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Multipole Radiation

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Multipole Radiation 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 0 j − 0j µ J (x ) eik x x A (x) = 0 d3x0 4π jx − x0j we expand j − 0j − · 0 ··· eik x x eik(r ^r x + ) = jx − x0j r − ^r · x0 + ··· 0 eikr e−ik^r·x = r 1 − 1 ^r · x0 r ( ) eikr 1 = (1 − ik^r · x0 + ···) 1 + ^r · x0 + ··· r r ( ( ) ) eikr 1 = 1 + − ik ^r · x0 + ··· 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 X 4π r0l = Y (θ; ') Y ∗ (θ0;'0) jx − x0j 2l + 1 rl+1 lm lm l;m j − 0j 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 p k jx − x0j = k r2 + x02 − 2x · x0 p 2 2 − · 0 < k rr + d 2rdx^ x^ 2d d2 < kr 1 − x^ · x^0 + r r2 ≪ 1 j − 0j eik x x ≈ 1 With this approximation, the wavelength dependence drops out and the potential becomes ˆ µ J (x0) A (x) = 0 d3x0 4π jx − x0j ˆ µ X 4π Y (θ; ') = 0 lm d3x0J (x0) r0lY ∗ (θ0;'0) 4π 2l + 1 rl+1 lm l;m The only time dependence is the sinusoidal oscillation, e−i!t, with the potential given in terms of the moments, ˆ ≡ 3 0 0 0l ∗ 0 0 Klm d x J (x ) r Ylm (θ ;' ) 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 j − 0j eik x x G (x; x0) = 4π jx − x0j in spherical harmonics, X 0 0 ∗ 0 0 G (x; x ) = glm (r; r ) Ylm (θ; ') Ylm (θ ;' ) 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 p1 r . They are discussed in Jackson, Section 9.6. The vector potential then takes the form X ˆ (1) 3 0 0 0 ∗ 0 0 A (x) = ikµ0 hl (kr) Ylm (θ; ') d x J (x ) jl (kr ) Ylm (θ ;' ) l;m The spherical Bessel function may be expanded in powers of kr to recover the previous near zone approxi- mations. 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 · 0 ≪ · 0 ≪ 1 · 0 ≪ · 0 In the far zone, we still have both both r ^r x 1 and ik^r x 1, but now r ^r x k^r x , so we expand x0 ∼ d only to zeroth order in r r first. With jx − x0j ≈ r the integrand becomes ikjx−x0j ik(r−^r·x0) e ≈ e jx − x0j r ikr e 0 = e−ik^r·x r The potential is then ˆ ikr µ e 0 A (x) = 0 d3x0J (x0) e−ik^r·x 4π r Now we use kx0 . kd ≪ 1 and expand the exponential. We may carry this power series in kx0 to order N 0 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 0 A (x) = 0 d3x0J (x0) e−ik^r·x 4π r ˆ µ eikr X (−ik)n = 0 d3x0J (x0) r0n (^r · ^r0)n 4π r n! n=0 Again, only the lowest nonvanishing moment of the current distribution ˆ d3x0J (x0) r0n (^r · ^r0)n ´ 0 dominates the radiation field. Notice that d3x0J (x0) e−ik^r·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 !, A (x; t) = A (x) e−i!t ϕ (x; t) = ϕ (x) e−i!t J (x; t) = J (x) e−i!t ρ (x; t) = ρ (x) e−i!t 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 = −rϕ − @t B = r × A However, for most configurations the vector potential alone is sufficient, since the Maxwell equation, r × − @D H @t = 0, shows that − r × q i!ϵ0E = H q ϵ0 µ0 Dividing by −i!ϵ0 = −ikcϵ0 = −ik and defining the impedence of free space, Z = , gives the µ0 ϵ0 electric field in the form iZ E = r × 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 0 0 1 3 0 0 ρ (x ; t ) 0 − 1 j − 0j ϕ (x; t) = d x dt 0 δ t t + x x 4πϵ0 jx − x j c ! d ≪ j − 0j ≈ with k = c . For r 1 we have x x r and this becomes ˆ ( ) 1 r ϕ (x; t) = d3x0ρ x0; 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 + r · J = 0, may be integrated over any region outside d, giving @t ˆ ( ) @ρ d3x + r · J = 0 ˆ @tˆ d d3xρ + d3xr · 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 = r × H k B = r × 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 (x0; t0) 1 A (x; t) = 0 d3x0 dt0 δ t0 − t + jx − x0j 4π jx − x0j c Then ˆ ˆ 0 ( ) µ J (x0) e−i!t 1 A (x) e−i!t = 0 d3x0 dt0 δ t0 − t + jx − x0j 4π jx − x0j c ˆ 0 0 −i! t− 1 jx−x j µ J (x ) e ( c ) = 0 d3x0 4π jx − x0j ! so that with k = c , we have ˆ 0 j − 0j µ J (x ) eik x x A (x) = 0 d3x0 4π jx − x0j A general approach to finding the vector potential, similar to the static case, is to expand the denominator, 1 X 4π r0l = Y (θ; ') Y ∗ (θ0;'0) jx − x0j 2l + 1 rl+1 lm lm l;m Then the vector potential is X ˆ µ0 4π 1 0 0 0 j − 0j ∗ 0 0 A (x) = Y (θ; ') d3x J (x ) r leik x x Y (θ ;' ) 4π 2l + 1 rl+1 lm lm l;m and since the integral is bounded ˆ 0 0 0 j − 0j ∗ 0 0 4 d3x J (x ) r leik x x Y (θ ;' ) < 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 r02 2r0 1=2 = 1 + − x^ · x^0 jx − x0j r r2 r ( ) ( ) ( )! 1 1 r02 2r0 3 r02 2r0 2 d3 = 1 − − x^ · x^0 + − x^ · x^0 + O r 2 r2 r 8 r2 r r3 ( ( ) ( )) 1 r0 1 r02 d3 = 1 + (x^ · x^0) − 1 − 3 (x^ · x^0)2 + O r r 2 r2 r3 and also expand the exponent, p jx − x0j = r2 + r02 − 2x · x0 r 2r0 r02 = r 1 − x^ · x^0 + r r2 ( ) ( ) ( )! 1 2r0 r02 1 1 2r0 r02 2 d3 = r 1 + − x^ · x^0 + − − x^ · x^0 + + O 2 r r2 2! 4 r r2 r3 ( ( ) ( )) r0 1 r02 d3 = r 1 − (x^ · x^0) + 1 − (x^ · x^0)2 + O r 2 r2 r3 5 so that ( ) ( ) 0 02 2 3 − 0 ikr−ikr r x^·x^0 + 1 ikr r 1− x^·x^0 +O d ikjx x j r ( ) 2 r2 ( ) r3 e = e ( ) 0 0 0 1 r 2 0 2 ikr −ikr x^·x^ ikr 1−(x^·x^ ) = e e ( )e 2 r2 ( )( ( ) ( ) ) 1 1 r0 = eikr 1 − ikr0 (x^ · x^0) − k2r02 (x^ · x^0)2 + ··· 1 + i (kr0) 1 − (x^ · x^0)2 + ··· 2 2 r 0 r0 and since both kr and r are small, ( ( ) ) 0 ( ) j − 0j 0 0 1 0 0 2 1 0 r 0 2 eik x x = eikr 1 − ikr (x^ · x^ ) − k2r 2 (x^ · x^ ) + ikr 1 − (x^ · x^ ) + ··· 2 2 r Substituting this into both the exponential and the denominator, ˆ 0 j − 0j µ J (x ) eik x x A (x) = 0 d3x0 4π jx − x0j ˆ ( ) ˆ µ eikr µ eikr 1 = 0 d3x0J (x0) + 0 − ik d3x0J (x0)(x^ · x0) 4π r 4π r r ˆ ( ( ) ) ( ) µ eikr 1 ik 3ik 3 d3 − 0 d3x0J (x0) r02 − + k2 + − (x^ · x^0)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.
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