focus on the charge pump BC REVIEWED

Focus on the charge pump BC PhL 3.20.14 Review 5.15.14. The very first part below seems OK and I refined it and put it into Section 6.5 (d). It is the idea that the function Ez(a,θ) tracks with n(θ), at least at large ω in Section 6.5. The tracking argument is not rigorous at low ω. But I think the cpbc is rigorous at all ω, so I think App D is OK down to ω = 0. The remainder of what is below is wrong because I confuse together two problems: [ In the rest of this doc, I seem to be confusing two different physical problems. One has a shorted end to the transmission line (where I expect to have symmetric Jz at ω = 0) and the other has Z = ∞ at the end at ω = 0, since it is after all supposed to be an infinite transmission line, and that is what Appendix D treats. For this reason, nothing said below seems too useful. ] I am still working on the low frequency limit anomaly of Appendix D. Here I will focus directly on the charge pump boundary condition which, together with divE = 0, seems to be causing the anomaly. Here is the BC in its two forms, where Er is "just under the surface" Er(a,θ) = (jω/σ) n(θ) . (D.2.24) Er(a,m) = (jω/σ) Nm . (D.2.25) Here is the div E = 0 equation at an arbitrary interior location, again in two forms. ∂r (r Er) + ∂θEθ + r ∂zEz = 0 . [1 + r∂r ] Er(r,m) + jmEθ(r,m) + r (-jβd)Ez(r,m) = 0 . (D.1.19) If we evaluate the second form at the surface, it says [1 + r∂r ] Er(a,m) + jmEθ(a,m) + a (-jβd)Ez(a,m) = 0 . (D.1.19) In the electrostatic capacitor model to which I "cling" strongly, Eθ(a,m) = 0 and div E=0 says [1 + r∂r ] Er(a,m) + a (-jβd)Ez(a,m) = 0 Ez(a,m) = (1/jaβd) [1 + r∂r ] Er(a,m) where Er(a,m) = (jω/σ) Nm This seems to provide a "mechanism" by which n(θ) propagates itself into Ez(a,m), but I only know the value of Er(a,m) = (jω/σ) Nm and not [∂rEr(r,m)]|r=a, so not clear what to do next. [ the above logic now appears in lines doc Section 6.5 (d). ] Now here is a problem I have. Write the following, where now "m" implies m > 0 always, Ez(a,m) = (1/jrβd) { Er(a,m) + [∂rEr(r,m)]|r=a } " non-symmetric part" Ez(a,0) = (1/jrβd) { Er(a,0) + [∂rEr(r,0)]|r=a } " symmetric part" Then look at the ratio of non-sym / sym : = Er(a,m) = (jω/σ) Nm => = ηm Er(a,0) = (jω/σ) N0 My problem is that I want this ratio to become very small as ω becomes very small, so that only the symmetric part survives in this limit. The βd factor cancels out in the ratio, so it cannot "help". It does seem to be a 0 / 0 type ratio as ω→0 . I do know that Er(r,m) = am x-1 Jm(x) + Jm+1(x) but of course if I install my known am and Km I will just reproduce the fact that this ratio does not get small. My problem is that when all the ingredients are mixed together (Helmholtz and divE = 0), I get a result which does not "make sense" for small ω. If instead I treat the coefficients as unknowns and do a small ω expansion, I will find that the ratio above is some function of Km,K0,am,a0 each of which can in general be a function of ω, so that by itself proves nothing. A big thing that does not "make sense" is the following. First consider curl E = [ r-1∂θEz - ∂zEθ] + [∂zEr - ∂rEz] + [ r-1∂r(rEθ) - r-1∂θEr ] so in partial waves curl E = [ r-1jmEz +jβdEθ] + [-jβdEr - ∂rEz] + [ r-1∂r(rEθ) - r-1jmEr ] and then [curl E]θ(r,m) = -jβdEr(r,m) - ∂rEz(r,m) Then Maxwell -jωBθ = [curl E]θ says -jωBθ(r,m) = -jβdEr(r,m) - ∂rEz(r,m) βd ≈ ω/vd ignoring loss When I assume βd = (ω/vd) and β = c = β', and when I install my known fields, I get for low ω, ω Bθ(r,m) = - (j/2a) ηm I Rdc m(m+1)(r/a)m-1 [ back to the B→∞ paradox ] The problem is that for very small ω, and for fixed I, Bθ(r,m) gets large without limit, which seems very unphysical. With these same β assumptions, we can write -jωBθ(r,m) ≈ - ∂rEz(r,m) very small ω and then it becomes clear that the problem is that ∂rEz(r,m) is failing to → 0 as ω→ 0. Since the small ω limit of Ez is (1/2) (m+1) ηm I Rdc (r/a)m, ∂rEz(r,m) is much the same and stays finite. Thus, the finiteness of ∂rEz(r,m) results in the infiniteness of Bθ(r,m) . So this is something I OUGHT to be able to understand and deal with. [ In the rest of this doc, I seem to be confusing two different physical problems. One has a shorted end to the transmission line (where I expect to have symmetric Jz at ω = 0) and the other has Z = ∞ at the end at ω = 0, since it is after all supposed to be an infinite transmission line, and that is what Appendix D treats. For this reason, nothing said below seems too useful. ] Now at the value f = 0.1 Hz, I have two infinite cylinders driven by Vac = 6cos(ωt) and I = very small due to the very large Z0. The ± n(θ) is constant and quite finite on the conductor pair for a long way down the line from the driving end. All this surface charge came from the AC battery driving source. On every full cycle, this entire n(θ) is pulled off the conductor surface and pushed onto the other, so to speak. So this brings me back to the "surface currents" issue. In the physical experiment just stated, one must ask: how does the n(θ) surface current "get" to where it is? It probably helps to have this little picture in mind, This definitely brings to mind the notion that it is n(θ,z) and this is decreasing in z due to the resistance elements. When we write n(θ,z) = ej(ωt-βz) n(θ,0), we will have |n| staying constant in z if we use our model that βd = (ω/vd) and then it looks like just a giant long capacitor with no resistance. But in this case we will have Z0 = 0 and then the capacitor is shorted out and the source drives an infinite current I. So whatever we do, we have to allow for finite resistance! I keep wanting to view the physical situation as two independent entities: [wrong!] (1) two long resistors with some I flowing in them (2) one long capacitor made from the surfaces of the two conductors. In this "model", I guess the capacitor charge is fed along the surface by surface currents. At DC there is no radial pumping current (let's say). But this "model" is intrinsically wrong! The current charging a downstream capacitor has to pass through some resistance before reaching that capacitor, and that current flows through the interior, and then it really is radially pumped out to the surface. So we must include in βd an imaginary part. I did the math on this in "including loss" doc and got βd = (ω/vd) - j [(Rdc1+ Rdc2)C ] (vd/2) = βd0 - jα α = [ (Rdc1+ Rdc2)C ] (vd/2) and there is the imaginary part of interest. e-jβz = e-jβz e-j(-jα)z = e-jβz e-αz So here then is the physical situation. You have some Rdc and C for your line. You have n(θ) at full value at the driving end, but it and everything else decays with e-αz which by the way is frequency independent! We can assume that α is small, but it is finite. In "DC limit of App D" I wrote this as βd = (1/vd)(ω-ωc) ωc = Rdc C vd2/2 and for the Belden 8281 I found that fc = 7.7 KHz. How does this change the above analysis? Go back to Er(a,θ) = (jω/σ) n(θ) . (D.2.24) Er(a,m) = (jω/σ) Nm . (D.2.25) Nm = ηmN0 What is N0 exactly? I do a little calculation below (D.2.30), but in this calculation I use my existing expression for Jz and Ez which may not be right if I am about to modify the theory. Maybe I should leave this as a TBD constant for a while. Let's just do that for the moment. Now the div E = 0 says this: ∂r (r Er) + ∂θEθ + r ∂zEz = 0 . [1 + r∂r ] Er(r,m) + jmEθ(r,m) + r (-jβd)Ez(r,m) = 0 . [1 + r∂r ] Er(a,m) + jmEθ(a,m) + r (-jβd)Ez(a,m) = 0 [1 + r∂r ] Er(a,m) + r (-jβd)Ez(a,m) = 0 Ez(a,m) = (1/jaβd) [1 + r∂r ] Er(a,m) where Er(a,m) = (jω/σ) Nm Now we DO have a difference because now I will insert βd = (1/vd)(ω-jωc) to get Ez(a,m) = [1 + r∂r ] Er(a,m) Er(a,m) = (jω/σ) Nm So this shows a significant modification in the case that we account for loss. Now when all is done, we are going to find that Er(r,m) has some form k(m,ω)(r/a)p near r = a, where p is some power like m or m±1 or something like that. Therefore I think we can say Er(r,m) = Er(a,m) (r/a)p for r near a ∂r Er(r,m) = Er(a,m) p(r/a)p-1(1/a) r ∂r Er(r,m) = Er(a,m) p(r/a)p-1 [r ∂r Er(r,m)]|r=a = Er(a,m) p and then our equation above becomes Ez(a,m) = (1+p) Er(a,m) Er(a,m) = (jω/σ) Nm and then we insert to get Ez(a,m) = (1+p(m)) (jω/σ) ηm N0 // I is inside N0 here As stated, I regard N0 as an unknown constant at this point, which I feel should be independent of ω. This does achieve the goal of causing Ez(a,m) → 0 as ω → 0, but unfortunately it achieves this goal both for m > 0 and m = 0. In fact we get = ηm This means then that you cannot have Ez(a,m) → 0 while Ez(a,m) → constant. For example, in the existing theory we have Ez(a,m) = (1/2) ηm I Rdc (m+1) Ez(a,0) = (1/2) I Rdc (1) and then in this theory we get = ηm which agrees with the above. Does this fix the Bθ problem? -jωBθ(r,m) = -jβdEr(r,m) - ∂rEz(r,m) jωBθ(r,m) = j (1/vd)(ω-ωc)Er(r,m) + ∂rEz(r,m) OK, now as ω→0 the first term is proportional to Er(r,m) which in the current theory → 0, never had an issue with that. The second term then gives jωBθ(r,m) ≈ ∂rEz(r,m) If we assume that Ez(r,m) = Ez(a,m) (r/a)s ∂rEz(r,m) = Ez(a,m) q (r/a)s-1 (1/a) ≈ Ez(a,m) (s/a) but from above, Ez(a,m) = (1+p(m)) (jω/σ) ηm N0 so we then have jωBθ(r,m) ≈ ∂rEz(r,m) = Ez(a,m) (s/a) = s (1+p(m)) (jω/σ) ηm N0 Then we can cancel the jω to get Bθ(r,m) = s (1+p(m)) (1/σ) ηm N0 and this then approaches a finite value instead of an infinite one, so in some sense the Bθ problem is somewhat solved by including the losses. dimensions: RHS = (1/σ) N0 = m/(sec m2) * sec * (ohm-m) * (Coul/m2) = = (ohm) * (Coul/m2) = henry/sec * Cou/m2 = amp-henry/m2 = correct For Bθ(r,m) I do expect to have an m = 0 contribution as well as m > 0 contributions since I expect the B field to be quite complicated even in the DC limit for two cylinders. I cannot find N0 until I know the full extent of Ez.