compute A transverse REVIEWED

Computation of At for "some situation" PhL 3.6.14 [ 5/14/14 final review: Everything in this doc now appears in Appendix M and Section 3.7, so no need to ever read this doc again. Here is where I first developed the theory. ] This appendix contains two big items: Early Appendix M proof and φ = constant proof. In this doc I make arguments for why 3.7Ax << Az , and then why Az ≈ (βd/ω)φ, and finally why the "capacitor model" for transmission line theory is justified. I add right now the point that Az ≈ (βd/ω)φ implies that Az = constant on conductor surfaces, and that in turn requires the strong skin effect limit. See red notes below. Section 1 below now appears in Appendix M (not yet installed). Added 5/12/14. Section 3 below I now realize was my original writeup of the set of Steps needed to justify φ = constant on the conductor surfaces. This is now part of lines doc Section 3.7. 1. Show At is small: taking a wild shot with the m=0 round wires case 1 2. How can one relate Az to something we know? 5 3. Restate previous section: Justification of the Capacitor Model of Transmission Line 9 1. Show At is small: taking a wild shot with the m=0 round wires case After pondering a few methods, I try using the Helmholtz integral to estimate the Ax,y size and I end up with some typical integral forms. I then assume wide spaced so only m = 0 contributes and I can then use the App D E fields for m=0. This puts some Bessel J1 function in my integrals, but I never use this result. I go back to having Jx inside the integrals and then I make my "strong cancellation" argument. My conclusion is then (1) At is small because Jt are small to begin with; (2) At are still smaller due to cancellations in the integration. This all exists now in Appendix M, not yet installed. I want to compute At for some simple situation in order to show that it can be neglected in this equation, Et = -tφ -jωAt What methods are available? There is an ODE system you could solve ( here x = t = transverse comp) (2 + βd2)Ax = - Σi=12 μiJx but I am unsure of the BC's. It seems better to use the Helmholtz integral A(x,ω) = Σi∫μiJi(x',ω) dV' R = |x - x'| . (1.5.9) For small β we quickly simplify this to say Ax1(x,ω) = ∫Jxi(x',ω) dV' R = |x - x'| . (1.5.9) Then we add the two conductors to get Ax(x,ω) = { !Syntax Error, IJx1(x',ω) dz' dx' dy' - !Syntax Error, IJx2(x',ω) dz' dx' dy' } Now what is the next step? For longitudinal we made the ansatz that Jz1(x,y,z) = b1(x,y) i1(z) A/m2 1/m2 A (4.7.3) but that makes no sense at all for the transverse components. We can instead take Jx1(x',ω) = e-jβdz Jx1(x',y',ω) but then again we are in the small βd regime so we ignore the phase. We can then do the dz' integration in the usual way using !Syntax Error, I = -2ln(s/Λ) and set Λ = 1 to get Ax(x,ω) = - { !Syntax Error, IJx1(x',ω) ln(s12) dx' dy' - !Syntax Error, IJx2(x',ω) ln(s22) dx' dy' } Ay(x,ω) = - { !Syntax Error, IJy1(x',ω) ln(s12) dx' dy' - !Syntax Error, IJy2(x',ω) ln(s22) dx' dy' } where s12 = (x-x'1)2 + (y-y'1)2 s22 = (x-x'2)2 + (y-y'2)2 Rewrite as Ax(x,ω) = - σ { !Syntax Error, IEx1(x',ω) ln(s12) dx1' dy1' - !Syntax Error, IEx2(x',ω) ln(s22) dx1' dy1' } Ay(x,ω) = - σ{ !Syntax Error, IEy1(x',ω) ln(s12) dx2' dy2' - !Syntax Error, IEy2(x',ω) ln(s22) dx2' dy2' } In the wide twinlead simplest case maybe we can say m = 0 in each wire where we know Ez(r,0) = - j (β'/βd) J0(x) // large a0 = 0 x = β'r β'2 = β2 - βd2 Er(r,0) = J1(x) . // small = (j/2) (aβd) I Rdc jEθ(r,0) = 0 Rdc = Here is a picture, In conductor 1 we have Ex1 = Er(r'1,0) cosθ'1 = J1(β'r'1) cosθ'1 Ey1 = Er(r'1,0) sin θ'1 = J1(β'r'1) sinθ'1 In conductor 2 we have Ex2 = Er(r'2,0) cosθ'2 = J1(β'r'2) cosθ'2 Ey2 = Er(r'2,0) sin θ'2 = J1(β'r'2) sinθ'2 Now cosθ'1 = x'1/r'1 sinθ'1 = y'1/r'1 cosθ'2 = (x'2-b)/r'2 sinθ'2 = y'2/r'2 So then Ex1 = J1(β'r'1) x'1/r'1 Ey1 = J1(β'r'1) y'1/r'1 Ex2 = J1(β'r'2) (x'2-b)/r'2 Ey2 = J1(β'r'2) y'2/r'2 s12 = (x-x'1)2 + (y-y'1)2 s22 = (x-x'2)2 + (y-y'2)2 Then we have Ax(x,ω) = - σ { !Syntax Error, I J1(β'r'1) x'1/r'1ln(s12) dx1' dy1' - !Syntax Error, I J1(β'r'2) (x'2-b)/r'2 ln(s22) dx1' dy1' } Ay(x,ω) = - σ{ !Syntax Error, I J1(β'r'1) y'1/r'1 ln(s12) dx2' dy2' - !Syntax Error, I J1(β'r'2) y'2/r'2 ln(s22) dx2' dy2' } Here is what one of these four integrals looks like: ∫ dx'1 dy'1 θ( x'12+ y'12 ≤ a1) J1(β') (x'1/r'1) ( ln [(x-x'1)2 + (y-y'1)2] If I change to r,θ this becomes a1 ∫ dr1 ∫dθ1 J1(β'r1) cosθ'1 ln [ (x-r'1cosθ1)2 + (y-r'1sinθ1)2] = a1 ∫ dr1 J1(β'r1) ∫dθ1 cosθ'1 ln [ x2+ y2+ r'12 - 2xr'1cosθ1 - 2yr'1sinθ1] Now I know the θ integral can be done because I ran into this same integral the last time I tried to do this calculation! See separate doc called "a certain integral...". The result is a bit messy, and this is just for the lowest moment m = 0. But look again at the above picture: Here the red arrows show the red radial Jr currents (for m = 0 there are no Jθ currents). And here is one of our integrals Ax(x,ω) = - { !Syntax Error, IJx1(x',ω) ln(s12) dx' dy' - !Syntax Error, IJx2(x',ω) ln(s22) dx' dy' } I argue there is a HUGE cancellation effect because for each positive contribution there is a nearly offsetting negative one from the source point inverted through the origin. It is a little different because the angle to (x,y) is not the same and R1 is not the same, but ln(s12) is a very slow variation for a slightly different s1. IN the higher partial waves we have Er and Eθ but the same idea is valid, you are going to get lots of cancellation in the integration. Maybe I can find a simpler parallel plate capacitor example or something like that where I can actually compute At. Barring that, here is a conclusion. Conclusion: Although I could not do the integrals explicitly to show it, I can see that in the two-cylinders case, in any partial wave one is going to have At be very small. This is so for two reasons: (1) the transverse currents are small to begin with. (2) there is massive cancellation which takes place during the integration. 2. How can one relate Az to something we know? Here I argue two ways that Az ≈ (βd/ω)φ which "connects Az to something in the capacitor world". I then apply my previous section notion that Ax << Az to conclude Ax << (βd/ω)φ . I then examine Ex = -∂xφ - jωAx . I estimate ∂xφ ~ (1/D)φ and then I show ωAx << (2π/D) φ which then implies that Ex ≈ -∂xφ for ω > 0 and this justifies the "capacitor approach" to transmission lines, and also the notion that Et = 0 on the conductor surfaces. I try with some examples to verify that ∂xφ ~ (1/D)φ (mixed success). I have now written Appendix M which writes up the previous section pretty well, I think, but is unable to say just "how small" At really is, because I cannot easily do the integrals. Here is an alternate idea: 1. Make the ansatz that (∂xAx+∂yAy) << ∂zAz . There is some dim logic to this ansatz since we have argued above that At is "small" . With this ansatz, we can write the King gauge condition, div A = - μεjωφ - μσφ = -jωμ(ε+σ/jω)φ = -jωμξφ = -j(β2/ω) φ King gauge (1.5.5) as : ∂zAz ≈ -j(βd2/ω) φ -jβdAz ≈ -j(βd2/ω) φ Az ≈ (βd/ω) φ vd = ω/k = ω/βd Az ≈ (1/vd) φ . I have gone this path several times before. For example, in "computation of Az for two cylinders.." I did it this way starting with a different equation, Ez = -∂zφ - ∂tAz Ez = jβdφ - jω Az -jωAz(x) = Ez(x) - jβdφ(x) Then ansatz that Ez can be neglected in the dielectric so then -jωAz(x) ≈ - jβdφ(x) ωAz(x) ≈ βdφ(x) Az = (βd/ω) φ and there we have it again. Main point: this is about the only connection I have between Az and anything else other than an intractable Helmholtz integral. Next, since we have the general feeling that Ax << Az, say, we can write Ax << (βd/ω) φ or perhaps even Ax < 10-3 (βd/ω) φ and now we have our first-ever connection between the size of Ax and something from the "capacitor" world. Next, I have been trying to deal with this question. In electrostatics, Ex = -∂xφ whereas in a transmission line we have Ex = -∂xφ - jωAx In the worst way I want to claim that the second term on the right can be neglected compared to -∂xφ and that would be the basis as maintaining the DC capacitor result as quasi-static into the transmission line world. Now suppose we dimly say that ∂xφ ~ (1/D)φ (2) below in next section where D is some vague transverse dimension of the line. Then I want to show that (1/D)φ >> ωAx or ωAx << (1/D)φ // want to demonstrate this] But what I know from above is that ωAx << βd φ or ωAx << (2π/λ) φ // what I think I know Is this going to fly, or bite me in the butt? We start λ >> D (1/λ) << (1/D) I know that ωAx << 2πφ (1/λ) But from just above (1/λ) << (1/D) therefore ωAx << 2πφ (1/λ) << 2πφ (1/D) and the conclusion is that ωAx << (2π/D) φ which is more or less what I wanted to demonstrate! Now go back to this dim idea ∂xφ ~ (1/D)φ One way to argue this is to ride only the y = 0 "center line" between the two conductors which are spaced by distance D, say. Then |∂xφ| ≡ = ~ (1/D) |φ| where φ is at some typical point in the capacitor near a plate! ____________________________________________________________________________ This example did not work too well but I keep it here: Example: For the two cylinders I found that φ(x,y) = ln[ ] Now replace x and y by dimensionless variables X = x/a and Y = y/a, and the above becomes In the region of strong fields for our capacitor, these X and Y coordinates are on the order of unity (ballpark), so the polynomial ratio is on the order of unity and we then have φ ≈ order of unity = ln (ratio) ∂xφ ≈ (4/a) * order of unity **************************************************** Try this a different way. We know for the two-cylinder cap that φ(ξ) = - V (10.8) And then ∂ξφ = - V Then ∂xφ = ∂φ/∂ξ * ∂ξ/∂x = ∂φ/∂ξ / ∂x/∂ξ But I know that x = a shξ/(chξ - cosu) and then ∂x/∂ξ = - a (chξcosu-1)/(chξ - cosu)2 and then ∂xφ = ∂φ/∂ξ / ∂x/∂ξ = (1/a) V * (chξ - cosu)2 /(chξcosu-1) = - (1/a)[ - V] * (chξ - cosu)2 /(chξcosu-1) = - (1/a)[ - V] (1/ξ) * (chξ - cosu)2 /(chξcosu-1) = - (1/a)[φ] (1/ξ) * (chξ - cosu)2 /(chξcosu-1) and then ∂xφ = φ * [ - (1/a) (1/ξ) * (chξ - cosu)2 /(chξcosu-1) ] ***************** 3. Restate previous section: Justification of the Capacitor Model of Transmission Line Here I recap the logic of the previous section. 1.By one of two methods, we argue first that Az ≈ (βd/ω) φ (1) 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 previous sections, that, |Ax| << |Az| // from previous sections => |Ax| << (βd/ω) |φ| // from (1) above in particular, |Ax| < 10-3 (βd/ω) |φ| from f = 0 Hz to 1000 GHz which we rewrite slightly, ω|Ax| << βd |φ| or |ωAx| << (2π/λ) |φ| = 2π |φ| (1/λ) (3) 4. Now consider 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.