14A scattering from dieletric without ignoring

Scattering from Dielectric Objects without ignoring radiated field PhL 1.28.03 In the conventional approach, we start with this equation A = -ik and we know that P(r') = E(r'), where E is the total electric field inside the medium. The coefficient already accounts for microscopic averaging and Jackson shows how you can relate it to , the factor for an individual atom or molecule in the material. So E is the total average macroscopic E field in the object of interest. Now we really have to say that E = E0 + Erad , where E0 is our incident plane way, and Erad is the field we are trying to calculate, the one we will get from A, then B, then E. So we have then a bit of a problem. The usual assumption is that Erad is much smaller than E0 so we can ignore it in the integral. When we do this, for example, in our sphere calculation, we get this result: E = k2 (4a3) f() (E0- E0r ) where f() is a form factor which is on the order of unity. It is -1/3 at central and drops to zero etc. In our particular problem, we have k = 2/ and a = 4 and = 1/2 , so ka = 16. Also, (n-1) = 2. So we can write the above as: Erad = (ka)2 4 eikr (a/r) f() (E0- E0r ) = (16)2 2 (n-1) eikr (a/r) f() (E0- E0r ) Unless n is exceedingly small, it is clear that our field is much larger than E0 at r = a, the surface of the sphere. But use of this equation is illegal there because in deriving the above, we assume r >> a in order to expand the exponential to first term. So we are not allowed to evaluate the above anywhere near the sphere, so we cannot know whether Erad << E0 inside the sphere! What can we do about this? (1) We can go back to (9.3) and try to do our calculation more correctly without assuming r >> d. But then we have J in the integral and we cannot do the parts trick to replace it with . I thought I was doing this approach in the Appendix calculation below, but the parts trick does not work with the propagator in there, so cannot get P inside the integral. (2) Looking at Jackson (6.27), why can't I associate J with the displacement current, set H = B, and get Jdisp = (/4) t E =( -i/4) E The reason you cannot do this is that the J driving A does not include this term. The displacement current is just the t E in the curl B equation. Notice that the above also exists outside our scattering object! (3) We can try to figure out what J is inside a dielectric. I know now what it is for a mechanical dipole, but don't know how to get to the macroscopic J in the dielectric material. I know for example that Ji(r) = -2ip (r - xi ) = the current due to a very closely spaced mechanical dipole I can put this into our equation for A to get A(r) = -2i ( Ei ) where Ri = | r - xi| where Ei is that internal macroscopic field that dipoles feel. In the world of page 116, we can replace Ei with (4/3) P and we then set P = E and then end up with A(r) = -2i ((4/3) E) where Ri = | r - xi | or something along these lines. The sum is due to the current delta functions. If we had a crystal, we might be able to actually do the sum, which is just over our scattering object. But in general, cannot do it, so cannot really know A, even in this toy model of nothing but dipoles. But it was a good shot. (4) I can try the full multipole expansion approach and use boundary conditions at the dielectric sphere surface. I think this might be Mie Scattering. I have done some probing here, and things are mighty complicated. The claim is that Mie even ferrets out the rainbows and multiple paths somehow. If you use the Debye expansion you sort by multiple paths. It is pretty mysterious how a general solution can contain all this phenomenological stuff. Another point is that in the multipole expansion of a plane wave, you don't just have dipole, so higher moments are important in the general solution. I think it might be fun to take a look at this stuff. All I did was compute the dipole term in the radiation zone of a dielectric sphere. I am not even sure it is the leading term. The subject is of great interest because it includes familiar objects like rainbows, coronas, glories, and so on. You can even download programs to do the Mie scattering computation on your PC! Here I thought I was computing A but I was using P inside even though cannot do parts on J to get it that way, so this entire calculation is meaningless. I have done a bit on this and this is where I get: A(r) = -2ik E0 dr' r'2 dx' where R = r - r' and where I selected the z' axis to point along the r direction, and where we compute that R = where x' = cos'. This looks like one of those famous undoable integrals, but it is in fact quite doable at least in x'. Think first that R = = a = r'2 + z2 b = - 2r'z Then set y = , dy = (1/) (b/2)dx' so that dx'/ = (2/b)dy which makes the integral quite trivial dx' = (2/b) * dy eiky = * { exp(ik) - exp(ik) } = * { exp(ik[r'-z]) - exp(ik[r'+z]) } and this leaves us with A(r) = -2ik E0 * dr' r' { exp(ik[r'-z]) - exp(ik[r'+z]) } = -2ik E0 * [ -2isin(kz) ] * dr' r' eikr' Maple tells us the inside integral from 0 to a is -(1/k)2 * { 1 + eika [ ika - 1] } So final result is then A(r) = -2ik E0 * [ -2isin(kz) ] * (-1) (1/k)2 * { 1 + eika [ ika - 1] } = (4i/k) E0 { 1 + eika [ ika - 1] } This is amazing! What have we done here? We started with Jackson (9.3)