continuity at conductor surface REVIEWED

Continuity at conductor surface PhL 3.4.14 [ 5.14.14 final review: Basically this is a test and vindication of the charge pump boundary condition that is so important to everything I do. It shows you can ignore "Debye currents" since they are really already included in the bulk volume current. ] This doc examines the connection between surface charge n and current densities Ji under the conductor surface using the continuity equation. Case 1 with the deep box gives a result not too useful. Case 2 develops the charge pump condition ignoring Debye surface currents. Case 3 includes these currents and shows that they don't affect the charge pump condition because they are negligible. Case 4 is nada. Continuity Analysis #1 : The deep gaussian box idea 1 Continuity Analysis #2 : Thin gaussian box at surface (ignoring Debye surface currents) 3 Continuity Analysis #3 : Thin gaussian box at surface (including Debye surface currents) 4 Continuity Analysis #4 : Gaussian box inside the skin layer 6 Continuity Analysis #1 : The deep gaussian box idea Here δ is normal skin depth. The conclusion here for the "deep box" shown in the figure is this: -jω n(x,0,z) = [∂xKx(x,0,z) + ∂zKz(x,0,z)]/(1+j) where Jr plays no role since the box is deep. This is probably correct for the deep box, but I don't think it proves anything one way or the other. First of all, I want a top quality picture so I can see what I am doing: The red thing is a Gaussian box with three pairs of faces. Assume there is some current J in the skin region and that all components drop off with δ. We take the box to extend 5δ div J = -jωρ ∂xJx + ∂yJy + ∂zJz = -jωρ ∫ J dS = -jω ∫ ρ dV Question 1: What is the contribution to ∫J dS from the top and bottom faces of the box? The area of the faces is dx dz. So we get [ arguments are always (x,y,z) ] top = dx dz Jy(x,ε,z) bottom = - dx dz Jy(x,-5δ,z) Now, the top is just above the surface and above the surface charge layer, and we assume the dielectric is non-conducting, so then Jy(x,ε,z) = 0. Meanwhile, the bottom is so far from the surface, Jy(x,-5δ,z) = 0. Thus, both these faces make no contribution ! Answer to Question 1 = 0 Question 2: What is the contribution to ∫ J dS from the left and right faces of the box? right = dz !Syntax Error, Idy Jx(x+dx,y,z) = dz δ Jx(x+dx,0,z) At this point, we model Jx as in (2.1.8) so that Jx(y) = Jx(0)e-(j+1)y/δ so then !Syntax Error, Idy' Jx(x+dx,-y',z) =!Syntax Error, Idy' Jx(x+dx,0,z)e-(j+1)y'/δ ≈ Jx(x+dx,0,z) !Syntax Error, I dy' e-(j+1)y'/δ = [δ/(j+1)] Jx(x+dx,0,z) So then right = dz [δ/(j+1)] Jx(x+dx,0,z) left = - dz [δ/(j+1)] Jx(x,0,z) right + left = dz [δ/(j+1)] Jx(x+dx,0,z) - dz [δ/(j+1)] Jx(x,0,z) = dz [δ/(j+1)] dx ∂x Jx(x,0,z) Answer to Question 2 = dz [δ/(j+1)] dx ∂xJx(x,0,z) Question 3: What is the contribution to ∫ J dS from the front and back faces of the box? Repeating question 2 where we change x to z, we get Answer to Question 3 = dx [δ/(j+1)] dz ∂zJz(x,0,z) Question 4: What is ∫ ρ dV ? Answer: ∫ ρ dV = n dx dz // since surface charge on surface. Then "continuity" says this: 0 + dz [δ/(j+1)] dx ∂xJx(x,0,z) + dx [δ/(j+1)] dz∂zJz(x,0,z) = -jω n dx dz or [δ/(j+1)] ∂xJx(x,0,z) + [δ/(j+1)] ∂zJz(x,0,z) = -jω n so that -jω n(x,0,z) = [δ/(j+1)] [∂xJx(x,0,z) + ∂zJz(x,0,z)] . This then gives the ridiculous result that as δ→ 0, the surface charge vanishes. I don't like that result. One problem is that 5δ is a pretty "deep box" and I know that there is strong phase variation of the transverse currents on the side walls which I have not incorporated. I assumed only a magnitude variation. So I guess my conclusion is that this result is not to be trusted and is in fact wrong. Notes added 3/6/14. 1. First, if you include the phase effect, the integral becomes δ/(1+j) instead of δ and so the nature of the result does not really change. [ 5.20.14: I have edited this fix into all of the above ] 2. I think the way to understand the above equation is to think of Kz = δJz and as you take δ → 0, Kz stays constant, but Jz → ∞. So making these two changes, the above would say -jω n(x,0,z) = [∂xKx(x,0,z) + ∂zKz(x,0,z)]/(1+j) // δ → 0 In the tail end of this limit, then, the surface currents remain finite and n remains finite and we are then not forced to the wrong conclusion that n = 0 for a perfect conductor. The reason Jz → ∞ is that the conductor is still constrained to carry some current i(z) and if you force it out to a skin sheath, Jz gets larger without limit. 3. What happens if we express things in terms of E? -jω n(x,0,z) = [δ/(j+1)] [∂xJx(x,0,z) + ∂zJz(x,0,z)] . -jω n(x,0,z) = [δ/(j+1)] σ [∂xEx(x,0,z) + ∂zEz(x,0,z)] . δ ≡ Write σ = 2/(ωμδ2) and then δσ = 2/(ωμδ) and then you get -jω n(x,0,z) = (2/ωμ) δ-1 [∂xEx(x,0,z) + ∂zEz(x,0,z)]/(j+1) How do we interpret THIS form then? Well, we know that Ei → 0 in the perfect conductor limit δ→ 0. Write for a round wire this approximation (wide spaced twin lead) Jz = (I/2πaδ) => Ez = Jz/σ = (I/2πa)(1/σδ) = (I/2πa) ωμδ/2 = (Iωμ/4πa) δ In this little model then we see explicitly that Ez → 0. Continuity Analysis #2 : Thin gaussian box at surface (ignoring Debye surface currents) Here, with a thin box, ignoring Debye surface currents, I obtain the usual CPBC used in Appendix D. Now we use a different box, The sides of the box approach 0 and so make no contribution because the transverse current is not infinite anywhere. Then for this box we get only a "bottom" contribution which is ∫ J dS = bottom = - Jy(x,-ε,z)dx dz and then continuity says δ λD -jω n dx dz = - Jy(x,-ε,z)dx dz or jω n = Jy(x,-ε,z) where ε is tiny, so this is the normal current "just below the surface". I have then replicated my "radial charge pumping boundary condition" from lines doc, which was: Jr(r=a-ε,φ) = jω n(φ) . (D.2.23) Continuity Analysis #3 : Thin gaussian box at surface (including Debye surface currents) Here we have a thin box that always catches the surface currents, giving the worrisome full boundary condition that -jω n = - Jy(x,-ε,z) + ∂xKx(x,0,z) + ∂zKz(x,0,z) Kx = δdebyeJx I then use Ohm's law J= σE only for transverse fields at the surface to rewrite this as (-jω/σ) n = - Ey(x,-ε,z) + δdebye [ ∂xEx(x,0,z) -j (2π/λ) Ez(x,0,z)] (*) I then assume that ∂xEx ~ DEx to get (-jω/σ) n ≈ - Ey(x,-ε,z) + (δdebye/D) Ex(x,0,z) -j (2π) (δdebye/λ) Ez(x,0,z)] (*) I then argue that both ratios are incredibly small so that last two terms may be neglected to give (-jω/σ) n ≈ - Ey(x,-ε,z) and then the CPBC is reconstituted with the conclusion that you can neglect Debye surface currents. Suppose there existed "Debye surface currents" Kx and Kz. Then no matter how thin I make the gaussian box in Analysis #2, I always capture these currents. Then: (∫ J dS) rightside = dz Kx(x+dx) (∫ J dS) leftside = - dz Kx(x) So these two faces then give contribution dz Kx(x+dx) - dz Kx(x) = dz dx ∂xKx(x,0,z) The front and back faces will contribute dx dz∂zKz(x,0,z) Then we have -jω n dx dz = - Jy(x,-ε,z)dx dz + dz dx ∂xKx(x,0,z) + dx dz∂zKz(x,0,z) or -jω n = - Jy(x,-ε,z) + ∂xKx(x,0,z) + ∂zKz(x,0,z) and this is then a "modified" charge pumping condition which now has to include the surface currents. Now if I take the point of view that the Debye stuff has some thickness δdebye, I can try to "squeeze out" the surface currents by making the gaussian red box super thin. But then I also squeeze out the charge, so that does not fly. Comments: In the model with some δdebye thickness we might say that Kx = δdebyeJx where Jx is the volume current inside the Debye layer and you might also say that Jx = σEx just as you would anywhere else inside the conductor. Then maybe we have -jω n = - Jy(x,-ε,z) + δdebye∂xJx(x,0,z) + δdebye∂zJz(x,0,z) -jω n = - σEy(x,-ε,z) + δdebye σ ∂xEx(x,0,z) + δdebyeσ ∂zEz(x,0,z) // replace Ji by Ei transverse (-jω/σ) n = - Ey(x,-ε,z) + δdebye [ ∂xEx(x,0,z) + ∂zEz(x,0,z)] // divide thru by σ (-jω/σ) n = - Ey(x,-ε,z) + δdebye [ ∂xEx(x,0,z) -j βdEz(x,0,z)] // ∂z → -j βd (-jω/σ) n = - Ey(x,-ε,z) + δdebye [ ∂xEx(x,0,z) -j (2π/λ) Ez(x,0,z)] (-jω/σ) n ≈ - Ey(x,-ε,z) + (δdebye/D) Ex(x,0,z) -j (2π) (δdebye/λ) Ez(x,0,z)] (*) You would certainly claim that (δdebye / λ) was awesomely small and that last term is 0, even though perhaps Ez is the largest of the three E field components. And maybe you could claim that ∂xEx is somehow on the order of D Ex where D is some transverse dimension of the transmission line or of the conductor, and then (δdebye / D) is super small as well. This offers a possible way to "get rid of" these Debye current terms. This argument then makes Kx and Kz go away again . Added 5.20.14. Here is a slightly different argument. Consider this picture where the Debye surface charge layer is greatly exaggerated in thickness. This is the layer that carries the Debye surface current as well. This picture shows a tiny slice of width dx of a piece of the conductor cross section at its surface. The total current in the z direction is the sum of that current in the Debye layer plus the current in the skin effect layer below the Debye layer, so we write I(D) = Jz(D) λD dx = σD Ez(D) λD dx I(δ-D) = Jz(δ-D)(δ-λD) dx = σ Ez(δ-D)(δ-λD) dx where we allow the Debye layer to have conductivity σD that differs slightly from the bulk σ. Now we use the fact that Ez, being a transverse field, is continuous at the lower Debye layer boundary, according to (1.1.41), so then Ez(D) = Ez(δ-D) = Ez Consider then the ratio of the currents carried in the two layers, = = ≈ 1 * As noted at the end of Appendix E, even at 100 GHz the ratio = 1/4000 for copper. Our conclusion then is that we may completely ignore the Debye layer from a current point of view because the current it carries is swamped by the current carried in the bulk conductor beneath the Debye layer. OK, that may be true, but the "thin gaussian box" is shown in red, and it includes only the Debye layer current. Let's go back then to (-jω/σ) n ≈ - Ey(x,-ε,z) + (λD /D) Ex(x,0,z) -j (2π) (λD /λ) Ez(x,0,z)] (*) Here we regard Ey as measured just below the Debye layer, while Ex and Ez are perhaps in the middle of the Debye layer. Since both Ex and Ey are transverse fields, we can regard all three field components as being evaluated just below the Debye layer. Since these fields are all inside a good conductor, they are all small, though we expect Ez >> Ex and Ey. Now consider the ratios for Belden 8281 (λD /D) ≈ 10-10m/ (400 x 10-6m) ≈ 10-10 / (0.4x10-3) ≈ 2.5 x 10-7 (λD /λ) =(λD /D)(D/λ) ≈ 2.5 x 10-7 (D/λ) << 2.5 x 10-7 since in the transmission line limit (D/λ) << 1. Based on these ratios, it seems reasonable to drop the last two surface current terms in (*) to arrive at the usual boundary condition (-jω/σ) n ≈ - Ey(x,-ε,z) [ I hope the presence of the charges in the Debye later does not heavily affect Ex or Ez there. That is sort of a plasma issue that I don't want to think about. You would think Ey was most affected. I expect them only to affect the normal Ey in the Debye later. Note: My elaborate argument about the ratio of currents supports the idea that the amount of current carried in the Debye later is much smaller than in the adjacent bulk skin effect layer. However, this argument does not seem to assist in obtaining the desired simple boundary condition. Going back to -jω n = - Jy(x,-ε,z) + ∂xKx(x,0,z) + ∂zKz(x,0,z) Kx = λDJx it is true that if we just ignore the surface currents and set Kx = Ky = 0, we do get the desired result, but just staring at this equation does not show why you can do that. Note added 3/6/14. Let's compare Analysis #3 to Analysis #1. In Case 1, as δ → 0 we had Jz → ∞ and Ez→ 0. The reason was that Jz has to carry the whole conductor current I. In Case 3 things are different. In Case 3, the Debye surface currents Kz does not have to carry all of I since most of I is carried by the normal skin sheath. Thus, in case 3 we have Ez does not change if you were to take δdebye → 0. This Ez is just the same Ez which lives in the normal skin sheath. That is why when we take δdebye → 0 in equation (*) above, only the first term - Ey(x,-ε,z) remains. In contrast, when we took δ → 0 in Case 1, it was Kz that remained constant and then we avoid the disaster of n → 0. Continuity Analysis #4 : Gaussian box inside the skin layer Not useful. Here we put a box tiny in all directions inside the conductor Since now div J = 0, we can just write ∂xJx(x,-ε,z) + ∂yJy(x,-ε,z) + ∂zJz(x,-ε,z) = 0 I am not sure this helps anywhere.