Assemble the Facts INSTALLED

Assemble the Facts in Sequence PhL 3.6.14 [5.11.14] A version of all this stuff is now installed in Section 3.7. I could not show φ = constant for the entire range of ω, just ω = 0 and high ω. I would have thought low ω proof would be easy, but I don't know how to do it. Perhaps it is not true, but I don't need this information. Final review 5/15/14. For the time being, I am going to be thinking always of the two cylinder problem rather than the general case. Later I can hopefully generalize. Fact 1: Et = 0 around a conductor cross section perimeter at constant z. [ for a while I was confused by using Et for both tangential and transverse E, but this is cleared up now in lines doc and Et is the transverse while Eθ is the tangential ] Here Et refers to E tangential to the conductor surface in the cross section picture. For the two-cylinder problem, this is Eθ with θ defined as in bipolar doc. In Section D.8 (a) we argue that this "quasi static" approximation is valid up to 1014 Hz which is 100,000 GHz, well beyond transmission lines of interest to us. Fact 2: In the King gauge, for a transmission line operating in the transmission line limit, the transverse vector potential is very small: |At| < 10-4 |Az| for f = DC to 1000 GHz. This is demonstrated in Appendix M, see (M.22). [ added this as a separate item ] Fact 3: φ = constant on the conductor cross section perimeter at all ω of interest. At ω = 0 ("electrostatics") we know that Et = 0 implies that scalar potential φ = constant on a conductor surface, since Et = -tφ and Et = 0 . For ω > 0, we have instead from (1.3.1) that Et = -tφ -jωAt . In order to continue to claim that φ = constant on a conductor cross section, we must show that -jωAt can be neglected relative to -tφ . Then we will have Et = -tφ and by Fact 1 we know Et = 0 for ω > 0 and then once again φ must to constant on each conductor cross section surface. We shall now demonstrate this fact in a series of 6 steps: 1. By either of two methods shown below, we argue first that Az ≈ (βd/ω) φ . // note that (βd/ω) = 1/vd (1) [ Note: Both methods are OK, and both methods implicitly assume the strong skin effect regime! The two methods appear in my recent Review of Pardox 1 and 2.doc ] Method A: Make the ansatz that ( ∂xAx+∂yAy) << (∂zAz) based on the Appendix M claim that |Ax,y| < 10-4 |Az| from DC to 1000 GHz. [ certainly a fudge! ] Then the King gauge says (in the dielectric), ∂zAz ≈ div A = -j (βd2/ω)φ . (1.5.5) As with all transmission line quantities, we assume Az(x,y,z) = e-jβz Az(x,y) so then -jβdAz(x,y) ≈ -j (βd2/ω)φ(x,y) => Az ≈ (βd/ω)φ Method A Revisited. Imagine as done below that ∂xAx+∂yAy ≈ (Ax + Ay)/D where D is some transverse dimension. Just simplify and write this as ∂xAx+∂yAy ≈ 2(Ax/D). Now to achieve Method A's goal, we have to show that ( ∂xAx+∂yAy) << (∂zAz) or (2/D) Ax << - jβdAz or (2/D) Ax << - j(2π/λ)Az or (λ/D) << (Az/Ax) // all should be absolute values or (D/λ) >> (Ax/Az) or (Ax/Az) << (D/λ) (*) Now we supposedly know from Appendix M that (Ax /Az) < 10-4 but this fact does not tell us anything about whether (Ax/Az) << (D/λ) . For example, we might have for some transmission line that (Ax/Az) = 10-4 but (D/λ) = 10-6 and then the << thing is false. So I guess I am arguing here that Method A is based on the claim that ( ∂xAx+∂yAy) << (∂zAz) but I know that this claim really can only be justified in the strong skin effect limit, and so Method A as stated above is an invalid argument. That is a good thing, since it does not require the skin limit. Method B: From (1.3.1) we have Ez = -∂zφ - ∂tAz. We know that Ez is extremely small inside and just outside a conductor surface (and vanishes for a perfect conductor), so ∂zφ ≈ - ∂tAz => -j βdφ ≈ -jωAz => βdφ ≈ ωAz => Az ≈ (βd/ω)φ This is a valid argument, since Ez extremely small requires strong skin limit! 2. We then argue that for φ evaluated near one of the conductors in the active region between the conductors, |∂xφ| ≈ ≈ ≈ (1/D) |φ| D = general transverse dimension (2) 3. We argue next, based on Appendix M and Fact 2 above : |Ax| << |Az| and then setting |Az| ≈ (βd/ω) |φ| according to item (1) above, we get |Ax| << (βd/ω) |φ| which we rewrite slightly, ω|Ax| << βd |φ| or |ωAx| << (2π/λ) |φ| = 2π |φ| (1/λ) (3) 4. Now consider the transmission lime limit inequality, λ >> D (1/λ) << (1/D) 2π |φ| (1/λ) << 2π |φ| (1/D) . (4) Thus from (3) and (4) and then finally (2) |ωAx| << 2π |φ| (1/λ) << 2π |φ| (1/D) ≈ 2π |∂xφ| . From this we conclude, ignoring the 2π factor, |ωAx| << |∂xφ| (5) 5. Finally, consider Ex = -∂xφ - jωAx . Based on (5) , we can ignore the last term and we get Ex ≈ -∂xφ (6) 6. Making this same argument for ∂yφ, we end up with Et ≈ - t φ which says that the transverse electric field Et for a transmission line operating at frequency ω > 0, but operating in the transmission line limit of λ >> D, is the same as the electrostatic electric field Et obtained from the ω = 0 DC capacitor problem. This is the basis of our "electro-quasi-static" approach to the transmission line analysis. Thus for any ω > 0 up to 1000 GHz the conductor cross section surfaces each have constant φ, just as they do at ω = 0. [ On 5/8/14 I think the above 6 step proof is OK. The argument requires (1) the transmission line limit and (2) the strong skin effect regime and (3) Appendix M to get the conclusion. ] Comment: If we are at low ω, then the strong skin is not true, and the above proof fails, though the conclusion may still be true. _________________________________________________________________________________ Commentary: So above I have supposedly nailed down the idea that φ = constant on a conductor accurate to 10-4 at any ω including low and high frequencies. This would seem to validate the V(z) portion of Ch 4. Above I argue by Method A and Method B that Az = (βd/ω)φ + 10-4*stuff. Question: if φ = constant on conductor, doesn't this mean Az= constant as well to this same 10-4 accurate at all ω ? So this would say Az = constant at DC, which I KNOW is not the case. Something is confused here, as usual. I think I have resolved this last issue.