Redo App D solutions for m = 0 REVIEWED

Redo App D solutions for m = 0 explicitly PhL 9.25.14 Here -- yet once again -- I verify that the "second summary" is valid for m = 0. I was having doubts midway below, but then the old results turned out to be valid, and I added a comment in lines doc pointing out that a0 = 0 and also ( + ) = 0, so Eθ(r,0) ≡ 0. The first summary is this First summary of the E field solutions (D.2.21) Ez(r,m) = - j (β'/k) Jm(x) x = β'r (D.1.27) Er(r,m) = am x-1 Jm(x) + Jm+1(x) β'2 = -k2 β' = jk (D.2.11) jEθ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) (D.2.15) I don't like seeing the am/m thing. Where did it come from? It comes from the fact that I use (D.26) to make a replacement, which equation says jmEθ(r,m) = stuff. Suppose I directly study these equations in the case m = 0 and see what happens then. So I am now going to do Section D.2 with m = 0 from the start (and for the moment I will forget ω=0 and just leave β' as its original self). I will now write new versions of various equations with m = 0: (a) The Ez Solution [r2∂r2 + r ∂r + (r2 β'2)] Ez(r,0) = 0 (D.2.1) [x2∂x2 + x ∂x + (x2)] Ez(x/β',0) = 0 . x = β'r (D.2.3) Ez(r,0) = Cz0 J0(β'r) (D.2.4) I think Y0 is logarithmically divergent just like ln r in the 2D thing and can be left out. So nothing is new here so far. (b) The Er Solution [r2∂r2 + r∂r - (1) + r2β'2] Er(r,0) = 0 (D.2.5) Equation is no longer coupled with Eθ. I think the solution to this equation is just Er(r,0) = Cr0 J1(β'r) // Schaum says yes Again, I would reject Y1(β'r) . So this makes for a much briefer Er solution section. Continuing, (c) The Eθ Solution The equation here is [r2∂r2 + r∂r - (m2+1) + r2(β2-k2)] Eθ(r,m) + 2jmEr(r,m) = 0 (D.1.18) which becomes [r2∂r2 + r∂r - (1) + r2β'2] Eθ(r,0) = 0 and this is the same as for Eθ, so I write Eθ(r,0) = Cθ0 J1(β'r) (c') The div E = 0 Condition // this section does not exist in App D But I then have the div E equation which has not been used at all, and it says ∂r [r Er(r,m)] + jmEθ(r,m) + r (-jk)Ez(r,m) = 0 or for m = 0 ∂r [r Er(r,0)] + r (-jk)Ez(r,0) = 0 and it then links Er and Ez together. Inserting my solutions then gives ∂r [r Cr0 J1(β'r) ] + r (-jk) Cz0 J0(β'r) = 0 x = β'r I think I can write this as Cr0 ∂x[x J1(x)] + Cz0 (x/β')(-jk)J0(x) = 0 Now Schaum p 137 24.21 with n = 1 says ∂x[x J1(x)] = x J0(x) Then my divE = 0 condition becomes Cr0 x J0(x)+ Cz0 (x/β')(-jk)J0(x) = 0 or x J0(x) [Cr0 - jkCz0/β' ] = 0 or Cr0 = j(k/β')Cz0 I feel I am treading on new territory here, I don't remember doing this before. Now recall that Czm = (1/2j)(β'/k) Km . (D.2.20) which is just my definition of the Km constants. I can then of course say Cz0 = (1/2j)(β'/k) K0 . (D.2.20) Then my div E = 0 result above says Cr0 = j(k/β')Cz0 = j(k/β')(1/2j)(β'/k) K0 = (1/2) K0 I then end up with Cz0 = (1/2j)(β'/k) K0 Cr0 = (1/2) K0 where both are related to the single unknown constant K0. And then Ez(r,0) = Cz0 J0(β'r) = (1/2j)(β'/k) K0 J0(β'r) = -j (β'/k) J0(β'r) Er(r,0) = Cr0 J1(β'r) = (1/2) K0 J1(β'r) = J1(β'r) Eθ(r,0) = Cθ0 J1(β'r) I can then restate my "first summary" for the E fields as First summary of the E field solutions (D.2.21) For m ≠ 0: Ez(r,m) = - j (β'/k) Jm(x) x = β'r (D.1.27) Er(r,m) = am x-1 Jm(x) + Jm+1(x) β'2 = β2 - k2 (D.2.11) jEθ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) (D.2.15) For m = 0: Ez(r,0) = -j (β'/k) J0(β'r) Er(r,0) = J1(β'r) Eθ(r,0) = Cθ0 J1(β'r) // but Cθ0 = 0 from below ! Observations: The Ez and Er solutions for m = 0 can be obtained from the m ≠0 solution set by setting m = 0 and assuming that a0 = 0 (as I do assume in App D). The Eθ solution can NOT be obtained in this manner! It is independent. I think in App D I would say that j Eθ(r,0) = J1(x), but I now see that is WRONG! So perhaps for the first time in a long time I have found a bug in App D. Go back and rereads this section. Done. The reason Eθ is independent is that the divE = 0 condition for m = 0 does not involve Eθ !!!! NOW let's apply the boundary conditions. Er(r=a,m) = (jω/σ) Nm (D.2.26) Eθ(r=a,m) = 0 (D.2.27) For m ≠0, things are unchanged, but for m = 0 I now get for the second, Eθ(r=a,0) = 0 => Cθ0 J1(β'a) = 0 => Cθ0 = 0 and now Cθ0 is GONE from the problem forever. The other BC says Er(r=a,0) = (jω/σ) N0 => J1(β'a) = (jω/σ) N0 and this tells us that K0 = 2 (jω/σ) N0 / J1(β'a) // which is the same as I had before by the way Now the new version of coefficients box (D.2.28) is this For m ≠ 0 : am = (jω/2σ) 2m Nm . (D.2.28) = (jω/2σ) Nm [ – ] (+ ) = (jω/2σ) Nm [ + ] For m = 0 : = (jω/σ) N0 Now I can insert these into the previous box to get: First for m ≠ 0: Ez(r,m) = - j (β'/k) Jm(x) = - j (β'/k) (jω/2σ) Nm [ – ] Jm(x) = - j (β'/k) (jω/2σ) Nm fm = (β'/k) (ω/2σ) Nm fm Er(r,m) = am x-1 Jm(x) + Jm+1(x) = (jω/2σ) 2m Nm x-1 Jm(x) + (jω/2σ) Nm [ – ] Jm+1(x) = (jω/2σ) Nm { 2m x-1 Jm(x) + [ – ] Jm+1(x) } At this point I have to use Schaum 24.17 which says 2m x-1Jm = Jm+1 + Jm-1 to get 1 2 3 4 = (jω/2σ) Nm { [Jm+1(x) + Jm-1(x)] + [ – ] Jm+1(x) } Terms 1 and 4 cancel giving = (jω/2σ) Nm { [Jm-1(x)] + [] Jm+1(x) } = (jω/2σ) Nm gm Now finally, jEθ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) = - (jω/2σ) 2m Nm x-1 Jm(x) + (jω/2σ) Nm [ + ]Jm+1(x) = (jω/2σ) Nm { - 2m x-1 Jm(x) + [ + ]Jm+1(x) } 1 2 3 4 = (jω/2σ) Nm { - [Jm+1(x) + Jm-1(x)] + [ + ]Jm+1(x) } Now terms 1 and 4 cancel and we get = (jω/2σ) Nm { - [Jm-1(x)] + [ ]Jm+1(x) } = (jω/2σ) Nm hm So here is my Second Summary for m ≠ 0: Ez(r,m) = (β'/k) (ω/2σ) Nm fm Er(r,m) = (jω/2σ) Nm gm jEθ(r,m) = (jω/2σ) Nm hm Now for m = 0, we can just copy the first two above, but we set the last to ≡ 0 !!! Ez(r,0) = (β'/k) (ω/2σ) N0 f0 Er(r,0) = (jω/2σ) N0 g0 Eθ(r,0) = 0 Now, if you go look at (D.2.31a) , since Ez(r,m) with m = 0 has not changed, the relation between I and N0 does not change!! So we can go ahead and write N0 = (k/2πωa) I Nm = ηm(k/2πωa) I Use these then to re-express the above 6 fields m ≠ 0: Ez(r,m) = (β'/k) (ω/2σ) ηm(k/2πωa) I fm = (β'a) (1/2σ) ηm(1/2πa2) I fm = (1/4) (β'a) ηm I Rdc fm Er(r,m) = (jω/2σ) ηm(k/2πωa) I gm = (j/2σ) ηm(ka/2πa2) I gm = (j/4) (ka) ηm I Rdc gm jEθ(r,m) = (jω/2σ) ηm(k/2πωa) I hm = (j/2σ) ηm(k/2πa) I hm = (j/4) (ka) ηm I Rdc hm Now for m = 0, we can just copy the first two above, but we set the last to ≡ 0 !!! Ez(r,0) = (β'/k) (ω/2σ) (k/2πωa) I f0 = (β'a) (1/2σ) (1/2πa2) I f0 = (1/4) (β'a) I Rdc f0 Er(r,0) = (jω/2σ) (k/2πωa) I g0 = (j/2σ) (ka/2πa2) I g0 = (j/4) (ka) I Rdc g0 Eθ(r,0) = 0 Here then is my "new" second summary m ≠ 0: Ez(r,m) = (1/4) (β'a) ηm I Rdc fm Er(r,m) = (j/4) (ka) ηm I Rdc gm Eθ(r,m) = (1/4) (ka) ηm I Rdc hm m = 0: Ez(r,0) = (1/4) (β'a) I Rdc f0 Er(r,0) = (j/4) (ka) I Rdc g0 Eθ(r,0) = 0 But let's now compare this to the existing box (D.2.33), Second summary of the E field solutions : Rdc = β'2 = β2 - k2 (D.2.33) 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 = [ - ] But we know that f0 = [ - ] = 2 g0 = [ + ] = 2 h0 = [ - ] = 0 D.6.1 Where have I written this down? They are all in D.6.1. Since h0 ≡ 0, I now see that the "second summary" is unchanged! It already gives Eθ(r,0) = 0 since h0 = 0. So the upshot is that there was some fuzziness with Eθ(r,0) when written in the general form, but in the end, there is no change to my Second Summary now that I have completely redone the m = 0 case separately. That would explain why these second summary fields solved the Helm equation as I have Maple show below (D.2.35). There I set both a0 = 0 and a0/0 = 0. I do think something is fishy with (D.2.21) so just now I added a clarifying note regarding m = 0 after that equation to show that Eθ(r,0) ≡ 0 from the git go.