altering the CPBC for radial Hall REVIEWED

Altering the CPBC for radial Hall PhL 6.25.14 An earlier doc alludes to pondering the radial Hall as the fall guy for low ω problems. Here I check it out and conclude that it is not the fall guy. I quote from below (page 12 or 30): Conclusion: If I have done the above correctly, then adding the radial Hall effect to the CPBC does not rescue that Jz asymmetry paradox! It was just another idea of many that did not pan out. I then carry on with 20 pages of flailing away about G = 0 and G > 0 cases and wondering about the magnetostatic problem. I think those situations are well handled now in lines doc. I think this is the cause of the asymmetry paradox. I think it very unlikely that this attempt will succeed [correct, it did not succeed] but it is worth a try because it is exactly at the right place in the machine where I am having trouble. The n(θ) shape is "getting into" Jz through this boundary condition. I know at at ω = 0 we have Jr = 0 and Er ≠ 0, so maybe near ω = 0 this idea will somehow help. The main issue here is trying to use the tensor magnetic Ohm's law in place of Jr = σEr. The BC I think is OK as is in terms of Jr. (a) obtain the appropriate magnetic Ohm's law for the round wire situation The magnetic Ohm's law discussion starts in Section N.4. There I was using a static version, so perhaps the dynamic version is this m = qE + qvxB - (m/τ) v . (N.1.5) mjωv = qE + qvxB - (m/τ) v . or m[jω +1/τ] v = qE + qvxB I know that 1/τ is about 1014 from (N.1.12). Even at 100 GHz we get ω = 2π x 1011 ~ 1013 and I am willing to ignore the jω term for my interests. Thus, (m/τ) v ≈ qE + qvxB This is the same as (N.4.1) and I have done all the solution work in Section N.7 with this result Jz = σ ( Ez + ωcτ Er) ωc ≡ (qB/m) Jr = σ (Er - ωcτ Ez) B = B Jθ = σEθ . σ = (nq2τ/m) (N.7.8) where this makes use of B(r) = B(r) which I will assume for the round wire! So this is our magnetic Ohm's law of interest even at some frequency ω. I know ωcτ is small, but I want to keep it for a while. My next step in radial Hall effect is to say Jr = 0 for the static situation, but I certainly DON'T want to say that here. Notice that if I set ωcτ = 0, then we just have the regular Ohm's law and they we are not fixing anything for Appendix D. The above equation is local to any point r inside the round wire. (b) What is the new boundary condition for Appendix D ? Jr(r=a-ε,θ) = jω n(θ) non-conducting dielectric (D.2.23) This then says [Er(a,θ) - ωc(a) τ Ez(a,θ)] = (jω/σ) n(θ) and this IS a different boundary condition. I guess I really need to add this: ωc(r) = (qB(r) /m) I could at this point install B(r) for a uniform Jz round wire. But no matter how the current is distributed in the wire, B(a) will be the same! Thus without any approximation we can use B(r) = (r/a) => B(a) = => ωc(a) = (q/m) ωc(a)τ = (qτ/m) = μ = mobility μ0 = the usual Now let's confirm that this is very small dim = tesla-1 henry/m amp / m = m2 amp-1henry-1 * henry/m amp / m = dimensionless as expected μ = (qτ/m) = 1.6 x 10-19 * 10-14 / [9.1 x 10-31] = 1.6/9.1 * 10-19-14+31 = 0.17 x 10-2 tesla-1 Then = 0.17 x 10-2 x 4π x 10-7 / 2π * (I/a) = 0.17 x 10-2 x 2 x 10-7* (I/a) = .34 x 10-9 I/a Example 1: I = 1000 amps and a = 10-2 m I/a = 105 = .34 x 10-4 Example 2: I = 1 amp and a = 400μ = 400x10-6 = 4 x 10-4 I/a = 1/ [4 x 10-4] ~ 103 So in either of these wildly different cases, we have << 1, so I will now simplify our new boundary condition and just say [Er(a,θ) - Ez(a,θ)] = (jω/σ) n(θ) Notice that the mixture on the left is the same at any ω of interest. I am not yet sure what affect this would have on the solution in Appendix D! Does this thing make any sense at DC? Assume uniform Ez. Then it says Er(a,θ) = Ez(a,θ) At DC this says Er(a,θ) = (1/σ) Jz(a,θ) = (1/σ) (I/πa2) Now we know that μ = qτ/m σ = (nq2τ/m) μ/σ = qτ/m * m/(nq2τ) = 1/nq so then Er(a,θ) = (1/σ) (I/πa2) = 1/nq (I/πa2) = = Ez of (N.7.12) So at DC the result of this BC agrees exactly with the radial Hall effect, I like it so far! (c) What is the impact on Appendix D? First write Er(a,m) - Ez(a,m) = (jω/σ) Nm Eθ(r=a,m) = 0 First summary of the E field solutions (D.2.21) Ez(r,m) = - j (β'/βd) Jm(x) x = β'r (D.1.27) Er(r,m) = am x-1 Jm(x) + Jm+1(x) β'2 = β2 - βd2 (D.2.11) jEθ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) (D.2.15) Things are going to be different now! I don't know how much different. Define = ωc(a)τ = κ just for short, a small dimensionless parameter! Then Er(a,m) - κ Ez(a,m) = (jω/σ) Nm Eθ(r=a,m) = 0 and we get am xa-1 Jm(xa) + Jm+1(xa) - κ [- j (β'/βd) Jm(xa)] = (jω/σ) Nm -am xa-1 Jm(xa) + ( + ) Jm+1(xa) where the new stuff is shown in red. Write again now am xa-1 Jm(xa) + Jm+1(xa) + κj (β'/βd) Jm(xa) = (jω/σ) Nm -am xa-1 Jm(xa) + ( + ) Jm+1(xa) [ Jm+1(xa) + κj (β'/βd) Jm(xa) ] + [xa-1 Jm(xa)] am = (jω/σ) Nm [Jm+1(xa)] + [-xa-1 Jm(xa) + Jm+1(xa)/m] am = 0 or [ Jm+1(xa) + κj (β'/βd) Jm(xa) ] + [xa-1 Jm(xa)] am = (jω/σ) Nm [m Jm+1(xa)] + [-m xa-1 Jm(xa) + Jm+1(xa)] am = 0 But right here we can use NIST (10.6.1) Bessel identity (2m/x)Jm(x) = Jm-1(x) + Jm+1(x). (m/x)Jm(x) = (Jm-1(x) + Jm+1(x))/2 which says -mxa-1 Jm(xa) + Jm+1(xa) = - (Jm-1(x) + Jm+1(x))/2 + Jm+1(xa) = - (Jm-1 - Jm+1)/2 = (Jm+1 - Jm-1)/2 so our two equations are then [ Jm+1(xa) + κj (β'/βd) Jm(xa) ] + [xa-1 Jm(xa)] am = (jω/σ) Nm [m Jm+1(xa)] + (1/2)[ Jm+1(xa) - Jm-1(xa)] am = 0 A+ B am = C D+ E am = 0 A = [Jm+1(xa) + κj (β'/βd) Jm(xa)] B = xa-1 Jm(xa) C = (jω/σ) Nm D = m Jm+1(xa) E = (1/2)[ Jm+1(xa) - Jm-1(xa)] Now let Maple solve the two equations: So our result is a bit complicated am = CD/den = (jω/σ) Nm mJm+1(xa) /den = - CE/den = - (jω/σ) Nm (1/2)[ Jm+1(xa) - Jm-1(xa)]/den = - (jω/2σ) Nm [ Jm+1(xa) - Jm-1(xa)]/den den = BD-EA = xa-1 Jm(xa) m Jm+1(xa) - (1/2)[ Jm+1(xa) - Jm-1(xa)] [Jm+1(xa) + κj (β'/βd) Jm(xa)] Note that if κ = 0 we get den = xa-1 Jm(xa) m Jm+1(xa) - (1/2)[ Jm+1(xa) - Jm-1(xa)] [Jm+1(xa)] = Jm+1(xa)[ xa-1 Jm(xa) m - (1/2)[ Jm+1(xa) - Jm-1(xa)] ] = (1/2) Jm+1(xa)[ 2mxa-1 Jm(xa) - Jm+1(xa) + Jm-1(xa) ] = (1/2) Jm+1(xa)[ Jm-1(x) + Jm+1(x) - Jm+1(xa) + Jm-1(xa) ] = (1/2) Jm+1(xa)[ 2Jm-1(xa) ] = Jm+1(xa)Jm-1(xa) In this case we get am = (jω/σ) Nm mJm+1(xa) /den = (jω/σ) Nm mJm+1(xa)/ [Jm+1(xa)Jm-1(xa)] = (jω/σ) Nm m/ Jm-1(xa)] and this does agree. Next, = - (jω/2σ) Nm [ Jm+1(xa) - Jm-1(xa)]/ [Jm+1(xa)Jm-1(xa)] = - (jω/2σ) Nm {1/ Jm-1(xa) - 1/ Jm+1(xa) } and this also agrees. Now without further ado, I can write Ez(r,m) = - j (β'/βd) Jm(x) = - j (β'/βd) Jm(x)[ - (jω/2σ) Nm [ Jm+1(xa) - Jm-1(xa)]/den] = j (β'/βd) Jm(x)[ (jω/2σ) Nm [ Jm+1(xa) - Jm-1(xa)]/den] = -(β'/βd) Jm(x)[ (ω/2σ) Nm [ Jm+1(xa) - Jm-1(xa)]/den] = -(β'/βd) Jm(x)[ (ω/2σ) Nm [ Jm+1(xa) - Jm-1(xa)] / { xa-1 Jm(xa) m Jm+1(xa) - (1/2)[ Jm+1(xa) - Jm-1(xa)] [Jm+1(xa) + κj (β'/βd) Jm(xa)] } Now for κ = 0 this reads Ez(r,m) = -(β'/βd) Jm(x)[ (ω/2σ) Nm [ Jm+1(xa) - Jm-1(xa)] / { xa-1 Jm(xa) m Jm+1(xa) - (1/2)[ Jm+1(xa) - Jm-1(xa)] [Jm+1(xa)] } = -(β'/βd) Jm(x)[ (ω/2σ) Nm [ Jm+1(xa) - Jm-1(xa)] (1/ Jm+1(xa)) / { xa-1 Jm(xa) m - (1/2)[ Jm+1(xa) - Jm-1(xa)] } = -(β'/βd) Jm(x)[ (2ω/2σ) Nm [ Jm+1(xa) - Jm-1(xa)] (1/ Jm+1(xa)) / { 2xa-1 Jm(xa) m - [ Jm+1(xa) - Jm-1(xa)] } = -(β'/βd) Jm(x)[ (2ω/2σ) Nm [ Jm+1(xa) - Jm-1(xa)] (1/ Jm+1(xa)) / { 2xa-1 Jm(xa) m - Jm+1(xa) + Jm-1(xa)] } = -(β'/βd) Jm(x)[ (2ω/2σ) Nm [ Jm+1(xa) - Jm-1(xa)] (1/ Jm+1(xa)) / { Jm-1(x) + Jm+1(x) - Jm+1(xa) + Jm-1(xa)] } = -(β'/βd) Jm(x)[ (2ω/2σ) Nm [ Jm+1(xa) - Jm-1(xa)] (1/ Jm+1(xa)) / {2Jm-1(x) } = -(β'/βd) Jm(x)[ (ω/2σ) Nm [ Jm+1(xa) - Jm-1(xa)] (1/ Jm+1(xa) Jm-1(x)) = -(β'/βd) Jm(x)[ (ω/2σ) Nm [1/ Jm-1(xa) - 1/ Jm+1(xa) ] = (β'/βd) Jm(x)(ω/2σ) Nm [1/ Jm+1(xa) - 1/ Jm-1(xa) ] = (β'/βd) (ω/2σ) Nm fm = (β'/βd) (ω/σ) ηm N0 fm = (β'/βd) (ω/2σ) ηm (βd/2πωa) I fm = β' (ω/2σ) ηm (1/2πωa) I fm Rdc = = β' ω πa2Rdc ηm (1/4πωa) I fm = (β'a) Rdc ηm (1/4) I fm which is correct. So having done all these tests, here is my new Ez when the radial Hall effect is included: Ez(r,m) = (β'/βd) Jm(x)(ω/2σ) Nm [ Jm-1(xa) - Jm+1(xa)] / { xa-1 Jm(xa) m Jm+1(xa) - (1/2)[ Jm+1(xa) - Jm-1(xa)] [Jm+1(xa) + κj (β'/βd) Jm(xa)] } = (β'/βd) (ω/2σ) Nm Jm(x) * Say it again Sam: Ez(r,m) = (β'/βd) (ω/σ) Nm Jm(x) * Now suppose I use the known fact that κ is small to write A = [Jm+1(xa) + κj (β'/βd) Jm(xa)] = Jm+1(xa) [ 1 + κj (β'/βd) Jm(xa)/ Jm+1(xa) ] Then the big ratio thing becomes, = * (1/Jm+1(xa)) The denominator of this ratio can be written { 2mxa-1 Jm(xa) - [ Jm+1(xa) - Jm-1(xa)] [ 1 + κj (β'/βd) Jm(xa)/ Jm+1(xa) ] = { 2mxa-1 Jm(xa) - Jm+1(xa) + Jm-1(xa)] + κj (β'/βd) [Jm(xa)/ Jm+1(xa)] [ Jm+1(xa) - Jm-1(xa)] = { Jm-1(xa) + Jm+1(xa) - Jm+1(xa) + Jm-1(xa)] + κj (β'/βd) [Jm(xa)/ Jm+1(xa)] [ Jm+1(xa) - Jm-1(xa)] = { 2Jm-1(xa) + κj (β'/βd) [Jm(xa)/ Jm+1(xa)] [ Jm+1(xa) - Jm-1(xa)] } = 2Jm-1(xa) { 1 + κj (β'/βd) [Jm(xa)/ Jm+1(xa)] [ Jm+1(xa) - Jm-1(xa)] / [2Jm-1(x) } = 2Jm-1(xa) { 1 + κj (β'/βd) [Jm(xa)/ 2Jm+1(xa) Jm-1(xa)] [ Jm+1(xa) - Jm-1(xa)] } Now write [ Jm+1(xa) - Jm-1(xa)]/ Jm+1(xa) Jm-1(xa) = 1/ Jm-1(xa) - 1/ Jm+1(xa) = - (1/Jm(xa)) [ - ] = - (1/Jm(xa)) fm(xa) = - fm(xa)/ Jm(xa) and then Jm(xa) [ Jm+1(xa) - Jm-1(xa)]/ Jm+1(xa) Jm-1(xa) = fm(xa) Then the denominator above can be written = 2Jm-1(xa) { 1 + (1/2) κj (β'/βd)fm(xa) } Then the Big Ratio becomes this (1/Jm+1(xa)) = (1/2)[ Jm+1(xa) - Jm-1(xa)]/ Jm+1(xa) Jm-1(xa) * = (1/2) { - fm(xa)/ Jm(xa)} * But now in the numerator Jm(xa) [ Jm+1(xa) - Jm-1(xa)]/ Jm+1(xa) Jm-1(xa) = fm(xa) so [ Jm+1(xa) - Jm-1(xa)]/ Jm+1(xa) Jm-1(xa) = fm(xa)/ Jm(xa) [ Jm+1(xa) - Jm-1(xa)] = [ Jm+1(xa) Jm-1(xa) / Jm(xa) ] fm(xa) So now the Big Ratio becomes = (1/2) { fm(xa)/ Jm(xa)} * = (1/2) { [fm(xa)/ Jm(xa)]2} * I seem now to get Ez(r,m) = (β'/βd) (ω/σ) Nm Jm(x) * (1/2) { [fm(xa)/ Jm(xa)]2} * But this seems to have the wrong limit for κ→0, since the correct limit from above is this = (β'/βd) Jm(x)(ω/2σ) Nm [1/ Jm+1(xa) - 1/ Jm-1(xa) ] = (β'/βd) Jm(x)(ω/2σ) Nm (Jm-1(xa) - Jm+1(xa))/[ Jm+1(xa) Jm-1(xa)] OK, I have had enough. This is really a task for Maple. The Big Question: Does having the small κ term present in the Ez solution somehow make us symmetric in the DC limit? Ez(r,m) = (β'/βd) (ω/σ) Nm Jm(x) * The κ term can only be useful if (β'/βd) is large, I would say. The verdict is still out on whether this can somehow yield Ez(r,m) ~ δm,0 (stuff) as ω → 0 Today it was all math. Tomorrow maybe I can ponder what this means. It just seems unlikely that having that little κ there can make any difference. June 26, 2014 I am going to switch to a Maple approach. First I set κ = 0 and have Maple solve the BC's for am and Km. That was quite easy, file is "solve Km and am v2.mws". I work first with κ = 0 hoping to reproduce old results. I find I used Q = (β'/βd') in the Ez expression, but this cancelled out in the ratio. Now if η1 ≠ 0, the ratio of the m=1 field to the m=0 field is as shown. If I make no assumption at all about β', then I cannot plot anything since x = β'r . But I can see that rat10 ≠ 0. Thus, if the m = 0 Ez component is present, then so is the m = 1 component, and this is my Paradox at ω = 0. Now I will rerun the program adding in the κ term! I now avoid doing all the manual algebra above and I just let Maple do it. Here are some results where again Q = (β'/βd') : Perhaps my hand computations agree with these results, they at least look vaguely similar. I then enter Ez as usual in terms of these constants and I compute this ratio rat10 for m=1 over m=1 Ez: Now here is the Question: is there some assumption I can make about Q at small ω which causes the above ratio to be zero? I know that at DC, the field Ez(r,m=1) has to be completely gone and I also know that the component Ez(r,m=0) has to be a constant, so the ratio has to go to 0. If I assume that Q → 0, I get this result : This ratio is NOT zero because the numerator is not zero for all x certainly. So that assumption about Q does not solve the problem. In fact, it just replicates the κ = 0 result shown earlier. If I assume that Q → ∞, I get this result : This result is independent of κ and is also NOT zero so this assumption does not resolve Paradox! The only other choice is that Q→Q0, some constant. They I get the full rat10 shown above with Q0. I would then need the numerator to be zero. That would require the following expression be 0 : since it appears in the numerator of the mess above, and no other numerator factors cancel. This would require that, as ω→ 0, Q κ = j J1(xa)/ J0(xa) which is to say (β'/βd') = j J1(xa)/ J0(xa) But this makes no sense at all. We might have(β'/βd') → 0, constant, or ∞ based on our choice of loss model. Certainly we can rule out the 0 and ∞ cases. The constant case would require that the constant ratio be proportional to 1/I and I no model which that gives that. Conclusion: If I have done the above correctly, then adding the radial Hall effect to the CPBC does not rescue that Jz asymmetry paradox! It was not another idea of many that did not pan out. Let's now go back to the modified CPBC : Er(a,θ) - Ez(a,θ) = (jω/σ) n(θ) κ = Here are arguments why this does NOT fix the Paradox. 1. The problem originally with Er(a,θ) = (jω/σ) n(θ) was that n(θ) "got into" Er and then into Ez . The new BC does not really stop n(θ) from getting into Er and Ez. You might imagine some delicate balance that caused a magic cancellation. but that would depend on I. 2. Also, since I → 0 in our limit ω→0, the BC is not really changed at all. 3. At any finite I, this is basically true for any I since κ << 1. One More Check. The modified BC is really this: [Er(a,θ) - (ωc(a)τ) Ez(a,θ)] = (jω/σ) n(θ) where (ωc(a)τ) = = κ so to be precise I should be writing the modified BC this way [Er(a,θ) - κ Ez(a,θ)] = (jω/σ) n(θ) [Er(a,θ) - κ Ez(a,θ)] = (jω/σ) n(θ)(1+κ2) [Er(a,m) - κ Ez(a,m)] = (jω/σ)Nm(1+κ2) If I do this in my Maple program, what happens? It just replaces N0 by N0(1+κ2) and then it cancels out in the ratio rat10 and so it makes no difference whatsoever. Another Mystery. The CPBC says Er(r=a,m) = (jω/σ) Nm Thus, it seems that you MUST get Er(r=a,m) = 0 when ω = 0 But the general solution at small ω is this Er(r,m) = (j/4) ηm I Rdc (aβd') [(r/a)m+1 + (r/a)m-1] Er(a,m) = (j/2) ηm I Rdc (aβd') so isn't this a conflict? Not if I = 0, the same escape hatch. But I also have I = 2πω (a/βd') N0 and then the above ways Er(a,m) = (j/2) ηm 2πω (a/βd') N0 Rdc (aβd') = (j/2) ηm 2πω a2N0 Rdc (1) = (j/2) Nm 2πω a2 Rdc and now our conflict goes away. I suppose this is the same as doing this I = V/Z0 Z0 = Then you get Er(a,m) = (j/2) ηm V Rdc (aβd') But why does this give a different looking result? Suppose I now use this App Q fact β'd = (1-j) The result is then Er(a,m) = (j/2) ηm V Rdc (a (1-j)) = (j/2) ηm V C ω (1-j)/ a Rdc = ( j/2) ηm V C ω e-jπ/4a Rdc = ( j/2) ηm V C ω a Rdc e-jπ/4e-jπ/4 = ( j/2) ηm V C ω a Rdc = ( j/2) ηm q ω a Rdc q = CV (2) Compare this to result (1) above Er(a,m) = (j/2) ηm 2πω a2N0 Rdc and recall that N0 = (1/2πa) q(0) (D.1.8) so this last result is then Er(a,m) = (j/2) ηm 2πω a2(1/2πa) q(0) Rdc = (j/2) ηm ω a q(0) Rdc (3) But this is the same as result (2), thank goodness. This shows that App Q really is right in this limit, otherwise we would not get the exact result. Maybe things are clearer if I get rid of I and make this replacement I = 2πω (a/βd') N0 = 2πω (a/βd') (1/2πa) q(0) = [ω q (1/βd') ] This replacement is valid at any ω. Then write Ez(r,m) = (1/4) ηm I Rdc (aβ') fm fm = [ - ] x = β'r Er(r,m) = (j/4) ηm I Rdc (aβd') gm gm = [ + ] xa = β'a Eθ(r,m) = (1/4) ηm I Rdc (aβd') hm hm = [ - ] Ez(r,m) = (1/4) ηm [ω (1/βd') q] Rdc (aβ') fm fm = [ - ] x = β'r Er(r,m) = (j/4) ηm [ω (1/βd') q] Rdc (aβd') gm gm = [ + ] xa = β'a Eθ(r,m) = (1/4) ηm [ω (1/βd') q] Rdc (aβd') hm hm = [ - ] or Ez(r,m) = (1/4) ηm [ωaq] (β'/βd') Rdc fm fm = [ - ] x = β'r Er(r,m) = (j/4) ηm [ωaq] Rdc gm gm = [ + ] xa = β'a Eθ(r,m) = (1/4) ηm [ωaq] Rdc hm hm = [ - ] (***) Where q is the charge/length at z = 0, a quantity we regard as fixed in our limit. But the above form is valid for any ω, just another way to write things that I have never used before. Now bring in these small β' results which are apropo for small ω fm = (r/a)m (m+1) (2/β'a) f0 = 4/(aβ') gm = (r/a)m+1 + (r/a)m-1 g0 = 2 (r/a) hm = (r/a)m+1 - (r/a)m-1 h0 = 0 to get For m > 0: Ez(r,m) = (1/2) ηm [ω q (1/βd') ]Rdc (r/a)m (m+1) Er(r,m) = (j/4) ηm [ω q a ]Rdc [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (1/4) ηm [ω q a ]Rdc [(r/a)m+1 - (r/a)m-1] For m= 0: Ez(r,0) = [ω q (1/βd') ]Rdc Er(r,0) = (j/2) [ω q a ]Rdc (r/a) Eθ(r,0) = 0 Now install this result for small ω, β'd = (1-j) = (1-j)/ = e-jπ/4 (1/βd') = ejπ/4 / and we then get For m > 0: Ez(r,m) = (1/2) ηm [ q ejπ/4 / ]Rdc (r/a)m (m+1) Er(r,m) = (j/4) ηm [ω q a ]Rdc [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (1/4) ηm [ω q a ]Rdc [(r/a)m+1 - (r/a)m-1] For m= 0: Ez(r,0) = [ q ejπ/4 / ]Rdc G = 0 Er(r,0) = (j/2) [ω q a ]Rdc (r/a) Eθ(r,0) = 0 Here we see ALL fields going to 0 in the limit ω→0. As the limit is approached, there exists Ez asymmetry as shown. Question: How does the above work out if G ≠ 0? First, here are my results for repaired Appendix D: Second summary of the E field solutions : Rdc = β"2 = β2 - β"d2 (D.9.25) (conducting dielectric) 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 = [ - ] Now write (see D.9.32 and below) I" = I = 2πω (a/βd) N0 = 2πω (a/βd") N0 But then we always have same as before N0 = (1/2πa) q(0) so that I" = 2πω (a/βd") (1/2πa) q(0) = ω (1/βd") q(0) = [ω q (1/βd")] Install to get something almost identical to (***) above, E"z(r,m) = (1/4) ηm [ω q a(β"/βd")] Rdcfm fm = [ - ] x = β"r E"r(r,m) = (j/4) ηm [ω q a] Rdc gm gm = [ + ] xa = β"a E"θ(r,m) = (1/4) ηm [ω q a] Rdc hm hm = [ - ] For the r and θ fields things are exactly the same! This is all for general ω so far. Now a big difference is going to be the Appendix Q limit which for G ≠ 0 is for ω→0 β"d = (ω/2) - j ≈ - j For β" we then get β"2 = β2 - β"d2 = - β"d2 = - [- j ]2 = - (-1) RG = RG β" → as ω → 0, something different! Then (β"/βd") = / [ - j ] = j We then get for ω → 0, E"z(r,m) = (j/4) ηm [ω q a] Rdc fm fm = [ - ] x = β"r E"r(r,m) = (j/4) ηm [ω q a] Rdc gm gm = [ + ] xa = β"a E"θ(r,m) = (1/4) ηm [ω q a] Rdc hm hm = [ - ] Even though ω→0, we can not in general use the small x arguments of Bessels. But if G is small, then β" is very small and we CAN. Then we get fm = (r/a)m (m+1) (2/β'a) f0 = 4/(aβ') gm = (r/a)m+1 + (r/a)m-1 g0 = 2 (r/a) hm = (r/a)m+1 - (r/a)m-1 h0 = 0 So we get these final results E"z(r,m) = (j/4) ηm [ω q a] Rdc fm fm = (r/a)m (m+1) (2/β"a) x = β"r E"r(r,m) = (j/4) ηm [ω q a] Rdc gm gm = [(r/a)m+1 + (r/a)m-1 ] xa = β"a E"θ(r,m) = (1/4) ηm [ω q a] Rdc hm hm = [(r/a)m+1 - (r/a)m-1 ] E"z(r,0) = (j/4) [ω q a] Rdc f0 f0 = 4/(aβ') x = β"r E"r(r,0) = (j/4) [ω q a] Rdc g0 g0 = 2 (r/a) xa = β"a E"θ(r,0) = (1/4) [ω q a] Rdc h0 h0 = 0 Install to get E"z(r,m) = (j/4) ηm [ω q ] (2/β") Rdc (r/a)m (m+1) x = β"r E"r(r,m) = (j/4) ηm [ω q a] Rdc [(r/a)m+1 + (r/a)m-1 ] xa = β"a E"θ(r,m) = (1/4) ηm [ω q a] Rdc [(r/a)m+1 - (r/a)m-1 ] E"z(r,0) = (j/4) [ω q ] Rdc 4/(β") x = β"r E"r(r,0) = (j/4) [ω q a] Rdc 2 (r/a) xa = β"a E"θ(r,0) = 0 Now use β" = to get E"z(r,m) = (j/4) ηm [ω q ] (2/) Rdc (r/a)m (m+1) x = β"r G ≠ 0 E"r(r,m) = (j/4) ηm [ω q a] Rdc [(r/a)m+1 + (r/a)m-1 ] xa = β"a E"θ(r,m) = (1/4) ηm [ω q a] Rdc hm = [(r/a)m+1 - (r/a)m-1 ] E"z(r,0) = (j/4) [ω q ] Rdc 4/ x = β"r E"r(r,0) = (j/4) [ω q a] Rdc 2 (r/a) xa = β"a E"θ(r,0) = 0 which we can compare with our earlier result for G = 0, For m > 0: Ez(r,m) = (1/2) ηm [ q ejπ/4 / ]Rdc (r/a)m (m+1) Er(r,m) = (j/4) ηm [ω q a ]Rdc [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (1/4) ηm [ω q a ]Rdc [(r/a)m+1 - (r/a)m-1] For m= 0: Ez(r,0) = [ q ejπ/4 / ]Rdc G = 0 Er(r,0) = (j/2) [ω q a ]Rdc (r/a) Eθ(r,0) = 0 We again find that all fields → 0, but the Ez fields now go with a full factor ω, whereas for G = 0 they only went to 0 as . In the limit in EITHER CASE, we find that Ez(r,m) / Ez(r,0) = (1/2) ηm (r/a)m (m+1) G = 0 or G ≠ 0 so AS the fields vanish, the Jz asymmetry remains. Question: What happens when you drive a line at very low ω? For small ω: Re(β'd) = Im(β'd) = - G = 0 and ω << (R/L) Thus, Ldecay = 1/α = 2/ As ω→0, the wavelength of course increases since λ = 2/ as well. And the decay distance also increases, so near the driving end, the decay seems to go away. The decay is due to R, and as R→0 it is the same result as if ω→0, there is no decay. For G≠0, we instead have For small ω: Re(β'd) = (ω/2) Im(β'd) = - ω << (R/L) and ω << (G/C) For this line, the decay distance is fixed by R and G as you see, so Ldecay = 1/α = 1/ but the wave's wavelength still goes to ∞. Question: Consider just the very first "section" of a transmission line which has some G > 0. Across this thing we have 1/G resistance and C capacitance in parallel. I talk about having a charge q at the start of the line, but if the time constant of ω is small enough, it seems that the capacitor has no effect and the charge q would just leak across the conductance and go away, leaving you with q= 0. So my assumption that there is some finite q at z = 0 is wrong in this case at low ω. Is this right? Or since V is across that 1/G resistor, there still is V there, and C charges up to Q = CV and then there IS a q sitting there. Yes, the driver keeps supplying new q. So above is wrong. Paradox B. Consider a line with G > 0. We know that as ω→0, we will have Z0 = and therefore some current I = V/Z0 = V will flow. The decay distance is Ldecay = 1/α = 1/ which could be very long. The paradox B is that the field equations above say E = 0 at DC. Then Ohm's law says you cannot have a current flowing. Maybe this is a hint as to my general problem. Resuming 6.27.14. I think something is wrong with my section D 9 (d) on generalizing the CPBC. I get down to (D.9.23) and I think the idea that Nm → (ξd/εd) Nm is the main change. Then looking at (D.2.28) agree that this is equivalent to doing Km → (ξd/εd) Km ≡ K'm and am → (ξd/εd) am ≡ a'm . I would argue right now that I should keep βd as a general complex parameter which is unchanged. It is a parameter of my theory. I can later evaluate it for special cases if I want. Looking at its derivation, I think D.2.31 says I = 2πω (a/βd) N0 → 2πω (a/βd) (ξd/εd) N0 and we can interpret this in 2 says. First, it is just an application of Nm → (ξd/εd) Nm . Second, it seems that I should increase due to G ≠ 0. But we are not really done yet. Well OK, I guess (D.8.25) is OK as it is stated, but the text βd' = βd where βd = ω only applies to the lossless case. I guess I would write that I' = 2πω (a/βd') (ξd/εd) N0 What is the meaning of N0 and n(θ) ? I think n(θ) is still the true surface charge as in (D.9.19). So then N0 is exactly what it was before. I think it would be clearer NOT to have I or I' appearing in my results. Let's back up and restate the un-modified equations first. My second summary was this Second summary of the E field solutions : Rdc = β'2 = β2 - βd2 (D.2.33) Ez(r,m) = (1/4) ηm I Rdc (aβ') fm fm = [ - ] x = β'r Er(r,m) = (j/4) ηm I Rdc (aβd) gm gm = [ + ] xa = β'a Eθ(r,m) = (1/4) ηm I Rdc (aβd) hm hm = [ - ] Dimension check: dimEz = amp ohm/m = volt/m = correct But suppose I replace at this early point I = 2πω (a/βd) N0, I then get an earlier form which I never wrote down, dim I = sec-1 m2 Cou/m2 = Cou/sec = amp = correct Second summary of the E field solutions : Rdc = β'2 = β2 - βd2 (D.2.33) Ez(r,m) = (1/4) ηm 2πω (a/βd) N0 Rdc (aβ') fm fm = [ - ] x = β'r Er(r,m) = (j/4) ηm 2πω (a/βd) N0 Rdc (aβd) gm gm = [ + ] xa = β'a Eθ(r,m) = (1/4) ηm 2πω (a/βd) N0 Rdc (aβd) hm hm = [ - ] dim Ez = sec-1 m m Cou/m2 ohm/m = sec-1 Cou ohm/m = amp ohm /m = volt/m = correct which I cosmetically rewrite as Second summary of the E field solutions : Rdc = β'2 = β2 - βd2 (D.2.33) Ez(r,m) = (1/4) ηm 2πa ω (1/βd) N0 Rdc (aβ') fm fm = [ - ] x = β'r Er(r,m) = (j/4) ηm 2πa2 ω N0 Rdc gm gm = [ + ] xa = β'a Eθ(r,m) = (1/4) ηm 2πa2 ω N0 Rdc hm hm = [ - ] DimEz = m sec-1 m Cou/m2 ohm/m = sec-1 Cou ohm/m = amp ohm/m = volt/m = correct DimEr = m2 sec-1Cou/m2ohm/m = sec-1Cou ohm/m = correct Now suppose at THIS point I turn on the conducting dielectric. I argued above that Nm → (ξd/εd) Nm is the fundamental change to make. So then the results are Second summary of the E field solutions : Rdc = β'2 = β2 - βd2 (D.2.33) dielectric conducts Ez(r,m) = (1/4) ηm 2πa ω (1/βd) (ξd/εd) N0 Rdc (aβ') fm fm = [ - ] Er(r,m) = (j/4) ηm 2πa2 ω (ξd/εd) N0 Rdc gm gm = [ + ] Eθ(r,m) = (1/4) ηm 2πa2 ω (ξd/εd) N0 Rdc hm hm = [ - ] where the following is still true N0 = (1/2πa) q(0) so rewrite as (D.1.8) Second summary of the E field solutions : Rdc = β'2 = β2 - βd2 (D.2.33) dielectric conducts Ez(r,m) = (1/4) ηm ω q (1/βd) (ξd/εd)Rdc (aβ') fm fm = [ - ] Er(r,m) = (j/4) ηm aω q (ξd/εd)Rdc gm gm = [ + ] Eθ(r,m) = (1/4) ηm aω q (ξd/εd)Rdc hm hm = [ - ] where q = q(0) is the charge per length at the start of the transmission line, and which I think is a good thing to think of as not changing when I turn on the conductivity of the dielectric. But now let's install the low ω xm factors fm = (r/a)m (m+1) (2/β'a) f0 = 4/(aβ') gm = (r/a)m+1 + (r/a)m-1 g0 = 2 (r/a) hm = (r/a)m+1 - (r/a)m-1 h0 = 0 For example, this results in: Ez(r,m) = (1/4) ηm ω q (1/βd) (ξd/εd)Rdc (aβ') [(r/a)m (m+1) (2/β'a)] Ez(r,0) = (1/4) ω q (1/βd) (ξd/εd)Rdc (aβ') [4/(aβ')] or Ez(r,m) = (1/4) ηm ω q (1/βd) (ξd/εd)Rdc [(r/a)m (m+1) 2] Ez(r,0) = (1/4) ω q (1/βd) (ξd/εd)Rdc [4] Dimension check: dimEz = sec-1 Cou/m m ohm/m = sec-1 Cou ohm/m = amp ohm/m = volt/m = correct Appendix Q says For small ω: Re(βd) = (ω/2) Im(βd) = - [ 1 + (ω2/8) ()2 ] ω << (R/L) and ω << (G/C) which fully incorporates dielectric conductivity. We then have close to ω = 0, βd = (ω/2) - j = - j { 1 + (ω/2) / [- j ] } = - j { 1 + j (ω/2) } I can then invert this since ω is small to get (1/βd) = [- j ]-1 { 1 - j (ω/2) } = (j/) { 1 - j (ω/2) } I then get Ez(r,m) = (1/4) ηm ω q (1/βd) (ξd/εd)Rdc [(r/a)m (m+1) 2] Ez(r,0) = (1/4) ω q (1/βd) (ξd/εd)Rdc [4] or Ez(r,m) = (1/4) ηm ω q (j/) { 1 - j (ω/2) } (ξd/εd)Rdc [(r/a)m (m+1) 2] Ez(r,0) = (1/4) ω q(j/) { 1 - j (ω/2) } (ξd/εd)Rdc [4] or Ez(r,m) = (1/4) ηm ω q (j/) (ξd/εd)Rdc [(r/a)m (m+1) 2] Ez(r,0) = (1/4) ω q (j/) (ξd/εd)Rdc [4] or Ez(r,m) = (1/4) ηm ω q (j/) ( 1 - jσd/εdω) Rdc [(r/a)m (m+1) 2] Ez(r,0) = (1/4) ω q (j/) ( 1 - jσd/εdω) Rdc [4] Now take the limit ω→0 and we get Ez(r,m) = (1/4) ηm q (j/) ( - jσd/εd) Rdc [(r/a)m (m+1) 2] Ez(r,0) = (1/4) q (j/) ( - jσd/εd) Rdc [4] or Ez(r,m) = (1/4) ηm q (1/) (σd/εd) Rdc [(r/a)m (m+1) 2] Ez(r,0) = (1/4) q (1/) (σd/εd) Rdc [4] Dim Ez = Cou/m m ohm-1/m m/far ohm/m = Cou/m 1/far = volt/m = correct Now we can write for a medium using (4.11.34), G = C (σd/εd ) => (σd/εd ) = G/C Then the results above become Ez(r,m) = (1/4) ηm q (1/) G/C Rdc [(r/a)m (m+1) 2] Ez(r,0) = (1/4) q (1/) G/C Rdc [4] or Ez(r,m) = (1/4) ηm (q/C) Rdc [(r/a)m (m+1) 2] Ez(r,0) = (1/4) (q/C) Rdc [4] or Ez(r,m) = (1/4) ηm V Rdc [(r/a)m (m+1) 2] Ez(r,0) = (1/4) V Rdc [4] or Ez(r,m) = (1/2) ηm V Rdc [(r/a)m (m+1)] Ez(r,0) = V Rdc where remember R is for the total transmission line. Dimension check: volt mho ohm/m = volt/m= correct Now let's consider this a different way. We know that as ω→0 we get Z0 = I = V/Z0 = V Jz = I/(πa2) = V/(πa2) Ez = Jz/σ = V/(σπa2) = V Rdc This agrees exactly with the Ez(r,0) term. But, we still have the Ez(r,m) term sitting there causing asymmetry. Comment: In my original work, I always had G = 0 and I got my asymmetric result for Jz and I argued that I → 0 so it did not matter for the infinite transmission line, you could sort of "sweep the asymmetry paradox under the rug". But now I have redone the analysis for G > 0, and this time you don't get I = 0, so this sweeping under the rug does not work. The uniform Jz school says one thing for Ez, while my theory says another thing for Ez, and they really do seem to disagree!! I cannot argue that it doesn't matter since I = 0. Note that in both cases G = 0 and G > 0, my Ez(r,m=0) term comes out exactly right!!! So the mystery continues with this better example. I now have a simple DC situation and both theories cannot be correct! Let's look at the other field components instead of just Ez . Go back to Second summary of the E field solutions : Rdc = β'2 = β2 - βd2 (D.2.33) dielectric conducts Ez(r,m) = (1/4) ηm ω q (1/βd) (ξd/εd)Rdc (aβ') fm fm = [ - ] Er(r,m) = (j/4) ηm aω q (ξd/εd)Rdc gm gm = [ + ] Eθ(r,m) = (1/4) ηm aω q (ξd/εd)Rdc hm hm = [ - ] and next we install fm = (r/a)m (m+1) (2/β'a) f0 = 4/(aβ') gm = (r/a)m+1 + (r/a)m-1 g0 = 2 (r/a) hm = (r/a)m+1 - (r/a)m-1 h0 = 0 to get Ez(r,m) = (1/4) ηm ω q (1/βd) (ξd/εd)Rdc (aβ') (r/a)m (m+1) (2/β'a) Er(r,m) = (j/4) ηm aω q (ξd/εd)Rdc [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (1/4) ηm aω q (ξd/εd)Rdc [(r/a)m+1 - (r/a)m-1] Ez(r,0) = (1/4) ω q (1/βd) (ξd/εd)Rdc (aβ') 4/(aβ') Er(r,0) = (j/4) aω q (ξd/εd)Rdc 2 (r/a) Eθ(r,0) = (1/4) aω q (ξd/εd)Rdc 0 or Ez(r,m) = (1/2) ηm ω q (1/βd) (ξd/εd)Rdc(r/a)m (m+1) Er(r,m) = (j/4) ηm aω q (ξd/εd)Rdc [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (1/4) ηm aω q (ξd/εd)Rdc [(r/a)m+1 - (r/a)m-1] Ez(r,0) = ω q (1/βd) (ξd/εd)Rdc Er(r,0) = (j/2) aω q (ξd/εd)Rdc(r/a) Eθ(r,0) = 0 Now for ω ≈0 we make the replacement, (ξd/εd) ≈ (- jσd/εdω) to get Ez(r,m) = (1/2) ηm q (1/βd) (- jσd/εd)Rdc(r/a)m (m+1) Er(r,m) = (j/4) ηm a q (- jσd/εd)Rdc [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (1/4) ηm a q (- jσd/εd)Rdc [(r/a)m+1 - (r/a)m-1] Ez(r,0) = q (1/βd) (- jσd/εd)Rdc Er(r,0) = (j/2) a q (- jσd/εd)Rdc(r/a) Eθ(r,0) = 0 or Ez(r,m) = (1/2) ηm q (1/βd) (- jσd/εd)Rdc(r/a)m (m+1) Er(r,m) = (j/4) ηm a q (- jσd/εd)Rdc [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (1/4) ηm a q (- jσd/εd)Rdc [(r/a)m+1 - (r/a)m-1] Ez(r,0) = q (1/βd) (- jσd/εd)Rdc Er(r,0) = (j/2) a q (- jσd/εd)Rdc(r/a) Eθ(r,0) = 0 Next as above we do this (1/βd) = (j/) which gives Ez(r,m) = (1/2) ηm q (j/) (- jσd/εd)Rdc(r/a)m (m+1) Er(r,m) = (j/4) ηm a q (- jσd/εd)Rdc [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (1/4) ηm a q (- jσd/εd)Rdc [(r/a)m+1 - (r/a)m-1] Ez(r,0) = q (j/) (- jσd/εd)Rdc Er(r,0) = (j/2) a q (- jσd/εd)Rdc(r/a) Eθ(r,0) = 0 The next step is to use σd/εd = G/C to get Ez(r,m) = (1/2) ηm q (j/) (- j G/C)Rdc(r/a)m (m+1) Er(r,m) = (j/4) ηm a q (- jG/C)Rdc [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (1/4) ηm a q (- j G/C)Rdc [(r/a)m+1 - (r/a)m-1] Ez(r,0) = q (j/) (- j G/C)Rdc Er(r,0) = (j/2) a q (- j G/C)Rdc(r/a) Eθ(r,0) = 0 or Ez(r,m) = (1/2) ηm q (1/) (G/C)Rdc(r/a)m (m+1) Er(r,m) = (1/4) ηm a q (G/C)Rdc [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (-j/4) ηm a q (G/C)Rdc [(r/a)m+1 - (r/a)m-1] Ez(r,0) = q (1/) ( G/C)Rdc Er(r,0) = (1/2) a q (G/C)Rdc(r/a) Eθ(r,0) = 0 or Ez(r,m) = (1/2) ηm (q/C) (1/) (G)Rdc(r/a)m (m+1) Er(r,m) = (1/4) ηm a (q/C) (G)Rdc [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (-j/4) ηm a (q/C) (G)Rdc [(r/a)m+1 - (r/a)m-1] Ez(r,0) = (q/C) (1/) ( G)Rdc Er(r,0) = (1/2) a (q/C) (G)Rdc(r/a) Eθ(r,0) = 0 or Ez(r,m) = (1/2) ηm V Rdc( r/a)m (m+1) Er(r,m) = (1/4) ηm a V (G) Rdc [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (-j/4) ηm a (V (G) Rdc [(r/a)m+1 - (r/a)m-1] Ez(r,0) = V Rdc Er(r,0) = (1/2) a V (G) Rdc(r/a) Eθ(r,0) = 0 Comment on the Radial Current Density. Somehow I expect that Er might be a function of θ just looking offhand at the problem. Outside the wires, you would expect E to be strongest between the conductors, so you would perhaps expect a larger current there! That then requires that some Er(r,m) terms be present! Question: Is there a simple way to solve the magnetostatics problem of the two cylinders? I can model the exterior problem using the ladder network and I know that for that problem, Z0 = => I = V/Z0 = V This is the total current going into the line at z = 0. I would expect this to decay since Im(β'd) = - = α so I then expect to have I(z) = V exp(-z ) But the exterior transmission line model tells me nothing about solution inside the wires. But I do know how to compute the exterior E fields for the two wires and from that I could compute Er just outside the wire surface and this will be basically Er(a,θ) = n(θ)/εd (confirmed). So the current density Jr(a,θ) just outside the wire is going to be asymmetric. If we apply this rule at the surface, as near D.2.23 div J = - jωρ -jω[∫V ρ dV] = ∫S J dS . (D.2.22) we get Jr being continuous through the surface! I already wrote this in D.9 (d) as Jr(a-α,θ) - Jr(a+α,θ) = 0 for ω = 0 Thus, just BELOW the surface, I know that Jr(a,θ) = σd Er(a,θ) = σd n(θ)/εd = (σd/εd) n(θ) (*) and there is your strong asymmetry. Does this agree with my theory ??? My theory above says: Ez(r,m) = (1/2) ηm V Rdc( r/a)m (m+1) Er(r,m) = (1/4) ηm a V (G) Rdc [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (-j/4) ηm a (V (G) Rdc [(r/a)m+1 - (r/a)m-1] Ez(r,0) = V Rdc Er(r,0) = (1/2) a V (G) Rdc(r/a) Eθ(r,0) = 0 which I can write at r = a as Ez(a,m) = (1/2) ηm V Rdc (m+1) Er(a,m) = (1/4) ηm a V (G) Rdc 2 Eθ(a,m) = (-j/4) ηm a (V (G) Rdc 0 Ez(a,0) = V Rdc Er(a,0) = (1/2) a V (G) Rdc Eθ(a,0) = 0 So the facts here are then Er(a,m) = ηm (1/2) a V G Rdc Er(a,0) = (1/2) a V G Rdc Using these two facts, I find that Er(a,θ) = !Syntax Error, I Er(r,m) ejmθ = (1/2) a V G Rdc !Syntax Error, I ηm ejmθ = (1/2) a V G Rdc/ N0 !Syntax Error, I Nm ejmθ = (1/2) a V G Rdc/ N0 n(θ) But N0= (1/2πa) q(0) so then Er(a,θ) = (1/2) a V G Rdc/ N0 n(θ) = (1/2) a V G Rdc n(θ) 2πa/q(0) = (1/C) G Rdc n(θ) πa2 = (G/C) (1/σ) n(θ) = (σd/εd) (1/σ) n(θ) and then Er(a,θ) = (σd/εd) n(θ) and this is the same as (*) above! So my quick little analysis of the Jr situation shows that my general theory model is correct in its statements of Er(a,m) and Er(a,0). This then lends credence to my full solution which is this Ez(r,m) = (1/2) ηm V Rdc(r/a)m (m+1) Er(r,m) = (1/4) ηm a V (G) Rdc [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (-j/4) ηm a (V (G) Rdc [(r/a)m+1 - (r/a)m-1] Ez(r,0) = V Rdc Er(r,0) = (1/2) a V (G) Rdc(r/a) Eθ(r,0) = 0 Remember that this is the exact solution of the Helmholtz equation and Maxwell equations with the limits of Appendix Q, so this HAS TO BE RIGHT! We really do end up with asym Ez in this case. In this limit, the CPBC just says that Jr is continuous at r = a, no big question about that.