The current asymmetry issue REVIEWED

The Current Asymmetry Issue PhL 3.2.14 [ Final review 5.14.14. Added summary below. (23 pages, long doc) ] Summary In Section 1 I quote the Appendix D E fields and examine values of β,β' and βd. In Section 2 I make δ graphs of Ez(r,m)/ Ez(a,m) which shows things about same for various m. In Section 3 I explain why all Ei field components have the same expo δ decay. In Section 4 I show that |fm| ≈ 2 for various m and so things don't depend much on m. Later in lines doc (6.5.22) I show that in fact fm = -2j in the extreme skin limit. Section 5 is Paradox #1: ηm implies Ez(a,θ) on surface, so how can Az be constant? Section 6 is early realization that Br = 0 on surface in skin regime and then Az = constant Section 7 is Paradox #2: King gauge says Az = (1/vd) φ so how explain Az ≠ constant at DC? Section 8 continues Section 7 on Br = 0 on surface in skin regime. Section 9 continues the study of Paradox #2 with the King gauge. Section 10 wonders whether φ = constant quasi-static is valid at all ω and questions low ω. This same questioning now exists in line doc Section 3.7 where I could only prove φ = constant for large ω. Section 11 resolves paradox #1 by saying φ = constant and Az = almost constant. I think now I would say that both are "almost" constant. Section 12 resolves paradox #2 with same notion that Az = almost constant. In Section 13 I show Az ≠ constant for DC twin lead with picture. I start thinking here about ω → 0 of the theory. I float the idea of a separate AzAC and AzDC but reject it. I then take low ω limits of the Appendix D fields. I find that the m=0 does the right thing, but to my surprise m > 0 fields don't vanish. This is a problem since I wanted to see Jz have no asymmetry at DC! So now I have current asymmetry down to ω = 0 and this violates my eddy current ideas. so a big new Paradox. This led to a whole new folder of docs on this subject! My current resolution is that the general theory does not apply at low ω, and of course the escape hatch idea that I = 0 since Z0 = ∞. But here is where I first discovered this problem. Comments added in red for each section on 3.4.14. A lot is going on in this doc. 1. Quoting the round wire E field partial wave results and β, βd and β' 1 2. Study of the Ratio Ez(r,m)/ Ez(a,m) and Er(r,m)/ Er(a,m) 4 3. Why do all components have same skin effect? 6 4. Comparison of fields for different m values 8 5. Paradox #1 Brewing 8 6. Review of H being tangent to the surface in skin regime: 10 7. Paradox #2 with the King Gauge 11 8. A Review of the Idea that Hn = 0 at a surface in the skin effect limit. 12 9. Review of Paradox #2 of the King Gauge 14 10. Residual Question on Quasi Static 16 11. Resolution of Paradox #1 (added 3/5/14) 16 12. Resolution of Paradox #2 (added 3/5/14) 16 13. But why is Az largely non-constant on conductor at DC? Paradox 3. 17 1. Quoting the round wire E field partial wave results and β, βd and β' Here I quote the results and then study the meaning of the various β one might use in these formulas Consider the E field solution of Appendix D, including the two assumed boundary conditions. Second summary of the E field solutions : Rdc = β'2 = β2 - βd2 (D.2.33) Ez(r,m) = (1/4) ηm I Rdc (aβ') [ - ] a = radius ηm ≡ Er(r,m) = (j/4) ηm I Rdc (aβd) [ + - ] x = β'r Eφ(r,m) = (1/4) ηm I Rdc (aβd) [ - + + ] xa = β'a What can be said about this partial wave solution in the Skin Effect Limit? In that limit, δ < (a/10), say. Comments about β First, recall β = ej3π/4 (/δ) and β2 = -2j/δ2 . (2.2.21) β2 = μεω2 - jωμσ = ω2μ ( ε - jσ/ω) ≈ -jωμσ . (1.5.1) In the conductor, we have assume in some sense that σ/ω >> ε. This is discussed below (2.2.2) and we conclude that for copper this approximation is valid up to 109 GHz, so that is really not an issue. So, we then have β = ej3π/4 (/δ) with δ < (a/10) |β| = (/δ) > 10 (/a) Notice that this is not particularly "huge" as I keep saying in lines doc, at least for this δ. My interest now is the radial dependence of the above E field solution components. Suppose I just make this exact identification for the moment, β = ej3π/4 10 (/a) Comments about βd Now βd2 = ω2μd ( εd - jσd/ω) If we include loss tangent stuff, then εd = εd' [1 - j tanL] but an easier way to think of this is βd2 = ω2μd ( ε'd - jσd,eff/ω) σd,eff = σdωε'd tanL The only real adjustment here is to realize that although you may have σd = 10-15, you will still have σd,eff = σdωε'd tanL ≈ ωε'd tanL ≈ ω * [2 *10-11] [ 2 *10-4 ] ≈ 8π f 10-15 So at f = 100GHz = 1011 we have σd,eff = 25 1011 10-15 = 25 10-4 = 0.25 * 10-2 So this is much larger than 10-15, but much smaller than copper's 107. Now back to βd : βd2 = ω2μd ( ε'd - j[σdωε'd tanL]/ω) ≈ ω2μd ( ε'd - j[ωε'd tanL]/ω) = ω2μd ( ε'd - jε'd tanL) = ω2μd ε'd ( 1 - j tanL) Another way to say this is shown in (3.3.7) ξ = ε'( 1 - j tanL) β2 = μεω2 - jωμσ = ω2μ ( ε - jσ/ω) = ω2μ ξ = ω2μ ε'( 1 - j tanL) So I am just making sure the result is correct. Here it is again: βd2 = ω2μd ε'd ( 1 - j tanL) Now consider ( 1 - j tanL)1/2 ≈ [1 - j (1/2)tanL] since tanL ~ 10-4 Then we get βd = ω [1 - j (1/2)tanL] = (ω/vd) [1 - j (1/2)tanL] and you might then say 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) Ratio of the two β's To summarize our main results β = ej3π/4 (/δ) ≈ ej3π/4 10 (/a) for δ = a/10 β = (j-1)(1/δ) βd = βd0 [1 - j (1/2)tanL] and then |β| = (/δ) ≈ 10 (/a) for δ = a/10 |βd| ≈ (ω/vd) = βd0 = 2π/λd Now the Transmission Line Limit requires λd > 10a lets say, so then |βd| = 2π/λd < 2π/(10a) So here we have a "reasonable" skin effect situation with a "reasonable" TL limit. In this case |β|/ |βd| = 10 (/a) * 10a/(2π) = 100 () * /(2π) = 100 / (π) = 22.5 Now this is the situation in which I want to study the E fields in the box above. Note added 3/6/14: I want to write this ratio in some other ways. 1. Here is one: |β|/ |βd| = (/δ) / 2π/λd = (λd/δ) δ ≡ which you can see is very small for normal values of δ. 2. Next, write λ = 2πv/ω and δ ≡ to get |β|/ |βd| = (λd/δ) = * 2πv/ω * / = v/ω = vd (1/) 3. Now we can write = 1/v and then = (1/)(1/v) so we get another form |β|/ |βd| = and |βd|/ |β| = This says that at low ω the second ratio is small, and at ω = 0 is zero, but as ω increases, the ratio increases. I think it increases to about 10-3 for copper at 1000 GHz. I think it is OK to approximate β' ≈ β given this factor of 22.5. Thus, the discussion of βd above will play no role, once we know it is small enough to be ignored in this context. 2. Study of the Ratio Ez(r,m)/ Ez(a,m) and Er(r,m)/ Er(a,m) I show that you get about the same skin effect decay in all the partial waves both for Ez and Er and obviously also for Eφ. So all components of the E field have the same skin effect. So what do the Bessel functions look like? It is quite complicated. Start with Ez and write Ez(r,m)/ Ez(a,m) = [ - ] / [ - ] x = βr xa = βa I am mainly interested in the abs value of this equation to see hopefully the skin effect at work. Rewrite as = = * = So this ratio is simpler than I thought it was. In (2.2.22) we had the m = 0 version of this ratio! E(r) = E(a) . (2.2.22) So let's try a Maple plot. The values m = 0,1,2,3,4,5 are almost identical, but the corner moves slightly to the right as m increases. The first plot shows this more dramatically. Fact: For Ez, all the partial waves exhibit the same characteristic skin effect decrease away from the surface, though higher m values give faster decays. Let's now try this for Er : = I don't see a simple way to reduce this. I wrote up the Maple and made the plots, and the two plots look exactly the same! Why is that? \ 3. Why do all components have same skin effect? I show that the reason for same skin effect of all components is that all have the same Bessel large argument behavior with same phase. I show that all square brackets in (D.2.33) above have same scale in skin effect, and therefore the relative size of the E field components are in fact determined by the leading factors and this is true for all m. Thus, the idea that Er and Eφ are scaled down from Ez by (βd/β) is valid in all partial waves m. Just confirming something I always presumed was true. Let's consider a single Bessel function like Jm(x) with x = βr and β = ej3π/4 10 (/a) For the range of r close to a, the argument is on the order of xa = βa = ej3π/4 10 (/a) a = ej3π/4 10 () = 14.1 ej3π/4 which perhaps is a "large argument" in this situation. Here is some math on that subject: [ Ref? ] Jm(x) = (2/πx)1/2 cos(x-mπ/2-π/4) β = (j-1)(1/δ) x = βr = (j-1)(r/δ) "Near" r = a, x is "large" so we can use this large x expansion. then cos = (1/2)exp[-j {x-mπ/2-π/4}] = (1/2)exp[-j {(j-1)(r/δ)-mπ/2-π/4} ] = (1/2)exp[+ (r/δ) + j(r/δ)+jmπ/2+jπ/4 ] = (1/2)e+r/δ ej[(r/δ)+mπ/2 + π/4] We ignore the other expo term since it blows down instead of up. So then Jm(x) = (2/πx)1/2 (1/2) e+r/δ ej[(r/δ)+mπ/2 + π/4] // assumes large x limit So this is the leading term for large x and it has weak m dependence. Now the coefficient in the next lowest term is proportional to m2 so very large m, that next term will play a role, and this explains why the Ez curves are different for large m values. Now if you ignore the difference in the m's, we have Er(r,m) = ~ [ + - ] ≈ = (2m/x) and then = (a/r) so for r ≈ a this explains the similarity in the curves more or less. I think I can assume that Eφ will have a similar decay pattern. We just change the sign of the first term. // Yes, curves look about the same. Fact: All the partial waves (all m) and all components exhibit the same skin depth behavior. Fact: The Relative Size between components is determined by the leading factors. We see that the square bracket quantities in (D.2.33) all look about the same in the Skin Depth Limit as functions of r coming in from the surface, that is what I showed above. Therefore, the relative size of components is determined by those leading factors for any m. Restating: Fact: For all m, Er and Eφ are both smaller than Ez by amount (βd/β) ≥ 22.5 for Skin Depth regime. As shown in (D.6.1), this is also true for m = 0. 4. Comparison of fields for different m values I show that the E field square bracket factors are more or less independent of m in terms of scale. Thus, the scale of contributions is really determined by the ηm moments. Now go back to Ez(r,m) = (1/4) ηm I Rdc (aβ') [ - ] If we ignore the ηm dependence here, and since curves versus r have same general shape, what can we say about |Ez(a,m)| as a function of m ? This says basically that the lowest 10 partial waves have about the same size at r = a, so you cannot in any way claim that somehow larger m waves are intrinsically weaker in our skin effect regime. 5. Paradox #1 Brewing Paradox #1 is this: Using the ηm obtained from n(θ), I can see from the general discussion above (without doing the calculation in detail) that Ez(r,θ) is going to have "strong θ dependence" similar to n(θ). Since Ez(θ) = +j βd φ(θ) - jωAz(θ) and since φ is constant on the periphery, it suggests that Az is therefore NOT constant on the periphery, and this then does not support the King approach in Ch 4 even in the strong skin effect regime. [ I agree on 5/7/14.] At the time, I was really torn between (a) Ez is constant on periphery; (b) Ez strongly varies on periphery something like n(θ). ALSO, if Az is not constant on the periphery, then H = Hθ is not true on the periphery, and I investigate this aspect in the next section. This seems to go against web pictures I have seen, for example. I found in "details of Chap 6 exact" that for the two cylinder problem, the moments are ηm ≡ Nm/ N0 = (-1)m exp(-|m|ξ1) and for ξ1 = 0.25 I got these numbers and I noted the alternating signs. The decay is rather slow, so there are lots of partial waves active, and I showed how this is needed to regenerate the surface charge function n(φ) in this case. Now if I use these ηm values in my Ez(r,m) expression, Ez(r,m) = (1/4) ηm I Rdc (aβ') [ - ] I find that the ηm decay slowly in magnitude, while the [..] factor does not decay with m near r = a. I also showed that n(φ)/N0 = 1 + 2 Σm=1∞ηm cos(mφ) = 1 + 2 Σm=1∞ (-1)m exp(-|m|ξ1)cos(mφ) Therefore, near φ = 0, cos(mφ) = 1 and we get cancellation between the partial waves, and so I expect the sum of partial waves Ez(a,φ) to be SMALL at φ = 0 due to these cancellations. On the other hand, at φ = π, all terms have the same sign, and I expect to have Ez = large. This case study suggests therefore that Ez(a,θ) varies going around the cross section perimeter if we are in the skin effect regime (where I could claim that [...] does not vary much with m ). Now go back to our old friend, E = - grad φ - ∂tA . (1.3.1) Ez = +j βd φ - jωAz I still maintain φ = constant around the perimeter. In the skin effect limit, I thought I was going to obtain the result that Az was also constant on the perimeter, and that would imply that Ez was constant on the perimeter. But in my case study, I find that Ez has large variation between φ = 0 and φ = π, so we do indeed have a Paradox here. 6. Review of H being tangent to the surface in skin regime: The gist here is that my Stokes and Gauss box rules with skin effect say that on the periphery, H = Hθ and this seems to agree with the idea of Az = constant on periphery. But previous section seems to say Az is not constant because Ez is not constant, so confusion continues. 1. Draw Stokes loop around skin layer and use Ampere integral form to conclude that Ht = Hθ = Jzδ = Kz and this is the result everyone quotes. It is based on H = 0 inside skin layer Claim can ignore the short portions of the Stokes loop. [ OK ] Note added 3/5/14. The above is based on curl H = ∂tD + J and C H ds = ∫S [∂tD+J] dS for a rectangular thin loop which is "sideways to the surface" so that dS = dS (always have a picture!!!). Thus, the displacement term involves jωDz = jωεEz. As in lines doc (3.4.1), we know we can always ignore this term compared to the conduction current term Jz = σEz. The short sides of the loop pick up Hn but only contribute if Hn = ∞. Note that we are NOT claiming Hn = 0 in this step, only Ht = Kz . 2. Draw Gaussian box but for magnetic. Conclude that Hn = 0, again based on H = 0 inside skin layer. Claim can ignore thin side surfaces of the box. [ true as δ → 0, or E ≡ 0 inside. ] 3. The conclusion is that, in the extreme skin effect regime, the H field is tangent to the surface is Ht = Kz . 4. Therefore, there is an H field line which is an exact circle going around a round conductor which is operating in a transmission line in the skin limit. [ In this limit, E ≈ 0, Hn ≈ 0 ] 5. There is no claim being made that |H| = Ht is constant going around the perimeter !!!! [ correct ] If there is current asymmetry, then we expect this to vary, and our example above strongly suggests there IS a current asymmetry. [ We expect Ht(θ) = Kz(θ) and Kz(θ) generally varies with θ . ] 6. Here is why I think H = Hθ on periphery means Az = constant on periphery: [ correct, see next ] Align a thin Stokes loop which is long in the z direction and such that the B flux through the loop is 0. The thin edges of the loop are along B field lines. Along the long loop edges, we must have Az1 = Az2 from our B = curlA Stokes law. Thus, even though the B field strength might vary along the B field lines in this area, Az is still a constant going along the field line. [ correct ] Similarly, in electrostatics, if you move along a φ = constant line in 2D, the E field might vary as you move along the equipotential line. [ correct analogy ] 7. Paradox #2 with the King Gauge King Gauge says Az(x) = constant * φ(x) if you ignore Ax and Ay. This conflicts with my DC plots of Az contours which do not match those of φ at surface. So it must be that at low frequency, you cannot ignore Ax and Ay. I give a simple reason why this might be so. The King gauge condition says (applying this in the dielectric right now) div A =-j(βd2/ω)φ If we ignore ∂xAx and ∂yAy then this says ∂zAz = -j(βd2/ω)φ and if Az has our usual wave form in z, this says -jβdAz = -j(βd2/ω)φ // βd0 ≈ βd = (ω/vd) ignoring loss tangent or Az(x) = (βd/ω)φ = (1/vd) φ(x) where I suppose x can be any point in the dielectric, including right at the conductor surfaces. This equation has several problems: Paradox 2: This says Az(x) = constant * φ(x) . This says that at some very low frequency ω, since circle has constant φ, circle must also have constant Az which I know is not true, having drawn the field lines [ It is correct that the B field lines do not match the periphery circles in my DC plots! ] Possible way out: At low frequency where there is little or no skin effect, there are Jr and Jφ currents inside the conductor, and these do create Ax and Ay, and things vary with x and y, and the way out would be that at low frequency, you cannot ignore ∂xAx + ∂yAy. As you move into strong skin effect, the ratio β'/β gets small, and then those Jr and Jφ go away! 8. A Review of the Idea that Hn = 0 at a surface in the skin effect limit. I show here that Hn = 0 for perfect conductor (δ → 0) , but not for an imperfect one. I am thinking of a piece of a round conductor and imagine that along the surface, the B field weakens as one moves away from the hot spot between the conductors. One could at least imagine this happening. So let's postulate that there exists an H field for which Hz = 0 (as apropo for a cross section like this), but might have some general Hx and Hy : [ ∫box H dS = 0 since div B = 0. ] This drawing shows a blowup of a piece of the round conductor surface in cross section. The red box has width dx and it has depth dz into the plane of paper. The height is skin depth δ. Assume now that there exists some Hx in the region of the box which varies with x. Then since div H = 0, Gauss's law applied to this box says (I add up all the outflows from the box) the total flux out of the box must be zero: Let's look at all surfaces of our gaussian box. We assume no action in the plane of paper, so the near and far sides which involve Hz we just ignore assuming Hz = 0. In terms of left and right, consider the right face. There we can say (think of y axis pointing down for the moment and x pointing to the left! ) ∫leftface H dS = - !Syntax Error, Idydz Hx(x,y) = - dz Hx(x,y=0) !Syntax Error, Idy e+y/δ = - dz Hx(x,y=0) !Syntax Error, Idy' e-y/δ y' = -y ≈ - dz Hx(x,y=0) !Syntax Error, Idy' e-y/δ = - dz Hx(x,y=0) δ ∫rightface H dS = + dz Hx(x+dx,y=0) δ // plus since face pointing right Meanwhile, the bottom face contributes nothing since Hn = 0 there and top face makes ∫topface H dS = dzdx Hy(x,y=0) We add the face contributions and set to 0 since div B = 0, - dz Hx(x,y=0) δ + dz Hx(x+dx,y=0) δ + dzdx Hy(x,y=0) = 0 Divide by dz to get - Hx(x,y=0) δ + Hx(x+dx,y=0) δ + dx Hy(x,y=0) = 0 Divide by dx to get δ [Hx(x+dx,y=0) - Hx(x,y=0)]/dx = - Hy(x,y=0) or δ ∂xHx(x,y=0) = - Hy(x,y=0) Now Hy = Hn (normal) at the surface, so we conclude that Hn(x,y=0) = - δ ∂xHx(x,y=0) Had we included the z-sides of the box, we would have gotten ∫frontface H dS = + dx Hz(z+dz,y=0) δ ∫backface H dS = - dx Hz(z,y=0) δ Then div H = 0 would have said - dz Hx(x,y=0) δ + dz Hx(x+dx,y=0) δ + dzdx Hy(x,y=0) + dx Hz(z+dz,y=0) δ - dx Hz(z,y=0) δ = 0 Divide by dz to get - Hx(x,y=0) δ + Hx(x+dx,y=0) δ + dx Hy(x,y=0) + dx δ [Hz(z+dz,y=0) - Hz(z,y=0)]/dz = 0 Then divide by dx to get δ [Hx(x+dx,y=0) - Hx(x,y=0)]/dx + Hy(x,y=0) + δ [Hz(z+dz,y=0) - Hz(z,y=0)]/dz = 0 or δ ∂x Hx(x,y=0) + Hy(x,y=0) + δ ∂zHz(x,y=0) = 0 so Hy(x,y=0) = - δ ∂xHx(x,y=0) - δ ∂zHz(x,y=0) or Hn(x,y=0) = - δ [∂xHx(x,y=0) + ∂zHz(x,y=0)] Now, if we go to the extreme skin effect limit where δ → 0, THEN we really do get Hn(x,y=0) = 0. Conclusion: Even if the H field at the surface varies in the transverse directions, in the extreme skin effect limit δ → 0 and we have Hn = 0 just outside the surface . This conclusion is not true of course as we approach the limit, only at the limit. Now recall Paradox #1 which consists of two logic threads. The first thread looks at the two cylinders case and explicitly calculates that Ez is not constant around the round conductor. The second thread is this: (a) skin limit => H tangential => H lines match surface => Az = constant on surface (b) φ = constant on surface as assumed in capacitor problem (c) Ez = -∂zφ + ∂tAz => Ez = jβdφ + jωAz (a) + (b) + (c) => Ez = constant on surface => Jz is symmetrical all around the conductor This paradox however has a Resolution as follows: If we have something like δ = a/10, we are in the skin effect regime, but not in the extreme limit. In this case, Ez varies with θ violently for our two-cylinder problem, but is always very small in magnitude. Thus we have jωAz(x) = - jβdφ(x) + Ez(x) and we conclude that Az(x) is very close to being constant on the surface, but not quite due to the small Ez. In terms of the Paradox statement above, what we are saying is that step (a) is not true with small but finite δ. 9. Review of Paradox #2 of the King Gauge Here I am wondering if ∂xAx + ∂yAy not small provides an escape hatch from this gauge paradox. I realize that I might be above to compute all of A for the two cylinder problem and this leads to a new doc on that subject. But there I find the calculation is too hard to do in a short time. Paradox #2 seems more fundamental and scary. Recall that if we ignore the At contributions, we get Az(x) = (1/vd) φ(x) (*) In our capacitor two-conductor analysis we have φ(x) constant around perimeter and we know that Az(x) is not constant on the perimeter, so (*) is blatantly violated in this example. [ correct ] So consider with the two-cylinder prototype example, E = - grad φ - ∂tA . (1.3.1) Ex = -∂xφ - ∂tAx Ex = -∂xφ - jωAx In my capacitor problem, I have ω = 0 and so this really does say Ex = -∂xφ and that is how the electrostatics capacitor problem works. But for ω > 0, we have the extra Ax term to worry about. We have to argue that ωAx ≈ 0 to justify the part of our capacitor problem transmission line solution which says that Et = -tφ (from which I get n(θ) etc etc). I do believe Ax really is small, but it is not zero because there are currents other than Jz . For example, inside both wires there is in fact a small Jx and then the King integral gives some small Ax. This is a "quasi-static" approximation somehow. So here is the question: if Ax is very small, you would think ∂xAx would also be small. Well I guess this has to be true: The King gauge condition says, div A = -j(β2/ω) φ Therefore ∂xAx + ∂yAy + ∂zAz = -j(β2/ω) φ ∂xAx + ∂yAy -jβd Az = -j(β2/ω) φ (∂xAx + ∂yAy)(x) = -j(β2/ω) φ(x) + -jβd Az(x) So this is a "formula" for the quantity on the left that we are wonder if is "small". Since in our example we think we know all about φ(x) So is there some way to check this equation for two-cylinders? Question: How can we compute Az for the two-cylinder problem in skin effect regime? I know that Ez = -∂zφ - ∂tAz Ez = jβdφ - jω Az -jωAz(x) = Ez(x) - jβdφ(x) OK, I think I know both φ and Ez for this problem. I know φ everywhere for sure. I get the moments ηm and then I know Ez from my Appendix D formula for Ez. So maybe I really need to go "work through" this problem so I at least have one example to look at. It will all assume that Eφ = 0 at the surface, however, but with small Ax that is the same as φ = constant on surface ring. 10. Residual Question on Quasi Static What happens to the quasi-static as ω increases from 0? Consider the two-cylinder problem. I compute φ using quasi-static capacitor. I compute static Et = -tφ and from that I compute n(θ) and the moments ηm. But as we ramp up ω above ω = 0, we then have Et = -tφ -jωAt It is incumbent upon "the writer" to show that as ω ramps up, one can neglect this extra term -jωAt. If one can show this, then the n(θ) distribution stays at its basic electrostatic form. I suppose it is possible that, as ω ramps up and we move away from electrostatics, φ itself changes away from its electrostatic functional form. I do think that, if we were to ramp up σ of the conductor to force more into the skin effect with ω = constant, then At → 0 because Jt → 0. I really do think King's stuff is valid for the skin effect or perfect conductor regime, but I have not yet assembled the pieces of this claim. [ but this is now all written up in lines doc Section 3.7 ] 11. Resolution of Paradox #1 (added 3/5/14) When this Paradox first arose, I was truly mystified, but in retrospect it is not so mysterious. My current view of things is that φ = constant exactly on the two cylinder surfaces, even for an imperfect conductor like copper, and even when one leaves the DC case and goes to ω > 0 perhaps up to the transmission line limit. Copper is a "good" conductor, not a "perfect" conductor. In this good conductor case, there is some Ez inside the conductor which is quite small in magnitude, though it does dramatically vary as a function of θ. But even at θ = π where it peaks, Ez is still small (since copper is so "good"). The way the paradox is resolved is this. Consider this equation evaluated at the conductor surface: Ez = j βd φ - jωAz I think of φ = constant and Az = "almost constant". Think of ωAz having magnitude 100 and Ez having magnitude 1. Then a 1% variation in Az around the conductor perimeter allows for small Ez to be present and of course to vary with θ. This 1% effect also means that the B field is not exactly tangent to the conductor surface and misses being so by about 1%. So one "take away" from this resolution is to keep in mind that Az is not exactly constant on the surface, but for a "good conductor", it is very close to being constant, and then King's development is justified in terms of W(z) depending only on z. I of course have to edit lines doc to clarify that fact. [ this resolution sounds good on 5/7/14 and the next day I reviewed it in detail in a separate doc which is called "review of paradox 1.doc" ]. 12. Resolution of Paradox #2 (added 3/5/14) Consider the King gauge condition again, which I write in a manner to isolate the transverse parts (we are in the dielectric here) (∂xAx + ∂yAy)(x) = -j(βd2/ω) φ(x) - ∂zAz(x) = -j(βd2/ω) φ(x) + jβdAz(x) The "paradox #2" is that I always want to say LHS = 0 because At ≈ 0 and then the RHS tells us that if φ is constant on the conductor surface, then Az must also be constant on the surface. The resolution here is related to the resolution of Paradox #1. For a "good" conductor, we know that although φ = constant, we have only that Az = "close to constant", and the non-constancy of Az(x) then gives (∂xAx + ∂yAy)(x) ≠ 0 going around the surface. So there is then no Paradox #2, nothing is being violated. The claim of course is that since Az is close to constant, the variation of (∂xAx + ∂yAy)(x) is small, but still exists, so one cannot willy-nilly just claim that (∂xAx + ∂yAy)(x) ≡ 0 as your approximation for a good conductor. This resolution is consistent with another argument. In the perfect conductor limit where δ = 0, we have only a surface current Kz on the conductor surface (skin + Debye if you like), and thus we have only the Az component ≠ 0 from the King integral: there are no transverse currents, so At ≡ 0 and Az is a perfect constant on the round conductor surface. But when we move to finite δ > 0, there is some depth to the skin current and there are then transverse currents and transverse E fields, and this then leads to transverse At ≠ 0 and this then leads to (∂xAx + ∂yAy)(x) ≠ 0 . So both Paradox #1 and Paradox #2 are resolve in the same way: Az = "almost constant" on the conductor surface for a "good" conductor. 13. But why is Az largely non-constant on conductor at DC? Paradox 3. In my two-cylinder plots of B I shows that the B lines are "way off" the conductor surfaces, so I presume that Az in this case is "highly non-constant" on the surfaces. But since presumably φ = constant at very low frequency on the surfaces, and presumably Ez is very small. So Paradox 3: how can φ not vary at all, and yet Az varies a lot, when I claim in Appendix M that we should have Az = constant to 1 part in 104 ? [ I don't claim that in Appendix M ! I claim that Ax << Az . So no Paradox 3 ] Hint #1. I presume that for ω near DC, At <<<< Az since there is no transverse current at all. So then you would think that Az(x) = (1/vd) φ(x) would be truer than ever! And we are very much in the TML. Support: I added some detail to "two cylinder calculations" where I actually compute Az directly from the Helm integral in the DC case. Here is a piece of the labeled plot I just now added: [ the code here is in Az lines 1.mws as an implicitplot, hence ratty but OK, I quote this now in lines doc. See doc just mentioned for more details. 5.14.14 ] For example, on the right black circle I show points with Az = 1.6, 1.0, 0.7, 0.5 . This shows that Az varies substantially on the black circle, just as you would think. This is not a 10-4 variation! Fact: since this is DC, I know that Ez = I/(σ πa12) inside the conductor. Thus consider the following where we are "just outside" a conductor. We know that Ez is still I/(σ πa12) just outside since Ez is a tangential field at the boundary. Ez = -∂zφ - ∂tAz Ez = jβdφ - jω Az βd = ω/vd Ez = j(ω/vd) φ - jω Az Think now of ω being very small but not 0. We then have I/(σ πa12) = j(ω/vd) φ - jω Az How do I explain that as ω → 0, we get I/(σ πa12) = 0 which is obviously wrong! Here is one possible explanation. In the time domain write Az(x,t) = Az(x,ω)ejωt + AzDC(x) where we allow a constant as part of the solution ∂tAz(x,t) = jω Az(x,ω)ejωt // constant does not appear here! Now when we say Az(x,ω) = (1/vd) φ(x,ω) we are referring only to the non-constant part of Az(x,t) ! We could write this as AzAC(x,ω) = (1/vd) φ(x,ω) and then this last equation can become true again to 10-4 accuracy! My plot shows AzDC(x). This explanation of course brings up a whole new raft of questions. I want to continue to believe that the quantity φ(x,ω) is independent of ω -- that is my Rock of Gibraltar. This then suggests that AzAC(x,ω) = AzAC(x) = (1/vd) φ(x) so then AzAC is also independent of ω to our 10-4 accuracy. And it is pretty small since vd is large. I have no direct way to compute AzAC(x,ω) . The Helmholtz integral with constant Jz gives me the other term AzDC(x). Perhaps I could compute AzAC(x,ω) by replacing Jz with Jz = (Jzincip - JzDC). I think in this tiny difference you would have to worry about all partial waves. I think I know that Ez(r,m) = (1/4) ηm I Rdc (aβ') [ - ] Jz(r,m) = σ (1/4) ηm I Rdc (aβ') [ - ] x = rβ' (*) From App M, we still have | βd/β| = and this is "always small" up to 1000 GHz, so β' = β = (j-1)(1/δ) with δ ≡ (M.8) βd = βd0 [1 - j (1/2)tanL] ≈ βd0 = (ω/vd) = 2π/λd (M.7) So then x = βr = (j-1)(r/δ) where δ is large at low ω so x is small. DC limits of the Appendix D Fields Question: What is the limit of Jz(r,m) shown in (*) as ω → 0 ? All Bessel arguments are small and in general we have x → 0. therefore Jm(x) = (x/2)m / m! for m = 0,1,2,..... But the rule J-m(x) = (-1)mJm(x) says then that J-m(x) = (-1)mJm(x) = (-x/2)m / m! for m = 0,1,2.... so be careful if negative m integers are encountered. Here I restrict to m = 0,1,2... : fm = [ - ] = [ - ] = [ (m+1) (x/2)m (xa/2)-m-1 - (1/m) (x/2)m(xa/2)-m+1] = (x/2)m (xa/2)-m [ (m+1) (xa/2)-1 - (1/m) (xa/2) ] = (x/xa)m [ (m+1) (xa/2)-1 - (1/m) (xa/2) ] = (x/xa)m [ (m+1) (xa/2)-1] since xa→ 0 For m = 0 we get [ - ] = 2 = twice the first term above = 2(x/xa)0 [ (0+1) (xa/2)-1 = 2 (xa/2)-1 = 2 (2/xa) = 4/xa So I think then that the low ω limit of Jz is this Jz(r,m) = σ (1/4) ηm I Rdc (aβ') (x/xa)m [ (m+1) (xa/2)-1 ] m ≠ 0 Jz(r,0) = σ (1/4) I Rdc (aβ') (x/xa)0 [2 (xa/2)-1 ] m = 0 = σ (1/4) I a β' (4/xa) = σ (1/4) I a β' (4/β'a) = σ I = I / (πa2) so the m = 0 partial wave DOES approach the DC current. The higher partial wave currents? Jz(r,m) = σ (1/4) ηm I Rdc (aβ') (r/a)m [ (m+1) (β'a/2)-1 ] m ≠ 0 = σ (1/4) ηm I Rdc (aβ') (r/a)m [ (m+1) (2/aβ') ] = σ (1/4) ηm I Rdc (r/a)m [ (m+1) (2) ] but β' → 0 so we get Jz(r,m) = σ (1/4) ηm I Rdc (r/a)m [ (m+1) (2)] = σ (1/4) ηm I (r/a)m [ (m+1) (2)] = (1/2) ηm I (r/a)m (m+1) This is a surprise to me. It suggests there is asymmetry at DC in the longitudinal current! I have checked the m=0 and general m results with a full pass. Paradox and Confusion: Suppose the conductor has q = 0 on the surface so that n(θ) ≡ 0. In this case Nm = (q/2π) (-1)m e-|mξ| and I would argue that all the moments vanish if q = 0, Nm= 0. But lines doc says N0 = (βd/2πωa) I = (1/2πa) I (βd/ω) = (1/2πa) I (1/vd) and this does NOT vanish at DC. Very Low Frequency Wave going down two fat closely spaced round conductors. From the above analysis, I seem to find this: Jz(r,0) = I / (πa2) Jz(r,m) = (1/2) ηm I (r/a)m (m+1) ηm = (-1)m e-|mξ| ******************************************************************************* What is Jz(r,-m)? Jz(r,m) = σ (1/4) ηm I Rdc (aβ') fm From elsewhere I know that fm ≡ [ - ] f-m = fm ηm ≡ Nm/ N0 = (-1)m exp(-|m|ξ1) η-m = ηm So then Jz(r,-m) = σ (1/4) η-m I Rdc (aβ') f-m = Jz(r,m) So for the mth partial wave we get Jz(r,θ) = Jz(r,m) ejmθ + Jz(r,-m) e-jmθ = Jz(r,m) 2cos(mθ) = ηm I (r/a)m (m+1) cos(mθ) in the DC limit Maybe as ω→0, the Helm for Ez becomes Laplace, and then these are atomic forms in polar coordinates. From bipolar doc we have ηm ≡ Nm/ N0 = (-1)m exp(-|m|ξ1) n(θ)/N0 = 1 + 2 Σm=1∞ηm cos(mθ) = 1 + 2 Σm=1∞ (-1)m exp(-|m|ξ1)cos(mθ) So here is the z current density at ω= 0 for the two-cylinders problem: Jz(r,θ) = I / (πa2) + Σm=1∞ ηm I (r/a)m (m+1) cos(mθ) = [ 1 + Σm=1∞ ηm (m+1) (r/a)m cos(mθ) ] = [ 1 + Σm=1∞ (-1)m (m+1) exp(-|m|ξ1) (r/a)m cos(mθ) ] Like n(θ), there are alternating sign coefficients so the current is strongly peaked where n(θ) is peaked. In fact we can see that Jz(a,θ) = [ 1 + Σm=1∞ (-1)m (m+1) exp(-|m|ξ1) cos(mθ) ] n(θ)/N0 = 1 + 2 Σm=1∞ (-1)m exp(-|m|ξ1)cos(mθ) Not quite the same however due to the (m+1) extra factor. [ at this point I did lots of edits on bipolar doc, including adding the ηm calculation as Appendix A ]