More on M and N functions

More on the M and N Functions PhL 2.16.03 1. About the M function In the previous paper on this topic, I made this identification for the "first term" multipole fields, Mm(r,,) = x [f(r) Ym r ] = - f(r) r x Ym = - f(r) Xm/C(,m) where Xm(,) = C(,m) * r x Ym(,) C(,m) = + 1 Carleton - 1 Mie (not sure what this means now) Jackson Sometimes, such as in Krugel, we found that Ym was replace with a linear combination of Ym within the same group. Let's now prove a few theorems concerning the M functions Theorem 1: If M x (c ) where c is either a constant vector or r, then M = - c x . Proof: x (c ) = x c + x c = x c = - c x Theorem 2: Let M x (c ). If c is either a constant vector or r, and if satisfies the wave equation (2 + k2) = 0, then so does M: (2 + k2) M = 0 Proof: If we apply 2 onto M, it moves through onto the c , and then we can apply our identity 2 (c ) = c2 + 2 c + 2 c where we think of a component of c at a time here, but write it as a vector. Now if c is a constant, the last two terms vanish and we get our result, 2 M x 2 (c ) = x (c 2) = - k2 x (c ) = - k2 M It turns out that this still works when c = r even though r is not a constant vector. Consider, 2 (r ) = r2 + 2 r + 2 r = - k2 (r ) + 0 + 2 where we have used the fact that 2 r = 0 since jjri= j j,i = 0. Obviously since r is linear, it cannot have a Laplacian. In the second term, we note that r = I , the identity matrix, which we might write in the form and then we would say that = (or do as components!) . So at this point we have 2 M x 2 (r ) = x [- k2 (r ) + 2 ] = - k2 M + 2 x = - k2 M QED Theorem 3: Let M x (r ) = - r x . If satisfies the wave-like equation 2 = f(r), then M satisfies the same equation, 2M = f(r) M . { This theorem is not true if we replace r with constant c.} Proof: 2 M = x 2 (r ) = x (r2) = x [(f(r) ) r ] = (f(r) ) x r + (f(r) ) x r = (f(r) ) x r = [ f(r) + f(r) ] x r = - r x f(r) + 0 = - f(r) r x = f(r) M by Theorem 1. Example of Theorem 3. Consider Xm = r x Ym so that = -Ym. We know that 2 Ym = - (+1)/r2 Ym so that f(r) = - (+1)/r2. Thus we know that 2 Xm = - (+1)/r2 Xm. We proved this fact in gory detail in our "vector harmonics.doc" paper. We might think of this as "the wave equation within a partial wave", loosely speaking. Summary of what we have so far: Define M x (r ) . Then: M = - r x , which is one of our usual VSH basis functions X , up to a constant M = 0 If satisfies a wave or wave-like equation , then so does M. Theorem 4. Suppose we define M f(r) Mo = f(r) x (r ) = - f(r) r x , where Mo = - r x and where: (1) = (,); (2) satisfies the wave-like equation 2 = g(r) where g(r) = -(+1)/r2 ; (3) f(r) is a linear combination of spherical Bessel functions such as j(kr), where k is the constant appearing in the Bessel equation. Then (2 + k2) M = 0. Corollary: If is any linear combination of spherical harmonics Ym within the manifold, theorem 4 is true. Proof of Theorem 4: From Theorem 2 we know that 2 Mo = g(r) Mo . So consider, 2 M = 2 [f(r) Mo ] = [2 f(r)] Mo + f(r) 2 Mo + 2 f(r) Mo = { 1/r2 r [ r2 r f(r)] + g(r) f(r) } Mo + 2 f '(r) r Mo Now comes a somewhat unobvious fact that I got caught on. Consider: - Mo = r x = r x ( 1/r + 1/r 1/S ) = - 1/S . So Mo is not even a function of r, and therefore r Mo = 0. Thus we get: 2 M = { 1/r2 r [ r2 r f(r)] -(+1)/r2 f(r) } Mo = { - k2 f(r) } Mo = - k2 M QED Since f(r) is a Bessel function as noted, we get the simplification of {...} shown on the right above. Comment: The above theorem shows an interesting result. We know that f(r) satisfies the scalar wave equation (where f(r) is a spherical Bessel of parameter k, and is a linear combination of spherical harmonics within ). If we are interested in solving the vector wave equation, then M = - f(r) r x works. It has the extra nice property that M = 0. Observation: The multipole field "main terms" for E and B are of exactly this form. Usually we have as a straight Ym , but it is OK to instead use something like = C [ Ym Y- m] if that is convenient. For the expansion of an polarized plane wave, this does turn out to be convenient, and m = 1. We are also satisfying E,B = 0 with this solution. 2. About the N function Now let's look at N defined by: N = x M where M satisfies the wave equation. Then N must also satisfy the wave equation, since 2 N = 2 x M = x (2 M) = x (-k2 M) = -k2 x M = -k2 N Notice also that N = 0. Usually, a multiplicative constant is used out in front, N = (1/k) x M since this removes a factor of 1/k in the "second terms" in the multipole expansions for E and B. We have just shown, then, that these "second terms" satisfy both the divergence condition and wave equation, so they represent a second solution to the wave equation mentioned earlier. So, we can now extend our earlier comment: Comment: We know that f(r) satisfies the scalar wave equation (where f(r) is a spherical Bessel of parameter k, and is a linear combination of spherical harmonics within ). If we are interested in solving the vector wave equation, then we know of two solutions: (1) M = - f(r) r x , and we also know that M = 0. (2) N = (1/k) x M = - (1/k) x [f(r) r x ], and N = 0. These are exactly the first and second terms we see in Jackson's multipole expansion formulas, apart from overall constants. We can think of M as M = constant x f(r) * X where X is one of Carleton's three VSH's. We know that we can explode N into a linear combination of the Y and Z VSH basis vectors. Only X of these three basis vectors satisfies the wave equation! However, the N linear combination of Y and Z also solves it! One other fact. With N scaled by the (1/k) factor as shown x N = k M So we really have this symmetrical situation: N = (1/k) x M M = (1/k) x N Also we know that (from inspection of M) r M = 0 so M is the "transverse" solution Of course we know this since M ~ X and we know that X and Z are transverse, Y radial. Note that , since N is a mixture of Y and Z, it has both transverse and radial pieces.