5. oblate bloid with p and q attempt 4

Oblate bloid with p and q attempt 4: success! PhL 2.14-21.10 The problem here is to find the Green's Function for an oblate hyperboloid. 1. The p and q functions. 1 2. The Smythian form, picture, and matching in the Neck 2 (a) get matching at the boundary between upper Vi and Vo 2 (b) the lower bloid form. 3 (c) Smythian form for the upper and lower bloids: 3 (d) Match the potential at the neck which is ζ = 0. 4 (e) Match the ζ derivative of the potential at the neck which is ζ = 0. 4 (f) Match the ξ derivative of the potential at the neck which is ζ = 0. 5 (g) Match the φ derivative of the potential at the neck which is ζ = 0. 6 (h) Status after doing all the neck matches: 6 (i) What are the advantages of our Smythian Form: 7 (j) Folding of the m sum. 7 (k) The spectrum of n. 7 3. Apply Gauss's Law to pillbox around point charge q. 8 (a) The Boiler Plate Warmup. 8 (b) Now we compute [ ∂ζVi - ∂ζVo ] 8 (c) compute the B coefficients: 9 (d) "pillbox" in the lower bloid. 10 (e) Tangential derivatives at the bloid surface ? 11 (f) Matching tangential derivatives on the spheroid surface? 11 (g) So here is how we now stand: 11 Final Result (ha ha) 13 (h) Discussion. 14 4. The Cone Limit 15 5. The On-Axis Limit 18 6. In-the-neck Limit 20 7. The Iris Limit 22 8. First put Green's charge in the neck, then go to the Iris Limit 23 (a) Potential on the iris and in the hole. 26 (b) Charge density on the iris // small correction made 3.8.10 27 (c) Comment: 28 9. A comparison between the real and complex bloid solutions. 28 10. Restatement of the General Bloid Solution: 29 Final Solution: 29 1. The p and q functions. The goal is to use these "real atom" functions in the ζ dimension, qnm(ζ) = (-i)2m (±i)n+1 Qnm(iζ) pnm(ζ) = [ (∓i)-m Pnm(iζ) + (±i)-mPnm(-iζ)] = (±i)-n 2cos[(n+m)π/2] Pnm(iζ) – (±i)-m/2 (-i)2m (2/π) sinπ(n+m) Qnm(iζ) These functions are real when n and m are and ζ are real. You must use the upper sign when Re(ζ) > 0, otherwise the lower sign. 2. The Smythian form, picture, and matching in the Neck . First, the picture We shall try a more general form than we used in Attempt #2. Let's first concentrate just on the upper bloid region and try this: Upper half bloid: Vo(ζ,ξ) = Σnm [ Anm qnm(ζ) ] Pnm(ξ) cos(mφ) Vi(ζ,ξ) = Σnm [Bnm pnm(ζ) + Cnm qnm(ζ) ] Pnm(ξ) cos(mφ) //upper ζ ≥ 0 We must have the q alone for the outside part due to large ζ decay, so this is about as general as we can get. (a) get matching at the boundary between upper Vi and Vo . Continuity at the boundary then tells us that Vo(ζ0,ξ) = Vi(ζ0,ξ) Σnm [Anm qnm(ζ0) ] Pnm(ξ) cos(mφ) = Σnm [Bnm pnm(ζ0) + Cnm qnm(ζ0) ] Pnm(ξ) cos(mφ) Matching terms tells us that Anm qnm(ζ0) = Bnm pnm(ζ0) + Cnm qnm(ζ0) So multiply and divide the upper form like this: Vo(ζ,ξ) = Σnm [ Anm qnm(ζ0)qnm(ζ)] Pnm(ξ) cos(mφ) / qnm(ζ0) = Σnm [Bnm pnm(ζ0) + Cnm qnm(ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) / qnm(ζ0) Then we have: Vo(ζ,ξ) = Σnm [Bnm pnm(ζ0) + Cnm qnm(ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) / qnm(ζ0) Vi(ζ,ξ) = Σnm [Bnm pnm(ζ) + Cnm qnm(ζ) ] Pnm(ξ) cos(mφ) Now define Bnm' = Bnm/ qnm(ζ0) and the same for C, then we have Vo(ζ,ξ) = Σnm [B'nm pnm(ζ0) + C'nm qnm(ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) Vi(ζ,ξ) = Σnm [B'nm pnm(ζ) + C'nm qnm(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) Now removed the primes to reduce clutter, and we arrive at this Symthian form: Upper half bloid: Vo(ζ,ξ) = Σnm [Bnm pnm(ζ0) + Cnm qnm(ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) Vi(ζ,ξ) = Σnm [Bnm pnm(ζ) + Cnm qnm(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 This is what it has to look like to match at the boundary between Vo and Vi . (b) the lower bloid form. Now, by the same argument, the lower bloid must have this form (new meaning for primes): Lower half bloid: Vo(ζ,ξ) = Σnm [B'nm pnm(-ζ0) + C'nm qnm(-ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) Vi(ζ,ξ) = Σnm [B'nm pnm(ζ) + C'nm qnm(ζ) ] qnm(-ζ0) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 The matching surface here has ζ = - ζ0, so have installed that value. (c) Smythian form for the upper and lower bloids: Upper half bloid: Vo(ζ,ξ) = Σnm [Bnm pnm(ζ0) + Cnm qnm(ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) Vi(ζ,ξ) = Σnm [Bnm pnm(ζ) + Cnm qnm(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 Lower half bloid: Vo(ζ,ξ) = Σnm [B'nm pnm(-ζ0) + C'nm qnm(-ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) Vi(ζ,ξ) = Σnm [B'nm pnm(ζ) + C'nm qnm(ζ) ] qnm(-ζ0) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 (d) Match the potential at the neck which is ζ = 0. We have at the neck: Vi(0+,ξ) = Σnm [Bnm pnm(0+) + Cnm qnm(0+) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper Vi(0- ,ξ) = Σnm [B'nm pnm(0-) + C'nm qnm(0-) ] qnm(-ζ0) Pnm(ξ) cos(mφ) //lower But our p and q functions are all perfectly symmetric, so rewrite the lower line Vi(0+,ξ) = Σnm [Bnm pnm(0+) + Cnm qnm(0+) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper Vi(0- ,ξ) = Σnm [B'nm pnm(0+) + C'nm qnm(0+) ] qnm(ζ0) Pnm(ξ) cos(mφ) //lower The condition is then: Bnm pnm(0+) + Cnm qnm(0+) = B'nm pnm(0+) + C'nm qnm(0+) (Bnm - B'nm) pnm(0+) = – ( Cnm - C'nm) qnm(0+) The good news is that we are not forced right here to have B = B' and C = C'. If we were, then due to the perfect symmetry of our P and Q functions, we would have the exact same potential on both upper and lower and we know this cannot be true since the Green's Charge is only in the upper. So, to coin a phrase, once again "we are still alive". (e) Match the ζ derivative of the potential at the neck which is ζ = 0. This is going to be the tricky one I think. We start with Vi(ζ,ξ) = Σnm [Bnm pnm(ζ) + Cnm qnm(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 Vi(ζ,ξ) = Σnm [B'nm pnm(ζ) + C'nm qnm(ζ) ] qnm(-ζ0) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 As usual, we pause to review the scale factors to make sure we don't trip up: h1 = c1 / q1, q2, q3 = ξ, ζ, φ h2 = c1 / h3 = c1 When we talk about the derivative normal to the neck disk plane, just above the plane we have our vector pointing up and we say ∂n = (1/h2) ∂ζ . Both n and ζ increase "going up". From above we know that h2 = c1ξ at the neck, where ξ is a positive number in the range (ξ0, 1). So ∂n = (c1ξ)-1 ∂ζ. Now what about just below the neck? As we move up, ζ gets less negative so dζ is positive, and dn is also positive, so we have the exact same rule: ∂n = (c1ξ)-1 ∂ζ . And of course ξ is the same on both sides of the neck. So we can ignore the factor (c1ξ)-1 in our match since it is the same on both sides, and then our matching condition is to make ∂ζVi be the same on both sides of the neck! We have: Vi(ζ,ξ) = Σnm [Bnm pnm(ζ) + Cnm qnm(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 Vi(ζ,ξ) = Σnm [B'nm pnm(ζ) + C'nm qnm(ζ) ] qnm(-ζ0) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 ∂ζ Vi(ζ,ξ) = Σnm [Bnm pnm '(ζ) + Cnm qnm '(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 ∂ζ Vi(ζ,ξ) = Σnm [B'nm pnm '(ζ) + C'nm qnm '(ζ) ] qnm(-ζ0) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 But we want this match to occur at ζ = 0. To allow for the possibility that derivatives are different above and below the plane of the neck, we write [ we also replace qnm(-ζ0) = qnm(ζ0) in the second line] ∂ζ Vi(0+,ξ) = Σnm [Bnm pnm '(0+) + Cnm qnm '(0+) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper ∂ζ Vi(0-,ξ) = Σnm [B'nm pnm '(0-) + C'nm qnm '(0-) ] qnm(ζ0) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 Since there is no surface charge layer in the neck, we require these to be the same. Therefore our matching condition is then Bnm pnm '(0+) + Cnm qnm '(0+) = B'nm pnm '(0-) + C'nm qnm '(0-) Feb 21, 2001: I now realize that, although pnm '(0±) = 0, this is NOT TRUE for qnm '(0±) as I originally thought. So starting here, things will be different. I will call this attempt 4! We have these results from "reality of P and Q" doc: pnm '(0±) = 0 qnm '(0±) = ∓2m [ Γ(1+n/2+m/2) / Γ(1/2+n/2-m/2)] Therefore our above derivative matching condition is this: + Cnm qnm '(0+) = + C'nm qnm '(0-) C'nm = – Cnm Wee move on to the next matching requirement: (f) Match the ξ derivative of the potential at the neck which is ζ = 0. Here we have ∂n = (1/h1) ∂ξ and h1 = c1ξ / which is a positive real number. This is a "radial" ρ-like derivative on the disk. In our picture on the right side of the center axis we have, as we move radially inward, we have dξ > 0, so we can think of as pointing inward. We just want to make sure this inward radial derivative is the same just above and just below the disk. The positive factor h1 can then be ignored. We then have, Vi(ζ,ξ) = Σnm [Bnm pnm(ζ) + Cnm qnm(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 Vi(ζ,ξ) = Σnm [B'nm pnm(ζ) + C'nm qnm(ζ) ] qnm(-ζ0) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 ∂ξVi(ζ,ξ) = Σnm [Bnm pnm(ζ) + Cnm qnm(ζ) ] qnm(ζ0) Pnm '(ξ) cos(mφ) //upper ζ ≥ 0 ∂ξ Vi(ζ,ξ) = Σnm [B'nm pnm(ζ) + C'nm qnm(ζ) ] qnm(-ζ0) Pnm'(ξ) cos(mφ) //lower ζ ≤ 0 ∂ξVi(0+,ξ) = Σnm [Bnm pnm(0+) + Cnm qnm(0+) ] qnm(ζ0) Pnm '(ξ) cos(mφ) //upper ∂ξ Vi(0-,ξ) = Σnm [B'nm pnm(0-) + C'nm qnm(0-) ] qnm(-ζ0) Pnm'(ξ) cos(mφ) //lower These will be equal if we set Bnm pnm(0+) + Cnm qnm(0+) = B'nm pnm(0-) + C'nm qnm(0-) But this is exactly the same condition we get for the potential match, so we get no new condition in this way, and we are still alive! (g) Match the φ derivative of the potential at the neck which is ζ = 0. This will work the same way as ξ, but instead of having Pnm '(ξ) replacing Pnm(ξ) we will have -msin(mφ) replacing cos(mφ). Our matching condition will then be: m(Bnm pnm(0+) + Cnm qnm(0+)) = m(B'nm pnm(0-) + C'nm qnm(0-)) which again gives us nothing new, whether or not m = 0. So we are STILL alive after doing all these matches at the neck. (h) Status after doing all the neck matches: We install the fact that C'nm = – Cnm to get Upper half bloid: Vo(ζ,ξ) = Σnm [Bnm pnm(ζ0) + Cnm qnm(ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) Vi(ζ,ξ) = Σnm [Bnm pnm(ζ) + Cnm qnm(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 Lower half bloid: Vo(ζ,ξ) = Σnm [B'nm pnm(-ζ0) – Cnm qnm(-ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) Vi(ζ,ξ) = Σnm [B'nm pnm(ζ) – Cnm qnm(ζ) ] qnm(-ζ0) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 with conditions: (Bnm - B'nm) pnm(0+) = – ( Cnm - C'nm) qnm(0+) C'nm = – Cnm Use the second to eliminate C' from the first (Bnm - B'nm) pnm(0+) = – ( 2Cnm ) qnm(0+) We still seem to have two sets of coefficients that are completely free. (i) What are the advantages of our Smythian Form: inside the upper bloid we are continuous at the boundary ζ = ζ0 inside the lower bloid we are continuous at the boundary ζ = -ζ0 ∂ζV is also continuous at ζ = ζ0 in the upper bloid so there is no surface charge out on that spheroid, except at the Green's charge. And the same thing will happen in the lower bloid. Going off to infinity in any direction is properly damped by the nature of qnm(ζ). The condition Pnm(ξ0) = 0 determines the n spectrum and makes V = 0 for upper and lower bloid. Since ξ is only in the limited range (ξ0, 1), we do not hit ξ = -1, and we then have no conflicting condition there requiring n to be integral. (j) Folding of the m sum. One imagines that the initial sums are m = (-∞,∞) all integers since this is the full spectrum of the φ ODE. However, each factor in the integrand has a symmetry property, to wit qn-m(ζ) = f(n,-m) qnm(ζ) m = integer, n = general, ζ = anything pn-m(ζ) = f(n,-m) pnm(ζ) Pn-m(ξ) = (-1)m f(n,-m)Pnm(ξ) // this (only!) is the on-the-cut version of P cos(-mφ) = cos(mφ) Therefore, we can take the sum -∞,-1 and "fold it over" into the positive sum by redefining the coefficients. We assume this was done right at the very start before we defined B,C,B',C' We use the short hand notation Σnm to stand for Σm=0∞Σn . (k) The spectrum of n. This is obtained from Pnm(ξ1) = 0 since ξ1 defines the surface of the entire metallic hyperboloid (=bloid). As shown elsewhere, (1) the eigenvalues of n are real and in general non integral; (2) the eigenvalues of n depend on m and ξ1 so we could write ni = ni(m, ξ1), i = 1,2,3... as a way to enumerate these zeros. The n sum therefore goes inside the m sum. We shall just write Σn with this spectrum understood. 3. Apply Gauss's Law to pillbox around point charge q. (a) The Boiler Plate Warmup. I show in "Smythe Greens oblate spheroid.doc" that, for a Green's point charge q located at point (ζ0, ξ0, φ0), Gauss's law says ∂ζVi - ∂ζVo = (q/ε) (h2/h3h1) δ(φ-φ0) δ(ξ - ξ0) where q1, q2, q3 = ξ, ζ, φ which is the key to these results, h1 = c1 / h2 = c1 / h3 = c1 h1h3 = c1 / c1 = c12 h2 = c1 / (1+ζ2) c1 h2 = h1h3 => (h2/h3h1) = [(1+ζ2) c1]-1 The charge density on the spheroid surface is given by [ this is Gauss's law for the pillbox ] σ/ε = ∂nVi - ∂nVo ∂n = (1/h2) ∂ζ => σ/ε = (1/h2)[ ∂ζVi - ∂ζVo ] Therefore, the h2 cancels when we write σ. Here are our two main results: ∂ζVi - ∂ζVo = (q/ε) (h2/h3h1) δ(φ-φ0) δ(ξ - ξ0) (**) σ/ε = (1/h2)[ ∂ζVi - ∂ζVo ] = (q/ε) (1/h3h1) δ(φ-φ0) δ(ξ - ξ0) dA2 = h3h1dq3dq1 = h3h1dφ dξ // area of pillbox side This then is the effective surface charge distribution "of the point charge q". In the doc last referenced, we took this double delta limit to the north pole which was an extra complication. Here we work as is. We can simplify ∂ζVi - ∂ζVo = (q/ε) (h2/h3h1) δ(φ-φ0) δ(ξ - ξ0) = (q/ε) [(1+ζ2) c1]-1 δ(φ-φ0) δ(ξ - ξ0) = (q/c1ε) δ(φ-φ0) δ(ξ - ξ0)/ (1+ζ02) (b) Now we compute [ ∂ζVi - ∂ζVo ] using our Smythian upper form above, which is this: Vo(ζ,ξ) = Σnm [Bnm pnm(ζ0) + Cnm qnm(ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) Vi(ζ,ξ) = Σnm [Bnm pnm(ζ) + Cnm qnm(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 ∂ζ Vo(ζ,ξ) = Σnm [Bnm pnm(ζ0) + Cnm qnm(ζ0)] qnm '(ζ) Pnm(ξ) cos(mφ) ∂ζ Vi(ζ,ξ) = Σnm [Bnm pnm '(ζ) + Cnm qnm '(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 ∂ζ Vi - ∂ζVo = Σnm Pnm(ξ) cos(mφ) * [Bnm pnm '(ζ) + Cnm qnm '(ζ) ] qnm(ζ0) – [Bnm pnm(ζ0) + Cnm qnm(ζ0)] qnm '(ζ) But we only care at ζ = ζ0 so we have ∂ζ Vi - ∂ζVo = Σnm Pnm(ξ) cos(mφ) * [Bnm pnm '(ζ0) + Cnm qnm '(ζ0) ] qnm(ζ0) – [Bnm pnm(ζ0) + Cnm qnm(ζ0)] qnm '(ζ0) The two C terms exactly cancel (meaning we will learn nothing about the C coefficients from this pillbox exercise) leaving us with: ∂ζ Vi - ∂ζVo = Σnm Pnm(ξ) cos(mφ) * Bnm pnm '(ζ0) qnm(ζ0) – Bnm pnm(ζ0) qnm '(ζ0) = – Σnm Pnm(ξ) cos(mφ) Bnm W[pnm(ζ0), qnm(ζ0)] For ζ0 > 0 we look this up in P and Q to find W[pnm(ζ), qnm(ζ)] = + 2cos[(n+m)π/2]f(n,m) * 1/(1+ζ2) so we have our result that ∂ζ Vi - ∂ζVo = – Σnm Pnm(ξ) cos(mφ) Bnm 2cos[(n+m)π/2]f(n,m) * 1/(1+ζ02) = - 2/(1+ζ02) Σnm Bnm cos[(n+m)π/2]f(n,m) Pnm(ξ) cos(mφ) Restate the final result: ∂ζ Vi - ∂ζVo = - 2/(1+ζ02) Σnm Bnm cos[(n+m)π/2]f(n,m) Pnm(ξ) cos(mφ) (c) compute the B coefficients: So combining part a and b above, we have (q/c1ε) δ(φ-φ0) δ(ξ - ξ0)/ (1+ζ02) = - 2/(1+ζ02) Σnm Bnm cos[(n+m)π/2]f(n,m) Pnm(ξ) cos(mφ) (q/c1ε) δ(φ-φ0) δ(ξ - ξ0) = - 2Σnm Bnm cos[(n+m)π/2]f(n,m) Pnm(ξ) cos(mφ) First, apply !Syntax Error, Idφ cos(m'φ) to both sides and use Schaum p 96 15.27 on the RHS, !Syntax Error, Idφ cos(m'φ) cos(mφ) = δm,m'2π/(2-δm,0) to get (first go to primes, then replaced m' by m, two steps), cos(mφ0) (q/c1ε) δ(ξ - ξ0) = - 4πΣn (2-δm,0)-1Bnm cos[(n+m)π/2]f(n,m) Pnm(ξ) But we assumed at the start in our cos(mφ) form that φ0 = 0 so we have (q/c1ε) δ(ξ - ξ0) = - 4π (2-δm,0)-1Σn Bnm cos[(n+m)π/2]f(n,m) Pnm(ξ) schk Now our next orthogonality is this (taken from "the Smyth method for Green's...", !Syntax Error, Idθ sinθ Pnm(cosθ) Pn'm(cosθ) = Knm(α) δn,n' Knm(α) = - sin2α /(2n +1) * ∂zPnm(cosα) ∂nPnm(cosα) schk from memory or !Syntax Error, Idξ Pnm(ξ) Pn'm(ξ) = knm(ξ1) δn,n' Knm(ξ1) = - (1-ξ12) /(2n +1) * ∂ξPnm(ξ1) ∂nPnm(ξ0) // ignore chg in fnctnl form of K So apply !Syntax Error, I Pn'm(ξ) to both sides of (*) to get (again, use n', then replace n' by n) (q/c1ε) Pnm(ξ0) = + 4π (2-δm,0)-1Bnm cos[(n+m)π/2]f(n,m) Knm(ξ1) (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) = + Bnm cos[(n+m)π/2]f(n,m) Knm(ξ1) Bnm = (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) sec[(n+m)π/2] f(n,-m)/ Knm(ξ1) So this tells us our B coefficients and of course they are REAL. (d) "pillbox" in the lower bloid. Suppose we just repeat the above steps in the lower bloid. Since qnm(-ζ0) = qnm(ζ0), the only differences we get would be these: 1. Replace B with B' and C with C'. 2. ζ < 0 in the lower bloid. 3. Since there is no point charge there, we get ∂ζ Vi - ∂ζVo = 0 at the image point of the Green charge. Then as before, C' cancels away and we learn nothing about it, but we would then find that 0 = + 2Σnm B'nm cos[(n+m)π/2]f(n,m) Pnm(ξ) cos(mφ) where I show a sign change because the Wronskian would have the other sign. But no matter, this tells us a very important fact: B'nm = 0 This is the condition such that the radial derivative ∂ζV be continuous at the boundary between Vi and V0 down in the lower bloid. (e) Tangential derivatives at the bloid surface ? This is a little new to me, first time I have considered it. We know that we get V = 0 on the bloid surface from the Pnm(ξ1) spectrum. We can get the charge σ from the ∂ξ derivatives. But are we sure that both the tangential derivatives at the bloid surface will be 0 (as they must be). Once again, Vi(ζ,ξ) = Σnm [Bnm pnm(ζ) + Cnm qnm(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 ∂ζ Vi(ζ,ξ) = Σnm [Bnm pnm '(ζ) + Cnm qnm '(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 ∂ζ Vi(ζ,ξ1) = Σnm [Bnm pnm '(ζ) + Cnm qnm '(ζ) ] qnm(ζ0) Pnm(ξ1) cos(mφ) //upper ζ ≥ 0 But Pnm(ξ1) = 0 so we are OK. Similarly, if we look at the φ tangential derivative, we get this same Pnm(ξ1) factor, so I think it will be OK as well. (f) Matching tangential derivatives on the spheroid surface? This is another aspect I have not considered. Again as have on the top Vo(ζ,ξ) = Σnm [Bnm pnm(ζ0) + Cnm qnm(ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) Vi(ζ,ξ) = Σnm [Bnm pnm(ζ) + Cnm qnm(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 ∂ξ Vo(ζ,ξ) = Σnm [Bnm pnm(ζ0) + Cnm qnm(ζ0)] qnm(ζ) Pnm '(ξ) cos(mφ) ∂ξ Vi(ζ,ξ) = Σnm [Bnm pnm(ζ) + Cnm qnm(ζ) ] qnm(ζ0) Pnm '(ξ) cos(mφ) //upper ζ ≥ 0 If we then set ζ= ζ0 for our spheroid, they match just fine. Same will happen for lower bloid. And the same argument shows that we have a match for the φ derivative which is the other tangent. (g) So here is how we now stand: Upper half bloid: Vo(ζ,ξ) = Σnm [Bnm pnm(ζ0) + Cnm qnm(ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) Vi(ζ,ξ) = Σnm [Bnm pnm(ζ) + Cnm qnm(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 Lower half bloid: Vo(ζ,ξ) = Σnm [– Cnm qnm(-ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) Vi(ζ,ξ) = Σnm [– Cnm qnm(ζ) ] qnm(-ζ0) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 with condition (Bnm) pnm(0+) = – ( 2Cnm ) qnm(0+) Cnm = (-1/2) (pnm(0+)/ qnm(0+) Bnm and with pillbox evaluation Bnm = (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) sec[(n+m)π/2] f(n,-m)/ Knm(ξ1) This almost looks like a solution to me!!! We know from reality of PQ doc that pnm(0±) = 2m+1 / [ Γ(1/2 - n/2 - m/2) Γ(1 + n/2 - m/2) ] qnm(0±) = (-1)m 2m-1 [ Γ(1/2 + n/2+ m/2) / Γ(1 + n/2- m/2)] So we find then that (pnm(0+)/ qnm(0+) = 2m+1 Γ(1 + n/2- m/2) / [(-1)m 2m-1 Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2) Γ(1 + n/2 - m/2) ] = 2+1 / [(-1)m 2-1 Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2) ] = 4 (-1)m / [ Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2) ] Therefore we have Cnm = (-1/2) (pnm(0+)/ qnm(0+) Bnm = = (-1/2) Bnm 4 (-1)m / [ Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2) ] = 2(-1)m+1 Bnm / [ Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2) ] Restate the tentative solution: Upper half bloid: Vo(ζ,ξ) = Σnm [Bnm pnm(ζ0) + Cnm qnm(ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) Vi(ζ,ξ) = Σnm [Bnm pnm(ζ) + Cnm qnm(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 Lower half bloid: Vo(ζ,ξ) = Σnm [– Cnm qnm(-ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) Vi(ζ,ξ) = Σnm [– Cnm qnm(ζ) ] qnm(-ζ0) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 Bnm = (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) sec[(n+m)π/2] f(n,-m)/ Knm(ξ1) Cnm =2(-1)m+1 Bnm / [ Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2) ] Let's define 1/ [ Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2) ] = g3(n,m) Then Cnm =2(-1)m+1 g3(n,m) Bnm and we write Upper half bloid: Vo(ζ,ξ) = Σnm Bnm [pnm(ζ0) – 2(-1)m g3(n,m)qnm(ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) Vi(ζ,ξ) = Σnm Bnm [pnm(ζ) – 2(-1)m g3(n,m)qnm(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 Lower half bloid: Vo(ζ,ξ) = Σnm Bnm [2(-1)m g3(n,m) qnm(-ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) Vi(ζ,ξ) = Σnm Bnm [2(-1)m g3(n,m) qnm(ζ) ] qnm(-ζ0) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 Bnm = -(2-δm,0) (q/[4πc1ε]) Pnm(ξ0) sec[(n+m)π/2] f(n,-m)/ Knm(ξ1) Notice now that the two lines for the lower bloid are the same! This is reasonable since there is nothing on the dividing line to divide the regions, so we can now give a Final Result (ha ha) Upper half bloid: Vo(ζ,ξ) = Σnm Bnm [pnm(ζ0) – 2(-1)m g3(n,m)qnm(ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) Vi(ζ,ξ) = Σnm Bnm [pnm(ζ) – 2(-1)m g3(n,m)qnm(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 Lower half bloid: V(ζ,ξ) = Σnm Bnm 2(-1)m g3(n,m) qnm(ζ0) qnm(ζ) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 We get the q function "all the way" in the lower bloid! Here is all supporting data: Bnm = (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) sec[(n+m)π/2] f(n,-m)/ Knm(ξ1) f(n,-m) = Γ(1+n-m)/ Γ(1+n+m) Knm(ξ1) = - (1-ξ12) /(2n +1) * ∂ξPnm(ξ1) ∂nPnm(ξ1) g3(n,m) = 1/ [ Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2) ] Below I find this result π sec((n+m)π/2) = Γ(1/2 + n/2+m/2) Γ(1/2 - n/2-m/2) = [1/ g3(n,m)] which allows us to write an alternate form for Bnm Bnm = (2-δm,0) (q/[4π2c1ε]) Pnm(ξ0) [1/g3(n,m)] f(n,-m)/ Knm(ξ1) We could write the solution even more densely as follows with min/max of (ζ,ζ0) : V(ζ,ξ) = Σnm Bnm [pnm(ζ<) – 2(-1)m g3(n,m) qnm(ζ<)] qnm(ζ>) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 V(ζ,ξ) = Σnm Bnm [ 2(-1)m g3(n,m) qnm(ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 (h) Discussion. I cannot think of any more boundary conditions. Everything seems right at the neck, at the bloid surface, at the Green's charge, and at infinity. Here is a comment from my attempt 3 after it failed: "Here is a very easy way to see the problem here: for the bloid, we know we are trying to come up with a solution which is different in the upper and lower regions. But our atoms p and q are the same in these two regions, and so can only generate symmetric solutions. The form is dead on arrival. " Why is this statement wrong? Here is our original Smythian form: Upper half bloid: Vo(ζ,ξ) = Σnm [Bnm pnm(ζ0) + Cnm qnm(ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) Vi(ζ,ξ) = Σnm [Bnm pnm(ζ) + Cnm qnm(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 Lower half bloid: Vo(ζ,ξ) = Σnm [B'nm pnm(ζ0) + C'nm qnm(ζ0)] qnm(ζ) Pnm(ξ) cos(mφ) Vi(ζ,ξ) = Σnm [B'nm pnm(ζ) + C'nm qnm(ζ) ] qnm(ζ0) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 It is true that both pnm(ζ) and qnm(ζ) are symmetric. It is true that the forms shown here are exactly the same for upper and lower, apart from primes on the coefficients. But it is NOT true that the coefficients are the same, so upper and lower are in fact not the same. Here is another statement: "So I need to go off and ponder this idea of my p and q functions supposedly being independent solutions of our ζ ODE, and yet both are even. But they are supposed to be a complete set like cos(mφ) and sin(mφ) to allow you to construct a general function f(φ), but in ζ I can only construct EVEN functions of ζ. " I guess this limitation of our real functions did not prevent us from solving the oblate bloid, as shown above. I am still not sure WHY these independent real solutions to the ζ flipped Legendre equation are symmetric, but I will ponder that. Feb 26. I have pondered it and I have all the answers ("Why are p and q symmetric.doc"). The two independent solutions of the flipped Legendre ζ equation can be and should be taken as the symmetric p and q functions. There exists another p' function, call it, which is odd, but of course it is a lincomb of p and q. Here is how you can make a non-even function out of p and q: f(ζ) = 3p(ζ) ± 2q(ζ) where the sign is sign(ζ), so that is not really the problem I thought it was. Even though p and q are even, f is not even. And of course g(ζ) = ± q(ζ) is an odd function. I am now ready to continue on taking limits of the above "Final Result". 4. The Cone Limit I took this limit in "Green's Function for an Oblate Bloid using cone method v3.doc" where I had my solution being unfortunately complex not real, but I think some of the data there will be useful here. Bateman's cone solution was this: r<a Vi = Σm=0∞ Σn Amn (r/a)n (-1)m Pnm(z) cos(mφ) r>a Ve = Σm=0∞ Σn Amn (r/a)-n-1 (-1)m Pnm(z) cos(mφ) ok a = his distance of the point charge from the cone origin q = 1 = his point charge value which we can write as a one-liner where r> and r< are max and min of r and a. Vcone = Σm=0∞ Σn Amn (r</a)n (r>/a)-n-1 (-1)m Pnm(z) cos(mφ) I have added (-1)m above because these are on the cut P functions and Smythe has that phase difference. The Amn are given by (Bateman P here as well) [ again, from above quoted doc] Amn = (2-δm,0) (2πεa)-1 (-1)m Pnm(ξ0) / [(2n+1)Kn(ξ1)] In the ξ coordinate, I am using entirely Bateman on-the-cut P functions. So much for the Smythe cone solution which I want to use as a check. My bloid result was this, for the distant region in the upper bloid V(ζ,ξ) = Σnm Bnm [pnm(ζ<) – 2(-1)m g3(n,m)qnm(ζ<)] qnm(ζ>) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 I am allowed now to think of both ζ< and ζ> as large, so will need large ζ limits of the p and q functions. And of course we have Bnm = (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) sec[(n+m)π/2] f(n,-m)/ Knm(ξ1) which needs no limits taken. Now I can see that for large ζ< , the p term will dominate the q term due to the usual large-z behavior rules, so I can replace my V above with this simpler one in the cone limit: V(ζ,ξ) = Σnm Bnm pnm(ζ<) qnm(ζ>) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 which looks promising. The limits I need are from P and Q doc, qnm(ζ) → [ Γ(1+n+m)/ Γ(n+3/2)] 2-n-1|ζ|-n-1 large |ζ| pnm(ζ) → cos[(n+m)π/2] (2n+1/) [Γ(n+1/2)/ Γ(1+n-m)] |ζ|n large |ζ| We also know that ζ = r/c1 and ζ0 = a/c1 => ζ> = r>/c1 and ζ< = r</c1 Then we have qnm(ζ>) → [ Γ(1+n+m)/ Γ(n+3/2)] 2-n-1(r>/c1)-n-1 pnm(ζ<) → cos[(n+m)π/2] (2n+1/) [Γ(n+1/2)/ Γ(1+n-m)] (r</c1)n so that pnm(ζ<) qnm(ζ>) = [ Γ(1+n+m)/ Γ(n+3/2)] 2-n-1(r>/c1)-n-1 * cos[(n+m)π/2] (2n+1/) [Γ(n+1/2)/ Γ(1+n-m)] (r</c1)n = [ Γ(1+n+m)/ Γ(n+3/2)] (r>/c1)-n-1 * cos[(n+m)π/2] [Γ(n+1/2)/ Γ(1+n-m)] (r</c1)n Now write (r>/c1)-n-1 (r</c1)n = (r>)-n-1 (r<)n c1 (r>/a)-n-1 (r</a)n = (r>)-n-1 (r<)n a => (r>/c1)-n-1 (r</c1)n = (r>/a)-n-1 (r</a)n * (c1/a) Then we have pnm(ζ<) qnm(ζ>) = [ f(n,m)/ Γ(n+3/2)] (r>/a)-n-1 * cos[(n+m)π/2] [Γ(n+1/2)] (r</a)n (c1/a) = (c1/a) f(n,m) cos[(n+m)π/2] [Γ(n+1/2)/ Γ(n+3/2)] (r</a)n (r>/a)-n-1 Now recall that Bnm = (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) sec[(n+m)π/2] f(n,-m)/ Knm(ξ1) so we can form this combination: Bnm pnm(ζ<) qnm(ζ>) = (c1/a) f(n,m) cos[(n+m)π/2] [Γ(n+1/2)/ Γ(n+3/2)] (r</a)n (r>/a)-n-1 * (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) sec[(n+m)π/2] f(n,-m)/ Knm(ξ1) = (c1/a) [Γ(n+1/2)/ Γ(n+3/2)] (r</a)n (r>/a)-n-1 * (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) / Knm(ξ1) But Γ(n+3/2) = (n+1/2) Γ(n+1/2) = Γ(n+1/2) (2n+1)/2 so we continue with the above = (c1/a) [2/(2n+1)] (r</a)n (r>/a)-n-1 * (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) / Knm(ξ1) = (2-δm,0) (q/[4πc1ε]) (c1/a) 2 (r</a)n (r>/a)-n-1 * Pnm(ξ0) / [(2n+1) Knm(ξ1) = (2-δm,0) (q/[2πaε]) (r</a)n (r>/a)-n-1 Pnm(ξ0) / [(2n+1) Knm(ξ1) = { q(-1)m } * { (2-δm,0) (1/[2πaε]) (r</a)n (r>/a)-n-1 (-1)m Pnm(ξ0) / [(2n+1) Knm(ξ1) = { q(-1)m } Amn (r</a)n (r>/a)-n-1 Therefore, my bloid limit is this: V(ζ,ξ) = Σnm Bnm pnm(ζ<) qnm(ζ>) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 = Σnm {q (-1)m } Amn (r</a)n (r>/a)-n-1 Pnm(ξ) cos(mφ) and we compare this to the Smythe cone result above which was (he has q = 1) Vcone = Σm=0∞ Σn Amn (r</a)n (r>/a)-n-1 (-1)m Pnm(z) cos(mφ) and we agree exactly! By the way, somehow this confirms the idea that I have only m ≥ 0 in my Σnm sum. If it were a full range sum, we would not match the cone limit. This has something to do with completeness of the m-spectrum, let's try to avoid worrying about that right now. I think m ≥ provides the complete spectrum and so that is the sum you need to use. Now what happens in the lower bloid? Recall our "final result" for the entire bloid: V(ζ,ξ) = Σnm Bnm [pnm(ζ<) – 2(-1)m g3(n,m)qnm(ζ<)] qnm(ζ>) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 V(ζ,ξ) = Σnm Bnm [ 2(-1)m g3(n,m) qnm(ζ<)] qnm(ζ>) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 In the upper region we threw out the q term relative to the p term for large arg. But in the lower bloid, all we have is the q(ζ) term. We in the lower bloid we want to evaluate this quantity: Bnm qnm(ζ<) qnm(ζ>) qnm(ζ>) → [ Γ(1+n+m)/ Γ(n+3/2)] 2-n-1(r>/c1)-n-1 qnm(ζ<) → [ Γ(1+n+m)/ Γ(n+3/2)] 2-n-1(r</c1)-n-1 so Vlower ~ Bnm qnm(ζ<) qnm(ζ>) ~ (r>/c1)-n-1 (r</c1)-n-1 which we can compare to what happens in the upper bloid Vupper ~ Bnm pnm(ζ<) qnm(ζ>) ~ (r>/c1)-n-1 (r</c1)+n So here is the ratio of potential strength within a partial wave n,m: Vlower/Vupper ~ (r</c1)-2n-1 So you get the idea that things are very weak in the lower bloid. As we truly go to the cone limit, the neck completely closes down, this means c1 → 0 and the above ratio → 0 for any n > -1/2 which range includes all n in our n spectrum. In this limit, the potential in the lower bloid is 0 everywhere. But in the upper bloid it is this, where Green's charge is at radial distance a from the cone apex at the origin: Vcone = Σm=0∞ Σn Amn (r</a)n (r>/a)-n-1 (-1)m Pnm(z) cos(mφ) Amn = (2-δm,0) (2πεa)-1 (-1)m Pnm(ξ0) / [(2n+1)Kn(ξ1)] Knm(ξ1) = - (1-ξ12) /(2n +1) * ∂ξPnm(ξ1) ∂nPnm(ξ1) 5. The On-Axis Limit Our starting position is this, V(ζ,ξ) = Σnm Bnm [pnm(ζ<) – 2(-1)m g3(n,m)qnm(ζ<)] qnm(ζ>) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 V(ζ,ξ) = Σnm Bnm [ 2(-1)m g3(n,m) qnm(ζ<)] qnm(ζ>) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 Bnm = (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) sec[(n+m)π/2] f(n,-m)/ Knm(ξ1) f(n,-m) = Γ(1+n-m)/ Γ(1+n+m) Knm(ξ1) = - (1-ξ12) /(2n +1) * ∂ξPnm(ξ1) ∂nPnm(ξ1) g3(n,m) = 1/ [ Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2) ] Putting the Green's charge on axis means ξ0 = 1. The only effect this has is on the Bnm . Consulting our leg prop doc we find that Pνμ(1) = δμ0 // μ = integers and general ν Therefore we have Bnm = (2-δm,0) (q/[4πc1ε]) Pnm(1) sec[(n+m)π/2] f(n,-m)/ Knm(ξ1) = (2-δm,0) (q/[4πc1ε]) δm0 sec[(n+m)π/2] f(n,-m)/ Knm(ξ1) = (1) (q/[4πc1ε]) δm0 sec[(n)π/2] 1/ Kn0(ξ1) = δm0 (q/[4πc1ε]) sec[nπ/2] / Kn0(ξ1) Bn0 = (q/[4πc1ε]) sec[nπ/2] / Kn0(ξ1) We have now the n-spectrum Pn(ξ1) = 0, and Kn0(ξ1) is still nothing simple I know of, so leave. Our solutions is then this for on-axis: V(ζ,ξ) = Σn Bn0 [pn0(ζ<) – 2 g3(n,0)qn0(ζ<)] qn0(ζ>) Pn(ξ) //upper ζ ≥ 0 V(ζ,ξ) = Σn Bn0 [ 2 g3(n,0)qn0(ζ<)] qn0(ζ>) Pn(ξ) //lower ζ ≤ 0 pn0(ζ<) = [Pn(iζ) + Pn(-iζ)] qn0(ζ) = (±i)n+1 Qn (iζ) Bn0 = (q/[4πc1ε]) sec[nπ/2] / Kn0(ξ1) Kn0(ξ1) = - (1-ξ12) /(2n +1) * ∂ξPn(ξ1) ∂nPn(ξ1) g3(n,0) = 1/ [ Γ(1/2 + n/2) Γ(1/2 - n/2 ) ] On scratch I show that Γ(1/2 + n/2) Γ(1/2 - n/2) = Γ(1+x)Γ(-x) x = n/2-1/2 = -π csc[ π(n/2-1/2)] = +π sec(nπ/2) so that Bn0 g3(n,0) = +[π sec(nπ/2)]-1 (q/[4πc1ε]) sec[nπ/2] / Kn0(ξ1) = (q/[4π2c1ε]) / Kn0(ξ1) So the on-axis result is pretty similar to the general result, except the m sum is gone and m = 0. 6. In-the-neck Limit Our starting position is this, V(ζ,ξ) = Σnm Bnm [pnm(ζ<) – 2(-1)m g3(n,m)qnm(ζ<)] qnm(ζ>) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 V(ζ,ξ) = Σnm Bnm [ 2(-1)m g3(n,m)qnm(ζ<)] qnm(ζ>) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 Bnm = (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) sec[(n+m)π/2] f(n,-m)/ Knm(ξ1) f(n,-m) = Γ(1+n-m)/ Γ(1+n+m) Knm(ξ1) = - (1-ξ12) /(2n +1) * ∂ξPnm(ξ1) ∂nPnm(ξ1) g3(n,m) = 1/ [ Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2) ] Putting the Green's charge in the neck means ζ0 = 0. Thus, ζ0 = 0 will always be ζ< and then ζ< will always be ζ, so we may write V(ζ,ξ) = Σnm Bnm [pnm(0) – 2(-1)m g3(n,m) qnm(0)] qnm(ζ) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 V(ζ,ξ) = Σnm Bnm [ 2(-1)m g3(n,m) qnm(0)] qnm(ζ) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 Our "reality of P and Q doc" tells us that pnm(0) = 2m+1 / [ Γ(1/2 - n/2 - m/2) Γ(1 + n/2 - m/2) ] qnm(0) = (-1)m 2m-1 [ Γ(1/2 + n/2+ m/2) / Γ(1 + n/2- m/2)] and now g3(n,m) qnm(0) = 1/ [ Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2) ] * (-1)m 2m-1 [ Γ(1/2 + n/2+ m/2) / Γ(1 + n/2- m/2)] = 1/ [Γ(1/2 - n/2 - m/2) ] * (-1)m 2m-1 [ 1 / Γ(1 + n/2- m/2)] = (-1)m 2m-1 1/ [Γ(1/2 - n/2 - m/2) Γ(1 + n/2- m/2)] And then [pnm(0) – 2(-1)m g3(n,m) qnm(0)] = (2m+1 – 2(-1)m (-1)m 2m-1) / [Γ(1/2 - n/2 - m/2) Γ(1 + n/2- m/2)] = (2m+1 – 2m) / [Γ(1/2 - n/2 - m/2) Γ(1 + n/2- m/2)] = 2m (2 – 1 ) / [Γ(1/2 - n/2 - m/2) Γ(1 + n/2- m/2)] = 2m / [Γ(1/2 - n/2 - m/2) Γ(1 + n/2- m/2)] Also, we can say 2(-1)m g3(n,m) qnm(0) = 2(-1)m(-1)m 2m-1 1/ [Γ(1/2 - n/2 - m/2) Γ(1 + n/2- m/2)] = 2m 1/ [Γ(1/2 - n/2 - m/2) Γ(1 + n/2- m/2)] If we now define h(n,m) = 1/[Γ(1/2 - n/2 - m/2) Γ(1 + n/2- m/2)] we have then showed that 2(-1)m g3(n,m) qnm(0) = 2m h(n,m) [pnm(0) – 2(-1)m g3(n,m) qnm(0)] = 2m h(n,m) so we can rewrite our result this way V(ζ,ξ) = Σnm Bnm [2m h(n,m)] qnm(ζ) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 V(ζ,ξ) = Σnm Bnm [2m h(n,m)] qnm(ζ) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 Bnm = (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) sec[(n+m)π/2] f(n,-m)/ Knm(ξ1) f(n,-m) = Γ(1+n-m)/ Γ(1+n+m) Knm(ξ1) = - (1-ξ12) /(2n +1) * ∂ξPnm(ξ1) ∂nPnm(ξ1) h(n,m) = 1/[Γ(1/2 - n/2 - m/2) Γ(1 + n/2- m/2)] qnm(ζ) ≡ (-i)2m (±i)n+1 Qnm(iζ) and we get the up/down symmetry that we expect in this situation !! So if Green's in the neck, you get the Q decay in both directions. There might be some further simplification of the constants but I don't think it adds much. I could now take the on-axis limit of the above neck result and I would then find that Bnm = δm0 (q/[4πc1ε]) sec[nπ/2] / Kn0(ξ1) and our result would then be V(ζ,ξ) = Σn Bn0 [h(n,0)] qn0(ζ) Pn(ξ) //upper and lower, in neck, on axis h(n,0) = 1/[Γ(1/2 - n/2) Γ(1 + n/2)] Bn0 = (q/[4πc1ε]) sec[nπ/2] / Kn0(ξ1) qn0(ζ) ≡ (±i)n+1 Qn(iζ) 7. The Iris Limit This is the limit I have long been seeking! Our starting position is this, V(ζ,ξ) = Σnm Bnm [pnm(ζ<) – 2(-1)m g3(n,m)qnm(ζ<)] qnm(ζ>) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 V(ζ,ξ) = Σnm Bnm [ 2(-1)m g3(n,m)qnm(ζ<)] qnm(ζ>) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 Bnm = (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) sec[(n+m)π/2] f(n,-m)/ Knm(ξ1) f(n,-m) = Γ(1+n-m)/ Γ(1+n+m) Knm(ξ1) = - (1-ξ12) /(2n +1) * ∂ξPnm(ξ1) ∂nPnm(ξ1) g3(n,m) = 1/ [ Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2) ] The Iris Limit means ζ1 = 0. I am not sure I have fully dealt with this limit properly in another doc, but I will look now. Found stuff in "Green's Function for an Oblate Bloid using cone method v3.doc". If we regard the n-spectrum as being complete for n values to the right of n = -1/2, we find this slightly strange n-spectrum, The non-redundant spectrum of n for a given m ≥ 0 value: m ≥0, even n = all positive odds + [evens in range (0,m-1)] n spectrum m= 0 n = all positive odds m > 0, odd n = all positive evens + [odds in range (0,m-1)] This spectrum comes from this fact: 0 = Pnm(0) = 2μ / [ Γ(1/2 - n/2 - m/2) Γ(1 + n/2 - m/2) ] = 2μh(n,m) Thus, for a given m, every n value in the n-spectrum causes h(n,m) = 0. I go on to show that 1/ Kn(0) = (2n+1) f(n,-m) So our solution now becomes V(ζ,ξ) = Σnm Bnm [pnm(ζ<) – 2(-1)m g3(n,m)qnm(ζ<)] qnm(ζ>) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 V(ζ,ξ) = Σnm Bnm [ 2(-1)m g3(n,m)qnm(ζ<)] qnm(ζ>) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 Bnm = (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) sec[(n+m)π/2] f(n,-m)/ Knm(0) f(n,-m) = Γ(1+n-m)/ Γ(1+n+m) 1/ Kn(0) = (2n+1) f(n,-m) g3(n,m) = 1/ [ Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2) ] So nothing really simplifies much except the n-spectrum and the K expression. 8. First put Green's charge in the neck, then go to the Iris Limit The in neck limit (ζ0 = 0) gives this from above V(ζ,ξ) = Σnm Bnm [2m h(n,m)] qnm(ζ) Pnm(ξ) cos(mφ) // upper and lower Bnm = (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) sec[(n+m)π/2] f(n,-m)/ Knm(ξ1) f(n,-m) = Γ(1+n-m)/ Γ(1+n+m) Knm(ξ1) = - (1-ξ12) /(2n +1) * ∂ξPnm(ξ1) ∂nPnm(ξ1) h(n,m) = 1/[Γ(1/2 - n/2 - m/2) Γ(1 + n/2- m/2)] qnm(ζ) ≡ (-i)2m (±i)n+1 Qnm(iζ) We then add our iris limit (ξ1= 0) to this and we get V(ζ,ξ) = Σnm Bnm [2m h(n,m)] qnm(ζ) Pnm(ξ) cos(mφ) // upper and lower Bnm = (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) sec[(n+m)π/2] f(n,-m)/ Knm(0) f(n,-m) = Γ(1+n-m)/ Γ(1+n+m) 1/ Kn(0) = (2n+1) f(n,-m) h(n,m) = 1/[Γ(1/2 - n/2 - m/2) Γ(1 + n/2- m/2)] qnm(ζ) ≡ (-i)2m (±i)n+1 Qnm(iζ) The spectrum of n was those n which made Pnm(x=0) = 2μ / [ Γ(1/2 - n/2 - m/2) Γ(1 + n/2 - m/2) ] = 2μ h(n,m) But this seems to say that at every point in the n spectrum, h(n,m) = 0, and therefore our entire potential collapses to 0. Again, this is for iris and Green's point in the neck somewhere. This time and last time I am just doing something wrong in my taking of this iris limit! Let's write out all the gammas and see if this is really true. First, copy down the following from above Γ(1/2 + n/2) Γ(1/2 - n/2) = Γ(1+x)Γ(-x) x = n/2-1/2 = -π csc[ π(n/2-1/2)] = +π sec(nπ/2) => π sec(nπ/2) = Γ(1/2 + n/2) Γ(1/2 - n/2) In this last equation, try replacing n → n+m : π sec((n+m)π/2) = Γ(1/2 + n/2+m/2) Γ(1/2 - n/2-m/2) Then consider sec[(n+m)π/2] h(n,m) = π-1 Γ(1/2 + n/2+m/2) Γ(1/2 - n/2-m/2) /[Γ(1/2 - n/2 - m/2) Γ(1 + n/2- m/2)] = π-1 Γ(1/2 + n/2+m/2) / Γ(1 + n/2- m/2) With this little preliminary, we now write: Bnm h(n,m) = (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) sec[(n+m)π/2] f(n,-m)/ Knm(0) * h(n,m) = (2-δm,0) (q/[4πc1ε]) (2n+1) Pnm(ξ0) { [f(n,-m)]2 sec[(n+m)π/2] h(n,m) } = (2-δm,0) (q/[4π2c1ε]) (2n+1) Pnm(ξ0) [f(n,-m)]2 Γ(1/2 + n/2+m/2) / Γ(1 + n/2- m/2) Now we have killed off one of the gamma in h(n,m). I know from the prev men doc that the denom gamma you see above kills off Bnm h(n,m) when n = m - even = 2,4,6.. (but not 0) so these terms will not survive in the n-sum. But the rest of the spectrum will. Those terms are these: n = -m + odd odd = 1,3,5... n + m = odd So here is our result written out: V(ζ,ξ) = Σnm Bnm [2m h(n,m)] qnm(ζ) Pnm(ξ) cos(mφ) = or V(ζ,ξ) = Σm=0∞ Σn=-m+odd 2m (2-δm,0) (q/[4π2c1ε]) [ Γ(1+n-m)/ Γ(1+n+m)]2 * (2n+1) [Γ(1/2+n/2+m/2) / Γ(1+n/2- m/2)] Pnm(ξ0) qnm(ζ) Pnm(ξ) cos(mφ) This is supposedly the potential everywhere for the bloid bent down to an iris, and the Green's charge in the hole! Green's is at (ξ0, ζ0 = 0, φo= 0), observation point is (ξ,ζ,φ). Let's try to simplify some of the gamma functions. Consider that 22x-1 Γ(x)Γ(x+1/2) = Γ(2x) => Γ(x)/ Γ(2x) = 21-2x / Γ(x+1/2) Try this with x = 1/2+n/2+m/2 so 2x = 1+n+m and x+1/2 = 1+n/2+m/2 and 1-2x = -n-m. Then we have Γ(1/2+n/2+m/2)/ Γ(1+n+m) = 2-n-m / Γ(1+n/2+m/2) and we negate m to get this further result: Γ(1/2+n/2-m/2)/ Γ(1+n-m) = 2-n+m / Γ(1+n/2-m/2) Therefore, f(n,-m) * [Γ(1/2+n/2+m/2) / Γ(1+n/2- m/2)] [ Γ(1+n-m)/ Γ(1+n+m)] * [Γ(1/2+n/2+m/2) / Γ(1+n/2- m/2)] = Γ(1/2+n/2+m/2) / Γ(1+n+m) * [Γ(1+n/2- m/2)/ Γ(1+n-m)]-1 = 2-n-m / Γ(1+n/2+m/2) * [2-n+m / Γ(1+n/2-m/2)]-1 = 2-2m / Γ(1+n/2+m/2) * [1 / Γ(1+n/2-m/2)]-1 = 2-2m Γ(1+n/2-m/2)/ Γ(1+n/2+m/2) = 2-2m f(n/2,-m/2) so this gives at least some reduction of our result above: V(ζ,ξ) = Σm=0∞ Σn=-m+odd 2m (2-δm,0) (q/[4π2c1ε]) [ Γ(1+n-m)/ Γ(1+n+m)]2 * (2n+1) [Γ(1/2+n/2+m/2) / Γ(1+n/2- m/2)] Pnm(ξ0) qnm(ζ) Pnm(ξ) cos(mφ) = Σm=0∞ Σn=-m+odd 2m (2-δm,0) (q/[4π2c1ε]) [ Γ(1+n-m)/ Γ(1+n+m)] 2-2m * (2n+1) [Γ(1+n/2-m/2)/ Γ(1+n/2+m/2)] Pnm(ξ0) qnm(ζ) Pnm(ξ) cos(mφ) or V(ζ,ξ) = Σm=0∞ Σn=-m+odd 2-m(2-δm,0) (q/[4π2c1ε]) f(n,-m) f(n/2,-m/2) (2n+1) * Pnm(ξ0) Pnm(ξ) qnm(ζ) cos(mφ) This result is then for Green's charge in the hole, and bloid folded down to an iris. (a) Potential on the iris and in the hole. Suppose now we try to set ξ = 0 so we are then right on the metal iris. Then factor Pnm(ξ) = 0 and we obtain the desired fact that V = 0 on the entire iris. Suppose we want to know the potential just in the hole itself. Then we set ζ = 0 and use qnm(0) = (-1)m 2m-1 [ Γ(1/2 + n/2+ m/2) / Γ(1 + n/2- m/2)] Using our same reduction above, we will get f(n,-m) qnm(0) = (-1)m 2m-1 Γ(1+n-m)/ Γ(1+n+m) * [ Γ(1/2 + n/2+ m/2) / Γ(1 + n/2- m/2)] = (-1)m 2m-1 * 2-2m Γ(1+n/2-m/2)/ Γ(1+n/2+m/2) = (-1)m 2-m-1 [Γ(1+n/2-m/2)/ Γ(1+n/2+m/2)] = (-1)m 2-m-1 f(n/2,-m/2) We then get V(ζ,ξ) = Σm=0∞ Σn=-m+odd 2-m(2-δm,0) (q/[4π2c1ε]) f(n,-m) f(n/2,-m/2) (2n+1) * Pnm(ξ0) Pnm(ξ) qnm(ζ) cos(mφ) = Σm=0∞ Σn=-m+odd 2-m(2-δm,0) (q/[4π2c1ε]) f(n,-m) [f(n/2,-m/2)]2 (2n+1) (-1)m 2-m-1 * Pnm(ξ0) Pnm(ξ) cos(mφ) = (q/[4πc1ε]) Σm=0∞ (-1)m 2-2m-1(2-δm,0) Σn=-m+odd (2n+1) f(n,-m) [f(n/2,-m/2)]2 * Pnm(ξ0) Pnm(ξ) cos(mφ) I suppose this could possibly be right. It has to blow up as ξ→ξ0 since then we are moving our observation point right to the Green's charge. Maybe this is reflected in that fact that, in this case, everything in the summand is positive (I think). (b) Charge density on the iris // small correction made 3.8.10 Now, suppose we are after the charge density on the iris. This will involve a ∂ξ derivative of the potential then evaluated at ξ = 0. So start with the general potential for the iris with charge in the hole at ξ0: V(ζ,ξ) = Σm=0∞ Σn=-m+odd 2-m(2-δm,0) (q/[4π2c1ε]) f(n,-m) f(n/2,-m/2) (2n+1) * Pnm(ξ0) Pnm(ξ) qnm(ζ) cos(mφ) ∂ξV(ζ,ξ) = Σm=0∞ Σn=-m+odd 2-m(2-δm,0) (q/[4π2c1ε]) f(n,-m) f(n/2,-m/2) (2n+1) * Pnm(ξ0) Pnm '(ξ) qnm(ζ) cos(mφ) and leg doc says [ recall that all ξ related P functions are on the cut functions ] Pνm ' (x=0) = – Pνm+1(x=0) m ≥ 0 and integer and we know that Pνμ(x=0) = 2μ / [ Γ(1/2 - ν/2 - μ/2) Γ(1 + ν/2 - μ/2) ] which we adjust to be Pνm ' (x=0) = – Pνm+1(x=0) m ≥ 0 and integer Pnm+1(x=0) = 2m+1 / [ Γ(1/2 - n/2 - (m+1)/2) Γ(1 + n/2 - (m+1)/2) ] = 2m+1 / [ Γ( - n/2 - m/2) Γ(1/2 + n/2 - m/2) ] => Pnm '(x=0) = - 2m+1 / [ Γ( - n/2 - m/2) Γ(1/2 + n/2 - m/2) ] so we then have ∂ξV(ζ,0) = Σm=0∞ Σn=-m+odd 2-m(2-δm,0) (q/[4π2c1ε]) f(n,-m) f(n/2,-m/2) (2n+1) * Pnm(ξ0) { -2m+1 / [ Γ( - n/2 - m/2) Γ(1/2 + n/2 - m/2) ]} qnm(ζ) cos(mφ) Now we have f(n/2,-m/2) / Γ(1/2 + n/2 - m/2) = [ Γ(1/2 + n/2 - m/2)/ Γ(1/2 + n/2 + m/2] / Γ(1/2 + n/2 - m/2) = 1/ Γ(1/2 + n/2 + m/2) so then ∂ξV(ζ,0) = Σm=0∞ Σn=-m+odd 2-m(2-δm,0) (q/[4π2c1ε]) f(n,-m) (2n+1) * Pnm(ξ0) { -2m+1 / [ Γ( - n/2 - m/2) Γ(1/2 + n/2 + m/2) ]} qnm(ζ) cos(mφ) = - 2 (q/[4πc1ε])Σm=0∞(2-δm,0) Σn=-m+odd f(n,-m) (2n+1) * Pnm(ξ0) { 1 / [ Γ(- n/2 - m/2) Γ(1/2 + n/2 + m/2) ]} qnm(ζ) cos(mφ) In this expression, we have: (ζ0, ξ0, φ0) = (0,ξ0,0) = location of the Green's point charge in the hole (ζ, ξ, φ) = (ζ, 0, φ) = location on the iris where we are looking at σ ξ0,ζ0=0,φ0=0 is the location of the charge in the hole, ζ labels the spheroid which marks your radial distance out on the iris from the hole center, and φ is your azimuth on the iris. The charge density on the iris will be this result combined with the hξ scale factor. With some work we could convert this thing to 2D polar coordinates on the iris. It is hard to imagine much simplification! At least the result is real! There is no obvious way to remove the P and q functions from this result! You could draw a nice iris picture and write the distance of the charge from the origin ρ0 and the distance of the observation point from the origin ρ (and angle φ), but those P and q functions will always be there. So we are far away from the Problem 38 result quoted below. The expansion above could in fact be correct, but it is not the form of the answer we want. (c) Comment: This brings a long voyage to an end. I got interested in the Green's for the bloid problem in the first place, because I somehow hoped that this final charge density formula would come out simple. But it is not simple, and it took about a month or more to get to this point! Let's look back now at a certain sequence of Smythe problems that got me motivated on this (Smythe page 203) So my path via the oblate bloid to get to this charge density gives the answer you see above, more or less, but obviously there is some much simpler method to do it. This problem in Smythe's problem list appears well before any problems which relate the oblate P and Q functions! Once you get the above solved, you can then do inversion and learn something about spherical bowls in the following problems. 9. A comparison between the real and complex bloid solutions. While I was writing this section just now, I realized why the Smythian form which led to the complex solution was invalid, resolving that mystery. I added a paragraph at the top of that complex-solution document explaining this detail. ( see "Green's Function for an Oblate Bloid using cone method v3.doc". ) 10. Restatement of the General Bloid Solution: V(ζ,ξ) = Σnm Bnm [pnm(ζ<) – 2(-1)m g3(n,m) qnm(ζ<)] qnm(ζ>) Pnm(ξ) cos(mφ) //upper ζ ≥ 0 V(ζ,ξ) = Σnm Bnm [ 2(-1)m g3(n,m) qnm(ζ<)] qnm(ζ>) Pnm(ξ) cos(mφ) //lower ζ ≤ 0 Bnm = (2-δm,0) (q/[4πc1ε]) Pnm(ξ0) sec[(n+m)π/2] f(n,-m)/ Knm(ξ1) f(n,-m) = Γ(1+n-m)/ Γ(1+n+m) Knm(ξ1) = - (1-ξ12) /(2n +1) * ∂ξPnm(ξ1) ∂nPnm(ξ1) g3(n,m) = 1/ [ Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2) ] π sec((n+m)π/2) = Γ(1/2 + n/2+m/2) Γ(1/2 - n/2-m/2) = [1/ g3(n,m)] Bnm = (2-δm,0) (q/[4π2c1ε]) Pnm(ξ0) [1/g3(n,m)] f(n,-m)/ Knm(ξ1) The summation is really this: Σnm = Σm=0∞ Σn=n(i,m) where n(i,m) is the spectrum of n with n > -1/2 Suppose we make this new definitions g4(n,m) = [ Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2) ] / 1/g3 Cnm = Bnm g3(n,m) = (2-δm,0) (q/[4π2c1ε]) Pnm(ξ0) f(n,-m)/ Knm(ξ1) Bnm = Cnm g4(n,m) Then our solution is this, where I now supply all supporting data: Final Solution: 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 Cnm = (2-δm,0) (q/[4π2c1ε]) Pnm(ξ0) f(n,-m)/ Knm(ξ1) g4(n,m) = Γ(1/2 + n/2+ m/2) Γ(1/2 - n/2 - m/2) f(n,-m) = Γ(1+n-m)/ Γ(1+n+m) Knm(ξ1) = - (1-ξ12) /(2n +1) * ∂ξPnm(ξ1) ∂nPnm(ξ1) Σnm = Σm=0∞ Σn=n(i,m) where Pn(i,m)(ξ1) = 0, i = 1,2,3... qnm(ζ) ≡ (-i)2m (±i)n+1 Qnm(iζ) Re(ζ) > 0 upper sign, Re(ζ) < 0 lower sign pnm(ζ) ≡ [ (∓i)-m Pnm(iζ) + (±i)-mPnm(-iζ)] Pnm(ξ) is Bateman on the cut, other P and Q are Bateman off the cut c1 = focal distance of the oblate hyperboloid potential of point charge is q/(4πε) location of Green's charge as shown below but with φ0 = 0; observation point (ζ,ξ,φ),