Homework Assignment #5 Solutions

Transcription

1 Physics 506 Winter 008 Textbook probems: Ch. 9: 9., 9., 9.4 Ch. 10: 10.1 Homework Assignment #5 Soutions 9. A spheric hoe of rdius in conducting medium cn serve s n eectromgnetic resonnt cvity. Assuming infinite conductivity, determine the trnscendent equtions for the chrcteristic frequencies ω m of the cvity for TE nd TM modes. Becuse of the spheric symmetry, it is ntur to describe the modes of the spheric cvity in terms of vector spheric wve expnsion. These wves f into either TE or TM modes, depending on whether r E = 0 or r H = 0, respectivey. The TE or mgnetic mutipoe modes re given by H = i k [j kr X m ], E = Z0 j kr X m 1 where we hve chosen the spheric Besse function j kr since it is regur t r = 0. FOr perfect conductor, we impose the boundry conditions H = 0 nd E = 0 t r =. More precisey, we demnd ˆr H = 0, r= ˆr E = 0 r= These re equivent to the condition j k = 0, nd eds to the quntiztion k nm = x n / where x n is the n-th zero of the spheric Besse function j. The TE nm frequencies re thus ω nm = x nc, j x n = 0, 1, m Ech frequency specified by nd n is + 1-fod degenerte, with zimuth quntum number beed by m. The TM or eectric mutipoe modes re simir, though the boundry conditions re somewht more invoved. The modes themseves re given by H = j kr X m, E = Z0 i k [j kr X m ] This time, the H = 0 boundry condition is utomtic, whie the E = 0 condition gives r [j kr X m ] = 0 r=

2 This vector quntity my be simpified using r V = r V V r V = r V 1 + r V = r r V r r V Using V = j kr X m with r X m = 0 gives r [j kr X m ] = r rj kr X m Hence the E = 0 boundry condition eds to the TM nm frequencies ω nm = y nc, d dx [xj x] = 0, 1, m z=yn The y n correspond to zeros of [xj x] or equiventy j x + xj x. b Ccute numeric vues for the wveength λ m in units of the rdius for the four owest modes for TE nd TM wves. The numeric vues for the wveengths re obtined from the zeros x n nd y n. For TE nm modes, the first four zeros of j x re x 11 = 4.494, x 1 = 5.765, x 1 = , x 1 = 7.75 Since k nm = x n / nd λ nm = π/k nm, we end up with λ nm / = π/x n or λ 11m = 1.98, λ 1m = 1.090, A these modes re + 1-fod degenerte. λ 1m = 0.899, For TM nm modes, the first four zeros of [xj x] re λ 1m = 0.81 y 11 =.747, y 1 =.870, y 1 = 4.974, y 41 = with corresponding wveengths λ 11m =.90, λ 1m = 1.6, λ 1m = 1.6, λ 14m = 1.06 Note tht the next mode, given by y 1 = is nery degenerte with y 41. c Ccute expicity the eectric nd mgnetic fieds inside the cvity for the owest TE nd owest TM mode. The owest TE nd TM modes both hve = 1. Thus we begin with n overview of = 1 vector spheric hrmonics X 1m = 1 LY1m

3 It is ntur to write the ngur momentum opertor L in terms of rising nd owering components L + = L x + il y, L = L x il y, L z Using L + Y m = + 1 mm + 1Y,m+1 for = 1 gives L Y m = + 1 mm 1Y,m 1 L z Y m = my m X + 11 = 0, Xz 11 = 1 Y 11, X 11 = Y 10 X 10 + = Y 11, X10 z = 0, X10 = Y 1, 1 X 1, 1 + = Y 10, X1, 1 z = 1 Y 1, 1, X1, 1 = 0 4 A vector with components V +, V, V z cn be converted to spheric coordintes V r, V θ, V φ ccording to V r = 1 V +e iφ + V e iφ sin θ + V z cos θ V θ = 1 V +e iφ + V e iφ cos θ V z sin θ V φ = i V +e iφ V e iφ Using the expicit form of the spheric hrmonics then gives X11 r = 0, X11 θ = 16π eiφ, X φ 11 = i 16π X10 r = 0, X10 θ = 0, X φ 10 = i 8π sin θ X1, 1 r = 0, X1, 1 θ = 16π e iφ, X φ 1, 1 = i 16π cos θeiφ cos θe iφ We re now redy to exmine the expicit eectric nd mgnetic fieds. From the expression 1 for TE nm modes, we hve E 11m = Z 0 j 1 kr X 1m, H11m = i Z 0 k E 11m The m = 0 mode is the most strightforwrd to write down E 110 = iz 0 8π j 1kr sin θ ˆφ H 110 = 1 j 1 kr cos θˆr [krj 0 kr j 1 kr] sin θˆθ kr 8π 5

4 Note tht we hve used the spheric Besse function identity j = j j Even more expicity, we hve j 1 = sin cos [j 1 ] = j 0 j 1 = 1 1 sin + cos The m = 1 mode is given by E 111 = Z 0 16π j 1kre iφ ˆθ + i cos θ ˆφ H 111 = 1 kr 16π eiφ j 1 kr sin θˆr [krj 0 kr j 1 kr]cos θˆθ + i ˆφ 6 whie the m = 1 mode is given by E 11, 1 = Z 0 16π j 1kre iφ ˆθ i cos θ ˆφ H 11, 1 = 1 kr 16π e iφ j 1 kr sin θˆr + [krj 0 kr j 1 kr]cos θˆθ i ˆφ 7 We now turn to the owest TM mode, which is the TM 11m mode with fieds given by H 11m = j 1 kr X 1m, E11m = iz 0 k H 11m It ought to be cer the the roes of E nd H re interchnged between the TE nd TM modes. In prticur, the TM 11m fieds my be obtined from the TE 11m fieds of 5, 6 nd 7 through the substitution E Z 0 H, Z0 H E This is essentiy the ction of eectric-mgnetic duity. Expicity, the TM 11m

5 modes correspond to H 110 = i 8π j 1kr sin θ ˆφ E 110 = Z 0 j 1 kr cos θˆr [krj 0 kr j 1 kr] sin θˆθ kr 8π H 111 = 16π j 1kre iφ ˆθ + i cos θ ˆφ E 111 = Z 0 kr 16π eiφ j 1 kr sin θˆr [krj 0 kr j 1 kr]cos θˆθ + i ˆφ H 11, 1 = 16π j 1kre iφ ˆθ i cos θ ˆφ E 11, 1 = Z 0 kr 16π e iφ j 1 kr sin θˆr + [krj 0 kr j 1 kr]cos θˆθ i ˆφ Note, however, tht the wvenumbers k nm re quntized differenty for the TE versus the TM modes. 9. The spheric resonnt cvity of Probem 9. hs nonpermebe ws of rge, but finite, conductivity. In the pproximtion tht the skin depth δ is sm compred to the cvity rdius, show tht the Q of the cvity, defined by eqution 8.86, is given by Q = δ Q = δ 1 where x m = /cω m for TM modes. + 1 x m for TE modes for TM modes In order to ccute the Q fctor, we need to obtin both the stored energy nd the power oss t the ws. We strt with the simper cse of TE modes, given by 1. The energy density for hrmonic fieds is u = ɛ 0 4 E + µ 0 4 H However, the energy is equy distributed between E nd H. Thus for TE modes we my immeditey write down u = ɛ 0 E = µ 0 j kr X m The stored energy is given by integrting this over the voume of the sphere U = µ 0 j kr X m r drdω = µ 0 j kr r dr 0

6 Note tht we hve used orthonormity of the vector spheric hrmonics to simpify the integr. We now use the normiztion integr for spheric Besse functions to obtin 0 j x m ρ/j x n ρ/ρ dρ = 1 [j x n ] δ mn U mn = µ 0 4 j x n 8 The power oss is given in terms of the tngenti mgnetic fied t the conducting surfce P = 1 ˆr H σδ d Using H = i/k j kr X m from 1 s we s the vector identity gives P mn = 1 1 d σδ r= kr dr rj kr X m r dω 1 = σδk [rj kr] 9 r= = 1 σδk j k + kj k = σδ j x n where in the st ine we mde use of the fct tht k = x n nd tht j x n = 0. Combining 8 nd 9 then gives the Q fctor for TE modes Q mn = ω U mn = µ 0σωδ P mn = δ where we mde use of the definition of the skin depth δ = /µ 0 σω. The ccution for TM modes is simir. However, the pproprite spheric Besse function normiztion integr needs to be modified for integrting to zeros of [xj x]. Here we simpy stte tht the pproprite normiztion integr my be written s j α m ρ/j α n ρ/ρ dρ = 1 pp αn [j α n ] δ mn where α n is the n-th positive zero of [x p j x] = 0 The fieds for the TM modes re given in, whie the chrcteristic frequencies re given in terms of zeros of [xj x]. We thus set p = 1 in the bove normiztion integr nd use the nottion y n to denote the n-th zero of [xj x] = 0. The expression for the TM stored energy then becomes U mn = µ 0 0 j kr r dr = µ y n j y n

7 The power oss is P mn = 1 σδ ˆr H d = 1 j kr ˆr X σδ m r dω = r= σδ j y mn As resut, the Q fctor for TM mn mode is Q mn = ω U mn = µ 0σωδ 1 P mn + 1 yn = 1 δ + 1 yn 9.4 Discuss the norm modes of oscition of perfecty conducting soid sphere of rdius in free spce. Determine the chrcteristic equtions for the eigenfrequencies for TE nd TM modes of oscition. Show tht the roots for ω wys hve negtive imginry prt, ssuming time dependence of e iωt. Setting up this perfecty conducting sphere probem is simir to wht we did for the spheric hoe probem. However, n importnt feture of the sphere in free spce is tht the voume of the resonnt cvity is unbounded ie it is of spce outside of the rdius. An importnt physic consequence of this is tht osciting eectromgnetic fieds wi rdite out to infinity. Since power is ost to infinity, these so-ced norm modes re ctuy unstbe in the sense tht they decy wy fter whie. Such modes re genery denoted qusi-norm modes, nd re described by compex frequency ω. For the eectric nd mgnetic fieds behve s ω = ω 0 i γ 10 E e iωt = e γt/ e iω 0t Hence the imginry prt of the qusi-norm mode frequency governs the decy of the fieds. Since energy is proportion to the squre of the fieds, the energy decys s e γt. Note tht γ 0 is essenti for this to mke sense. If γ were negtive, then the mode woud grow exponentiy with time. Cery this woud viote energy considertions. In fct, so ong s rdition is emitted nd reches infinity, the mode must necessriy decy. In this cse, we my rgue tht γ is stricty positive. In terms of the frequency ω in 10, energy conservtion then demnds tht ω wys hs negtive imginry prt. In order to ctuy work out the qusi-norm mode frequencies, we note tht TE modes re given by the nog of 1 for the exterior probem with outgoing rdition H = i k h 1 kr X m, E = Z0 h 1 kr X m

8 Here we hve used physic outgoing rdition boundry conditions to seect the first spheric Hnke function h 1. The TE boundry conditions re identic to wht we found bove, nmey ˆr H = 0, ˆr E = 0 r= r= This corresponds to the eqution h 1 k = 0 TE modes Unike in the cse of the spheric Besse functions j nd n, the spheric Hnke functions do not dmit ny re zeros. One wy to see this is to note tht h 1 is defined s the compex combintion h 1 = j + in If were re, then the ony wy for h 1 to vnish is if both re nd imginry prts [ie j nd n ] were to simutneousy vnish for the sme. However, it is esy to see tht the zeros of j nd n never coincide. Therefore, the zeros of h 1 re wys compex. In fct, h 1 hs precisey zeros in the compex pne. To show this, we note tht h 1 my be written s compex poynomi in 1/ times the outgoing spheric wve fctor e i /. In prticur h 1 +1 ei s + s! i = i s! s! Ignoring the irregur point t infinity, the zeros of h 1 then correspond to the zeros of the poynomi s! P = s!s! is s=0 Since this is poynomi of degree, it dmits precisey compex zeros. In fct, it cn be shown tht these zeros hve negtive imginry prt, nd pproximtey ie ong n rc in the ower hf compex pne. The zeros of h 1 re potted for sm vues of s 4 s=0 =1 = = =4 =

9 The TE nm frequencies re thus ω nm = x nc, h1 x n = 0, 1, m, n = 1,,..., where x n denotes the n-th zero of the spheric Hnke function h 1. The TM modes my so be worked out in simir fshion. In prticur, the nog of for the exterior probem is H = h 1 kr X m, E = Z0 i k [h 1 kr X m ] This time, the conducting sphere boundry conditions ed to d dx [xh1 x] = 0 x=k TM modes This time, there re + 1 zeros, which so pproximtey ie ong rcs in the ower hf compex pne 4 =1 = = =4 = Hence the TM nm frequencies re ω nm = y nc, d dx [xh1 x] = 0, 1, m, n = 1,,..., +1 x=yn b Ccute the eigenfrequencies for the = 1 nd = TE nd TM modes. Tbute the wveength defined in terms of the re prt of the frequency in units of the rdius nd the decy time defined s the time tken for the energy to f to e 1 of its initi vue in units of the trnsit time /c for ech of the modes. For = 1 nd =, the spheric Hnke functions re expicity h 1 1 = ei 1 + i, h 1 iei = 1 + i

10 The zeros of h 1 = 1, re then x 11 = i x 1 = i, x = i whie the zeros of [h 1 ] = 1, re y 11 = i, y 1 = i y i, y i, y i Since the compex frequencies re given by these zeros mutipied by c/, we end up with Mode nm λ/ τ//c TE 11m 1/ TE 1m 4π/ 1/ TM 11m 4π/ 1 TM 1m 0.1 TM m where the wveength λ nd the energy decy time τ is given by ω = πc λ i τ 10.1 Show tht for rbitrry initi poriztion, the scttering cross section of perfecty conducting sphere of rdius, summed over outgoing poriztions, is given in the ong-wveength imit by [ 5 dω ɛ 0, ˆn 0, ˆn = k ɛ 0 ˆn 1 ] 4 ˆn ˆn 0 ɛ 0 ˆn 0 ˆn where ˆn 0 nd ˆn re the directions of the incident nd scttered rditions, respectivey, whie ɛ 0 is the perhps compex unit poriztion vector of the incident rdition ɛ 0 ɛ 0 = 1; ˆn 0 ɛ 0 = 0. If poriztions re specified, the conducting sphere scttering cross section is given by dω ˆn, ɛ; ˆn 0, ɛ 0 = k 4 6 ɛ ɛ 0 1 ˆn ɛ ˆn 0 ɛ 0 11 Wht we woud ike to do is to sum this over both orthogon outgoing poriztions. One wy to do this is to introduce iner poriztion bsis trnsverse to the outgoing direction ˆn. To do so, we first ssume the scttering is not in

11 the forwrd direction. Then the incoming direction ˆn 0 my be used to define orthogon poriztions ɛ 1 = ˆn ˆn 0 sin θ, ɛ = ˆn ɛ 1 = ˆnˆn ˆn 0 ˆn 0 sin θ where θ is the nge between ˆn nd ˆn 0. In prticur, we my write sin θ = 1 ˆn ˆn 0. In this cse, the cross section summed over outgoing poriztions becomes dω ˆn; ˆn 0, ɛ 0 = = = = k ˆn ˆn 0 [ ˆn ˆn0 ɛ 0 1 ˆn ˆn ˆn 0 ˆn 0 ɛ 0 + ˆnˆn ˆn 0 ˆn 0 ɛ 0 1 ˆn ˆnˆn ˆn 0 ˆn 0 ˆn 0 ɛ 0 ] k ˆn ˆn 0 [ ˆn ˆn0 ɛ 0 1 ˆnˆn ˆn 0 ˆn 0 ˆn 0 ɛ 0 + ˆn ˆn 0 ˆn ɛ 0 1 ˆn 0 ˆn ˆn 0 ɛ 0 ] k ˆn ˆn 0 [ ˆn ˆn0 ɛ 0 1 ˆn ˆn 0ˆn ˆn 0 ɛ 0 + ˆn ˆn 0 ˆn ɛ 0 1 ˆn ɛ 0 ] k ˆn ˆn 0 [ ˆn ˆn0 ɛ ˆn ˆn 0 + ˆn ɛ 0 1 ˆn ˆn 0 ] Note tht we hve used trnsversity of the initi poriztion, ˆn 0 ɛ 0 = 0. To proceed, we expnd the squres nd rewrite the bove s dω ˆn; ˆn 0, ɛ 0 = k ˆn ˆn 0 [ 5 4 ˆn ˆn 0 ˆn ˆn 0 ɛ 0 + ˆn ɛ 0 1 ˆn ˆn ˆn ˆn 0 ɛ 0 + ˆn ɛ 0 ] 1 The second ine cnces the denomintor. However the first ine needs bit of work. We now use the fct tht ɛ 0 is unit poriztion vector orthogon to ˆn 0. As resut, the three vectors ˆn 0, ɛ 0, ˆn 0 ɛ 0 1 form normized right-hnded coordinte bsis spnning the three-dimension spce. There is sight subtety if ɛ 0 is compex, though the end resut is oky, provided we re crefu with mgnitude squres. The components of ˆn expnded in this bsis re ˆn ˆn 0, ˆn ɛ 0, ˆn ˆn 0 ɛ 0 nd since ˆn is unit vector, the sum of the squres of these components must be one. In other words ˆn ˆn 0 + ˆn ɛ 0 + ˆn ˆn 0 ɛ 0 = 1

12 where we hve been crefu bout compex quntities. Using this resut, we see tht the denomintor in 1 cn be competey eiminted, resuting in dω ˆn; ˆn 0, ɛ 0 = k 4 6 [ 5 4 ˆn ˆn ˆn ˆn 0 ɛ 0 ˆn ɛ 0 ] 14 b If the incident rdition is inery porized, show tht the cross section is [ 5 dω ɛ 0, ˆn 0, ˆn = k cos θ cos θ ] 8 sin θ cos φ where ˆn ˆn 0 = cos θ nd the zimuth nge φ is mesured from the direction of the iner poriztion. As stted, the scttering nge θ is given by ˆn ˆn 0 = cos θ. The zimuth nge φ is the one between ˆn nd ɛ 0, mesured in the pn perpendicur to ˆn 0. Wht this mens is tht, using the bsis vectors 1 with ɛ 0 re, the components of ˆn cn be written s or terntivey ˆn = ˆn 0 cos θ + ɛ 0 sin θ cos φ + ˆn 0 ɛ 0 sin θ sin φ ˆn ˆn 0 = cos θ, ˆn ɛ 0 = sin θ cos φ, ˆn ˆn 0 ɛ 0 = sin θ sin φ Substituting this into 14 gives dω θ, φ = k4 6 [ 5 4 cos θ 1 4 sin θ sin φ sin θ cos φ] = k 4 6 [ 5 4 cos θ 1 8 sin θ1 cos φ 1 sin θ1 + cos φ] = k 4 6 [ cos θ cos θ 8 sin θ cos φ] c Wht is the rtio of scttered intensities t θ = π/, φ = 0 nd θ = π/, φ = π/? Expin physicy in terms of the induced mutipoes nd their rdition ptterns. At θ = π/, we hve Hence dω π/, φ = k4 6 [ cos φ] dω π/, 0 = 1 4 k4 6, dω π/, π/ = k4 6 Scttering t 90 is firy esy to understnd physicy. For φ = 0, the scttered wve is ined up with the incident poriztion ɛ 0. Since the poriztion is given by the eectric fied vector, this indictes tht the induced eectric dipoe of the sphere is ined up with the direction of the scttered wve. Since the rdition must be trnsverse, no dipoe rdition cn be emitted on xis, nd in this cse the scttering must be purey mgnetic dipoe in nture. On the other hnd, for φ = π/, the scttered wve is ined up with the incident mgnetic fied, nd hence the scttering must be purey eectric dipoe in nture. This demonstrtes tht the mximum strength of mgnetic dipoe scttering is qurter tht of eectric dipoe scttering. This is in fct evident by the fctor of 1/ in the mgnetic dipoe term in the cross section expression 11.