Multipole Radiation

March 17, 2014

1 Zones

We will see that the problem of harmonic radiation divides into three approximate regions, depending on the relative magnitudes of the distance of the observation point, r, and the wavelength, λ. We assume throughout that the extent of the source is small compared to the both of these, d ≪ λ and d ≪ r. We consider the cases:

d ≪ r ≪ λ (static zone) d ≪ r ∼ λ (induction zone) d ≪ λ ≪ r (radiation zone)

or simply the near, intermediate, and far zones. The behavior of the fields is quite different, depending on the zone. The near and far zones allow us to use different approximations. Since d is always much smaller than r d d or λ, we may always expand in powers of r and/or λ . Starting from

ˆ ′ | − ′| µ J (x ) eik x x A (x) = 0 d3x′ 4π |x − x′|

we expand

| − ′| − · ′ ··· eik x x eik(r ˆr x + ) = |x − x′| r − ˆr · x′ + ··· ′ eikr e−ikˆr·x = r 1 − 1 ˆr · x′ r ( ) eikr 1 = (1 − ikˆr · x′ + ···) 1 + ˆr · x′ + ··· r r ( ( ) ) eikr 1 = 1 + − ik ˆr · x′ + ··· r r

We can work from this expansion, or we can restrict to a particular zone to simplify the expansion. There is another possible expansion, in terms of spherical harmonics. For the denominator,

1 ∑ 4π r′l = Y (θ, φ) Y ∗ (θ′, φ′) |x − x′| 2l + 1 rl+1 lm lm l,m

| − ′| This is most useful in the near zone, where the factor of eik x x is close to unity.

1 1.1 The near zone 1.1.1 Near zone In the near zone, r ≪ λ so √ k |x − x′| = k r2 + x′2 − 2x · x′ √ 2 2 − · ′ < k √r + d 2rdxˆ xˆ 2d d2 < kr 1 − xˆ · xˆ′ + r r2 ≪ 1 | − ′| eik x x ≈ 1

With this approximation, the wavelength dependence drops out and the potential becomes ˆ µ J (x′) A (x) = 0 d3x′ 4π |x − x′| ˆ µ ∑ 4π Y (θ, φ) = 0 lm d3x′J (x′) r′lY ∗ (θ′, φ′) 4π 2l + 1 rl+1 lm l,m

The only time dependence is the sinusoidal oscillation, e−iωt, with the potential given in terms of the moments, ˆ ≡ 3 ′ ′ ′l ∗ ′ ′ Klm d x J (x ) r Ylm (θ , φ )

of the current distribution. These are just like the multipole moments for the electrostatic potential, but with the charge density replaced by the current density. The lowest nonvanishing term in the series will d dominate the ﬁeld, since increasing l decreases the potential by powers of order r .

1.2 Intermediate zone In the intermediate zone, with r ∼ λ, an exact expansion of the Green function is useful. This is found by expanding | − ′| eik x x G (x, x′) = 4π |x − x′| in spherical harmonics, ∑ ′ ′ ∗ ′ ′ G (x, x ) = glm (r, r ) Ylm (θ, φ) Ylm (θ , φ ) l,m and solving the rest of the Helmholtz equation for the radial function. The result is spherical Bessel functions, (1) (2) jl (x) , nl (x), and the related spherical Henkel functions, hl , hl , which are essentially Bessel function times √1 r . They are discussed in Jackson, Section 9.6. The vector potential then takes the form ∑ ˆ (1) 3 ′ ′ ′ ∗ ′ ′ A (x) = ikµ0 hl (kr) Ylm (θ, φ) d x J (x ) jl (kr ) Ylm (θ , φ ) l,m

The spherical Bessel function may be expanded in powers of kr to recover the previous near zone approximations. There are also applicable approximation methods. Although r and λ are of the same order of magnitude, both are much greater than d, so we may expand in a double power series. This gives the multipole expansion in subsequent sections.

2 1.3 Far zone 1.3.1 Truncation of the series in r 1 · ′ ≪ · ′ ≪ 1 · ′ ≪ · ′ In the far zone, we still have both both r ˆr x 1 and ikˆr x 1, but now r ˆr x kˆr x , so we expand x′ ∼ d only to zeroth order in r r ﬁrst. With |x − x′| ≈ r the integrand becomes

ik|x−x′| ik(r−ˆr·x′) e ≈ e |x − x′| r ikr e ′ = e−ikˆr·x r

The potential is then ˆ ikr µ e ′ A (x) = 0 d3x′J (x′) e−ikˆr·x 4π r Now we use kx′ ≲ kd ≪ 1 and expand the exponential. We may carry this power series in kx′ to order N ′ N d as long as we can still neglect (kx ) relative to r , d ≪ (kd)N r The expansion is then ˆ ikr µ e ′ A (x) = 0 d3x′J (x′) e−ikˆr·x 4π r ˆ µ eikr ∑ (−ik)n = 0 d3x′J (x′) r′n (ˆr · ˆr′)n 4π r n! n=0 Again, only the lowest nonvanishing moment of the current distribution ˆ d3x′J (x′) r′n (ˆr · ˆr′)n

´ ′ dominates the radiation ﬁeld. Notice that d3x′J (x′) e−ikˆr·x is just the Fourier transform of the current density.

2 Multipole expansion of time dependent electromagnetic fields 2.1 The fields in terms of the potentials Consider a localized, oscillating source, located in otherwise empty space. Let the source fields be confined in a region d ≪ λ where λ is the wavelength of the radiation, and let the time dependence be harmonic, with frequency ω,

A (x, t) = A (x) e−iωt ϕ (x, t) = ϕ (x) e−iωt J (x, t) = J (x) e−iωt ρ (x, t) = ρ (x) e−iωt

We are interested in the ﬁeld at distances r ≫ d.

3 In general, we have the fields in terms of the scalar and vector potentials, ∂A E = −∇ϕ − ∂t B = ∇ × A However, for most configurations the vector potential alone is sufficient, since the Maxwell equation, ∇ × − ∂D H ∂t = 0, shows that − ∇ × √ iωϵ0E = H √ ϵ0 µ0 Dividing by −iωϵ0 = −ikcϵ0 = −ik and defining the impedence of free space, Z = , gives the µ0 ϵ0 electric field in the form iZ E = ∇ × H k 2.2 Electric monopole There is one exception to this conclusion: the case of vanishing magnetic field. In this case, we may use the scalar potential satisfying ρ □ϕ = − ϵ0 with solution ( ) ˆ ˆ ′ ′ 1 3 ′ ′ ρ (x , t ) ′ − 1 | − ′| ϕ (x, t) = d x dt ′ δ t t + x x 4πϵ0 |x − x | c ω d ≪ | − ′| ≈ with k = c . For r 1 we have x x r and this becomes ˆ ( ) 1 r ϕ (x, t) = d3x′ρ x′, t − 4πϵ0(r ) c q t − r = tot c 4πϵ0r r that is, just the Coulomb potential for the total charge at time t = c . But for an isolated system, the continuity equation, ∂ρ + ∇ · J = 0, may be integrated over any region outside d, giving ∂t ˆ ( ) ∂ρ d3x + ∇ · J = 0 ˆ ∂tˆ d d3xρ + d3x∇ · J = 0 dt ˆ ˛ d d3xρ = − d2xn · J dt dq tot = 0 ( dt) − r Therefore, the total charge cannot change, so qtot t c = qtot = constant, and the potential is independent of time, q ϕ (x, t) = tot 4πϵ0r As a result, there is no electric monopole radiation. Thus, when the conserved electric charge is confined to a bounded region, all radiation effects follow from the resulting currents, via the vector potential, with the fields following from iZ E = ∇ × H k B = ∇ × A We first consider general results for A (x, t), then some cases which hold only in certain regions.

4 2.3 The vector potential We now examine the vector potential. 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 ˆ ˆ ( ) µ J (x′, t′) 1 A (x, t) = 0 d3x′ dt′ δ t′ − t + |x − x′| 4π |x − x′| c Then ˆ ˆ ′ ( ) µ J (x′) e−iωt 1 A (x) e−iωt = 0 d3x′ dt′ δ t′ − t + |x − x′| 4π |x − x′| c ˆ ′ ′ −iω t− 1 |x−x | µ J (x ) e ( c ) = 0 d3x′ 4π |x − x′| ω so that with k = c , we have ˆ ′ | − ′| µ J (x ) eik x x A (x) = 0 d3x′ 4π |x − x′| A general approach to finding the vector potential, similar to the static case, is to expand the denominator, 1 ∑ 4π r′l = Y (θ, φ) Y ∗ (θ′, φ′) |x − x′| 2l + 1 rl+1 lm lm l,m Then the vector potential is ∑ ˆ µ0 4π 1 ′ ′ ′ | − ′| ∗ ′ ′ A (x) = Y (θ, φ) d3x J (x ) r leik x x Y (θ , φ ) 4π 2l + 1 rl+1 lm lm l,m and since the integral is bounded ˆ ′ ′ ′ | − ′| ∗ ′ ′ 4 d3x J (x ) r leik x x Y (θ , φ ) < πdl+3J lm 3 max ( ) d l+1 the terms fall off with increasing r as r . We may therefore approximate the vector potential by the lowest nonvanishing term. d An easier way to keep track of increasing powers of r is to simply expand the denominator, ( )− 1 1 r′2 2r′ 1/2 = 1 + − xˆ · xˆ′ |x − x′| r r2 r ( ( ) ( ) ( )) 1 1 r′2 2r′ 3 r′2 2r′ 2 d3 = 1 − − xˆ · xˆ′ + − xˆ · xˆ′ + O r 2 r2 r 8 r2 r r3 ( ( ) ( )) 1 r′ 1 r′2 d3 = 1 + (xˆ · xˆ′) − 1 − 3 (xˆ · xˆ′)2 + O r r 2 r2 r3 and also expand the exponent, √ |x − x′| = r2 + r′2 − 2x · x′ √ 2r′ r′2 = r 1 − xˆ · xˆ′ + r r2 ( ( ) ( ) ( )) 1 2r′ r′2 1 1 2r′ r′2 2 d3 = r 1 + − xˆ · xˆ′ + − − xˆ · xˆ′ + + O 2 r r2 2! 4 r r2 r3 ( ( ) ( )) r′ 1 r′2 d3 = r 1 − (xˆ · xˆ′) + 1 − (xˆ · xˆ′)2 + O r 2 r2 r3

5 so that ( ) ( ) ′ ′2 2 3 − ′ ikr−ikr r xˆ·xˆ′ + 1 ikr r 1− xˆ·xˆ′ +O d ik|x x | r ( ) 2 r2 ( ) r3 e = e ( ) ′ ′ ′ 1 r 2 ′ 2 ikr −ikr xˆ·xˆ ikr 1−(xˆ·xˆ ) = e e ( )e 2 r2 ( )( ( ) ( ) ) 1 1 r′ = eikr 1 − ikr′ (xˆ · xˆ′) − k2r′2 (xˆ · xˆ′)2 + ··· 1 + i (kr′) 1 − (xˆ · xˆ′)2 + ··· 2 2 r

′ r′ and since both kr and r are small, ( ( ) ) ′ ( ) | − ′| ′ ′ 1 ′ ′ 2 1 ′ r ′ 2 eik x x = eikr 1 − ikr (xˆ · xˆ ) − k2r 2 (xˆ · xˆ ) + ikr 1 − (xˆ · xˆ ) + ··· 2 2 r

Substituting this into both the exponential and the denominator,

ˆ ′ | − ′| µ J (x ) eik x x A (x) = 0 d3x′ 4π |x − x′| ˆ ( ) ˆ µ eikr µ eikr 1 = 0 d3x′J (x′) + 0 − ik d3x′J (x′)(xˆ · x′) 4π r 4π r r ˆ ( ( ) ) ( ) µ eikr 1 ik 3ik 3 d3 − 0 d3x′J (x′) r′2 − + k2 + − (xˆ · xˆ′)2 + O 8π r r2 r r r2 r3 ( ) ( ) 1 1 d 1 1 d2 1 kd2 1 2 2 The ﬁrst integral is of order r , the second is of order max r r , r kd and the third of order max r r , r r , r k d , d where both r and kd are small. We show below that the ﬁrst of these gives the electric dipole radiation, while the second gives both electric quadrupole and magnetic dipole radiation. The third and higher terms give higher electric and magnetic multipoles.

ˆ ˆ µ eikr µ eikr A (x) = 0 d3x′J (x′) + 0 (1 − ikr) d3x′J (x′)(ˆr · x′) 4π r 4π r2 ( ) ˆ µ eikr 1 + 0 1 − ikr − k2r2 d3x′J (x′)(ˆr · x′)2 + ··· 4π r3 2

3 Electric dipole radiation

The dominant term in the multipole expansion of A (x) is ˆ µ eikr A (x) = 0 d3x′J (x′) 4π r From the continuity equation, ∂ρ 0 = + ∇ · J ∂t = −iωρ (x) + ∇ · J

′ and a cute trick. Since the current vanishes at inﬁnity, we may write a vanishing total divergence of xjJ (x ): ˆ ˆ ( ) ( ) 3 ′∇ · ′ ′ 2 ′ · ′ ′ d x xjJ (x ) = d x ˆr xjJ (x ) = 0

6 Then ˆ ′ ( ) 3 ′∇ · ′ ′ 0 = d x xjJ (x ) ˆ ∑ ( ) 3 ′∇′ ′ ′ = d x i xjJi (x ) i ∑ ˆ ( ) 3 ′ ∇′ ′ ′ ′ ∇′ ′ = d x ixjJi (x ) + xj iJi (x ) i ∑ ˆ ( ) 3 ′ ′ ′ ∇′ ′ = d x δijJi (x ) + xj iJi (x ) ˆi ( ) ′ 3 ′ ′ ′ ∇ · = d x Jj (x ) + xj J

and we have ˆ ˆ ′ 3 ′ ′ − 3 ′ ′ ∇ · d x Jj (x ) = d x xj J ˆ − 3 ′ ′ ′ = iω d x xjρ (x )

This integral is the electric dipole moment, ˆ p = d3x′ x′ρ (x′)

and the vector potential is iωµ eikr A (x) = − 0 p 4π r The magnetic ﬁeld is the curl of this, 1 H (x) = ∇ × A µ 0 ( ) iω eikr = − ∇ × p 4π r ( ) iω eikr = − ∇ × p 4π r ( ) iω ∂ eikr = − ˆr × p 4π ∂r r ( ) iω ikeikr eikr = − − ˆr × p 4π r r2 ( ) ωk 1 eikr = 1 − ˆr × p 4π ikr r ( ) k2c 1 eikr = 1 − ˆr × p 4π ikr r and is therefore transverse to the radial direction. For the electric ﬁeld, iZ E = ∇ × H k (( ) ) iZkc 1 eikr = ∇ × 1 − ˆr × p 4π ikr r

7 Using ∇ × (a × b) = a (∇ · b) − b (∇ · a) + (b · ∇) a − (a · ∇) b this becomes [ [( ) ] ( [( ) ])] iZkc 1 eikr 1 eikr E = (p · ∇) 1 − ˆr − p ∇ · 1 − ˆr 4π ikr r ikr r

We easily compute the ﬁrst term in the brackets using the identity f (r) ∂f (a · ∇)(ˆrf (r)) = [a − (a · ˆr) ˆr] + (a · ˆr) ˆr r ∂r Thus, the ﬁrst term is [( ) ] 1 eikr f (r) ∂f (p · ∇) 1 − ˆr = [p − (p · ˆr) ˆr] + (p · ˆr) ˆr ikr r r ∂r ( ) [ ( ) ( ) ] 1 1 eikr 1 eikr 1 eikr 1 ikeikr = 1 − [p − (p · ˆr) ˆr] + (p · ˆr) ˆr − 1 − + 1 − r ikr r ikr2 r ikr r2 ikr r [( ) ( )] eikr 1 1 2 2 = − [p − (p · ˆr) ˆr] + (p · ˆr) ˆr ik − + r r ikr2 r ikr2

For the second term, we use 2 ∂f ∇ · (ˆrf (r)) = f + r ∂r so that [( ) ] ( ) (( ) ) 1 eikr 2 1 eikr ∂ 1 eikr ∇ · 1 − ˆr = 1 − + 1 − ikr r r ikr r ∂r ikr r ( ) ( ) ( ) 2 1 eikr 1 eikr 1 eikr 1 ikeikr = 1 − + − 1 − + 1 − r ikr r ikr2 r ikr r2 ikr r ( ) eikr 2 2 1 1 1 1 = − + − + + ik − r r ikr2 ikr2 r ikr2 r ikeikr = r √ √ Combining these results, and using Zc = µ0 1 = 1 , ϵ0 µ0ϵ0 ϵ0 [ [( ) ( )] ( )] iZkc eikr 1 1 2 2 ikeikr E = − [p − (p · ˆr) ˆr] + (p · ˆr) ˆr ik − + − p 4π r r ikr2 r ikr2 r [( ) ( ) ] iZkc eikr 1 1 2 2 = − [p − (p · ˆr) ˆr] − − (p · ˆr) ˆr − ik (p − (p · ˆr) ˆr) 4π r r ikr2 r ikr2 [( ( )) ( ) ] iZkc eikr 1 1 2 1 = −ik + 1 − (p − (p · ˆr) ˆr) − 1 − (p · ˆr) ˆr 4π r r ikr r ikr [ ( ) ( ) ] iZkc eikr 1 1 2 1 = −ik (p − (p · ˆr) ˆr) + 1 − (p − (p · ˆr) ˆr) − 1 − (p · ˆr) ˆr 4π r r ikr r ikr [ ( ) ] 1 eikr ik 1 = k2 (p − (p · ˆr) ˆr) + 1 − (p − 3 (p · ˆr) ˆr) 4πϵ0 r r ikr Finally, noting that ˆr × (p × ˆr) = p − (p · ˆr) ˆr

8 we write the elecrtric field as [( ) ( ) ] 1 eikr−iωt ik 1 2ik 2 2 − ˆ × × ˆ − − · ˆ ˆ E (x, t) = k + 2 r (p r) 2 (p r) r 4πϵ0 r r r r r where the first term is transverse and the second is not. The electric and magnetic fields for an oscillating dipole field are therefore, ( ) k2c 1 eikr−iωt H (x, t) = 1 − ˆr × p 4π ikr r [ ] 1 eikr−iωt 1 ( ) 2 2 2 − ˆ × × ˆ − − · ˆ ˆ E (x, t) = 2 k r + ikr 1 r (p r) 2 (ikr 1) (p r) r 4πϵ0 r r r In the radiation zone, kr ≫ 1, these simplify to

k2c eikr−iωt H (x, t) = ˆr × p 4π r k2 eikr−iωt E (x, t) = ˆr × (p × ˆr) 4πϵ0 r = Z0H × ˆr while in the near zone, kr ≪ 1, ikc 1 H (x, t) = e−iωtˆr × p 4π r2 ikr 1 −iωtˆ × = 3 e r p Z 4πϵ0r 1 −iωt · ˆ ˆ − E (x, t) = 3 e (3 (p r) r p) 4πϵ0r Notice that in the near zone, the electric field is just e−iωt times a static dipole field, while kr 1 p H = 3 Z 4πϵ0 r 1 p | ˆ · ˆ ˆ − ˆ| E = 3 3 (p r) r p 4πϵ0 r so that H ≪ E, while the spatial oscillation is negligible. In the far zone, by contrast, there is a transverse wave travelling radially outward with the electric and magnetic fields comparable.

4 Electric quadrupole and magnetic dipole radiation

To go to higher multipoles, we generalize our expression for multipole integrals of the current.

4.1 The trick, in general Does the trick for expressing the moments of the current work for higher multipoles? Not quite! We show here 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 charge density, ˆ ˆ 3 ′ ′ − 3 ′ ′ ′ d x Jj (x ) = iω d x xjρ (x )

9 Now, suppose we know the ﬁrst n − 1 moments of a current distribution in terms of moments of the charge density, and want to ﬁnd the nth, ˆ ≡ 3 jk1...kn d x xk1 . . . xkn J (x)

Then consider the (vanishing) integral of a total divergence, ˆ ( ) 3 ∇ · 0 = d x xk1 xk2 . . . xkn+1 J (x) ˆ ∑ ( ) 3 ∇ = d x i xk1 xk2 . . . xkn+1 Ji i ˆ ∑ [( ) ] 3 ∇ = d x δik1 xk2 . . . xkn+1 + xk1 δik2 . . . xkn+1 + ... + xk1 xk2 . . . δikn+1 Ji + xk1 xk2 . . . xkn+1 iJi ˆi [( ) ] 3 = d x Jk1 xk2 . . . xkn+1 + xk1 Jk2 . . . xkn+1 + ... + xk1 xk2 ...Jkn+1 + iωxk1 xk2 . . . xkn+1 ρ

Each of the ﬁrst n + 1 integrals, ˆ 3 d x Jk1 xk2 . . . 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 the current. 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 ˆ 3 − d x (Jixk Jkxi) xk2 . . . xkn This does not vanish, but it can be expressed in terms of the magnetic moment density. Recall 1 M = x × J 2 1 ∑ M = ε x J i 2 ijk j k j,k ∑ 1 ∑ M ε = ε ε x J i imn 2 imn ijk j k i i,j,k 1 ∑ = (δ δ − δ δ ) x J 2 mj nk mk nj j k j,k 1 = (xmJn − xnJm) 2 ∑ xmJn = xnJm + 2 Miεimn i Consider the second term in our expression for the moments, ˆ ( ) 3 d x xk1 Jk2 . . . xkn+1

We can now turn it into the same form as the ﬁrst term, ˆ ˆ (( ) ) ( ) ∑ 3 3 M d x xk1 Jk2 . . . xkn+1 = d x xk2 Jk1 + 2 iεik1k2 . . . xkn+1 i ˆ ∑ ˆ 3 3 M = d xJk1 xk2 . . . xkn+1 + 2 d x iεik1k2 xk3 . . . xkn+1 i

10 The ﬁrst term on the right is now the same as the ﬁrst term in our expression. Repeating this for each of the n + 1 terms involving Jk gives ˆ ∑ ˆ ∑ ˆ ˆ 3 3 M 3 M 3 0 = d x (n + 1) Jk1 xk2 . . . xkn+1 + 2 d x iεik1k2 xk3 . . . xkn+1 + ... + 2 d x iεik1kn+1 xk2 xk3 . . . xkn+1 + iω d xxk1 xk2 . . . xkn+1 ρ i i and therefore, ˆ 2 ∑ [ ] iω j = − d3x M ε x . . . x + ... + M ε x x . . . x − p k2...kn+1 n + 1 i ik1k2 k3 kn+1 i ik1kn+1 k2 k3 kn+1 n + 1 k1...kn+1 i 2n! ∑ iω j = − ε m − p k2...kn+1 n + 1 ik1(k2 ik3...kn+1) n + 1 k1...kn+1 i where ˆ 3 M mk1...kn = d x xk1 . . . xkn

are the higher magnetic moments. The parentheses notation means symmetrization. For example, 1 T ≡ (T + T + T + T + T + T ) (ijk) 3! ijk jki kij jik kji ikj This shows that there are two types of moments that we will encounter: electric multipole moments built from the charge density, and magnetic multipole moments built from the magnetic moment density. We see this explicitly in the calculation of the electric quadrupole term of our general expansion.

4.2 Electric quadrupole and magnetic dipole radiation If the electric dipole term is absent, the dominant term in our expansion of the vector potential is ( ) ˆ µ eikr 1 A (x) = 0 − ik d3x′J (x′)(xˆ · x′) 4π r r Now we need the next higher moment of the current, ˆ d3x′J (x′)(ˆr · 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 vector identity. Starting with the magnetic moment density 1 M = (x × J) 2 we take a second curl 1 ˆr × M = ˆr × (x′ × J) 2 1 = (x′ (ˆr · J) − J (ˆr · x′)) 2 so that J (ˆr · x′) = x′ (ˆr · J) − 2ˆr × M In components, this is ∑ ∑ ∑ ′ ′ − M rˆk (Jixk) = rˆkxiJk 2 εijkrˆj k k k j,k

11 Therefore, the ith component of the current moment is ( ) [ˆ ] ˆ ∑ 3 ′ ′ · ′ 3 ′ ′ d x J (x )(ˆr x ) = d x rˆkJixk i k ˆ ( ) 1 ∑ 1 ∑ = d3x′ rˆ J x′ + rˆ J x′ 2 k i k 2 k i k  k k  ˆ 1 ∑ 1 ∑ ∑ = d3x′  rˆ J x′ +  rˆ x′ J − 2 ε rˆ M  2 k i k 2 k i k ijk j k k k j,k   ˆ ∑ 1 ∑ = d3x′  rˆ (J x′ + x′ J ) − ε rˆ M  k 2 i k i k ijk j k kˆ ˆ j,k iω ∑ ∑ = − rˆ d3x′ (x′ x′ ρ) + d3x′ ε rˆ M 2 k i k ijk k j k  j,k  ˆ ˆ ∑ ∑ iω = rˆ d3x′  ε M − d3x′ (x′ x′ ρ) k ijk j 2 i k k j

Notice that ˆ ∑ ( ) 3 ′ ′2 fi = rˆk d x r ρδik kˆ 3 ′ ′2 =r ˆi d x r ρ has vanishing curl, ˆ ∇ × f = (∇ × ˆr) d3x′ r′2ρ = 0

This means that this term may be added to the vector potential without affecting the fields (Jackson never mentions this). This allows us to define the quadrupole moment of the charge distribution as the traceless matrix ˆ ( ) 3 ′ ′ ′ − ′2 ′ Qik = d x 3xixk δikr ρ (x ) ∑ Qkk = 0 k without changing the fields. Then, with the magnetic dipole moment equal to ˆ m = d3xM and setting ∑ ≡ [Q (ˆr)]i rˆkQik k the vector potential becomes ( ) ˆ µ eikr 1 A (x) = 0 − ik d3x′J (x′)(ˆr · x′) 4π r r

12 ( )( ) µ ikeikr 1 iω = 0 1 − ˆr × m + Q (ˆr) 4π r ikr 6 ( )( ) µ ikeikr 1 iω = 0 1 − Q (ˆr) + ˆr × m 4π r ikr 6 For comparison, here is the divergence trick applied to the present case. Consider ˆ 3 0 = d x∇ · (xjxkJ (x)) ∑ ˆ 3 = d x∇i (xjxkJi (x)) i ∑ ˆ 3 = d x [(δijxk + δikxj) Ji + xjxk∇iJi] î 3 = d x [Jjxk + Jkxj + iωxjxkρ] ˆ 3 = d x [2Jjxk + (Jkxj − Jjxk) + iωxjxkρ] ˆ ˆ ˆ 3 3 3 = 2 d x Jjxk + d x (Jkxj − Jjxk) + iω d x ρ xjxk so that finally, ˆ ˆ ˆ 1 iω d3x J x = − d3x (J x − J x ) − d3x ρ x x j k 2 k j j k 2 j k Now using the definition of the magnetic moment density, 1 M = (x × J) 2 and noting that 1 ∑ [ˆr × M] = ε rˆ (x′ × J) i 2 ijk j k jk 1 ∑ = ε ε rˆ x′ J 2 ijk kmn j m n jkmn 1 ∑ = (δ δ − δ δ )r ˆ x′ J 2 im jn in jm j m n jmn 1 ∑ ( ) = rˆ x′ J − x′ J 2 j i j j i j Returning to the expression for the vector potential ( ) ˆ µ eikr 1 A (x) = 0 − ik d3x′J (x′)(ˆr · x′) 4π r r ( ) ˆ µ eikr 1 ∑ A = 0 − ik rˆ d3x′J x′ i 4π r r k i k ( ) k ( ˆ ˆ ) µ eikr 1 ∑ 1 iω = 0 − ik rˆ − d3x′ (J x′ − J x′ ) − d3x′ ρ x′ x′ 4π r r k 2 k i i k 2 i k k ( ) ( ˆ ˆ ) µ eikr 1 iω ∑ = 0 − ik − d3x′ ([ˆr × M] ) − rˆ d3x′ ρ x′ x′ 4π r r i 2 k i k k

13 ( )( ) µ ikeikr 1 iω A (x) = 0 1 − Q (ˆr) + ˆr × m 4π r ikr 6 so the result is just as above.

4.2.1 Magnetic dipole The fields are now found in the usual way. The magnetic dipole potential is µ ikeikr A (x) = 0 ˆr × m 4π r Notice that the magnetic field for the electric dipole had exactly this form, iµ µ ikeikr−iωt 0 H (x, t) = 0 ˆr × p kc e 4π r −→ iZ0 ∇ × once we replace p m. Since the electric field in that case was given by Ee (x, t) = k He, we can write the curl of A immediately. We have 1 H (x, t) = ∇ × A (x) m µ m 0 ( )

1 iµ0 = ∇ × He (x, t) µ0 kc p−→m i k | = Ee (x, t) p−→m kc iZ0 [ ( ) ] 1 1 eikr−iωt ik 1 = k2ˆr × (m × ˆr) + 1 − (m − 3 (m · ˆr) ˆr) cZ 4πε r r ikr 0 [ ( ) ] 1 eikr−iωt ik 1 = k2ˆr × (m × ˆr) + 1 − (m − 3 (m · ˆr) ˆr) 4π r r ikr

We can resort to this sort of magic again, because we know that this form of Ee (x, t) was achieved by taking the curl of He, and the Maxwell equations for harmonic sources tell us that iωB = ∇ × E ( )m m

iµ0 iω∇ × He (x, t) = ∇ × Em kc p−→m − | µ0He (x, t) p−→m = Em Notice that dropping the curl on both sides is not quite allowed, since the right and left sides could diﬀer by a gradient, but the answer here is correct. The electric ﬁeld for magnetic dipole radiation is correctly given by

Em = −µ0He (x, t)| −→ ( ( p m ) ) 2 ikr−iωt k c 1 e = −µ0 1 − ˆr × p 4π ikr r −→ ( ) p m Z 1 eikr−iωt = − 0 k2 1 − ˆr × m 4π ikr r This gives the magnetic dipole ﬁelds as [ ( ) ] 1 eikr−iωt ik 1 H = k2ˆr × (m × ˆr) + 1 − (m − 3 (m · ˆr) ˆr) 4π r r ikr ( ) Z 1 eikr−iωt E = − 0 k2 1 − ˆr × m 4π ikr r

14 in complete analogy to the electric dipole ﬁeld, but with magnetic and electric parts interchanged. In the radiation zone, these become 1 eikr−iωt H = k2 [ˆr × (m × ˆr)] 4π r Z eikr−iωt E = − 0 k2 ˆr × m 4π r so they are once again transverse and comparable in magnitude. In the near zone, e−iωt H = (3 (m · ˆr) ˆr − m) 4πr3 iZ kr e−iωt E = − 0 ˆr × m 4π r3 so the electric ﬁeld is much weaker than the magnetic, which takes the form of a dipole.

4.2.2 Electric quadrupole For the quadrupole fields, we begin with the quadrupole piece of the vector potential ( ) iωµ ikeikr 1 A (x) = 0 1 − Q (ˆr) 24π r ikr where writing ∑ ≡ [Q (ˆr)]i rˆkQik k allows us to write the potential in vector form. The magnetic field is then 1 H (x) = ∇ × A (x) µ0 1 Keeping only terms of order r , this gives [ ] 1 iωµ ikeikr H (x) = ∇ × 0 Q (ˆr) µ 24π r 0 [ ] 1 −iωk2µ eikr = 0 ˆr × Q (ˆr) µ0 24π r −ick3 eikr = ˆr × Q (ˆr) 24π r [ ( ) ] 1 iωµ0 ik ∇ ikr × since the only derivative term that does not increase the power of r is 24π r e Q (ˆr) . The electric field is then E (x) = Z H × ˆr 0 ( ) −ick3 eikr = Z ˆr × Q (ˆr) × ˆr 0 24π r ik3 eikr = − [ˆr × Q (ˆr)] × ˆr 24πϵ0 r The fields in the radiation zone are therefore ik3 eikr−iωt H (x, t) = − ˆr × Q (ˆr) 24πϵ0Z r ik3 eikr−iωt E (x, t) = − [ˆr × Q (ˆr)] × ˆr 24πϵ0 r

15 Near ﬁeld? With ( ) iωµ ikeikr 1 A (x) = 0 1 − Q (ˆr) 24π r ikr

the magnetic ﬁeld is then 1 H (x) = ∇ × A (x) µ 0 ( ( ) ) 1 iωµ ikeikr 1 = ∇ × 0 1 − Q (ˆr) µ 24π r ikr 0 ( ( ) ) k2c ∂ eikr 1 H (x) = − ε 1 − Q (ˆr) i 24π ijk ∂x r ikr k j ( ( ) ) k2c ∑ ∂ eikr 1 = − Qkmεijk 1 − rˆm 24π ∂xj r ikr j,k,m ( ( ) ( ) ( )( )) k2c ∑ 1 1 ikrˆ eikr 1 eikr 1 eikr 1 1 1 = − Q ε − rˆ eikr 1 − rˆ + j 1 − rˆ + rˆ rˆ + 1 − δ − rˆ rˆ 24π km ijk r2 j ikr m r ikr m r ikr2 j m r ikr r jm r j m j,k,m (( ) ( ) ) k2c eikr ∑ 2 1 = − Q ε ikr − 2 + rˆ rˆ + 1 − (δ − rˆ rˆ ) 24π r2 km ijk ikr j m ikr jm j m j,k,m   ( ) ( ) k2c eikr 2 1 ∑ = −  ikr − 2 + [ˆr × Q] + 1 −  Q ε − [ˆr × Q]  24π r2 ikr i ikr kj ijk i j,k ( ) k2c eikr 3 = − ikr − 3 + [ˆr × Q] 24π r2 ikr i that is, ( ) k2 eikr 3 − − ˆ × H (x) = 2 ikr 3 + r Q 24πϵ0Z r ikr with the electric ﬁeld given by iZ E = ∇ × H k ( ( )) ik ∂ eikr 3 = − ikr − 3 + ˆr × (ˆr × Q) 24πϵ ∂r r2 ikr 0 ( ( ) ) ik eikr 3 − ikr − 3 + ∇ × (ˆr × Q) 24πϵ r2 ikr 0 ( ) ik eikr 9 = − −k2r2 − 4ikr + 9 − ˆr × (ˆr × Q) 24πϵ r3 ikr 0 ( ( ) ) ik eikr 3 + ikr − 3 + ((Q · ˆr) ˆr + 2Q) 24πϵ r3 ikr 0 ( ) ik eikr 9 = − −k2r2 − 4ikr + 9 − ((Q · ˆr) ˆr − Q) 24πϵ r3 ikr 0 ( ( ) ) ik eikr 3 − · ˆ ˆ + 3 ikr 3 + ((Q r) r + 2Q) 24πϵ0 r ikr

16 ( ) ik eikr 12 = − −k2r2 − 5ikr + 12 − (Q · ˆr) ˆr 24πϵ r3 ikr 0 ( ) ikr ik e − 2 2 − − 3 + 3 k r 2ikr + 3 Q 24πϵ0 r ikr where we use ( ) ∑ r r [∇ × (ˆr × Q)] = ε ∇ ε l Q n i ijk j klm r mn r jklm ( ) ∑ r δ δ r r r r = (δ δ − δ δ ) Q l jn + lj n − l n j il jm im jl mn r2 r2 r4 jklm r 1 rˆ 1 = Q i + Q − Q rˆ i − 3Q nn r2 r i j j r i r 1 = − ((Q · ˆr) ˆr + 2Q) r In the radiation zone this is dominated by the ﬁrst term, ik3 eikr H (x) = − ˆr × Q 24πϵ0Z r ik3 eikr E (x) = − (Q − (Q · ˆr) ˆr) 24πϵ0 r ik3 eikr = ˆr × (ˆr × Q (ˆr)) 24πϵ0 r and in the near zone by ikr eikr ˆ × H (x) = 4 r Q 8πϵ0Z r 1 eikr − − · ˆ ˆ E (x) = 4 (Q 4 (Q r) r) 8πϵ0 r 5 Radiated power

The energy per unit area carried by an electromagnetic wave is given by the Poynting vector, S = E × H For a plane wave, we have

E = E0 cos (k · x − ωt) 1 √ 1 H = µε k × E cos (k · x − ωt) µ k 0 So × S = E H ( ) 1 √ 1 = E cos (k · x − ωt) × µε k × E cos (k · x − ωt) 0 µ k 0 √ ε 1 = E × (k × E ) cos2 (k · x − ωt) µ k 0 0 √ ε 1 = E2k cos2 (k · x − ωt) µ k 0

17 with real part √ ε 1 S = E2k cos2 (k · x − ωt) µ k 0

with time average ( √ ) 1 ε S = E2 kˆ 2 µ 0 For a complex representation of the wave,

i(k·x−ωt) E = E0e 1 √ 1 H = µε k × E ei(k·x−ωt) µ k 0 we may write the same quantity as ( ) 1 1 1 √ Re (E × H∗) = Re E ei(k·x−ωt) × µεkˆ × E∗e−i(k·x−ωt) 2 2 0 µ 0 √ 1 ε = |E |2 kˆ 2 µ 0

so the time-averaged energy ﬂow per unit area per unit time is 1 S = Re (E × H∗) 2 Now consider the average power carried oﬀ by electric dipole, electric quadrupole, and magnetic dipole radiation.

5.1 Electric dipole The radiation zone ﬁelds were found above to be k2c eikr−iωt H (x, t) = ˆr × p 4π r k2 eikr−iωt E (x, t) = ˆr × (p × ˆr) 4πε0 r so that dP = ˆr · S dA 1 = Re (ˆr · (E × H∗)) 2 ( ( )) 1 k2 1 k2c 1 = Re ˆr · [ˆr × (p × ˆr)] × [ˆr × p] 2 4πε0 r 4π r ck4 1 = |p|2 sin2 θ 32π2ε r2 √0 c2k4 µ ε 0 0 1 | |2 2 = 2 2 p sin θ 32π ε0 r c2Z 1 = 0 k4 |p|2 sin2 θ 32π2 r2

18 This is the power per unit area. Since the area element at large distances is dA = r2dΩ, where Ω is the solid angle, we may write the diﬀerential power radiated per unit solid angle using dP 1 dP = dA r2 dΩ so that dP c2Z = 0 k4 |p|2 sin2 θ dΩ 32π2 5.2 Magnetic dipole The radiation zone ﬁelds for magnetic dipole radiation are

k2 eikr−iωt H (x, t) = ˆr × (m × ˆr) 4π r Z k2 eikr−iωt E (x, t) = − 0 ˆr × m 4π r so the result is the same as for the electric dipole with the substitution p −→ m/c,

dP Z = 0 k4 |m|2 sin2 θ dΩ 32π2 5.3 Electric quadrupole moment For electric quadrupole radiation the ﬁelds are given by

−ick3 eikr H (x, t) = ˆr × Q (ˆr) 24π r ik3 eikr E (x, t) = − [ˆr × Q (ˆr)] × ˆr 24πε0 r giving an average power per unit solid angle of

dP r2 = |Re (E × H∗)| dΩ 2 [( ) ( )] 2 3 ikr 3 −ikr r ik e ick e = Re − [ˆr × Q (ˆr)] × ˆr × ˆr × Q (ˆr) 2 24πε0 r 24π r 1 k3 ck3 = |([ˆr × Q (ˆr)] × ˆr) × (ˆr × Q (ˆr))| 2 24πε0 24π ck6 | ˆ × ˆ × ˆ × ˆ × ˆ | = 2 ([r Q (r)] r) (r Q (r)) 1152π ε0 Z c2 = 0 k6 |[ˆr × Q (ˆr)] × ˆr|2 1152π2 Notice that the power radiated by the quadrupole moment depends on k6, whereas the power radiated by the dipole moments both go as k4. This pattern continues for higher moments.