solution to-9.18.pdf

7/30/2019 solution to-9.18.pdf

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Multipole Radiation

Notice that throughout, we use r and n interchangably as the unit radial vector.Consider a localized, oscillating source, located in otherwise empty space. We know that the solution for

the vector potential (e.g. using the Green function for the outer boundary at infinity) is

A (x, t) =

0

4

d3

x

dtJ (x, t)

|x x| t t +1

c |x x|Let the source fields be confined in a region d where is the wavelength of the radiation, and let thetime dependence be harmonic, with frequency ,

A (x, t) = A (x) eit

J (x, t) = J (x) eit

(x, t) = (x) eit

Then

A (x) eit =04

d3x

dt

J (x) eit

|x x|

t t + 1c

|x x|

=04

d3

xJ (x) ei(t

1c

|xx

|)

|x x|so that with k = c , we have

A (x) =04

d3x

J (x) eik|xx||x x|

The electric and magnetic fields follow immediately. We know that B =A, so

H =1

0A

while the Maxwell equation, H Dt = 0, shows thati (0E) = H

Dividing byi0 = ikc0

= ik000

= ik

00

and defining the impedence of free space, Z =

00

, gives the electric field in the form

E =iZ

kH

Now we consider the radiation. The problem divides into three approximate regions, depending on the

length scales d and . We assume d . If r is the distance of the observation from the source, we willconsider

d r (static zone)d r (induction zone)d r (radiation zone)

or simply the near, intermediate, and far zones.

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Near zone

For the near zone,

kr =2r

1

implieseikr

1

and the potential becomes

A (x) =04

limkr0

d3x

J (x)

|x x|=

04

limkr0

l,m

4

2l + 1

Ylm (, )

rl+1

d3xJ (x) rlYlm (

, )

and the only time dependence is the sinusoidal oscillation, eit. The spatial integration depends on thedetails of the source.

Far zone

The exponential becomes important in the radiation zone, kr 1. Setting x = rn = rr, we have|x x| =

r2 + x2 2rn x

= r

1 +

x2

r2 2

rn x

and since x2 r2, we may drop the quadratic term and approximate the square root,

|x x| r

1 2rn x

r

1 1rn x

= r n xThe potential is then

A (x) =04

d3x

J (x) eik(rnx)

r n xFor the lowest order approximation, we neglect the x term in the denominator,

A (x) =04

eikr

r

d3x

J (x) eiknx

1 1r n x

=04

eikr

r

d3xJ (x) eiknx

1 +

1

rn x

=

0

4

eikr

r

d

3

xJ (x) eikn

x

where the last step follows because d implies 1r n x kn x.For higher orders in the kx kd 1, note that powers of kx decrease rapidly in magnitude. We can

carry a power series in kx to higher order N in kx kd as long as we can still neglect dr ,d

r (kd)N

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The expansion is then

A (x) =04

eikr

r

d3xJ (x) eiknx

=04

eikr

r n=0(ik)n

n!

d3xJ (x) (n x)n

and because each term is smaller than the last by a factor on the order of kd, it is only the lowest nonvanishingmoment of the current distribution

d3xJ (x) (n x)n

that dominates the radiation field.The radiation zone solution is characteristic of radiation. Returning to lowest order

A (x) =04

eikr

r

d3xJ (x) eiknx

we note that the integral contributes only angular dependence of the field,

d3

xJ (x) eikn

x

= f(, )

so the waveform is

A (x) =04

eikr

rf(, )

The potential is therefore a harmonic, radially-expanding waveform, with amplitude decreasing as 1r ,

A (x, t) ei(krt)

r

The magnetic field is given by

B (x) = A (x)

= 04 eikr

r

d3xJ (x) eiknx

=04

eikr

r

d3xJ (x) eiknx e

ikr

r

d3xJ (x) eiknx

The first term is in the direction

n

d3xJ (x) eiknx

and therefore transverse. Since the gradient includes a term

eikr

r ikne

ikr

r

the magnetic field will fall off as 1

r

.For the second term

d3xJ (x) eiknx =

d3xJ (x) eik(xx)/r

the gradient gives

eik(xx)/r = eik(xx

)/r

ikx

r+

ikn xr

n

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The second term is also transverse, so we need only consider

d3x (x J) eiknx

If we think of the exponential as giving a modified source,

J = Jeiknx

then this integral is just twice the magnetic dipole moment of that source. We know that the resultingmagnetic field may be written as

B =04

3n (n m) m

r3

Although this does have a radial component, the magnitude falls off as the cube of the distance, and istherefore negligible. The magnetic field is therefore transverse to the radial direction of propagation.

The electric field is also dominated by the

eikr ikneikr

term, falling off as 1r , and therefore being transverse to the radial direction.

Intermediate zone

In the intermediate zone, r , an exact expansion of the Green function is required. This is found byexpanding

G (x,x) =eik|xx|

4 |x x|in spherical harmonics,

G (x,x) =l,m

glm (r, r) Ylm (, ) Y

lm (

, )

and solving the rest of the Helmholtz equation for the radial function. The result is spherical Bessel functions,

jl (x) , nl (x), and the related spherical Henkel functions, h

(1)

l , h

(2)

l , which are essentially Bessel function times1r

. They are discussed in Jackson, Section 9.6. The vector potential then takes the form

A (x) = ik0l,m

h(1)l (kr) Ylm (, )

d3xJ (x)jl (kr

) Ylm (, )

The spherical Bessel function may be expanded in powers of kr to recover the previous approximations.

Explicit multipoles: n = 0 and n = 1

For higher multipoles (n > 0), we require the vector potential in order to get both magnetic and electricfields. We can then find both fields using

H = 10A

E =iZ

kH

Since there is no magnetic monopole field, we may use the scalar potential to demonstrate the absence ofmonopole radiation.

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Monopole field

The lowest order far field is the electric monopole field. For this it is easiest to use the solution for the scalarpotential, in terms of charge density,

(x, t) =1

40

d3x

dt

(x, t)

|x

x

|

t t + 1

c|x x|

=1

40r

d3x

dt (x, t)

t t + 1

c|x x|

However, all charge is confined to a central region, and total charge is conserved. This means that the spatialintegral is independent of time,

qtot =

d3x (x, t)

so

(x, t) =qtot

40r

The electric field is therefore a static Coulomb field; there is no radiation.

Dipole field

If the first nonvanishing term in the multipole expansion,

A (x) =04

eikr

r

d3xJ (x) eiknx

=04

eikr

r

n=0

(ik)nn!

d3xJ (x) (n x)n

is the lowest (n = 0), then we have

A (x) =04

eikr

r

d3xJ (x)

From the continuity equation,

0 =

t+ J

= i (x) + Jand a cute trick. Since the current vanishes at infinity, we may write a vanishing total divergence of xjJ (x

):

d3x xjJ (x) =

d2xn xjJ (x)= 0

Then

0 =

d3x

xjJ (x)=

i

d3xi

xjJi (x

)

=i

d3x

ixjJi (x) + xj

iJi (x)

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=i

d3x

ijJi (x

) + xj

iJi (x)

=

d3x

Jj (x

) + xj J

and we have

d3xJj (x) =

d3x xj

J

= i

d3x xj (x)

This integral is the electric dipole moment,

p =

d3x x (x)

and the vector potential is

A (x) =

i0

4

eikr

r

p

The magnetic field is the curl of this,

H (x) =1

0A

= i4

eikr

rp

= i4

eikr

r p

= i4

r

eikr

r

r p

= i4

ike

ikr

r e

ikr

r2r p

=k

4

1 1

ikr

eikr

rr p

=k2c

4

1 1

ikr

eikr

rr p

and is therefore transverse to the radial direction. For the electric field,

E =iZ

kH

=iZkc

4

1

1

ikreikr

rr

p

Using (a b) = a ( b) b ( a) + (b )a (a )b

this becomes

E =iZkc

4

(p )

1 1

ikr

eikr

rr

p

1 1

ikr

eikr

rr

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We easily compute the first term in the brackets using the identity

(a ) (nf(r)) = f(r)r

[a (a n)n] + (a n)nfr

Thus, the first term is

(p ) 1 1ikr eikr

r r

=

f(r)

r [p (p r) r] + (p r) rf

r

=1

r

1 1

ikr

eikr

r[p (p r) r] + (p r) r

1

ikr2eikr

r

1 1ikr

eikr

r2+

1 1

ikr

=eikr

r

1

r 1

ikr2

[p (p r) r] + (p r) r

ik 2

r+

2

ikr2

For the second term, we use

(rf(r)) = 2r

f +f

rso that

1 1

ikreikr

rr =

2

r 1 1

ikreikr

r+

r 1 1

ikreikr

r =

2

r

1 1

ikr

eikr

r+

1

ikr2eikr

r

1 1ikr

eikr

r2+

1 1

ikr

ikeikr

r

=eikr

r

2

r 2

ikr2+

1

ikr2 1

r+

1

ikr2+ ik 1

r

=ikeikr

r

Combining these results, and using Zc =

00

1

00= 10 ,

E =iZkc

4

eikr

r

1

r 1

ikr2

[p (p r) r] + (p r) r

ik 2

r+

2

ikr2

p

ikeikr

r

= iZkc4

eikrr

1r

1ikr2

[p (p r) r]

2r

2ikr2

(p r) r ik (p (p r) r)

=iZkc

4

eikr

r

ik + 1

r

1 1

ikr

(p (p r) r) 2

r

1 1

ikr

(p r) r

=iZkc

4

eikr

r

ik (p (p r) r) + 1

r

1 1

ikr

(p (p r) r) 2

r

1 1

ikr

(p r) r

=1

40

eikr

r

k2 (p (p r) r) + ik

r

1 1

ikr

(p 3 (p r) r)

Finally, noting thatr (p r) = p (p r) r

we see that this is equivalent to the expression in Jackson. Notice that the first term is transverse, but notthe second.

The electric and magnetic fields for an oscillating dipole field are therefore,

H (x, t) =k2c

4

1 1

ikr

eikrit

rr p

E (x, t) =1

40

eikrit

r

k2r (p r) + ik

r

1 1

ikr

(p 3 (p r) r)

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In the radiation zone, kr 1, these simplify to

H (x, t) =k2c

4

eikrit

rr p

E (x, t) =k2

40

eikrit

rr (p r)

= Z0H rwhile in the near zone, kr 1,

H (x, t) =ikc

4

1

r2eikrit r p

=ikrc

4

1

r2eikritr p

E (x, t) =1

40r3eikrit (3 (p r) r p)

Notice that in the near zone, the electric field is just eit times a static dipole field, while

H =kr

Z

1

40

p

r3

E =1

40

p

r3|3 (p r) r p|

so that H E.

An aside: the trick, in general

Does the trick for expressing the moments of the current work for higher multipoles? Not quite! We showhere that two distinct distributions are required: the charge density and the magnetic moment density.

First, we know how to find the zeroth moment of the current in terms of the first moment of the chargedensity,

d3xJj (x) = i d3x xj (x)

Now, suppose we know the first n 1 moments of a current distribution in terms of moments of the chargedensity, and want to find the nth,

jk1...kn

d3x xk1 . . . xknJ (x)

Then consider the (vanishing) integral of a total divergence,

0 =

d3x xk1xk2 . . . xkn+1J (x)

= i

d3xi xk1xk2 . . . xkn+1Ji

=i

d3x

ik1xk2 . . . xkn+1 + xk1ik2 . . . xkn+1 + . . . + xk1xk2 . . . ikn+1

Ji + xk1xk2 . . . xkn+1iJi

=

d3x

Jk1xk2 . . . xkn+1 + xk1Jk2 . . . xkn+1 + . . . + xk1xk2 . . . J kn+1

+ ixk1xk2 . . . xkn+1

=

d3x

Jk1xk2 . . . xkn+1 + xk1Jk2 . . . xkn+1 + . . . + xk1xk2 . . . J kn+1

+ ixk1xk2 . . . xkn+1

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Each of the first n + 1 integrals, d3x Jk1xk2 . . . xkn+1

is a component of jk1...kn , since there are n factors of the coordinates, xk2 . . . xkn+1. The problem is that

we have expressed the (n + 1)st

moment of the charge density in terms of the symmetrized moments of thecurrent. We cannot solve for jk1...kn unless we also know the antisymmetric parts. Since the products of the

coordinates are necessarily symmetric (i.e., xk2 . . . xkn+1 is the same regardless of the order of the ki indices),the only antisymmetric piece is

d3x (Jixk Jkxi) xk2 . . . xknThis does not vanish, but it can be expressed in terms of the magnetic moment density. Recall

M =1

2x J

Mi = 12

j,k

ijkxjJk

iMiimn = 1

2 i,j,kimnijkxjJk

=1

2

j,k

(mjnk mknj) xjJk

=1

2(xmJn xnJm)

xmJn = xnJm + 2i

Miimn

Consider the second term in our expression for the moments,

d3x

xk1Jk2 . . . xkn+1

We can now turn it into the same form as the first term,d3x

xk1Jk2 . . . xkn+1

=

d3x

xk2Jk1 + 2

i

Miik1k2

. . . xkn+1

=

d3xJk1xk2 . . . xkn+1 + 2

i

d3xMiik1k2xk3 . . . xkn+1

The first term on the right is now the same as the first term in our expression. Repeating this for each ofthe n + 1 terms involving Jk gives

0 =

d3x (n + 1) Jk1xk2 . . . xkn+1 + 2

i

d3xMiik1k2xk3 . . . xkn+1 + . . . + 2

i

d3xMiik1kn+1xk2xk3 . . . xk

and therefore,

jk2...kn+1 = 2

n + 1

i

d3x

Miik1k2xk3 . . . xkn+1 + . . . + Miik1kn+1xk2xk3 . . . xkn+1 in + 1pk1...kn+1jk2...kn+1 =

2n!

n + 1

i

ik1(k2mik3...kn+1) i

n + 1pk1...kn+1

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where

mk1...kn =

d3xMxk1 . . . xkn

are the higher magnetic moments. The parentheses notation means symmetrization. For example,

T(ijk)

1

3!

(Tijk + Tjki + Tkij + Tjik + Tkji + Tikj)

This shows that there are two types of moments that we will encounter: electric multipole moments builtfrom the charge density, and magnetic multipole moments built from the magnetic moment density. We seethis explicitly in the calculation of the electric quadrupole term of our general expansion.

Electric quadrupole and magnetic dipole radiation

Return to the exact expression for the vector potential, and expand to the next order:

A (x) =04

eikr

r

d3x

J (x) eiknx

1 1r n x

=

04

eikr

r

d3

xJ (x) (1 ikn x)1 + 1r n x=

04

eikr

r

d3xJ (x) +

04

eikr

r

d3xJ (x)

ikn x + 1

rn x

We know that the first term on the right leads to electric dipole radiation. Suppose this term vanishes, sothat the radiation is described by the next order terms:

A (x) =04

eikr

r

1

r ik

d3xJ (x) (n x)

Now we need the next higher moment of the current,

d

3

xJ (x) (n x)We have seen that we can express the nth moment of the charge distribution in terms of symmetrized(n 1)st moments of the current. In order to get the symmetrized moments of J we may use a vectoridentity. Starting with the magnetic moment density

M =1

2(x J)

we take a second curl

rM = 12r (x J)

=1

2(x (r

J)

J (r

x))

so thatJ (r x) = x (r J) 2rM

In components, this is k

rk (Jixk) =

k

rkxiJk 2

j,k

ijk rjMk

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Therefore, the ith component of the current moment isd3xJ (x) (n x)

i

=

d3x

k

rkJixk

= d3x1

2k

rkJixk +

1

2k

rkJixk

=

d3x

1

2

k

rkJixk +

1

2

k

rkxiJk 2

j,k

ijk rjMk

=

d3x

k

rk1

2(Jix

k + x

iJk)

j,k

ijk rjMk

= i2

k

rk

d3x (xix

k) +

d3x

j,k

ijk rkMj

= k

rk

d3x

j

ijkMj i2

d3x (xix

k)

Notice that

fi =k

rk

d3x

r2ik

= ri

d3x r2

has vanishing curl,

f = ( r)

d3x r2

= 0

This means that this term may be added to the vector potential without affecting the fields (Jackson nevermentions this). This allows us to define the quadrupole moment of the charge distribution as the tracelessmatrix

Qik =

d3x

3xix

k ikr2

(x)

k

Qkk = 0

without changing the fields. Then, with the magnetic dipole moment equal to

m =

d3xM

and setting

[Q (r)]i k

rkQik

the vector potential becomes

A (x) =04

eikr

r

1

r ik

d3xJ (x) (r x)

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12/16

=04

ikeikr

r

1 1

ikr

r m + i

6Q (r)

=04

ikeikr

r

1 1

ikr

i

6Q (r) + rm

For comparison, here is the divergence trick applied to the present case. Consider

0 =

d3x (xjxkJ (x))

=i

d3xi (xjxkJi (x))

=i

d3x [(ijxk + ikxj) Ji + xjxkiJi]

=

d3x [Jjxk + Jkxj + ixjxk]

=

d3x [2Jjxk + (Jkxj Jjxk) + ixjxk]

= 2

d

3

x Jjxk +

d

3

x (Jkxj Jjxk) + i d3

x xjxk

so that finally, d3x Jjxk = 1

2

d3x (Jkxj Jjxk) i

2

d3x xjxk

Now using the definition of the magnetic moment density,

M =1

2(x J)

and noting that

[r M]i =1

2

jk

ijk rj (x J)k

= 12

jkmn

ijkkmnrjxmJn

=1

2

jmn

(imjn injm) rjxmJn

=1

2

j

rj

xiJj xjJi

Returning to the expression for the vector potential

A (x) =04

eikr

r

1

r ik

d3xJ (x) (r x)

Ai = 04 e

ikr

r

1r ik

krk

d3xJixk

=04

eikr

r

1

r ik

k

rk

1

2

d3x (Jkx

i Jixk)

i

2

d3x xix

k

=04

eikr

r

1

r ik

d3x ([rM]i) i

2

k

rk

d3x xix

k

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13/16

A (x) =04

ikeikr

r

1 1

ikr

i

6Q (r) + rm

so the result is just as above.The fields are now found in the usual way. The magnetic dipole potential is

A (x) =

0

4

ikeikr

r rmNotice that the magnetic field for the electric dipole had exactly this form,

i0kc

He (x, t) =04

ikeikrit

rr p

once we replace p m. Since the electric field in that case was given by Ee (x, t) = iZ0k He, we canwrite the curl ofA immediately. We have

Hm (x, t) =1

0Am (x)

=1

0

i0kc

He (x, t)pm=

i

kc

k

iZ0Ee (x, t)|pm

=1

cZ

1

40

eikrit

r

k2r (m r) + ik

r

1 1

ikr

(m 3 (m r) r)

=1

4

eikrit

r

k2r (m r) + ik

r

1 1

ikr

(m 3 (m r) r)

We can resort to this sort of magic again, because we know that this form ofEe (x, t) was achieved by takingthe curl ofHe, and the Maxwell equations for harmonic sources tell us that

iBm = Em

i i0kc He (x, t)pm

= Em

0He (x, t)|pm = EmNotice that dropping the curl on both sides is not quite allowed, since the right and left sides could differ bya gradient, but the answer here is correct. The electric field for magnetic dipole radiation is correctly givenby

Em = 0He (x, t)|pm= 0

k2c

4

1 1

ikr

eikrit

rr p

pm

=

Z0

4

k21 1

ikr e

ikrit

r

r

m

This gives the magnetic dipole fields as

H =1

4

eikrit

r

k2r (m r) + ik

r

1 1

ikr

(m 3 (m r) r)

E = Z04

k2

1 1ikr

eikrit

rrm

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in complete analogy to the electric dipole field, but with magnetic and electric parts interchanged. In theradiation zone, these become

H =1

4k2

eikrit

r[r (m r)]

E =

Z0

4k2

eikrit

rr

m

For the quadrupole fields, we begin with the quadrupole piece of the vector potential

A (x) =i024

ikeikr

r

1 1

ikr

Q (r)

where writing

[Q (r)]i k

rkQik

allows us to write the potential in vector form. The magnetic field is then

H (x, t) =1

0A (x)

Keeping only terms of order 1r , this gives

H (x, t) =1

0

i024

ikeikr

rQ (r)

=1

0

ik2024

eikr

rrQ (r)

=ick3

24

eikr

rrQ (r)

since the only derivative term that does not increase the power of 1r isi024

ikr

eikr

Q (r). The electricfield is then

E = Z0H r= Z0

ick324

eikr

rrQ (r)

r

= ik3

240

eikr

r[rQ (r)] r

Radiated power

The energy per unit area carried by an electromagnetic wave is given by the Poynting vector,

S = EH

For a plane wave, we have

E = E0 cos(k x t)H =

1

1

kkE0 cos(k x t)

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So

S = EH= E0 cos(k x t)

1

1

kk E0 cos(k x t)

=

1

kE0 (kE0)cos2

(k x t)=

1

kE20k cos

2 (k x t)

with real part

S =

1

kE20k cos

2 (k x t)

with time average

S =

1

2

E20

k

For a complex representation of the wave,

E = E0ei(kxt)

H =1

1

kk E0ei(kxt)

we may write the same quantity as

1

2Re (EH) = 1

2Re

E0e

i(kxt) 1

k E0ei(kxt)

=1

2

|E0|2 k

so the time-averaged energy flow per unit area per unit time is

S =1

2Re (EH)

Now consider the average power carried off by electric dipole, electric quadrupole, and magnetic dipoleradiation.

Electric dipole

The radiation zone fields were found above to be

H (x, t) =k2c

4

eikrit

rr p

E (x, t) =

k2

40

eikrit

r r (p r)so that

dP

dA= r S

=1

2Re (r (EH))

15


16/16

=1

2Re

k2

40

1

rr

[r (p r)] k2c

4

1

r[r p]

=ck4

3220

1

r2|p|2 sin2

=c2k4

00

3220

1

r2|p|2 sin2

=c2Z0322

1

r2k4 |p|2 sin2

This is the power per unit area. Since the area element at large distances is dA = r2d, where is the solidangle, we may write the differential power radiated per unit solid angle using

dP

dA=

1

r2dP

d

so that

dP

d=

c2Z0322

k4 |p|2 sin2

Magnetic dipole

The radiation zone fields for magnetic dipole radiation are

H (x, t) =k2

4

eikrit

rr (m r)

E (x, t) = Z0k2

4

eikrit

rrm

so the result is the same as for the electric dipole with the substitution p m/c,dP

d=

Z0322

k4 |m|2 sin2

Electric quadrupole moment

For electric quadrupole radiation the fields are given by

H (x, t) =ick3

24

eikr

rrQ (r)

E (x, t) = ik3

240

eikr

r[rQ (r)] r

giving an average power per unit solid angle of

dP

d=

r2

2|Re (EH)|

=r2

2

Re

ik3

240

eikr

r[rQ (r)] r

ick3

24

eikr

rrQ (r)

=

1

2

k3

240

ck3

24|([rQ (r)] r) (rQ (r))|

=ck6

115220|([rQ (r)] r) (rQ (r))|

=Z0c

2

11522k6 |[rQ (r)] r|2

Notice that the power radiated by the quadrupole moment depends on k6, whereas the power radiatedby the dipole moments both go as k4. This pattern continues for higher moments.

16

Documents

solution to-9.18.pdf