Multipole Radiation. March 17, 2014

Transcription

1 Multpole Radaton Mach 7, 04 Zones We wll see that the poblem of hamonc adaton dvdes nto thee appoxmate egons, dependng on the elatve magntudes of the dstance of the obsevaton pont,, and the wavelength, λ. We assume thoughout that the extent of the souce s small compaed to the both of these, d λ and d. We consde the cases: d λ statc zone d λ nducton zone d λ adaton zone o smply the nea, ntemedate, and fa zones. The behavo of the felds s qute dffeent, dependng on the zone. The nea and fa zones allow us to use dffeent appoxmatons. Snce d s always much smalle than o λ, we may always expand n powes of d and/o d λ. Statng fom we expand A x = µ 0 d 3 x J x e x x x x e x x x x e x + = x + = e = e = e e x x + x + x + + x + We can wo fom ths expanson, o we can estct to a patcula zone to smplfy the expanson. Thee s anothe possble expanson, n tems of sphecal hamoncs. Fo the denomnato, x x = l,m l + l+ Y lm θ, φ Ylm θ, φ l Ths s most useful n the nea zone, whee the facto of e x x s close to unty.

2 . The nea zone.. Nea zone In the nea zone, λ so x x = + x x x e x x < + d dx x < d x x + d Wth ths appoxmaton, the wavelength dependence dops out and the potental becomes A x = µ 0 d 3 x J x x x = µ 0 Y lm θ, φ l + l+ d 3 x J x l Ylm θ, φ l,m The only tme dependence s the snusodal oscllaton, e ωt, wth the potental gven n tems of the moments, K lm d 3 x J x l Ylm θ, φ of the cuent dstbuton. These ae just le the multpole moments fo the electostatc potental, but wth the chage densty eplaced by the cuent densty. The lowest nonvanshng tem n the sees wll domnate the feld, snce nceasng l deceases the potental by powes of ode d.. Intemedate zone In the ntemedate zone, wth λ, an exact expanson of the Geen functon s useful. Ths s found by expandng G x, x = e x x x x n sphecal hamoncs, G x, x = l,m g lm, Y lm θ, φ Y lm θ, φ and solvng the est of the Helmholtz equaton fo the adal functon. The esult s sphecal Bessel functons, j l x, n l x, and the elated sphecal Henel functons, h l, h l, whch ae essentally Bessel functon tmes. They ae dscussed n Jacson, Secton 9.6. The vecto potental then taes the fom A x = µ 0 l,m h l Y lm θ, φ d 3 x J x j l Y lm θ, φ The sphecal Bessel functon may be expanded n powes of to ecove the pevous nea zone appoxmatons. Thee ae also applcable appoxmaton methods. Although and λ ae of the same ode of magntude, both ae much geate than d, so we may expand n a double powe sees. Ths gves the multpole expanson n subsequent sectons.

3 .3 Fa zone.3. Tuncaton of the sees n In the fa zone, we stll have both both x and x, but now x x, so we expand only to zeoth ode n x d fst. Wth x x the ntegand becomes The potental s then e x x x x A x = µ 0 e e x = e e x d 3 x J x e x Now we use x d and expand the exponental. We may cay ths powe sees n x to ode N as long as we can stll neglect x N elatve to d, The expanson s then d dn A x = µ 0 e d 3 x J x e x = µ 0 e n d 3 x J x n n n! n=0 Agan, only the lowest nonvanshng moment of the cuent dstbuton d 3 x J x n n domnates the adaton feld. Notce that d 3 x J x e x densty. s just the Foue tansfom of the cuent Multpole expanson of tme dependent electomagnetc felds. The felds n tems of the potentals Consde a localzed, oscllatng souce, located n othewse empty space. Let the souce felds be confned n a egon d λ whee λ s the wavelength of the adaton, and let the tme dependence be hamonc, wth fequency ω, We ae nteested n the feld at dstances d. A x, t = A x e ωt ϕ x, t = ϕ x e ωt J x, t = J x e ωt ρ x, t = ρ x e ωt 3

4 In geneal, we have the felds n tems of the scala and vecto potentals, E = ϕ A t B = A Howeve, fo most confguatons the vecto potental alone s suffcent, snce the Maxwell equaton, H D t = 0, shows that ωϵ 0 E = H Dvdng by ωϵ 0 = cϵ 0 = ϵ0 µ 0 and defnng the mpedence of fee space, Z = electc feld n the fom E = Z H µ0 ϵ 0, gves the. Electc monopole Thee s one excepton to ths concluson: the case of vanshng magnetc feld. In ths case, we may use the scala potental satsfyng wth soluton ϕ x, t = ϵ 0 ϕ = ρ ϵ 0 d 3 x dt ρ x, t x x δ t t + c x x wth = ω c. Fo d we have x x and ths becomes ϕ x, t = d 3 x ρ x, t ϵ 0 c = q tot t c ϵ 0 that s, just the Coulomb potental fo the total chage at tme t = c. But fo an solated system, the contnuty equaton, ρ t + J = 0, may be ntegated ove any egon outsde d, gvng ρ d 3 x t + J = 0 d d 3 xρ + d 3 x J = 0 dt d d 3 xρ = d xn J dt dq tot = 0 dt Theefoe, the total chage cannot change, so q tot t c = qtot = constant, and the potental s ndependent of tme, ϕ x, t = q tot ϵ 0 As a esult, thee s no electc monopole adaton. Thus, when the conseved electc chage s confned to a bounded egon, all adaton effects follow fom the esultng cuents, va the vecto potental, wth the felds followng fom E = Z H B = A We fst consde geneal esults fo A x, t, then some cases whch hold only n cetan egons. 4

5 .3 The vecto potental We now examne the vecto potental. Consde a localzed, oscllatng souce, located n othewse empty space. We now that the soluton fo the vecto potental e.g. usng the Geen functon fo the oute bounday at nfnty s Then so that wth = ω c, we have A x, t = µ 0 A x e ωt = µ 0 = µ 0 d 3 x d 3 x dt J x, t x x δ t t + c x x dt J x e ωt x x δ t t + c x x d 3 x J x e ωt c x x x x A x = µ 0 d 3 x J x e x x x x A geneal appoach to fndng the vecto potental, smla to the statc case, s to expand the denomnato, Then the vecto potental s A x = µ 0 x x l,m = l + l+ Y lm θ, φ Ylm θ, φ l,m l l + l+ Y lm θ, φ d 3 x J x l e x x Y lm θ, φ and snce the ntegal s bounded d 3 x J x l e x x Y lm θ, φ < 4 3 πdl+3 J max the tems fall off wth nceasng as d l+. We may theefoe appoxmate the vecto potental by the lowest nonvanshng tem. An ease way to eep tac of nceasng powes of d s to smply expand the denomnato, x x and also expand the exponent, / = + x x = x x = + x x 3 x x + O x x = + x x = x x + x x + x x d 3 + O d 3 3! 4 x x + = + = x x + x x + O 5 d d 3 + O 3

6 so that e x x = e x x + x x d +O 3 3 = e e x x e x x = e x x x x + + x x + and snce both and ae small, e x x = e x x x x + x x + Substtutng ths nto both the exponental and the denomnato, A x = µ 0 = µ 0 e µ 0 e 8π d 3 x J x e x x x x d 3 x J x + µ 0 e d 3 x J x x x d 3 x J x x x + O d 3 The fst ntegal s of ode, the second s of ode max d, d and the thd of ode max, d, whee both d and d ae small. We show below that the fst of these gves the electc dpole adaton, 3 d, d whle the second gves both electc quadupole and magnetc dpole adaton. The thd and hghe tems gve hghe electc and magnetc multpoles. A x = µ 0 e d 3 x J x + µ 0 e d 3 x J x x + µ 0 e 3 d 3 x J x x + 3 Electc dpole adaton The domnant tem n the multpole expanson of A x s Fom the contnuty equaton, A x = µ 0 e d 3 x J x 0 = ρ t + J = ωρ x + J and a cute tc. Snce the cuent vanshes at nfnty, we may wte a vanshng total dvegence of x j J x : d 3 x x jj x = d x x jj x = 0 6

7 Then 0 = d 3 x x jj x = d 3 x x j J x = = = d 3 x J j x + x j J d 3 x x jj x + x j J x d 3 x δ j J x + x j J x and we have d 3 x J j x = d 3 x x j J = ω d 3 x x jρ x Ths ntegal s the electc dpole moment, and the vecto potental s p = d 3 x x ρ x The magnetc feld s the cul of ths, A x = ωµ 0 e p H x = A µ 0 = ω e p = ω e p = ω e = ω e = ω e = c p e p p e p and s theefoe tansvese to the adal decton. Fo the electc feld, E = Z H = Zc e p 7

8 Usng ths becomes E = Zc a b = a b b a + b a a b [ [ p ] [ e p ]] e We easly compute the fst tem n the bacets usng the dentty Thus, the fst tem s [ p ] e Fo the second tem, we use so that [ ] e a f = f [a a ] + a f = f [p p ] + p f = e [ = e [p p ] + p [p p ] + p f = f + f = e + = e = e = e e + e [ e + e + ] e e + ] e Combnng these esults, and usng Zc = µ0 ϵ 0 µ 0 ϵ 0 = ϵ 0, E = Zc = Zc e = Zc e = Zc e = e ϵ 0 [ [ e [ [ + [p p ] + p + [p p ] [ p p + [ p p + ] p ] e ] p p p p p ] p p p ] p 3 p ] p Fnally, notng that p = p p 8

9 we wte the electc feld as E x, t = e ωt ϵ 0 [ + p ] p whee the fst tem s tansvese and the second s not. The electc and magnetc felds fo an oscllatng dpole feld ae theefoe, H x, t = c E x, t = e ωt ϵ 0 In the adaton zone,, these smplfy to whle n the nea zone,, e ωt p [ + p ] p H x, t = c e ωt p E x, t = e ωt p ϵ 0 = Z 0 H H x, t = c e ωt p = Z ϵ 0 3 e ωt p E x, t = ϵ 0 3 e ωt 3 p p Notce that n the nea zone, the electc feld s just e ωt tmes a statc dpole feld, whle H = p Z ϵ 0 3 p E = 3 p p ϵ 0 3 so that H E, whle the spatal oscllaton s neglgble. In the fa zone, by contast, thee s a tansvese wave tavellng adally outwad wth the electc and magnetc felds compaable. 4 Electc quadupole and magnetc dpole adaton To go to hghe multpoles, we genealze ou expesson fo multpole ntegals of the cuent. 4. The tc, n geneal Does the tc fo expessng the moments of the cuent wo fo hghe multpoles? Not qute! We show hee that two dstnct dstbutons ae equed: the chage densty and the magnetc moment densty. Fst, we now how to fnd the zeoth moment of the cuent n tems of the fst moment of the chage densty, d 3 x J j x = ω d 3 x x jρ x 9

10 Now, suppose we now the fst n moments of a cuent dstbuton n tems of moments of the chage densty, and want to fnd the n th, d 3 x x... x n J x j... n Then consde the vanshng ntegal of a total dvegence, 0 = d 3 x x x... x n+ J x = d 3 x x x... x n+ J = d 3 x [ ] δ x... x n+ + x δ... x n x x... δ n+ J + x x... x n+ J = d 3 x [ J x... x n+ + x J... x n x x... J n+ + ωx x... x n+ ρ ] Each of the fst n + ntegals, d 3 x J x... x n+ s a component of j... n, snce thee ae n factos of the coodnates, x... x n+. The poblem s that we have expessed the n + st moment of the chage densty n tems of the symmetzed moments of the cuent. We cannot solve fo j... n unless we also now the antsymmetc pats. Snce the poducts of the coodnates ae necessaly symmetc.e., x... x n+ s the same egadless of the ode of the ndces, the only antsymmetc pece s d 3 x J x J x x... x n Ths does not vansh, but t can be expessed n tems of the magnetc moment densty. Recall M = x J M = ε j x j J j, M ε mn = ε mn ε j x j J,j, = δ mj δ n δ m δ nj x j J j, = x mj n x n J m x m J n = x n J m + M ε mn Consde the second tem n ou expesson fo the moments, d 3 x x J... x n+ We can now tun t nto the same fom as the fst tem, d 3 x x J... x n+ = d 3 x x J + M ε... x n+ = d 3 xj x... x n+ + d 3 xm ε x 3... x n+ 0

11 The fst tem on the ght s now the same as the fst tem n ou expesson. Repeatng ths fo each of the n + tems nvolvng J gves 0 = d 3 x n + J x... x n+ + d 3 xm ε x 3... x n d 3 xm ε n+ x x 3... x n+ + ω and theefoe, j... n+ = n + j... n+ = n! n + whee d 3 x [ ] M ε x 3... x n M ε n+ x x 3... x n+ ε m 3... n+ ω n + p... n+ m... n = d 3 xmx... x n ae the hghe magnetc moments. The paentheses notaton means symmetzaton. Fo example, T j 3! T j + T j + T j + T j + T j + T j ω n + p... n+ Ths shows that thee ae two types of moments that we wll encounte: electc multpole moments bult fom the chage densty, and magnetc multpole moments bult fom the magnetc moment densty. We see ths explctly n the calculaton of the electc quadupole tem of ou geneal expanson. 4. Electc quadupole and magnetc dpole adaton If the electc dpole tem s absent, the domnant tem n ou expanson of the vecto potental s A x = µ 0 e d 3 x J x x x Now we need the next hghe moment of the cuent, d 3 x J x x We have seen that we can expess the n th moment of the chage dstbuton n tems of symmetzed n st moments of the cuent. In ode to get the symmetzed moments of J we may use a vecto dentty. Statng wth the magnetc moment densty we tae a second cul M = x J M = x J = x J J x so that In components, ths s J x = x J M J x = x J ε j j M j,

12 Theefoe, the th component of the cuent moment s [ ] d 3 x J x x = d 3 x J x = d 3 x J x + J x = d 3 x J x + x J ε j j M j, = d 3 x J x + x J ε j j M j, = ω d 3 x x x ρ + d 3 x ε j M j j, = d 3 x ε j M j ω d 3 x x x ρ j Notce that f = d 3 x ρδ = d 3 x ρ has vanshng cul, f = = 0 d 3 x ρ Ths means that ths tem may be added to the vecto potental wthout affectng the felds Jacson neve mentons ths. Ths allows us to defne the quadupole moment of the chage dstbuton as the taceless matx Q = d 3 x 3x x δ ρ x Q = 0 wthout changng the felds. Then, wth the magnetc dpole moment equal to m = d 3 xm and settng [Q ] Q the vecto potental becomes A x = µ 0 e d 3 x J x x

13 = µ 0 e = µ 0 e m + ω6 Q ω Q + m 6 Fo compason, hee s the dvegence tc appled to the pesent case. Consde 0 = d 3 x x j x J x = d 3 x x j x J x = d 3 x [δ j x + δ x j J + x j x J ] = d 3 x [J j x + J x j + ωx j x ρ] = d 3 x [J j x + J x j J j x + ωx j x ρ] = d 3 x J j x + d 3 x J x j J j x + ω d 3 x ρ x j x so that fnally, d 3 x J j x = d 3 x J x j J j x ω d 3 x ρ x j x Now usng the defnton of the magnetc moment densty, and notng that M = x J [ M] = ε j j x J = j jmn ε j ε mn j x mj n = δ m δ jn δ n δ jm j x mj n jmn = j x J j x jj Retunng to the expesson fo the vecto potental A x = µ 0 e d 3 x J x x A = µ 0 e d 3 x J x = µ 0 e d 3 x J x J x ω d 3 x ρ x x = µ 0 e d 3 x [ M] ω d 3 x ρ x x j 3

14 A x = µ 0 e ω Q + m 6 so the esult s just as above. 4.. Magnetc dpole The felds ae now found n the usual way. The magnetc dpole potental s e A x = µ 0 m Notce that the magnetc feld fo the electc dpole had exactly ths fom, µ 0 c H e x, t = µ 0 e ωt once we eplace p m. Snce the electc feld n that case was gven by E e x, t = Z 0 H e, we can wte the cul of A mmedately. We have H m x, t = A m x µ 0 = µ 0 µ 0 c H e x, t = E e x, t c Z p m 0 = e ωt cz ε 0 = e ωt p m [ m + [ m + p ] m 3 m m 3 m We can esot to ths sot of magc agan, because we now that ths fom of E e x, t was acheved by tang the cul of H e, and the Maxwell equatons fo hamonc souces tell us that µ 0 ω c H e x, t ωb m = E m p m = E m µ 0 H e x, t p m = E m Notce that doppng the cul on both sdes s not qute allowed, snce the ght and left sdes could dffe by a gadent, but the answe hee s coect. The electc feld fo magnetc dpole adaton s coectly gven by Ths gves the magnetc dpole felds as E m = µ 0 H e x, t p m c = µ 0 e ωt p = Z 0 e ωt m H = e ωt E = Z 0 [ m + e ωt 4 m p m ] m 3 m ]

15 n complete analogy to the electc dpole feld, but wth magnetc and electc pats ntechanged. In the adaton zone, these become H = e ωt [ m ] E = Z 0 e ωt m so they ae once agan tansvese and compaable n magntude. In the nea zone, H = e ωt 3 m m 3 E = Z 0 e ωt 3 m so the electc feld s much weae than the magnetc, whch taes the fom of a dpole. 4.. Electc quadupole Fo the quadupole felds, we begn wth the quadupole pece of the vecto potental A x = ωµ 0 e Q whee wtng [Q ] Q allows us to wte the potental n vecto fom. The magnetc feld s then Keepng only tems of ode, ths gves H x = µ 0 A x H x = µ 0 [ ωµ0 = µ 0 [ ω µ 0 = c3 e ] Q ] Q e e Q snce the only devatve tem that does not ncease the powe of s [ ωµ 0 feld s then E x = Z 0 H The felds n the adaton zone ae theefoe = Z 0 c 3 e Q = 3 e [ Q ] ϵ 0 H x, t = 3 ϵ 0 Z e ωt Q E x, t = 3 e ωt [ Q ] ϵ 0 5 e Q ]. The electc

16 Nea feld? Wth A x = ωµ 0 e Q the magnetc feld s then H x = A x µ 0 = ωµ0 e µ 0 H x = c ε e j x j = c e Q m ε j x j = c j,,m j,,m = c e = c e = c e Q Q m Q m ε j je j,,m Q m ε j m + je j m + [ Q] + j, [ Q] m + e δ jm j m Q j ε j [ Q] j m + e δ that s, H x = e ϵ 0 Z Q wth the electc feld gven by E = Z H = e ϵ 0 e ϵ 0 = e ϵ ϵ 0 e 3 = e ϵ ϵ 0 e Q Q Q Q + Q Q Q Q + Q 6

17 whee we use = e ϵ e ϵ 0 3 [ Q] = jlm = jlm l ε j j ε lm Q n mn Q Q δ l δ jm δ m δ jl Q mn l δ jn = Q nn + Q Q j j 3Q = Q + Q + δ lj n l n j 4 In the adaton zone ths s domnated by the fst tem, and n the nea zone by H x = 3 e ϵ 0 Z Q E x = 3 e Q Q ϵ 0 3 e = Q ϵ 0 H x = e 8πϵ 0 Z 4 Q E x = e 8πϵ 0 4 Q 4 Q 5 Radated powe The enegy pe unt aea caed by an electomagnetc wave s gven by the Poyntng vecto, S = E H Fo a plane wave, we have So E = E 0 cos x ωt H = µε µ E 0 cos x ωt S = E H = E 0 cos x ωt µε µ E 0 cos x ωt ε = µ E 0 E 0 cos x ωt ε = µ E 0 cos x ωt 7

18 wth eal pat wth tme aveage S = Fo a complex epesentaton of the wave, ε µ E 0 cos x ωt S = ε µ E 0 we may wte the same quantty as Re E H = Re E = E 0 e x ωt H = µε µ E 0e x ωt = E 0 e x ωt µε E µ 0 e x ωt ε µ E 0 so the tme-aveaged enegy flow pe unt aea pe unt tme s S = Re E H Now consde the aveage powe caed off by electc dpole, electc quadupole, and magnetc dpole adaton. 5. Electc dpole The adaton zone felds wee found above to be H x, t = c E x, t = e ωt p e ωt p ε 0 so that dp da = S = Re E H = Re ε 0 [ p ] c = c 4 3π ε 0 p sn θ = c 4 µ 0 ε 0 3π ε 0 p sn θ = c Z 0 3π 4 p sn θ [ p] 8

19 Ths s the powe pe unt aea. Snce the aea element at lage dstances s da = dω, whee Ω s the sold angle, we may wte the dffeental powe adated pe unt sold angle usng so that dp da = dp dω dp dω = c Z 0 3π 4 p sn θ 5. Magnetc dpole The adaton zone felds fo magnetc dpole adaton ae e ωt H x, t = E x, t = Z 0 e ωt m m so the esult s the same as fo the electc dpole wth the substtuton p m/c, 5.3 Electc quadupole moment dp dω = Z 0 3π 4 m sn θ Fo electc quadupole adaton the felds ae gven by H x, t = c3 gvng an aveage powe pe unt sold angle of dp dω = Re E H = e Q E x, t = 3 e [ Q ] ε 0 [ Re 3 e ε 0 c 3 [ Q ] = 3 c 3 [ Q ] Q ε 0 c 6 = 5π [ Q ] Q ε 0 Z 0 c = 5π 6 [ Q ] e Q ] Notce that the powe adated by the quadupole moment depends on 6, wheeas the powe adated by the dpole moments both go as 4. Ths patten contnues fo hghe moments. 9