Review of Pardox 1 and 2 REVIEWED

Working on Paradox #1 and Paradox #2 PhL 5.8.14 [ Final review 5.4.14. I am still totally happy with the resolutions of these paradoxes, and it is all written up now in lines doc. ] As of May 8, 2014 I am totally happy with the resolution of paradox 1 outlined here. It is just one of those little things you check that is OK. Always looking for trouble! 5/12/14. Things here are all resolved in lines doc. Statement of the Paradox 1. 1 Claimed Resolution of Paradox 1. 1 Numerical Example of Paradox 1. 2 1. What is φ for this example? 2 2. What is Ez for this example? 3 3. What do we know so far with our numerical example? 8 Is there a more general way to show this result? 9 Paradox #2. 10 Statement of the Paradox 1. This "paradox" is discussed in "the current asymmetry issue.doc". The paradox is the following: We know that for any pair of closely spaced conductors, Jz and hence Ez = Jz/σ will be highly asymmetrical across each conductor cross section. Ez will be largest in the "active region" close to the other conductor. This asymmetry exists for all values of ω, even very low values. At one time I was unhappy about this being true at very low ω until I learned that the ω→0 limit of the transmission line theory has I = 0 so there is no current at all. But basically for any ω > 0, the asymmetry seems to be constant in some sense, though that claim needs work. At higher ω the skin effect pushes the asymmetry out to the boundary sheath only. Here then is the paradox: We have this known transmission line equation, Ez = j βd φ - jωAz and we know that φ is constant on the boundary, but Ez varies greatly as noted above. Looking at the equation w have jωAz = j βd φ - Ez (*) and this suggests that Az also varies greatly on the boundary. At large ω, we were hoping that in the skin effect regime Az would NOT vary greatly but would be nearly constant on the boundary in line with King's theory. So the paradox #1 is this: how can we have Az = constant on the boundary in the skin effect regime? This is required for the Chapter 4 King theory to be valid. That seems inconsistent with the equation just quoted, where Ez varies and φ = constant on the boundary. Claimed Resolution of Paradox 1. My claimed resolution is this: Ez is very small since we are dealing with good conductors. So when we look at the above equation, jωAz = j βd φ - Ez the Ez term, though highly variable on the boundary, is always much smaller than the j βd φ term, and therefore the right side and hence Az is fairly constant on the boundary. Numerical Example of Paradox 1. I want to see some numbers to prove this resolution really is true -- an example. The example I seem to know the most about is the two cylinders transmission line. I would like to compute φ, Ez and Az separately for this example, and then see if the above resolution is viable. 1. What is φ for this example? Looking at Chapter 6, I see in (6.3.1) that φt(x) = ln(s22/s12) with some BC's, and I know that φ(x,y,z) = q(z) φt(x,y) Unfortunately, I don't state φt(x) more "specifically" in Ch 6, I concentrate there on K which controls the line parameters. But I think I do treat this subject in my bipolar.doc starting there with Section 10. If the two cylinders have ξ1 and ξ2 as labels, I find that φ(ξ) = - V (10.8) and later I write this as φ(x,y,z) = q(z) φt(x,y) Ref [5] (5.1.1) and comparison with (10.19a) shows that (using also (4.5) above), φt(x,y) = ln [] = ln[ ] = -2ξ (10.20) and there we have things in (x,y) as well as ξ coordinates. So I think I know all about "φ for this example". This is in the dielectric of course, and if I then approach the surface, I get φ(surface 1) = - V (10.8) which is of course just some finite fraction of the applied voltage V. Now "closely spaced conductors" means that ξ2 > 0 is small, and ξ1< 0 is also small as in Fig (10.2) of bipolar doc. But this fact does not really affect the fraction much. For a symmetric case, the fraction will always be 1/2 no matter how closely the cylinders are spaced. Thus, in terms of "scale", we might just say |φ(surface 1)| ≈ V/2 and then we have a handle on one of our three quantities in (*). But our equation is this: Ez = j βd φ - jωAz so what is βd ? From "very low freq limit" non v2 we have βd = (ω/vd) ≈ (1/3) x 10-8 ω If we want a/δ = 10, and since I will use a = 1mm = 1000 μ below, we have from lines doc table (2.3.9) for δ = 100 μ that f = 500 KHz let's say so ω = 2πf = 6*1/2*106 = 3 x 106 so then, βd = (1/3) x 10-8 * 3 x 106 = 10-2 m-1 We then get βd φ = 10-2 * V/2 = (1/200) V volts/m 2. What is Ez for this example? Appendix A of bipolar doc I think get's us started on this question. It shows that n1(ξ1,θ) = , (10.28) ηm ≡ Nm/N0 = (-1)m e-|mξ| . (A.12) n1(ξ1,θ) = (q/2π)[ 1 + 2 !Syntax Error, I (-1)m e-m|ξ| cos(mθ) ] (A.13) so at least we know the charge distribution on the conductor surface (and of course it is strongly peaked in the active region between the conductors. Now my plan at this point has been to use the above ηm values in lines doc (D.2.33) which says Ez(r,m) = (1/4) ηm I Rdc (aβ') [ - ] ηm ≡ Nm/N0 = (-1)m e-|mξ| . (A.12) so this seems to be a very explicit statement of the partial wave Ez(r,m). I can write this then as Ez(r,m) = (1/4) ηm I Rdc (aβ') fm fm = [ - ] where fm appears in low frequency folder "very low frequency limit of Appendix D fields v.2.doc". There I take the low ω limit and find that fm = (2/aβ) (m+1)(r/a)m m > 0 f0 = (4/aβ) m = 0 β = e3πj/4 ω1/2 but I am not interested in that limit right now, I do know that β'2 = β2 - βd2 and these parameters are treated in detail in the non-v,2 version of the just-quoted doc: β' ≈ = e3πj/4 ω1/2 = β [ this is still OK at any reasonable ω ] Now where have I already computed Ez(r,θ) from these partial wave amplitudes?? I do it in the non v2 doc just mentioned and I get for low ω only, Ez(r,θ) = I Rdc [ 1 + !Syntax Error, I (-1)m e-m|ξ| (r/a)m [ (m +1)] cos(mθ) ] where β' has been dealt with and ends up being in the (r/a)m factor. Also, this is using the low-ω limit of fm. I also plot this function Ez(r,θ), where I used ξ1 = - 2 which in fact means things are fairly widely spaced. And so there is the famous picture of the Ez asymmetry if I have done things right! I remember being surprised at getting such a strong asymmetry at low ω. The more general result from non-v2 is this [ from "Computation of Az for two cylinders case..".] Ez(r,θ) = I Rdc [ (1/4) xa f0 + (1/2) xa!Syntax Error, I ηm fm(x,xa) cos(mθ) ] or Ez(r,θ) = I Rdc (1/4) xa [ f0 + 2 !Syntax Error, I ηm fm(x,xa) cos(mθ) ] (*) where ηm ≡ Nm/N0 = (-1)m e-|mξ| fm = [ - ] all m x = β'r xa = β'a I have a plot of this function in "compute Ez for 2cyl.mws" in the overhaul folder. There I enter just [...] portion of (*) and I add perhaps 20 terms (easy to vary). What can we say about the "scale" of this Ez ? First of all, somewhere I have plotted fm and have discussed the plots. Seems not to be in the low freq folder. I am having much trouble locating my doc having these plots! It has to be an mws file, so I will now look through every single such file in all my related folders. bipolar folder: examined 10 mws files, not there overhaul folder: YES, the plots are in "current asym studies.mws". So we have now the right folder. The doc discussion is in "the current asymmetry issue.doc". What I show is that as you vary m, the Ez(r,m) curve is always vary similar. I want now to do some new plotting of fm just to see what it does in terms of scale. Recall that β = (j-1)/δ from box (2.2.30) Then xa = βa = (j-1)(a/δ) If we are in the skin effect regime, this xa is "large" so maybe we can use some large-x limits on Bessel functions to see what to expect. Let's first rewrite the above as β = ej3π/4 (/δ) = (j-1)/ * (/δ) = (j-1)/δ Then xa = βa = ej3π/4 (a/δ) => za = a/δ = large Then we can use from lines doc Jm(ej3π/4z) = Mm(z) ejθ(z) . (2.3.3) Mm(z) ≈ [ 1 - + O(1/z2) ] θm(z) ≈ (z/) + (π/2) [ m - 1/4 ] + + O(1/z2) . (2.3.5) Let's try to express m in terms of these M things, were x = ej3π/4z and xa = ej3π/4za = Mm(z) ejθ(z) / [Mm+1(za) ejθ(za)] = ej[θ(z)- θ(z)] If z is large, keep only the leading term in M and we then have [ I am setting x = xa now ] = ej[θ(za)- θ(za)] ≈ ej[θ(za)- θ(za)] = expj [{ (za/) + (π/2) [ m - 1/4 ] } - { (za/) + (π/2) [ m+1 - 1/4 ] } ok = expj [{+ (π/2) 0 } - { + (π/2) [1] } = expj [{-π/2}] = -j If we now consider the second term in fm which is - , we repeat the above, but there are two changes. The first is the leading - sign. The second is that m+1 → m-1 so we will then find ≈ - expj [{+π/2}] = -j Thus, the two terms together give fm ≈ -2j in the limit (a/δ) is very large. [ I have verified this in Maple with an example, so we really do seem to get -2j ] so we expect | fm | ≈ 2. at r = a, in the significant skin effect regime Here are some Maple plots confirming this rough estimate from "plots of fm.mws" : You see that the peak value at r = a on the right is about | fm | ≈ 2, and everywhere else we have a lesser magnitude. Now go back to Ez(r,θ) = I Rdc [ 1 + (1/2) xa!Syntax Error, I ηm fm(x,xa) cos(mθ) ] where ηm ≡ Nm/N0 = (-1)m e-|mξ| fm = [ - ] x = β'r xa = β'a where we were asking about the "scale" of Ez. We have seen now something about the "scale" of fm. So we then have xa = βa = ej3π/4 (a/δ) with magnitude on the order of maybe 14 for ratio a/δ = 10. ηm ≈ 1 or less cos(mθ) ≈ 1 or less fm ≈ 2 or less Now if we assume a large ξ1 meaning cylinders not closely spaced, we will have only the leading term in the series more or less! So I think we can fairly reach this conclusion which screamed out at the start: scale(Ez) ≈ I Rdc. = I * 1/(πa2σ) Suppose I = 1 amp and a = 1 mm, then so we have Ez = 5.5 mV/m, fine! Notes added 5/13/14 while writing Section 6.5. Assuming the result above is correct that fm = -2j in our large x limit, and assuming Jz(r,θ) = Jdc (βa/4) [ f0(r) + 2 Σm=1∞ (-1)m e-m|ξ| fm(r) cos(mθ) ] (6.5.15) we find in this limit that Jz(a,θ) = Jdc (βa/4)(-2j) [ 1+ 2 Σm=1∞ (-1)m e-m|ξ|cos(mθ) ] (6.5.15) = Jdc (βa/4)(-2j) n1(ξ1,θ)(2π/q) = Jdc (βa/4)(-2j) (2π/q) n1(ξ1,θ) = Jdc (βa)(-j) (π/q) n1(ξ1,θ) = (-j)(I/πa2)(βa) (π/q) n1(ξ1,θ) = (-j)(I/a)(β) (1/q) n1(ξ1,θ) = (-j)(I/a)(β/q) n1(ξ1,θ) q = 2πε = (-j)(I/aV)(β/2πε) (ξ2 - ξ1) n1(ξ1,θ) = (-j)(I/V)(β/2πεa) (ξ2 - ξ1) n1(ξ1,θ) = (-j)(β/2πεaZ0) (ξ2 - ξ1) n1(ξ1,θ) Now what is Z0 for large ω? Box in lines doc shows Z0 = (1/4π) K 1/Z0 = 4π (1/K) Then we have Jz(a,θ) = (-j)(β/2πεa)[ 4π (1/K)] (ξ2 - ξ1) n1(ξ1,θ) But we know below (6.5.3) that K = 2(ξ2-ξ1) so we then have Jz(a,θ) = (-j)(β/2πεa)[ 4π (1/K)] (1/2) 2(ξ2 - ξ1) n1(ξ1,θ) = (-j)(β/2πεa)[ 4π ] (1/2) n1(ξ1,θ) = (-j)(β/πεa)[ π ] n1(ξ1,θ) = (-j)(β/εa)[ ] n1(ξ1,θ) = (-j)(β/a)[ 1/] n1(ξ1,θ) = (-j)(β/a)vd n1(ξ1,θ) = (-j) (vd/a) ej3π/4 (/δ) n1(ξ1,θ) = (-j) ej3π/4 (vd/a) (a/δ) n1(ξ1,θ)(1/a) = e-jπ/2 ej3π/4 (vd/a) (a/δ) n1(ξ1,θ)(1/a) = e-j2π/4 ej3π/4 (vd/a) (a/δ) n1(ξ1,θ)(1/a) = e+jπ/4 (vd/a) (a/δ) (1/a) n1(ξ1,θ) and this is fairly close to my intuitive derivation of this fact !!! That result is this Jz(a,θ) = ( vd /a) [1/2 + (1+j)a/δ ] (1/a) n(θ) and if I take large a/δ it becomes Jz(a,θ) = ( vd /a) [(1+j)a/δ ] (1/a) n(θ) = ( vd /a) [(1/)(1+j)a/δ ] (1/a) n(θ) = ( vd /a) [ejπ/4a/δ ] (1/a) n(θ) = ejπ/4 ( vd /a) [a/δ ] (1/a) n(θ) and we are now in full agreement. 3. What do we know so far with our numerical example? We are considering the equation, Ez = j βd φ - jωAz and we are in the skin effect regime where maybe a/δ = 10. We have found that scale(φ) = V/2 scale(βd φ) = V/200 = (1/2)V x 10-2 volts/m scale(Ez) = 5 x 10-3 volts/m Suppose our signal is V = 1 volt. Then we get scale(βd φ) = V/200 = 0.5 x 10-2 = 5 x 10-3 volts/m scale(Ez) = 5 x 10-3 volts/m So this is throwing mud at my "claimed resolution of the problem". My claim was that Ez was very small compared to βd φ, but here in my example they are about the same!!! Thus, as Ez does its variation around the boundary, ωAz also does a large variation, blowing a giant hole in my argument. But: I assumed V = 1 volt and I = 1 amp which implies that Z0 = 1 ohm. But if I am talking a normal coaxial cable, I know that Z0 = 50 ohms, say. For twin lead and using chapter 6 I have Z0 = (K /) 30Ω K = 2 ch-1 [ (b2/2a2) - 1] So let a = 1 mm as above, and assume we are moderately spaced so b = 3. Then K = 2 ch-1[ 9/2 -1] = 2 ch-1(3.5) = 3.8 and then Z0 = 3.8 * 30 = 114 ohms. Remember that TV twinlead is 300 ohms. So for this case, let's say Z0 = 100 ohms. Then if V = 1 volt, I am forced to take I = 10-2 amps and then scale(Ez) = 5 x 10-3 volts/m * 10-2 = 5 x 10-5 volts/m scale(βd φ) = V/200 = 0.5 x 10-2 = 5 x 10-3 volts/m Now things are recovered a bit. The Ez term is now only 1/100th of the βd φ term, so then the variation in Az might only be 1% which I can live with! Ez = j βd φ - jωAz Is there a more general way to show this result? Well, here is one method. Suppose we are in extreme skin effect so Az = constant on surfaces. In this case, I think the solutions for φt and Azt are the same, since they have the same ODE and the same BC's. Here are some items from Chapter 5: φ(x,y,z) = q(z) φt(x,y) (5.1.1) [ t2 + (β2 - kφ2)] φt(x,y) = 0 (5.1.8) φt(C1) = K1 φt(C2) = K2 K1 - K2 = K (5.1.10) Az(x,y,z) = i(z) Azt(x,y) (5.2.1) [ t2 + (β2 - kA2)] Azt(x,y) = 0 (5.2.8) Azt(C1) = W1 Azt(C2) = W2 W1 - W2 = K (5.2.10) For the "lossless" case where the Helm parm is zero, I think we find that Azt(x,y) = φt(x,y) // both are dimensionless Then we must have Az(x,y,z) = φ(x,y,z) = [ - V ] or Az = [ - V ] = με [ - V ] = (1/vd)2 [ - V ] But i(z) = q(z) vd ( draw a picture, both sides are Cou/sec) so this really says Az = (1/vd) [ - V ] = (1/vd) φ OK, now let's take a look at our famous little equation, Ez = j βd φ - jωAz = jω(φ/vd) - jωAz = jω [(φ/vd) - Az ] = jω [Az - Az ] = 0 I found this simple result elsewhere. If you assume the extreme skin effect limit, then Az really does hug the boundary, then you really do get Azt(x,y) = φt(x,y) , and you really do get Ez= 0. Extreme skin effect means σ → ∞ and that means Ez= 0. So when we are in strong skin effect, Az is not quite given by what I write above, and Ez is not quite zero, and we then get something like our numerical example. Paradox #2. This is DIRECTLY RELATED to Paradox #1. When you write out the King gauge condition, div A = - μεjωφ - μσφ = -jωμ(ε+σ/jω)φ = -jωμξφ = -j(β2/ω) φ King gauge (1.5.5) in the dielectric, one way to express this is div A = -j βd (βd/ω) φ = -j βd (1/vd) φ Writing out the details, this says (∂xAx + ∂yAy) + (∂zAz) = -j βd(1/vd) φ Using the usual transmission line ∂z → -jβd this says (∂xAx + ∂yAy) - jβd Az = -j βd (1/vd) φ or (∂xAx + ∂yAy) = jβd [ Az - (φ/vd) ] This is the same RHS we see in the Ez expression above. In the extreme skin effect limit, we expect as previously to get Az = (φ/vd) and the result is then (∂xAx + ∂yAy) = 0. This agrees with the fact that in this limit, all current flows in a pure skin, so there can be no transverse current perp z. But for just a strong skin effect, we will have some small (∂xAx + ∂yAy) ≠ 0 just as we have some small Ez ≠ 0 in the previous section. Why is this a "paradox"? Well, only if you insist that (∂xAx + ∂yAy) ≡ 0 and then you find that the King gauge forces you to have Az = (φ/vd) and that was a paradox away from the strong skin limit (such as near DC) where we know Az does not track the conductor boundary.