Redo Sec 4 with averaging eveywhere REVIEWED

Redo Sec 4 with averaging everywhere PhL 9.14.14 I think this is the first doc where I attempt to maintain (∂xAx12(x) + ∂yAy12(x)) in the Chapter 4 analysis leading to the transmission line equations. I made some mistakes below, such as T(z) missing from one of the transmission line pair. I am also trying to support the general notion of averaging better. All this stuff ended up in Appendix S where I think it is done OK. The doc ends with a lot of scratch work, and one item starts to show that "superposition" does not seem to fix that Jz asymmetry problem. Reviewed 10.8.14. I think the very first thing would be a new version of (4.4.1), V(x1,x2) ≡ φ12(x1) - φ12(x2) = q(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' α1(x1',y1') – !Syntax Error, Idx2' dy2' α2(x2',y2') } – q(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' α1(x1',y1') – !Syntax Error, Idx2' dy2' α2(x2',y2') } (4.4.1) Continuing on, we would then get to V(x1,x2) = q(z) {!Syntax Error, Idx1' dy1' α1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' α2(x2',y2') ln(s222/s122) } s212 = (x2-x1')2 + (y2-y1')2 s222 = (x2-x2')2 + (y2-y2')2 (4.4.6) s112 = (x1-x1')2 + (y1-y1')2 s122 = (x1-x2')2 + (y1-y2')2 . Next would come = = {!Syntax Error, Idx1' dy1' α1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' α2(x2',y2') ln(s222/s122) } = K(x1,x2) and V(x1,x2) = q(z) K(x1,x2) (4.4.7) Now I guess I can decompose C' into its real and imaginary parts and give them names as shown, C'(x1,x2) = C(x1,x2) + G(x1,x2)/(jω) (4.4.9) And next of course C(x1,x2) = 4πεd/K(x1,x2) capacitance per unit length of the transmission line G(x1,x2) = 4πσd/K(x1,x2) conductance per unit length of the transmission line so C(x1,x2) /G(x1,x2) = εd/σd . (4.4.10) I don't have to interpret anything, things like C(x1,x2) are just functions with dimensions you know. Le(x1,x2) = (μd/4π)K(x1,x2) . (4.4.11) Z0(x1,x2) = = (1/4π) K(x1,x2) (4.4.14) OK, and so ends Section 4.4. I then do the two examples, and theory resumes in 4.7. But I think we can skip down to Section 4.10 where we talk about W between the two conductors. W(x1,x2) ≡ Az12(x1) - Az12(x2) = i(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' b1(x1',y1') – !Syntax Error, Idx2' dy2' b2(x2',y2') } – i(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' b1(x1',y1') – !Syntax Error, Idx2' dy2' b2(x2',y2') } (4.10.1) and then later W(x1,x2) = i(z) {!Syntax Error, Idx1' dy1' b1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' b2(x2',y2') ln(s222/s122) } (4.10.4) W(x1,x2) = Le(x1,x2) i(z) . (4.10.7) Le(x1,x2) = = KL(x1,x2) (4.10.8) where KL is the following dimensionless real number, KL(x1,x2) ≡ !Syntax Error, Idx1' dy1' b1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' b2(x2',y2') ln(s222/s122) . (4.10.9) K(x1,x2) ≡ !Syntax Error, Idx1' dy1' α1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' α2(x2',y2') ln(s222/s122) . (4.4.8) Thus ends Section 4.10 and we start into Section 4.11. E = - grad φ - ∂tA (1.3.1) div A = - μdεd ∂tφ - μσφ . // the King gauge (1.3.18) Both the above equations are exact. But then my next step is this Ez(x) = - ∂zφ(x) - jωAz(x) ∂zAz(x) + (∂xAx + ∂yAy) = - j (βd2/ωφ(x) . (4.11.3) The first above is still exact, the second is NOT exact and I will have more to say about it later on (next section of this doc). I will nevertheless just carry on with this inexactness being allowed. [But I am now going back and adding the missing stuff in blue.] Next comes Ez(x) = - ∂zφ12(x) - jωAz12(x) (4.11.4a) ∂zAz12(x) + (∂xAx12(x) + ∂yAy12(x)) = - j (βd2/ωφ12(x) . (4.11.4b) These are implied by the previous pair, so no new assumptions. Then Ez(x1) - Ez(x2) = -∂z[φ12(x1) - φ12(x2)] - jω[Az12(x1) - Az12(x2)] (4.11.5a) ∂z[Az12(x1) - Az12(x2)] = - j (β2/ω[φ12(x1) - φ12(x2)] . (4.11.5b) ∂z[Az12(x1) - Az12(x2)] + (∂xAx12(x1) - ∂xAx12(x2) ] + (∂yAy12(x1) - ∂yAy12(x2) ] = - j (β2/ω[φ12(x1) - φ12(x2)] Again, there are no new assumptions, I just did a subtraction. I then define V(x1,x2) ≡ φ12(x1) - φ12(x2) (4.4.1) W(x1,x2) ≡ Az12(x1) - Az12(x2) (4.10.1) and this leads to Ez(x1) - Ez(x2) = - ∂zV(x1,x2) - jωW(x1,x2) (4.11.6a) ∂zW(x1,x2) = - j (βd2/ωV(x1,x2) . (4.11.6b) ∂zW(x1,x2) = (∂xAx12(x1) - ∂xAx12(x2) ] + (∂yAy12(x1) - ∂yAy12(x2) ] - j (βd2/ωV(x1,x2) Suppose for shorthand I write this as ∂zW(x1,x2) = T(x1,x2) - j (βd2/ωV(x1,x2) T = "Transverse derivatives" If I double average this equation, I get something like this ∂zW(z) = T(z) - j (βd2/ωV(z) I then average the first equation of the previous pair to get [<Ez(x1) >C1 - <Ez(x2) >C2] = - ∂z[<φ12(x1) >C1 - <φ12(x2) >C2] - jω [<Az12(x1) >C1 - <Az12(x2) >C2 ] . We now redefine V and W to be the averages appearing in these equations, along with Ez1 and Ez2 : Ez1(z) ≡ <Ez(x1) >C1 = (1/P1) ∫C1 ds1 Ez(x1) Ez2(z) ≡ <Ez(x2) >C2 = (1/P2) ∫C2 ds2 Ez(x2) V(z) ≡ <φ12(x1) >C1 - <φ12(x2) >C2 = <V(x1,x2)>C1,C2 W(z) ≡ <Az12(x1) >C1 - <Az12(x2) >C2 = <W(x1,x2)>C1,C2 (4.11.7) with this result [Ez1(z) - Ez2(z)] = - ∂z V(z) - jω W(z) . (4.11.8) I think the above equation with its averages is exact! Meanwhile, the surface impedances on C1 and C2 are defined by (see C.2.1) , Ez1(x1) = Zs1(x1) i1(z) Ez2(x2) = Zs2(x2) i2(z) (C.2.1) which we average in the same way to obtain Ez1(z) = Zs1 i1(z) Zs1 ≡ (1/P1) ∫C1 ds1 Zs1(x1) Ez2(z) = Zs2 i2(z) Zs2 ≡ (1/P2) ∫C2 ds2 Zs2(x2) . (4.11.9) There will be some location on C1 where Ez1(x1) and thus Zs1(x1) will be maximal (for example on the walls of the gap in Fig 4.12). Referring to this value as Zs1,max we can define p1 ≡ (Zs1/Zs1,max)P1 p2 ≡ (Zs2/Zs2,max)P2 (4.11.10) Then using i(z) = i1(z) = -i2(z) and (4.11.9), rewrite (4.11.8) and (4.11.6b) as [Zs1 + Zs2] i(z) = - ∂z V(z) - jω W(z) ∂zW(z) = - j (βd2/ωV(z) + T(z) (4.11.11) Again, the first equation here is exact in terms of my averages, but the second is NOT exact for the reasons mentioned above going back to (4.11.3). I now back up to W(x1,x2) = Le(x1,x2) i(z) . (4.10.7) which do a double average of the functions to get W(z) = Le i(z) and then this is also exact, given the averages. Then install this to get (Zs1 + Zs2) i(z) = - ∂zV(z) - jω Le i(z) Le ∂z i(z) = - j (βd2/ωV(z) + T(z) which we then rearrange as ∂zV(z) = - [ Zs1+ Zs1+ jωLe] i(z) ∂z i(z) = - [ jβd2/(ωLe)] V(z) + T(z)/Le (4.11.14a) Again, the second equation is not exact (but I made it exact in blue). These are the classical transmission line equations. They are usually written in this form: [ ∂/∂z = d/dz] = - z i(z) = - y V(z) + T(z)/Le with z = R + jωL y = G +jωC . (4.11.14b) Zs1 ≡ (1/P1) ∫C1 ds1 Zs1(x1) Zs2 ≡ (1/P2) ∫C2 ds2 Zs2(x2) V(z) ≡ <φ12(x1) >C1 - <φ12(x2) >C2 = <V(x1,x2)>C1,C2 W(z) ≡ <Az12(x1) >C1 - <Az12(x2) >C2 = <W(x1,x2)>C1,C2 (4.11.7) Le ≡ < Le(x1,x2)>C1,C2 Thus, we end up with R,L,C,G which are functions of the above averages. I don't have any great interpretations for the averages, but they are just averages of the functions shown. So we have z = R + jωL = Zs1 + Zs2 + jωLe (4.11.17) y = G + jωC = jβd2/(ωLe) (4.11.18) where the Z's and Le are the averaged values. But what about the K = KL business? Despite the averaging done above, I think there are still some loose ends that I have to deal with. Go back to: = = K (4.4.7) Le = = KL (4.10.8) Here we have to regard this as saying < >C1,C2 = 1/q(z) * <V(x1,x2)>C1,C2 = < K(x1,x2)>C1,C2 and then I define 1/C' ≡ < >C1,C2 so 1/C' = V(z)/q(z) = K Question: Is this true: < >C1,C2 = 1 / [ <C'(x1,x2)>C1,C2 ] The answer is no. For example, ∫ dx 1/x ≠ 1/ ∫dx x over some interval. So be careful! I think it is still OK to define C' as I have done it 2 lines above. In the same way everything here is an average Le = = KL // this average seems clear There is more to do. Consider some next equations: ∂z i(z) = - [ jωC'] V(z) + (1/Le)T(z) . [ I omitted this! ] (4.11.20) ∂z i(z) = - [ jβd2/(ωLe)] V(z) + T(z)/Le [ updated] (4.11.14a) Rewrite them both in this manner which is more suitable for talking about averaging, - V(z) = ∂z i(z) -V(z) = ∂z i(z) Le Now set these equal to find that = Le We could then average this equation to get the same result with 1/C' and Le being the averages discussed above. THEN we can say (using the averaged values), LeC' = βd2/ω2 = μdξd . // see (1.5.1a) regarding βd2 (4.11.21) y = G + jωC = jβd2/(ωLe) = j (βd2/ω2) (ω/Le) = j (LeC') (ω/Le) = jωC' . (4.11.22) We NOW have z = R + jωL = Zs1 + Zs2 + jωLe (4.11.17) y = G + jωC = jωC' (4.11.18) where everything the Z's and Le and C' are the averages above. y = G + jωC = jωC' = jω(ξd/εd)C = jω (1 - jσd/εdω)C = jωC + (σd/εd)C (4.11.24) Once we have C' by doing the average of 1/C'(x1,x2), we can use C' as a constant. For example, C'/C = qc/qs = (ξd/εd) . (4.11.23) (1/C) = (ξd/εd) (1/C') (1/C(x1,x2)) = (ξd/εd) (1/C'(x1,x2)) Now apply the average and we get (1/C) = (ξd/εd) (1/C') C'/C = qc/qs = (ξd/εd) where C' and C are well defined quantities. More equations that are still true in averages. LeC' = μdξd // I just did this above, both are averages (4.11.27) LeC = μdεd = 1/vd2 // just using result above (4.11.28) Recall now that in terms of averages, Le = KL . (4.11.29) Now go back to this result, in terms of averages, LeC' = βd2/ω2 = μdξd write as Le = μdξd but this says KL = μdξd K and finally we have our KL = K in terms of averages. Conclusion: I think my averaging idea is in fact exact and unambiguous, and in itself, it puts no frequency range limitation on the resulting transmission line equations. Possible Break But now we go back to the assumption made earlier. In (4.11.3) I write div A = - j (βd2/ωφ . // the King gauge, see (1.5.1a) and (1.5.5) re βd2 (4.11.2) ∂zAz(x) = - j (βd2/ωφ(x) . (4.11.3) so I am assuming we can ignore the transverse At component derivatives. In new App M I claim this is justified for the non-derivatives all the way down to DC for the At but I say nothing about derivatives. Now look back at this: => φ/vd - Az = Ez/(jω) (3.7.6) Inside a good conductor Ez is small to begin with, and in the limit δ → 0 (ω→∞) the right side of (3.7.6) is small due to this fact and due to the 1/ω factor. Therefore, Az ≈ φ/vd small or extreme skin effect (3.7.7) Now I would guess that (3.7.7) is NOT valid as ω→ 0 since in that case Ez/(jω) becomes very large despite the fact that Ez is very small! So this might be a key fact. This is my Step 1 result. Note for manana: There is something afoot here. Before I derive the TL equations (with averaging), I do get to the steps shown above as (4.11.2) and then (4.22.3). This is then the source of the second equation needed to get the TL equations! Thus, I really am assuming that ω is larger than some amount!!! I think I have finally found something! I will pursue this tomorrow. It is now manana. I have three arguments for the same conclusion: Argument A. We have to go back to Chapter 3 and review this logic sequence: 1. If ω is large, then Az ≈ φ/vd as shown in (3.7.7), using the strict Ez = -∂zφ - jωAz relation. 2. If | ∂xAx+∂yAy| << |∂zAz|, then Az ≈ φ/vd as shown below (3.7.8), this using the King gauge. From these two facts I conclude that If ω is small, then this is NOT true: | ∂xAx+∂yAy| << |∂zAz|, Argument B. I reach this same conclusion roughly in doc " App M on the subject of derivatives of Ax and Az" by considering the Helmholtz integrals. Argument C. Write the above condition replacing ∂z → -jk : | ∂xAx+∂yAy| << |kAz|, Large ω means large k and that makes the above be true. Small ω means small k and that makes it be NOT true. In a cruder sense, we might write (1/D) |Ax| << (1/λ) |Az| ? Then this would be true of λ is not too large relative to D, which means moderate ω. But for low ω, λ is very large and then this is NOT true. In the extreme transmission line limit, λ is very large, ω is very low, and then this is NOT true. OK, let's now back up to the derivation of the TL equations. OK, I have included the omitted transverse terms in the above math and I end up with this = - z i(z) = - y V(z) + T(z)/Le with z = R + jωL y = G +jωC . (4.11.14b) where blue shows the now-extra term. Diff to get ∂z2V(z) = - z ∂z i(z) ∂2i(z) = - y ∂zV(z) + (1/Le)∂zT(z) We then get ∂z2V(z) = -z [- y V(z) + T(z)/Le] ∂2i(z) = - y [-z i(z) ] + (1/Le)∂zT(z) Then the transmission line 2nd order equations become [∂z2 - zy ] V(z) = (1/Le)T(z) [∂z2 - zy ] i(z) = (1/Le)∂zT(z) Without the blue right sides, these equations have the solution e-jkz where k2 = - zy . With the blue right side, the solutions do NOT have this simple exponential form!!! Thus, the ansatz of Appendix D is not realized. In this case, the solutions have to be integrals over k. This is then a REASON why I get anomalous results as ω → 0. OK, what implications does this have? 1. For "large ω", the ansatz form e-jkz is justified and k2 = - zy. Thus, in the solutions of Appendix D, we set k = -j . 2. For "small ω", since the TL equations are no longer valid, the ansatz form e-jkz is not justified. In this case, the general form of an Appendix D partial wave solution is this Ez(r,m,z) = (1/4) ηm I Rdc∫dk C(k) fm(r) e-jkz Er(r,m,z) = (j/4) ηm I Rdc∫dk C(k) gm(r) e-jkz Eθ(r,m.z) = (1/4) ηm I Rdc∫dk C(k) hm(r) e-jkz where a coefficient function C(k) determines the distribution of k in the solution. For large ω then we must have C(k) → δ(k - [-j]) For small ω, C(k) must be determined by solving the full problem including the matching boundary conditions on all fields (or potentials) at the two conductor surfaces, a problem we have not attempted. Note: We know the solutions for a given value of k, namely. Ez(r,m) = (1/4) ηm I Rdc (aβ') fm fm = [ - ] x = β'r Er(r,m) = (j/4) ηm I Rdc (ak) gm gm = [ + ] xa = β'a Eθ(r,m) = (1/4) ηm I Rdc (ak) hm hm = [ - ] This is the solution for some value k = k1. The solution has the same form for k = k2. I guess I could do this linear combination to form a third solution: Ez(r,m)(3) = Am Ez(r,m; k1) + Bm Ez(r,m; k2) Er(r,m)(3) = Am Er(r,m; k1) + Bm Er(r,m; k2) Eθ(r,m)(3) = Am Eθ(r,m; k1) + Bm Eθ(r,m; k2) which really says Ei(r,m)(3) = Am Ei(r,m; k1) + Bm Ei(r,m; k2) If I now look at the 4 equations of (D.1.20), what happens? We know that the z equation says [∂r2 + (1/r) ∂r + (1/r2) ∂θ2 + ∂z2 + β2 ] ej(ωt-k1z)Ez(r,θ;k1) = 0 . (1) [∂r2 + (1/r) ∂r + (1/r2) ∂θ2 + ∂z2 + β2 ] ej(ωt-k2z)Ez(r,θ;k2) = 0 . (1) Then I suppose [∂r2 + (1/r) ∂r + (1/r2) ∂θ2 + ∂z2 + β2 ] Ez(r,θ)(3) = [∂r2 + (1/r) ∂r + (1/r2) ∂θ2 + ∂z2 + β2 ] { Am Ei(r,m; k1) + Bm Ei(r,m; k2) 3. Thus, for small ω, the fact that our k = -j Appendix D solution set has the two anomalies already noted can be explained as being due to the fact that the z dependence is not given by e-jkz for low ω, but rather must be a linear combination as shown with C(k) which must be determined. It is a pretty complicated explanation, but I have been looking for an explanation for a long time and have tried everything I could think of. So let's now go back to the critical step in Chapter 3. There I started with this φ/vd - Az = Ez/(jω) (3.7.6) I argued that for large ω, this says Az ≈ φ/vd small or extreme skin effect (3.7.7) I then show that this same result obtains from the King gauge condition , (∂xAx+∂yAy) - jβdAz = -j (βd2/ω)φ . (3.7.8) if we neglect the transverse terms. I conclude that Fact: The transverse derivative terms can be neglected if | φ/vd - Az | >> |Ez/(jω)| . This is all inside a conductor where Ez is small. You might crudely say V/vd >> Ez/ω Do I have an estimate for Ez ? It is Jz/σ so condition is V/vd >> Jz/(σω) In a ballpark sense, maybe Jz = I/area so we then have V/vd >> I/(σω * area) ωZ0 >> vd / (σπa2) ω >> [vd / (σπa2)] (1/Z0) Here I am already thinking in the low ω range with this area statement. dim(ω) = m/sec * m-2 * ohm-m * 1/ohm = 1/sec = OK Now at low ω maybe write Fact 8: The small ω limit for Z0(ω), assuming ωd = 0, is given by (Q.8.6) Re(Z0) ≈ 1/ + (1/2) [Ldc/] + O(ω3/2) Im(Z0) ≈ - 1/ + (1/2) [Ldc/] + O(ω3/2) So the general idea is Z0 = (1/)(1-j) |Z0| = (1/) Then our inequality is ω >> [vd / (σπa2)] * / or >> [vd / (σπa2)] / But for a twinlead each of radius a, perhaps Rdc = 2 σ/(πa2) and we then have ω >> [vd / (σπa2)]2 (C/Rdc) Now we can estimate Rdc = 2 * 1/(σπa2) so 1/Rdc = (1/2) σπa2 and then we get ω >> [vd2 / (σ2π2a4)] (C) (1/2) σπa2 or ω >> [vd2 / (σπa2)] (C) (1/2) or ω >> [Cvd2 / (2πσa2)] Another dimension check dim(RHS) = far/m * m2/sec2 * m-2 * ohm-m = far * 1/sec2 * ohm = far-ohm/sec2 = sec-1 OK Now we know that vd2 = 1/(μdεd) so we have ω >> [C/ (2π μdεdσa2)] We also know that C = 4πεd/K so then we have ω >> [4πεd / (2π K μdεdσa2)] = 2 / (K μdσa2)] or ωμdσ >> (2/Ka2) Now δ ≡ , and if we assume the same μ we have δ2 = 2/(ωμσ) => ωμdσ = 2/δ2 and then 2/δ2 >> (2/Ka2) δ2 << Ka2 (δ/a) << This is how ALL inequalities end up, something about the ratio δ/a being small. Plan A. Let's write a new section 4.11 (g) which discusses averaging being OK at any ω, but the TL equations are not valid for low ω . Perhaps that is a good place to insert this new stuff. Or should I just rewrite ALL of Chapter 4 in the full x1, x2 notation? That will really mess things up a lot. Maybe I have a section reviewing the averaging idea, and then another section on the low ω implication. 9/16/14. So exactly how does this explanation remove my two paradoxes? We now have Ez(r,m,z) = (1/4) ηm I Rdc∫dk C(k) fm(r) e-jkz Er(r,m,z) = (j/4) ηm I Rdc∫dk C(k) gm(r) e-jkz Eθ(r,m.z) = (1/4) ηm I Rdc∫dk C(k) hm(r) e-jkz It would then seem that Ez(r,m,z) / Ez(r,0,z) = ηm ∫dk C(k) fm(r) e-jkz / ∫dk C(k) f0(r) e-jkz or = ηm There could be a wide spectrum of k values in C(k), so I cannot take any particular limit of fm. Is it possible that the upper integral on the right side could be 0 ? In fact, the low ω requirement really is ∫dk C(k) fm(r) e-jkz = 0 for all m, all r, all z, as ω→ 0 Now C(k) can have either sign, but how can this possibly be true as a function of z ? F(m,β,r,z ) = 0 as ω → 0 ??