Exterior solution attempted rewrite of section 2_5 REVIEWED

Attempted rewrite of Section 2.5 now 2.6 I think PhL 11.27.13 Solving the Wave Equations 1 Look now at just the exterior wave equations. 3 Does the exterior solution satisfy Maxwell's Equations? 4 Plan A: Ignore the Maxwell violations and complete the solution 7 Plan B. Try adding in an Er radial field. 9 Plan C: Have all 6 fields active -- but this is Appendix D! 12 Solving the Wave Equations First, write the two wave equations of interest (2 + β12)E(x,ω) = 0 (2 + β12) B(x,ω) = 0 // region 1 r>a (2 + β22)E(x,ω) = 0 (2 + β22) B(x,ω) = 0 // region 2 r<a Next, assume that in both regions E(x,ω) = Ez(x,ω) => only Ez B(x,ω) = Bθ(x,ω) => only Bθ Next, use the vector Laplacian in cylindrical coordinates 2E = [2Er -2 r-2∂θEθ - r-2Er] + [2Eθ +2 r-2∂θEr - r-2Eθ] + 2Ez = + 2Ez 2B = [2Br -2 r-2∂θBθ - r-2Br] + [2Bθ +2 r-2∂θBr - r-2Bθ] + 2Bz = [ -2 r-2∂θBθ] + [2Bθ - r-2Bθ] = [2Bθ - r-2Bθ] so that in region 1 we get (2 + β12)E(x,ω) = 0 2E(x,ω) + β12 E(x,ω) = 0 2Ez(x,ω) + β12 Ez(x,ω) = 0 (2 + β12)Ez(x,ω) = 0 And for the B field also in region 1, (2 + β12) B(x,ω) = 0 2B(x,ω) + β12 B(x,ω) = 0 [2Bθ - r-2Bθ] + β12 Bθ(x,ω) = 0 (2 - r-2 + β12) Bθ(x,ω) At this point then our four wave equations are (2 + β12)Ez(x,ω) = 0 (2 - r-2 + β12) Bθ(x,ω) = 0 // region 1 (2 + β22)Ez(x,ω) = 0 (2 - r-2 + β22) Bθ(x,ω) = 0 // region 2 Next, assume that all fields have exp(-jβ1z) z-dependence, Ez(x,ω) = exp(-jβ1z)Ez(x,y,ω) Bθ(x,ω) = exp(-jβ1z)Bθ(x,y,ω) Then the four equations become, since ∂z2 → -β12, (22D)Ez(x,y,ω) = 0 (22D - r-2) Bθ(x,y,ω) = 0 // region 1 (22D + [β22-β12]) Ez(x,y,ω) = 0 (22D - r-2 + [β22-β12]) Bθ(x,y,ω) = 0 // region 2 Now invoke azimuthal symmetry to say Ez(x,y,ω) = E(r) Bθ(x,y,ω) = B(r) so the four equations are then (22D)E(r) = 0 (22D - r-2) B(r) = 0 // region 1 (22D + [β22-β12]) E(r) = 0 (22D - r-2 + [β22-β12]) B(r) = 0 // region 2 Now when acting on a function only of r we know that 22Dg(r) = r-1∂r(r∂rg(r)) = r-1[ r ∂r2g(r) + ∂rg(r)] = g"(r) + r-1g'(r) Therefore our four equations become E"(r) + r-1E'(r) = 0 B"(r) + r-1B'(r) - r-2 B(r) = 0 // region 1 E"(r) + r-1E'(r) + [β22-β12] E(r) = 0 // region 2 B"(r) + r-1B'(r) + [β22-β12] B(r) - r-2 B(r) = 0 Multiply by r2 to get r2E"(r) + r E'(r) = 0 r2B"(r) + r B'(r) - B(r) = 0 // region 1 r2 E"(r) + rE'(r) + r2 [β22-β12] E(r) = 0 r2 B"(r) + rB'(r) + r2 [β22-β12]B(r) - B(r) = 0 // region 2 Look now at just the exterior wave equations. OK, let's stop the train and look at the region 1 equations r2E"(r) + r E'(r) = 0 r2B"(r) + r B'(r) - B(r) = 0 // region 1 Here is what Maple has to say: Thus we conclude that E(r) = A + B lnr // as expected since it is just Laplace azisym B(r) = Cr + Dr-1 // not sure how to relate this to ang mom or whatever I did check by hand that the B(r) works. So in the context of the round wire problem we really have E(r) = A + B lnr B(r) = Dr-1 // region 1 Does the exterior solution satisfy Maxwell's Equations? Let's try a few consistency checks. (1) What about div E = 0? That reads 0 = div E = r-1∂r(rEr) + r-1∂θEθ + ∂zEz = ∂zEz = -jβ1Ez = -jβ1[A + B lnr] so then 0 = -jβ1[A + B lnr] That does not look good! I know from the boundary match at r = a that A+Bln(a) = E(a) there which is some large value because Jz and Ez are both large in the wire just below the surface in the skin effect situation, and Ez is a tangential field with then is continuous through the surface. (2) What about div B = 0? 0 = div B = r-1∂r(rBr) + r-1∂θBθ + ∂zBz = 0 So this one is OK. (3) What about curl E = - ∂tB ? curl E = - ∂tB E(x,ω) curl E = [ r-1∂θEz - ∂zEθ] + [∂zEr - ∂rEz] + [ r-1∂r(rEθ) - r-1∂θEr ] = [ r-1∂θEz] + [- ∂rEz] = [- ∂rEz] The test here is then to see if - ∂rEz = -jω Bθ LHS = - ∂rEz = -∂rE(r) = -∂r[A+Blnr] = -B(1/r) RHS = -jωBθ = -jωB(r) = -jω[Dr-1] The forms agree, and moreover we conclude that B = jωD ! (4) Now let's try the other curl equation (in region 1 there is no free current J ) curl H = ∂tD + J => curl B = μ1 jω ε1E So curl B = [ r-1∂θBz - ∂zBθ] + [∂zBr - ∂rBz] + [ r-1∂r(rBθ) - r-1∂θBr ] = [∂zBθ] + [ r-1∂r(rBθ)] = [∂zB(r)] + [ r-1∂r(rB(r))] = [ r-1∂r(rB(r))] The test here then is to see if r-1∂r(rB(r)) = μ1 jω ε1E(r) LHS = r-1∂r(rB(r)) = r-1∂r(r Dr-1) = 0 RHS = μ1 jω ε1E(r) = μ1 jω ε1[A + Blnr] 0 = μ1 jω ε1[A + Blnr] We have already shown that B = jωD so we cannot just set A = B = 0 to satisfy this equation. This is the paradox I ended up yesterday (which was 11.27.13) using the integral forms, but the curl equations are simpler to deal with, paradox obtained faster. Perhaps this second equation is roughly valid for small ω. So let's back up and show how the two curl equations make the Helmholtz equation. Well, the derivations are all laid out in detail in Section 1.2 of lines. We find there that, let's say region 1 only, (2 - μ1ε1 ∂t2)E = μ1∂tJ + (1/ε) grad ρ1 (1.2.1) (2 - μ1ε1 ∂t2)B = - μ1 curl J (1.2.2) or (2+β12)E = μ1∂tJ + (1/ε) grad ρ1 (2+β12)B = - μ1 curl J But in region 1 (away from boundaries) we have ρ = 0 and J = 0 so we get (2+β12)E = 0 (2+β12)B = 0 So there is nothing wrong with these wave equations, they are exact in region 1. So the problem is likely with our various ansatz assumptions. Recall they were 4 (the azisym form is not an assumption once you get to the f(x,y) forms ) E(x,ω) = Ez(x,ω) => only Ez B(x,ω) = Bθ(x,ω) => only Bθ Ez(x,ω) = exp(-jβ1z)Ez(x,y,ω) Bθ(x,ω) = exp(-jβ1z)Bθ(x,y,ω) I know that the first equation are not correct in an active transmission line because there are radial currents pumping out the surface charge. But in the current wire, there is no surface charge. Possible Conclusion #1: Maxwell does not support an exact solution of my form! If that conclusion is correct, you cannot have a wave going down a single wire as I have tried to make it out. I have never heard of a single wire transmission line. Is there some limit in which my solution is approximately valid? In a wide spaced twinlead we might have the azisym situation for one of the wires. But we know there are charge densities active on the surfaces in that case. How are things above affected? Pause to rewrite some of Chapter 2: I now realize that you can do this trick just below the wire surface, 2πaHθ = I => 2πaBθ/μ2 = I => B(a) = [μ2I/(2πa)] I added this to Chapter 2 first section, changed the box and ending 2.2.xx equation numbers Plan A: Ignore the Maxwell violations and complete the solution If I assume that ω is "small" in some sense, then my two Max violations can be ignored. But I don't know "how small" unless maybe I do some work. So let's go ahead and accept these violations for the moment and finish our solution. From the new box (2.2.31) we have the interior solution shown interior solution: region 2 E(r) = [μ2I/(2πa)] (-jω/β2) B(a) = [μ2I/(2πa)] B(r) = [μ2I/(2πa)] E(a) = [μ2I/(2πa)] (-jω/β2) exterior solution: region 1 E(r) = A' + B' ln(r/a) B(r) = D'r-1 = (B'/jω)r-1 since above we found that B' = jωD' Matching boundary conditions (assuming for the moment that μ1= μ2)for the fields then gives [μ2I/(2πa)] (-jω/β2) = A' match Ez (1/μ2) [μ2I/(2πa)] = (1/μ1)(B'/jω)a-1 match Hθ The two constants are found to be A' = [μ2I/(2πa)] (-jω/β2) B' = [μ2I/(2πa)] (jωa) Thus, the exterior fields are E(r) = [μ2I/(2πa)] (-jω/β2) + [μ2I/(2πa)] (jωa) ln(r/a) = [μ2I/(2πa)] { (-jω/β2) + (jωa) ln(r/a) } = [μ2I/(2πa)](jω) { (-1/β2) + a ln(r/a) } = [μ2I/(2πa)](jω) { (-1/β2) + a ln(r/a) } B(r) = (B'/jω)r-1 = (1/jω)B' r-1 = (1/jω) [μ2I/(2πa)] (jωa) r-1 = [μ2I/(2πa)] a r-1 = [μ2I/(2πa)] (a/r) = [μ1I/(2πr)] This our result for B(r) comes out being the simple Ampere's law result, so in effect we have just neglected the displacement current in region 1. To summarize Region 1 B(r) = [μ2I/(2πa)] (a/r) B(a+) = [μ2I/(2πa)] E(r) = [μ2I/(2πa)] (jω) { (-1/β2) + a ln(r/a) } E(a) = [μ2I/(2πa)] (jω) { (-1/β2) Region 2 B(r) = [μ2I/(2πa)] B(a-) = [μ2I/(2πa)] E(r) = [μ2I/(2πa)] (-jω/β2) E(a) = [μ2I/(2πa)] (-jω/β2) This is then my complete solution given the "violations" mentioned above. Can I now estimate the size of the violations in region 1? Are there similar violations in region 2? The quantities I need to be small are these -jβ1[A' + B' ln(r/a)] μ1 jω ε1[A' + B' ln(r/a)] That is to say, I want both these quantities to be 0 in fact. But I just no "scale" for doing this. You need a larger picture somehow and then maybe you can throw out small terms relative to large terms. Plan B. Try adding in an Er radial field. Certainly there is a transverse E field and that does not appear in the above. That might be the major issue. Perhaps near one of the wires write E(x,ω) = Ez(x,ω) + Er(x,ω) What happens to everything above if I add this extra term in the E world and make no change in B 2E = [2Er -2 r-2∂θEθ - r-2Er] + [2Eθ +2 r-2∂θEr - r-2Eθ] + 2Ez = [2Er - r-2Er] + [2 r-2∂θEr] + 2Ez = [2Er - r-2Er] + 2Ez Then (2 + β12)E(x,ω) = 0 2E(x,ω) + β12 E(x,ω) = 0 [2Er - r-2Er] + 2Ez + β12 Ez + β12Er(x,ω) = 0 This then leads to two equations instead of one: 2Er(x,ω) + β12Er(x,ω) - r-2Er(x,ω) = 0 2Ez(x,ω) + β12Ez(x,ω) = 0 or (2 + β12 - r-2)Er(x,ω) = 0 (2 + β12 )Ez(x,ω) = 0 Now let's make the ansatz that both components have the same z behavior, Ez(x,ω) = exp(-jβ1z)Ez(x,y,ω) = exp(-jβ1z)E(r) Er(x,ω) = exp(-jβ1z) Er(r) The two equations then become (22D - r-2) Er(r) = 0 (22D ) E(r) = 0 The second equation is the same as obtained last time without the Er field, but now we have added in the Er field and its governing equation. From above, I know we have these two solution types: E(r) = A + B lnr Er(r) = Fr-1 Pause just one moment to check divergence equations. 0 = div E = r-1∂r(rEr) + r-1∂θEθ + ∂zEz = r-1∂r(rEr) + ∂zEz = r-1∂r(rEr) - jβ1Ez = r-1∂r(rEr(r)) - jβ1E(r) = 0 - jβ1F r-1 so entirely in the E world we have a violation in region 1 since no charge there! But now I might claim that Er is very small so F is small, so this is then a small violation. OK, I like that. Now onto the curl equations. Let's assume that B still has this form B(r) = Dr-1 B(x,ω) = Bθ(x,ω) Then here goes for the curl E equation: curl E = - ∂tB = -jωB curl E = [ r-1∂θEz] + [∂zEr - ∂rEz] + [- r-1∂θEr ] = + [-jβ1Er - ∂rEz] The curl E equation then has LHS = [-jβ1Er - ∂rEz] RHS = -jωBθ so we must then have -jβ1Er - ∂rEz = -jωBθ or -jβ1Er(r) - ∂rE(r) = -jωB(r) Now use the forms above to get -jβ1[Fr-1] - ∂r[A + B lnr] = -jω [Dr-1] -jβ1Fr-1 - Br-1 = -jω Dr-1 -jβ1F - B = -jω D This is fairly encouraging, maybe we can make this work. Then here goes for the curl B equation. Since B has not changed, we still have curl B = [ r-1∂r(rB(r))] and so the curl equation curl B = μ1 jω ε1E now says [ r-1∂r(rB(r))] = μ1 jω ε1[Ez(x,ω) + Er(x,ω) ] which then says [ r-1∂r(rB(r))] = μ1 jω ε1[E(r) + Er(r) ] which gives two equations r-1∂r(rB(r)) = μ1 jω ε1E(r) 0 = μ1 jω ε1 Er(r) We can dismiss the second violation by arguing small Er, just as we did in the div E = 0 case. The first equation then says r-1∂r(r Dr-1) = μ1 jω ε1[A + B lnr] 0 = μ1 jω ε1[A + B lnr] Therefore, adding in the extra Er field component does not repair the problem we had without it! Unless it says that both A and B must be very small and we then can ignore the violation as with other examples above. But here there is no Er field sitting there which we can claim is small! Now consider what happens when we match the BC at r = a. We get A + B ln(a) = E(a) I know that E(a) is large from the inside analysis, so I don't thing both A and B can be small! So the problem persists! Statement of Problem: In the curl equation curl B = μ1 jω ε1E, we cannot get a balance in the z component. The displacement current is required to feed the curl of B. There IS non-zero a displacement current for sure since we know Ez exists in region 1, but the curl B is always 0 since Bθ = D/r . So something is then wrong with B! Plan C: Have all 6 fields active -- but this is Appendix D! If I draw the known twin-lead pattern, I see clearly that there is some Br component! It must be this extra component that fixes the curl equation. I guess I could try to add in such a term. What would such an extra term do? I have to start ALL over I guess First, write the two wave equations of interest (2 + β12)E(x,ω) = 0 (2 + β12) B(x,ω) = 0 // region 1 r>a (2 + β22)E(x,ω) = 0 (2 + β22) B(x,ω) = 0 // region 2 r<a Let's just assume that ALL components exist and see what happens. E(x,ω) = Er(x,ω) + Eθ(x,ω) + Ez(x,ω) B(x,ω) = Br(x,ω) + Bθ(x,ω) + Bz(x,ω) Next, use the vector Laplacian in cylindrical coordinates 2E = [2Er -2 r-2∂θEθ - r-2Er] + [2Eθ +2 r-2∂θEr - r-2Eθ] + 2Ez 2B = [2Br -2 r-2∂θBθ - r-2Br] + [2Bθ +2 r-2∂θBr - r-2Bθ] + 2Bz The region 1 Helmholtz equations are then [2Er -2 r-2∂θEθ - r-2Er] + [2Eθ +2 r-2∂θEr - r-2Eθ] + 2Ez = -β12 [Er(x,ω) + Eθ(x,ω) + Ez(x,ω) ] [2Br -2 r-2∂θBθ - r-2Br] + [2Bθ +2 r-2∂θBr - r-2Bθ] + 2Bz = -β12 [Br(x,ω) + Bθ(x,ω) + Bz(x,ω) ] This gives us the following 6 equations: 2Er -2 r-2∂θEθ - r-2Er = -β12Er(x,ω) 2Eθ +2 r-2∂θEr - r-2Eθ = -β12 Eθ(x,ω) 2Ez = -β12 Ez(x,ω) 2Br -2 r-2∂θBθ - r-2Br = -β12Br(x,ω) 2Bθ +2 r-2∂θBr - r-2Bθ = -β12 Bθ(x,ω) 2Bz = -β12 Bz(x,ω) Now haven't I done all this somewhere before? Some appendix? It is Appendix D where I do the "general case" including non-azisym! I assume the exp(-jβdz) dependence for all components in (D.1.9) where I do only E and ignore B. I use the cylin coords vector Laplacian explicitly. I then first look at the z equation which is [2E]z + β2 Ez = 0 . Here β is generic, I don't pick any region yet. I end up with this radial equation [r2∂r2 + r ∂r - m2 +r2( β2- βd2)] Ez(r,m) = 0 I then do this for each of the three component equations and I end up with this summary: Summary of Examples: [2E]z + β2 Ez = 0 : [r2∂r2 + r ∂r - m2 + r2 ( β2- βd2)] Ez(r,m) = 0 (D.1.16) [2E]r + β2 Er = 0 : [r2∂r2 + r∂r - (m2+1) + r2(β2-βd2)] Er(r,m) - 2jm Eφ(r,m) = 0 (D.1.17) [2E]φ + β2 Eφ = 0 : [r2∂r2 + r∂r - (m2+1) + r2(β2-βd2)] Eφ(r,m) + 2jmEr(r,m) = 0 (D.1.18) div E = 0 : ∂r [r Er(r,m)] + jmEφ(r,m) -jβd r Ez(r,m) = 0 (D.1.19) Again, this is all done in light of the full partial wave expansion. Glad I did that! I think I go on to completely solve the problem ! Wow. And matching boundary conditions too. Here are my results, E Fields Inside a Round Wire x = β'r β'2 = β2- βd2 ≈ β2 β = ω ξ =[ε + σ/(jω)] ≈ σ/(jω) conductor xa = β'a βd = ω ξd =[εd + σd/(jω)] ≈ εd dielectric E(r,φz,t) = ej(ωt-βz) E(r,φ) (D.1.2) E = Er + Eφ + Ez (for any arguments) E(r,φ) = E(r,0) + 2!Syntax Error, I[Re{E(r,m)}cos(mφ) - Im{E(r,m)} sin(mφ)] = real (D.1.4) where: Ez(r,m) = (1/4) ηm I Rdc [ - ] a = radius ηm ≡ Ez(r,0) = (1/2) I Rdc [] // = (2.2.22) for βd = 0 η0 = 1 Er(r,m) = (j/4) (aβd) ηm I Rdc [ + - ] Er(r,0) = (j/2) (aβd) I Rdc [ ] Eφ(r,m) = (1/4) (aβd) ηm I Rdc [ - + + ] Eφ(r,0) = 0 (D.4.6) Good heavens! But do I know these solutions satisfy the two curl equations, or is my z dependence ansatz too strong? Well, I only have the E solution! Where is the B solution? Maybe I get that from the curl E equation! My entire effort here is "inside the wire only". Hmmm. Comments: I have been ignoring this Appendix D for a long time, just a painful messy thing. But now I see that it is basically a complete solution to the round wire problem, as claimed. I even then have a special section on the outside of the wire, which I have been struggling with here in this doc. I found: β' = βd2 - βd2 = 0 so that: [2E]z + β2 Ez = 0 : [r2∂r2 + r ∂r - m2] Ez(r,m) = 0 (D.1.16)o [2E]r + β2 Er = 0 : [r2∂r2 + r∂r - (m2+1)] Er(r,m) - 2jm Eφ(r,m) = 0 (D.1.17)o [2E]φ + β2 Eφ = 0 : [r2∂r2 + r∂r - (m2+1)] Eφ(r,m) + 2jmEr(r,m) = 0 (D.1.18)o div E = 0 : ∂r [r Er(r,m)] + jmEφ(r,m) -jβd r Ez(r,m) = 0 (D.1.19)o OK, I am ending this doc now, I have seen enough.