Rethink Debye Surface Currents at low w REVIEWED

Rethink Debye Surface Currents at low ω PhL 8.16.14 Reviewing these notes on 9.10.14 I am reading through Section D.9 and taking notes here on how things are different. (a) The notion of Debye Surface Currents I think this section is still completely correct. At ω → 0 you would just have δ → ∞ and everything is still valid. Fact 1, the Corollary, and Fact 2 are still true as ω→ 0. (b) The role of Debye Surface Currents in the boundary condition I think things are OK down through this equation -jω n(θ) = - Jr(a,θ) -jk [σ Ez(a,θ) λD] = - σEr(a,θ) -jk [σ Ez(a,θ) λD] = - σEr(a,θ)[ 1 - jkλD ] . (D.9.7) where I continue to dispense with Eθ effects. At this point, my discussion goes into ansatz mode. I assume the second term is small, and with this assumption, the fields I obtain USING the traditional CPBC result in ~ | | which should really be β' in the ratio, but I was always thinking k << β so β = β'. Let that lie for now. I then conclude the ansatz with a self-consistency argument to show that indeed, the second term is small in (D.9.7). My end up condition is (δ/λd) >> 1 which is of course true also as ω→0. But perhaps this is just a self-fulfilling prophecy. IF I assume the second term is small, then I get the App D solutions, and then the second term is indeed small. Look at the logic here" Step 1: Derive fact that -jω n(θ) = - σEr(a,θ)[ 1 - jkλD ] Step 2: Assume 2nd term can be neglected to get -jω n(θ) = - σEr(a,θ). Step 3: Run through Appendix D with this as the CPBC and thereby obtain solutions (D.2.33) Step 4: Observe that for these (D.2.33) solutions, one has ~ | | Step 5: Using this field ratio, verify that the "second term" in Step 1 can be neglected. I have then proven the following: "If the 2nd term can be neglected, then it can be neglected" ! [ I agree, the argument seems circular, I need to do something to improve this ******** ] Ouch. What I really need is some OTHER criterion by which this second term can be neglected. Something to keep in mind in the Debye layer J = σE - D grad ρ . dim(D) = m2/sec (E.1) Now, go back to earlier statement -jω n(θ) = - Jr(a- λD,θ) -jk [σ Ez(a,θ) λD] = - Jr(a- λD,θ) -jk λd Jz(a,θ) = - Jr(a- λD,θ) -jk λd JzD(a,θ) = - Jr(a- λD,θ) -j k KzD(a,θ) This is the equation I really need to ponder! When is the inside radial current just below the surface much larger than the Debye current? I need some kind of handle or estimate for Jr which does NOT assume the CPBC. Question. What do I know about divE ? [ perhaps a digression not useful ] [ I am trying to find a way to compare the size of the two terms in n(θ) above. ] Below the Debye layer there is no free charge, so there I claim that div E = 0. To the extent that Eθ is zero in this region (still relatively close to the surface), div E = 0 tells me that ∂r (r Er(a- λD,θ)) - jka Ez(a- λD,θ) ≈ 0 r ≈ a below Debye layer But in the Debye layer I claim that ρ(r) = ρ(0) e-(a-r)/λ Because λd is so small, you can regard this as a 1D problem with the above result in terms of r. Then I guess we have this result then: In the Debye layer just below the surface, we have [ from divE = ρ/ε0 ] ∂r (r Er(a,θ)) - jka Ez(a,θ) ≈ r ρ(0) e-(a-r)/λ /ε0 ≈ a ρ(0)/ε0 = a e ne/ε0 [ ne = net ] The RHS is in general a huge number! Stealing a result from below λD e ne/ε0 = 10,000 volt/m a e ne/ε0 = (a/λd) λD e ne/ε0 = (a/λd) 10,000 volt/m = perhaps 1012 volt/m This seems ridiculous. Dimension check: LHS = E = volt/m RHS = m * Cou * m-3 * m/far = volt/m So we have shown that ∂r (r Er(a,θ)) - jka Ez(a,θ) ≈ a e ne/ε0 Now define f ≡ - Jr(a,θ) -jk [σ Ez(a,θ) λD] [ the two terms of interest ] which is the RHS of the "true" boundary condition, and I want to see when the second term may be neglected. Write f ≡ - Jr(a,θ) -jk [σ Ez(a,θ) λD] = - Jr(a,θ) -jkλDσ Ez(a,θ) = - Jr(a,θ) + σ(λD/a) [-jkaEz(a,θ)] = - Jr(a,θ) + σ(λD/a) [e ane/ε0 - ∂r (r Er(a,θ))] =σ { - Er(a,θ) + (λD/a) [e a ne/ε0 - ∂r (r Er(a,θ))] } so then f/σ = - Er(a,θ) + (λD/a) [e a ne/ε0 - ∂r (r Er(a,θ))] and again I want to know when the second term can be neglected. I have managed so far to eliminate Ez from the question, but now I need to know SOMETHING about Er, but without the App D solutions, I really don't know a thing about Er ! I can see that if Er(a,θ) ≡ 0, then you can NOT neglect the second term. Looking at the Helm equation (D.1.17) is not too helpful because it is 2nd order in ∂r. Write again f/σ = - Er(a,θ) + (λD/a) [a e ne/ε0] - (λD/a) ∂r (r Er(a,θ)) [ nothing seems useful in the above really ] Plan A. Perhaps I can argue that (λD/a) << 1 so the last term can be neglected. Then we have f/σ = - Er(a,θ) + λD e ne/ε0 How large is this second term? What are its dimensions? dim(λD ene/ε0) = m * Cou * m-3 * m/farad = volt/m Here are some numbers for copper λD = 5.55 x 10-11 m e = - 1.6 x 10-19 Coul ne = 1022 m-3 [ net ne inside the Debye layer ] ε0 = 8.8541877 x 10-12 farad/m Maple gives for λD ene/ε0 : So we then have f/σ = - Er(a,θ) + 10,000 volts / m An unexpected result. In this plan, the second term would always dominate, and you would never have the CPBC as presented in App D. [ so something is not right here ] Plan B: Regroup. In the above I have come up with four distinct equations: Equation #1: -jω n(θ) = - Jr(a- λD,θ) -jk [σ Ez(a,θ) λD] = - Jr(a- λD,θ) -jk λd Jz(a,θ) = - Jr(a- λD,θ) -jk λd JzD(a,θ) = - Jr(a- λD,θ) -j k KzD(a,θ) // involves Jr below Debye layer OK Equation #2: ∂r (r Er(a- λD,θ)) - jka Ez(a- λD,θ) ≈ 0 r ≈ a Er below Debye layer Equation #3: ∂r (r Er(a,θ)) - jka Ez(a,θ) ≈ a e ne/ε0 ~ 1012 Er in the Debye layer Equation #4: Ez(a- λD,θ) = Ez(a,θ) Since I am interested in Equation #1 as a modified CPBC, equation #3 is not very useful so perhaps just chuck that from the list of useful equations. I can rewrite #1 and #2 and #4 this way : -jω n(θ) = - σEr(a- λD,θ) -jk [σ Ez(a,θ) λD] #1 + Ohms Law outside Debye layer ∂r (r Er(a- λD,θ)) - jka Ez(a,θ) ≈ 0 r ≈ a #2 and #4 [ both OK ] At this point, one thing I could do (though probably not useful) is eliminate Ez(a,θ) to get -jω n(θ) = - σEr(a- λD,θ) - jka Ez(a,θ) [σ (λD/a)] = - σEr(a- λD,θ) - ∂r (r Er(a- λD,θ)) [σ (λD/a)] or -jω n(θ)/σ = - Er(a- λD,θ) - (λD/a) ∂r (rEr(a- λD,θ)) or jω n(θ)/σ = Er(a- λD,θ) + (λD/a) ∂r (rEr(a- λD,θ)) or jω n(θ)/σ = Er(a-λD,θ) + (λD/a) Er(a- λD,θ) + (λD/a) a ∂rEr(a- λD,θ) Now we know that (λD/a) << 1 for a practical transmission line a value, so this says jω n(θ)/σ = Er(a-λD,θ) + λd ∂rEr(a- λD,θ) [ seems OK ] This is the first time in this doc I have stated this claimed equation. Only the Er field below the Debye layer is involved here! I want to know the relative size of the two terms shown. At high ω, I might guess this to be true below the Debye layer and toward wire center, Er(r,θ) = Er(a- λD,θ) e(r-a)/δ and then roughly we get ∂r Er(a- λD,θ) ≈ (1/δ) Er(a- λD,θ) Then one gets jω n(θ)/σ = Er(a-λD,θ) + (λd/δ) Er(a- λD,θ) = [ 1 + (λd/δ) ] Er(a- λD,θ) Even at 100 GHz we know that (λd/δ) ≈1/4000 and the first term dominates. Below this, the first term dominates even more, and this is how I have used Appendix D. So maybe this is a better argument than the one I now present in Section D (9) (b). [ taking note of that ] This really says nothing about what happens at low ω [agreed] . If you just say δ → ∞ in the above formula as a representation of low ω, you get the same result as for high ω. Let's go back to jω n(θ)/σ = Er(a-λD,θ) + λd ∂rEr(a- λD,θ) radial pump Debye term Another form jω n(θ) = Jr(a- λD,θ) -jk [Jz(a,θ) λD] #1 + Ohms Law outside Debye layer [ so, to this point I have learned nothing new about these two terms ] Concern: Have I ever said i(z) = dq(z)/dt ? I hope not. In my notation, i(z) is the total current in a conductor, whereas q(z) is the charge per unit length all of which lies on the surface. So ∂tq(z) is only the current feeding the surface charge, not the total current! I will collect some evidence: 1. Below Fig 3.6a I say this: i(z) = q(z) vd, (4.11.19a) where I am quoting from a future location in lines doc. Going there I see first ∂z i(z) = -jωq(z) . (4.11.19) which seems to say that ∂zi(z,t) = -∂tq(z,t) assuming no dielectric current out box sides. This is supposed to come from div J = - jωρ -jω[∫V ρ dV] = ∫S J dS . (1.1.35) Why doesn't capacitor current appear here? Because symbol J is conduction current only! So Ok, if the total wire current i(z) is decreasing in z, it has to be going somewhere, and the only place is q(z). I think then that (4.11.19) is OK. Now look at my larger context, ∂z i(z) = -jωq(z) . (4.11.19) Jumping the gun again, if we again use ∂z → -jk with k = (ω/v), we arrive at the intuitive relation i(z) = q(z) v (4.11.19a) which just says the charge per unit length is [in effect, see D.9 (c)] traveling down the line at phase velocity v. In the lossless case v = vd (dielectric speed of light), whereas more generally v is complex. Now saying k = ω/vφ can be regarded as a definition of the complex phase velocity vφ. Then with that definition, we can say ∂zi(z) = -jωq(z) => -jk i(z) = -jωq(z) => k i(z) = ωq(z) => i(z) = (ω/k)q(z) and finally that i(z) = vφ q(z). This is just a definition. The more interesting claim is just that i(z) = (ω/k)q(z) which is a relationship between the TOTAL current in the conductor to the surface charge. I don't think I have said anything wrong yet, and I have not said i(z) = dq(z)/dt so far. [ so this little "concern" turned up nothing in lines doc that is wrong ] Plan C. Let's go back to an earlier stage: -jω n(θ) = - Jr(a- λD,θ) -jk [σ Ez(a,θ) λD] where I think the second term is the Debye current contribution. I want a solution where this second term "carries the whole load". Something like this for very low ω -jω n(θ) = -jk [σ Ez(a,θ) λD] Since n(θ) is a function of θ, how are you going to avoid having Ez(a,θ) depend on θ? Well. I was hoping that JzD = σ Ez would depend on θ only in the Debye layer, and then below that layer I wanted to see Jz= σEz being a constant in θ! But then how do I get Ez being continuous at the "Debye boundary"? Maybe my model of a hard boundary is no good. [ at least a new idea just presented: maybe you can have Jz be uniform inside the round wire, while at the same time the Debye current on the surface is non-uniform. ] (next day Aug 17, 2014) Plan D. Have been working on scratch this AM. Here is a possible logic thread. At high ω, strong skin effect, current constrained to sheath, Z0 is low and some constant value. Then i(z) = V(z)/Z0 = some finite value. Perhaps V = 10 volts and Z0 = 75 ohms so i(z) = 1/7 amp. The sheath is δ thick, so for symmetric round wire you could say that 2πa δ Jz = i(z), which says Jz(sheath) = i(z)/[2πaδ]. For some large ω there is some small δ and there is then some Jz(sheath) value. You cannot have Jz(sheath) be 10 times this value! The Debye surface current is swamped by the δ sheath current as I show somewhere, so we can ignore the Debye layer. Perhaps vφ = c, speed of light. Now with all this background, we ask if n(θ) can just move along the Debye layer and no Jr is necessary to make n(θ) "go". Then Jz = neevφ = neec in order to make that work. [ Why do I put phase velocity into the J = nev formula, that seems wrong. It should be a drift velocity. I think therefore Plan D is no good. ] What is this value? Jz = neec = 1022* 1.6 x 10-19* 3x108 = 5 x 1011 amps/m2. The required Ez field to do this is then Ez = Jz/σ = 1012/ 108 = 104 volts/m. But our limit is Jz(sheath) = i(z)/[2πaδ] which for a = Belden = 10-3 m and δ = 0.21 x 10-6 at 100 GHz so then Jz(sheath) = i(z)/[2πaδ] = (1/7) / [ 6 x 10-3 x 0.21 x 10-6 ] = 109/8 ~ 108 amp/m2 Since Jz(sheath) = 108 amp/m2 and Ez = Jz/σ = 108/ 108 = 1 volt/m, our attempt to make drift velocity be "c" would require Jz = 1012 amp/m2 and Ez = 104 volt/m, so that is why you cannot "feed n(θ)" by the Debye mechanism at 100 GHz. So why could you maybe "feed n(θ)" by Debye at low ω? In this case, perhaps Jz = i(z)/(πa2) and there is some corresponding Ez in the wire. Now Jz = neevφ is small because vφ is very small. In this case, we have Z0 = 1/ [ 1-j] = 1/ e-jπ/4 from App Q. Then i(z) = V(z)/Z0 ~ V(z) e+jπ/4 and this is getting very small. So then Jz = i(z)/(πa2) = (1/πa2) V(z) e+jπ/4 which is getting small as . The required Debye solution would need Jz = neevφ with vφ = ω1/2 [ same error as above I think ]according to my last doc, so we then have Jz = neevφ = nee / . If we were to equate these values Jz = neevφ = nee / required Debye value of JzD Jz = i(z)/(πa2) = (1/πa2) V(z) e+jπ/4 value determined by uniform Jz assumption These have the same ω dependence, which is encouraging. But to be equal we would need nee / = (1/πa2) V(z) e+jπ/4 or nee/ = (1/πa2) V(z) e+jπ/4 or nee = (1/πa2) V(z) e+jπ/4 = (1/πa2) V(z) C e+jπ/4/ = (1/πa2) q(z) e+jπ/4/ LHS = Coul/m3 RHS = m-2 * volt * far/m = Coul/m3 OK after 8 fixes So what does it mean to require that nee = (1/πa2) q(z) e+jπ/4/ ? q(z) = (πa2 ) nee e-jπ/4 nee = total 3D free charge density in the Debye layer n(θ) = nee λD q(z) = (2πa) a nee (1/) e-jπ/4 q(z) = (2πa) (a/λd) n(θ) (1/) e-jπ/4 q(z) = q(z) (a/λd) (1/) e-jπ/4 hmmmm.. problem here Well, I am playing in the right city at least, but things just remain fuzzy all the time, now noon. [ so in Plan D I wonder if at low ω the Debye layer could carry all the current, but I reach no conclusion] Comments (9.10.14) I read all the above stuff today, it is a lot of failing around. Since ne is always small compared to the large number of carrier electrons, maybe the surface charge has no effect at all on the total Jz current. If Ez is the same in the Debye later as below the Debye layer, Jz will be about the same in both regions. This would be true at any ω. OK, but this does not answer the question of how the z directed Debye current contributes to n(θ) as compared to the radial Jr current. So everything is inclusive.