Physics 506 Winter 2006 Homework Assignment #6 Solutions

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1 Physics 506 Winter 006 Homework Assignment #6 Soutions Textbook probems: Ch. 10: 10., 10.3, 10.7, Eectromagnetic radiation with eiptic poarization, described (in the notation of Section 7. by the poarization vector, ɛ = r ( ɛ + + re iα ɛ is scattered by a perfecty conducting sphere of radius a. Generaize the ampitude in the scattering cross section (10.71, which appies for r = 0 or r =, and cacuate the cross section for scattering in the ong-waveength imit. Show that [ dσ 5 dω = k4 a 6 8 (1 + cos θ cos θ 3 ( ] r r sin θ cos(φ α Compare with Probem Using a spherica wave expansion with the above poarization vector, we may write the incident pane wave as E = i π( r [ j (kr X,1 + 1 k j (kr X,1 + re iα (j (kr X, 1 1 k j (kr X, 1 ] (and a simiar expression for H. The scattered wave then takes the normaized form π( + 1 E sc = 1 i [ α+ 1 + r (h (1 (kr X,1 + β +( k h (1 (kr X,1 + re iα (α (h (1 (kr X, 1 β ( h (1 (kr X k, 1 ] Note that, since the incident wave has both ɛ + and ɛ (coherenty, the scattered eectric fied (essentiay the scattering ampitude has a coherent sum of positive and negative heicities. The scattering cross section may be written as dσ sc dω = π [ k (1 + r + 1 α+ ( X,1 + iβ + (ˆn X,1 + re iα (α ( X, 1 iβ (ˆn X, 1 ]

2 This is the generaization of (10.63, and eads to a tota scattering cross section π σ sc = k (1 + r ( + 1[ α + ( + β + ( + r ( α ( + β ( ] In the ong waveength imit, we ony need to worry about the = 1 terms in the above. The partia wave coefficients α ± ( and β ± ( are those for a perfecty conducting sphere, and are unchanged by the eiptica poarization. For = 1, we use the Jackson resut As a resut, we obtain dσ sc dω α ± (1 = 1 β ±(1 i 3 (ka3 π 3k (1 + r (ka6 X1,1 iˆn X 1,1 + re iα ( X 1, 1 + iˆn X 1, 1 (1 We now work out the expicit functiona forms of X 1,1 and X 1, 1. This is most straightforward in spherica coordinates where Using X 1,±1 = 1 LY1,±1 L = r ( i ˆr = i ˆθ 1 sin θ φ ˆφ θ as we as Y 1,±1 = 3/8π sin θe ±iφ, we find ( 3 X 1,±1 = i ˆθ 1 16π sin θ φ ˆφ θ This aso yieds ˆn X 1,±1 = ˆr X 1,±1 = Inserting these expressions into (1 gives sin θe ±iφ = 3 16π ( ˆφ iˆθ cos θe ±iφ dσ sc dω = (ka 6 [ˆθ(1 ] cos θ + i ˆφ(cos θ e iφ 8k (1 + r 3 16π (ˆθ ± i ˆφ cos θe ±iφ + re i(α φ[ˆθ(1 ] cos θ i ˆφ(cos θ (ka 6 = ˆθ(1 cos θ(1 + re i(α φ 8k (1 + r = + i ˆφ(cos θ (1 re i(α φ (ka 6 8k (1 + r [ (1 cos θ (1 + r + r cos(α φ + (cos θ (1 + r r cos(α φ ]

3 Mutipying this out and rearranging terms gives the fina resut [ dσ sc 5 dω = k4 a 6 8 (1 + cos θ cos θ 3 ( ] r r sin θ cos(φ α ( Aternativey, we coud take the resut of 10.1a [ dσ 5 dω = k4 a 6 4 ˆɛ 0 ˆn 1 ] 4 ˆn ˆn 0 ˆɛ 0 ˆn 0 ˆn and substitute in the poarization ˆɛ 0 = r (ˆɛ + + re iαˆɛ We take expicity ˆn 0 = ẑ, ˆɛ ± = 1 (ˆx ± iŷ, ˆn 0 ˆɛ ± = iˆɛ ± as we as so that dσ dω = k4 a 6 [ 5 4 sin θ = k 4 a 6 [ 5 4 cos θ 1 sin θ ˆn = sin θ(ˆx cos φ + ŷ sin φ + cos θẑ (1 + r eiφ + re i(α φ sin θ ( 1 + r 1 + r 1 8 sin θ ( 1 r = k 4 a 6 [ 5 8 ( sin θ cos θ 3 4 cos(α φ ] cos(α φ 1 + r ( r 1 + r This is identica to the = 1 partia wave resut (. ] 8(1 + r eiφ re i(α φ cos θ sin θ cos(α φ 10.3 A soid uniform sphere of radius R and conductivity σ acts as a scatterer of a panewave beam of unpoarized radiation of frequency ω, with ωr/c 1. The conductivity is arge enough that the skin depth δ is sma compared to R. a Justify and use a magnetostatic scaar potentia to determine the magnetic fied around the sphere, assuming the conductivity is infinite. (Remember that ω 0. We first note that for harmonic fieds (ω 0 both the magnetic fied and eectric fied must vanish inside a perfect conductor. Furthermore, there are no source currents outside the soid sphere. As a resut of J = 0, and since we are in the ong waveength imit kr 1 (so we may work with a quasi-static magnetic fied ]

4 with B 0, we may use a magnetostatic scaar potentia B = Φ M, at east in the vicinity (but aways outside of the sphere. Immediatey outside the sphere, we may take a Legendre expansion Φ M = B 0 z + α r +1 P (cos θ = B 0 rp 1 (cos θ + α r +1 P (cos θ Note that we have taken the incident magnetic fied to point aong the z direction. (Since eectromagnetic waves are transverse, this means the incident wave is actuay traveing in the x-y pane. We now use the fact that the perpendicuar magnetic fied must vanish at the surface r = R of the conducting sphere. This gives 0 = B r r=r = Φ M r = B 0 P 1 (cos θ + r=r ( + 1α R + P (cos θ Since the Legendre poynomias form an orthogona set, this indicates that a α must vanish for 1, whie α 1 = 1 B 0R 3 This gives ( Φ M = B 0 r + R3 r P 1 (cos θ = B 0 z (1 + R3 r 3 The resuting magnetic fied is B = Φ M = B 0 [ ẑ R3 ] 3ˆr(ˆr ẑ ẑ r 3 (3 The second term is ceary that of a magnetic dipoe of strength m = πr3 µ 0 B0 ( B 0 = B 0 ẑ This agrees with the conducting sphere resut of ( When combined with the eectric dipoe term, this gives the ong waveength scattering cross section of ( b Use the technique of Section 8.1 to determine the absorption cross section of the sphere. Show that it varies as (ω 1/ provided σ is independent of frequency. We start with the power oss cacuation P oss = 1 σδ ˆn H da

5 where ˆn H = 1 µ 0 ˆr B = B 0 µ 0 ] [1 + R3 r 3 r=r ˆr ẑ = 3B 0 µ 0 sin θ ˆφ We have used B given in (3, and evauated the fied at the surface of the conductor. Integrating this over the sphere gives P oss = 1 9 B 0 σδ 4µ sin θ R d cos θ dφ = 3π B 0 R 0 σδµ 0 For normaization, note that the incident fux is I 0 = 1 Z 0 E 0 = This gives an absorption cross section c B 0 = Z 0 µ 0 /ɛ 0 µ B 0 0 σ abs = P oss I 0 = 6πR σδz 0 Using δ = /µ 0 σω gives σ abs = 6πR ɛ0 ω σ which is ceary proportiona to (ω 1/ provide σ is independent of frequency Discuss the scattering of a pane wave of eectromagnetic radiation by a nonpermeabe, dieectric sphere of radius a and dieectric constant ɛ r. a By finding the fieds inside the sphere and matching to the incident pus scattered wave ouside the sphere, determine without any restriction on ka the mutipoe coefficients in the scattered wave. Define suitabe phase shifts for the probem. For the spherica wave anaysis, we start with the outside soution, which is a combination of the incident and scattered wave E = H = 1 Z 0 i 4π( + 1[(j (kr + 1 α ±(h (1 (kr X,±1 ± 1 k (j (kr + 1 β ±(h ( (kr X,±1 ] i 4π( + 1[ i k (j (kr + 1 α ±(h (1 (kr X,±1 i(j (kr + 1 β ±(h (1 (kr X,±1 ] Inside the dieectric sphere, we have no sources, and ony a modified dieectric constant ɛ r. As a resut, the waves inside the sphere must be ordinary spherica waves, however with modified wave number k = ω µ 0 ɛ = (ω µ 0 ɛ 0 ɛ r = k ɛ r (4

6 Defining aso Z = µ0 ɛ = Z 0 ɛr the spherica waves inside the dieectric sphere may be parametrized by E = i 4π( + 1[a M,± (j (k r X,±1 ± 1 k a E,±( j (k r X,±1 ] H = 1 i 4π( + 1[ i Z k a M,±( j (k r X,±1 ia E,± (j (k r X,±1 ] (5 Note that the choice of constants was made to simpify the comparison with (4. Since we have a dieectric boundary, we now perform matching between the inside and outside fieds. Note that this is the spherica generaization of matching pane waves incident on a fat dieectric boundary. There are four matching conditions, namey continuity of B, E, D and H. For the perpendicuar fieds, note that whie ˆr X m = 0 ˆr f (kr X m = ˆr f (kr X m = i L f (kr X m = if (kr L X m = i ( + 1f (kry m This indicates that ony the cur terms in E and H survive in the perpendicuar direction. For the parae fieds, on the other hand, both terms contribute. In particuar ˆr X m 0 and ˆr ( f (kr X m = (f (krˆr X m 1 r f (kr( X m ˆr(ˆr X m (ˆr f (kr X m = 1 d r dr (rf (kr X m where we have used ˆr X m = 0. Matching ineary independent terms in the inside (5 and outside (4 soutions gives B : a M,± (j (x = j (x + 1 α ±(h (1 (x H : ɛr a E,± (j (x = j (x + 1 β ±(h (1 (x a M,± ( d dx x j (x = d dx x(j (x + 1 α ±(h (1 (x D : ɛr a E,± (j (x = j (x + 1 β ±(h (1 (x E : a M,± (j (x = j (x + 1 α ±(h (1 (x a E,± ( d dx x j (x = d ɛ r dx x(j (x + 1 β ±(h (1 (x

7 where we have defined x = ka, x = k a = x ɛ r We note that two of the six equations are redundant (this aso happened in the case of pane waves refecting and refracting off of a pane dieectric boundary. This aows us to sove four equations for four unknowns α, β, a E and a M. Since we are ony directy interested in the mutipoe coefficients α and β, we eiminate a E and a M from the above to obtain the soution α ± ( + 1 = h( (x d dx x j (x j (x d h (1 (x d dx x j (x j (x d dx xh( dx xh(1 β ± ( + 1 = h( (x d dx x j (x ɛ r j (x d h (1 (x d dx x j (x ɛ r j (x d (x (x dx xh( dx xh(1 We now note that (at east for rea ɛ r the above expressions are of the form of a ratio of a compex quantity divided by its compex conjugate. This indicates that the fractions have unit magnitude, and can be written in terms of rea phase shifts α ± ( = 1 = e iδ, β ± ( + 1 = e iδ Noting that gives e iδ = a ib a + ib tan δ = a b tan δ = j (x d dx x j (x j (x d dx xj (x n (x d dx x j (x j (x d dx xn (x tan δ = j (x d dx x j (x ɛ r j (x d dx xj (x n (x d dx x j (x ɛ r j (x d dx xn (x With a bit of simpification, these can be rewritten in the form (x (x (6 tan δ = xj (x B j (x xn (x B n (x tan δ = xj (x B j (x xn (x B n (x (7 where the coefficients B and B are B = x j (x j (x, B = 1 ɛ r ( x j (x j (x + 1 ɛ r (8 and may be thought of as parametrizing the matching conditions at the boundary of the dieectric sphere. Note that the expressions for tan δ and B are identica

8 to that from the quantum mechanica scattering probem. The presence of the primed quantities is the resut of vector waves as opposed to scaar waves. b Consider the ong-waveength imit (ka 1 and determine expicity the differentia and tota scattering cross sections. Compare your resuts with those of Section 10.1.B. For ka 1 ony the owest ( = 1 phase shift is important. In this case, we may approximate the spherica Besse functions j 1 (x = x x ( , n 1(x = 1 x (1 + x + Substituting this directy into (6, and keeping ony the owest non-trivia terms gives tan δ 1 = 1 45 x3 (x x = 1 45 (ɛ r 1(ka 5 tan δ 1 = ɛ r 1 3 ɛ r + x3 = ɛ r 1 3 ɛ r + (ka3 The mutipoe expansions are then approximated by α ± (1 = e iδ 1 1 iδ 1 = i 45 (ɛ r 1(ka 5 β ± (1 = e iδ 1 1 iδ 1 = 4i 3 ɛ r 1 ɛ r + (ka3 We see that ony the β 1 (eectric dipoe coefficient dominates at ow energies. The scattering cross section is dσ dω = π k + 1[α± ( X,±1 ± iβ ± (ˆn X,±1 ] π 4i ɛ r 1 k 3 3 ɛ r + (ka3ˆn X 1,±1 = 16π ( ɛr 1 6k (ka 6 X ɛ r + 1,±1 ( = 1 k4 a 6 ɛr 1 (1 + cos θ ɛ r + This agrees perfecty with the dipoe approximation. c In the imit ɛ r compare your resuts to those for the perfecty conducting sphere. When we take ɛ r we are taking x = ɛ r x. In this case, we use a arge argument approximation to the spherica Besse function j (x j (x 1 x sin(x π

9 This resuts in the asymptotic forms of the coefficients (8 B x cot(x π 1 B 1 ɛr x cot(x π 1 1 Substituting this into (7 gives tan δ = j (x n (x, tan δ = xj (x + j (x xn (x + n (x = d dx xj (x d dx xn (x which reproduce exacty the perfecty conducting sphere phase shifts The aperture or apertures in a perfecty conducting pane screen can be viewed as the ocation of effective sources that produce radiation (the diffracted fieds. An aperture whose dimensions are sma compared with a waveength acts as a source of dipoe radiation with the contributions of other mutipoes being negigibe. a Beginning with ( show that the effective eectric and magnetic dipoe moments can be expressed in terms of integras of the tangentia eectric fied in the aperture as foows: p = ɛˆn ( x E tan da m = iωµ (ˆn E tan da where E tan is the exact tangentia eectric fied in the aperture, ˆn is the norma to the pane screen, directed into the region of interest, and the integration is over the area of the openings. The diffraction resut ( states In the radiation zone, we may take E( x = 1 π (ˆn E eikr apertures R da (9 e ikr R eikr r e i k x Furthermore, for a sma aperture (ong waveength imit, we may expand the second exponentia e ikr R eikr r (1 i k x

10 Inserting this into (9 and noting that we may use the repacement i k in the radiation zone, we obtain the expansion E = i e ikr π r k (ˆn E(1 i k x da (10 We start with the first term in the expansion E 1 = i e ikr π r k ˆn E da which may be compared with the eectric fied of magnetic dipoe radiation (in the radiation zone E = Z 0 4π r ˆk m This aows us to read off the effective magnetic dipoe moment m = ˆn E ikz da = ˆn E 0 iωµ da (11 0 k eikr The effective eectric dipoe moment is somewhat trickier to extract. It is reated to the second term in (10, which we write as E = 1 e ikr π r k (ˆn E( k x da (1 Since we have a fat screen, the norma vector ˆn is constant. Furthermore, the outgoing momentum vector k is unreated to the integration coordinates (which ine on the screen. Thus these two vectors may be pued out of the integra. This means, we need to evauate the integra (given in components E i x j da where the indices i and j ony ie in the screen directions (ie i, j = 1, if we take ˆn = ẑ. We now show that E i x j da = 1 δ ij E x da (13 where we reemphasize that i and j ie in the screen directions ony. Perhaps the most direct way to prove this is to write E i x j in tensor form ( E x E1 x = 1 E 1 x E x 1 E x = 1 ( E1 x 1 + E x 0 0 E 1 x 1 + E x + 1 ( E1 x 1 E x E 1 x E x 1 E 1 x 1 + E x = 1 δ ij( E x + 1 [E ix j (ˆn x i (ˆn E j ] (14

11 The second term vanishes when integrated over the openings. This is because we may use E = 0 in a source-free region. Then 0 = x ix j ˆn ( E da = ɛ kmˆn k x ix j E m da = ɛ kmˆn k (x ix je m da = ˆn k (ɛ ikm x j + ɛ jkm x ie m da = [x i(ˆn E j + x j(ˆn E i ] da (15 Note that the surface term arising from the integration by parts vanishes because it is proportiona to E, which must vanish on the boundaries of the openings. Substituting in expicit components ij = 11, 1, and then proves that the integra of E x 1, E 1 x 1 E x, and E 1 x vanish, as needed to remove the second term from (14. This can aso be seen directy by taking a cross product of (15 with ˆn in the ith component to get [(ˆn x i (ˆn E j x je i ] da = 0 In any case, the resut is simpy (13, which may be substituted into (1 to obtain E = 1 e ikr 4π r k (ˆn k x E da = 1 e ikr 4π r k ( k ˆn x E da Comparing this with the radiation patter for eectric dipoe radiation E = k e ikr 4πɛ 0 r ˆk (ˆk p gives an effective eectric dipoe moment p = ɛ 0ˆn x E da Note the curious fact that the magnetic dipoe term comes from the owest order in the expansion of (9, whie the eectric dipoe term comes from the next order. This is backwards from what happens for a conventiona source given by a specified current density. b Show that the expression for the magnetic moment can be transformed into m = µ x(ˆn B da

12 Be carefu about possibe contributions from the edge of the aperture where some components of the fieds are singuar if the screen is infinitesimay thick. To reate the eectric fied to the magnetic fied, we may use Faraday s equation for harmonic fieds E iω B = 0 to write ˆn ( E = iω(ˆn B Mutipying this by a vector x and integrating gives iω x (ˆn B da = x [ˆn ( E] da = x ɛ ijkˆn i j E k da = j ( x ɛ ijkˆn ie k da = ˆn E da Note that for integration by parts, we use the fact that ˆn is a constant surface norma vector and that E vanishes at the edges of the aperture. More precisey, the generaization of Stokes theorem indicates that the surface term is of the form x ( E d so the eectric fied contribution indeed arises ony from the parae component to the edge of the aperture. Finay, substituting this integrated reation between E and B into (11 gives m = iωµ 0 ˆn E da = µ 0 x (ˆn B da