where ω 0 is the angular frequency of rotation. Using this, we first examine the time-dependent multipole moments

Transcription

1 9. A common textbook example of a adiating system (see Poblem 9.2) is a configuation of chages fixed elative to each othe but in otation. The chage density is obviously a function of time, but it is not in the fom of (9.). a) Show that fo otating chages one altenative is to calculate eal time-dependent multipole moments using ρ( x, t) diectly and then compute the multipole moments fo a given hamonic fequency with the convention of (9.) by inspection o Fouie decomposition of the time-dependent moments. Note that cae must be taken when calculating q lm (t) to fom linea combinations that ae eal befoe making the connection. Fo a otating set of chages, whee the otation is along the z axis, the chage density may be witten as ρ = ρ(, θ, φ ω t) whee ω is the angula fequency of otation. Using this, we fist examine the time-dependent multipole moments q lm (t) = l Ylm(θ, φ)ρ(, θ, φ ω t)d 3 x = l Ylm(θ, φ + ω t)ρ(, θ, φ )d 3 x

2 whee in the second line we have made the substitution φ = φ + ω t. We now note that the azimuthal behavio of the spheical hamonics goes as Hence Y lm (θ, φ) e imφ Y lm (θ, φ + ω t) = Y lm (θ, φ )e imω t This allows us to isolate the time dependence of q lm (t) as q lm (t) = q lm e imω t (8) whee q lm is the static multipole moment calculated in the body fame [ie with ρ(, θ, φ)]. This expession is almost of the analogous fom as (9.), in the sense that the hamonic time dependence is given by a complex exponential. One inteesting diffeence, howeve, is that (9.) involves a pue fequency ω of the fom ρ( x, t) = ρ( x )e iωt while (8) involves a diffeent fequency ω m = mω fo each diffeent value of m. This demonstates that a otating set of chages geneally adiates at the fundamental fequency ω as well as all highe hamonics. Anothe impotant diffeence, howeve, is that m < components of (8) appea to have a negative fequency. This is somewhat atificial, since the hamonic pesciption we ae using is to take the eal pat of the complex time-dependent quantities. In paticula R(e imωt ) = R(e +imωt ) = cos(mωt) This indicates that modes q lm (t) and q l, m (t) adiate at the same fequency mω. To avoid negative fequencies, we may use the identity Y l, m (θ, φ) = ( ) m Y lm(θ, φ) to show that q l, m = ( ) m q l,m. This allows us to ewite (8) as q lm e imω t m > q lm (t) = q l m = ( ) m [q l m e i m ωt ] m < Note that the m = tem has zeo fequency, and hence does not adiate. At this stage, we still have not specified what to do with the m < multipoles. We note, howeve, that since ultimately we only cae about the eal pats of these complex

3 expessions, it does not matte much whethe we take a complex conjugate o not. Hence we can dop the complex conjugate in the m < expession above. In this case, both q lm (t) and q l, m (t) can be expessed using q l m, at least up to a possible minus sign. To see how to deal with this sign, we note that q lm (t) is essentially the coefficient of Y lm (θ, φ) in the spheical hamonic expansion. The poduct q lm (t)y lm (θ, φ) then has a simple m m behavio q l, m (t)y l, m (θ, φ) = [q lm (t)y lm (θ, φ)] Linealy supeposing the +m and m moments then gives q lm (t)y lm (θ, φ) + q l, m (t)y l, m (θ, φ) = R[2q lm Y lm (θ, φ)e imω t ] This demonstates that, when summing ove all multipoles fo adiation, it is sufficient to sum ove the positive fequency modes only while including an exta facto of two. In paticula, we can take { qlm eff 2qlm m > = q l m = with fequencies mω (9) b) Conside a chage density ρ( x, t) that is peiodic in time with peiod T = 2π/ω. By making a Fouie seies expansion, show that it can be witten as whee ρ( x, t) = ρ ( x ) + ρ n ( x ) = T R[2ρ n ( x )e inωt ] n= T ρ( x, t)e inω t dt This shows explicitly how to establish connection with (9.). Recall that the complex Fouie seies in the time vaiable t may be witten as ρ( x, t) = ρ n ( x ) = T n= T ρ n ( x )e inω t ρ( x, t)e inω t dt Assuming that ρ( x, t) is eal (as it ought to be) we note that ρ n ( x ) = ρ n ( x )

4 Hence ρ( x, t) = ρ ( x ) + = ρ ( x ) + = ρ ( x ) + [ρ n ( x )e inωt + ρ n ( x )e inωt ] n= [ρ n ( x )e inωt + (ρ n ( x )e inωt ) ] n= R[2ρ n ( x )e inωt ] n= Taking the eal pat of a complex time hamonic quantity is of couse what we want to make connection to (9.). In paticula, we show that the peiodic in time chage distibution ρ( x, t) may be teated as a collection of hamonic chage densities ρ eff n ( x ) = { 2ρn ( x ) n > ρ ( x ) n = Of couse, ρ ( x ) is static and does not adiate. between this and (9). with fequencies nω () Note the similaity in fom c) Fo a single chage q otating about the oigin in the x-y plane in a cicle of adius R at constant angula speed ω, calculate the l = and l = multipole moments by the methods of pats a and b and compae. In method b expess the chage density ρ n ( x ) in cylindical coodinates. Ae thee highe multipoles, fo example, quadupole? At what fequencies? Fo a single otating chage q, the time dependent chage density may be witten in spheical coodinates as ρ( x, t) = q R 2 δ( R)δ(cos θ)δ(φ ω t) We stat with the method of pat a. Hee we calculate the body-centic multipole moments q lm = l Ylm(θ, φ)ρ(, θ, φ) 2 dd cos θdφ The l = and l = moments ae = qr l Ylm(π/2, ) = qr l 2l + (l m)! 4π (l + m)! P l m () () q = 4π q, so that, accoding to (9), we have q eff = 4π q, 3 q = 8π qr qeff = 3 2π qr

5 Fo the method of pat b, we stat by calculating the n-th Fouie mode ρ n ( x ) = ω 2π = ω 2π = q 2π/ω 2π/ω ρ( x, t)e inω t dt q R 2 δ( R)δ(cos θ)δ(φ ω t)e inω t dt δ( R)δ(cos θ)einφ 2πR2 The multipole moments calculated fom ρ n ( x) ae q lm [ρ n ] = l Ylm(θ, φ)ρ n (, θ, φ) 2 dd cos θdφ = q 2πR 2 = qr l 2π l Y lm(θ, φ)δ( R)δ(cos θ)e inφ 2 dd cos θdφ 2π Y lm(π/2, φ)e inφ dφ = qr l δ mn Ylm(π/2, ) = qr l 2l + (l m)! δ mn 4π (l + m)! P l m () Note that the moments calculated fom ρ n ( x ) have m = n, but othewise agee with (). Since the effective chage density ρ n ( x ) is doubled fo n > accoding to (), the effective moments q lm [ρ n ] ae doubled as well. This effective doubling is consistent acoss pats a and b. Finally, we note fom () that all highe multipoles ae pesent, so long as P m l () is non-vanishing. By paity, this happens wheneve l + m is even. Thus the l-th multipole will adiate at fequences lω, (l 2)ω, (l 4)ω, Two halves of a spheical metallic shell of adius R and infinite conductivity ae sepaated by a vey small insulating gap. An altenating potential is applied between the two halves of the sphee so that the potentials ae ±V cos ωt. In the long-wavelength limit, find the adiation fields, the angula distibution of adiated powe, and the total adiated powe fom the sphee. In the long wavelength limit, the electic dipole appoximation ought to be easonable. In this case, we may fist wok out the multipole expansion of the souce, and then extact the dipole tem. Fo this poblem, the souce is essentially a hamonically (e iωt ) vaying vesion of the electostatic poblem with hemisphees at opposite potential. The long wavelength limit is also equivalent to the low fequency limit. Thus it is valid to think of the souce as a quasi-static object. Using azimuthal symmety, the potential then admits an expansion in Legende polynomials Φ(, θ) = l α l ( R ) l+ P l (cos θ)

6 whee α l = 2l + 2 Φ(R, cos θ)p l (cos θ)d cos θ Fo hemisphees at opposite potential ±V (times e iωt, which is to be undestood), the expansion coefficients ae α l = (2l + )V P l (x) dx odd l only The dipole tem is dominant, and it is easy to see that α = 3 2V. This gives ise to a dipole potential of the fom Φ l= = 3 2 V ( R Compaing this with the dipole expession ) 2 P (cos θ) = 3 2 V R2 z 3 Φ = p 4πɛ 3 allows us to ead off an electic dipole moment p = 4πɛ ( 3 2 V R2 ẑ) = 6πɛ V R 2 ẑ Woking in the adiation zone, this electic dipole gives and H = ck2 (ˆ p )eik 4π = ck2 4π 6πɛ V R 2 eik E = Z ˆ H = 3 eik V (kr)2 sin θˆθ 2 The angula distibution of dipole adiation gives sin θ ˆφ = 3 2 Z eik V (kr)2 sin θ ˆφ dp dω = c2 Z 32π 2 k4 p 2 sin 2 θ = c2 Z 32π 2 k4 36π 2 ɛ 2 V 2 R 4 sin 2 θ = 9 8 Z V 2 (kr) 4 sin 2 θ and the total adiated powe is P = 3πZ V 2 (kr) 4

7 4. Poblem 9.5 Points a): Fo A(x), copy Eqns of Jackson. Fo Φ(x), wite the analogue of Eq. 9.3 fo Φ(x), Φ(x) = ( ) exp(ik) 4πɛ ik ρ(x )ˆn x d 3 x = ( ) exp(ik) 4πɛ ik ˆn p b): B = A = iµ ( ω 4π p exp(ik) [( ) = iµ ω 4π = iµ ω 4π = ck2 µ 4π exp(ik) ( ˆn exp(ik) exp(ik) ) p + exp(ik) [ik ]) 2 p ] ( p) = [ ] (ˆn p) (5) ik One way to obtain E is E = t A Φ = exp(ik) 4πɛ { [ ( ( ) )] 2 k 2 p + ˆ ˆ p ik + 2 [ ( ˆθ ˆθ p ik ) ( )] [ ( ˆφ ˆφ p ik ) ( )] } whee we have used θ (ˆ p) = p ˆθ and φ (ˆ p) = (p ˆφ) sin θ. The esult simplifies to E = exp(ik) k 2 {p p ˆ} ( ) exp(ik) {ˆθpθ 4πɛ 4πɛ 2 ik + ˆφp } φ 2ˆp Noting that ˆ = ˆn, the fist culy backet equals (ˆn p) ˆn and the second p 3ˆn(ˆn p) we find the final esult, E = 4πɛ { k 2 (ˆn p) exp(ik) ( + [3ˆn(ˆn p) p] 3 ik ) } 2 exp(ik) (6)

8 9. Thee chages ae located along the z axis, a chage +2q at the oigin, and chages q at z = ±a cos ωt. Detemine the lowest nonvanishing multipole moments, the angula distibution of adiation, and the total powe adiated. Assume that ka. We stat by specifying the chage and cuent densities ρ = q[2δ(z) δ(z a cos ωt) δ(z + a cos ωt)]δ(x)δ(y) J = ẑqaω sin ωt[δ(z a cos ωt) δ(z + a cos ωt)]δ(x)δ(y) (6) It should be clea that these moving chages do not diectly coespond to time hamonic souces of the fom ρe iωt, Je iωt Thus we must use some of the techniques discussed in poblem 9. in Fouie decomposing the souce chage and cuent distibutions. Essentially we find it easiest to take the appoach of 9.a, which is to compute the time-dependent multipole moments fist befoe Fouie decomposing in fequency. Assuming that ka, we may diectly compute the fist few multipole moments. Woking with Catesian tensos, we have p(t) = x ρ d 3 x = q(a cos ωt a cos ωt) = and m(t) = 2 x J d 3 x = In fact, all magnetic multipole moments vanish since the chages ae undegoing linea motion. The electic quadupole moment is non-vanishing, howeve Q ij (t) = (3x i x j 2 δ ij )ρ(t) d 3 x = qa 2 cos 2 ωt(3δ i3 δ j3 δ ij ) The non-vanishing moments ae then Q 33 (t) = 2Q (t) = 2Q 22 (t) = 4qa 2 cos 2 ωt Note that this may be witten as Q 33 (t) = 2qa 2 [ + cos(2ωt)] = R[ 2qa 2 ( + e 2iωt )] Since the zeo fequency tem does not adiate, this indicates that we may assume a hamonic quadupole moment Q 33 = 2Q = 2Q 22 = 2qa 2 (7)

9 which oscillates with angula fequency 2ω. The angula distibution of adiation is then given by dp dω = c2 Z k 6 52π 2 Q 33 2 sin 2 θ cos 2 θ = Z q 2 28π 2 (ck)2 (ka) 4 sin 2 θ cos 2 θ Using ck = 2ω (since the hamonic fequency is 2ω), we find dp dω = Z q 2 ω 2 32π 2 (ka)4 sin 2 θ cos 2 θ (8) Integating this ove the solid angle gives a total powe P = Z q 2 ω 2 6π (ka)4 (9) Altenatively, we may apply the multipole expansion fomalism to wite down all multipole coefficients. Using the ka appoximation, these expansion coefficients ae given by a E (l, m) ckl+2 l + Q lm i(2l + )!! l () a M (l, m) ikl+2 l + M lm (2l + )!! l whee Q lm = l Ylmρ d 3 x, M lm = l + l Y lm ( J ) d 3 x To poceed, we convet the chage and cuent densities (6) to spheical coodinates ρ = q ( ) δ() δ( a cos ωt)[δ(cos θ ) + δ(cos θ + )] 2π 2 J = ˆ qaω sin ωt δ( a cos ωt)[δ(cos θ ) + δ(cos θ + )] 2π2 Fo the magnetic multipoles, we see that since J ˆ the coss poduct vanishes, J =. Thus all magnetic multipoles vanish a M (l, m) = We ae thus left with the electic multipoles Q lm (t) = q l Y 2π lm( δ() δ( a cos ωt) [δ(cos θ ) + δ(cos θ + )] ) dd cos θdφ ( = qδ m, 2δl, Y [Yl(, ) + Yl(π, )](a cos ωt) l)

10 Using then gives Y = 2l + 4π, [Y l(, ) + Y l (π, )] = 4π [P l() + P l ( )] 2l + Q l (t) = 2q 4π (a cos ωt)l l = 2, 4, 6,... A Fouie decomposition gives both positive and negative fequency modes 2l + ( a ) l ( Q l (t) = 2q e iωt + e iωt) l 4π 2 2l + ( a ) l l ( ) l = 2q e i(l 2k)ωt l = 2, 4, 6,... 4π 2 k k= Howeve, since the eal pat of e ±inωt does not cae about the sign of inωt we may goup such tems togethe to eliminate negative fequencies 2l + ( a ) l Q l (t) = 4q R 4π 2 2 ( ) l + ( l/2 n=2,4,...,l l (l n)/2 ) e inωt whee l = 2, 4, 6,.... Note that the zeo fequency mode does not adiate, and hence may be ignoed. In geneal, the l-th mode adiates at fequencies lω, (l 2)ω, (l 4)ω,.... The lowest non-vanishing moment is the electic quadupole moment Q 2 = qa 2 5 4π and its hamonic fequency is 2ω. Note, in paticula, that this agees with (7) when conveted to a Catesian tenso. Using (), this yields a multipole coefficient a E (2, ) = iqk(ck)(ka) 2 () 2π We now tun to the angula distibution of the adiation. Fo a pue multipole of ode (l, m), the angula distibution of adiated powe is dp (l, m) dω = Z 2k 2 l(l + ) a(l, m) 2[ 2 (l m)(l + m + ) Y l,m (l + m)(l m + ) Y l,m 2 + m 2 Y lm 2]

11 This simplifies consideably fo m = dp (l, ) dω = Z 2k 2 a(l, ) 2 Y l, 2 Since the lowest multipole is the electic quadupole, we substitute in l = 2 and m = to obtain dp dω = Z 2k 2 a E(2, ) 2 5 8π sin2 θ cos 2 θ (2) Using () then gives dp dω = Z q 2 28π 2 (ck)2 (ka) 4 sin 2 θ cos 2 θ = Z q 2 ω 2 32π 2 (ka)4 sin 2 θ cos 2 θ Note that we have used ck = 2ω, since the hamonic fequency is 2ω. The total adiated powe is given by P = Z 2k 2 [ a E (l, m) 2 + a M (l, m) 2 ] l,m Fo the electic quadupole, this gives P = Z 2k 2 a E(2, ) 2 = Z q 2 ω 2 6π (ka)4 (3) The next non-vanishing multipole would be l = 4, which adiates at fequencies 2ω and 4ω. Howeve, this will be subdominant, so long as ka. Note that the angula distibution (2) and the total adiated powe (3) agee with those found ealie, namely (8) and (9).