electrodynamics flooved

8/13/2019 electrodynamics flooved

1/140

-8

-6

-4

-2

0

2

4

6

8

-8 -6 -4 -2 0 2 4 6 8

rotating dipole: t = 2


2/140


3/140

Contents

1 Maxwells electrodynamics 11.1 Dynamical variables of electromagnetism 11.2 Maxwells equations and the Lorentz force 31.3 Conservation of charge 41.4 Conservation of energy 41.5 Conservation of momentum 6

1.6 Junction conditions 71.7 Potentials 101.8 Greens function for Poissons equation 121.9 Greens function for the wave equation 141.10 Summary 161.11 Problems 17

2 Electrostatics 212.1 Equations of electrostatics 212.2 Point charge 212.3 Dipole 222.4 Boundary-value problems: Greens theorem 242.5 Laplaces equation in spherical coordinates 28

2.6 Greens function in the absence of boundaries 302.7 Dirichlet Greens function for a spherical inner boundary 342.8 Point charge outside a grounded, spherical conductor 352.9 Ring of charge outside a grounded, spherical conductor 382.10 Multipole expansion of the electric field 402.11 Multipolar fields 422.12 Problems 45

3 Magnetostatics 473.1 Equations of magnetostatics 473.2 Circular current loop 473.3 Spinning charged sphere 493.4 Multipole expansion of the magnetic field 51

3.5 Problems 53

4 Electromagnetic waves in matter 554.1 Macroscopic form of Maxwells equations 55

4.1.1 Microscopic and macroscopic quantities 554.1.2 Macroscopic smoothing 564.1.3 Macroscopic averaging of the charge density 574.1.4 Macroscopic averaging of the current density 604.1.5 Summary macroscopic Maxwell equations 61

4.2 Maxwells equations in the frequency domain 624.3 Dielectric constant 63

i


4/140

ii Contents

4.4 Propagation of plane, monochromatic waves 654.5 Propagation of wave packets 68

4.5.1 Description of wave packets 684.5.2 Propagation without dispersion 694.5.3 Propagation with dispersion 694.5.4 Gaussian wave packet 71

4.6 Problems 73

5 Electromagnetic radiation from slowly moving sources 755.1 Equations of electrodynamics 755.2 Plane waves 765.3 Spherical waves the oscillating dipole 78

5.3.1 Plane versus spherical waves 785.3.2 Potentials of an oscillating dipole 795.3.3 Wave-zone fields of an oscillating dipole 815.3.4 Poynting vector of an oscillating dipole 835.3.5 Summary oscillating dipole 84

5.4 Electric dipole radiation 84

5.4.1 Slow-motion approximation; near and wave zones 855.4.2 Scalar potential 855.4.3 Vector potential 875.4.4 Wave-zone fields 885.4.5 Energy radiated 905.4.6 Summary electric dipole radiation 90

5.5 Centre-fed linear antenna 915.6 Classical atom 945.7 Magnetic-dipole and electric-quadrupole radiation 95

5.7.1 Wave-zone fields 965.7.2 Charge-conservation identities 985.7.3 Vector potential in the wave zone 995.7.4 Radiated power (magnetic-dipole radiation) 100

5.7.5 Radiated power (electric-quadrupole radiation) 1015.7.6 Total radiated power 1025.7.7 Angular integrations 103

5.8 Pulsar spin-down 1035.9 Problems 105

6 Electrodynamics of point charges 1096.1 Lorentz transformations 1096.2 Fields of a uniformly moving charge 1126.3 Fields of an arbitrarily moving charge 114

6.3.1 Potentials 1156.3.2 Fields 1166.3.3 Uniform motion 120

6.3.4 Summary 1216.4 Radiation from an accelerated charge 122

6.4.1 Angular profile of radiated power 1226.4.2 Slow motion: Larmor formula 1236.4.3 Linear motion 1246.4.4 Circular motion 1266.4.5 Synchrotron radiation 1276.4.6 Extremely relativistic motion 131

6.5 Radiation reaction 1326.6 Problems 135


5/140

Chapter 1

Maxwells electrodynamics

1.1 Dynamical variables of electromagnetism

The classical theory of the electromagnetic field, as formulated by Maxwell, involvesthe vector fields

E(t,x) electric field at position x and time t (1.1.1)and

B(t,x) magnetic field at position x and time t. (1.1.2)We have two vectors at each position of space and at each moment of time. Thedynamical system is therefore much more complicated than in mechanics, in whichthere is afinite numberof degrees of freedom. Here the number of degrees of freedomis infinite.

The electric and magnetic fields are produced by charges and currents. In aclassical theory these are best described in terms of a fluid picture in which thecharge and current distributions are imagined to be continuous (and not made ofpointlike charge carriers). Although this is not a true picture of reality, these con-

tinuous distributions fit naturally within a classical treatment of electrodynamics.This will be our point of view here, but we shall see that the formalism is robustenough to also allow for a description in terms of point particles.

An element of charge is a macroscopically small (but microscopically large)portion of matter that contains a net charge. An element of charge is locatedat position x and has a volume dV . It moves with a velocityv that depends ontime and on position. The volume of a charge element must be sufficiently largethat it contains a macroscopic number of elementary charges, but sufficiently smallthat the density of charge within the volume is uniform to a high degree of accuracy.

In the mathematical description, an element of charge at position x is idealized asthe pointx itself. The quantities that describe the charge and current distributionsare

(t,x) density of charge at position x and time t, (1.1.3)v(t,x) velocity of an element of charge at positionx and time t,(1.1.4)j(t,x) current density at positionx and time t. (1.1.5)

We shall now establish the important relation

j= v. (1.1.6)

The current density j is defined by the statement

j da current crossing an element of area da, (1.1.7)

1


6/140

2 Maxwells electrodynamics

(v t)cos

Figure 1.1: Point charges hitting a small surface.

so that j is the current per unit area. The current flowing across a surface Sis thenS

j da.

On the other hand, current is defined as the quantity of charge crossing a surfaceper unit time. Let us then calculate the current associated with a distribution ofcharge with density and velocity v.

Figure 1.1 shows a number of charged particles approaching with speed va smallsurface of area a; there is an angle between the direction of the velocity vectorv and the normal nto the surface. The particles within the dashed box will all hit

the surface within a time t. The volume of this box is (vt cos )a= v nta.The current crossing the surface is then calculated as

current = (charge crossing the surface)/t

= (total charge)

volume of box

total volume

1

t

= (density)(volume of box)/t

= (v nta)/t= v (na)= v a.

Comparing this with the expression given previously, current = j

x, we find that

indeed,v is the current density.As was mentioned previously, the fluid description still allows for the existence

of point charges. For these, the density is zero everywhere except at the positionof a charge, where it is infinite. This situation can be described by a -function.Suppose that we have a point charge qat a position r. Its charge density can bewritten as

(x) = q(x r),where(x) (x)(y)(z) is a three-dimensional-function. If the charge is movingwith velocity v , then

j(x) = qv(x r)


7/140

1.2 Maxwells equations and the Lorentz force 3

is the current density.More generally, let us have a collection of point charges qA at positions rA(t),

moving with velocities vA(t) = drA/dt. Then the charge and current densities ofthe charge distribution are given by

(t,x) = A

qA

x rA(t) (1.1.8)and

j(t,x) =A

qAvA(t)x rA(t)

. (1.1.9)

EachqA is the integral of(t,x) over a volumeVA that encloses this charge but noother:

qA=

VA

(t,x) d3x, (1.1.10)

whered3x dxdydz. Thetotal chargeof the distribution is obtained by integratingthe density over all space:

Q (t,x) d3x= A

qA. (1.1.11)

1.2 Maxwells equations and the Lorentz force

The four Maxwell equations determine the electromagnetic field once the chargeand current distributions are specified. They are given by

E = 10

, (1.2.1)

B = 0, (1.2.2)

E =

B

t

, (1.2.3)

B = 0j+00 Et

. (1.2.4)

Here, 0 and 0 are constants, and = (x, y, z) is the gradient operatorof vector calculus (with the obvious notationxf = f/x for any function f).The Maxwell equations state that the electric field is produced by charges andtime-varying magnetic fields, while the magnetic field is produced by currents andtime-varying electric fields. Maxwells equations can also be presented in integralform, by invoking the Gauss and Stokes theorems of vector calculus.

The Lorentz-force law determines the motion of the charges once the electro-magnetic field is specified. Let

f(t,x)

force density at position x and time t, (1.2.5)

where the force density is defined to be the net force acting on an element of chargeat x divided by the volume of this charge element. The statement of the Lorentz-force law is then

f=E+ j B= E+ v B. (1.2.6)The net force Facting on a volume Vof the charge distribution is the integral ofthe force density over this volume:

F(t, V) =

V

f(t,x) d3x. (1.2.7)


8/140


For a single point charge we have = q(xr),j = qv(xr), and the total forcebecomes

F =qE(r) + v B(r).

Here the fields are evaluated at the charges position. This is the usual expressionfor the Lorentz force, but the definition of Eq. (1.2.6) is more general.

Taken together, the Maxwell equations and the Lorentz-force law determine thebehaviour of the charge and current distributions, and the evolution of the electricand magnetic fields. Those five equations summarize the complete theoryof classicalelectrodynamics. Every conceivable phenomenon involving electromagnetism can bepredicted from them.

1.3 Conservation of charge

One of the most fundamental consequences of Maxwells equations is that charge islocally conserved: charge can move around but it cannot be destroyed nor created.

Consider a volume V bounded by a two-dimensional surface S. Charge conser-vation means that the rate of decrease of charge within V must be equal to the

total current flowing across S:

ddt

V

d3x=

S

j da. (1.3.1)

Local charge conservation means that this statement must be true for any volumeV, however small or large. We may use Gauss theorem to turn Eq. (1.3.1) into adifferential statement. For any smooth vector field u withinV we have

V

ud3x=S

u da.

The right-hand side of Eq. (1.3.1) can thus be written as

V j d3x, while the

left-hand side is equal to V(/t) d3x. Equality of the two sides for arbitraryvolumes V implies

t+ j = 0. (1.3.2)

This is the differential statement of local charge conservation, and we would like toprove that this comes as a consequence of Maxwells equations.

To establish this we first differentiate Eq. (1.2.1) with respect to time to obtain

t =0 E

t,

where we have exchanged the order with which we take derivatives of the electricfield. If we now use Eq. (1.2.4) to eliminate the electric field, we get

t

= 10

( B) j.

The first term on the right-hand side vanishes identically (the divergence of a curl isalways zero) and we arrive at Eq. (1.3.2). We have therefore established that localcharge conservation is a consequence of Maxwells equations.

1.4 Conservation of energy

[The material presented in this section is also covered in Sec. 6.7 of Jacksons text.]


9/140

1.4 Conservation of energy 5

Other conservation statements follow from Maxwells equations and the Lorentz-force law. In this section we formulate and derive a statement of energy conser-vation. In the next section we will consider the conservation of linear momentum.It is also possible to prove conservation of angular momentum, but we shall notpursue this here.

Conservation of energy is one of the most fundamental principle of physics, andMaxwells electrodynamics must be compatible with it. A statement of local energyconservation can be patterned after our previous statement of charge conservation.Let

(t,x) electromagnetic field energy density, (1.4.1)S(t,x) da field energy crossing an element of

surfaceda per unit time. (1.4.2)

So is analogous to charge density, and S(which is known as Poyntings vector),is analogous to current density.

If electromagnetic field energy were locally conserved, we would write the state-ment

ddt

V

d3x=S

S da,

which is analogous to Eq. (1.3.1). But we should not expect field energy to beconserved, and this statement is not correct. The reason is that the field does workon the charge distribution, and this takes energy away from the field. This energygoes to the charge distribution, and total energy is conserved. A correct statementof energy conservation would therefore be (field energy leaving Vper unit time) =(field energy crossingSper unit time) + (work done on charges within V per unittime). To figure out what the work term should be, consider an element of chargewithin V. It moves with a velocity v and the net force acting on it is

fdV =(E+ v B) dV.

While the element undergoes a displacement dx, the force does a quantity of workequal to f dxdV . The work done per unit time is then f v dV = E v dV =j EdV. Integrating this over Vgives the total work done on the charges containedin V, per unit time. The correct statement of energy conservation must thereforehave the form

ddt

V

d3x=

S

S da +V

j Ed3x. (1.4.3)

In differential form, this is

t+ S= j E. (1.4.4)

This last equation must be derivable from Maxwells equations, which have notyet been involved. And indeed, the derivation should provide expressions for thequantities and S. What we have at this stage is an educated guess for a correctstatement of energy conservation, but Eq. (1.4.4) has not yet been derived nor thethe quantities and S properly defined in terms of field variables. Now the hardwork begins.

We start with the right-hand side of Eq. (1.4.4) and eliminate j in favour of fieldvariables using Eq. (1.2.4). This gives

j E= 10

E ( B) +0E Et

.


10/140


We use a vector-calculus identity to replace E(B) with B(E)(EB),and in this we replace EwithB/t, using Eq. (1.2.3). All of this gives us

j E = 10

B Bt

+0E Et

+ 1

0 (E B)

=

t1

2 0E2

+

1

20B2

+ 10E B.We have arrived at an equation of the same form as Eq. (1.4.4), and this shows thatwe do indeed have energy conservation as a consequence of Maxwells equations andthe Lorentz-force law.

The calculation also gives us definitions for the fields energy density,

12

0E2 +

1

20B2, (1.4.5)

and for the fields energy flux (Poynting) vector,

S 10

E B. (1.4.6)

It is an important consequence of Maxwells theory that the electromagnetic fieldcarries its own energy.

1.5 Conservation of momentum

[The material presented in this section is also covered in Sec. 6.7 of Jacksons text.]We should expect a statement of momentum conservation to take a form similar

to our previous statement of energy conservation. In that case we had a scalarquantity representing the density of energy, a vectorial quantity Sa representingthe flux of energy, and the conservation statement took the form of

t

+

aSa = work term.

We have introduced an explicit component notation for vectors, and summationover a repeated index is understood.

For momentum conservation we will need a vectorial quantity a to representthe density of momentum, and a tensorial quantity Tab to represent the flux ofmomentum (one index for the momentum component, another index for the fluxdirection). So let

a(t,x) density ofa-component of field momentum, (1.5.1)Tab(t,x) dab a-component of field momentum crossing

an element of surface daper unit time. (1.5.2)

The minus sign in front ofTab is introduced by convention.

The statement of momentum conservation is (field momentum leavingVper unittime) = (field momentum crossing Sper unit time) + (rate at which momentum iscommunicated to the charges). Because a rate of change of momentum is a force,we have the integral statement

ddt

V

a d3x=

S

Tab dab+

V

fa d3x, (1.5.3)

or the equivalent differential statement

at

+ bTab = fa= Ea+ (j B)a. (1.5.4)


11/140

1.6 Junction conditions 7

We now would like to derive this from Maxwells equations, and discover the iden-tities of the vector a and the tensorTab.

We first use Eqs. (1.2.1) and (1.2.4) to eliminate and j from Eq. (1.5.4):

f = (0 E)E+ 1

0 B 0 E

t B= 0( E)E+ 1

0( B)B + 1

0( B) B

0 t

(E B) +0E Bt

,

where to go from the first to the second line we have inserted a term involving B = 0 and allowed the time derivative to operate on E B. The next step isto use Eq. (1.2.3) to replace B/t with E. Collecting terms, this gives

f= 0 t

(EB) + 0

( E)EE (E)+ 10

( B)BB (B).

To proceed we invoke the vector-calculus identity 12(u

u) = (u

)u+u

(

u),

where u stands for any vector field. We use this to clean up the quantities withinsquare brackets. For example,

( E)E E ( E) = ( E)E+ (E )E 12(E E).

In components, the right-hand side reads

(bEb)Ea+ (Ebb)Ea 12aE2 = EabEb+EbbEa 1

2abbE2

= b

EaEb 12

abE2

.

Doing the same for the bracketed terms involving the magnetic field, we arrive at

fa = 0 t

(E B)a+0b

EaEb 12

abE2

+

1

0b

BaBb 1

2abB

2

.

This has the form of Eq. (1.5.4) and we conclude that momentum conservation doesindeed follow from Maxwells equations and the Lorentz-force law.

Our calculation also tells us that the fields momentum density is given by

a 0(E B)a, (1.5.5)

and that the fields momentum flux tensor is

Tab = 0EaEb 1

2abE

2

+ 1

0BaBb 1

2abB

2

. (1.5.6)Notice that the momentum density is proportional to the Poynting vector: =00S. This means that the fields momentum points in the same direction as theflow of energy, and that these quantities are related by a constant factor of00.

1.6 Junction conditions

[The material presented in this section is also covered in Sec. I.5 of Jacksons text.]Maxwells equations determine how the electric and magnetic fields must be

joined at an interface. The interface might be just a mathematical boundary (in


12/140


which case nothing special should happen), or it might support a surface layerof charge and/or current. Our task in this section is to formulate these junctionconditions.

To begin, we keep things simple by supposing that the interface is the x-y planeatz = 0 a nice, flat surface. If the interface supports a charge distribution, thenthe charge density must have the form

(t,x) = +(t,x)(z) +(t,x)(z) +(t,x,y)(z), (1.6.1)

where + is the charge density above the interface (in the region z > 0), thecharge density below the interface (in the regionz 0 and 0 otherwise, and the Dirac -function, which is such that

f(z)(z) dz =

f(0) for any smooth function f(z). These distributions (also known as generalizedfunctions) are related by the identities

d

dz(z) = (z),

d

dz(z) = (z). (1.6.2)

As for any distributional identity, these relations can be established by integratingboth sides against a test function f(z). (Test functions are required to be smoothand to fall off sufficiently rapidly as z .) For example,

d(z)

dz f(z) dz =

(z)df(z)

dz dz

= 0

df

= f(0),

and this shows that d(z)/dz is indeed distributionally equal to(z). You can showsimilarly that f(z)(z) = f(0)(z) and f(z)(z) = f(0)(z)

f(0)(z) are valid

distributional identities (a prime indicates differentiation with respect to z).If the interface at z = 0 also supports a current distribution, then the current

density must have the form

j(t,x) = j+(t,x)(z) +j(t,x)(z) + K(t,x,y)(z), (1.6.3)

where j+ is the current density above the interface, j the current density belowthe interface, andKis the surface density of current (charge per unit time and unitlength) supported by the interface.

We now would like to determine how EandB behave across the interface. Wecan express the fields as

E(t,x) = E+(t,x)(z) + E(t,x)(

z),

(1.6.4)

B(t,x) = B+(t,x)(z) + B(t,x)(z),

in an obvious notation; for example, E+ is the electric field above the interface. Wewill see that these fields are solutions to Maxwells equations with sources given byEqs. (1.6.1) and (1.6.3). To prove this we need simply substitute Eq. (1.6.4) intoEqs. (1.2.1)(1.2.4). Differentiating the coefficients of(z) is straightforward, butwe must also differentiate the step functions. For this we use Eq. (1.6.2) and write

(z) = (z)z,


13/140

1.6 Junction conditions 9

where z is a unit vector that points in the z direction.A straightforward calculation gives

E = E+(z) + E(z) + z E+ E(z), E = E+(z) + E(z) +z E+ E(z), B = B+(z) + B(z) +z B+ B(z), B = B+(z) + B(z) + z B+ B(z).

Substituting Eq. (1.6.1) and the expression for E into the first of Maxwellsequations, E= /0, reveals that E is produced by and that the surfacecharge distribution creates a discontinuity in the normal component of the electricfield:

z E+ Ez=0 = 10 . (1.6.5)Substituting the expression for Binto the second of Maxwells equations, B =0, reveals thatB satisfies this equation on both sides of the interface and that thenormal component of the magnetic field must be continuous:

z B+ Bz=0 = 0. (1.6.6)Substituting Eq. (1.6.4) and the expression for Einto the third of Maxwellsequations, E= B/t, reveals that Eand Bsatisfy this equation on bothsides of the interface and that the tangential (x and y) components of the electricmust also be continuous:

z E+ Ez=0 = 0. (1.6.7)Finally, substituting Eqs. (1.6.3), (1.6.4), and the expression for B into thefourth of Maxwells equations, B = 0j +00E/t, reveals that B isproduced byjand that the surface current creates a discontinuity in the tangential

components of the magnetic field:

z B+ Bz=0 = 0K. (1.6.8)Equations (1.6.5)(1.6.8) tell us how the fields E and B are to be joined at aplanar interface.

It is not difficult to generalize this discussion to an interface of arbitrary shape.All that is required is to change the argument of the step and functions from zto s(x), where s is the distance from the interface in the direction normal to theinterface; this is positive above the interface and negative below the interface. Thens= n, the interfaces unit normal, replaces z in the preceding equations. IfE+,B+ are the fields just above the interface, and E, B the fields just below theinterface, then the general junction conditions are

n E+ E = 10

, (1.6.9)

n B+ B = 0, (1.6.10)n E+ E = 0, (1.6.11)n B+ B = 0K. (1.6.12)

Here, is the surface charge density on the general interface, and Kis the surfacecurrent density. It is understood that the normal vector points from the minus sideto the plus side of the interface.


14/140


1.7 Potentials

[The material presented in this section is also covered in Secs. 6.2 and 6.3 of Jack-sons text.]

The Maxwell equation B= 0 and the mathematical identity (A) = 0imply that the magnetic field can be expressed as the curl of a vector potentialA(t,x):

B= A. (1.7.1)Suppose that B is expressed in this form. Then the other inhomogeneous Maxwellequation, E+B/t = 0, can be cast in the form

E+

A

t

= 0.

This, together with the mathematical identity () = 0, imply that E+A/tcan be expressed as the divergence of a scalar potential(t,x):

E= At

; (1.7.2)

the minus sign in front of is conventional.Introducing the scalar and vector potentials eliminates two of the four Maxwell

equations. The remaining two equations will give us equations to be satisfied by thepotentials. Solving these is often much simpler than solving the original equations.An issue that arises is whether the scalar and vector potentials are unique: WhileEand B can both be obtained uniquely from and A, is the converse true? Theanswer is in the negative the potentials are not unique.

Consider the following gauge transformationof the potentials:

new = ft

, (1.7.3)

A Anew= A + f, (1.7.4)

in which f(t,x) is an arbitrary function of space and time. We wish to show thatthis transformation leaves the fields unchanged:

Enew = E, Bnew= B . (1.7.5)

This implies that the potentials are not unique: they can be redefined at will by agauge transformation. The new electric field is given by

Enew = Anewt

new

= t

A + f

f

t

= E tf+

f

t;

the last two terms cancel out and we have established the invariance of the electricfield. On the other hand, the new magnetic field is given by

Bnew = Anew= A + f= B + (f);

the last term is identically zero and we have established the invariance of the mag-netic field. We will use the gauge invarianceof the electromagnetic field to simplifythe equations to be satisfied by the scalar and vector potentials.


15/140

1.7 Potentials 11

To derive these we start with Eq. (1.2.1) and cast it in the form

1

0= E=

A

t

=

t A 2,

or

10= 00

2

t2 2 t A +00 t

, (1.7.6)

where, for reasons that will become clear, we have added and subtracted a term00

2/t2. Similarly, we re-express Eq. (1.2.4) in terms of the potentials:

0j = B 00 Et

= ( A) 00 t

A

t

= ( A) 2A +00 2A

t2 +00

t,

where we have used a vector-calculus identity to eliminate

(

A). We have

obtained

0j= 002A

t2 2A +

A +00

t

. (1.7.7)

Equations (1.7.6) and (1.7.7) govern the behaviour of the potentials; they are equiv-alent to the two Maxwell equations that remain after imposing Eqs. (1.7.1) and(1.7.2).

These equations would be much simplified if we could demand that the potentialssatisfy the supplementary condition

A +00 t

= 0. (1.7.8)

This is known as the Lorenz gauge condition, and as we shall see below, it can

always be imposed by redefining the potentials according to Eqs. (1.7.3) and (1.7.4)with a specific choice of function f(t,x). When Eq. (1.7.8) holds we observe thatthe equations for and A decouple from one another, and that they both take theform of a wave equation:

1c2

2

t2+ 2

(t,x) = 1

0(t,x), (1.7.9)

1c2

2

t2+ 2

A(t,x) = 0j(t,x), (1.7.10)

where

c 100

(1.7.11)

is the speed with which the waves propagate; this is numerically equal to the speedof light in vacuum.

To see that the Lorenz gauge condition can always be imposed, imagine thatwe are given potentials old and Aold that do not satisfy the gauge condition. Weknow that we can transform them according to old = old f/t andAold A= Aold +f, and we ask whether it is possible to find a function f(t,x)such that the new potentials and A will satisfy Eq. (1.7.8). The answer is in theaffirmative: If Eq. (1.7.8) is true then

0 = A +00 t


16/140


= Aold+ f +00 t

old f

t

= Aold+00 oldt

00 2f

t2 + 2f,

and we see that the Lorenz gauge condition is satisfied iff(t,x) is a solution to the

wave equation 1

c22

t2+ 2

f= Aold 00 old

t .

Such an equation always admits a solution, and we conclude that the Lorenz gaugecondition can alwaysbe imposed. (Notice that we do not need to solve the waveequation for f; all we need to know is that solutions exist.)

To summarize, we have found that the original set of four Maxwell equations forthe fields Eand Bcan be reduced to the two wave equations (1.7.9) and (1.7.10) forthe potentials and A, supplemented by the Lorenz gauge condition of Eq. (1.7.8).Once solutions to these equations have been found, the fields can be constructedwith Eqs. (1.7.1) and (1.7.2). Introducing the potentials has therefore dramaticallyreduced the complexity of the equations, and correspondingly increased the ease offinding solutions.

In time-independent situations, the fields and potentials no longer depend ont,and the foregoing equations reduce to

2(x) = 10

(x), (1.7.12)

2A(x) = 0j(x), (1.7.13)as well as E = and B = A. The Lorenz gauge condition reduces to A = 0 (also known as the Coulomb gauge condition), and the potentials nowsatisfy Poissons equation.

1.8 Greens function for Poissons equation[The material presented in this section is also covered in Sec. 1.7 of Jacksons text.]

We have seen in the preceding section that the task of solving Maxwells equa-tions can be reduced to the simpler task of solving two wave equations. For time-independent situations, the wave equation becomes Poissons equation. In this andthe next section we will develop some of the mathematical tools needed to solvethese equations.

We begin in this section with the mathematical problem of solving a genericPoisson equation of the form

2(x) = 4f(x), (1.8.1)where(x) is the potential, f(x) a prescribed source function, and where a factor

of 4 was inserted for later convenience.To construct the general solution to this equation we shall first find a Greens

functionG(x,x) that satisfies a specialized form of Poissons equation:

2G(x,x) = 4(x x), (1.8.2)where (xx) is a three-dimensional Dirac -function; the source point x isarbitrary. It is easy to see that if we have such a Greens function at our disposal,then the general solution to Eq. (1.8.1) can be expressed as

(x) = 0(x) +

G(x,x)f(x) d3x, (1.8.3)


17/140

1.8 Greens function for Poissons equation 13

where0(x) is a solution to the inhomogeneous (Laplace) equation,

20(x) = 0. (1.8.4)

This assertion is proved by substituting Eq. (1.8.3) into the left-hand side of Eq. (1.8.1),and using Eqs. (1.8.2) and (1.8.4) to show that the result is indeed equal to

4f(x).

In Eq. (1.8.3), the role of the integral is to account for the source term inEq. (1.8.1). The role of0(x) is to enforce boundary conditionsthat we might wishto impose on the potential (x). In the absence of such boundary conditions,0can simply be set equal to zero. Methods to solve boundary-value problems will beintroduced in the next chapter.

To construct the Greens function we first argue that G(x,x) can depend onlyon the difference xx; this follows from the fact that the source term depends onlyon x x, and the requirement that the Greens function must be invariant undera translation of the coordinate system. We then express G(x,x) as the Fouriertransform of a function G(k):

G(x,x) = 1

(2)3 G(k)eik(xx

) d3k, (1.8.5)

wherek is a vector in reciprocal space; this expression incorporates our assumptionthat G(x,x) depends only on xx. Recalling that the three-dimensional -function can be represented as

(x x) = 1(2)3

eik(xx

) d3k, (1.8.6)

we see that Eq. (1.8.2) implies k2G(k) = 4, so that Eq. (1.8.6) becomes

G(x,x) = 4

(2)3

eik(xx

)

k2 d3k. (1.8.7)

We must now evaluate this integral.To do this it is convenient to switch to spherical coordinates in reciprocal space,defining a radius k and angles and by the relations kx = k sin cos , ky =k sin sin , and kz = k cos . In these coordinates the volume element is d

3k =k2 sin dkdd. For convenience we orient the coordinate system so that the kzaxis points in the direction ofRx x. We then have k R= kR cos , whereR |R| is the length of the vector R. All this gives

G(x,x) = 1

22

0

dk

0

d

20

d eikR cos sin .

Integration over d is trivial, and the integral can easily be evaluated by using= cos as an integration variable:

0

eikR cos sin d=

1

1

eikR d=2sin kR

kR .

We now have

G(x,x) = 2

0

sin kR

kR dk=

2

R

0

sin

d,

where we have switched to = kR. The remaining integral can easily be evaluatedby contour integration (or simply by looking it up in tables of integrals); it is equalto/2.


18/140


Our final result is therefore

G(x,x) = 1

|x x| ; (1.8.8)

the Greens function for Poissons equation is simply the reciprocal of the distance

between x and x

. The general solution to Poissons equation is then

(x) = 0(x) +

f(x)

|x x| d3x. (1.8.9)

The meaning of this equation is that apart from the term 0, the potential at x isbuilt by summing over source contributions from all relevant points x and dividingby the distance between x and x. Notice that in Eq. (1.8.8) we have recovered afamiliar result: the potential of a point charge at x goes like 1/R.

1.9 Greens function for the wave equation

[The material presented in this section is also covered in Sec. 6.4 of Jacksons text.]

We now turn to the mathematical problem of solving the wave equation,

(t,x) = 4f(t,x), (1.9.1)for a time-dependent potential produced by a prescribed source f; we have intro-duced the wave, or dAlembertian, differential operator

1c2

2

t2+ 2. (1.9.2)

For this purpose we seek a Greens function G(t,x; t,x) that satisfies

G(t,x; t,x) = 4(t t)(x x). (1.9.3)In terms of this the general solution to Eq. (1.9.2) can be expressed as

(t,x) = 0(t,x) +

G(t,x; t,x)f(t,x) dtd3x, (1.9.4)

where0(t,x) is a solution to the homogeneous wave equation,

0(t,x) = 0. (1.9.5)

That the potential of Eq. (1.9.4) does indeed solve Eq. (1.9.1) can be verified bydirect substitution.

To construct the Greens function we Fourier transform it with respect to time,

G(t,x; t,x) = 1

2 G(;x,x)ei(tt

) d, (1.9.6)

and we represent the time -function as

(t t) = 12

ei(tt

) d. (1.9.7)

Substituting these expressions into Eq. (1.9.3) yields2 + (/c)2G(;x,x) = 4(x x), (1.9.8)which is a generalized form of Eq. (1.8.2). From this comparison we learn thatG(0;x,x) = 1/|x x|.


19/140

1.9 Greens function for the wave equation 15

At this stage we might proceed as in Sec. 1.8 and Fourier transform G(;x,x)with respect to the spatial variables. This would eventually lead to

G(;x,x) = 4

(2)3

eik(xx

)

k2 (/c)2 d3k,

which is a generalized form of Eq. (1.8.7). This integral, however, is not defined,because of the singularity at k2 = (/c)2. We shall therefore reject this method ofsolution. (There are ways of regularizing the integral so as to obtain a meaningfulanswer. One method involves deforming the contour of the integration so as toavoid the poles. This is described in Sec. 12.11 of Jacksons text.)

We can anticipate that for = 0, G will be of the form

G(;x,x) =g(, |x x|)

|x x| , (1.9.9)

with g representing a function that stays nonsingular when the second argument,R |x x|, approaches zero. That G should depend on the spatial variablesthrough R only can be motivated on the grounds that three-dimensional space isboth homogeneous (so that G can only depend on the vector R

x

x) and

isotropic (so that only the length of the vector matters, and not its direction). ThatG should behave as 1/RwhenR is small is justified by the following discussion.

Take Eq. (1.9.8) and integrate both sides over a sphere of small radius centeredat x. Since2G= G, we can use Gauss theorem to get

R=

G Rda+ (/c)2

R


20/140


21/140

1.11 Problems 17

and 1

c22

t2+ 2

A(t,x) = 0j(t,x) (1.10.2)

when the Lorenz gauge condition

A +

1

c2

t = 0 (1.10.3)

is imposed. The speed of propagation of the waves, c = 1/

00, is numericallyequal to the speed of light in vacuum.

The retarded solutions to the wave equations are

(t,x) = 0(t,x) + 1

40

t |x x|/c,x|x x| d

3x (1.10.4)

and

A(t,x) = A0(t,x) +04

j

t |x x|/c,x|x x| d

3x. (1.10.5)

In static situations, the time dependence of the sources and potentials can bedropped.

Once the scalar and vector potentials have been obtained, the electric and mag-netic fields are recovered by straightforward differential operations:

E= At

, B= A. (1.10.6)This efficient reformulation of the equations of electrodynamics is completely equiva-lent to the original presentation of Maxwells equations. It will be the starting pointof most of our subsequent investigations.

1.11 Problems

1. We have seen that the charge density of a point charge q located at r(t) isgiven by

(x) = qx r(t),

and that its current density is

j(x) = qv(t)x r(t),

where v = dr/dt is the charges velocity. Prove that and j satisfy thestatement of local charge conservation,

t+ j = 0.

2. In this problem we explore the formulation of electrodynamics in theCoulombgauge, an alternative to the Lorenz gauge adopted in the text. The Coulombgauge is especially useful in the formulation of a quantum theory of electro-

dynamics.a) Prove that the Coulomb gauge condition, A= 0, can always be imposed

on the vector potential. Then show that in this gauge, the scalar potentialsatisfies Poissons equation, 2 = /0, so that it can be expressed as

(t,x) = 1

40

(t,x)

|x x| d3x.

Show also that the vector potential satisfies the wave equation

A= 0j+00t

.


22/140


b) Define the longitudinal current by

jl(t,x) = 14

j(t,x)|x x| d

3x,

and show that this satisfies

j l = 0. Then prove that the wave

equation for the vector potential can be rewritten as A= 0jt, wherejt j jl is the transverse current.

c) Prove that the transverse current can also be defined by

jt(t,x) = 1

4

j(t,x)

|x x| d3x

,

and that it satisfies jt= 0.d) The labels longitudinal and transverse can be made more meaningful

by formulating the results of part b) and c) in reciprocal space instead ofordinary space. For this purpose, introduce the Fourier transform j(t,k)of the current density, such that

j(t,x) = 1

(2)3

j(t,k)eikx d3k.

Then prove that the Fourier transform of the longitudinal current is

jl(t,k) =k j(t,k)k,

where k = k/|k| is a unit vector aligned with the reciprocal positionvector k; thus, the longitudinal current has a component in the directionofk only. Similarly, prove that the Fourier transform of the transversecurrent is

jt(t,k) = k j(t,k) k;thus, the transverse current has components in the directions orthogo-nal to k only. For a concrete illustration, assume that k points in thedirection of thez axis. Then show that jl= jzz andjt = jxx +jyy.

3. Prove that

G(x, x) = 1

2ikeik|xx

|

is a solution to d2

dx2+k2

G(x, x) = (x x),

where k is a constant. This shows thatG(x, x) is a Greens function for theinhomogeneous Helmholtz equation in one dimension, (d2/dx2 +k2)(x) =f(x), where is the potential and f the source.

4. The differential equationd2x

dt2 +2x= f(t)

governs the motion of a simple harmonic oscillator of unit mass and naturalfrequency driven by an arbitrary external force f(t). It is supposed thatthe force starts acting at t= 0, and that prior to t= 0 the oscillator was atrest, so that x(0) = x(0) = 0, with an overdot indicating differentiation withrespect to t.


23/140

1.11 Problems 19

Find the retarded Greens functionG(t, t) for this differential equation, whichmust be a solution to

d2G

dt2 +2G= (t t).

Then find the solution x(t) to the differential equation (in the form of an

integral) that satisfies the stated initial conditions.

Check your results by verifying that iff(t) = (t)cos(t), so that the oscillatoris driven at resonance, then x(t) = (2)1t sin t.

5. In this problem (adapted from Jacksons problem 6.1) we construct solutionsto the inhomogeneous wave equation = 4fin a few simple situations.A particular solution to this equation is

(t,x) =

G(t,x; t,x)f(t,x) dtd3x,

where

G(t,x; t,x

) = (t

t

|x

x

|/c)

|x x|is the retarded Greens function.

a) Take the source of the wave to be a momentary point source, describedby f(t,x) = (x)(y)(z)(t). Show that the solution to the waveequation in this case is given by

(t ,x,y,z) = (t r/c)

r ,

where r

x2 +y2 +z2. The wave is therefore a concentrated wave-front expanding outward at the speed of light.

b) Now take the source of the wave to be a line source described by f(t,x) =(x)(y)(t). (Notice that this also describes a point source in a spaceof two dimensions.) Show that the solution to the wave equation in thiscase is given by

(t,x,y) =2c(t /c)

(ct)2 2 ,

where

x2 +y2 and(s) is the Heaviside step function. Notice thatthe wave is no longer well localized, but that the wavefront still travelsoutward at the speed of light.

c) Finally, take the source to be a sheet source described by f(t,x) =

(x

)(t

). (Notice that this also describes a point source in a spaceof a single dimension.) Show that the solution to the wave equation inthis case is given by

(t, x) = 2c(t |x|/c).

Notice that apart from a sharp cutoff at|x| =ct, the wave is now com-pletely uniform.

6. In the quantum version of Maxwells theory, the electromagnetic interactionis mediated by a massless photon. In this problem we consider a modified


24/140


classical theory of electromagnetism that would lead, upon quantization, to amassivephoton. It is based on the following set of field equations:

E+2 = 10

,

B = 0,

E = Bt

,

B +2A = 0j+00 Et

,

where is a new constant related to the photon mass m by = mc/, andwhere the fieldsEand B are related in the usual way to the potentials andA. The field equations are supplemented by the same Lorentz-force law,

f=E+j B,

as in the original theory.

a) Prove that the modified theory enforces charge conservationprovidedthatthe potentials are linked by

A +00 t

= 0.

The Lorenz gauge condition must therefore always be imposed in themodified theory.

b) Find the modified wave equations that are satisfied by the potentials and A.

c) Verify that in modified electrostatics, the scalar potential outside a pointcharge qat x = 0 is given by the Yukawa form

= q40e

r

r .

d) Prove that the modified theory enforces energy conservation by derivingan equation of the form

t+ S= j E.

Find the new expressions for and S; they should reduce to the oldexpressions when 0. In the original theory the potentials have nodirect physical meaning; is this true also in the modified theory?


25/140

Chapter 2

Electrostatics

2.1 Equations of electrostatics

For time-independent situations, the equations of electromagnetism decouple intoa set of equations for the electric field alone electrostatics and another set for

the magnetic field alone magnetostatics. The equations of magnetostatics willbe considered in Chapter 3. The topic of this chapter is electrostatics.The equations of electrostatics are Poissons equation for the scalar potential,

2(x) = 10

(x), (2.1.1)

and the relation between potential and field,

E= . (2.1.2)

The general solution to Poissons equation is

(x) = 0(x) + 1

40 (x

)

|x x|d3x, (2.1.3)

where 0(x) satisfies Laplaces equation,

20(x) = 0. (2.1.4)

The role of 0(x) is to enforce boundary conditions that we might wish to imposeon the scalar potential.

2.2 Point charge

The simplest situation involves a point charge q located at a fixed point b. Thecharge density is

(x) = q(x b), (2.2.1)and in the absence of boundaries, Eq. (2.1.3) gives

(x) = 1

40

q

|x b| . (2.2.2)

To compute the electric field we first calculate the gradient ofR |x b|, thedistance between the field point x and the charge. We have R2 = (x bx)2 +(y by)2 + (z bz)2 and differentiating both sides with respect to, say, x gives2RR/x = 2(x bx), or R/x = (x bx)/R. Derivatives of R with respect

21


26/140

22 Electrostatics

+q

p

qx

z

y

Figure 2.1: Geometry of a dipole.

to y and z can be computed in a similar way, and we have established the useful

vectorial relation|x b| = x b|x b| . (2.2.3)

Notice that R is a unit vector that points in the direction ofR x b.The calculation of the electric field involves R1 = R2R, or

1

|x b| = x b|x b|3 . (2.2.4)

This gives

E(x) = q

40

x b|x b|3 . (2.2.5)

The electric field goes asq/R2 and points in the direction of the vector R= x b.This well-known result is known as Coulombs law.

2.3 Dipole

Another elementary situation is that of two equal charges, one positive, the othernegative, separated by a distance d 2. We align the charges along the directionof the unit vector p, and place the origin of the coordinate system at the middlepoint; see Fig. 2.1.

The charge density of this distribution is given by

(x) = q(x p) q(x + p), (2.3.1)and the potential is

(x) = q

40

1

|x p| 1

|x + p|

. (2.3.2)

This is an exact expression. We can simplify it if we take the observation point xto be at a large distance away from the dipole,

r |x| . (2.3.3)


27/140

2.3 Dipole 23

We can then approximate

|x p| =

(x p) (x p)=

r2 2 p x +2

= r1 p x/r2 +O(2/r2),

or1

|x p| =1

r

1 p x/r2 +O(2/r2)

.

Keeping terms of order only, the potential becomes

= 1

40

(2q p) xr3

.

The vectorp 2q pis thedipole momentof the charge distribution. In general,for a collection of several charges, this is defined as

p= A

qAxA, (2.3.4)

where xA is the position vector of the charge qA. In the present situation thedefinition impliesp = (+q)( p) + (q)( p) = 2q p, as was stated previously. Thepotential of a dipole is then

(x) = 1

40

p xr3

, (2.3.5)

whenr satisfies Eq. (2.3.3). It should be noticed that here, the potential falls off as1/r2, faster than in the case of a single point charge. This has to do with the factthat here, the charge distribution has a vanishing total charge.

It is instructive to rederive this result by directly expanding the charge density

in powers of. For any functionfof a vector x we have f(x + p) =f(x) + p f(x) +O(2), and we can extend this result to Diracs distribution:

(x p) = (x) p (x) +O(2).

The charge density of Eq. (2.3.1) then becomes (x) = 2q p (x) +O(2), or

(x) = p (x). (2.3.6)

This is the charge density of a point dipole.Substituting this into Eq. (2.1.3) and setting 0 = 0 yields

(x) = pa40

a(x)

|x

x

|d3x,

where summation over the vectorial index a is understood. The standard procedurewhen dealing with derivatives of a -function is to integrate by parts. The integralis then

(x)a1

|x x| d3x =

(x)

(x x)a|x x|3 d

3x = xar3

,

and we have recovered Eq. (2.3.5). Notice that in these manipulations we haveinserted a result analogous to Eq. (2.2.4); the minus sign does not appear becausewe are now differentiating with respect to the primed variables.


28/140

24 Electrostatics

S

V

n

n

in

outS

Figure 2.2: An inner boundary Sin on which is specified, an outer boundarySout on which is specified, and the region V in between that contains a charge

distribution. The outward unit normal to both boundaries is denoted n.

From the potential of Eq. (2.3.5) we can calculate the electric field of a pointdipole. Taking a derivative with respect to x gives

x = 140

x(p x)r3

3(p x)r4

xr

.

We havex(p x) = px and according to Eq. (2.2.3),xr= x/r. So

x = 140

pxr3

3(p x)xr5

,

and a similar calculation can be carried out for the y and z components of.Introducing the unit vector

r= x/r= (x/r,y/r, z/r), (2.3.7)

we find that the electric field is given by

E(x) = 1

40

3(p r)r pr3

. (2.3.8)

This is an exact expression for the electric field of a point dipole, for which thecharge density is given by Eq. (2.3.6). For a physical dipole of finite size, thisexpression is only an approximationof the actual electric field; the approximationis good forr .

2.4 Boundary-value problems: Greens theorem

[The material presented in this section is also covered in Secs. 1.8 and 1.10 ofJacksons text.]

After the warmup exercise of the preceding two sections we are ready to facethe more serious challenge of solving boundary-value problems. A typical situationin electrostatics features a charge distribution (x) between two boundaries onwhich the potential (x) is specified (see Fig. 2.2) the potential is then said tosatisfy Dirichlet boundary conditions. A concrete situation might involve an inner


29/140

2.4 Boundary-value problems: Greens theorem 25

S

V

in

outSda

da

Figure 2.3: The union of the inner boundarySin, the outer boundarySout, and thenarrow channel between them forms a closed surface Sthat encloses the regionV.

boundary Sin that is also a grounded conducting surface (on which = 0) andan outer boundary Sout that is pushed to infinity (so that = 0 there also). Wewould like to derive an expression for (x) in the region between the boundaries;this should account for both the charge density and the boundary data. To arriveat this expression we will need to introduce a new type of Greens function, whichwe will call aDirichlet Greens functionand denoteGD(x,x

); its properties will beidentified along the way. To get there we will make use of Greens identity, whichis essentially an application of Gauss theorem.

Gauss theorem states that for a volume V enclosed by a surface S,

V bd3x= Sb da,

where b is any vector field defined within V and da is an outwardsurface elementon S. The theorem, as stated, says nothing about a region V bounded by twoboundaries Sin and Sout, the situation that interests us. But we can make oursituation fit the formulation of the theorem by digging a narrow channel fromSout to Sin, as depicted in Fig. 2.3. The union ofSin, Sout, and the channel is aclosed surface, and integrating b over both sides of the channel produces a zeroresult because dapoints in opposite directions. We therefore have

V

bd3x=Sout

b da +Sin

b da,

and noting the opposite orientations ofdaand non Sin (see Fig. 2.3), we write this

as V

b d3x =Sout

b nda Sin

b nda

SoutSin

b nda.

It is important to notice that the unit vector npoints outof both Sin andSout.We now choose the vector b to be

b= ,


30/140

26 Electrostatics

where(x) and (x) are two arbitrary functions. We have b= 2 2and Gauss theorem gives

V

2 2 d3x=

SoutSin

nda. (2.4.1)

This is Greens identity. For convenience we write it in the equivalent formV

2 2 d3x =

SoutSin

nda,

where all fields are now expressed in terms of the new variables x.We now pick

(x) (x) scalar potentialand

(x) GD(x,x) Dirichlet Greens function,where the properties of the Dirichlet Greens function will be described in detailbelow. For now we assume that it satisfies the equation

2GD(x,x) = 4(x x), (2.4.2)while the scalar potential satisfies Poissons equation, 2(x) = (x)/0. Mak-ing these substitutions in the equation following Eq. (2.4.1) leads to

4(x) + 10

V

GD(x,x)(x) d3x =

SoutSin

(x)GD(x,x

)

GD(x,x)(x)

nda,

or

(x) = 1

40 V GD(x,x)(x) d3x

14

SoutSin

(x)GD(x,x

) GD(x,x)(x)

nda.

Apart from a remaining simplification, this is the kind of expression we were seeking.The volume integral takes care of the charge distribution within V , and the twosurface integrals take care of the boundary conditions. But there is one problem:While the value of the scalar potential is specified on the boundaries, so that thefunctions (x) are known inside the surface integrals, we are given no informationabout n (x), its normal derivative, which also appears within the surfaceintegrals. It is not clear, therefore, how we might go about evaluating these integrals:the problem does not seem to be well posed.

To sidestep this problem we demand that the Dirichlet Greens function satisfythe condition

GD(x,x) = 0 when x is on the boundaries. (2.4.3)

Then the boundary terms involvingn (x) simply go away, because they aremultiplied by GD(x,x

) which vanishes on the boundaries. With this property wearrive at

(x) = 1

40

V

GD(x,x)(x) d3x

14

SoutSin

(x)n GD(x,x)da. (2.4.4)


31/140

2.4 Boundary-value problems: Greens theorem 27

This is our final expression for the scalar potential in the region V . The problemof finding this potential has been reduced to that of finding a Dirichlet Greensfunction that satisfies Eqs. (2.4.2) and (2.4.3). Notice that this is a specialized formof our original problem: GD(x,x

) is the potential produced by a point charge ofstrength 40 located at x, and its boundary values on Sin andSout are zero. If wecan solve this problem and obtain the Dirichlet Greens function, then Eq. (2.4.4)gives us the means to calculate (x) in very general circumstances.

We already know that GD(x,x) satisfies Eq. (2.4.2) and vanishes when x lies

on Sin and Sout. We now use Greens identity to show that the Dirichlet Greensfunction is symmetric in its arguments:

GD(x,x) = GD(x,x

). (2.4.5)

This implies that it is also a solution to the standard equation satisfied by a Greensfunction,

2GD(x,x) = 4(x x), (2.4.6)which is identical to Eq. (1.8.2). To establish Eq. (2.4.5) we take y to be theintegration variables in Greens identity,

V

2y 2y

d3y =

SoutSin

y y

nday,and we pick(y) GD(x,y) and(y) GD(x,y). Then according to Eq. (2.4.2)we have2y =4(x y) and2y =4(x y). The left-hand side ofGreens identity gives4[GD(x,x) GD(x,x)] after integration over d3y. Theright-hand side, on the other hand, gives zero because both and are zero wheny is on Sin andSout. This produces Eq. (2.4.5).

To summarize, we have found that the Dirichlet Greens function satisfies Eq. (2.4.6),

2GD(x,x) = 4(x x),possesses the symmetry property of Eq. (2.4.5),

GD(x,x) = GD(x,x

),

and satisfies the boundary conditions of Eq. (2.4.3),

GD(x,x) = 0 when x or x is on the boundaries.

Notice that the boundary conditions have been generalized in accordance to Eq. (2.4.5):the Dirichlet Greens function vanishes whenever x or x happens to lie on theboundaries. Once the Dirichlet Greens function has been found, the potentialproduced by a charge distribution (x) in a region V between two boundaries Sinand Sout on which (x) is specified can be obtained by evaluating the integralsof Eq. (2.4.4). The task of finding a suitable Dirichlet Greens function can onlybe completed once the boundaries Sin and Sout are fully specified; the form of the

Greens function will depend on the shapes and locations of these boundaries. Inthe absence of boundaries, GD(x,x) reduces to the Greens function constructedin Sec. 1.8, G(x,x) = |x x|1.

In the following sections we will consider a restricted class of boundary-valueproblems, for which

there is only an inner boundary Sin (Sout is pushed to infinity and does notneed to be considered);

the inner boundary is spherical (Sin is described by the statement|x| = R,whereR is the surfaces radius);


32/140

28 Electrostatics

the inner boundary is the surface of a grounded conductor ( vanishes onSin).

With these restrictions (introduced only for simplicity), Eq. (2.4.4) reduces to

(x) = 1

40|x|>R

GD(x,x)(x) d3x. (2.4.7)

In these situations, the electric field vanishes inside the conductor and there is aninduced distribution of charge on the surface. The surface charge density is givenby Eq. (1.6.9),

= 0E n, (2.4.8)where Eis the electric field just aboveSin. This is given by E= (|x| =R). Toproceed we will need to find a concrete expression for the Dirichlet Greens function.For the spherical boundary considered here, this is done by solving Eq. (2.4.6) inspherical coordinates.

2.5 Laplaces equation in spherical coordinates

[The material presented in this section is also covered in Secs. 3.1, 3.2, and 3.5 ofJacksons text.]

When formulated in spherical coordinates (r,,), Laplaces equation takes theform

2= 1r2

r

r2

r

+

1

r2 sin

sin

+

1

r2 sin2

2

2 = 0. (2.5.1)

To solve this equation we separate the variables according to

(r,,) = R(r)Y(, ), (2.5.2)

and we obtain the decoupled equations

d

dr

r2

dR

dr

= (+ 1)R (2.5.3)

and1

sin

sin

Y

+

1

sin2

2Y

2 = (+ 1)Y, (2.5.4)

where(+ 1) is a separation constant.The solutions to the angular equation (2.5.4) are the spherical harmonics Ym(, ),

which are labeled by the two integers and m. While ranges from zero to infinity,mis limited to the values, + 1, , 1, . Form 0,

Ym(, ) =2+ 1

4( m)!(+m)! Pm (cos )eim, (2.5.5)

wherePm (cos ) are the associated Legendre polynomials. Form


33/140

2.5 Laplaces equation in spherical coordinates 29

Y11 =

3

8 sin ei,

Y10 =

3

4 cos ,

Y22 =

1

415

2 sin2

e2i

,

Y21 =

15

8 sin cos ei,

Y20 = 1

2

5

4(3 cos2 1).

Two fundamental properties of the spherical harmonics are that they are orthonor-mal functions, and that they form a complete setof functions of the angles and. The statement of orthonormality is

Ym(, )Ym(, ) d = mm, (2.5.7)

whered = sin ddis an element of solid angle. The statement of completenessis that anyfunction f(, ) can be represented as a sum over spherical harmonics:

f(, ) ==0

m=

fmYm(, ) (2.5.8)

for some coefficients fm. By virtue of Eq. (2.5.7), these can in fact be calculatedas

fm=

f(, )Ym(, ) d. (2.5.9)

Equation (2.5.8) means that the spherical harmonics form a complete set ofbasisfunctionson the sphere.

It is interesting to see what happens when Eq. (2.5.9) is substituted into Eq. (2.5.8).To avoid confusion we change the variables of integration to and:

f(, ) =

m

Ym(, )

f(, )Ym(

, ) d

=

f(, )

m

Ym(, )Ym(, )

d.

The quantity within the large square brackets is such that when it is multipliedby f(, ) and integrated over the primed angles, it returns f(, ). This musttherefore be a product of two -functions, one for and the other for . Moreprecisely stated,

=0

m=

Ym(, )Ym(, ) =

( )( )sin

, (2.5.10)

where the factor of 1/ sin was inserted to compensate for the factor of sin ind (the -function is enforcing the condition = ). Equation (2.5.10) is knownas the completeness relation for the spherical harmonics. This is analogous to awell-known identity,

12

eikx

1

2eikx

dk= (x x),


34/140

30 Electrostatics

in which the integral over dk replaces the discrete summation over and m; thebasis functions (2)1/2eikx are then analogous to the spherical harmonics.

The solutions to the radial equation (2.5.3) are power laws, R r or Rr(+1). The general solution is

Rm(r) = amr +bmr

(+1), (2.5.11)

wheream andbm are constants.The general solution to Laplaces equation is obtained by combining Eqs. (2.5.2),

(2.5.5), (2.5.11) and summing over all possible values of and m:

(r,,) ==0

m=

amr

+bmr(+1)

Ym(, ). (2.5.12)

The coefficients am and bm are determined by the boundary conditions imposedon(r,,). If, for example, is to be well behaved atr = 0, then bm 0. If, onthe other hand, is to vanish at r =, then am 0. From these observationswe deduce that the only solution to Laplaces equation that is well behaved at the

origin and vanishes at infinity is the trivial solution = 0.

2.6 Greens function in the absence of boundaries

[The material presented in this section is also covered in Secs. 3.6, and 3.9 of Jack-sons text.]

Keeping in mind that our ultimate goal is to obtain the Dirichlet Greens functionGD(x,x) for a spherical inner boundary, in this section we attempt somethingsimpler and solve

2G(x,x) = 4(x x) (2.6.1)in the absence of boundaries. We already know the answer: In Sec. 1.8 we foundthat

G(x,x) = 1|x x| . (2.6.2)

We will obtain an alternative expression for this, one that can be generalized toaccount for the presence of spherical boundaries.

The-function on the right-hand side of Eq. (2.6.1) can be represented as

(x x) = (r r)

r2( )( )

sin

= (r r)

r2

m

Ym(, )Ym(, ), (2.6.3)

where we have involved Eq. (2.5.10). The factor of 1/(r2 sin ) was inserted be-

cause in spherical coordinates, the volume element is d3x= r2 sin drdd, and theJacobian factor ofr2 sin must be compensated for. We then have, for example,

f(x) =

f(x)(x x) d3x

=

f(r,,)

(r r)r2

( )( )sin

r2 sin drdd

=

f(r,,)(r r)( )( ) drdd

= f(r, , ),


35/140

2.6 Greens function in the absence of boundaries 31

as we should.Inspired by Eq. (2.6.3) we expand the Greens function as

G(x,x) =m

gm(r, r)Ym(

, )Ym(, ), (2.6.4)

using the spherical harmonicsYm(, ) for angular functions and gm(r, r

)Y

m(

,

)for radial functions. For convenience we have made the radial function dependonx through the variables r , , and; these are treated as constant parameterswhen substituting G(x,x) into Eq. (2.6.1). Applying the Laplacian operator ofEq. (2.5.1) on the Greens function yields

2G(x,x) =m

1

r2d

dr

r2

dgmdr

(+ 1)

r2 gm

Ym(

, )Ym(, ),

and setting this equal to4(x x) as expressed in Eq. (2.6.3) produces anordinary differential equation for gm(r, r):

d

dr

r2

dgmdr

(+ 1)gm= 4(r r). (2.6.5)

This equation implies that the radial function depends only on , and not on m:gm(r, r

) g(r, r). To simplify the notation we will now omit the label onthe radial function.

To solve Eq. (2.6.5) we first observe that when r =r, the right-hand side of theequation is zero andg is a solution to Eq. (2.5.3). We letg(r, r) be the solution to the right ofr (forr > r ).We then assume that g(r, r) can be obtained by joining the solutions at r = r:

g(r, r) = g(r, r)(r r), (2.6.6)

where(r r) and (r r) are step functions. Differentiating this equation withrespect to r gives

g(r, r) = g(r, r

)(r

r) + g>(r, r) g(r, r) g(r

, r) g(r

, r) gdr

(+ 1)g>

(r r)

+r2

dg>dr

(r, r) dg(r, r) g


36/140


37/140

2.6 Greens function in the absence of boundaries 33

and such integrals typically cannot be evaluated directly: they are just too compli-cated.

Let us consider an example to illustrate the power of the addition theorem. Wewant to calculate the electrostatic potential both inside and outside a sphericaldistribution of charge with density (x) =(r). There are no boundaries in thisproblem, but the distribution is confined to a sphere of radius R. After substitutingEq. (2.6.12) into Eq. (2.1.3) and writingd3x =r2drd, we obtain

(x) = 1

40

m

4

2+ 1Ym(, )

R0

r

(r)r2 dr

Ym(, ) d.

To evaluate the angular integral we multiply and divide by (4 )1/2 Y00(, ): Ym(

, ) d =

4

Ym(

, )Y00(, ) d =

4,0m,0,

where we have involved Eq. (2.5.7). At this stage we happily realize that the infinitesum over and m involves but a single term. We therefore have

(x) = 140

4

4Y00(, ) R

0

r2

r>(r) dr = 1

404

R

0

r2

r>(r) dr,

and as we might have expected from the spherical symmetry of the problem, thepotential depends onr only.

To evaluate the radial integral we must be careful with the meaning of r>max(r, r). Suppose first thatx is outsidethe charge distribution, so that r > R.Thenr > r, r>= r, and we have

4

R0

r2

r>(r) dr =

1

r4

R0

(r)r2 dr =Q

r,

where Q

(x) d3x = 4 R0 (r)r2 dr is the total chargeof the distribution.

In this case we haveout(x) =

1

40

Q

r, (2.6.13)

the same result as in Eq. (2.2.2). Suppose next that x is insidethe charge distri-bution, so that r < R. Then part of the integration covers the interval 0r < rand the remaining part covers r r R. In the first part r is smaller thanr andr> = r; in the second partr

is larger than r and r> = r. So

4

R0

r2

r>(r) dr =

1

r4

r0

(r)r2 dr + 4

Rr

(r)r dr.

In the first term we recognize

q(r) |x|r

(x) d3x = 4 r0

(r)r2 dr, (2.6.14)

the charge enclosed by a sphere of radius r; ifr > R then q(r) = Q. The internalpotential can thus be expressed as

in(x) = 1

40

q(r)

r + 4

Rr

(r)r dr

. (2.6.15)

While the expression (2.6.15) for the internal potential depends on the detailsof the charge distribution the particular form of the function (r) a nicer


38/140

34 Electrostatics

conclusion applies to the electric field. The field is obtained by differentiating inwith respect to r , and

dindr

= 1

40

q

r2+

1

r

dq

dr 4r

.

But since dq/dr = 4r2 according to Eq. (2.6.14), we see that the last two termscancel out. The electric field is then

E(x) = 1

40

q(r)

r2 r, (2.6.16)

both inside and outside the charge distribution; recall that when r > R, q(r) be-comes Q, the total charge. We have recovered the well-known result that for anyspherical distribution of charge, the electric field at a radiusr depends only on thecharge q(r) enclosed by a sphere of that radius; and the field is the same as if allthis charge were concentrated in a point at r = 0.

2.7 Dirichlet Greens function for a spherical

inner boundary

[The material presented in this section is also covered in Sec. 3.9 of Jacksons text.]We now return to the task of finding the Dirichlet Greens function GD(x,x) for

the specific situation described near the end of Sec. 2.4, in which we have a sphericalinner boundary at|x| = R and no outer boundary. It will be a simple matter tomodify the treatment presented in Sec. 2.6 to account for the inner boundary.

As in Eq. (2.6.4), the Dirichlet Greens function can be expanded as

GD(x,x) =

m

g(r, r)Ym(

, )Ym(, ). (2.7.1)

As in Eq. (2.6.6), the radial function can be expressed as

g(r, r) = g(r, r)(r r), (2.7.2)where g< and g> are solutions to Eq. (2.5.3) that must also satisfy the matchingconditions of Eqs. (2.6.7) and (2.6.8),

g>(r, r) gdr

(r, r) dg

r<

R2+1

r+1 is the greater ofr and r (noticethat r rr). We see that g(r, r) vanishes on the inner boundary: Whenr = R it must be that r < r, so that r< = r = R, and the terms within thebrackets vanish; on the other hand, when r = R it must be that r < r, so thatr< = r =R, and the same conclusion applies.

The Dirichlet Greens function is obtained by substituting Eq. (2.7.8) into Eq. (2.7.1):

GD(x,x) =

m

4

2+ 1

1

r+1>

r<

R2+1

r+1R

GD(x,x)(x) d3x, (2.8.1)

and the Dirichlet Greens function was obtained as

GD(x,x) =

m

4

2+ 1

1

r+1>

r<

R2+1

r+1 Rfrom the centre ofthe conducting sphere. The charges position vector is z0 = z0z and its sphericalcoordinates arer0 = z0, 0 = 0, and 0 is undetermined. The charge density is

(x) = q(x z0). (2.8.4)


40/140

36 Electrostatics

Substituting this into Eq. (2.8.1) gives

(x) = q

40GD(x, z0)

= q

40m

4

2+ 1 r

R2+1

r

+1

< r

+1

>Ym(0, 0)Ym(, ),

where r< now stands for the lesser ofr and z0, while r> stands for the greater ofthe two. It is a property of the spherical harmonics that

Ym(0, 0) =

2+ 1

4 m,0,

so that the potential is actually independent of0 and involves a single sum over. It is another property of the spherical harmonics that for m = 0, they reduce toordinary Legendre polynomials,

Y,0(, ) = 2+ 14

P(cos ).

After cleaning up the algebra, the final expression for the potential is

(x) = q

40

=0

r

R2+1/z+10

r+1

P(cos ). (2.8.5)

We see that the potential depends on r and only; the fact that it does not dependon reflects the axial symmetry of the problem.

Before moving on we pause and recall the main properties of the Legendrepolynomials. These are ordinary polynomials of order ; if is even the functionP(x) contains only even terms, while if is odd it contains only odd terms. Forexample,

P0(x) = 1,

P1(x) = x,

P2(x) = 1

2(3x2 1),

P3(x) = 1

2(5x3 3x),

P4(x) = 1

8(35x4 30x2 + 3).

The Legendre polynomials are orthogonal functions,

1

1

P(x)P(x) dx= 2

2+ 1, (2.8.6)

and they possess the special values

P2n(0) = (1)n (2n 1)!!(2n)!!

, P2n+1(0) = 0, P(1) = 1, (2.8.7)

where the double factorial notation means (2n)!! = (2n)(2n 2)(2n 4) (2) or(2n 1)!! = (2n 1)(2n 3)(2n 5) (1).

The potential of Eq. (2.8.5) is expressed as an infinite sum over . We willevaluate this sum in a moment, but for now we calculate the charge density ()


41/140

2.8 Point charge outside a grounded, spherical conductor 37

0

5

10

15

20

25

30

0.5 1 1.5 2 2.5 3

theta

Figure 2.4: Surface charge induced by a point charge above a spherical conductor.Plotted is 4R2/q as a function of for z0/R = 1.3 (upper curve), z0/R = 1.6(middle curve), and z0/R = 1.9 (lower curve). The charge density is peaked at= 0 and its width increases with increasing values ofz0/R: the closer the chargeis to the surface the narrower is the charge distribution.

on the surface of the conductor. For r close to R we haver < z0, r< = r, r> = z0,and Eq. (2.8.5) becomes

(x) = q

40 1

z+10 r R

2+1

r+1 P(cos ).Let f(r) = r R2+1/r+1 be the function ofr that appears inside the brackets.Its derivative is f(r) = r1 + (+ 1)R2+1/r+2 and evaluating this at r = Rgives f(R) = (2+ 1)R1. From Eq. (2.8.3) we then obtain

= q4

(2+ 1)R1

z+10P(cos ),

or

() = q

4R2

=0

(2+ 1)(R/z0)P(cos ), (2.8.8)

whereq

q(R/z0) (2.8.9)

will be shown presently to be the total induced chargeon the surface of the con-ductor. Plots of the charge density are presented in Fig. 2.4 for selected values ofR/z0.

Thatq is truly the total surface charge can be shown by evaluating the integralof the charge density over the surface of the conductor; we should recover q =

da. To evaluate this we let da = R2 d = R2 sin dd, which reduces toda= 2R2 sin d after integration over . Then

da=1

2q

=0

(2+ 1)(R/z0)

0

P(cos )sin d.


42/140

38 Electrostatics

The integral over is equal to11 P(x)P0(x) dx = 2,0 by virtue of Eq. (2.8.6),

and we obtain

da= q, as required.The potential of Eq. (2.8.5) can be decomposed as

(x) = q(x) + q(x), (2.8.10)

whereq(x) =

q

40

=0

r

P(cos ) (2.8.11)

is the potential that would be obtained in the absence of a boundary, and

q(x) = q

40

=0

(z0)

r+1P(cos ) (2.8.12)

is the modification introduced by the boundary condition (r = R) = 0; hereq = q(R/z0) was defined in Eq. (2.8.9) and

z0 R2/z0. (2.8.13)Notice that z 0/R= R/z0 < 1, so that z

0 < R

r.

The sums overcan now be evaluated. For q we already know that Eq. (2.8.11)is equivalent to

q(x) = 1

40

q

|x z0| , (2.8.14)because q is the potential produced by a charge q at a position z0 = z0z in theabsence of boundaries. By analogy, because z0 < r we have that Eq. (2.8.12) isequivalent to

q(x) = 1

40

q

|x z0|, (2.8.15)

wherez0 = z0z. This is the potential of a fictitious image chargeq

at a position z0inside the conductor. While in reality the chargeq is distributed on the surface ofthe conductor, the potential q is the same as if all this charge were concentrated

at z

0.This observation is the basis for the method of images, which consists of findingsolutions for the potential in the presence of conducting surfaces by guessing thecorrect positions of the correct number of image charges. It is a powerful, but tricky,method.

2.9 Ring of charge outside a grounded, spherical

conductor

In this section we consider a uniformly charged ring of radius a > R. If the ring isplaced in the x-y plane, its charge density is described by

(x

) =

q

2a2 (r

a)(

2 ), (2.9.1)

whereq=

(x) d3x is the rings total charge. The -function inr indicates thatthe ring is infinitely thin, while the -function in tells us that it is placed in thex-y plane, as was previously stated.

Substituting Eq. (2.9.1) into Eq. (2.8.1) along with Eq. (2.8.2) gives

(x) = 1

40

q

2a2

m

4

2+ 1Ym(, )

0

r

R2+1

r+1< r+1>

(r a)r2 dr

Ym(, )( 2 ) d,


43/140

2.9 Ring of charge outside a grounded, spherical conductor 39

where d = sin dd, r< = min(r, r), and r> = max(r, r

). To perform the

integration we recall thatYm eim

, so that integrating gives zero unless m= 0.So 2

0

Ym(, ) d = 2Y,0(

)m,0.

Integration over

then gives

2Y,0(2 )m,0 = 2

2+ 1

4 P(0)m,0.

On the other hand, the radial integration gives

a2

r

R2+1

r+1< r+1>

where r< now stands for the lesser of r and a, while r> stands for the greater ofthe two. With these results the potential becomes

(x) =

q

40

4

2+ 1 Y,0() r R2+1

r+1< r+1>

2+ 14 P(0).

Invoking once more the relation between Y,0() and P(cos ), we arrive at

(x) = q

40

=0

P(0)

r

R2+1/a+1

r+1

P(cos ). (2.9.2)

Once more the potential is independent of, and is expressed as an infinite sumover . Once again we will see that the sum over can be evaluated.

We first calculate the charge density on the surface of the conductor. For r closetoR and therefore smaller thana, we haver< = r,r> = a, and Eq. (2.9.2) becomes

(x) = q40

=0

P(0) f(r)a+1 P(cos ),

where the function f(r) = r R2+1/r+1 was previously encountered in Sec. 2.8.We recall that f(R) = (2+ 1)R1, and Eq. (2.8.3) gives

= q4

(2+ 1)P(0)R1

a+1P(cos ),

or

() = q

4R2

=0

(2+ 1)P(0)(R/a)P(cos ), (2.9.3)

whereq q(R/a) (2.9.4)

is the total charge on the conducting surface. Plots of the surface charge densityare presented in Fig. 2.5 for selected values ofR/a.

As in the previous section the potential of Eq. (2.9.2) can be decomposed as

(x) = q(x) + q(x), (2.9.5)

where q is the potential of a ring of total charge qand radius a, while q is thepotential of a fictitious image ring of total charge q and radius a = R2/a. (Youshould work through the details and make sure that these statements are true.)


44/140

40 Electrostatics

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5 3

theta

Figure 2.5: Surface charge induced by a ring of charge outside a spherical conduc-tor. Plotted is 4R2/q as a function of for a/R = 1.3 (upper curve), a/R = 1.6(middle curve), anda/R= 1.9 (lower curve). The charge density is peaked at = 2and its width increases with increasing values ofa/R: the closer the ring is to thesurface the narrower is the charge distribution.

2.10 Multipole expansion of the electric field

[The material presented in this section is also covered in Sec. 4.1 of Jacksons text.]We now leave boundary-value problems behind, and derive the fact that in the

absence of boundaries, the electrostatic potential outside a bounded distribution of

charge can be characterized by a (potentially infinite) number ofmultipole moments.Among them are

Q

(x) d3x total charge monopole moment, (2.10.1)

pa

(x)xa d3x dipole moment vector, (2.10.2)

Qab

(x)

3xaxb r2ab

d3x quadrupole moment tensor.(2.10.3)

Additional multipole moments can be constructed in a similar way, up to an infi-nite number of tensorial indices. But to proceed along this road quickly becomescumbersome, and we shall find a better way of packaging the components of these

multipole-moment tensors. Notice that the quadrupole moment is a symmetric andtracefree tensor: Qba = Qab and Qaa = 0 (summation over the repeated index isunderstood); it therefore possesses 5 independent components.

Consider the complex quantities

qm

(x)rYm(, ) d3x. (2.10.4)

By virtue of Eq. (2.5.6), they satisfy the identity

q,m= (1)mqm. (2.10.5)


45/140

2.10 Multipole expansion of the electric field 41

Because m ranges from to , for a given there are 2+ 1 independent realquantities within the qms. For = 0 we have only one, and we will see that q00 isproportional to the total charge. For= 1 we have three independent quantities,and we will see that theq1ms give the components of the dipole moment vector. For = 2 we have five independent quantities, and we will see that the q2ms give thecomponents of the quadrupole moment tensor. Thus, the quantities of Eq. (2.10.4)give a convenient packaging of the multipole moments, to all orders.

To evaluate Eq. (2.10.4) for = 0 we recall thatY00 = (4)1/2, so that

q00 = 1

4Q. (2.10.6)

We see that indeed, q00 is directly related to the total charge. For = 1 we haverY11 = (3/8)1/2(r sin )ei = (3/8)1/2(x iy) andrY10 = (3/4)1/2r cos =(3/4)1/2z. This gives

q11 =

3

8 px ipy

, q10 =

3

4pz, (2.10.7)

and we see that indeed, the q1ms are directly related to the dipole moment vec-tor. For = 2 we have r2Y22 = (15/32)

1/2(r sin ei)2 = (15/32)1/2(x iy)2,r2Y21 =(15/8)1/2(r sin ei)(r cos ) =(15/8)1/2(xiy)z, and r2Y20 =(5/16)1/2r2(3cos2 1) = (5/16)1/2(3z2 r2). This gives

q22 = 1

12

15

2

Qxx Qyy 2iQxy

,

q21 = 16

15

2

Qxz iQyz

,

q20 = 1

2

5

4Qzz , (2.10.8)

and we see that indeed, the q2ms are directly related to the quadrupole momenttensor. For higher values of it is best to stick with the definition of Eq. (2.10.4).

How are the multipole moments qmrelated to the potential? The answer comesfrom Eq. (2.1.3) in which we set 0(x) = 0 (because we are no longer consideringboundaries) and substitute Eq. (2.6.12):

(x) = 1

40

m

4

2+ 1Ym(, )

(x)

r

Ym(, ) d3x.

Because we are looking at the potential outsidethe charge distribution, r is largerthanr and we can set r = r. This gives

(x) =

1

40m

4

2+ 1

Ym(, )

r+1

(x

)r

Y

m(

,

) d

3

x

.

We recognize the integral over the primed variables as the definition of the qms,and we arrive at

(r,,) = 1

40

=0

m=

4

2+ 1qm

Ym(, )

r+1 . (2.10.9)

This is an expansion of the potential in inverse powers of r, and the multipolemoments play the role of expansion coefficients.


46/140

42 Electrostatics

The leading term in Eq. (2.10.9) comes from the monopole momentq00, or totalcharge Q:

monopole(x) = 1

40

Q

r. (2.10.10)

The next term comes from the dipole moment. Using Eq. (2.10.7) and recalling the

explicit expressions for the spherical harmonics of degree = 1, we have

1m=1

q1,mY1,m= 3

4

pxx+pyy+pzz

and

dipole(x) = 1

40

p xr3

, (2.10.11)

as in Eq. (2.3.5). The next term in the expansion of the electrostatic potential comesfrom the quadrupole moment. Using Eq. (2.10.8) and the explicit expressions forthe spherical harmonics of degree = 2, it is easy to show that

quadrupole(x) = 140

12

Qab xaxbr5

. (2.10.12)

In a situation in which Q= 0, the electrostatic potential far outside the chargedistribution is well approximated by monopole; in this case the potential is charac-terized by a single quantity, the total charge Q. When the total charge vanishes,however, the potential is approximated by dipole, and in this case we need threequantities, the components of p, the dipole moment vector. When this vanishesalso, the potential is more complicated and is approximated by quadrupole; in thiscase the characterization of the potential involves the five independent componentsof the quadrupole moment tensor.

It is important to be aware of the fact that the multipole moments qm dependon the choice of origin for the coordinate system. For example, a point charge q

at x = 0 has a charge density described by (x) = q(x) and multipole momentsgiven by

qm=

q/

4 = 0

0 otherwise .

As expected, only the monopole moment is nonzero. But the same point chargeplaced at another position x0 possesses a very different set of multipole moments.In this case the density is (x) = q(x x0) and Eq. (2.10.4) gives

qm= qr0Ym(0, 0),

where (r0, 0, 0) are the spherical coordinates of the point x0. We have agreementonly for q00, the lowest nonvanishing multipole moment. This is a special case of

a general result: Only the lowest nonvanishing multipole moments have values thatare independent of the choice of origin for the coordinate system; all other momentschange under a translation of the coordinates. We shall not provide a proof of thisstatement.

2.11 Multipolar fields

To conclude this chapter we construct plots of electric field lines that would beproduced by a pure multipole of order . We shall consider only axisymmetricsituations and setm = 0.


47/140

2.11 Multipolar fields 43

The potential for such a multipole is obtained from Eq. (2.10.9), in which weset [4/(2+ 1)]q0Y0(, ) = qP(cos ). This gives

(r, ) = q40

P(cos )

r+1 , (2.11.1)

and the corresponding electric field is

E(r, ) = q40

1

r+2

(+ 1)P(cos )r + sin P

(cos )

, (2.11.2)

in which a prime indicates differentiation with respect to the argument.

The electric field lines are represented by parametric curves x() for which isan arbitrary parameter. These curves are integral curves of the vector fieldE(x),which means that the tangent vector to the curves is defined to be everywhere equalto E. Thus, the field lines are determined by solving the differential equations

dx

electrodynamics flooved

Documents