1. Green's Function for an Oblate Bloid non-cone-method

Green's Function for an Oblate Hyperboloid PhL 1.27.10 Summary 2.28.10. This is perhaps my first "bloid attempt" dated 1.27.10. I use the complex P and Q functions, the "right" coordinate parameterization where an entire bloid has a single value of ξ (though I thought I was using the wrong one), and I use a hyperboloid instead of the later spheroid as the "dividing surface" between two regions. The Smythian form is this: V(ζ,ξ,φ) = Σnm Enm [Pnm(ξ1) Qnm(ξ<) – Qnm(ξ1) Pnm(ξ<) ] Pnm(ξ>) Qnm(jζ) eimφ V(ζ,ξ,φ) = Σnm Fnm [Pnm(ξ1) Qnm(ξ<) – Qnm(ξ1) Pnm(ξ<) ] Pnm(ξ>) Qnm(jζ) eimφ This approach requires orthogonal Q functions to isolate the coefficients, something like this: !Syntax Error, Idζ Qnμ(iζ) Qn'μ(iζ)* = Hn(ξ1) δn,n' I was able to formulate a S-L problem in ξ with Q(0) = 0 and interval (0,i∞) which produced a spectrum for n and produced the above orthogonality. But then I found that the resulting n spectrum was forcing V = 0 in the neck [ since Q(0) = 0 ] which I know is wrong (but good for the jigger problem). I tried some rescue attempts, but did not succeed. I tried (-i∞,i∞) as an alternate interval which gave just a continuous n spectrum, then I got side-tracked on those Qnμ(iζ)OR type functions (" Idea #2"). I then gave up on this approach. In retrospect, my attempt #4 presumably-successful solution to this problem was V(ζ,ξ,φ) = Σnm Cnm [g4(n,m) pnm(ζ<) – 2(-1)m qnm(ζ<)] qnm(ζ>) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 V(ζ,ξ,φ) = Σnm Cnm [ 2(-1)m qnm(ζ<)] qnm(ζ>) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 I suspect now that the approach of this current doc might still be workable if you start with the real p and q functions, but it would be a very different looking solution since it would have ξ> instead of ζ> etc. It would be difficult to verify this result in the cone limit, since the cone solution is based on the cone method rather than the method given here. But it still might be viable. The orthogonality of the Q would come out reminiscent of the messy orthogonality of P's in the cone method. At this point, I decided to use the cone method to get V = 0 on the bloid, then I went with the spheroid as the dividing Green's surface. 1. The Smythian Form. 2 2. Compute Normal Derivatives and take difference, use Wronskian. 3 3. Apply Gauss's Law to pillbox around point charge q. 4 4. Use orthogonality twice to compute the coefficients Enm : first the φ step. 5 5. The ζ direction as a Sturm Liouville Problem: asking the questions 5 6. The ζ direction as a Sturm Liouville Problem: answering the questions! 6 7. Continuing with our oblate bloid solution : the final answer. 8 8. Physical Interpretation (looking for trouble). 11 Matching of Potential and Its Gradients in the Hole. 12 Where do we go from here? 15 7A. Continuing with our oblate bloid solution. 15 Idea #1: Suppose we try the SL problem in ζ with the range (-i∞, i∞). 15 Idea #2. Go through the cut 16 Idea #3: The odd-ν idea to get V continuity at ζ = 0 16 Idea #4 Why not use the following Q function: Qnm(j|ζ|) ? 18 Our Green's point charge sits inside the one-sheet hyperboloid: Vi is on the "inside" meaning here only that it is the side close to the grounded surface. I choose this bad convention just because it matches the spheroid problem I just finished doing. 1. The Smythian Form. We seek a linear combination of oblate atoms that causes V = 0 on the bloid marked by parameter ξ1 Vo(ζ,ξ,φ) = Σnm Enm [Pnm(ξ1) Qnm(ξ0) – Qnm(ξ1) Pnm(ξ0) ] Pnm(ξ) Qnm(jζ) eimφ Vi(ζ,ξ,φ) = Σnm Enm [Pnm(ξ1) Qnm(ξ) – Qnm(ξ1) Pnm(ξ) ] Pnm(ξ0) Qnm(jζ)eimφ These forms have these properties: when ζ→ ∞, which can happen in either region, potential decays. when ξ = ξ1, we have Vi = 0 as it must be on the bloid surface when ξ = ξ0, Vi = Vo everywhere away from the point charge. we don't get either V = 0 at ξ0. (simpler forms violate this) So, I think we have it! There are no other choices for the ζ function, or for the φ function. We can write this Smythian form in the usual manner as a single expression V(ζ,ξ,φ) = Σnm Enm [Pnm(ξ1) Qnm(ξ<) – Qnm(ξ1) Pnm(ξ<) ] Pnm(ξ>) Qnm(jζ) eimφ ξ> = max(ξ,ξ0) ξ< = min(ξ,ξ0) Note added 2.14.10. When I was doing this, I had just come from the oblate spheroid problem, so I think I was assuming m = all integers, and n = 0,1,2... I am not doing the "cone method" here, but am causing V = 0 on the bloid using "the spheroid method", you might say, built into the Smythian form. Notice that the picture does not say which way I am parameterizing the coordinates in the lower region. At this point, my SL problems are in the φ and ξ spaces, causing m and n to quantize. That fact really says that I was assuming ξ goes to -1 in the lower region, and that in turn implies the parameterization type, which is that the entire spheroid gets the same ζ, and upper and lower have opposite ξ. 2. Compute Normal Derivatives and take difference, use Wronskian. Our derivative of interest is now ∂ξ since this is normal to our new pillbox. So we have (note where the primes go) ∂ξVo(ζ,ξ,φ) = Σnm Enm [Pnm(ξ1) Qnm(ξ0) – Qnm(ξ1) Pnm(ξ0) ] Pnm '(ξ) Qnm(jζ) eimφ ∂ξVi(ζ,ξ,φ) = Σnm Enm [Pnm(ξ1) Qnm '(ξ) – Qnm(ξ1) Pnm '(ξ) ] Pnm(ξ0) Qnm(jζ)eimφ Then we shall need the difference: ∂ξVi(ζ,ξ) – ∂ξVo(ζ,ξ) = Σnm Enm Qnm(jζ) eimφ * { [Pnm(ξ1) Qnm '(ξ) – Qnm(ξ1) Pnm '(ξ) ] Pnm(ξ0) – [Pnm(ξ1) Qnm(ξ0) – Qnm(ξ1) Pnm(ξ0) ] Pnm '(ξ) } But we are only going to want this quantity at ξ = ξ0 which marks the bloid holding our point charge, so set ξ = ξ0 ∂ξVi(ζ,ξ) – ∂ξVo(ζ,ξ) = Σnm Enm Qnm(jζ) eimφ * { [Pnm(ξ1) Qnm '(ξ0) – Qnm(ξ1) Pnm '(ξ0) ] Pnm(ξ0) – [Pnm(ξ1) Qnm(ξ0) – Qnm(ξ1) Pnm(ξ0) ] Pnm '(ξ0) } ∂ξVi(ζ,ξ) – ∂ξVo(ζ,ξ) = Σnm Enm Qnm(jζ) eimφ * { [Pnm(ξ1) Qnm '(ξ0) Pnm(ξ0) – Qnm(ξ1) Pnm '(ξ0) Pnm(ξ0) ] – [Pnm(ξ1) Qnm(ξ0) Pnm '(ξ0) – Qnm(ξ1) Pnm(ξ0) Pnm '(ξ0) ] } ∂ξVi(ζ,ξ) – ∂ξVo(ζ,ξ) = Σnm Enm Qnm(jζ) eimφ * { Pnm(ξ1) Qnm '(ξ0) Pnm(ξ0) – Qnm(ξ1) Pnm '(ξ0) Pnm(ξ0) – Pnm(ξ1) Qnm(ξ0) Pnm '(ξ0) + Qnm(ξ1) Pnm(ξ0) Pnm '(ξ0) } This time, it is the second and fourth terms which cancel, leaving us with ∂ξVi(ζ,ξ) – ∂ξVo(ζ,ξ) = Σnm Enm Qnm(jζ) eimφ * Pnm(ξ1) [Qnm '(ξ0) Pnm(ξ0) – Qnm(ξ0) Pnm '(ξ0) ] = Σnm Enm Qnm(jζ) eimφ * Pnm(ξ1) W[Pnm(ξ0), Qnm(ξ0) ] Bateman tells us that W[ Pnm(z), Qnm(z)] = (-1)m (n+m)! / (n-m)! * 1/(1-z2) So we then have ∂ξVi(ζ,ξ) – ∂ξVo(ζ,ξ) = (1-ξ02)-1 Σnm (-1)m [(n+m)! / (n-m)!] Enm Qnm(jζ) Pnm(ξ1) eimφ (*) which I would again say is "not too bad". 3. Apply Gauss's Law to pillbox around point charge q. In the Exterior Green's case, we got this result: ∂ζVi - ∂ζVo = (q/ε) (h2/h3h1) δ(φ-φ0) δ(ξ - ξ0) q1, q2, q3 = ξ, ζ, φ We now want to swap 1↔ 2 in order to get [ we maintain q1, q2, q3 = ξ, ζ, φ ] ∂ξVi - ∂ξVo = (q/ε) (h1/h3h2) δ(φ-φ0) δ(ζ - ζ0) where h1 = c1 / h2 = c1 / h3 = c1 h2h3 = c1 / * c1 = c12 h1 = c1 / (1-ξ2) c1 h1 = h2h3 => (h1/h3h2) = [(1-ξ2) c1]-1 The charge density on the bloid surface is given by [ this is Gauss's law for the pillbox ] σ/ε = ∂nVi - ∂nVo ∂n = (1/h1) ∂ξ => σ/ε = (1/h1)[ ∂ξVi - ∂ξVo ] Therefore, the h1 cancels when we write σ. Here are our two main results: ∂ξVi - ∂ξVo = (q/ε) (h1/h3h2) δ(φ-φ0) δ(ζ - ζ0) = q/[εc1 (1-ξ2)] δ(φ-φ0) δ(ζ - ζ0) (**) σ/ε = (1/h1)[ ∂ξVi - ∂ξVo ] = (q/ε) (1/h3h2) δ(φ-φ0) δ(φ-φ0) δ(ζ - ζ0) dA2 = h3h2dq3dq2 = h3h2dφ dζ // area of pillbox side This then is the effective surface charge distribution "of the point charge q". So we now know from (*) and (**) that (q/ε) (h1/h3h2) δ(φ-φ0) δ(ζ-ζ0) = (1-ξ02)-1 Σnm (-1)m [(n+m)! / (n-m)!] Enm Qnm(jζ) Pnm(ξ1) eimφ To review: we computed the normal difference across the bloid at the point charge location and set this ~ equal first to the surface charge density that charge represents, and second to our Smythian form derivative difference, and we got the above! Now it should be an easy matter to computer the Enm. In my form, we have both positive and negative m integers in the m sum, whereas Smythe uses cos(m[φ-φm]). 4. Use orthogonality twice to compute the coefficients Enm : first the φ step. We first apply ∫dφ e-imφ to both sides of the above and use the azimuthal orthogonality relation, ∫dφ e-imφ' eimφ = 2π δm',m to get (q/ε) (h1/h3h2) e-im'φ0 δ(ζ-ζ0) = (1-ξ02)-1 Σnm (-1)m [(n+m)! / (n-m)!] Enm Qnm(jζ) Pnm(ξ1) 2π δm',m = [ 2π / (1-ξ02)] Σn (-1)m' [(n+m')! / (n-m')!] Enm' Pnm'(ξ1) Qnm'(jζ) Now suddenly things are different! We need some kind of Q orthogonality in order to continue! So we pause for a digression, and hopefully after that we can continue here where we left off. 5. The ζ direction as a Sturm Liouville Problem: asking the questions The ODE of interest is the Legendre equation, our solutions are Pνμ(iζ) and Qνμ(iζ) . Because we have a full azimuth in our Green's problem, we know μ = m (integers either sign including 0). In the "polar" ξ department we have ξ = 1 and ξ = -1 both included in our "range" of solution, so I think that forces ν = n (positive integers including 0). [ This is wrong: my ξ range is only (0,1) in this problem's set up ]. In the Smythe cone problem, which we did in sphericals, we had ν = the zeros of P. We could consider doing that here in ξ, but that would seem to violate our rule that we need n = integer in ξ to stay finite at ξ = -1, which is a very regular location it seems in our solution. So let's for the moment continue to assume Pnm(iζ) and Qnm(iζ). Our next item is that the "interval" for our ODE is in z = i0 to z = i∞. We know (finally) that both P and Q are finite at the 0 end of things, and only Q is finite at the ∞ end of things. So if we consider this interval (i0,i∞), we might be able to argue that the Qnm(iζ) are not only solutions, but are orthogonal eigenfunctions somehow. Q is certainly oscillatory in this range (example below). So let's just approach this as an isolated SL problem and ask about such orthogonal solutions. Knowing nothing about the other dimensions, we would be looking at the general solution Qνμ(iζ) with no quantization. If we think of μ as a fixed number, maybe we expect to see something like this: !Syntax Error, Idζ Qνμ(iζ) Qν'μ(iζ)* = Kνμ δ(ν-ν') Firstly, I want to know: what is the "general theory" on this type of question? Is the ODE still self-adjoint when we replace z = iζ ? Well, here is the normal Legendre ODE L = (1-z2) D2 - 2z D + [ n(n+1) - m2/(1-z2)] Lu = 0 Bateman p 121 Let's rewrite this to show it has the Stak self-adjoint form L = -D(pD) + q Lz = ∂z[(1-z2)∂z] + [ n(n+1) - m2/(1-z2)] => p = - (1-z2) q = [ n(n+1) - m2/(1-z2)] Now if we take z = iζ, then dz = i dζ and ∂z = -i∂ζ and it all becomes Lζ = (-i∂ζ )[(1+ζ2) (-i∂ζ )] + [ n(n+1) - m2/(1+ξ2)] = – ∂ζ[(1+ζ2)∂ζ] + [ n(n+1) - m2/(1+ξ2)] => p = + (1+ζ2) q = [ n(n+1) - m2/(1+ξ2)] So this is manifestly formally self-adjoint. What about the boundary conditions at the endpoints. They are supposed to be homogeneous or finite in a SL problem, I think. At the 0 end, something like Au(0) + Bu'(0) = 0 is required in an ODE. This is where my understanding is lacking I think. 6. The ζ direction as a Sturm Liouville Problem: answering the questions! I had to go off on a long side trip which involved these documents: Properties of Legendre functions.doc math / ODE / Legendre When are ODE solutions orthogonal.doc math / ODE Attempting the oblate Q integral.doc math / Integrals * GR Here is what I learned: (a) the orthogonality relation conjectured above does in fact exist and is valid, but in this form: ( uμν, uμν') = !Syntax Error, Idζ Qνμ(iζ)* Qν'μ(iζ) = Kνμ δν,ν' (b) for a given value of μ, there is a finite discrete spectrum with these eigenvalues ν = μ- N N = 2,4,6... but stops at last N such that ν > - 1/2 (c) the spectrum is created by our imposition of the homogeneous BC at iζ = 0+: uμν(0+) = Qνμ(0+) = 0 I showed that there was no other homo BC that is different than this. It is the only one. [ What about using the derivative Q' instead? ] (d) the truncation of the spectrum at ν > -1/2 is caused by the requirement that uμν be of finite s norm at the high endpoint. (e) An exact and general formula for these eigenfunctions is this: Qμ-Nμ(z) = 2μ-N [ Γ(κ) Γ(κ+μ)/ Γ(2κ) ] (z +1)-μ/2+N-1 (z -1)-μ/2 F(1-N, κ, 2κ; 2/(1+z) ) where κ = κ ≡ 1+μ-N = 1+ν N = 2,4,6... The F function is a finite series in 2/(1+z) having N terms, since its "a" parameter is a negative integer. So the general form of the eigenfunction is this: Qμ-Nμ(z) = AμN (z +1)-μ/2+N-1 (z -1)-μ/2 Σn=0N-1 anμ [2/(1+z)]n a0μ = 1 As you see, there are no logs. We just have some power factors and a finite inverse series. In the special case that μ = integer, one can use the following easy formula to compute the functions: Qm-Nm(z) = (z2-1)m/2∂mQm-N(z) One can obtain the Qi(z) functions from Maple or Maxima, respectively: LegendreQ(ν,μ,z) assoc_legendre_q(n,m,z) (f) In order to compute the normalization constants Kνμ, you have to write out the function explicitly as shown above in (e) and then do the integral. !Syntax Error, Idζ | Qνμ(iζ)|2 = Kνμ The integrals involve nothing but elementary functions. I found, for example, that K24 = 72π. (g) In our application here we have μ = integer m. Although all integers m are part of the φ spectrum, only certain m are allowed in order to even have a ζ spectrum! And for each allowed m, there are only m/2 allowed values of n. Here is a list of the eigenfunctions, ordered by increasing integer m. Notice there are no eigenfunctions for negative m or even for m = 0 or m = 1. This does seem unusual, but it is true. m ≤ 1 none! m=2 Q02(iξ) m=3 Q13(iξ) m=4 Q24(iξ) , Q04(iξ) m=5 Q35(iξ) , Q15(iξ) m=6 Q46(iξ) , Q26(iξ), Q06(iξ) m=7 Q57(iξ) , Q37(iξ), Q17(iξ) etc. // in general, we have n = m-2, m-4.... (1 or 0) (h) Two of these functions that I calculated using the methods above: Q24(z) = 48z/(z2-1)2 Q04(z) = 24z(z2+1)/ (z2-1)2 Notice that for z = iξ, these particular functions are pure imaginary. (i) An easy way to the functions when μ is an integer m is to use the formula Qm-Nm(z) = (z2-1)m/2∂mQm-N(z) For general μ, one must use the formula shown in (e) above. I conjectured that these functions were "oscillatory" above, and here is a plot of the two functions shown above: plot(Im(LegendreQ(0,4,I*z)),z=0..10); left plot(Im(LegendreQ(2,4,I*z)),z=0..10); right You can see how they might be orthogonal. I verified this analytically and numerically. I computed the norm factor K24 and got 72π, but analytically and numerically ! 7. Continuing with our oblate bloid solution : the final answer. In section 4 above, we left off with this result: (q/ε) (h1/h3h2) e-im'φ0 δ(ζ-ζ0) = [ 2π / (1-ξ02)] Σn (-1)m' [(n+m')! / (n-m')!] Enm' Pnm'(ξ1) Qnm'(jζ) We now multiply both sides by Qn'm'(jζ)* and integrate !Syntax Error, Idζ. But first we have a new observation: the values of n which appear in the n sum on the right are not "all positive integers", but are only those integers that are "in the spectrum of the ζ problem". Only then can we do our "inversion". If we find a solution, since it is unique, we have it, and then this assumption is justified de facto. So we use the integral result shown above, which we restate here as !Syntax Error, Idζ Qn'm'(iζ)* Qnm'(iζ) = Kn'm' δn',n We then find that (q/ε) (h1/h3h2) e-im'φ0 Qn'm'(jζ0)* = [ 2π / (1-ξ02)] Σn (-1)m' [(n+m')! / (n-m')!] Enm' Pnm'(ξ1) Kn'm δn',n = [ 2π / (1-ξ02)] (-1)m' [(n'+m')! / (n'-m')!] En'm' Pnm'(ξ1) Kn'm' and we now remove all primes to get (q/ε) (h1/h3h2) e-imφ0 Qnm(jζ0)* = [ 2π / (1-ξ02)] (-1)m [(n+m)! /(n-m)!] Enm Pnm(ξ1) Knm The scale factors on the LHS are evaluated at ξ, ζ, φ = ξ0, ζ0, φ0 which is the location of our point charge around which we put the pillbox. In fact, we have (h1/h3h2) = [(1-ξ02) c1]-1 and the factor (1-ξ02)-1 pleasantly cancels on both side so we get: (q/c1ε)e-imφ0 Qnm(jζ0)* = 2π (-1)m [(n+m)! /(n-m)!] Enm Pnm(ξ1) Knm We now know our coefficients Enm ( this is the whole point of this exercise ! ) Enm = (q/[2πc1ε Knm]) e-imφ0(-1)m [(n-m)! /(n+m)!] Qnm(jζ0)* / Pnm(ξ1) We can now copy down our Smythian form V(ζ,ξ,φ) = Σnm Enm [Pnm(ξ1) Qnm(ξ<) – Qnm(ξ1) Pnm(ξ<) ] Pnm(ξ>) Qnm(jζ) eimφ ξ> = max(ξ,ξ0) ξ< = min(ξ,ξ0) and we then install our hard-won coefficients: V(ζ,ξ,φ) = (q/[2πc1ε]) Σnm (1/Knm] (-1)m [(n-m)! /(n+m)!] eim(φ-φ0) Qnm(jζ0)* / Pnm(ξ1) [Pnm(ξ1) Qnm(ξ<) – Qnm(ξ1) Pnm(ξ<) ] Pnm(ξ>) Qnm(jζ) V(ζ,ξ,φ) = (q/[2πc1ε]) Σnm (1/Knm) (-1)m [(n-m)! /(n+m)!] eim(φ-φ0) * [Qnm(ξ<) – { Qnm(ξ1)/Pnm(ξ1)}Pnm(ξ<) ] Pnm(ξ>) Qnm(jζ) [Qnm(jζ0)]* = g(ζ,ξ,φ | ζ0,ξ0,φ0) for hyperboloid ξ> = max(ξ,ξ0) ξ< = min(ξ,ξ0) This is our final answer in closed form! The Knm are the Q function orthogonality constants that we have to go compute if we really want to use this formula. Note that the ζ0 Q has a * on it. And note that the summation only includes spectral points. So we might write it as Σnm = Σm=2∞ Σn=(0/1)m-2 where the lower endpoint is 0 if m is even, and 1 if m is odd. And here again is our picture. Comment: For me, this is an astounding result. You put your Green's point charge at an arbitrary location on the inside of this infinite grounded one-sheeted oblate hyperboloid, some complicated charge is induced on the bloid, and that charge plus the point charge produces the potential shown above! As quick checks on our result, we let ξ → ξ1 and we expect to get V = 0. In this case, ξ> = max(ξ,ξ0) = ξ0 ξ< = min(ξ,ξ0) = ξ = ξ1 V(ζ,ξ,φ) = (q/[2πc1ε]) Σnm (1/Knm) (-1)m [(n-m)! /(n+m)!] eim(φ-φ0) * [Qnm(ξ1) – { Qnm(ξ1)/Pnm(ξ1)}Pnm(ξ1) ] Pnm(ξ0) Qnm(jζ) [Qnm(jζ0)]* = 0 so at least that limit works. I am now of course going to be very interested in taking limits of this result, maybe looking at the charge distribution, and so on. 8. Physical Interpretation (looking for trouble). Right off the bat, I see trouble. In the ζ SL problem, I required Qnm(0+) = 0 to get quantization of the n for given m. But, looking at the above solution, that implies that V = 0 when ζ = 0. But ζ = 0 is the super flat spheroid that is the interior disk in this limit. So my "solution" to the problem is producing V = 0 everywhere in the hole! I know that cannot be right. Just put the charge in the hole a sit one inch away from it, potential cannot possibly be zero in the hole. This arose from the way I treated the ζ SL problem. I arbitrarily required Qnm(0+) = 0 just to get a spectrum, with no physical connection to things. So something has to be fixed! Basically, I eliminated all (n,m) loci which could make V vary on the ζ=0 disk. If I don't do this, then we have full capability with the Pnm(ξ) eimφ combination to obtain an arbitrary on-disk value. As I examine my picture, I see another related issue. In my "choice" of oblate coordinates, the upper and lower bloid halves have different ξ labels! We run (1→0) on the top, then (0→-1) on the bottom. This reminds me of the Smythe hole in plate problem set up. If I keep it the way I have it, then I have not really met the V=0 BC on the entire bloid. AND, there is a discontinuity in the potential as you cross the center disk, which is just free space. It certainly would seem better for THIS problem to split the spheroids into two halves and have ζ > 0 on the upper half and ζ < 0 on the lower half, as Smythe did for his hole in plate problem. Then ξ is the label for a full top + bottom bloid. How do we justify these two possible "choices" of coordinates? Certainly if we keep ξ > 0, our P and Q functions are still oblate "atoms", and our products still solve Laplace. That is a key point. And making ζ < 0 does not stop Qnm(jζ) from being a solution of its separated equation. This alternate setup is MUCH more appropriate for the geometry of this problem! Notice that Qnm(z) decays for large |z| in all directions, so going up and down would give decay, and in fact we need decay in a large solid angle both up and down in this problem. So somehow, if we adopt this new coordinate sub-choice for oblate, we have these Smythian forms For ζ ≥ 0 and for ξ in (0,1) only Vo(ζ,ξ,φ) = Σnm Enm [Pnm(ξ1) Qnm(ξ0) – Qnm(ξ1) Pnm(ξ0) ] Pnm(ξ) Qnm(jζ) eimφ Vi(ζ,ξ,φ) = Σnm Enm [Pnm(ξ1) Qnm(ξ) – Qnm(ξ1) Pnm(ξ) ] Pnm(ξ0) Qnm(jζ)eimφ For ζ ≤ 0 and for ξ in (0,1) only Vo(ζ,ξ,φ) = Σnm Fnm [Pnm(ξ1) Qnm(ξ0) – Qnm(ξ1) Pnm(ξ0) ] Pnm(ξ) Qnm(jζ) eimφ Vi(ζ,ξ,φ) = Σnm Fnm [Pnm(ξ1) Qnm(ξ) – Qnm(ξ1) Pnm(ξ) ] Pnm(ξ0) Qnm(jζ)eimφ where we now have E and F coefficients to worry about. Certainly we would not claim the potential to be up/down symmetric by setting E = F. [ But setting E = F does not make things symmetric if the Q(jζ) function is not symmetric.] Here is our corresponding picture: So I am sort of repeating my own hole-in-plate approach here. We then have a new BC to impose, which is that the potential and its gradient be continuous going across the hole (assuming charge not in the hole). By using Qnm(jζ) in both upper and lower regions, I have assured continuity of V since Q is continuous through the origin (I will discuss the cut there soon). Matching of Potential and Its Gradients in the Hole. Matching the potential gives Enm Qnm(j0) = Fnm Qnm(j0). So either Qnm(j0) = 0, or Enm = Fnm. So this brings us to the gradient. It has three directions of interest of course. (1) The ξ direction gradient. This is gradient in the direction in the hole. We have already computed things for this direction: So we now have: ∂n = (1/h1) ∂ξ and h1 = c1 / = c1ξ / in the hole. For ζ ≥ 0 and for ξ in (0,1) only ∂ξVo(ζ,ξ,φ) = Σnm Enm [Pnm(ξ1) Qnm(ξ0) – Qnm(ξ1) Pnm(ξ0) ] Pnm '(ξ) Qnm(jζ) eimφ ∂ξVi(ζ,ξ,φ) = Σnm Enm [Pnm(ξ1) Qnm '(ξ) – Qnm(ξ1) Pnm '(ξ) ] Pnm(ξ0) Qnm(jζ)eimφ For ζ ≤ 0 and for ξ in (0,1) only ∂ξVo(ζ,ξ,φ) = Σnm Fnm [Pnm(ξ1) Qnm(ξ0) – Qnm(ξ1) Pnm(ξ0) ] Pnm '(ξ) Qnm(jζ) eimφ ∂ξVi(ζ,ξ,φ) = Σnm Fnm [Pnm(ξ1) Qnm '(ξ) – Qnm(ξ1) Pnm '(ξ) ] Pnm(ξ0) Qnm(jζ)eimφ Now if we require that things match up in the hole, we set ζ = 0 and repeat the above four lines: For ζ ≥ 0 and for ξ in (0,1) only ∂ξVo(ζ,ξ,φ) = Σnm Enm [Pnm(ξ1) Qnm(ξ0) – Qnm(ξ1) Pnm(ξ0) ] Pnm '(ξ) Qnm(0) eimφ ∂ξVi(ζ,ξ,φ) = Σnm Enm [Pnm(ξ1) Qnm '(ξ) – Qnm(ξ1) Pnm '(ξ) ] Pnm(ξ0) Qnm(0)eimφ For ζ ≤ 0 and for ξ in (0,1) only ∂ξVo(ζ,ξ,φ) = Σnm Fnm [Pnm(ξ1) Qnm(ξ0) – Qnm(ξ1) Pnm(ξ0) ] Pnm '(ξ) Qnm(0) eimφ ∂ξVi(ζ,ξ,φ) = Σnm Fnm [Pnm(ξ1) Qnm '(ξ) – Qnm(ξ1) Pnm '(ξ) ] Pnm(ξ0) Qnm(0)eimφ where we are using the Qnm(0)0,0R (principle sheet, z=1 cut pulled right) as discussed in a separate doc. In order to have things match for all φ and for all ξ, it certainly seems that we will need: Enm Qnm(0) = Fnm Qnm(0) which we write this way to allow for the possibility that Qnm(0) might = 0 for certain m,n values. In fact we know that Qnm(0)0,0R = 2m-1 e+iπ(-n-1)/2 [ Γ(1/2 + n/2+ m/2) / Γ(1 + n/2- m/2)] We know what this means. For n = m - N, N = 2,4,6... we will have Qnm(0) = 0, so for these values, our matching condition here tells us nothing about the relationship between Enm and Fnm. For all other values, the two must be equal. Let's put this on hold and consider the other gradients. (2) The φ direction gradient. Let's just do the derivative on all four of our Smythian form lines, AND lets set ζ = 0 for being in the hole: For ζ ≥ 0 and for ξ in (0,1) only ∂φVo(ζ,ξ,φ) = Σnm im Enm [Pnm(ξ1) Qnm(ξ0) – Qnm(ξ1) Pnm(ξ0) ] Pnm(ξ) Qnm(0) eimφ ∂φVi(ζ,ξ,φ) = Σnm im Enm [Pnm(ξ1) Qnm(ξ) – Qnm(ξ1) Pnm(ξ) ] Pnm(ξ0) Qnm(0)eimφ For ζ ≤ 0 and for ξ in (0,1) only ∂φVo(ζ,ξ,φ) = Σnm im Fnm [Pnm(ξ1) Qnm(ξ0) – Qnm(ξ1) Pnm(ξ0) ] Pnm(ξ) Qnm(0) eimφ ∂φVi(ζ,ξ,φ) = Σnm im Fnm [Pnm(ξ1) Qnm(ξ) – Qnm(ξ1) Pnm(ξ) ] Pnm(ξ0) Qnm(0)eimφ The related scale factor is hφ = h3 = c1 and ∂n = (1/h3) ∂φ . In the hole, ζ = 0 and this just becomes hφ = h3 = c1 which is thankfully not discontinuous. It is the same both above and below the hole plane. So, if we require that ∂φVo and be the same on both sides of the hole plane (which makes ∂φVo be the same, where n points in the φ direction), for all values of ξ and φ, it seems that we need to have Enm m Qnm(0) = Fnm mQnm(0) which is quite similar to our last result, except here we have an added bit of information. If m = 0, we know for sure there is no relationship forced here between the Enm and the Fnm (that is to say, no relation between the En0 and the Fn0 coefficients, at least no relation forced by this gradient condition. ) So this too we shall now put on hold and do the one remaining gradient: (3) The ζ direction gradient. First, here is the action of derivative ∂ζ on things For ζ ≥ 0 and for ξ in (0,1) only ∂ζ Vo(ζ,ξ,φ) = Σnm j Enm [Pnm(ξ1) Qnm(ξ0) – Qnm(ξ1) Pnm(ξ0) ] Pnm(ξ) Qnm '(0) eimφ ∂ζ Vi(ζ,ξ,φ) = Σnm j Enm [Pnm(ξ1) Qnm(ξ) – Qnm(ξ1) Pnm(ξ) ] Pnm(ξ0) Qnm '(jζ)eimφ For ζ ≤ 0 and for ξ in (0,1) only ∂ζ Vo(ζ,ξ,φ) = Σnm j Fnm [Pnm(ξ1) Qnm(ξ0) – Qnm(ξ1) Pnm(ξ0) ] Pnm(ξ) Qnm '(0) eimφ ∂ζ Vi(ζ,ξ,φ) = Σnm j Fnm [Pnm(ξ1) Qnm(ξ) – Qnm(ξ1) Pnm(ξ) ] Pnm(ξ0) Qnm '(0)eimφ Again, for Q on the 0,0R sheet, we know Qnm(z) is continuous as we move across the location the cut used to be in the 0,0L analysis. Nothing special happens at z = 0, so the function and all derivatives are going to be continuous there, so it is the same Qnm '(0) we have in the above and below formulas. Again doing the comparison, we conclude that Enm Qnm '(0) = Fnm Qnm '(0) Again we can write ∂n = (1/h2) ∂ζ where h2 = c1 / = c1 ξ in the hole. (4) Summary of gradient matching in the hole: potential match: Enm Qnm(j0) = Fnm Qnm(j0) ξ gradient match: Enm Qnm(0) = Fnm Qnm(0) ∂n = (1/h1) ∂ξ h1 = c1ξ / φ gradient match: Enm m Qnm(0) = Fnm mQnm(0) ∂n = (1/h3) ∂φ h3 = c1 ζ gradient match: Enm Qnm '(0) = Fnm Qnm '(0) ∂n = (1/h2) ∂ζ h2 = c1 ξ Where do we go from here? At this point, my inclination is to just have one set of constants Enm and go back to the original Smythian form: Vo(ζ,ξ,φ) = Σnm Enm [Pnm(ξ1) Qnm(ξ0) – Qnm(ξ1) Pnm(ξ0) ] Pnm(ξ) Qnm(jζ) eimφ Vi(ζ,ξ,φ) = Σnm Enm [Pnm(ξ1) Qnm(ξ) – Qnm(ξ1) Pnm(ξ) ] Pnm(ξ0) Qnm(jζ)eimφ and now we have our upper/lower asymmetry handled by the innate asymmetry of the Qnm(jζ)0,0R functions, my prototype being Q1(iζ) = -1 + ζ cot-1(ζ). Now at least I don't have any mystery about this asymmetry. What about quantum numbers? The φ gives all integer m as usual. Then we get all "the usual" integer n from the ξ dimension. And in the ζ dimension, I do not require Qnm(0) = 0, so I get no quantization there, and then any Qnm(iζ) is viable as long as Re(n) > -1/2, [ wrong ] and this includes everything in my spectrum. So the entire presentation above goes through unaltered until we reach this point, which I will now call section 7A: 7A. Continuing with our oblate bloid solution. In section 4 above, we left off with this result: (q/ε) (h1/h3h2) e-im'φ0 δ(ζ-ζ0) = [ 2π / (1-ξ02)] Σn (-1)m' [(n+m')! / (n-m')!] Enm' Pnm'(ξ1) Qnm'(jζ) I am just dead in the water unless I can come up with some new orthogonality stuff! This solution method is now on permanent hold until this barrier can be overcome. Idea #1: Suppose we try the SL problem in ζ with the range (-i∞, i∞). In this SL problem, both endpoints are singular and both are limit point, and we know that Qνμ(jζ) are the only possible solutions. We no longer care anything about Qνμ (0) in particular. The spectrum for ν is then the entire continuous real axis from (-1/2,∞). We then expect or at least hope for a result I mentioned long ago, but which takes this new form: !Syntax Error, Idζ Qnμ(iζ) Qn'μ(iζ)* = Cnμ δ(n-n') n in (-1/2,∞) Suppose this were true, then we have some trouble as follows. Here is our form (q/ε) (h1/h3h2) e-im'φ0 δ(ζ-ζ0) = [ 2π / (1-ξ02)] Σn (-1)m' [(n+m')! / (n-m')!] Enm' Pnm'(ξ1) Qnm'(jζ) and we want to apply !Syntax Error, Idζ Qn'm'(iζ)* to both sides and we get (q/ε) (h1/h3h2) e-im'φ0 Qn'm'(iζ)* = [ 2π / (1-ξ02)] Σn (-1)m' [(n+m')! / (n-m')!] Enm' Pnm'(ξ1) Cnm' δ(n-n') and then we have a "measure problem" on the RHS. We would then have to assume that Σn is really an integral over n, but then that violates the Pnm(ξ) rule which said we need n = integer to keep this from blowing up at ξ = -1 which is a loci inside our solution space. So, as usual, I am between a rock and a hard place, with no one do advise me but myself and the web and the few books I have. Idea #2. Go through the cut Qνμ(z)0,0L = 2ν [ Γ(1+ν)Γ(1+ν+μ)/ Γ(2+2ν)] z-ν-1 // Bateman p 132 (36) I also fiddled around and found that Q1(z)0,1L = Q1(z)0,0L - iπz which makes me think that we are NOT going to get z-ν-1 going for Qnm'(jζ)0,0R. So how should I go after the large z asymptotic behavior going south? All my Bateman results are north only. Let's look at a simple case: Q1(z)0,1L = 2-2 Γ(2)/Γ(5/2) z-2 F(3/2, 1; 5/2; 1/z2) // Bateman p 122 (5) Well, I went off and studied this whole subject in the Legendre section, see Maze doc for a summary. It did not give me a way out. At this point I wrote the Maze document, and I am now resuming here for a while. Idea #3: The odd-ν idea to get V continuity at ζ = 0 This during a 4 AM Smith's walk today Feb 1. For the ζ SL homo BC, instead of taking Q(j0+) = 0 [ which is the jigger problem ], do this instead: (1) Considering [ Note that we are talking about the Qνμ(z=0±)0,0L function here , and we are talking the parameterization where each spheroid has two ζ values, so ζ < 0 in the lower bloid region. This is the preferred way to go. Qnm(jζ) decays on all rays.] Qνμ(z=0±) = 2μ-1 e±iπ(-ν-1)/2 [ Γ(1/2 + ν/2+ μ/2) / Γ(1 + ν/2- μ/2)] // p 134 (40) limit the spectrum to ν = values such that e±iπ(-ν-1)/2 = the same for + and -. This would cause Qnm(jζ) to be continuous at ζ = 0, a condition we certainly want to have. This despite the presence of the cut. This seems to limit us to the following values for ν: (ν+1)/2 = integer N. Then e±iπ(-ν-1)/2 = e±iπN = (-1)N which is clearly the same for + and -. So that says ν = 2N+1 N = 0,1,2... or ν = 1,3,5,7 or ν = odd. So then we are not forcing Qνμ(0) = 0, so we are staying away from the jigger problem. (2) Then require that Qνμ '(z=0±) = 0 as the homo BC condition. This means the slope is zero coming in from both + and - directions, so the slope of Q is then continuous as well as Q being continuous. Having the slope continuous is consistent with there being no charge lying on the hole disk! I only set the slope to 0 because then we have a homo BC for the SL problem. Otherwise I would just say Qνμ '(z=0±) = same for + and - and does not have to be 0. But you have to ask, in an asymmetric problem like this, why would you expect to have Qνμ '(z=0±) = 0 ? Looking at ∂ζ Vo(ζ,ξ,φ) = Σnm j Enm [Pnm(ξ1) Qnm(ξ0) – Qnm(ξ1) Pnm(ξ0) ] Pnm(ξ) Qnm '(jζ) eimφ we would conclude that ∂ζ Vo(ζ,ξ,φ) = 0 at the hole plane. But for the Green's charge just above the hole plane, we know this cannot be true on the entire Vo part of the hole plane. It is true that we are getting into the region of how we parameterize the coordinates, but if we stay at ζ = +iε and wander just above and over the hole disk, it does seem this is forcing ∂ζVo(ζ=0+,ξ,φ) = 0. The close point charge shows this cannot be right, so this idea is no good. "Wrong Way, Dead End" . But maybe we can take idea (1) without idea (2). (2') Want to have Qνμ '(z=0+) = Qνμ '(z=0-). Could figure out what this implies spectrally from Legendre properties. But how would we bring this into the SL problem as a homo BC? Comments: In this idea, we let go of Qνμ(0) = 0 as a homo BC and let Qνμ(0) come out whatever it is. It is the same for ± because we select odd ν only. But for gradient we need Qνμ '(z=0+) = Qνμ '(z=0-) but this is not in the form of a homo BC for the ζ SL problem. I later learned that Qνm '(0±) = ∓ i Qνm+1 (0±) m ≥ 0 and integer Qν-m ' (0±) = ∓ i Qνm+1 (0±) Γ(ν-m+1)/Γ(ν+m+1) m ≥ 0 and integer With ν = odd only, these would say Qνm '(0±) = ∓ i Qνm+1 (0) m ≥ 0 and integer Qν-m ' (0±) = ∓ i Qνm+1 (0) Γ(ν-m+1)/Γ(ν+m+1) m ≥ 0 and integer and then our (2') condition becomes (for μ=m>0) Qνμ '(z=0+) = Qνμ '(z=0-) - i Qνm+1 (0) = + iQνm+1 (0) and we are then forced back to Qνm+1(0) = 0 again, which is what we were trying to "get away from", since this forces Vo= 0 on the hole plane (ie, this is the jigger problem). So Idea #4 fails. Idea #4 Why not use the following Q function: Qnm(j|ζ|) ? Above the hole this is Qnm(jζ), and below the hole it is Qnm(-jζ). In both regions this is a Laplace atom (see Bateman p 122 (6) area). Granted, one must be careful at ζ = 0. But this lets us use Qnm(jζ)0,0L on both top and bottom, so we get decay of the Q function both ways. The cost is the extra burden of requiring continuity at ζ = 0. We have seen Jackson use e-k|z| in his cylindrical coordinate work for example. Let's track through what this would do. We start with our Smythian form. (where I have now added abs value on ζ) Vo(ζ,ξ,φ) = Σnm Enm [Pnm(ξ1) Qnm(ξ0) – Qnm(ξ1) Pnm(ξ0) ] Pnm(ξ) Qnm(j|ζ|) eimφ Vi(ζ,ξ,φ) = Σnm Enm [Pnm(ξ1) Qnm(ξ) – Qnm(ξ1) Pnm(ξ) ] Pnm(ξ0) Qnm(j|ζ|)eimφ Now, which coordinates are we using? If we use "spheroid has two labels ±ζ", then the ξ is (0,1) only. Our Qnm(j|ζ|) decays whether ζ is + or -, that is, whether we are above or below the hole plane. In this system, a point in the hole plane has an unambiguous single value of ξ. This is our system: In this scheme, as we approach the hole ζ = 0 from either above or below, we have Qnm(j0+) appearing in our formulas for both Vo and Vi. Therefore, we have achieved continuity of V in the hole. But we also need to have continuity of Qnm '(j|ζ|) at the hole plane so that we don't have any charge appearing on that plane. Now the only way that f(|ζ|) can have a continuous slope at ζ = 0 is if that slope is zero (draw a picture), because this f is a mirror function. Thus, we need to have Qnm '(j0+) = 0. According to our Leg props doc we have [ I show both signs here, but we are only interested in the upper sign ] Qνm '(0±) = ∓ i Qνm+1 (0±) m ≥ 0 and integer Qν-m ' (0±) = ∓ i Qνm+1 (0±) Γ(ν-m+1)/Γ(ν+m+1) m ≥ 0 and integer Qνm ' (0±) = ∓ i Qν-m+1 (0±) Γ(ν+m+1)/Γ(ν-m+1) m < 0 and integer Let's first look at the m > 0 case. In order to have Qnm '(j0+) = 0, we need Qνm+1 (0+) = 0. But our Theorem 1 says Qνm+1 (0+) = 0 only when m+1>0 and n = m-1,m-3.... -m-1. So these are the allowed n values when m ≥ 0 which make Qnm '(j0+) = 0 . Now consider m < 0. Let's define m1 = -m. Then we have m1 > 0 and from the last line above, Qν-m1 ' (0±) = ∓ i Qνm1+1 (0±) Γ(ν-m1+1)/Γ(ν+m1+1) m1 ≥ 0 and integer = ∓ i Qνm1+1 (0±) f(ν,-m1) = ∓ i Qνm1+1 (0±) f(ν,m) There are now two possible ways to have a zero. The first is when Qνm1+1 (0±) has a zero, and this occurs when ν = m1-1,m1-3.... -m1-1. The second is when f(ν,m) has a zero. But Theorem 3 says for m<0, f(ν,m) has no zeros. So the only zeros when m < 0 are for ν = m1-1,m1-3.... -m1-1. BUT, be careful, because f(ν,m) has poles when ν = |m|-1, |m|-2...- |m| = m1-1,m1-2.... -m1. Only ν = -m1-1 avoids these poles. So we have to eliminate Q functions which blow up, and only ν = -m1-1 = m-1 survives for m < 0. We can combine our results to say this: Qnm '(j0+) = 0 and is finite when: m > 0: n = m-1,m-3.... -m-1 m < 0: n = m-1 Now let's impose our requirement that our Q function decay for large ξ. Since Q ~ |ζ|-n-1 we have to rule out all n values which are < 0. This kills off the second line above, and leaves only part of the first line. So here is our new spectrum Qnm '(j0+) = 0 and is finite, and Qnm (jξ) is finite s-norm, when: m > 0: n = m-1,m-3.... 0/1 whichever one you end on We now observe that Vo contains the product Pnm(ξ) Qnm(j|ζ|). We now call upon, Theorem 2: If m is an integer, (a) Pnm(ξ) = finite for all complex n, including all integer n. (b) Pnm(ξ) = 0 only for m > 0 and n = m-1,m-2, ... -m. [ n integer and -m ≤ n < m ] We see that the set of n values that make P = 0 includes all of our allowed Q spectrum!!! This means that every single one of our "allowed" Q functions is killed off by its partner P function. We now have Vo(ζ,ξ,φ) = Σnm Enm [Pnm(ξ1) Qnm(ξ0) – Qnm(ξ1) Pnm(ξ0) ] Pnm(ξ) Qnm(j|ζ|) eimφ = 0 !!! Therefore, our whole plan crashes to the ground (as I expected and almost hoped it would). So I now return to the Maze document to summarize this last finding.