App D actions 1 v 2 REVIEWED

App D Actions 1 v 2 PhL 11.30.13 This document supersedes App D Actions 1 in which I made a math error. So no need ever to look at the older App D Actions 1 ever again. [ I threw that doc out! ] This doc addresses a whole battery of ugly questions and I think as of 12/5/13 they have all been resolved so this doc is now "fully reviewed". See TOC below for subject matter. 1. This is done now for general m and for m = 0 in Appendix D Section D.4 2. I showed this explicitly and it is reported now in Appendix D section D.5. 3. βd2 << |β|2 is now addressed in Appendix D just below equation (D.2.2). 4. I only assume βd2 << |β|2 very late in Appendix D in (D.4.12) for the B fields only 5.6. The Eφ(r=0) = 0 idea is rejected as a real BC, it will just be true. I explain Eφ = 0 at the surface now at the very start of Appendix D below (D.1.2), so that BC is now understood. 7,8. Resolving the above BC then resolved these questions. 1. Compute the Appendix D Magnetic Field B for m = 0. 1 2. Do these E and B fields satisfy Maxwell's equations? 3 3. What exactly does the condition βd2 << |β|2 imply? 6 4. Did I assume βd2 << |β|2 somewhere in my Appendix D development? Answer is no! 9 5. My strange Eφ boundary condition. 10 6. Would it be better to say maybe that Eφ(r=0,m) = 0 ? 12 7. So now I am stuck with a new question: how to compute am and Km ? 13 8. What should I do about boundary conditions then for m > 0 ? 15 1. Compute the Appendix D Magnetic Field B for m = 0. I did not do this in App D and maybe I should, but I will try to do it right here for the first time. I comment on this at the end of App D and perhaps will adjust that comment. It seems the easiest method is this: curl E = - jωB B = (-1/jω) curl E Notes before starting: In this section, write β' where things should be β'. Assume Eφ = Eθ ≡ 0 for this m=0 analysis, since can select Cφ0 = 0 as shown below Everything here is for inside the wire only Here are the Appendix D m = 0 results for the E field Ez(r,0) = (1/2) (aβ') I Rdc [ J0(β'r)/ J1(β'a) ] Er(r,0) = (j/2) (aβd) I Rdc [J1(β'r)/ J1(β'a)] Eφ(r,0) = Cφ0 J1(β'r) = 0 β'2 = β2- βd2 => β'2+ βd2 = β2 Remember that β has a strange and large phase, but think of βd as real. Rdc = β2 ≈ -jωμσ and therefore (-jω/β) ≈ β/(μσ) So we use (quote from the web just to back up mine from my TK doc) curl E = [ r-1∂θEz - ∂zEθ] + [∂zEr - ∂rEz] + [ r-1∂r(rEθ) - r-1∂θEr ] = [ r-1∂θEz] + [∂zEr - ∂rEz] + [- r-1∂θEr ] // Eθ ≡ 0 = [∂zEr - ∂rEz] = [-jβdEr - ∂rEz] = [-jβd{Er} + {- ∂rEz}] // ok with Ez = (1/2) (aβ') I Rdc [ J0(β'r)/ J1(β'a) ] // sign agrees with Chap 2 and Matick Er = (j/2) (aβd) I Rdc [J1(β'r)/ J1(β'a)] . // I have no verification of this sign! First, compute the second term: Ez = (1/2) (aβ') I Rdc [ J0(β'r)/ J1(β'a) ] ∂rEz = (1/2) (aβ') I Rdc β' [ J0'(β'r)/ J1(β'a) ] = - (1/2) (aβ') I Rdc β' [ J1(β'r)/ J1(β'a) ] Then we have [curl E]θ = -jβd { (j/2) (aβd) I Rdc [J1(β'r)/ J1(β'a)]} + {(1/2) (aβ') I Rdc β' [ J1(β'r)/ J1(β'a) ]} = + βd (1/2) (aβd) I Rdc [J1(β'r)/ J1(β'a)] + (1/2) (aβ'2) I Rdc [ J1(β'r)/ J1(β'a) ] = (1/2) aβd2 I Rdc [J1(β'r)/ J1(β'a)] + (1/2)(aβ'2 I Rdc [ J1(β'r)/ J1(β'a) ] = (1/2) a I Rdc [J1(β'r)/ J1(β'a)] { βd2 + β'2 } = (1/2) a β2 I Rdc [J1(β'r)/ J1(β'a)] Therefore, B = (-1/jω) curl E = [(-1/jω) (1/2) a β2 I Rdc [J1(β'r)/ J1(β'a)]] // exact How does this compare to the Chapter 2 result? Bθ = (-1/jω) (1/2) a β2 I Rdc [J1(β'r)/ J1(β'a)] = (-1/jω) (1/2) a β2 I [J1(β'r)/ J1(β'a)] ≈ (-1/jω) (1/2) a (-jωμσ) I [J1(β'r)/ J1(β'a)] // here assume β2 ≈ -jωμσ ≈ ( (μ) I [J1(β'r)/ J1(β'a)] ≈ [μI/(2πa)] [J1(βr)/ J1(βa)] // This is exactly the Chapter 2 result! Recall that inside Chapter 2 we make the assumption that β2 ≈ -jωμσ , which says β2 ≈ β'2 and all that stuff, so our exact App D result → Chapter 2 result when this approx is added. Comments: The B field is entirely in the direction which is certainly a simplifying fact. This result is exact with the assumptions I have made about z dependence. So here are the m = 0 field results a la Appendix D: Ez(r,0) = (1/2) (aβ') I Rdc [J0(β'r)/ J1(β'a)] Er(r,0) = (j/2) (aβd) I Rdc [J1(β'r)/ J1(β'a)] Bθ(r,0) = (-1/jω)(a/2) I Rdc [J1(β'r)/ J1(β'a)] (βd2 + β'2) The other three field components are each identically 0. 2. Do these E and B fields satisfy Maxwell's equations? It would be good the check them all, one at time! Trust but verify. (1) div B = 0? div B = r-1∂r(rBr) + r-1∂θBθ + ∂zBz = 0 because there is no Br or Bz , and because Bθ does not depend on θ. One down! (2) div E = 0? div E = r-1∂r(rEr) + r-1∂θEθ + ∂zEz = r-1∂r(rEr) + ∂zEz = r-1∂r(rEr) -jβd Ez Compute the first term rEr = (j/2) (aβd) I Rdc(1/ J1(β'a)) [ rJ1(β'r) ] ∂r(rEr) = (j/2) (aβd) I Rdc(1/ J1(β'a)) [ rβ'J1'(β'r) + J1(β'r) ] r-1∂r(rEr) = (j/2) (aβd) I Rdc(1/ J1(β'a)) [ β'J1'(β'r) + r-1J1(β'r) ] Therefore div E = {(j/2) (aβd) I Rdc(1/ J1(β'a)) [ β'J1'(β'r) + r-1J1(β'r) ]} -jβd { (1/2) (aβ') I Rdc [J0(β'r)/ J1(β'a)]} = (j/2) (aβd) I Rdc(1/ J1(β'a)) { J1'(β'r) + r-1J1(β'r) - β'J0(β'r) } = (j/2) (aβd) I Rdc(1/ J1(β'a) β' { J1'(β'r) + (rβ')-1J1(β'r) - J0(β'r) } = (j/2) (aβd) I Rdc(1/ J1(β'a) (β'/x) { xJ1'(x) + J1(x) - xJ0(x) } Now here is what Spiegel 24.19 says xJ1'(x) = xJ0(x) - J1(x) = 0 => xJ1'(x) + J1(x) - xJ0(x) = 0 and therefore we have shown that div E = 0, hurray for once! Since this was built into App D, it is certainly an expected result. (3) curl E = - ∂tB or curl E = -jωB This is the equation we used to compute B in the first place, so it will be OK. But let's do it anyway just to make sure: curl E = [(1/2) a I Rdc [J1(β'r)/ J1(β'a)] (βd2 + β'2)] from above -jωB = (-jω) [(-1/jω) (1/2) a I Rdc [J1(β'r)/ J1(β'a)] (βd2 + β'2)] and yes we are OK. (4) curl H = ∂tD + J or curl B = jωμε E + μσE = (μσ + jωμε) E or curl B = μ(σ+jωε) E This is the most dangerous one (my experience of failures), I hope it works. Start on the left, curl B = [ r-1∂θBz - ∂zBθ] + [∂zBr - ∂rBz] + [ r-1∂r(rBθ) - r-1∂θBr ] = [- ∂zBθ] + [ r-1∂r(rBθ)] = [+jβdBθ] + [ r-1∂r(rBθ)] Compute the second term: Bθ = (-1/jω)(a/2) I Rdc [J1(β'r)/ J1(β'a)] (β2) rBθ = (-1/jω)(a/2) I Rdc(1/ J1(β'a)) (β2) [r J1(β'r) ] ∂r(rBθ) = (-1/jω)(a/2) I Rdc(1/ J1(β'a)) (β2) [ r β'J1'(β'r) + J1(β'r) ] r-1∂r(rBθ) = (-1/jω)(a/2) I Rdc(1/ J1(β'a)) (β2) [ β'J1'(β'r) +r-1 J1(β'r) ] Therefore we have [curl B]z = (-1/jω)(a/2) I Rdc(1/ J1(β'a)) (β2) [ β'J1'(β'r) +r-1 J1(β'r) ] Our Maxwell says this is supposed to be equal to μ(σ+jωε)Ez = μ(σ+jωε) (1/2) (aβ') I Rdc [ J0(β'r)/ J1(β'a) ] It does not look promising. We now compare the two sides (-1/jω)(a/2) I Rdc(1/ J1(β'a)) (β2) [ β'J1'(β'r) +r-1 J1(β'r) ] = μ(σ+jωε) (1/2) (aβ') I Rdc [ J0(β'r)/ J1(β'a) ] ? (-1/jω) (β2) [ β'J1'(β'r) +r-1 J1(β'r) ] = μ(σ+jωε) (β') J0(β'r) ] ? (-1/jω) (β2) [ β'J1'(β'r) + r-1 J1(β'r) ] = μ(σ+jωε) (β') J0(β'r) ] ? Now we know that β2 = ω2μξ = ω2μ [ε + σ/(jω)] = ω2μ/(jω) * [jωε + σ] = -jω μ(σ+jωε) which says μ(σ+jωε) = (-1/jω) β2 So now our test is this: (-1/jω) (β2) [ β'J1'(β'r) + r-1 J1(β'r) ] = (-1/jω) β2 (β') J0(β'r) ] ? [ β'J1'(β'r) + r-1 J1(β'r) ] = (β') J0(β'r) ] ? β' [J1'(β'r) + (rβ')-1 J1(β'r) ] = (β') J0(β'r) ] ? [J1'(β'r) + (rβ')-1 J1(β'r) ] = (J0(β'r) ] ? [J1'(x) +x-1 J1(x) ] = J0(x) ? xJ1'(x) + J1(x) = xJ0(x) ? This is the same Spiegel 24.19 that I used above, so success for the component of curl B = μ(σ+jωε) E. Now let's examine the component of this Maxwell equation: [curl B]r = +jβdBθ = (jβd) (-1/jω)(a/2) I Rdc [J1(β'r)/ J1(β'a)] (β2) = - (βd) (1/ω)(a/2) I Rdc [J1(β'r)/ J1(β'a)] (β2) Max ways this is supposed to equal μ(σ+jωε) Er = μ(σ+jωε) (j/2) (aβd) I Rdc [J1(β'r)/ J1(β'a)] So we then have another equation to test out: - (βd) (1/ω)(a/2) I Rdc [J1(β'r)/ J1(β'a)] (β2) = μ(σ+jωε) (j/2) (aβd) I Rdc [J1(β'r)/ J1(β'a)] ? - (βd) (1/ω)(a/2) (β2) = μ(σ+jωε) (j/2) (aβd) ? - (βd) (1/ω)(a/2) (β2) = μ(σ+jωε) (j/2) (aβd) ? - (1/ω) (β2) = μ(σ+jωε) j ? Recall from above that μ(σ+jωε) = (-1/jω) β2 so we continue - (1/ω) (β2) = (-1/jω) β2j ? - (1/ω) (β2) = (-1/ω) β2 ? and again agreement is exact so success for the component of curl B = μ(σ+jωε) E. Conclusions on Verification of the four Maxwell equations. (1) div B = 0 exactly with no assumptions. (2) div E = 0 exactly with no assumptions. (3) curl E = -jωB because that is how we compute B (4) curl B = μ(σ+jωε) E with no assumptions Conclusion: The ansatz ej(ωt-βz) yields a solution of the round wire interior which satisfies all four of Maxwell's equations with no approximations. It works for all frequencies 3. What exactly does the condition βd2 << |β|2 imply? This assumption is not made in Appendix D nor above, but is made in Chapter 2. βd2 << |β|2 βd2 = ω2μdεd // I assume all is real to keep it simple β2 = ω2μξ = ω2μ[ε + σ/(jω)] |β2|2 = (ω2μ)2[ε - j σ/(ω)] [ε + j σ/(ω)] = (ω2μ)2 [ ε2 + σ2/ω2] |β2| = ω2μ So condition is βd2 << |β|2 ω2μdεd << ω2μ μdεd << μ (μdεd)2 << μ2(ε2+σ2/ω2) (μdεd)2 - (με)2 << μ2 σ2/ω2 [(μdεd)2 - (με)2]-1 >> ω2/ μ2 σ2 ω2 << μ2 σ2 [(μdεd)2 - (με)2]-1 ω2 << ω2 << Now typically we will have μ = μd = μ0 in which case this says ω2 << At this point I don't know what to say about ε . This result seems wrong to me. Well what is the largest that the denominator might be, because this controls the bound. Perhaps ε = ε0 and εd = 2.3 ε0 so then ω2 << = 1/(2.3)2 * σ2/ε02 = 0.19 σ2/ε02 ω << 0.43 σ/ε0 // so this undoes the squaring I did above, regarding << ω << 0.43 [5.81 x 107 mho/m]/ [8.8541877 x 10-12 farad/m] = 0.43 (5.8/8.6) 1019 mho/m * m/farad // mho/F = sec-1 ≈ 0.3 x 1019 sec-1 2πf << 0.3 x 1019 f << (1/2π) 0.3 x 1019 = (1/2π)30 x 1017 = 4.8 x 1017 = 4.8 x 108 GHz So as long as we keep the frequency below 1/2 billion GHz, our condition is met. Let's start over on the above like this: β2 = ω2μξ = ω2μ[ε + σ/(jω)] I argue below (2.2.3) that as long as f << one billion GHz, we can neglect the ε term. Then we have β2 ≈ ω2μ[σ/(jω)] |β2| ≈ ωμσ // sec-1* henry/m * mho/m = sec-1 ohm-sec * mho/m = 1/m2 OK Then my I get βd2 << |β|2 ω2μdεd << ωμσ ωμdεd << μσ ω << μσ/( μdεd) Now let us suppose the μd = μ. This condition is then ω << σ/(εd) Let's now evaluate both sides for copper, assuming roughly that εd = ε0. RHS = σ/ε0 = [5.81 x 107 mho/m] / [8.8541877 x 10-12 farad/m] = (5.8/8.6) 1019 mho/m * m/farad // mho/F = sec-1 = (5.8/8.6) 1019 sec-1 So again we get 2πf << (5.8/8.6) 1019 f << (58/8.6*2π) 1018 = 1.07 x 1018 ≈ 109 GHz But this is the same value and the same calculation as the first. Conclusion: Assuming μ = μd = μ0 and ε = ε0 and εd = 2.3 ε0 we conclude that if f << 5 x 108 GHz // note: typically f << 100GHz then we will have βd2 << |β|2 4. Did I assume βd2 << |β|2 somewhere in my Appendix D development? Answer is no! Question: What assumptions did I make in doing the Appendix D solution? D.1 (a) Only assumption here is ansatz of ej(ωt-βz) for everything. (b) no new assumptions here, so Summary equations are exact (c) no new assumptions here (d) Here I note that β'2 = β2 - βd2 and I say we could ignore the distinction between β and β', but I continue to use β' which maintains that distinction! So this is a little warning. In Chapter 2 everything is done with β, but App D shows that it is really β'2 which appears in most equations. D.2 (a) no new assumptions made (b) no new assumptions made, but m = 0 gets treated as a special case (c) no new assumptions made D.3 that third H equation, no new assumptions made D.4. no new assumptions made, and here I have the summary box! So to answer the question: the only assumption make is ej(ωt-βz) . I do not assume β' = β, I do not assume that βd << β or anything like that. So maybe ej(ωt-βz) is not an exactly realizable ansatz! [ and that is my conclusion #2 above ] Appendix D summary (just for fun): Here is what App D does: (1) makes the e-jβz ansatz for the z dependence of all functions (2) does a PWE to replace azimuthal variable φ by m. (3) writes out the Vector Helmholtz Equation for E in each coordinate (4) writes out the div E = 0 equation as well (5) derives a charge pumping boundary condition at r = a which relates Er there to Nm which is the amount of charge in that mode. [ but this condition is only used later for constant eval on am and Km ] (6) solves first for Ez(r,m) and shows it has the form Czm Jm(β'r) which agrees with Chap 2. (7) finds the Er solution using both its Helm and using div E = 0. Get solution for Er(r,m) as (D.2.11). (8) finds the Eφ(r,m) solution as (D.2 15). At this points, we have constants am and Km . 5. My strange Eφ boundary condition. I then DO make a new ansatz assumption which is that Eφ= 0 at r = a. How do I justify that? I really have no justification other than "it would be nice". This ansatz is used in part to determine those constant. I then end up with the "complete solution". Comments on Appendix D: I have no external verification for this complete solution The Ez does agree with my Chapter 2, but THAT had no justification either, maybe a little Matick. Yes, Matick confirms my expression for Ez= E(r) exactly, so that is worth a lot to me. my BC that Eφ= 0 at r = a seems totally out of the blue I think the charge-pumping other BC is basically OK. everything is suspect when m = 0, so that needs a solid review Even when I make the assumption Eφ= 0 at r = a, I still get this weird solution for m ≠ 1. Eφ(r,m) = (1/4) (aβd) ηm I Rdc [ - + + ] So it may be zero on the surface, but it is swirling around inside. Could the above expression be 0 and I failed to notice it? Is it possible that + = or Jm+1(x) [ 1/Jm+1(xa) + 1/Jm-1(xa) ] = 2mx-1Jm(x)/ Jm-1(xa) or Jm+1(x) [Jm-1(xa) + Jm+1(xa) ]/[ Jm+1(xa) Jm-1(xa) ] = 2mx-1Jm(x)/ Jm-1(xa) or Jm-1(xa)Jm+1(x) [Jm-1(xa) + Jm+1(xa) ] = Jm+1(xa) Jm-1(xa) 2mx-1Jm(x) The only Bessel rule I could use is this (only one with no derivatives) Jm+1(x) = (2m/x)Jm(x) - Jm-1(x) => Jm+1(x) + Jm-1(x) = (2m/x)Jm(x) Jm+1(xa) = (2m/xa)Jm(xa) - Jm-1(xa) => Jm+1(xa) + Jm-1(xa) = (2m/xa)Jm(xa) Using the lower right we get Jm-1(xa)Jm+1(x) [(2m/xa)Jm(xa) ] = Jm+1(xa) Jm-1(xa) 2mx-1Jm(x) Jm+1(x) [(1/xa)Jm(xa) ] = Jm+1(xa) x-1Jm(x) Jm+1(x) Jm(xa)/xa = Jm+1(xa) Jm(x)/x This is clearly not true for general x, only when x = xa. So no, and it is true that Eφ does swirl around inside the wire. And I just happen to force it to be 0 on the surface. Again, no justification whatsoever!!! Here is 100% confirmation from Maple (assuming my Appendix D result for Eφi is correct: This shows how to use the alias idea to simplify everything the fact that Maple does really know about Jm relationships (the 0 after simplify) that Eφ(r,m) ≠ 0 . I test m = 3 use x = 2 and xa = 2.4 and I DON'T get 0. 6. Would it be better to say maybe that Eφ(r=0,m) = 0 ? How can you have an E field NOT obeying this condition? Consider: curl E = - ∂tB C E ds = -∂t[∫S B dS] (1.1.36) Apply to tiny loop of radius r centered at r = 0: Eφ 2πr = -jω (πr2)Bz => Eφ = -jω(1/2) r Bz → 0 Also, what possible direction can Eφ point in at the origin of a polar system? There is no vortex line there. So yes, I think Eφ(r=0,m) = 0 is a much better choice. But, I see that my existing solution meets this condition already: jEφ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) I can see at once that a0 = 0. Then we have jEφ(r,0) = ( ) J1(x) ~ ( ) x → 0 jEφ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) ≈ am x-1(x/2)m/m! + ( + ) (x/2)m+1/(m+1)! for all m ≥ 0 in fact ≈ am(1/2) (2/x) (x/2)m/m! + ( + ) (x/2)m+1/(m+1)! ≈ am(1/2) (x/2)m-1/m! + ( + ) (x/2)m+1/(m+1)! m > 0 For m = 1 this says, as x→ 0 jEφ(r,1) = a1(1/2) + ( + ) (x/2)2/(2)! = a1(1/2) For m > 1 we get jEφ(r,m) = 0 since both terms vanish for any coefficients. So consideration of this limit r → 0 gives a very strange result : jEφ(r→0,m) = 0 for all m except for m = 1 in which the limit is a1/2. By my argument above, I think this requires that a1 = 0. But the point is this: requiring jEφ(r→0,m) = 0 does not nail down anything other than maybe a1 , so it is not a very valuable BC. 7. So now I am stuck with a new question: how to compute am and Km ? Question: Starting at equation (D.2.19), assume no errors were made above it, how do you determine the coefficients am and Km ? Basically since I don't like my (D.2.19) BC, the tail end of this Appendix D has now fallen apart. All of App D is OK except for this final part. Review: Km came from the Ez solution which is just some constant in effect times the J solution of this equation, just a scaling factor. Constant am came from the fact that there is an arbitrary homo solution to the Er equation that you have to add in! That is then where (D.2.11) comes from. Here are my solutions without using any BC's (except I can see that a0 = 0 is required in last below) Km ≡ 2j (βd/β') Czm . (D.2.5) Ez(r,m) = Czm Jm(β'r) = Km (β'/βd)(1/2j) Jm(β'r) (D.1.27) Er(r,m) = am x-1 Jm(x) + Jm+1(x) . (D.2.11) jEφ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) x = β'r . (D.2.15) I do show above (D.4.1) that I = -j(β'βd)-1 2πσ [ xa J1(xa) ] = -j(β'βd)-1 πσ K0 [ xa J1(xa) ] = -j(1/βd) (1/β') πσ K0 [β'a J1(β'a) ] = -j(1/βd) πaσ K0 J1(β'a) ] so this does determine K0 and I have K0 = jI βd/(πσa) * 1/ J1(β'a) So I think for the m=0 fields I get Ez(r,0) = { jI βd/(πσa) * 1/ J1(β'a)} (β'/βd)(1/2j) J0(β'r) = { I /(2πσa) } (β') J0(β'r)/ J1(β'a) = (1/2) (aβ') I Rdc [ J0(β'r)/ J1(β'a) ] which compare to chapter 2 E(r) = [μI/(2πa)] (-jω/β) Now use approximations that β2 = -jωμσ (as used in Ch 2) and β' = β to write β'2 = -jωμσ => β' = (1/β')[ -jωμσ] => (1/β') = β' / [ -jωμσ] => (-jω/β') = β' / [μσ] and using these facts, rewrite the Ch 2 result as E(r) = [μI/(2πa)] (-jω/β') = [μI/(2πa)] β' / [μσ] = [I/(2πaσ)] β' and thus the Chapter 2 result agrees with the Appendix D result, and to get this agreement I assumed no boundary conditions at all except relating K0 to I and setting a0 = 0. Thus, the boundary condition part of Appendix D is NOT NEEDED to get a justification of the Chapter 2 result. What about the other two fields for m = 0? Er(r,0) = J1(x) . (D.2.11) = (1/2) {jI βd/(πσa) * 1/ J1(β'a)}J1(β'r) = (1/2) {jI βd/(πσa) [J1(β'r)/ J1(β'a)] = (1/2) {jI (aβd)/(πσa2) [J1(β'r)/ J1(β'a)] = (j/2) (aβd) I Rdc [J1(β'r)/ J1(β'a)] where Rdc is the resistance per unit length of the wire. Agrees with my original box D.4.6. Next we have jEφ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) ∂(r,0) = Cφ0 J1(β'r) (D.2.17) where in this last, Cφ0 is another unknown constant, it is not related to any other constants at this point. 8. What should I do about boundary conditions then for m > 0 ? When I write Er(r=a,m) = (jω/σ) Nm (D.2.18) which implies am xa-1 Jm(xa) + Jm+1(xa) = (jω/σ) Nm (1) I am merely trading into the new set of unknown constants Nm which are the charge moments which are unknown. So in a way, this BC doesn't really do anything for us. Perhaps we really have to tie in with the exterior solution in order to nail down Km and am . This is something I could work on. Another task would be to think again about those Maxwell equations being satisfied, where I do everything in terms of the am and Km constants. Once again here is my solution for E(r.m) Ez(r,m) = Czm Jm(β'r) = Km (β'/βd)(1/2j) Jm(β'r) (D.1.27) Er(r,m) = am x-1 Jm(x) + Jm+1(x) . (D.2.11) jEφ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) x = β'r . (D.2.15)