Green's Function for Two Circles REVIEWED

2D Green's Function for Two Circles PhL 1.10.14 This is early work which is now all installed into Chapter 6, the guy from Perga and all that stuff. I think I just started over in Chapter 6 and ignored what is below. 1. First, I plan to use 2D bipolar coordinates, M&S picture of this is page 64 Fig 2.09. horizontal circles are labeled by u, vertical circles by v. This is extruded into a cylindrical coordinate system on page 81 system E4C where they are then called bi-cylindrical coordinates with that same Fig 2.09 mentioned. But here they write x = ashη/(chη-cosψ) y = asinψ/(chη-cosψ) z = z g11 = g22 = a2/(chη-cosψ)2 g33 = 1 h1 = h2 = a/(chη-cosψ) h3 = 1 Then on page 89 M&S give the differential operators including 2 = [(chη-cosψ)2/a2] (∂η2 + ∂ψ2) + ∂z2 From this I then know that 22D = [(chη-cosψ)2/a2] (∂η2 + ∂ψ2) Now I have to figure out how the various thing's are labeled! This really is the same as toroidal if you set φ = 0, and I have lots of data on that. 2. Using my "toroidal coordinates.doc" notes I will switch now to my usual coordinates, I then go on to get the Cartesian equations of the circles. I next change the meaning of θ to the following and after this my equations become x = ashξ/(chξ+cosθ)  hξ = hθ = a/(chξ+cosθ) -y/x = sinθ/shξ -y = asinθ/(chξ+cosθ) ξ = tanh-1[2ax/(a2+ x2+ y2)] θ = tan-1[ -2ay/(a2-x2-y2)] r2 = a2 (sh2ξ + sin2θ) / (chξ+cosθ)2 tanφ ≡ y/x = -sinθ/shξ x2 + (y - acotθ)2 = a2/sin2θ yc = a cot(θ) R = a/|sinθ| I then change from θ to u like so and here are the new equations. x = ashξ/(chξ–cosu)  hξ = hu = a/(chξ–cosu) y/x = sinu/shξ y = asinu/( chξ–cosu) ξ = tanh-1[2ax/(a2+ x2+ y2)] tanu = [ -2ay/(a2-x2-y2)] r2 = a2 (sh2ξ + sin2u) / (chξ-cosu)2 tanφ ≡ y/x = sinu/shξ x2 + (y- acotu)2 = a2/sin2u yc = a cot(u) R = a/|sinu| (x - acothξ)2 + y2 = a2/sh2ξ xc = acothξ R = a/|shξ| THESE equations then match those of M&S and then I presume that 2 = [(chξ-cosu)2/a2] (∂ξ2 + ∂u2) + ∂z2 22D = [(chξ-cosu)2/a2] (∂ξ2 + ∂u2) The Apollonius horizontal circles are labeled by ξ (the toroid label). So my capacitor circles are going to be surfaces of constant ξ. Values are negative for left side circles, positive for right side. Then I guess my little lines doc Chapter 5 equation becomes [(chξ-cosu)2/a2] (∂ξ2 + ∂u2)φt(ξ,u) = 0 φt(ξ1) = K/2 φt(ξ2) = -K/2 Now here then is the Green's function problem [(chξ-cosu)2/a2] (∂ξ2 + ∂u2)g(ξ,u|ξ',u') = δ(ξ-ξ')δ(u-u') (1/J) ?? Let's start in Cartesians -(∂x2 + ∂y)2g(x,y|x',y') = δ(x-x')δ(y-y') If we wanted to go to polar coordinates, we would have h1 = r and h2 = 1 ( think) and then dV = rdrdθ = J drdθ = dxdy x = rcosθ J = r y = rsinθ δ(x-x')δ(y-y') = (1/r)δ(r-r')δ(θ-θ') = δ(r-r')δ(θ-θ')(1/J) Now start again in Cartesians and go to bipolar coordinates. In this case J = [a/(chξ–cosu)]2 = h1h2, and so we get -[(chξ-cosu)2/a2] (∂ξ2 + ∂u2)g(ξ,u|ξ',u') = δ(ξ-ξ')δ(u-u') [(chξ-cosu)2/a2] so the factor cancels on both sides and we end up with -(∂ξ2 + ∂u2)g(ξ,u|ξ',u') = δ(ξ-ξ')δ(u-u') Do I believe this, it seems just too easy. It is identical to 2D Cartesian! In lines Appendix I I show that in 2D, -2[ln(1/R)/2π] = δ(r-r') R = | r - r' | (I.1.4) Now I suppose in bipolar coordinates this says [(chξ-cosu)2/a2] (∂ξ2 + ∂u2) { - ln(1/R)/2π } = δ(ξ-ξ')δ(u-u') [(chξ-cosu)2/a2] or - (∂ξ2 + ∂u2) { ln(1/R)/2π } = δ(ξ-ξ')δ(u-u') where you have to write our R in terms of the bipolar coordinates. x = ashξ/(chξ–cosu)  y = asinu/( chξ–cosu) x' = ashξ'/(chξ'–cosu')  y' = asinu'/( chξ'–cosu') R2 = (x-x')2 + (y-y')2 = [ashξ/(chξ–cosu) - ashξ'/(chξ'–cosu')]2 + [asinu/( chξ–cosu) - asinu' /(chξ'–cosu')]2 = a2(chξ–cosu)-2(chξ'–cosu')-2 { [shξ(chξ'–cosu') - shξ'(chξ–cosu)]2 + [ sinu(chξ'–cosu')-sinu'(chξ–cosu)]2 Maple says this is a big mess. What am I doing wrong here? I looked on the web and found some very messy Green's Functions in the presence of two circles, so maybe I don't really want to go down this path.