scattering from a dielectric sphere

Scattering from the dielectric sphere PhL 2.5.03 Assume 1 in the outer region, and 2 inside the sphere. Outer region = 1, inner = 2. We start off as on page 569. We use 16.139 for the plane wave coming in (with the B' and k' mods), and we assume 16.141 for the scattered fields with these same mods. The new feature in this problem compared to the conducting sphere is that we have to model the fields inside the sphere. It seems to me that 16.131 is exactly what we want in this case: all modes must be finite at r=0 because there is nothing singular there. We don't expect to have infinite E or B fields at the origin! Now here are the boundary conditions at the sphere surface: 1) 1E1 = 2E2 // the normal field has a discontinuity due to pol charge 2) x E1 = x E2 // the tangential field is same since no surface currents. 3) B1 = B2 // the normal field is same since no magnetic charge on surface 4) x B1 = x B2 // the tangential field is same since no surface currents. Comments on conditions 2 and 4. These come from little contour loops which are done at the surface. We will find that things are continuous across the boundary as long as a little area loop of the curl X gives 0, where X = E or B depending on which condition you have. Since the loop is of super thin area, the only way to get something non-zero is to have some kind of infinite surface term like a surface current. We do expect to have such a thin surface charge of this nature. Looking at 6.112, we expect the surface charge to be -P. If we follow through on this with our pill box, we get condition 1. Now 6.112 also shows a surface current which is dP/dt. What effect does this have? It tells us that x B = 4ikP. This in turn tells us that (B1 - B2) = 4ik PdA where the integral is over the loop surface. Half the loop is in material 1, the other half in material 2. In either half, we have P = E. Unless at least one of the tangential E fields is infinite at the surface, this integral goes to 0 as the surface shrinks, and we conclude that condition 4 is OK after all. Condition 2 is valid similarly provided the tangential B field is finite. Comments on conditions 3 and 4. There are conditions on the true B field, but not on our special B' field that we need to use in the multipole theory in the presence of and ! Let's now replace 3 and 4 with corrected versions for the B' fields! ( see separate document on this subject) 3) B'1 = B'2 4) x B'1 = x B'2 . So, we are going to have 4 "unknowns" in this problem: 2 coefficients in the scattered field equations, and 2 coefficients in the "inside field" equations. But we know that each curl condition above will yield two separate conditions, so we would appear to have 6 conditions on our 4 unknowns. In the conducting sphere case we had 3 equations in 2 unknowns, but two of the equations were the same, maybe that will happen here as well. Now let's write out the expansions: Einc = (1/2) i [ 2j(k1r) X,1 2/k1 x { j(k1r) X,1} ] B'inc = (1/2) i [ 2i j(k1r) X,1 - 2i/k1 x { j(k1r) X,1} ] Esc = (1/2) i [ () h(1)(k1r) X,1 ()/k1 x { h(1)(k1r) X,1} ] B'sc= (1/2) i [ i () h(1)(k1r) X,1 - i ()/k1 x { h(1)(k1r) X,1} ] E2 = (1/2) i [ ' () j(k2r) X,1 ' ()/k2 x { j(k2r) X,1} ] B'2= (1/2) i [ i ' () j(k2r) X,1 - i' ()/k2 x { j(k2r) X,1} ] _________________________________________________________________________ Boundary Condition (1): 1 (Einc + Esc) = 2E2 We know that the first expansion term gives nothing, and in the second expansion term we get only the first term of 16.143. If we ignore common factors, we get, 1 { 2/k1 j(k1r) ()/k1 h(1)(k1r) } = 2 { ' ()/k2 j(k2r) } _________________________________________________________________________ Boundary Condition (3): (B'inc + B'sc) = B'2 This is done in similar fashion, but with the B fields, and we get - 2i/k1 j(k1r) - i ()/k1 h(1)(k1r) = - i' ()/k2 j(k2r) _________________________________________________________________________ Boundary Condition (2): x (Einc + Esc) = x E2 Here we pick up the two kinds of terms and we get two conditions, to wit, x Xm terms: 2j(k1r) + () h(1)(k1r) = ' () j(k2r) - Xm /r terms: 2/k1 r[r j(k1r)] ()/k1 r[r h(1)(k1r)] = ' ()/k2 r[r j(k2r) ] _________________________________________________________________________ Boundary Condition (4): x (B'inc + B'sc) = x B'2 Same idea as in the previous boundary condition: x Xm terms: 2i j(k1r) i () h(1)(k1r) = i ' () j(k2r) - Xm /r terms: - 2i/k1 r[r j(k1r)] - i ()/k1 r[r h(1)(k1r)]= - i' ()/k2 r[r j(k2r)] Summary of Boundary Conditions: (1) 1 { 2/k1 j(k1r) ()/k1 h(1)(k1r) } = 2 { ' ()/k2 j(k2r) } (2a) 2j(k1r) + () h(1)(k1r) = ' () j(k2r) (2b) 2/k1 r[r j(k1r)] ()/k1 r[r h(1)(k1r)] = ' ()/k2 r[r j(k2r) ] (3) - 2i/k1 j(k1r) - i ()/k1 h(1)(k1r) = - i' ()/k2 j(k2r) (4a) 2i j(k1r) i () h(1)(k1r) = i ' () j(k2r) (4b) - 2i/k1 r[r j(k1r)] - i ()/k1 r[r h(1)(k1r)]= - i' ()/k2 r[r j(k2r)] We seem to have 6 conditions on our 4 coefficients, which is a little scary. I can see from the above that (1) and (4a) completely determine () and ' (). Similarly, it would seem that (2a) and (3) completely determine () and ' (). This means we have (2b) and (4b) left over, and these could mean we are overdetermined, but I presume they will in fact just be redundant. Let's go ahead and compute the 's as just described and see what we get: (1)*ik1 2i1 j(k1r) 1i () h(1)(k1r) = 2 i ' ()(k1/k2) j(k2r) (4a) * 1 / 2i1 j(k1r) i 1 () h(1)(k1r) = i1' () j(k2r)(/) Add these to get, 0 + 0 = 2 i ' ()(k1/k2) j(k2r) i1' () j(k2r) (/) or 0 = i ' ()j(k2r) { 2(k1/k2) 1 (/)} But k1 = k so that (k1/k2) = /. So {..} factor is then { 2(k1/k2) 1 (/)} = { 2(/) 1 (/)} Now to get a good result, we have to assume that 1 = 2 otherwise we have magnetization M and I may not have done things right up to this point. If we assume that, then we get: {...} = { 2 1 } = { } = 0!!! This is a very welcome result! It just says that our two equations (1) and (4a) are not independent. Let's go back and look again at those equations: (1)*ik1 2i1 j(k1r) 1i () h(1)(k1r) = 2 i ' ()(k1/k2) j(k2r) (4a) * 1 / 2i1 j(k1r) i 1 () h(1)(k1r) = i1' () j(k2r)(/) The RHS of the lower equation has factor , and the upper equation RHS has the exact same factor. so finally I have broken through this logjam!!! The trick was to be sure to state boundary conditions 3 and 4 in terms of the B' field! So now let's take one of these equations along with (2b) and try again to get a solution for the 's! (1)*k1 * /1 * r[r h(1)(k1r)] 2j(k1r) r[r h(1)(k1r)] + ()h(1)(k1r) r[r h(1)(k1r)] = (2/1) ' ()(k1/k2 ) j(k2r) r[r h(1)(k1r)] (2b)*k1* * h(1)(k1r) 2r[r j(k1r)] h(1)(k1r) + ()r[r h(1)(k1r)] h(1)(k1r)= ' ()(k1/k2 )r[r j(k2r) ] h(1)(k1r) Now finally we can subtract to cancel the second term. In doing this, if we do out the two derivative terms in the first term, one of those two terms cancels when we subtract, leaving us with 2r{ j(k1r) r[h(1)(k1r)] - h(1)(k1r) r[j(k1r)] } = ' ()(k1/k2 ){ (2/1) j(k2r) r[r h(1)(k1r)] - r[r j(k2r) ] h(1)(k1r) } On the LHS we can multiply by k1 / k1 to get LHS = 2r k1 { j(k1r) rk1[h(1)(k1r)] - h(1)(k1r) rk1[j(k1r)] } = 2r k1 W [j(k1r), h(1)(k1r) ] From Jackson page 541 we can write the LHS as 2r k1 W[j(k1r), h(1)(k1r)] = 2r k1 * i (k1r)-2 = 2i/( k1r) We are left with 2i/( k1r) = ' ()(k1/k2 ){ (2/1) j(k2r) r[r h(1)(k1r)] - r[r j(k2r) ] h(1)(k1r) } If we had (2/1) = 1, the RHS would also make a the same Wronskian, and we would have that ' () = 2. This says that the incident wave just comes through and becomes the area 2 wave, at least for the electric (?) component. So this is a good check on the result I think. So here is our result, using the fact that (k1/k2 ) = , [ 2i/( k1r)] = ' (){j(k2r) r[r h(1)(k1r)] - r[r j(k2r) ] h(1)(k1r) } ' () = It ain't pretty, but I think it might be right! Now we need to get something as well for regular . Let's just start over and make the RHS's cancel: (1) /2*r[r j(k2r)] 2(1/2)/k1 j(k1r) r[r j(k2r)] ()(1/2)/k1 h(1)(k1r) r[r j(k2r)] = ' ()/k2 j(k2r) r[r j(k2r)] (2b)* j(k2r) 2/k1 j(k2r)r[r j(k1r)] ()/k1 j(k2r)r[r h(1)(k1r)] = ' ()/k2 j(k2r)r[r j(k2r) ] Subtract and the RHS cancels, so we have 2(1/2)/k1 j(k1r) r[r j(k2r)] - 2/k1 j(k2r)r[r j(k1r)] + ()(1/2)/k1 h(1)(k1r) r[r j(k2r)] - ()/k1 j(k2r)r[r h(1)(k1r)] = 0 Mult by k1 and group a little 2{ (1/2) j(k1r) r[r j(k2r)] - j(k2r)r[r j(k1r)] } = () { j(k2r)r[r h(1)(k1r)] -(1/2)/k1 h(1)(k1r) r[r j(k2r)] } Now kill the overall sign and we get () = Maybe it's time to power up Maple a bit to do this algebra. Well, OK, let's go after the 's so we can put this problem to bed! I will pick 2 of the 3 equations that seem to be independent: (2a) 2j(k1r) + () h(1)(k1r) = ' () j(k2r) (4b) - 2i/k1 r[r j(k1r)] - i ()/k1 r[r h(1)(k1r)]= - i' ()/k2 r[r j(k2r)] Now adjust to cancel the RHS (2a) *r[r j(k2r)] *i/ k2 2ij(k1r) r[r j(k2r)] / k2 + () i h(1)(k1r) r[r j(k2r)] / k2 = i ' ()/ k2 j(k2r) r[r j(k2r)] (4b)/* j(k2r) - 2i/k1 j(k2r)r[r j(k1r)] - i ()/k1 j(k2r)r[r h(1)(k1r)]= - i' ()/k2 j(k2r)r[r j(k2r)] Add and RHS cancels and we get 2ij(k1r) r[r j(k2r)] / k2 + 2i/k1 j(k2r)r[r j(k1r)] = - i ()/k1 j(k2r)r[r h(1)(k1r)] - () i h(1)(k1r) r[r j(k2r)] / k2 Mult all by k1 and group a bit, 2i { (k1/ k2 ) j(k1r) r[r j(k2r)] + j(k2r)r[r j(k1r)] } = - i () { j(k2r)r[r h(1)(k1r)] + (k1/ k2 ) h(1)(k1r) r[r j(k2r)] } But (k1/ k2 ) = so that common factor can be completely removed along with an i 2{ j(k1r) r[r j(k2r)] + j(k2r)r[r j(k1r)] } = - () { j(k2r)r[r h(1)(k1r)] + h(1)(k1r) r[r j(k2r)] } which gives for our result - () = One more to go! (2a) 2j(k1r) + () h(1)(k1r) = ' () j(k2r) (4b) - 2i/k1 r[r j(k1r)] - i ()/k1 r[r h(1)(k1r)]= - i' ()/k2 r[r j(k2r)] (2a) **i*r[r h(1)(k1r)] 2ij(k1r) r[r h(1)(k1r)] + i ()h(1)(k1r) r[r h(1)(k1r)] = i' () j(k2r) r[r h(1)(k1r)] (4b)*k1 * h(1)(k1r) - 2i h(1)(k1r) r[r j(k1r)] - i ()h(1)(k1r) r[r h(1)(k1r)] = - i' () ( k1/ k2 ) h(1)(k1r) r[r j(k2r)] No add to cancel the terms: 2ij(k1r) r[r h(1)(k1r)] - 2i h(1)(k1r) r[r j(k1r)] = i' () j(k2r) r[r h(1)(k1r)] - i' () ( k1/ k2 ) h(1)(k1r) r[r j(k2r)] Group a little, divide by 2i { j(k1r) r[r h(1)(k1r)] - h(1)(k1r) r[r j(k1r)] } = i' ()j(k2r) r[r h(1)(k1r)] - i(/)' () ( k1/ k2 ) h(1)(k1r) r[r j(k2r)] On the right side, the k and factors cancel leaving us with 2i { j(k1r) r[r h(1)(k1r)] - h(1)(k1r) r[r j(k1r)] } = i' () { j(k2r) r[r h(1)(k1r)] - h(1)(k1r) r[r j(k2r)] } Only the LHS gives a Wronskian LHS = 2i r k1 W[j(k1r), h(1)(k1r) ] = 2i r k1 * i (k1r)-2 = -2/(k1r) leaving us with -2/(k1r) = i' () { j(k2r) r[r h(1)(k1r)] - h(1)(k1r) r[r j(k2r)] } so that we get this result, ' () = Now let's grab all the results of this problem: Einc = (1/2) i [ 2j(k1r) X,1 2/k1 x { j(k1r) X,1} ] B'inc = (1/2) i [ 2i j(k1r) X,1 - 2i/k1 x { j(k1r) X,1} ] Esc = (1/2) i [ () h(1)(k1r) X,1 ()/k1 x { h(1)(k1r) X,1} ] B'sc= (1/2) i [ i () h(1)(k1r) X,1 - i ()/k1 x { h(1)(k1r) X,1} ] E2 = (1/2) i [ ' () j(k2r) X,1 ' ()/k2 x { j(k2r) X,1} ] B'2= (1/2) i [ i ' () j(k2r) X,1 - i' ()/k2 x { j(k2r) X,1} ] ' () = ' () = - () = () = Of course everywhere I have r, I mean it to be evaluated at r = a, the sphere radius. The next step I think would be to verify these results and fix them using Maple. Then the trick is to come up with phase shifts, if that is possible. In Problem 16.12 Jackson hints that phase shifts are possible At some point, I could go look up the Mie and Debye 1908 paper. I would of course expect this subject to be summarized in many later works, but I have not found such a review on the web. I did want to do this for myself before reviewing someone else's results. Maybe Born and Wolf will have the results as Jackson hints. I wonder if there are advanced texts that are much heavier duty that even Jackson's text.