King gauge problem continued REVIEWED

King Gauge Problem Continued PhL 11.4.13 The struggle continues here from the previous doc, and I draw the picture you see below which ended up being the final picture. I wrote down the King condition and wondered which regions it applied to. I then got the idea of using the region 1 King gauge in all three regions and suddenly King Problem #2 seemed to have a solution in terms of my unified wave equation for A over entire region R. But at this point, all regions had to have the same μ -- but better than a poke in the eye with a sharp stick! I then assumed different μ's and still got a unified equation! " I am just astounded" he says. I then wrote up the general plan which is now installed (same picture) into Section 1.5 of lines doc. In the original doc of this title, I pursued this same question at length. Here just since some time has gone by, I will give it another shot starting "afresh", though I am sure I will end up at the same place. Start with and make the immediate simplifying assumption that μ1 = μ2 = μ3 = μ . If we consider the magnetostatic analog to the Jackson point charge near boundary electrostatic problem, we would maybe have Az = (μ/4π)(JzdV/R) as the replacement for the electrostatic problem φ = (1/4πε)(1/r). Then we might argue that Az(x) does not even see the boundary, and so the lines of B would not diffract at the boundary. Perhaps this will somehow help things in what follows. It seems to at least remove one complicating factor. I lines doc section 1.6 I started out like so: The 3 pre-gauge A equations are (2 + β0(i)2) A(i) = grad [μiεi jωφ(i) + (divA(i)) ] - μiJ(i) . (1.3.3) If we decide to have three region-specific King gauges div A(i) = - μεijω φ(i) - μσi φ(i) we then get in each region (2 + β(i)2) A(i)(x,ω) = 0 since all currents are conduction currents and have been absorbed and this puts me in a bad position. Before going on, since σ is so violently different, the gauge condition is violently different in each region, and the (A,φ)(i) in each region have these violently different gauges, so I see no sense of continuity of these potential functions at the boundaries! That seems like a bad idea if you are trying to unify the regions somehow. So let's try having just one gauge condition for the entire region. Plan H: Suppose we assume only the region 1 gauge condition for ALL regions. Then there is only one set of functions (A,φ) which give (E,B) everywhere, I like that. We then have, thinking only of region 1: div A = - με1jω φ - μσ1 φ // using the common μ in all regions. (global King gauge) (2 + β12) A(x,ω) = 0 // since conduction current has been absorbed into β1 and we are then stuck with our famous homo equation with no source inside region 1, which is correct. What is the equation for A in region 2? This comes from (1.3.3) converted to ω space, pre-gauge: (2 + β0(2)2) A = grad [με2 jωφ + (divA) ] - μJ where now J is some current in region 2 and I want to try to leave it "as is". Now we have to use the global King gauge to deal with div A. RHS = grad [με2 jωφ + (- με1jω φ - μσ1 φ) ] - μJ = με2 jω gradφ - με1jω gradφ - μσ1 gradφ - μJ ok = μ(jω[ε2-ε1] - σ1) gradφ - μJ ok Now what do we do? I cannot eliminate φ without using E = - grad φ - ∂tA and that then brings in both A and E. But I don't want to eliminate E with J = σE. But go ahead with E = - grad φ - ∂tA : RHS = μ(jω[ε2-ε1] - σ1) (-E -jωA) - μJ // in region 2 keep in mind Perhaps since in conductor I could just set E = 0 ?? That is, σ2 = ∞. Try it So here I am making an approximation that E = 0 inside the region 2 conductor: RHS = μ(jω[ε2-ε1] - σ1) (-jωA) - μJ ok Then I end up with (2 + β0(2)2) A = μ(jω[ε2-ε1] - σ1) (-jωA) - μJ ok // region 2 (2 + ω2με2) A = μ(jω[ε2-ε1] - σ1) (-jωA) - μJ ok [2 + ω2με2 + μ(jω[ε2-ε1] - σ1) (+jω) ] A = - μJ ok [2 + ω2με2 - ω2μ[ε2-ε1] - jωμσ1) ] A = - μJ ok [2 + ω2με1 - jωμσ1) ] A = - μJ ok [2 + β(1)2 ] A = - μJ ok Wow! Is that really true??? If so, I think we might have made progress here. I conclude that this is the PDE for A in region 2 [2 + β(1)2 ] A = - μJ // for region 2. where β(1)2 = ω2μξ1. In other words, although we are in region 2, the propagator is that of region 1 !!! I then have these equations: (2 + β12) A(x,ω) = 0 // region 1 (2 + β12) A(x,ω) = - μJ // region2 (2 + β12) A(x,ω) = - μJ // region3 This is definitely the first time I have got these equations! I then have this unified equation (2 + β12) A(x,ω) = - μJ // region R where J includes only the currents in region 2 and region 3. Then I get a solution even for FAT conductors! The ingredients to get this result are μ = same in all regions E = 0 on conductors region 1 King gauge applies to all regions This is a very big breakthrough for me. Plan H version 2: Let's just see now what happens if we don't assume all the μ are the same. div A = - μ1ε1jω φ - μ1σ1 φ // (global King gauge) (2 + β12) A(x,ω) = 0 // equation for region 1 with conduction current absorbed β12 = ω2μ1ξ1 What is the equation for A in region 2? This comes from (1.3.3) converted to ω space, pre-gauge: (2 + ω2μ2ε2) A = grad [μ2ε2 jωφ + (divA) ] - μ2J Now ignoring the - μ2J term, the RHS is RHS = grad [μ2ε2 jωφ + (divA) ] = grad [μ2ε2 jωφ + (- μ1ε1jω φ - μ1σ1 φ ) ] = grad [μ2ε2 jωφ - μ1ε1jω φ - μ1σ1 φ ] = [μ2ε2 jω - μ1ε1jω - μ1σ1 ] grad φ Now make the replacement E = - grad φ - ∂tA to get = [μ2ε2 jω - μ1ε1jω - μ1σ1 ] (-E -jωA) Then set E = 0 since inside a conductor to get = - [μ2ε2 jω - μ1ε1jω - μ1σ1 ] (jωA) = - [ jω(μ2ε2 - μ1ε1) - μ1σ1 ] (jωA) = - [ -ω2(μ2ε2 - μ1ε1) - jωμ1σ1 ] A Then our region 2 equation is (2 + ω2μ2ε2) A = - [ -ω2(μ2ε2 - μ1ε1) - jωμ1σ1 ] A - μ2J Swing the first term RHS over to get (2 + ω2μ2ε2 + [ -ω2(μ2ε2 - μ1ε1) - jωμ1σ1 ] ) A = - μ2J (2 + ω2μ2ε2 -ω2μ2ε2 +ω2 μ1ε1 - jωμ1σ1 ] ) A = - μ2J (2 +ω2 μ1ε1 - jωμ1σ1 ] ) A = - μ2J (2 + β12 ) A = - μ2J // region 2 I am just astounded. Now only two assumptions were made E = 0 on conductors region 1 King gauge applies to all regions Now I might define J(x) = θ2(x)J2(x) + θ3(x)J3(x) Then we have this single equation which covers the entire region R: (2 + β12 ) A = - μ2J I keep being amazed by this. So now I will start a major rewrite of parts of Chapter 1! No more need for that surface current stuff. Lorenz Gauge Plan H v 2 First do A: Region 1: (2 - μ1ε1 ∂t2)A = - μ1J (1.3.5) divA = - μ1ε1 ∂tφ . // Lorenz Gauge (1.3.6) Region 2: Start with the pre-gauge equation (2 - μ2ε2 ∂t2) A = grad [μ2ε2 ∂tφ + divA ] - μ2J = grad [μ2ε2 ∂tφ + (- μ1ε1 ∂tφ) ] - μ2J = [μ2ε2- μ1ε1]∂tgrad φ - μ2J Then replace gradφ = -E - ∂tA and set E = 0 to get = [μ2ε2- μ1ε1]∂t(-E-∂tA) - μ2J = - [μ2ε2- μ1ε1]∂t2A - μ2J then we have (2 - μ2ε2 ∂t2) A = - [μ2ε2- μ1ε1]∂t2A - μ2J (2 - μ2ε2 ∂t2+ [μ2ε2- μ1ε1]∂t2) A = - μ2J (2 - μ1ε1 ∂t2) A = - μ2J2 So carrying out the program H here we get (2 - μ1ε1 ∂t2)A = - μ1J1 region 1 (2 - μ1ε1 ∂t2)A = - μ2J2 region 2 (2 - μ1ε1 ∂t2)A = - μ3J3 region 3 (2 - μ1ε1 ∂t2)A = - Σi=1NμiJi Now do the φ equations. Region 1: 2φ + ∂t[div A] = -ρ/ε1 Region 2: 2φ + ∂t[div A] = -ρ/ε2 2φ + ∂t[- μ1ε1 ∂tφ] = -ρ/ε2 (2 - μ1ε1 ∂t2)φ = -ρ/ε2 thus we get (2 - μ1ε1 ∂t2)φ = -ρ/ε1 region 1 (2 - μ1ε1 ∂t2)φ = -ρ/ε2 region 2 (2 - μ1ε1 ∂t2)φ = -ρ/ε3 region 3 (2 - μ1ε1 ∂t2)φ = -ρ/ε1 all of R Plan H v 2 for the fields E and B From (1.2.1) and (1.2.2) we can write (2 - μ1ε1 ∂t2)E = μ1∂tJ1 + (1/ε1) grad ρ1 = μ1∂tJ1 + (1/ε1) grad ρs // region 1 (2 - μ2ε2 ∂t2)E = μ2∂tJ2 + (1/ε2) grad ρ2 = μ2∂tJ2 // region 2 (2 - μ3ε3 ∂t2)E = μ3∂tJ3 + (1/ε3) grad ρ3 = μ3∂tJ3 // region 3 (2 - μ1ε1 ∂t2)B = - μ1 curl J1 // region 1 (2 - μ2ε2 ∂t2)B = - μ2 curl J2 // region 2 (2 - μ3ε3 ∂t2)B = - μ3 curl J3 // region 3 Now maybe I can simplify this somewhat. If J1 = σ1E then the first becomes (2 - μ1ε1 ∂t2)E = μ1σ1∂tE + (1/ε1) grad ρs // region 1 (2 - μ1ε1 ∂t2 - μ1σ1∂t)E = + (1/ε1) grad ρs // region 1 And also curl J1 = ε1curl E = ε1(-∂tB) so that (2 - μ1ε1 ∂t2)B = - μ1 curl J1 (2 - μ1ε1 ∂t2)B = - μ1 ε1(-∂tB) (2 - μ1ε1 ∂t2 - μ1σ1∂t)B = 0 So we can rewrite the 6 equations as (2 - μ1ε1 ∂t2 - μ1σ1∂t)E = + (1/ε1) grad ρs // region 1 (2 - μ2ε2 ∂t2)E = μ2∂tJ2 + (1/ε2) grad ρ2 = μ2∂tJ2 // region 2 (2 - μ3ε3 ∂t2)E = μ3∂tJ3 + (1/ε3) grad ρ3 = μ3∂tJ3 // region 3 (2 - μ1ε1 ∂t2 - μ1σ1∂t)B = 0 // region 1 (2 - μ2ε2 ∂t2)B = - μ2 curl J2 // region 2 (2 - μ3ε3 ∂t2)B = - μ3 curl J3 // region 3