idea on Jm and the homo solutions v2 REVIEWED

Crazy idea on Jm and the homo solutions PhL 1.15.14 Again, this general idea is now installed as Section B.6 (b) Conjectured Theorem: If you add Jm to your Jc, the extra Jm integral gives exactly the right homo solutions to make the Az boundary conditions match. Attempted Proof: Start with Appendix C.4's starting point Az(x) = ∫ dV' [μ2Jcz(x') + μ0Jmz (x')] R = |x - x'| ω = 0 (1.5.9) = Az(c) + Az(m) I will show later that one can add an arbitrary surface current and that only changes the homo solutions, and I choose the current shown above. 1. I know that the "conduction current term" has this property at the boundary between μ1 and μ2: [Az(c)]1 = [Az(c)]2 1 = dielectric = outside wire [∂n Az(c)]1 = [∂n Az(c)]2 Therefore I know that [∂n Az(c)]1 – [∂n Az(c)]2 = [-][∂n Az(c)] This is the quantity that we want the Jm term to cancel out. 2. The "magnetization current term" is this Az(m) = μ0 ∫ dV' Jmz (x')] where Jmz (x') is confined to the conductor surface and this has the usual (-∞,∞) dz' integral. The mag current density is in the form of a 3D density of amps/m2. However, with the usual set of assumptions I can replace this 3D integral and propagator with as 2D integral and propagator as follows: Az(m) = μ0 ∫ dA' Jmz (x')] [ - ln(R) ] Now since we know that Jmz (x') is confined to the 2D perimeter in this 2D picture, we then write Az(m) = μ0 ∫C ds' Kmz (x')] [ - ln(R) ] where we have now "taken out the delta function" so K is amps/m. I showed in Appendix B that Kz = - ( - ) Hθ where Hθ > 0 and Kz ≤ 0. so we then have Az(m) = - (μ2-μ1) ∫C ds' Hθ(x')] [ - ln(R) ] = ∫C ds' [(μ1-μ2) Hθ(x')] [ - ln(R) ] = ∫C ds' a(x')] E(x|x') . a(x') = (μ1-μ2) Hθ(x') 3. Now recall from Stakgold these facts u(x) = ∫σ dSξ a(ξ) E(x|ξ) u(s) = ∫σ dSξ a(ξ) E(s|ξ) ∂νu(x) = ∫σ dSξ a(ξ) ∂νE(x|ξ) ∂ν = normal to surface σ = I write as ∂n ∂νu(s) = [∂νu(x)]x=s+ = ∫σ dSξ a(ξ) ∂νE(x|ξ) - a(s)/2 // the extra term! where σ is a surface of interest. We take now this surface to be our perimeter and then E is the 2D E. Thus we expect to find that [∂nAz(m)(x)]1 = ∫C ds' [(μ1-μ2) Hθ(x')] ∂nE(x|x') - [(μ1-μ2) Hθ(x)]/2 (**) [∂nAz(m)(x)]2 = ∫C ds' [(μ1-μ2) Hθ(x')] ∂nE(x|x') + [(μ1-μ2) Hθ(x)]/2 Now compute our "correction term" for the difference boundary condition, [∂nAz(m)(x)]1 – [∂nAz(m)(x)]2 = {∫C ds' [(μ1-μ2) Hθ(x')] ∂nE(x|x') [- ] - [(μ1-μ2) Hθ(x)] [+ ]/2 = (μ1-μ2) [- ] ∫C ds' Hθ(x')] ∂nE(x|x') - (1/2) (μ1-μ2) [+ ] Hθ(x) This then is our candidate correction term which we want to cancel the original term. I am not sure how that is going to happen! The original was this: [∂n Az(c)]1 – [∂n Az(c)]2 = [- ][∂n Az(c)] Thus this is what I WANT to be true (add above two equations and hope for 0) (μ1-μ2) [- ] ∫C ds' Hθ(x')] ∂nE(x|x') - (1/2) (μ1-μ2) [+ ] Hθ(x) + [- ][∂nAz(c)]= 0 The μ factors don't look right, but let's plough ahead . Write [- ] = (μ2- μ1)/(μ1μ2) = - (μ1-μ2) /(μ1μ2) on the far right only, then cancel one factor (μ1-μ2) to get [- ] ∫C ds' Hθ(x')] ∂nE(x|x') - (1/2) [+ ] Hθ(x) - [∂n Az(c)] = 0 Further simplify to (μ2- μ1) ∫C ds' Hθ(x')] ∂nE(x|x') - (1/2) (μ2+μ1) Hθ(x) - [∂n Az(c)] = 0 (*) ?? I have obtained the above equation in my Section B.6 writeup. Once again, I am trying to show that (*) is true, I have not shown it is true! I do know from (**) that ∫C ds' [(μ1-μ2) Hθ(x')] ∂nE(x|x') = [∂nAz(m)(x)]1 + [(μ1-μ2) Hθ(x)]/2 agree so then here is what I want to show is true: -[∂nAz(m)(x)]1 - [(μ1-μ2) Hθ(x)]/2 - (1/2) (μ2+μ1) Hθ(x) - [∂n Az(c)] = 0 ?? or [∂nAz(m)(x)]1 + [(μ1-μ2) Hθ(x)]/2 + (1/2) (μ2+μ1) Hθ(x) + [∂n Az(c)] = 0 ?? or [∂nAz(m)(x)]1 + μ1 Hθ(x) + [∂n Az(c)] = 0 ?? Now how is Hθ related to Az ? If we define a little local cylindrical coordinate system, then we know that B = curl A = [ r-1∂θAz - ∂zAθ] + [∂zAr - ∂rAz] + [ r-1∂r(rAθ) - r-1∂θAr = [ r-1∂θAz] + [- ∂rAz] For a circular boundary this would be just B = [- ∂rAz] since Az is constant on the surface. Let us now conjecture that for a general perimeter we have B = [- ∂nAz ] where is a local tangent vector as I often use. Then we could claim that Bθ = - ∂nAz . What Az is this? It should be the "true one" where BC's are met, so this says Bθ = - ∂nAz(c) - ∂nAz(m) But is this inside or outside? I guess we have to say Bθ1 = - [∂nAz]1 Bθ2 = - [∂nAz]2 Then we can write H(θ) = Bθ1/μ1 = Bθ2/μ2 take your choice So I choose 1 and outside to get H(θ) = Bθ1/μ1 = -(1/μ1) [∂nAz(c) + [∂nAz(m)]1] or μ1 H(θ) = Bθ1 = - [∂nAz(c) + [∂nAz(m)]1] Now try this in our ?? equation: [∂nAz(m)(x)]1 + μ1 Hθ(x) + [∂n Az(c)] = 0 ?? [∂nAz(m)(x)]1 - [∂nAz(c) + [∂nAz(m)]1] + [∂n Az(c)] = 0 ?? 0 = 0 !!!!! Did I just go in a circle, or have I really succeeded here? Time is 8:13 AM 1.16.14.