Debye Surface Currents REVIEWED

Debye Surface Currents PhL 5.19.14 Today is 10.8.14 and I don't really feel like reviewing the stuff below. I think I have put this topic to bed in Appendix D.9 (b). But if it comes back, below is a review of my various pieces of work on this topic. I thought this issue was resolved, but now it is back. It has come and gone several times already. Part I: Review of Past Work on Debye Surface Currents. 1 1. bull by the horns doc Section 2 1 2. Continuity at conductor surface doc 1 3. Computation of Az for the two cylinders.... (Section 11) 1 4. The Surface Currents Issue 2 5. Work done in Appendix A below in this document. 2 6. Bug to deal with. 3 Appendix A 4 Part I: Review of Past Work on Debye Surface Currents. 1. bull by the horns doc Section 2 Here I consider the cpbc situation including surface currents, and if I include the surface currents in the two directions, the boundary condition becomes: -jωn(θ)a = -Jr(a-ε)a + (∂zKz)a + (∂θKθ) In this doc I am convinced that Kz and Kθ are important and theory needs to be completely redone. I note that a free surface current Kz also appears in (1/μ2) (∂nAz)2 - (1/μ1) (∂nAz)1 = Kzfree which then seems to wreck my earlier work regarding Az and μ boundaries. Argghhhh! it says. This led me to do a web search on such surface currents. I thought I had a hit with Pozar, but he was a red herring. I then review what sections of lines doc are threatened by the addition of Kz and Kθ. End of doc. 2. Continuity at conductor surface doc This seems to be a very careful analysis of the subject of Debye surface currents, probably the best thing I have on the subject, and I added a little to it just now. It argues strongly that the surface currents can be neglected in doing the cpbc. Case 3 has my best direct argument for the validity of the cpbc !!! 3. Computation of Az for the two cylinders.... (Section 11) This is not really about Debye surface currents since "deep box", but the ideas are worth noting. If I take the deep box of the previous section and now define Kz as the TOTAL surface current over depth δ (not just the Debye), and if I assume Eθ = 0 over the entire δ depth so the two θ sides of the box give no contribution since Jθ= 0, I find these results Kz(φ,z) = (j+1)vd n(φ,z) Kz(θ,z) = δ Jz(a,θ,z) -jω n(θ) = [δ/(j+1)] [-jβd Jz(a,θ)] The first and last are really the same as they say this Jz(a,θ) = (j+1) (vd/δ) n(θ,z) and this is perhaps my earliest derivation of the fact that " Jz(a,θ) at the surface tracks n(θ)". This thing comes from div J = -∂tρ evaluated for the "deep box" assuming Eθ = 0 on box sides. Later in lines doc Section 6.5 (d) I derive a different result using both div E = 0 and the cpbc and I get Ez(a,θ) = (-jω/σ) (β/βd) n(θ) but right now I see this is the same thing: Jz(a,θ) = (-jω) (β/βd) n(θ) = (-jω) [ej3π/4 (/δ) ] / (ω/vd) n(θ) = (-j) vd [ej3π/4 (/δ) ] n(θ) = (-j) vd (j-1)/ * (/δ) ] n(θ) = (1+j) vd (1/δ) ] n(θ) = (1+j) (vd/δ) n(θ) which is the same! So this is perhaps a better way to get this result! 4. The Surface Currents Issue This is an earlier summary of the Debye and other issues. Sections 1,7,8,10 involve Debye. In Section 1 I look again on the web for surface currents but find nothing on Debye. In Section 7 is an EXELLENT summary of why I can ignore Debye surface currents. It is based on the idea that such a small fraction of the current is in the Debye layer. Still, I think I item 2 Case 3 above is the best direct argument for validity of the cpbc. In Section 8 I talk about "where does n(θ) come from" which is still a very active question for me. The arguments here are relevant, but they don't do what I want them to do and I will work on that soon. Section 10 looks at this perhaps reasonable hybrid cpbc version Jr(a,θ) + jβdKz(θ) = jωn(θ) each term = amps/m2 Er(a,m) + jβd δdebyeEz(a,m) = j(ω/σ)Nm In this form, it is hard to arm-wave the Kz term away. I then use this form to recompute am and Km in a very long algebra session. But in the end, I argue that (β' δdebye) << 1 and I then end up with the exact same Appendix D values I now have for am and Km ! 5. Work done in Appendix A below in this document. As in earlier items above, I consider a box at the surface and get (same as just above) -jω n = Jr - j(ω/vd) KzDebye But then if I set KzDebye = vDn(θ), I get Jr = 0. I then have BC's Jr = 0 and Jθ = 0 and I show that this forces Appendix D to have am = 0 and Km = 0 so the inside of the wire is totally quiet! I then try to rescue things by bringing back the azimuthal current so then [ with the cancelling terms above ] 0= Jr + ∂θKθDebye which makes a horrible world of azimuthal surface currents swirling around. The problem here which I have to write up somehow is this: KzDebye ≠ vdn(θ) Kz = vdn(θ) where Kz includes the full δ thickness including the Debye I think by the way that (4.11.19a) claiming i(z) = q(z)vd is correct, an equation having a similar form. TBC is now 1:45 May 20, 2014 6. Bug to deal with. I want to talk about the Debye layer, but it seems to have a carrier density a lot different from the normal bulk conductor. Recall that the conductivity is given by σ = (ncq2τ/m) so you would imagine that nc is a lot larger in the Debye layer than elsewhere where it is only 1 charge per atom. Larger conductivity weakens my argument of ignoring the current of this layer. In Appendix E I write that ρ(x) = ρ(0) e-x/λ as you move away from the surface. This stuff has to integrate to n(θ) so we get the familiar type of result, n(θ) = ρ(0) λD ρ(0) = n(θ) / λD = probably a very LARGE number. nc(0) = ρ(0)/e = n(θ) / (eλD) Looking at the numbers in Appendix N, can I learn just how big this free charge stuff is compared to the normal carrier density in the conductor? A typical value for n(θ) we can get from our two cylinder thing n(θ)a = 2πε ~ εV n(θ) ~ εV/a dim = farad/m * volts/m = farad-volts/m2 = Cou/m2 correct How big is this for say a = 1 mm and V = 10 volts n(θ) = 9x10-12 * 10/ 10-3 = 90 x 10-9 ~ 10-7 Cou/m2 How about parallel plate capacitor of area A and separation s. Then V=Es, εE = n so n = εV/s which has the same form. Separation 1 mm gives the above. Then from the above nc(0) = n(θ) / (eλD) dim = Cou/m2 / (Cou-m) = 1/m3OK = 10-7 / [1.6 x 10-19 * 10-10 ] = 1/1.6 * 10-7+29 = 0.7 x 1022 carriers/m3 This is large, but it is still much smaller than n = 8.5 x 1028 electrons/m3 // for copper Conclusion: In a practical situation, the volume density of free charge in the Debye layer 1022 is negligible compared to the density of conduction carriers 1028. Therefore, although these free charges are available as extra current carriers, they basically have no significant effect. This is consistent with my general claim that KzDebye is very small compared to any volume current in the conductor. I think this says I can set σ = σd in my presentation argument *************************************************** Appendix A I am still worried about surface currents flowing out the front and back sides of the box: If I compress the box right to the math surface, I exclude the Debye currents, but then when I measure Jr just under the surface, I would have to use J = σE - D grad ρ Jr = σEr - D a ∂rρ so then I lose the Jr = σEr connection. So just including the Debye stuff in the bulk maybe doesn't work. I think the idea of continuous Ez at the surface says the surface charge sees the same Ez as inside the volume, say at 10A in from the surface. So maybe Jzbulk = σbulk Ez JzDebye = σDebye Ez KzDebye = λD JzDebye amps/m Let's say the box is λD thick which is on the order of 1 Angstrom. The z current through the front and back sides of the box could then be [ JzDebye(z+dz) - JzDebye(z)] λD adθ Then -jω[∫V ρ dV] = ∫S J dS -jω dq = JrdA + [ JzDebye(z+dz) - JzDebye(z)] λD adθ dA = adθdz or -jω n dA = JrdA + [ JzDebye(z+dz) - JzDebye(z)] λD adθ or -jω n adθdz = Jr adθdz + [ JzDebye(z+dz) - JzDebye(z)] λD adθ or -jω n = Jr + [ JzDebye(z+dz) - JzDebye(z)] (1/dz)λD or -jω n = Jr + ∂zJzDebye(z) λD or -jω n = Jr -jβdJzDebye(z) λD or -jω n = Jr -j(ω/vd)JzDebye(z) λD or -jω n = Jr - j(ω/vd) KzDebye Not sure what to conclude there. How about instead KzDebye = vDn(θ) = m/sec * Coul/m2 = amp/m I know the size of vD and n(θ) so I know Kz. Is there really such a current? If the pattern n(θ) really moves down the surface in the z direction at vD, then I guess yes. But then I get -jω n = Jr - j(ω/vd) KzDebye = Jr - jωn(θ) and then I end up with Jr = 0 !!! That would certainly wreak havoc on lines doe! [ this makes one suspicious of the equation KzDebye = vDn(θ) ] There must be some way to find out what really happens at the surface! 1. n(θ) must move down the wire at vD like everything else including V(z), E, B, q and so on. 2. If this is so, then this movement uses up divE = -jωρ and then Jr = 0. 3. This is a very different boundary condition on the round wire than what I was using. Where would this new boundary condition lead? 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) If I now assume that Er(a,m) = 0 => am xa-1 Jm(xa) + Jm+1(xa) = 0 jEθ(a,m) = 0 => - am xa-1 Jm(xa) + ( + ) Jm+1(xa) = 0 I can add these equations to get Jm+1(xa) + ( + ) Jm+1(xa) = 0 which says Km + = 0 and I still retain a0 = 0 as in (D.2.16) Now go back to the first equation: am xa-1 Jm(xa) + Jm+1(xa) = 0 m xa-1 Jm(xa) + Jm+1(xa) = 0 -Km m xa-1 Jm(xa) + Jm+1(xa) = 0 This equation then requires that Km = 0, so contradiction! Look back at the two conditions again, interpret as Aam + BKm = 0 Cam + DKm = 0 Interpret again as Ax + By = 0 Cx + Dy = 0 This is two straight lines through the origin. Unless both lines are the same, the only solution is x = y = 0, and that is what is happening above. Fact: If you assume that Er and Eθ both vanish on the cylinder boundary, you are forced to have E = 0 everywhere inside. [ correct ] You can think of this as coming from div E = 0 just inside the surface. Does this say Eθ ≠ 0 at the surface? That there are azimuthal surface currents after all? But then the boundary condition would change again: -jω[∫V ρ dV] = ∫S J dS dA = adθdz -jω dq = JrdA + [ JzDebye(z+dz) - JzDebye(z)] λD adθ + [Jθ(θ+dθ)-Jθ(θ)] λDdz or -jω n dA = JrdA + [ JzDebye(z+dz) - JzDebye(z)] λD adθ + [Jθ(θ+dθ)-Jθ(θ)] λDdz or -jω n adθdz = Jr adθdz + [ JzDebye(z+dz) - JzDebye(z)] λD adθ + [Jθ(θ+dθ)-Jθ(θ)] λDdz or -jω n = Jr + [ JzDebye(z+dz) - JzDebye(z)] (1/dz)λD + [Jθ(θ+dθ)-Jθ(θ)](1/adθ) λD or -jω n = Jr + ∂zJzDebye(z) λD + (1/a)∂θJθDebye(z) λD or -jω n = Jr -jβdJzDebye(z) λD + (1/a)∂θJθDebye(z) λD or -jω n = Jr -j(ω/vd)JzDebye(z) λD + (1/a)∂θJθDebye(z) λD or -jω n = Jr - j(ω/vd) KzDebye + (1/a)∂θKθDebye Now what? What about the idea that KzDebye = vDn(θ) ? If "the pattern just moves down the line", then this must still be true, and then -jω n = Jr - j(ω/vd) [vDn(θ)] + (1/a)∂θKθDebye -jω n = Jr - jω n + (1/a)∂θKθDebye 0= Jr + (1/a)∂θKθDebye and then we are back with a condition on Jr again. Then I have Jr "feeding" the azimuthal current instead of feeding n(θ). Very twisted!