more on surface currents REVIEWED

More on Surface Current PhL 11.16.13 This is early work on the mag surface current issue. This eventually led to the new BC on ∂nAz to be included in Chapter 1, and this BC later became important for many reasons! I also first discovered the little formula for Kz surface current right here that became so important later on for the general theory. 1. I think I may have a bug. First, in (1.1.9g)m I make this claim, derived before your eyes: curl B = μ0(curlH + curlM) = μ0(∂tD + Jc) + μ0Jm = μ0(∂tD + Jc + Jm) line 1 " B sees all currents" (1.1.9g)m [ now (1.1.24) ] But then in (1.1.24) [ this is (1.1.1)] I state basically that curl H = ∂tD + Jc => curl (B/μ) = ∂tD + Jc => curl B = μ ∂tD + μJc line 2 [ this seems correct ] (1.1.24) Suppose we are doing statics so ∂tD = 0. Then I have from the above two lines, curl B = μ0 Jc + μ0Jm // line 1 curl B = μJc // line 2 [ both OK ] How can these both be valid? If they are valid, we are claiming that μJc = μ0Jc + μ0Jm Jc (μ-μ0) = μ0Jm Jm = ( μ/μ0 - 1)Jc [ this still seems right on 2/9/14, inside a medium of μ ] This says mag current is determined in this way by conduction current? Is this really true? Well, let's test it Jm = ( μ/μ0 - 1)Jc curl M = ( μ/μ0 - 1) curl H M = ( μ/μ0 - 1)H χm H = ( μ/μ0 - 1)H χm = ( μ/μ0 - 1) 1 + χm = μ/μ0 and this is correct !!! So OK, there is no bug after all. [ I think it is all correct, away from boundaries ] 2. I want to redo the H BC if there is a surface current? I can work with either of the forms above: curl B = μ0(∂tD + Jc + Jm) curl B = μ ∂tD + μJc The second form is not useful since I won't be able to "capture" the surface current Jm with it, so use the first equation (even though both are valid) Now consider this surface geometry Suppose there is a surface current flowing toward the viewer, Jm(x) = δ(y)K(x) = Jm(x) Start with the result above curl B = μ0(∂tD + Jc + Jm) curl B = μ0(∂tD + Jc + Jm) B ds = ∫S [μ0(∂tD + Jc + Jm)] dA Then LHS = (B2x-B1x)L And if the D and Jc are non-singular at the boundary, we just have RHS = μ0 ∫SJm dA = μ0 ∫Sδ(y)K(x) dA = μ0L Kz This seems to say that μ0Kz = B2x-B1 How does this compare with my Appendix B claim? μ0Kz = (Bx2 - Bx1) [ this is the current Appendix B claim ] Question: Why did I get μ1 on the left in Appendix B ? [ an earlier version of App B ] Answer: In App B I quoted (1.1.24) correctly, but in (1.1.24) the J there does not include Jm. since the rule is originating from curl H = ∂tD + J which does not see Jm. But then in App B I wrongly say that this J does include Jm. Question: If the correct μ is μ0 in the K formula, how does this alter later sections of App B? Now consider: B = curl A => Bx = ∂yAz - ∂zAy Then we have shown that μ0Kz = (Bx2 - Bx1) = (∂yAz2 - ∂zAy2) - (∂yAz1 - ∂zAy1) = [∂yAz2 - ∂yAz1] - [∂zAy2 - ∂zAy1] Since y is the normal direction, if we could somehow ignore the second term this says μ0Kz = ∂nAz2 - ∂nAz1 and this at least resembles something I quoted long ago from the web // I finally made sense of this early finding Now suppose we knew that A = Az in both regions. Then my result would say μ0Kz = ∂nAz2 - ∂nAz1 μ0Kz = (∂nAz2) - (∂nAz1) Now ∂n = 0, so this says μ0K = ∂nA1- ∂nA2 and this agrees exactly with the quoted result. Can we show that in general [∂zAy2 - ∂zAy1] = 0 ? That is to say ∂z(Ay2- Ay1) = 0 ? If we are in some situation where everything varies slowly in the z direction (transmission line limit? ), then we could say that ∂zAy2 ≈0 and ∂zAy1 ≈ 0 separately, and then this is true. Another case would be where we argue that we can neglect Ay. Is it possible that Ay2 - Ay1 = 0 all the time? In our King gauge for all of region R we have div A = - μ1ε1 ∂tφ - μ1σ1φ // applied to all of R . (1.3.18) div A = - μ1ε1jω φ - μ1σ1φ // applied to all of R . (1.3.18) div A = [- μ1ε1jω - μ1σ1]φ = -jωμ1ξ1φ // applied to all of R . (1.3.18) If we apply the divergence theorem we get div E = ρ/ε (1/ε)∫V ρ dV = ∫S E dS if ε is constant in space (1.1.19) div A = -jωμ1ξ1φ ∫V -jωμ1ξ1φ dV = ∫S A dS -jωμ1ξ1 [∫V φ dV] = ∫S A dS Now if φ is non-singular and continuous at the boundary, then ∫V φ dV = 0. Then we really can conclude that Ay2- Ay1 = 0 and Ay1 = Ay2 Wow. This has to be added to the arsenal somewhere! It is King gauge specific, or Lorenz gauge, either one. Coulomb gauge as well! What about continuity of Az ? In that case we want curl A appearing somewhere, so try this curl A = B A ds = [∫S B dA] The left side will then pick up A1t - At2. Then if B is finite at the surface, you have it! Wow!