A study of (R)-1 expansions in toroidal coordinates

A study of 1/R expansions in toroidal coordinates PhL 5.22.10 I have done this now for spherical and cylindrical coordinates, so let's try to replicate things here for the toroidal case. From "Toroidal coordinates.doc" here are the main atomic combinations times: {Pmn-1/2(chμ), Qmn-1/2(chμ)} e±inη e±imφ n,m = integers As usual, I assume oscillatory in φ. Then η will be the other oscillatory, and μ the expo coordinate. Overview (added 10.10.10, 4 pages) 1 0. The 1/R expansion in toroidal coordinates: the Smythian Form 5 1. Use the pillbox condition to find the Amn coefficients. 7 2. Development of some Toroidal Sum Rules 14 3. Derivation of the Selvaggi/Snow 1/R Expansion 16 4. Development of a new toroidal 1/R expansion and the addition theorem. 18 5. Something I did not do in this doc: the "other" Smythian form. 21 Overview (added 10.10.10, 4 pages) In Section 0 part (a), as my Smythian form candidate for 1/R, I make this choice of toroidal atoms: f(r | r') = Σnm Anm Pmn-1/2(chμ<) Qmn-1/2(chμ>) einη eimφ n,m = all integers I then show that the above Smythian form exhibits the right behavior in the limits μ→ 0 (huge donut radius) and μ→ ∞ (donut shrinks to wire ring at focal distance). In part (b) after some fiddling I am able to write R = |r-r'| in the toroidal coordinates (not atoms) r = μ,η,φ and r' = μ',η',φ' as follows: ( and I have verification from a certain Wilson pdf on this) R2 = 2 a2 { chγ - cos(η-η')} / [(chμ-cosη) (chμ'-cosη')] 1/R = (1/a) / where chγ ≡ chμ chμ' - shμ shμ' cos(φ-φ') In Section 1 I use the pillbox method to evaluate Anm . Note that φ and η are oscillatory, so the pillbox is in the μ direction, which is normal to a donut. I find, after doing two trig orthogonalities and after using the P,Q Wronskian and doing much general labor, that Anm = (q/πa) (-1)m f(n-1/2,-m) e-inη' e-imφ' and upon insertion into the above Smythian form we then find that (setting q = 1) f(r | r') = 1/R = (1/πa) Σnm (-1)m f(n-1/2,-m) Pmn-1/2(chu<) Qmn-1/2(chu>) ein(η-η') eim(φ-φ') I then examined the symmetry under n and m reflection of pieces of the integrand and found that: (a) The factor f(n-1/2,-m) → +f when n→ –n, and f(n-1/2,-m) → –f when m→ –m. (b) Each Legendre function is symmetric under n → –n . For P this is obvious, and for Q it is only true since we have ν = n-1/2 and n = integer. Therefore, the entire summand is symmetric under n → -n. (c) The entire summand is also symmetric under m → –m, this takes some work to show. Since the summand is even under reflection of both n and m independently, we know that f is real. Here is a quick proof: f = Σnm hnm ein(η-η') eim(φ-φ') = Σnm hnm [ cos(nx)+isin(nx)] [ cos(my)+isin(my)] Im(f) = Σnm hnm cos(nx) sin(my) + Σnm hnm sin(nx) cos(my) The first sum vanishes because hnm is symmetric in m, while sin(my) is antisymmetric, so the summand as a whole is antisymmetric in m, and sum is over all integers m, positive and negative. The second term vanishes for the same reason with respect to index n. Things had to come out this way, I just wanted to verify it. Once we know the above symmetries, we can rewrite our 1/R expansion in several other ways. With full n and m sums, we can replace either or both exponentials with cosines (I show this). Here are some rewrites: f(n-1/2,-m) = Γ(n-m+1/2)/ Γ(n+m+1/2) 1/R = (1/πa) Σnm (-1)m f(n-1/2,-m)Pmn-1/2(chu<)Qmn-1/2(chu>) cos[n(η-η')] cos[m(φ-φ')] 1/R = (1/πa) Σn=0∞ Σm=0∞ εn εm (-1)m f(n-1/2,-m)Pmn-1/2(chu<)Qmn-1/2(chu>) cos[n(η-η')] cos[m(φ-φ')] I find verification of this last 1/R form in MF, but they have the typo (-i)m or perhaps they have a different definition of the Q function which seems quite possible since MF are very "shy" of using the Bateman forms and tend to use their own notations. MF was 1953, the exact same year of Bateman. I then have a second verification from the Wilson 2005 PDF of this last result, and agreement is perfect. In Section 2 I first select the Green's charge location r' on the z axis. This means μ' = 0 since on infinite horizontal circle (donut), and then η' picks out a vertical circle (bowl) to give us some point on the z axis. For such a point r', we find (using the simpler cylindrical coordinates with ρ' = 0) that R2 = ρ2 +(z-z')2 ρ = a shμ/(chμ - cosη) z = a sinη/(chμ - cosη) z' = a sinη'/(1 - cosη') If we examine our last 1/R expansion above in the limit μ' → 0, I show that it becomes 1/R = (1/πa) Σn=0∞εn Qn-1/2(chμ) cos[n(η-η')] which then gives us this little hybrid [ ie, cylindrical/toroidal ] sum rule, 1/ = (1/πa) Σn=0∞εn Qn-1/2(chμ) cos[n(η-η')] We can run our point down the positive z axis to z' = 0 by taking η' → π where then 1/ = (1/πa) Σn=0∞εn Qn-1/2(chμ) cos[n(η-π)] and if we express the LHS in toroidals and shuffle factors around, this becomes (π/) (1/) = Σn=0∞εn Qn-1/2(chμ) cos[n(η-π)] (*) = Σn=0∞εn (-1)n Qn-1/2(chμ) cos(nη) This is essentially a specialized case of our toroidal 1/R expansion when r' = 0. Analogous equations from our cylindrical world are these: 1/ = !Syntax Error, Idk exp(-k|z|) J0(kρ) = (2/π) !Syntax Error, Idk cos(kz) K0(kρ) Notice that the toroidal world Smythian form has Σn whereas the cylindrical has ∫dk . And notice that the non-trig atom in the toroidal world is the ring function Qn-1/2(chμ), whereas it is a Bessel function Z0(kρ) in the cylindrical world. If we replace η → η+π in (*) we get this simpler result (π/) (1/) = Σn=0∞εn Qn-1/2(chμ) cos(nη) (**) which happens to appear on page 1304 of MF, but MF are missing the εn factor! (errata!) // should contain εn If we set a = coshμ, b = 1 and η = x, we can interpret the above (**) as the expansion direction of our little Fourier Cosine analysis ( see doc just on this subject) 1/ = (1/π) Σn=0∞ εn Qn-1/2(a/b) cos(nx) // expansion !Syntax Error, Idx cos(nx)/ = Qn-1/2(a/b) // projection In Section 3 I use specific "arbitrary" values for μ and η in the 1/R expansion (*) above, chμ = [ρ2 + ρ'2 + (z-z')2]/(2ρρ') ≡ B η = φ-φ' and I quickly arrive at this result, where r and r' are arbitrary points expressed in cylindricals, 1/R = 1/(π Σn=0∞εn Qn-1/2(B) cos[n(φ-φ')] which is what I call the Selvaggi/Snow 1/R expansion. This thing is really expressed in cylindrical coordinates, but the toroidal atom Qn-1/2(B) appears in the sum, so it is a hybrid animal. By comparing this general 1/R expansion with those in my cylindrical doc, we conclude that Qn-1/2{[ρ2 + ρ'2 + (z-z')2]/(2ρρ')} = π !Syntax Error, Idk exp(-k|z-z'|) Jn(kρ) Jn(kρ') = 2!Syntax Error, I dk cos[k(z-z')] In(kρ<) Kn(kρ>) If z = z', we get this interesting special case: Qn-1/2{[ρ2 + ρ'2 ]/(2ρρ')} = π !Syntax Error, Idk Jn(kρ) Jn(kρ') = 2!Syntax Error, I dk In(kρ<) Kn(kρ>) We could perhaps regard these as some "integral addition theorems" that live in the hybrid world of the Snow expansion. In Section 4 I first come up with a more general version of the toroidal sum rule which is this: 1 / = (/π) Σn=0∞εn Qn-1/2(chγ) cos[n(η-η')] 4.5 where one can install any chγ one wants. Using chγ = B and other values shown in Section 3, this leads to the same Selvaggi 1/R expansion we got before. That is, η' = π, and so on. Next, I develop this general 1/R toroidal expansion 1/R = (1/πa) Σn=0∞εn Qn-1/2(chγ) cos[n(η-η')] 4.6 = (1/a) / where chγ ≡ chμ chμ' - shμ shμ' cos(φ-φ') This thing is somewhat analogous to the following spherical coordinates 1/R expansion 1/R = Σn (r<)n(r>)-n-1 Pn(cosγ) // note m = 0 where cosγ ≡ cosθcosθ' + sinθsinθ'cos(φ-φ') In cylindrical coordinates, the analogous items I think are these (these also have m = 0) 1/R = !Syntax Error, Idk exp(-k|z-z'|) J0(k) 1/R = (2/π) !Syntax Error, Idk cos[k(z-z')] K0(k) where you have a single "atom" with a complicated argument. I then compare my new 1/R toroidal expansion to the general Smythian one in Section 1 and this leads do the following toroidal addition theorem, Qn-1/2(chμ chμ' - shμ shμ' cos(φ-φ')) = Σm=0∞ εm cos[m(φ-φ')] (-1)m f(n-1/2,-m)Pmn-1/2(chμ<)Qmn-1/2(chμ>) We may compare this with other addition theorems we have encountered: Pn[cosθcosθ' + sinθsinθ'cos(φ-φ')] = Σm=0∞ εm cos[m(φ-φ')] f(n,-m) Pnm(cosθ) Pnm(cosθ')] J0(k) = Σm=0∞εm cos[m(φ-φ')] Jm(kρ) Jm(kρ') K0(k) = Σm=0∞ εm cos[m(φ-φ')] Im(kρ<) Km(kρ>) All three coordinate systems have an azimuthal coordinate φ, and all three of these addition theorem types thus have a similar "look" to them. I do not have direct verification for my toroidal addition theorem, but I think I have seen it somewhere! As with everything else, I expect it has a group theoretic interpretation. In Section 5 I conjecture briefly on what the 1/R expansion might look like had I taken the alternate Smythian form starting position where η is expo. The result would look something like this: 1/R = (1/πa) // just a conjecture! !Syntax Error, Idτ Σm=0∞ εm (-1)m f(iτ-1/2,-m)Pmiτ-1/2(chu<)Qmiτ-1/2(chu>) e-τ|η-η'| cos[m(φ-φ')] ________________________________________________________________________________ 0. The 1/R expansion in toroidal coordinates: the Smythian Form (a) Smythian Form. The potential of a point charge at the origin is q/R with R = r. If we move the point charge to some other origin r', the potential is q/R where R = |r - r'|. In either case, the potential is a solution of the Laplace equation (away from the point charge) and can therefore be written in terms of the Laplace "atoms" of whatever coordinate system we want to use. For example, if we think of R as f(r) we know we can expand it this way in toroidals (the notation here means that r' is a parameter, r is the variable) f(r | r') = Σnm Anm Pmn-1/2(chμ<) Qmn-1/2(chμ>) einη eimφ n,m = integers Our m is the usual azimuthal quantum number. As for n and cos(nη) stuff, we know that if we let η range from (0,2π), we create all the partial circles of the vertical circle family (bowls in toroidals) as shown in my Maple graphs of the above doc. I think these partial circles will "repeat" as you go (2π,4π) and so on, and that is what causes n to also be quantized to be an integer. Very similar to the reason for m to do this. And the atomic forms I have seen all say this same thing. An explanation of the choice for the P and Q appearance is given below after we review the coordinate parameterizations. Here μ is the radial coordinate, and we know from our doc on toroidals that as you move away from the focus on the μ coordinate, μ → 0. See picture in that doc. So we really want Pmn-1/2(chμ) for this limit since P is happy at argument 1. Conversely, as we move the other way toward the focus with our horizontal circles, μ → ∞ and chμ → ∞ and we need Qmn-1/2(chμ) in this case. The Green's charge is at μ' somewhere. So we want Pmn-1/2(chu<) Qmn-1/2(chu>), and this is then how I made it appear in our Smythe atomic form above. (b) expressing R in toroidal coordinates. Notice that we are parameterizing our two points r and r' this way r = a (shμ cosφ, shμ sinφ, sinη)/(chμ-cosη) r' = a (shμ' cosφ', shμ' sinφ', sinη')/(chμ'-cosη') r2 = a2 ( sh2μ + sin2η) / (chμ-cosη)2 r'2 = a2 ( sh2μ' + sin2η') / (chμ'-cosη')2 rr' = a2 [ shμ shμ' cos(φ-φ') + sinη sinη'] / [(chμ-cosη)( chμ'-cosη')] which is not all that pleasant a result. We then have R2 = |r-r'|2 = r2 + r'2 - 2 rr' = a big mess. Here is what Maple says, if we let μ = μ1 η = η1 μ' = μ2 η' = η2 R2 = a2N/[(chμ-cosη)2( chμ'-cosη')2] with numerator N given by : This does not seem very useful, does it! Another approach would be to start with cylindricals and write R2 = ρ2 + ρ'2 + (z-z')2 - 2ρρ'cos(φ-φ') then install ρ = a shμ /(chμ-cosη) z = a sinη /(chμ-cosη) ρ' = a shμ' /(chμ'-cosη') z' = a sinη' /(chμ'-cosη') but this does not seem to buy much, it has to come out being the same mess. Looking at section (g1) of the toroidal notes, I just note in passing that we can invert the above two pairs of equations as follows: thμ = 2aρ / (a2+ρ2+z2) tanη = - 2az/(a2- ρ2-z2) thμ' = 2aρ' / (a2+ρ'2+z'2) tanη' = - 2az'/(a2- ρ'2-z'2) Well, I have shown on scratch paper that the following is true (R2/2a2) (chμ-cosη) (chμ'-cosη') = [chμ chμ' - shμ shμ' cos(φ-φ')] - cos(η-η') where I must have done this using just the above R2 written out in cylindrical coordinates and then using the replacements shown above for ρ,ρ',z.z', so this does give an expression for R which is not TOO bad, R2 = 2 a2 {[chμ chμ' - shμ shμ' cos(φ-φ')] - cos(η-η')} / [(chμ-cosη) (chμ'-cosη')] We can see that the numerator [..] is at smallest chμ chμ' - shμ shμ' = ch(μ-μ') and so is ≥ 1, so we can represent it by another ch. So let's define, in analogy with cosγ from spherical coordinates, chγ = chμ chμ' - shμ shμ' cos(φ-φ') Then we have ( I add the first line just to have it here for comparison) R2 = ρ2 + ρ'2 + (z-z')2 - 2ρρ'cos(φ-φ') R2 = 2 a2 { chγ - cos(η-η')} / [(chμ-cosη) (chμ'-cosη')] 1/R = (1/a) / This result is verified in the Wilson paper which has this notation 1. Use the pillbox condition to find the Amn coefficients. The general pillbox condition is this (Jackson units) ∂q1Vi - ∂q1Vo = 4πq (h1/h2h3) δ(q2-q2') δ(q3-q3') where q1 is the dimension perp to our partition of space into two regions at the point charge. For toroidals, we want q1 = μ, the radial coordinate. The Green's charge sits at some μ' with η' and φ'. So I will use the term "outer" to mean the smaller μ value, sort of inverted from the usual situation, see long paragraph just above. We then have ( selecting where the prime goes on the P and Q) Vi = Σnm Anm Pmn-1/2(chu') Qmn-1/2(chu) einη eimφ μ > μ' Vo = Σnm Anm Pmn-1/2(chu) Qmn-1/2(chu') einη eimφ μ < μ' So q1 = μ. We can then take q2 = η and q3 = φ and use these facts h1 = hμ = a/ [chμ - cosη] h2 = hη = a/ [chμ - cosη] h3 = hφ = a shμ/ [chμ - cosη] which come from my toroidal doc, and which agrees with M&M page 191. Our pillbox condition is then this: (note that (h1/h2h3) = 1/h3 ) ∂μVi - ∂μVo = 4πq [chμ - cosη]/(a shμ) δ(η-η') δ(φ-φ') We then need to compute the derivatives. Things are a little different now compared to previous coordinate systems because we have μ dependence in the leading factor. We need to know ∂μ = (1/2) shμ / We then have: Vi = Σnm Anm Pmn-1/2(chu') Qmn-1/2(chu) einη eimφ μ > μ' Vo = Σnm Anm Pmn-1/2(chu) Qmn-1/2(chu') einη eimφ μ < μ' ∂μVi = {(1/2) shμ / } Σnm Anm Pmn-1/2(chu') Qmn-1/2(chu) einη eimφ + Σnm Anm Pmn-1/2(chu') Qmn-1/2(chu)' shμ einη eimφ and to get ∂μVo we just switch P ↔ Q. When we subtract the first two terms, they will contribute this amount to the difference: {(1/2) shμ / } Σnm Anm einη eimφ ( Pmn-1/2(chu') Qmn-1/2(chu) - Qmn-1/2(chu') Pmn-1/2(chu)} But when we evaluate at μ = μ', this contribution will (thankfully) be 0 so we forget about it. The contribution of the other two terms is this: Σnm Anm shμ einη eimφ W[Pmn-1/2(chu), Qmn-1/2(chu)] For "regular" (off the cut) P and Q functions, Leg prop doc tells us that W[ Pνμ(z), Qνμ(z)] = eiμπ (1-z2)-1[Γ(1+μ+ν)/ Γ(1-μ+ν) ] or W[ Pνm(z), Qνm(z)] = (-1)m(1-z2)-1 f(ν,m) or W[ Pνm(chu), Qνm(chu)] = (-1)m f(ν,m) (1-ch2μ)-1 = (-1)m f(ν,m) 1/sh2μ so at this point then we seem to have (ν = n-1/2) ∂μVi - ∂μVo = Σnm Anm shμ einη eimφ { 1/sh2μ }(-1)m f(n-1/2,m) = Σnm (-1)m f(n-1/2,m)Anm einη eimφ / shμ so our pillbox condition is then 4πq [chμ - cosη]/(a shμ) δ(η-η') δ(φ-φ') = Σnm (-1)m f(n-1/2,m)Anm einη eimφ / shμ and we get cancellation of the shμ factor (this seems to happen in all coordinate systems) and we are left with 4π(q/a) δ(η-η') δ(φ-φ') = Σnm (-1)m f(n-1/2,m)Anm einη eimφ and it should now be a simple matter to extract the Anm coefficients. For the moment, let's define (-1)m f(n-1/2,m)Anm = Bnm so we have then 4π(q/a) δ(η-η') δ(φ-φ') = Σnm Bnm einη eimφ First, apply ∫dφ e-im'φ to both sides and use (1/2π) !Syntax Error, I dφ eimφ e-im'φ = δmm' // orthogonality The RHS becomes Σnm Bnm einη ∫dφ e-im'φ eimφ = Σnm Bnm einη {2πδmm'} = 2π Σn Bnm einη and the LHS becomes 4π(q/a) δ(η-η') e-im'φ' so we then have (replace m' with m) 4π(q/a) δ(η-η') e-imφ' = 2π Σn Bnm einη (2q/a) δ(η-η') e-imφ' = Σn Bnm einη Now apply ∫dη e-in'η to both sides and use (1/2π) !Syntax Error, I dη einη e-in'η = δnn' // orthogonality The RHS becomes Σn Bnm ∫dη e-in'η einη = Σn Bnm {2πδnn'} = 2π BAn'm and the LHS becomes (2q/a) e-in'η' e-imφ' so we have (replace n' with n) (2q/a) e-inη' e-imφ'= 2π Bnm Bnm = (q/πa) e-inη' e-imφ' and thus Anm = Bnm (-1)m f(n-1/2,-m) = (q/πa) (-1)m f(n-1/2,-m) e-inη' e-imφ' So we now have our coefficients! I set μ = μ' because all this was done at the point charge. As expected, Anm is a function of the coordinates of r' . We can install these into our expansion above V(r | r') = q/R = Σnm Anm Pmn-1/2(chu<) Qmn-1/2(chu>) einη eimφ = (q/πa) Σnm (-1)m f(n-1/2,-m)Pmn-1/2(chu<) Qmn-1/2(chu>) ein(η-η') eim(φ-φ') where we have f(n-1/2,-m) = Γ(n-m+1/2)/ Γ(n+m+1/2) Before going another step, we want to look at the symmetry of this factor under inversion of n or m: f(n-1/2,-m)Pmn-1/2(chu<) Qmn-1/2(chu>) (*) (a) We know what the f factor does if m changes sign, it goes upside down. But what happens if n changes sign? We have to use the reflection formula which says Γ(z) Γ(1-z) = π/sin(πz). Setting z = n-m+1/2 we find that 1-z = 1-( n-m+1/2) = 1-n+m-1/2 = -n+m+1/2, so Γ(n-m+1/2) Γ(-n+m+1/2) = π/sin(π[n-m+1/2]) = π/cosπ(n-m) = π (-1)n-m so that Γ(n-m+1/2) = π (-1)n-m / Γ(-n+m+1/2) We then find that f(n-1/2,-m) = Γ(n-m+1/2)/ Γ(n+m+1/2) = { π (-1)n-m / Γ(-n+m+1/2)}/ { π (-1)n+m / Γ(-n-m+1/2)} = { 1 / Γ(-n+m+1/2)}/ { 1 / Γ(-n-m+1/2)} = Γ(-n-m+1/2)/ Γ(-n+m+1/2) = f(-n-1/2,-m) So the f factor is simply symmetric when n → -n, and it flips over when m → -m. (b) In terms of the n reflection, our Leg prop doc shows that P-ν-1μ(z) = Pνμ(z) => P-n-1/2μ(z) = Pn-1/2μ(z) The Q is less obvious. We start with: Q-ν-1μ(z) sinπ(ν-μ) = Qνμ(z) sinπ(ν+μ) – π eiπμ cosπν Pνμ(z) and set ν = n-1/2 and μ = m to get Q-n-1/2m(z) sinπ(n-1/2-m) = Qn-1/2m(z) sinπ(n-1/2+m) – π eiπm cosπ(n-1/2) Pn-1/2μ(z) As we have seen elsewhere, the last term vanishes since cosπ(n-1/2) = ±sin(πn) = 0. We are left with Q-n-1/2m(z) cosπ(n-m) = Qn-1/2m(z) cosπ(n+m) But we know that cosπN = (-1)N for so we then have Q-n-1/2m(z) (-1)n-m = Qn-1/2m(z) (-1)n+m Q-n-1/2m(z) (-1)-m = Qn-1/2m(z) (-1)m Q-n-1/2m(z) = Qn-1/2m(z) So both our P and Q functions are simply symmetric when n → -n. This means that our entire factor quoted above in (*) is symmetric under n inversion. (c) As for the m folding, from Leg prop doc we have f(ν,m) Pν-m(z) = Pνm(z) f(ν,m) Qν-m(z) = Qνm(z) where ν = n-1/2 as usual. Looking at our (*) factor we have [ replace u with μ everwhere ] F(m) ≡ f(ν,-m)Pmν(chu<) Qmν(chu>) F(-m) = f(ν,m)P-mν(chu<) Q-mν(chu>) = f(ν,m) Pmν(chu<) Qmν(chu>) f(ν,-m) f(ν,-m) = f(ν,-m) Pmν(chu<) Qmν(chu>) = F(m) so our (*) factor is also simply symmetric under m → -m. These symmetries have to come out this way in order that the imaginary part of the double sum be zero! But we have explicitly shown this to be true in (a), (b), (c) here. So, we can now write our result in various ways. We start with the original way found above (n and m are over all integers) 1/R = (1/πa) Σnm (-1)m f(n-1/2,-m)Pmn-1/2(chu<)Qmn-1/2(chu>) ein(η-η')eim(φ-φ') If we look at the contribution to this sum from sin[n(η-η')], we know it is zero because the factor (*) noted above is symmetric under n → -n. Thus we conclude that 1/R = (1/πa) Σnm (-1)m f(n-1/2,-m)Pmn-1/2(chu<)Qmn-1/2(chu>) cos[n(η-η')]eim(φ-φ') If we now ask about the contribution to this last sum from sin[m(φ-φ')], it is also zero since the factor is symmetric in m → -m. Thus, we get this rewrite as well, and this form is then manifestly real: 1/R = (1/πa) Σnm (-1)m f(n-1/2,-m)Pmn-1/2(chu<)Qmn-1/2(chu>) cos[n(η-η')] cos[m(φ-φ')] Of course another way would be to start with the other factor, so we would get 1/R = (1/πa) Σnm (-1)m f(n-1/2,-m)Pmn-1/2(chu<)Qmn-1/2(chu>) ein(η-η') cos[m(φ-φ')] Now due to our two symmetries just noted, we can reflect both the n and m sums (independently) and in the usual way we get this result: 1/R = (1/πa) Σn=0∞ Σm=0∞ εn εm (-1)m f(n-1/2,-m)Pmn-1/2(chu<)Qmn-1/2(chu>) cos[n(η-η')] cos[m(φ-φ')] where f(n-1/2,-m) = Γ(n-m+1/2)/ Γ(n+m+1/2) As noted in the toroidal doc, this result is actually stated in M&F, which I again quote: (a=1) and my "suspicion" is now confirmed, MF have a typo: their (-i)m should be (-1)m . And they are using a = 1 for the focal distance, so that my result shows the correct dimensions. [ or they have a different phase definition of Q and maybe P and Q for all I know, recall CS and NCS stuff and eimμ in Bateman Q tables.] So I am very happy about getting this result!! It gives me a waypoint on the way to the next section. Here is a second "verification" of the above result from a pdf paper "wilson toroid greens.doc" which appears in Electromagnetics in the ripe year 2005 (!) : where These authors are using ξ in place of μ, but continue to use η for the bowl labels. The second equality here agrees exactly with my result above. The first equality is also correct and has been commented on in the previous section. 2. Development of some Toroidal Sum Rules Now, what happens if we put our point charge on the z axis? This was our starting point in both the spherical and cylindrical systems. Then r' = a (shμ' cosφ', shμ' sinφ', sinη')/(chμ'-cosη') = (0,0 ,z') which clearly means μ' = 0 so you are on a "big" horizontal circle (fat toroid) that is coming down the z' axis and then your z' location is picked off by an η' circle (sphere). Certainly in this case μ' = μ< and so we have Pmn-1/2(chu<) = Pmn-1/2(chu') = Pmn-1/2(1) = δm0 so the RHS of our equation 1/R = (1/πa) Σn=0∞ Σm=0∞ εn εm (-1)m f(n-1/2,-m)Pmn-1/2(chμ<)Qmn-1/2(chμ>) cos[n(η-η')] cos[m(φ-φ')] becomes ( azimuthal dependence is being removed) 1/R = (1/πa) Σn=0∞εn Qn-1/2(chμ) cos[n(η-η')] When I write out R on the LHS, however, it is still a big mess even with μ' = 0. However, I could write out this R in cylindrical coordinates which are a little less opaque. In general we have R2 = |r-r'|2 = r2 + r'2 - 2 rr' = ρ2 + z2 + ρ'2 + z'2 - 2ρρ'cos(φ-φ') -2 zz' and for r' on the z axis this becomes ρ' = 0 so R2 = ρ2 +(z-z')2 ρ = a shμ/(chμ - cosη) z = a sinη/(chμ - cosη) z' = a sinη'/(1 - cosη') so we are then saying that, for r' on the z axis, we have 1/R = 1/ = (1/πa) Σn=0∞εn Qn-1/2(chμ) cos[n(η-η')] We might go further (as I did in the cylindrical 1/R doc) and set z' = 0. If we look carefully at our toroidal pictures, we see that this means η' = π (not 0), so cosη' = -1 and sinη' = 0 and we then get R2 = ρ2 +z2 ρ = a shμ/(chμ - cosη) z = a sinη/(chμ - cosη) And then our 1/R equality becomes 1/R = 1/ = (1/πa) Σn=0∞εn Qn-1/2(chμ) cos[n(η-π)] which is just our toroidal 1/R expansion where we put the Green's charge at the origin. We can of course then write = [ a/(chμ - cosη) ] so 1/R = 1 / = [(chμ - cosη)/a] 1/ Our equality then becomes [(chμ - cosη)/a] 1/ = (1/πa) Σn=0∞εn Qn-1/2(chμ) cos[n(η-π)] π / = Σn=0∞εn Qn-1/2(chμ) cos[n(η-π)] (π/) [ (chμ-cosη) / (sh2μ + sin2η)]1/2 = Σn=0∞εn Qn-1/2(chμ) cos[n(η-π)] (*) Now process the LHS a bit: LHS = (π/) [ (chμ-cosη) / (ch2μ - 1 + sin2η)]1/2 = (π/) [ (chμ-cosη) / (ch2μ - cos2η)]1/2 = (π/) [ (chμ-cosη) / {(chμ - cosη) (chμ + cosη)}]1/2 = (π/) [ 1/ (chμ + cosη)]1/2 so we get this simpler result (π/) [ 1/ (chμ + cosη)]1/2 = Σn=0∞εn Qn-1/2(chμ) cos[n(η-π)] Every time we shift cos(x) by π, we change its sign, so this can be written as (π/) [ 1/ (chμ + cosη)]1/2 = Σn=0∞εn (-1)n Qn-1/2(chμ) cos(nη) (π/) 1/ = Σn=0∞εn (-1)n Qn-1/2(chμ) cos(nη) (*) Once again, apart from an overall constant factor, this "toroidal sum rule" is just a special case of our general toroidal 1/R expansion where we have put the Green's charge at the origin, r' = 0. NOTE: I continue my "addition theorem development" in Section 4 below. 3. Derivation of the Selvaggi/Snow 1/R Expansion This "sum rule" (*) is valid for any values of μ and η we want, so let's emphasize this fact by rewriting the sum rule this way (π/) [ 1/(B + cosα)]1/2 = Σn=0∞εn Qn-1/2(B) cos[n(α-π)] where B is any positive real number and α is any angle. Let's make one further change, by defining β ≡ α - π cos(α) = cos(π+β) = -cosβ Our generic sum rule now becomes (π/) [ 1/(B - cosβ)]1/2 = Σn=0∞εn Qn-1/2(B) cos(nβ) or 1/D ≡ 1/ = (/π) Σn=0∞εn Qn-1/2(B) cos(nβ) (**) which, by the way, we can interpret as just the Fourier Series expansion of 1/, as I deal with in a ragged doc elsewhere. Now, forgetting about all earlier coordinates, let's consider two new arbitrary points r and r' in cylindrical coordinates. We know of course that R2 = ρ2 + ρ'2 + (z-z')2 - 2ρρ'cos(φ-φ') Let's now apply our formula (**) above where we blindly select B = [ρ2 + ρ'2 + (z-z')2]/(2ρρ') β = (φ-φ') We then find that D2 = B - cosβ = [ρ2 + ρ'2 + (z-z')2]/(2ρρ') - cos(φ-φ') = [ρ2 + ρ'2 + (z-z')2 - 2ρρ' cos(φ-φ')]/ (2ρρ') = R2/2ρρ' => 1/D = (1/R) Then (**) becomes (1/R) = (/π) Σn=0∞εn Qn-1/2(B) cos(n (φ-φ')) or 1/R = 1/(π Σn=0∞εn Qn-1/2(B) cos[n(φ-φ')] (***) where B = [ρ2 + ρ'2 + (z-z')2]/(2ρρ') and this is precisely the "Selvaggi" 1/R expansion. I think there is a simple geometric interpretation here, but I cannot quite nail it down, so I will defer that to some future time. The question I cannot answer is this: what is the interpretation of B ? Ignoring this question, we can conclude that the Selvaggi expansion has been derived from the toroidal sum rule, and is some kind of geometric transformation or application of it. In the sum rule, the trig angle is really η which is not an azimuthal angle at all. But in the Selvaggi thing, we really have a difference of azimuths. So what should we call this Selvaggi thing? It seems more of a cylindrical coordinates 1/R expansion, but it contains the toroidal Q function atom! The Selvaggi paper doesn't apply any name to this expansion. Now one thing we can do at this point is compare our Selvaggi 1/R expansion to our cylindrical general 1/R expansion which is either of these forms: 1/R = Σm=0∞εm cos[m(φ-φ')] !Syntax Error, Idk exp(-k|z-z'|) Jm(kρ) Jm(kρ') 1/R = (2/π) Σm=0∞ εm cos[m(φ-φ')]!Syntax Error, I dk cos[k(z-z')] Im(kρ<) Km(kρ>) 1/R = 1/(π) Σm=0∞εm cos[m(φ-φ')] Qm-1/2{[ρ2 + ρ'2 + (z-z')2]/(2ρρ')} // Selvaggi where we changed the summation variable to m in the Selvaggi form. We can now make a term by term match to conclude that the following equations must be true, with B = [ρ2 + ρ'2 + (z-z')2]/(2ρρ'): 1/(π) Qm-1/2(B) = !Syntax Error, Idk exp(-k|z-z'|) Jm(kρ) Jm(kρ') 1/(π) Qm-1/2(B) = (2/π) !Syntax Error, I dk cos[k(z-z')] Im(kρ<) Km(kρ>) Let's rewrite these as Qm-1/2{[ρ2 + ρ'2 + (z-z')2]/(2ρρ')} = π !Syntax Error, Idk exp(-k|z-z'|) Jm(kρ) Jm(kρ') Qm-1/2{[ρ2 + ρ'2 + (z-z')2]/(2ρρ')} = 2!Syntax Error, I dk cos[k(z-z')] Im(kρ<) Km(kρ>) I guess you could call these things "addition theorems" where the sum is an integral over k. They can be compared to the cylindrical and spherical addition theorems which are these J0(k) = Σm=0∞εm cos[m(φ-φ')] Jm(kρ) Jm(kρ') K0(k) = Σm=0∞ εm cos[m(φ-φ')] Im(kρ<) Km(kρ>) Pn[cosθcosθ' + sinθsinθ'cos(φ-φ')] = Σm=0∞ εm cos[m(φ-φ')] f(n,-m) Pnm(cosθ) Pnm(cosθ')] Somehow the roles of m and k are swapped. In the "normal" addition theorems we are summing over m which is associated with the azimuth φ, and the quantity k (or n for spherical) is fixed. But in our Q addition theorems, we are instead summing over k, and m is fixed. 4. Development of a new toroidal 1/R expansion and the addition theorem. My earlier attempt here led to a big mess which I have deleted and replaced it with what follows. Let's now give it one more go using the "new" form I found for 1/R. Start again with 1/ = (1/πa) Σn=0∞εn Qn-1/2(chμ) cos[n(η-η')] ρ = a shμ/(chμ - cosη) z = a sinη/(chμ - cosη) z' = a sinη'/(1 - cosη') where we have μ' = 0 since we are on the z axis. Our general 1/R formula is this: 1/R = (1/a) / 4.1 where chγ ≡ chμ chμ' - shμ shμ' cos(φ-φ') 4.2 and we want to set μ' = 0 so chμ' = 1 and chγ = chμ, so that we have 1/R = (1/a) / 4.3 We have then shown that (1/a) / = (1/πa) Σn=0∞εn Qn-1/2(chμ) cos[n(η-η')] We can cancel factors on both sides to find that 1 / = (/π) Σn=0∞εn Qn-1/2(chμ) cos[n(η-η')] 4.4 and we have then done pretty will in "exposing" chμ on the LHS. The sum rule shown here must be value for any chμ we like. So we could install the full chγ on both sides to get 1 / = (/π) Σn=0∞εn Qn-1/2(chγ) cos[n(η-η')] 4.5 Now multiply both sides by (1/a) . The LHS becomes general 1/R and we have then shown that 1/R = (1/πa) Σn=0∞εn Qn-1/2(chγ) cos[n(η-η')] 4.6 = (1/a) / where chγ ≡ chμ chμ' - shμ shμ' cos(φ-φ') The first line above is the addition theorem I have been looking for. I now want to make sure earlier results in previous sections are still OK. Suppose in 4.5 se set η' = π. It then reads 1 / = (/π) Σn=0∞εn (-1)nQn-1/2(chγ) cos(nη) 4.7 and this agrees with the second (*) in Section 2 above. Now what about Selvaggi? We can derive it exactly as before, so all is well, but I am wondering if I don't now have some "interpretation" of things. Is B = [ρ2 + ρ'2 + (z-z')2]/(2ρρ') some special case of the quantity chγ ? I know from above that B = D2 + cos(φ-φ') = R2/2ρρ' + cos(φ-φ') chγ ≡ chμ chμ' - shμ shμ' cos(φ-φ') The only way these two could be the same is if cos(φ-φ') = 0 in which case we would have B = R2/(2ρρ') R2 = 2 a2 {chμ chμ' - cos(η-η')} / [(chμ-cosη) (chμ'-cosη')] chγ ≡ chμ chμ' We can then write B = R2/(2ρρ') = a2 {chμ chμ' - cos(η-η')} / [(chμ-cosη) (chμ'-cosη')(ρρ')] But we know that ρ = a shμ /(chμ-cosη) => (chμ-cosη) ρ = a shμ ρ' = a shμ' /(chμ'-cosη') => (chμ'-cosη') ρ' = a shμ' so we then have B = a2 {chμ chμ' - cos(η-η')} / [ a2 shμ shμ'] = {chμ chμ' - cos(η-η')}/ [ shμ shμ'] But this does not in general give B = chγ. I am just not finding any useful "interpretation" here. What about addition theorems? We have shown the following 1/R expansion 1/R = (1/πa) Σn=0∞εn Qn-1/2(chγ) cos[n(η-η')] 4.6 = (1/a) / where chγ ≡ chμ chμ' - shμ shμ' cos(φ-φ') We usually set this equal to some other 1/R expansion and try to compare terms. Our toroidal form was this: 1/R = (1/πa) Σn=0∞ Σm=0∞ εn εm (-1)m f(n-1/2,-m)Pmn-1/2(chu<)Qmn-1/2(chu>) cos[n(η-η')] cos[m(φ-φ')] If we set these things equal we obtain: (1/πa) Σn=0∞εn Qn-1/2(chγ) cos[n(η-η')] = (1/πa) Σn=0∞ Σm=0∞ εn εm (-1)m f(n-1/2,-m)Pmn-1/2(chu<)Qmn-1/2(chu>) cos[n(η-η')] cos[m(φ-φ')] or Σn=0∞εn Qn-1/2(chγ) cos[n(η-η')] = Σn=0∞ Σm=0∞ εn εm (-1)m f(n-1/2,-m)Pmn-1/2(chu<)Qmn-1/2(chu>) cos[n(η-η')] cos[m(φ-φ')] or Qn-1/2(chγ) = Σm=0∞ εm (-1)m f(n-1/2,-m)Pmn-1/2(chu<)Qmn-1/2(chu>) cos[m(φ-φ')] where chγ ≡ chμ chμ' - shμ shμ' cos(φ-φ') So THIS is what we would call "the toroidal addition theorem". We could for fun consider the special case of this where we put r' back on the z axis with μ' = 0 and we know Pmn-1/2(chμ<) = Pmn-1/2(chμ')Pmn-1/2(1) = δm0 and chγ = chμ so we end up then with Qn-1/2(chμ) = ε0 (-1)0 f(n-1/2,0) Q0n-1/2(chu>) cos(0) = Qn-1/2(chμ) so things check out OK. Next topic: is there some kind of transform to worry about? In our atomic approach here our two SL problems were the trivial ones involving φ and η. But if you were to use e-nη then you get those cone functions and they do have a SL problem, as noted elsewhere, called Mehler-Foch. 5. Something I did not do in this doc: the "other" Smythian form. I could have taken the "other" Smythian form at the start, which might be something like this: f(r | r') = !Syntax Error, Idτ Σm Aτm Pmiτ-1/2(chμ<) Qmiτ-1/2(chμ>) e-τ|η| eimφ where now μ is oscillatory and η is expo, the reverse of our previous case. One would then do the pillbox in the η direction to deduce the Aτm coefficients. More clarity is needed regarding the exponential, I am just doing a symbolic outline here. The pillbox would require Mehler-Fock orthogonality as well as the usual trig φ orthogonality. I have never done this, and I have never seen it done, so it would be "advanced" work that perhaps I don't need to do right now. Based on Green's symmetry, I would expect the result to have this general form: 1/R = (1/πa) !Syntax Error, Idτ Σm=0∞ εm (-1)m f(iτ-1/2,-m)Pmiτ-1/2(chu<)Qmiτ-1/2(chu>) e-τ|η-η'| cos[m(φ-φ')] but I expect there to be extra factors not shown. Perhaps I could derive this using contour integration methods to get the sum into the τ integral, and then start with the already known result. I would then obtain perhaps a sum rule and an addition theorem and something like the Selvaggi/Snow expansion. I dimly recall maybe seeing one or two papers using this kind of stuff.