standing wave version of Appendix D REVIEWED

Standing wave version of Appendix D PhL 6.12.14 Reread Aug 7, 2014: here is a summary ( red comments added today) I successfully develop a "standing wave version" of Appendix D in the Plan C section. Both BC's, the divE = 0 and the three Helm equations are all satisfied by this E field solution. Basically the solution is the same as appears in regular Appendix D, except symbol I is no longer the total current in the line. All quantities exhibit reasonable standing waves. The problem is that this new solution, being of the same form as the old solution, has Jz asymmetry, so it does not solve THAT problem. Thus, although I seem to have a successful "reflection solution" with a shorting bar at z = L, I still get Jz asymmetry when G = 0. The first section Plan A just shows that you cannot assume the same trig z function for all E components. Section Plan B just explains why this does not work and suggests Plan C. Plan D is wrong. With the new loss model, β' = 0 when G = 0 and ω = 0, so Jz = constant is in fact a viable solution to the Helm equation, and in fact is just the m = 0 term. Plan E comes up with an interesting but complicated loss model and I think my k(ω) model is easier. Plan A Trying a different App D starting form -- NO GO 1 Plan B -- Why Plan A is No Go 2 Plan C Standing wave version of Appendix D 3 Plan D (a question) : Does the solution Jz = constant [ Ez = constant ] satisfy Helmholtz? 8 Plan E: An interesting though complicated Loss Model 9 Plan F: Review fact that Jz asymmetry exists in the Regular App D solution. 13 [ When this appendix was written, what is now k in Appendix D was called βd ] Plan A Trying a different App D starting form -- NO GO If you superpose traveling waves of the same amplitude, you end up with a standing wave as Maple animate well demonstrates. In this case you can maintain V(L) = 0 all the time as you would have a shorted line ending. Basically, to find the E field solutions for this case, you have to redo Appendix D making this replacement right at the very start: E(r,θz,t) = ej(ωt-βz) E(r,θ) traveling wave E(r,θz,t) = ejωt sin(βdz) E(r,θ) standing wave There is a big difference in these two forms. Notice that ∂z2 E(r,θz,t) = - βd2 E(r,θz,t) same as before but ∂z E(r,θz,t) = βd cot(βdz) E(r,θz,t) Proof of this last line: ∂z E(r,θz,t) = ejωt E(r,θ)βd cos(βdz) = ejωt E(r,θ)βd cos(βdz)/ sin(βdz) * sin(βdz) = βd cot(βdz) E(r,θz,t) So we then have to replace the rule ∂z → -jβd with the new rule ∂z → βd cot(βdz) which I am sure is a very significant difference! [ agreed ] Comment: I went down this path long ago I think, but don't know right now where it is. So what then happens in Appendix D? (1) Since the Helmholtz equations all have ∂z2 ONLY, they are unchanged. (2) The div E = 0 gets the replacement -j → cot(βdz) for the ∂z term. We then have ∂r (r Er) + ∂θEθ + r ∂zEz = 0 . Applying the conversion rules gives ∂r [r Er(r,m,z,t)] + jmEθ(r,m,z,t) + r (cot(βdz)βd)Ez(r,m,z,t) = 0 [1 + r∂r ] Er(r,m,z,t) + jmEθ(r,m,z,t) + r (cot(βdz)βd)Ez(r,m,z,t) = 0 . (D.1.19) Now install: Er(r,m,z,t) = ejωt sin(βdz) Er(r,m) etc to get sin(βdz)∂r [r Er(r,m)] + jm sin(βdz)Eθ(r,m) + r (cos(βdz)βd)Ez(r,m) = 0 But this is a contradiction by saying that that functions are of (r,m) but you see functions of z entering the equation! Again I think this is saying: "The assumed form E(r,θz,t) = ejωt sin(βdz) E(r,θ) E(r,mz,t) = ejωt sin(βdz) E(r,m) cannot solve div E = 0 ". [ agreed ] Plan B -- Why Plan A is No Go (a) I think I have an explanation of why the above form fails. What I have written down in App D as solutions for a +z wave are really these Ei(r,m;βd) For a -z wave, the solutions are NOT THE SAME. They are in fact Ei(r,m; -βd) which in general is not the same function. Now when you superpose you get ej(ωt-βz) Ei(r,m;βd) + ej(ωt+βz) Ei(r,m; -βd) and this is NOT of the simple standing wave form Ei(r,m) ejωt 2 cos(βdz) and that is because the two functions are DIFFERENT (at least for some components). [ agreed ] (b) Suppose you superpose two opposite-directed waves of exactly the same amplitude as I have just done in writing ej(ωt-βz) Ei(r,m;βd) + ej(ωt+βz) Ei(r,m; -βd) It seems to me that each wave would have the same function n(θ). Then you might get n(θ,z) = ej(ωt-βz) n(θ) + ej(ωt+βz) n(θ) = ejωt 2 cos(βdz) n(θ) [ ok ] and then you CAN get a standing wave form for the charge density and thus for the external problem! For example you would then expect to have a standing wave form for V(z) since this is based on n. Plan C Standing wave version of Appendix D 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) = ej(ωt-β[z-L]) Ei(r,m;βd) - ej(ωt+β[z-L]) Ei(r,m; -βd) I expect these two waves to have n(θ) and -n(θ), so I expect to get n(θ,z) = ej(ωt-β[z-L]) n(θ) + ej(ωt+β[z-L]) [-n(θ)] = - ejωt 2jsin(βd[z-L]) n(θ) and THIS is a charge density that I can then force to always be 0 at z = L, as might be correct for a shorting bar at that location! [ seems good ] [ BUT Fatal Flaw noted at this point: If βd→0 as ω→0, then according to the line above, n(θ,z) → 0 and thus q(z)→0. Since q = CV, that implies V(z) = 0 for all z. But I seek a DC solution with some V(0) ≠ 0 to drive DC current through the shorted line! So Plan C is dead meat right here. ] [ Possible rescue: If βd → A as ω→0, where A is small, then we would get n(θ,z) = - ejωt 2jA[z-L] n(θ) and in turn q(z) = - ejωt 2jA[z-L] q0 and in turn V(z) = - ejωt 2jA[z-L] q0/C and this IS in fact a linearly decreasing V(z) which vanishes at z = L, more what I want to see happen. For small G ≠ 0, this is exactly what happens since then A = -j and then V(z) = - ejωt 2 [z-L] q0/C. But I am looking for my non-vanishing DC solution V(z) in particular when G = 0. ] Now, how do I "access" the work of Appendix D and make it bear on this standing wave situation? I could first take hold of the "first summary" Ez(r,m; βd) = - j (β'/βd) Jm(x) x = β'r (D.1.27) Er(r,m; βd) = am(βd) x-1 Jm(x) + Jm+1(x) β'2 = β2 - βd2 (D.2.11) jEθ(r,m; βd) = - am(βd) x-1 Jm(x) + (+ ) Jm+1(x) (D.2.15) with the idea that Km and am might be functions of βd and in particular its sign. Then for each component the total field is given by Ei,tot(r,m,z,t) = ej(ωt-β[z-L]) Ei(r,m;βd) - ej(ωt+β[z-L]) Ei(r,m; -βd) Ei,tot(r,θ,z,t) = ej(ωt-β[z-L]) Ei(r,θ;βd) - ej(ωt+β[z-L]) Ei(r,θ; -βd) Next, we must "face up to" the two boundary conditions. This is where I always collapse. How about this idea: do the CPBC Gaussian box at a particular value of z : Er,tot(r=a-ε,θ,z,t) = (jω/σ) n(θ,z) = -(jω/σ) ejωt 2jsin(βd[z-L]) n(θ) In order for this to be viable, we must then have this form for the non z,t Er,tot: Er,tot(r=a-ε,θ,z,t) = -2jsin(βd[z-L]) ejωt Er,tot(r=a-ε,θ) (*) This then in turn says Er,tot(r=a-ε,θ) = (jω/σ) n(θ) (**) which at least "looks like" my original BC. [ so far so good ] Now suppose we assume that Er(r,m; -βd) = + Er(r,m; βd) for the r component only. Then we get Er,tot(r,θ,z,t) = ej(ωt-β[z-L]) Er(r,θ;βd) - ej(ωt+β[z-L]) Er(r,θ; -βd) = ej(ωt-β[z-L]) Er(r,θ;βd) - ej(ωt+β[z-L]) Er(r,θ; βd) = [ej(ωt-β[z-L]) - ej(ωt+β[z-L])] Er(r,θ; βd) = -2jsin(βd[z-L]) ejωt Er(r,θ; βd) [ good to have Er = 0 at z = L; gold brick idea ] and this is then consistent with the form I assumed in (*) above if I say Er,tot(r,θ) = Er(r,θ; βd) Then our CPBC is (**) which looks exactly the same as it did in the single-wave Appendix D, namely: Er(r=a,θ; βd) = (jω/σ) n(θ) For the first time EVER, I now at least have some kind of standing wave form for one of the internal E field components of the round wire. Now, I will make the ansatz that Eθ has this same form, so Eθ(r,m; -βd) = + Eθ(r,m; βd) Eθ,tot(r,θ,z,t) = -2jsin(βd[z-L]) ejωt Eθ(r,θ; βd) [ same gold brick idea ] Eθ,tot(r,θ) = Eθ(r,θ; βd) In this case, the other BC ALSO looks the same Eθ(r=a,θ; βd) = 0 I now make this third ansatz, which is that Ez gets a minus sign, so Ez(r,m; -βd) = - Ez(r,m; βd) Ez,tot(r,θ,z,t) = 2cos(βd[z-L]) ejωt Ez(r,θ; βd) . [ different sign from Er and Eθ ] Summary of the three -βd equations: Er(r,m; -βd) = + Er(r,m; βd) Eθ(r,m; -βd) = + Eθ(r,m; βd) Ez(r,m; -βd) = - Ez(r,m; βd) . Summary of the resulting three total field-component expressions: Er,tot(r,θ,z,t) = -2jsin(βd[z-L]) ejωt Er(r,θ; βd) Eθ,tot(r,θ,z,t) = -2jsin(βd[z-L]) ejωt Eθ(r,θ; βd) Ez,tot(r,θ,z,t) = 2cos(βd[z-L]) ejωt Ez(r,θ; βd) Reminder: Since ∂z2 = -βd2 for each of the new Ei,tot components, the three Helm equations are the same for these functions as they were for the old App D functions. I am hopeful now that the div E = 0 equation, which was problematical in Plan A, is now going to be OK. We have the raw cyl coordinates form, ∂r (r Er,tot(r,θ,z,t)) + ∂θEθ,tot(r,θ,z,t) + r ∂zEz,tot(r,θ,z,t) = 0 which then says ∂r (r (-)2jsin(βd[z-L]) ejωt Er(r,θ; βd)) + ∂θ 2 (-)jsin(βd[z-L]) ejωt Eθ(r,θ; βd) + r ∂z2cos(βd[z-L]) ejωt Ez(r,θ; βd) = 0 or ∂r (r (-)2jsin(βd[z-L]) ejωt Er(r,θ; βd)) + ∂θ (-)2jsin(βd[z-L]) ejωt Eθ(r,θ; βd) - r βd ∂z2sin (βd[z-L]) ejωt Ez(r,θ; βd) = 0 or -∂r (r j) ejωt Er(r,θ; βd)) - ∂θ j ejωt Eθ(r,θ; βd) - r βd ejωt Ez(r,θ; βd) = 0 or -j ∂r(r Er(r,θ; βd)) - j ∂θ Eθ(r,θ; βd) - r βd Ez(r,θ; βd) = 0 or ∂r(r Er(r,θ; βd)) + ∂θ Eθ(r,θ; βd) -j r βd Ez(r,θ; βd) = 0 Now this equation is at least consistent with itself since there are no functions of z appearing (recall in Plan A there was a cot thing). Also, this equation is identical to (D.1.19) !!! [ good ] At this point then we have this situation: The three Helm equations are the same The div E equation is the same The two BC's are the same. [ seems very promising at this point ] I think this means that our solutions must be the same TO THIS POINT: [ agreed ] Ez(r,m;βd) = - j (β'/βd) Jm(x) x = β'r (D.1.27) Er(r,m;βd) = am x-1 Jm(x) + Jm+1(x) β'2 = β2 - βd2 (D.2.11) jEθ(r,m;βd) = - am x-1 Jm(x) + ( + ) Jm+1(x) (D.2.15) am = (jω/2σ) 2m Nm . a0 = 0 (D.2.28) = (jω/2σ) Nm [ – ] = (jω/σ) N0 (+ ) = (jω/2σ) Nm [ + ] ( + ) = 0 The next question is this: What happens to the normalization condition relating I and N0 which appears in Appendix D below (D.2.30) ? It starts off this way: Itot = !Syntax Error, Idθ !Syntax Error, Ir dr Jz,tot(r,θ,z,t) = !Syntax Error, Idθ !Syntax Error, Ir dr { σ !Syntax Error, I Ez,tot(r,m,z,t) ejmθ } = σ !Syntax Error, I !Syntax Error, Ir dr Ez,tot(r,m,z,t) !Syntax Error, Idθ ejmθ = 2π σ!Syntax Error, Ir dr Ez,tot(r,0,z,t) [ ok ] But now we want to use, Ez,tot(r,m,z,t) = 2cos(βd[z-L]) ejωt Ez(r,m; βd) and then Itot(z) = 2π σ!Syntax Error, Ir dr 2cos(βd[z-L]) ejωt Ez(r,0; βd) = [ 2cos(βd[z-L]) ejωt ] 2πσ !Syntax Error, Ir dr Ez(r,0; βd) Apart from the red factor, things proceed exactly as below (D.2.30) and we get Itot(z,t) = [ 2cos(βd[z-L]) ejωt ] 2πω (a/βd) N0 We could at this point just define I ≡ 2πω (a/βd) N0 [ this has the exact same form as above (D.2.31) ] and then Itot(z,t) = [ 2cos(βd[z-L]) ejωt ] I I ≡ 2πω (a/βd) N0 So then I is no longer "the magnitude" of the current in the wire, but it is the coefficient you see here of the standing wave of the current. I think then that this leads to the following solutions: Ez(r,m;βd) = (1/4) ηm I Rdc (aβ') fm fm = [ - ] x = β'r Er(r,m;βd) = (j/4) ηm I Rdc (aβd) gm gm = [ + ] xa = β'a Eθ(r,m;βd) = (1/4) ηm I Rdc (aβd) hm hm = [ - ] where I has the interpretation just stated above. The problem now is that each of these solution fields has the wrong "symmetry" for negation of βd. OK, ignoring this detail (which could of course be fatal) I will rewrite the solutions as Ez(r,m;βd) = sign(βd) (1/4) ηm I Rdc (aβ') fm fm = [ - ] x = β'r Er(r,m;βd) = sign(βd) (j/4) ηm I Rdc (aβd) gm gm = [ + ] xa = β'a Eθ(r,m;βd) = sign(βd) (1/4) ηm I Rdc (aβd) hm hm = [ - ] where I have to just assume that βd is real and ignore the loss business. [ so far so good ] So MAYBE I now have a standing wave model for a shorting bar at z = L. The following quantities vanish at z = L: (all the sine quantities) n(θ,z) Er,tot(r,θ,z,t) Eθ,tot(r,θ,z,t) V (since based on n ) However, the following quantities are a maximum at z = L Ez,tot(r,θ,z,t) Itot(z,t) This does at least SEEM right for a shorting bar. The transmission line ends in a gold plane which does not allow n(θ,z) to exist at z = L. The current is always a maximum, as is Jz current density. So this was my "idea" which motivated the standing wave approach of Plan C. [ it all seems very promising to this point ] So let us then look at the current density Jz,tot : Jz,tot(r,m,z,t) = 2σ cos(βd[z-L]) ejωt Ez(r,m; βd) = 2σ cos(βd[z-L]) ejωt sign(βd) (1/4) ηm I Rdc (aβ') fm(β',r,a) At every point z, we still have our asymmetry problem! The ηm are the same as they have always been, and if we do a low ω limit, we will find moments of Jz for m ≠ 0 and we will have current asymmetry. Conclusion: This "shorting bar" model does not seem to "solve" the Jz asymmetry problem. Nevertheless, I think it was a valiant attempt, perhaps one of my best so far. [ I agree, this was indeed a carefully done and "valiant" attempt, and I agree with the conclusion.] Plan D (a question) : Does the solution Jz = constant [ Ez = constant ] satisfy Helmholtz? This is not so much a "plan" as an "explanation" of what is happening. 1. If you start off with the Helm equation with β2, things make sense in this regard: If you take ω → 0, you find that β→0 and then Helm is just vector Laplace 2E = 0 and then Ez = constant is a viable solution for inside a conductor. Ez = constant also satisfies div E = 0. 2. However, once you make the ANSATZ that field components go like ej(ωt-βz), the situation changes. Now the Helm that gets used contains β'2, for example here from below (D.1.14), [ agreed ] [∂r2 + (1/r) ∂r + (1/r2) ∂θ2 + β'2 ] ej(ωt-βz)Ez(r,θ) = 0 . (1) In order to still allow a DC solution, you would need β' = 0. If you ignore the loss model, then you do in fact get β→0 and βd → 0 which allows β'→0 and then you could have a DC solution for Ez. But ignoring the loss model means perfect conductors which means Ez ≡ 0 and that is not what we want. [ for my "modern" loss model with G = 0, you do in fact get all three β → 0 as ω → 0 even when there are losses. See current (D.11.2). ] 3. When I use my loss model of Section D.11, I have to assume the transmission line equations, and that in turn means I have to assume ω > ω0 for some min ω0, and that in turn rules out the DC limit. [ but I have disowned that old loss model with its ωc except perhaps for very small loss. ] 4. Thus, once I make the ej(ωt-βz) ansatz along with the D.11 loss model, I have barricaded the Appendix D solution from reaching a DC limit! It is an impossible pathway. You can see directly from the equation (1) above that Ez = constant only solves (1) if β' = 0. So there is the explanation. No matter what you do, reflections, large Z0, whatever, the Jz= 0 solution will always be illegal in the assumptions context. [ I disagree then with this conclusion. For ω = 0 we have β' = 0 and constant is a viable solution. ] Plan E: An interesting though complicated Loss Model Is there some other loss model that is valid down to ω = 0? Assuming G = 0 as I always do in Appendix D, all loss is due to σ in the conductor. We know that J = σE (at least for our frequency range!), so I ought to be able to compute the loss per unit length of transmission line. I have often argued that Jz is the predominant current, and that is certainly true near DC. So maybe I can just use Jz = σEz to compute the loss. I sort of do this right now in Appendix P where I write dP = (dI)2(dR) = [Jz(r,θ)dA]2 ρ dz / dA = ρ Jz(r,θ)2 dA dz . (P.10.3) which is the loss in a "tiny resistor" inside the medium. If I integrate over the conductor cross section P = ∫dP = dz ∫dA ρ Jz(r,θ)2 . (P.10.4) and this is the Ohmic power loss in a coin-shaped resistor of thickness dz. I might write this as dP/dz = (1/σ) ∫dA Jz(r,θ)2 Before continuing, however, there are several complications! From Chapter 2 and just generally we know that Ez and Jz are complex, not simply real. But imagine a tiny resistor and P = I2R. The phase of the current I does not really affect the real power burn in the resistor. So I guess I should modify the above stuff to read dP = |dI| 2(dR) = |Jz(r,θ)dA|2 ρ dz / dA = ρ |Jz(r,θ)|2 dA dz . (P.10.3) P = ∫dP = dz ∫dA ρ |Jz(r,θ)|2 . (P.10.4) dP/dz = (1/σ) ∫dA |Jz(r,θ)|2 Even though tiny resistors at different radii have different phase current, you still just add up the little real power dissipations! For a round wire we then have dP/dz = (1/σ) !Syntax Error, Irdr !Syntax Error, I dθ |Jz(r,θ)|2 I will check later, but I think this leads to (up to a constant I would have to determine) dP/dz = (1/σ) Σm =-∞∞ !Syntax Error, Irdr |Jz(r,m)|2 and then each partial wave contributes to the dissipation. This can also be written dP/dz = σ Σm!Syntax Error, Irdr |Ez(r,m)|2 Now what happens if I go ahead and use the Appendix D solution? Ez(r,m) = (1/4) ηm I Rdc (aβ') fm(r) Then we have dP/dz = σ [ (1/4) I Rdc (a|β'|)]2 Σm |ηm|2 !Syntax Error, Irdr |fm(r)|2 I am hacking a completely new first-time path now, I have never done this before. The integral might be doable ! fm = Jm(x) [ - ] !Syntax Error, Irdr |fm|2 x = β'r dx = β' dr !Syntax Error, Irdr |fm|2 = !Syntax Error, I(x/β')(dx/β') |fm(x)|2 = (1/β')2 !Syntax Error, Idx x |fm(x)|2 = | - |2 (1/β')2 !Syntax Error, Idx x |Jm(x)|2 What does this integral mean? Well consider: Wow, just what the doctor ordered with α = 1 and p = m, so ∫ dx x |Jm(x)|2 = (x2/2) { |Jm(x)|2 - Jm-1(x)Jm+1(x) } I already know the small argument stuff Jm(x) = (x/2)m / m! for m = 0,1,2,..... J-1(x) = - (x/2) for m = -1 . (D.11.16) So for m = 1,2,3... I get RHSm(small x) = (x2/2) { [(x/2)m / m!]2 - [(x/2)m-1 / (m-1)!] [(x/2)m+1 / (m+1)!] } and then as x→ 0 this is all zero. For m = 0 we get RHS0(small x) = (x2/2) { |J0(x)|2 - J-1(x)J1(x) } = (x2/2) { 1 - (-x/2)(x/2) } and as x→ 0 this is also 0. This we may conclude that !Syntax Error, Idx x |Jm(x)|2 = (xa2/2) { |Jm(xa)|2 - Jm-1(xa)Jm+1(xa) } and so !Syntax Error, Irdr |fm|2 = | - |2 (1/β')2 (xa2/2) { |Jm(xa)|2 - Jm-1(xa)Jm+1(xa) } Although messy, it is in closed form. Now xa/β' = β'a/β' = a, so can simplify !Syntax Error, Irdr |fm|2 = (1/2) a2 | - |2 { |Jm(xa)|2 - Jm-1(xa)Jm+1(xa) } and then dP/dz = σ [ (1/4) I Rdc (a|β'|)]2 Σm |ηm|2 * a2 | - |2 { |Jm(xa)|2 - Jm-1(xa)Jm+1(xa) } = σ (1/16) I2 1/(π2a4σ2) a2|β'|2 Σm |ηm|2 * a2 | - |2 { |Jm(xa)|2 - Jm-1(xa)Jm+1(xa) } = (1/16) I2 1/(π2σ) |β'|2 Σm |ηm|2 * | - |2 { |Jm(xa)|2 - Jm-1(xa)Jm+1(xa) } Dimension check: I2 / σ * |β'|2 = amp2 [ohm-m] m-2 = amp-volt /m = watt/m = correct Amazingly, I have a closed-form expression for the power loss per meter of this conductor. [ if no math errors, seems OK. Note that Jr and Jθ contributions are ignored. ] Comments: This formula gives the rate at which power is consumed inside a round conductor that is part of a transmission line. The total power at location z on the transmission line is the energy stored both inside and outside the conductors, and I would guess this was mostly outside. Without using the transmission line equations (not valid at low ω), I don't have a simple expression for this power, so I cannot then get a simple expression for the βd imaginary part or decay exponent. Perhaps I could just say that P = I2Z0 ? Then we have dP/dz = (1/16) (P/Z0) 1/(π2σ) |β'|2 Σm |ηm|2 * | - |2 { |Jm(xa)|2 - Jm-1(xa)Jm+1(xa) } or dP/P = (1/16) (dz /Z0) 1/(π2σ) |β'|2 Σm |ηm|2 * | - |2 { |Jm(xa)|2 - Jm-1(xa)Jm+1(xa) } ln P = (1/16) (z /Z0) 1/(π2σ) |β'|2 Σm |ηm|2 * | - |2 { |Jm(xa)|2 - Jm-1(xa)Jm+1(xa) } = -2α z P(z) = P(0) e-2αz So we seem to have -2α = (1/16) (1/Z0) 1/(π2σ) |β'|2 Σm |ηm|2 * | - |2 { |Jm(xa)|2 - Jm-1(xa)Jm+1(xa) } where xa = β'a with β'2 = β2 - βd2 = β2 - (βd0 - jα)2 so we have some kind of complicated equation which determines α. If α is assumed << βd0 then we are done and the above is our result for α. [ but I now have a much simpler way to compute α using the k(ω) stuff of Appendix Q ] Plan F: Review fact that Jz asymmetry exists in the Regular App D solution. Let's go back to lossless βd and let's go back to this result below (D.1.14): [∂r2 + (1/r) ∂r + (1/r2) ∂θ2 – βd2 + β2 ] Ez(r,θ) = 0 . (2) This is a PDE and I don't know much about general PDE solutions. In the lossless model, as ω→0 we get β→0 and βd→ 0 so this reads [∂r2 + (1/r) ∂r + (1/r2) ∂θ2 ] Ez(r,θ) = 0 . (2) [ agreed ] I think this is the 2D Laplace equation, and it has general solutions of this form rm cos(mθ) [∂r2 + (1/r) ∂r] rm cos(mθ) = [ m(m-1)rm-2 cos(mθ) + mrm-2 cos(mθ) ] = m2rm-2cos(mθ) (1/r2) ∂θ2 rm cos(mθ) = - rm-2 m2 cos(mθ) The m = 0 solution form is just a constant, and that too is a viable solution. [ yes ] But somehow as ω→0, my low ω solutions are Ez(r,θ) = (1/4) I Rdc [ 4 + 4 Σm=1∞ (r/a)m (m+1) ηm cos(mθ) ] Er(r,θ) = (j/4) I Rdc (aβd) [2 (r/a) + 2 Σm=1∞ [(r/a)m+1 + (r/a)m-1] ηm cos(mθ) ] Eθ(r,θ) = (1/4) I Rdc (aβd) [2 Σm=1∞ [(r/a)m+1 - (r/a)m-1] ηm cos(mθ) ] . (D.11.18) and then as βd → 0 this says Ez(r,θ) = (1/4) I Rdc [ 4 + 4 Σm=1∞ (r/a)m (m+1) ηm cos(mθ) ] Er(r,θ) = 0 Eθ(r,θ) = 0 (D.11.18) which just seems immensely strange to me. Yes, it IS a solution of the Helm → Laplace. What about div E = 0 equation ∂r (r Er) + ∂θEθ + r ∂zEz = 0 ? Each of the three terms is 0 in this limit. The CPBC in this limit Er(r=a,m) = (jω/σ) Nm . (D.2.25) says that 0 = 0 // since Er = 0 on the left and since ω= 0 on the right The other BC just says 0 = 0 as well. But somehow the "ghost" of the CPBC is shaping my solution even in this ω→0 limit. Plan G So what IS the ω = 0 solution obtained in Plan C ? First of all, recall this summary Er,tot(r,θ,z,t) = -2jsin(βd[z-L]) ejωt Er(r,θ; βd) Eθ,tot(r,θ,z,t) = -2jsin(βd[z-L]) ejωt Eθ(r,θ; βd) Ez,tot(r,θ,z,t) = 2cos(βd[z-L]) ejωt Ez(r,θ; βd) Assuming the functions on the right are non-singular in βd we then have Er,tot(r,θ,z,t) = 0 Eθ,tot(r,θ,z,t) = 0 Ez,tot(r,θ,z,t) = 2 ejωt Ez(r,θ; βd = 0) // this is the ω = 0 solution Itot(z,t) = [ 2 ejωt ] I I ≡ 2πω (a/βd) N0 Ez(r,m;βd) = (1/4) ηm I Rdc (aβ') fm So we end up somehow with NO Er and NO Eθ, but somehow Ez is still asymmetric! Not clear yet what happens to I in this limit. If N0 = <n(θ)> as in (D.1.8), it would seem that I = 2πω (a/βd) N0 = 2πω (a/k) N0 // in new notation But we have from Appendix Q: Fact 4: In the low frequency limit with G = 0 , (Q.4) Re(k) ≈ ω1/2 + ω3/2 Im(k) ≈ - ω1/2 + ω3/2 ω << R/L so it seems that we might end up with I = 2πω (a/k) N0 = 2πω <n(θ)> = stuff * ω1/2 → 0 SO! We still end up with our I→0 situation in the DC limit for this reflection scenario! It does not yield a DC limit where there is a finite current, as we need to be doing Appendix P eddy currents. TBC.