scattering from a dielectric sphere_confused

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 suspect? // 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. 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) = /. Unless it happens that the 1 and 2 parameters are the same, this last { .. } is non-zero and we are forced to conclude that ' () = 0! If the 1 and 2 are the same, then this equation tells us nothing about ' (). Assuming then that ' () = 0, we can solve either equation for the 's: 2i1 j(k1a) 1 i () h(1)(k1a) = 0 => () = -2 j(k1a)/ h(1)(k1a) Note that this seems to be the () solution in the Jackson conducting sphere problem. Now if these are the correct solution and ', then what does equation (2b) have to say about things! ? (2b) 2/k1 r[r j(k1r)] ()/k1 r[r h(1)(k1r)] = ' ()/k2 r[r j(k2r) ] => 2r[r j(k1r)] + ()r[r h(1)(k1r)] = ' ()(k1/k2 )r[r j(k2r) ] => 2r[r j(k1r)] + ()r[r h(1)(k1r)] = 0 => 2rr[j(k1r)] + ()rr[h(1)(k1r)] + 2 j(k1r) + ()h(1)(k1r) = 0 => 2rr[j(k1r)] + {-2 j(k1a)/ h(1)(k1a) } rr[h(1)(k1r)] + 2 j(k1r) +{-2 j(k1a)/ h(1)(k1a)} h(1)(k1r) = 0 The last two terms cancel, so this requires that => 2rr[j(k1r)] + {-2 j(k1a)/ h(1)(k1a) } rr[h(1)(k1r)] = 0 => r[j(k1r)] + {- j(k1a)/ h(1)(k1a) } r[h(1)(k1r)] = 0 => h(1)(k1a) r[j(k1r)] - j(k1a) r[h(1)(k1r)] = 0 => Wronskian (h(1)(k1a), j(k1a)) = 0 But the Wronskian W[h(1)(k1a), j(k1a)] = -i (k1a)-2 according to Jackson page 541, so we definitely have a conflict, there is no point in continuing until this conflict is resolved. Comments: I have very carefully gone over every item in the above document, I have not discovered what my mistake is that leads to this inconsistency. Undaunted, let's go ahead and try to compute the coefficients in a similar manner: (2a) *i 2ij(k1r) + i () h(1)(k1r) = i' () j(k2r) (3)*k1 - 2i j(k1r) - i () h(1)(k1r) = - i' ()(k1/k2) j(k2r) Add and as before, the first two terms cancel. We then get 0 + 0 = i' () j(k2r) { 1 - (k1/k2) } As before, if the mediums have different k's, then we conclude that ' () = 0 and then 2ij(k1r) + i () h(1)(k1r) = 0 => () = -2 j(k1r)/ h(1)(k1r) and I am sure then that equation (4b) will lead to another Wronskian violation! So our entire solution to this problem (ignoring our "little problem") is this: ' () = ' () = 0 () = () = -2 j(k1a)/ h(1)(k1a) This solution says there is no E or B field whatsoever inside the sphere, which certainly seems unlikely! If no E field in there, then no P, and then no surface charge . Certainly in the large limit we would expect our incoming plane wave to enter the sphere by Snell's law, etc. Conclusion: my assumed form for the field inside the sphere must be wrong! ?