details of Chap 6 exact solution REVIEWED

Details of the Chapter 6 Exact Solution PhL 2/15/14 updated: 2/25/14 Title refers to the two-cylinders geometry. [ Final review 5.14.14. Here is a little summary: ] Summary Section 1 shows the extreme skin effect ideal that Azt = φt and thus Az (1/v)φ and I go around in a circle ending up with Ez = 0! I also realize that i(z) = v q(z) which is now in lines doc. Section 2 suggests that we still have Azt = φt even if not low-loss, since transverse equations are the same. But on the other hand, this makes Az ≠ constant, so probably this fact is useless. Very short. Section 3 addresses The Zs Paradox which is now all resolved in lines doc Ch 4. At the time I even wondered if somehow Zs might really be constant for any conductor pair. Section 4 uses Appendix D Ez expression to show that there will be some moments ηm (I don't know what they are at this point) and Zs(θ) really does depend on θ for the two cylinders. Section 5 calculates the two-cylinder moments ηm I guess for the first time! Section 6 summarizes all my disasters resulting in my pulling lines doc off the web circa 2/20. There really work several big problems! Reviewed this on 3/5/14. Two weeks have gone by, working every day, since I write this doc. About 8 days since I pulled lines doc from the web. Things are not resolved, but I claim to have made significant progress to that end. The TOC is the outline here. Reviewed again on 5/12/14. The main problems are now all cleared up in into lines doc! The title of this doc only refers to item 5 below where I compute the ηm moments. This doc is really about why I had to pull lines doc from the web due to several big problems. Doing this now due to a paradox noted in the edit log [ concerns Zs(θ)! ], but probably this is a good thing to do anyway! After all, this is the only non-trivial problem my method can solve! 1. Comparing PDE's for φ and Az : for truly lossless, I think φt and Azt are the same. 1 2. What happens to φ and Az when losses are "turned on"? 4 3. Review of the Chapter 4 Logic Flow which leads to "the Zs Paradox" 4 4. Why I think Zs(θ) varies strongly with θ for the two-cylinder problem. 6 5. Calculate the moments ηm for the two-cylinder problem! 8 6. Status Report and Pull lines doc from the Web 12 1. Comparing PDE's for φ and Az : for truly lossless, I think φt and Azt are the same. [ 5/12/14 I think yes, these solutions are exactly the same for a perfect conductor ] OK, so what do we know about the solution of fat round conductors from Chapter 6 ?? 1. The scalar potential is this: φ(x,y,z) = q(z) φt(x,y) φt(x) = ln(s22/s12) = 2 ln(s2/s1) φt(C1) = 2 ln(s2/s1)|C1 = -2B1 φt(C2) = 2 ln(s2/s1)|C2 = -2B2 . (6.3.1) 2. But what is the vector potential Az ? I have not made an ansatz for this as I have for φ. Could it have exactly the same form? I claim in (5.3.14) that Az has the exact same asymptotic form. Suppose that were the case. Then I would have Az(x,y,z) = i(z) Azt(x,y) Azt(x) = ln(s22/s12) = 2 ln(s2/s1) It is true that the same type of BC applies, φt(C1) = K1 φt(C2) = K2 K1 - K2 = K (5.1.10) Azt(C1) = W1 Azt(C2) = W2 W1 - W2 = K (5.2.10) If the solution is exactly the same, then W1 = K1 and W2 = K2. Is this really the solution? How does this fit with the gauge condition, ∂zAz(x) = - j (β2/ωφ(x) (4.11.3) -jk Az(x) = - j (β2/ωφ(x) k Az(x) = (β2/ωφ(x) ? But in low-loss we are supposed to have k2 = β2 from (5.3.8) so that seems to say k Az(x) = (k2/ωφ(x) Az(x) = (k/ωφ(x) = (1/v)φ(x) v = ω/k = phase velocity in dielectric // low-loss Question: What are the dimensions of things!!! Look at (1.3.1) A = vector potential (tesla-m = amp-henry/m = volt-sec/m) E = volt/m φ = scalar potential (volts) B = tesla . so they have different dimensions, the difference is velocity. Fine. Now go back a bit φ(x,y,z) = q(z) φt(x,y) = [Coul/m]/[ farad/m] = volt OK Az(x,y,z) = i(z) Azt(x,y) = henry/m * amp = ohm-sec m-1 amp = volt sec/m OK As I already state, φt and Azt are both dimensionless. So then we would have as above, and everything is correct. So now I am sure that Azt is the same as φt exactly! 3. What is Ez ? E = - grad φ - ∂tA . dim: volt/m and sec-1 volt sec/m A = vector potential (tesla-m = amp-henry/m = volt-sec/m) E = volt/m φ = scalar potential (volts) B = tesla . E = - grad φ - jωA Ez(x) = - ∂zφ(x) - jωAz(x) Ez(x) = +jk φ(x) - jωAz(x) RHS = m-1 volt and sec-1 volt sec/m OK Ez(x) = +jk [ q(z) φt(x,y)] - jω[ i(z) Azt(x,y)] Ez(x) = +jk [ q(z) 2 ln(s2/s1)] - jω[ i(z) 2 ln(s2/s1)] Ez(x) = {+(k/ε) q(z) - (ωμ) i(z)} 2j ln(s2/s1)](1/4π) Ez(x) = {+(k) q(z) - (ωεμ) i(z)} 2j ln(s2/s1)](1/4πε) What do I make of this result?? Well recall that με = 1/v2 but also v = ω/k so 1/v2 = k2/ω2 Ez(x) = {+k q(z) - (ω/v2) i(z)} 2j ln(s2/s1)](1/4πε) Ez(x) = {+k q(z) - (ω/v2) i(z)} 2j ln(s2/s1)](1/4πε) How are q(z) and i(z) related to each other? [ draw a cylinder picture, charge/sec across area A ] i(z) = v q(z) C/sec = m/sec * C/m How come this is not written down anywhere in lines doc??? I seem only to be interested in i(z) and V(z). This is a bad equation to omit! He omits it as well. So let's install it and see what happens: [ On 5/12/14 I added this above equation as (4.11.19a) to lines doc. ] Beware what v this actually is! Here it is the loss-less v. Ez(x) = {+k q(z) - (ω/v2) v q(z)} 2j ln(s2/s1)](1/4πε) Ez(x) = {+k q(z) - k q(z)} 2j ln(s2/s1)](1/4πε) Ez(x) = 0 Interpretation? I have examined the exact 2 round wires solution. This solution assumes lossless so that we have the capacitor problem and potential theory in 2D. I compute Ez from φ and Az and I find that Ez = 0 everywhere!!! Not just in the wires, but in the dielectric. That of course says Ez = 0 at the conductor surfaces, and that in turn says Zs = 0 at the conductors, and that is consistent with "lossless", around the circle we go! [ it all makes sense, really ] Misnomer: lossless really means Zs= 0, it does not mean there is no loss in the dielectric, so this is a bad term to use and that should be repaired. [ On 5/12/14 I added a comment about low loss at the start of Section 5.4 ] ok to here 2/25/14 Note added 3/5/14. Consider the current in one of the conductors this way i(z) = dQ/dz * dz/dt Now dQ/dz = q(z), the charge per unit length. And dz/dt says how fast this charge is moving by, which is basically vd . Therefore, i(z) = q(z) vd which is the result just found above. This seems to be a solid and general result for any shaped conductor! You can argue about the precision of vd regarding losses. Fact: All of Chapter 6 operates in the loss-less limit of Zs = 0, so you cannot use anything there to check on asymmetric Zs !! All of Chapter 5 after the point of setting Zs = 0 to get the Laplace equation is also lossless. The whole 2D capacitor problem is based on lossless. Only in Chapter 4 do I have mention Zs ≠ 0. So right now, that is the only place I can look for any information on this subject. 2. What happens to φ and Az when losses are "turned on"? Note added 3/5/14. I do have a certain inconsistency in today's thinking on this subject. With lossless, both the φ and Az equations become Laplace equations and they both have the same solution which is the Chapter 6 solution. But when you "turn on the loss", you add a certain Helmholtz term to both the φ and the Az equations as shown in lines doc here: [ t2 + (β2 - kA2)] Azt(x,y) = 0 (5.2.8) [ t2 + (β2 - kφ2)] φt(x,y) = 0 (5.1.8) where I later claim that kφ2 = kA2 ≡ k2 = -zy = - (R+jωL)(G+jωC) (5.3.4) but OK, this does assume Chapter 4 stuff which right now is in question. But if the above three equations survive, then you would argue that the Helmholtz term is the SAME for both equations, and so the solutions should somehow stay equal, though different from the Laplace solution. So why do I assume φ stays the same, but Az "changes slightly"? I am just noting this inconsistency for future reference. So this question is unresolved. 3. Review of the Chapter 4 Logic Flow which leads to "the Zs Paradox" [ This was the reason I pulled lines doc from the web! Call it the Zs paradox. It even caused me to doubt the fact that Jz is asymmetric on the conductor perimeter, so in the next section I went and computed Jz for the round wire case and found that it is indeed asymmetric (proximity effect). It was not until 5/11/14 that I figured out how to resolve the Zs paradox and end up with meaningful Zs constants in the transmission line equations, in a manner compatible with my earlier discussions of active perimeters and such. ] Making another pass through Chapter 4, examining assumptions. assumption of separation of variables such as (4.1.2) assumption of balanced line as in (4.1.4) transmission line limit same as small β as in Section 4.3 assumption (3.8.10) that φ is constant at a given z. assumption that transverse components Ax and Ay can be neglected so that A = Az. (4..7.1) assumption about μ time dependence is ejωt It is with this long list of assumptions that we enter Section 4.11. Notice that "low loss" has NOT been assumed at this point [ except I claim φ(Ci) = constant ], we have K and KL separate. Everything we know at this point I think appears in box (4.11.1). There are 6 equations in that box. Note that i(z) and q(z) do appear. Also NOT assumed is the wave dependence on z. With the above listed assumptions, King gauge, (4.11.2) is undeniable. The first line of (4.11.3) is nothing new, while the second line is new and uses the assumption that we can ignore Ax and Ay AND their derivatives. So a big assumption then going into (4.11.3). [ I now know on 3/5/14 that this assumption is the same as extreme skin depth limit .] Then (4.11.4) is nothing but a rewrite. Then (4.11.5) is the evaluate and subtract thing for selected x1 and x2 points. Then (4.11.7) I would write this way: [ I am allowing that W = W(z) only here, for time being! ] Ez12(x1) - Ez12(x2) = - ∂zV(z) - jωW(z) ∂zW(z) = - j (β2/ωV(z) . (4.11.7) Then Ez1(x1) = Zs1(x1) i1(z) Ez2(x2) = Zs2(x2) i2(z). (4.11.8) so at this point both Z and E could vary azimuthally around a round conductor, say. For example, due to the form shown, if Ez1(x1) is small at some azimuth, then Zs1(x1) is also small at that azimuth. Now let's maintain this notation and write (4.11.9): [Zs1(x1) + Zs2(x2)] i(z) = - ∂zV(z) - jωW(z) ∂zW(z) = - j (β2/ωV(z) . (4.11.9) [ the Zs paradox is visible above in 1st line. Right side function only of z, left side of x1 and x2 ] Our set of assumptions above has then led us to this conclusion: [Zs1(x1) + Zs2(x2)] = value for any x1 and any x2 in the same z plane. In other words, the claim is that [Zs1(x1) + Zs2(x2)] = f(z) This is not really a claim, it is a fact resulting from all our work above including the assumptions made. But I think we can obtain more. We can keep x2 fixed and let x1 vary to some new x1'. Then we have [Zs1(x1') + Zs2(x2)] = f(z) Subtract to get Zs1(x1) = Zs1(x1') but this in turn implies that, since the above is true for all points on the conductor perimeter. Zs1(x1) = f1(z) The same then is true for the other, so we have Zs2(x2) = f2(z) and these are stronger claims than just for the sum. Now go back to Ez1(x1) = Zs1(x1) i1(z) Ez2(x2) = Zs2(x2) i2(z). (4.11.8) These then read Ez1(x1) = f1(z) i1(z) Ez2(x2) = f2(z) i2(z). (4.11.8) and we then conclude that Ez1(x1) = Ez1(z) Ez2(x2) = Ez2(z) This is a very dramatic result. We know from box (1.1.50) that Ez is continuous at the surface. This tells us that Jz(x1) = Jz(z) The current "just below the surface" is constant around the entire conductor! This seems extremely strange to me. It makes no comments about currents deeper down in the conductor. This conclusion is extremely problematic for me. It says for example that all higher partial wave Ez fields must vanish, which is nonsense. [ Keep feet on the ground. Both Ez and Jz do vary in θ very strongly, they are not constant around the perimeter! And so Zs = Zs(θ) with this same strong variation. This paradoxical result arises due to earlier confusions which I will write up soon! ] 4. Why I think Zs(θ) varies strongly with θ for the two-cylinder problem. Exercise #1 For Chapter 6, I could in theory compute the moments ηm from the charge distribution, and from that I could in theory compute Ez along the surface of one of the conductors. Here is my formula from (D.2.33) for Ez : Ez(r,m) = (1/4) ηm I Rdc (aβ') [ - ] Setting r = a then gives Ez(a,m) = (1/4) ηm I Rdc (aβ') [ - ] xa = β'a β'2 = β2 - βd2 Note that, [ - ] = Jm(xa) [ - ] = Jm(xa) = Jm(xa) so this is not identically zero. Ez is "small" because Rdc is very small. Well, I already did this, and here is the result: Zs(φ) = (1/I) !Syntax Error, I Ez(a,m) ejmφ // (D.1.3a) = (1/4) Rdc !Syntax Error, I ηm [ - ] ejmφ // (D.2.33) where, (D.2.35) ηm = Nm/N0 = !Syntax Error, Idφ n(φ) e-jmφ // (D.1.5b) and (D.2.31) So this will give for sure a non-uniform Zs(φ) around the wire. This is a calculation I really could do using the Chapter 6 solutions since they determine surface charge n(φ). Back up Again. Here are some steps: E = - grad φ - jωA Ez(x) = - ∂zφ(x) - jωAz(x) Ez12(x) = - ∂zφ12(x) - jωAz12(x) Ez12(x1) - Ez12(x2) = - ∂zV(z) - jωW(z) I can ignore the 2nd equation in each pair, my problem is with the 1st equation of each pair! Idea #1. I think several of my "mysteries" are tied together. For example, consider this situation, and exaggerate it every more by putting the conductors VERY close together so there is only a tiny active area. Since the current is so localized in the heavy dark region, I think a B field line will look like that shown in red, and this is NOT parallel to the conductor, so W(z) is NOT a constant on this surface as King always likes to say it is! This is really my "first proof" and here you see it totally falling apart. So this gets me back to the EB = 0 business! Probably B is only parallel to the conductor surface "in the active region". s12 = (x+d)2 + y2 s22 = (x-d)2 + y2 (6.2.3) φt(x) = ln(s22/s12) = ln [] ∂xφt = So OK, this should be the E field [ignoring Az!], and I expect it to be large when d is small. Update Feb 25, 2014. [ updated again 3/8/14 after bipolar doc edits made ] 5. Calculate the moments ηm for the two-cylinder problem! [ this whole calculation is now Appendix A in bipolar doc ] In bipolar doc I show that the angular charge densities on the two cylinders are given by n1(ξ1,θ) = dQ1(ξ1,θ)/[dθdz] = n2(ξ2,θ) = dQ2(ξ2,θ)/[dθdz] = - (10.28) and each one is independent of the other one, what could be simpler! We know that ∫ n1(ξ1,θ) dθ = q Now from lines doc the moments are given by Nm = (1/2π) !Syntax Error, Idθ n1(ξ1,θ) e-jmθ In my case, n1 is an even function of θ so only the cos(mθ) contributes and then Nm = (1/2π) |shξ1| 2 !Syntax Error, Idθ = (1/2π) 2 [|shξ1|!Syntax Error, Idθ ] Maple does not know this integral. It looks familiar, however. Consider |shξ1| !Syntax Error, Idθ = |thξ1| !Syntax Error, Idθ Let a = 1/chξ1 = sechξ1 which is then a < 1 . 1 - a2 = 1 - sech2ξ1= th2ξ1 Then = thξ1 so we get !Syntax Error, Idθ = (π/ thξ1) [ ]m But = = shξ1- chξ1 = - e-ξ1 // says Maple So our integral is then !Syntax Error, Idθ = (π/ thξ1) (-1)m exp(-mξ1) Then we have, Nm = (1/2π) 2 [|shξ1|!Syntax Error, Idθ ] = (1/2π) 2 [ |thξ1| (π/ thξ1) (-1)m exp(-mξ1) ] = sign(ξ1) (-1)m exp(-mξ1) But the integral is really even in m from the very start, so the result is really Nm = sign(ξ1) (-1)m exp(-|m|ξ1) [ this result is superseded by that shown in Appendix A: Nm = (-1)m exp(-|mξ1|) ] Good. Now if ξ1 is very large, the m = 0 moment is sign(ξ1), but all the higher moments basically vanish, just as I wanted. Note added: But the above says N0 = sign(ξ1) which seems to conflict a bit with N0 as it appears in lines doc which is N0 = (βd/2πωa) I . (D.2.31) Now suppose ξ1 = 0.25 as in my bipolar doc example, then ηm ≡ Nm/ N0 = (-1)m exp(-|m|ξ1) η0 = 1 η1 = - exp(-.25) = - 0.778 η2 = + 0.607 etc. Maple says, I was wrong in lines doc! I thought this would have a strong m = 1 only, but it is a slowly converging series involving a lot of moments. Let's try to reconstruct n(θ) from all the moments. n(φ) = !Syntax Error, I Nm ejmφ (D.1.5a) = N0 + Σm=-∞-1Nm ejmφ + Σm=1∞Nm ejmφ = N0 + Σm=1∞N-m e-jmφ + Σm=1∞Nm ejmφ = N0 + Σm=1∞Nm e-jmφ + Σm=1∞Nm ejmφ = N0 + Σm=1∞Nm (e-jmφ + ejmφ ) = N0 + Σm=1∞Nm 2cos(mφ) so n(φ)/N0 = 1 + 2 Σm=1∞ηm cos(mφ) = 1 + 2 Σm=1∞ (-1)m exp(-|m|ξ1)cos(mφ) and this looks exactly right for that value of ξ1 = 0.25. The reason there are many is that the alternating sines cancel out near θ = 0 and are additive at θ = π. Conclusions: I think everything makes complete sense: (1) If conductors are closely spaced and charge is localized, there will be lots of moments with alternating signs to give a peak like that shown above. (2) If conductors are more widely spaced (larger ξ) then the m = 1 will be the main correction to m = 0 I could compute Z(φ) using the above formula, it will be some complicated mess, but it won't be constant around the conductor, that is for sure. It will have φ dependence. [ It will be Zs(θ) = Ez(θ)/ i(z) ] The moments control Jz at the surface, which is Ez at the surface, and so I think you can compute the internal Jz from the moments without too much work using lines doc. Ez(r,m) = (1/4) ηm I Rdc (aβ') [ - ] = Km,a ηm Ez(r,φ) =!Syntax Error, I Ez(r,m) ejmφ = Σm Km,a ηm ejmφ It is not totally obvious that Ez(r,φ) will also have a peak at φ = π, but that will probably be true. 6. Status Report and Pull lines doc from the Web I have multiple problems now that I think are all related, and they are big enough that I decided to pull my lines doc paper from the web for the time being until I can effect repairs. Here is a list of some of the problems: (1) I have a logical failure in Chapter 4 as described above [the Zs paradox], and I suspect that the problem is either that W(z) on conductor cross sections is invalid, or that I need some approximation hierarchy for what to do after I do the loss-less analysis. [ both V(z) and W(z) are slightly invalid ] (2) If the B field lines are not tangent to the conductor surface, then I think W(z) cannot be constant on the surface, so this whole underpinning of lines doc wobbles. [ correct, so must stick to large ω only ] (3) This all ties in with the E B = 0 problem which I did a half-baked rescue of. [ As of 5.12.14 I have resolved the above three problems and installed into lines doc. ] (4) I had the general feeling that the asymmetry of current in a conductor matched that asymmetry of the charge density on the surface and may have even stated that in lines doc. I suspect that this association of the two asymmetries is only valid at "high frequency" where the skin effect is active. Probably at low frequency you could have a huge charge density asymmetry and a completely uniform Ez in the round wire or other conductor. [ I will investigate this item ] (5) I am still unhappy about the question of "where does the surface charge come from"? I made that into a reader exercise because I could not answer it! [ I will investigate this item at some point ] So my current version of lines doc is then only a "working platform" which at least organizes certain pieces of EM information, but it is not ready for prime time because there are too many fudged and unanswered questions. I kept sweeping things under the rug to get it done, and now those things have come back to roost. But I do think I have the tools to approach the issues, and in the end I think I can get lines doc into something worth while. Right now it is worthless and embarrassing. In particular, I think I can compute a lot of things for the parallel cylinders based on the bipolar coordinates doc.