Physics 506 Winter 2008

Homework Assignment #4 Solutions

Textbook problems: Ch. 9: 9.6, 9.11, 9.16, 9.17

9.6 a) Starting from the general expression (9.2) for ~A and the corresponding expressionfor , expand both R = |~x~x | and t = tR/c to first order in |~x |/r to obtainthe electric dipole potentials for arbitrary time variation

(~x, t) =1

4pi0

[1r2~n ~pret + 1

cr~n ~pret

t

]~A(~x, t) =

04pir

~prett

where ~pret = ~p(t = t r/c) is the dipole moment evaluated at the retarded timemeasured from the origin.

We start with the scalar potential, which is given by

(~x ) =1

4pi0

(~x , t |~x ~x |/c)

|~x ~x | d3x (1)

We now use the expansion

|~x ~x | r n ~x

as well as

t = t |~x ~x|

c t r

c+n ~x c

= tret +n ~x c

where tret = t r/c. Since is a function of time t, we make the expansion

(~x , t) = (~x , tret) +n ~x c

(~x , tret)t

+

= ret +n ~x c

rett

+

As a result, the expansion of (1) becomes

(~x ) =1

4pi0r

[ret + n ~x

(1rret +

1c

rett

)+

]d3x

=1

4pi0r

[Q+ n

(1r~pret +

1c

~prett

)+

]

where we have used the expressions for charge and electric dipole moment

Q =ret d

3x, ~pret =~x ret d3x

Note that, by charge conservation, Q is independent of time, so the subscript Qretis superfluous. Dropping the static Coulomb potential (which does not radiate)then gives

(~x ) 14pi0

[1r2n ~pret + 1

crn ~pret

t

](2)

For the vector potential, the expansion is even simpler. We only need to keep thelowest order behavior

~A(~x ) =04pi

~J(~x , t |~x ~x |/c)|~x ~x | d

3x =04pir

[~Jret +

]d3x

Using integration by parts, we note thatJret i d

3x =

xixj

Jret j d3x =

xi(~ ~Jret) d3x =

xirett

d3x =pret it

Hence~A(~x ) 0

4pir~prett

(3)

b) Calculate the dipole electric and magnetic fields directly from these potentialsand show that

~B(~x, t) =04pi

[ 1cr2

~n ~prett 1c2r

~n 2~prett2

]~E(~x, t) =

14pi0

{(1 +

r

c

t

)[3~n(~n ~pret) ~pret

r3

]+

1c2r

~n(~n

2~prett2

)}

For the magnetic field, we use ~B = ~ ~A, where the vector potential is givenby (3). It is important to note that the electric dipole ~pret in (3) is actually afunction of retarded time

~pret = ~p(t r/c)Application of the chain rule then gives

~pretr

= 1c

~prett

Since ~r = n, the magnetic field turns out to be

~B =04pi

~(

1r

~prett

)=04pin

( 1r2~prett 1cr

2~prett2

)(4)

The expression for the electric field is a bit more involved. Using (2) and (3), weobtain

~E = ~ ~A

t

= 14pi0

~(~x

r3 ~pret + ~x

cr2 ~prett

) 0

4pir2~prett2

= 14pi0

[~pretr3 3~x(~x ~pret)

r5 ~xcr4

(~x ~pret

t

)+

1cr2

~prett

2~xcr4

(~x ~pret

t

) ~xc2r3

(~x

2~prett2

)] 1

4pi01c2r

2~prett2

= 14pi0

[~pret 3n(n ~pret)

r3+

1cr2

t(~pret 3n(n ~pret))

+1c2r

2

t2(~pret n(n ~pret))

]=

14pi0

[(1 +

r

c

t

)3n(n ~pret) ~pret

r3+

1c2r

n(n

2~prett2

)]

(5)

c) Show explicitly how you can go back and forth between these results and theharmonic fields of (9.18) by the substitutions i /t and ~peikrit ~pret(t).

Making the substitution

~pret ~peikr and t i

the magnetic field (4) becomes

~H =1

4pin

( 1r2

(i)~p 1cr

(2)~p)eikr

=2

4picr(n ~p )

(1 c

ir

)eikr =

ck2

4pi(n ~p )e

ikr

r

(1 1

ikr

)while the electric field (5) becomes

~E =1

4pi0

[(1 +

r

c(i)

) 3n(n ~p ) ~pr3

+1c2r

(2)n (n ~p )]eikr

=1

4pi0

[eikr

r3(1 ikr)(3n(n ~p ) ~p ) k2 e

ikr

rn (n ~p )

]To go in the other direction, we simply read these equations backwards.

9.11 Three charges are located along the z axis, a charge +2q at the origin, and charges qat z = a cost. Determine the lowest nonvanishing multipole moments, the angulardistribution of radiation, and the total power radiated. Assume that ka 1.

We start by specifying the charge and current densities

= q[2(z) (z a cost) (z + a cost)](x)(y)~J = zqa sint[(z a cost) (z + a cost)](x)(y) (6)

It should be clear that these moving charges do not directly correspond to timeharmonic sources of the form

eit, ~Jeit

Thus we must use some of the techniques discussed in problem 9.1 in Fourierdecomposing the source charge and current distributions. Essentially we find iteasiest to take the approach of 9.1a, which is to compute the time-dependentmultipole moments first before Fourier decomposing in frequency.

Assuming that ka 1, we may directly compute the first few multipole moments.Working with Cartesian tensors, we have

~p(t) =~x d3x = q(a cost a cost) = 0

and~m(t) =

12

~x ~J d3x = 0

In fact, all magnetic multipole moments vanish since the charges are undergoinglinear motion. The electric quadrupole moment is non-vanishing, however

Qij(t) =

(3xixj r2ij)(t) d3x = qa2 cos2 t(3i3j3 ij)

The non-vanishing moments are then

Q33(t) = 2Q11(t) = 2Q22(t) = 4qa2 cos2 t

Note that this may be written as

Q33(t) = 2qa2[1 + cos(2t)] =

which oscillates with angular frequency 2. The angular distribution of radiationis then given by

dP

d=c2Z0k

6

512pi2|Q33|2 sin2 cos2 = Z0q

2

128pi2(ck)2(ka)4 sin2 cos2

Using ck = 2 (since the harmonic frequency is 2), we find

dP

d=Z0q

22

32pi2(ka)4 sin2 cos2 (8)

Integrating this over the solid angle gives a total power

P =Z0q

22

60pi(ka)4 (9)

Alternatively, we may apply the multipole expansion formalism to write downall multipole coefficients. Using the ka 1 approximation, these expansioncoefficients are given by

aE(l,m) ckl+2

i(2l + 1)!!

l + 1lQlm

aM (l,m) ikl+2

(2l + 1)!!

l + 1lMlm

(10)

where

Qlm =rlY lm d

3x, Mlm = 1l + 1

rlY lm~ (~r ~J ) d3x

To proceed, we convert the charge and current densities (6) to spherical coordi-nates

=q

2pir2((r) (r a cost)[(cos 1) + (cos + 1)])

~J = rqa

2pir2sint (r a cost)[(cos 1) + (cos + 1)]

For the magnetic multipoles, we see that since ~J r the cross product vanishes,~r ~J = 0. Thus all magnetic multipoles vanish

aM (l,m) = 0

We are thus left with the electric multipoles

Qlm(t) =q

2pi

rlY lm

((r) (r a cost) [(cos 1) + (cos + 1)])drd cos d

= qm,0(2l,0Y 00 [Y l0(0, 0) + Y l0(pi, 0)](a cost)l

)

Using

Y00 =

1

4pi, [Yl0(0, 0) + Yl0(pi, 0)] =

2l + 1

4pi[Pl(1) + Pl(1)]

then gives

Ql0(t) = 2q

2l + 14pi

(a cost)l l = 2, 4, 6, . . .

A Fourier decomposition gives both positive and negative frequency modes

Ql0(t) = 2q

2l + 14pi

(a2

)l (eit + eit

)l= 2q

2l + 1

4pi

(a2

)l lk=0

(l

k

)ei(l2k)t l = 2, 4, 6, . . .

However, since the real part of eint does not care about the sign of int wemay group such terms together to eliminate negative frequencies

Ql0(t) = 4q

2l + 14pi

(a2

)l