Case Study, A for round doc REVIEWED

A Case Study PhL 11.13.19 Fully reviewed. Here I am going to look at the A solution for the round wire with uniform current in two ways, which I will call the Problem I Way and the Problem II Way. This is all done at ω = 0 and in the 2D way Problem I Way. [ this method is supportable ] The problem is described separately for regions 1 and 2, there is no common PDE, there is no image current on the boundary. (2 + β12) AI = 0 x in region 1 (2 + β12) AI = - μ2 Jc2 x in region 2 (AzI)2 = (AzI)1 (1/μ2) (∂nAzI)2 = (1/μ1) (∂nAzI)1 To solve this problem I make these assumptions: 1. The solution in region 2 is the particular solution plus a constant d. Thus AI,region2 = (1/2π) ∫dV' [μ2Jc(x')] ln(1/R) + d = W + d where constant d is a homo solution to (2 + β12) u = 0 . 2. The solution in region 1 is a homo solution of this form: AI,region1 = A lnr 3. I then evaluate the integral W. Just for fun, I do it for both region 1 and region 2, but I only need it right now for region 2 where r < a. W = (1/2π) ∫dV' [μ2Jc(x')] ln(1/R) = -(1/4π) μ2∫dV'[Jcz(x')] ln(R2) } = -(1/4π) μ2Jcz∫dV' ln(R2) } = -(1/4π) μ2[I/πa2] // using ∫dV' ln(R2) from App B = -(1/4π) μ2[I/πa2] 2πa2 = -(I/2π) μ2 The relevant part is in region 2 where I use the integral above, where then = -(I/2π) μ2 [ lna + (1/2)(r2/a2-1) ] // integral evaluated in region 2 4. Here then is what the solutions look like in the two regions at this point: AI,region1(r) = A lnr AI,region2(r) = d - (I/2π) μ2 [ lna + (1/2)(r2/a2-1) ] Notice these facts which will be used in the next little section AI,region1(a) = A lna AI,region2(a) = d - (I/2π) μ2 lna ∂r AI,region1(r) = A/r ∂r AI,region2(r) = - (I/2π) μ2 * (1/2a2)2r = - (I/2πa2) μ2 r ∂r AI,region1(a) = A/a ∂r AI,region2(a) = - (I/2πa2) μ2 a 4. I then match the boundary conditions AI,region2(a) = AI,region1(a) (1/μ2) ∂rAI,region2(a) = (1/μ1) ∂rAI,region1(a) which conditions then say d - (I/2π) μ2 lna = A lna (1/μ2) [- (I/2πa2) μ2 a] = (1/μ1)[ A/a] or d - (I/2π) μ2 lna = A lna - (I/2π) = (1/μ1)A The solution to these 2 equations in 2 unknowns is A = -μ1 (I/2π) d = (I/2π) μ2 lna -μ1 (I/2π) lna = (I/2π)(μ2-μ1)lna Installing these constants, we then obtain our final solution AI,region2(r) = W+d = -(I/2π) μ2 [ lna + (1/2)(r2/a2-1) ] + (I/2π)(μ2-μ1)lna AI,region1(r) = Alnr = -μ1 (I/2π)lnr or AI,region2(r) = -(I/2π) μ2 lna -(I/4π) μ2(r2/a2-1) + (I/2π)(μ2-μ1)lna AI,region1(r) = -μ1 (I/2π)lnr or AI,region2(r) = -(I/2π) μ1 lna - (I/4π) μ2(r2/a2-1) AI,region1(r) = -μ1 (I/2π)lnr or AI,region2(r) = -(I/2π) [ μ1 lna + (1/2)μ2(r2/a2-1) AI,region1(r) = -μ1 (I/2π)lnr Comment: This is the problem solution obtained by treating each region separately and making assumptions about the assumed forms then finding the constants. [ By the way, this solution happens to agree with the Appendix B Version 2 solution. ] Problem II Way. [ this method works, but is not supportable] The problem is described uniformly for region 1 + 2 = region R, but an image surface current is added at the boundary whose value matches the known mag current there. (2 + β12) AII = -μ0Jm - μ2Jc2 x in region R // image current present! AII,region2(a) = AII,region1(a) (1/μ2) ∂rAII,region2(a) - (1/μ1) ∂rAII,region1(a) = Kzimage = [(μ1-μ2)/μ0] I/(2πa) (*) wrong! Question: Below do I ever make use of equation (*) ? Answer: No, I never use this, and moreover, it is not true! [ correct] See comments in Plan C of Disaster Log 2. To solve this problem I make these assumptions: 1. The entire solution is the particular solution of the uniform PDE AII,regionR = (1/2π) ∫dV' [μ2Jc(x') + μ0Jm(x')] ln(1/R) = W + Z where W is the same integral we saw in the Problem I analysis and which we happened to evaluate in both regions 1 and 2, to wit, W = -(I/2π) μ2 . The new integral here is then Z = (1/2π) ∫dV'Jzm(x')] ln(1/R) = (1/2π) μ0 ∫dS' Kzm(x') ln(1/R) = (1/2π) μ0 [(μ1-μ2)/μ0] I/(2πa) ∫dS' ln(1/R) = (1/2π) (μ1-μ2) I/(2πa) ∫dS' ln(1/R) = - (1/2π) (μ1-μ2) I/(2πa)(1/2) { ∫dS' ln(R2) } = - (1/2π) (μ1-μ2) I/(2πa)(1/2) 4πa // {..} from Appendix B = - (I/2π) (μ1-μ2) = - (I/2π) (μ1-μ2) Then the unified solution over R is this AII,regionR = W + Z = -(I/2π) μ2 - (I/2π) (μ1-μ2) = -(I/2π) [ μ2 + (μ1-μ2) ] We then write out the two region solutions separately AII,regionR (r>a) = -(I/2π) [ μ2lnr + (μ1-μ2)lnr ] = -(I/2π) μ1lnr AII,regionR (r<a) = -(I/2π)[ μ2(lna + (1/2)(r2/a2-1)) + (μ1-μ2)lna ] = -(I/2π)[ μ2lna + μ2 (1/2)(r2/a2-1) + (μ1-μ2)lna ] = -(I/2π)[ μ1lna + μ2 (1/2)(r2/a2-1) ] and once again AII,regionR (r>a) = -(I/2π) μ1lnr region 1 AII,regionR (r<a) = -(I/2π)[ μ1lna + μ2 (1/2)(r2/a2-1) ] region 2 [ Note: these agree with the "correct" solutions obtained in the Disaster Log 2 "Resume 11.21.13. ] Thus, I at least have A METHOD of solving this problem and getting the correct answer!] Comparison of the solutions done the two ways AI,region1(r) = -μ1 (I/2π)lnr AI,region2(r) = -(I/2π) [ μ1 lna + (1/2)μ2(r2/a2-1) AII,region1(r) = -μ1 (I/2π)lnr AII,region2(r) = -(I/2π)[ μ1lna + μ2 (1/2)(r2/a2-1) ] We observe that the solution to this problem is EXACTLY THE SAME using either method of computation: I. Smythian form segregated method in each region. Simple homo form is selected for region 1, and a simple homo form constant adder is selected for region 2, and these are sufficient to solve the problem. Each region has a homo contribution. II. Unified region R solution but an image current has been added at the boundary. There is no homo contribution at all. This is my Prototype Example of the two Ways. The question is: WHY do these two Ways give the exact same answer? Is it just a coincidence for this very simple geometry, or is it a more general fact? [ I cannot prove it as general fact, so "cloy" ] Comment: At r = 0 we get AII,region2(0) = -(I/2π)[ μ1lna + μ2 (1/2)( -1) ] = -(I/2π)[ μ1lna - μ2/2 ] AII,region2(a) = -(I/2π)[ μ1lna + μ2 (1/2)(12-1) ] = -(I/2π)[ μ1lna ] ] Problem III Way ? Blue shows the original method curl H = ∂tD + J // Maxwell (1.1.1) (1/μ) curl curl A = με ∂tE + J // H = B/μ , B = curl A from (1.3.1), and D = εE grad divA - 2A = με ∂t[- grad φ - ∂tA] + μJ // vector identity and E = - grad φ - ∂tA (2 - με ∂t2) A = grad [με ∂tφ + divA ] - μJ [ = Jackson (6.11) ] (1.3.3) (2 - με ∂t2)A = - μJ . // apply same gauge choice divA = - με ∂tφ Black shows a possible different path: curl B = μ0(∂tD + Jc + Jm) // Maxwell (1.1.24) grad divA - 2A = μ0(∂tεE + Jc + Jm) grad divA - 2A = μ0(ε∂t[- grad φ - ∂tA] + Jc + Jm) grad divA - 2A = - μ0ε ∂tgrad φ - μ0ε∂t2A + μ0Jc + μ0Jm - grad divA + 2A = μ0ε ∂tgrad φ + μ0ε∂t2A - μ0Jc - μ0Jm 2A - μ0ε∂t2A = μ0ε ∂tgrad φ + grad divA - μ0Jc - μ0Jm (2A - μ0ε∂t2) A = grad [divA + μ0ε∂tφ] - μ0Jc - μ0Jm Still have not made gauge choice, but I can see that this is NOT the wave operator I want to see. This is totally at odds with the King wave operator, so this Way III is just not useful. Here I am trying to "expose" the mag current so it gets picked up. [ I pursue this more in Disaster 2, Plan E1. It led I think to me finding my big bug, but it is a lousy gauge to work in. ]