the mu paradox REVIEWED

The Mu Paradox PhL 11.6.13 This nano-detail could cost me a month! We shall see. I resolved the paradox on 11.7.13, it did not take month, just a day. I am adding comments in red below. // All done. There really is a big paradox if you don't account for surface currents! Statement of the Paradox. On the One Hand: My current derivation of this equation [ surface currents missing ] (2 - μ1ε1 ∂t2 - μ1σ1) A = - Σi=2NμiJi all of region R (1.3.23) clearly has the μi associated with the μ inside conductor i and these μi ≠ μ1 in the dielectric. [ correct] This result occurs when I use the "region-1 King gauge" inside both dielectric and inside all conductors: div A = - μ1ε1 ∂tφ - μ1σ1φ // applies to all of R (1.3.18) If I then work through the W(z) section of Chapter 4, I end up with the wire having the following external inductance Le = {!Syntax Error, Idx' dy' b1(x',y') ln(s2/s1) -!Syntax Error, Idx' dy' b2(x',y') ln(s2/s1) } (4.4.8) where μ is that for the conductor, not the dielectric. The integral is dimensionless and the bi are dimensionless transverse currents which integrate to 1 across each conductor. Continuing, I end up finding that Le = (μ/2π)K where K is a geometric integral and I thus end up with μ for the conductor inside this formula for Le. [ if you omit surface currents, the conclusion is correct]. On the Other Hand: In appendix C I compute Le by computing the stored energy in the dielectric, which of course has μ1 . The resulting formula is Le = ln . (C.3.8) where μ1 is for the medium in which that energy was stored. External verification? Which is the correct answer? Matick p 97 computes Lint and the μ appearing there is that of the wire. Matick p 315 (8.11) then gives the external inductance where μ is for the dielectric. I looked on the web a bit, this is surely the right answer. [ yes, c.3.8 with μ1 = diel is correct.] So something is wrong with my development [yes] , and I already have an idea or two and it has to do with my choice of gauge [no, the gauge is OK]. Consider these two equations div A = - μ1ε1 ∂tφ - μ1σ1φ // applies to all of R (1.3.18) E = - grad φ - ∂tA B = curl A. Plan A. Proof that A in the region-1-King-gauge is completely continuous at a boundary [ I think this proof is OK, but it does not really bear on the paradox at hand. ] I want to study things at the boundary between conductor and dielectric. Work in the ω domain and we then have for the first equation div A = -jωμ1ξ1φ = -j(β12/ω) φ // King gauge Now do an integral form for div A at the 1/2 boundary: ∫ div A dV = ∫ A dS LHS = ∫ -j(β12/ω) φ dV = -j(β12/ω) ∫ φ dV which seems rather strange. If φ is continuous at the boundary which I certainly think is true, then we have ∫ φ dV = sA φ where pillbox has area A and height s. Meanwhile,. RHS = ∫ A dS = An1 A - An2A and therefore -j(β12/ω) sA φ = (An1- An2)A -j(β12/ω) s φ = (An1- An2) . Then take s→0 and conclude that An1 = An2 [ derivation seems OK, makes use of the gauge condition ] which says the normal A is continuous at the 1,2 boundary. In the T line app, this A is not of much interest since T line uses a transverse A (Az), but at least we have a little result here. This is in my special gauge. Notice that if the gauge were different in the two regions, then I would get LHS = ∫ div A dV = { [-j(β12/ω) φ] - [-j(β22/ω) φ] }/2 *V and then An would NOT be continuous at the boundary. [ good reason to use one gauge everywhere] Now look at B = curl A and see what that has to say: ∫S ( x A) dS = A ds = ∫S B dS Apply this to the usual little loop at a boundary of width s and height L. Then (region 1 on right) A ds = (At1- At2)L ∫S B dS = (Bt1+Bt2)/2 * sL // Ft points at viewer [ Note: Bt1 is discontinuous at the boundary, but it is no infinite, so argument is OK ] Then (Bt1+Bt2)/2 * sL = (At1- At2)L (Bt1+Bt2)/2 * s = (At1- At2) Then as s→0 we seem to get At1 = At2 [ derivation seems OK, gauge not used for this part ] I seem to conclude that all components of A are continuous at such a boundary. [ yes ] It is true that the Bt field jumps, (1/μ1)B1t = (1/μ2)B2t but that does not affect the above discussion. Fact: In the region-1-King gauge, A is continuous at a dielectric/conductor boundary. [ yes ] I regard that as a good thing really, since it is like φ. Now consider E = - grad φ - jωA If A is continuous at a boundary, then this is consistent with E taking a jump because φ changes slope. Corollary: When I compute W(z) as ΔAz in Chapter 4, there will be no issue with Az taking a sudden jump at the conductor surfaces, so that is not going to "explain" my paradox. [ correct ] Plan B. Maybe setting E = 0 inside the conductor the way I did it is wrong. [ no, this was not it ] I have this (2 - μ2ε2 ∂t2) A = [ (μ2ε2 - μ1ε1) ∂t - μ1σ1) ] (-E-∂tA) - μ2J // using (1.3.1) Now what happens if I set E = J/σ2 which is exactly correct rather than E = 0. This will create following new term on the right of my final equation [ (μ2ε2 - μ1ε1) ∂t - μ1σ1) ] (-J/σ2) which is a big mess, but then I would say σ2→ ∞ and this entire term vanishes and I am back. Plan C. OK, let's try to do the single round wire all by itself using my two equations and see what happens. [ in other words, try the case n = 2 ] Here is our picture of the wire cross section, And here are the two equations in my special gauge: [ surface current missing for region 1 ] (2 - μ1ε1 ∂t2 - μ1σ1) A = 0 region 1 (2 - μ1ε1 ∂t2 - μ1σ1∂t ) A = - μ2J2 region 2 time domain (2+β12) A = 0 region 1 (2+β12) A = - μ2J2 region 2 ω domain I would then claim a common equation for both regions (2+β12) A = - μ2J2 region 1 union region 2 Now I claim somehow that we can neglect At since Jt ≈ 0 and we then have (2+β12) Az = - μ2Jz2 region 1 union region 2 and we then get to treat this as a scalar wave equation like φ [correct] and we get Az(x,ω) = + ∫ Jz2 (x',ω)dV' R = |x - x'| . (1.5.9) where I suppose x can be anywhere, but the integral is only over region 2. At low ω this should be Az(x,ω) = + ∫ Jz2 (x',ω) dx'dy'dz' R = |x - x'| I can now jump to Appendix C.4 and result this discussion. There I deal with the divergence issues of this integral. I get this temp result Az(r) = - !Syntax Error, Ir' dr'!Syntax Error, Idθ ln (/Λ ) (C.4.4) This integral I actually do in my "verifications doc" with the solution reviewed as a "reader exercise". Now this is very interesting. In that document I start with Az(x,ω) = R = |x - x'| (1.5.9)z which has a "certain μ" that you can plainly see. I then boil this down to Az(r) = - !Syntax Error, Ir' dr'!Syntax Error, Idθ ln (/Λ ) (C.4.4) I then show that !Syntax Error, Idθ ln ( ) = π ln [ ] . where the important abs value is visible. I obtain these results Az(r) = - (μ1Jz/2) a2 ln(r) r > a // outside wire (*) B(r) = + (μ1Jza2/2)/r H(r) = B(r)/μ1 = (Jza2/2)/r = (I/πa2) a2/(2r) = (I/2π) r-1 Az(r) = - (μ2Jz/4) r2 r < a // inside wire B(r) = (μ2Jz/2) r H(r) = B(r)/μ2 = (Jz/2) r = (I/2πa2) r I have assigned the μ's above as they must be in order to agree with results below, BUT, it seems that you have to have μ1 = μ2 = μ since Az(r) came from a single initial equation!! So how can the μ's be the same and be different at the same time? [ this is the paradox ] In that same appendix I do this for internal inductance (2 = conductor) dU = (1/2) μ2H2 dV . (C.3.1) Ui = (1/2) (Li dz) I2 . (C.3.2) Ui = ∫inside (1/2) μ2H2dV, H(r) = (I/2πa2) r r ≤ a . (C.3.3) Ui = (μ2dz I2/16π) = (1/2) [μ2/8π] dz I2 Li = μ2/8π So there is μ2 properly sitting in the internal L formula, all seems clean here. Now here is my external L calculation: H(r) = (I/2π) r-1 . (C.3.6) dU(r) = (1/2) μ1H2dV Le = ln Just for the record here, H1t = H2t or (1/μ1)B1t = (1/μ2)B2t (1.1.26) Bn1 = Bn2 or μ1Hn1 = μ2 Hn2 (1.1.29) So here is the Big Mystery: To get Li and Le to come out right, you have to have μ1 and μ2 separately. But my unified Az(r) from which both results derive can only have a single μ sitting in front! [ yes, it is a mystery of you leave off the surface current ] Idea: Maybe there is a surface magnetization current [ ding! ]that I have ignored, analogous to the polarization charge at the boundary between two dielectric media? Since I keep saying low ω, why not just set ω = 0 and β1 = 0 and then it is just a Poisson type problem. Yes, let's just do the magnetostatics problem [ that was the correct decision ] and see what falls out. I will power this up in a separate doc.