Show that linear combinations solve Appendix D REVIEWED

Show that linear combinations solve Appendix D PhL 9.16.14 (1) I show that you can superpose solutions of different k, discretely or continuously. (2) I do the Smythian Form analysis in Part C and clarify the m=0 case in Part D. (3) I show that Jz asym does not go away if you superpose. (4) Much of this stuff is quoted now in Section 7. 2 (5) I never resolved the paradox regarding superposition mentioned in "low freq discussion". Reviewed on 10.8.14. Preliminary Question: 1 Part A: First shot at the superposition of k idea (discrete sum of two solutions) 2 Part B: Generalization from 2-term superposition to continuous superposition 4 Part C: Smythian Approach to the Continuous Superposition 5 Part D: Repeat Part C with focus on the m = 0 partial wave 13 Part E: The Jz Asymmetry Paradox is still present with a Superposed Solution 17 Reflection Example: 18 Eliminate the two terms m and -m possible escape hatch. 18 I thought this would be obvious, but after fiddling a bit, it seems less obvious. Preliminary Question: If you consider the single-k solution and the Helm equation and the div E = 0 equation and the two boundary conditions, does this system determine the SCALE of the solutions or not? Answer: Here is the App D solution where I is replaced by 2πω (a/k) N0 Ez(r,m) = (1/4) ηm 2πω (a/k) N0 Rdc (aβ') fm fm = [ - ] (**) Er(r,m) = (j/4) ηm 2πω (a/k) N0 Rdc (ak) gm gm = [ + ] Eθ(r,m) = (1/4) ηm 2πω (a/k) N0 Rdc (ak) hm hm = [ - ] We know that N0 = q/(2πa) and q = CV so you can replace N0 = [C/2πa] V. Then each Ei is proportional to V. Since you can apply any V you want to a TL, you can have any scale you want for the solution set. Alternative Answer: Here is the App D solution in its usual form, Ez(r,m) = (1/4) ηm I Rdc (aβ') fm fm = [ - ] Er(r,m) = (j/4) ηm I Rdc (ak) gm gm = [ + ] Eθ(r,m) = (1/4) ηm I Rdc (ak) hm hm = [ - ] We know that I = V/Z0 where Z0 = . Once again, we end up with each Ei being proportional to V, and so yes we have that same scaling idea. Comparing the two Forms It must be true that I = 2πω (a/k) N0 = 2πa (ω/k) [C/2πa] V = V/Z0 . Then it must be true that 2πa (ω/k) [C/2πa] = 1/Z0 or (ω/k) [C] = 1/Z0 or k/Z0 = ωC Is this true? We can install k and Z0 to get k/Z0 = -j * = -j (G+jωC) But in App D for the CP BC I assume that G = 0, so then k/Z0 = -j (0+jωC) = -j2ωC = ωC and so the two forms do in fact compare OK. I show in separate paradox doc that if you allow G > 0, then you have a new N0(new) = N0(old) * (ξd/εd) requiring that k/Z0 = ωC' which is also true, just another internal check of the theory that works! Conclusion: Solution Ei are proportional to V, and you could double V to double all solutions. Part A: First shot at the superposition of k idea (discrete sum of two solutions) Here I show that you can superpose two solutions of Helm to get a new Helm solution. This is done in θ space, so there is no m label on the coefficients C(k,ω) and C(k',ω). I then treat the n(θ) boundary condition incorrectly which gives the incorrect requirement that C(k,ω) + C(k',ω) = 1. In m space, the superposition has θ replaced by m, but the coefficients of course do not depend on m since they are the same coefficients used in θ space. 1. Suppose I know that (2 + β2) E(r,θz,t; k) = 0 single-k solution for k (2 + β2) E(r,θz,t; k') = 0 single-k solution for k' Then create this linear combination solution E(r,θz,t)(3) = C(k,ω)E(r,θz,t; k) + C(k',ω) E(r,θz,t; k') // dim(C) = 1 for discrete sum We then find that (2 + β2) E(r,θz,t)(3) = C(k) (2 + β2) E(r,θz,t; k) + C(k') (2 + β2) E(r,θz,t; k') = C(k,ω) 0 + C(k',ω) 0 = 0 Thus, our linear combination is also a solution of the full vector Helmholtz equation. The two simple solutions have these forms, E(r,θz,t; k) = ej(ωt-kz) E(r,θ; k) E(r,θz,t; k') = ej(ωt-k'z) E(r,θ; k') and then E(r,θz,t)(3) = C(k,ω) ej(ωt-kz) E(r,θ; k) + C(k',ω) ej(ωt-k'z) E(r,θ; k') 2. Go to partial waves: I think each of the three functions E(r,θz,t; k) E(r,θz,t; k') E(r,θz,t)(3) goes to partial waves in the obvious manner. Thus we end up with E(r,mz,t)(3) = C(k,ω) E(r,mz,t; k) + C(k',ω) E(r,mz,t; k') . Notice that the coefficients do not depend on m, they just come through from the θ-space forms. 3. What happens to the boundary conditions? The new boundary conditions must be in θ space Er(r=a-ε,θ)(3) = (jω/σ) n(θ) Eθ(r=a,θ)(3) = 0 But if we write out the expansions, we end up with C(k,ω) Er(r=a,θz,t; k) + C(k',ω) Er(r=a,θz,t; k') = (jω/σ) n(θ) C(k,ω) Eθ(r=a,θz,t; k) + C(k',ω) Eθ(r=a,θz,t; k') = 0 The above is wrong because we really need some n(θ,z) = n(θ) r(z) on the right. I do this better below. That is, for a single-k you could cancel e-jkz on both dies, but not for mult k. Note: the fields all depend on ω as well, I just don't display that fact. We can prove the second one by assuming that the single-k solutions each satisty the BC, but the first BC is unclear! I think n(θ) is determined by the electrostatic problem (especially at very low ω), so you cannot adjust n(θ). We then end up assuming that each of the single-k solutions satisfies the same BC Er(r=a,θz,t; k) = (jω/σ) n(θ) Er(r=a,θz,t; k') = (jω/σ) n(θ) then we are forced to have Er(r=a-ε,θ)(3) = C(k,ω) Er(r=a,θz,t; k) + C(k',ω) Er(r=a,θz,t; k') = C(k,ω) (jω/σ) n(θ) + C(k',ω) (jω/σ) n(θ) = [ C(k,ω) + C(k',ω)] (jω/σ) n(θ) The only way to rescue things is to require that C(k,ω) + C(k',ω) = 1 wrong for reason shown in red above If I insist on this restriction, then E(r,mz,t)(3) does indeed satisfy both boundary conditions as well as the vector Helmholtz equation! I think the derived B fields would be the obvious linear combinations, and the verification of Maxwell's equations would then go through as before. 4. What happens to quantity I when there is a linear combination solution? The simple form N0 = (k/2πωa) no longer applies. So rather than updating I right now, let's just not use I in our solution forms for the single-k solutions. That is to say, replace I by 2πω (a/k) N0 in those solutions, which then read: Ez(r,m) = (1/4) ηm 2πω (a/k) N0 Rdc (aβ') fm fm = [ - ] Er(r,m) = (j/4) ηm 2πω (a/k) N0 Rdc (ak) gm gm = [ + ] Eθ(r,m) = (1/4) ηm 2πω (a/k) N0 Rdc (ak) hm hm = [ - ] Part B: Generalization from 2-term superposition to continuous superposition Here I just state the candidate continuous superposition called E(3). This is an explicit integral with arbitrary coefficient C(k,ω). As happened in Part A, in going from θ to m space, the C(k,ω) continue to have no m index. I then show that the double expansion Σm ∫dk for Ez(r,θz)(3) has the exact Smythian atomic superposition form you would expect with a certain specific cm(k,ω) = [stuff(m)] C(k.ω). 5. Generalization of the results: Here we must mimic the above sum of two terms to get: E(r,θz,t)(3) = ∫dk C(k,ω) E(r,θz,t; k) = ∫dk C(k,ω) ej(ωt-kz) E(r,θ; k) E(r,mz,t)(3) = ∫dk C(k,ω) E(r,mz,t; k) = ∫dk C(k,ω) ej(ωt-kz) E(r,m; k) ∫dk C(k,ω) = 1 in order to meet the CP BC. this line wrong, ignore it Note: Now dim(C) = m, since we require dim[dk C] = 1. More specifically [ note that I is not used ] using boxed result (**) above Ez(r,mz,t)(3) = ∫dk C(k,ω) [- ]ej(ωt-kz) (1/4) ηm 2πω (a/k) N0 Rdc (aβ') Jm(β'r) or Ez(r,mz)(3) = (1/4) Nm 2πωa2 Rdc ∫dk C(k,ω) [- ]e-jkz (β'/k) Jm(β'r) = ∫ dk cm(k,ω) e-jkz Jm(β'r) Added: dim [ (1/4) Nm 2πωa2 Rdc] = Cou/m2 m2/sec ohm/m = amp-ohm/m = volt/m OK and going back to θ space Ez(r,θz)(3) = Σm ∫dk cm(k,ω) { e-jkz Jm(β'r) ejmθ } Note Added: dim[dk cm] = (1/m) dim[cm] = V/m so must have dim[cm] = V. which does have my general Smythian structure with m and k dependence of the coefficient. And cm(k,ω) = (1/4) Nm 2πωa2 Rdc [- ] (β'/k) ] C(k,ω) = (1/2) Nm (ω/σ) [- ] (β'/k) C(k,ω) Dimensional analysis: We know that dim(cm) = V. Look now at the RHS and see if OK or not OK: dim(RHS) = Cou/m2 * sec-1 ohm-m * m = Cou * sec-1 ohm = amp-ohm = volt OK Part C: Smythian Approach to the Continuous Superposition As an alternate approach, here I start with the Smythian form with unknown coefficient dm(k,ω) and I try to determine that coefficient. I apply the Helm equations to this Smythian form (for m ≠ 0) and I show that one must have dm(k,ω) = Czm = (1/2j)(β'/k) Km where Km has to be determined from the BC's. A new feature is that the CP BC now has this form Er(r=a-ε,θ,z)(3) = (jω/σ) n(θ,z) where n(θ,z) = n(θ)r(z) and rk is the FT of r(z) (in the single-k world, we could just cancel e-jkz on both sides). Dealing then with the two BC's I find this new value for Km (which has an extra factor rk) Km = 2 (jω/2σ) Nm rk [ – ] // rk is the new feature (new factor) so dm(k,ω) = (1/2j)(β'/k) Km = (1/2j)(β'/k) 2 (jω/2σ) Nm rk [ – ] = (1/2)(β'/k)(ω/σ) Nm rk [ – ] So the superposition coefficient rk is directly related to the z shape of the surface charge along the line. Setting dm here to cm of the previous section, I find that C(k,ω) = rk and I show this earlier as well. Thus, we now have an interpretation of the arbitrary coefficient C(k,ω) 5a. Most General Form for Appendix D. Now suppose from the very start that we tried this as a most general form for Ez(r,θz)(3) dm(k,ω) and I try to determine that coefficient. [ since dim(dk) = m-1, we must have dim[dm] = volts ] Ez(r,θz)(3) = Σm ∫dk dm(k,ω) { e-jkz Jm(β'r) ejmθ } where dm(k,ω) is completely arbitrary. Since { e-jkz Jm(β'r)ejmθ } is the proper atomic form for the scalar Ez Helmholtz equation, I would argue that this expression for Ez(r,mz)(3) is the most general form possible for a solution of this particular scalar Helmholtz equation. [ agreed ] What is the m-space version of this equation? Can write Σm Ez(r,m,z)(3) ejmθ = Ez(r,θz)(3) from which I would conclude that Ez(r,m,z)(3) = ∫dk dm(k,ω) e-jkz Jm(β'r) If we track this through the Appendix D steps, what happens? I have no "most general form" for the other two components. Here are the four equations [∂r2 + (1/r) ∂r + (1/r2) ∂θ2 + ∂z2 + β2 ] Ez(r,θz)(3) = 0 [∂r2 + (1/r)∂r + (1/r2)∂θ2 + ∂z2 - (1/r2) + β2] Er(r,θz)(3) - (2/r2) ∂θEθ(r,θz)(3) = 0 [∂r2 + (1/r)∂r + (1/r2)∂θ2 + ∂z2 - (1/r2) + β2] Eθ(r,θz)(3) + (2/r2) ∂θEr(r,θz)(3)= 0 ∂r (r Er(r,θz)(3)) + ∂θ Eθ(r,θz)(3) + r ∂z Ez(r,θz)(3) = 0 I can surely convert all this to m-space to get [∂r2 + (1/r) ∂r - (1/r2) m2 + ∂z2 + β2 ] Ez(r,mz)(3) = 0 [∂r2 + (1/r)∂r - (1/r2)m2 + ∂z2 - (1/r2) + β2] Er(r,mz)(3) - (2/r2) jmEθ(r,mz)(3) = 0 / 2nd eq [∂r2 + (1/r)∂r - (1/r2)m2 + ∂z2 - (1/r2) + β2] Eθ(r,mz)(3) + (2/r2) jmEr(r,mz)(3)= 0 ∂r (r Er(r,mz)(3)) + jm Eθ(r,mz)(3) + r ∂z Ez(r,mz)(3) = 0 I would then rewrite the last equation as -jm Eθ(r,mz)(3) = ∂r (r Er(r,mz)(3)) + r ∂z Ez(r,mz)(3) and then I could sub this into the 2nd equation to get [∂r2 + (1/r)∂r - (1/r2)m2 + ∂z2 - (1/r2) + β2] Er(r,mz)(3) + (2/r2) [∂r (r Er(r,mz)(3)) + r ∂z Ez(r,mz)(3)] = 0 which can be rewritten, [∂r2 + (1/r)∂r - (1/r2)m2 + ∂z2 - (1/r2) + β2 + (2/r2) [∂r (r .. )] Er(r,mz)(3) = -(2/r) ∂z Ez(r,mz)(3) or [r2∂r2 + r∂r -m2 + r2∂z2 - 1 + β2r2 + (2) [∂r (r .. )] ] Er(r,mz)(3) = -2r ∂z Ez(r,mz)(3) or [r2∂r2 + r∂r -(m2+1) + r2∂z2 + β2r2 + (2) (r∂r + 1) ] Er(r,mz)(3) = ... or [r2∂r2 + r∂r -(m2+1) + r2∂z2 + β2r2 + 2r∂r + 2] Er(r,mz)(3) or [r2∂r2 + 3r∂r -(m2-1) + r2∂z2 + β2r2 ] Er(r,mz)(3) or [r2∂r2 + 3r∂r +(1-m2) + r2∂z2 + β2r2 ] Er(r,mz)(3) = -2r ∂z Ez(r,mz)(3) I could at this point install the most general form for Ez to get [r2∂r2 + 3r∂r +(1-m2) + r2∂z2 + β2r2] Er(r,mz)(3) = -2r ∂z ∫dk dm(k,ω) e-jkz Jm(β'r) = -2r ∫dk dm(k,ω) (-jk) e-jkz Jm(β'r) = +2j r ∫dk dm(k,ω) k e-jkz Jm(β'r) Now of course we want to solve this for Er(r,mz)(3) . It seems that just to have a chance in terms of z, you would have to make this assumption about Er(r,mz)(3) Er(r,mz)(3) = ∫dk e-jkz Fm(r) where we then want to solve for Fm(r). OK, if we put this in, the above equation reads [r2∂r2 + 3r∂r +(1-m2) + r2∂z2 + β2r2]∫dk e-jkz Fm(r) = +2j r ∫dk dm(k,ω) k e-jkz Jm(β'r) ∫dk e-jkz [r2∂r2 + 3r∂r +(1-m2) + r2(-k2) + β2r2] Fm(r) = ∫dk e-jkz 2j r dm(k,ω) k Jm(β'r) This would then require, using completeness of the e-jkz, that [r2∂r2 + 3r∂r +(1-m2) + β'2r2] Fm(r) = 2j r dm(k,ω) k Jm(β'r) This is the same as (D.2.7) if I set dm(k,ω) = Czm. I can then make use of Appendix D to solve this equation for Fm(r) and the answer will be Fm(r) = am x-1 Jm(β'r) + Jm+1(β'r) // but don't yet know constants am and Km At this point I then know that Er(r,mz)(3) = ∫dk e-jkz Fm(r) = ∫dk e-jkz [am x-1 Jm(β'r) + Jm+1(β'r)] [ Note: since dim(dk) = m-1, the am and Km here will differ in dimensions from App D coeffs.] I think the same thing will happen when I solve for Eθ, so the answer is going to be First summary of the E field solutions (D.2.21) Ez(r,m,z)(3) = ∫dk e-jkz [- j (β'/k) Jm(x)] x = β'r (D.1.27) Er(r,m,z)(3) = ∫dk e-jkz [am x-1 Jm(x) + Jm+1(x)] β'2 = β2 - k2 (D.2.11) jEθ(r,m,z)(3) = ∫dk e-jkz [- am x-1 Jm(x) + ( + ) Jm+1(x)] (D.2.15) where dm(k,ω) = Czm = (1/2j)(β'/k) Km . Note however that coefficients am and Km are not yet determined because we have not yet considered the boundary conditions. So the above solutions are like (D.2.21) "first summary" So let's look at the BC's. Things are suddenly very different now because of the z dependence. I guess we have to say, Er(r=a-ε,θ,z)(3) = (jω/σ) n(θ,z) // new idea of n(θ,z) Er(r=a-ε,m,z)(3) = (jω/σ) Nm(z) // dim[n(θ,z)] = dim[Nm(z)] = Cou/m2 Dim checks: dim(n) = Cou/m2 and dim(ω/σ) = sec-1 m-ohm so dim [(jω/σ) n(θ,z)] = sec-1 m-ohm * Cou/m2 = ohm amp /m = volt/m so OK. This then reads, ∫dk e-jkz [am x-1 Jm(xa) + Jm+1(xa)] = (jω/σ) Nm(z) In order to have a chance, you have to be able to write Nm(z) = ∫dk e-jkz Nm(k) and then you find that am x-1 Jm(xa) + Jm+1(xa) = (jω/σ) Nm(k) . The other boundary condition says jEθ(r=a,m,z)(3) = 0 which then gives - am xa-1 Jm(xa) + ( + ) Jm+1(xa) = 0 . But these last two equations are just (1) and (2) after (D.2.27) where Nm → Nm(k). We then get (D.2.28) where Nm has this new meaning, whatever it is. So if we solve the above two equations for the coefficients, we will get (D.2.28) with Nm → Nm(k) [ which below we see is Nm rk ] Pause. How do we relate this all back to the electrostatic n(θ) business? Maybe write n(θ,z) = n(θ) r(z) // dim r(z) = 1 Now just go to m-space to get: Nm(z) = Nm r(z) // dim[Nm] = dim[Nm(z)] = Cou/m2 and then for any z, we at least know the θ dependence of n(θ,z) is the electrostatic value. I guess this is an ansatz of sorts, but for a given z slice, it has to be true Then Nm(z) = Nm r(z) r(z,ω) to be general // dims OK Now let Nm(k) be the standard FT of Nm(z), Nm(z) = ∫dk e-jkz Nm(k) dim Nm(k) = Cou/m k like t Nm(k) = (1/2π) ∫dz e+jkz Nm(z) // consistent with (1.6.8) z like ω = (1/2π) ∫dz e+jkz Nm r(z) = Nm [(1/2π) ∫dz e+jkz r(z) ] = Nm rk // dim rk = m . // Cou/m = Cou/m2 * m OK Of course I don't know r(z) or rk . The App D equations apply, but I have to do this: Nm → Nm(k) = Nm rk = Nm/N0 N0 rk = ηm N0 rk ηm = Nm / N0 = Nm(k)/ N0(k) = unchanged In particular, with this change we get am = (jω/2σ) 2m Nm rk = (jω/2σ) Nm [ – ] rk This in turn tells us that dm(k,ω) = Czm = (1/2j)(β'/k) Km = (1/2j)(β'/k) (jω/σ) Nm [ – ] rk = (1/2)(β'/k) (ω/σ) Nm [ – ] rk // no factor of a Bug? The first line, Nm → Nm(k) = Nm rk replaces something that was Cou/m2 with something that is Cou/m. Is that OK? That will change the dims of am and Km ! OK, then Ez(r,m,z)(3) = ∫dk e-jkz (1/4) ηm 2πω (a/k) N0 rk Rdc (aβ') fm fm = [ - ] Er(r,m,z)(3) = ∫dk e-jkz (j/4) ηm 2πω (a/k) N0 rk Rdc (ak) gm gm = [ + ] Eθ(r,m,z)(3) = ∫dk e-jkz (1/4) ηm 2πω (a/k) N0 rk Rdc (ak) hm hm = [ - ] Bug? How can you take the original single-k Ei expression and multiply by rk which has dim = m and still end up with dimEz = volts/m ? The reason is that you have also added dk which has dim = m-1. But now we are stuck with this new unknown rk . So I guess just think of it as what I used to call C(k), a generic weighting function (apart from a constant, see below). So we now have Ei(r,m,z)(3) = ∫dk e-jkz C(k,ω) Ei(r,m) and I am then back to my original method (but without the condition ∫ dk C(k,ω) = 1). Comparing to the above for Ez says C(k,ω) = rk // both have dimensions of meters Fact: If you assume the most general form for Ez(r,m,z)(3) Ez(r,m,z)(3) = ∫dk dm(k,ω) e-jkz Jm(β'r) then in order to solve the Helm equations and the BC's, you end up with just linear combinations of the App D Ei form for all three components Ei(r,m,z)(3) = ∫dk e-jkz r(k,ω) Ei(r,m) where the Ei(r,m) are the exact Appendix D solutions. The CP BC is then Er(r=a-ε,m,z)(3) = (jω/σ) Nm(z) = (jω/σ) Nm r(z) Er(r=a-ε,θ,z)(3) = (jω/σ) n(θ) r(z) n(θ,z) = n(θ) r(z) where r(z) is the way charge density varies in z as this disturbance moves on the transmission line. r(z) = ∫dk e-jkz r(k) = ∫dk e-jkz C(k,ω) As before, we would argue that at large ω we have C(k,ω) → δ(k - k1) k1 = the usual function in which case r(z) = e-jk1z and we are back to the regular Appendix D stuff. Conclusion: I think k superposition of Appendix D is reasonable. Added question: How is rk related to C(k,ω) of the previous section? Answer: In the above I show that dm(k,ω) = (1/2j)(β'/k) Kmnew = (1/2j)(β'/k) 2 (jω/2σ) Nm [ – ] rk = (1/2)(β'/k)(ω/σ) Nm [ – ] rk But in the previous section I showed that cm(k,ω) = (1/2) Nm (ω/σ) [- ] (β'/k) C(k,ω) But we must have dm = cm since both expansions are the same, which requires, (1/2) Nm (ω/σ) [- ] (β'/k) C(k,ω) = (1/2)(β'/k)(ω/σ) Nm rk [ – ] or C(k,ω) = rk in agreement with earlier. review is OK to here. Part D: Repeat Part C with focus on the m = 0 partial wave I was concerned that something might be wrong with my m = 0 App D analysis, so here I do it all as a special case. The result is that the App D m=0 results are correct. So Part C is really valid for all m. I do the single-k case and the continuous superposition as well here. I just copy paste and edit from section 5a as far as is possible. A few lines down for m = 0 says Ez(r,m,z)(3) = ∫dk dm(k,ω) e-jkz Jm(β'r) so Ez(r,0,z)(3) = ∫dk d0(k,ω) e-jkz J0(β'r) I then write out the 4 equations in config space, and I then convert to m space to get [∂r2 + (1/r) ∂r - (1/r2) m2 + ∂z2 + β2 ] Ez(r,mz)(3) = 0 [∂r2 + (1/r)∂r - (1/r2)m2 + ∂z2 - (1/r2) + β2] Er(r,mz)(3) - (2/r2) jmEθ(r,mz)(3) = 0 [∂r2 + (1/r)∂r - (1/r2)m2 + ∂z2 - (1/r2) + β2] Eθ(r,mz)(3) + (2/r2) jmEr(r,mz)(3)= 0 ∂r (r Er(r,mz)(3)) + jm Eθ(r,mz)(3) + r ∂z Ez(r,mz)(3) = 0 Now set m = 0 everywhere to get [∂r2 + (1/r) ∂r + ∂z2 + β2 ] Ez(r,0z)(3) = 0 [∂r2 + (1/r)∂r + ∂z2 - (1/r2) + β2] Er(r,0z)(3) = 0 [∂r2 + (1/r)∂r + ∂z2 - (1/r2) + β2] Eθ(r,0z)(3) = 0 ∂r (r Er(r,0z)(3)) + r ∂z Ez(r,0z)(3) = 0 Hmm! Very different from m ≠ 0. The Helm equations are all separated when m = 0. Let's go ahead and use the ansatz of e-jkz to get ∂z → -jk, then we have [∂r2 + (1/r) ∂r -k2 + β2 ] Ez(r,0z)(3) = 0 [∂r2 + (1/r)∂r -k2 - (1/r2) + β2] Er(r,0z)(3) = 0 [∂r2 + (1/r)∂r -k2 - (1/r2) + β2] Eθ(r,0z)(3) = 0 ∂r (r Er(r,0z)(3)) - r jk Ez(r,0z)(3) = 0 or [∂r2 + (1/r) ∂r + β'2 ] Ez(r,0z)(3) = 0 [∂r2 + (1/r)∂r - (1/r2) + β'2] Er(r,0z)(3) = 0 [∂r2 + (1/r)∂r - (1/r2) + β'2] Eθ(r,0z)(3) = 0 ∂r (r Er(r,0z)(3)) - r jk Ez(r,0z)(3) = 0 or [r2∂r2 + r ∂r + r2β'2 ] Ez(r,0 = 0 solution is J0(β'r) [r2∂r2 + r∂r -1 + r2β'2] Er(r,0 = 0 solution is J1(β'r) [r2∂r2 + r∂r - 1 + r2β'2] Eθ(r,0 = 0 solution is J1(β'r) ∂r (r Er(r,0)) - r jk Ez(r,0) = 0 relates solutions I am back to the single-k world right now, just want to see what happens here. So Ez(r,0Az J0(β'r) // Az = Az (k,ω) I suppose Er(r,0Ar J1(β'r) Eθ(r,0Aθ J1(β'r) Now what happens when you stuff the first two forms into the div E = 0 above? ∂r (r Ar J1(β'r)) - r jk Az J0(β'r) = 0 Change to x = β'r so dx = β'dr and ∂x = (1/β')∂r and then ∂r = β' ∂x so β' ∂x ((x/β') Ar J1(x)) - (x/β') jk Az J0(β'r) = 0 Ar ∂x (x J1(x)) - Az(jk/β') x J0(x) = 0 I know from Spiegel 24.21 that ∂x (x J1(x)) = x J0(x) so the above says Ar x J0(x) - Az(jk/β') x J0(x) = 0 [ Ar - Az(jk/β') ] xJ0(x) = 0 and so we find that the coefficients must be related by Ar = (jk/β') Az . // this then is from div E = 0 At this point, our solutions take this form Ez(r,0Az J0(β'r) Er(r,0 (jk/β') Az J1(β'r) Eθ(r,0Aθ J1(β'r) It is then time for the BC's. We write Er(r=a,0) = (jω/σ) N0 Eθ(r=a,0) = 0 The first says (jk/β') Az J1(β'a) = (jω/σ) N0 The second says Aθ J1(β'a) = 0 => Aθ = 0 Wow, I think this is all new stuff. So we then find that (jk/β') Az J1(β'a) = (jω/σ) N0 (k/β') Az J1(β'a) = (ω/σ) N0 Az = (β'/k) (ω/σ) N0 / J1(β'a) Ar = (jk/β') Az = (jk/β') (β'/k) (ω/σ) N0 / J1(β'a) = j (ω/σ) N0 / J1(β'a) The resulting m = 0 fields are Ez(r,0Az J0(β'r) = (β'/k) (ω/σ) N0 J0(β'r)/ J1(β'a) Er(r,0Ar J1(β'r) = j (ω/σ) N0 J1(β'r)/ J1(β'a) (**) Eθ(r,0) = 0 Is this different from my existing Appendix D for m = 0, or is it the same? Here are my existing App D solutions for general m Ez(r,m) = (1/4) ηm 2πω (a/k) N0 Rdc (aβ') fm fm = [ - ] Er(r,m) = (j/4) ηm 2πω (a/k) N0 Rdc (ak) gm gm = [ + ] Eθ(r,m) = (1/4) ηm 2πω (a/k) N0 Rdc (ak) hm hm = [ - ] I know from below (D.11.5) that f0 = 2 g0 = 2 h0 = 0 Thus, my Appendix D existing solutions are Ez(r,0) = (1/4) 2πω (a/k) N0 Rdc (aβ') 2 Er(r,0) = (j/4) 2πω (a/k) N0 Rdc (ak) 2 Eθ(r,0) = 0 or Ez(r,0) = πωa2 (β'/k) N0 Rdc Er(r,0) = (j) πωa2 N0 Rdc Eθ(r,0) = 0 or Ez(r,0) = πωa2 (β'/k) N0 1/(σπa2) Er(r,0) = (j) πωa2 N0 1/(σπa2) Eθ(r,0) = 0 or Ez(r,0) = (ω/σ) (β'/k) N0 Er(r,0) = j (ω/σ) N0 Eθ(r,0) = 0 These agree exactly with the results found more directly in (**) above. So I guess I don't have some major m = 0 malfunction in Appendix D. Then the superposition work of Part C is valid for m = 0. Part E: The Jz Asymmetry Paradox is still present with a Superposed Solution 6. The Jz Asymmetry Paradox. First, in the single-k scenario we have = ηm where k has some value As ω→ 0, we get that β' = jkr which we can insert into the ratio above's RHS. Now matter what you assume about k as ω→0, the ratio is non-zero, and that is the "single-k Jz symmetry paradox". With superposition we have instead from above, Ez(r,m,z)(3) = (1/4) ηm Rdc∫dk 2πω (a/k) N0C(k,ω) (aβ') fm(r) e-jkz Er(r,m,z)(3) = (j/4) ηm Rdc∫dk 2πω (a/k) N0C(k,ω) (ak) gm(r) e-jkz Eθ(r,m,z)(3) = (1/4) ηm Rdc∫dk 2πω (a/k) N0C(k,ω) (ak) hm(r) e-jkz or Ez(r,m,z)(3) = (1/4) Nm Rdc2πωa∫dk (1/k) C(k,ω) (aβ') fm(r) e-jkz Er(r,m,z)(3) = (j/4) Nm Rdc2πωa ∫dk (1/k) C(k,ω) (ak) gm(r) e-jkz Eθ(r,m,z)(3) = (1/4) Nm Rdc2πωa ∫dk (1/k) C(k,ω) (ak) hm(r) e-jkz Now take this ratio and cancel common factors, = ηm (**) fm = [ - ] Comment: Note that C(k,ω) itself has no m subscript, and cannot have one. Note: If I assume that C(k,ω) = δ(k-k) (so to speak) the above becomes = ηm fm = [ - ] which agrees with my new Chapter 7 start. This thing has the Jz asym problem as described in Chapter 7. But now let's return to the continuous superposition and see if that fixes the problem. My paradox-removal requirement is that the ratio (**) → 0 as ω → 0. In that limit, β → 0 so our requirement then is this: [ β' = lim = = jk, so β'/k = j ∫dk C(k,0) j fm(jkr) e-jkz = 0 m ≠ 0 F(r,m,z) = 0 ∫dk C(k,0) j f0(jkr) e-jkz ≠ 0 m = 0 The first line requires that C(k,0) = 0 as the only way really to get F(r,m,z) to vanish for arbitrary values of r, m and z. But this makes the second line above be not true! Note: in the superposition scenario, we don't think about what k does as ω→ 0. Here k is just an integration variable. We do know that C(k,0) = rk at ω = 0 which is the FT of r(z). Conclusion: Thus the Jz paradox does NOT go away with superposition of different k solutions. No matter how you superpose, you still have Jz asymmetry. Reflection Example: In the reflection scenario you have C(k,ω) = A(k1)δ(k-k) + B(k1) δ(k+k) so to speak This is just a special case of superposition. We would then have to have A(k1) fm(jk1r) e-jkz + B(k1) fm(-jk1r) e+jkz = 0 m ≠ 0 A(k1)/k1 f0(jk1r) e-jkz + B(k1)/k1 f0(-jk1r) e+jkz ≠ 0 m = 0 There is no way to make the first line be true unless both A(k1) = 0 and B(k1) = 0. That is why I found long ago that reflection does not "fix" the Jz asym problem. Eliminate the two terms m and -m possible escape hatch. 7. Let's eliminate the usual m and -m escape hatch. The two terms in the expansion are terms(m,θ) = Ez(r,m,z)(3) ejmθ + Ez(r,-m,z)(3) e-jmθ for some m > 0 Ez(r,m,z)(3) = (1/4) ηm I Rdc∫dk C(k,ω) fm(r) e-jkz η-m = ηm f-m = fm terms(m,θ) = [ (1/4) Nm Rdc2πωa ∫dk/k * β'C(k,ω) fm(r) e-jkz ] ejmθ + [ (1/4) N-m Rdc2πωa ∫dk/k* β'C(k,ω) f-m(r) e-jkz ] e-jmθ = [ (1/4) Nm Rdc2πωa ∫dk/k* β'C(k,ω) fm(r) e-jkz ] [ejmθ + e-jmθ] = [ (1/4) Nm Rdc2πωa ∫dk/k* β'C(k,ω)) fm(r) e-jkz ] 2cos(mθ) Now consider terms(m=0,θ) = Ez(r,0,z)(3) e0 = Ez(r,0,z)(3) // only one term for m = 0 Then if we include both the m and -m terms together we have = ηm = ηm 2 cos(mθ) As ω→ 0 we set = jk to get = ηm 2 cos(mθ) But this is the exact same ratio found above with an extra 2 cos(mθ) on the outside, and we already showed that said ratio is not 0 ! Conclusion: Allowing superposition of k values does NOT resolve the Jz asymmetry paradox! This enhances my Appendix D mystery! There is no possible superposition of solutions that makes this asymmetry go away!! It has nothing to do with the specific function k(ω) of Appendix Q, for example. 8. The Infinite Br Paradox. Start with curl E(r,θ,z) = [ r-1∂θEz - ∂zEθ] + [∂zEr - ∂rEz] + [ r-1∂r(rEθ) - r-1∂θEr ] and in particular [curl E(r,θ,z)]r = r-1∂θEz - ∂zEθ Now apply this to our superposed solution, so it says [curl E(r,θ,z)(3)]r = r-1∂θEz(r,θ,z)(3) - ∂zEθ(r,θ,z)(3) Now install the superposed solutions E(r,θz,t)(3) = ∫dk C(k,ω) ej(ωt-kz) E(r,θ; k) E(r,θz)(3) = ∫dk C(k,ω) e-jkz E(r,θ; k) Ei(r,θz)(3) = ∫dk C(k,ω) e-jkz Ei(r,θ; k) We then get [curl E(r,θ,z)(3)]r = r-1∂θ ∫dk C(k,ω) e-jkz Ez(r,θ; k) - ∂z ∫dk C(k,ω) e-jkz Eθ(r,θ; k) I think it is OK to go to partial waves and write [curl E(r,m,z)(3)]r = r-1jm ∫dk C(k,ω) e-jkz Ez(r,m; k) - ∂z ∫dk C(k,ω) e-jkz Eθ(r,m; k) Now install the known single-k solutions, Ez(r,m) = (1/4) ηm 2πω (a/k) N0 Rdc (aβ') fm fm = [ - ] Er(r,m) = (j/4) ηm 2πω (a/k) N0 Rdc (ak) gm gm = [ + ] Eθ(r,m) = (1/4) ηm 2πω (a/k) N0 Rdc (ak) hm hm = [ - ] to get [curl E(r,m,z)(3)]r = = r-1jm ∫dk C(k,ω) e-jkz {(1/4) ηm 2πω (a/k) N0 Rdc (a) fm(r) } - ∂z ∫dk C(k,ω) e-jkz {(1/4) ηm 2πω (a/k) N0 Rdc (ak) hm(r) } = r-1jm Nm 2πω Rdc(1/4)a2 ∫dk C(k,ω) e-jkz { (1/k) () fm(r) } - Nm 2πω Rdc(1/4)a2 ∂z ∫dk C(k,ω) e-jkz { hm(r) } Divide both sides by (jω) to get Br(ω) = r-1m Nm 2π Rdc(1/4)a2 ∫dk C(k,ω) e-jkz { (1/k) () fm(r) } + jNm 2π Rdc(1/4)a2 ∫dk C(k,ω) (-jk)e-jkz { hm(r) } Now we want to take ω→ 0. This gives, Br(0; r,z,m) = r-1m Nm 2π Rdc(1/4)a2 ∫dk C(k,0) e-jkz { (1/k) (jk) fm(jkr) } + jNm 2π Rdc(1/4)a2 ∫dk C(k,0) (-jk)e-jkz { hm(jkr) } But now there is nothing trying to blow up the right hand side! In this superposition model, there is no connection between k and ω, and that is where the problem arose before. In the large ω limit, we have instead I think that C(k,ω) = δ(k-k1) so to speak where k1 is our formula. Then Br(ω) = r-1m Nm 2π Rdc(1/4)a2 e-jkz { (1/k) () fm(r) } + jNm 2π Rdc(1/4)a2 (-jk)e-jkz { hm(r) } and NOW you have to worry about k = k(ω). Rewrite for small ω, Br(ω) = r-1m Nm 2π Rdc(1/4)a2 e-jkz { (1/k) (jk) fm(jkr) } + jNm 2π Rdc(1/4)a2 (-jk)e-jkz { hm(jkr) } So OK, where is the problem now?? fm = (r/a)|m| (|m|+1) (2/β'a) f0 = 4/(aβ') gm = (r/a)|m|+1 + (r/a)|m|-1 g0 = 2 (r/a) hm = (r/a)|m|+1 - (r/a)|m|-1 h0 = 0 . (D.11.6) where now β' = jk as ω → 0 so we then have Br(ω) = r-1m Nm 2π Rdc(1/4)a2 e-jkz { (1/k) (jk) (r/a)|m| (|m|+1) (2/jka) } (r/a)|m| (|m|+1) + jNm 2π Rdc(1/4)a2 (-jk)e-jkz { (r/a)|m|+1 - (r/a)|m|-1 } The second term has just k and goes then as and so is not a problem. The first term now has 1/k and goes as 1/ and THAT is the problem, as always. Conclusion: The superposition solution of k values can make the infinite Br paradox go away.