Plan C Revised (reflection) REVIEWED

Plan C Revised PhL 8.8.14 At the end of "why superposition doesn't work" I realized that you need to add an overall coefficient that is a function of k. So let's step through Plan C of "standing wave version of Appendix D". In this valiant document, I construct a pretty good "reflection scenario" model which I think correctly handles the transmission line with a shorting bar. Lots of things come out right, but the Jz asymmetry in the DC limit does not go away, now for finite DC current i(z). Basically, as I always suspected and as I now have confirmed, the Jz asymmetry present in the Appendix D solution simply propagates into the reflection scenario. If I really want to solve this problem, I have to solve it in Appendix D. There is another problem with the reflection solution at DC, which is that curl E ≠ 0 and I don't know why [ see Section III.10 below), but it is perhaps related to the Jz asymmetry problem of Appendix D. So any further work on this subject should I think be done with the Appendix D single-wave solution and not with this derived reflection scenario solution. Note added 10.8.14. Notice the comment above about curl E ≠ 0 at DC. Another way to say this is that B is blowing up at ω = 0 since curl E = -jωB, and that is exactly what I much later found as a problem in Chapter 7. Now I know why I was having this curl E problem back in August! [converted from βd to k on 8.18.14 ] Part I: General transmission line quantities (outside the conductors) 1 Part II: Superposition inside one of the conductors. 6 Part III. Redo Part II in a simpler manner. 24 1. Show that div Etot(r,θ,z) = 0 and that the three Helmholtz equations are satisfied 24 2. Show that (**) satisfies the Eθ,tot(r=a,m) = 0 boundary condition: 25 3. The Reflection Boundary Conditions at z = L 25 4. Determination of the constant A(k) 26 5. Final form for the reflection scenario E fields inside the conductor 27 6. Show that (*) satisfies the charge pump boundary condition at z. 27 7. Summary for the reflection scenario fields inside a round wire conductor of a TL with G = 0: 29 8. Current Jz is asymmetric in the general solution and then also at DC 29 9. What is the limit as ω → 0 of the E fields? 30 10. Does the DC solution satisfy curl E = 0 ? No !!! 31 (a) Review Appendix D solutions for curl E = 0 at DC. 31 (b) Study curl E = 0 for DC again in the reflection scenario. 33 Part IV. So what is wrong? 38 Appendix. What IS the function that keeps appearing? 39 End 43 Part I: General transmission line quantities (outside the conductors) Let's now try this variation of Plan B. Let's have the reflected -z wave have opposite sign instead of the same sign, and at the same time replace z → z-L where L will be the "reflection point". Then we get Ei,tot(r,m,z,t) = Ai(k) [ ej(ωt-k[z-L]) Ei(r,m;k) - ej(ωt+k[z-L]) Ei(r,m; -k)] I have added an arbitrary scaling function Ai(k) but we shall not use the above for a while. [ I am assuming G = 0 here so there is no confusion about the meaning of n(θ), and also y = jωC. It is in fact with G = 0 that I have my must mysterious results! ] I expect these two waves to have n(θ) and -n(θ), so I expect to get n(θ,z) ejωt = B(k) {ej(ωt-k[z-L]) n(θ) + ej(ωt+k[z-L]) [-n(θ)] } // ansatz = - ejωt 2j B(k) sin(k[z-L]) n(θ) where n(θ) is the surface charge for the +z going infinite line, and where I have added a TBD scaling constant which is a function of k. Then n(θ,z) = - 2j B(k) sin(k[z-L]) n(θ) and THIS is a charge density that will always be 0 at z = L, as might be correct for a shorting bar at that location! Now let's assume at at z = 0 we have some n(θ,z=0). Then we can write n(θ,z=0) = - 2j B(k) sin(k[-L]) n(θ) or n(θ,z=0) = 2j B(k) sin(kL) n(θ) and we can solve for B: B(k) = [n(θ,z=0)/ n(θ)] / [2j sin(kL)] Installing this then gives n(θ,z) = - 2j B(k) sin(k[z-L]) n(θ) = - 2j { [n(θ,z=0)/ n(θ)] / [2j sin(kL)]} sin(k[z-L]) n(θ) = - { [n(θ,z=0)] / [ sin(kL)]} sin(k[z-L]) = - n(θ,z=0) = + n(θ,z=0) (*) This is then a standing wave pattern for n(θ,z) which has a maximum for example at z = 0. For large k there might be several peaks with this same max value, and in between nodes where n = 0. If we take the limit k → 0 we get n(θ,z) = n(θ,z=0) { (L-z)/L } = n(θ,z=0) { 1 - z/L } which I would call a reasonable limit. It just scales down linearly from the driving end to the shorting bar and z = L. Comment: Everything seems reasonable about this n(θ,z) function (*) 1) it vanishes at z = L where the capacitor is shorted 2) it is maximal at the driving point 3) it exhibits the expected standing waves 4) it gives the right DC limit! If we were to integrate this over θ, we end up with q(z), the charge density per unit length. Then q(z) = q(0) // general form q(z) = q(0) { 1 - z/L } // limit k → 0 Now since q(z) = CV(z), we have V(z) = q(z)/C and V(0) = q(0)/C, so we then have shown that V(z) = V(0) // general form V(z) = V(0) { 1 - z/L } // limit k → 0 This is a very reasonable form since it does all the things I want! What about the current i(z)? If we go to some general location z on our line (say to the left of the shorting bar), we have this TL equation, = - z i(z) z = R + jωL // assumption that TL equations are valid! We can solve this for i(z) to get general result i(z) = (-1/z) ∂zV(z) = (-1/z) V(0) [ 1/sin(kL)] ∂z sin(k[L-z]) = (+1/z) V(0) [ 1/sin(kL)] ∂z sin(k[z-L]) = (+1/z) V(0) [ 1/sin(kL)] k cos(k[z-L]) = (k/z) V(0) Then as k → 0 we get i(z) = (k/z) V(0) / (kL) = V(0)/ ( Lz) which is a constant, independent of z. But at ω = 0, we can write z = R where R is the total resistance of both conductors of the line per unit length. Then we get i(z) = V(0)/ ( LR) R = R1+ R2 So here is a summary: i(z) = (k/z) V(0) // general form i(z) = V(0)/ ( LR) // limit k → 0 This also does just what I want: It is a max at z = L, and perhaps has other maxima for large k of the same size, with nodes in between. We can look at our two special locations: i(0) = (k/z) V(0) cot(kL) // general form i(L) = (k/z) V(0) // general form Thus we can also write i(z) = i(L) cos(k[z-L]) // general form i(0) = i(L) cos(kL) If we wanted, we could go on to define the impedance Z(z) in this way Z(z) = V(z)/i(z) = V(0) / [ (k/z) V(0) ] = (z/k) tan(k[L-z]) // DC limit here is Z(z) = z (L-z) = R(L-z) Z(L) = 0 //which is correct for our shorting bar Z(0) = (z/k) tan(kL) Just for fun, let's check the other TL equation which says this: = - y V(z) so that V(z) = (-1/y) ∂zi(z) = (-1/y) ∂z[i(L) cos(k[z-L])] = (-1/y) i(L) ∂z cos(k[z-L]) = (+k/y) i(L) sin(k[z-L]) = (+k/y) (k/z) V(0) sin(k[z-L]) = - V(0) sin(k[z-L]) // since k2 = k2 = [ -j]2 = - yz = V(0) and yes, this does agree with the result found earlier! Status: At this point, I think I know all these objects for my reflection superposition model: V(z), q(z), i(z), Z(z) ∫dθ n(θ,z) = q(z) = a ∫dθ n(θ,z) n(θ,z) = a n(θ,z) At z = 0, we imagine driving with some V(0) and there is then some q(0) = CV(0) per unit length. I know at this point z = 0 that n(θ,0) = (1/a) where a is the radius of my conductor of interest and where ξ1 is its bipolar coordinate label. So I guess I know n(θ,z) as well as all the other quantities listed. So the quantities I claim to "know about" are those associated "in general" with a transmission line. Summary of Transmission Line Quantities General form for any ω: DC limit where ω=0 k = k(ω) = -j n(θ,z) = n(θ,z=0) n(θ,z) = n(θ,z=0) { 1 - z/L } q(z) = q(0) q(z) = q(0) { 1 - z/L } V(z) = V(0) V(z) = V(0) { 1 - z/L } i(z) = (k/z) V(0) i(z) = V(0)/ ( LR) Z(z) = (z/k) tan(k[L-z]) Z(z) = R(L-z) I have only assumed the transmission line equations for the line, and G = 0. The DC limits are all exactly what I expect them to be, so this is certainly encouraging and suggests that a reflection model might actually truly model the situation. I next want to try to find out what is going on INSIDE one of the round conductors. Part II: Superposition inside one of the conductors. Please don't start off on the wrong foot here! At the start of Part I I wrote, Ei,tot(r,m,z,t) = Ai(k) ej(ωt-k[z-L]) Ei(r,m;k) + Ci(k) ej(ωt+k[z-L]) Ei(r,m; -k)] where the idea is that Ei(r,m;k) is precisely a regular Appendix D E field solution. I am not really sure about the sign between the two terms so I just make two separate constants (functions of k) for now. Claim: Each of the two terms shown above separately satisfies the following equations: the three Helmholtz equations for E yes the div E = 0 equation defer the two boundary conditions defer Proof: We know the Helms are true for the first term for any Ai(k) because that is what App D shows. It is then true as well for the second term since we just take k → - k everywhere and call upon "symmetry" for a wave going to the left instead of to the right. We shall deal with the last two items below. I think that I want Er = 0 and Eθ = 0 at z = L. I am now imagining that the "shorting bar" is in fact a reflecting plane or "brick" of something like solid silver that conducts really well, I assume it is a perfect conductor. Thus, just inside this brick, I expect to have Er = 0 and Eθ = 0 . Since these are both tangential to the shorting plane, they are continuous through the boundary so then Er = 0 and Eθ = 0 at z = L-ε. In contrast, I want Ez to be maximal at z = L because Jz = σEz and i(z) is maximal at z = L. Now consider the above equation with index i = r (radial), Er,tot(r,m,z,t) = Ar(k) ej(ωt-k[z-L]) Er(r,m;k) + Cr(k) ej(ωt+k[z-L]) Er(r,m; -k)] Recall now from Appendix D that Ez(r,m;k) = (1/4) ηm I Rdc (aβ') fm fm = [ - ] x = β'r Er(r,m;k) = (j/4) ηm I Rdc (ak) gm gm = [ + ] xa = β'a Eθ(r,m;k) = (1/4) ηm I Rdc (ak) hm hm = [ - ] where β'2 = β2 - k2. I can see then that gm is invariant under negation of k. And I can see that (ak) negates. But what is the meanings of I in this expression? I am going to take a shot in the dark here (based on looking at the Ez solution). I will say I = I(k) is the current going to the right in the infinite line solution, and this negates if you instead make the wave go to the left. This will then cause Ez to have the opposite sign in the left going wave. So therefore I will assume that I(k) negates when you negate k. Therefore, Er (and also Eθ) do not change sign. We then have Er,tot(r,m,z,t) = Ar(k) ej(ωt-k[z-L]) Er(r,m;k) + Cr(k) ej(ωt+k[z-L]) Er(r,m; +k)] Now, if we want to have Er,tot(r,m,z=L,t) = 0 at the surface of the silver plane/brick terminator, we have to take this form Er,tot(r,m,z,t) = Ar(k) { ej(ωt-k[z-L]) Er(r,m;k) - ej(ωt+k[z-L]) Er(r,m; +k)] = Ar(k) ejωt Er(r,m;k) { 2j sin(k[L-z]) } ie Cr(k) = - Ar(k) The exact same argument applies to Eθ , so we will also require that Cθ(k) = - Aθ(k) What about Ez? Using the logic above, I argue that Ez(r,m;-k) = - Ez(r,m;k) and therefore Ez,tot(r,m,z,t) = Az(k) ej(ωt-k[z-L]) Ez(r,m;k) - Cz(k) ej(ωt+k[z-L]) Ez(r,m; +k)] = ejωt Ez(r,m;k) { Az(k) ej(-k[z-L]) - Cz(k) ej(+k[z-L]) } I want this to be maximal at z = L. As another shot in the dark, I will assume Cz(k) = - Az(k) and then I get Ez,tot(r,m,z,t) = ejωt Az(k) Ez(r,m;k) [ 2 cos(k[L-z]) ] Jz,tot(r,m,z,t) = ejωt σAz(k) Ez(r,m;k) [ 2 cos(k[L-z]) ] or Ez,tot(r,m,z) = Az(k) Ez(r,m;k) [ 2 cos(k[L-z]) ] Jz,tot(r,m,z) = σAz(k) Ez(r,m;k) [ 2 cos(k[L-z]) ] At least this has the property that it is maximal at z = L and so I then expect i(z) to be maximal at z = L since this is essentially an integral of Ez. Recall from lines doc that Jz,tot(r,θ) =!Syntax Error, I Jz,tot(r,m) ejmθ // expansion (D.1.3a) When we integrate this over the cross section area of our round wire, the dθ integral kills off all but the m = 0 term, so we have ∫dθ Jz,tot(r,θ) = ∫dθ Jz,tot(r,m=0) = 2π Jz,tot(r,m=0) We then get i(z) = ∫dA Jz,tot(r,θ) = ∫r dr 2π Jztot(r,m=0) = ∫r dr 2π σAz(k) Ez(r,0;k) [ 2 cos(k[L-z]) ] = 2π σAz(k) [ 2 cos(k[L-z]) ] ∫r dr Ez(r,m=0;k) This same integral appears in lines doc from which I quote [ see below (D.2.30) ], 2π σ!Syntax Error, Ir dr Ez(r,m=0) = 2πω (a/k) N0 Thus I get i(z) = Az(k) [ 2 cos(k[L-z]) ] { 2π σ!Syntax Error, Ir dr Ez(r,m=0) } = Az(k) [ 2 cos(k[L-z]) ] { 2πω (a/k) N0 } But in my work from Part I I found that i(z) = (k/z) V(0) Comparison then says (k/z) V(0) = Az(k) [ 2 cos(k[L-z]) ] { 2πω (a/k) N0 } or (k/z) V(0) = Az(k) [ 2 ] { 2πω (a/k) N0 } or (k2/z) V(0) = Az(k) 4πωaN0 and we can then solve to get Az(k) = (k2/z) V(0) But the Appendix D symbols I and N0 are related by N0 = (k/2πωa) I (D.2.31) or N0/2 = (k/4πωa) I or 1/2 = (k/4πωaN0) I or 1/(2Ik) = (1/4πωaN0) = and then we can write Az(k) = (k2/z) V(0) = (k/z) V(0) and we have then evaluated our coefficient. We can then insert it into our expression above to get Ez,tot(r,m,z) = Az(k) Ez(r,m;k) [ 2 cos(k[L-z]) ] = 2 cos(k[L-z]) { (k/z) V(0) } { (1/4) ηm I Rdc (aβ') fm } = cos(k[L-z]) { (k/z) V(0) } { (1/4) ηm Rdc (aβ') fm(β'r) } and this then is our final expression for Ez,tot(r,m,z) in our superposition model. Status to this point: Ez,tot(r,m,z) = { (k/z) V(0) } { (1/4) ηm Rdc (aβ') fm(β'r) } Er,tot(r,m,z) = Ar(k) Er(r,m;k) { 2j sin(k[L-z]) } Eθ,tot(r,m,z) = Aθ(k) Eθ(r,m;k) { 2j sin(k[L-z]) } Now what about the Appendix D boundary conditions? This one is obvious: Eθ,tot(r=a,m,z) = Aθ(k) Eθ(r=a,m;k) { 2j sin(k[L-z]) } = 0 since Eθ(r=a,m;k) = 0 The other boundary condition would be this: Er,tot(r=a-ε,θ,z) = (jω/σ) n(θ,z) In m-space this says Er,tot(r=a,m,z) = (jω/σ) Nm (z) From Part I we have that n(θ,z) = n(θ,z=0) Nm(z) = Nm(0) where n(θ,z=0) = (1/a) and Nm(0) is the transform of this which I computed somewhere. Thus our boundary condition requires that Ar(k) Er(r=a,m;k) { 2j sin(k[L-z]) } = (jω/σ) Nm(0) I am relieved to see that same sin factor on both sides, so we then have Ar(k) Er(r=a,m;k) { 2j } = (jω/σ) Nm(0) 1 / sin(kL) or Ar(k) Er(r=a,m;k) = (ω/2σ) Nm(0) 1 / sin(kL) But recall now that Er,tot(r,m,z) = Ar(k) Er(r,m;k) { 2j sin(k[L-z]) } Inserting the previous then gives Er,tot(r,m,z) = Ar(k) Er(r,m;k) { 2j sin(k[L-z]) } = [ Er(r,m;k) / Er(a,m;k) ] Ar(k) Er(a,m;k){ 2j sin(k[L-z]) } = [ Er(r,m;k) / Er(a,m;k) ] (ω/2σ) Nm(0) 1 / sin(kL) * { 2j sin(k[L-z]) } = (jω/2σ) Nm(0) This expression then satisfies the boundary condition, Er,tot(r=a-ε,θ,z) = (jω/σ) Nm(z) at every point z along the line. Now Er(r,m;k) = (j/4) ηm I Rdc (ak) gm(r) Er(a,m;k) = (j/4) ηm I Rdc (ak) 2 and therefore = gm(r)/2 and we then have Er,tot(r,m,z) = [gm(r)/2 ] (jω/2σ) Nm(0) This is mildly encouraging (though of course I know better). So far we have Ez,tot(r,m,z) = { (k/z) V(0) } { (1/4) ηm Rdc (aβ') fm(β'r) } Er,tot(r,m,z) = [gm(r)/2 ] (jω/2σ) Nm(0) Eθ,tot(r,m,z) = Aθ(k) Eθ(r,m;k) { 2j sin(k[L-z]) } I think Aθ(k) will be determined now by the div Etot = 0 equation. ∂r (r Er,tot) + ∂θEθ,tot + r ∂zEztot = 0 . In partial waves this says ∂r (r Er,tot(r,m,z)) + jm Eθ,tot(r,m,z) + r ∂zEz,tot(r,m,z) = 0 . I think it will be easiest to write each total field in terms of its App D field. From above, Ez,tot(r,m,z) = Az(k) Ez(r,m;k) [ 2 cos(k[L-z]) ] Eθ,tot(r,m,z) = Aθ(k) Eθ(r,m;k) { 2j sin(k[L-z]) } Er,tot(r,m,z) = Ar(k) Er(r,m;k) { 2j sin(k[L-z]) } Then divEtot = 0 says Ar(k) { 2j sin(k[L-z]) } ∂r (r Er(r,m;k) ) + Aθ(k) { 2j sin(k[L-z]) } jm Eθ(r,m;k) + Az(k) Ez(r,m;k) r ∂z{ 2 cos(k[L-z]) } = 0 or Ar(k) { 2j sin(k[L-z]) } ∂r (r Er(r,m;k) ) + Aθ(k) { 2j sin(k[L-z]) } jm Eθ(r,m;k) + Az(k) Ez(r,m;k) r { 2k sin(k[L-z]) } = 0 At least we have a chance here, and things simplify to Ar(k) { 2j } ∂r (r Er(r,m;k) ) + Aθ(k) { 2j } jm Eθ(r,m;k) + Az(k) Ez(r,m;k) r { 2k } = 0 Divide through by 2j to get Ar(k) ∂r (r Er(r,m;k) ) + Aθ(k) jm Eθ(r,m;k) - j k Az(k) Ez(r,m;k) r = 0 I know from App D that ∂r [r Er(r,m;k)] + jmEθ(r,m,k) + r (-jk)Ez(r,m,k) = 0 I can use this to eliminate the messy r derivative so we then have Ar(k) { - jmEθ(r,m,k) + r (jk)Ez(r,m,k) } + Aθ(k) jm Eθ(r,m;k) - j k Az(k) Ez(r,m;k) r = 0 which we regroup to get Eθ(r,m,k) { -jm Ar(k) + Aθ(k) jm } + Ez(r,m,k) { r (jk) Ar(k) - r j k Az(k) } = 0. Rewrite this requirement once again Eθ(r,m,k) jm { - Ar(k) + Aθ(k) } + Ez(r,m,k) (jkr){ Ar(k) - Az(k) } = 0. Each term has to separately be zero for this to work. This then requires that Ar(k) = Aθ(k) = Az(k) We can then select Aθ(k) = Ar(k) to get the first term to vanish. But what about the second term? We already determined the coefficients and I doubt they are going to "work out". We found that Ar(k) Er(r=a,m;k) = (ω/2σ) Nm(0) / sin(kL) correctly copied from above and Az(k) = (k2/z) V(0) I would like to show that Ar(k) - Az(k) = 0 ? Ar(k) Er(r=a,m;k) - Az(k) Er(r=a,m;k) = 0 ? (ω/2σ) Nm(0) 1 / sin(kL) - (k2/z) V(0) Er(r=a,m;k) = 0 ? (ω/2σ) Nm(0) - (k2/z) V(0) Er(r=a,m;k) = 0 ? Now I know that Er(r=a,m;k) = (j/4) ηm I Rdc (ak) gm(a) = (j/2) ηm I Rdc (ak) So have to show that (ω/2σ) Nm(0) - (k2/z) V(0) (j/2) ηm I Rdc (ak) = 0 ? This looks very unlikely! But recall that N0 = (k/2πωa) I so that I/N0 = 2πωa/k and then (ω/2σ) Nm(0) - (k2/z) V(0) (j/2) ηm 2πωa/k Rdc (ak) = 0 ? (ω/2σ) Nm(0) - (k/z) V(0) (j/4) ηm Rdc (ak) = 0 ? (ω/2σ) Nm(0) - (k/z) V(0) (j/4) ηm (1/πa2σ) (ak) = 0 ? ω Nm(0) - (k2/z) V(0) ηm (j/2πa) = 0 ? How can this possibly be true?? Nm(0) corresponds to n(θ,z=0) = (1/a) which I guess means that Nm(0) is Nm at z = 0, so assume that and continue ω N0 ηm - (k2/z) V(0) ηm (j/2πa) = 0 ? ω N0 - (k2/z) V(0) (j/2πa) = 0 ? Now we know the k2 = k2 = -zy so we then have ω N0 - (- y) V(0) (j/2πa) = 0 ? ω N0 + (jωC) V(0) (j/2πa) = 0 ? // assume G = 0 N0 + (jC) V(0) (j/2πa) = 0 ? N0 - C V(0) (1/2πa) = 0 ? N0 - q(0) (1/2πa) = 0 ? But recall that N0 = (1/2πa) q(0) = <n(θ)> (D.1.8) so amazingly it works! Maybe this is obvious from the start. Recall that Ei,tot(r,m,z,t) = Ai(k) ej(ωt-k[z-L]) Ei(r,m;k) + Ci(k) ej(ωt+k[z-L]) Ei(r,m; -k) We then want to show that ∂r (r Er,tot(r,m,z)) + jm Eθ,tot(r,m,z) + r ∂zEz,tot(r,m,z) = 0 . Write this out as ∂r (r { Ar(k) ej(ωt-k[z-L]) Er(r,m;k) + Cr(k) ej(ωt+k[z-L]) Er(r,m; -k)}) + jm { Aθ(k) ej(ωt-k[z-L]) Eθ(r,m;k) + Cθ(k) ej(ωt+k[z-L]) Eθ(r,m; -k) } + r ∂z{ Az(k) ej(ωt-k[z-L]) Ez(r,m;k) + Cz(k) ej(ωt+k[z-L]) Ez(r,m; -k)} = 0 Now do the z derivatives in the last line , and simplify the first line, to get Ar(k) ej(ωt-k[z-L])∂r (r Er(r,m;k) + Cr(k) ej(ωt+k[z-L]) ∂r (r Er(r,m; -k)) } + jm { Aθ(k) ej(ωt-k[z-L]) Eθ(r,m;k) + Cθ(k) ej(ωt+k[z-L]) Eθ(r,m; -k) } + r jk{ - Az(k) ej(ωt-k[z-L]) Ez(r,m;k) + Cz(k) ej(ωt+k[z-L]) Ez(r,m; -k)} = 0 Now in the first line make these replacements using App D. ∂r (r Er(r,m;k) = - jmEθ(r,m;k) + r (jk)Ez(r,m;k) ∂r (r Er(r,m;-k) = - jmEθ(r,m;-k) - r (jk)Ez(r,m;-k) Then we get Ar(k) ej(ωt-k[z-L]) [- jmEθ(r,m;k) + r (jk)Ez(r,m;k)] + Cr(k) ej(ωt+k[z-L])[ - jmEθ(r,m;-k) - r (jk)Ez(r,m;-k)] + jm { Aθ(k) ej(ωt-k[z-L]) Eθ(r,m;k) + Cθ(k) ej(ωt+k[z-L]) Eθ(r,m; -k) } + r jk{ - Az(k) ej(ωt-k[z-L]) Ez(r,m;k) + Cz(k) ej(ωt+k[z-L]) Ez(r,m; -k)} = 0 Now the two functions of z have to have the same coefficients, so this then says Ar(k) ej(ωt-k[z-L]) [- jmEθ(r,m;k) + r (jk)Ez(r,m;k)] + jm { Aθ(k) ej(ωt-k[z-L]) Eθ(r,m;k) - r jk{ Az(k) ej(ωt-k[z-L]) Ez(r,m;k) = 0 and Cr(k) ej(ωt+k[z-L]) [ - jmEθ(r,m;-k) - r (jk)Ez(r,m;-k)] + jm { Cθ(k) ej(ωt+k[z-L]) Eθ(r,m; -k)} + r jk { Cz(k) ej(ωt+k[z-L]) Ez(r,m; -k) } = 0 In each of these we then cancel the z function to get instead these two simpler equations Ar(k) [- jmEθ(r,m;k) + r (jk)Ez(r,m;k)] + jm { Aθ(k) Eθ(r,m;k) - r jk{ Az(k) Ez(r,m;k) = 0 Cr(k) [ - jmEθ(r,m;-k) - r (jk)Ez(r,m;-k)] + jm { Cθ(k) Eθ(r,m; -k)} + r jk { Cz(k) Ez(r,m; -k) } = 0 Now in each of these, we regroup in terms of Eθ and Eθ : Eθ(r,m;k) jm { - Ar(k) + Aθ(k) } + Ez(r,m;k) (jrk){ Ar(k) - Az(k)} = 0 Eθ(r,m;-k) jm { - Cr(k) + Cθ(k) } + Ez(r,m;-k) (jrk){ -Cr(k) + Cz(k)} = 0 OK, now having done this, you only get it to work if the following four equations are true: Ar(k) = Aθ(k) Ar(k) = Az(k) Cr(k) = Cθ(k) Cr(k) = Cz(k) But we can reduce this two the following Ar(k) = Aθ(k) = Az(k) ≡ A(k) Cr(k) = Cθ(k) = Cz(k) ≡ C(k) Now in my previous long calculation above, I did find that the three Ai were equal. I also found that Cz(k) = - Az(k) Cr(k) = - Ar(k) Cθ(k) = - Aθ(k) So in retrospect, I could have just started with this form Ei,tot(r,m,z,t) = A(k) { ej(ωt-k[z-L]) Ei(r,m;k) - ej(ωt+k[z-L]) Ei(r,m; -k) } or Etot(r,m,z,t) = A(k) { ej(ωt-k[z-L]) E(r,m;k) - ej(ωt+k[z-L]) E(r,m; -k) } I would then want to show that div [e-jkz E(r,m;k) ] = 0 div [e+jkz E(r,m;-k) ] = 0 (*) Recall again that we know that r div F (r,θ,z) = ∂r (r Fr(r,m,z)) + jm Fθ(r,m,z) + r ∂z Fz(r,m,z) Then r div [e-jβz E(r,m;k) ] = ∂r (r e-jkz Er(r,m;k)) + jm e-jkz Eθ(r,m;k) + r ∂z e-jkz Ez(r,m;k) = e-jkz { ∂r (r Er(r,m;k)) + jm Eθ(r,m;k) + r (-jk) Ez(r,m;k) } But recall from App D that r div E(r,m;k) = ∂r [r Er(r,m;k)] + jmEθ(r,m;k) + r (-jk)Ez(r,m;k) = 0 Thus we have shown that r div [e-jkz E(r,m;k) ] = e-jkz r div E(r,m;k) = 0 The verification for the second line of (*) is then obvious. Summary of results for Part II: For r = r,θ,z we have these E fields: Ei,tot(r,m,z,t) = A(k) { ej(ωt-k[z-L]) Ei(r,m;k) - ej(ωt+k[z-L]) Ei(r,m; -k) } A(k) = (k2/z) V(0) = (k/z) V(0) = (-y) V(0) = (-y) V(0) // dimensionless I had these three symmetry relations, Er(r,m; -k) = + Er(r,m; k) Eθ(r,m; -k) = + Eθ(r,m; k) Ez(r,m; -k) = – Ez(r,m; k) We then obtain Er,tot(r,m,z,t) = A(k) { ej(ωt-k[z-L]) Er(r,m;k) - ej(ωt+k[z-L]) Er(r,m; k) } Eθ,tot(r,m,z,t) = A(k) { ej(ωt-k[z-L]) Eθ(r,m;k) - ej(ωt+k[z-L]) Eθ(r,m; k) } Ez,tot(r,m,z,t) = A(k) { ej(ωt-k[z-L]) Ez(r,m;k) + ej(ωt+k[z-L]) Ez(r,m; k) } which we can write as Er,tot(r,m,z,t) = A(k) Er(r,m;k) ejωt{ e-jk[z-L]) - e+jk[z-L]) } Eθ,tot(r,m,z,t) = A(k) Eθ(r,m;k) ejωt{ e-jk[z-L]) - e+jk[z-L]) } Ez,tot(r,m,z,t) = A(k) Ez(r,m;k) ejωt{ e-jk[z-L]) + e+jk[z-L]) } and again as Er,tot(r,m,z,t) = A(k) Er(r,m;k) ejωt{ -2j sin(k[z-L]) } Eθ,tot(r,m,z,t) = A(k) Eθ(r,m;k) ejωt{ -2j sin(k[z-L]) } Ez,tot(r,m,z,t) = A(k) Ez(r,m;k) ejωt{ 2 cos(k[z-L]) } We can then insert the factor A(k) in the form A(k) = (-y) V(0) = - (yV(0)/2Ik) 1/sin(kL) to get Er,tot(r,m,z,t) = - (yV(0)/2Ik) Er(r,m;k) ejωt{ -2j sin(k[z-L]) } / sin(kL) Eθ,tot(r,m,z,t) = - (yV(0)/2Ik) Eθ(r,m;k) ejωt{ -2j sin(k[z-L]) } / sin(kL) Ez,tot(r,m,z,t) = - (yV(0)/2Ik) Ez(r,m;k) ejωt{ 2 cos(k[z-L]) } / sin(kL) or Er,tot(r,m,z,t) = +j (yV(0)/Ik) Er(r,m;k) ejωt{ sin(k[z-L]) } / sin(kL) Eθ,tot(r,m,z,t) = +j (yV(0)/Ik) Eθ(r,m;k) ejωt{ sin(k[z-L]) } / sin(kL) Ez,tot(r,m,z,t) = - (yV(0)/Ik) Ez(r,m;k) ejωt{ cos(k[z-L]) } / sin(kL) or Er,tot(r,m,z) = +j (yV(0)/Ik) Er(r,m;k) { sin(k[z-L]) } / sin(kL) Eθ,tot(r,m,z) = +j (yV(0)/Ik) Eθ(r,m;k) { sin(k[z-L]) } / sin(kL) Ez,tot(r,m,z) = - (yV(0)/Ik) Ez(r,m;k) { cos(k[z-L]) } / sin(kL) where Ez(r,m;k) = (1/4) ηm I Rdc (aβ') fm fm = [ - ] x = β'r Er(r,m;k) = (j/4) ηm I Rdc (ak) gm gm = [ + ] xa = β'a Eθ(r,m;k) = (1/4) ηm I Rdc (ak) hm hm = [ - ] So now install these fields to get Er,tot(r,m,z) = +j (yV(0)/Ik) (j/4) ηm I Rdc (ak) gm { sin(k[z-L]) } / sin(kL) Eθ,tot(r,m,z) = +j (yV(0)/Ik) (1/4) ηm I Rdc (ak) hm { sin(k[z-L]) } / sin(kL) Ez,tot(r,m,z) = - (yV(0)/Ik) (1/4) ηm I Rdc (aβ') fm { cos(k[z-L]) } / sin(kL) We then get cancellation of the I factors, and so Er,tot(r,m,z) = +j yV(0) (j/4) ηm Rdc (ak) gm sin(k[z-L]) / ksin(kL) Eθ,tot(r,m,z) = +j yV(0) (1/4) ηm Rdc (ak) hm sin(k[z-L]) / ksin(kL) Ez,tot(r,m,z) = - yV(0) (1/4) ηm Rdc (aβ') fm cos(k[z-L]) / ksin(kL) So, in my model for the reflection scenario with a shorting plane at z = L, the above are my predictions for the E fields inside a round wire of radius a which is one of the two conductors for a transmission line that is made from two round wires. We could set y = jωC in the case that G = 0. Question: do we still have asymmetry for Jz as ω → 0 ? Start with this: Ez,tot(r,m,z) = - yV(0) (1/4) ηm Rdc (aβ') fm cos(k[z-L]) / ksin(kL) so that Jz,tot(r,m,z) = - σ yV(0) (1/4) ηm Rdc (aβ') fm cos(k[z-L]) / ksin(kL) As ω → 0, we know that k → 0 and β' → 0. This last fact allows us to use, fm = (r/a)|m| (|m|+1) (2/β'a) f0 = 4/(aβ') We then have Jz,tot(r,m,z) = - σ yV(0) (1/4) ηm Rdc(aβ') (r/a)|m| (|m|+1) (2/β'a)cos(k[z-L]) / ksin(kL) m ≠ 0 Jz,tot(r,0,z) = - σ yV(0) (1/4) Rdc(aβ') 4/(aβ') cos(k[z-L]) / ksin(kL) m = 0 or Jz,tot(r,m,z) = - σ yV(0) (1/4) ηm Rdc (r/a)|m| (|m|+1) (2)cos(k[z-L]) / ksin(kL) m ≠ 0 Jz,tot(r,0,z) = - σ yV(0) (1/4) Rdc 4 cos(k[z-L]) / ksin(kL) m = 0 or Jz,tot(r,m,z) = - σ yV(0) (1/2) ηm Rdc (r/a)|m| (|m|+1) cos(k[z-L]) / ksin(kL) m ≠ 0 Jz,tot(r,0,z) = - σ yV(0) Rdc cos(k[z-L]) / ksin(kL) m = 0 or Jz,tot(r,m,z) = - yV(0) (1/2) ηm (1/πa2) (r/a)|m| (|m|+1) cos(k[z-L]) / sin(kL) m ≠ 0 Jz,tot(r,0,z) = - yV(0) (1/πa2) cos(k[z-L]) / sin(kL) m = 0 or Jz,tot(r,m,z) = - jωC V(0) (1/2) ηm (1/πa2) (r/a)|m| (|m|+1) cos(k[z-L]) / ksin(kL) m ≠ 0 Jz,tot(r,0,z) = - jωC V(0) (1/πa2) cos(k[z-L]) / ksin(kL) m = 0 Now what happens when we take the limit k → 0 corresponding to ω → 0 ? Jz,tot(r,m,z) = - jωC V(0) (1/2) ηm (1/πa2) (r/a)|m| (|m|+1) 1 / (k2L) m ≠ 0 Jz,tot(r,0,z) = - jωC V(0) (1/πa2) 1 / (k2L) m = 0 We need to know the way in which k → 0. From Section D.11, for G = 0, k2 = RCω(-j) = -jωRC In this case ω/k2 = ω/[ -jωRC] = j/(RC) We then get Jz,tot(r,m,z) = - j[j/(RC)]C V(0) (1/2) ηm (1/πa2) (r/a)|m| (|m|+1) 1 / L m ≠ 0 Jz,tot(r,0,z) = - j[j/(RC)]C V(0) (1/πa2) 1 / L m = 0 or Jz,tot(r,m,z) = [1/(R)] V(0) (1/2) ηm (1/πa2) (r/a)|m| (|m|+1) 1 / L m ≠ 0 Jz,tot(r,0,z) = [1/(R)] V(0) (1/πa2) 1 / L m = 0 or Jz,tot(r,m,z) = V(0) (1/2) ηm (1/πa2) (r/a)|m| (|m|+1) 1 / (RL) m ≠ 0 Jz,tot(r,0,z) = V(0) (1/πa2) 1 / (RL) m = 0 Recall now from Part I that i(z) = V(0)/ ( LR) R = R1+ R2 So the rewrite again as Jz,tot(r,m,z) = i(z) (1/2) ηm (1/πa2) (r/a)|m| (|m|+1) m ≠ 0 Jz,tot(r,0,z) = i(z) (1/πa2) m = 0 For m = 0, this is exactly the result I expect. But the asymmetry continues to persist at DC!! I think this is the first time I have obtained this asymmetry in a situation where the current i(z) is finite as ω → 0. So this then is a more serious "issue" ! Is it really true that Ji = σ Ei since B fields are present? My magnetic Ohm's law discussion concludes that you can in general ignore the magnetic distortion of ohm's law as long as B << 569 T. But you cannot ignore it in the Hall effect, as I point out at the very end. So maybe you cannot ignore it here?? What is going on with the other two field components? For Er we start with Er,tot(r,m,z) = +j yV(0) (j/4) ηm Rdc (ak) gm sin(k[z-L]) / ksin(kL) Then write this out for the two cases: Er,tot(r,m,z) = +j yV(0) (j/4) ηm Rdc (ak) gm sin(k[z-L]) / ksin(kL) m≠0 Er,tot(r,0,z) = +j yV(0) (j/4) Rdc (ak) g0 sin(k[z-L]) / ksin(kL) m=0 Now go to the ω→0 limit where β' and k go to zero. We then get from App D.11, gm = (r/a)|m|+1 + (r/a)|m|-1 g0 = 2 (r/a) So then Er,tot(r,m,z) = +j yV(0) (j/4) ηm Rdc (ak) [(r/a)|m|+1 + (r/a)|m|-1] sin(k[z-L]) / ksin(kL) m≠0 Er,tot(r,0,z) = +j yV(0) (j/4) Rdc (ak) [2 (r/a)] sin(k[z-L]) / ksin(kL) m=0 But at the same time we have to do the small k limit on other things, so we have Er,tot(r,m,z) = +j yV(0) (j/4) ηm Rdc (ak) [(r/a)|m|+1 + (r/a)|m|-1] (k[z-L]) / k(kL) m≠0 Er,tot(r,0,z) = +j yV(0) (j/4) Rdc (ak) [2 (r/a)] (k[z-L]) / k(kL) m=0 or Er,tot(r,m,z) = +j yV(0) (j/4) ηm Rdc (ak) [(r/a)|m|+1 + (r/a)|m|-1] ([z-L]) / k(L) m≠0 Er,tot(r,0,z) = +j yV(0) (j/4) Rdc (ak) [2 (r/a)] ([z-L]) / k(L) m=0 or Er,tot(r,m,z) = +j yV(0) (j/4) ηm Rdc (a) [(r/a)|m|+1 + (r/a)|m|-1] ([z-L]) / (L) m≠0 Er,tot(r,0,z) = +j yV(0) (j/4) Rdc (a) [2 (r/a)] ([z-L]) / (L) m=0 or Er,tot(r,m,z) = +j yV(0) (j/4) ηm Rdc (a) [(r/a)|m|+1 + (r/a)|m|-1] ([z-L]) / (L) m≠0 Er,tot(r,0,z) = +j yV(0) (j/4) Rdc (a) [2 (r/a)] ([z-L]) / (L) m=0 Assuming Ohm's law is valid, this becomes Jr,tot(r,m,z) = +j yV(0) (j/4) ηm (1/πa) [(r/a)|m|+1 + (r/a)|m|-1] ([z-L]) / (L) m≠0 Jr,tot(r,0,z) = +j yV(0) (j/4) (1/πa) [2 (r/a)] ([z-L]) / (L) m=0 But now use y = jωC Jr,tot(r,m,z) = -ωC V(0) (j/4) ηm (1/πa) [(r/a)|m|+1 + (r/a)|m|-1] ([z-L]) / (L) m≠0 Jr,tot(r,0,z) = -ωC V(0) (j/4) (1/πa) [2 (r/a)] ([z-L]) / (L) m=0 Now since we are doing ω → 0, we end up with Jr,tot(r,m,z) = 0 m≠0 Jr,tot(r,0,z) = 0 m=0 so there is then no radial current. This seems consistent with the fact that the surface charge is not changing in this DC limit. We might as well do Eθ here as well: For Eθ we start with Eθ,tot(r,m,z) = +j yV(0) (1/4) ηm Rdc (ak) hm sin(k[z-L]) / ksin(kL) which we then write as Eθ,tot(r,m,z) = +j yV(0) (1/4) ηm Rdc (ak) hm sin(k[z-L]) / ksin(kL) m ≠0 Eθ,tot(r,m,z) = +j yV(0) (1/4) Rdc (ak) h0 sin(k[z-L]) / ksin(kL) m = 0 We then use App D.11 results that for small β', hm = (r/a)|m|+1 - (r/a)|m|-1 h0 = 0 Then we get Eθ,tot(r,m,z) = +j yV(0) (1/4) ηm Rdc (ak) [(r/a)|m|+1 - (r/a)|m|-1] sin(k[z-L]) / ksin(kL) m ≠0 Eθ,tot(r,0,z) = +j yV(0) (1/4) Rdc (ak) h0 sin(k[z-L]) / ksin(kL) m = 0 Now do the other limits to get Eθ,tot(r,m,z) = +j yV(0) (1/4) ηm Rdc (a) [(r/a)|m|+1 - (r/a)|m|-1] ([z-L]) / (L) m ≠0 Eθ,tot(r,0,z) = +j yV(0) (1/4) Rdc (a) h0 ([z-L]) / (L) m = 0 Now we safely set h0 = 0 to get Eθ,tot(r,m,z) = +j yV(0) (1/4) ηm Rdc (a) [(r/a)|m|+1 - (r/a)|m|-1] ([z-L]) / (L) m ≠0 Eθ,tot(r,0,z) = 0 m = 0 Finally use again y = jωC to get Eθ,tot(r,m,z) = 0 m ≠0 Eθ,tot(r,0,z) = 0 m = 0 and then the same for Jθ. So, assuming Ohm's law is valid, here is our DC limit for all field results inside the round wire: Jr,tot(r,m,z) = 0 m≠0 Jr,tot(r,0,z) = 0 m=0 Jθ,tot(r,m,z) = 0 m ≠0 Jθ,tot(r,0,z) = 0 m = 0 Jz,tot(r,m,z) = i(z) (1/2) ηm (1/πa2) (r/a)|m| (|m|+1) m ≠ 0 Jz,tot(r,0,z) = i(z) (1/πa2) m = 0 So what IS IT that is making Jz be asymmetric like this? Do we still have div E = 0 inside the round wire?? It seems that , after doing ω→0 and then doing div E, div E = ∂zEz = stuff * ∂z [i(z)] = 0 so div E = 0 is OK even with this asymmetry. How on earth can Jz be asymmetric here? What mechanism could make that happen? There is no radial current, there is no azimuthal current, so the mechanisms I am used to based on ηm seem to no longer apply. I think the two App D boundary conditions are met. Part III. Redo Part II in a simpler manner. We start off with this ansatz for the total field Etot in the reflection scenario which matches the scenario of Part I above which was mainly for "outside" the conductors; Ei,tot(r,m,z,t) = A(k) { ej(ωt-k[z-L]) Ei(r,m;k) - ej(ωt+k[z-L]) Ei(r,m; -k) } or Etot(r,m,z,t) = A(k) { ej(ωt-k[z-L]) E(r,m;k) - ej(ωt+k[z-L]) E(r,m; -k) } This is the superposition of right and left going waves, where Ei(r,m;k) are the solutions of Appendix D found for a wave going in the +z direction on an infinite transmission line. 1. Show that div Etot(r,θ,z) = 0 and that the three Helmholtz equations are satisfied Consider: Etot(r,θ,z,t) = A(k) { ej(ωt-k[z-L]) E(r,θ;k) - ej(ωt+k[z-L]) E(r,θ; -k) } = A(k) { ejkL [ej(ωt-kz) E(r,θ;k)] - e-jkL [ej(ωt+kz) E(r,θ; -k)] } (*) Recall now from App D [ where E(r,θ) are the specific Appendix D solutions ] E(r,θz,t; k) = ej(ωt-kz) E(r,θ) . (D.1.1) so we then have Etot(r,θ,z,t) = A(k) { ejkL E(r,θz,t; k) - e-jkL E(r,θz,t; -k) } (**) We showed in App D that div E(r,θz,t; k) = 0, and this is also true for -k , so each term in (**) has zero divergence, so the sum of the two terms then has zero divergence, QED. We also showed in App D that (2 + β2) E(r,θz,t; k) = 0 where β is for the conductor. Looking at (**) we see that each term separately satisfies the three Helmholtz equations, so the sum does as well. 2. Show that (**) satisfies the Eθ,tot(r=a,m) = 0 boundary condition: Our ansatz expression above for i = θ reads: Eθ,tot(r,m,z,t) = A(k) { ej(ωt-k[z-L]) Eθ(r,m;k) - ej(ωt+k[z-L]) Eθ(r,m; -k) } Therefore Eθ,tot(a,m,z,t) = A(k) { ej(ωt-k[z-L]) Eθ(a,m;k) - ej(ωt+k[z-L]) Eθ(a,m; -k) } But we know from App D that Eθ(a,m;k) = 0, and thus Eθ(a,m; -k) = 0, so Eθ,tot(a,m,z,t)= 0, QED. 3. The Reflection Boundary Conditions at z = L First, if we just look at the Appendix D solutions for E(r,θ;k), Ez(r,m;k) = (1/4) ηm I Rdc (aβ') fm fm = [ - ] x = β'r Er(r,m;k) = (j/4) ηm I Rdc (ak) gm gm = [ + ] xa = β'a Eθ(r,m;k) = (1/4) ηm I Rdc (ak) hm hm = [ - ] we know that Ez = σJz must change sign if we make the right-moving wave move to the left. None of the factors shown above in Ez changes sign under k→ - k except I. Obviously the current changes sign if you change the wave direction. Formally, I(-k) = - I(k) = - I. Looking then at Er and Eθ, the extra visible factor k causes Er and Ez not to change sign. We then get these symmetry relations, Er(r,m; -k) = + Er(r,m; k) Eθ(r,m; -k) = + Eθ(r,m; k) Ez(r,m; -k) = – Ez(r,m; k) When there are used in (**) we find Er,tot(r,m,z,t) = A(k) { ej(ωt-k[z-L]) Er(r,m;k) - ej(ωt+k[z-L]) Er(r,m; k) } Eθ,tot(r,m,z,t) = A(k) { ej(ωt-k[z-L]) Eθ(r,m;k) - ej(ωt+k[z-L]) Eθ(r,m; k) } Ez,tot(r,m,z,t) = A(k) { ej(ωt-k[z-L]) Ez(r,m;k) + ej(ωt+k[z-L]) Ez(r,m; k) } which then simplifies to Er,tot(r,m,z,t) = A(k) Er(r,m;k) ejωt{ -2j sin(k[z-L]) } Eθ,tot(r,m,z,t) = A(k) Eθ(r,m;k) ejωt{ -2j sin(k[z-L]) } Ez,tot(r,m,z,t) = A(k) Ez(r,m;k) ejωt{ 2 cos(k[z-L]) } Notice these facts at z = L, the reflection point: Er,tot(r,m,z,t) = 0 Eθ,tot(r,m,z,t) = 0 Ez,tot(r,m,z,t) = A(k) Ez(r,m;k) ejωt{ 2 } We are then happy to see that Er and Eθ vanish at z = L because these are transverse fields at the shorting block, and inside the shorting block (perfect conductor) these fields must vanish. We see also that Ez takes its max value at z = L so we expect current Jz to be max at z = L where voltage is minimum 0. These results just fall out of our Appendix D solutions for the single wave. 4. Determination of the constant A(k) Assuming Ohm's law is valid, we know from 4" above that Jz,tot(r,m,z,t) = A(k) Jz(r,m;k) ejωt{ 2 cos(k[z-L]) } or in the θ domain Jz,tot(r,θ,z,t) = A(k) Jz(r,θ;k) ejωt{ 2 cos(k[z-L]) } If we integrate over the wire cross section, the above says i(z) = A(k) I ejωt { 2 cos(k[z-L]) } where we use the fact below D.2.30 which says Jz(r,θ;k) integrates to I. We could write i(z) = Itot(z). But we found in Part I that i(z) = (k/z) V(0) ejωt // assumption! where I have reinserted the time factor. We can compare these two expressions for i(z) to obtain coefficient A(k), A(k) I ejωt { 2 cos(k[z-L]) } = (k/z) V(0) ejωt A(k) I { 2 } = (k/z) V(0) A(k) = (k/z) V(0) There it is. 5. Final form for the reflection scenario E fields inside the conductor If we now install this coefficient into our "tot" field results and at the same time insert the E solutions of Appendix D quoted just above, here is what we get Er,tot(r,m,z,t) = (k/z) V(0) (j/4) ηm I Rdc (ak) gm ejωt{ -2j sin(k[z-L]) } Eθ,tot(r,m,z,t) = (k/z) V(0) (1/4) ηm I Rdc (ak) hm ejωt{ -2j sin(k[z-L]) } Ez,tot(r,m,z,t) = (k/z) V(0) (1/4) ηm I Rdc (aβ') fm ejωt{ 2 cos(k[z-L]) } Notice that the factors of I cancel on each line and we end up with [ note k/k multiplied by ] Er,tot(r,m,z) = +j yV(0) (j/4) ηm Rdc (ak) gm sin(k[z-L]) / ksin(kL) Eθ,tot(r,m,z) = +j yV(0) (1/4) ηm Rdc (ak) hm sin(k[z-L]) / ksin(kL) Ez,tot(r,m,z) = - yV(0) (1/4) ηm Rdc (aβ') fm cos(k[z-L]) / ksin(kL) Here we added a k in the denominator on each line in order to get (k2/z) = - y since k2 = -yz . Since we have assumed that G = 0, we can write this as Er,tot(r,m,z) = +j (jωC)V(0) (j/4) ηm Rdc (ak) gm sin(k[z-L]) / ksin(kL) Eθ,tot(r,m,z) = +j (jωC)V(0) (1/4) ηm Rdc (ak) hm sin(k[z-L]) / ksin(kL) Ez,tot(r,m,z) = - (jωC)V(0) (1/4) ηm Rdc (aβ') fm cos(k[z-L]) / ksin(kL) (*) 6. Show that (*) satisfies the charge pump boundary condition at z. That boundary condition (recall G = 0) says Er,tot(r=a-ε,θ,z) = (jω/σ) Nm(z) Now recall from Part I that n(θ,z) = n(θ,z=0) // assumption which in m space says Nm(z) = Nm(0) where Nm(0) is the charge density moment at z = 0, the start of the line. We can define relative moments for this as ηm(0) = Nm(0)/N0(0) But these moments are really determined by the "capacitor problem" and are independent of z and are the same as the moments which appear in Appendix D for the single wave. So write ηm(0) = ηm and then Nm(z) = N0(0) ηm Now recall (D.1.8), N0 = (1/2πa) q(0) = <n(θ)> (D.1.8) If we apply this to our reflection scenario quantities, we get N0(0) = (1/2πa) q(0) = <ntot(θ,0)> Thus (using only the left equality above) we may write Nm(z) = (1/2πa) q(0) ηm . Our charge pump boundary condition requires that Er,tot(r=a-ε,θ,z) = (jω/σ) Nm(z) or inserting on both sides, +j (jωC)V(0) (j/4) ηm Rdc (ak) gm(a) sin(k[z-L]) / ksin(kL) = (jω/σ) (1/2πa) q(0) ηm ? Is this condition met? Looking again at our App D solution box quoted above, we see that gm(r=a) = 2. Then cancelling obvious factors the above becomes [ note that the z-L order is different in the two sines ] -j (C)V(0) (j/4) Rdc (ak) 2 / k = (1/σ) (1/2πa) q(0) ? Now use Rdc = 1/(σπa2) and C V(0) = q(0) to get -j (j/4) 1/(σπa2) (ak) 2 / k = (1/σ) (1/2πa) ? -j (j/4) 2 = (1/2) ? (1/2) = (1/2) ? yes! Thus, we have shown that our reflection scenario solutions satisfy the charge pump boundary condition, QED. 7. Summary for the reflection scenario fields inside a round wire conductor of a TL with G = 0: Er,tot(r,m,z) = +j (jωC)V(0) (j/4) ηm Rdc (ak) gm(r) sin(k[z-L]) / ksin(kL) Eθ,tot(r,m,z) = +j (jωC)V(0) (1/4) ηm Rdc (ak) hm(r) sin(k[z-L]) / ksin(kL) Ez,tot(r,m,z) = - (jωC)V(0) (1/4) ηm Rdc (aβ') fm(r) cos(k[z-L]) / ksin(kL) These solutions satisfy all of the following: the Helmholtz equation (2 + β2)E = 0 div E = 0 the Eθ = 0 boundary condition the charge pump boundary condition Er = Eθ = 0 at z = L, the shorting block location Ez = maximum at the shorting block, so Jz = max and i(z) = max there V(z) = 0 at z = L 8. Current Jz is asymmetric in the general solution and then also at DC Without going any further, we can see that Ez,tot(r,m,z) / Ez,tot(r,0,z) = ηm fm(r)/f0(r) fm(r) = [ - ] f0(r) = 2 x = β'r xa = β'a For general values of r and k = k, it is clear that the ratio fm(r)/f0(r) is non-vanishing and that therefore the current density Jz is asymmetric in θ inside the wire. What is very surprising is that this asymmetric continues to exist in the DC limit ω → 0. To see this, we recall from App D.11 that fm(r) = (r/a)m (m+1) (2/β'a) f0(r) = 4/(aβ') and then in this limit we have Ez,tot(r,m,z) / Ez,tot(r,0,z) = ηm fm(r)/f0(r) = (r/a)m (m+1) (2/β'a) / 4/(aβ') = (r/a)m (m+1)/2 and there is your DC current asymmetry! I now have this problematic situation in the shorted-bar situation where the current i(z) does NOT vanish at DC. Here then is the Big Question: What physically is causing this asymmetry at DC ? It is not eddy currents since eddy currents go away at ω = 0. 9. What is the limit as ω → 0 of the E fields? As a first step, take k very small, and I will do this in phases so you can see the details: Er,tot(r,m,z) = +j (jωC)V(0) (j/4) ηm Rdc (ak) gm(r) sin(k[z-L]) / ksin(kL) Eθ,tot(r,m,z) = +j (jωC)V(0) (1/4) ηm Rdc (ak) hm(r) sin(k[z-L]) / ksin(kL) Ez,tot(r,m,z) = - (jωC)V(0) (1/4) ηm Rdc (aβ') fm(r) cos(k[z-L]) / ksin(kL) Er,tot(r,m,z) = +j (jωC)V(0) (j/4) ηm Rdc (ak) gm(r) (k[z-L]) / k(kL) Eθ,tot(r,m,z) = +j (jωC)V(0) (1/4) ηm Rdc (ak) hm(r) (k[z-L]) / k(kL) Ez,tot(r,m,z) = - yV(0) (1/4) ηm Rdc (aβ') fm(r) 1 / k(kL) Er,tot(r,m,z) = +j (jωC)V(0) (j/4) ηm Rdc (a) gm(r) ([z-L]) / (L) Eθ,tot(r,m,z) = +j (jωC)V(0) (1/4) ηm Rdc (a) hm(r) ([z-L]) / (L) Ez,tot(r,m,z) = - (y/k2) V(0) (1/4) ηm Rdc (aβ') fm(r) 1 / (L) Er,tot(r,m,z) = +j (y)V(0) (j/4) ηm (1/σπa2) (a) gm(r) (z-L)/L Eθ,tot(r,m,z) = +j (y)V(0) (1/4) ηm (1/σπa2) (a) hm(r) (z-L)/L Ez,tot(r,m,z) = (1/z)V(0) (1/4) ηm (1/σπa2) (aβ') fm(r)/L Er,tot(r,m,z) = +j (y)V(0) (j/4) ηm (1/σπa) [gm(r)] (z-L)/L Eθ,tot(r,m,z) = +j (y)V(0) (1/4) ηm (1/σπa) [hm(r)] (z-L)/L Ez,tot(r,m,z) = (1/z)V(0) (1/4) ηm (1/σπa2) [fm(r)] (aβ') /L Next, here are limits for the three fm type factors from App D.11 for m ≥ 0 fm = (r/a)m (m+1) (2/β'a) f0 = 4/(aβ') gm = (r/a)m+1 + (r/a)m-1 g0 = 2 (r/a) hm = (r/a)m+1 - (r/a)m-1 h0 = 0 We then have to break things into two groups. Er,tot(r,m,z) = +j (jωC)V(0) (j/4) ηm (1/σπa) [(r/a)m+1 + (r/a)m-1] (z-L)/L Eθ,tot(r,m,z) = +j (jωC)V(0) (1/4) ηm (1/σπa) [(r/a)m+1 - (r/a)m-1] (z-L)/L Ez,tot(r,m,z) = (1/z)V(0) (1/4) ηm (1/σπa2) [(r/a)m (m+1) (2)] /L m > 0 Er,tot(r,0,z) = +j (jωC)V(0) (j/2) (1/σπa) (r/a) (z-L)/L Eθ,tot(r,0,z) = 0 Ez,tot(r,0,z) = (1/z)V(0) (1/σπa2) /L m = 0 The final step is the explicit ω = 0 so we get Er,tot(r,m,z) = 0 Eθ,tot(r,m,z) = 0 Ez,tot(r,m,z) = (1/R)V(0) (1/4) ηm (1/σπa2) [(r/a)m (m+1) (2)] /L m > 0 Er,tot(r,0,z) = 0 Eθ,tot(r,0,z) = 0 Ez,tot(r,0,z) = (1/R)V(0) (1/σπa2) /L m = 0 10. Does the DC solution satisfy curl E = 0 ? No !!! Start with general result in cyl coordinates curl E(r,θ,z) = [ r-1∂θEz - ∂zEθ] + [∂zEr - ∂rEz] + [ r-1∂r(rEθ) - r-1∂θEr ] Then go into partial waves to get curl E(r,m,z) = [ r-1jmEz - ∂zEθ] + [∂zEr - ∂rEz] + [ r-1∂r(rEθ) - r-1jmEr ] At DC, none of the fields varies with z so ∂z = 0 (I think this is legal). [ Or sayAlso Er = Eθ = 0 so get curl E(r,m,z) = [ r-1jmEz] + [- ∂rEz] But this is clearly NOT 0, so something is wrong in my machinery!!! After many months I finally have an indication that something is wrong! (a) Review Appendix D solutions for curl E = 0 at DC. In Appendix D, I use the curl E equation to compute B when ω ≠ 0 : curl E = -jωB . This reminds me of a problem I once thought I had, which was that B must then →∞ as ω → 0 which is physically wrong and would be repaired by curl E → 0. So maybe let's look at this entirely in App D terms and forget the reflection scenario for a while. Question 1: What happens if I look at the E field solutions before BC's are added. Those solutions are First summary of the E field solutions (D.2.21) Ez(r,m) = - j (β'/k) Jm(x) x = β'r (D.1.27) Er(r,m) = am x-1 Jm(x) + Jm+1(x) β'2 = β2 - k2 (D.2.11) jEθ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) (D.2.15) Now from above we have, where I replace ∂z → - jk as apropos for App D, curl E(r,m,z) = [ r-1jmEz+jkEθ] + [+jk Er - ∂rEz] + [ r-1∂r(rEθ) - r-1jmEr ] The three component equations are then r-1jmEz+jkEθ = 0 +jk Er - ∂rEz = 0 r-1∂r(rEθ) - r-1jmEr = 0 at DC ? I presume the E fields are all finite as the limit is approached. I know that k→ 0 at DC so we then have mEz = 0 ∂rEz = 0 ∂r(rEθ) - jmEr = 0 at DC ? Assuming Km and am are non-singular in this DC limit [ can check from (D.2.28) ] I see from (D.2.21) [ also assuming that all Bessel combinations are finite ] that β'/k = / [ ω1/2 e-jπ/4] = / [ e-jπ/4] for G = 0 and this result is a constant. On the other hand, I know that Jm(x) = (x/2)m / m! x = β'r so that as β'→0, x→0 and then Jm(x) → 0 except for m = 0. Thus, for general Km constant, we really do have mEz = 0 for all m in the DC limit, so the first equation above is OK. Maybe curl E = 0 for all coefficients! Question 2: Let's repeat the above with our full solutions where BC's are included. Let's now look at the solutions with BC's and in the DC approaching limit. Ez(r,m) = (1/2) ηm I Rdc (r/a)|m| (|m|+1) Er(r,m) = (j/4) ηm I Rdc (ak) [(r/a)|m|+1 + (r/a)|m|-1] (D.11.7) Eθ(r,m) = (1/4) ηm I Rdc (ak) [(r/a)|m|+1 - (r/a)|m|-1] Ez(r,0) = I Rdc Er(r,0) = (j/2) I Rdc (ak) (r/a) Eθ(r,0) = 0 // low ω E fields Go back to our three test equations to see if curl E = 0: mEz = 0 ∂rEz = 0 ∂r(rEθ) - jmEr = 0 at DC ? The first one is OK for m = 0. For m ≠ 0 we need ( 1/2) ηm I Rdc (r/a)|m| (|m|+1) → 0 but this only works in the case I→0 ! One might use I = 2πω (a/k) N0 of (D.2.31) to replace I. Then we have I = 2πω (a/k) N0 = 2πω (a/ [ ω1/2 e-jπ/4]) N0 = 2πω1/2 (a/ [ e-jπ/4]) N0 and then as ω → 0, we really do have I→ 0 and then mEz = 0 is rescued. By the same argument, we will find that ∂rEz = 0. The ∂r does not create anything singular, and then I = 0. Same argument for the last line. Conclusion: The solutions of App D do respect curl E = 0 in the DC limit! It just took me a little while to verify that above. (b) Study curl E = 0 for DC again in the reflection scenario. Somehow maybe my reflection scenario wrecks this fact by having a finite resulting current? Let's look at the general form of the reflection solution, Etot(r,m,z,t) = A(k) { ej(ωt-β[z-L]) E(r,m;k) - ej(ωt+β[z-L]) E(r,m; -k) } where A(k) = (k/z) V(0) In the DC limit we see that A(k) → (V(0)/ z) (1/2IL) so this is immediately suspicious since it appears to blow up at DC! But let's for now take the curl of {} and see what happens. Maybe it is better to work with these forms instead, Er,tot(r,m,z) = A(k) Er(r,m;k) { -2j sin(k[z-L]) } Eθ,tot(r,m,z) = A(k) Eθ(r,m;k) { -2j sin(k[z-L]) } Ez,tot(r,m,z) = A(k) Ez(r,m;k) { 2 cos(k[z-L]) } . Then curl E (r,θ,z) = [ r-1∂θEz(r,θ,z) - ∂zEθ] + [∂zEr - ∂rEz] + [ r-1∂r(rEθ) - r-1∂θ Er ] so curl E (r,m,z) = [ r-1jmEz(r,m,z) - ∂zEθ] + [∂zEr - ∂rEz] + [ r-1∂r(rEθ) - r-1jm Er ] where imagine "tot" in each subscript. Our three equations are r-1jmEz,tot - ∂zEθ,tot = 0 ∂zEr,tot - ∂rEz,tot = 0 r-1∂r(rEθ,tot) - r-1jmEr,tot = 0 for DC ? Let's study just the first r component equation : r-1jm{Ez,tot} = ∂z{Eθ,tot} ? r-1jm{ A(k) Ez(r,m;k) { 2 cos(k[z-L]) } = ∂z{ A(k) Eθ(r,m;k) { -2j sin(k[z-L]) } ? r-1jm{ A(k) Ez(r,m;k) { 2 cos(k[z-L]) } = { A(k) Eθ(r,m;k) { -2j kcos(k[z-L]) } ? A(k) r-1jm{ Ez(r,m;k) { 1 } = A(k){ Eθ(r,m;k) { -j k} ? A(k) r-1m{ Ez(r,m;k) } = - A(k)k {Eθ(r,m;k)} ? Let's now install the App D solutions: A(k) r-1m{ (1/4) ηm I Rdc (aβ') fm } = - A(k)k {(1/4) ηm I Rdc (ak) hm } ? Now we showed above that A(k) I = (k/z) V(0) so the above becomes (k/z) V(0) r-1m{ (1/4) ηm Rdc (aβ') fm } = - (k/z) V(0) {(1/4) ηm Rdc (ak) hm } The ratio k/sin is finite even at ω = 0, so divide it out and other things to get (1/z) r-1m{ (1/4) (β') fm } = - (1/z) {(1/4) (k) hm } ? z is also finite, so get rid of it as well r-1m{ (β') fm } = - { (k) hm } ? Now look at the DC limits of fm and hm : fm = (r/a)|m| (|m|+1) (2/β'a) f0 = 4/(aβ') gm = (r/a)|m|+1 + (r/a)|m|-1 g0 = 2 (r/a) hm = (r/a)|m|+1 - (r/a)|m|-1 h0 = 0 . (D.11.6) For m > 0 we then want to check r-1m{ (β') (r/a)|m| (|m|+1) (2/β'a)} = - { (k) [(r/a)|m|+1 - (r/a)|m|-1]} ? r-1m{ (r/a)|m| (|m|+1) (2/a)} = - { (k) [(r/a)|m|+1 - (r/a)|m|-1]} ? r-1m{ (r/a)|m| (|m|+1) (2/a)} = 0 ? This is clearly NOT true, and therefore we do not have curl Etot = 0 at DC! Let's now repeat what we just did this time using the DC limits earlier on: A(k) r-1m{ Ez(r,m;k) } = - A(k)k {Eθ(r,m;k)} ? A(k)r-1m{ (1/2) ηm I Rdc (r/a)|m| (|m|+1)} = - A(k)k { (1/4) ηm I Rdc (ak) [(r/a)|m|+1 - (r/a)|m|-1]} A(k)r-1m{ I (r/a)|m| (|m|+1)} = - A(k) k { (1/2) I (ak) [(r/a)|m|+1 - (r/a)|m|-1]} Now use the fact that A(k)I → (V(0)/ z) (1/2L) so we want to test (V(0)/ z) (1/2L)r-1m{ (r/a)|m| (|m|+1)} = - (V(0)/ z) (1/2L) k { (1/2) (ak) [(r/a)|m|+1 - (r/a)|m|-1]} r-1m{ (r/a)|m| (|m|+1)} = - k { (1/2) (ak) [(r/a)|m|+1 - (r/a)|m|-1]} ? r-1m{ (r/a)|m| (|m|+1)} = 0 ? Same result: This is NOT an equality, so curl E = 0 does not obtain in the DC limit for the reflection scenario. The problem is caused by the fact that A(k)I → constant. Thus, I have found a FLAW in my little constructed reflection scenario! For ω > 0, it seems OK because then as usual you compute B from curl E, and then you have div B = div curl E = 0 and then you have all 4 Maxwell's satisfied! div Etot = 0 by direct computation I did above in (1). the Helmholtz equation (2+β2)E = 0 as verified in (1) the two boundary conditions as verified in (2) and (6) Here is a verification of the curl B equation: (imagine all have tot subscripts) B = (-1/jω) curl E // this is how I would compute B from E curl B = (-1/jω) curl curl E = (-1/jω) [ ( div E) - 2E ] = (+1/jω) 2E = (+1/jω) (- β2 E) // using Helmholtz = ( jβ2/ω) E But this IS the correct curl B equation since (assuming Ohm's law) curl B = μ J + μ jωεE = μ(σ + jωε) E = μ(jω)( ε - jσ/ω) E = jω μξ E = j (β2/ω) E . // see (1.5.1c) Notice also that then, computing B as above, div B = div (-1/jω) curl E = (-1/jω) div curl E = 0 How about the other Helmholtz equation? Can we show it is also respected? curl B = j (β2/ω) E / / known true from above curl curl B = j (β2/ω) curl E // apply curl to both sides [ ( div B) - 2B ] = j (β2/ω) curl E // vector ident [- 2B ] = j (β2/ω) [-jωB] // div B = 0 from above. [- 2B ] = (β2) [B] (2 + β2) B = 0 Thus, for ω > 0 I can extend the above list to div Etot = 0 by direct computation I did above in (1). the Helmholtz equations (2+β2)Etot = 0 as verified in (1) the two boundary conditions as verified in (2) and (6) curl Btot = μ J + μ jωε Etot div Btot = 0 curl Etot = -jωBtot the Helmholtz equations (2+β2)Btot = 0 How about continuity? Inside we have that div E = 0 so therefore div J = 0. At the boundary, the CPBC assures that div J = -jωρ in the interaction with the surface charge and G = 0. Thus, for ω > 0 my round wire transmission line reflection scenario solution satisfies : all four Maxwell equations both E and B Helmholtz equations the two boundary conditions continuity inside and at the surface I would have thought that uniqueness would imply that this then is the solution to the problem!!! It just seems very unlikely that this solution would fail in the limit ω→ 0 if it is OK for any ω > 0 ! It really makes me think I have made a mistake somewhere. Let's just recap the calculation above. I get [curl Etot (r,m,z)]r = r-1jmEz,tot - ∂zEθ,tot = r-1jm{ A(k) Ez(r,m;k) { 2 cos(k[z-L]) } - { A(k) Eθ(r,m;k) { -2j kcos(k[z-L]) } = 2 A(k) cos(k[z-L]) { r-1jm Ez(r,m;k) - Eθ(r,m;k) (-jk) } = 2 A(k) cos(k[z-L]) { r-1jm (1/4) ηm I Rdc (aβ') fm - (1/4) ηm I Rdc (ak) hm (-jk) } = 2 [ A(k) I ] cos(k[z-L]) ηmRdc (1/4) { r-1jm (aβ') fm + (ak) hm jk) } = 2 [ A(k) I ] cos(k[z-L]) ηmRdc (j/4) { r-1m (aβ') fm + (ak2) hm ) } = 2 [(k/z) V(0) ] cos(k[z-L]) ηmRdc (j/4) { r-1m (aβ') fm + (ak2) hm ) } Now we take the DC limit to find = 2 [(1/z) V(0) ] cos(k[z-L]) ηmRdc (j/4) { r-1m (aβ') fm + (ak2) hm ) } = 2 [(1/z) V(0) ] cos(k[z-L]) ηmRdc (j/4) { r-1m (aβ') (r/a)|m| (|m|+1) (2/β'a)+ (ak2) 0] ) } = 2 [(1/z) V(0) ] cos(k[z-L]) ηmRdc (j/4) { r-1m (aβ') (r/a)|m| (|m|+1) (2/β'a)) } = 2 [(1/z) V(0) ] ηmRdc (j/4) { r-1m (r/a)|m| (|m|+1) (2)) } m > 0 Thus, we conclude that for m > 0 [curl Etot (r,m,z)]r = 2 [(1/z) V(0) ] ηmRdc (j/4) { r-1m (r/a)|m| (|m|+1) (2)) } Note: As shown just below, we must have curl E = ω * [stuff] where stuff is finite near ω = 0. But the above expression does NOT have this form! Big trouble here in River City, my friend, yes sir. This DC result is a finite value which is a function of r and z and m. It is not 0. One implication is this. Recall that B = (-1/jω) curl E so Br(r,m) = (-1/jω) {2 [(1/z) V(0) ] ηmRdc (j/4) { r-1m (r/a)|m| (|m|+1) (2)) }} As ω→0. we get Br = ∞ everywhere which is real nonsense! In a magnetostatic situation, one must have curl E = 0 as in electrostatics. We know this from curl E = jωB with a finite B field. So what is causing Br,tot to be infinite at DC ? Part IV. So what is wrong? I reread Part III today 8.12.14 and it seems fine, but I was looking for assumptions made. I first checked for the 100th time the limit of fm and of course it is the same as it was the previous 99 times. Idea 1. One thing I did notice is that when it comes to determining A(k) in section (4), I have to make use of my Part I expression for i(z) which is a standing wave. But this i(z) was obtained from the TL equations, and I make a big issues of the fact that I think these TL equations are invalid for small ω. They are valid in the network model all the way down to ω = 0, but I don't think they are valid in the true physics model! But I sort of rescue the TL equations in Chapter 4 by doing "averaging" of various things. As part of this rescue, I get at one point ∂zV(z) = - [ Zs1+ Zs1+ jωLe] i(z) ∂z i(z) = - [ jk2/(ωLe)] V(z) . (4.11.14a) In these equations which are basically the TL equations, these quantities are all averages: V(z) Zsi Le whereas i(z) is not an average, it is the total current in the conductor. The first equation here is basically what we used above to compute i(z). It just reads ∂zV(z) = - z i(z) and this is the only equation I used in Part I. So this really seems reasonable and is probably NOT the source of my problem! Idea 2. When all is said and done, the Jz asymmetry present in the Appendix D solutions just passes through into the reflection scenario. In this doc, I think I finally constructed a solid reflection scenario, getting all the details and limits right. But I think "the problem" exists right in that Appendix D solution for the single direction TL wave for an infinite line. The problem is present there, and of course it just moves into the reflection solution pretty much unaltered. The big difference is that whereas I→0 in appendix D, we have i(z) → constant ≠ 0 in the reflection scenario, so now the asymmetry is a more painful violation of our eddy current intuition. I was tempted to ignore the Jz asymmetry present in the Appendix D solution because I→0. But deep down, I was always expecting to see the Jz asymmetry gradually fade away for small ω, before the ω→0 final limit was reached, but it does not fade away, it approaches a constant value. In Section D.11 (a) I try to justify this asymmetry failure of the theory. I argue that the TL equations are exact for ω all the way down to 0 in the network model, but they are not exact in the physics model. In retrospect, the averaging was supposed to repair this problem, so the argument at best is "weak". I am just looking for something to explain the asymmetry anomaly! Perhaps I was hoping that somehow the reflection construction would make the asymmetry go away, but it did not do that! The only way to find out was to do the scenario in detail, and that is what this document has done! Conclusion: If I really want to solve the Jz asymmetry mystery, I need to do it directly in the App D solution and not worry about the reflection scenario. *************************************************************************** Appendix. What IS the function that keeps appearing? [ This stuff is not very interesting since I can just plot the series. ] Ez(r,θ) = I Rdc { 1 + Σm=1∞ (r/a)m (m+1) ηm cos(mθ) } ηm ≡ Nm/N0 = (-1)m e-|mξ| = I Rdc { 1 + Σm=1∞ (r/a)m (m+1) (-1)m e-m|ξ| cos(mθ) } ≡ I Rdc S For shorthand, lets write ξ ≡ |ξ1| so then we have S = 1 + Σm=1∞ (r/a)m (m+1) (-1)m e-mξ cos(mθ) So I will now attempt to evaluate this sum. But first simplify cos(mθ) = (ejmθ + e-jmθ)/2 S = 1 + Σm=1∞ (r/a)m (m+1) (-1)m e-mξ (ejmθ + e-jmθ)/2 = 1 + { (1/2) Σm=1∞ (r/a)m (m+1) (-1)m e-mξ ejmθ + c.c. } Next, write (r/a)m = emln(r/a) so we then have S = 1 + { (1/2) Σm=1∞ emln(r/a) (m+1) (-1)m e-mξ ejmθ + c.c. } = 1 + { (1/2) Σm=1∞ (m+1) (-1)m e-m[ln(a/r)+ξ-jθ] + c.c. } Let x ≡ ln(a/r)+ξ-jθ Then we have' = 1 + { (1/2) Σm=1∞ (m+1) (-1)m e-mx + c.c. } Now the expo is as simple as it can get. Now define S1 = Σm=1∞ (-1)m e-mx S2 ≡ - ∂xS1 = Σm=1∞ (-1)m m e-mx Then we can write S = 1 + (1/2) [(S1 + S2) + c.c. ] Now call upon GR7 page 27, which says tanh(x/2) = 1 + 2 Σm=1∞ (-1)m e-mx or Σm=1∞ (-1)m e-mx = (1/2)[ tanh(x/2) - 1] Then S1 = Σm=1∞ (-1)m e-mx = (1/2)[ tanh(x/2) - 1] S2 = - ∂xS1 = -(1/2) (1/2) sech2(x/2) = -(1/4) sech2(x/2) We then know that S = 1 + (1/2) { (1/2)[ tanh(x/2) - 1] - (1/4) sech2(x/2) + c.c. } = 1 + (1/4) { [ tanh(x/2) - 1] - (1/2) sech2(x/2) + c.c. } = 1 + (1/2) Re { [ tanh(x/2) - 1] - (1/2) sech2(x/2) } = 1 + (1/2) Ref Now let y ≡ x/2 Then we have S = 1 + (1/2) Re { [ tanh(y)- 1] - (1/2) sech2(y) } where y = x/2 = [ ln(a/r)+ξ-jθ]/2 = [ (1/2)(ln(a/r) + ξ)] + j [ -(1/2)θ ] = a + jb a = (1/2)(ln(a/r) + ξ) b = - θ/2 This is the first time I have ever computed this result. Maybe Maple can simplify this result? This result is more complicated than I want to deal with even though the denominators are the same apart for powers. But at least I know there is a closed form. However, for plotting purposes it is a lot easier to just use the series for S up to some finite number mmax of terms. The plot of Ez as this sum S S = 1 + Σm=1∞ (r/a)m (m+1) (-1)m e-m|ξ| cos(mθ) has "the usual form" : which shows strong asymmetry. What is new is that I am now getting this asymmetry at ω = 0 !!! End