large r behavior of the bowl potential v2 REVIEWED

Finding the large r behavior of the bowl potential v2 PhL 12.5.15 This is incorporated into Section 2.5 item 3 of bowl doc, all is well. The potential is this V(ξ,u) = (V0/π) { π - cot-1[] + cot-1[] } (1) In the large r region we know from (1.3.16) that ≈ a/r ξ << 1 u = 2π+ε with |ε| << 1 For our purposes, we can approximate many expressions in the above potential cos(u/2) ≈ cos(2π/2) = cos(π) = -1 cos(2u0-u) ≈ cos(2u0-2π) = cos(2u0) chξ - cos(2u0-u) ≈ 1 - cos(2u0) = 2sin2u0 = |sinu0| cos(u0-u/2) ≈ cos(u0-π) = -cosu0 Then the ratios appearing in *** may be approximated as u0 = β π-u0 ≈ = - r/a ≈ = R/r R = a |sinu0| ≈ = - σ cotu0 σ ≡ sign(sinu0) The resulting potential is then V(ξ,u) = (V0/π) { π - cot-1[- r/a] + (R/r) cot-1[ - σ cotu0] } The first two terms are π - cot-1[- r/a] = cot-1[r/a] = tan-1[a/r] ≈ a/r = R |sinu0| (1/r) The last term depends on σ. σ = -1: cot-1[cotu0] = u0 σ = +1: cot-1[-cotu0] = π -cot-1[cotu0] = π - u0 Taking the case σ = +1 which means 0 ≤ u0≤ π we find V(ξ,u) = (V0/π) { R |sinu0| (1/r) + (R/r)(π-u0)] } = (V0R/π) { sinu0 + (π-u0) } (1/r) As shown in the upper part of Fig (1.3.18), in this case one has bowl angle β = π-u0 and so V(ξ,u) = (V0R/π) { sinβ + β } (1/r) For σ = -1 the potential is instead V(ξ,u) = (V0/π) { R |sinu0| (1/r) + (R/r) u0 } = (V0R/π) { |sinu0| + u0 } As shown in the lower part of Fig (1.3.18), in this case one has bowl angle β = u0 and so again, V(ξ,u) = (V0R/π) { sinβ + β } (1/r) Writing V = Q/r = CV0/r one reads off the bowl capacitance, C = (R/π) { sinβ + β } 1. Find region where r is large Consider r = a / (chξ - cosu) (2) To get large r, we must have small ξ and cosu close to 1. This happens in range near u = 2π. Let u = 2π+ε so cosu = cosε ≈ 1-ε2/2 . Also sinu = sinε ≈ ε numer = a= a denom = chξ - cosu = 1+ξ2/2 - [ 1-ε2/2] = (ξ2 + ε2)/2 Thus r = a / (chξ - cosu) = a / [(ξ2 + ε2)/2] = 2a / Now note that z/ρ = sinu/shξ = ε/ξ ≡ 1/f could be any size!!! ξ = fε Then = = ε so that r = 2a / = [2a/] (1/ε) (3) 1/r = ε / 2a (4) Thus, r is large where ε is small. 2. Evaluate the potential Now go through things as in the previous paradox document: V(ξ,u) = (V0/π) { π - cot-1[] + cot-1[] } (1) R1 R2 R3 cos(u/2) = cos(π+ε/2) = cos(ε/2+π) = -cos(ε/2) ≈ - [ 1 - (ε/2)2/2] = ε2/8 - 1 cos(u) = cos(2π+ε) = cosε ≈ 1 - ε2/2 chξ - cosu = (1 + ξ2/2 - 1 + ε2/2) = (ξ2+ε2)/2 = ( f2ε2+ε2)/2 = (1/2)(f2+1)ε2 ≈ (1/) ε 1/ = 1/[ (1/) ε ] = /( ε ) cos(2u0-u) = cos(2u0- [2π+ε]) = cos(2u0- ε - 2π) = cos(2u0 - ε) ≈ cos(2u0) - ε [ -sin(2u0) ] = cos(2u0) + ε sin(2u0) chξ - cos(2u0-u) = 1+ξ2/2 - cos(2u0 - ε) = ξ2/2 + [1-cos(2u0 - ε)] = ξ2/2 + 2sin2(u0 - ε/2) but sin(u0- ε/2) ≈ sin(u0) - (ε/2) cos(u0) sin2(u0- ε/2) ≈ sin2(u0) - ε sinu0cosu0 Therefore chξ - cos(2u0-u) ≈ ξ2/2 + 2sin2(u0 - ε/2) = (f2/2)ε2 + 2 [ sin2(u0) - ε sinu0cosu0] = 2sin2u0 - 2ε sinu0cosu0 + order(ε2) which I will ignore ≈ 2sin2u0 - 2ε sinu0cosu0 = 2sinu0 ( sinu0 - 2εcosu0) = 2sin2u0 - 4ε sinu0cosu0 = 2sin2u0[ 1 - 2εcotu0] [chξ - cos(2u0-u)]-1 = [ 2sin2u0[ 1 - 2εcotu0] ]-1 = (1 + 2εcotu0) /(2sin2u0) 1/ = (1 + εcotu0)/(|sinu0|) = |sinu0| = |sinu0| (1 - εcotu0) the above agrees with my old v3 doc 1/ = 1/ [|sinu0| (1 - εcotu0)] = (1+εcotu0)/(|sinu0|) cos(u0- u/2) = cos(u0 - π - ε/2) = - cos(u0-ε/2) ≈ - [ cos(u0) + (ε/2) sinu0] So here is a summary of the above results cos(u/2) = - 1 + ε2/8 ok cos(u) = 1 - ε2/2 ok chξ - cosu = (1/2)(f2+1)ε2 ok = (1/) ε ok cos(2u0-u) = cos(2u0) + ε sin(2u0) ok 1/ = (1+εcotu0)/(|sinu0|) ok cos(u0- u/2) = - [ cos(u0) + (ε/2) sinu0] . ok in v3 doc I get |cos(u0- u/2)| = | | = | cos(u0) | so this does not confirm my sign here. Now write again V(ξ,u) = (V0/π) { π - cot-1[] + cot-1[] } (1) R1 R2 R3 R1 = = = = - 2/[ε] + (1/4) ε ok R2 = = (1/) ε * (1+εcotu0)/(|sinu0|) ≈ (1/) ε / (|sinu0|) + order(ε2) ok R3 = = - [ cos(u0) + (ε/2) sinu0] * (1+εcotu0)/(|sinu0|) = - [ cos(u0) + (ε/2) sinu0] [(1+εcotu0]/|sinu0| = - sign(sinu0) [ cot(u0) + (ε/2)] [1+εcotu0] = - sign(sinu0) { cotu0 + ε [ 1/2 + cot2u0]} + order(ε2) ok To summarize: R1 = - 2/[ε] + (1/4) ε ok R2 = (1/) ε / (|sinu0|) ok R3 = - sign(sinu0) { cotu0 + ε [ 1/2 + cot2u0]} ok The potential is then V(ξ,u) = (V0/π) { π - cot-1[R1] + R2 cot-1[R3] } So let's now look at the individual terms. First π - cot-1[R1] = cot-1[-R1] = cot-1 [ 2/[ε] - (1/4) ε] As ε→0, the first term becomes +∞ and the second term does not matter at all, So π - cot-1[R1] = cot-1 (+∞) = 0 that term is completely gone!!! (wrong!!!!) STOP!!! Consider this alternative evaluation of the first term! π - cot-1[R1] = cot-1[-R1] = cot-1 [ 2/[ε] - (1/4) ε] ≈ cot-12/[ε] ≈ tan-1 ( ε /2) ≈ ε /2 and there suddenly we have a new linear ε term that was not there before! So I will add this "new term" in red below. Next, consider cot-1[R3] = cot-1[- sign(sinu0) { cotu0 + ε [ 1/2 + cot2u0]} ] Let's now assume that 0 ≤ u0 < π . This is the normal range to describe any "upper bowl" over the full range from full sphere to disk! Then sign(sinu0) = +1 and we have cot-1[R3] = cot-1[- { cotu0 + ε [ 1/2 + cot2u0]} ] = π - cot-1[ { cotu0 + ε [ 1/2 + cot2u0]} ] There is a correction term here. We have cot-1(a + bε) = cot-1(a) +bε [ -1/(1+a2) ] Now set a = cotu0 and b = (1/2 + cot2u0) // did a check on this, it seems right so then cot-1[R3] = π - cot-1[ { cotu0 + ε [ 1/2 + cot2u0]} ] = π - { cot-1(cotu0) + (1/2 + cot2u0) (-1/(1+cot2u0)) ε } OK, I computed this extra term, but I can see that it is not going to do anything, so we have cot-1[R3] = (π - u0) Next we have R2 cot-1[R3] = R2 (π - u0) = (1/) ε / (|sinu0|) * (π - u0) So my final result is V(ξ,u) = (V0/π) { π - cot-1[R1] + R2 cot-1[R3] } = (V0/π) { 0 + ε /2 + (1/) ε / (|sinu0|) * (π - u0) } = (V0/π) { ε /2 + (ε/2) / (|sinu0|) * (π - u0) } doing corrected from here on, absorbing the red correction term : = (V0/π) ε /2 { 1 + 1 / (|sinu0|) * (π - u0) } = (V0/π) ε /2 { |sinu0| + (π - u0) }/ |sinu0| = (V0/π) [ (ε /2a)] { |sinu0| + (π - u0) } * (a/|sinu0| ) = (V0/π) [ 1/r] { |sinu0| + (π - u0) } * R = (RV0/π) [ 1/r] { |sinu0| + (π - u0) } So my conclusion is that V(ξ,u) = (RV0/π){ (π-u0) + |sinu0| } *(1/r) This is the correct answer now that I have the missing term. To get the correct answer, note that C = (R/π) { (π- u0) + sin(u0) } Q = CV0 V = Q/r = (CV0)/r = (V0R/π) { (π- u0) + sin(u0) }/ r whereas I got (RV0/π){ (π-u0) + |sinu0| } *(1/r) So as in my last attempt, I get the first term but I have somehow lost the second term!!! Note: I see looking back at "capacitance of the toroidal bowl v1.doc" that I had this same missing term problem there on 1/25/11. I penciled in the missing term in red at that time, but I don't seem to show where it should come from! I state in the intro that I never resolved the mystery. Is a resolution in the v2.doc or v3.doc of this series? Well, I try this again in v2.doc and there I get a different kind of wrong answer! There I am just missing a π. Don't think the secret is there. In the version 3 I actually get it right for the first time. So what IS the secret here??? I will have to study that document and compare to this. 3. Where is my missing term??? I have now later found the missing term. So how did I miss it below in my careful "analysis"? The missing term is this ΔV = (V0R/π)sin(u0) / r = (V0/π)a / r = (V0/π)a ε / 2a = (V0/π) (ε/2) Where might this come from?? I have V(ξ,u) = (V0/π) { π - cot-1[R1] + R2 cot-1[R3] } So let's start the search: Fact: The following missing term must be hiding somewhere π - cot-1[R1] + R2 cot-1[R3] must include hidden term (ε/2) This is a linear term in ε, so make sure you don't throw out linear terms! Recall R1 = - 2/[] (1/ε) + (1/4) ε + order(ε3) < 0 always!!!! R2 = (1/) ε / (|sinu0|) + order(ε2) R3 = - { cotu0 + ε [ 1/2 + cot2u0]} + order(ε2) Warning: cotu0 > 0 only for 0 ≤ u0< π/2, so I will just assume that for now. Write this as R1 = -A/ε + Bε A,B > 0 A = 2/[], B = (1/4) R2 = Cε C > 0 C = / (2|sinu0|) R3 = -D - εE D,E > 0 D = cotu0, E = [ 1/2 + cot2u0] Then we have π - cot-1[R1] + R2 cot-1[R3] = π - cot-1[ -A/ε + Bε] + Cε cot-1[ -D - εE] = cot-1[A/ε - Bε] + Cε cot-1[ -D - εE] The missing term is in one of these two terms, or maybe in both! We know that ∂xcot-1(x) = -1(1+x2) cot-1 (x+α) = cot-1 (x) - α/(1+x2) α = small For the first term, let x = A/ε and α = -Bε. Then cot-1[A/ε - Bε] = cot-1 (A/ε) + Bε / (1+(A/ε)2) = cot-1 (A/ε) + B ε3 / (ε2+A2) ≈ cot-1 (A/ε) + B ε3/A2 // this is all OK Now as ε→ 0, the first term is zero, and the correction term is order ε3!!! [ wrong!!! ] That cannot be it. Assuming I made no algebra mistakes prior. Here is what you then failed to do cot-1 (A/ε) = tan-1(ε/A) ≈ ε/A and there is the missing linear term!!! Fact: Them missing term is NOT in the first term of V. Now for the second term cot-1[ -D - εE] = π - cot-1[ D + εE] = π - { cot-1(D) -εE/(1+D2) } = π - cot-1(D)-εE/(1+D2) D = cotu0 = π - u0 - ε [ 1/2 + cot2u0] / (1+cot2u0) = π-u0 - εQ = π - u0 - ε [ 1/2 + cot2u0] * sin2u0 I am suspicious of the 1/2, but I checked it and could not find anything wrong with it. So here at least is an "extra term" of sorts. We then have R2 cot-1[R3] = Cε[ π-u0 - εQ ] = Cε(π-u0) + Kε2 But an ε2 term in the second term cannot supply my missing term which has to be linear in ε. Fact: Them missing term is NOT in the scond term of V. So for the fourth time, I cannot see where this term is lurking! The missing term within {.....} must have the form Kε, that is, it must be linear in ε. Conclusion: There must be an algebra error earlier on of a major sort. Comment: I don't really know that my V(ξ,u) is correct. I know that in Maple it gives what looks like a 1/r drop off, but I don't know if the constant is correct or not. So do I have any confirmation on any of my potential forms?? V(ξ,u) = V0 !Syntax Error, Idτ Piτ-1/2(chξ) . (2.4.13) V(ξ,u) = (V0/π) { cot-1[] + cot-1[] } I have no Lebedev result to look at. Check his book again. No, it is not there. He has a section on toroidal coordinates. How about Smythe? No, he gets the capacitance. I claim that if I use the first form above I get the right capacitance. Maybe I should process that section now to make sure my claim is correct!