toroid charge density calculation REVIEWED

Section between ************ lines is installed, do not edit here! The stuff below the lower ****** involves integrating things over the toroid area. It also involves doing the two integrals which now appear in Appendix H.4. // 12.31.15 ************************************************************************* 10.5 Surface charge density on a toroid The toroid surface charge density may be obtained from the potential in this manner, σ = + (1/4π) (1/hξ) [∂ξV(ξ,u)]|ξ=ξ0 1/hξ = (chξ0 - cosu)/a . (10.5.1) An explanation is given below (4.1.2), and the sign here is + because ξ decreases moving outward from the toroid surface. The toroid potential was found in (10.1.11) to be V(ξ,u) = V0 Σn=0∞ εn Pn-1/2(chξ) cos(nu) . (10.1.11) (10.5.2) = V0 Σn=0∞ εn cos(nu) [ Pn-1/2(chξ) ] Therefore (10.5.10 reads, σ = + V Σn=0∞εn cos(nu) * ∂ξ [ Pn-1/2(chξ) ] |ξ=ξ0 (10.5.3) We need then to compute, ∂ξ [ Pn-1/2(chξ) ] = [ P'n-1/2(chξ) shξ ] + [(1/2) (1/) * shξ ] Pn-1/2(chξ) = shξ [ P'n-1/2(chξ) + (1/2) (1/) Pn-1/2(chξ) ] |ξ=ξ0 = shξ0 [ P'n-1/2(chξ0) + (1/2)(1/) Pn-1/2(chξ0) ] , (10.5.4) where P'ν(z) means ∂zPν(z). Then using 1/hξ = (chξ0 - cosu)/a , (1/hξ)∂ξ [ Pn-1/2(chξ) ] |ξ=ξ0 = (shξ0/a) (chξ0 - cosu) [ P'n-1/2(chξ0) + (1/2)(1/)Pn-1/2(chξ0) ] . = (shξ0/a) [ (chξ0 - cosu)3/2 P'n-1/2(chξ0) + (1/2)Pn-1/2(chξ0) ] . (10.5.5) Inserting this into (10.5.3) gives σ = + V0 Σn=0∞ εn cos(nu) * (shξ0/a) * [ (chξ0 - cosu)3/2 P'n-1/2(chξ0) + (1/2)Pn-1/2(chξ0) ] . (10.5.6) We now write this as the sum of the two terms σ = σ1 + σ2 where σ1 = + V0 Σn=0∞ εn cos(nu) * (shξ0/a) (chξ0 - cosu)3/2 P'n-1/2(chξ0) σ2 = V0 Σn=0∞ εn cos(nu) * (shξ0/a) * * Pn-1/2(chξ0) ] which we then reorganize to get σ1 = V0 (shξ0/a) (chξ0 - cosu)3/2 Σn=0∞ εn P'n-1/2(chξ0) cos(nu) σ2 = V0 (shξ0/a) Σn=0∞ εn Qn-1/2(chξ0) cos(nu) . (10.5.7) Recall (10.1.8a) with a = chξ0 and b = 1 and x = u, 1 = Σn=0∞ εn Qn-1/2( chξ0) cos(nu) . (10.5.8) This "sum rule" greatly simplifies σ2 so that now, σ2 = V0 (shξ0/a) = V0 (shξ0/a) { } . (10.5.9) Reconstruct the sum σ = σ2 + σ1 to get σ(u; ξ0) = V0 (shξ0/a) { + (chξ0 - cosu)3/2 Σn=0∞ εn P'n-1/2(chξ0) cos(nu) } . Next, extract a factor π/ from {...} and replace (shξ0/a) = 1/R from (1.3.7) to get, σ(u; ξ0) = [ + (chξ0 - cosu)3/2 Σn=0∞ εn P'n-1/2(chξ0) cos(nu) ] . (10.5.10) This is our final result for the surface charge density on a toroid having label ξ0 and tube radius R held at potential V0. We have scanned all our known sources, but cannot find verification of this result, so we shall be extra attentive to doing "checks". First Check: The thin-wire toroid limit In the thin wire limit the parameter ξ0 gets large as shown in Fig (1.2.3). The distance ρc to the tube center line approaches a, and the tube radius R approaches 0. Since chξ0 gets large, we invoke the large-x limits of the P and Q functions from Appendix H. Consider first the terms in the sum in (10.5.10) which have n > 0. With x = chξ0, one finds P'n-1/2(x) → xn-3/2 (H.6.3) → x-2n (H.6.7) (chξ0 - cosu)3/2 → x3/2 . Therefore, the nth sum term goes as x3/2 xn-3/2 x-2n = x-n . (10.5.11) For large x, these terms all decay away, relative to the constant term 1/2 appearing in (10.5.10). The n = 0 term has different behavior. Again from Appendix H, P'-1/2(x) → x-3/2 [ 1 - (1/2)ln(8x)] (H.6.16) → = (π2/2) 1/ln(8x) (H.6.15) (chξ0 - cosu)3/2 → x3/2 . For large x = chξ0 the n = 0 term in (10.5.1) is then * x3/2 * 1 * x-3/2 [ 1 - (1/2)ln(8x)] * (π2/2) 1/ln(8x) = [ 1 - (1/2)ln(8x)] (1/ln(8x)) = 1/ln(8x) - 1/2 (10.5.12) This -1/2 cancels the +1/2 appearing in (10.5.10) and we end up with σ(u; ξ0) ≈ . (10.5.13) We keep in mind that R → 0 in our limit, but we maintain R for a while longer. Meanwhile, x = chξ0 ≈ shξ0 = a/R // (1.3.7) so ln(8x) = ln(8a/R) = ln(a) + ln(8/R) ≈ ln(8/R) as R → 0 Then we find σ(u; ξ0) ≈ (10.5.14) This charge density is uniform in u, as one would expect since a piece of the thin ring thinks it is a piece of straight wire with uniform σ. The curvature radius ρc = a is huge compared to the wire radius R. Thus, to find the total charge on the toroid, we multiply σ by the area of a toroid, A= 4π2Rρc // toroid area to get Q = σA = 4π2Rρc = V0 πρc = CV0 Setting ρc = 1 as in Section 10.4 below, one gets C = (10.5.15) which agrees with (10.2.12) below as the capacitance of the toroid in the thin-wire limit. We therefore regard our σ result (10.5.10) as being correct in the thin-wire limit. Second Check: Integrating the surface charge This task is anything but simple and is carried out in Appendix L with support from other Appendices. The integration of course is performed directly in toroidal coordinates. We outline the main steps here. 1. Start with the charge density (10.5.10), σ(u; ξ0) = [ + (chξ0 - cosu)3/2 Σn=0∞ εn P'n-1/2(chξ0) cos(nu) ] . (L.2.1) 2. Integrate over the toroidal surface to get the total charge Q, Q = a2 shξ0 { π (chξ0)(1/sh3ξ0) + Σn=0∞ εn P'n-1/2(chξ0) } . (L.2.6) 3. Use the Wronskian 1/(1-z2) = Pν(z) Q'ν(z) - P'ν(z) Qν(z) to rewrite the above sum as Σn=0∞ εn P'n-1/2(chξ0) = (1/shξ0)2 Σn=0∞ εn (L.2.9) + Σn=0∞ εn Qn-1/2(chξ0) Q'n-1/2(chξ0) . (L.2.10) 4. Evaluate the second sum using these two facts, where the second is the derivative of the first, Σn=0∞ εn [Qn-1/2(z)]2 = (π2/2) (K.4.1) Σn=0∞ εn Qn-1/2(z) Q'n-1/2(z) = -(π2/4) z (z2-1)-3/2 . (K.4.2) The QQ' sum term in item 3 exactly cancels the first term in item 2 above, giving this result Q = V0 Σn=0∞ εn (10.5.16) which implies that the toroid has capacitance C = Σn=0∞ εn . (10.5.17) This agrees with the result (10.2.3) which was found completely independently by taking the far-away limit of the potential (10.1.11). ************************************************************************* Question: Suppose I try the degenerate toroid limit here. What will happen? I have chξ0 = ρc/R and this does approach 1. We have above, σ(u; ξ0) = [ + (chξ0 - cosu)3/2 Σn=0∞ εn P'n-1/2(chξ0) cos(nu) ] . I need to know about P' and so I need to know about P near 1. I recently did this somewhere. Pn-1/2(z) ≈ 1 + (1/2) [n2 - 1/4](z-1) P'n-1/2(z) = (1/2) [n2 - 1/4] valid for all n Let's try to build the limit! σ(u; ξ0) = [ + (1 - cosu)3/2 Σn=0∞ εn P'n-1/2(chξ0) cos(nu) ] . ≈ - ln(x-1) valid for all n so we seem to get σ(u; ξ0) = - [ + (1 - cosu)3/2 Σn=0∞ εn (1/2) [n2 - 1/4] ln(x-1) cos(nu) ] . = - [ + (1 - cosu)3/2(1/2) ln(x-1) Σn=0∞ εn [n2 - 1/4] cos(nu) ] Does that n sum do anything interesting? Ask Maple. The sum approaches 8.28 and seems to converge. It does not go to 0. So how do I explain a divergent σ in the horn toroid limit? Well at least when u = 0 we get (1 - cosu) = 0 and perhaps we can argue this kills the divergence. But for any other u, the series is finite and ln(x-1) blows up! Well, here is my defense. The limit is not uniformly convergent z → 1 here as well as in the T(z) case, so you cannot take the z→1 limit through the infinite sum, and that action is what creates the ln(x-1) problem which does not really exist. It is wrong just as saying T(1) = 0 is wrong. Probably numerically you would find that there is some finite limit of σ at the horn toroid limit. It would be very good to plot σ for something near this case to see what happens. I have done really no plots and this needs doing tomorrow! Not sure where to put the check that integral is right. ********************************** The good news: this exactly agrees with what I published in the original bowl doc! As a check on this result, we integrate it over the toroid using this differential area, dA = (hudu)(hφdφ) = a/(chξ0 - cosu) * a shξ0/(chξ0 - cosu) * dudφ = a2 shξ0 (chξ0 - cosu)–2dudφ Insert test: Test this to see if you can generate the toroid area!! ∫∫dA = 2π a2 shξ0 !Syntax Error, Idu (chξ0 - cosu)–2 = 2πa2 shξ0 2π = 4π2a2 / chξ0 The result is supposed to be 4π2Rρc and I know that ρc = a/thξ and Rt = a/shξ so 4π2Rρc = 4π2 a/shξ0 * a/thξ0 = 4π2a2/ STOP, something is wrong. Then Q = ∫∫ σ dA = a2 shξ0 [ !Syntax Error, Idφ ] [ !Syntax Error, Idu (chξ0 - cosu)-2 σ(u; ξ0) ] = 2πa2 shξ0!Syntax Error, Idu (chξ0 - cosu)-2 σ(u; ξ0) We are then faced with these two integrals !Syntax Error, Idu cos(nu)(chξ0 - cosu)-2 !Syntax Error, Idu cos(nu) (chξ0 - cosu)-1/2 = 2 Qn-1/2(chξ0) // from (K.3.8) On scratch I show that !Syntax Error, Idu cos(nu)(a - cosu)-2 This does not look good, but at least I have a tentative result for this integral. Another integral !Syntax Error, Idu (z - cosu)–2 = 2π // wrong Out of steam at 8:30 PM. Next day 12.19.15. Let's try a simpler case. Compute the circumference of the toroid. ds = (hudu) = a/(chξ0 - cosu) * du circ = a!Syntax Error, Idu (α - cosu)–1 = a !Syntax Error, Idu α = chξ0 b = α-1 = ab !Syntax Error, Idu Let's go with this integral from GR7 Use a = -b, n = 0 and result is !Syntax Error, Idu = so then circ = = = = 2πa/shξ0 But R = a/shξ0 so we get 2πR which is the correct answer. Idea for area integral. Start with this integral Check this with Maple numeric: a = 0.7 n = 0 ok a = 0.7 n = 1 ok a = - 0.7 n = 1 ok and replace a with a = 1/b everywhere , where ab = 1. We have three pieces to compute: J = !Syntax Error, Idx cos(nx) = !Syntax Error, Idx cos(nx) = b !Syntax Error, Idx cos(nx) = b = ( - b) . Then we know that J = b!Syntax Error, Idx cos(nx) = b ( - b)n or I(b) ≡ !Syntax Error, Idx cos(nx) = ( - b)n b>1 b = |b| I think that if b < 1, you have to replace → - if I draw a phasor diagram. Then I(b) ≡ !Syntax Error, Idx cos(nx) = - (- - b)n b<-1 = - (-1)n (+ b)n = (-1)n+1 (+ b)n b <-1 b = -|b| I can combine these to get I(b) = (- |b| )n b > 1 I(b) =(-1)n+1 (- |b| )n b < -1 Combine again to get I(b) = (sign b)n+1 (- |b| )n This now checks out in Maple: b = 1.7 n = 2 b = -1.7 n = 2 b = 1.7 n = 3 b = -1.7 n = 3 all four are now OK Now compute ∂bI(b) = - !Syntax Error, Idx cos(nx)(b+cosx)-2 = ∂b [( - b)n ] and this at least gives a path for computing our integral of interest. Now for n = 0 we get I(b) ≡ !Syntax Error, Idx = ∂bI(b) = - !Syntax Error, Idx (b+cosx)-2 = ∂b [] = π ∂b(b2-1)-1/2 = π (-1/2) (b2-1)-3/2 2b = -πb (b2-1)-3/2 and therefore we have a new integral to add to the list !Syntax Error, Idx (b+cosx)-2 = π|b| (b2-1)-3/2 . |b| > 1 Notice that both sides are positive, and that there is no pole inside the implied contour. Application: compute the toroid area: dA = (hudu)(hφdφ) = a/(chξ0 - cosu) * a shξ0/(chξ0 - cosu) * dudφ = a2 shξ0 (chξ0 - cosu)–2dudφ = a2 shξ0 (-chξ0 + cosx)–2dxdφ u = x The dφ integral is 2π so we end up with A = 2π a2 shξ0 2 !Syntax Error, Idx (b+cosx)-2 where b = -chξ0 |b| > 1 check = 2π a2 shξ0 2 π|b| (b2-1)-3/2 = 4π2a2shξ0 b (b2-1)-3/2 = 4π2a2shξ0 [chξ0] (sh2ξ0)-3/2 = 4π2a2shξ0 [chξ0] (shξ0)-3 = 4π2a2 chξ0/sh2ξ0 = 4π2 (a/shξ0)(a/thξ0) = 4π2Rρc which is the correct result for toroid area. Maybe put these integrals into H.4 ! What other integrals might I need to check the total charge? σ(u; ξ0) = [ + (chξ0 - cosu)3/2 Σn=0∞ εn P'n-1/2(chξ0) cos(nu) ] I will need two integrals. Recall from above that, after the dφ integration dA = 2π a2 shξ0 (chξ0 - cosu)–2du So I will need these two integrals: J1 = !Syntax Error, Idu cos(nu)(chξ0 - cosu)–2 J2 = !Syntax Error, Idu cos(nu)(chξ0 - cosu)–2(chξ0 - cosu)3/2 (1/2)!Syntax Error, Idu cos(nu) (chξ0 - cosu)-1/2 = Qn-1/2(chξ0) // from (K.3.8) So only the first integral remains. But I have a path to do that integral from above, where recall. ∂bI(b) = - !Syntax Error, Idx cos(nx)(b+cosx)-2 = ∂b [( - b)n ] where b = - chξ0 That is to say, J1 = !Syntax Error, Idu cos(nu)(chξ0 - cosu)–2 = !Syntax Error, Idx cos(nx)(-chξ0 + cosx)–2 = - { ∂b [( - b)n ] } |b = -chξ0 Now I need to preserve analyticity with signs. So here is what Maple says But now I have to consider the two cases b > 1 and b < -1. The result is correct as is for b > 1. I think this is the result for b < - 1 f = (b - n)(--b)n π / [ - 3 ] = (-1)n+1 (b - n)(+b)n π (b2-1)-3/2 b < - 1 f = (b + n)(-b)n π(b2-1)-3/2 b > 1 So here is side by side: f = π (b + n)(-b)n (b2-1)-3/2 b > 1 |b| = b f = (-1)n+1π (b - n)(+b)n (b2-1)-3/2 b < - 1 Now rewrite the second as f = π (b + n)(-b)n (b2-1)-3/2 b > 1 |b| = b f = (-1)n π (-b + n)(+b)n (b2-1)-3/2 b < - 1 |b| = -b I can combine these to get f = (sign(b))n π (|b| + n)(-|b|)n (b2-1)-3/2 So here is my claim now: !Syntax Error, Idx cos(nx)(b+cosx)-2 = (sign(b))n π (|b| + n)(-|b|)n (b2-1)-3/2 Take it out for a Maple spin please: b = 1.7 n = 2 ok b = -1.7 n = 2 ok b = 1.7 n = 3 ok b = -1.7 n = 3 ok NOW lets take b = -chξ, so sign(b) = -1. The result is then !Syntax Error, Idx cos(nx)(-chξ+cosx)-2 = (-1)n π (chξ + nshξ)(shξ - chξ)n (1/sh3ξ) ξ > 0 Now shξ - chξ = -e-ξ so write as = (-1)n π (chξ + nshξ)(-e-ξ)n (1/sh3ξ) ξ > 0 = π (chξ + nshξ)e-nξ (1/sh3ξ) ξ > 0 Now do a Maple test on this result ξ = 1.7 n = 2 ok ξ = 1.7 n = 3 ok So here is my conclusion: !Syntax Error, Idx cos(nx)(-chξ+cosx)-2 = π (chξ + nshξ)e-nξ (1/sh3ξ) ξ > 0 !Syntax Error, Idu cos(nu)(chξ-cosu)-2 = π (chξ + nshξ)e-nξ (1/sh3ξ) ξ > 0 Now go back to the "two integrals needed" point: σ(u; ξ0) = [ + (chξ0 - cosu)3/2 Σn=0∞ εn P'n-1/2(chξ0) cos(nu) ] So I will need these two integrals: J1 = !Syntax Error, Idu cos(nu)(chξ0 - cosu)–2 = π (chξ0 + nshξ0)e-nξ0 (1/sh3ξ0) J2 = !Syntax Error, Idu cos(nu)(chξ0 - cosu)–2(chξ0 - cosu)3/2 = Qn-1/2(chξ0) // from (K.3.8) So now I have both the required integrals, so maybe now I can compute the total charge on the toroid. dA = 2π a2 shξ0 (chξ0 - cosu)–2du after dφ integration Then ∫σ(u; ξ0) dA = 2π a2 shξ0 ∫σ(u; ξ0) (chξ0 - cosu)–2du = 2π a2 shξ0 { !Syntax Error, Idu (chξ0 - cosu)–2 + Σn=0∞ εn P'n-1/2(chξ0) !Syntax Error, Idu cos(nu) (chξ0 - cosu)–1/2 } Now recall that !Syntax Error, Idu (chξ0 - cosu)–2 = 2π (chξ)(1/sh3ξ) since n = 0 !Syntax Error, Idu cos(nu) (chξ0 - cosu)–1/2 = 2 Qn-1/2(chξ0) Here then is my claimed answer (which looks very wrong) Q = 2π a2 shξ0 { 2π (chξ)(1/sh3ξ) + Σn=0∞ εn P'n-1/2(chξ0) 2 Qn-1/2(chξ0) } or Q = a2 shξ0 { 2π (chξ0)(1/sh3ξ0) + Σn=0∞ εn P'n-1/2(chξ0) } But this looks nothing like the Snow capacitance formula I expect from Q = CV0. I so know this however, 1/(1-z2) = Pν(z) Qν'(z) - Pν'(z) Qν(z) That could get rid of P', but then you are stuck with a Q'. Problem: The σ is going to contain P' no matter what! So there must be some way to get rid of it! Look at the properties. Here are some derivative formulas I might use to get rid of P' These say (1-x2)P'ν(x) = (-ν-1)Pν+1(x) + (ν+1)xPν(x) (1-x2)P'ν(x) = ν Pν-1(x) - νxPν(x) I write again with ν = n-1/2 (1-x2)P'n-1/2(x) = - (n+1/2)Pn+1/2(x) + (n+1/2)xPn-1/2(x) (1-x2)P'n-1/2(x) = (n-1/2) Pn-3/2(x) - (n-1/2) xPn-1/2(x) and says Bateman (1-x2)Q'n-1/2(x) = - (n+1/2)Qn+1/2(x) + (n+1/2)xQn-1/2(x) (1-x2)Q'n-1/2(x) = (n-1/2) Qn-3/2(x) - (n-1/2) xQn-1/2(x) Now my charge formula seems to have this action, Σn=0∞ εn P'n-1/2(chξ0) whereas the capacitance has this action Σn=0∞ εn Just for fun, use the wronskian in the first sum 1/(1-z2) = Pν(z) Qν'(z) - Pν'(z) Qν(z) Σn=0∞ εn P'n-1/2(chξ0) = Σn=0∞ εn P'n-1/2(chξ0)Qn-1/2(chξ0) = Σn=0∞ εn [ Pn-1/2(chξ0)Q'n-1/2(chξ0) - 1/(1-z2) ] This just doesn't do anything for me. So something is wrong somewhere!! Suppose for some reason this is true: Σn=0∞ εn [ Pn-1/2(chξ0)Q'n-1/2(chξ0) ] = f(ξ0) = a simple function Then I would have Σn=0∞ εn P'n-1/2(chξ0) = f(ξ0) - 1/(1-z2) * Σn=0∞ εn = f(ξ0) - (1/shξ0)2 Σn=0∞ εn . Then I would have Q = a2 shξ0 { 2π (chξ0)(1/sh3ξ0) + Σn=0∞ εn P'n-1/2(chξ0) } = a2 shξ0 { 2π (chξ0)(1/sh3ξ0) + [ f(ξ0) + (1/shξ0)2 Σn=0∞ εn ] } Now suppose f(ξ0) cancels the first term. Then we get Q = a2 shξ0 { + [ + (1/shξ0)2 Σn=0∞ εn ] } = a2(1/ shξ0) Σn=0∞ εn And then I would get C = a2(1/ shξ0) Σn=0∞ εn Now R = a/shξ0 so we then get C = a Σn=0∞ εn = Σn=0∞ εn My earlier calculation was C = (2a/π ) Σn=0∞ εn and then this makes things work! So I have at least a theory. I have then to show that Σn=0∞ εn [ Pn-1/2(chξ0)Q'n-1/2(chξ0) ] = f(ξ0) 2π (chξ0)(1/sh3ξ0) + [ f(ξ0) ] = 0 π (chξ0)(1/sh3ξ0) + f(ξ0) = 0 - (π2/4) (chξ0)(1/sh3ξ0) = f(ξ0) So I would have to show that Σn=0∞ εn Qn-1/2(chξ0)Q'n-1/2(chξ0) = - (π2/4) (chξ0)(1/sh3ξ0) Good luck!!! Note that f(x) f'(x) = (1/2) ∂x (f(x))2 Then the above would read Σn=0∞ εn Qn-1/2(z)Q'n-1/2(z) = - (π2/4) z (z2-1)-3/2 (1/2) Σn=0∞ εn ∂z Q2n-1/2(z) = - (π2/4) z (z2-1)-3/2 (1/2)∂z [ Σn=0∞ εn [Qn-1/2(z)]2 ] = - (π2/4) z (z2-1)-3/2 ∂z [ Σn=0∞ εn [Qn-1/2(z)]2 ] = - (π2/2) z (z2-1)-3/2 // hope for result At least this looks like a simpler series to add up. So what is g(z) = Σn=0∞ εn [Qn-1/2(z)]2 ? The only tool I really have is this: Qn-1/2(z) = (1/) !Syntax Error, Idt cos(nt)/ Then I could say Σn=0∞ εn [Qn-1/2(z)]2 = (1/2) Σn=0∞ εn !Syntax Error, Idt cos(nt)/!Syntax Error, Idt' cos(nt')/ = (1/2)!Syntax Error, Idt !Syntax Error, Idt' Σn=0∞ εn cos(nt)cos(nt') But recall from Appendix K that Σn=0∞ (εn/π) cos(nθ)cos(nθ) = δ(θ-θ') Σn=0∞ (εn/π) cos(nt)cos(nt') = δ(t-t') Then we get Σn=0∞ εn [Qn-1/2(z)]2 = (1/2)!Syntax Error, Idt !Syntax Error, Idt' π δ(t-t') = π/2 !Syntax Error, Idt But this is an integral I think I know from above. Above I showed that !Syntax Error, Idx cos(nx) = (sign b)n+1 (- |b| )n and for n = 0 this says !Syntax Error, Idx = (sign b) Now write !Syntax Error, Idt = - !Syntax Error, Idt = - [(-1) ] = = π/shξ Then I have shown that Σn=0∞ εn [Qn-1/2(z)]2 = (π/2) !Syntax Error, Idt = (π/2) = (π2/2) = (π2/2)(z2- 1)-1/2 which would be astonishing to say the least. Maple: A plot confirms that Σn=0∞ εn [Qn-1/2(z)]2 = (π2/2)(z2- 1)-1/2 !!!!! I was hoping to get ∂z [ Σn=0∞ εn [Qn-1/2(z)]2 ] = - (π2/2) z (z2-1)-3/2 Is this what I get? Well here is what I got ∂z [ Σn=0∞ εn [Qn-1/2(z)]2 ] = ∂z (π2/2)(z2- 1)-1/2 = (π2/2) ∂z (z2- 1)-1/2 = -(π2/2)(1/2)(z2- 1)-3/2 2z = - (π2/2) z(z2- 1)-3/2 Done!!!! I have confirmed that the σ integrates to Q !! Let's try a Maple check on our Q sum! This Q series is so simple, it must be in the books. Here is one AS series that is nice No, not in AS or GR7 or Bate. Why can't I do the capacitance sum that way? It's the denominator! What do you do with 1/P ? I don't know any integral reps for that. Now that I know how this all works, maybe there is a better way to write σ. ************************************************************ Charge density in terms of θ instead of u. I know from (L.1.4) that dA = a2 shξ0 (chξ0 - cosu)–2dudφ The charge in this area is dQ = σdA = σ [ a2 shξ0 (chξ0 - cosu)–2] du dφ = σ [ R2 sh3ξ0 (chξ0 - cosu)–2] du dφ Suppose I want to change from toroidal parameter u to the "circle angle θ" shown Bipolar (7.1). Then dQ = σ [ a2 shξ0 ] [ ] dθ dφ Now I can use certain Bipolar facts, = = = Then dQ = σ [ a2 shξ0 ] [ ] dθ dφ = σ [ a2 shξ0 ()2] [ ] dθ dφ = σ a2 (chξ+cosθ) dθ dφ = σ a2 (chξ+cosθ) dθ dφ = σ R2 (chξ+cosθ) dθ dφ and then the thing you want to plot is dQ/dφ = σ a2 (chξ0+cosθ) dθ = σ R2 (chξ0+cosθ) dθ and then this is a task for Maple. What happens in the thin wire limit ? chξ0 is large so we get dQ/dφ = σ a2 chξ0/sh2ξ0 dθ ≈ σ/shξ0 But we know from (10.5.14) that σ(u; ξ0) ≈ = a constant for fixed small R Then dQ/dφ ≈ a2 (1/shξ0)dθ As expected, there is no θ dependence going around the circle. Thus Q per dφ = a2(1/shξ0) 2π The total charge on the right is then Q = a2(1/shξ0) 2π * 2π which says the capacitance is C = a2(1/shξ0) 2π * 2π = a2(π/R)(1/shξ0) ≈ ≈ aπ a = ρc = 1 C(R) ≈ π / ln(8/R) ≡ C0(R) . OK, so here are the plotting instructions: dQ/dφ = σ a2 (chξ0+cosθ) dθ = σ R2 (chξ0+cosθ) dθ But now we have to convert σ: σ(u; ξ0) = [ + (chξ0 - cosu)3/2 Σn=0∞ εn P'n-1/2(chξ0) cos(nu) ] . Look up in Bipolar, = Now what about cos(nu) ? That could be a problem. sinu = cosu = First time I realized this difficulty. Let's try this idea cosu + i sinu = eiu = [chξcosθ+1 + i |shξ| sinθ]/[chξ+cosθ} Plan B: Let's just do it all numerically somehow. New Maple mws file please. OK, I am entering this now. What do I do about P' ? Pν'(z) = ∂zPν(z) = dPν(z)/dz = dPν(chξ)/d(chξ) = dPν(chξ)/[shξ dξ] = (1/shξ) ∂ξ Pν(chξ) ≡ dPν(chξ) So then I define dP(ν,ξ) = (1/shξ) ∂ξ P(ν,ξ) What will Maple do when I try this? Well, luckily it knows how to do this, so I just enter it as shown. dP := (nu,xi) -> (1/shξ) diff(P(nu,xi),xi):