Iris Green's by Dirichlet method Attempt #1

Iris Green's by Dirichlet method Attempt #1 PhL 4.24.10 This section is a continuation of what is in "Finding Green's Functions by the Dirichlet Method.doc" which I have stored in the Stakgold folder at the moment. There I deal with four problems which all come out OK, but this 5th one is much harder so we put it into a separate document and we will now have "attempts" in the usual manner. This was my "first cut" attempt to "summit" the iris mountain. Contents: For overview, see meta review doc. Problem 5: Point charge and an Iris 1 Smythian Form and the Dual Integral Equations with coefficients Am(k) 4 The Trantor Dual Integral Equation Model 7 Expression for the charge density on the iris sticky math surface in terms of Am(k). 8 Trying to work it backwards from the Smythe Form for σ. 8 Now go to the special case in which d = 0 (point charge in the hole). 9 Back to the Charge Distribution 9 Making use of the Hankel Transform to get a Smythe-σ based result for Am(k) 11 Try to show that the Smythe-implied Am(k) satisfies the dual integral equations. 13 Appendix A: Compute !Syntax Error, Idk Jm(ρk) Jm(axk) 16 Problem 5: Point charge and an Iris Can the Dirichlet method maybe solve this problem I have been long suffering with? Here is my picture We use cylindrical coordinates. The surface is the iris plus a Great Sphere, I guess it has two cavities. We could just consider the z ≥ 0 side. The prescribed potential on the iris is this (use r = ρ interchangeably) V(r',θ') = -q / r1 r12 = r'2 + b2 - 2br' cosθ' but rewrite in simpler form V(r,θ) = -q/ r ≥ a only We don't have any prescribed potential in the hole region in this problem, which makes it different from the previous examples where the surfaces did not have any holes in them. So we don't have a pure play Dirichlet problem here. We don't know the potential in the hole. We do know there is no charge density in the hole, but it is not clear yet how to make use of that fact. Our big problem is finding a Smythian form for this problem. We want the prescribed plane z = 0 to be in the non-oscillatory dimension, so we might try this for our Dirichlet solution (this would then be the potential due do the induced charge) V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-k|z|Am(k) Jm(kρ) where we are using certain cylindrical atoms. Then our Dirichlet boundary condition is this V(0,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk Am(k) Jm(kρ) = -q/ ρ > a only Comment: I think I need to move the Green's charge out of the hole and into one of the side cavities, so the problem can have a real 3D life. This will make it similar to the previous examples I have done. And then we can talk maybe about ∂V/∂z and the charge density on the iris and the lack of it in the hole. Then if we do all this, maybe we can take a limit of the point charge approaching the hole. I am mildly optimistic for some strange reason. Start over with Green's off the z = 0 plane: We now allow the points b,r,r' to have arbitrary z components, so all three points are lifted out of the z = 0 plane where they were above. Think of this as a top view with z axis out of paper and x axis to the right. Then we have the following coordinates, where no label means cylindrical: ( if d = 0 and c<a, point charge is in the hole) b = (c, 0, d)Car = (c, 0,d) b2 = c2 + d2 br = cx+dz r = (x,y,z)Car = (ρ,θ,z) r2 = ρ2 + z2 br' = cx'+dz' r' = (x',y',z')Car = (ρ',θ',z') r'2 = ρ'2 + z'2 rr' = xx'+yy'+zz' x = ρcosθ y = ρsinθ rr' = ρ ρ'(cosθ cosθ' + sinθ sinθ') + zz' x' = ρ'cosθ' y' = ρ'sinθ' = ρ ρ' cos(θ-θ') + zz' What is the potential at point r' due to the point charge at b? It is V = +q/|b-r'| |b-r'|2 = b2 + r'2 - 2 br' = b2 + r'2 - 2(cx'+dz') = b2 + ρ'2 + z'2 - 2cρ'cosθ' - 2dz' |b-r|2 = b2 + r2 - 2 br = b2 + r2 - 2(cx+dz) = b2 + ρ2 + z2 - 2cρcosθ - 2dz I am unsure which coordinates to use, Cartesian or Cylindrical, so keeping both alive. In particular, then, for r on the z = 0 plane we have |b-r|2 = b2 + ρ2 - 2cρcosθ // notice c, not b, so not law of cosines We know then that the potential at point r on the z = 0 plane due to our Green's charge is this: V(ρ,θ,0) = +q/ This is of course true on the iris and in the hole. For our Dirichlet problem, we will want V(ρ,θ,0) = - q/ // only on the iris, so ρ ≥ a only When we later add back the point charge potential, the total potential will be V = 0 on the iris, as desired. Now, the point charge +q does not create any charge density in the hole. So if we assume in our Dirichlet problem that there is no such hole charge, then that fact will be true in the total solution. So let's add a Neumann condition to our "Dirichlet problem" saying no surface charge in the hole. The question now is this: how is surface charge in the hole related to our Dirichlet problem potential V? I want to say this, where ρ1 is now a volume charge density and σ a surface charge density on z = 0. -2V = ρ1 = σ δ(z) But I don't know exactly what to do with such a distributional equation. Instead, let's put a flat pillbox centered on the z = 0 plane. The sides are tiny, so contribute nothing to the EdA integral of Gauss's law, so only the round top and bottom of the pillbox make contributions. We have ∫EdA = 4πqenclosed in world where V = q/r EdA = -∂zV(z+ε)dA top surface EdA = -∂zV(z-ε)(-dA) bottom surface [ -∂zV(z+ε) + ∂zV(z-ε)] dA = 4πσdA ∂zV(z+ε)- ∂zV(z-ε) = -4πσ where σ for the iris is the sum of the surface densities on both sides, and in the hole σ = 0. I have gone through all this detail just to verify what I know is true: the condition that there is σ = 0 in the hole is stated in terms of the potential in this way: ∂zV(z+iε) = ∂zV(z-iε), in other words, ∂zV is continuous as you move through the hole. Because our Green's charge is above the hole, we no longer have the symmetry that Jackson had in his disk problem, so we cannot say ∂zV = 0 in the hole as he did, except in the limit that d→ 0. So I am reaching this conclusion: the "Neumann part" of my "Dirichlet problem specification" is merely that ∂zV = continuous on both sides of the hole. It is certainly not clear at this point how this fact can be "exploited" to get something useful. But let's now summarize our "Dirichlet problem" for V: (1) V(ρ,θ,0) = - q/ ρ ≥ a (2) ∂zV(ρ,θ,0+) = ∂zV(ρ,θ,0-) 0 < ρ <a Remember that this V is "the potential of the induced charge on the iris" and has no singularities in the hole even if the point charge is moved there, because the point charge is not part of this Dirichlet problem. Smythian Form and the Dual Integral Equations with coefficients Am(k) At this point, we need some kind of Smythian form, and that is going to be our next big problem. Consider first this shot in the dark, V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-|k|zAm(k) Jm(kρ) Several comments. First, this is a valid atomic form. Second, even though point b is off the z=0 plane, we still have ±θ symmetry so cos(mθ) is still correct. Third, the Dirichlet and "Neumann" specifications are at z = 0 and this is associated with the non-oscillatory coordinate z. Fourth, the potential is symmetric in z, and this is what I (now) expect. See argument at the end of Attempt #2. The basic idea is that on the central boundary (the iris), for the Dirichlet potential V above we have a single value of V everywhere on the central plane, and there is only a single math layer of sticky induced charge σ = σ++ σ-, so each cavity sees the exact same BC so V must be symmetric. I first though this was wrong, which is why I started Attempt #2, but I now think it is right. So all these observations support the above form. We expect that V→0 for large ρ as well, but somehow that has to "come out in the wash" since we are oscillatory in ρ. [ we do know that Jm(kρ) → 1/ ] So here is the Dirichlet boundary condition part of our "Dirichlet problem": (1) Σm=0∞ cos(mθ) !Syntax Error, Idk Am(k) Jm(kρ) = - q/ ρ ≥ a Looking into the Neumann part, we see that ∂z e-k|z| = -k ∂z|z| = -k (±1) for z 0 This says then that ∂zV(ρ,θ,z) = ∓ Σm=0∞ cos(mθ) !Syntax Error, Idk k e-k|z|Am(k) Jm(kρ) and in particular ∂zV(ρ,θ,0±) = ∓ Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) But our Neumann condition is that ∂zV(ρ,θ,0+) = ∂zV(ρ,θ,0-) for ρ < a so we must have Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) = 0 0 < ρ < a which is very encouraging. This is similar to Jackson's m=0 only result 3.172. So here we are with our two conditions: (1) Σm=0∞ cos(mθ) !Syntax Error, Idk Am(k) Jm(kρ) = - q/ ρ ≥ a (2) Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) = 0 0 < ρ < a Our next problem is to express RHS(1) as a cosine series and then do our partial wave analysis. This is again the subject of a whole paper I wrote and the result was this: "Consider the Fourier Analysis of the function 1/ on (0,2π) , εn = 2-δn,0 Neumann factor, f(x) = 1/ = a0/2 + Σn=1∞ an cos(nx) = (1/2) Σn=0∞ εnancos(nx) // expansion I = πan = !Syntax Error, Idx cos(nx)/ // projection an = I/π = (2/π ) Σs=0n (-1)s (2n, 2n-2s) B[ s+1/2, n-s+1/2 ] F [s+1/2, 1/2; n+1; -2b/(a-b) ] ao= 4/[(a-b)] K(k)/π a1= [4/ ] K(k)/π k2 = -2b/(a-b) a2 = [4/ ] [(a/b) K(k) + (1 - (a/b) E(k) ] /π So in the RHS of (1) we say that ( change to α and β to avoid confusion on two meanings for b) α = b2+ ρ2 β = 2cρ - q/ = (-q) (1/2) Σn=0∞ εnan(α,β) cos(nθ) = (-q/2) Σn=0∞ εnan(α,β) cos(nθ) am(α,β) = (2/π ) Σs=0m (-1)s (2m, 2m-2s) B[ s+1/2, m-s+1/2 ] F [s+1/2, 1/2; m+1; -2β/(α-β) ] So our two Dirichlet/Neumann conditions are now these: (1) Σm=0∞ cos(mθ) !Syntax Error, Idk Am(k) Jm(kρ) = (-q/2) Σm=0∞ εmam(α,β) cos(mθ) ρ ≥ a (2) Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) = 0 0 < ρ < a We now do our partial wave analysis and write these as (1) !Syntax Error, Idk Am(k) Jm(kρ) = (-q/2) εmam(α,β) ρ ≥ a α = b2+ ρ2 β = 2cρ (2) !Syntax Error, Idk k Am(k) Jm(kρ) = 0 0 < ρ < a This is the closest I have ever been to having a dual integral equation for the iris Green's problem! The difference from Jackson p 91 (3.173) is that the RHS of (1) is a function of ρ so is not a constant, as it was in his application where constant was called V. Let's now look more closely at am(α,β). α-β = b2 + ρ2 - 2cρ -2β/(α-β) = -4cρ/( b2 + ρ2 - 2cρ) So let's make this definition Gm(ρ) ≡ (-q/2) εmam(α,β) = some known but messy function of ρ, b, and c (and q and m) Our dual integral equations are then these: (1) !Syntax Error, Idk Am(k) Jm(kρ) = Gm(ρ) ρ ≥ a (2) !Syntax Error, Idk k Am(k) Jm(kρ) = 0 0 < ρ < a Our goal is now to make these equations somehow fit into one of the molds Bateman shows on page 76. The first step is to define x = ρ/a so make the ranges look right, and to reverse the order of the equations: (2) !Syntax Error, Idk k Am(k) Jm(kax) = 0 0 < x < 1 (1) !Syntax Error, Idk Am(k) Jm(kax) = Gm(ax) x ≥ 1 Next, let ka = y so we have (2) !Syntax Error, Idy/a y/a Am(y/a) Jm(xy) = 0 0 < x < 1 (1) !Syntax Error, Idy/a Am(y/a) Jm(xy) = Gm(ax) x ≥ 1 (2) !Syntax Error, Idy y Am(y/a) Jm(xy) = 0 0 < x < 1 (1) !Syntax Error, Idy Am(y/a) Jm(xy) = a Gm(ax) x ≥ 1 The Trantor Dual Integral Equation Model This looks pretty close to the Tranter 1951 forms shown in (79) of Bateman 2 p 76 and we can read off: Φ(y) = Am(y/a) ν = m f(x) = 0 F(x) = a Gm(ax) Bateman claims we have a closed form solution for Φ(y). Here are more details: F(x) = a Gm(ax) = (-aq/2) εmam(α,β) where ρ = ax α = b2+ a2x2 β = 2cax Am(k) = Φ(ak) The solution proceeds in three sequential parts, each involving integrals. First H(y) ≡ F(1) Jm+1(y) + y !Syntax Error, Idx x F(x)Jm(xy) = (-aq/2) εmam(a2+b2,2ca) Jm+1(y) + (-aq/2) εm y !Syntax Error, Idx x am(b2+ a2x2, 2cax) Jm(xy) At least we can write it down! The next thing we need is this, which uses H(y) from above, and has a double integral L(t) = (2/π) t-2m !Syntax Error, I dx / [ - !Syntax Error, Idy y H(y) Jm(xy) ] = – (2/π) t-2m !Syntax Error, I dx / !Syntax Error, Idy y H(y) Jm(xy) Then once we get H(y) and L(t), our result is this, which involves one more integral, Φ(y) = H(y) + !Syntax Error, Idx x am(b2+ a2x2, 2cax) Jm(xy) A simpler model? Let's go back to our last step before doing Trantor above (2) !Syntax Error, Idy y Am(y/a) Jm(xy) = 0 x < 1 (1) !Syntax Error, Idy Am(y/a) Jm(xy) = a Gm(ax) x ≥ 1 We would like to get this into the mold of Bateman (77) which Jackson used. The problem is that in our iris problem, the RHS of 0 occurs in the hole which is x < 1, whereas in the disk problem RHS = 0 occurs in the iris region which matches (77). I don't see how to transform the iris into the disk mold. I think then we are stuck with the messy Tranter solution. Maybe some work has been done in this poorly developed field since 1951. Expression for the charge density on the iris sticky math surface in terms of Am(k). First, above we found that ∂zV(z+ε)- ∂zV(z-ε) = -4πσ where σ was the sum of the charge on the two sides. But in our Dirichlet "induced charge problem" there is no distinction between the sides and all we have is a thin math layer holding this σ. We found above that: ∂zV(ρ,θ,0±) = ∓ Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) so we conclude that 4πσ = 2 Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ σ(θ,ρ) = (1/2π) Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) Needless to say, we need to know the Am(k) for all m in order to know the total σ. Trying to work it backwards from the Smythe Form for σ. Recall that Smythe claims the answer is this: σ(ρ',θ') = - (q/2πR2) / R2 = ρ'2 + b2 - 2bρ' cosθ' which I prefer to write as σ(ρ,θ) = - (q/2πR2) / R2 = ρ2 + b2 - 2bρ cosθ Assuming Smythe is right, and assuming my work here is right, then we should have σ(θ,ρ) = (1/2π) Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) = - (q/2π) / [ ( ρ2 + b2 - 2bρ cosθ) ] Of course Smythe is assuming the point charge is in the hole, so let's digress on that for a moment, Now go to the special case in which d = 0 (point charge in the hole). Where does d even appear? Well, it appears in exactly one place: b2 = c2 + d2 . So in the special case, we can just interpret b as its old meaning, distance from point charge to origin. This means that nothing at all changes in our result, which is expressed in terms of b. There is, however, one simplification which is that the surface charge σ shown above will be split into σ/2 on each side of the iris. Big deal. Back to the Charge Distribution Our first order of business is to do a Fourier Series expansion for 1/(a-bcos(x)) : εn = 2-δn,0 f(x) = 1/(a+bcosx) = a0/2 + Σn=1∞ an cos(nx) = (1/2) Σn=0∞εnancos(nx) // expansion I = πan = !Syntax Error, Idx cos(nx)/ (a+bcosx) // projection where I am for now cribbing this from Schaum p 131 with L = π. Again, this is a Fourier Series integral, not a Fourier Transform one, so we just have to find it in GR somewhere. Let's define α = b/a and write this as I = (1/a) !Syntax Error, Idx cos(nx)/ (1+α cosx) = (2/a) !Syntax Error, Idx cos(nx)/ (1+α cosx) α = b/a < 1 GR page 391 tells us that [ I don't really understand the superscript 6, but the n ≥ 0 has been added since my 4th edition. Maybe this is saying that the correction was first made, preface is very unclear. ] So I then have I = (π/a) 1/ [ (- 1) / α ]n an = (1/a) 1/ [ (- 1) / α ]n So the upshot is this: 1/(a+bcosx) = (1/2) Σn=0∞εnancos(nx) an = (1/a) 1/ [ (- 1) / α ]n Now back to our Smythe claim which was this: [ valid for ρ > a, by the way, since that is the iris ] σ(θ,ρ) = (1/2π) Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) = - (q/2π) / [ ( ρ2 + b2 - 2bρ cosθ) ] = - (q/2π) [/] 1/(A + B cosθ) A = ρ2 + b2 B = -2bρ = - (q/2π) [/](1/2) Σm=0∞εmamcos(mθ) = - (q/4π) [/] Σm=0∞εmamcos(mθ) where am = (1/a) ( 1/) [ (- 1) / α ]m where α = B/A = -2bρ/(ρ2+ b2) If we then do our θ PWA on the σ equation, we get this result: (1/2π) !Syntax Error, Idk k Am(k) Jm(kρ) = (q/4π) [/]εmam ρ > a !Syntax Error, Idk k Am(k) Jm(kρ) = (q/2a) [/ ] εm ( 1/) [ (- 1) / α ]m where α = -2bρ/(ρ2+ b2) Comments: This says that I could at least verify Smythe's amazing claim if I could obtain the above result for this object !Syntax Error, Idk k Am(k) Jm(kρ) ρ > a This does NOT mean I have to compute all the Am(k), I just have to compute this integral. It is "interesting to note" that I earlier found this result: !Syntax Error, Idk k Am(k) Jm(kρ) = 0 ρ < a So let's go back to our Smythian form, which is valid everywhere. V(z,ρ,θ) = Σm=0∞ cos(mθ) !Syntax Error, Idk e-|k|zAm(k) Jm(kρ) Then we have ∂zV(ρ,θ,0±) = ∓ Σm=0∞ cos(mθ) !Syntax Error, Idk k Am(k) Jm(kρ) as noted earlier. This result should be valid for all ρ. But I am begging the question here because I don't know ∂zV(ρ,θ,0±) on the iris and this is basically the charge density there which I don't know. So I guess I don't see any shortcut for getting the Smythe result without actually computing the Am(k) using the Tranter method outlined above, and that would be a complete nightmare, although it is at least a prescription. It might be easier to use the oblate hyperboloid result! Making use of the Hankel Transform to get a Smythe-σ based result for Am(k) Suppose we define our integral of interest to be a function name like so, f(ρ) ≡ !Syntax Error, Idk k Jm(kρ) Am(k) What would a Hankel Transform have to say about this thing? Fν(μ) = !Syntax Error, Idx xJν(μx)f(x) f(x) = !Syntax Error, Idμ μJν(μx) Fν(μ) δ(x-x')/x = !Syntax Error, Idμ μJν(μx') Jν(μx) δ(μ-μ')/μ = !Syntax Error, Idx xJν(μx) Jν(μ'x) Transcribe this to say Am(k) = !Syntax Error, Idρ ρJm(kρ)f(ρ) f(ρ) = !Syntax Error, Idk kJm(kρ) Am(k) Now Smythe is making this claim: f(ρ) = !Syntax Error, Idk k Am(k) Jm(kρ) = (q/2a) [/ ] εm ( 1/) [ (- 1) / α ]m where α = -2bρ/(ρ2+ b2) ρ > a and I am making this other claim f(ρ) = 0 ρ < a So if we accept Smythe's result, we could compute the Am(k) from it as follows: Am(k) = !Syntax Error, Idρ ρJm(kρ)f(ρ) α = -2bρ/(ρ2+ b2) = (q εm /2a) !Syntax Error, Idρ ρ Jm(kρ) [1 / ] ( 1/) [ (- 1) / α ]m Maybe we could find this in the Bateman transform tables in the last volume? But the integration range is not right. Now here is an interesting fact: 1 - α2 = (ρ2-b2)2/ (ρ2+b2)2 = (ρ2-b2)/ (ρ2+b2) and (- 1) = (-2b2)/(ρ2+ b2) since in the integration we have ρ > a > b. We also have (- 1) = (-2b2)/(ρ2+ b2) [ (- 1) / α ] = (b/ρ) so we can rewrite the above as Am(k) = (q εm /2a) !Syntax Error, Idρ ρ Jm(kρ) [1 / ] [(ρ2+b2)/ (ρ2-b2)] (b/ρ)m = (q εm bm /2a) !Syntax Error, Idρ ρ1-m Jm(kρ) [1 / ] [(ρ2+b2)/ (ρ2-b2)] so this is not quite as horrible as I first thought, though pretty horrible still. Suppose we define ρ = a ρ' dρ = a dρ' we then get = (q εm bm /2a) !Syntax Error, I a dρ' (aρ') 1-m Jm(kaρ') [1 / ] [((aρ')2+b2)/ ((aρ')2-b2)] = (q εm bm a1-m/2a) !Syntax Error, I dρ' ρ' 1-m Jm(kaρ') [1 / ] [(ρ'2+(b/a)2)/ (ρ'2-(b/a)2)] = (q εm bm a-m/2) !Syntax Error, I dx x 1-m Jm(kax) [1 / ] [(x2+(b/a)2)/ (x2-(b/a)2)] = (q εm /2) (b/a)m !Syntax Error, I dx x 1-m Jm(kax) [1 / ] [(x2+α2)/ (x2-α2)] α = b/a < 1 So we wonder if we can find this integral somewhere: !Syntax Error, I dx x 1-m Jm(βx) [1 / ] [(x2+α2)/ (x2-α2)] β = ka α = b/a < 1 Off hand I don't see this in GR7. Well, the Smythe result and the solution of this iris problem still eludes me. Maybe review this Attempt #1 tomorrow and see if something obvious has been overlooked. 9:15 PM. ( a little more added 4.28.10): Try to show that the Smythe-implied Am(k) satisfies the dual integral equations. Above we found that ( note that b = c in our Smythe case) Am(k) = (q εm /2) (b/a)m !Syntax Error, I dx x 1-m Jm(kax) [1 / ] [(x2+α2)/ (x2-α2)] α = b/a < 1 An interesting fact is that k appears at only one location in this equation! Can we show that this Am(k) in fact solves our dual integral equations? The first thing you would need to show would be this: (2) !Syntax Error, Idk k Am(k) Jm(kρ) = 0 0 < ρ < a This would require the following integral lookup: !Syntax Error, Idk k Jm(ρk) Jm(axk) But we know this integral from our Transforms, it is the Hankel thing, so !Syntax Error, Idk k Jm(ρk) Jm(axk) = δ(ρ-ax)/ax Then we get, with α = b/a < 1, !Syntax Error, Idk k Am(k) Jm(kρ) = (q εm /2) (b/a)m !Syntax Error, I dx x 1-m [1 / ] [(x2+α2)/ (x2-α2)] δ(ρ-ax)/ax The delta only gets a hit if ρ = ax which means ρ = something larger than a. But for ρ < a we get NO HIT and therefore we get our desired 0 for a result. Fascinating. Maybe I just went around in a circle. The no doubt harder thing to show is this: (1) Σm=0∞ cos(mθ) !Syntax Error, Idk Am(k) Jm(kρ) = - q/ ρ ≥ a which we write out as Σm=0∞ cos(mθ) !Syntax Error, Idk Am(k) Jm(kρ) = Σm=0∞ cos(mθ) !Syntax Error, I Jm(kρ) [(q εm /2) (b/a)m !Syntax Error, I dx x 1-m Jm(kax) [1 / ] [(x2+α2)/ (x2-α2)] = (q/2) Σm=0∞ cos(mθ) εm (b/a)m !Syntax Error, I dx x 1-m[1 / ] [(x2+α2)/ (x2-α2)] α = b/a !Syntax Error, I Jm(kρ) Jm(kax) Now we need the integral without the k factor, !Syntax Error, Idk Jm(ρk) Jm(axk) = (1/π) (ρax)-1/2Qm-1/2[ (ax/2ρ) + (ρ/2ax)] See Appendix A below for the evaluation, the result is as shown above. Our task is now to show that (q/2) Σm=0∞ cos(mθ) εm (b/a)m !Syntax Error, I dx x 1-m[1 / ] [(x2+α2)/ (x2-α2)] (1/π) (ρax)-1/2Qm-1/2[ (ax/2ρ) + (ρ/2ax)] = - q/ ρ ≥ a Rewrite as ( I now set c = b on the RHS) (1/2π) (ρa)-1/2Σm=0∞ cos(mθ) εm (b/a)m !Syntax Error, I dx x 1/2-m[1 / ] [(x2+α2)/ (x2-α2)] Qm-1/2[ (ax/2ρ) + (ρ/2ax)] = - 1/ ρ ≥ a I know how to PWA the RHS, - 1/ = - (1/2) Σm=0∞ εmam(α,β) cos(mθ) So here is what we would have to show: (1/π) (ρa)-1/2 (b/a)m !Syntax Error, I dx x 1/2-m[1 / ] [(x2+(b/a)2)/ (x2-(b/a)2)] Qm-1/2[ (ax/2ρ) + (ρ/2ax)] = - am(α,β) where am(α,β) = (2/π ) Σs=0m (-1)s (2m, 2m-2s) B[ s+1/2, m-s+1/2 ] F [s+1/2, 1/2; m+1; -2β/(α-β) ] α = b2+ ρ2 β = 2bρ α-β = b2 + ρ2 - 2bρ = (ρ-b)2 -2β/(α-β) = -4bρ/( b2 + ρ2 - 2bρ) = -4bρ/(ρ-b)2 So we would have to show this for ρ > a (1/π) (ρa)-1/2 (b/a)m !Syntax Error, I dx x 1/2-m[1 / ] [(x2+(b/a)2)/ (x2-(b/a)2)] Qm-1/2[ (ax/2ρ) + (ρ/2ax)] = - (2/π(ρ-b)) Σs=0m (-1)s (2m, 2m-2s) B[ s+1/2, m-s+1/2 ] F [s+1/2, 1/2; m+1; -4bρ/(ρ-b)2 ] I wonder if this is true for m = 0 ? (1/π) (ρa)-1/2 !Syntax Error, I dx x 1/2[1 / ] [(x2+(b/a)2)/ (x2-(b/a)2)] Q-1/2[ (ax/2ρ) + (ρ/2ax)] = - (2/π(ρ-b))(0, 0) B[ 1/2, 1/2 ] F [1/2, 1/2; 1; -4bρ/(ρ-b)2 ] = - (2/π(ρ-b))π F [1/2, 1/2; 1; -4bρ/(ρ-b)2 ] = - (2/(ρ-b)) F [1/2, 1/2; 1; -4bρ/(ρ-b)2 ] or (1/π) (ρa)-1/2 !Syntax Error, I dx x 1/2[1 / ] [(x2+(b/a)2)/ (x2-(b/a)2)] Q-1/2[ (x2 + (ρ/a)2)/(2x(ρ/a))] = - (2/(ρ-b)) F [1/2, 1/2; 1; -4bρ/(ρ-b)2 ] OK, enough. Even if I were able to show this is true, I would still no very little about the solution method for this problem! I really think this has a toroidal aspect to it that might make things clearer. Appendix A: Compute !Syntax Error, Idk Jm(ρk) Jm(axk) GR7 p 660 has the following candidate, which is just a Bateman ET II contribution, If we set μ = ν = m this becomes !Syntax Error, Idk Jm(ρk) Jm(axk) = (ax)m (ρ)-m-1 Γ(m+1/2)/ [ m! Γ(1/2)] F(m+1/2, 1/2, m+1; (ax/ρ)2) I suspect this is a P or Q function, and I suspect it will be Qm-1/2(something) based on later knowledge. So look through Bateman thinking μ = 0 and ν = m-1/2. We will need third argument = m+1 = ν+3/2 and this leads us to page 135 Bateman. First two arguments need to be a = m+1/2 = ν+1 b = 1/2 and our attention is redirected to page 137 (45) with μ = 0 and ν = m-1/2 so that 1+ν+μ = 1+ν = m+1/2. So here then is what (45) tells us: Qm-1/2(z) = Γ(m+1/2) ( z + )-m-1/2/ Γ(m+1) F(1/2, m+1/2, m+1; ξ) ξ = (ax/ρ)2 where ξ = (z - )/(z + ) // = s-1/s = s-2 => s = 1/ Sneak a peak at geometry/ellipses.doc to conclude that z = (1/2) [ + 1/] and in our case we have ξ = (ax/ρ)2 which says z = (1/2) [ (ax/ρ) + (ρ/ax) ] . We can then solve the above for F to get F(1/2, m+1/2, m+1; ξ) = { Γ(m+1)/ [ Γ(m+1/2) ( z + )-m-1/2] } Qm-1/2(z) = { m! / [ Γ(m+1/2) s-m-1/2]} Qm-1/2(z) = sm+1/2 m! / [ Γ(m+1/2) ] Qm-1/2(z) = ξ-m/2-1/4 m! / [ Γ(m+1/2) ] * Qm-1/2(z) = (ax/ρ)-m-1/2 m! / [ Γ(m+1/2) ] * Qm-1/2[(ax/2ρ) + (ρ/2ax)] So the upshot of all this is that !Syntax Error, Idk Jm(ρk) Jm(axk) = (ax)m (ρ)-m-1 Γ(m+1/2)/ [ m! Γ(1/2)] F(m+1/2, 1/2, m+1; (ax/ρ)2) = (ax)m (ρ)-m-1 Γ(m+1/2)/ [ m! Γ(1/2)] * (ax/ρ)-m-1/2 m! / [ Γ(m+1/2) ] * Qm-1/2[(ax/2ρ) + (ρ/2ax)] = (ax)m (ρ)-m-1 / [ Γ(1/2)] * (ax/ρ)-m-1/2 / [] * Qm-1/2[(ax/2ρ) + (ρ/2ax)] = (ax)m (ρ)-m-1 / [] * (ax/ρ)-m-1/2 / [] * Qm-1/2[(ax/2ρ) + (ρ/2ax)] = (1/π) (ax)m (ρ)-m-1 * (ax/ρ)-m-1/2 * Qm-1/2[(ax/2ρ) + (ρ/2ax)] = (1/π) (ax)-1/2 ρ-1/2 Qm-1/2[(ax/2ρ) + (ρ/2ax)] = (1/π) (axρ)-1/2 Qm-1/2[(ax/2ρ) + (ρ/2ax)] I now realize this ought to be in Bateman ET II Russian, lets go find it and see what goofs I made. Here it is: So set α = 0 and he is saying this (replace integration variable x with k) !Syntax Error, Idk [ 1] Jm(βk) Jm(ρk) = (1/π) Qm-1/2[ β2 + ρ2)/2βρ] !Syntax Error, Idk Jm(βk) Jm(ρk) = (1/π) Qm-1/2[ β2 + ρ2)/2βρ] !Syntax Error, Idk Jm(βk) Jm(ρk) = (1/π) Qm-1/2[ β2 + ρ2)/2βρ] Then set β = ax to get !Syntax Error, Idk Jm(axk) Jm(ρk) = (1/π) Qm-1/2[ (ax)2 + ρ2)/2axρ] = (1/π) (ρax)-1/2Qm-1/2[ (ax/2ρ) + (ρ/2ax)] This agrees with my evaluation.