try to show phi = constant at low w REVIEWED

Try to show φ = constant for low ω PhL 5.12.14 Review 5.14.14: This is a very recent doc (2 days old) written after Section 3.7 was updated. In that section now I allow as you might say that φ ≈ constant in the low ω regime based on the little picture where Az is "ballpark constant" at DC. But I obtain here no silver bullet to prove what ought to be easy to prove: that φ = constant applies at ω = 0 up at least to some low ω1 . Part I of 2 I tried to show this for about an hour today and I got nowhere, so for now I give up. It seems odd that I can show that φ = constant for ω = 0 and ω = strong and extreme skin effect, but I can't show it for low frequencies. That just seems wrong to me. Basically we have to show (ignore abs values in this doc) that on the surface at least, ωAt << tφ In terms of my little coordinate system of Section 3.7, this says I have to show ωAx << ∂xφ and same for y You might think the gauge condition would play a role in the proof. I do claim in Step 2 that |∂xφ| ≈ ≈ ≈ (1/D) |φ| ballpark which at least gets rid of the ∂x problem. But then I have to show that ωAx << φ/D How do I get an estimate for Ax ? I do think I know that Ax << Az. Suppose I could show that ωAz << φ/D Again, I need some estimate for Az. In Appendix G I compute Az for a DC round wire and find Az(a) = - [Iμ1/2π] ln(a) at the surface. So then I would have to show that ω Iμ1/2π] ln(a) << V/D say This requires a relation between I and V, but I don't have that yet! Also, by using Ax << Az I am giving up a lot of space relative to <<, so that is not a good pathway. Can Appendix M help? It says that Ax(x) = - Σi μi∫Jx(x',y') ln(s2) dx' dy' I just don't see at this point how to relate any current like Jx to V. Maybe one could do an "after the fact" ansatz thing here. Let's go back to showing ω Iμ1/2π] ln(a) << V/D This then becomes ω μ ln(a) << 2π Z0/a But as ω increases, this gets into trouble, and my estimate for Az is DC only. ************************ I can use the equation Ez = jβdφ - jωAz = j(ω/vd)φ - jωAz = jω [ φ/vd - Az ] . βd = (ω/vd) (3.7.6) or φ/vd - Az = Ez/(jω) Maybe I need some better words at this point. I know in a round wire that Ez= Jz/σ ≈ (I/area*σ) low ω but on the other hand skin effect boosts Jz so that Jz = I / [δ 2πa ] so then in that limit Ez = I / [δ 2πa σ ] δ ≡ δ2 = 2/(ωμσ) Ez = I / [ 2πa σ ] = = Ez/ω = const / ω1/2 But this is all for strong and extreme skin. As noted above, for low ω I get Ez = Jz/σ ≈ (I/πa2σ) Ez/ω = (I/πa2σ)(1/ω) Then I would have to show that the RHS is small here: φ/vd - Az = Ez/(jω) ≈ (I/πa2)(1/σω) But that is only true as (1/ωσ) → ∞. But (1/ωσ) = μδ2/2 so I am back to needing δ → 0 to make this work. Part II of 2 Again, I am trying to show that I can ignore the last term Et = -tφ - jωAt (3.7.2) I do know that |At| < 10-4 |Az| so can I find some upper bound for ωAz ? I guess I can continue with the assumption of Step 2 that |tφ| ≈ (1/D)|φ| . Question: Do I have some real world numbers for Az anywhere? The way I think of φ ~ V, what can I say about the general size of Az? How about this argument. I know that φ(x,y,z) = q(z) φt(x,y) (5.1.1) Az(x,y,z) = i(z) Azt(x,y) . (5.2.1) and I know that [ t2 + (β2-k2)] φt(x,y) = 0 φt(C1) = K1 φt(C2) = K2 K1- K2 = K (5.3.10) [ t2 + (β2-k2)] Azt(x,y) = 0 Azt(C1) = W1 Azt(C2) = W2 W1- W2 = K (5.3.11) So if would seem that both φt and Azt have the same scale, which is K, and I do know a few things about K. So then I get something like this: φ(x,y,z) ≈ q(z) K (5.1.1) Az(x,y,z) = i(z) K (5.2.1) It the first one realistic? It is in fact exact for a difference, so it says V = q(z) K or C = V/q = K and this is the result I always use, so yes it is realistic. Then I should be able to say Az(x,y,z) = i(z) K = q(z) K 4πε = vd φ(x,y,z) 4πε = vd με φ(x,y,z) = vd (1/vd)2 φ(x,y,z) = φ/vd But this is the result (3.7.7) that I claimed was only valid for skin effect. But now I am claiming this is ballpark reasonable at ANY frequency all the way down to DC, for a lossless line. Of course then lossless means skin effect! Question: Are φt(x,y) and Azt(x,y) exactly the same? Well, maybe at extreme skin effect where then the boundary conditions on Azt really are constant on both those conductors. But the answer is NO at lower frequency where we KNOW that Az is not constant on the conductors.