Disaster Repair Log 2 REVIEWED

Disaster Repair Log 2 PhL 11.17.13 On 11.21.13 6 PM I finished a complete review of this document. Holy Cow. 1. Background for the Paradox 2 2. So here is the Paradox: 2 3. Look Harder now at the Region 1 and Region 1 wave equation derivations: 3 4. Vague Idea for Resolving the Paradox. 6 November 19, 2013 11 Creation of Appendix B Version 2. 11 Comparing Problem I and Problem II 13 Plan A: Superposition? 17 Plan B: The Stakgold v function idea. 18 Plan C: Boundary Conditions on A 19 Plan D: Ansatz to simulate correct BC's and Stak surface layer limits 20 Plan D1: Try again: 21 11.20.13 Back At It Again: The new μ0 gauge idea. 23 Plan E1: Question: What do you have to do to cause Jm to honestly appear as a source in a wave equation for Az ? This is my μ0 gauge approach. 23 Plan E2: φ in the same gauge 26 Plan E3. Do the DC round wire in this new gauge. Use Appendix B as a guide. 28 Plan E4. Adding homo solutions to new gauge solution 31 Plan E4. Show directly that the correct Az satisfies a certain wave equation 36 Plan E5: Gauge Violation? 38 Redo the μ0 gauge solution as a two-region solution with BC's. 40 There is the little annoying fact about the integral and Maple 43 Resume 11.21.13. 46 (a) What are the "right" Az expressions for the round wire? 46 (b) Do these correct Az solutions solve the μ0 gauge wave equation [as I perceived it] 47 (c) So let's continue the reverse trace debug with a narrowed range 49 Bug Found !!! // there was indeed a Loose Bolt 52 Planar Scenarios 52 Dethreading the Nov 16 Disaster after Bug Found 55 Attempt to solve the μ0 gauge wave equation using the correct Jmz 56 Death Knell for the μ0 Gauge 63 Reminder of the Just-Fine two region Case Study solution to the round wire problem. 63 The one-region Case Study solution to the round wire problem. 64 1. Background for the Paradox Here is my A wave equation derivation: [ reviewed] 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) Here I have not used a gauge yet, nor have I talked about "regions". But what I do know is that the J shown here does NOT include Jm mag currents, because H does not "see" such currents so these currents don't exist in (1.1.1) [ correct] . Thus, there is no equation of the form (1.3.3) in any "region" which is going to include such Jm currents [correct] . Fine. I think in the way I already did things, here is how I end up. (2 - μ1ε1 ∂t2 - μ1σ1∂t ) A = 0 region 1 (2 - μ1ε1 ∂t2 - μ1σ1∂t ) A = - μ2J2 region 2 [ both correct ] This is the King gauge for region 1 applied in both region 1 and region 2, so I end up with unified result [be careful when you say that!] (2 - μ1ε1 ∂t2 - μ1σ1∂t ) A = - μ2J2 region R [ correct and viable ] where J2 is the conduction current in the conductor. Going to ω space, and then going to DC, this says (2 + β2)A(x,ω) = - μ2J2(x,ω) [correct] What could be simpler! We know that at ω = 0 β2 = 0 so this says in 2D A(x,ω=0) = +(μ2/4π)∫dV' J2(x',ω=0) (1/R) R = | x - x' | (*) [ particular sol ] Now, I think J2 has a meaning independent of μ2 of the conductor. It is amps/m2 inside. Since this formula is supposed to be valid for all of region R, it is valid inside and outside of the conductor. [ yes ] 2. So here is the Paradox: On the one hand: (1) A(x,0) shown above (at ω = 0) does not depend on μ1 in any way (the dielectric's perm. value) (2) B = curl A, therefore B cannot depend on μ1. On the other hand: Outside the conductor, I know from the Maxwell curl H(x,ω) = J(x,ω) in ω space I know that H(r,ω) = I(ω)/2πr r > a I(ω) = ∫dx'dy' J2(x',ω) and therefore B(r,ω) = μ1H(r,ω) = μ1 I/2πr r > a and we see that in fact B is in fact a function of μ1 for r > a. So doing things the first way gives B not a function of μ1, while doing things the second way gives a B which is a function of μ1. This is the Paradox. Both these ways cannot be correct, and I know the second way is correct, so there is something wrong with the first way. [ correct ] Fact: Since B = curl A, in order for B to depend on μ1, we must have A depend on μ1, and that is true no matter what gauge you use, and no matter what ω you work at in ω-space. [ correct ] Paradox Resolution: It is true that the particular solution (*) does not depend on μ1. But that does not say that possibly required homogeneous adder solutions won't depend on μ1, and in fact they do! These adders might be needed to satisfy boundary conditions. 3. Look Harder now at the Region 1 and Region 1 wave equation derivations: Region 1: [ reviewed ] We first obtain the wave equation for A in region 1. Start with (1.3.3) which gives the wave equation for A in region 1 before any gauge choice is made: (2 - μ1ε1 ∂t2) A = grad [μ1ε1 ∂tφ + divA ] - μ1J . // region 1 (1.3.3) Now insert the King gauge (1.3.18) to get (2 - μ1ε1 ∂t2) A = grad [μ1ε1 ∂tφ + (- μ1ε1 ∂tφ - μ1σ1φ) ] - μ1J // note cancelling of terms! = - μ1σ1 grad φ - μ1J = - μ1σ1 (-E -∂tA) - μ1J . // from (1.3.1) Now in region 1 we have relation J = σ1E where J is the conduction current in region 1, and this is exactly the J which appears in the above equation. Therefore, = - μ1σ1 (-J/σ1 -∂tA) - μ1J = + μ1J + μ1σ1 ∂tA - μ1J = μ1σ1 ∂tA (2 - μ1ε1 ∂t2) A = μ1σ1 ∂tA which says (2 - μ1ε1 ∂t2 - μ1σ1∂t) A(x,t) = 0 x in region 1 I just do NOT see anything wrong with the above region 1 derivation. [ there IS nothing wrong ] Region 2: Start with are pre-Gauge wave equation now in region 2 (2 - μ2ε2 ∂t2) A = grad [μ2ε2 ∂tφ + divA ] - μ2J . // region 2 (1.3.3) So far all I have done is insert μ,ε,σ appropriate for this new region. Next, I use the exact same gauge used for region . div A = - μ1ε1 ∂tφ - μ1σ1φ Is there anything wrong with deciding to do this? [ no ] It is the same gauge used in region 1, but now I apply it to x values in region 2. I really think this is OK to do. We then get (2 - μ2ε2 ∂t2) A = grad [μ2ε2 ∂tφ + (- μ1ε1 ∂tφ - μ1σ1φ) ] - μ2J . // region 2 = [μ2ε2 ∂t + (- μ1ε1 ∂t - μ1σ1) ]grad φ - μ2J // region 2 = [μ2ε2 ∂t + (- μ1ε1 ∂t - μ1σ1) ] (-E -∂tA) - μ2J // region 2 = [μ2ε2 ∂t + (- μ1ε1 ∂t - μ1σ1) ] (-E) - μ2J (*) + [μ2ε2 ∂t + (- μ1ε1 ∂t - μ1σ1) ]( -∂tA) = [μ2ε2 ∂t + (- μ1ε1 ∂t - μ1σ1) ] (-J/σ2) - μ2J - [μ2ε2 ∂t + (- μ1ε1 ∂t - μ1σ1) ]( ∂tA) = [μ2ε2 ∂t + (- μ1ε1 ∂t - μ1σ1) ] (-J/σ2) - μ2J - [μ2ε2 ∂t2 + (- μ1ε1 ∂t2 - μ1σ1∂t) ] A = [μ2ε2 ∂t + (- μ1ε1 ∂t - μ1σ1) ] (-J/σ2) - μ2J + ( -μ2ε2 ∂t2 + μ1ε1 ∂t2 + μ1σ1∂t) ] A Now look at the second line, which has 3 terms. The first term cancels the corresponding term on the LHS. We then move the two other terms to the LHS to get (2 - μ1ε1 ∂t2 - μ1σ1∂t) A = [μ2ε2 ∂t + (- μ1ε1 ∂t - μ1σ1) ] (-J/σ2) - μ2J Now we simplify the RHS [μ2ε2 ∂t + (- μ1ε1 ∂t - μ1σ1) ] (-J/σ2) - μ2J = [(μ2ε2-μ1ε1) ∂t - μ1σ1 ] (-J/σ2) - μ2J = [- (μ2ε2-μ1ε1) ∂t + μ1σ1 ] (J/σ2) - μ2J = [- {(μ2ε2-μ1ε1)/σ2} ∂t + μ1σ1/σ2 ] J - μ2J = [- {(μ2ε2-μ1ε1)/σ2} ∂t + μ1σ1/σ2 - μ2 ] J This seems to be something new and a little unexpected. Previously at step (*) I just set E = 0 saying it was inside a conductor. Maybe that was the wrong thing to do. But let's go to ω space (2 + β12) A = [- {(μ2ε2-μ1ε1)/σ2} jω + μ1σ1/σ2 - μ2 ] J At this point you could say that σ2 is very large [ correct ] and so you then really to get (2 + β12) A = - μ2 J // region 2 [ correct ] as long as ω is not TOO high. J is the conductors conduction current. So here then are my results: (2 + β12) A = 0 x in region 1 [ both correct ] (2 + β12) A = - μ2 J x in region 2 Understanding that J exists only in region 2 of region R, we unify to get (2 + β12) A = - μ2 J x in region R [ unified is correct and viable ] Then we go do DC ω = 0 to get (2) A = - μ2 J x in region R and then we are back to the Paradox, because μ1 does not appear in this equation! [ But as noted, μ1 may appear in homo adder solutions.] 4. Vague Idea for Resolving the Paradox. [reviewed] The general idea is this: I treat things like ε,μ,σ as "constants" in space, but in the region of a boundary, they are NOT constants. They have different values on the two sides of the boundary, so you have to be a little careful in such a region. You cannot for example pass them through a curl operator right at the boundary. But if you stay cleanly in one region or the other, you are OK. [ This was just a straw in the wind idea. But really I already am careful which region I am in, and this vague idea is not the source of the paradox. ] curl H = ∂tD + J // Maxwell (1.1.1) in region 1 H = B/μ1 in region 1 (1/μ1) curl curl A = μ1ε1 ∂tE + J // H = B/μ , B = curl A from (1.3.1), and D = εE grad divA - 2A = μ1ε1 ∂t[- grad φ - ∂tA] + μ1J // vector identity and E = - grad φ - ∂tA (2 - μ1ε1 ∂t2) A = grad [μ1ε1 ∂tφ + divA ] - μ1J [ = Jackson (6.11) ] (1.3.3) I can do this in region 2 as well. BUT I don't think I can just take the union of these two equations. (2 + β2) A = 0 x in region 1 away from the boundary [ correct as usual ] (2 + β2) A = - μ2 J x in region 2 away from the boundary But region R includes regions 1 and 2 and the boundary region. So that is my vague idea. Let's roll out this picture Then I would claim (2 + β2) A = 0 for y > ε region 1 (2 + β2) A = - μ2 J for y < ε region 2 Now in region 2 I might be able to say (ω = 0) A(x,ω=0) = +(μ2/4π)∫R2 dV' J2(x',ω=0) (1/R) for y < ε region 2 My idea here is this claim: You cannot simply take the union of these two equations, because then you have included the boundary region that you don't know much about. [ Again, just a straw. In Stak we do this multi-region stuff all the time, but there are BC's to deal with at the boundary. At this point, I was not very cognizant of what BC's on Az might look like! I think that came later. Just grasping. ] So suppose go try to solve the problem for region 2. This gives some result and I can take ε→0 and approach the surface from below. In fact, if I do this for the round conductor, I will get this: ( I can just look at my 2D Section B.6 result! ) Az = - [I/(4π2a2)] { μ2∫dV' ln(R2) } (B.6.4) ∫dV' ln(R2) = 2π a2lna + π(r2-a2) r < a inside Then Az(r) = - [I/(4π2a2)] μ2{ 2π a2lna + π(r2-a2) } + Az(r)homo2 r ≤ a where I now allow some homogeneous solution inside the wire which is region 2. [ I am slowly waking up here, this is the key idea : there will be homo adders ]. The limit then gives Az(a-) = - [I/(4π2a2)] μ2{ 2π a2lna } + Az(a)homo Now perhaps the next step would be to examine the boundary condition for Az at this boundary and maybe for some derivatives of Az as well. From my newly reworked Section 1.1 I have this result (1/μ2) (∂nAz)2 - (1/μ1) (∂nAz)1 = Kzfree But I don't think there is any free surface current on this boundary [correct] , so this would then read (1/μ2) (∂rAz)2 = (1/μ1) (∂rAz)1 . [ correct, and other BC is Az1 = Az2 ] Now let's go back to region 2 Az = - [I/(4π2a2)] μ2{ 2π a2lna + π(r2-a2) } + Az(r)homo2 r ≤ a (∂rAz)2 = - [I/(4π2a2)] μ2 * 2πr + (∂rAzhomo2)2 → - [I/(4π2a2)] μ2 * 2πa + (∂rAzhomo2)|r=a = - [I/(2πa)] μ2 + (∂rAzhomo2)|r=a. So I then claim that (1/μ1) (∂rAz)1 = (1/μ2) { - [I/(2πa)] μ2 + (∂rAzhomo2)|r=a } = - [I/(2πa)] + (1/μ2) (∂rAzhomo2)|r=a. So maybe what I have for region 1 is a Neumann problem, or at least some kind of BV problem. [ Well, it is not a Neumann problem because ∂rAz is not specified on a closed boundary, nor is it a mixed problem. But like any Stak potential theory problem with multiple regions, you have to decide what your BC's are at boundaries between regions and then make sure they are met. Title of Stak book ! ] Now let's take a look at region 1 again with β1 = 0. (2 + β12) A = 0 for y > ε region 1 Now I have already done this in App B, so I have Az(r) = Azhomo1(r) I guess I won't get far without some general atomic form which allows coefficient match at r = a. I now jump to the 2D model of things and write: Azhomo1(r) = A lnr + Br + Cr-1 + D for example from Stak II p 92 [ Error: for azisym function, r and r-1 are not atoms, so B = C = 0, as I do later anyway. ] Now in region 1 we can go do large r and I think we can set B = 0 at least. So let's just try Azhomo1(r) = A lnr +Cr-1 + D Azhomo2(r) = α lnr + br + cr-1 +d = br + d // I guess [ Not "I guess" but true since r and r-1 are not azisym atoms. ] OK, then we have (∂rAzhomo)1 = [ A(1/r) - Cr-2 ]|a=r = A(1/a) - Ca-2 and (∂rAz)1 = (∂rAzhomo)1 = A(1/a) - Ca-2 Meanwhile (∂rAzhomo)2 = b and Az(r) = - [I/(4π2a2)] μ2{ 2π a2lna + π(r2-a2) } + Az(r)homo2 r ≤ a => (∂rAz)2 = - [I/(4π2a2)] μ22πr|r=a + (∂rAzhomo)2 = - [I/(2πa)] μ2 + b Then the boundary condition says (1/μ2) (∂rAz)2 = (1/μ1) (∂rAz)1 [ correct since no free boundary current.] (1/μ2){ - [I/(2πa)] μ2 + b } = (1/μ1){ A(1/a) - Ca-2} Meanwhile, I conjecture another BC which is that Az is continuous so we would then have (Az)2 = (Az)1 [ correct ] LHS = (Az)2 = - [I/(4π2a2)] μ2{ 2π a2lna + π(r2-a2) } + Az(r)homo2 = - [I/(4π2a2)] μ22π a2lna + ba + d = - [I/2π] μ2 lna + ba + d RHS = (Az)1 = Azhomo1(r) = A lna +Ca-1 + D So this BC says - [I/2π] μ2 lna + ba + d = A lna +Ca-1 + D Here then are the two BC's: (1/μ2){ - [I/(2πa)] μ2 + b } = (1/μ1){ A(1/a) - Ca-2} - [I/2π] μ2 lna + ba + d = A lna +Ca-1 + D and we have these unknowns: A, C, b, d, D I don't think there are any other BC's, so maybe I can set some of these to 0. Let's just try on the outside that C = D = b = 0 and see what happens - I/(2π) = (1/μ1){ A } => A = - (1/2π)μ1I - [I/2π] μ2 lna + d = A lna => d = +[I/2π] μ2 lna + A lna = +[I/2π] μ2 lna - (1/2π)μ1I lna = [I/2π] (μ2- μ1) lna where only A and d are left and I solved for them already on the right. Then our solutions are Azhomo1(r) = A lnr = - (1/2π)μ1I lnr Azhomo2(r) = [I/2π] (μ2- μ1) lna And then the full solution is Az1 = Azhomo1(r) = - (1/2π)μ1I lnr Az2 = - [I/(4π2a2)] μ2{ 2π a2lna + π(r2-a2) } + [I/2π] (μ2- μ1) lna = - [I/(4π2a2)] μ22π a2lna - [I/(4π2a2)] μ2 π(r2-a2) + [I/2π] (μ2- μ1) lna = - [I/(2π)] μ2 lna - [I/(4πa2)] μ2 (r2-a2) + [I/2π] (μ2- μ1) lna = -[I/2π]μ1 lna - [I/(4πa2)] μ2 (r2-a2) Say it again Az(r) = - (1/2π)μ1I lnr r > a Az(r) = -[I/2π]μ1 lna - [I/(4πa2)] μ2 (r2-a2) r < a Now how does this compare with my Appendix B result with the surface current thing? There I got Az(r>a) = -[I/(2π)] { μ1ln(r) } // which agrees! Az(r<a) =-[I/(2π)] { μ1 ln(a) + μ2(r2-a2)/(2a2) ) } // which agrees Wow! I have resolved the paradox, and now have a lot to ponder. Why does the assumption of a surface magnetization current give the same answer??? [ OK, THIS section has the main and correct idea. I just had trouble accepting it. I was hoping those King Helmholtz integrals didn't need adder homo solutions I guess. ] November 19, 2013 Creation of Appendix B Version 2. I made a copy of Appendix B and am now editing it, making corrections. This results in a slightly different formula for Kzm. Section B.1: I now make clear that Kzfree = 0 and only Kzm exists at the boundary and that is why you are allowed to say Hx2 = Hx1 [true]. Then I use the Maxwell equation version curl B = μ0(jωD + Jc + Jm) in which form we can say that "B sees all currents" and they all have a common μ0 factor. I derive this equation carefully in Section 1.1 as (1.1.24) [correct] . This is different from what I had before for curl B. I then end up with a different result equation which is now Kzm = [(μ2-μ1)/μ0] Hx2 = - [(μ2-μ1)/μ0] Hθ . My result for the simple round wire (B.1.11) is now different from what it was before. [ The new Kzm expression is the correct one. But I never realized Jm exists everywhere, not just on the boundary. ] [ Section B.1 is interesting about the surface mag current, but I am not sure it plays a role in the solution later in the section. ] Section B.2. My derivation of H(x,y) = - ∫d2x' ln(R2) curl' J(x') which is unchanged. Section B.3. Unchanged except the formula for Kzm now has the new form which still includes the factor (μ2-μ2). [ I think my General Method fails because including Jm in the wave equation is wrong in the first place, so why have a fancy General Method to find the surface part of Jm ? ] Section B.4: This is the simple-method way to compute Hθ for round wire and then I use the new Kzm formula and I restate Kzm for the round wire. Only thing changed here is formula for Kzm. [ but does anybody care about this surface current? ] Section B.5. This is my very long and detailed calculation of Hθ for the round wire with uniform current which makes use of my fancy formula H(x,y) = ∫d2x' [ ln(1/R) ] curl' J(x'). I compute the curl J thing for the wire and insert it. I renamed the integral C instead of K to avoid confusion. This section makes no mention of surface current and is therefore completely unchanged except for name C. [ again, I don't think anybody really cares about this surface current computation method. Perhaps it is interesting that you can compute the surface current even if we have no use for it! ] Section B.6: The big change here is that I assume without justification that 22DA(x) = -μ0Jm(x) -μ2Jc(x) . [ wrong and unjustifiable starting point ! ] where I have μ0 in the first term whereas I used to have μ1. Later I will try to justify this equation! [ I never did ] There are two issues: (1) why μ0 for Jm ? (2) why should you be including Jm in the first place? Given this new equation, I use Appendix I to write down the particular solution (B.6.1), nothing new here. But now when I insert the new form for Kzm and account for the new presence of μ0, these two changes cancel and I get the exact same result for Az that I got before. This then yields the expected H and B results for the round wire, both inside and outside! So luckily, most of this Section B.6 is the same including the Maple plot. Section B.7 is my reader exercise. Only change I made is μ1→μ0 at the start. This concludes my major edit of Appendix B in its separate Version 2 doc. [ This Appendix B has to be completely redone, even its purpose now is not clear. ] Here then is the major issue: I want to justify this fact 2A(x) = -μ0Jm(x) -μ2Jc(x) for all of region R Somehow this is the proper "union" of region 1 and region 2. What it says is that A "sees" both Jc and Jm and that is why B will see both as well. [ call it coincidence. ] I think a similar situation arises in my Jackson ε problem, but I have not cleaned that up yet. So here is the Theorem I want to somehow derive: Theorem: We know that separately in our two regions we have (2 + β12) A = 0 x in region 1 (2 + β12) A = - μ2 Jc2 x in region 2 We know that there is a magnetization current Jm on the boundary where [ ignoring other Jm] Kzm = - [(μ2-μ1)/μ0] Hθ = - [(μ2-μ1)/μ0] I/(2πa) amp/m (B.4.4) Kzm = Kzm (B.4.5) Jm = Kzmδ(r-a) amp/m2 . (B.4.6) We know we can find the solution in the two regions by using Smythian forms in each region and matching boundary conditions to find the unknown assumed Smythian form constants, and we know that in this method we have a homo solution in each region as part of the Smythian form. In fact, in region 1 the entire solution is obviously a homo solution since (2 + β12) A = 0. [ all correct ] The claim of the Theorem is that there is an alternate solution method [yes, that is the claim] . In that alternate method, you take the union of the two regions to form region R, and you then have this unified equation for region R (2 + β12) A = -μ0Jm - μ2Jc2 x in region R where now A is simply the particular integral of the above equation, where we have ADDED the mag current on the right side with a μ0 multiplier. The resulting A is valid everywhere in R, which means it is valid in both region 1 and region 2. To "prove" the theorem, we have to explain why this unified region R problem is the same problem! I think this explanation may involve the use of the word "image". [ perhaps ] Viewed from either side, if we add an "image current" Kz = Kzm on the boundary, we get the same solution A that we get using the Smythian method for both regions outlined above. [ I don't think this idea pans out. ] Comparing Problem I and Problem II [ reviewed ] Suppose we claim this Theorem as an ansatz [ OK, can do that] , and then see where it leads. Let's call this situation with the image current to be Problem II. We want to somehow show that Problem II is the same as Problem I where there is no image current and where we solve the regions separately using Smythian forms [ I think I am sort of on the right track here in comparing these two problems (ie, with and without the added-in surface charge). But I could not drive it to any useful conclusion. ] (1) First, if we stay away from the boundary, we know that Problem 2 is described by (2 + β12) AII = 0 x in region 1 PDE set II (2 + β12) AII = - μ2 Jc2 x in region 2 And we know that these equations are the same as those for Problem 1 which we can write as (2 + β12) AI = 0 x in region 1 PDE set I (2 + β12) AI = - μ2 Jc2 x in region 2 (2) If we then focus on the boundary region, we know for Problem 2 (AzII)2 = (AzII)1 BC set II (1/μ2) (∂nAzII)2 - (1/μ1) (∂nAzII)1 = Kzimage whereas for Problem 1 we have instead (AzI)2 = (AzI)1 BC set I (1/μ2) (∂nAzI)2 - (1/μ1) (∂nAzI)1 = 0 Summary Problem I: (2 + β12) AI = 0 x in region 1 PDE set I (2 + β12) AI = - μ2 Jc2 x in region 2 (AzI)2 = (AzI)1 BC set I (1/μ2) (∂nAzI)2 - (1/μ1) (∂nAzI)1 = 0 Problem II (2 + β12) AII = -μ0Jm - μ2 Jc2 x in region R // image current present! (2 + β12) AII = 0 x in region 1 PDE set II (2 + β12) AII = - μ2 Jc2 x in region 2 (AzII)2 = (AzII)1 BC set II (1/μ2) (∂nAzII)2 - (1/μ1) (∂nAzII)1 = Kzimage AII = ∫R dV' [-μ0Jm - μ2 Jc2] e-jβR/R = the particular solution (3) Now we might conjecture that, for the entire region R, we can write AII - AI = Ahomo where the following is true (2 + β12) Ahomo = 0 x in region 1 (2 + β12) Ahomo = 0 x in region 2 This would then allow PDE set I and PDE set II to both be satisfied even though AI ≠ AII. (4) Now consider (1/μ2) (∂nAzII)2 - (1/μ1) (∂nAzII)1 = Kzimage (1/μ2) (∂nAzI)2 - (1/μ1) (∂nAzI)1 = 0 If we subtract these equations (and also the first equations) we find (Azhomo)2 = (Azhomo)1 (1/μ2) (∂nAzhomo)2 - (1/μ1) (∂nAzhomo)1 = Kzimage So far then we seem to have identified the following description of this homo solution (2 + β12) Ahomo = 0 x in region 1 (2 + β12) Ahomo = 0 x in region 2 (Azhomo)2 = (Azhomo)1 (1/μ2) (∂nAzhomo)2 - (1/μ1) (∂nAzhomo)1 = Kzimage = [(μ2-μ1)/μ0] Hx2 = (Bx2 - Bx1)/μ0 Suppose we could find a solution Azhomo that solves the above stated boundary value problem. Then this Azhomo expresses the difference between the Problem I and Problem II solutions. If we could show that this homo solution created zero B field, then maybe we could say the problems are "the same". But it seems to me that a surface charge Kzimage is not going to have a 0 B field! But go back to the picture If there is a Kzimage and if we know that A = Az(y,z) , then I show in (1.1.46) that Bhomo = Bxhomo = (∂yAzhomo) so I conclude that such a Kzimage really does create a Bhomo field which is non-zero. But then (Azhomo)2 = (Azhomo)1 (1/μ2) (∂yAzhomo)2 - (1/μ1) (∂yAzhomo)1 = Kzimage = [(μ2-μ1)/μ0] Hx2 = (Bhomox2 - Bhomox1)/μ0 so (Azhomo)2 = (Azhomo)1 (1/μ2) Bxhomo2 - (1/μ1) Bxhomo1 = (Bhomox2 - Bhomox1)/μ0 which just says Bxhomo2 is some multiple of Bxhomo1 and does not say both are 0. Conclusion so Far: I wanted to show that Problem I and Problem II were "the same problem", but using the thread above, I end up with Problem I and Problem II having different B fields, so how can they be the same problem? Idea during Review: Can you show directly somehow that A = (1/2π) ∫dV' [μ0Jm(x') + μ2Jc(x')] ln(1/R) satisfies the two BC's? That is to say, apply ∂r to this Helm integral and somehow compute ∂rAz on the two sides. If you could show that, then you know that the first Jm term only creates homo solutions on the two sides, so then you could say it is creating exactly the right homo terms needed to meet the BC's ! I will pursue this a bit a little later, though it is not necessary./ ] Lets look again at the Smythian solution for Problem I found yesterday. In region 1 we have (2 + β12) AzI = 0 x in region 1 AzI = Az(r) = Azhomo1(r) = A lnr A = to be determined In region 2 we have (2 + β12) AzI = = - μ2 Jc2 x in region 2 AzI,part = - [I/(4π2a2)] { μ2∫dV' ln(R2) } = - [I/(4π2a2)] μ2{ 2π a2lna + π(r2-a2) } AzI = AzI,part + AzI,homo2 AzI,homo2 = d d = to be determined There is no free surface charge at the boundary, so our two boundary conditions are (AzI)2 = (AzI)1 (1/μ2) (∂nAzI)2 - (1/μ1) (∂nAzI)1 = 0 From these two conditions we learn that A = - (1/2π)μ1I d = [I/2π] (μ2- μ1) lna . Therefore we have Region 1: AzI = Az(r) = Azhomo1(r) = A lnr = - (1/2π)μ1I lnr Region 2: AzI = AzI,part + AzI,homo2 = - [I/(4π2a2)] μ2{ 2π a2lna + π(r2-a2) } + d = - [I/(4π2a2)] μ2{ 2π a2lna + π(r2-a2) } + [I/2π] (μ2- μ1) lna = -[I/2π]μ1 lna - [I/(4πa2)] μ2 (r2-a2) and once again our Problem I solution is : Region 1: AzI = - (1/2π)μ1I lnr Region 2: AzI = -[I/2π]μ1 lna - [I/(4πa2)] μ2 (r2-a2) Compare this to the Problem II solution In Region R = 1 + 2 we have (2 + β12) AII = -μ0Jm - μ2 Jc2 x in region R // image current present! (AzII)2 = (AzII)1 BC set II (1/μ2) (∂nAzII)2 - (1/μ1) (∂nAzII)1 = Kzimage = [(μ1-μ2)/μ0] I/(2πa) Now the - μ2 Jc2 source gives the exact same particular solution as obtained in Problem I, so AzII,parta = - [I/(4π2a2)] { μ2∫dV' ln(R2) } = - [I/(4π2a2)] μ2{ 2π a2lna + π(r2-a2) } But then we have part b of this particular solution which is the -μ0Jm contribution AzII,partb = - [I/(4π2a2)] {(μ1-μ2) (a/2) 4πa = - [I/(2π)] (μ1-μ2) = - [I/(2π)] (μ1-μ2) ln(r>). Each of these solutions is defined over all R. When we add these solutions we get AII = AzII,parta + AzII,partb = - [I/(4π2a2)] μ2{ 2π a2lna + π(r2-a2) } - [I/(2π)] (μ1-μ2) ln(r>) = - [I/(2π)] μ2{ lna + (1/2)(r2/a2-1) } - [I/(2π)] (μ1-μ2) ln(r>) = - [I/(2π)] { μ2 lna - (μ1-μ2) ln(r>) + (μ2/2)(r2/a2-1) } which we can write as AII = - [I/(2π)] { μ2 lna - (μ1-μ2) ln(r) + (μ2/2)(r2/a2-1) } r>a AII = - [I/(2π)] { μ2 lna - (μ1-μ2) ln(a) + (μ2/2)(r2/a2-1) } r>a Pause for Back and Fill I have "carefully" written up the two solution methods for the round wire problem in "Case Study, A for a round wire.doc". I show there all the calculation except the two integrals which are done in Appendix B. I show that the two methods give exactly the same solution. But I don't know WHY they give the same solution. Plan A: Superposition? Suppose we had nothing but a real surface charge Kz and magically we could set Jc = 0 while maintaining that Kz . What would be the solution in that case? It would be the Z integral in my Case Study doc. AII,regionR = - (I/2π) (μ1-μ2) The two media still have μ1 and μ2 , and Kz still vanishes when μ1 = μ2. [ notice that the solution is a homo solution on both sides! ] Plan B: The Stakgold v function idea. [ Well, this is more a method to compute true Green's Functions, it does not really help here, but it was OK to review it. I am only interested in using the free-space Green's Function. ] In this idea, you have some boundaries and a Green's function g which vanishes on same. Then you define v = g - E = full Green's minus free-space Green's. Then you ask: what do I know about the function v? One thing you know about v is that on the surface, its value is given by v = -E ! And you know E, so you then know v = -E on the surface. But then you are in a Dirichlet situation with a known v on the surface. Let's put more flesh on these bones. We have these Green problems with point charge at x : -2g = δ(x-ξ) with g = 0 on σ, g = full Green's inside region R -2E = δ(x-ξ) with E = 0 at ∞ and symmetric, E = free-space Green's Here is the regular Dirichlet solution: -2u = q => u = ∫R dV g q + ∫σ dSξ f ∂ng f(ξ) = u(ξ) = prescribed Now what do we know about v ? v = g - E -2v = 0 inside region R vx(ξ) = - E(ξ|x) and this is known if we know E and we know σ Now, how do you solve the following problem: -2v = 0 in region R vx(ξ) = - E(ξ|x) on boundary σ or region R It seems to me this is a simple Laplace Dirichlet problem. To solve such a problem, we need the full Green g such that -2g = δ(x-ξ) with g = 0 on σ, g = full Green's inside region R Then we can solve -2v = 0 by saying v = ∫dV g (q=0) + ∫dSξ vx(ξ) ∂ng or vx(x) = ∫σdSξ vx(ξ) ∂ng or vx(x) = ∫σdSξ [- E(ξ|x)] ∂ng Now in the Green problem -2g = δ(x-ξ) with g = 0 on σ, g = full Green's inside region R we know that ∂ng(x|ξ) is the charge induced on surface σ. But I don't know where to go with this idea. I think Stak wanted you to take this system -2v = 0 in region R vx(ξ) = - E(ξ|x) on boundary σ or region R and solve it somehow without using the Green's Function method, to get function v. Then you would have calculated g as g = E + v. Now with this warmup, go find this in the Stak notes empire. It relates to Section 6.7 (1). I read this a bit, it relates to integral equations for ∂ng and Stak refers to -∂ng as function I and we then get an integral equation of the form (6.97) when you take s→ x. If you could really solve this integral equation for I, then we have a second equation (6.96) which then tells you g. So in theory, this is jus one of Stak's methods of computing g functions. I don't see it having any bearing on what I am trying to do today. But I am grasping at any straw I can remotely think of. Plan C: Boundary Conditions on A This "plan" is in the form of a question. I have shown that in Method I and Method II, I get exactly the same A fields on the two sides of the boundary. Here are those A fields: AI,region1(r) = -μ1 (I/2π)lnr AI,region2(r) = -(I/2π) [ μ1 lna + (1/2)μ2(r2/a2-1) Method I AII,region1(r) = -μ1 (I/2π)lnr AII,region2(r) = -(I/2π)[ μ1lna + μ2 (1/2)(r2/a2-1) ] Method II Well, if that is the case, how can it be that there is a "jump" of Kz in Method II, but the jump is 0 in Method I?? Answer: Well, as I look back at the Method II solution, I see that although I wrote down a "jump by Kz " statement as a BC, I never used this BC and in fact the BC is wrong. There is in fact no "jump" and that is in agreement with the Method I solution. This strengthens the argument that my surface charge Kz is in fact not a "free" one, because if it were free, we would have to have a jump by Kz. [ OK, here I just corrected an error and after correction, things make sense. ] Plan D: Ansatz to simulate correct BC's and Stak surface layer limits [ There is something to this I think, and I may be back on this tack later if I try to directly show that the adder Jm makes the full particular integral satisfy BC's. I ignore this section for the time being. ] Can I say that adding Jm is just an ansatz, which then gives something that "simulates" the right boundary conditions? Az(r) = (1/2π) ∫dV' [μ2Jc(x') + μ0Jm(x')] ln(1/R) ansatz adder Az(r) = (1/2π) ∫dV' μ2Jcz(x') ln(1/R) + (1/2π) ∫dA' μ0 Kz(x')] ln(1/R) Az(r) = -(1/4π) ∫dV' μ2Jcz(x') ln(R2) - (1/4π) ∫dA' μ0 Kz(x')] ln(R2) In 2D I guess we have R2 = r2 + r'2 - 2rr'cos(θ-θ') ∂rR2 = 2r - 2r'cos(θ-θ') ∂rlnR2 =(1/R2) ∂rR2 = 2[r-r'cos(θ-θ')]/[ r2 + r'2 - 2rr'cos(θ-θ') ] = smooth function Now suddenly and amazingly, I am reminded of all Stak's painstaking work with those "layers". From my meta meta notes on Ch 6 we have this: u(s) = ∫σ dSξ a(ξ) E(s|ξ) (**) u = Ea Fred 1 ∂νu(x) = ∫σ dSξ a(ξ) ∂νE(x|ξ) value for x not on σ (***) limx→s+[∂νu(x)] = ∫σ dSξ a(ξ) ∂νE(x|ξ) - a(s)/2 "the extra term pops out" I think moreover that if you approach from the other side you get limx→s-[∂νu(x)] = ∫σ dSξ a(ξ) ∂νE(x|ξ) + a(s)/2 and that then jump [∂νu(x)] = a(s) ignoring sign Now look at my Az(r) equation and write it this way Az(r) = (1/2π) ∫dV' μ2Jcz(x') ln(1/R) + (1/2π) ∫dA' μ0 Kz(x') ln(1/R) Az(r) = μ2 ∫dV' Jcz(x') E(x|x') + μ0∫σ dξ Kz(ξ) E(x|ξ) I suspect that the volume integral here does not create that extra term, so I think we can say ∂rAz(r) = μ2 ∫dV' Jcz(x') ∂rE(x|x') + μ0∫σ dξ Kz(ξ) ∂rE(x|ξ) limr→a+ [∂rAz(r)] = μ2 [∫dV' Jcz(x') ∂rE(x|x')]r=a + μ0[∫σ dξ Kz(ξ) ∂rE(x|ξ)]r=a – μ0Kz(s)/2 limr→a- [∂rAz(r)] = μ2 [∫dV' Jcz(x') ∂rE(x|x')]r=a + μ0[∫σ dξ Kz(ξ) ∂rE(x|ξ)]r=a + μ0Kz(s)/2 and then limr→a+ [∂rAz(r)] – limr→a- [∂rAz(r)] = μ0Kz(s) = (μ1-μ2) I/(2πa) This looks good but I am missing the (1/μ1) and (1/μ2) factors. But I think I am definitely ON to something here. The idea is that by doing the ansatz of adding that surface current, you do in fact "simulate" the boundary conditions and therefore that is WHY Method 2 works! Maybe I have to generalize this Stak idea to a boundary with μ1 on one side and μ2 on the other side. BUT, the boundary condition I want to simulate is THIS: (1/μ1) limr→a+ [∂rAz(r)] – (1/μ2) limr→a- [∂rAz(r)] = 0 !!! So not only am I messing the factors, I am getting Kz instead of getting 0 . Oops. Plan D1: Try again: Az(r) = μ2 ∫dV' Jcz(x') E(x|x') + μ0∫σ dξ Kz(ξ) E(x|ξ) Az(r) = μ2 ∫dV' Jcz(x') E(x|x') + μ0∫σ dξ Kz(ξ) E(x|ξ) Suppose you DO get some kickout from the first term as well. Let's just wing it and see what might happen ∂rAz(r) = μ2 ∫dV' Jcz(x') ∂r E(x|x') + μ0∫σ dξ Kz(ξ) ∂r E(x|ξ) limr→a+ [∂rAz(r)] = μ2 ∫dV' Jcz(x') {∂r E(x|x')}r=a - (1/2)μ2 ∫dr' Jcz(x') + μ0∫σ dξ Kz(ξ) {∂rE(x|x')}r=a - (1/2)μ0 Kz(a) limr→a- [∂rAz(r)] = μ2 ∫dV' Jcz(x') {∂r E(x|x')}r=a- + (1/2)μ2 ∫dr' Jcz(x') + μ0∫σ dξ Kz(ξ) {∂rE(x|x')}r=a + (1/2)μ0 Kz(a) limr→a+ [∂rAz(r)] = μ2 ∫dV' Jcz(x') {∂r E(x|x')}r=a - (1/2)μ2 ∫dr' Jcz(x') + μ0∫σ dξ Kz(ξ) {∂rE(x|x')}r=a - (1/2)μ0 Kz(a) = (main stuff) - (1/2)μ2 ∫dr' Jcz(x') - (1/2)μ0 Kz(a) = (main stuff) - (1/2)μ2 a (I/2πa2) - (1/2)(μ1- μ2) I/(2πa) ?? = (main stuff) - (1/2)μ1 a (I/2πa2) ?? And somehow maybe we can arrange limr→a- [∂rAz(r)] = μ2 ∫dV' Jcz(x') {∂r E(x|x')}r=a - (1/2)μ2 ∫dr' Jcz(x') + μ0∫σ dξ Kz(ξ) {∂rE(x|x')}r=a + (1/2)μ0 Kz(a) where only one of the signs changes for some reason. Then we have.. No, this is never going to give the BC I am seeking which is this: (1/μ2) (∂rAz)2 = (1/μ1) (∂rAz)1 Somehow there must be a way to show that this result can be obtained from Az(r) = μ2 ∫dV' Jcz(x') E(x|x') + μ0∫σ dξ Kz(ξ) E(x|ξ) Kz = - [(μ2-μ1)/μ0] I/(2πa) I obtain the result if I compute all the integrals, but I want to show it is true without doing the integrals! Simpler: Az(r) = μ2 (I/πa2) ∫dV' E(x|x') + μ0 (μ1-μ2) I/(2πa) ∫σ dξ E(x|ξ) At least it has both μ1 and μ2 present somewhere. Maybe it just a coincidence with the round wire that it works. Status at 7:30 PM 11.19.13. I am beginning to think it is just a coincidence that adding μ0Jm to the particular integral for the full range R just happens to give the exact correct answer. That would be an evil connivance of nature to lead me off on several days of trying to show it was a general rule. After several days, I can think of no "justification" for adding this Jm to the particular integral. If I go ahead and add it, then it just happens to give the exact right answer for the round wire. A huge Red Herring for me. I am unable to show that adding this as an ansatz "simulates" the correct BC's. This simulation fact only comes out at the very end when everything is evaluated. That was a nice idea, a hopeful one, but it just did not pan out, even with Stak's fancy help. If this is correct that it is just a coincidence, then only Method I is valid. The reader is then instructed to do a Smythian fit solution to both regions and match the BC's. Signoff 8 PM. [ review complete to here ] 11.20.13 Back At It Again: The new μ0 gauge idea. [ Here I start over doing wave equations with curl B = μ0(∂tD + Jc + Jm) which exposes Jm. When I did this, I was not aware that Jm exists everywhere and not just on the wire surface. I was surprised that I could in fact come up with a full set of A and φ wave equations by this method, using the King μ0 gauge as I now call it. That is to say, the φ operator is the same as the A operator. In retrospect, this is a lousy gauge because you have to then know all about Jm a priori in order to do anything! ] Plan E1: Question: What do you have to do to cause Jm to honestly appear as a source in a wave equation for Az ? This is my μ0 gauge approach. I have taken this path at least twice, but not in either of these Disaster logs. I see it is in the main edit log, so I will just copy that here. Right now we could be in either region 1 or region 2: curl B = μ0(∂tD + Jc + Jm) " B sees all currents" (1.1.9g)m curl curl A = μ0(∂tD + Jc + Jm) grad divA - 2A = μ0(∂tD + Jc + Jm) grad divA - 2A = μ0(ε ∂tE + Jc + Jm) grad divA - 2A = μ0(ε ∂t[- grad φ - ∂tA] + Jc + Jm) grad divA - 2A = μ0ε ∂t[- grad φ - ∂tA] + μ0Jc + μ0Jm grad divA - 2A = - μ0ε ∂tgrad φ - μ0ε ∂t2A + μ0Jc + μ0Jm grad divA - 2A = - μ0ε grad ∂t φ - μ0ε ∂t2A + μ0Jc + μ0Jm -grad divA + 2A = + μ0ε grad ∂t φ + μ0ε ∂t2A - μ0Jc - μ0Jm 2A - μ0ε ∂t2A = grad divA + μ0ε grad ∂t φ - μ0Jc - μ0Jm 2A - μ0ε ∂t2A = grad [divA + μ0ε ∂t φ] - μ0(Jc+Jm) (1.3.3) OK, once "price" you pay with this method is that you get μ0ε on that ∂t2A part of the operator. I accept that as just fine for the moment. Now before stating the gauge condition, let's apply this to region 1 (dielectric which conducts). As a first possible choice, suppose we decide to absorb the conduction of region 1 into the operator somehow. Then we do these steps: 2A - μ0ε1 ∂t2A = grad [divA + μ0ε1 ∂t φ] - μ0(Jc1+Jm1) region 1 2A - μ0ε1 ∂t2A = grad [divA + μ0ε1 ∂t φ] - μ0σ1E - μ0Jm1 2A - μ0ε1 ∂t2A = grad [divA + μ0ε1 ∂t φ] - μ0σ1[- grad φ - ∂tA] - μ0Jm1 2A - μ0ε1 ∂t2A - μ0σ1∂tA = grad [divA + μ0ε1 ∂t φ] + μ0σ1grad φ - μ0Jm1 2A - μ0ε1 ∂t2A - μ0σ1∂tA = grad [divA + μ0ε1 ∂t φ + μ0σ1φ] - μ0Jm1 OK, now this then is our choice of gauge divA = - μ0ε1 ∂t φ - μ0σ1φ so it is a King gauge but with μ0 in place of the former μ1. A certain niceness in that since maybe it is more applicable to both the μ1 and μ2 regions? Well, just continue on with this gauge to get 2A - μ0ε1 ∂t2A - μ0σ1∂tA = - μ0Jm1 (2 - μ0ε1 ∂t2 - μ0σ1∂t)A = - μ0Jm1 region 1 Here I have just assumed there is some surface mag current Jm1 which lies in region 1. I will return to that question later. Notice that, in order to have Jc1 disappear, we needed μ0Jc1 to appear in the sources. So that is half the story. Now let's do region 2. Start off with 1.3.3 but in region 2: 2A - μ0ε2 ∂t2A = grad [divA + μ0ε2 ∂t φ] - μ0Jc2 - μ0Jm2 (1.3.3) Now the safest option it seems to me is to use the same gauge here, so that we at least have a CHANCE of having some kind of unified equation for all of R with a uniform gauge. Also, in region 2 I want to leave the current Jc "as is". Then we can do steps. First step is to insert the above gauge, 2A - μ0ε2 ∂t2A = grad [- μ0ε1 ∂t φ - μ0σ1φ + μ0ε2 ∂t φ] - μ0Jc2 - μ0Jm2 2A - μ0ε2 ∂t2A = grad [ μ0(ε2-ε1) ∂tφ - μ0σ1φ] - μ0Jc2 - μ0Jm2 2A - μ0ε2 ∂t2A = [ μ0(ε2-ε1) ∂t - μ0σ1] gradφ - μ0Jc2 - μ0Jm2 2A - μ0ε2 ∂t2A = [ μ0(ε2-ε1) ∂t - μ0σ1] [-E - ∂tA] - μ0Jc2 - μ0Jm2 Now region 2 is our good conductor so try setting E = 0 to get 2A - μ0ε2 ∂t2A = [ μ0(ε2-ε1) ∂t - μ0σ1] [ - ∂tA] - μ0Jc2 - μ0Jm2 2A - μ0ε2 ∂t2A = - [ μ0(ε2-ε1) ∂t - μ0σ1] [ ∂tA] - μ0Jc2 - μ0Jm2 2A - μ0ε2 ∂t2A = - [ μ0(ε2-ε1) ∂t2A - μ0σ1 ∂tA] - μ0Jc2 - μ0Jm2 2A - μ0ε2 ∂t2A = - μ0(ε2-ε1) ∂t2A + μ0σ1 ∂tA] - μ0Jc2 - μ0Jm2 Now notice that the μ0ε2 ∂t2A cancels on both sides which we like to see, and then we can move the μ0ε1 term to the left side, giving 2A - μ0ε1 ∂t2A = + μ0σ1 ∂tA - μ0Jc2 - μ0Jm2 2A - μ0ε1 ∂t2A - μ0σ1 ∂tA = - μ0Jc2 - μ0Jm2 (2 - μ0ε1 ∂t2 - μ0σ1 ∂t)A = - μ0Jc2 - μ0Jm2 region 2 Here then are the results for this pathway: (2 - μ0ε1 ∂t2 - μ0σ1∂t)A = - μ0Jm1 region 1 (2 - μ0ε1 ∂t2 - μ0σ1 ∂t)A = - μ0Jc2 - μ0Jm2 region 2 Comments on this approach: 1) the gauge is exactly the same in both regions 2) the wave operator is exactly the same in both regions 3) the surface [ part of the ] mag current Jm appears as a bona fide source in each region 4) current Jc2 has a multiplier of μ0 as does the surface current [ not very nice ] 5) the gauge is DIFFERENT from that I was using before, since has μ0 Now it certainly seems a logical step at this point to combine the two regions. Maybe this is wrong, but if we do it, then we can combine the Jm's into a single Jm. We then don't have to decide how much of Jm is in each region, only the total Jm counts. We then get (2 - μ0ε1 ∂t2 - μ0σ1 ∂t)A = - μ0Jc2 - μ0Jm region R where it is understood that Jm only exists on the boundary and Jc2 exists only in region 2. [ With the full interpretation of Jm, this is a perfectly fine wave equation for region R! ] Plan E2: φ in the same gauge Just as an aside, with this gauge, what does the φ wave equation look like? It might be completely different for all I know right now. So go back to the source in lines doc. Start in a generic unspecified region E = - grad φ - ∂tA // (1.3.1) [= Jackson (6.9)] div E = - div grad φ - ∂t (div A) // take div of both sides ρ/ε = - 2φ - ∂t (div A) 2φ = - ∂t (div A) - ρ/ε Note: ρ includes both free and polarization charge!!! Now use the same King gauge we used for the magnetic stuff, divA = - μ0ε1 ∂t φ - μ0σ1φ and we then get 2φ = - ∂t (- μ0ε1 ∂t φ - μ0σ1φ) - ρ/ε 2φ = ∂t ( μ0ε1 ∂t φ + μ0σ1φ) - ρ/ε 2φ = ( μ0ε1 ∂t2 φ + μ0σ1∂t φ) - ρ/ε 2φ = μ0ε1 ∂t2 φ + μ0σ1∂t φ - ρ/ε 2φ - μ0ε1 ∂t2 φ- μ0σ1∂t φ = - ρ/ε (2 - μ0ε1 ∂t2- μ0σ1∂t) φ = - ρ/ε Now let's talk regions. But nothing depends on region except ρ and ε! Things are a little different in the electric case, but let's just move forward: (2 - μ0ε1 ∂t2- μ0σ1∂t) φ = - ρ1/ε1 region 1 (2 - μ0ε1 ∂t2- μ0σ1∂t) φ = - ρ2/ε2 region 2 The very happy news so far is that the wave operator here is the same as for the A equations. Now let's take the step if we like and unify into a single equation (2 - μ0ε1 ∂t2- μ0σ1∂t) φ = - ρ1/ε1- ρ2/ε2 region R Now probably we would say at this point that all the charge density lies on the outside surface of the conductor, which says ρ2 = 0. That is how we always "draw it". Unlike the Jm case, we seem to get some ambiguity here depending on where you imagine ρ to lie. Where does polarization charge fit into this picture? Let's defer this question for a moment, other than to say both ρ1 and ρ2 would include any polarization charges in their respective regions. Comments on the result: 1) the gauge is exactly the same in both regions 2) the wave operator is exactly the same in both regions and it is the exact same as in the A equation 3) the gauge is DIFFERENT from that I was using before, since has μ0 So here is a possibly valid summary of our results: (2 - μ0ε1 ∂t2 - μ0σ1∂t)A = - μ0Jm1 region 1 (2 - μ0ε1 ∂t2 - μ0σ1∂t)A = - μ0Jc2 - μ0Jm2 region 2 (2 - μ0ε1 ∂t2- μ0σ1∂t) φ = - ρ1/ε1 region 1 (2 - μ0ε1 ∂t2- μ0σ1∂t) φ = - ρ2/ε2 region 2 (2 - μ0ε1 ∂t2 - μ0σ1∂t)A = - μ0Jc2 - μ0Jm region R (2 - μ0ε1 ∂t2- μ0σ1∂t) φ = - ρ1/ε1- ρ2/ε2 region R This is mildly encouraging in that things are fairly uniform. But I fear there would be problems. First, let's take all these equations to ω space and write β012(ω) = μ0ε1ω2 - μ0σ1jω = ω2μ0 (ε1-jσ1/ω) = ω2μ0ξ1  ξ1 ≡ ε1 - jσ1/ω β012(ω) = μ0ε1ω2 - μ0σ1jω = ω(μ0ε1ω - j μ0σ1) There is no question but that β012(ω) = 0 when ω = 0. So we then have these Helmholtz equations. (2 + β012)A = - μ0Jm1 region 1 (2 + β012)A = - μ0Jc2 - μ0Jm2 region 2 (2 + β012)φ = - ρ1/ε1 region 1 (2 + β012)φ = - ρ2/ε2 region 2 (2 + β012)A = - μ0Jc2 - μ0Jm region R (2 + β012)φ = - ρ1/ε1- ρ2/ε2 region R Now at DC, we have β01 = 0 and then we have in particular this equation 2A = - μ0Jc2 - μ0Jm region R divA = - μ0ε1jωφ - μ0σ1φ = -μ0(ε1 - jσ1/ω)jωφ = -μ0 ξ1jωφ I would like now to do the round wire computation in this new gauge and see what happens! Plan E3. Do the DC round wire in this new gauge. Use Appendix B as a guide. [ This section is completely wrong because I am including only the surface part of Jm in a calculation that requires all of Jm! ] Start off with these modified equations: 22DA(x) = -μ0Jm(x) -μ0Jc(x) . (B.6.1a) Az = -(1/4π) { μ0∫dS' [Kzm(x')] ln(R2) + μ0∫dV'[Jcz(x')] ln(R2) } + H. (B.6.1b) We now insert the uniform prescribed current Jcz and our previously computed surface current Kz, Jcz(x') = Jcz = I/(πa2) // uniform in round wire (B.6.2) Kzm(x') = Kzm = - [(μ2-μ1)/μ0] I/(2πa) // from (B.4.4) (B.6.3) to get Az = - [I/(4π2a2)] { (μ1-μ2) (a/2) ∫dS' ln(R2) + μ0∫dV' ln(R2) } + H. (B.6.4) It seems that everything stays the same except for this μ2 → μ0 in the volume term, so I should be able to jump far ahead in the calculation: ∫dV' ln(R2) = (B.6.14) ∫dS' ln(R2) = 4πa . (B.6.15) The above integrals are "just geometry". We then get Az = - [I/(4π2a2)] {(μ1-μ2) (a/2) ∫dS' ln(R2) + μ0∫dV' ln(R2) } + H. (B.6.4) = - [I/(4π2a2)] {(μ1-μ2) (a/2) 4πa + μ0} + H (B.6.16) Then write these out for the two separate regions : Az(r>a) = -[I/(4π2a2)] { μ1-μ2) (a/2) 4πa ln(r) + μ0 2πa2 ln(r) } = -[I/(4π2a2)] { μ1-μ2) 2πa2 ln(r) + μ0 2πa2 ln(r) } = -[I/(2π)] (μ1-μ2+μ0) ln(r) + H(r>a) region 1,r>a (B.6.17) Az(r<a) = -[I/(4π2a2)] { (μ1-μ2) (a/2) 4πa ln(a) + μ0( 2πa2ln(a) + π(r2-a2) ) } = -[I/(4πa2)] { (μ1-μ2) 2a2 ln(a) + μ02a2ln(a) + μ0(r2-a2) ) } = -[I/(4πa2)] { (μ1-μ2+μ0) 2a2 ln(a) + μ0(r2-a2) ) } = -[I/(4πa2)] { (μ1-μ2+μ0) 2a2 ln(a) + 2a2μ0(r2-a2)/(2a2) ) } = -[I/(2π)] { (μ1-μ2+μ0) ln(a) + μ0(r2-a2)/(2a2) ) } + H(r<a) region 2, r < a (B.6.18) There results are now different because we are in a different gauge! The next step is to compute the magnetic field B = curl A given our function Az(r). We find from the usual cylindrical coordinates curl formula (B.5.2), B = curl A = - ∂rAz(r) . (B.6.20) At the boundary r = 0 both expressions give Az(r=a) = = -[I/(2π)] { (μ1-μ2+μ0) ln(a) (B.6.19) First, for r > a one finds, -∂rAz(r) = ∂r[[I/(2π)] (μ1-μ2+μ0) ln(r) = [I/(2π)] (μ1-μ2+μ0) (1/r) r > a (B.6.21) and then for r < a, -∂rAz(r) = I/(2π) ∂r {(μ1-μ2+μ0) ln(a) + μ0(r2-a2)/(2a2) ) } = (I μ0/2π) r/a2 r < a (B.6.22) Thus B = [I/(2π)] (μ1-μ2+μ0) (1/r) for r > a which is in the dielectric medium with μ1 B = (I μ0/2π) r/a2 for r < a which is in the conductor medium with μ2 . (B.6.23) Using B = μH in each region we then get H = [I/(2π)] (μ1-μ2+μ0)/μ1 (1/r) for r > a => Hθ = (I/2πr) H = (I μ0/2π)/μ2 r/a2 for r < a => Hθ = (Ir/2πa2) (B.6.24) These are NOT my standard results. Wow. This new "honest gauge" has given a different result! I KNOW that the standard results are right for arbitrary μ1 and μ2 so there is a LOOSE BOLT somewhere in the works that is messing up everything I do. [ As now expected, if you omit the remaining bulk pieces of Jm, you get the wrong answer! ] Let's now look back at the "simple derivation" of these results in Section B.4 [ I keep wanting to verify even the simplest things. That is what you do when you don't know where to go! ] For a round wire of radius a with uniform Jz (as would be the DC case ω = 0) , geometric symmetry makes the calculation of H very easy. One need only apply Ampere's Law separately for a point r outside the wire, and for another point r inside the wire. For the outside case one finds 2πr Hθ(r) = I => Hθ(r) = I/(2πr) r ≥ a . (B.4.1) And then for the inside case the "current enclosed" is determined by a simple area fraction. 2πr Hθ(r) = I (πr2/πa2) => Hθ(r) = I r/(2πa2) r ≤ a . (B.4.2) Ampere's Law is curl H = Jc and is not supposed to "see" the mag charge on the surface. Then we have this integral form, curl H = jωεE + J C H ds = ∫S [jωεE +J] dS (1.1.36) Hmmm. But we are at DC so ω = 0 and then curl H = J C H ds = ∫S J dS (1.1.36) and then we do what is done above and reach the usual conclusions. New Paradox: I have come up with a new gauge for A in which Jm appears in the A wave equation in an "honest way". But in this new gauge, when I compute H or B for a round wire, I get the wrong answer! There is indeed a loose bolt somewhere. My model flunks the simplest n = 2 test you can think of. [ The loose bolt in this case is the omission of the extra Jm stuff in the bulk regions. I never fully did the problem with that extra Jm stuff with success, but I am pretty sure it would succeed. But knowledge a priori of Jm requires knowing H or B, and that requires the problem to already be solved! Catch 22. ] The problem originates right here: 22DA(x) = -μ0Jm(x) - μ0Jc(x) . (B.6.1a) If we replace μ0 → μ2 for the conductor of region 2, then everything comes out right. But the above derivation shows that it really is μ0 that appears in this equation. [ And I know at least in BC terms why you get the right answer with μ0 → μ2 for the round wire, but I have not yet been able to prove this in any general sense. ] Are we not allowed to merge the regions into a single region R? Is that the fallacy? [ no it is not ] Plan E4. Adding homo solutions to new gauge solution [ The plan here was this: still not knowing about the bulk Jm contributions, I use the μ0 gauge to try and compute A. I accept that I might have to add homo solutions to make the BC's work. I add in such homo solutions, and I tune them to make BC's match. But after doing this, I get the wrong answers for the round wire H field. I now think this can never work because you have to include the bulk Jm terms in the calculation, and if you omit them, homo solutions probably won't help. ] Start off with these modified equations: 22DA(x) = -μ0Jm(x) -μ0Jc(x) . (B.6.1a) Az = -(1/4π) { μ0∫dS' [Kzm(x')] ln(R2) + μ0∫dV'[Jcz(x')] ln(R2) } . (B.6.1b) Question: I have just blindly assumed that the particular solution is THE solution. But how do I know that some homo solution might not be needed to be added in to get the correct full solution? Is this my loose bolt in this case? OK, let's go back to the derivation and just add H everywhere. If I do this, I end up with this final result for Az Az(r>a) = -[I/(2π)] (μ1-μ2+μ0) ln(r) + H1(r>a) region 1, r>a (B.6.17) Az(r<a) = -[I/(2π)] { (μ1-μ2+μ0) ln(a) + μ0(r2-a2)/(2a2) ) } + H2(r<a) region 2, r <a (B.6.18) Now maybe I have to select the homo function H such that some known BC's are met. Here are the BC's I know have to be met Az(a)1 = Az(a)2 the vector potential is continuous at the surface (1/μ1) ∂rAz(a)1 = (1/μ2) ∂rAz(a)2 there is no Kzfree at the boundary This second condition says that the slope of Az still jumps, and the reason is that there is as surface mag charge at the boundary which makes it jump. This fact is embedded in the fact that μ1 ≠ μ2. Now recall from above: First, for r > a one finds, -∂rAz(r) = ∂r[[I/(2π)] (μ1-μ2+μ0) ln(r) - ∂r H1(r) = [I/(2π)] (μ1-μ2+μ0) (1/r) - ∂r H1(r) r > a (B.6.21) and then for r < a, -∂rAz(r) = I/(2π) ∂r {(μ1-μ2+μ0) ln(a) + μ0(r2-a2)/(2a2) ) }- ∂r H2(r) = (I μ0/2π) r/a2 - ∂r H2(r) r < a (B.6.22) So now write down the two BC's: Az(a)1 = Az(a)2 -[I/(2π)] (μ1-μ2+μ0) ln(r) + H1(r>a) = -[I/(2π)] { (μ1-μ2+μ0) ln(a) + μ0(r2-a2)/(2a2) ) } + H2(r<a) or -[I/(2π)] (μ1-μ2+μ0) ln(a) + H1(a) = -[I/(2π)] { (μ1-μ2+μ0) ln(a) } + H2(a) To make this work, we just have H1(a) = H2(a) Next, consider -(1/μ1) ∂rAz(a)1 = -(1/μ2) ∂rAz(a)2 which says (1/μ1){ [I/(2π)] (μ1-μ2+μ0) (1/r) - ∂r H1(r) } = (1/μ2) {(I μ0/2π) r/a2 - ∂r H2(r)} or (1/μ1){ [I/(2π)] (μ1-μ2+μ0) (1/a) - ∂r H1(a) } = (1/μ2) {(I μ0/2π)1/a - ∂r H2(a)} Let's rescale the homo function so that H = [I/(2π)] (1/a) h Then our condition is (1/μ1){ [I/(2π)] (μ1-μ2+μ0) (1/a) - [I/(2π)] (1/a)∂r h1(a) } = (1/μ2) {(I μ0/2π)1/a - [I/(2π)] (1/a)∂r h2(a)} Then we can cancel lots of stuff to get (1/μ1){ (μ1-μ2+μ0) - ∂r h1(a) } = (1/μ2) {( μ0) - ∂r h2(a)} or (1/μ1){ (μ1-μ2+μ0) - ∂r h1(a) } = (1/μ2) {( μ0) - ∂r h2(a)} or (1/μ2) ∂r h2(a) - (1/μ1) ∂r h1(a) = (1/μ2) ( μ0) - (1/μ1)(μ1-μ2+μ0) = [μ1μ0 - (μ1-μ2+μ0)μ2 ]/(μ1μ2) = [μ1μ0 - (μ1 μ2-μ22+μ0 μ2) ]/(μ1μ2) = [ μ1μ0 - μ1 μ2+μ22-μ0 μ2) ]/(μ1μ2) = [ (μ2- μ1)(μ2- μ0) ]/(μ1μ2) Now multiply both sides by μ1μ2 to get μ1∂r h2(a) - μ2 ∂r h1(a) = (μ2- μ1)(μ2- μ0) So here then are our two conditions on the h functions h1(a) = h2(a) μ1∂rh2(a) - μ2 ∂rh1(a) = (μ2- μ1)(μ2- μ0) So we have to find functions that make this work! Wow. the atoms are the 2D Laplace atoms which include of course ln(r) and constant and linear r and 1/r. We need to pick Smythian forms which are reasonable for the range of r. Thus we might pick h1(r) = Clnr // since this does the right thing as r → ∞ h2(r) = B // which is OK at r = 0 Then ∂rh1 = C/r ∂rh2 = 0 and then the conditions are Clna = B - μ2(C/a) = (μ2- μ1)(μ2- μ0) Then solve the second to get C = (μ1- μ2)(μ2- μ0)(a/μ2) B = C ln a = (μ1- μ2)(μ2- μ0)(a/μ2) ln a OK, then we have h1(r) = Clnr h2(r) = Clna and then H1 = [I/(2π)] (1/a) Clnr H2 = [I/(2π)] (1/a) Clna Then go back to Az(r>a) = -[I/(2π)] (μ1-μ2+μ0) ln(r) + H1(r>a) Az(r<a) = -[I/(2π)] { (μ1-μ2+μ0) ln(a) + μ0(r2-a2)/(2a2) ) } + H2(r<a) and add in the homo components to get Az(r>a) = -[I/(2π)] (μ1-μ2+μ0) ln(r) + [I/(2π)] (1/a) C lnr Az(r<a) = -[I/(2π)] { (μ1-μ2+μ0) ln(a) + μ0(r2-a2)/(2a2) ) } + [I/(2π)] (1/a) Clna where C = (μ1- μ2)(μ2- μ0)(a/μ2) STOP. As μ2→ 0, C → ∞ and so Az(r>a) → ∞, but in the correct answer Az(r>a) =-[I/(2π)] { μ1ln(r) } is independent of μ2, so something is wrong above this point. Try to get simpler forms: Az(r>a) = -[I/(2π)] (μ1-μ2+μ0) ln(r) + [I/(2π)] (1/a) (μ1- μ2)(μ2- μ0)(a/μ2) lnr = -[I/(2π)] ln(r) { (μ1-μ2+μ0) - (μ1- μ2)(μ2- μ0)(1/μ2) } = -[I/(2π)] ln(r) { μ2(μ1-μ2+μ0) + (μ2- μ1)(μ2- μ0) }/μ2 = -[I/(2π)] ln(r) { μ2μ1- μ2μ2+ μ2μ0 + μ1μ0 - μ1 μ2+μ22-μ0 μ2}/μ2 = -[I/(2π)] μ0 (μ1/μ2) ln(r) STOP here, this is the wrong result! Then Az(r<a) = -[I/(2π)] { (μ1-μ2+μ0) ln(a) + μ0(r2-a2)/(2a2) ) } + [I/(2π)] (1/a) Clna = -[I/(2π)] { (μ1-μ2+μ0) ln(a) + μ0(r2-a2)/(2a2) ) } + [I/(2π)] (1/a) (μ1- μ2)(μ2- μ0)(a/μ2) lna = -[I/(2π)]{ (μ1-μ2+μ0) ln(a) + μ0(r2-a2)/(2a2) - (μ1- μ2)(μ2- μ0)(1/μ2) lna } = -[I/(2π)]{ + μ0(r2-a2)/(2a2) + (μ1-μ2+μ0) ln(a) - (μ1- μ2)(μ2- μ0)(1/μ2) lna } = -[I/(2π)]{ + μ0(r2-a2)/(2a2) + [ (μ1-μ2+μ0) - (μ1- μ2)(μ2- μ0)(1/μ2)] lna } = -[I/(2π)]{ + μ0(r2-a2)/(2a2) + [ (μ1-μ2+μ0)μ2 + (μ2- μ1)(μ2- μ0)] lna/μ2 } = -[I/(2π)]{ + μ0(r2-a2)/(2a2) + μ0(μ1/μ2) lna } So when the homo pieces are added, our new solution Az is this Az(r>a) = -[I/(2π)] μ0 (μ1/μ2) lnr // these meet BC at r = a Az(r<a) = -[I/(2π)]{μ0(r2-a2)/(2a2) + μ0(μ1/μ2) lna } From these we find ∂r Az(r>a)1 = -[I/(2π)] μ0 (μ1/μ2)(1/r) ∂r Az(r<a)2 = -[I/(2π)] μ0 r /(a2) Then ∂r Az(a) = -[I/(2π)] μ0 (μ1/μ2)(1/a) region 1 ∂r Az(a) = -[I/(2π)] μ0 1/a And then (1/μ1) ∂r Az(a)1 = -[I/(2π)] μ0 (1/μ2)(1/a) (1/μ2) ∂r Az(a)2 = -[I/(2π)] μ0 1/a(1/μ2) // 2nd BC is also met at r = a which then verifies our matching boundary conditions. Now what about the fields? B = curl A = - ∂rAz(r) . (B.6.20) B1 = - ∂rAz(r)1 = [I/(2π)] μ0 (μ1/μ2)(1/r) B2 = - ∂rAz(r)2 = [I/(2π)] μ0 r /(a2) and then H1 = B1/μ1 = [I/(2π)] μ0 (1/μ2)(1/r) compare Hθ(r) = I/(2πr) H2 = B2/μ2 = [I/(2π)] μ0(1/μ2) r /(a2) compare Hθ(r) = I r/(2πa2) The result is CLOSE, but we have an extra (μ0/μ2) factor in both equations! So if I did not make an algebraic error, this says that adding the homo solutions did NOT fix up my paradox. [ Again, the above computation ignored bulk Jm contributions so is doomed to fail. ] Plan E4. Show directly that the correct Az satisfies a certain wave equation Collect More Evidence. I did the following exercise. Start with this equation showing a μ2 : 22DAz(x) = -μ0Jmz(x) -μ2Jcz(x) = -[ (μ1-μ2)(I/2πa)] δ(a-r) - μ2(I/πa2) θ(r < a) Then if you take the Appendix B solutions Az(r>a) == -[I/(2π)] { μ1ln(r) } Az(r<a) = = -[I/(2π)] { μ1 ln(a) + μ2(r2-a2)/(2a2) ) } and if you write 22DAz = (1/r)∂r(rAz) then you find that the Az solution does in fact satisfy the equation for r > a and for r < a. [ yes, this is true.] I did this on the whiteboard and won't repeat it here, it is pretty simple. Now write the equation as Az(x) = -[ (μ1-μ2)(I/2πa)] δ(a-r) - μ2(I/πa2) θ(r < a) (*) [ This is a useful general idea, to use the divergence theorem to deal with a delta function. ] Then integrate both sides over all 2D space ∫dV Az(x) = ∫dV { -[ (μ1-μ2)(I/2πa)] δ(a-r) - μ2(I/πa2) θ(r < a) } = ∫rdr∫dθ { -[ (μ1-μ2)(I/2πa)] δ(a-r) - μ2(I/πa2) θ(r < a) } = 2π !Syntax Error, Irdr { -[ (μ1-μ2)(I/2πa)] δ(a-r) - μ2(I/πa2) θ(r < a) } = 2π !Syntax Error, Irdr { [ (μ2-μ1)(I/2πa)] δ(a-r) - μ2(I/πa2) θ(r < a) } = 2π [ (μ2-μ1)(I/2πa)] !Syntax Error, Irdr δ(a-r) - 2π μ2(I/πa2) !Syntax Error, Irdr θ(r < a) = 2π [ (μ2-μ1)(I/2πa)] a - 2π μ2(I/πa2) !Syntax Error, Irdr = 2π [ (μ2-μ1)(I/2πa)] a - 2π μ2(I/πa2) (a2/2) = [ (μ2-μ1)I] - μ2(I} = - μ1I // the δ(a-r) made a contribution here So we find then that, by evaluating the right side above, we have ∫dV div2DAz(x) = - μ1I Now we want to evaluate the left side to see what it comes out being. Use the divergence theorem over all space ( maybe a large cylinder) which says ∫dV div2DAz(x) = ∫ds Az(x) = C ds (∂rAz) where the C curve is a great circle. On this great circle, the r>a solution for Az should apply so we get C ds (∂rAz) = C ds ∂r[-[I/(2π)] { μ1ln(r) }] = -[I/(2π)]μ1C ds ∂r ln(r) = -[I/(2π)]μ1C ds (1/r) = -[I/(2π)]μ1∫rdθ (1/r) = -[I/(2π)]μ1 2π = -[I]μ1 Thus, our solution is verified not only for r > a and for r < a, but the delta function term δ(r-a) is also verified as we show by integrating both sides of our equation (*) over all space. [ Well, I only used the exterior solution to get the -[I]μ1, so maybe I only have verified the r>a solution by this method. I could maybe redo the above for a volume inside the wire and then verify the r<a . ] Conclusion: By direct calculation using just the PDE (with μ2) and no Helmholtz integrals or any of that stuff -- by direct calculation I have shown that the Appendix B solution for Az does in fact solve 22DAz(x) = -μ0Jmz(x) -μ2Jcz(x) Therefore, that second term must have a μ2 and cannot have a μ0 despite earlier work which shows μ0 to be the constant! I further verify this by doing the Helmholtz integral and getting the right answer with no homo terms added, and I show that with no homo terms, the BC's both are satisfied. [ everything in this Conclusion is correct! I did show directly that my known-good solution satisfies the PDE with μ2 and that "extra" Jm surface-only thing present. ] So, the Grand Master Debugger is slowly collecting evidence. Many fingers are now pointing to the requirement for μ2 going with the current Jc2. [ and that is how it ends up! ] Plan E5: Gauge Violation? [ An interesting thing to examine. ] In my most recent derivation of a potential wave equation, I had this gauge condition divA = - μ0ε1jωφ - μ0σ1φ so at ω = 0 that seems to say divA = - μ0σ1φ . But if I assume that A = Az(x,y) , then div A = ∂zAz = 0, so I am not respecting the gauge! So just making that assumption violates the gauge unless it happens that φ = 0, which might be the case in a purely magnetic problem. Suddenly my little world has an internal inconsistency. [ I think that is correct. If you assume A = Az(x,y), you always get div A =0. If you are in some Lorenz or King gauge, you have to assume φ = 0 to be consistent. The other solution is either Az(x,y,z) or Ax and/or Ay ≠0. ] Let's go back then to our full derived 3D μ0-gauge equations which were (2 + β012)A = - μ0Jm1 region 1 (2 + β012)A = - μ0Jc2 - μ0Jm2 region 2 (2 + β012)φ = - ρ1/ε1 region 1 (2 + β012)φ = - ρ2/ε2 region 2 (2 + β012)A = - μ0Jc2 - μ0Jm region R (2 + β012)φ = - ρ1/ε1- ρ2/ε2 region R divA = - μ0ε1jωφ - μ0σ1φ [ Reminder: I think this is all correct if you include all Jm in Jm! ] Now I just showed that Ax and Ay have to exist or we have to have Az(x,y,z) with change in z, otherwise the gauge condition is not satisfied. If we have a simple infinite wire with some ρ on the outside surface and with some DC current, then Az is not going to depend on z, so Ax and Ay must then exist. I think I can continue to insist the Jc2 is entirely in the z direction as is Jm1, so we then have (2 + β012)A = - μ0Jc2 - μ0Jm region R meaning (2 + β012)Az = - μ0Jc2z - μ0Jmz region R (2 + β012)Ax = 0 (2 + β012)Ay = 0 and this then allows homo solutions for Ax and Ay which in turn will allow us to satisfy the gauge condition. But the z equation is unaltered, I still claim that Az does not depend on z, so we then have (22D + β012)Az(r) = - μ0Jc2z(r) - μ0Jmz(r) region R So what is new here is that there are some other components Ax and Ay floating around which are non-zero! But since B = curl A these extra components must somehow not alter B from what the "standard solution" gives for B. [ and assuming φ = 0 is always another way out. ] General Dialog I am surprised that in my μ0 gauge, I could not get a viable solution for Az by doing the particular integral and then adding homo terms to meet the BC's. How can this be explained? (1) the wave equation extended to region R is wrong (2 - μ0ε1 ∂t2 - μ0σ1 ∂t)A = - μ0Jc2 - μ0Jm2 region 2 (2) I did something wrong adding the homo terms. However, after I added them, my solution did at least meet the BC's as I verified from the solution. (3) But maybe the BC's themselves are not correct (4) my Kz expression is wrong. (5) there is something else wrong that I am just not aware of, am too blind to see at this time. [ Number (5) wins! I was blind to the fact that Jm was not just a surface current! ] Redo the μ0 gauge solution as a two-region solution with BC's. Perhaps once again that unifying R is to blame [ wrong] . So let's look again at our separated solutions which I will take to be these: (2 - μ0ε1 ∂t2 - μ0σ1∂t)A = - μ0Jm1 region 1 (2 - μ0ε1 ∂t2 - μ0σ1 ∂t)A = - μ0Jc2 - μ0Jm2 region 2 I will decide to throw all the surface current into region 1, so we then have [ but when bulk Jm is present you cannot do this! ] (2 - μ0ε1 ∂t2 - μ0σ1∂t)A = - μ0Jm region 1 (2 - μ0ε1 ∂t2 - μ0σ1 ∂t)A = - μ0Jc region 2 and then the Helmholtz is this (2 + β012)A = - μ0Jm region 1 (2 + β012)A = - μ0Jc2 region 2 β012(ω) = μ0ε1ω2 - μ0σ1jω = ω(μ0ε1ω - j μ0σ1) // = 0 at DC So at DC, here are our equations 2Az = - μ0Jm region 1 2Az = - μ0Jc2 region 2 We assume that Az = Az(r) so no z dependence, and then 22DAz = - μ0Jm region 1 22DAz = - μ0Jc2 region 2 [ By ignoring bulk Jm in the μ0 gauge you doom the problem to fail and solution to be wrong! ] Region 1 Solution The particular integral here is this (we are only doing r > a) Az = -(1/4π) { μ0∫dS' [Kzm(x')] ln(R2) = +(1/4π)(μ2-μ1) I/(2πa) ∫dS' ln(R2) = (1/4π)(μ2-μ1) I/(2πa) * 4πa lnr // from B..6.15 = (μ2-μ1) I/(2π) lnr // which is a homo solution for r > a, as expected I will add more homo solution of the form Azh1 = C lnr since this seems right for large r. Then Az(r>a) = [(μ2-μ1) I/(2π) + C] ln(r). Region 2 Solution The particular integral here is this (we are only doing r < a) Az(r) = -(1/4π) μ0∫dV'[Jcz(x')] ln(R2) = -(1/4π) μ0 I/(πa2) ∫dV' ln(R2) = -(1/4π) μ0 I/(πa2) [ 2πa2ln(a) + π(r2-a2) ] // from B..6.14 = - (I/2π) μ0 [ ln(a) + (r2-a2)/2a2 ] STOP: (then resume) Isn't this a homo solution? No, the only atoms for axial symmetry are A + B lnr, so r2 is not an atom, so this is not a homo solution, no problem. In fact, 22D (r2) = 4 so 22D (Az(r)) = - (I/2π) μ0 (1/2a2) 4 = - (I/π) μ0 (1/a2) = -μ0 I/(πa2) = -μ0Jc, all is fine. [ I was a little late realizing that only A and Blnr are legal atoms for this azisym situation. ] I will add a homo solution of the form Azh2 = B since this seems right for small r. Then Az(r<a) = - (I/2π) μ0 [ ln(a) + (r2-a2)/2a2 ] + B. So far then we have these solutions for the two regions Az(r>a) = [(μ2-μ1) I/(2π) + C] ln(r). Az(r<a) = - (I/2π) μ0 [ ln(a) + (r2-a2)/2a2 ] + B. ∂rAz(r>a) = [(μ2-μ1) I/(2π) + C]/r // made error here earlier! ∂r Az(r<a) = - (I/2π) μ0 r/a2 We now write these four objects at r = a: Az(a)1 = [(μ2-μ1) I/(2π) + C] ln(a). Az(a)2 = - (I/2π) μ0 [ ln(a) ] + B. ∂rAz(a)1 = [(μ2-μ1) I/(2π) + C]/a // ditto ∂rAz(a)2 = - (I/2π) μ0 (1/a) Our boundary conditions are supposedly Az(a)1 = Az(a)2 ∂rAz(a)1/μ1 = ∂rAz(a)2/μ2 which we then write out as [(μ2-μ1) I/(2π) + C] ln(a) = - (I/2π) μ0 ln(a) + B (1/μ1)[(μ2-μ1) I/(2π) + C]/a = - (I/2π) μ0 (1/a)(1/μ2) // ditto Solve the second equation for C in steps (1/μ1)[(μ2-μ1) I/(2π) + C] = - (I/2π) μ0 (1/μ2) [(μ2-μ1) I/(2π) + C] = - (I/2π) μ0 (μ1/μ2) C = - (I/2π) μ0 (μ1/μ2) - (1/μ1)[(μ2-μ1) I/(2π) = - (I/2π) [μ0 (μ1/μ2) + (1/μ1)(μ2-μ1) ] = - (I/2π) [μ0 μ12 + (μ22-μ1μ2) ]/(μ1μ2) = - (I/2π) [μ0 μ12 + μ2(μ2-μ1) ]/(μ1μ2) // new and ugly, never needed Then the first equation may be solved for B B = [(μ2-μ1) I/(2π) + C] ln(a) + (I/2π) μ0 ln(a) Well, what we really want here is this fact, a form of our solution for C, [(μ2-μ1) I/(2π) + C] = - (I/2π) μ0 (μ1/μ2) and then we find B = - (I/2π) μ0 (μ1/μ2)ln(a) + (I/2π) μ0 ln(a) = (I/2π) ln(a) [ - μ0 (μ1/μ2) + μ0] = (I/2π) ln(a)μ0 [ - (μ1/μ2) + 1 ] // ditto Then with these constants determined, we end up with ∂rAz(r>a) = [(μ2-μ1) I/(2π) + C]/r = - (I/2π) μ0 (μ1/μ2)/r ∂rAz(r<a) = - (I/2π) μ0 r/a2 and then the magnetic B fields are B = curl A = - ∂rAz(r) . (B.6.20) B1 = - ∂rAz(r)1 = (I/2π) μ0 (μ1/μ2)/r // did not change B2 = - ∂rAz(r)2 = (I/2π) μ0 r/a2 These results are incorrect, so there is some malfunction in this method. I thought maybe maintaining the separate regions would fix things up rather than a unified R, but that is not the case. [Again: the malfunction is omitting bulk Jm contributions. ] Comment: The first time through I got the C expression wrong, but the end result did not change and it is still wrong. By the way, in cylindricals we could redo our slope BC this way Assume A = Az(r) curl A = [ r-1∂θAz - ∂zAθ] + [∂zAr - ∂rAz] + [ r-1∂r(rAθ) - r-1∂θAr ] = [- ∂rAz] = Bθ where Bθ = - ∂rAz Then lines (1.1.44) says (1/μ2)Bθ2 - (1/μ1)Bθ1 = Kzfree or -(1/μ2) ∂rAz2 + (1/μ1) ∂rAz1 = Kzfree and that is where our slope BC comes from. I just assume Az itself is continuous (gulp). [ I think this is pretty well nailed down, except maybe sign of Kz , and installed in Section 1.1 ] So this is a Major Problem. I claim to be teaching how to solve problems in this area, but my solution of the simplest possible problem is failing!! I have spent perhaps 4 whole days trying to find out why with no success. And before that maybe 90 days on Lines, and before that maybe 1 year on Stak! I need to develop some debugger probing tools. [ yes you do! ] The bug is deep in the Machine here [ yes it was] . This is going to go well into January 2014, I can see that now [ wrong ]. If four days does not nail it, it is something very serious and hard to find. I'm glad no one is having to pay me for this flailing work. [ but the fifth day was golden. An argument to support "persistence". ] There is the little annoying fact about the integral and Maple [ an interesting side line. Maple does it weird, but I know GR7 is correct as is my sum method. ] [ Again, every single item is on the suspect list, including this integral, since I am thrashing!] In Appendix B I quote this result, This integral may be evaluated using GR7 p 531 4.224, but when I tell Maple to do this, here is what I get The simple integral says that the integral should be π ln( (1/2) ( 3 + ) ) Wolfram Alpha can't do it unless you pay them. Hmmm. I think Stak has a series for this log thing that I used somewhere. Page 104 (6.22) says this ln(1 + α2-2α cosx) = -2 Σn=1∞αn cos(nx)/n valid for |α| < 1. Then write, using α = r'/r < 1 ln(r2 + r'2-2rr' cosx) = ln[(r2)(1 + α2 - 2α cosx)] = ln(r2) + ln(1 + α2-2α cosx) = 2ln(r) -2 Σn=1∞αn cos(nx)/n And then we have Q(r',r) = !Syntax Error, Idx ln [r'2 +r2-2rr' cosx)] = !Syntax Error, Idx { 2ln(r) -2 Σn=1∞αn cos(nx)/n } = 2πln(r) - 2 Σn=1∞αn/n (!Syntax Error, Idx cos(nx) ) = 2πln(r) - 2 Σn=1∞αn/n sin(πn)/n = 2π ln(r) valid for r' < r You could swap r↔r' in the above few lines to obtain Q(r',r) = !Syntax Error, Idx ln [r'2 +r2-2rr' cosx)] = 2πln(r') valid for r' > r So we end up then with Q(r',r) = 2π and this gives the same result as the GR7 quoted integral. So I don't know what Maple is thinking. So I guess I will cross off worrying about the GR7 integral and Maple's problem with it. Here is another Maple attempt which shows conclusively that Maple gets this integral wrong: In the above explicit example, Maple's numerical integration agrees with the GR7 value. I think Maple has gone off on some strange branch of the analytic function ln(3 +2cos(z)) which is not the obvious branch. I will not pursue this further! Resume 11.21.13. Signal tracing day. [ Amen ] I will start this time at the end of the chain and trace backwards to look for the malfunction. (a) What are the "right" Az expressions for the round wire? We know the correct B fields for the round wire B = (I μ1/2π) (1/r) for r > a which is in the dielectric medium with μ1 B = (I μ2/2π) (r/a2) for r < a which is in the conductor medium with μ2 . (B.6.23) With our assumption of A = Az(r) we also know that B = curl A = - ∂rAz(r) . (B.6.20) So we then have - ∂rAz(r)1 = (I μ1/2π) (1/r) - ∂rAz(r)2 = (I μ2/2π) (r/a2) Now integrate both these equations to get Az(r)1 = -(I μ1/2π) ln(r) + C1 Az(r)2 = -(I μ2/4π) (r2/a2) + C2 So these are the forms that the potential must have. The potentials already meet the slope condition by inspection. And I think C1 = 0 for correct large r behavior. Then value match says -(I μ2/4π) (r2/a2) + C2 = -(I μ1/2π) ln(r) -(I μ2/4π) + C2 = -(I μ1/2π) ln(a) C2 = -(I μ1/2π) ln(a) +(I μ2/4π) = -(I/2π)[ μ1lna - μ2/2] The potential must then be this Az(r)1 = -(I μ1/2π) ln(r) Az(r)2 = -(I μ2/4π) (r2/a2) - (I/4π)[ 2μ1lna - μ2] = -(I/4π) [ μ2(r2/a2) -μ2 + 2μ1lna ] = -(I/4π) [ μ2{(r2/a2) -1} + 2μ1lna ] Once again Az(r)1 = -(I μ1/2π) ln(r) Az(r)2 = -(I/4π) [ μ2{(r2/a2) -1} + 2μ1lna ] Let's now verify by tracing forward again - ∂rAz(r)1 = (I μ1/2π) 1/r = (I μ1/2π) 1/r = correct - ∂rAz(r)2 = (I/4π) [ μ2{(2r/a2) } ] = (I μ2/2π) (r/a2) = correct Now the gauge condition says div A = ∂zAz = 0 which trivially true, so not violating this at ω = 0. Conclusion: given the requirement A→ constant * lnr for large r, and given the known B fields, and given the Az assumption, the potentials MUST have this form for the round wire [ correct ] Az(r)1 = -(I μ1/2π) ln(r) Az(r)2 = -(I/4π) [ μ2{(r2/a2) -1} + 2μ1lna ] // both agree with original App B calc and these do meet both our boundary conditions. (b) Do these correct Az solutions solve the μ0 gauge wave equation [as I perceived it] Now that we know the required Az fields, let's check to see if they satisfy our wave equations: 22DAz = - μ0Jm region 1 [ I imagined Jm only on boundary] 22DAz = - μ0Jc2 region 2 We can write these as r-1∂r(r∂rAz) = -μ0 {- [(μ2-μ1)/μ0] [ I/(2πa)}δ(r-a) region 1 r-1∂r(r∂rAz) = -μ0 {I/2πa2} region 2 Now compute left sides: First quote from a few lines above - ∂rAz(r)1 = (I μ1/2π) 1/r = correct - ∂rAz(r)2 = (I μ2/2π) (r/a2) = correct Then r∂rAz(r)1 = -(I μ1/2π) r∂rAz(r)2 = -(I μ2/2π) (r2/a2) ∂r(r∂rAz(r)1) = 0 ∂r(r∂rAz(r)2) = -(I μ2/π) (r/a2) r-1∂r(r∂rAz(r)1) = 0 r-1∂r(r∂rAz(r)2) = - (I μ2/π) (1/a2) = -μ2 (I/πa2) Looking at region 2, we have a gaping contradiction. Fact: The known-correct Az potential in region 2 does NOT satisfy this wave equation: 22DAz = - μ0Jc2 [ a major signal tracing discovery ] Instead, it satisfies THIS wave equation 22DAz = - μ2Jc2 // NOTE WELL !!! We have a forward path to the wave equation which is like this curl B = μ0(∂tD + Jc + Jm) first step 22DAz = - μ0Jc2 last step region 2 Somewhere along this path, something is going wrong! [ yup!] This is our signal tracer debug method. Let's check the very first step! Start with the known correct fields B = (I μ1/2π) (1/r) for r > a which is in the dielectric medium with μ1 B = (I μ2/2π) (r/a2) for r < a which is in the conductor medium with μ2 . (B.6.23) Then use curl B = [ r-1∂θBz - ∂zBθ] + [∂zBr - ∂rBz] + [ r-1∂r(rBθ) - r-1∂θBr ] = [ - ∂zBθ] + [ r-1∂r(rBθ) ] = [ r-1∂r{rBθ} ] Then curlB1 = [ r-1∂r{r (I μ1/2π) (1/r)} ] = 0 curlB2 = [ r-1∂r{r (I μ2/2π) (r/a2)} ] = (I μ2/2πa2) [ r-1∂r{r2} ] = (I μ2/2πa2) [ r-12r ] = (I μ2/πa2) So our very first step in region 2 says curl B = μ0(∂tD + Jc + Jm) [curl B]z = μ0Jcz = μ0I/(πa2) Conclusion: The very first step is wrong, signal tracing procedure is concluded. [ a breakthrough! ] Statement of Conclusion: The known B fields for the round wire in region 2 (inside wire) do NOT satisfy this Maxwell equation curl B = μ0(∂tD + Jc + Jm) [ with Jm as I was imagining it, just a surface thing ] (c) So let's continue the reverse trace debug with a narrowed range We start with curl H = ∂tD + Jc // a true Maxwell equation (1.1.1), very first of lines doc! We then claim that B = μ0(H+M) = μ0(H+ χm2 H) = μ0(1+χm2)H = μ2H which looks OK. Then we can say, since μ0 is surely a constant, curl B = μ0(curl H+curl M) We then insert curl H = ∂tD + Jc along with Jm = curl M to get curl B = μ0([∂tD + Jc]+[Jm]) = μ0( ∂tD + Jc + Jm) So our problem lies somewhere right in these last few lines. Let's simplify this step sequence even more. Just set ∂tD = 0 for ω = 0, Then ( green mean step verified, purple means step fails to verify. I colored them after doing each below. ) Step 1: curl H = Jc in all of region R = 1 + 2 Step 2: B = μ0(H+M) in all of region R = 1 + 2 We note in passing that the entire wire interior has M ≠ 0 as indicated by my App B picture. The curl of M is 0 in the bulk region and only non-zero at the boundary. Step 3: curl B = μ0(curl H+curl M) // no question about this step Step 4: curl M = Jm = isolated to boundary region // this is where the error was! We know insert step 1 and step 4 into step 3 to get Step 5: curl B = μ0(Jc + Jm) we showed above this is violated. In region 2, we have shown that Step 5 is violated by the known B fields of a round wire. Verify Step 1. Let's verify that Step 1 is at least valid for our round wire. The known fields are H1 = (I /2π) (1/r) for r > a => Hθ = (I/2πr) H2 = (I /2π) (r/a2) for r < a => Hθ = (Ir/2πa2) (B.6.24) curl H = [ r-1∂θHz - ∂zHθ] + [∂zHr - ∂rHz] + [ r-1∂r(rHθ) - r-1∂θHr ] = [ - ∂zHθ] + [ r-1∂r(rHθ) ] = [ r-1∂r{rHθ} ] curl H1 = 0 curl H2 = [ r-1∂r{r(I /2π) (r/a2)} ] = (I /2πa2 ) r-1∂r{r2} = (I /2πa2 ) r-12r = (I /πa2 ) Jc2 = (I /πa2 ) Thus I have just verified that in region 2 we have curl H2 = Jc2, so Step 1 is OK. Verify Step 2. Next, let's check Step 2 which claims that B = μ0(H+M) I think I know B and H, but what is M? Well, we think that M = χm H = [ μ2/μ0 - 1] H M2 = [ μ2/μ0 - 1] H2 = [ μ2/μ0 - 1] (I /2π) (r/a2) = [ μ2/μ0 - 1] (I /2πa2) r Note that this magnetization is circumferential, not constant, and not longitudinal. We then want to verify this equation B2 = μ0(H2+M2) LHS = μ2H2 = μ2 (Ir/2πa2) RHS = μ0 { (Ir/2πa2) + [ μ2/μ0 - 1] (I /2πa2) r } = μ0 (I /2πa2){ (r + [ μ2/μ0 - 1] r } = μ0 (I /2πa2) r{ 1 + [ μ2/μ0 - 1] } = μ0 (I /2πa2) r{ μ2/μ0 } = (I /2πa2) μ2r Thus we have LHS = RHS and so this Step 2 is verified in region 2, it goes green. Verify Step 3. We have shown in Step 2 that B2 = μ0(H2+M2) in the bulk region 2. It seems a no brainer that we would then have curl B2 = μ0(curl H2+curl M2) Since everything is a function only of r and nothing depends on z, we write this as [ r-1∂r{rBθ2} ] = μ0 ( r-1∂r{rHθ2} + r-1∂r{rMθ2} ) (*) where from above Hθ2 = (Ir/2πa2) Bθ2 = μ2 Hθ2 = μ2 (Ir/2πa2) Mθ2 = [ μ2/μ0 - 1] Hθ2 = [ μ2/μ0 - 1] (Ir/2πa2) Looking now at (*) we have LHS = r-1∂r(r μ2 (Ir/2πa2)) = μ2(I/2πa2) r-1∂r(r2) = μ2(I/πa2) RHS = μ0 { r-1∂r(r (Ir/2πa2)) + r-1∂r{r [ μ2/μ0 - 1] (Ir/2πa2) } = μ0(I/2πa2) r-1∂r { r r + [ μ2/μ0 - 1]r r} = μ0(I/2πa2) r-1∂r {r2 + [ μ2/μ0 - 1]r2} = μ0(I/2πa2) r-1∂r { μ2/μ0]r2} = μ0(I/2πa2) r-1{ μ2/μ0]2r} = (I/πa2) { μ2} and we have then verified Step 3 with our known fields, so it goes green. Verify Step 4. Let's compute Jm2 = curl M2 = [ r-1∂r{rMθ2} ] = [ r-1∂r{r [ μ2/μ0 - 1] (Ir/2πa2)} ] = [ μ2/μ0 - 1] (I/2πa2) [ r-1∂r{rr } ] = [ μ2/μ0 - 1] (I/2πa2) [ 2 ] = [ μ2/μ0 - 1] (I/πa2) Bug Found !!! // there was indeed a Loose Bolt DING !!!!!!! [ maybe 11.21.13 at 2 PM circa ] I was thinking that Jm existed only on the boundary of the round wire!!! That is the way things work at a planar boundary between μ1 and μ2 materials in scenarios #2 and #3 below. But here you see that Jm2 is non-vanishing everywhere inside the round wire!! (outside as well) That was a big mistake in my interpretation of things. Jm2 is in fact a constant everywhere inside the wire, it is not 0 as I thought. This is a big game changer. [ yes it is! ] Planar Scenarios Look again at the planar boundary situation: We imagine current in the z direction in region 2. We have here M = χm H = [ μ2/μ0 - 1] H Mx1 = χm1 Hx = [ μ1/μ0 - 1] Hx1 Mx2 = χm2 Hx = [ μ2/μ0 - 1] Hx2 Hx2 = Hx1 Scenario #1 The physical situation here is supposed to be uniform Jz in the bottom infinite half space. If we put our little loop entirely in the lower half space, we would find that curl H = J => [Hx(y+2s) - Hx(y)] L = JzL2s so that [Hx(y+2s) - Hx(y)]/2s = Jz = d Hx/dy =? Hx(y) = Jzy + Hx(y=0) So the H field in the conductor is not constant, but is linear in y. Then Mx2 = [ μ2/μ0 - 1] Hx2 = [ μ2/μ0 - 1] [ Hx(0) + Jzy ] So the magnetization in this scenario is NOT constant (since of course M = χm H ). Then the magnetization current is given by Jm = curl M = (∂yMz - ∂zMy) + (∂zMx - ∂xMz) + (∂xMy - ∂yMx) = (- ∂yMx) = (- ∂y{ [ μ2/μ0 - 1] [ Hx(0) + Jzy ]}) = - [ μ2/μ0 - 1]Jz So even in this planar boundary case, we find that Jm is NOT a surface current, it is a constant throughout region 2, just as in the round wire. So where did this "surface current" fixation come from? Scenario #2 Well, suppose in this same scenario instead of having a uniform Jz, suppose we have Jz existing only as a surface current at some y = -y0 . That is a sheet down somewhere in the conductor. Then for loops outside that sheet, you find that Hx is constant in y. For a loop around the sheet we would get Hx(-y0+ s) - Hx(-y0- s) = Kz where Kz is our sheet current. So what we have in THIS scenario is Hx(y) = Hx(0) for y in the range y = 0 down to y = -y0 Hx(y) = Hx(0) - Kz for y < - y0 ignoring sign of Kz Then in this situation we have region 2 divided into two constant H regions, which means that we have then two constant M regions down there. So with a surface current as our source, we STILL end up with bulk magnetizations in this planar boundary case. Now lets go up into region 1. We know that Hx continues across the boundary, and if we assume there is no current in region 1, it stays constant as you go up, using our loop rule. Then we have Hx(y) = Hx(0) for y in the range y > 0 Hx(y) = Hx(0) for y in the range y = 0 down to y = -y0 Hx(y) = Hx(0) - Kz for y < - y0 ignoring sign of Kz Now we then have, since Mx = [ μi/μ0 - 1] Hx we have Mx(y) = [ μ1/μ0 - 1] Hx(0) for y in the range y > 0 Mx(y) = [ μ2/μ0 - 1] Hx(0) for y in the range y = 0 down to y = -y0 Mx(y) = [ μ2/μ0 - 1] ( Hx(0) - Kz ) for y < - y0 ignoring sign of Kz So now we have three distinct regions of constant magnetization. Then the mag current is this: Mx(y) = [ μ1/μ0 - 1] Hx(0) θ(y>0) + [ μ2/μ0 - 1] Hx(0) θ(y<0) θ(y>-y0) + [ μ2/μ0 - 1] ( Hx(0) - Kz ) θ(y<-y0) Jm = (- ∂yMx) - Jmz = ∂yMx = Now ∂y θ(y>0) = ∂yθ(y) = δ(y) ∂y [θ(y<0) θ(y>-y0)] = δ(-y0) - δ(y) ∂y [θ(y<-y0)] = - δ(-y0) So we find that - Jmz = [ μ1/μ0 - 1] Hx(0) δ(y) + [ μ2/μ0 - 1] Hx(0) { δ(-y0) - δ(y)} + [ μ2/μ0 - 1] ( Hx(0) - Kz )[ - δ(-y0)] = [ μ1/μ0 - 1] Hx(0) δ(y) - [ μ2/μ0 - 1] Hx(0) δ(y) + [ μ2/μ0 - 1] Hx(0) { δ(-y0) - [ μ2/μ0 - 1] ( Hx(0) - Kz ) δ(-y0) } = [μ1/μ0 - μ2/μ0] Hx(0) δ(y) + Kz δ(-y0) = (μ1-μ2)/μ0 Hx(0) δ(y) + Kz δ(-y0) so then Jmz = (μ2-μ1)/μ0 Hx(0) δ(y) - Kz δ(-y0) Thus, in this case we have TWO sheets of magnetization current! One sheet is at the boundary between μ1 and μ2, while the other is at that sheet source current. So at y = -y0 we have both a real current sheet with Kz and we have a magnetization current sheet with -Kz. So to some extent, THIS is the situation perhaps analogous to the "Jackson problem" with a point charge and ε1 and ε2 where we get a surface polarization charge on the boundary. Scenario #3 I suppose we could take a "point current" Jzd3x at say x = 0 and y = -y0. That is really a wire in the z direction down there which seems an OK thing to have. This might be the true analog to the Jackson problem. Then B&B p 142 suggest dH = (1/4πr3) J d3x x r and we have some H field which is trying to go around the conductor. Since there will be a sudden change in M all along the boundary (Hx continuous through the boundary), we will have a magnetization current all along the boundary, but it won't be constant. it will be something like the Jackson polarization charge density. So this would I guess make a very good "exercise" to perhaps include in an appendix with the Jackson problem. It would have the same image type solution. I suspect up in region 1 we would have perfectly circular magnetic field lines. But save all this for the future, right now I want to de-thread from my Disaster! It is 8:30 AM on 11.21.13, Dethreading the Nov 16 Disaster after Bug Found OK, back to the round wire. What does Jm really look like here? We have for this problem: Mθ1 = [ μ1/μ0 - 1] Hθ1 = [ μ1/μ0 - 1] (I/2πr) Mθ2 = [ μ2/μ0 - 1] Hθ2 = [ μ2/μ0 - 1] (Ir/2πa2) You can see that at r = a, there will be a sudden jump in this thing, so there will be a surface current component. We can write Mθ = [ μ1/μ0 - 1] (I/2πr) θ(r>a) + [ μ2/μ0 - 1] (Ir/2πa2)θ(r<a) ∂rMθ = [ μ1/μ0 - 1] (I/2πr) δ(r-a) + [ μ2/μ0 - 1] (Ir/2πa2)[ -δ(r-a)] = [ μ1/μ0 - μ2/μ0] (I/2πr) δ(r-a) [ OUCH! On 2.7.14 I realized I did the ∂rM wrong above because I have omitted two terms!!! ] [ The error is now corrected in Appendix G, we end up with only one bulk term, same δ term ] And then Jm = curl M = [ r-1∂r{rMθ} ] so that [ I think this little calculation is done correctly including all signs ] Jmz = r-1∂r{rMθ} = r-1[ Mθ + r∂rMθ] = r-1Mθ + ∂rMθ = r-1[ μ1/μ0 - 1] (I/2πr) θ(r>a) + [ μ2/μ0 - 1] (Ir/2πa2)θ(r<a)] + [ μ1 - μ2] (I/2πrμ0) δ(r-a) = [ μ1/μ0 - 1] (I/2πr2) θ(r>a) + [ μ2/μ0 - 1] (I/2πa2)θ(r<a) + [ μ1 - μ2] (I/2πaμ0) δ(r-a) So in fact there are three distinct regions for Jm : [ WRONG as of 2.7.14 ] Jmz = [ μ2/μ0 - 1] (I/2πa2) inside the wire, it is a constant in this region Jmz = [ μ1/μ0 - 1] (I/2πr2) outside the wire, it decays with 1/r2 Jmz = [ μ1 - μ2] (I/2πaμ0) δ(a) on the surface of the wire Now, the third term I used to think was the only place Jmz existed, and I wrote in App B Kzm = - [(μ2-μ1)/μ0] [ I/(2πa)] round wire of radius a and μ2, dielectric μ1 (B.1.11) and this agrees exactly with my third component above [including sign]. So I had that part right! But I did not have the other two parts right! Attempt to solve the μ0 gauge wave equation using the correct Jmz [ I raced through this long calculation and probably made many errors. I suspect the divergence stuff is really there though and has to be dealt with. In the end I did not get the correct result, but I think if you did it carefully you would get the correct result. But I have no interest now in this μ0 gauge! ] So let's now go back to our putative wave equation (from far above) 2Az = - μ0Jc2z - μ0Jmz region R Jc2z(r) = (I/πa2)θ(r<a) Jmz(r) = [ μ1/μ0 - 1] (I/2πr2) θ(r>a) + [ μ2/μ0 - 1] (I/2πa2)θ(r<a) + [ μ1 - μ2] (I/2πaμ0) δ(a) This is a whole new world now! Let's compute the Helmholtz integral particular in this strange gauge which exposes the magnetization currents: Az = +μ0 ∫dV' { [ μ1/μ0 - 1] (I/2πr'2) θ(r'>a) + [ μ2/μ0 - 1] (I/2πa2)θ(r'<a) } [ ln(1/R)/2π] + μ0 ∫dS' [ μ1 - μ2] (I/2πaμ0) [ ln(1/R)/2π] + ∫dV' (I/πa2)θ(r'<a) [ ln(1/R)/2π] We might as well replace ln(1/R) by -(1/2 ln(R2) right at the start to get Az = -(μ0/2) ∫dV' { [ μ1/μ0 - 1] (I/2πr'2) θ(r'>a) + [ μ2/μ0 - 1] (I/2πa2)θ(r'<a) } [ ln(R2)/2π] -(μ0/2) ∫dS' [ μ1 - μ2] (I/2πaμ0) [ ln(R2)/2π] - (1/2) ∫dV' (I/πa2)θ(r'<a) [ ln(R2)/2π] Now pull out the 2π from the propagator Az = -(μ0/4π) ∫dV' { [ μ1/μ0 - 1] (I/2πr'2) θ(r'>a) + [ μ2/μ0 - 1] (I/2πa2)θ(r'<a) } [ ln(R2)] -(μ0/4π) ∫dS' [ μ1 - μ2] (I/2πaμ0) [ ln(R2)] - (μ0/4π)μ0 ∫dV' (I/πa2)θ(r'<a) [ ln(R2)] Now move the obvious factor to the left side to get (include I in this) (-4π/μ0I) Az = ∫dV' { [ μ1/μ0 - 1] (1/2πr'2) θ(r'>a) + [ μ2/μ0 - 1] (1/2πa2)θ(r'<a) } [ ln(R2)] + ∫dS' [ μ1 - μ2] (1/2πaμ0) [ ln(R2)] + ∫dV' (1/πa2)θ(r'<a) [ ln(R2)] Now break into four separate terms (-4π/μ0I) Az = ∫dV' [ μ1/μ0 - 1] (1/2πr'2) θ(r'>a) [ ln(R2)] + ∫dV' [ μ2/μ0 - 1] (1/2πa2) θ(r'<a) [ ln(R2)] + ∫dS' [ μ1 - μ2] (1/2πaμ0) [ ln(R2)] + ∫dV' (1/πa2)θ(r'<a) [ ln(R2)] Now take out constants in each term (-4π/μ0I) Az = [ μ1/μ0 - 1] (1/2π) ∫dV' (1/r'2) θ(r'>a) [ ln(R2)] + [ μ2/μ0 - 1] (1/2πa2) ∫dV' θ(r'<a) [ ln(R2)] + [ μ1 - μ2] (1/2πaμ0)∫dS' [ ln(R2)] + (2/2πa2)∫dV' θ(r'<a) [ ln(R2)] Notice that we can combine the 2nd and 4th terms to get a single term with [ μ2/μ0 + 1]. So (-4π/μ0I) Az = [ μ1/μ0 - 1] (1/2π) ∫dV' (1/r'2) θ(r'>a) [ ln(R2)] + [ μ2/μ0 + 1] (1/2πa2) ∫dV' θ(r'<a) [ ln(R2)] + [ μ1 - μ2] (1/2πaμ0)∫dS' [ ln(R2)] We now have three integrals to calculate. Let's write things out in more detail using ∫dV' = !Syntax Error, Ir'dr'!Syntax Error, Idθ' ∫dS' = a!Syntax Error, Idθ' Then we have (-4π/μ0I) Az = [ μ1/μ0 - 1] (1/2π) !Syntax Error, Ir'dr'!Syntax Error, Idθ'(1/r'2) θ(r'>a) [ ln(R2)] + [ μ2/μ0 + 1] (1/2πa2) !Syntax Error, Ir'dr'!Syntax Error, Idθ'θ(r'<a) [ ln(R2)] + [ μ1 - μ2] (1/2πaμ0) a!Syntax Error, Idθ' [ ln(R2)] Now remove θ's (-4π/μ0I) Az = [ μ1/μ0 - 1] (1/2π) !Syntax Error, Ir'dr'!Syntax Error, Idθ'(1/r'2) [ ln(R2)] + [ μ2/μ0 + 1] (1/2πa2) !Syntax Error, Ir'dr'!Syntax Error, Idθ' [ ln(R2)] + [ μ1 - μ2] (1/2πaμ0) a!Syntax Error, Idθ' [ ln(R2)] // but here r' = a inside R2 and reorder things (-4π/μ0I) Az = [ μ1/μ0 - 1] (1/2π) !Syntax Error, Ir'-1dr'!Syntax Error, Idθ' [ ln(R2)] + [ μ2/μ0 + 1] (1/2πa2) !Syntax Error, Ir'dr'!Syntax Error, Idθ' [ ln(R2)] + [ μ1 - μ2] (1/2πaμ0) a!Syntax Error, Idθ' [ ln(R2)] // r' = a Now write, with advanced knowledge, !Syntax Error, Idθ' [ ln(R2)] = 2 !Syntax Error, Idθ' [ ln(R2)] = 2 Q(r',r) Q(r',r) ≡ !Syntax Error, Idx ln [r'2 + r2 - 2rr' cosx] Thus we have (-4π/μ0I) Az = [ μ1/μ0 - 1] (1/π) !Syntax Error, Ir'-1dr' Q(r',r) + [ μ2/μ0 + 1] (1/πa2) !Syntax Error, Ir'dr' Q(r',r) + [ μ1 - μ2] (1/πaμ0) a Q(a,r) Now from App B we know that Q(r',r) = 2π = 2π ln(r') θ(r'>r) + 2π ln(r)θ(r'<r) So we have two integrals to compute. First !Syntax Error, Ir'-1dr' Q(r',r) = !Syntax Error, Ir'-1dr' [2π ln(r') θ(r'>r) + 2π ln(r)θ(r'<r)] = 2π { !Syntax Error, Ir'-1dr' [ln(r') θ(r'>r) + ln(r)θ(r'<r)] } = 2π { !Syntax Error, Ir'-1dr' ln(r') θ(r'>r) + !Syntax Error, Ir'-1dr' ln(r)θ(r'<r) } = 2π { !Syntax Error, Ir'-1dr' ln(r') θ(r'>r) + ln(r)!Syntax Error, Ir'-1dr'θ(r'<r) } = 2π { !Syntax Error, Ir'-1dr'ln(r') + ln(r)!Syntax Error, Ir'-1dr' } Now separate the two cases r > a region 1 outside r < a region 2 inside Then [!Syntax Error, Ir'-1dr' Q(r',r) ]1 = 2π { !Syntax Error, Ir'-1dr' ln(r') + ln(r)!Syntax Error, Ir'-1dr'] } We have Maple do this integration to avoid errors. So my surprise, we get a divergence on the first integral so I put in a cutoff W. Then we have [!Syntax Error, Ir'-1dr' Q(r',r) ]1 = 2π {(1/2)ln(W2) - ln(r) +[ ln(r)]2 - lna lnr } This was not what I was expecting. Meanwhile for r < a [!Syntax Error, Ir'-1dr' Q(r',r) ]2 = 2π { !Syntax Error, Ir'-1dr' ln(r') + ln(r)!Syntax Error, Ir'-1dr' } = 2π { !Syntax Error, Ir'-1dr'ln(r') + 0 } We then have [!Syntax Error, Ir'-1dr' Q(r',r) ]2 = 2π { (1/2)ln(W2) - ln(a) } So we summarize our results for this painful integral: [!Syntax Error, Ir'-1dr' Q(r',r) ]1 = 2π {(1/2)ln(W2) - ln(r) +[ ln(r)]2 - lna lnr } r > a [!Syntax Error, Ir'-1dr' Q(r',r) ]2 = 2π { (1/2)ln(W2) - ln(a) } r < a Now we have our second integral to deal with !Syntax Error, Ir'dr' Q(r',r) = 2π !Syntax Error, Ir'dr' [ ln(r') θ(r'>r) + ln(r)θ(r'<r)] = 2π !Syntax Error, Ir'dr' ln(r') θ(r'>r) + 2π !Syntax Error, Ir'dr' ln(r)θ(r'<r) = 2π{ !Syntax Error, Ir'dr' ln(r') θ(r'>r) + ln(r) !Syntax Error, Ir'dr'θ(r'<r) } = 2π{ !Syntax Error, Ir'dr' ln(r') + ln(r) !Syntax Error, Ir'dr' } So again we do our two regions [!Syntax Error, Ir'dr' Q(r',r)]1 = 2π{ 0 + ln(r) !Syntax Error, Ir'dr' } = 2π ln(r) (1/2)a2 = πa2 lnr r>a [!Syntax Error, Ir'dr' Q(r',r)]2 = 2π{ !Syntax Error, Ir'dr' ln(r') + ln(r) !Syntax Error, Ir'dr' } r<a So we summarize these as [!Syntax Error, Ir'dr' Q(r',r)]1 = πa2 lnr r>a [!Syntax Error, Ir'dr' Q(r',r)]2 = 2π [ (a2/2)lna + (1/4)(r2-a2)] r <a Finally, we have to consider Q(a,r) = 2π ln(a) θ(a>r) + 2π ln(r)θ(a<r) [Q(a,r)]1 = 2π ln(r) [Q(a,r)]2 = 2π ln(a) OK, we now put the pieces together (-4π/μ0I) Az1 = [ μ1/μ0 - 1] (1/π) [!Syntax Error, Ir'-1dr' Q(r',r)]1 + [ μ2/μ0 + 1] (1/πa2) [!Syntax Error, Ir'dr' Q(r',r) ]1 + [ μ1 - μ2] (1/πaμ0) a [Q(r',a)]1 r > a (-4π/μ0I) Az1 = [ μ1/μ0 - 1] (1/π) 2π {(1/2)ln(W2) - ln(r) +[ ln(r)]2 - lna lnr } + [ μ2/μ0 + 1] (1/πa2) πa2 lnr + [ μ1 - μ2] (1/πaμ0) a2π ln(r) r > a (-4π/μ0I) Az1 = [ μ1/μ0 - 1] {ln(W2) - 2ln(r) +2[ ln(r)]2 - 2lna lnr } + [ μ2/μ0 + 1] lnr + [ μ1 - μ2] (1/μ0) 2 ln(r) r > a = [ μ1/μ0 - 1] {ln(W2) +2[ ln(r)]2 } + ln(r) { -2 [ μ1/μ0 - 1] - 2 lna [ μ1/μ0 - 1] + [ μ2/μ0 + 1] + 2 [ μ1 - μ2] (1/μ0) } = [ μ1/μ0 - 1] {ln(W2) +2[ ln(r)]2 } + ln(r) { - 2μ1/μ0 +2 - 2 lna [ μ1/μ0 - 1] + μ2/μ0 + 1 + 2 [ μ1/μ0 - μ2/μ0] } = [ μ1/μ0 - 1] {ln(W2) +2[ ln(r)]2 } + ln(r) { - 2μ1/μ0 +3 - 2 lna [ μ1/μ0 - 1] + μ2/μ0+ 2 μ1/μ0 - 2μ2/μ0] } = [ μ1/μ0 - 1] {ln(W2) +2[ ln(r)]2 } + ln(r) { +3 - 2 lna [ μ1/μ0 - 1] - μ2/μ0] } It's ugly and certainly wrong, but leave for the moment. Next (-4π/μ0I) Az2 = [ μ1/μ0 - 1] (1/π) [!Syntax Error, Ir'-1dr' Q(r',r)]2 + [ μ2/μ0 + 1] (1/πa2) [!Syntax Error, Ir'dr' Q(r',r) ]2 + [ μ1 - μ2] (1/πaμ0) a [Q(r',a)]2 r < a (-4π/μ0I) Az2 = [ μ1/μ0 - 1] (1/π) 2π { (1/2)ln(W2) - ln(a) } + [ μ2/μ0 + 1] (1/πa2) 2π [ (a2/2)lna + (1/4)(r2-a2)] + [ μ1 - μ2] (1/πaμ0) a 2π ln(a) r < a (-4π/μ0I) Az2 = [ μ1/μ0 - 1] { ln(W2) - 2ln(a) } + [ μ2/μ0 + 1] [ lna + (1/2a2)(r2-a2)] + [ μ1 - μ2] (1/μ0) 2 ln(a) r < a = [ μ1/μ0 - 1] ln(W2) - 2ln(a) [ μ1/μ0 - 1] + [ μ2/μ0 + 1] [ lna + (1/2a2)(r2-a2)] + [ μ1 - μ2] (1/μ0) 2 ln(a) = [ μ1/μ0 - 1] ln(W2) + [ μ2/μ0 + 1] (1/2a2)(r2-a2) + + lna { -2 [ μ1/μ0 - 1] + [ μ2/μ0 + 1] + [ μ1 - μ2] (1/μ0) } = [ μ1/μ0 - 1] ln(W2) + [ μ2/μ0 + 1] (1/2a2)(r2-a2) + + lna { -2 μ1/μ0 +2 + μ2/μ0 + 1 + μ1/μ0 - μ2/μ0 } = [ μ1/μ0 - 1] ln(W2) + [ μ2/μ0 + 1] (1/2a2)(r2-a2) + + lna { - μ1/μ0 + 3 } So including various no doubt algebra mistakes, here is the result: (-4π/μ0I) Az1 = { +3 - 2 lna [ μ1/μ0 - 1] - μ2/μ0] } ln(r) r > a (-4π/μ0I) Az2 = [ μ1/μ0 - 1] ln(W2) + [ μ2/μ0 + 1] (1/2a2)(r2-a2) + lna { - μ1/μ0 + 3 } r<a These are of course wrong, but they have some semblance of truth in them. Recall the correct results, Az(r)1 = -(I μ1/2π) ln(r) + C1 Az(r)2 = -(I μ2/4π) (r2/a2) + C2 I think errors resulted in failure to get cancellations, as suggested by the "3" for example, and the fact that μ1 and μ2 both appear in each equation (cancellation failures). I will assume that with enough corrective hours I could make things work out here. But this is more important: Death Knell for the μ0 Gauge I only considered the μ0 gauge because I wanted to "expose" the magnetization current Jm. I thought at that time Jm was something very simple, just the surface Kz thing, and then I was getting what looked like a simple problem to solve : 22DAz = - μ0Jc2z - μ0Jmz region R where Jm was localized to the boundary Kz thing. Since that interpretation was wrong about Jm, naturally this equation is not satisfied by the correct solution for Az. In retrospect, since Jm is much more complicated than I thought, and also unknown at the start, this is a terrible gauge in which to work!! In the messy calculation just finished above, I had to assume the solution in order to know H and thus to know Jm so I could put it on the right side! So certainly this is the end of the μ0 gauge for me, though it was instructive and led me to my misunderstanding of Jm. [ Sayonara to the μ0 gauge ] Reminder of the Just-Fine two region Case Study solution to the round wire problem. This μ0 gauge was a huge distraction used as a probe into things. If I then go back to the μ1 gauge, I still have this situation, 22D Az = - μ2 Jcz x in region R the King μ1 gauge I now appreciate this equation much more because all that messy mag current Jm is completely hidden from sight, we do not include it in Jcz. Remember that Jm exists in both bulk regions 1 and 2, and on the boundary, so it is very ugly. Also notice that μ2 goes with Jc2z which I know is very helpful. In the μ0 gauge we have - μ0 Jcz which caused headaches galore. Now remember that I successfully solved the above wave equation in "Case Study" and got the exact right answer. Here is how that solution went 1. Treat things as two regions, so have this 22D Az = 0 region 1 22D Az = - μ2 Jcz region 2 2. Compute the particular integral for region 2 and add a homo function to this solution in the form of a simple constant d based on the idea that the other atom lnr is not allowed in region 2. Az2 = (1/2π) ∫dV' [μ2Jcz(x')] ln(1/R) + d = W + d 3. In region 1, assume a homo solution of this form Az1 = A lnr We could have added a second atom so perhaps Az1 = A lnr + A' , but we would have found A' = 0 when we did the BC match. 4. Compute the W integral to find W = -(I/2π) μ2 but we only make use of this result in region 2 (r<a) where it showed up above. 5. The solutions at this point are (where A and d are to be determined) A1 = A lnr A2 = - (I/2π) μ2 [ lna + (1/2)(r2/a2-1) ] + d 6. Match both function and derivative BC's at r = a to find that A = -μ1 (I/2π) d = (I/2π)(μ2-μ1)lna 7. Install these to get the final results which we know are the correct results! A1 = -(I/2π) [ μ1 lna + (1/2)μ2(r2/a2-1) A2 = -μ1 (I/2π)lnr So basically "all is well" with the round wire and the King Helmholtz method, BUT you have to play with homo atoms in addition to the particular integral, something King did not mention but he would say was an obvious fact of life. The one-region Case Study solution to the round wire problem. What happens if we reconsider the above wave equation in the single-region R view 22D Az = - μ2 Jcz x in region R the King μ1 gauge You could compute the particular integral just as done above for all of R to get Az2 = (1/2π) ∫dV' [μ2Jcz(x')] ln(1/R) + Azh = W + Azh // for all R W = -(I/2π) μ2 Your task is then to come up with a homo adder function Az to make the BC's work at the boundary. You might conjecture that Az1 = -(I/2π) μ2 ln(r) + C ln(r) r > a Az2 = -(I/2π) μ2 [ ln(a) + (r2/a2-1)/2 ] + d r < a If you were then to match the boundary conditions and thereby compute d and C, you would then arrive at the correct solution. No matter how you cut it, the BC's are really what determine the solution! So let's do this explicitly. We have Az1 = [-(I/2π) μ2 + C ] ln(r) r > a Az2 = -(I/2π) μ2 [ ln(a) + (r2/a2-1)/2 ] + d r < a -∂rAz1 = [(I/2π) μ2 - C ]1/r r > a -∂rAz2 = (I/2π) μ2 [ (r/a2) ] r < a Then evaluate both at the boundary Az1 = [-(I/2π) μ2 + C ] ln(a) r > a Az2 = -(I/2π) μ2 [ ln(a) ] + d r < a -∂rAz1 = [(I/2π) μ2 - C ]1/a r > a -∂rAz2 = (I/2π) μ2 [ (1/a) ] r < a then we find [-(I/2π) μ2 + C ] ln(a) = -(I/2π) μ2 [ ln(a) ] + d (1/μ1)[(I/2π) μ2 - C ]1/a = (1/μ2) (I/2π) μ2 [ (1/a) ] [(I/2π) μ2 - C ] ln(a) = (I/2π) μ2 [ ln(a) ] - d [(I/2π) μ2 - C ] = (I/2π)μ1 C = (I/2π) μ2 - (I/2π)μ1 = (I/2π)(μ2-μ1) d = (I/2π) μ2 [ ln(a) ] - [(I/2π) μ2 - C ] ln(a) = (I/2π) μ2 [ ln(a) ] - (I/2π)μ1 ln(a) = (I/2π) ln(a)(μ2- μ1) Thus our values are C = (I/2π)(μ2-μ1) d = (I/2π) (μ2- μ1) ln(a) and then our solution is Az1 = [-(I/2π) μ2 + C ] ln(r) r > a Az2 = -(I/2π) μ2 [ ln(a) + (r2/a2-1)/2 ] + d r < a Az1 = [-(I/2π) μ2 + (I/2π)(μ2-μ1) ] ln(r) r > a Az2 = -(I/2π) μ2 [ ln(a) + (r2/a2-1)/2 ] +(I/2π) (μ2- μ1) ln(a) r < a Az1 = [-(I/2π) μ1 ] ln(r) r > a Az2 = -(I/2π) μ2 ln(a) -(I/2π) μ2 (r2/a2-1)/2 +(I/2π) (μ2- μ1) ln(a) = -(I/2π) μ1 ln(a) -(I/2π) μ2 (r2/a2-1)/2 r < a = -(I/2π)[ μ1 ln(a) + μ2 (r2/a2-1)/2 ] r < a and this is the "correct" solution as found in Appendix B. [ reviewed to here ] There does remain this old mystery Which I am not sure I am worried about. It was presented as Method II in my Case Study doc. I found that IF you assumed a wave equation of this form 22D A = -μ0Km δ(r-a) - μ2Jc2 THEN the particular integral happens to give the exact right solution with no homo terms needed. This was precisely the subject of Appendix B Version 2. I fretted for a long time trying to find why this is true. Notice that you are only using "some" of the full mag current Jm in this equation, and that with no justification whatsoever. Somehow it happens that adding -μ0Km δ(r-a) must generate exactly the precise homo adder solution which appears in the one-region case study above. In other words, it must happen that this is true: Z = (1/2π) ∫dV' [μ0 Kz δ(r-a)] ln(1/R) = d = (I/2π)(μ2-μ1)lna r < a Z = (1/2π) ∫dV' [μ0Kz δ(r-a)] ln(1/R) = C lnr = (I/2π)(μ2-μ1) lnr r > a I can quickly check on this using App D stuff. Z = (1/2π) ∫dV' [μ0Kz δ(r-a)] ln(1/R) = - (1/2π)(1/2) ∫dV' [μ0Kz δ(r-a)] ln(R2) = - (1/2π)(1/2) [(μ1-μ2) I/(2πa) ∫dV' δ(r-a) ln(R2) = - (1/2π)(1/2) [(μ1-μ2) I/(2πa) ∫dr' r' dθ δ(r-a) ln(R2) = - (1/2π)(1/2) [(μ1-μ2) I/(2πa) a ∫ dθ ln(R2) = - (1/2π)(1/2) [(μ1-μ2) I/(2πa) a 2 Q(a,r) = - (1/2π)(1/2) [(μ1-μ2) I/(2πa) a 2 * 2π = (I/2π) [(μ2-μa) and this then all checks out right. So the cloying question remains: WHY does adding source -μ0Kz δ(r-a) to the right side of the wave equation generate these exactly correct homo adder solutions? Here is the problem with just this term 22D Az = -μ0Kz δ(r-a) all R 22D Az = 0 r>a 22D Az = 0 r<a Whatever the solution is, it will be a homo solution in the two separate regions, so we know THAT much. And we expect continuity of Az at the boundary, and the solution shows that. Maybe this added current somehow cancels the mag current, causing B and hence A to "see nothing there" and then you get the regular jump on ∂rAz with no μi factors. I do not think it is worth much to come up with a precise interpretation here, and I will let it lay. It certainly is a special case I would say! I did show earlier that Jmz = [ μ1/μ0 - 1] (I/2πr2) θ(r>a) + [ μ2/μ0 - 1] (I/2πa2)θ(r<a) + [μ1 - μ2] (I/2πaμ0) δ(a) I think this Disaster of Nov 16 has come to an end, I am now going to back thread through all related docs with red text to make sure I have not missed something. [ I went through this entire 67 page doc in much detail and still may pursue the "cloying question" a bit more. ]