Computation of Az for two cylinders case REVIEWED

Computation of Az for two cylinders case PhL 3.3.14 [ 5/14/14 final review done. No need to study this doc other than to read the summary. There are no field plots in this doc! ] [ Lots of Chapter 6.5 data in this doc. Proximity effect! A LOT happened in this doc. ] Summary: In Sections 1-8 I do the two-cylinder analysis which is now all written up in lines doc in Section 6.5, so no need to ponder these sections any more. Here is where I developed it. Section 9 just shows you can write Kz(φ) = δ Jz(a,φ) if you want the extreme skin current as a surface current Kz. Section 10 tries to compute Az from the Helmholtz integral and leads to a messy sinθ cosθ integral that is so ugly that I wrote a whole separate doc about it. This section is then a failure. Section 11 relates n(θ) to Jz (or Kz) and this is in lines doc now in Section 6.5. I have this here in a simpler form Jz(θ,z) = (vd/δ) n(θ,z) which now appears fancier in (6.5.21), but basically same idea. I refer to this equation as a "new continuity fact". In Section 12 I use -jωAz(x) = Ez(x) - jβdφ(x) with my newly found Ez(x) = Jz/σ to produce an expression for Az at the conductor surface, using also φ from the two-cylinder problem. I then get some expression for Az(a,θ) which is a constant plus a θ dependent term. Section 13 digresses on notational problems in bipolar.doc which I go fix Section 14 continues to study the new Az expression. I show that the θ-dependent 2nd term is negligible in the extreme skin regime, so Az= constant on perimeter and King is justified! This was a big deal for me when I first did it. I then ponder paradoxes 1,2 and 3 which now are all resolved. Section 15 on "hierarchy" is about different levels of a "good conductor" and this is now embodied in my low, strong and extreme δ limits of (3.7.3). which is non-constant on the surface, though I seem to ignore that fact. Recall the title of this doc "finding Az for two cylinders". In retrospect 1. Formula to use 1 2. Bipolar doc solution for φ 1 3. Review of the moments ηm 2 4. Use moments to compute Ez(r,φ) 2 5. Symmetry of fm and value of f0. 2 6. Final formula for Ez(r,φ). 3 7. What value of β' should be used? 4 8. Maple program to compute Ez(r,θ). 5 9. How to get surface current Kz from Jz 6 10. Attempt computation of Az 6 11. Getting Jz directly from n(φ) ? 10 12. Implications of this new continuity fact. 12 13. Dimension problems in bipolar doc 14 14. Resume on new Az expression 14 Reviewed this earlier, and then again on 5/9/14. There is a lot of good stuff here. Things happened here for the first time. 1. Formula to use Statement of the relationship between Az(x) and Ez(x) and φ(x) for wave solution. I will try to use this formula: E = - grad φ - ∂tA (1.3.1) Ez = -∂zφ - ∂tAz Ez = jβdφ - jω Az -jωAz(x) = Ez(x) - jβdφ(x) 2. Bipolar doc solution for φ Statement φ(x) for two-cylinders in bipolar coordinates. My bipolar doc has these results: φ(ξ) = - V (10.8) V2 = - V = - V < 0 V1 = - V = + V > 0 . (10.9) φ(ξ) = - V = - ln [] = - ln [] (10.8) (10.18) So I have the solution potential φ in lots of different forms. 3. Review of the moments ηm Statement of the charge moments for the two-cylinder problem. I need to put this somewhere! In "details of Chap 6" I compute from the above φ first E, then surface charge n, and then the moments and I find that ηm ≡ Nm/ N0 = (-1)m exp(-|m|ξ1) n(φ)/N0 = 1 + 2 Σm=1∞ηm cos(mφ) = 1 + 2 Σm=1∞ (-1)m exp(-|m|ξ1)cos(mφ) and I plotted this thing for ξ1 = 0.25 and it looked good. 4. Use moments to compute Ez(r,φ) Compute Ez from ηm for two-cylinder problem. Using these moments, I am able then to compute Ez 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 So I can insert my explicit moments to get Ez(r,m) = (1/4) I Rdc (aβ') { (-1)m exp(-|m|ξ1) [ - ] } Ez(r,φ) =!Syntax Error, I Ez(r,m) ejmφ = !Syntax Error, I ejmφ (1/4) I Rdc (aβ') { (-1)m exp(-|m|ξ1) [ - ] } 5. Symmetry of fm and value of f0. Show that f-m = fm. Question: Define: fm ≡ [ - ] Then f-m = [ - ] But Spiegel p 136 tells me that J-m(x) = (-1)mJm(x) m = integer which I recall from dim past. Then we get J-m(x) = (-1)m Jm(x) J-m+1 = (-1)m+1 Jm-1 J-m-1 = (-1)m+1 Jm+1 Then f-m = [ - ] = - [ - ] = fm which is a good result f-m = fm Also, while we are at it f0 = [ - ] = [ + ] = 2 6. Final formula for Ez(r,φ). State simplified sum for Ez in terms of the moments. Now go back to Ez(r,φ) =!Syntax Error, I Ez(r,m) ejmφ = !Syntax Error, I ejmφ (1/4) I Rdc (aβ') { (-1)m exp(-|m|ξ1) [ - ] } = (1/4) I Rdc (aβ') !Syntax Error, I ejmφ { (-1)m exp(-|m|ξ1) fm } To reflect the negative part of the sum, write Σm=-∞∞ ejmφ (-1)m exp(-|m|ξ1) fm = f0 + Σm=-∞-1 ejmφ (-1)m exp(-|m|ξ1) fm + Σm=1∞ ejmφ (-1)m exp(-|m|ξ1) fm = f0 + Σm=1∞ ejmφ (-1)m exp(-|m|ξ1) fm + Σm=1∞ e-jmφ (-1)m exp(-|m|ξ1) f-m = f0 + Σm=1∞ (-1)m exp(-|m|ξ1) fm [ejmφ + e-jmφ] = f0 + 2 Σm=1∞ (-1)m exp(-|m|ξ1) fm cos(mφ) . Then we have Ez(r,φ) = (1/4) I Rdc (aβ') [f0 + 2 Σm=1∞ (-1)m exp(-|m|ξ1) fm cos(mφ) ] = (1/4) I Rdc (aβ') { [2 ] + 2 Σm=1∞ (-1)m exp(-|m|ξ1) [ - ] cos(mφ) } = (1/4) I Rdc (aβ') { [2 ] + 2 Σm=1∞ ηm [ - ] cos(mφ) } = I Rdc { (xa/4) [2 ] + (1/2) xa Σm=1∞ ηm [ - ] cos(mφ) } = I Rdc { (xa/4) f0 + (1/2) xa Σm=1∞ ηm fm cos(mφ) } This is something I could plot! The r dependence is in x = β'r. 7. What value of β' should be used? Claim can set β' = β = (j-1) 10 (1/a) for 1/10th skin depth. From "current asymmetry issue" I have β = ej3π/4 (/δ) βd = βd0 [1 - j (1/2)tanL] βd0 = (ω/vd) The main observation is that βd is mainly real with a small negative imaginary part. Also, |βd| ≈ | βd0| = (ω/vd) I shows that if we are in the skin effect regime of δ = a/10 or better, then βd is about 1/22.5 of β in abs value. I could put in the exact stuff, but for the moment, I will assume β' = β. So then β' = β = ej3π/4 10 (/a) = (j-1) 10 (1/a) where I will work directly with δ = a/10 as a specific example. Perhaps make this a/N with variable N. 8. Maple program to compute Ez(r,θ). Maple plots of the sum for Ez derived above. See that Jz is indeed "asymmetric". The program was quick to write, I keep 10 terms, set δ = a/10 and here is a plot of the result, Here I keep 20 terms and plot Ez(a,θ), It sure looks like n(φ) !!!! Is there some more direct way to get from n(φ) to Ez(a,φ) ? See Section 11 below. You see that most current (Jz = σEz) is near the surface which is r = 10 as expected, and you see that the current is strongly peaked at φ = π, as I have always thought !! Recall how φ = θ is defined: Jz(r,φ) = σ (1/4) I Rdc (aβ') [f0 + 2 Σm=1∞ (-1)m exp(-|m|ξ1) fm cos(mφ) ] So angle θ = φ, and I am dealing with the left conductor where φ is CCW which I don't think is a problem. Now I should be able to use this Jz expression to compute Az. I have to assume this in each of the two conductors. This is likely to be a heavy duty numeric integration. Maybe I should first go extreme skin effect and try to make it a 1D integration. The above says Jz(r,φ) = σ (1/4) I Rdc (aβ') [f0 + 2 Σm=1∞ (-1)m exp(-|m|ξ1) fm cos(mφ) ] fm ≡ [ - ] f0 = 2 How do I turn the expo decay surface layer into a true surface current? [ Note added 5/9/14. I think I dropped the ball right here. I had just computed Ez above to be Ez(r,φ) == (1/4) I Rdc (aβ') { [2 ] + 2 Σm=1∞ ηm [ - ] cos(mφ) } where [] = f0. Perhaps nicer to write this as in my review of paradox a and 2, Ez(r,θ) = I Rdc (1/4) xa [ f0 + 2 !Syntax Error, I ηm fm(x,xa) cos(mθ) ] xa = β'a So I laboriously computed this Ez(r,θ) and I failed to continue on to get Az from: Ez = jβdφ - jω Az => jωAz = jβdφ - Ez where Ez as above and φ as φ(ξ) = - V = - ln [] = - ln [] (10.8) (10.18) which I could of course write n (r,θ) coordinates for the left wire, say, to match Ez's coordinates. But then we have I appearing in Ez and we have V appearing in φ and since V/I = Z0 this brings in Z0 which has a hazy evaluation although a clear definition here as shown. But I think what I would find if I carried through here is that Ez, though highly variable (as the plot above shows), is still always quite small (in the strong skin limit) and so Az would then basically follow φ according to jβdφ - jω Az = 0 or Az = (1/vd)φ. This subject is explored more in Review of Paradox 1 and 2.doc. Instead of computing Az as just described, below I start with Jz and try to get Az using the Helm integral! ] 9. How to get surface current Kz from Jz How to represent skin depth current Jz as a purely surface current Kz. Suppose Jz(r) = e-r/δ. Since δ is very small, we just treat this as a 1D problem and say K = !Syntax Error, I e-r/δ dr = δ Jz(0) So in our case this translates to Kz(φ) = δ Jz(a,φ) and there is our surface current which is a function of φ. 10. Attempt computation of Az Here I try to compute Az for the two-cylinder problem using the King Helmholtz integral method as developed in lines Ch 4, making use of a pure surface current in the extreme skin depth limit, and I end up with this rather intractable result: - Az12(x) = i(z) { !Syntax Error, I dθ1 a Kz(π -θ1) ln[x2 + y2 + a12 - 2xa1cosθ1 - 2ya1sinθ1] -!Syntax Error, I dθ2 a Kz(θ2) ln[(x-b)2 + y2 + a12 - 2(x-b) a1cosθ1 - 2ya1sinθ1] } Kz(θ) = δ Jz(a,θ) Jz(a,θ) = σ (1/4) I Rdc (aβ') [f0 + 2 Σm=1∞ (-1)m exp(-|m|ξ1) fm cos(mθ) ] with x→xa The required integral here is then !Syntax Error, Idθ cos(mθ) ln [A + Bsinθ + Ccosθ] I think I can do this by converting all to expo notation and then letting z = ejθ and doing it as a contour integral, after first doing parts. I am used to the King method giving horrible integrals. I show in a separate doc how to do this integral as a contour integral in Maple and for each m we get a simple analytic function for this integral, so in theory I can actually compute Az(x,y) at any point for the two cylinder problem, but I have not carried out that task. Analytically this requires some fancy dθ integrals Now consider from lines doc, Az12(x) = i(z) !Syntax Error, Idz'{ !Syntax Error, Idx1' dy1' b1(x1',y1') – !Syntax Error, Idx2' dy2' b2(x2',y2') } (4.9.1) R12 = (x-x1')2 + (y-y1')2 + (z-z')2 = s12 + (z-z')2 s12 = (x-x1')2 + (y-y1')2 (4.8.1) R22 = (x-x2')2 + (y-y2')2 + (z-z')2 = s22 + (z-z')2 s22 = (x-x2')2 + (y-y2') . Let's replace !Syntax Error, Idx1' dy1' b1(x1',y1') = !Syntax Error, I dθ1' a Kz(θ'1) Then we have Az12(x) = !Syntax Error, Idz' i(z') { !Syntax Error, I dθ1' a Kz(θ'1) -!Syntax Error, I dθ2' a Kz(θ'2) } where we have to be a little careful now. Recall again, The angle θ'1 is the θ shown on the left, and θ'2 is the θ shown on the right. Let's rewrite things this way Az12(x) = !Syntax Error, Idz' i(z') { !Syntax Error, I dθ1 a Kz(π -θ1) -!Syntax Error, I dθ2 a Kz(θ2) } where I am now using this only to show the two angles θ1 and θ2 But I need now a different picture in order to see R1 and R2 . But luckily I have this one Fig 4.1 where x is at some arbitrary location. I know that !Syntax Error, Idz' = - 2 ln(s1/Λ) (J.10) So also taking out i(z) in the usual way, we get Az12(x) = i(z) { !Syntax Error, I dθ1 a Kz(π -θ1) [- 2 ln(s1/Λ)] -!Syntax Error, I dθ2 a Kz(θ2) [- 2 ln(s2/Λ)] } where s12 = (x-x1')2 + (y-y1')2 s22 = (x-x2')2 + (y-y2')2 But now I have to draw a NEW transverse picture in order to understand these distances: Now x'1 = a1 cosθ1 y'1 = a1 sinθ1 x'2 = a2 cosθ2 + b y'2 = a2 sinθ2 So our distances are quite messy s12 = (x-x1')2 + (y-y1')2 = (x- a1 cosθ1)2 + (y- a1 sinθ1)2 s22 = (x-x2')2 + (y-y2')2 = (x- a2 cosθ2 - b)2 + (y- a2 sinθ2)2 So we then have - Az12(x) = i(z) { !Syntax Error, I dθ1 a Kz(π -θ1) ln(s1/Λ)2 -!Syntax Error, I dθ2 a Kz(θ2) ln(s2/Λ)2 } I know I can just set Λ = 1, so we still have this huge mess - Az12(x) = i(z) { !Syntax Error, I dθ1 a Kz(π -θ1) ln[(x- a1 cosθ1)2 + (y- a1 sinθ1)2] -!Syntax Error, I dθ2 a Kz(θ2) ln[(x- a2 cosθ2 - b)2 + (y- a2 sinθ2)2 ] } Kz(φ) = δ Jz(a,φ) Jz(a,φ) = σ (1/4) I Rdc (aβ') [f0 + 2 Σm=1∞ (-1)m exp(-|m|ξ1) fm cos(mφ) ] with x→xa Now maybe we can simplify a bit, s12 = (x-x1')2 + (y-y1')2 = (x- a1 cosθ1)2 + (y- a1 sinθ1)2 = x2 + a12cosθ12 - 2xa1cosθ1 + y2 + a12sinθ12 - 2ya1sinθ1 = x2 + a12 - 2xa1cosθ1 + y2 - 2ya1sinθ1 = x2 + y2 + a12 - 2xa1cosθ1 - 2ya1sinθ1 s22 = (x-x2')2 + (y-y2')2 = (x- a2 cosθ2 - b)2 + (y- a2 sinθ2)2 just replace x by x-b everywhere and 1→2 so s22 = (x-b)2 + y2 + a12 - 2(x-b) a1cosθ1 - 2ya1sinθ1 The general form is s2 = A + Bsinθ + Ccosθ so to do this stuff analytically we would need this integral !Syntax Error, Idθ cos(mθ) ln [A + Bsinθ + Ccosθ] I don't offhand see this in GR7, but Maple can do the integral for m = 1,2,3 and results are of a poly nature with increasing complexity of the result as m increases. I think a numerical approach would be better. Another approach would be to try this all directly in bipolar coordinates! Actually, I think that is going to work pretty well. 11. Getting Jz directly from n(φ) ? [I added the (j+1) factors below on 5.20.14] This is the big new realization for working in the skin depth limit: Jz(φ,z) = (ω/βd) n(φ,z)/δ = (vd/δ) n(φ,z) dim OK Kz(φ,z) = vd n(φ,z) dim OK This shows that indeed Jz is asymmetrical on the two-cylinder cross section since n is asymmetrical. Consider a Gaussian box where z is perpendicular to paper Imagine some Jz(r,φ) which characterize by Kz(φ,z) = δ Jz(a,φ,z) where I now show a z argument. [ Kz is not the Debye surface current, it is the total surf current ] The front and back sides (other sides added in later) of the above box give ∫J dS = [ Jz(φ,z+dz) - Jz(φ,z)] dw δ = -jω Qenclosed = -jω n(φ,z) dw dz [perhaps I am using the average values of Jz on front and back; well, as shown in "continuity at conductor surfaces", the integrals of Jz over the front and back sides creates 1/(j+1) due to the assumed skin depth model of Jx(y) = Jx(0)e-(j+1)y/δ so then [ Jz(φ,z+dz) - Jz(φ,z)] dw [δ/(j+1)] = -jω n(φ,z) dw dz ∂zJz(φ,z) [δ/(j+1)] = -jω n(φ,z) -jβdJz(φ,z) [δ/(j+1)] = -jω n(φ,z) βdJz(φ,z) [δ/(j+1)] = ω n(φ,z) Jz(φ,z) = (ω/βd) n(φ,z)(j+1)/δ = (j+1)(vd/δ) n(φ,z) dim OK Kz(φ,z) = (j+1)vd n(φ,z) dim OK Something seems fishy here. We know that n "really is on the surface itself" whereas Jz has some thickness to it, but you see the derivation here! This then explains why Jz(φ,z) looks like n(φ,z) when I plot it above. This simple result could be a major breakthrough for The Dim Wit. But what about the two other pairs of box faces? The top and bottom faces involve Er (area dwdz) while the left and right involve Eθ (area δ dz). If we include these faces, we then have ∫J dS = [ Jz(θ,z+dz) - Jz(θ,z)] dw [δ/(j+1)] + [ 0 - 0] dwdz + [ Jθ(θ+dθ,z) - Jθ(θ,z)] [δ/(j+1)] dz [ this is wrong because the top of the box is outside the conductor in the non-conducting dielectric and so the term shown in red above should be 0. Bottom of box is below skin layer so ≈ 0 there as indicated. ] This is then set equal to -jω n(φ,z) dw dz so we get -jω n(θ,z) dw dz = [ Jz(θ,z+dz) - Jz(θ,z)] dw [δ/(j+1)] + [0 - 0] dwdz + [ Jθ(θ+dθ,z) - Jθ(θ,z)] [δ/(j+1)] dz where write dw = adθ say. -jω n(θ,z) adθ dz = [ Jz(θ,z+dz) - Jz(θ,z)] adθ [δ/(j+1)] + [0 - 0] adθ dz + [ Jθ(θ+dθ,z) - Jθ(θ,z)] [δ/(j+1)] dz Now divide by dθ -jω n(θ,z) a dz = [ Jz(θ,z+dz) - Jz(θ,z)] a [δ/(j+1)] + [0 - 0] a dz + [ Jθ(θ+dθ,z) - Jθ(θ,z)]/dθ [δ/(j+1)] dz Then divide by dz -jω n(θ,z) a = [ Jz(θ,z+dz) - Jz(θ,z)]/dz a [δ/(j+1)] + [0 - 0] a + [ Jθ(θ+dθ,z) - Jθ(θ,z)]/dθ [δ/(j+1)] Then write as -jω n(θ,z) a = ∂zJz a [δ/(j+1)] + 0 a + ∂θJθ [δ/(j+1)] -jω n(θ,z) = ∂zJz[δ/(j+1)] + 0 + ∂θJθ (j+1)(δ/a) -jω n(θ,z) = -jβd Jz [δ/(j+1)] + 0 + ∂θJθ (j+1)(δ/a) = 0 + [δ/(j+1)] [∂θJθ (1/a) -jβd Jz ] Note added 3/6/14. The corrected version of the above is (but I have already corrected it in place) -jω n(θ,z) = [δ/(j+1)] [∂θJθ (1/a) -jβd Jz ] This is the cylinder version of Case 1 "deep box" from "continuity at conductor surface", which says: -jω n(x,0,z) = [δ/(j+1)] [∂xJx(x,0,z) + ∂zJz(x,0,z)] . Now, I have claimed that Eθ = 0 on the surface based on the idea that φ = constant on ring AND on the assumption that Ax = Ay = 0 (valid roughly for strong skin effect). This was one of my two Appendix D boundary conditions. If I assume this is true here not just at the surface but on average over depth δ (remember we have a deep box), then we must have ∂θJθ = 0, and then I get this simple result -jω n(θ) = [δ/(j+1)] [-jβd Jz(a,θ)] => Jz(a,θ) = (j+1)(ω/δβd) n(θ) = (j+1)(vd/δ) n(θ,z) This is the result I originally obtained, and then later decided was not valid, and now I think it is valid again! Typical of how things progress with me! [ 5.14.14. I fancied up the work above in lines doc, but the above result is basically correct. ] Question: Given the above, does that not alter my charge pumping boundary condition? Suppose in the above picture I add a certain Jr coming up from the bottom. Then I would have ∫J dS = [ Jz(φ,z+dz) - Jz(φ,z)] dw [δ/(j+1)] - Jr dw dz = -jω Qenclosed = -jω n(φ,z) dw dz This is then set equal to -jω n(φ,z) dw dz so we get -jω n(φ,z) dw dz = and then [ Jz(φ,z+dz) - Jz(φ,z)] dw [δ/(j+1)] - Jr dw dz = -jω n(φ,z) dw dz [ Jz(φ,z+dz) - Jz(φ,z)] [δ/(j+1)] - Jr dz = -jω n(φ,z) dz ∂zJz(φ,z) [δ/(j+1)] - Jr = -jω n(φ,z) -jβd Jz(φ,z) [δ/(j+1)] - Jr(a-δ,φ,z) = -jω n(φ,z) Now what? This is my new charge pumping condition! But, (βd δ) = 2π δ/λ = 2π(δ/λ) = very small So maybe this scales down the large Jz and makes it be a similar size to Jr and then we must keep both terms! But something is still wrong with this box! If I pancake it down a whole lot, that should exclude all of the Jz current but I don't see that happening here. Well, the Jr appearing in my last equations is really Jr(a-δ,φ,z) and this is still far from the surface. I think my charge pump condition then is still OK. Note added 3/6/14. I show in "continuity at conductor surface" Case 2 and Case 3 that the charge pumping BC survives all this swirling around, with or without Debye surface currents! So the last conclusion just stated here is correct, 12. Implications of this new continuity fact. Review of this section: Since Jz(φ,z) = (j+1)(vd/δ) n(φ,z) from previous section in skin limit, I know Ez(φ,z) = (j+1)(vd/δσ) n(φ,z) on the surface. Then from the E,φ,A equation I can compute Az everywhere on the surface and I end up with (using bipolar computation of n(θ) ) -jωAz(a,θ) = V1 [ jβd + (j+1)(vd/σδ) ε/(R1ξ1) ] which I will further process in Section 14 below. I have shown that in the skin effect limit, Jz(φ,z) = (j+1)(vd/δ) n(φ,z) dim OK from above and when I replace the z argument with the r argument this says, Jz(a,φ) = (j+1) (vd/δ) n(φ) = σ Ez(a,φ) Therefore Ez(a,φ) = (j+1) (vd/σδ) n(φ) But now go back to this equation, where we now take x to be a point on the conductor surface -jωAz(x) = Ez(x) - jβdφ(x) -jωAz(a,φ) = Ez(a,φ) - jβdφ(a,φ) But φ(a,φ) = V1 since I assume we are talking conductor C1. Then -jωAz(a,φ) = Ez(a,φ) - jβd V1 or -jωAz(a,φ) = (j+1) (vd/σδ) n(φ) - jβd V1 Dimension check: dim(LHS) = (1/sec)* (volt-sec/m) = volt/m from (1.3.1) dim(RHS2) = (1/m)*volt = OK dim(RHS1) = (m/sec)*(ohm-m)*(1/m) * (Cou/m2) = (1/sec)*(ohm) * (Cou/m) = Amp-ohm/m = V/m so OK So for the first time ever, I have an expression for Az(a,φ) on the surface of conductor C1 . I also know dQ1(ξ1,θ) = [dz dθ] . // dim RHS = (cou/m) m = cou, // (10.27) But φ(ξ) = - ξ , (10.16) so V1 = φ(ξ1) = - ξ1 => = -ε V1/ξ1 and then dQ1(ξ1,θ) = -ε dz dθ V1/ξ1 // dim(RHS) = (farad/m) m volt = coul But dQ1(ξ1,θ) = n1(θ) dA = n1(θ)R1dθ dz // 10.27 which says n1(θ)R1dθ dz = -ε dz dθ V1/ξ1 n1(θ)R1= -ε V1/ξ1 and then -jωAz(a,θ) = (j+1) (vd/σδ) n(θ) - jβd V1 = - (j+1) (vd/σδ) ε V1/(R1ξ1) - jβd V1 or jωAz(a,θ) = V1 [ jβd + (j+1) (vd/σδ) ε/(R1ξ1) ] So this is another first! I wonder what the relative sizes of the two terms might be? Dimension check: dim[first term] = dim(β) = 1/m dim[2nd term] = m /sec* m-ohm * 1/m * (farad/m)*(1/m) = 1 /sec* ohm * 1/m * (farad) = 1 /sec* sec * 1/m = 1/m 13. Dimension problems in bipolar doc I realized just above that I had some bad notation (wrong in fact) in bipolar doc, and I will fix that soon, [ DONE] the changes are listed in a separate working doc. It has to do with n versus Q. Let's just review a few equations: φ(ξ) = - ξ , (10.16) LHS = volts q = Coul/m ε = farad/m RHS = (Coul/m) /(farad/m) = Coul / Farad = volt so OK Next look at (10.11) n(ξ1,u) = ε // surface charge density on the left conductor C1 (10.11) RHS = (farad/m)*volts = Coul/m = linear surface charge, not 2D surface charge. WRONG I see I have to make some fixes to bipolar doc on this issue. I have n standing for both linear charge and for 2D charge density. // Wrong, no fixes were necessary, all dimensions are correct, and variable n is indeed charge/m2. I studied this carefully in a separate working doc. My mistake is on the above line where I forgot that h contains a! so RHS = (farad/m)*volts*(1/m) = Coul/m2 = correct for surface charge. OK, I will create an edit list for updating bipolar with cleaner notation. 14. Resume on new Az expression Here I simplify my Section 12 result for Az on the surfaces of the two cylinders. The result is the following: Az(R1,θ) = V1(1/vd) [ 1 - j (j+1) (δ/2R1) * (1/ξ1) ] The second term has the θ dependence. You see that in the skin depth limit (δ/2R1) << 1 and so the second term becomes small and we get - Az(R1,θ) = constant, consistent with King's assumption. jωAz(a,θ) = V1 [ jβd + (j+1)(vd/σδ) ε/(R1ξ1) ] vd = ω/k = ω/βd jAz(a,θ) = V1 [ j(βd/ω) + (j+1) (vd/ωσδ) ε/(R1ξ1) ] vd = ω/k = ω/βd jAz(a,θ) = V1 [ j (1/vd) + (j+1) (vd/ω)(1/(σδ)) ε/(R1ξ1) ] jAz(a,θ) = V1 [ j (1/vd) + (j+1) (1/(βdσδ)) ε/(R1ξ1) ] jAz(R1,θ) = V1 [ j (1/vd) + (j+1)ε/(βdσδR1ξ1) * ] δ ≡ The second term is the Ez term, by the way, while the first term comes from the φ term. Again, I wonder about the relative size of these two terms! Another dim check V1 / vd = volts sec/m = correct V1 ε / βd σ δ R1 = volt * (farad/m)*m*ohm-m*(1/m2) = volt * (farad/m)*ohm = volt sec/m So at least both sides have the same dimensions Now recall that R1 = a/|shξ1|, but that does not buy me much and then two meanings for a! Here is the ratio of the two term coefficients rat = ( first/second) = first * (1/second) = δ2 = 2/(ωμσ) ωμσ = 2/δ2 rat2 = Now replace δ2 = to get rat2 = = = = = = = 2μωσR12 = 2R12 (μωσ) = 2R12 2/δ2 = 4 R12 / δ2 = (2R1/δ)2 Amazing. Then rat = (2R1/δ) => (2nd coeff) = (first coeff) (1/rat) = ( 1/vd) (δ/2R1) so it would seem that jAz(R1,θ) = V1 [ j (1/vd) + (j+1)ε/(βdσδR1ξ1) * ] = V1 [ j (1/vd) + (j+1) ( 1/vd) (δ/2R1) * (1/ξ1) ] = V1(1/vd) [ j + (j+1) (δ/2R1) * (1/ξ1) ] Az(R1,θ) = V1(1/vd) [ 1 - j (j+1) (δ/2R1) * (1/ξ1) ] (*) Now the relative sizes of terms is clearer. The second term is largest when cosθ = -1 so we have Az(R1,π) = V1(1/vd) [ 1 - j (j+1) (δ/2R1) * (1/ξ1) ] Now for a very excellent conductor, and for some fixed ξ1 , we argue that (δ/2R1) << 1 and then the second term goes away in (*) and we end up with Az(R1,θ) = V1(1/vd) which is then a constant around the perimeter!!! I am surprised at this result. [ 5/14/14: but in lines doc this is now stated as Step 1 (3.7.5) for extreme skin effect, so I was basically on track.] [ So I have shown that for the extreme skin regime, the second term is tiny and then Az = constant on the surface. This is correct. ] Go back a bit to -jωAz(a,φ) = Ez(a,φ) - jβd V1 The claim is that, in the extreme skin depth regime relative to wire radius, although Ez(a,φ) has a rather violent shape going around the perimeter of the circle and is definitely not constant, its overall scale is so small that we can neglect this term relative to the last term so we get -jωAz(a,φ) ≈ -jβd V1 Az(a,φ) ≈ -(βd/ω) V1 = (1/vd) V1 and then King's claim that Az is constant on surface comes back to life! [ correct ] What about the two Paradoxes? They are on the white board, and I think I see how they resolve!! But each one will need a write up. [ I have now done this write-up in "the current asymmetry issue" ] Paradox 3. If Az is constant round the perimeter, we know that a B field line must match the circle, and then there is no normal B component. But it seems that B strength would be strongest where Jz was strongest, so you would think the B strength varied, and therefore there is going to be some Hn which seems a contradiction. But perhaps this resolves the same as Paradox 1: [ I am assuming in the above paragraph that "B varies around circle => Hn ≠ 0". But in the extreme skin effect limit, I show in "the current asymmetry issue Section 6 that even if B varies on the circle, you still have Hn = 0.] (1) Az is very close to constant on the circle, but not exact. Therefore, the B contour around the circle is not exact, so the actual contour is a slightly distorted circle, like the red B field line shown here, I am just guessing this is now it is distorted. So the point is that yes, there really is some Hn component where we are distorted. But I claim the largest Hn should occur where ∂θHθ is largest. Well, OK, that gradient is 0 at the max point, so I think we are in like flint here. A lot of stuff happened today!!! 15. Approximation Hierarchy (a) When dealing with a "perfect conductor" one is in the extreme skin depth limit and Ez = 0 inside the conductor. In this case the EAφ equation just says -jωAz(x) = Ez(x) - jβdφ(x) = - jβdφ(x) Strictly this applies inside the conductor, but I think there would also be no Ez field outside the conductor either. In this limit then we have -jωAz(x) = - jβdφ(x) ωAz(x) = βdφ(x) Az(x) = (βd/ω) φ(x) βd0 = (ω/vd) Az(x) = (1/vd) φ(x) This says that Az(x) is just a scaled copy of φ(x) and this is King-compatible. (b) Recall the King gauge condition β2 = μεω2 - jωμσ = ω2μ ( ε - jσ/ω) = ω2μ ξ ξ ≡ ε - jσ/ω . (1.5.1) div A = - μεjωφ - μσφ = -jωμ(ε+σ/jω)φ = -jωμξφ = -j(β2/ω) φ King gauge (1.5.5) In the dielectric, and ignoring transverse Ai, this last says ∂zAz = -j(β2/ω) φ or -jβdAz = -j(βd2/ω) φ or Az = (βd/ω) φ = (1/vd) φ(x) This is the same as my part (a) result! Conclusion: The assumption of a perfect conductor (extreme skin effect) results (EAφ equation) in Az(x) = (1/vd) φ(x) everywhere in space since Ez= 0 in this case. This is the same result as obtained from the gauge condition where you throw out the transverse contributions ∂xAx + ∂yAy. Perhaps the extreme skin effect limit squashes both Eφ and Er since β→∞, and then there are no transverse currents, and then there is no transverse Ai ! I think this makes some sense. [ 5.14.14: All of this conclusion is now to be found in lines doc in various places. ]