,

Psh = c"0E20 a2: (7.106)

Therefore, the total power reradiated by the sphere is

P = Pill + Psh = c"0E20 2a2; (7.107)

and the total scattering cross-section in the short wavelength limit is

= 2a2; ka 1: (7.108)

This is consistent with the asymptotic (ka!1) value of the general cross-section derived earlier,

(ka) =2

k2

1Xl=1

(2l + 1)hjAlj2 + jBlj2

i; (7.109)

where

Al = jl(ka)

h(1)l (ka)

; Bl = d

da[ajl(ka)]

d

da[ah

(1)l (ka)]

: (7.110)

The summation over l in Eq. (7.109) requires up to l ' O(ka) for su¢ cient accuracy and thus becomescumbersome in optical regime where ka can be huge. Also, it should be noted that the assumption of ideally

conducting sphere ("e¤ =1; e¤ = 0) entirely breaks down in high frequency (short wavelength) regime.The e¤ective relative permittivity of ordinary metals in optical frequency regime is in general complex and

remains of the order of unity. (The relative magnetic permeability in the regime may be assumed to be

unity.)

7.4 Scattering by a Knife Edge

In scattering by a knife edge, there is no geometrical scale size to speak of except for k where is the radial

distance from the edge tip. In short wavelength optical regime, assuming an incident electric eld along

the plane above the edge may provide the lowest order approximation. In rigorous approach, the boundary

conditions for the electromagnetic elds must be incorporated. We assume an observing point at (; ): The

19

Figure 7-6: Geometry for analyzing scattering by a conducting knife edge. Top: special case of normalincidence. Bottom: general incidence angle.

distance between a point on the y axis and the observing point is

0 =p( sin y0)2 + 2 cos2 ' y0 sin+ 1

2

y02

: (7.111)

Therefore, for small ; the path di¤erence between and 0 is

1

2

y02

;

and corresponding phase di¤erence is1

2

ky02

: (7.112)

The electric eld at (x0; y) is thus given by

E(y) =E0p2

Z qky

1exp

i

2t2dt: (7.113)

The function, shown in Fig. (??),

F (x) =

Z x

1exp

i

2t2dt; (7.114)

is known by Fresnels integral and its special values are:

F (1) = 0; F (0) = 1

2(1 + i); F (1) = 1 + i:

20

At large positive y (well above the edge); the incident eld E0 is recovered as expected. Well below the edge

in the shadow region, y ! 1; the eld vanishes also as expected. At y = 0; the eld is 12E0 and intensity

is 14 of the incident intensity. Fig. (7-8) shows the intensity as a function of the normalized vertical distance

y ! yq

2x0:

F (x) =

Z x

1exp

i

2t2dt

­0.2

0

0.2

0.4

0.6

0.8

1.2

­0.2 0.2 0.4 0.6 0.8 1 1.2

Figure 7-7: Fresnel integral F (x) =R x1 exp

i2 t

2dt: F (1) = 0; F (1) = 1 + i:

0

0.2

0.4

0.6

0.8

1

1.2

­4 ­2 2 4 6y

Figure 7-8: Intensity I=I0 as a function of normalized vertical distance yq

2x0. I(y = 0) = 1

4I0; Imax '1:37I0:

The analysis above is rather primitive because no consideration was given to the boundary conditions

of the electric and magnetic elds. For incident electric eld parallel to the conducting plate, E0 = E0ez

(E-mode), the total eld (incident + scattered) must vanish at the surface of the conductor plate. If the

21

incident wave approaches the plate at an angle 0; the total wave eld is given by

Ez(; ) =1 i2E0e

ik cos(0)Z 2pk= cos

02

1ei

2 t

2

dt

1 i2E0e

ik cos(+0)

Z 2pk= sin

+02

1ei

2 t

2

dt: (7.115)

If the incident wave is so polarized that its magnetic eld is parallel to the plate, H0 = H0ez (H-mode), the

boundary condition is

E _@Hz@

= 0 at = 2;3

2:

This yields

Hz(; ) =1 i2H0e

ik cos(0)Z 2pk= cos

02

1ei

2 t

2

dt

+1 i2H0e

ik cos(+0)

Z 2pk= sin

+02

1ei

2 t

2

dt: (7.116)

The reader should verify that the solutions (rst formulated by Sommerfeld) satisfy the Helmholtz equation

and the respective boundary conditions.

22

Problems

7.1 An electric eld E(; ) = E(; )e + E(; )e is specied on the surface of a sphere of radius a:

Determine the radiation electric eld.

7.2 A magnetic eld B(; ) = B(; )e + B(; )e is specied on the surface of a sphere of radius a:

Determine the radiation magnetic eld.

7.3 A plane wave is incident normal to a long conducting cylinder having a square cross-section of side

a:The electric eld is axial (Ez only) and the incident wave falls normal to one of the rectangular

surfaces. Derive an integral equation for the surface current density Jz(r): (The integral equation

can be solved numerically following the procedure developed for nding the capacitance of a square

conducting plate.)

7.4 Repeat the preceding problem for H mode (Hz only).

7.5 A point light source is placed at a distance z0 on the axis of an opaque disk of radius a ( ): Determine

the light intensity along the axis behind the disk as a function of distance z:

7.6 A point light source is placed at a distance z0 on the axis of a circular opening of radius a in an opaque

screen. Determine the light intensity behind the disk as a function of axial distance z:

7.7 Show that in Fresnel di¤raction of incident wave normal to the plate, the maximum intensity occurs

at y ' 1:2px0=2 and is given by Imax ' 1:37I0 where I0 is the intensity of the incident wave.

7.8 Show that near a knife edge, an E-mode axial electric eld

Ez(; ) = Ap sin

2

;

and corresponding magnetic eld

H =A

2i!0

1p

cos

2ex + sin

2ey

;

satisfy the boundary conditions. The x component of the magnetic eld is discontinuous at = 0

and = 2: Interpret this peculiarity. Show that the intensity of a wave scattered by a knife edge is

insensitive to the observing angle : (A knife edge appears shiny regardless of observing angle.)

7.9 The transmission cross section of a small (ka 1) circular aperture in a conducting plate is

T =64

27k4a6;

where a is the hole radius. Show that the magnitude of the electric eld in the aperture should be of

the order of

Ea ' kaE0;

where E0 is the incident eld.

Note: For incident eld polarized in the x direction, Bouwkamp found the electric eld component in

the aperture responsible for di¤raction,

Ey(x; y) = 8k

3

x2 + 2y2 2a2pa2 x2 y2

E0:

23

(Ex (x; y) is an odd function of both x and y:) Integration over the aperture yieldsZEy (x; y) dS =

ZEy (r; ) rdrd =

16

3ka3E0;

which is consistent with the e¤ective magnetic dipole,

m = 2 83a3H0:

At the edge of a thin conductor, the electric eld parallel to the edge vanishes in a manner Ek =p;

while the normal component diverges as E? = 1=p where is the distance from the edge. For

example, the electric eld at the rim of a charged conducting disk behaves as

E? _1q

a2 (a )2=

1p2a

:

Such properties can be exploited to nd electric eld when the boundary involves sharp conducting

edge.)

7.10 In low frequency limit, scatterimg by a conducting sphere may be analyzed using dipole approximation.

Relevant dipole moments are

p = 4"0a3E0; m = 2a3H0;

where E0 and H0 are the elds of the incident plane wave. Show that the di¤erential scattering cross

section isd

d= k4a6

"1 1

2cos

2sin2 +

cos 1

2

2cos2

#;

the total scattering cross section is

=10

3k4a6;

and that the force exerted on the sphere is

Fz =14

3k4a6"0E

20 ; (N).

24

Chapter 8

Radiation by Moving Charges

8.1 Introduction

The problem of radiation of electromagnetic waves by a single charged particle moving at an

arbitrary velocity had correctly been formulated independently by Lienard and Wiechert before

the advent of the special relativity theory. This is because once emitted from a charged particle,

electromagnetic waves propagate at the speed c irrespective of the velocity of the charged particle,

just as sound waves propagate at a speed independent of the source velocity. (The major nding

made by Einstein was that electromagnetic waves still propagate at the speed c regardless of the

observers velocity in contrast to the case of sound waves.)

The scalar and vector potentials due to a moving charge can be found rigorously using the

Greens function for the wave equation. Then radiation electromagnetic elds can readily be cal-

culated. In nonrelativistic regime, the radiation power only depends on the acceleration of charged

particles. As the velocity approaches c; however, signicant increase in the radiation power occurs.

Furthermore, in highly relativistic limit, radiation occurs primarily along the direction of velocity

within an angular spread of order ' 1= about the velocity irrespective of the direction of ac-

celeration. Here = 1=p1 2 is the relativity factor with = v=c: Hence radiation frequency

is subject to strong Doppler shift. For example, in synchrotron radiation due to highly relativistic

electron beam bent or undulated by a magnetic eld, radiation even in hard x-ray regime can be

created.

In material medium, radiation processes without acceleration on charged particles are possible.

If the velocity of a charged particle exceeds the velocity of electromagnetic waves in the medium

v >1

p"0

;

where " is the permittivity, Cherenkov radiation occurs. Furthermore, if a charged particle crosses a

boundary of two dielectric media, the transition radiation occurs even if the condition for Cherenkov

radiation is not met. Transition radiation is due to sudden change in the normalized velocity from

1 = vp"10 to 2 = v

p"20 which may be regarded as an e¤ective acceleration even though the

1

particle velicty v remains constant.

8.2 Lienard-Wiechert Potentials

The charge and current densities of a moving point charge are singular and described by

(r; t) = e[r rp(t)]; (8.1)

J(r; t) = ev(t)[r rp(t)]; (8.2)

where e is the charge, rp(t) is the instantaneous location of the charge and v(t) = drp(t)=dt is the

instantaneous velocity of the charge which may be changing with time. Exploiting the Greens

function for the wave equation,

G(r r0; t t0) = 1

4 jr r0jt t0 jr r

0jc

; (8.3)

we can write down solutions for the inhomogeneous wave equations,r2 1

c2@2

@t2

=

"0; (8.4)

r2 1

c2@2

@t2

A = 0J; (8.5)

in the form

(r; t) =e

4"0

ZdV 0

Zdt0

1

jr r0j[r0rp(t0)][f(t0)]; (8.6)

A(r; t) =0e

4

ZdV 0

Zdt0

v(t0)

jr r0j[r0rp(t0)][f(t0)];

where

f(t0) = t0 t+ jr r0j

c: (8.7)

The volume integrations can be carried out immediately with the results

(r; t) =e

4"0

Z1

jr rp(t0)j[f(t0)]dt0; (8.8)

A(r; t) =0e

4

Zv(t0)

jr rp(t0)j[f(t0)]dt0; (8.9)

where f(t0) is now

f(t0) = t0 t+ jr rp(t0)j

c: (8.10)

2

The integral involving the delta function can be simplied asZg(t0)[f(t0)]dt0 =

Zg(t0)[f(t0)]

1df(t0)dt0

df=

g(t0)

jdf=dt0j

f=0

; (8.11)

where t0 is now understood as a solution for t0 satisfying f(t0) = 0 or

t0 t+ jr rp(t0)j

c= 0: (8.12)

This is in general an implicit equation for t0: The time derivative of f(t0) is

df

dt0= 1 n(t

0) vp(t0)c

= 1 n(t0) (t0); (8.13)

where n(t0) is the unit vector along the relative distance r rp(t0) and = v=c: The observing

time t and time t0 are related through

[1 n(t0) (t0)]dt0 = dt;

ordt

dt0= 1 n(t0) (t0):

After performing time integration, we nally obtain

(r; t) =e

4"0

1

1 n(t0) (t0)1

jr rp(t0)j=

e

4"0

1

(t0) jr rp(t0)j; (8.14)

A(r; t) =0e

4

1

1 n(t0) (t0)v(t0)

jr rp(t0)j=0e

4

v(t0)

(t0) jr rp(t0)j; (8.15)

where

(t0) = 1 n(t0) (t0): (8.16)

These retarded potentials, called Lienard-Wiechert potentials, had been formulated in 1898. They

are applicable to arbitrary velocity of the charged particle. Retarded nature of the potentials clearly

appears in the condition that all time varying quantities, rp(t0);v(t0);n(t0); must be evaluated at

t0; not at the observing time t because of nite propagation speed of electromagnetic disturbance.

Having found the retarded potentials, we are now ready to calculate the electromagnetic elds

due to a moving point charge. The electric eld is to be found from

E(r; t) = r @A@t; (8.17)

where spatial and time derivatives pertain to r (the coordinates of the observing location) and

3

t (observing time). Since the potentials and A are implicit functions of r and t; it is more

convenient to use the original integral representations, Eqs. (8.8) and (8.9), respectively, for proper

di¤erentiation with respect to r and t: For example, the spatial derivative of the scalar potential

can be performed as follows. Letting R(t0) = jr rp(t0)j ; and introducing a unit vector in thedirection r rp(t0);

n(t0) =r rp(t0)jr rp(t0)j

; (8.18)

we nd

r =e

4"0

Zr

1

jr rp(t0)j[f(t0)]

dt0

=e

4"0

Zn@

@R

1

R[f(t0)]

dt0

= e

4"0

Z n

R2[f(t0)] n

cR

d

df[f(t0)]

dt0

= e

4"0

n

R2+1

c

d

dt0

nR

; (8.19)

where use is made of the integration by parts,Zg(t0)

d

df[f(t0)]dt0

=

Zg(t0)

1

df

dt0

d

dt0[f(t0)]dt0 (8.20)

=

1

df=dt0d

dt0

g (t0)

df=dt0

f(t0)=0

(8.21)

= 1

(t0)

d

dt0

g (t0)

(t0)

f(t0)=0

: (8.22)

Similarly,@A

@t=

e

4"0

1

c

d

dt0

R

; (8.23)

and the electric eld becomes

E(r; t) =e

4"0

n

R2+1

c

d

dt0

n R

f(t0)=0

: (8.24)

To proceed further, we need concrete expressions for the derivatives,

dn

dt0and

d

dt0

1

R

:

The unit vector n along the distance vector R = r rp(t0) changes its direction only through the

4

velocity component perpendicular to R;

dn = v?Rdt0; (8.25)

as can be seen in Fig. 8-1. This yields

Figure 8-1: The change in the unit vector n is caused by the perpendicular velocity v?:

dn

dt0= v?

R: (8.26)

Also,

d

dt0

1

R

= 1

(R)2d

dt0[(1 n )R]

= 1

(R)2

hv? (n _)R c(1 n )n

i= c

(R)2

2 n 1

cR(n _)

: (8.27)

Substitution of Eqs. (8.26) and (8.27) to Eq. (8.24) gives

E(r; t) =e

4"0

"n (R)2

n (R)2

2 n 1

cR(n _)

_

c2R

#

=e

4"0

1 2

3R2(n ) + 1

c3Rn [(n ) _]

f(t0)=0

: (8.28)

The rst term in the RHS,

ECoulomb =e

4"0

1 2

3R2(n )

f(t0)=0

; (8.29)

5

is the Coulomb eld corrected for relativistic e¤ects. It is proportional to 1=R2 and thus does not

contribute to radiation of energy. The second term,

Erad =e

4"0c

1

3Rn [(n ) _]

f(t0)=0

; (8.30)

contains acceleration _ and is proportional to 1=R: This is the desired radiation electric eld due

to a moving charged particle.

The magnetic eld can be calculated in a similar manner from

B = rA=

1

c[nE]f(t0)=0 : (8.31)

Derivation of this result is left for an exercise.

8.3 Radiation from a Charge under Linear Acceleration

If the acceleration is parallel (or anti-parallel) to the velocity, _ = 0; the radiation electric eldreduces to

Erad =e

4"0

"n (n _)

c3R

#f(t0)=0

: (8.32)

The angular distribution of radiation power at the observing time t is

dP (t)

d= c"0 jEradj2R2

=1

4"0

e2

4c

"n (n _)

c3

#2f(t0)=0

: (8.33)

Denoting the angle between _ and n by ; we have [n (n _)]2 = _2sin2 ; and thus

dP (t)

d=

1

4"0

e2

4c

"_2sin2

(1 cos )6

#f(t0)=0

: (8.34)

However, the power P (t) in the above formulation is the rate of energy radiation at t; the observing

time, which is not necessarily equal to the energy loss rate of the charge at the retarded time t0

determined from f(t0) = 0: To nd the radiation power at the retarded time P (t0) ; let us consider

the amount of di¤erential energy dE=d radiated during the time interval between t0 and t0+dt0: Bydenition dE = P (t0)dt0: The radiation energy dE is sandwitched between two eccentric sphericalsurfaces with a volume

dV = R2cdt0(1 cos ):

6

Therefore, the di¤erential radiation energy is

dEd

=1

2"0 jEradj2R2cdt0(1 cos );

and

dP (t0)

d=

1

4"0

e2

4c

"_2sin2

(1 cos )6 (1 cos )#f(t0)=0

=1

4"0

e2

4c

"_2sin2

(1 cos )5

#f(t0)=0

: (8.35)

In nonrelativistic limit jj 1; the radiation occurs predominantly in the direction perpendicular

to the acceleration = =2. The total radiation power in this case is

P ' 1

4"0

e2 _2

4c

Zsin2 d

=1

4"0

2e2 _2

3c; 1: (8.36)

This is the well known Larmors formula for radiation power due to a nonrelativistic charge. Since

in nonrelativistic limit,

n [(n ) _] ' n (n _); (8.37)

Larmors formula is applicable for acceleration in arbitrary direction relative to the velocity.

For arbitrary magnitude of the velocity ; the radiation power can be found from

P (t0) =1

4"0

e2 _2

4c

Zsin2

(1 cos )5d

=1

4"0

e2 _2

4c2

Z

0

sin3

(1 cos )5d

=1

4"0

2e2 _2

3c 6; (8.38)

where

=1p1 2

; (8.39)

is the relativity factor and the following integral is used,Z

0

sin3

(1 cos )5d =Z 1

1

1 x2(1 x)5dx =

4

3

1

(1 2)3: (8.40)

7

The angular dependence of the radiation intensity,

sin2

(1 cos )6 ; (8.41)

peaks at angle 0 where

cos 0 =

p1 + 242 1

4: (8.42)

In highly relativistic limit 1; the angle 0 becomes of order

0 '1

1; (8.43)

which indicates a very sharp pencil or beam of radiation along the direction of the velocity : (This

is also the case for acceleration perpendicular to the velocity as shown in the following section.)

Angular distribution of radiation intensity I ()) for = 0; 0:2; 0:9 and 0:999 is shown below for a

common acceleration. Note that the radiation intensity rapidly increases with = 1=p1 2:

­1­0.8­0.6­0.4­0.2

0

0.20.40.60.8

1

­1 ­0.8 ­0.4 0.2 0.4 0.6 0.8 1

= 0 ( = 1:0):

­1.0 ­0.5 0.5

­1.0

­0.5

0.5

1.0

x

y

= 0:2 ( = 1:02):

2000 4000 6000 8000 10000 12000 14000

­2000

0

2000

x

y

= 0:9 ( = 2:3):

2e+8 4e+8 6e+8 8e+8 1e+9 1.2e+9­9e+7

9e+7

x

y

= 0:99 ( = 7:09):

8

Polar plot of I() for several factors but with common parallel acceleration.

For linear acceleration, the momentum change of the charged particle is

dp

dt=d

dt

mvp1 2

!= m 3 _v: (8.44)

Therefore, the radiation power can be rewritten as

P (t0) =1

4"0

2e2

3c3m2

dp

dt0

2=

1

4"0

2e4

3c3m2E2ac

=1

4"0

2e2

3c3m2

dEdx0

2; (8.45)

where Eac is an external acceleration electric eld and

dEdx0;

is the energy gradient of a linear accelerator which is at most of the order of 100MeV/m in practice.

The radiation loss in linear accelerators is negligibly small compared with energy gain. This is one

of the advantages of high energy linear accelerators. Note that the radiation power due to linear

acceleration is independent of the particle energy or the relativity factor :

8.4 Radiation from a Charge in Circular Motion

In circular motion, the acceleration is perpendicular to the velocity. Here we consider highly

relativistic motion of a charged particle with ' 1, for nonrelativistic case has already been

discussed in Chapter 4. A geometry convenient for analysis to follow is shown in Fig.8-2. A

particle undergoes circular motion with an orbit radius in the x z plane and it passes theorigin at t0 = 0: At that instant, the acceleration is in the x direction while the velocity is in the z

direction,_ =

_ ex; = ez:

The radiation electric eld can then be written down in terms of cartesian components,

E(r; t) =e

4"0

1

c3Rn [(n ) _]

=e

4"0

_c3R

[( cos ) cose + (1 cos ) sine] ; (8.46)

9

Figure 8-2: Particle undergoing circular motion in the x z plane with radius and frequency !0:At t = 0; the particle passes the origin.

and the angular distribution of radiation power is given by

dP (t0)

d=

1

4"0

e2 _24c

1

(1 cos )5

(1 cos )2 1

2sin2 cos2

: (8.47)

The total radiation power is

P (t0) =1

4"0

e2 _24c

Z1

(1 cos )5

(1 cos )2 1

2sin2 cos2

d

=1

4"0

e2 _24c

Z

0sin d

Z 2

0d

1

(1 cos )5

(1 cos )2 1

2sin2 cos2

=

1

4"0

2e2j _vj2

3c3 4: (8.48)

Relevant integrals are: Z 1

1

dx

(1 x)3 =2

(1 2)2= 2 4;

Z 1

1

1 x2(1 x)5dx =

4

3

1

(1 2)3=4

3 6:

In highly relativistic case, the acceleration may be approximated by

j _vj = v2

' c2

: (8.49)

10

Then, the radiation power in terms of the orbit radius is

P (t0) ' 1

4"0

2e2c 4

32: (8.50)

To maintain the radiation loss in a circular accelerator at a tolerable level, the orbit radius must

be increased as the particle energy mc2 increases. Note that the radiation power is a sensitive

function of the particle energy in contrast to the case of linear acceleration.

As an example, let us consider the Betatron, the well known inductive electron accelerator

invented by Kerst. In the Betatron, the electron cyclotron orbit is maintained constant,

mc2

= ecB (t) = ecB0 sin!t ' ecB0!t;

where only the initial phase of the sinusoidal magnetic eld is useful for acceleration, !t 1. Then

the rate of electron energy gain is

d

dt

mc2

= ecB0! = const.

Equating this to the radiation energy loss,

Pt0=

1

4"0

2

3c3

c2

2 4;

we nd

max '3!eB0

3

2mc2re

1=4; (8.51)

where

re =e2

4"0mc2= 2:8 1015 m; (8.52)

is the classical radius of electron. If = 50 cm, ! = 2 60 rad/s, and B0 = 0:5 T (5 kG), the

upper limit of is about 400 and the maximum electron energy attainable is approximately 200

MeV.

8.5 Fourier Spectrum of Radiation Fields

The formulae such as Eqs. (8.45) and (8.50) only tell us the total radiation power integrated

over the frequency. Radiation eld emitted by highly relativistic particle is hardly monochromatic

but cnsists of broad frequency spectrum. Knowing such frequency spectrum of radiation is of

practical importance for identifying radiation source. A typical example is synchrotron radiation

due to highly relativistic electrons bent by, or trapped in, a magnetic eld. In nonrelativistic limit,

the radiation elds all have a single frequency component corresponding to the classical electron

cyclotron frequency !c = eB=m: However, as the relativity factor increases, the radiation elds

11

consist of harmonics of the fundamental frequency !c = eB= m. In highly relativistic case 1;

the frequency spectrum becomes almost continuous peaking at the frequency ! ' 3!c = 2eB=m:Frequency spectrum of the radiation electric eld can be formulated by directly applying Fourier

transformation on the eld in Eq. (8.24),

E(r; !) =

Z 1

1E(r; t)ei!tdt

=1

4"0

e

c

Z 1

1

"n [(n ) _]

3R

#f(t0)=0

ei!tdt: (8.53)

Changing the variable from t to t0 and assuming r rp; we obtain

E(r; !) =1

4"0

e

c

ei!r=c

r

Z 1

1

n [(n ) _]

2exp

i!t0 1

cn rp(t0)dt0: (8.54)

The unit vector n(t0) may be regarded constant since r rp and thus approximated by n ' r=r:Then,

n [(n ) _]

2' d

dt0

n (n )

; (8.55)

and Eq. (8.53) can be integrated by parts,

E(r; !) = i!

4"0

e

c

ei!r=c

r

Z 1

1n (n ) exp

i!t0 1

cn rp(t0)dt0: (8.56)

Any time derivatives contained in the physical electric eld are merely multiplied by i! in theFourier space and the disappearance of the acceleration _ is not surprising. Amazing fact about

Eq. (8.56) is that it is applicable to radiation elds which do not require particle acceleration

such as Cherenkov and transition radiation provided a proper velocity of electromagnetic waves in

dielectrics is substituted for c:

The radiation energy (not power) associated with the electric eld is

c"0r2

Z ZjE(r; t)j2 dtd; (J). (8.57)

However, since the electric eld E(r; t) and its Fourier transform E(r; !) are related through

Parsevals theorem:ZjE(r; t)j2 dt = 1

2

ZjE(r; !)j2 d!; (8.58)

the radiation energy can be written in terms of the Fourier transform E(r; !) as

1

2c"0r

2

Z ZjE(r; !)j2 dd!: (8.59)

12

The quantitydI(!)

d 1

2c"0r

2 jE(r; !)j2 ; (8.60)

can therefore be identied as the radiation energy per unit solid angle per unit frequency. Substi-

tuting Eq. (8.56), we nd

dI(!)

d=

1

4"0

e2

82c!2Z n (n ) exp

i!t0 1

cn rp(t0)dt02 ; 1 < ! <1; (8.61)

or

dI(!)

d=

1

4"0

e2

42c!2Z n (n ) exp

i!t0 1

cn rp(t0)dt02 ; 0 < ! <1: (8.62)

Let us work on a few examples.

8.6 Synchrotron Radiation I

Synchrotron radiation is due to highly relativistic electrons trapped in a magnetic eld. The

radiation beam rotates together with an electron and is directed along the direction of the velocity

with an angular spread of order ' 1= 1: If the orbiting frequency is !0 = eB= me; the

radiation beam shines a detector for a duration

t0 '

!0=

1

!0; (8.63)

as seen by the electron at the retarded time t0. Since

t

t0= 1 n = 1 ' 1

2 2; (8.64)

which is entirely due to Doppler e¤ect, the pulse width detected is of order

t ' t0

2 2=

1

2 3!0: (8.65)

Therefore, synchrotron radiation is dominated by frequency components in the range

! ' 3!0 = 2eB

me: (8.66)

In ultrarelativistic case, the frequency spectrum of synchrotron radiation can extend to very high

frequencies even for a modest magnetic eld.

We calculate the amount of energy radiated in one period of cyclotron motion T = 2=!0: Since

the radiation power is constant, the radiated energy is

E = 1

T

Z 1

0I (!) d! =

!02

Z 1

0I (!) d!: (8.67)

13

The time integration in the energy spectrum,

dI (!)

d=

1

4"0

e2

42c!2

Z T=2

T=2n (n ) exp

i!t0 1

cn rp(t0)dt0

2

; 0 < ! <1; (8.68)

can be extended from 1 to 1 since the characteristic frequency of the radiation eld is much

higher than the fundamental frequency ! !0;

dI (!)

d=

1

4"0

e2

42c!2Z 1

1n (n ) exp

i!t0 1

cn rp(t0)dt02 ; 0 < ! <1: (8.69)

To perform the integration, we assume the trajectory shown in Fig. (8-2) in which an electron

passes the origin at t = 0: The vector n is assumed to be in the y z plane since the radiationprole is essentially symmetric about the z axis. The trajectory is described by

rpt0=

1 cos!0t0

ex + sin!0tez

; (8.70)

and the velocity is

=!0

c(sin!0tex + cos!0tez) : (8.71)

Then,1

cn rp(t0) =

csin (!0t) cos : (8.72)

Since

n (n ) = ? = sin!0tex + cos!0t sin e?; (8.73)

where

e?= n ex; (8.74)

is a unit vector perpendicular to both n and x axis, and the radiation lasts for a very short time

and is limited within a small angle ; Eq. (8.73) reduces to

? ' !0tex + e?: (8.75)

Within the same order of accuracy, the phase function !(t n rp=c) can be approximated by

!t n rp

c

= !

t sin!0t cos

c

' !

2

1

2+ 2

t+

c2

32t3; (8.76)

where v has been approximated by c but 1 by

1 ' 1

2 2:

14

Then, Z 1

1? exp

hi!t n rp

c

idt

=

Z 1

1(!0tex e?) exp

i!

2

1

2+ 2

t+

c2

32t3dt

= 2i!0

Z 1

0t sin

!

2

1

2+ 2

t+

c2

32t3dt ex

2Z 1

0cos

!

2

1

2+ 2

t+

c2

32t3dt e?: (8.77)

The integrals reduce to the modied Bessel functions of fractional orders (or the Airys functions),Z 1

0cos

3

2x

+

1

33d =

1p3K1=3 (x) ; (8.78)

Z 1

0 sin

3

2x

+

1

33d =

1p3K2=3 (x) ; (8.79)

where

= !0tp1 + 22

; x =!

3 3!0

1 + 22

3=2: (8.80)

Then

2i!0

Z 1

0t sin

!

2

1

2+ 2

t+

c2

32t3dt = 2i

1 2+ 2

!0

1p3K2=3 (x) ;

2

Z 1

0cos

!

2

1

2+ 2

t+

c2

32t3dt = 2

q1 2+ 2

!0

1p3K2=3 (x) ;

and for I (!) =d; we obtain

dI (!)

d=

1

4"0

e2

32c

!2

!20

1 2+ 2

h1 2+ 2

K22=3 (x) +

2K21=3 (x)

i=

1

4"0

3e2 2

2c!21 + 2

n1 + 2

K22=3[!

1 + 2

3=2] + 2K2

1=3[!1 + 2

3=2]o;

where = and ! is the normalized frequency,

! =!

3 3!0: (8.81)

The modied Bessel functions K2=3 (x) and K1=3 (x) both diverge at x ! 0: However, xK2=3 (x)

and xK1=3 (x) are well behaving and vanish at small x: Fig. (8-3) shows 2x2K22=3 (x) and x

2K21=3 (x)

which represent radiation intensities associated with electric eld polarization along ex (that is, in

the particle orbit plane) and e?; respectively.

The energy spectrum I (!) emitted during one revolution (T = 2=!0) can be found by inte-

15

0 1 2 3 4 50.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

x

y

Figure 8-3: 2x2K22=3 (x) (solid line) and x

2K21=3 (x) (dotted line). The factor of 2 in 2x

2K22=3 (x)

assumes ' 1= :

grating dI (!) =d over the solid angle,

I (!)

=1

4"0

3e2 2

2c!2Z

0sin 0d0

Z 2

0d01 + 2

n1 + 2

K22=3[!

1 + 2

3=2] + 2K2

1=3[!1 + 2

3=2]o

=1

4"0

6e2 2

c!2Z =2

=2cos d

1 + 2

n1 + 2

K22=3[!

1 + 2

3=2] + 2K2

1=3[!1 + 2

3=2]o

' 1

4"0

6e2

c!2Z 1

1

1 + 2

n1 + 2

K22=3[!

1 + 2

3=2] + 2K2

1=3[!1 + 2

3=2]od

=1

4"0

6e2

cf (!) ; (8.82)

where 0 = 2 is the polar angle from the axis of electron revolution (y axis), 0 is the azimuthal

angle about the y axis, and the function f (!) is dened by

f (!) = 2!2Z 1

0

1 + 2

n1 + 2

K22=3[!

1 + 2

3=2] + 2K2

1=3[!1 + 2

3=2]od; (8.83)

and shown in Fig. (8-4) (linear scale) and Fig. (8-5) (log-log scale). In Fig. (8-5), the straight line

in the low frequency regime! 3!0

has a slope of 1=3; and indicates I (!) _ !1=3: f (!) peaks

at ! ' 0:14; or ! ' 0:42 3!0; and its peak value is about 0.83. In high frequency regime ! 3!0;

the spectrum decays exponentially. The energy radiated per revolution can be calculated as

16

0

0.2

0.4

0.6

0.8

1 2 3 4 5

Figure 8-4: The function f (x) = f!=3 3!0

plotted in linear scale.

E =Z 1

0I (!) d! ' 1

4"0

6e2

c3 3!0

Z 1

0f (!) d!; (8.84)

where the integral numerically evaluated is approximatelyZ 1

0f (!) d! ' 0:713:

Then

E =Z 1

0I (!) d! =

1

4"0

4:08e2 4!0c

; (J)

and the radiation power is

P = E !02=

1

4"0

0:65e2c 4

2; (8.85)

which agrees reasonably well with Eq. (8.50),

P =1

4"0

2e2c 4

32' 1

4"0

0:667e2c 4

2; (W).

The discrepancy may be attributed to the various approximations made in the analysis.

8.7 Synchrotron Radiation II

An alternative approach to nding the frequency spectrum of synchrotron radiation is to apply

discrete Fourier analysis in terms of harmonics of the fundamental frequency !0 directly to radiation

17

.1e­2

.1e­1

.1

.1e­3 .1e­2 .1e­1 .1 1.

Figure 8-5: Log-log plot of the function f (x) = f!=3 3!0

:

elds. If a charge e is in circular motion with a constant angular frequency !0, the radiation eld

contains higher harmonics of !0 and can be Fourier decomposed as follows. Let us recall the

Lienard-Wiechert vector potential,

A(r; t) =04

ev(t0)

(1 n )R(t0)

f(t0)=0

; (8.86)

where R = jr rp(t0)j and the subscript f(t0) = 0 indicates that all time dependent quantities

should be evaluated at the retarded time t0 determined from the implicit equation for t0,

f(t0) = t0 t+ jr rp(t0)j

c= 0: (8.87)

The vector potential can be Fourier decomposed as

A(r; t) =Xl

Al(r)eil!0t; (8.88)

where

Al(r) =0e

4

1

T

Z T

0

ev(t0)

(1 n )Reil!0tdt; (8.89)

with T = 2=!0 being the period of the circular motion. Changing the integration variable from t

to t0 by notingdt

dt0= 1 n ; (8.90)

18

leads to

Al(r) '0e

4r

eiklr

T

Z T

0v(t0)ei(l!0t

0klnrp)dt0; (8.91)

where kl = l!0=c. Note that the period T remains unchanged through the transformation. Let the

particle trajectory be

rp(t0) = (cos!0t

0ex + sin!0t0ey); (8.92)

v(t0) = !0( sin!0t0ex + cos!0t0ey): (8.93)

Since all radiation elds rotate with the charge, the observing point can be chosen at arbitrary

azimuthal angle and we choose = =2; so that n = (1; ; = =2): Then

n rp(t0) = sin sin!0t0: (8.94)

The velocity in the spherical coordinates is

v(t0) = !0(sin cos!0t0er + cos cos!0t

0e + sin!0t0e): (8.95)

Thus, the component of Al(r) is given by

Al =0e

4reiklr

!0Tcos

Z T

0cos!0t

0eil!0(t0 sin sin!0t0=c)dt0: (8.96)

Letting x = !0t0; and noting = !0=c; we can rewrite this as

Al = eiklr0e!04r

cos

2

Z 2

0cosxeil(x sin sinx)dx: (8.97)

The integral reduces to Z 2

0cosxeil(x sin sinx)dx

= [Jl+1(l sin ) + Jl1(l sin )]

=2

sin Jl(l sin );

and thus nally,

Al =0ec

4reiklr cot Jl(l sin ): (8.98)

Similarly,

Al = i0e!04r

eiklrJ 0l (l sin ); (8.99)

where use has been made of the recurrence formula of the Bessel functions,

Jl1(x) Jl+1(x) = 2J 0l (x): (8.100)

19

The far-eld radiation magnetic eld can be found from

Hl 'i

0kl Al; (8.101)

which yields

Hl =e!0kl4r

eiklrJ 0l (l sin ); (8.102)

Hl = ieckl4r

eiklr cot Jl(l sin ): (8.103)

The radiation power associated with the l-th harmonic is

Pl = c0r2

ZjHlj2d

=1

80

(el!0)2

c

Z

0

2J 02l (l sin ) + cot

2 J2l (l sin )sin d: (8.104)

Since Pl = Pl, the total power is

P =

1Xl=1

Pl; (8.105)

where Pl is now

Pl =1

40

(el!0)2

c

Z

0

2J 02l (l sin ) + cot

2 J2l (l sin )sin d; l 1: (8.106)

In nonrelativistic limit 1; the l = 1 term is dominant. For x 1;

J1(x) '1

2x; J 01(x) '

1

2: (8.107)

Then the lowest order radiation power agrees with the Larmors formula,

P1 ' 1

40

(e!0)2

c

2

4

Z

0(1 + cos2 ) sin d

=1

40

2

3

e2a2

c3; (8.108)

where a = v2= is the acceleration.

The integral in Eq. (8.105) cannot be reduced to elementary functions. However, the total

power given in Eq. (8.105) should reduce to Eq. (8.50),

P =1

40

2

3

e2a2

c3 4 =

1

40

2

3

e2a2

c31

(1 2)2: (8.109)

To show this, we modify the integral by noting

J2l (x) =1

Z

0J0(2x sin ) cos(2l)d; (8.110)

20

J 02l (x) =1

Z

0J0(2x sin )

cos 2 l2

x2

cos(2l)d: (8.111)

Then

2J 02l (l sin ) + cot2 J2l (l sin ) =

1

Z

0J0(2l sin sinx)(

2 cos 2x 1) cos(2lx)dx: (8.112)

Furthermore, Z

0J0(2l sinx sin ) sin d =

sin(2l sinx)

l sinx; (8.113)

and the power Pl reduces to

Pl =1

40

(e!0l)2

c

Z

0

sin(2l sinx)

l sinx(2 cos 2x 1) cos(2lx)dx: (8.114)

Noting Z

0sinx sin(2l sinx) cos(2lx)dx

=1

2

Z

0[sin(2l + 1)x+ sin(1 2l)x] sin(2l sinx)dx

=

2[J2l+1(2lx) J2l1(2l)]

= J 02l(2l) (8.115)

and Z

0

sin(2l sinx) cos(2lx)

sinxdx

= 2l

Z

0

Z

0cos(2l sinx) cos(2lx)dxd

= 2l

Z

0J2l(2l)d; (8.116)

the power Pl can be rewritten as

Pl =1

40

2(e!0l)2

cl

2J 02l(2l) l(1 2)

Z

0J2l(2l)d

; (8.117)

and the total power is

P =

1Xl=1

Pl:

Relevant sum formula of the Bessel functions is

1Xl=1

J2l(2lx) =x2

2(1 x2) : (8.118)

21

Di¤erentiating by x; Xl

2lJ 02l(2lx) =x

(1 x2)2 : (8.119)

Also, Xl

l2Z

0J2l(2lx)dx =

3

6(1 2)3: (8.120)

This is one of Kapteyn series formulae. (See for example, Mathematical Formulae (in Japanese),

(Iwanami, Tokyo, 1960), vol. 3, p. 212.) Then, nally, the total radiation power becomes

P =1

40

2e2a2

3c3 4; a =

v2

(8.121)

which is consistent with the known radiation power from a charge undergoing circular motion.

Analytic expression for the radiation power Pl can be found by exploiting following approxima-

tion,

J2l(2l) '1pl1=3

Ai

l2=3

2

!; (8.122)

where Ai(x) is the Airy function dened by

Ai(x) =1p

Z 1

0cos

1

3t3 + xt

dt: (8.123)

f(x) =1p

Z 20

0cos

1

3t3 + xt

dt

For large x 1; the function takes the form

Ai(x) ' 1

2x1=4exp

23x3=2

; x 1: (8.124)

Also, Ai0(0) = 0:4587: Using these approximations, we nd the following approximate formulae,

Pl '1

4"0 0:5175e

2!20cl1=3; 1 l 3; (8.125)

Pl '1

4"0

e2!202pc

sl

exp

23

l

3

; l 3: (8.126)

Note that the radiation power increases with l in the manner Pl _ l1=3 up to l ' 3 beyond whichPl decays exponentially. This is consistent with the analysis in the preceding section.

8.8 Free Electron Laser

In a synchrotron radiation source, many beamlines can be installed by bending an electron beam.

In straight sections of race track, no radiation occurs. However, by inserting a device called wiggler,

22

Figure 8-6: In a wiggler, an electron beam is modulated by a periodic magnetic eld. Electronsacquire spatially oscillating perpendicular displacement x(z) and velocity vx(z) which together withthe radiation magnetic eld BRy produces a ponderomotive force vx(z)BRy(z) directed in the zdirection. The force acts to cause electron bunching required for amplication of coherent radiation.

high intensity radiation can be extracted. A wiggler consists of periodically alternating magnets

and gives an electron beam periodic kick perpendicular to both the beam velocity and magnetic

eld. Electrons receive kicks at an interval

t0 =wc; (8.127)

where w is the wavelengthof the periodic wiggler structure. Because of Doppler shift, this time

interval is shortened by a factor1

1 ' 2 2 for a stationary detector in front of the beam,

t ' w2 2c

: (8.128)

Therefore, the wavelength of resultant radiation is approximately given by

' w2 2

w; (8.129)

and the frequency by

! ' 4 2c

w: (8.130)

The intensity of free electron laser can be orders of magnitude higher than that of synchrotron

radiation because of coherent amplication through the periodic structure. Electrons tend to be

bunched in the wiggler as the electron beam travels through the periodic structure. In contrast, no

collective interaction between electrons and electromagnetic waves exists in synchrotron radiation.

In electron bunching, the magnetic ponderomotive force plays a major role. Let us assume a

23

periodic wiggler magnetic eld in y direction,

By = B0 cos

2

wz

= B0 cos kwz; kw =

2

w: (8.131)

The Lorentz force is

F = ev B;

or

Fx = evzB0 cos kwz:

Then electron acquires a velocity vx in x direction,

vx =eB0m kw

sin kwz; (8.132)

and a resultant ponderomotive force is

evx BR; (8.133)

where BR is the radiation magnetic eld propagating in the form ei(kz!t) along the beam. Since

the acceleration due to the wiggler magnetic eld is in x direction, the radiation electric eld is

predominantly in x direction and radiation magnetic eld BR is in y direction. Then the pondero-

motive force directed in z direction is proportional to ei[(k+kw)z!t] and propagates at a velocity

!

k + kw:

When this propagation velocity matches the electron beam velocity c, strong interaction between

the radiation eld and electron motion takes place and electrons tend to be bunched. This results

in positive feedback for wave amplication. From the condition

!

k + kw' c; or k

k + kw= ;

we readily recover

k =

1 kw ' 2 2kw:

8.9 Radiation Accompanying Decay

Equation (8.61) for the angular distribution of radiation energy can be applied to cases in which

particle acceleration is not involved explicitly. In decay, an energetic electron (or positron) is

suddenly released from a nucleus together with neutrino. The situation is equivalent to sudden

acceleration of an electron. The duration of acceleration t is limited by the uncertainty principle

24

t mc2 & ~: Therefore, the upper limit of the frequency spectrum should be of the order of

!max ' mc2

~: (8.134)

In decay, the maximum value of is of order of 30.

Let us assume that an electron suddenly acquires a velocity = v=c and then travels at a

constant velocity. The integration in Eq. (8.61) is limited from t = 0 to 1;

dI(!)

d=

1

4"0

e2

42c!22 sin2

Z 1

0ei!(1 cos )tdt

2=

1

4"0

e2

42c

2 sin2

(1 cos )2 ; (8.135)

where is the angle between the velocity and the unit vector n: Integration over the solid angle

yields

I(!) =1

4"0

e2

c

1

ln

1 +

1

2= const.; ! . !max =

mc2

~: (8.136)

The frequency spectrum is at up to !max: Therefore, the total energy radiated through decay

is approximately given by

E ' 1

4"0

e2

c

1

ln

1 +

1

2 mc2

~' 1

ln(4 2) 2

mc2; (8.137)

where the dimensionless quantity ,

=1

4"0

e2

c~' 1

137; (8.138)

is the ne structure constant. The energy emitted as radiation through decay is a small fraction

of the electron energy.

8.10 Cherenkov Radiation

Cherenkov radiation occurs when a charged particle travels faster than electromagnetic waves in

a material medium. It does not require acceleration of charges and the basic mechanism is very

similar to that of sound shock waves in gases. As in the case of decay, we assume a charge

travelling along a straight line at a velocity = v=c(!); where

c(!) =1p"(!)0

;

25

is the velocity of electromagnetic waves in a dielectric having a permittivity "(!): Eq. (8.61) should

be modied as follows after taking into account the proper denition of c(!);

dI(!)

d=04

e2

42c(!)!2v2 sin2

Z 1

1ei!(1 cos )tdt

2 : (8.139)

Note that the integration limits are from 1 to 1: The time integral is singular,Z 1

1ei!(1 cos )tdt = 2[!(1 cos )];

and the condition for radiation is

cos =1

< 1;

or

(!) =v

c(!)> 1: (8.140)

Then,dI(!)

d=04

e2

(2)2c(!)!2v2 sin2 2[!(1 cos )]: (8.141)

Square of a delta function is not integrable and the radiation energy simply diverges. This is merely

due to the assumption that the charge is radiating forever from t = 1 to 1 which is of course

unphysical. It is more appropriate to consider a radiation power rather than energy. For this

purpose, we consider a thin slab of the dielectric of thickness dz. The transit time over the distance

dz is T = dz=v and we calculate energy radiated during that time,

dI(!)

d=04

e2

42c(!)!2v2 sin2

Z T=2

T=2ei!(1 cos )tdt

2

: (8.142)

The integral can be carried out easily,Z T=2

T=2ei!(1 cos )tdt =

2

!(1 cos ) sin(1 cos )T!

2

:

ThusdI(!)

d=04

e2!2

42c(!)sin2

sin

2(dz)2; (8.143)

where

=1

2(1 cos )!T: (8.144)

26

In high frequency regime !T 1; the function (sin=)2 may be approximated by a delta

function (): Integration over the solid angle yields

dI(!)

dz=

1

4"0

e

c0

2!

1 1

2

=

1

4"0

e

c0

2!

1 1

"(!)0v2

; (8.145)

where c0 = 1=p"00 is the speed of light in vacuum. The rate of energy loss due to Cherenkov

emission is given by

dEdz=

1

4"0

e

c0

2 Z 1

0!

1 1

"(!)0v2

d!; 0 < ! <1: (8.146)

The result obtained is meaningful only if

> 1; or v > c(!) =1p"(!)0

; (8.147)

which is the condition for Cherenkov radiation. The instantaneous radiation power can be estimated

from

P =d

dt

ZI(!)d!

=1

4"0

e

c0

2v

Z!

1 1

"(!)0v2

d!: (8.148)

In order to nd the eld proles emitted through Cherenkov radiation, we start from the wave

equations for the potentials in a material medium,r2 0~"

@2

@t2

~"(r; t) = free; (8.149)

r2 0~"

@2

@t2

A(r; t) = 0J; (8.150)

where ~" is the dielectric operator containing time derivative,

~" = ~"

@

@t

: (8.151)

For a charged particle e travelling at a constant velocity v; the charge density and current density

are described by

= e(r vt); (8.152)

J = ev(r vt): (8.153)

27

Then, after Fourier-Laplace transformation, the Fourier potentials can readily be found,

(k; !) =2e

"(!)

k2 !2

c2(!)

(! k v); (8.154)

A(k; !) =2e0v

k2 !2

c2(!)

(! k v); (8.155)

where, as before,

c2(!) =1

"(!)0; (8.156)

and the transformation Z Z(r vt)ei(!tkr)dV dt = 2(! k v); (8.157)

is substituted. Since the physical electric eld is

E(r; t) = r(r; t) @

@tA(r; t); (8.158)

the Fourier component of the electric eld is given by

E(k; !) = ik(k; !) + i!A(k; !)

= 2iek !

c2(!)v

"(!)

k2 !2

c2(!)

(! k v): (8.159)

Similarly, the Fourier-Laplace component of the magnetic eld is

B(k; !) = ikA(k; !)

= 2ie0k v

k2 !2

c2(!)

(! k v): (8.160)

The physical electromagnetic elds can then be found through inverse transformations,

E(r; t) =1

(2)4

Zd3k

Zd!E(k; !)ei(kr!t); (8.161)

B(r; t) =1

(2)4

Zd3k

Zd!B(k; !)ei(kr!t): (8.162)

To proceed further, we assume that the charged particle is travelling along the z axis at a

constant velocity v: The system is symmetric about the axis and we may assume an observing

point in the x z plane without loss of generality. We denote the cylindrical coordinates of the

28

Figure 8-7: Geometry for Fourier inverse transform.

observing point by (; = 0; z) and spherical Fourier coordinates by k =(k; ; ): Then,

k r = k(z cos + sin cos):

Because of the cylindrical symmetry, radiation of energy is expected in the radial direction ; and

the relevant Poynting vector is

S = (EH)

= Ez(r; t)H(r; t): (8.163)

The energy radiated per unit length along the particle trajectory (z axis) can be calculated from

dEdz

= 2ZEz(r; t)H

(r; t)dt

= ZEz(r; !)H

(r; !)d!; (8.164)

where Ez(r; !) is the Laplace transform of the electric eld,

Ez(r; !) =1

(2)3

Zd3kEz(k; !)e

ikr; (8.165)

and H(r; !) is the Laplace transform of the magnetic eld,

H(r; !) =1

(2)3

Zd3kH(k; !)e

ikr: (8.166)

29

The Laplace transform of the axial electric eld Ez(r; !) can be calculated as follows:

Ez(r; !) =1

(2)3

Zd3kEz(k; !)e

ikr

= ie

(2)2

Z 1

0k2dk

Z

0sin d

Z 2

0d

k cos !vc2(!)

"(!)k2 !2

c2(!)

(! kv cos )eik(z cos + sin cos)= i0e!

(2)2

Z 1

1=d

12 1

2 1

Z 2

0exp

"i!

c(!)

(z

+

s1 1

()2cos

)#d

= i0e!2

exp

i!z

c(!)

Z 1

1=d

12 1

2 1J0!

c(!)

q2 1

2

; (8.167)

where

=kc(!)

!; =

v

c(!): (8.168)

Letting

2 = 2 1

2; (8.169)

we nally obtain

Ez(r; !) =i0e!

2exp

i!z

c(!)

1 1

2

Z 1

0

J0

!c(!)

2 1 + 1

2

d

=ie02

! exp

i!z

c(!)

1 1

2

K0(); (8.170)

where

= !

c (!)

s1

2 1; (8.171)

and use is made of the integral representation of the modied Bessel function K0(ax);

K0(ax) =

Z 1

0

tJ0(at)

t2 + x2dt: (8.172)

In the asymptotic regime jj 1; K0() approachesr

2e: (8.173)

For this to be propagating radially outward in the form eik; we must choose

= !

c (!)

s1

2 1

= i !

c (!)

s1 1

2; > 1: (8.174)

30

With this choice for ; the Laplace transform of the axial electric eld becomes proportional to

exp

"i!

c(!)

s1 1

2+

z

!#; (8.175)

and the radial and axial wavenumbers can be identied as

k =!

c(!)

s1 1

2; kz =

!

v; (8.176)

respectively. Cherenkov radiation is conned in a cone characterized by an angle ;

sin =c(!)

v; (8.177)

as shown in Fig. 8-8.

Figure 8-8: Cherenkov cone. Radiation elds are conned in the cone.

The Laplace transform of the azimuthal magnetic eld is given by

H(r; !) =1

(2)3

Zd3kH(k; !)e

ikr

= ie

(2)2

Zk2dk

Z

0sin d

Z 2

0d

kv sin

k2 !2

c2(!)

(! kv cos )

eik(z cos + sin cos)

=e

2K1() exp

i!

c(!)z

; (8.178)

31

where the following integral Z 1

0

x2J1(bx)

x2 + a2dx = aK1(ab); (8.179)

is noted. (Calculation steps are left for an exercise.) Substituting Ez(r; !) and H(r; !) into Eq.

(8.164), we nd

dEdz= i0e

2

(2)2

Z 1

1!

1 1

2

K0()K

1 ()d!: (8.180)

In the asymptotic region jj 1; this reduces to

dEdz=

1

4"0

e2

c20

Z 1

0!

1 1

2

d!; (8.181)

in agreement with the earlier result, Eq. (8.146). Eq. (8.180) can be used even when Cherenkov

condition is not satised. In this case, energy loss is through near eld Coulomb interaction between

a charged particle and ions and electrons in molecules in the dielectric media. We will return to

this problem in Section 8.12.

8.11 Transition Radiation

Transition radiation occurs when a charge crosses a boundary of two dielectric media. No accelera-

tion is required, nor is it necessary for charge to move faster than the speed of light as in Cherenkov

radiation. In this respect, transition radiation is a least demanding radiation mechanism. Radia-

tion emitted from a charge approaching a conductor is an extreme case of transition radiation with

an innite permittivity, and may be regarded as the inverse process of radiation accompanying

decay. Disappearance, rather than creation, of charge is responsible for transition radiation.

We rst consider a simple case: a charge e approaching normally a conducting plate at a velocity

v (> 0). On impact, the charge is assumed to come to rest. A conducting plate is mathematically

equivalent to an innitely permissive dielectric plate. An image charge e moving in the oppositedirection in the conducting plate can be introduced so that the current density is

Jz(r; t) = ev [(z + vt) + (z vt)] (x)(y); 1 < t < 0; (8.182)

where at t = 0 (or z = 0) the particle is brought to rest. Its Laplace transform is

Jz(r; !) =

Z 0

1Jz(r;t)e

i!tdt

= e expi!vjzj(x)(y): (8.183)

Let the observing point be at P located at (r; ): (The system is symmetric about the axis and thus

32

Figure 8-9: Radiation from a charge impinging on a metal surface. Sudden deceleration at themetal surface is the inverse process of radiation accompanying beta decay.

is ignorable.) The vector potential at P can be calculated in the usual manner,

Az(r; !) =04

ZJz(r

0; !)

jr r0j eikjrr0jdV 0 ' 0

4reikr

ZJz(r

0;!)eikr0dV 0

= 0e4r

eikrZ 1

0

hexp

i!vz0eikz

0 cos + expi!vz0eikz

0 cos idz0

= i0e

4reikr

1

k

1

1 + cos +

1

1 cos

; z > 0; (8.184)

and the magnetic eld from

H(r; !) ' ikAz sin =0

=e

4reikr

1

1 + cos +

1

1 cos

sin : (8.185)

33

The angular distribution of radiation energy is thus given by

dI(!)

d=

1

2r2c0 jH(r; !)j2

=e22c0323

1

1 + cos +

1

1 cos

2sin2

=1

4"0

e2v2

22c3sin2

(1 2 cos2 )2; 1 < ! <1: (8.186)

Integration over the solid angle in the region z > 0 (0 < < =2) yields

I(!) =1

4"0

e2

2c

1

1 + 2

ln1 +

1 2; ! > 0:

Evidently, the radiation energyR10 I(!)d! diverges. This is due to the assumption of a perfect

conductor. In practice, metals cannot be regarded as perfect conductor. The condition that the

surface impedance of metal

Z =

ri!0

i!"0 + ;

be su¢ ciently small compared with the free space impedance Z0 =p0="0 imposes an upper limit

of the frequency, !"0 and a cuto¤ emerges in the integralR !max0 I(!)d!:

Figure 8-10: Transition radiation emitted by a charge passing through a dielectric boundary.

We now analyze the case of a dielectric slab having a relative permittivity "r = "="0: In this

34

case, the particle continues to travel after passing the boundary and the current density is now

Jz(r; t) = ev(x)(y)(z + vt); 1 < t <1: (8.187)

Its Laplace transform is

Jz(r; !) = ev(x)(y)ei!v t: (8.188)

The contribution to the radiation elds from the region z > 0 consists of two parts, one directly

from the charge (as in free space) and the other via reection at the dielectric boundary. Denoting

the magnetic reection coe¢ cient by in the Fresnels formulae,

="r cos

p"r sin 2

"r cos +p"r sin 2

; (8.189)

and following the same procedure as in the case of conductor plate, we nd

H1(r; !) 'ev

4creikr

1

1 + cos +

1 cos

sin ; z > 0: (8.190)

The contribution from the region z < 0 involves refraction at the boundary, and thus additional

retardation because of the longer path length. In Fig.8-10, for z < 0, we observe

l = r + z0 tan 00 sin ; z0 < 0

l0 = z0= cos 00;sin 0

sin 00=p";

and thus

kl +pkl0 = kr kz

p sin 2: (8.191)

Then the contribution from the region z < 0 to the integral becomes

H2(r; !) = ev

4cr(1 ) 1

1 + p"r sin 2

eikr sin (8.192)

Note that the factor 1 here indicates the eld amplitude transmitted into the air region. Thetotal magnetic eld is H1 +H2; and the angular distribution of the radiation energy is given by

dI(!; )

d=r2c02

jH(r;!)j2 =0e

2v2

323cA2 sin 2; (8.193)

where

A =1

1 + cos +

1 cos 1

1 + p"r sin 2

: (8.194)

In the case of ideal conductor "r !1; = 1; we recover

dI(!; )

d=

1

40

e2v2

22c3

sin

1 2 cos2

2.

35

When "r = 1; = 0; radiation evidently disappears.

The factor A in Eq. (8.194) does not fully agree with that in the original work by Frank and

Ginzburg,

A0 =1

1 + cos +

1 cos 1 +

"r(1 + p"r sin 2)

; (8.195)

although this too vanishes when "r ! 1 and reduces to the case of conducting medium when

"r !1:

8.12 Energy Loss of Charged Particles Moving in Dielectrics

The formula derived in Eq. (8.180) yields a physically meaning energy loss rate even when the

Cherenkov condition is not satised, < 1: A charged particle moving in a dielectric medium

collides with atoms and lose its energy through Coulomb interaction with electrons in atoms.

Electrons in an atom are bounded. However, they do respond to electromagnetic disturbance and

absorb energy through the resonance "(!) = 0, where

"(!) = "0

1

!2p!2 !20

!: (8.196)

Resonance of the type1

x x0; (8.197)

can be handled mathematically by introducing an imaginary part,

1

x x0= P

1

x x0 i(x x0); (8.198)

where P stands for the principal part. This is justiable because the imaginary part of the function

1

x x0 + i"=

x x0(x x0)2 + "2

i "

(x x0)2 + "2; (8.199)

remains nite even in the limit "! 0;

lim"!0

Z"

(x x0)2 + "2dx = : (8.200)

Physically, the resonance leads to absorption of wave energy by charged particles in a material

medium dielectrics, plasmas, etc.

The characteristic scale length of interaction between a charged particle and atoms in a dielec-

tric is evidently of the order of atomic size which indicates that the interaction is of near-eld,

nonradiating nature dominated by longitudinal (electrostatic) elds. As we will see, the major

contribution to the energy loss occurs through the pole of the dielectric function, "(!) = 0:

In the near-eld region 1; the modied Bessel functions K0();K1() may be approx-

36

imated by

K0() ' ln2

E ; (8.201)

K1() '1

; (8.202)

where E = 0:5772 is the Eulers constant. The real part of Eq. (8.180) becomes

dEdz' 0e

2

(2)2Re

Z 1

1i!

1 1

"(!)0v2

hln2

+ E

id!; (8.203)

with

=!

c(!)

s1

2 1; < 1: (8.204)

The dielectric function "(!) is in the form

"(!) = "0

1

!2p!2 !20

!:

Therefore, the integration can be carried out by evaluating the pole contribution at "(!) = 0; which

occurs at

! = q!2p + !

20; (8.205)

and exploiting Plemeljs formula,

lim!0

1

x a+ i = P1

x a i(x a);

where P indicates the principal part of the singular function 1=(x a): The result is

dEdz' 1

4"0

e!pv

2ln

0@ 1:12v

q!2p + !

20

1A : (8.206)

As the minimum distance ; the intermolecular distance may be substituted because the shell

electrons e¤ectively shield the electric eld of the charge well inside the atom.

In a plasma, !0 is evidently zero (because electrons in a plasma are free). Then,

dEdz' 1

4"0

e!pv

2ln

v

min!p

; (8.207)

where

min '3

4n

1=3;

is the average distance between ions. For a Maxwellian electron distribution with a temperature

37

Te; the average energy loss rate may be estimated from

dEdt' 1

4"0

(e!p)2

3vTeln

8p2

3n3D

!; (8.208)

where

D =vTe!p; (8.209)

is the Debye shielding length.

8.13 Bremsstrahlung

Whenever a charged particle collides with another charged particle, electromagnetic radiation occurs

due to acceleration by Coulomb force. Collisions between like particles (e.g., electron-electron)

emit quadrupole radiation while collisions between unlike particles (e.g., electron-ion) emit dipole

radiation. Bremsstrahlung is due to collisions between electrons and ions and provides a basic

mechanism for x-ray production.

Let an electron approach an ion having a charge Ze with a velocity v ( c) and impact

parameter b: The acceleration due to Coulomb force is of the order of

a ' 1

4"0

Ze2

mb2; (8.210)

and consequent radiation power can be estimated from the Larmors formula,

P ' 1

4"0

2e2a2

3c3: (8.211)

The total energy radiated can in principle be found by integrating the power over time along the

electron trajectory. However, in experiments, one is seldom interested in measuring radiation power

or energy associated with a single electron. What is more relevant is the radiation associated with

a beam of electrons impinging on an ion. In this case, some electrons have impact parameters

vanishingly small. However, the impact parameter has a lower bound imposed by the uncertainty

principle,

bmin '~p; (8.212)

where p is the electron momentum. For an impact parameter b; the time duration in which the

acceleration is signicant is

t ' b

v: (8.213)

38

Therefore, energy radiated by a single electron is

E = Pt

' 1

(4"0)32

3c3

Ze3

m

21

b3v: (8.214)

For an electron beam having a density n; the radiation power can thus be estimated from

P = nev

Z 1

bmin

E2bdb

=1

(4"0)34

3

Ze3

m

2nemv

c3~: (8.215)

If the ion density is ni; the quantity Pni denes the power density,

1

(4"0)34

3

Ze3

m

2nenimv

c3~; (W/m3): (8.216)

The frequency spectrum of radiation energy may qualitatively be found as follows. Since the

characteristic time of acceleration

=b

v; (8.217)

is short, radiation occurs as an impulse and the spectrum is at in the region 0 < ! < v=b; and

vanishes for ! > v=b: Since the minimum impact parameter is

bmin '~mv

; (8.218)

the upper limit of the frequency spectrum extends to

!max 'mv2

~; or ~! . mv2: (8.219)

This is essentially a statement of energy conservation, that is, the maximum photon energy emit-

ted during bremsstrahlung is limited by the incident electron kinetic energy, which is reasonable.

Therefore, the frequency spectrum of bremsstrahlung is

I(!; b) =

8>>><>>>:1

(4"0)32

3

(Ze3)2

mc2mv2b2; 0 < ! <

v

b;

0; ! >v

b:

(8.220)

I(!; b) has dimensions of J/frequency. It is convenient to introduce a radiation cross-section (!)

39

dened by

(!) =

Z bmax

bmin

I(!; b)2bdb

=1

(4"0)34

3

(Ze3)2

mc2mv2ln

bmaxbmin

=

1

(4"0)34

3

(Ze3)2

mc2mv2ln

mv2

~!

: (8.221)

The integral over the frequency, Z(!)d!; (8.222)

evidently diverges at the lower end ! ! 0. To remedy this di¢ culty, Bethe and Heitler recognized

that if the velocity v is understood as the mean value of the initial and nal velocities, i.e., before

and after emission of a photon,

v =1

2(vinitial + vnal)

=1p2m

pE +

pE~!

; (8.223)

the integral remains nite,

Z(!)d! =

1

(4"0)34

3

(Ze3)2

mc2mv2

Z !max

0ln

0B@pE +

pE~!

2~!

1CA d!=

1

(4"0)34

3

(Ze3)2

mc3~

Z 1

0ln

1 +

p1 xpx

dx; x =

~!E ;

=1

(4"0)34

3

(Ze3)2

mc3~: (8.224)

The integral in the intermediate step is unity. Multiplying by the ion density ni; we thus obtain

the bremsstrahlung rate per unit length,

dE

dz= ni

Z(!)d!

=1

(4"0)34

3

(Ze3)2nim2c3~

; (8.225)

and radiation power density,

P=V =1

(4"0)34

3

(Ze3)2neniv

mc3~; (W m3): (8.226)

This agrees with the earlier qualitative estimate in Eq. (8.216).

Relativistic correction to the classical bremsstrahlung formulae can be readily found if we move

40

to the electron frame wherein the electron velocity is nonrelativistic. Since the energy and frequency

are Lorentz transformed in the same manner, and the transverse dimensions are Lorentz invariant,

it follows that the radiation cross-section (!) is Lorentz invariant,

lab(!lab) = 0(!0);

where the primed quantities are those in the electron frame. The frequencies !lab and !0 are related

through the relativistic Doppler shift,

!0 = !lab(1 cos ); !lab = !0(1 + cos 0); (8.227)

where and 0 are the angles with respect to the electron velocity in each frame. Since in the

electron frame, the radiation is conned in a small angle about 0 = =2; we have

!0 ' !lab : (8.228)

The collision time is shortened by the factor since the transverse eld is intensied by the same

factor through Lorentz transformation. Therefore, the maximum impact parameter is modied as

bmax = v

!0= 2v

!lab; (8.229)

and the radiation cross-section in the laboratory frame becomes

(!) = 0(!0) =1

(4"0)34

3

e4(Ze)2

m2c3v2ln

2mv2

~!

: (8.230)

It is noted that the minimum impact parameter remains unchanged through the transformation

because it is essentially the Compton length based on the uncertainty principle.

8.14 Radiation due to Electron-Electron Collision

In this case the dipole radiation is absent because in the center of mass frame, two electrons stay

at opposite positions, r1 = r2; and the dipole moment identically vanishes.. The lowest orderradiation process is that due to electric quadrupole. (The magnetic dipole moment also vanishes.)

The quadrupole moment tensor is

Qij =1

2exixj ; (8.231)

where xi is the i-th component of the relative distance 2r: In Chapter 5, a general formula for the

quadrupole radiation power has been derived. Noting

d3

dt3(xixj) =

:::xixj + 3

::xi:xj + 3

:xi::xj + xi

:::xj ; (8.232)

41

Figure 8-11: Colliding electrons in the center of mass frame. The impact parameter is b:

::xi =

1

4"0

2e2

mr3xi; (8.233)

:::xi =

1

4"0

2e2

m

xir3 3xivr

r4

; (8.234)

we ndd3

dt3(xixj) =

1

4"0

2e2

mr3

4(vixj + vjxi)

6vrxixjr

; (8.235)

:::Q =

1

4"0

e3

mr3

4(vixj + vjxi)

6vrxixjr

(8.236)

Substituting this into the quadrupole radiation power,

P =1

4"0

1

60c5

243Xij

:::Q2

ij X

i

:::Qii

!235 ; (8.237)

we obtain, after somewhat lengthy calculations,

P =1

4"0

2

15c5

e3

m

2 v2 + 11v2r4

; (8.238)

where v is the component of the velocity related to the initial angular momentum bv0 = r(t)v(t)

with v0 the velocity at r !1: Energy conservation reads

1

2mv20 =

1

2mv2 +

1

4"0

e2

2r: (8.239)

42

Then

v2 + 11v2 = v20

4e2

4"0mr+ 11

(bv0)2

r2; (8.240)

and the total radiation energy can be found by integrating the radiation power over time along the

trajectory,

E = 1

4"0

2

15c5

e3

m

2 Z 1

1

1

r4

v20

4e2

4"0mr+ 11

(bv0)2

r2

dt: (8.241)

This can be converted to an integral over the distance r by noting

dt =dr

vr=

drsv20

(bv0)2

r2 4e2

4"0mr

; (8.242)

E = 1

4"0

2

15c5

e3

m

2 2

Z 1

rmin

1

r4

v20 4e2

4"0mr+ 11

(bv0)2

r2sv20

(bv0)2

r2 4e2

4"0mr

dr; (J) (8.243)

where rmin is the distance of the closest approach,

rmin =1

2v20

0@4e2m+

s4e2

m

2+ 4(bv0)2

1A : (8.244)

However, the radiation energy by a single electron pair is of no practical interest. What is more

relevant is the radiation power emitted by an electron beam impinging on a single electron which

can be evaluated from

P = 2nv0

Z 1

0E(b)bdb

=1

4"0

4

15c5

e3

m

22nv0

Z 1

0bdb

Z 1

rmin

1

r4

v20 4e2

4"0mr+ 11

(bv0)2

r2sv20

(bv0)2

r2 4e2

4"0mr

dr; (W). (8.245)

The double integral reduces to

Z 1

0bdb

Z 1

rmin

dr1

r4

v20 4e2

4"0mr+ 11

(bv0)2

r2sv20

(bv0)2

r2 4e2

4"0mr

=25

3v0

Z 1

rmin

1

r5

r2 4re

2

mv20

3=2dr =

5

6

mv30e2: (8.246)

43

Therefore,

P =1

4"0

4

9

ne4v40mc5

; (W). (8.247)

This denes a radiation cross-section,

=P

1

2nmv30

=8

9

v

cr2e ; (m

2) (8.248)

where 12nmv

30 is the energy ux density of the beam and

re =1

4"0

e2

mc2= 2:85 1015 m;

is the classical radius of electron.

44

Problems

8.1 A charged particle with charge e and mass m undergoes motion in external electric and

magnetic elds E and B: Show that the radiation power is given by

P (t0) =1

4"0

2e2

3c3 6a2 ( a)2

=

1

4"0

2e2

3m2c3 2(E+ v B)2 ( E)2

;

where a is the acceleration. Since a2 ( a)2 is positive denite, so is (E+ v B)2 ( E)2.

Note: In an instantaneous rest frame of the particle, v = = 0 and = 1: The radiation

power in that frame is thus with respect to the proper time,

dE 0d

= P () =1

4"0

2e2

3m2c3E02;

where E0 is the electric eld also in that frame. Since the electric eld is Lorentz transformed

as

E0k = Ek; E0? = (E? + v B);

we readily see that

P () = P (t0);

that is, the proper radiation power is Lorentz invariant. This is not surprising because both

time and energy are Lorentz transformed in a similar manner.

8.2 Carry out the integration in the radiation power due to acceleration perpendicular to the

velocity

P (t0) =1

4"0

e2 _?24c

Z1

(1 cos )5

(1 cos )2 1

2sin2 cos2

d

to obtain

P (t0) =1

4"0

2e2 _23c

4:

Plot the angular distribution of the radiation intensity

f() =1

(1 cos )6

(1 cos )2 1

2sin2 cos2

;

for = 0 and =2 in a highly relativistic case.

8.3 A nonrelativistic electron with velocity v collides head-on with a heavy negative ion having

a charge Ze:Find the total energy radiated. What is the dominant frequency component?

45

8.4 A highly relativistic electron passes by a proton at an impact parameter b:Assuming that

the electron trajectory is una¤ected by the proton and radiation, show that the total energy

radiated is

E = 1

4"0

e6

12m2ec3b3v

4 2

1 2' 1

4"0

e6 2

4m2ec4b3:

What is the dominant frequency component contained in the radiation eld?

8.5 An electron having an initial energy E(t = 0) = 0mc2 undergoes cyclotron motion in a

magnetic eld B: Show that the rate of energy loss to radiation is given by

dE(t)dt

=1

4"0

2e4B2

m4c5E2 m2c4

:

Integrate this and nd an expression for E(t):

8.6 An electron is placed in a plane wave E0ei(kx!t): If ~! mc2; electron recoil may be ignored.

Show that the radiation power is

P = I08

3r2e ;

where I0 = c"0E20 (W/m2) is the intensity of the incident wave and

re =e2

4"0mc2= 2:8 1015 (m),

is the classical radius of electron.

=8

3r2e ;

is known as Thomson scattering cross section.

Note: For ~! mc2; Klein and Nishina, solving the Dirac equation to take into account

electron recoil, obtained

' r2emc2

~!

1

2+ ln

2~!mc2

:

8.7 The light from Crab nebula is believed to be synchrotron radiation emitted by highly energetic

electrons trapped in a weak intergalactic magnetic eld. If the electron energy is 2 1011 eVand B = 107 T, what is the dominant wavelength component?

8.8 Perform numerically the integration in the radiation power of l-th harmonic of synchrotron

radiation,

Pl =1

40

(e!0)2l

c

Z

0

sin(2l sinx)

sinx(2 cos 2x 1) cos(2lx)dx;

for the case = 0:99 ( ' 7) and plot the result as a function of l:

8.9 Estimate the bremsstrahlung cooling rate (in W/m3) of a hydrogen plasma for an electron

temperature of 10 keV and plasma density of n = 1020 m3: If the plasma is conned by a

magnetic eld of 5 T, what is the cooling rate due to cyclotron radiation if reabsorption by

46

the plasma is ignored? (Cyclotron radiation can be reabsorbed by a plasma if its density is

high enough and its contribution to energy loss may be ignored.)

8.10 Assuming a permittivity in the form

" (!) =

1

!2p!2 !20

!"0;

carry out the integration in the rate of Cherenkov radiation,

dEdz=

1

4"0

e2

c20

Z 1

0!

1 1

2

d!:

8.11 Two identical charges are moving along a circular orbit of radius a at an angular velocity !:

Assume nonrelativistic motion, !a c: Angular separation of the charges is 0: If 0 = 0;

the problem reduces to a charge 2e rotating at ! which radiates as a dipole. If 0 = ; the

dipole moment vanishes and higher order quadrupole radiation should be considered. At what

angular separation 0 does the dipole radiation power become equal to quadrupole radiation

power? Is it necessary to consider magnetic dipole radiation?

8.12 Four identical charges are placed along a circle of radius a with equal angular spacing =2:

If the system rotates about the axis at angular velocity !; what is the lowest order radiation

power? What would happen if the number of charges N increases?

8.13 Generalize the quadrupole radiation in electron-electron collision to relativistic velocity.

47

Chapter 9

Conned and Guided Waves

9.1 Introduction

Electromagnetic waves can be conned in a volume surrounded by a conductor wall as in microwave

cavities and also guided to propagate in conducting tubes and dielectric bers. Parallel wire

transmission lines and coaxial cables are also used in transmitting low frequency waves. In this

chapter, electromagnetic waves in cavities and waveguides will be discussed. In waveguides, axial

electric and magnetic elds Ez and Hz satisfy scalar wave equation and can be readily solved by

incorporating the boundary conditions. Transverse elds E? and H? can then be found in the

form of linear combination of the axial elds. In conductor waveguides, electromagnetic waves can

be classied into modes without axial magnetic eld (TE mode) and modes without axial magnetic

eld (TM mode). The axial phase velocity of electromagnetic waves in conductor waveguides with

a smooth wall is always larger than c: In slow waveguides with a periodically structured wall, the

phase velocity can be adjusted close to c as needed in linear accelerators. In dielectric bers, there

can be no pure TE or TM modes because evanescent elds outside the ber do a¤ect mode structure

through the boundary condition at the ber surface. Optical bers used in communication all have

gradual radial () change in the index of refraction, n(); and poses a problem of wave propagation

in a nonuniform medium.

1

9.2 Spherical Cavity

Electromagnetic waves in a spherical cavity surrounded by a conducting wall may be analyzed in

terms of the TE and TM spherical eigenvectors formulated in Chapter 6,

E = i![rr+r (rr )]; (9.1)

B = r (rr) + k2rr ; (9.2)

where and both satisfy the Helmholtz equation

@2

@r2+2

r

@

@r+1

r21

sin

@

@

sin

@

@

+

1

r2 sin2

@2

@2+ k2

0@ (r)

(r)

1A = 0: (9.3)

Here k = !=c: Solutions for and are in the form of standing (non-propagating) wave,

(r; ; ; t); (r; ; ; t) = jl(kr)Pml (cos )

8<: sinm

cosm

9=; ei!t: (9.4)

Note that nl(kr) is discarded because the elds should remain nite at r = 0: As in mechanical

standing waves wherein the sum of potential energy and kinetic energy is constant, in electromag-

netic standing waves the sum of electric energy and magnetic energy is constant. This is guaranteed

because the magnetic eld is out of phase with the electric eld by =2 as seen from Eqs. (9.1) and

(9.2). For TE modes without radial component, the boundary condition that E and E vanish at

the inner wall r = a yields

jl(ka) = 0: (9.5)

Then the resonance frequency is given by

!ls =cxlsa; (9.6)

2

where xls is the s-th root of jl(x) = 0. Some roots are listed below.

l = 0 1 2 3

s = 1 4:493 5:763 6:988

2 2 7:725 9:095 10:417

3 3 10:904 12:323 13:698

4 4 14:066 15:515 16:924

For TM modes, the radial component of the magnetic eld should vanish at r = a: This yields

d

dr[rjl(kr)] = 0 at r = a; (9.7)

and the resonance frequency of TM mode is

!ls =cx0lsa; (9.8)

where x0ls is the s-th root ofddx [xjl(x)] = 0:

l = 0 1 2 3

s = 1 =2 2:744 3:870 4:973

2 3=2 6:177 7:443 8:722

3 5=2 9:317 10:713 12:064

4 7=2 12:486 13:921 15:413

The smallest eigenvalue is that of l = 1 TM mode,

!TM;l=1 ' 2:744c

a; ka = 2:744:

The Q value (cavity quality factor) is dened by

Q = !time averaged energy stored in a cavity

time averaged rate of energy absorption by the wall

= !

RV02 H

2dVq!02

RS12(nH)2dS

=2

RV H

2dVRS(nH)2dS

; (9.9)

3

where the magnetic eld is assumed to be in the form of standing wave H(r)ei!t; and

=

s2

!0=

1pf0

; (9.10)

is the skin depth of the wall. Note that in the case of standing wave, the sum of electric and

magnetic energies is constant and equal to either

02

ZVH2dV;

or"02

ZVE2dV:

Time averaged dissipation rate at the wall is

1

2

r!02

(nH)2; (W/m2);

where the factor 1=2 is for rms (time average) value.

For the lowest order TM mode with l = 1; the azimuthal magnetic eld is in the form

H = Cj1 (kr) sin ; (9.11)

where C is a constant. Then, for ka = 2:744;

QTM;l=1 =a

1 j0 (ka) j2 (ka)

j21 (ka)

' 0:734

a

; (9.12)

where use is made of the integral,

Zx2j2l (x) dx =

x3

2

j2l (x) jl1 (x) jl+1 (x)

: (9.13)

Alternative expression is

QTM;l=1 = 0:734a

= 0:32

: (9.14)

The atmospheric region between the ionosphere and ground earth can be regarded as a thin

4

cavity for electromagnetic waves. The simplest standing wave is the one which is azimuthally

symmetric (@=@ = 0) : The component of the wave equation

r2 + k2

B = 0; k = !=c; (9.15)

yields r2 1

r2 sin2 + k2

B = 0; (9.16)

provided Br = B = 0: Solution for B (r; ) is

B (r; ) =ul (r)

rP 1l (cos ) ; (9.17)

where ul (r) satisesd2uldr2

+

k2 l (l + 1)

r2

ul (r) = 0: (9.18)

Corresponding electric eld is

Er = ic2

!rl (l + 1)

ul (r)

rPl (cos ) ; (9.19)

E = ic2

!r

dul (r)

drP 1l (cos ) : (9.20)

In the narrow region between the ionosphere and ground, we may approximate

l (l + 1)

r2' l (l + 1)

a2; (9.21)

where a is the earth radius. Then

ul (r) = cos

rk2 l (l + 1)

a2(r a)

!; (9.22)

and the condition E = 0 at r = a yields a resonance frequency,

! =pl (l + 1)

c

a: (9.23)

The fundamental resonance (known as Schumann resonance) frequency is approximately 10 Hz.

(Observed frequency is approximately 8 Hz.) The Schumann mode can be excited by lightning

5

discharges and is being continuously monitored as one of the most important electromagnetic

phenomenon of global scale. Note that the resonance frequency is essentially the inverse of the

transit time around the globe, = 2a=c:

9.3 Perturbative Change in Cavity Resonance Frequency

For a given eigenmode, the eigenvalue k2 of a cavity mode can be found from

k20 =

RV (rE)

2dV 2HS(nE) (rE)dSR

V E2dV

; (9.24)

subject to the condition that r E = 0 in the volume and nE = 0 on the wall surface. Note

that the normal vector n is directed away from the cavity volume. The surface integral evidently

vanishes for exact eigenmodes but is retained here for the purpose of variational calculation because

trial functions may not exactly satisfy the boundary conditions. If the cavity surface is only slightly

deformed, the unperturbed electric eld may still be used in the majority of region except at the

deformation. Let us assume a small dent of volume V and surface area S: The new eigenvalue

to order V and S is

k2 =

RVV (rE)

2dV 2HS0+S(nE) (rE)dSR

Vv E2dV

:

=

RV (rE)

2dV RV (rE)

2dV 2HS0+S(nE) (rE)dSR

V E2dV

RV E

2dV

'RV (rE)

2dVRV E

2dV

1 +

RV E

2dVRV E

2dV

RV (rE)

2dVRV E

2dV 2

HS(nE) (rE)dSR

V E2dV

= k20

1 +

RV E

2dVRV E

2dV

RV (rE)

2dVRV E

2dV 2

HS(nE) (rE)dSR

V E2dV

; (9.25)

6

where S0 is the unperturbed surface on which nE = 0 and S is the surface of V facing the

cavity interior. The surface integral can be converted into volume integral as

IS(nE) (rE)dS =

IS(n0E) (rE)dS

= ISn0 [E (rE)]dS

= ZVr [E (rE)]dV

= ZV(rE)2dV + k20

ZVE2dV (9.26)

where n0= n is the normal vector directed away from V: Then, the change in the eigenvalue is

k2 =

RV [(rE)

2 k20E2]dVRV E

2dV: (9.27)

Since the electric and magnetic energy are identical in the cavity, we nally nd

k2 ' k20

RV H

2dVRV H

2dVRV E

2dVRV E

2dV

; (9.28)

where the rst term indicates the fraction of magnetic energy removed and second term the fraction

of electric energy removed from the cavity. Corresponding change in the resonance frequency is

!

!0' 1

2

RV H

2dVRV H

2dVRV E

2dVRV E

2dV

: (9.29)

This result indicates that if a dent is made at a position on the inner wall where the magnetic energy

is concentrated, the eigenvalue (and thus the frequency) increases through an e¤ective decrease in

the inductance. If a dent is made where the electric energy is dominant, the resonance frequency

decreases through an e¤ective increase in the capacitance. This is analogous to a perturbation in

inductance and capacitance. If a conductor is inserted into an inductor, the inductance decreases

and if a conductor is inserted in a capacitor, the capacitance increases.

If a small object is placed in the volume of a cavity, it will perturb the electromagnetic elds.

A resultant frequency shift may be analyzed in terms of dipole approximation for a body whose

dimension is su¢ ciently small compared with the wavelength. For example, a conductor sphere of

7

Figure 9-1: A dent on a cavity wall with volume V and area S: S0 is the unperurbed area onwhich nE0 = 0 with E0 the unperturbed eld.

radius a induces an electric dipole moment

p = 4"0a3E(rs); (9.30)

and magnetic dipole moment

m = 2a3H(rs); (9.31)

where E(rs) and H(rs) are the elds at the location of the sphere. The perturbed electric energy

is

Ee = p E(rs) = 4"0a3E2(rs); (9.32)

and perturbed magnetic energy is

Em = m B (rs) = +20a3H2(rs): (9.33)

Then the magnetic energy removed is

20a3H2(rs):

and the frequency change is given by

!

!0= 2"0E

2 (rs) + 0H2 (rs)

"0RV E

2dVa3: (9.34)

8

9.4 TE and TM Modes in Conductor Waveguides

In an air-lled waveguide, the electromagnetic elds satisfy the vector wave equations,

r2 1

c2@2

@t2

E = 0; (9.35)

r2 1

c2@2

@t2

H = 0: (9.36)

Since we are interested in waves propagating along the waveguide extended in the z-direction, the

z and time dependence can be singled out in the form

E(r; t) = E(r?)ei(kzz!t); H(r; t) = H(r?)e

i(kzz!t); (9.37)

where r? indicates the coordinates transverse to the z-axis. The axial components Ez and Hz both

satisfy scalar wave equation, r2 1

c2@2

@t2

Ez = 0; (9.38)

r2 1

c2@2

@t2

Hz = 0; (9.39)

Finding solutions to the axial elds Ez and Hz is su¢ cient to determine electromagnetic elds in

a waveguide, for the Maxwells equations

rE = i!0H; rH = i!"0E; (9.40)

yield transverse components entirely in terms of the axial components as follows,

E? =i

(!=c)2 k2z(kzr?Ez + !0r?Hz ez) ; (9.41)

H? =i

(!=c)2 k2z(kzr?Hz !"0r?Ez ez) : (9.42)

Modes having no axial electric elds, Ez = 0; are called Transverse Electric (TE) modes and modes

having no axial magnetic elds, Hz = 0; are called Transverse Magnetic (TM) modes. As for

radiation elds, these two modes form basic eigenfunctions.

9

9.5 Rectangular Waveguides

Figure 9-2: Rectangular waveguide.

We assume a conductor tube having a cross-section a b: The conductor walls are assumed to

be ideally conducting, that is, the skin-depth

==

p2

!0=

1pf0

; (9.43)

is su¢ ciently small and Ohmic dissipation is negligible. Later, we will relax this assumption in

evaluating the damping factor of electromagnetic waves conned in a waveguide. However, for

analyzing mode structure, the assumption of ideal conductor does not introduce signicant errors.

For TE modes, it is su¢ cient to nd the axial magnetic eld Hz(r; t) = Hz(x; y)ei(kzz!t) which

obeys the following Helmholtz equation,

@2

@x2+

@2

@y2+!c

2 k2z

Hz(x; y) = 0: (9.44)

The boundary condition for Hz(x; y) can be found from the vanishing tangential components of the

electric eld at the wall of the waveguide,

r?Hz(x; y) = 0 at x = 0; a and y = 0; b: (9.45)

10

This condition is satised if we choose

Hz(x; y) = H0 cosmaxcosnby; (9.46)

where m and n are integers. Either m or n can be zero but not simultaneously. Substitution into

Eq. (9.44) yields a dispersion relation,

!2 = c2m

a

2+nb

2+ k2z

; (9.47)

which determines the axial wavenumber kz for a given wave frequency ! and given dimensions of

the waveguide. The minimum frequency allowed for wave propagation (kz > 0) in a waveguide

occurs at kz = 0 and is given by

!cmn = c

rma

2+nb

2: (9.48)

This is called the cuto¤ frequency of the TEmn mode. If a > b; the smallest cuto¤ frequency is

that of the TE10 mode (m = 1; n = 0),

!c10 =c

a; or fc10 =

c

2a(Hz): (9.49)

The TE10 mode is most commonly used in microwave communication.

The phase and group velocities along the wave guide are given by

!

kz=

cp1 (!cmn=!)2

> c; (9.50)

d!

dkz= cp1 (!cmn=!)2 < c: (9.51)

In the waveguide, waves propagate along a zig-zag path being reected at the walls. The velocity

along the zig-zag path is simply c; for the total wavenumber is

k =

rma

2+nb

2+ k2z : (9.52)

The phase velocity along the z-axis larger than c is due to a kz which is smaller than k: The group

velocity along the z-axis corresponds to the axial component of c.

11

For TM modes, the axial electric eld Ez(x; y) satisfying the Helmholtz equation@2

@x2+

@2

@y2+!c

2 k2z

Ez(x; y) = 0; (9.53)

and the boundary conditions

Ez = 0 at x = 0; a and y = 0; b (9.54)

is given by

Ez(x; y) = E0 sinmaxsinnby; (9.55)

where m and n are nonzero integers. The cuto¤ frequency of the TM mode is

!cmn = c

rma

2+nb

2; m 1; n 1: (9.56)

9.6 TE10 Mode

In this section, some properties of the TE10 mode will be studied. We assume an electric eld in

the form

Ey(x; z; t) = E0 sinaxei(kzz!t): (9.57)

Corresponding magnetic eld can be found from

H =1

i!0rE

= kz!0

E0 sinaxei(kzz!t)ex i

=a

!0cosaxei(kzz!t)ez: (9.58)

The magnetic eld prole seen from top is shown below for the case kz = =a: Also shown is the

prole of the surface current on the lower wall,

Js= nH: (9.59)

12

x 10.80.60.40.20

z

2

1.5

1

0.5

0

Figure 9-3: Magnetic eld prole of the TE10 mode in a rectangular waveguide.

The power transmitted in the waveguide can be found by integrating the Poynting vector in the

axial direction given by

Sz = EyHx =

kz!0

E20 sin2ax; (9.60)

P =

Z a

0dx

Z b

0dy Sz(x) (9.61)

=kz!0

E20

Z a

0sin2

axZ b

0dy (9.62)

=kz!0

E20ab

2; (peak value). (9.63)

The rms power is

Prms =kz!0

E20ab

4(9.64)

=ab

4

r"00

p1 (!c10=!)2E20 (9.65)

=ab

4

E20ZTE10

; (W) (9.66)

13

x 10.80.60.40.20

z

2

1.5

1

0.5

0

Figure 9-4: Surface current density pattern on the upper inner surface.

where

ZTE10 =

p0="0p

1 (!c10=!)2; (9.67)

is the impedance of the TE10 mode.

In practice, the nite (not innite) conductivity of the wall causes dissipation of electromagnetic

energy and the power decays as the wave propagates along the waveguide. To evaluate the damping

factor, we recall that the Poynting ux normal to a conductor wall is

S? = Z jHtj2 =ri!0

jHtj2 ; (9.68)

where

Zs =

ri!0

; (9.69)

is the surface impedance of the conductor and Hs is the tangential component of the magnetic eld

at the wall surface. The real part of S? indicates net dissipation,

Re

ri!0

jHtj2 =r!02

jHtj2 : (9.70)

14

For the TE10 mode, the tangential component of the magnetic eld at the side walls at x = 0 and

a is

Hz(x = 0; a) = i=a

!0E0e

i(kzz!t): (9.71)

Therefore, the time averaged loss rate per unit length of the side walls is

dP1dz

= 2

r!02

=a

!0

2 E202b: (9.72)

Similarly, on the top and bottom walls,

dP2dz

= 2

r!02

"=a

!0

2+

kz!0

2# E204a

=1

2

"00

r!02

E20a; (9.73)

and the total loss rate is

dPdz

="00

r!02

b!c!

2+a

2

E20 : (9.74)

The power damping factor is given by

1P

dP

dz= 2

r!"02

a+ 2b(!c=!)2

abp1 (!c=!)2

2; (9.75)

and the eld intensity decays as E0ez along the waveguide. For example, the damping factor in a

copper waveguide ( ' 6 107 S/m) having a cross-section 2 1 cm2 excited at a frequency f = 9

GHz is approximately 2 = 0:04/m. The microwave power e-folds in a distance of 1=2 = 25 m.

Example 1 Rectangular Cavities

Electromagnetic elds conned in a rectangular wave guide closed at z = 0 and L form complete

standing waves in all directions, x:y; z: We assume L > a > b without loss of generality. It is

su¢ cient to consider TM modes because after suitable coordinates changes, TM modes can be

recovered from TE modes. The axial electric eld Ez(r; t) must vanish at x = 0; a and at y = 0; b

and thus may be assumed to be

Elmnz (r; y) = E0 sinmaxsinnbycos

l

Lz

; (9.76)

15

where coslL zis chosen so that x and y components of the electric eld vanish at z = 0 and L:

The resonance frequency of (l;m; n) is thus given by

!lmn = c

sma

2+nb

2+

l

L

2; (9.77)

where m;n are nonzero integers and l = 0; 1; 2; The lowest resonance frequency is

!011 = c

ra

2+b

2: (9.78)

9.7 Excitation of Rectangular TE Modes

Electromagnetic waves in waveguides can be excited by a current source placed in the guide. To

illustrate the general methodology, we consider a thin vertical current placed at x = a=2 in a

rectangular waveguide. The y component of the wave equation,

r2 1

c2@2

@t2

E =

1

"0r+ 0

@J

@t: (9.79)

is r2 1

c2@2

@t2

Ey = 0

@Jy@t

; (9.80)

where the charge density has been ignored. Since the transverse electric eld of TE modes can

be generated by the axial magnetic eld Hz as

E? =0k2?

@

@tr?Hz ez; (9.81)

we obtain

Ey = 0k2?

@

@t

@Hz@x

: (9.82)

Substitution into Eq. (9.82) yields

r2 1

c2@2

@t2

1

k2?

@Hz@x

= Jy = I(t)x a

2

(z): (9.83)

16

Let us assume TEm0 modes are excited by the current. Then the axial magnetic eld may be

expanded as

Hz(x; z; t) =

1Xm=1

An(z; t) cosmax; (9.84)

where Am(z; t) is an expansion function and the eigenfunction for TEmn mode is recalled,

Hmnz (x; y) = cos

maxcosnby: (9.85)

Multiplying Eq. (9.83) by

sin

m0

ax

;

and integrating over the cross section area a b; we nd

r2z

1

c2@2

@t2 k2m0

Am0(z; t) =

2m

a2I(t)(z) sin

m2

; (9.86)

with k2m0 = (m=a)2 :

If the current is an impulse, I(t) = q0(t) where q0 is the amount of charge transferred, the

Laplace transform of Am0(z; t) obeysd2

dz2 s2

c2 k2m0

Am0(z; s) =

2m

a2q0(z) sin

m2

; (9.87)

which can be solved as

Am0(z; s) =mq0a2

1p(s=c)2 + k2m0

exp

q(s=c)2 + k2m0 jzj

sinm2

: (9.88)

Inverse Laplace transformation yields

Am0(z; t) =

8<:cq0aJ0

ma

p(ct)2 z2

sinm2

; ct > jzj

0; ct < jzj(9.89)

Note that the pulse front propagates at the speed c: Of course, the pulse shape is severely distorted

as it propagates over a large distance because of the dispersion of the electromagnetic modes

conned in a waveguide.

For a steady oscillating current I(t) = I0ei!t; we may assume Am0(z; t) = A(z)ei!t; where

17

Am0(z) satises d2

dz2+!2

c2 k2m0

Am0(z) =

2m

a2I0(z) sin

m2

: (9.90)

Notingd2

dz2eikjzj = 2ik(z) k2eikjzj;

we nd the following steady state solution,

Am0(z) =mI0

ia2p(!=c)2 k2m0

sinm2

eip(!=c)2k2m0jzj: (9.91)

9.8 Circular Waveguide

Figure 9-5: Circular waveguide. The lower gure shows qualitatively the electric and magnetic eldproles.

Let us now consider electromagnetic waves conned in a conductor tube of radius a: The

Helmholtz equation for the axial electric and magnetic elds is

@2

@2+1

@

@+1

2@2

@2+!c

2 k2z

0@ Ez

Hz

1A = 0: (9.92)

18

General solutions bounded at the axis = 0 are

Ez(; ); Hz(; ) = Jm(k)eim; (9.93)

where m is an integer and

k2 =!c

2 k2z > 0: (9.94)

(The case k2 < 0 will be considered later for slow wave circular waveguides.) For TE modes, the

boundary condition H( = a) = 0 requires

d

dJm(k) = 0 at = a; (9.95)

and for TM modes, Hz( = a) = 0 requires Jm(ka) = 0: In the table, roots of J 0m(x) = 0 relevant

to TE modes and roots of Jm(x) = 0 relevant to TM modes are shown. [Note the special case

J 00(x) = J1(x):]

Roots of J 0m(x) = 0

m = 0 1 2

1st 3.832 1.841 3.053

2nd 7.016 5.331 6.706

3rd 10.173 8.536 9.969

Roots of Jm(x) = 0

m = 0 1 2

1st 2.405 3.832 5.136

2nd 5.520 7.016 8.417

3rd 8.654 10.173 11.620

The dispersion relation of the TE modes is

ka = a

r!c

2 k2z = x0mn; (9.96)

or

!2 = c2

"x0mna

2+ k2z

#; (9.97)

and the cuto¤ frequency is

!c =c

ax0mn: (9.98)

The smallest root is the rst root of J 01(x) = 0; x11 ' 1:841: This is the fundamental mode in a

circular waveguide and corresponds to the TE10 mode in a rectangular waveguide. Through gradual

tapering, the rectangular TE10 mode can be converted to circular TE11 mode.

19

Example 2 Cylindrical Cavity

Consider TE modes in a circular waveguide closed at z = 0; L:The axial magnetic eld may be

assumed to be

Hz(r) = H0Jm(k)eim sin

l

Lz

;

where

k =

r!c

2 k2z :

From

E? =i

(!=c)2 k2z(kzr?Ez + !0r?Hz ez)

=i!0

(!=c)2 k2zr?Hz ez;

we nd the transverse electric eld,

E? =i!0

(!=c)2 k2z

im

Hze

@Hz@

e

:

From the condition that E( = a) = 0; it follows that

kmnl =

s!mnlc

2l

L

2=x0mna;

where x0mn is the n-th root of J0m(x) = 0: The resonance frequency is

!mnl = c

sx0mna

2+

l

L

2:

The transverse magnetic eld is

H? =ikz

(!=c)2 k2zr?Hz

=ikz

(!=c)2 k2z

@Hz@

e +im

Hze

:

The TE0n1 mode is commonly used in practical applications because the surface current on the end

walls is purely azimuthal (J only as can be seen from Js = nH) and thus a thin circular gap

20

does not signicantly perturb the elds (and resonance frequency). The resonance frequency can

be controlled by varying the length L and a small but nite gap cannot be avoided in moveable

plungers. The TE0n1 is evidently independent of and thus degeneracy problems between cosm

and sinm modes can also be avoided.

9.9 TM Mode in Circular Slow Waveguide

Waveguides used in linear electron accelerators must accommodate TM modes having a phase

velocity close to c. Waveguides with smooth inner walls can only accommodate modes having

phase velocities larger than c and thus cannot be used for this purpose. Modes must be TM

because TE modes have no electric eld in the axial direction needed to accelerate electrons.

Figure 9-6: Circular slow waveguide of radius a with diaphragms of radius b. The spacing betweendiaphragms is much smaller than the axial wavelength z:

Slow waveguides have conductor diaphragms placed periodically along the axis as shown in

Fig.9-7. The purpose of the diaphragms is to increase the capacitance per unit length of the

waveguide which contributes to slowing down the phase velocity of electromagnetic waves. We

consider modes symmetric about the axis, m = 0: The waveguide has a radius a and diaphragms

have holes of radius b: In the region < b; the Helmholtz equation for the axial electric eld is

@2

@2+1

@

@+!c

2 k2z

Ez() = 0; < b: (9.99)

Since we are interested in modes having an axial phase velocity slightly smaller than c; that is,

!

kz. c; (9.100)

21

the quantity (!=c)2 k2z must be negative, and thus solution for Ez() may be assumed to be

Ez() = AI0(k); < b (9.101)

where

k2 = k2z !c

2> 0: (9.102)

In the diaphragm region b < < a; the electric elds lines are essentially straight provided the

axial period of the diaphragms is su¢ ciently smaller than the axial wavelength. We assume that

this condition is met. Then, the wave equation in the region b < < a may be approximated by

@2

@2+1

@

@+!c

2Ez() = 0; b < < a: (9.103)

General solutions are

Ez() = BJ0

!c+ CN0

!c; b < < a: (9.104)

The boundary conditions are:

Ez( = a) = 0; (9.105)

and

Ez and H be continuous at = b: (9.106)

These boundary conditions yield

BJ0

!ca+ CN0

!ca= 0; (9.107)

AI0(kb) = BJ0

!cb+ CN0

!cb; (9.108)

A

kI1(kb) =

c

!

hBJ1

!cb+ CN1

!cbi; (9.109)

where J 00(x) = J1(x); N 00(x) = N1(x); I 00(x) = I1(x) are noted. Eqs. (9.107) through (9.109)

give the following dispersion relation

ck

!

I0(kb)

I1(kb)=J0

!caN0

!cb J0

!cbN0

!ca

J1

!cbN0

!ca J0

!caN1

!cb : (9.110)

22

For the purpose of accelerating highly relativistic electrons, the axial phase velocity !=kz must be

close to c or k ' 0: Then I0(kb) ' 1; I1(kb) ' kb=2; and the LHS of Eq. (9.110) reduces to 2c=!b:

For a given rf frequency ! and the size of the waveguide a; the dispersion relation

2c

!b'J0

!caN0

!cb J0

!cbN0

!ca

J1

!cbN0

!ca J0

!caN1

!cb ; (9.111)

can be solved numerically to determine the aspect ratio a=b of a slow wave circular waveguide.

Fig.9-7 shows the function

f(x) =2a

xbJ0 (x)N0

b

ax

J0

b

ax

N0 (x)

J1

b

ax

N0 (x) J0 (x)N1

b

ax

; x = !a

c; (9.112)

when a=b = 2:5: The rst root occurs at x ' 3:89 and for a given rf frequency !; the outer radius

a can thus be determined.

x 43.93.83.73.63.5

1

0.5

0

­0.5

­1

Figure 9-7: Root of f(x) = 0 when a=b = 2:5:

9.10 Dielectric Waveguides

An optical ber can conne light waves because of total reection at the surface. In contrast to

conductor waveguides, light waves in optical waveguides cannot be pure TE or TM modes. This

is because electromagnetic elds outside, as well as inside, the optical ber must be considered

23

simultaneously. Although the outer elds are evanescent (otherwise waves cannot be conned), the

elds near the surface do a¤ect those inside.

We rst consider a simple case of step change in the index of refraction,

n() =

8<: n; < a;

1; > a:

Such an optical ber is of no practical interest, for bers used in optical communication all have

graded index of refraction with a gradual change with the radius : The axial electric eld Ez(r; t)

satises the wave equations in both regions,

@2

@2+1

@

@+1

2@2

@2+

@2

@z2+ 0"!

2

E<z (r) = 0; < a; (9.113)

@2

@2+1

@

@+1

2@2

@2+

@2

@z2+ 0"0!

2

E>z (r) = 0; > a: (9.114)

The azimuthal dependence may be assumed to be eim and the axial dependence eikzz;

E(r) = E()eim+ikzz:

Then, d2

d2+1

d

d m2

2+ 0"!

2 k2zE<z () = 0; < a; (9.115)

d2

d2+1

d

d m2

2+ 0"0!

2 k2zE>z () = 0; > a; (9.116)

which admit the following bounded solutions,

E<z () = AJm(k1); < a; (9.117)

E>z () = BKm(k2); > a: (9.118)

Here

k1 =p0"!

2 k2z ; k2 =pk2z 0"0!2: (9.119)

Note that the outer eld should be evanescent for the waveguide to conne light waves.

24

Similarly, the axial magnetic eld Hz(r) may be assumed to be

H<z () = CJm(k1); < a; (9.120)

H>z () = DKm(k2); > a: (9.121)

The transverse elds E? and H? can then be calculated by referring to Eqs. (9.41) and (9.42).

The azimuthal components of the elds are

E< () =ik21

kzim

AJm(k1) !0k1CJ 0m(k1)

; (9.122)

E> () =i

k22

kzim

BJm(k2) !0k2DK 0

m(k2)

; (9.123)

H< () =

i

k21

kzim

CJm(k1) + !"k1AJ

0m(k1)

; (9.124)

and

H> () =

ik21

kzim

DKm(k2) + !"0k2BK

0m(k1)

: (9.125)

The continuity of Ez; E;Hz and H yields

AJm(k1a) = BKm(k2a); (9.126)

CJm(k1a) = DKm(k2a); (9.127)

1

k21

kzim

aAJm(k1a) !0k1CJ 0m(k1a)

=1

k22

kzim

aBKm(k2a) !0k2DK 0

m(k2a)

; (9.128)

1

k21

kzim

aCJm(k1a) + !"k1AJ

0m(k1a)

= 1

k22

kzim

aDKm(k2a) + !"0k2BK

0m(k1a)

: (9.129)

Then the determinantal dispersion relation is

1

k21+1

k22

2mkza

2=

!

k1c1

2J 0m(k1a)Jm(k1a)

2+

!

k2c2

2K 0m(k2a)

Km(k2a)

2+!2

k1k2

1

c21+1

c22

J 0m(k1a)K

0m(k2a)

Jm(k1a)Km(k2a): (9.130)

Figure 9-8 shows the dispersion relation, namely, axial wavenumber kz normalized by k0 = !=c

as a function of the normalized frequency k0a = !a=c when m = 1 and n = 1:1: It can be seen that

25

a cuto¤ occurs at !a=c ' 1:386: Fig.9-9 shows the case when m = 2: The cuto¤ frequency increases

to !a=c ' 5:43:

x 302520151050

1.1

1.08

1.06

1.04

1.02

1

Figure 9-8: ckz=! vs. !a=c when n = 1:1; m = 1: The cuto¤ frequency is !ca=c ' 1:386:

x 302520151050

1.1

1.08

1.06

1.04

1.02

1

Figure 9-9: ckz=! vs. !a=c when n = 1:1;m = 2: The cuto¤ frequency is !ca=c ' 5:43:

26

9.11 Graded Index Fibers

In optical bers used in practical communication, the index of refraction is designed to have gradual,

rather than step, variation with the radius. Quadratic variation is commonly employed,

n() = n01 22

; = constant (9.131)

because the electromagnetic elds are then well conned with a Gaussian prole ea22 . The

corresponding permittivity is

"() = "0n20

1 22

2: (9.132)

Since

r ["(r)E] = "(r)r E+E r" = 0; (9.133)

the wave equation

rrE = !20"(r)E; (9.134)

reduces to

r2E+ !20"(r)E+rE r""

= 0: (9.135)

If the change in the permittivity is small, the last term can be ignored in the lowest order approx-

imation, and we obtain a simple wave equation with an inhomogeneous permittivity,

r2E+ !20"(r)E ' 0: (9.136)

In the cylindrical geometry, a cartesian component of the transverse electric eld satises

@2

@2+1

@

@+1

2@2

@2+

@2

@z2+ !20"()

Ei = 0: (9.137)

For weak variation of "();

"() = "0n20(1 22)2 ' "0n

20

1 222

;

27

and axially symmetric mode @=@ = 0; Eq. (9.137) reduces to

d2

d2+1

d

d k2z + k20

1 222

E() = 0; (9.138)

where

k20 = !2"00n20: (9.139)

Assuming

E() = E0ea22 ;

we nd

a2 =1p2k0; (9.140)

k2z = k20 4p2k0: (9.141)

The electric eld is conned with a Gaussian prole in the radial direction. The e-folding radial

distance is

w =1

a=

4p2pk0

; (9.142)

which is called beam radius. A constant beam radius is maintained only for appropriate injection

of light wave at the input end. If not, the beam radius varies with the axial distance accompanied

by periodic focusing and defocusing as intuitively expected from the picture of repeated total

reections.

28

Problems

1. A waveguide has a semicircular cross section of radius a:Determine the lowest cuto¤ frequency.

2. Prove Eqs. (??) and (??).

3. Each corner of a rectangular waveguide having a cross section a b is smoothed as shown

with a; b: Find the change in the cuto¤ frequency of the TE10 mode.

4. The cuto¤ frequency of a ridge waveguide shown may be estimated from equivalent capaci-

tance and inductance,C

l' "0

d

g(F/m); L l = 1

20ab (H m)

!c =1pLC

= c

r2g

abd:

Explain.

5. Thin diaphragms along the wide walls of a rectangular waveguide act as an e¤ective ca-

pacitance, while diaphragms along the narrow walls act as an e¤ective inductance. Give

qualitative explanations.

6. In the magic T shown, TE10 mode entering port 3 (or 4) is equally divided between ports

1 and 2 but does not come out of port 4 (or 3). Explain in terms of cuto¤ of higher order

modes.

7. Two waveguides are coupled through two small holes drilled in the wider walls separated by

a quarter wavelength in each direction as shown. TE10 mode entering port 1 comes out of

ports 2 and 4, but not port 3. Explain.

8. Show that the dispersion relation of TM modes in a coaxial cable having inner and outer radii

a and b is given by

Jm(ka)Nn(kb) Jm(kb)Nm(ka) = 0;

where

k2 =!c

2 k2z :

When a = 1 mm, b = 4 mm, what are cuto¤ frequencies? Consider m = 0 and 1:

29

9. Repeat the preceding problem for TE modes to show that the dispersion relation in this case

is

J 0m(ka)N0n(kb) J 0m(kb)N 0

m(ka) = 0:

10. Wave propagation along a conducting helix with a pitch angle is expected to be characterized

by the axial propagation speed,!

kz' c sin ;

because electromagnetic waves are guided along the spiral path of the helix. Show that an

approximate dispersion relation is given by

!c

2cot2 = k2

I0(ka)K0(ka)

I1(ka)K1(ka);

where

k2 = k2z !c

2:

30

Chapter 10

Electromagnetism and Relativity

10.1 Introduction

If I am moving at a large velocity along a light wave, what propagation velocity should I measure?

This was a question young Einstein asked himself and in 1905, he published a monumental paper

on special relativity which formulated how to transform coordinates, velocity and electromagnetic

elds between two inertial frames. Einstein postulated that:

1. all physical laws remain intact in any inertial frames, and

2. the light velocity c is invariant against inertial coordinate transformation.

Postulate 1 means that, for example, the Maxwells equations in a moving frame remain formally

identical to those in the laboratory frame, provided the spatial coordinates, time, and electro-

magnetic elds are appropriately transformed. Electromagnetic elds appear and disappear as we

change observing frame. For example, if a charge is moving in the laboratory frame, we observe

both electric and magnetic elds. In the frame of the moving charge, the current and magnetic eld

evidently disappear. However, the electromagnetic elds (primed quantities) in the charge frame

are subject to the transformation

E0k = Ek; E0? = (E? +V B?) ; (10.1)

B0k = Bk; B0? = (B? V E?) ; (10.2)

where k and ? are relative to the direction of the velocity V: Since in this example, Bk = 0

and B? = V E? in the laboratory frame, the magmatic eld in the frame of the moving chargeindeed vanishes consistent with our intuition. The pertinent static Maxwells equations are satised

in both frames

laboratory frame: r E =

"0; rB = 0J (10.3)

in the frame of moving charge: r0 E0 = 0

"0; r0 B0 = 0; (J0 = 0): (10.4)

1

Here the primed operators and quantities are those in the moving frame which are subject to the

Lorentz transformation.

As shown in Chapter 8, electromagnetic elds due to a charged particle moving at an arbitrary

velocity can be correctly formulated by the Lienard-Wiechert potentials which had been discovered

prior to the theory of relativity. Electromagnetic disturbances propagate at the velocity c regardless

of the velocity of the charge, just as sound waves emitted by a moving source propagate at a

sound velocity independent of the source velocity. A major di¤erence between sound waves and

electromagnetic waves occurs for stationary source and moving observer. For sound waves, if an

observer is approaching a source at a velocity VO; the apparent sound velocity becomes cs + VO

because both cs and VO are well dened with respect to the medium of sound waves, namely, air. In

electromagnetic waves that can propagate in vacuum, there is no preferred inertial frame to dene

velocities and a moving observer will measure the same propagation velocity c regardless of the

relative velocity between two inertial frames. Of course, the frequency and wavelength are Doppler

shifted but the product = 0 0 = c or the ratio !=k = !0=k0 = c remains invariant.

10.2 CGS-ESU System

In this Chapter, the CGS-ESU (Electro-Static Unit) unit system is used so that electromagnetic

elds E (statvolt/cm ' 300 volt/cm = 3 104 volt/m) and B (gauss = 104 T) have the same

dimensions. In CGS-ESU, the Coulombs law is adopted to connect the mechanical world and

electromagnetic world. (Recall that in SI, the magnetic force is adopted to dene 1 ampere current

which in CGS-USU is 3 109 stat-ampere.) If two equal charges separated by 1 cm exert a force

of 1 dyne (= erg/cm = 107J/102 m = 105 N) on each other, the charge is dened to be 1 ESU

' 3 109 C. The Coulombs law in CGS-ESU system is

Coulombs law: F =q1q2r2

(dyne). (10.5)

The electronic charge is e = 4:8 1010 ESU (= 1:6 1019 C). The potentials and A also

have common dimensions (statvolt) in CGS-ESU. This is particularly convenient in theoretical

electrodynamics because the elds E (polar vector) and B (axial or pseudo vector) are in fact

components of a unied 4 4 eld tensor and the potentials , A form a four vector (; A):The

Maxwells equations in this unit system are:

r E = 4; rE = 1c

@B

@t; (10.6)

r B = 0; rB = 4

cJ+

1

c

@E

@t; (10.7)

and the relationships between the elds and potentials are

E = r 1c

@A

@t; B = rA: (10.8)

2

The wave equations for the potentials are modied asr2 1

c2@2

@t2

= 4; (10.9)

r2 1

c2@2

@t2

A = 4

cJ; (10.10)

subject to the Lorentz gauge,

r A+ 1c

@

@t= 0: (10.11)

Electromagnetic force in the CGS-ESU system is

F (dyne) = e

E+

1

cv B

; f (dyne/cm3) = E+

1

cJB; (10.12)

the Poynting ux is

S =c

4EB; (erg cm2 sec1); (10.13)

the ux of momentum is1

4EB; (dyne cm2); (10.14)

and the momentum density is1

4cEB; (dyne sec cm3): (10.15)

Electromagnetic energy density is

1

8

E2 +B2

; (erg cm3): (10.16)

The vacuum impedance for a plane wave is unity (dimensionless),

B =c

!kE; jBj = jEj : (10.17)

In CGS-ESU system, macroscopic proportional constants inevitably have unfamiliar units. For

example, the capacitance has dimensions of length (cm) as seen from its denition,

[C] =[q]

[]= length. (10.18)

The conductivity relates the electric eld and current density, J = E; and has dimensions of

frequency, sec1; since

[] =[J ]

[E]=[q] cm2 sec1

[q] cm2= sec1 :

3

10.3 Lorentz Transformation

The null result of Michelson-Morleys extensive interference experiments to detect the ether velocity

was explained by Lorentz who assumed that a moving object contracts in the direction of its velocity

by a factor ;

L0 =1

L0 =

q1 2L0: (10.19)

This was followed by the nding by Lorentz and Poincaré that if the spatial coordinates, time and

electromagnetic elds are all transformed according to what is known as Lorentz transformation,

the Maxwells equations remain intact. If a relative velocity V is assumed in the x direction,

the laboratory coordinates (ct; x; y; z) and coordinates in the moving frame (ct0; x0; y0; z0) may be

assumed to be related through a linear transformation,

t0 = 0 (t aV x)x0 = 1(x V t)y0 = y; z0 = z;

provided that the two coordinate systems coincide at t = 0: The invariance of the coordinates

perpendicular to the relative velocity, y = y0 and z = z0; follows from isotropy of space which is

implicitly assumed. For light pulse emitted at t = t0 = 0 and x = x0 = 0 in the positive x direction

(same direction as V ),

x0 = ct0; x = ct;

which yield

c = 1 0

c V1 acV ; (10.20)

while for light pulse emitted in the negative x direction

x0 = ct; x = ct;

or

c = 1 0

c+ V

1 + acV(10.21)

From Eqs. (10.20) and (10.21), we nd

0 = 1 = and a =1

c2: (10.22)

To determine ; consider a light pulse emitted along the y0 axis (x0 = 0; that is, x = V t) in the

moving frame,

y0 = ct0 = c

t 1

c2V x

= c

1 V 2

c2

t:

4

In the laboratory frame, light propagation is tilted due to the relative motion between the two

coordinates, (ct)2 = y2 + (V t)2 or

y =pc2 V 2t:

Since y0 = y; we nd

=1r

1 V 2

c2

: (10.23)

Desired transformation between (x; t) and (x0; t0) is

x0 = (x V t);

t0 =

t V

c2x

;

and the four dimensional coordinates in the laboratory frame (ct; x; y; z) and those in the moving

frame (ct0; x0; y0; z0) are related through266664ct0

x0

y0

z0

377775 =266664

Vc 0 0

Vc 0 0

0 0 1 0

0 0 0 1

377775266664ct

x

y

z

377775 (10.24)

Its inverse transformation can be found by replacing V with V;266664ct

x

y

z

377775 =266664

Vc 0 0

Vc 0 0

0 0 1 0

0 0 0 1

377775266664ct0

x0

y0

z0

377775 (10.25)

Graphically, Lorentz transformation may be visualized as contraction in the (ct; x) plane with a

quasi angle dened by

cosh = ; sinh = ; tanh = ;

as illustrated in Fig. (10-1). Note that the coordinates (ct0; x0) are not orthonormal if those in the

laboratory frame are so chosen. Since

tanh( 1 + 2) =tanh 1 + tanh 21 + tanh 1 tanh 2

=1 + 21 + 12

;

sum of two velocities cannot exceed c:

Time dilation and length contraction can be visualized as follows. A clock stationary at x = 0

in the laboratory frame moves along the vertical ct axis (x = 0) as time elapses. 1 second in the

laboratory frame appears as second in the moving frame. If seen from the moving frame, the

clock is moving and a moving clock ticks slower. If a clock is stationanry at x0 = 0 in the moving

5

Figure 10-1: Graphical representation of Lorentz transformation for the case = 0:5: Hyperboliccurves show (ct)2 x2 = 1 (interval of time-like events) and (ct)2 x2 = 1 (length of space-likeobject).

frame, it travelsalong the t0 axis (x0 = 0). In the laboratory frame, 1 second in the moving frameappears as second. In both cases, a moving clock ticks slower.

Likewise, a stick one meter long in the moving frame is contarcted by a factor if seen from

the laboratory frame. Note that length measurements should be done for a common time in both

frames.

It is convenient to introduce a metric tensor dened by

gij = gij =

2666641 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

377775 (10.26)

A contravariant vector (vector in ordinary sense) xi = (x0; x1; x2; x3) = (ct; x; y; z) can be converted

to a covariant vector through

xi = gijxj = (x0;x1;x2;x3) (10.27)

so that

s2 = xixi: (10.28)

6

Note that we follow Einsteins convention, that is, repeated subscripts and superscripts mean

summation is to be taken,

gijxj =

3Xj=0

gijxj ; (10.29)

xixi = x0x0 + x1x1 + x

2x3 + x4x4: (10.30)

s2 can be either positive or negative. The Lorentz transformation for the coordinates266664ct0

x0

y0

z0

377775 =266664

0 0

0 0

0 0 1 0

0 0 0 1

377775266664ct

x

y

z

377775can be written in the form

x0i = Lijxj ; (10.31)

where Lij is the Lorentz transformation mixed tensor,

Lij =

266664 0 0

0 0

0 0 1 0

0 0 0 1

377775 (10.32)

Its inverse tensor is

Lij1

=

266664 0 0

0 0

0 0 1 0

0 0 0 1

377775 (10.33)

A four dimensional vector that transforms according to the Lorentz transformation law is called

a four-vector. The positionvector xi evidently forms a four vector. For an object moving at a

velocity v in the laboratory frame, the velocity four vector is

vi =dxi

d= (c; vxvy; vz) ; (10.34)

where

d =1

dt =

q1 2dt; (10.35)

is the proper time as measured in the moving frame of the object. This is the celebrated time

dilation e¤ect. The magnitude of the four velocity is constant,

vivi = c2: (10.36)

7

The momentum four vector is dened by

pi =

Ec;p

=

Ec; px; py; pz

; (10.37)

where

E = mc2; (10.38)

is the energy and p is the momentum of mass of a particle having a mass m;

p = mv: (10.39)

The magnitude of this four vector is also invariant,

pipi =

E

c

2 p2 = 2(1 2)(mc)2 = (mc)2: (10.40)

Example 1 How are the velocity and acceleration transformed?

For a relative velocity V between two inertial frames, spatial coordinates are transformed as

r0k = (rk Vt); r0? = r?; (10.41)

where k and ? indicate components parallel and perpendicular to the velocity V: Since the time istransformed as

dt0 =

1 v V

c2

dt; (10.42)

the velocity in the direction of the relative velocity is transformed as

v0k =dr0kdt0

=vkV1 vV

c2

; (10.43)

and the normal component as

v0? =dr0?dt0

=v?

1 vV

c2

: (10.44)

For example, if two particles are approaching each with a velocity v relative to the laboratory frame,

the relative velocity in the frame of either particle is

2v

1 + (v=c)2;

which cannot exceed c:Note that 2v itself can exceed c:However, a relative velocity between two

objects is meaningful only if it is measured in rest frame of either object. 2v pertains to an

observer in the laboratory frame in which both objects are moving. Therefore, 2v itself is not a

very meaningful velocity. As an example, consider head on collision of two protons. We assume

each proton has a velocity v = 0:9c relative to the laboratory frame. In the rest frame of either

8

proton, the two protons approach with a relative velocity

V =2v

1 + (v=c)2= 0:9945c;

and the kinetic energy available for nuclear interaction is

( 1)mc2 = 8 GeV.

To nd transformation of acceleration, we note the acceleration parallel to the relative velocity

is transformed as

a0k =dv0kdt0

=ak

31 vV

c2

3 ; (10.45)

and the normal component as

a0? =dv0?dt0

=a?

21 vV

c2

2 + v?

21 vV

c2

3 a Vc2 : (10.46)

Note that transformation of the normal acceleration involves parallel acceleration as well.

The inverse transformation for the acceleration is

ak =dvkdt

=a0k

31 + v0V

c2

3 (10.47)

a? =a0?

21 + v0V

c2

2 v0?

21 + v0V

c2

3 a0Vc2 : (10.48)

In an instantaneously rest frame of a particle, v0 = V; and v? = 0: Then

a0k = 31

1 V 2

c2

3ak = 3ak; (10.49)

a0? = 2a?: (10.50)

The current density J and charge density form the following four vector

J i = (c;J) = (c; Jx; Jy; Jz); (10.51)

where

(v) = en(v) = en0q

1vc

2 ; (10.52)

J = en(v)v; (10.53)

9

n(v) =n0q

1vc

2 ; (10.54)

is the density of charged particles corrected for length contraction in the direction of the velocity v

and n0 is the charge density in the rest frame, v = 0: The magnitude of the current four vector is

J iJi = (c0)2 = const. (10.55)

where 0 is the proper charge density in the rest frame of the charge.

In Lorentz gauge, the potentials satisfy the decoupled wave equations,r2 1

c2@2

@t2

= 4; (10.56)

r2 1

c2@2

@t2

A = 4

cJ: (10.57)

Therefore, a resultant four vector potential is

Ai = (;A) = (; Ax; Ay; Az);

which satises the wave equation 1

c2@2

@t2r2

Ai =

4

cJi:

Noting1

c2@2

@t2r2 = @

@xi@

@xi= gii@i@i = @i@i;

the wave equation can readily be Lorentz transformed as

@0j@0jA0i =

4

cJ 0i

since @i@i is Lorentz invariant. The electromagnetic wave equation is thus Lorentz invariant which

guarantees the constancy of the wave propagation velocity c.

10.4 Transformation of Electromagnetic Fields

The electric and magnetic elds, E and B; do not form four vectors. This is due to di¤erent

vectorial nature of the respective elds. The electric eld is a polar vector (or true vector) because

it changes the sign if coordinates are reversed, r! r: In contrast, the magnetic eld

B(r) =1

c

Z(r r0) J(r0)

jr r0j3dV 0;

10

remains unchanged against coordinate inversion since both r r0 and J(r) change sign. The mag-netic eld is an axial vector (or pseudo vector). Rather they are components of an antisymmetric

pseudo tensor,

Bij =

264 0 Bz By

Bz 0 BxBy Bx 0

375The conventional magnetic Lorentz force

fm =1

cJB

now takes a form

f im =1

cJjB

ij ; (i; j = 1; 2; 3)

where

Jj = (Jx;Jy;Jz)

is the covariant current density. Combining the electric eld and magnetic eld into a single eld

tensor

F ij =

2666640 Ex Ey EzEx 0 Bz By

Ey Bz 0 BxEz By Bx 0

377775 ;the conventional electromagnetic force

f = E+1

cJB

can be rewritten as

f i =1

cF ijJj ; i = 1; 2; 3; j = 0; 1; 2; 3:

The component f0

f0 =1

cE J

indicates the work done by the electromagnetic eld. A resultant force four vector is

f j =

1

cE J; f

:

Using the eld tensor in Eq. (), we now reformulate Maxwells equations as follows. Since

E = r 1c

@A

@t= rA0 1

c

@A

@t; B = rA;

the eld tensor can be written as

F ij = @iAj @jAi

11

where @i is the contravariant derivative

@i =@

@xi=

1

c

@

@t; @

@x; @

@y; @

@z

:

For example,

F ii = @iAi @iAi = 0; (no summation here)

F 01 = @0A1 @1A0 = 1

c

@Ax@t

+@

@x= Ex;

F 12 = @1A2 @2A1 = @Ay@x

+@Ax@y

= Bz;

and so on. Di¤erentiating F ij = @iAj @jAi with respect to xi covariantly, we obtain

@iFij = @i@

iAj @i@jAi = @i@iAj @j@iAi:

The rst term in the RHS is

@i@iAj =

1

c2@2

@t2r2

Aj =

4

cJ j ;

while the second term vanishes because of our choice of Lorentz gauge,

@iAi =

1

c

@

@t+r A = 0:

Thus

@iFij =

4

cJ j :

For j = 0; noting J0 = c; we recover Gausslaw,

@iFij = r E = 4:

For j = 1; 2; 3; we also recover j-th component of generalized Amperes law,

rB = 4

cJ+

1

c

@E

@t:

Another identity satised by F ij is

@iF jk + @jF ki + @kF ij = 0;

as can be readily checked by substituting F ij = @iAj @jAi: When i = 0; j = 1; k = 2; Eq. ()

yields

@0F 12 + @1F 20 + @2F 01 = 0;

12

or1

c

@

@t(Bz)

@

@xEy

@

@y(Ex) = 0;

that is, we recover Faradays law

(rE)z = 1

c

@Bz@t

:

Furthermore, for i = 1; j = 2; k = 3; we recover

r B = 0:

In order to nd how the eld tensor F ij is Lorentz transformed, let us consider an arbitrary

contravariant vector Bj dened by

Bj = F ijAi:

After Lorentz transformation, this becomes

B0j = F 0ijA0i;

where

B0j = LjkBk; A0i = L k

i Ak:

Then,

LjkBk = F 0ijL m

i Am

LjkFlkAl = F 0ijL l

i Al

Since Al is arbitrary, we obtain

F 0ijL mi = LjnF

mn:

Multiplying both sides by Lrm and noting

L mi Lrm = ri ;

we nd

F 0ij = LimLjnF

mn = LimFmnLnj :

For example,

F 001 = L0mL1nF

mn

= L00L11F

01 + L01L10F

10

= 2Ex + 22Ex= Ex = F 01;

13

that is, the electric eld parallel to the relative velocity V is invariant. For F 002; we nd

F 002 = L00L22F

02 + L01L22F

12

= (Ey Bz);

and so on. The overall result is

F 0ij =

2666640 Ex (Ey Bz) (Ez + By)Ex 0 (Bz Ey) (By + Ez)

(Ey Bz) (Bz Ey) 0 Bx (Ez + By) (By + Ez) Bx 0

377775For a relative velocity in an arbitrary direction, the electromagnetic elds are transformed according

to

E0k = Ek; E0?= (E?+ B?);B0k = Bk; B0?= (B? E?);

where k and ? indicate components parallel and perpendicular to the relative velocity V:The invariance of Bx; the magnetic eld parallel to V; can be seen from the following observation.

Bx appears only in F 23 = F 32: Since the component F ij transforms similar to the coordinatesxi and xj ; Bx trarnsforms as y and z which are invariant. Therefore, Bx does not change through

Lorentz transformation. Likewise, F 02 = Ey transforms as x0 = ct and x2 = y;

E0y = (Ey Bz);

F 03 = Ez asE0z = (Ez + By);

and so on. F 01 = Ex = F 01 is invariant since the Lorentz transformation corresponds torotation in the (ct; x) plane and F 00 = F 11 = 0; F 01 = F 10 form an antisymmetric tensor.

Example 2 A charge e is moving at a velocity V along the x axis. Find the electric eld and

compare it with the eld expected from the Lienard-Wirchert potentials.

In the frame of the moving charge, the scalar potential and electric eld are

0 =e

r0; E0 =

er0

r03;

where

r0 =px02 + y02 + z02:

The vector potential in the moving frame is zero, A0 = 0: The x component of the electric eld is

14

invariant,

Ex = E0x

=ex0

r03

= e (x V t)

[ 2(x V t)2 + y2 + z2]3=2

= e(1 2) x V t(x V t)2 + (1 2)(y2 + z2)

3=2 :Similarly,

Ey = E0y = e(1 2) y(x V t)2 + (1 2)(y2 + z2)

3=2 ;Ez = E0z = e(1 2) z

(x V t)2 + (1 2)(y2 + z2)3=2 :

Therefore, the electric eld in the laboratory frame is

E = e(1 2) (x V t)ex + y + z(x V t)2 + (1 2)(y2 + z2)

3=2 :Equivalence of this expression to the Coulomb eld emerging from the Lienard-Wiechert potentials,

E = e(1 2) n 3R2

ret;

can be readily proven by noting

n 1 n

ret=

(x V t)ex + y + z(x V t)2 + (1 2)(y2 + z2)

1=2 ;where retmeans the retarded time. Denoting the angle between the x axis and the position

vector R = (x V t)ex + y + z by ; we nd

E =eR

R31 2

(1 2 sin2 )3=2:

The Coulomb eld is radialwith respect to the present location of the charge. At = 0;

Ek =e

R2(1 2) = e

R21

2;

and at = =2;

E? =e

R2 :

In highly relativistic case, the eld is dominated by components perpendicular to the velocity.

15

Angular dependence of the electric eld for the case = 0:9 is shown below in polar plot.

1 2

(1 2 sin2 )3=2

­2 ­1 1 2

­3

­2

­1

1

2

3

Example 3 Show that the quantities E2 B2 and E B are invariant in Lorentz transformation.

Since

E2 = E2k + E2?;

and

E0k = Ek; E0? = (E?+ B?);

we nd

E02 = E2k + 2(E?+ B?)2:

Similarly,

B02 = B2k + 2(B? E?)2:

Then

E02 B02 = E2k B2k +

2(E?+ B?)2 2(B? E?)2

= E2k B2k +

2(1 2)(E2? B2?) + 2 2[E?( B?) +B?( E?)]= E2 B2:

16

For E B;

E0B0 = E0kB0k +E

0? B0?

= EkBk + 2(E?+ B?) (B? E?)

= EkBk + 2(1 2)E? B?

= E B:

Note that

(AB) (CD) = (A C)(B D) (A D)(B C):

The invariance of E B means that if E and B are normal to each other in one reference frame,

they remain so in any other frames. If E or B is zero in one reference frame, in other frames they

are normal to each other. Furthermore, if the elds in the laboratory frame are E and B; there

exists a frame moving at a velocity

V

1 (V=c)2 = cEBE2 +B2

;

wherein electric and magnetic elds are parallel to each other.

10.5 Energy and Momentum Tensor

As shown in Chapter 1 and 3, the electromagnetic force

f = E+1

cJB;

can be expressed as the divergence of the Maxwells stress tensor T ij ;

fi = @jT ij = 1

4@j

1

2(E2 +B2)ij EiEj BiBj

; (i; j = 1; 2; 3):

The time component of the force four vector was

f0 =1

cJ E:

Since

J E = 1

8

@

@t(E2 +B2)r S;

where S is the Poynting vector

S =c

4EB;

f0 can be written as

f0 = 1

8c

@

@t(E2 +B2) 1

4r (EB) ;

17

and a resultant four dimensional Maxwells stress tensor is

T =

26666418 (E

2 +B2) 14 (EB)

x 14 (EB)

y 14 (EB)

z

14 (EB)

x T 11 T 12 T 13

14 (EB)

y T 21(= T 12) T 22 T 23

14 (EB)

z T 31(= T 13) T 32 = (T 23) T 33

377775The quaintly

1

cS =

1

4EB;

is the momentum ux density and thus

1

4cEB;

is the momentum density of electromagnetic elds. The angular momentum density is accordingly

given by1

4cr (EB);

and the total electromagnetic angular momentum is

1

4c

Zr (EB)dV:

In four dimensional form, the angular momentum tensor can thus be dened by

Kij =1

c

Z(xiT

ik xjT ik)dk;

where dk is the area element in the four dimensional (hyper) space having the dimensions of

volume (cm3):

10.6 Relativistic Mechanics

In terms of the velocity four vector

vi = (c;v) = (c; vx; vy; vz);

and the eld tensor

F ij =

2666640 Ex Ey EzEx 0 Bz By

Ey Bz 0 BxEz By Bx 0

377775

18

the force four vector to act on a charged particle having a charge e can be formulated as

F i =e

cF ijvj ;

where

vj = (c;v) = (c;vx;vy;vz);

is the covariant four velocity. F i reduces to

F i = (e E;F);

where

F = e(E+ B);

is the space component of the electromagnetic force. Note that

F = e E;

reecting the fact that the magnetic eld does not do any work on charged particles.

The magnitude of the velocity four vector is constant,

vivi = 2(1 2)c2 = c2:

The corresponding momentum four vector

pi =

Ec;p

;

also has a constant magnitude,

pipi =

Ec

2 p2 = (mc)2;

where

E2 = (cp)2 + (mc2)2;

is the energy of the particle.

The Lagrangian in nonrelativistic mechanics,

L =1

2mv2 + e( A );

can be readily generalized as

L = mc2q1 2 + e( A )

= mc2q1 2 + v (P mv) e;

19

where

P =@L

@v= mv +

e

cA = p+

e

cA;

is the canonical momentum. The equation of motion can be derived from Lagrange equation

d

dt

@L

@v

= rL:

Noting

r(v A) = vrA+Arv + v rA+Ar v= vrA+ v rA;

since the velocity should be xed in carrying out spatial di¤erentiation, we nd

d

dt(p+

e

cA) =

e

c(vrA+ v rA)er:

NotingdA

dt=@A

@t+ vrA; rA = B; E = r 1

c

@A

@t;

we recover the familiar equation of motion for a charged particle,

dp

dt=

d

dt( mv) = e

E+

1

cv B

:

Since the momentum p and energy E are related through

p =v

c2E ;

the acceleration a = dv=dt can be readily found,

a =e

m [E+ B ( E)] : (10.58)

Example 4 Analyze the motion of an electron in the Coulomb eld of a heavy ion carrying acharge Ze:

Since the electric eld is static, the energy of electron is conserved,

E = cpp2 + (mc)2 Ze2

r= E0 (const.) (10.59)

By suitable choice of coordinates, the problem can be made two dimensional so that electron motion

is conned in the plane (r; ): The momentum can be decomposed into

p2 = p2r +

Lr

2; (10.60)

20

where L = rp; the angular momentum, is also conserved. Then,

c

sp2r +

Lr

2+ (mc)2 Ze2

r= E0. (10.61)

In nonrelativistic limit v c, we have

1

2m

sp2r +

Lr

2 Ze2

r= const. (10.62)

In the limit r ! 0; the LHS diverges. (Note that in nonrelativistic limit, the momentum is bounded.)

This means that in nonrelativistic limit, the electron cannot approach the ion indenitely. In

relativistic case, however, the LHS of Eq. () remains nite when r ! 0 provided pr !1:The electron trajectory can be found by di¤erentiating Eq. (10.61) with respect to time,

d

dt(m _r) L2

m

1

r3+Ze2

r2= 0: (10.63)

L = const. means

m r2 _ = L = const.

Then time derivative can be converted into angular derivative through

d

dt=

Lm r2

d

d(10.64)

and Eq. () reduces tod2

d21

r+1 2

1r=

Ze2

L2c2E0; (10.65)

where

=Ze2

cL: (10.66)

When < 1; the solution for r() is quasi-oscillatory,

r() =b

1 + a cos(p1 2)

; (10.67)

where a is an integration constant determined by the initial condition and

b =(cL)

2 (Ze2)2Ze2E0

: (10.68)

Sincep1 2 is in general an irrational number, the oscillation is quasi-periodic without closed

orbits. The electron cannot approach the ion indenitely. If > 1; or Ze2 > L; cos(p1 2) in

Eq. (10.67) becomes cosh(p2 1) and in this case the electron can collapse on the ion.

21

Example 5 Find the power radiated by a charge explicitly in terms of the external electric andmagnetic elds.

In nonrelativistic limit, the radiation power is given by the Larmors formula,

P =2

3

(ea)2

c3: (10.69)

This is still applicable in an instantaneous rest frame of the charged particle,

P =2

3

(ea0)2

c3; (10.70)

where a0 is the acceleration in that frame. According to Example 1, the acceleration in the labora-

tory frame a and that in the instantaneous rest frame of the charge a0 are related through

a0 = 3ak + 2a?; (10.71)

as can be found by choosing v = V: Since

ak =e

m 3Ek; a? =

e

m (E? + B); (10.72)

we nd for the radiation power

P =2

3

(ea0)2

c3

=2

3

e4

m2c3

hE2k +

2(E?+ B)2i

=2

3

e4

m2c3 2(E+ B)2 ( E)2

: (10.73)

If the electric eld is parallel to the velocity and there is no magnetic eld as in linear accelerators,

the power becomes independent of the particle energy,

P =2

3

e4

m2c3E2k :

In the absence of the electric eld, we obtain

P =2

3

e4

m2c3 2( B)2:

22

10.7 Radiation Damping

Consider an electron subject to an acceleration a for a duration T: In nonrelativistic limit, the

electron acquires a kinetic energy of order

Ekin =1

2m(aT )2;

while the amount of energy radiated is

Erad =2

3

(ea)2

c3T:

It is reasonable to conjecture that the radiation energy should not exceed the kinetic energy, Erad <

Ekin for otherwise, one must wonder where the radiation energy comes from. This imposes a limit

on T;

T & e2

mc3= ' 1023 sec,

where is approximately the transit time of light over the classical electron radius,

re =e2

mc2= 2:85 1013 cm.

(To realize such a time scale in cyclotron motion in a magnetic eld, we need an unrealistically

large magnetic eld, B ' 1012 T.) In this time scale, radiation reaction on dynamics of charged

particle is expected to become signicant. However, the uncertainty principle imposes a more strict

constraint,

Et & ~:

If we choose the self energy of the electron for E = mc2; we nd

t & ~mc2

=~ce2

e2

mc3= 137 ;

wheree2

~c=

1

137;

is the ne structure constant. Therefore, in nonrelativistic cases, quantum mechanical e¤ects

become important well before recoil force due to radiation need to be considered.

However, in radiation from a charge undergoing harmonic motion, radiation reaction is well

dened and has been observed experimentally. Recoil force exerted by radiation may be estimated

from energy balance, Z T

0F vdt = 2

3

e2

c3

Z T

0a2dt

=2

3

e2

c3

Z T

0vd2v

dt2dt:

23

Then

F =2

3

e2

c3d2v

dt2;

and the equation of motion of harmonic oscillator is modied as

x ...x + !20x = eE

mei!t:

Solution for x(t) is

x(t) =eE

m

ei!t

!2 + i!3 !20;

which remains bounded at the resonance ! ' !0 due to radiation damping. The shift in the

resonance frequency is given by

! ' 58!30

2:

24

Mathematical Formulae

1. Vector Formulae

Bold characters are vector functions and f is a scalar function.

A (BC) = C (AB) = B (CA)

A (BC) = B(A C)C(A B)

(AB) (CD) = (A C)(B D) (A D)(B C)

r (fA) = rf A+ fr A

r (fA) = rf A+ frA

r(A B) = A (rB) +B (rA) + (B r)A+ (A r)B

r (AB) = B (rA)A (rB)

r (AB) = Ar BBr A+ (B r)A (A r)B

rrf 0

r (rA) 0

r (rA) = r(r A)r2A

r r = 3; r = position vector

r r = 0

r2 1

jr r0j = 4(r r0)

Substantive derivative:df

dt=@f

@t+ (vr)f; v = velocity

Substantive derivative:dA

dt=@A

@t+ (vr)A

Substantive derivative:dv

dt=@v

@t+ (vr)v =@v

@t+1

2rv2 + (r v) v

Gausstheorem:ZVr AdV =

ISA dS

StokestheoremZSrA dS =

ICA dlI

S(rA) dS = 0 for closed surfaceZ

VrAdV =

ISdSA =

ISA dS

1

2. Delta Function Zf(t)(t t0)dt = f(t0)

(at) =1

jaj(t);Zg(t)[f(t)]dt =

g(t0)dfdtt=t0

where f(t0) = 0

(r r0) (3-D delta function)= (x x0)(y y0)(z z0) (Cartesian)

=(r r0)rr0

( 0)sin

( 0) (spherical)

=(r r0)rr0

(cos cos 0)( 0) (spherical)

=( 0)

( 0)(z z0) (cylindrical)

3. Curvilinear Coordinates

Let ui(x; y; z) (i = 1; 2; 3) be a system of curvilinear coordinates. The metric coe¢ cients are

hi =

s@x

@ui

2+

@y

@ui

2+

@z

@ui

2;

and the length segments in each direction are hiduiei (ei unit vector).Area elements

dSi = hjhkdujdukej ekVolume element

dV = h1h2h3du1du2du3

Gradient

rf =3Xi=1

1

hi

@f

@ui

Divergence

r A =1

h1h2h3

@

@u1(h2h3A1) +

@

@u2(h3h1A2) +

@

@u3(h1h2A3)

Curl

rA =

e1h2h3

e2h3h1

e3h1h2

@

@u1

@

@u2

@

@u3

h1A1 h2A2 h3A3

Scalar Laplacian

r2f = r rf = 1

h1h2h3

@

@u1

h2h3h1

@f

@u1

+

@

@u2

h3h1h2

@f

@u2

+

@

@u3

h1h2h3

@f

@u3

2

Vector Laplacian (denition)

r2A r(r A)r (rA)

Spherical Coordinates (r; ; )

Transformation

x = r sin cos

y = r sin sin

z = r cos

Metric coe¢ cientshr = 1; h = r; h = r sin :

Transformation of the unit vectors24 eree

35 =24 sin cos sin sin cos cos cos cos sin sin sin cos 0

3524 exeyez

3524 exeyez

35 =24 sin cos cos cos sinsin sin cos sin coscos sin 0

3524 eree

35For example,

ex = rx = r(r sin cos)= sin coser + cos cose sine

Derivative of the unit vectors

@er@

= e;@er@

= sin e;@e@

= er;@e@

= cos e;@e@

= sin er cos e:

Gradient

rf = @f

@rer +

1

r

@f

@e +

1

r sin

@f

@e

Divergence

r A =1

r2@

@r(r2Ar) +

1

r sin

@

@(sin A) +

1

r sin

@A@

Curl

rA =

err2 sin

er sin

er

@

@r

@

@

@

@

Ar rA r sin A

Scalar Laplacian

r2f = @2f

@r2+2

r

@f

@r+

1

r2 sin

@

@

sin

@f

@

+

1

r2 sin2

@2f

@2

3

Vector Laplacian

r2A r(r A)r (rA)

=

r2Ar

2

r2Ar

2

r2@A@

2 cot r2

A 2

r2 sin

@A@

er

+

r2A

1

r2 sin2 A +

2

r2@Ar@

2 cos

r2 sin2

@A@

e

+

r2A

1

r2 sin2 A +

2

r2 sin

@Ar@

+2 cos

r2 sin2

@A@

e

Note that in non-cartesian coordinates,

(r2A)i 6= r2Ai

Cylindrical Coordinates (; ; z)

Transformation

x = cos

y = sin

z = z

Metric coe¢ cientsh = 1; h = ; hz = 1

Derivatives of the unit vectors

@e@

= e;@e@

= e

Gradient

rf = @f

@e +

1

@f

@e +

@f

@zez

Divergence

r A =1

@

@(A) +

1

@A@

+@Az@z

Curl

rA =

e

eez

@

@

@

@

@

@z

A A Az

Scalar Laplacian

r2f = @2f

@2+1

@f

@+1

2@2f

@2+@2f

@z2

4

Vector Laplacian

r2A r(r A)r (rA)

=

r2A

2

2@A@

1

2A

e

+

r2A +

2

2@A@

1

2A

e +r2Azez

4. Special Functions

Bessel Functions Zm(x) = Jm(x); Nm(x) [or Ym(x) in some books]

J0(x); J1(x)

20151050

1

0.75

0.5

0.25

0

­0.25 x

y

x

y

J0 (x) (solid line) and J1 (x) (dashed line).

Y0(x); Y1(x)

2015105

1

0.5

0

­0.5

­1

x

y

x

y

Y0 (x) (solid line) and Y1 (x) (dashed line).

Di¤erential equation d2

d2+1

d

d+ k2 m

2

2

Zm(k) = 0

WronskianJm(x)N

0m(x) J 0m(x)Nm(x) =

2

x

5

Series representation of Jm(x)

Jm(x) =x2

m 1Xn=0

(1)n(m+ n)!n!

x2

2n; Jm(x) = (1)mJm(x)

Jm (x) 'x2

mfor small x 1

J0 (x) ' 1 14x2 + 1

64x4 for small x 1

J1 (x) ' 12x

116x

3 for small x 1

Generating functions

exp

x

2

t 1

t

=

1Xm=1

Jm(x)tm

eix sin =1X

m=1Jm(x)e

im

n-th root of J0(x) = 0; x0n; and value of J1(x0n)

x0n 2:40483 5:52008 8:65373 11:79153 14:93092 18:07106J1(x0n) 0:51915 0:34026 0:27145 0:23246 0:20655 0:18773

n-th root of J1(x) = 0; x1n; and value of J0(x1n)

x1n 3:83171 7:01559 10:17347 13:32369 16:47063 19:61586J0(x1n) 0:40276 0:30012 0:24970 0:21836 0:19647 0:18006

For small x ( 1);

Jm(x) 'x2

m; Nm(x) '

2

lnx

2+ E

; E = 0:5772 (Eulers constant).

Hankel functions of the rst and second kinds

H(1;2)m (x) = Jm(x) iNm(x)

Asymptotic forms at large x 1

Jm(x) 'r2

xcos

x 2m+ 1

4

Nm(x) 'r2

xsin

x 2m+ 1

4

H(1;2)m (x) '

r2

xexp

ix 2m+ 1

4

Recurrence formulae

Z 00(x) = Z1(x)

Z 0m(x) =1

2[Zm1(x) Zm+1(x)]

Zm(x) =x

2m[Zm+1(x) + Zm1(x)]

6

d

dx[xmZm(x)] = x

mZm1(x)

Integral representations (There are many. A few are listed.)

Jm(x) =1

2

Z 2

0cos(m x sin )d; J0(x) =

2

Z 1

0

cos(xt)p1 t2

dt

N0(x) = 2

Z 1

1

cos(xt)p1 t2

dt

Nm(x) = 2m+1xmp(12 m)

Z 1

1

cosxt

(t2 1)m+(1=2)dt

Integrals Z 1

0Jm(ax)dx =

1

a;

Z 1

0Nm(ax)dx =

1

atan

m2

Z 1

0J0(ax)J0(bx)dx =

2

bK(a=b); K : complete elliptic integral of the rst kind

Z 1

0J0(ax)J1(bx)dx =

8<:1=b; b > a > 01=2b; a = b > 00; a > b > 0

(step function)

Z 1

0xJ1(ax)J1(bx)dx =

(a b)a

; (derivative of the above with respect to a)

In fact for any integer m; Z 1

0xJm(ax)Jm(bx)dx =

(a b)aZ 1

0J+x(ax)Jx(ax)dx = J+(2a); + > 1Z 1

0x1J(ax)dx =

21+

2

a

2 + 1

; : gamma functionZ 1

0

J(ax)J(bx)

x2 y2 dx =i

2J(by)H

(1) (ay)Z 1

0eaxJ0(bx)dx =

1pa2 + b2Z 1

0eaxJ(bx)dx =

(pa2 + b2 a)

bpa2 + b2Z 1

0eaxJ(bx)J(cx)dx =

1

pbcQ 1

2

a2 + b2 + c2

2bc

; Q : Legendre function of the 2nd kindZ 1

0ea

2x2J2(px)dx =

p

2aexp

b2

8a2

I

b2

8a2

Z 1

0ea

2x2x2J0(bx)dx =1

2a2exp

b2

4a2

Z 1

0ea

2x2J(px)J(qx)dx =1

2a2exp

p

2 + q2

4a2

I

pq2a2

7

Z 1

0sin(ax)J(bx)dx =

(1pb2a2 sin

sin1(a=b)

; b > a

bpa2b2(a+

pa2b2) cos

2

; a < b

Z 1

0sin(ax)J0(bx)dx =

8<: 0; b > a1p

a2 b2; a > b

Sum

J0(x) + 21Xn=1

J2n(x) = 1

1Xn=1

J2n(x) = 1

1Xn=0

n2J2n(x) =x2

4

J0(x) + 2

1Xn=1

(1)nJ2n(x) = cosx

1Xn=0

(1)nJ2n+1(x) =1

2sinx

1Xn=0

(2n+ 1)J2n+1(x) =x

2

1Xn=1

1

n2J2n(2nx) =

x2

2

1Xn=1

J2n(2nx) =x2

2(1 x2)1Xn=1

n2J2n(2nx) =x2(1 + x2)

2(1 x2)4

1Xn=1

n2Z x

0J2n(2nt)dt =

x3

6(1 x2)3

Spherical Bessel Functions zl(x) = jl(x); nl(x)

Spherical Bessel functions are elementary functions. Some low order forms are:

j0(x) =sinx

x; j1(x) =

sinx x cosxx2

; j2(x) =(3 x2) sinx 3x cosx

x3

n0(x) = cosx

x; n1(x) =

cosx+ x sinx

x2; n2(x) =

(3 x2) cosx+ 3x sinxx3

Denition

jl(x) r

2xJl+ 1

2(x); nl(x)

r

2xNl+ 1

2(x)

8

h(1;2)l (x) = jl(x) inl(x)

Di¤erential equation d2

dr2+2

r

d

dr+ k2 l(l + 1)

r2

zl(kr) = 0

Wronskianjl(x)n

0l(x) j0l(x)nl(x) =

1

x2

For x 1;

jl(x) 'xl

(2l + 1)!!; nl(x) '

(2l 1)!!xl+1

For x 1;

jl(x) '1

xcos

x l + 1

2

; nl(x) '

1

xsin

x l + 1

2

h(1)l (x) ' (i)

l+1 eix

x; h

(2)l (x) ' i

l+1 eix

x

Recurrence formulae(2l + 1)zl(x) = x[zl1(x) zl+1(x)]

(2l + 1)dzl(x)

dx= lzl1(x) (l + 1)zl+1(x)

d

dx[xl+1zl(x)] = x

l+1zl1(x)

d

dx[xlzl(x)] = xlzl+1(x)

Modied Bessel Functions Im(x); Km(x)

I0(x); I1(x)

53.752.51.250

25

20

15

10

5

0

x

y

x

y

I0(x) (solid line) and I1 (x) (dashed line).

K0(x);K1(x)

9

32.521.510.50

10

7.5

5

2.5

0

x

y

x

y

K0 (x) (solid line) and K1 (x) (dashed line).

Denition

Im(x) = eim=2Jm(ix)

=x2

m 1X0

(x=2)2n

n!(m+ n)!

Km(x) = (1)m+1Im(x) + ln

x

2

+(1)m2

1Xk0

(x=2)m+2k

k!(m+ k)!

"kXn=1

1

n+

k+mXn=1

1

n

#

+1

2

m1Xr=0

(1)r (m r 1)!r!

x2

2rmDi¤erential equation

d2

d2+1

d

d+ k2 +

m2

2

Im(k)Km(k)

= 0

WronskianI 0m(x)Km(x) Im(x)K 0

m(x) =1

x

Series representation of Im(x)

Im(x) =x2

m 1Xn=0

1

m!(m+ n)!

x2

2nFor x 1

I0(x) ' 1 +1

4x2 +

I1(x) 'x

2+1

16x3 +

Recurrence formulae

Im1(x) Im+1(x) =2m

xIm(x); Im1(x) Im+1(x) = 2I 0m(x)

Km1(x)Km+1(x) = 2m

xKm(x); Km1(x) +Km+1(x) = 2K 0

m(x)

10

I 00(x) = I1(x); K 00(x) = K1(x)

Integral representation

K0(x) =

Z 1

0

tJ0(xt)

t2 + 1dt =

Z 1

0

cosxtpt2 + 1

dt

K1(x) = K 00(x) =

Z 1

0

t2J1(xt)

1 + t2dt

K1=3

2x3=2

33=2

!=

3px

Z 1

0cos(t3 + xt)dt; (Airys integral)

Legendre Functions Pml (x); Qml (x)

Di¤erential equation(1 x2) d

2

dx2 2x d

dx+ l(l + 1) m2

1 x2

Pml (x)Qml (x)

= 0

Pl(x) =1

2ll!

dl

dxl(x2 1)l (Rodriguesformula)

Ql(x) =1

2Pl(x) ln

1 + x

1 x Wl1(x); x real and jxj 1

Ql(z) =1

2Pl(z) ln

z + 1

z 1 Wl1(z) for general complex z

W1(x) = 0; W0(x) = 1; W1(x) =3

2x; W2(x) =

5

2x2 2

3;

Special values (m = 0)Pl(1) = 1; Pl(1) = (1)l

Pl(0) = 0 for odd l; Pl(0) = (1)l=2(l 1)!!l!!

for even l

P 0l (1) =l(l + 1)

2; P 0l (0) = (l + 1)Pl+1(0)

Denition of Pml (x); Qml (x) in terms of Pl(x); Ql(x) (x real, jxj 1)

Pml (x) =1 x2

m=2 dmdxm

Pl(x); Qml (x) =1 x2

m=2 dmdxm

Ql(x)

For general complex argument z

Pml (z) =z2 1

m=2 dmdzm

Pl(z); Qml (z) =z2 1

m=2 dmdzm

Ql(z)

Orthogonality of Pml (x) Z 1

1

Pml (x)Pm0l (x)

1 x2 dx =1

m

(l +m)!

(l m)!mm0

Z 1

1Pml (x)P

ml0 (x)dx =

2

2l + 1

(l +m)!

(l m)!ll0

11

Wronskianm = 0; Pl(x)Q

0l(x) P 0l (x)Ql(x) =

1

1 x2

Pml (x)dQml (x)

dx dP

ml (x)

dxQml (x) =

22m

1 x2[(l +m+ 1)=2][(l +m)=2 + 1]

[(l m+ 1)=2][(l m)=2 + 1]

=1

1 x2(l +m)!

(l m)! ; 0 m l

Generating function

(cos + i sin cos)l = Pl(cos ) + 2lX

m=1

iml!

(l +m)!Pml (cos ) cos(m)

Low order forms of Pml (x); (x = cos )m = 0

P0(x) = 1; P1(x) = x; P2(x) =3x2 12

; P3(x) =5x3 3x

2;

l = 1; m = 1

P 11 (x) =p1 x2 = sin

l = 2; m = 1; 2

P 12 (x) = 3p1 x2x = 3 sin cos = 3

2sin 2

P 22 (x) = 3(1 x2) = 3 sin2 = 3

2(1 cos 2)

l = 3; m = 1; 2; 3

P 13 (x) =3

2

p1 x2(5x2 1) = 3

8(sin + 5 sin 3)

P 23 (x) = 15(1 x2)x = 15

4(cos cos 3)

P 33 (x) = 15(1 x2)3=2 = 15 sin3 = 15

4(2 sin 2 + 7 sin 4)

Qml (x) (associated Legendre function of the second kind)

Special case z = i sinh

Q0(i sinh ) = i cot1(sinh ); Q1(i sinh ) = sinh cot1(sinh ) 1

Q11(i sinh ) = cosh cot1(sinh ) tanh

Spherical harmonic function Ylm(; )

Positive m : Ylm(; ) = (1)ms2l + 1

4

(l m)!(l +m)!

Pml (cos )eim for 0 m l

12

Negative m : Yl;m(; ) = (1)mY lm(; ) for l m 0

General form

Ylm(; ) = (1)m+jmj2

s2l + 1

4

(l jmj)!(l + jmj)!P

jmjl (cos )eim; for l m l

Y00(; ) =1p4

Y10(; ) =

r3

4cos

Y1;1(; ) = r3

8sin ei

Y20(; ) =

r5

4

1

2(3 cos2 1)

Y2;1(; ) = r15

8sin cos ei

Y2;2(; ) =1

4

r15

2sin2 ei2

Orthogonality of Ylm(; )IYlm(; )Y

l0m0(; )d = ll0mm0 ; d = sin dd

Toroidal functions Pl 12(cosh ); Ql 1

2(cosh ) satisfy

d2

d2+ coth

d

d l2 + 1

4m2cosech2

F () = 0

Integral representations

Pml 1

2

(cosh ) =(1)m(2l 1)!!

2m+1(2l 2m 1)!!

Z

0

cosm

(cosh + cos sinh )l+12

d

Qml 1

2

(cosh ) =(1)m(2l 1)!!

2m+1(2l 2m 1)!!

Z 1

0

coshmt

(cosh + cosh t sinh )l+12

dt

Gamma Function

Denition

(z) =

Z 1

0ettz1dt

Properties

(z + 1) = z (z) ; (z) (1 z) =

sin (z); z + 1

2

12 z

=

cos (z)

If z is a positive integer, z = n; (n+ 1) = n!

13

Special values

(1) = 1; 12

=p;

n+ 1

2

=(2n 1)!!

2np

(x)

52.50­2.5­5

5

2.5

0

­2.5

­5

x

y

x

y

Gamma function (x) :

Elliptic Integrals K (k2) and E (k2)

Complete Elliptic Integrals of the First Kind K(k2) and Second Kind Ek2

Denitions

Kk2=

Z =2

0

1p1 k2 sin2

d; Ek2=

Z =2

0

p1 k2 sin2 d; 0 k2 1

Special values

K(0) = E (0) =

2; lim"!0

K (1 ") = ln4p"

; E (1) = 1

K (x) =

Z =2

0

1p1 x sin2

d

E (x) =

Z =2

0

p1 x sin2 d

14

10.750.50.250

3

2.5

2

1.5

1

0.5

0

x

y

x

y

Kk2(thick line) and E

k2(thin line).

Relationship between K and E

E(k2) =1 k2

ddk

kK

k2

Integrals that can be reduced to the elliptic integralsZ =2

0

sin2 p1 k2 sin2

d =1

k2Kk2 E

k2

Z =2

0

cos2 p1 k2 sin2

d =1

k2Ek2 (1 k2)K

k2

Z =2

0

1

(1 k2 sin2 )3=2d =

1

1 k2Ek2=d

dk

kK

k2

Z =2

0

sin2

(1 k2 sin2 )3=2d =

1

k2 (1 k2)Ek2 (1 k2)K

k2

Z =2

0

cos2

(1 k2 sin2 )3=2d =

1

k2Kk2 (1 k2)E

k2

Series Expansion of Elementary Functions1

1 + x= 1 x+ x2 x3 + ; jxj < 1

ex = 1 + x+1

2!x2 +

1

3!x3 +

cosx = 1 1

2!x2 +

1

4!x4

sinx = x 1

3!x3 +

1

5!x5

tanx = x+1

3x3 +

2

15x5 +

17

315x7 +

15

coshx = 1 +1

2!x2 +

1

4!x4+

sinhx = x+1

3!x3 +

1

5!x5 +

tanhx = x 13x3 +

2

15x5 17

315x7 +

ln(1 + x) = x 12x2 +

1

3x3 ; jxj < 1

Innite Products

1Yn=1

1 +

x2

n2

=sinhx

x

1Yn=1

1 x

2

n2

=sinx

x

1Yn=1

1 +

x2

(2n 1)2

= cosh

x

2

1Yn=1

1 x2

(2n 1)2

= cos

x

2

1Yn=1

1 +

x2

(a 2n)2

=coshx cos a1 cos a

1Yn=1

1 x2

(a 2n)2

=cosx cos a1 cos a

16

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