Green's Function for two concentric rings in 3D

Green's Function for two concentric rings in 3D PhL 3.9.10 When I started this doc, I thought I knew the solution to the one ring problem and I wanted to see then what the two-ring solution might be in hopes of maybe assembling a flat disk from a set rings. I then realized that I didn't even know the one-ring solution, and so then I started my little one-ring doc and more or less abandoned this document. Overview (perhaps written 4.21.10) 1 1. The setup. 2 2. How do you solve the integral equations? 4 3. What about a "force" approach to finding the charge density? 6 4. What about the horrible cos(nθ) expansion? 6 2A. How do you solve the integral equations? 7 2B. How do you solve the integral equations? 9 __________________________________________________________________________________ Overview (1/2 page, perhaps written 4.21.10) In Section 1 I "set up" the two-ring problem in terms of a pair of coupled 1D integral equations each of which says that the potential ON one of the two rings is 0. At section end, I simplify to one ring only so there is just one integral equation. (Only later in the one-ring doc did I learn that the intra-ring integrals in these integral equations in fact diverge. ) In Section 2 I try doing formal expansions of the various 1/R factors in these integral equations in Pn(cosθ )functions, mainly because the expansions are pretty simple. Since these are not representation functions of SO(2), the integral equations fail to formally diagonalize and I end up with a digital integral equation such that the integral over θ is now replaced by a sum over m, and I am nowhere man. In Section 3 I think about a "force" approach instead of a potential one, and quickly give up on that. In Section 4 I consider expanding all the 1/R factors in cos(nθ) functions, which actually is a more correct thing to do in this problem. But I cringe at the horrible Fn coefficients which arise, and at this time I am unaware that I have a simple O(2) diagonalization problem. So I quickly give up on this avenue. [ These are the Fn things that I much later found were Qn-1/2 functions. ] Then Section 2A is a fruitless rehash of Section 2 above with a few alterations. Section 2B is a similar effort. I now know that even the one-ring problem has "no solution" in the sense of infinitesimally thin wires, so it is a big waste of time to think about problems with more than one such wire! But of course I did not know that when I started this doc. I was somehow hoping to "build up" the disk from some wires, and that was my original vague motivation to think about a two-wire problem. I was sort of following Smythe's advice somewhere. ___________________________________________________________________________________ 1. The setup. We start by picking an observation point on the inner ring at location r1 = (a,θ). Point charge q is inside both rings at location c = c . Integration points are called ra = (a,θ') and rb = (b,θ'). The inner ring has a linear charge density σa(θ') and ds = b dθ' as shown in the figure, and similarly for the other ring. The potential at our selected point r1 must be zero since it lies on a grounded metal ring, so we write 0 = V(r1) = q/|r1-c| + ∫dθ' [ aσa(θ')/|ra-r1| + bσb(θ')/|rb-r1| ] for all r1 on ring a Similarly for a point r2 on the outer ring we have 0 = V(r2) = q/|r2-c| + ∫dθ' [ aσa(θ')/|ra-r2| + bσb(θ')/|rb-r2| ] for all r2 on ring b So we seem to have two integral equations (in 1 dimension) with two unknowns. We can write distances in this way | p - q|2 = p2 + q2 - 2pqcos(pq) so then we have | r1-c |2 = a2+ c2-2accos(θ) | r2-c |2 = b2+c2-2bc cos(θ) // when we observe from here, also θ1 say | ra-r1 |2 = a2 + a2 - 2a2 cos(θ-θ') = 2a2[1-cos(θ-θ')] | ra-r2 |2 = a2+b2-2abcos(θ-θ') | rb-r1 |2 = a2+b2-2abcos(θ-θ') | rb-r2 |2 = b2 + b2 - 2b2 cos(θ-θ') = 2b2[1-cos(θ-θ')] = 4 b2 sin2(θ-θ') So here are our two integral equations: 0 = q/ + ∫dθ' [ aσa(θ')/ + bσb(θ')/ ] 0 = q/ + ∫dθ' [ aσa(θ')/ + bσb(θ')/ ] Let's process these just a bit to get a more dimensionless form: 0 = q/ + ∫dθ' [ σa(θ')/ + bσb(θ')/ ] 0 = q/ + ∫dθ' [ aσa(θ')/ + σb(θ')/ ] Write a2+ c2-2accos(θ) = c2[ 1+α2-2α cos(θ)] α = a/c a2+ b2-2abccos(θ) = c2[α2+β2-2αβ cos(θ)] β = b/c b2+c2-2bc cos(θ) = c2[1+β2-2β cos(θ)] Then we have these equations: 0 = (q/c)/ + ∫dθ' [ σa(θ')/ + βσb(θ')/ ] 0 = (q/c)/ + ∫dθ' [ ασa(θ')/ + σb(θ')/ ] where now all terms have dimensions charge/length in form q/c or σ . Let's simplify again using α2+β2-2αβ cos(θ-θ') = α2[1+(β/α)2- 2(β/α) cos(θ-θ')] to get 0 = (q/c)/ + ∫dθ' [ σa(θ')/ + σb(θ')/ ] 0 = (q/c)/ + ∫dθ' [σa(θ')/ + σb(θ')/ ] Then make the change θ" = θ-θ' to get 0 = (q/c)/ + ∫dθ" [ σa(θ-θ")/ + σb(θ-θ")/ ] 0 = (q/c)/ + ∫dθ" [σa(θ-θ")/ + σb(θ-θ")/ ] and again changing back to single prime 0 = (q/c)/ + ∫dθ' [ σa(θ-θ')/ + σb(θ-θ')/ ] 0 = (q/c)/ + ∫dθ' [σa(θ-θ')/ + σb(θ-θ')/ ] I am not sure which pair will be best for solution. If there is only one ring of radius a, we can set σb = 0 in the above and we don't have the second equation which says the potential is 0 on the outer ring. So we have just this integral equation in this case: 0 = (q/c)/ + ∫dθ' [ σa(θ')/ or ∫dθ' σa(θ') / |sin[(θ-θ')/2] | = – (2q/c)/ α = a/c 2. How do you solve the integral equations? I don't know! One obvious starting point is to expand σa(θ') in some carefully selected manner. I know now that you do not want to expand on cos(nθ) functions since things are a mess. Everything seems to point to Legendre polynomials. The famous generating function is shown page 146 Schaum 1/ = Σn=0∞ Pn(cosθ) βn If we set β = 1 this becomes 1/ = Σn=0∞ Pn(cosθ) where I have to wonder about convergence. But suppose we formally expand all 1/radicals in our two equations in this manner. Looking at our first pair of coupled equations above, 0 = (q/c)/ + ∫dθ' [ σa(θ')/ + βσb(θ')/ ] 0 = (q/c)/ + ∫dθ' [ ασa(θ')/ + σb(θ')/ ] we see that we will then find ourselves dealing with things like this: ∫dθ' σa(θ') Pn(cos[θ-θ']) We know that the Legendre function addition theorem will come into play here. First cos(θ-θ') = cosθcosθ' + sinθsinθ' We can then use the form shown Bateman p 168-9 which is this: ψ = 1 Pn(cos[θ-θ']) = Pn(cosθcosθ' + sinθsinθ') = Pn(cosθ) Pn(cosθ') + 2 Σm=1∞ (-1)m Pn-m(cosθ) Pnm(cosθ') Then our integral above is this: ∫dθ' σa(θ') Pn(cos[θ-θ']) = ∫dθ' σa(θ') { Pn(cosθ) Pn(cosθ') + 2 Σm=1∞ (-1)m Pn-m(cosθ) Pnm(cosθ') } = Pn(cosθ) ∫dθ' σa(θ') Pn(cosθ') + 2 Σm=1∞ (-1)m Pn-m(cosθ) ∫dθ' σa(θ') Pnm(cosθ') So let's just define the following coefficients: σanm ≡ ∫dθ' σa(θ') Pnm(cosθ') so we then have ∫dθ' σa(θ') Pn(cos[θ-θ']) = Pn(cosθ) σan0 + 2 Σm=1∞ (-1)m Pn-m(cosθ) σanm (*) Our integral equations are then going to have this sum over m sitting in them. We can balance the n sums term by term (maybe) and get rid of the n sum. All we then accomplish is replacement of the integral equations with sum over m equations. Suppose we assume that σb = 0 and we ignore the second equation which said that the potential on ring b was zero. This reduces us to the "one ring problem" of radius a. We then have this integral equation to start with. 0 = (q/c)/ + ∫dθ' σa(θ')/ Let's now actually insert our expansions quoted above, namely 1/ = Σn=0∞ Pn(cos[θ-θ']) 1/ = Σn=0∞ Pn(cosθ) αn α = a/c We then get 0 = (q/c) Σn=0∞ Pn(cosθ) αn + Σn=0∞ ∫dθ' σa(θ') Pn(cos[θ-θ']) We then insert our result (*) above and get: 0 = (q/c) Σn=0∞ Pn(cosθ) αn + Σn=0∞ { Pn(cosθ) σan0 + 2 Σm=1∞ (-1)m Pn-m(cosθ) σanm } or 0 = Σn=0∞ Pn(cosθ) [(q/c) αn + σan0] + 2 Σn=0∞ Σm=1∞ (-1)m Pn-m(cosθ) σanm This result certainly is clumsy and suggests that I took a wrong term. The first term is a reasonable expansion on the Pn but the second term is a double expansion on Pnm so we have not expanded on a complete set of basis functions, but on functions from different complete sets! Very ugly. You have to decide what complete set of functions you are aiming for, but here I just let the math pull me along and it dumped me in a ravine. Below the **** line I show other attempts to deal with things that don't help much. So I conclude that even the "two-ring" problem is too hard for me to do! Ouch! 3. What about a "force" approach to finding the charge density? Consider the point r1 in our picture: Presumably the tangential force on a piece of charge at r1 is zero. Since the charge density even on the a ring is not azimuthally symmetric, we will have a force integral that is just as ugly if not uglier than that in the potential approach. We have 1/r2 so we get rid of the radicals, maybe that is a plus. But then we have the vector mess to deal with. I cannot imagine this is a simpler approach! 4. What about the horrible cos(nθ) expansion? Here is one of our integral equations above 0 = (q/c)/ + ∫dθ' [ σa(θ')/ + σb(θ')/ ] Suppose we go ahead and use this mess Coefficients of 1/R = 1/|r-r'| = (1/r) 1/ a=1+f2 b = 2f f = r'/r = (1/π) Σn Fn(f) cos(nθ) Fn(f) = (2/ | 1-f ]) Σs=0n (-1)s (2n, 2n-2s) B[ s+1/2, n-s+1/2 ] F [s+1/2, 1/2; n+1; -4f/(1-f)2 ] How would we apply this to one of our radicals? | r1-c |2 = a2+ c2-2accos(θ) = c2[ 1+α2-2α cos(θ)] α = a/c 1/R = 1/| c-r1| = (1/c) 1/ a = 1+α2 b = 2α α = a/c f = α So using this, we can expand our outside term. But what about one of the integral terms: | ra-r1 |2 = a2 + a2 - 2a2 cos(θ-θ') = a2[2-2cos(θ-θ')] 1/R = 1/| ra-r1 | = (1/a) 1/ a = 2 b = 2 f = 1 In this case, our F formula diverges. In any event, even if things were nice here, we still have Σs=0n sum left over so things do not diagonalize. ******************************************************************************** These two sections are alternate versions of section 2 above, but neither was more successful! 2A. How do you solve the integral equations? Here I tried a slightly different threading through the same math, where I added a sine factor(red) to make the required end-game integral look better. But no cigar. I don't know! One obvious starting point is to expand σa(θ') in some carefully selected manner. I know now that you do not want to expand on cos(nθ) functions since things are a mess. Everything seems to point to Legendre polynomials. The famous generating function is shown page 146 Schaum 1/ = Σn=0∞ Pn(cosθ) βn If we set β = 1 this becomes 1/ = Σn=0∞ Pn(cosθ) where I have to wonder about convergence. But suppose we formally expand all 1/radicals in our two equations in this manner. We will then find ourselves dealing with things like this: ∫dθ' σa(θ') Pn(cos[θ-θ']) Suppose we simply expand the charge densities in the (almost) obvious manner: [σa(θ')/sinθ'] = Σn=0∞ σan Pn(cosθ') Then we will be stuck with this Σn=0∞ σan ∫dθ' sinθ' Pn(cosθ') Pn(cos[θ-θ']) We know that the Legendre function addition theorem will come into play here. First cos(θ-θ') = cosθcosθ' + sinθsinθ' We can then use the form shown Bateman p 168-9 which is this: ψ = 1 Pn(cosθcosθ' + sinθsinθ') = Pn(cosθ) Pn(cosθ') + 2 Σm=1∞ (-1)m Pn-m(cosθ) Pnm(cosθ') Then our integral above is this: ∫dθ' sinθ' Pn(cosθ') Pn(cos[θ-θ']) = ∫d(cosθ') Pn(cosθ') { Pn(cosθ) Pn(cosθ') + 2 Σm=1∞ (-1)m Pn-m(cosθ) Pnm(cosθ') } = ∫d(cosθ') Pn(cosθ') Pn(cosθ') + ∫d(cosθ') Pn(cosθ') 2 Σm=1∞ (-1)m Pn-m(cosθ) Pnm(cosθ') = Pn(cosθ) ∫d(cosθ') Pn(cosθ') Pn(cosθ') + 2 Σm=1∞ (-1)m Pn-m(cosθ) ∫d(cosθ') Pn(cosθ') Pnm(cosθ') = Pn(z) ∫dz' Pn(z') Pn(z') + 2 Σm=1∞ (-1)m Pn-m(z) ∫dz' Pn(z') Pnm(z') If we just assume these were Bateman on-cut functions, we can use this rule !Syntax Error, Idz Pnm(z)Pkm(z) = δn,k (n+1/2)-1 [(n+m)! / (n-m)! ] = δn,k (n+1/2)-1 f(n,m) but this does not tell us how to do the last integral! If I could cause an extra factor (1-z'2)-1 to be in the integrand, then the entire summation term vanishes, but the first term then diverges. 2B. How do you solve the integral equations? Here I try yet another threading through the same math, but don't get anything very useful. I had in mind the integral p 171 (20) of Bateman. I don't know! One obvious starting point is to expand σa(θ') in some carefully selected manner. I know now that you do not want to expand on cos(nθ) functions since things are a mess. Everything seems to point to Legendre polynomials. The famous generating function is shown page 146 Schaum 1/ = Σn=0∞ Pn(cosθ) βn If we set β = 1 this becomes 1/ = Σn=0∞ Pn(cosθ) where I have to wonder about convergence. But suppose we formally expand all 1/radicals in our two equations in this manner. Looking at our first pair of coupled equations above, 0 = (q/c)/ + ∫dθ' [ σa(θ')/ + βσb(θ')/ ] 0 = (q/c)/ + ∫dθ' [ ασa(θ')/ + σb(θ')/ ] we see that we will then find ourselves dealing with things like this: ∫dθ' σa(θ') Pn(cos[θ-θ']) Suppose we simply expand the charge densities in the (almost) obvious manner: σa(θ') = Σn=0∞ σan Pn(cosθ') Then we will be stuck with this little item to deal with: Σn=0∞ σan ∫dθ' Pn(cosθ') Pn(cos[θ-θ']) We know that the Legendre function addition theorem will come into play here. First cos(θ-θ') = cosθcosθ' + sinθsinθ' We can then use the form shown Bateman p 168-9 which is this: ψ = 0 Pn(cosθcosθ' + sinθsinθ') = Pn(cosθ) Pn(cosθ') + 2 Σm=1∞ (-1)m Pn-m(cosθ) Pnm(cosθ') Then our integral above is this: ∫dθ' Pn(cosθ') Pn(cos[θ-θ']) = ∫dθ' Pn(cosθ') { Pn(cosθ) Pn(cosθ') + 2 Σm=1∞ (-1)m Pn-m(cosθ) Pnm(cosθ') } = Pn(cosθ) ∫dθ' Pn(cosθ') Pn(cosθ') + ∫dθ' Pn(cosθ') 2 Σm=1∞ (-1)m Pn-m(cosθ) Pnm(cosθ') = Pn(cosθ) ∫dθ' Pn(cosθ') Pn(cosθ') + 2 Σm=1∞ (-1)m Pn-m(cosθ) ∫dθ' Pn(cosθ') Pnm(cosθ') = Pn(z) ∫dθ' Pn(z') Pn(z') + 2 Σm=1∞ (-1)m Pn-m(z) ∫dθ' Pn(z') Pnm(z') Now write z' = cosθ' so that dz' = -sinθ' dθ' = - dθ' and I think we then get ∫dθ' = ∫dz' / with the usual limits, so we continue the above as = Pn(z) ∫dz' Pn(z') Pn(z')/ + 2 Σm=1∞ (-1)m Pn-m(z) ∫dz' Pn(z') Pnm(z') / These are assumed all to be Bateman on the cut functions, so we can now use Bateman p 171 (20) which says that this last integral vanishes except when m = 0, but m = 0 is not in the sum, so the entire second piece vanishes and we have only the first piece.