multipole

The Multipole Expansion PhL 1.29.03 Personal history: When I took Jackson's class perhaps in 1971, we did multipoles for statics, but we never got to the famous Chapter 16 for dynamics. Now with Mie scattering being an interest, I am more motivated to delve into this subject in more detail. Chapter 2 Statics Review In statics, you have the wave-equation with k=0 ( = ) with as the driving source [ known as Poisson], as in (1.28), and the general solution is (1.17) where 1/R is the Green's function or propagator (think exp(ikR)/R with k=0). This is the "particular solution", and you can add solutions to 2 = 0 [ known as Laplace ] , as in (1.36), in order to find a complete solution that matches some boundary conditions. The two boundary surface terms can be interpreted as from an effective surface charge and "dipole layer", page 15. In the boundary conditions, you are either specifying the normal E field at the surface (Neumann), or you specify itself (Dirichlet). If you specify either one on a closed boundary, you get a unique and correct solution, but if you try to specify both (Cauchy) , it is "too much" and the only solution is 0. Specifying any of these three on an open (partial) surface is "not enough" to get a solution. This is shown in the first column of the table on page 17. Notice the second column which applies to "hyperbolic" ODE's like the wave equation. In this case, specifying anything on a closed surface is too much, and the only chance you have is doing Cauchy on an open partial surface. In Kirchhoff scalar diffraction theory (field ) applied to an aperture, you are in effect trying to specify and /n on a closed surface (in the hole, you are assuming you have the incident field unaltered, on the screen you assume 0, on the great sphere you assume 0). Thus you are trying to do Cauchy on the wave equation, and that is known to only give a 0 solution. The third column applies to ODE's like the heat equation which I have never really studied! Jackson does not prove these things, but refers us to famous historic sources like Morse & Feshbach, and Sommerfeld himself. On page 18 Jackson starts into the problem of "finding the Green's function" for a particular set of boundary conditions. He notes that the 1/R thing is only the particular solution, and to this you can add a solution F of Laplace, and this is the part that lets you satisfy the BC's. In Dirichlet, you interpret F as the potential caused by charges a point source induces on conductors. We then move into Chapter 2 and start looking at some real problems. For a point charge near a plane, you can "simulate" the boundary conditions you know must exist at the plane by using an image point charge to replace the plane. In this problem, =G is the solution, and = constant is required on the metal plane surface, and we know how to make an image charge to get G=0. In this case, the potential of the image charge simulates the potential that is in fact created by the surface charge on the metal surface. Example: The Problem of a Point Charge Near a Sphere I never realized until just now the significance of this problem and how it relates to something I recently was wondering about. Here is the logic flow. First, Jackson puts a point charge q distance y from the center of a grounded sphere of radius a, and puts it outside so y > a. He shows that you can get = G = 0 on the sphere by assuming one image charge inside the sphere whose size is -q(a/y) and whose distance from sphere center is a2/y . The potential solution is then given by the first 2 terms of (2.8), and you can then go and compute whatever you want. In this solution, the grounded sphere brings in from ground a total charge of -q(a/y), as Gauss's law tells you. Again, we see that the image charge is simulating what the surface charge distribution really does. Suppose you add to this solution a sphere with total charge Q1 where Q1 = Q - [-q(a/y)] . The total solution then has Q on the sphere and solution (2.8) with the third term. This is then the solution you get by putting a point charge outside a charged insulated sphere. But why is this interesting? The potential is the sum of three terms, and if you do , you get the electric field (F(x)) at all points in space, and it is a function of a, y, and Q. However, if you want to know the force on the point charge q, you only consider the last two terms in (2.8)and set x=y, and this then gives (2.9). The point charge feels a force from the Q at sphere center, and from the image charge which lies under the sphere surface below the point charge. The distance between the point charge and the image charge is given by y - a2/y . As the point charge approaches the surface, the image charge approaches from the other side and it's force dominates. The image charge always has a sign opposite q. If Q and q have opposite signs, then q is attracted by the central Q and by the image charge. The more interesting case is when Q and q have the same sign. Then q is repelled by the central Q, but attracted by the image charge. The two forces are equal when q lies about (a/2)above the sphere surface in the case Q>>q. For an electron over a macroscopic sphere of some reasonable charge, this is a very small distance. As q approaches the surface, the image charge wins out (the reforming surface charge, that is to say) and the force becomes infinite right at the surface. So here is the point: an individual electron on a charged sphere is held on by an infinite restoring force if it tries to leave. No doubt if you work with a non-idealized metal surface, you will find the force is finite, and a finite energy can get an electron over the equilibrium hump and this is called the work function. Not too long ago I made a bogus argument I not think about why electrons did not leave a charged metal object in which I referred to the 1 eV chemical idea. I thought the electron was held by its binding stability energy. That might be true, but this image charge force seems much more significant. This whole discussion could have been tried with regard to a charge over a flat metal surface, but there it does not work so easily. You could imagine an insulated large round plate that is grounded and then you get the image charge solution assuming very large radius. But then you think of that plate as having charge Q in addition and you compute of this, things have log divergence, not very nice. You have to put in a cutoff radius R, but I think the whole thing can be made to work. For radius R of the plate, I think this is the potential the point charge sees from the image and the plate: (y) = -q/(2y) + (Q+q)/(R2) [ - y] If I set '(y) = 0 with large Q, I find that y = Ris the equilibrium point. Yes, it is quite ugly. The problem here is that distances do matter, you cannot just ignore far away things and say that locally the surface is flat and we will use the "flat solution". Jackson solution diverges for a as well. So there sphere is a very instructive case indeed that shows clearly the principle involved. In other cases, the same general idea will probably be true, but much harder to compute. Now we resume our static review. Jackson does a conducting sphere in a uniform E field, uses image charges at to simulate the E field and finds that the sphere acts as a dipole object. We skipped and I now skip the method of inversion and we get to page 41. Here, he reminds us of what G must be for a charge near a conducting sphere (charge + image), and reminds us that since this makes G=0 on the sphere, this is also the G for the general problem of some defined on the sphere, where you set q=1. So we can use this G to solve fancier Dirichlet problems where is prescribed on a spherical surface. The solution is given in (1.44) in this case. The first term is 0 if you are outside the sphere, and the second term is what we want, and we get (2.25). Jackson then applies this to a capacitor made of two hemispheres and gets (2.26). Aha! Here in fact is a non-doable integral and he admits it! He can get the result on-axis, and he can power-series expand things and he gets something that looks like multipole, but that idea has not been set up yet.