more on the mu fix REVIEWED

More on the μ fix PhL 2.5.14 This is an earlier version of what is now called the Jm theorem. The essential proof succeeds right here. I then rewrote it a bit, and it is all installed. In Lines Appendix B, I somewhat lose the thread of how the surface current Jm is supposed to "repair" things for a transmission line, so maybe I can bail it out here. The main problem is that I did not treat more than one conductor so the result is hazy. The starting point should really be an equation like Az(x) = ∫ Jz1(x1')d3x1' + ∫ Jz2(x2')d3x2' R1 = |x - x1'| R2 = |x - x2'| where this is the total potential due to all conductors integrations. Then we consider the boundary condition at the surface first of conductor C1 . How do we do the repair? If I trace through B.7 I would start going just 1/R factors. Then B.6.1 would say -Az(x) = ∫ dA2' [μ2 Jcz2(x2') + μ0 Jmz2(x2')] + ∫ dA3' [μ3 Jcz3(x3') + μ0 Jmz3(x3')] where the dielectric has μ1 to be consistent with Appendix B, so the two conductors are now 2 and 3. I then have to restate the theorem more carefully. This form should fix the boundary conditions at both conductor surfaces. I would then have (1) Preliminaries with no change. Then I guess for section (2) I would have to have a break up into four pieces instead of 2: Az(c2)(x) = ∫ [μ2 Jcz2(x2')] dV2' . R = |x - x2'| Az(c3)(x) = ∫ [μ3 Jcz3(x3')] dV3' . R = |x - x3'| Az(m2)(x) = ∫ [μ0 Jmz2(x2')] dV2' . R = |x - x2'| Az(m3)(x) = ∫ [μ0 Jmz3(x3')] dV3' . R = |x - x3'| Az(x) = Az(c2)(x) + Az(m2)(x) + Az(c3)(x) + Az(m3)(x) Now I would specialize so that x → s which is a point on C2. The claim is that each Az(ci)(x) is smooth at such a point and we have Az(ci)(s+) = Az(ci)(s-) ∂nAz(ci)(s+) = ∂nAz(ci)(s-) . i = 2, 3 (B.6.7) Then for the sum of the two (c) terms I would write ∂n[Az(c2)(s+) + Az(c3)(s+)] – ∂n[Az(c2)(s-) + Az(c3)(s-)] = + [ - ] ∂n[Az(c2)(s) + Az(c3)(s)]. (B.6.8) where I could have used either s+ or s- on the right, since each is continuous at s on C1. So this is the fancier version of (B.6.8). The goal is then to show that ∂n[Az(m2)(s+) + Az(m3)(s+)] – ∂n[Az(m2)(s-) + Az(m3)(s-)] = - [ - ] ∂n[Az(c2)(s) + Az(c3)(s)] . (B.6.9) So it then remains to demonstrate this new and fancier (B.6.9). (3) Verification of (B.6.9). Now suppose I just regard this whole section as a applying to C2 only, so imagine little "2" subscripts as needed. We just march down through this section thinking this way. In (B.6.18) we would then have ∂nAz(m2)(s±) = C2 ds2' [(μ1-μ2) Hθ2(x')] ∂nE2(s|x2') ∓ [(μ1-μ2) Hθ2(s)]/2 . ∂nAz(m3)(s±) = C3 ds3' [(μ1-μ3) Hθ3(x')] ∂nE2(s|x3') (B.6.18) The second line does not have the Stakgold "extra term" because as x → s, since s lies on C2 and not on C3, we are not approaching the monopole later Hθ3 on C3 so this singularity effect does not occur. Notice that each line is an integral over its own conductor with its own Hθ value. The big equation after (B.6.18) is now even bigger: ∂n[Az(m2)(s+) + Az(m3)(s+)] – ∂n[Az(m2)(s-) + Az(m3)(s-)] = { C2 ds2' [(μ1-μ2) Hθ2(x2')] ∂nE2(s|x2') - [(μ1-μ2) Hθ2(s)]/2 } – { C2 ds2' [(μ1-μ2) Hθ2(x2')] ∂nE2(s|x2') + [(μ1-μ2) Hθ2(s)]/2 } + { C3 ds3' [(μ1-μ3) Hθ3(x3')] ∂nE2(s|x3') } – { C3 ds3' [(μ1-μ3) Hθ3(x3')] ∂nE2(s|x3') } = (μ1-μ2) [ - ] C2 ds2' Hθ2(x2') ∂nE2(s|x2') – [ + ] (μ1-μ2) Hθ2(s)/2 + [ - ] (μ1-μ3) C3 ds3' Hθ3(x3')] ∂nE2(s|x3') where now we have an extra term on the last line which was not present before. Our task of showing that (B.6.9) is true then boils down to showing that the last expression above is equal to - [ - ] ∂n[Az(c2)(s) + Az(c3)(s)] . I smell trouble brewing. I will first write this big equation: (μ1-μ2) [ - ] C2 ds2' Hθ2(x2') ∂nE2(s|x2') – [ + ] (μ1-μ2) Hθ2(s)/2 + [ - ] (μ1-μ3) C3 ds3' Hθ3(x3')] ∂nE2(s|x3')= - [ - ] ∂n[Az(c2)(s) + Az(c3)(s)] ? Cancelling (μ1-μ2) factors gives [ - ] C2 ds2' Hθ2(x2') ∂nE2(s|x2') – [ + ] Hθ2(s)/2 - (μ1-μ3) C3 ds3' Hθ3(x3')] ∂nE2(s|x3') = + ∂n[Az(c2)(s) + Az(c3)(s)] ? or (μ2-μ1) C2 ds2' Hθ2(x2') ∂nE2(s|x2') – (μ2+μ1) Hθ2(s)/2 - (μ1-μ3) C3 ds3' Hθ3(x3')] ∂nE2(s|x3') = + ∂n[Az(c2)(s) + Az(c3)(s)] ? (B.6.19) Now we want to replace both integrals using (B.6.18) where we make the s+ choice ∂nAz(m2)(s+) = C2 ds2' [(μ1-μ2) Hθ2(x2')] ∂nE2(s|x2') - (1/2)[(μ1-μ2) Hθ2(s) ∂nAz(m3)(s±) = C3 ds3' [(μ1-μ3) Hθ3(x2')] ∂nE2(s|x3') (B.6.18)+ Our equation of interest (B.6.19) then becomes - ∂nAz(m2)(s+) - (1/2)[(μ1-μ2) Hθ2(s) – (1/2)(μ2+μ1) Hθ2(s) - ∂nAz(m3)(s±) = + ∂n[Az(c2)(s) + Az(c3)(s)] ? or - ∂nAz(m2)(s+) - μ1Hθ2(s) - ∂nAz(m3)(s±) = + ∂n[Az(c2)(s) + Az(c3)(s)] ? or ∂n[Az(c2)(s) + Az(c3)(s) + Az(m2)(s+) + ∂nAz(m3)(s±) ] = - μ1Hθ2(s) or ∂nAz(s) = - μ1Hθ2(s) ? But this equation is (B.6.23) in the language C → C2 on which point s lies. Thus, we can erase all the question marks!!!!!!!! ******************************************************************* Is there a faster way? Let's go back to the point where I have the "fancier B.6.9 ∂n[Az(m2)(s+) + Az(m3)(s+)] – ∂n[Az(m2)(s-) + Az(m3)(s-)] = - [ - ] ∂n[Az(c2)(s) + Az(c3)(s)] . (B.6.9) I don't think I can claim that the "2" part of this is true on its own. The reason is the last step above where you have to add all four elements to get Hθ2. The philosophy is that we "force" J in the conductors, and that forces Hθ2 for example. Then the total Az has to satisfy ∂nAz(s) = - μ1Hθ2(s) . OK, I first redid equation numbers in parts (1), (2) and (3) . Now I add section (c) and I will do this in a separate doc so I can copy from here. .