reflection scenario REVIEWED

The Reflection Scenario PhL 5.16.14 This is 30 long pages on this subject. Among other things, I tried rewriting all of Appendix D in the reflection framework. Although I don't see the conclusion here, somewhere I concluded that even when you do this reflection stuff, you still end up with Jz asymmetry as ω→ 0. It is embedded in the two waves you superpose, so it does not go away. (10.8.14) 1. Setup 1 2. The meaning of ∂z when acting on V at low ω 3 3. What is the current i(z) in the wire segment? 4 4. What is the impedance seen from the driving end? 6 5. What is the current density Jz in the wire segment? 7 6. What does div E = 0 have to say? 7 7. What does the cpbc have to say? 7 8. What would Appendix D1 look like? 7 Resume this development on 6.9.14. 12 Resume this development on 6.10.14. 14 Resume this development on 6.19.14. 16 I started this in "conclusions on low freq" and will start over here. 1. Setup Transmission line segment runs from z = 0 to z = L. The line is driven at z = 0 and is shorted at z = L. don't ever confuse L with inductance in the following please. We model the wave action for V(z) on this segment by superposing waves of amplitude A going to the right and B going to the left. Thus, V(z)/V = A ej(ωt-βz) + B ej(ωt+βz) The boundary conditions at the two ends are: A+B = 1 at the left driving end Ae-jβL + Be+jβL = 0 at the shorting bar point Maple says So Maple tells us that, A = e+jβL/ [2jsin(βdL)] B = -e-jβL/ [2jsin(βdL)] Therefore the solution V(z) is given by V(z) = [ ej(ωt-βz+βL) - ej(ωt+βz-βL) ] = ejωt 2j sin(-βdz + βdL) = ejωt sin(βd(L-z)) Ignoring the time phasor as we normally do, this says V(z) = V = standing wave pattern βd = ω/v As ω→ 0 we get V(z) = V Here are some pictures of V(z) at various values of ω using L = 10 units and vd = 1 unit so βd = ω : Above is ω = 1 Above is ω = 1/2 Above is ω = 1/5 Above is ω = 1/5 Above is ω = 1/10 Above is ω = 1/100 2. The meaning of ∂z when acting on V at low ω For a normal wave we know that ∂z → -jβd = -j(ω/vd). For our standing wave above, the result is different: [ here vd = 1 ] As ω → 0 we can see that ∂zV(z) → - (V/L) cos[ω(L-z)] = - V/L or ∂z → -1/L and this then is just the negative slope in the last picture above. So we can say ∂z → -1/L for very low values of ω > 0 as well as ω = 0. Now, certain other quantities have this same behavior. For example n(θ,z) = n(θ,0) so charge sloshes back and forth at ω maximally at each antinode, and not at all at each node. In the limit though the only node is at z = L, and the max sloshing is at z = 0. ok to here 6/5/14 3. What is the current i(z) in the wire segment? Go back to V(z) again: V(z) = A ej(ωt-βz) + B ej(ωt+βz) where we superpose the two wave potentials. Perhaps we can say current in right going wave is iR(z) = A ej(ωt-βz)/ Z0 current in left going wave is iL(z) = B ej(ωt-βz)/ Z0 This would say i(z) = V(z)/Z0 which I think is wrong because I expect i(z) to be large at z = L because there is a short, whereas V(z) = 0 there! On the other and maybe expect i(z) = 0 at L since V(z) = 0. I am not sure of that. What about Chapter 4.11 equations? As long as I don't replace ∂z I think they still apply! Thus, I end up with all these applicable equations = - z i(z) = - y V(z) " the classics" with z = R + jωL y = G +jωC . (4.11.15) In particular, this says that i(z) = (-1/z) ∂zV(z) So then we get V(z) = V i(z) = (-1/z)[ -Vβd ] = (Vβd/z) where derivative is from Maple Now for small βd this becomes i(z) = (VβdL/Lz) ≈ (V/Lz) cos(0) = V/(Lz) and for small ω this is a constant, independent of z, as I would expect. What happens then with the other classic TLE: = - y V(z) => V(z) = (-1/y) i(z) = (-1/z) ∂zV(z) V(z) = (-1/y) (-1/z) ∂z2V(z) = (1/yz) [ -V βd2 ] according to Maple, So I then get V(z) = (-βd2/yz) V(z) But if I go down to (5.3.4) I find kφ2 = kA2 ≡ k2 = -zy = - (R+jωL)(G+jωC) (5.3.4) and in the low loss worked we set k2 = βd2 and then we have βd2 ≈ -zy and then everything is consistent (whew!! ). So we have answered the question "what is the current in the wire" i(z) = (Vβd/z) // for any ω = V/(Lz) = constant // for small ω 4. What is the impedance seen from the driving end? At some point z we have shown that V(z) = V i(z) = (Vβd/z) Therefore Z(z) = V(z)/i(z) = (z/βd) tan[ βd(L-z) ] ohms // any ω = (z/βd)[ βd(L-z)] = z (L-z) ohms // small ω Fascinating. It has nodes of 0 (where V(z) = 0) and antinodes of ∞ (where i(z) = 0). In particular Z(0) = V(0)/i(0) = (z/βd) tan[ βdL ] // any ω = (z/βd)(βdL) = zL // small ω = RL // ω = 0 Notice that this is completely different from Z0 of the transmission line. Z0 = = ∞ at DC 5. What is the current density Jz in the wire segment? I can only assume it has the same z dependence as i(z), so then Jz(r,θ,z) = f(r,θ) (Vβd/z) // general ω = f(r,θ) V/(Lz) // small ω Ez(r,θ,z) = f(r,θ) (1/σ) V/(Lz) // small ω Maybe similar forms for Er and Eθ. 6. What does div E = 0 have to say? ∂r (r Er(r,θ,z)) + ∂θEθ(r,θ,z) + r ∂zEz(r,θ,z) = 0 . ∂r (r Er(r,θ,z)) + r ∂zEz(r,θ,z) ≈ 0 // near r = a " near surface" Now for small ω I get ∂zEz(r,θ) ≈ 0 so then ∂r (r Er(r,θ,z)) = 0 r Er(r,θ,z) = g(θ,z), some unknown function Er(r,θ,z) = g(θ,z) / r But this says we must have g(θ,z) = 0, and we conclude that Er = for small ω. 7. What does the cpbc have to say? Er(a,θ,z)) = (jω/σ) n(θ,z)) Both sides go to 0 as ω→0. So here we end up for very small ω with div E saying nothing at all about asymmetry or even Ez! I think in this "scenario", as ω→ 0 the asymmetry goes away! A detailed analysis would require a whole new Appendix D1 which is geared to this physical experiment. But maybe there is still some way to use superposition and then make use of existing Appendix D results. 8. What would Appendix D1 look like? Try this: E(r,θz,t) = ej(ωt-βz) ER(r,θ) + ej(ωt+βz) EL(r,θ) (D.1.1)' ∂z E(r,θz,t) = -jβd ej(ωt-βz) ER(r,θ) + jβd ej(ωt+βz) EL(r,θ) = (-jβd)[ ej(ωt-βz) ER(r,θ) - ej(ωt+βz) EL(r,θ) ] ∂z2 E(r,θz,t) = (-jβd)2 E(r,θz,t) = - βd2 E(r,θz,t) so the second derivative is the same as before, the first derivative is different. The only place the first derivative appears is in the div E = 0 equation section. ∂r (r Er) + ∂θEθ + r ∂zEz = 0 . What does this now say? ∂r (r [ej(ωt-βz) ERr(r,θ) + ej(ωt+βz) ELr(r,θ)]) + ∂θ ([ej(ωt-βz) ERθ(r,θ) + ej(ωt+βz) ELθ(r,θ)]) + r ∂z([ej(ωt-βz) ERθ(r,θ) + ej(ωt+βz) ELθ(r,θ)]) = 0 or ej(ωt-βz) { ∂r(r ERr(r,θ)) + ∂θ ERθ(r,θ) + r(-jβd) ERθ(r,θ) } + ej(ωt+βz) { ∂r(r ELr(r,θ)) + ∂θ ELθ(r,θ) + r(+jβd) ELθ(r,θ) } = 0 So now I think we end up with TWO divE = 0 equations to worry about ∂r(r ERr(r,θ)) + ∂θ ERθ(r,θ) + r(-jβd) ERθ(r,θ) = 0 ∂r(r ELr(r,θ)) + ∂θ ELθ(r,θ) + r(+jβd) ELθ(r,θ) = 0 Giving each the usual θ expansion, we then get ∂r(r ERr(r,θ)) +jm ERθ(r,θ) + r(-jβd) ERθ(r,θ) = 0 ∂r(r ELr(r,θ)) +jm ELθ(r,θ) + r(+jβd) ELθ(r,θ) = 0 I think we then end up with this summary starting position: The Three Helmholtz Equations and the div E = 0 equation (in partial waves) (D.1.20) [2E]z + β2 Ez = 0 : [r2∂r2 + r ∂r - m2 + r2 ( β2- βd2)] Ez(r,m) = 0 (D.1.15) [2E]r + β2 Er = 0 : [r2∂r2 + r∂r - (m2+1) + r2(β2-βd2)] Er(r,m) - 2jm Eθ(r,m) = 0 (D.1.17) [2E]θ + β2 Eθ = 0 : [r2∂r2 + r∂r - (m2+1) + r2(β2-βd2)] Eθ(r,m) + 2jmEr(r,m) = 0 (D.1.18) div E = 0 : ∂r(r ERr(r,m)) +jm ERθ(r,m) + r(-jβd) ERθ(r,m) = 0 (D.1.19) ∂r(r ELr(r,m)) +jm ELθ(r,m) + r(+jβd) ELθ(r,m) = 0 E(r,θz,t) = ej(ωt-βz) ER(r,θ) + ej(ωt+βz) EL(r,θ) which is not THAT much more complicated than we had. What we really have is this in fact: The Three Helmholtz Equations and the div E = 0 equation (in partial waves) (D.1.20) [2E]z + β2 Ez = 0 : [r2∂r2 + r ∂r - m2 + r2 ( β2- βd2)] ERz(r,m) = 0 (D.1.15) [r2∂r2 + r ∂r - m2 + r2 ( β2- βd2)] ELz(r,m) = 0 (D.1.15) [2E]r + β2 Er = 0 : [r2∂r2 + r∂r - (m2+1) + r2(β2-βd2)] ERr(r,m) - 2jm ERθ(r,m) = 0 (D.1.17) [r2∂r2 + r∂r - (m2+1) + r2(β2-βd2)] ELr(r,m) - 2jm ELθ(r,m) = 0 (D.1.17) [2E]θ + β2 Eθ = 0 : [r2∂r2 + r∂r - (m2+1) + r2(β2-βd2)] ERθ(r,m) + 2jmERr(r,m) = 0 (D.1.18) [r2∂r2 + r∂r - (m2+1) + r2(β2-βd2)] ELθ(r,m) + 2jmELr(r,m) = 0 (D.1.18) div E = 0 : ∂r(r ERr(r,m)) +jm ERθ(r,m) + r(-jβd) ERθ(r,m) = 0 (D.1.19) ∂r(r ELr(r,m)) +jm ELθ(r,m) + r(+jβd) ELθ(r,m) = 0 E(r,θz,t) = ej(ωt-βz) ER(r,θ) + ej(ωt+βz) EL(r,θ) This appears to be two completely independent sets of four equations each. (a) The Ez Solution The solutions to the Ez equation will be as before: Ez(r,m) = Czm Jm(β'r) (D.2.4) where Czm is at this point unknown. Ah, but I now see that it will be two equations! ERz(r,m) = CRzm Jm(β'r) (D.2.4) ELz(r,m) = CLzm Jm(β'r) (D.2.4) (b) The Er Solution The Er equation is exactly as in Appendix D [r2∂r2 + r∂r - (m2+1) + r2β'2] Er(r,m) - 2jm Eθ(r,m) = 0 (D.2.5) I am inclined to treat this now as two different equations, matching z dependence [r2∂r2 + r∂r - (m2+1) + r2β'2] ERr(r,m) - 2jm ERθ(r,m) = 0 (D.2.5) [r2∂r2 + r∂r - (m2+1) + r2β'2] ELr(r,m) - 2jm ELθ(r,m) = 0 (D.2.5) and now I have two div E equations of the form: -jmERθ(r,m) = [1 + r∂r ] ERr(r,m) - r (jβd)ERz(r,m) . (D.2.6) -jmELθ(r,m) = [1 + r∂r ] ELr(r,m) + r (jβd)ELz(r,m) . (D.2.6) which I can insert to get these two Er equations: [r2∂r2 + 3r∂r + (1-m2) + r2 β'2] ERr(r,m) = 2r (jβd)ERz(r,m) . (D.2.7) [r2∂r2 + 3r∂r + (1-m2) + r2 β'2] ELr(r,m) = - 2r (jβd)ELz(r,m) . (D.2.7) I then insert the Ez solutions to get [r2∂r2 + 3r∂r + (1-m2) + r2 β'2] ERr(r,m) = 2r (jβd) CRzm Jm(β'r) = KRm β'r Jm(β'r) [r2∂r2 + 3r∂r + (1-m2) + r2 β'2] ELr(r,m) = - 2r (jβd) CLzm Jm(β'r) = KLm β'r Jm(β'r) where KRm ≡ + 2j (βd/β') CRzm KLm ≡ - 2j (βd/β') CLzm I think we will end up then with the following Er solutions: ErR(r,m) = aRm x-1 Jm(x) + Jm+1(x) . (D.2.15) ErL(r,m) = aLm x-1 Jm(x) + Jm+1(x) . (D.2.15) (c) The Eθ Solution I think I first get these results jmERθ(r,m) = - aRm Jm'(x) + [ - Jm+1(x) - x Jm+1'(x) + x Jm(x) ] jmELθ(r,m) = - aLm Jm'(x) + [ - Jm+1(x) - x Jm+1'(x) - x Jm(x) ] // note - sign The sign difference then makes a difference. Use next -xJm+1' = -xJm + (m+1)Jm+1 Jm' = -Jm+1 + (m/x)Jm and then you get jmERθ(r,m) = - aRm [-Jm+1 + (m/x)Jm] + [ - Jm+1(x) + { (m+1)Jm+1 -xJm } + x Jm(x) ] jmELθ(r,m) = - aLm [-Jm+1 + (m/x)Jm] + [ - Jm+1(x) + { (m+1)Jm+1 -xJm } - x Jm(x) ] or jmERθ(r,m) = - aRm [-Jm+1 + (m/x)Jm] + [ - Jm+1(x) + (m+1)Jm+1 ] jmELθ(r,m) = - aLm [-Jm+1 + (m/x)Jm] + [ - Jm+1(x) + (m+1)Jm+1 - 2x Jm(x) ] so maybe easier to just keep the original forms. First summary of E field solutions: First write ERz(r,m) = CRzm Jm(β'r) KRm/2 ≡ + j (βd/β') CRzm ELz(r,m) = CLzm Jm(β'r) KLm /2≡ -2j (βd/β') CLzm so that ERz(r,m) = - j (β'/βd) Jm(x) ELz(r,m) = + j (β'/βd) Jm(x) ErR(r,m) = aRm x-1 Jm(x) + Jm+1(x) . (D.2.15) ErL(r,m) = aLm x-1 Jm(x) + Jm+1(x) . (D.2.15) jmERθ(r,m) = - aRm Jm'(x) + [ - Jm+1(x) - x Jm+1'(x) + x Jm(x) ] jmELθ(r,m) = - aLm Jm'(x) + [ - Jm+1(x) - x Jm+1'(x) - x Jm(x) ] // note - sign Now of course there are four unknown constants instead of 2. If I were doing App D1, I would then check that these solutions really solve the four equations claimed! The cpbc ? What is it? Er(a-ε,θ) = (jω/σ) n(θ) . (D.2.24) But I think I have to state this as ERr(a-ε,θ) = (jω/σ) nR(θ) . (D.2.24) ELr(a-ε,θ) = (jω/σ) nL(θ) . (D.2.24) where now each wave is associated with its own charge density, and it must then be that n(θ,z) = ej(ωt-βz) nR(θ) + ej(ωt+βz) nL(θ) I think nR and nL have the same shape, but different amplitudes. I have pushed far enough into this Appendix D1 wilderness. Basically you cannot just superpose two Appendix D solutions, because more info is needed, such as V(z) which has no place in the discussion of a single round wire in an Appendix D. I gave it a valiant try, don't think it is meaningful. But the reflection scenario I think is reasonable. Resume this development on 6.9.14. I basically have two sets of four equations each. The sets are independent, but they will be linked together by the boundary conditions and by the reflection conditions I assume. The two boundary conditions ought to be these: ERr(a-ε,θ,z) + ELr(a-ε,θ,z) = (jω/σ) [nR(θ,z) + nL(θ,z)] (D.2.24) ERθ(a,θ,z) + ELθ(a,θ,z) = 0 Suppose I imagine a right and left directed ni component. Then we really have two separate conditions. ERr(a-ε,θ) = (jω/σ) nR(θ) ELr(a-ε,θ) = (jω/σ) nL(θ) If we meet these two separate conditions, they we satisfy the first equation above. I think the same would have to apply to the Eθ condition as well, so ERθ(a,θ,z) = 0 ELθ(a,θ,z) = 0 But then each directional wave has the overall general solution I already know: ERz(r,m) = (1/4) ηRm IR Rdc (aβ') fm fm = [ - ] x = β'r ERr(r,m) = (j/4) ηRm IR Rdc (aβd) gm gm = [ + ] xa = β'a ERθ(r,m) = (1/4) ηRm IR Rdc (aβd) hm hm = [ - ] ELz(r,m) = (1/4) ηLm IL Rdc (aβ') fm fm = [ - ] x = β'r ELr(r,m) = - (j/4) ηLm IL Rdc (aβd) gm gm = [ + ] xa = β'a ELθ(r,m) = - (1/4) ηLm IL Rdc (aβd) hm hm = [ - ] Back up. Here is a wave going to the right: ERz(r,m) = (1/4) ηRm IR Rdc (aβ') fm fm = [ - ] x = β'r ERr(r,m) = (j/4) ηRm IR Rdc (aβd) gm gm = [ + ] xa = β'a ERθ(r,m) = (1/4) ηRm IR Rdc (aβd) hm hm = [ - ] where nRm are moments of nR(θ), and IR is the total current in the wire, integral of JRz or Erz. What happens if I do IR → - IR ? Well, Ez changes sign. We still have exp(-jβdz) so I think βd has to also change sign to represent current going the opposite direction. . Then ERr gets a double sign change and so stays the same, and same for ERθ . Resume this development on 6.10.14. I think for either of the two waves, the ηm are the same and are determined by the capacitance of the line and the electrostatic n(θ) which must be independent of direction. Even if the waves have different amplitude in the two directions, ηm is normalized to the current. So we then have this slightly simpler form ERz(r,m) = (1/4) ηm IR Rdc (aβ') fm fm = [ - ] x = β'r ERr(r,m) = (j/4) ηm IR Rdc (aβd) gm gm = [ + ] xa = β'a ERθ(r,m) = (1/4) ηm IR Rdc (aβd) hm hm = [ - ] ELz(r,m) = (1/4) ηm IL Rdc (aβ') fm fm = [ - ] x = β'r ELr(r,m) = - (j/4) ηm IL Rdc (aβd) gm gm = [ + ] xa = β'a ELθ(r,m) = - (1/4) ηm IL Rdc (aβd) hm hm = [ - ] Notice the two minus signs in the last two lines, since βd is negated for the L wave. Example: Suppose the reflected traveling wave has IL = - IR ? Then: ERz(r,m) = (1/4) ηm IR Rdc (aβ') fm fm = [ - ] x = β'r ERr(r,m) = (j/4) ηm IR Rdc (aβd) gm gm = [ + ] xa = β'a ERθ(r,m) = (1/4) ηm IR Rdc (aβd) hm hm = [ - ] ELz(r,m) = - (1/4)ηm IR Rdc (aβ') fm fm = [ - ] x = β'r ELr(r,m) = + (j/4) ηm IR Rdc (aβd) gm gm = [ + ] xa = β'a ELθ(r,m) = + (1/4) ηm IR Rdc (aβd) hm hm = [ - ] Fact: The exponential ej(ωt-βz) or ej(ωt+βz) which you tack onto Ei(r,θ) is the same as what you should tack onto Ei(r,m). Therefore in this example we have Ez(r,m,z,t) = (1/4) ηm IR Rdc (aβ') fm [ ej(ωt-βz) - ej(ωt+βz) ] = (1/4) ηm IR Rdc (aβ') fm ejωt [ 2j sin(-βdz)] and there (finally!) is our standing wave situation with full nodes and maxima. If we picked some other ratio of IR to IL, we would get some kind of SWR situation where the maxima are not 100%. Fact: No matter what ratio you take for IR and IL,you are going to get Ez(r,m,z,t) NOT vanishing for m = 1,2,3... because neither ηm nor fm vanish. Thus, you will never get uniform Jz. So I don't think that the notion of "reflection" resolves this uniform Jz issue. Recall that for low ω we have Ez(r,m) = (1/4) ηm I Rdc (aβ') fm fm = (r/a)m (m+1) (2/β'a) f0 = 4/(aβ') = (1/2) ηm I Rdc (r/a)m (m+1) for m ≠ 0 and thus in the above example Ez(r,m,z,t) = (1/2) ηm I Rdc (r/a)m (m+1) [ 2j sin(-βdz)] m ≠ 0 Question: Is this two-wave superposition idea non-viable due to the BC's? Since Max's are linear, we should be able to superpose solutions to get new solutions. But I must admit the BC situation is very fuzzy to me. Go back to n(θ,z) = ej(ωt-βz) nR(θ) + ej(ωt+βz) nL(θ) Perhaps since this is something like n(θ,z) = n(θ) [ Aej(ωt-βz) + Bej(ωt+βz) ] Perhaps even n(θ,z) = n(θ) [ 2j sin(-βdz)] and there is your standing wave n(θ) deal. I wrote above the BC this way (cpbc) ERr(a-ε,θ,z) + ELr(a-ε,θ,z) = (jω/σ) [nR(θ,z) + nL(θ,z)] (D.2.24) so maybe this just says, after you cancel the [ 2j sin(-βdz)] factor on both sides, ERr(a-ε,θ) + ELr(a-ε,θ) = (jω/σ) [An(θ) + Bn(θ)] (D.2.24) where A and B are some constants. β2 ≈ - jωμσ β = ej3π/4 (/δ) δ ≡ New Starting Point: Resume this development on 6.19.14. Had a new idea this AM for resolving the DC limit mystery for the shorted transmission line of length L. Idea of superposing AC and DC solutions. But I cannot find a way top justify doing this! Idea: Perhaps the solution should be a superposition of these two solutions: (a) the reflected wave situation as described above (asym Jz) (b) a DC solution determined by R and G (sym Jz) Each solution seems to have its own world and is computable. As ω→0, the solution (a) gradually goes away to nothing and you are then left with just solution (b). Then you can see the total Jz gradually go from asym to uniform! [ that is the motivation for doing this ] With this as motivation, let's reconsider the reflection solution allowing as its Jz will always be asym. To this end, let's start with this assumption for the two waves of the reflection Let's also assume that the right-going wave has factor ej(ωt-β[z-L]). This just adds a factor e+jβL to the initial App D ansatz. It does not change ∂z or ∂t so I don't think it changes the solutions above AT ALL. Only when you assemble the full solution physical will you get this factor. Assume that for the moment. Then if we add the two waves shown above Ez we get Ez(r,m,z,t) = (1/4) ηm IR Rdc (aβ') fm { ej(ωt-β[z-L]) - ej(ωt+β[z-L]) } = (1/4) ηm IR Rdc (aβ') fm ejωt { 2j sin(-βd[z-L]) } // wrong choice and this is our standing wave solution which is the AC part of the problem. It vanishes at z = L. I am not sure this is the component we want to vanish however. If the termination is a "gold block", then in that black the transverse fields will be 0 for example, and this will then force the transverse fields just to the left of the block to be 0. So maybe start over with this: ERz(r,m) = (1/4) ηm I Rdc (aβ') fm fm = [ - ] x = β'r ERr(r,m) = (j/4) ηm I Rdc (aβd) gm gm = [ + ] xa = β'a ERθ(r,m) = (1/4) ηm I Rdc (aβd) hm hm = [ - ] ELz(r,m) = (1/4) ηm I Rdc (aβ') fm fm = [ - ] x = β'r ELr(r,m) = - (j/4) ηm I Rdc (aβd) gm gm = [ + ] xa = β'a ELθ(r,m) = - (1/4) ηm I Rdc (aβd) hm hm = [ - ] where the minus signs in the last two equations arise because βd changes sign for the left going wave, but we are writing βd as if it were positive everywhere. THEN suppose we add the r and θ components which are the transverse components to get Er(r,m,z,t) = (j/4) ηm I Rdc (aβd) gm { ej(ωt-β[z-L]) - ej(ωt+β[z-L]) } = (j/4) ηm I Rdc (aβd) gm { 2j sin(-βd[z-L]) } ej(ωt) So get rid of the time to define Er(r,m,z) =(j/4) ηm I Rdc (aβd) gm { 2j sin(-βd[z-L]) } So then we have Ez(r,m,z) = (1/4) ηm I Rdc (aβ') fm { 2 cos(-βd[z-L]) } Er(r,m,z) = (j/4) ηm I Rdc (aβd) gm { 2j sin(-βd[z-L]) } Eθ(r,m,z) = (1/4) ηm I Rdc (aβd) hm { 2j sin(-βd[z-L]) } // this is the AC solution! Ez is a max at the gold block surface and this can be associated with surface charges on that surface which affect Ez only. The transverse fields vanish at z = L as desired. So this is then the "reflection AC solution" for our shorted pair of cylindrical conductors! We also have I = V/Z0 and if G = 0, we get I→0 as ω→0 as we well know, since Z0 → ∞. One more line: Jz(r,m,z) = σ (1/4) ηm I Rdc (aβ') fm { 2 cos(-βd[z-L]) } ≡ JzAC Now what is the DC solution? Pretty simple IDC = V/RDC RDC = Rdc*L*2 since two conductors, each length L, and Rdc is resistance per unit length. So IDC = V/(2LRdc) Rdc = 1/(σπa2) JzDC = IDC / (πa2) = (V/2L)(σπa2) Now JzDC = IDC / (πa2) so we then have JzDC = (V/2L)(σπa2) / (πa2) = σ (V/2L) EzDC = (V/2L) Of course EzDC = (V/2)/L since the voltage drop in each conductor is V/2. Then we are just getting JzDC = σ EzDC How do I express this DC solution in partial waves so I can add apples to apples? JzDC(r,m,z) = (1/2π) !Syntax Error, Idθ [σ EzDC] e-jmθ = [σ EzDC] δm,0 = (σV/2L) δm,0 where it happens there is So now I can add these two solutions to get Jztot(r,m,z) = JzAC + JzDC = σ (1/4) ηm I Rdc (aβ') fm { 2 cos(-βd[z-L]) } + σ (V/2L) δm,0 = σ [ (1/4) ηm (V/Z0) Rdc (aβ') fm { 2 cos(-βd[z-L]) } + (V/2L) δm,0 ] = σ V [ (1/4) ηm (1/Z0) Rdc (aβ') fm { 2 cos(-βd[z-L]) } + (1/2L) δm,0 ] = σ V [ (1/4) ηm (Rdc/Z0) (aβ') fm { 2 cos(-βd[z-L]) } + (1/2L) δm,0 ] In the above, I make the replacement I = V/Z0 without justification. This was valid for the single wave. Note that dim(Rdc/Z0) = m-1 since Rdc is per length and Z0 is not. dim(σ V ) = ohm-1 m-1 volts = volt/ohm m-1 = amp/m So here is my proposed final solution to the problem of the shorted two cylinders: Jz(r,m,z) = σ V [ (1/4) ηm (Rdc/Z0) (aβ') fm { 2 cos(-βd[z-L]) } + (1/2L) δm,0 ] = σ V [ (1/4) ηm (1/Z0) 1/(σπa2) (aβ') fm { 2 cos(-βd[z-L]) } + (1/2L) δm,0 ] = V [ (1/4) ηm (1/Z0) 1/(πa2) (aβ') fm { 2 cos(-βd[z-L]) } + (σ/2L) δm,0 ] = V [ (1/4) ηm (1/Z0) 1/(πa2) (aβ') fm { 2 cos(-βd[z-L]) } + (σ/2L) δm,0 ] Now Z0 = 1/Z0 = = assuming G = 0. Then Jz(r,m,z) = V [ (1/4) ηm 1/(πa2) (aβ') fm { 2 cos(-βd[z-L]) } + (σ/2L) δm,0 ] Now what about β' ?? First, β2 = μεω2 - jωμσ = ω2μ ( ε - jσ/ω) = ω2μ ξ ξ ≡ ε - jσ/ω . (1.5.1) So write β2 = ωμ(εω - jσ) ≈ -jσωμ for reasonable high ω and also as ω → 0. Meanwhile, βd2 = [(ω-jωc)/vd]2 at general ω = [(0-jωc)/vd]2 = - (ωc/vd)2 at low ω Then we get β'2 = β2 - βd2 = -jσωμ - [(ω-jωc)/vd]2 at general ω = -jσωμ + (ωc/vd)2 at low ω = (ωc/vd)2 as ω → 0 Now as I claimed in App D, β' is going to be small, so we can use fm = (x/xa)m (m+1) (2/xa) m > 0 f0 = 4/xa m = 0 But recall that these cancel, so I don't need all the β stuff above. For low ω we have then Jz(r,m,z) = V [ (1/4) ηm 1/(πa2) xa (x/xa)m (m+1) (2/xa){ 2 cos(-βd[z-L]) } ] m>0 = V [ (1/4) ηm 1/(πa2) xa 4/xa { 2 cos(-βd[z-L]) } + (σ/2L) ] m=0 = V [ (1/4) ηm 1/(πa2) (x/xa)m (m+1) (2){ 2 cos(-βd[z-L]) } ] m>0 = V [ (1/4) ηm 1/(πa2) 4 { 2 cos(-βd[z-L]) } + (σ/2L) ] m=0 = V [ ηm 1/(πa2) (x/xa)m (m+1){cos(βd[z-L]) } ] m>0 = V [ ηm 1/(πa2) { 2 cos(βd[z-L]) } + (σ/2L) ] m=0 = V [ δm,0 (σ/2L) + ηm 1/(πa2) cos(βd[z-L]) ] = V [ δm,0 (σ/2L) + ηm 1/(πa2) cos(βd[z-L]) ] Now I still have βd sitting in there. If I ignore the loss model, then cos(βd[z-L]) ≈ 1 and then Jz(r,m,z) = V [ (σ/2L) δm,0 + ηm 1/(πa2) ] and then at low ω we further get Jz(r,m,z) = V [ (σ/2L) δm,0 + ηm 1/(πa2) ] DC + AC super = V [ (σ/2L) δm,0 + qm ] // assume q-m = qm Now (finally after 4 months!) you see that Jz → constant as ω → 0. In the θ domain I guess this says Jz(r,θ,z) = Σm=-∞∞ V [ (σ/2L) δm,0 + qm ] ejmθ = V [ (σ/2L) + q0 ] + 2 Σm=1∞ V qm cos(mθ) = V [(σ/2L) + 1/(πa2) 2 ] + 2 V 1/(πa2) Σm=1∞ cos(mθ) ηm (r/a)m (m+1) So now I have to justify why I think I am allowed to just brute-force add in this DC solution to the AC solution. The entire section above is very poor. ________________________________________________________________________________ What is the meaning of symbol I ? What is the total current in the wire with reflection running? Here I consider only the AC solution. From above, for the reflecting wire we have Ez(r,m,z) = (1/2) ηm I Rdc (aβ') fm cos(βd[z-L]) Jz(r,m,z) = σ (1/2) ηm I Rdc (aβ') fm cos(βd[z-L]) The total current in a wire ought to be the integral of Jz(r,θ,z). So let's first compute Jz(r,θ,z) : Jz(r,θ,z) = Σm=-∞∞ Jz(r,m,z)ejmθ = Jz(r,0,z) + 2Σm=1∞ Jz(r,m,z) cos(mθ) = σ (1/2) η0 I Rdc (aβ') f0 cos(βd[z-L]) + 2 σ (1/2) I Rdc (aβ') cos(βd[z-L]) Σm=1∞ ηm fm cos(mθ) Now if I integrate this over θ, all the m≠0 terms vanish, so in fact ∫dθ Jz(r,θ,z) = 2π σ (1/2) η0 I Rdc (aβ') f0 cos(βd[z-L]) Finally then, since dA = rdrdθ, we have (η0 = 1) Itot(z) = 2π σ (1/2) I Rdc (aβ') cos(βd[z-L]) !Syntax Error, Idr r f0(r) I already did this integral somewhere, but I will do it again, !Syntax Error, Idr r f0(r) = !Syntax Error, Idr r { 2 = [ 2/ J1(xa) ] !Syntax Error, Idr r J0(β'r) x = β'r dx = β'dr = [ 2/ J1(xa) ] !Syntax Error, I[dx/β'] [x/β'] J0(x) = [ 2/ J1(xa) ] (1/β')2 !Syntax Error, Idx x J0(x) But quoting from GR7 on this, we have !Syntax Error, Idx x J0(x) = x J1(x)|xa0 = xaJ1(xa) - 0 J1(0) = xaJ1(xa) Therefore !Syntax Error, Idr r f0(r) = [ 2/ J1(xa) ] (1/β')2 * xaJ1(xa) = 2xa/β'2 = 2(β'a)/β'2 = 2 a/β' Then the total current is this: Itot(z) = 2π σ (1/2) I Rdc (aβ') cos(βd[z-L]) !Syntax Error, Idr r f0(r) = 2π σ (1/2) I Rdc (aβ') cos(βd[z-L]) 2 a/β' = 2π σ I Rdc a2 cos(βd[z-L]) = 2π σ I (1/πa2σ) a2 cos(βd[z-L]) = 2 I cos(βd[z-L]) // I think this is correct which even looks reasonable, it is a standing wave. Then Itot(0) = 2 I cos(βdL) This is the total current in the wire at the left end where we are presumably applying V ejωt . What happens as ω→0? Assuming βd = ω/vd (without the loss model), we just get Itot(z; ω) = 2 I cos(βd[z-L]) Itot(z; ω=0) = 2 I So! In this gold block reflection scenario, there IS current all along the wire and it is 2I when ω = 0. But at DC, I know that the current is going to be Itot(z; ω=0) = V(0)/RDC with RDC = Rdc*L*2 so Itot(z; ω=0) = V(0)/[2LRdc] But since this is 2I, I now have a meaning for the symbol I : I = (1/4) V(0)/(LRdc) // OK if you require Itot = DC solution at ω= 0 (**) and then I know that Itot(z; ω) = 2 I cos(βd[z-L]) = (1/2) V(0)/(LRdc) cos(βd[z-L]) So let's go back now and insert this value for I in previous reflection equations: Ez(r,m,z) = (1/2) ηm [(1/4) V/(LRdc)] Rdc (aβ') fm cos(βd[z-L]) Jz(r,m,z) = σ (1/2) ηm[(1/4) V/(LRdc)] Rdc (aβ') fm cos(βd[z-L]) or Ez(r,m,z) = (1/8) ηm (V/L) (aβ') fm cos(βd[z-L]) Jz(r,m,z) = σ (1/8) ηm (V/L) (aβ') fm(r) cos(βd[z-L]) Jz(r,θ,z) = σ (1/8) η0 (V/L) (aβ') f0 cos(βd[z-L]) + 2 σ (1/8) (V/L) (aβ') cos(βd[z-L]) Σm=1∞ ηm fm cos(mθ) = σ (1/8) (V/L) cos(βd[z-L]) (aβ') [ f0(r) + 2 Σm=1∞ ηm fm(r) cos(mθ)] So here we have an asymmetric Jz with a finite magnitude (no Z0 anymore) and it is clearly asymmetric. If I now put in the low ω values for the fm I get Jz(r,θ,z) = σ (1/8) (V/L) cos(βd[z-L]) (aβ') [ f0(r) + 2 Σm=1∞ ηm fm(r) cos(mθ)] = σ (1/8) (V/L) cos(βd[z-L]) xa [4/xa + 2 Σm=1∞ ηm(x/xa)m (m+1) (2/xa) cos(mθ)] = σ (1/8) (V/L) cos(βd[z-L]) [4 + 4 Σm=1∞ ηm(x/xa)m (m+1) cos(mθ)] = σ (1/2) (V/L) cos(βd[z-L]) [1 + Σm=1∞ ηm(r/a)m (m+1) cos(mθ)] and there staring you in the face is the asym Jz in the reflection scenario! And there is no Z0 to give a rescue any more! In the last lines above, I have replaced symbol I by (**) . I then have my usual asym Jz Sisyphus paradox and at the same time, I no longer have Z0 so this paradox exists at ω = 0 while current remains finite! _______________________________________________ Try to compute the series that appears above. // just an aside. Recall that for the twin lead we have, ηm ≡ Nm/N0 = (-1)m e-|mξ| . Bipolar (A.12) (6.5.4) so we have then this strange function g(θ) = 1 + Σm=1∞ (-1)m e-|mξ| (r/a)m (m+1) cos(mθ) = 1 + Σm=1∞ ejmx (m+1) cos(mθ) Lets try first this sum S1 = Σm=1∞ (-1)m e-|mξ| (r/a)mcos(mθ) = Σm=1∞ (-1)m e-m|ξ| em ln(r/a) cos(mθ) = Σm=1∞ ejmπ e-m|ξ| em ln(r/a) cos(mθ) = Σm=1∞ ejm[π–|ξ1|+ln(r/a)] cos(mθ) = Σm=1∞ ejmx cos(mθ) x = π–|ξ1|+ln(r/a) = Σm=1∞ ejmx (1/2)[ ejmθ + e-jmθ ] = (1/2)[ Σm=1∞ ejm(x+θ) + Σm=1∞ ejm(x-θ) ] Let y = ej(x+θ) and z = ej(x-θ) so then we have = (1/2)[ Σm=1∞ ym + Σm=1∞ zm ] = (1/2)[ y/(1-y) + z/(1-z) ] ≡ s1 = S1 Now the other sum is s2 = Σm=1∞ ejmx m cos(mθ) = Σm=1∞ ejmx d/dθ [ sin(mθ)] = d/dθ [Σm=1∞ ejmx sin(mθ)] = d/dθ [Σm=1∞ ejmx (1/2j)[ ejmθ - e-jmθ ]] = (1/2j) d/dθ [Σm=1∞ ejm(x+θ) - Σm=1∞ ejm(x-θ)] = (1/2j) d/dθ [Σm=1∞ ym - Σm=1∞ zm] = (1/2j) d/dθ [y/(1-y) - z/(1-z)] = (1/j) d/dθ { (1/2) [y/(1-y) - z/(1-z)] } = (1/j) d/dθ s2 Maple gets an answer, but it ain't purdy, lots of terms. ________________________________________________________________________________ Does the superposed wave solution meet the charge pumping boundary condition? I am now looking for reasons to disqualify the reflection solution, since I see it is giving asym Jz I think the charge density looks like this: n(θ,z,t) = n(θ) [ ej(ωt-β[z-L]) - ej(ωt+β[z-L]) ] = n(θ) 2j sin(-βd(z-L))ejωt I choose the minus sign because n is associated with Jr and Er have this minus sign. Also, I like n = 0 at the shorting point where the cap is shorted out. So n(θ,z) = n(θ) 2j sin(-βd(z-L)) which I think says Nm(z) = Nm 2j sin(-βd(z-L)) Meanwhile from above we had Er(r,m,z) = (j/4) ηm I Rdc (aβd) gm { 2j sin(-βd[z-L]) } Now gm = [ + ] and so gm(r=a) = 2. Thus, I get Er(a,m,z) = (j/4) ηm I Rdc (aβd) 2{ 2j sin(-βd[z-L]) } Now the CPBC is supposed to say (using a Gaussian box! ) Er(r=a,m) = (jω/σ) Nm or I guess at location z, Er(r=a,m,z) = (jω/σ) Nm(z) This would require, based on the above, that (j/4) ηm I Rdc (aβd) 2{ 2j sin(-βd[z-L]) } = (jω/σ) Nm 2j sin(-βd(z-L)) or (j/4) ηm I Rdc (aβd) 2 = (jω/σ) Nm or (1/2) ηm I Rdc (aβd) = (ω/σ) ηmN0 or I Rdc (aβd) = (ω/σ) 2N0 If I solve for I, I get I = (ω/σ) 2N0 1/Rdc (1/aβd) = (ω/σ) 2N0πa2σ (1/aβd) = (ω) 2N0π(a/βd) = 2πω(a/βd) N0 which is the same as (D.2.31). Now I found above that, if I require this theory to produce the correct DC limit, I = (1/4) V/(LRdc) so I would have to be able to show (in order to have the CPBC be respected) that (1/4) V/(LRdc) = 2πω(a/βd) N0 or (1/4) V/(LRdc) = 2πω(a/βd) N0 = 2πω(a/βd) (1/2πa) q(0) = ω(1/βd) q(0) = q(0) vd All would be well if this really determined q(0) to be q(0) = (1/4) V/(LRdcvd) But in fact q(0) is determined by the capacitance, so we really have q(0) = CV so I think this says that the CPBC is NOT respected by my reflection solution! Conclusion: The Charge Pump Boundary Condition is NOT met by my reflection superposition if I require the model to have the correct DC limit. In the next section I show that the driving V is related to N0 by V = (1/2j) sin(βdL) N0 and this is definitely not consistent with the strange claim above. ________________________________________________________________________________ What do we know about V(z) ?? This is all useless with lots of algebra errors. Let's assume as I did above that n(θ,z) = n(θ) 2j sin(-βd(z-L)) But n(θ) = !Syntax Error, I Nm ejmθ ∫n(θ)dθ = N0 The integral of n(θ) around the surface is going to be ( q is per unit length) q(z) = ∫n(θ,z)adθ = 2j a sin(-βd(z-L)) ∫n(θ)dθ = 2j a sin(-βd(z-L)) 2π N0 where this N0 is not the same as App D since n(θ) is not the same. So here a statement that q(z) is also a standing wave, based closely on n(θ). If cap is C, we then have V(z) = q(z)/C = 4πj a sin(-βd(z-L)) N0/C So now we have V(z) also doing a standing wave. V(L) = 0 by design. And V = V(0) = 4πj a sin(-βd(-L)) N0/C = 4πj a sin(βdL) N0/C So this relates the applied V and N0 ! From above I had (**) Itot(z; ω) = (1/2) V(0)/(LRdc) cos(βd[z-L]) Itot(0; ω) = (1/2) V(0)/(LRdc) cos(βdL) This suggests the following impedance looking in at the left end 1/ Z(0) = Itot(0; ω)/V(0) = (1/2) 1/(LRdc) cos(βdL) or Z(0) = 2 LRdc sec(βdL) In fact, 1/ Z(z) = Itot(z; ω)/V(z) = (1/2) V(0)/(LRdc) cos(βd[z-L]) / 2j a sin(-βd(z-L)) N0/C = (1/4j) V(0)/(LRdc) * C/N0 * cos(βd(z-L))/ a sin(-βd(z-L)) = - (1/4j) V(0)/(LRdc) * C/N0 * cos(βd(z-L))/ a sin(βd(z-L)) = - (1/4j) V(0)/(LRdc) * C/N0 * cot(βd(z-L)) = - (1/4j) 2j a sin(βdL) N0/C /(LRdc) * C/N0 * cot(βd(z-L)) = - (1/2) a sin(βdL) 1 /(LRdc) * cot(βd(z-L)) Z(z) = -2LRdc csc(βdL) tan(βd(z-L)) so this thing has poles periodically along the way. As a check, set z = 0 to get Z(0) = -2LRdc csc(βdL) tan(βd(-L)) = 2LRdc csc(βdL) tan(βdL) = 2LRdc sec(βdL) which agrees with above. ________________________________________________________________________________ Now try to redo this with G present! Recall first from Section D.9, Second summary of the E field solutions : Rdc = β'2 = β2 - β'd2 (D.9.25) I' = V/Z0 ξd = εd + σd/jω βd' = βd where βd = ω ( I' = I where I was for σd = 0 ) E'z(r,m) = (1/4) ηm I' Rdc (aβ') fm fm = [ - ] x = β'r E'r(r,m) = (j/4) ηm I' Rdc (aβ'd) gm gm = [ + ] xa = β'a E'θ(r,m) = (1/4) ηm I' Rdc (aβ'd) hm hm = [ - ] Note that β does not change at all (inside conductor). Once I use I' = V/Z0 , I' is gone so I can forget about I'. Maybe better to rewrite the above as Second summary of the E field solutions : Rdc = β'2 = β2 - β'd2 (D.9.25) I' = V/Z0 ξd = εd + σd/jω βd' = βd where βd = ω ( I' = I where I was for σd = 0 ) E'z(r,m) = (1/4) ηm I' Rdc (aβ') fm fm = [ - ] x = β'r E'r(r,m) = (j/4) ηm I' Rdc (aβd) gm gm = [ + ] xa = β'a E'θ(r,m) = (1/4) ηm I' Rdc (aβd) hm hm = [ - ] This does alter β' but β' cancels out in my limit. The transverse fields have extra factor , but I was only working with the Ez field which really has no difference at all other than that fact that I' got larger! So I think the result for the AC solution is exactly the same except G ≠ 0. I think this DC calculation is correct. What about the DC solution? I still have the shorting bar in place at location L. It is not obvious what the solution is here, so I turn to the Network appendix. // Well this took quite a while. If I set z = 0 at the short and have z increase to the left, I find that R(z) = tanh(z) is the resistance looking into the ladder from the left end. If the ladder has length L, then R(L) = tanh(L) If L >> 1, then the limit is just R(L) = Let's check now the limit G → 0. Then tanhx ≈ x so we get R(L) ≈ (L) = 2RL which agrees with earlier RDC = Rdc*L*2 So in the G case we get JzDC = IDC / (πa2) where IDC = V/RDC where RDC = tanh(L) or JzDC = V/[ πa2RDC] = V/[ πa2 tanh(L)]