Multipole Radiation. February 29, The electromagnetic field of an isolated, oscillating source

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1 Multipole Radiation Febuay 29, 26 The electomagnetic field of an isolated, oscillating souce Conside a localized, oscillating souce, located in othewise empty space. We know that the solution fo the vecto potential e.g. using the Geen function fo the oute bounday at infinity is A x, t µ d 3 x dt J x, t x x δ t t + c x x Let the souce fields be confined in a egion d λ whee λ is the wavelength of the adiation, and let the time dependence be hamonic, with fequency ω, Then so that with k ω c, we have A x e iωt µ µ A x, t A x e iωt J x, t J x e iωt ρ x, t ρ x e iωt d 3 x dt J x e iωt x x δ t t + c x x d 3 x J x e iωt c x x x x A x µ d 3 x J x e ik x x x x These consideations mean that the entie souce oscillates vey nealy coheently. The chaacteistic time of one oscillation of the souce, λ c, is much longe than the time it takes light to coss the souce, d c. The electic and magnetic fields follow immediately. We know that B A, so while the Maxwell equation, H D t Dividing by H µ A, shows that iω ɛ E H iωɛ ikcɛ ikɛ µ ɛ ɛ ik µ

2 and defining the impedence of fee space, Z µ ɛ, gives the electic field in the fom E iz k H Now we conside the adiation. The poblem divides into thee appoximate egions, depending on the length scales d and λ. We assume d λ. If is the distance of the obsevation fom the souce, we will conside o simply the nea, intemediate, and fa zones. d λ static zone d λ induction zone d λ adiation zone. Nea zone Fo the nea zone, implies and the potential becomes A x µ lim k µ lim k l,m k 2π λ e d 3 x J x x x 2l + Y lm θ, ϕ l+ d 3 x J x l Y lm θ, ϕ and the only time dependence is the sinusoidal oscillation, e iωt. The spatial integation depends stongly on the details of the souce..2 Fa zone The exponential becomes impotant in the adiation zone, k. Setting x, we have x x 2 + x 2 2 x + x x and since x 2 2, we may dop the quadatic tem and appoximate the squae oot, x x 2 x x x The potential is then A x µ d 3 x J x e ik x x 2

3 Fo the lowest ode appoximation, we neglect the x tem in the denominato, A x µ e µ e µ e d 3 x J x e ik x x d 3 x J x e ik x + x d 3 x J x e ik x whee the last step follows because d λ implies x k x. Fo highe odes in kx kd, note that powes of kx decease apidly in magnitude. We can cay a powe seies in kx to highe ode N in kx kd as long as we can still neglect d, The expansion is then d kdn A x µ e d 3 x J x e ik x µ e ik n d 3 x J x x n n! n and because each tem is smalle than the last by a facto on the ode of kd, it is only the lowest nonvanishing moment of the cuent distibution d 3 x J x x n that dominates the adiation field. The adiation zone solution is chaacteistic of adiation. Retuning to lowest ode A x µ e d 3 x J x e ik x we note that the integal contibutes only angula dependence of the field, d 3 x J x e ik x f θ, ϕ so the wavefom is A x µ e f θ, ϕ The potential is theefoe a hamonic, adially-expanding wavefom, with amplitude deceasing as, The magnetic field is given by A x, t eik ωt B x A x µ µ e [ e d 3 x J x e ik x d 3 x J x e ik x e d 3 x J x e ik x ] 3

4 The fist tem is in the diection d 3 x J x e ik x and theefoe tansvese. Since the gadient includes a tem e the magnetic field will fall off as. Fo the second tem d 3 x J x e ik x the gadient gives ik e e ikx x / e ikx x / ikx + The second tem is also tansvese. Fo the emaining tem, d 3 x x J e ik x we may think of the exponential as giving a modified souce, J Je ik x d 3 x J x e ikx x / ik x Then this integal is just twice the magnetic dipole moment of that modified souce. We know that the esulting magnetic field may be witten as B µ [ ] 3 m m 3 Although this does have a adial component, the magnitude falls off as the cube of the distance, and is theefoe negligible. The magnetic field is theefoe tansvese to the adial diection of popagation. The electic field is also dominated by the e ikne tem, falling off as, and is theefoe also tansvese to the adial popagation..3 Intemediate zone In the intemediate zone, λ, an exact expansion of the Geen function is equied. This is found by expanding G x, x eik x x x x in spheical hamonics, G x, x l,m g lm, Y lm θ, ϕ Y lm θ, ϕ and solving the est of the Helmholtz equation fo the adial function. The esult is spheical Bessel functions, j l x, n l x, and the elated spheical Henkel functions, h l, h 2 l, which ae essentially Bessel function times. They ae discussed in Jackson, Section 9.6. The vecto potential then takes the fom A x ikµ l,m h l k Y lm θ, ϕ d 3 x J x j l k Y lm θ, ϕ The spheical Bessel function may be expanded in powes of k to ecove the pevious appoximations. 4

5 2 Explicit multipoles: n and n Fo highe multipoles n >, we equie the vecto potential in ode to get both magnetic and electic fields. We can then find both fields using H µ A E iz k H Since thee is no magnetic monopole field, we may use the scala potential to demonstate the absence of monopole adiation. 2. Monopole field The lowest ode fa field is the electic monopole field. Fo this it is easiest to use the solution fo the scala potential, in tems of chage density, φ x, t d 3 x dt ρ x, t ε x x δ t t + c x x d 3 x dt ρ x, t δ t t + ε c x x Howeve, all chage is confined to a cental egion, and total chage is conseved. This means that the spatial integal is independent of time, q tot d 3 x ρ x, t and the time integal of the delta function simply gives one. Theefoe, φ x, t q tot ε The electic field is a static Coulomb field; thee is no adiation. 2.2 Dipole field If the fist nonvanishing tem in the multipole expansion, A x µ e d 3 x J x e ik x µ e ik n d 3 x J x x n n! n is the lowest n, then we have A x µ e d 3 x J x Fom the continuity equation, we have ρ t + J iωρ x + J 5

6 Now, since the cuent vanishes at infinity, the integal of the divegence of x j J x must vanish: d 3 x x jj x d 2 x x jj x Then d 3 x x jj x d 3 x i x j J i x i d 3 x ix j i i d 3 x J j x + x j J J i x + x j ij i x d 3 x δ ij J i x + x j ij i x whee the fist integal is the one we equie. Using the continuity equation to eplace the divegence, we have d 3 x J j x d 3 x x j J iω d 3 x x jρ x This integal is the electic dipole moment, and the vecto potential is p d 3 x x ρ x The magnetic field is the cul of this, A x iωµ e p H x A µ iω e p iω e p iω e iω ike 2 ωk e k2 c 6 p e p p e p

7 This is tansvese to the adially popagating wave. Fo the electic field, Using with a e Fo the fist tem, so that with we have E izkc E iz k H izkc e p a b a b b a + b a a b and b p, this becomes [ p e [ ] e [ ] [ e + p ]] e f f + f 2 f + f 2 e 2 + ik e 2 + ik + e 2 ik e 2 e 2 e + ik e 2 e ik ike We compute the second tem in the backets fo E using the identity a f a f + f a a f + f a Since is constant in the diection, a depends only on the angula deivatives, a a θ sin θ cos ϕi + sin θ sin ϕj + cos θk + θ sin θ a ϕ sin θ cos ϕi + sin θ sin ϕj + cos θk ϕ a θ cos θ cos ϕi + cos θ sin ϕj sin θk + sin θ a ϕ sin θ sin ϕi + sin θ cos ϕj a θ cos θρ sin θk + a ϕ ϕ a θθ + aϕ ϕ a a a 7

8 Thus, the second tem is [ p ] e p p ik ik e e 2 p e p p e p p e p Combining these esults, and using Zc µ ɛ µ ɛ ɛ, E ik ɛ ik ɛ ik e ɛ e ɛ [ [ p [ ike p + [ [ ik 2 + k2 ] [ e + p ik ik p + p + ik p + k 2 3ik ]] e ] e 2 p ] p ] p Reaanging tems, E e [ k 2 p p + ik ɛ ] p 3 p Finally, noting that p p p we see that this is equivalent to the expession in Jackson. Notice that the fist tem is tansvese, but not the second. The electic and magnetic fields fo an oscillating dipole field ae theefoe, H x, t k2 c E x, t e iωt ɛ In the adiation zone, k, these simplify to while in the nea zone, k, e iωt p [ k 2 p + ik H x, t k2 c E x, t e iωt k 2 e iωt ɛ Z H p ] p 3 p p H x, t ikc 2 e iωt p E x, t ɛ 3 e iωt 3 p p 8

9 Notice that in the nea zone, the electic field is just e iωt times a static dipole field, while H k p Z ɛ 3 p E 3 p p ɛ 3 so that H E. 9