Exterior Solution REVIEWED

Exterior Solution PhL 12.6.13 (12/12/13) In my original work here, I assumed you could have m ≠ 0 partial waves in the coaxial cable picture where r = ∞ is part of the picture. I later realized this is wrong as shown in pictures below. In the original work I ended up with a contraction. I had these 3 "boundary conditions" ξ Er(a,m) = ξd Erd(a,m) (-jσ/ω)Er(a,m) = εd Erd(a,m) (-jσ/ω) (jω/σ) Nm = εd Erd(a,m) where "d" means dielectric or exterior solution. For the interior values I had Ezd(a,m) = - j (β'/βd) Jm(x) where = (jω/2σ) Nm [ – ] Erd(a,m) = Nm/εd jEφd(a,m) = 0 where the last is questionable but OK for the moment. But for the exterior values I had Ezd(r,m) = Bzmr-m Erd(r,m) = -(1/2)(m-1)-1 jβd Bzm r1-m + Dmr-m-1 jEφd(r,m) = (1/2) (m-1)-1 jβd Bzm r1-m + Dm r-m-1 This was forcing 3 conditions on the two constants Bzm and Dm, an impossible situation. This entire document is a total mess and useless due to my starting error, but I will keep it with that understand. Parts of it might be salvaged. I have (I think) successfully computed the E and B fields inside a round wire, this is all written up in Appendix D sections D.1 through D.6. I wanted to add a short section describing the fields outside the wire just to finish things off. However, when I went to do this, I ended up with three boundary conditions to determine only two available constants, so I came into yet another problem. First, here is the method I used: Ideas for an exterior solution writeup for Appendix B? 1. I think of the exterior solution as entirely driven by the interior solution and its current I. Unlike typical situations, you don't really have to consider both at once to get a solution to either mainly because β is much larger than βd. [ suspect this is not true] 2. I already know all the interior fields, and therefore I know them all on the r=a boundary. I know that both E and B satisfy (2 + βd2)E = 0 and (2 + βd2)B = 0 outside the wire. There is no gauge situation here, these are the E and B fields. [ok ] 3. I can assume that the exterior fields have the same exp(***) wave dependence that the interior solutions do because they must match at the boundary. So then E(x,y,z,ω) = e-jβdz E(x,y,ω) for the exterior region. Therefore I end up with (22D + 0) E(x,y,ω) = 0. ok 4. This is still a vector 2 operator so must be careful in cylindrical coordinates. What then have is basically the main line of Appendix D except we replace β2 by βd2, and that changes the nature of ALL the component equations obtained with the vector 2. If you look at the summary (D.1.20) the exterior solution situation will be this [ I think all this is ok ] EXTERIOR The Three Helmholtz Equations and the div E = 0 equation (in partial waves) (D.1.20) [2E]z + βd2 Ez = 0 : [r2∂r2 + r ∂r - m2] Ez(r,m) = 0 (D.1.15)o [2E]r + βd2 Er = 0 : [r2∂r2 + r∂r - (m2+1)] Er(r,m) - 2jm Eφ(r,m) = 0 (D.1.17)o [2E]φ + βd2 Eφ = 0 : [r2∂r2 + r∂r - (m2+1)] Eφ(r,m) + 2jmEr(r,m) = 0 (D.1.18)o div E = 0 : ∂r [r Er(r,m)] + jmEφ(r,m) -jβd r Ez(r,m) = 0 (D.1.19) The last equation stays the same, but in the three previous we no longer even have Bessel equations! So I then have a completely new "solutions" section. 5. This is exactly the point I reached in "edit beta_d to k in App D.doc" Here is what I said there next: The differential operators appearing in the above equations are no longer Bessel-style operators, they are Euler-style operators. Euler ODE's have the general form [ r2∂r2 + a r ∂r + b] f(r) = 0, and the solutions have this form from p 45 of Polyanin's excellent ODE compendium, For Ez(r,m) the equation (D.1.15)o shown just above is in fact an Euler equation which has a = 1 and b = -m2 so μ = m and the solution forms are these (r ≥a outside the wire), Ez(r,m) = Azmrm + Bzmr-m m > 0 Ez(r,0) = Czln(r) + Dz m = 0 ok [ I verified that this really is the solution with Maple ] . This Ez Helmholtz equation is the same as the radial equation one obtains when solving the 2D Laplace equation 2u = 0 in separated polar coordinates, see for example Stakgold Vol II p 92 (6.7), and so what I see here are the familiar "atoms" of that problem. 6. I have to then do some work to solve the other equations, same as in main Appendix D, since they are coupled. So I have to trace through how all that goes. For example go down to (D.2.5) for the Er solution and just take it from there. I can get rid of the r+m power certainly [ wrong, see below] . So basically I just mimic the little sections of the main line. I will do that tomorrow. Fri Dec 6, 2013 My physical picture is a coaxial cable with some very large radius call it R. Like an electric dipole at the origin surrounded by a large Great Sphere of radius R, I expect fields to drop off in r, not to be constant or to blow up, except maybe in the m = 0 case. Note Added 12/12/13: The coaxial cable can only support the m = 0 wave, so it is the wrong "picture" for thinking about these boundary conditions. You need to use an off-center coax, or a twin lead or something like that. Here is a twin lead picture with the cylin coord system on the left conductor. This system will have m ≠ 0 components, especially m = 1 I think: You can talk about a Smythian Form within the red cylinder, but not outside because it hits the other transmission line conductor. So you could use atomic forms within the red cylinder. But this does not include r = ∞, so r = ∞ boundary conditions are not relevant! On the right you could use a Smythian Form outside the blue cylinder, and then the r = ∞ BC's would apply. But in this case you would have to think of that blue cylinder as the place where BC's apply, and this is not r = a! So either way, things don't work as described below. Continuing the above with D.2 (a). There it says Ez(r,m) = Czm Jm(β'r), here we have Ez(r,m) = Bzmr-m at least for m > 0. [ But in the red cylinder, it would be the full Ez(r,m) = Azmrm + Bzmr-m ] Then we go to (b) [r2∂r2 + r∂r - (m2+1)] Er(r,m) - 2jm Eφ(r,m) = 0 -jmEφ(r,m) = [1 + r∂r ] Er(r,m) - r (jβd)Ez(r,m) It is the same as in the main doc but we have to set β' = 0. We get to [r2∂r2 + 3r∂r + (1-m2)] Er(r,m) = 2r (jβd)Ez(r,m) . (D.2.7) [r2∂r2 + 3r∂r + (1-m2)] Er(r,m) = 2r (jβd) Bzmr-m . (D.2.7) Then I have to solve this somehow. Restate [r2∂r2 + 3r∂r + (1-m2)] Er(r,m) = 2(jβd) Bzmr1-m . (D.2.7) [r2∂r2 + 3r∂r + (1-m2)] Er(r,m) = 2(jβd) Bzmr1-m . (D.2.7) Maple gives the solution as (this must be particular plus homo) [ exterior solution 1.mws ] Er(r,m) = -(1/2)(m-1)-1jβd Bzm r1-m + Cm rm-1 + Dmr-m-1 and I verified that this does solve the ODE. [ But if we include Azmrm + Bzmr-m in Ez(r,m) we then have [r2∂r2 + 3r∂r + (1-m2)] Er(r,m) = 2(jβd) Azmr1+m + 2(jβd) Bzmr1-m (D.2.7) and this will have some other particular solution plus the same homo solution. I ran this in exterior solution 2. mws and here is the result where the first term is so then here is the solution Er(r,m) = -(j/2)(m-1)-1 βdBm r1-m - 2j βd(m2-1)-1Am r1+m + Cm rm-1 + Dmr-m-1 where there are now four constants. This is the same as the original result except for the Am term. [ What about m = 1? I tell Maple to solve [r2∂r2 + 3r∂r] Er(r,1) = 2(jβd) Bz1r0 and it gives Er(r,m=1) = j βd Bzm [ ln(r) - 1/2] + E + F/r2 This seems to conflict with my dipole line charge result which gives Er = 1/r3. ] [ What about m = 0 ? I tell Maple to solve [r2∂r2 + 3r∂r + 1] Er(r,0) = 2(jβd) Bz0r and it gives Er(r,m=0) = (1/2)jβd Bz0 r + G/r + H ln(r)/r I guess I would then set Bz0 = 0 and H = 0 and get Er(r,m=0) = G/r which is analogous to the line charge which has potential ln(r) and field 1/r. ] So it seems you might have to deal with m = 0 and m = ±1 separately. Now I have already assumed m > 0. I will then assume that Cm = 0 since that term is constant or blows up [ but that is wrong for our red cylinder, have to keep all terms] . Then we have Er(r,m) = -(1/2)(m-1)-1jβd Bzm r1-m + Dm r-m-1 Er(r,m) = -(j/2)(m-1)-1 βdBm r1-m - 2j βd(m2-1)-1Am r1+m + Cm rm-1 + Dmr-m-1 // new and this is then like Er(r,m) = am x-1 Jm(x) + Jm+1(x) . (D.2.15) where there are two unknown constants. I continue then to the Eφ solution -jmEφ(r,m) = [1 + r∂r ] Er(r,m) - r (jβd)Ez(r,m) = [1 + r∂r ] { -(1/2)(m-1)-1jβd Bzm r1-m + Dm r-m-1} - r(jβd) Bzm r-m = -(1/2)(m-1)-1jβd Bzm [1 + r∂r ] r1-m + Dm [1 + r∂r ] r-m-1 - (jβd) Bzm r-m+1 -jmEφ(r,m) = [1 + r∂r ] Er(r,m) - r (jβd)Ez(r,m) = [1 + r∂r ] { -(j/2)(m-1)-1 βdBm r1-m - 2j βd(m2-1)-1Am r1+m + Cm rm-1 + Dmr-m-1 } - r (jβd){ Amrm + Bmr-m } = [1 + r∂r ] { -(j/2)(m-1)-1 βdBm r1-m - 2j βd(m2-1)-1Am r1+m + Cm rm-1 + Dmr-m-1 } - (jβd){ Amrm+1 + Bmr-m+1 } Now [1 + r∂r ] r1-m = r1-m + r(1-m)r-m = r1-m + (1-m)r1-m = (2-m) r1-m [1 + r∂r ] r-m-1 = r-m-1 + r (-m-1)r-m-2 = r-m-1 + (-m-1)r-m-1 = -mr-m-1 [1 + r∂r ] rs = rs + rsrs-1 = (1+s)rs So then -jmEφ(r,m) = -(1/2)(m-1)-1jβd Bzm (2-m) r1-m - Dm mr-m-1 - (jβd) Bzmr-m+1 = jβd Bzm { -(1/2)(m-1)-1 (2-m) - 1 } r1-m - Dm mr-m-1 = jβd Bzm (1/2){ -(m-1)-1 (2-m) - 2 } r1-m - Dm mr-m-1 = jβd Bzm (1/2)(m-1)-1{ - (2-m) - 2(m-1) } r1-m - Dm mr-m-1 = jβd Bzm (1/2)(m-1)-1{ -2+m -2m+2 } r1-m - Dm mr-m-1 = jβd Bzm (1/2)(m-1)-1{ -m } r1-m - Dm mr-m-1 -jmEφ(r,m) = { -(j/2)(m-1)-1 βdBm (2-m)r1-m - 2j βd(m2-1)-1Am (2+m)r1+m + Cm m rm-1 + Dm(-m)r-m-1 } - (jβd){ Amrm+1 + Bmr-m+1 } = -(j/2)(m-1)-1 βdBm (2-m)r1-m - 2j βd(m2-1)-1Am (2+m)r1+m + Cm m rm-1 + Dm(-m)r-m-1 - (jβd)Am rm+1 - (jβd) Bmr-m+1 = r1-m { -(j/2)(m-1)-1(2-m) βdBm - (jβd) Bm } + r1+m { - 2j βd(m2-1)-1Am (2+m) - (jβd)Am } + Cm m rm-1 + Dm(-m)r-m-1 = r1-m j βdBm(1/2) { - (m-1)-1(2-m) - 2 } + r1+m j βdAm { - 2 (m2-1)-1 (2+m) - 1 } + Cm m rm-1 + Dm(-m)r-m-1 = r1-m j βdBm(1/2) { - (2-m) - 2(m-1) }/(m-1) + r1+m j βdAm { - 2 (2+m) - (m2-1) }/(m2-1) + Cm m rm-1 + Dm(-m)r-m-1 = r1-m j βdBm(1/2) { - 2+m -2m + 2 }/(m-1) + r1+m j βdAm { -4 - 2m - m2+1) }/(m2-1) + Cm m rm-1 + Dm(-m)r-m-1 = r1-m j βdBm(1/2) { -m }/(m-1) + r1+m j βdAm { -3 - 2m - m2) }/(m2-1) + Cm m rm-1 + Dm(-m)r-m-1 = - r1-m j βdBm(1/2) { m }/(m-1) - r1+m j βdAm { 3 + 2m + m2 }/(m2-1) + Cm m rm-1 + Dm(-m)r-m-1 and the above agrees with Maple. Then -jmEφ(r,m) = - r1-m j βdBm(1/2) { m }/(m-1) - r1+m j βdAm { 3 + 2m + m2 }/(m2-1) + Cm m rm-1 + Dm(-m)r-m-1 jmEφ(r,m) = r1-m j βdBm(1/2) { m }/(m-1) + r1+m j βdAm { 3 + 2m + m2 }/(m2-1) - Cm m rm-1 + Dm mr-m-1 jEφ(r,m) = r1-m j βdBm(1/2) { }/(m-1) + r1+m j βdAm { 3 + 2m + m2 }/[m(m2-1)] - Cm rm-1 + Dm r-m-1 jEφ(r,m) = j βdBm(1/2) r1-m + j βdAm r1+m - Cm rm-1 + Dm r-m-1 and then -jEφ(r,m) = - jβd Bzm (1/2)(m-1)-1 r1-m - Dm r-m-1 = - (1/2) (m-1)-1 jβd Bzm r1-m - Dm r-m-1 so that jEφ(r,m) = (1/2) (m-1)-1 jβd Bzm r1-m + Dm r-m-1 which is a problem for m = 1, otherwise seems reasonable. All needs Maple checking! [ But now Eφ will have 4 constants and not two constants! ] So I end up with (for m >1) Ezd(r,m) = Bzmr-m Erd(r,m) = -(1/2)(m-1)-1 jβd Bzm r1-m + Dmr-m-1 jEφd(r,m) = (1/2) (m-1)-1 jβd Bzm r1-m + Dm r-m-1 Then we come to the boundary conditions. From the First Summary of the inside we know First summary of the E field solutions r = a (D.2.21) Ez(a,m) = - j (β'/βd) Jm(xa) x = β'r (D.1.27) Er(a,m) = am xa-1 Jm(xa) + Jm+1(xa) . (D.2.11) jEφ(a,m) = - am xa-1 Jm(xa) + ( + ) Jm+1(xa) . (D.2.15) We then go look up the BC's to find: Ez(a,m) = Ezd(a,m) ξ Er(a,m) = ξd Erd(a,m) // = ξ (jω/σ) Nm according to (D.2.27) jEφ(a,m) = jEφd(a,m) // = 0 according to (D.2.26) BUT, I assumed no current in the dielectric in order to get the first boundary condition at the wire surface in App D, so that means ξd = εd. And I know that inside the metal ξ ≈ (-jσ/ω). So later I can insert these values, but keep the two ξ for the time being. That then says - j (β'/βd) Jm(xa) = Bzma-m ξ (jω/σ) Nm = ξd[-(1/2)(m-1)-1jβd Bzm a1-m + Dma-m-1 ] third equation 0 = jβd Bzm (1/2)(m-1)-1a1-m + Dm a-m-1 second equation This seems a little strange because we then seem to have 3 conditions to determine Bzm and Dm . [ but in the red cylinder situation we have 3 conditions to determine 4 constants Bzm, Azm, Dm and Cm. ] Let's go ahead and solve the first for Bzm Bzm = - j (β'/βd) am (jω/2σ) Nm [ – ] Jm(xa) = (β'/βd) am(ω/2σ) Nm [ – ] Jm(xa) Then we put this into the 2nd equation to get - Dm a-m-1 = jβd Bzm (1/2)(m-1)-1a1-m = jβd (β'/βd) am(ω/2σ) Nm [ – ] Jm(xa) (1/2)(m-1)-1a1-m = jβ' am (ω/2σ) (1/2)(m-1)-1a1-m Nm [ – ] Jm(xa) = jβ' a (ω/4σ) (m-1)-1Nm [ – ] Jm(xa) So then Dm = - am+2 jβ' (ω/4σ) (m-1)-1Nm [ – ] Jm(xa) What happens when we insert these into the third equation? RHS = ξd[-(1/2)(m-1)-1jβd a1-m (β'/βd) am(ω/2σ) Nm [ – ] Jm(xa) - ξd a-m-1 am+2 jβ' (ω/4σ) (m-1)-1Nm [ – ] Jm(xa) = - j ξd[(m-1)-1 aβ' (ω/4σ) Nm [ – ] Jm(xa) - j ξd a β' (ω/4σ) (m-1)-1Nm [ – ] Jm(xa) = -j ξd (ω/4σ) Nm aβ' (m-1)-1 [ – ] Jm(xa) { 1 + 1 } = -j ξd (ω/2σ) Nm aβ' (m-1)-1 [ – ] Jm(xa) whereas LHS = j 2 ξ (ω/2σ) Nm so things don't seem to be working very well. But consider [ – ] Jm(xa) = [ Jm-1(xa) - Jm+1(xa)] Jm(xa) / [Jm+1(xa) Jm-1(xa)] = (2m/xa) Jm(xa) Jm(xa) / [Jm+1(xa) Jm-1(xa)] but this does not further simplify. So we have a problem. Comments on the above method: I consider the exterior dielectric wave equation ( 2 + βd2 ) E(r,φz,t) = 0, (D.1.1) and as before I make this ansatz E(r,φz,t) = ej(ωt-βz) E(r,φ) . (D.1.2) and I make the same partial wave expansion, E(r,φ) =!Syntax Error, I E(r,m) ejmφ . This leads to the three Helmholtz equations in cylindricals plus the div E = 0 equation [2E]z + βd2 Ez = 0 : [r2∂r2 + r ∂r - m2] Ez(r,m) = 0 (D.1.15)o [2E]r + βd2 Er = 0 : [r2∂r2 + r∂r - (m2+1)] Er(r,m) - 2jm Eφ(r,m) = 0 (D.1.17)o [2E]φ + βd2 Eφ = 0 : [r2∂r2 + r∂r - (m2+1)] Eφ(r,m) + 2jmEr(r,m) = 0 (D.1.18)o div E = 0 : ∂r [r Er(r,m)] + jmEφ(r,m) -jβd r Ez(r,m) = 0 (D.1.19) These equations are the same as the inside equations but I have set β' = 0 since β'2 = βHelm2 - βd2. But they are no longer Bessel equations. I solve these equations thinking for m > 1 and keeping only terms which don't blow up for large r and I get Ezd(r,m) = Bzmr-m Erd(r,m) = -(1/2)(m-1)-1 jβd Bzm r1-m + Dmr-m-1 jEφd(r,m) = (1/2) (m-1)-1 jβd Bzm r1-m + Dm r-m-1 [ but for red cylinder, these are different and also have Azm and Cm ] The first equation is homo and I keep only the decaying solution of this homo. To get the second equation, I use the same trick I used in App D: I look at the Er equation and see jm Eφ(r,m) sitting in it. I then replace this by a lin com of Er and Ez from the div E = 0 equation. This then gives an equation which has only Er and Ez, namely [r2∂r2 + 3r∂r + (1-m2)] Er(r,m) = 2r (jβd)Ez(r,m) . I then install my solution for Ez(r,m) into this to get a driven equation [r2∂r2 + 3r∂r + (1-m2)] Er(r,m) = 2r (jβd) Bzmr-m . I then solve this using Maple and after throwing out positive r powers the solution is Er(r,m) = -(1/2)(m-1)-1jβd Bzm r-m+1 + Dmr-m-1 where only one of the two homo solution coefficients Dm appears. In this equation, the last term is the sort of m-pole expected power decay (m = 2 is octupole and is r-3 in 2D). The first term arises from the admixture of the Ez solution into the Er solution caused by the Helmholtz equation and it has a lesser decay by two powers of r. When I tie this in, one can think of the Ez solution in the dielectric as being driven by the Ez in the wire just below the surface, and that is where this two-powers-less strange term is coming from. I then take the divE = 0 and it provides a solution for Eφ as follows, jEφ(r,m) = (1/2) (m-1)-1 jβd Bzm r-m+1 + Dm r-m-1 where the first term involves that same admixture of the Ez solution. So I then have the three solutions shown above which I repeat here: Ezd(r,m) = Bzmr-m Erd(r,m) = -(1/2)(m-1)-1 jβd Bzm r1-m + Dmr-m-1 jEφd(r,m) = (1/2) (m-1)-1 jβd Bzm r1-m + Dm r-m-1 My conundrum now arises because I am aware of three boundary conditions for these fields, but there are only two undetermined constants. [ conundrum is resolved! ] Thus boundary conditions are (really ξd = εd since I assumed in my second inside BC that σd = 0, and also ξ = (-jσ/ω) inside the wire. Ez(a,m) = Ezd(a,m) // [tangential] ξ Er(a,m) = ξd Erd(a,m) // = ξ (jω/σ) Nm according to (D.2.27) jEφ(a,m) = jEφd(a,m) // = 0 according to (D.2.26) [tangential] I then insert these facts: (1) Eφ = 0 at the boundary, just one of my inside BC's. (2) Er = (jω/σ) Nm at the boundary, another of my inside BC's (3 ) insert ξ = (-jσ/ω) and ξd = εd Doing this, Er BC becomes ξ Er(a,m) = ξd Erd(a,m) (-jσ/ω)Er(a,m) = εd Erd(a,m) (-jσ/ω) (jω/σ) Nm = εd Erd(a,m) Nm = εd Erd(a,m) so my three exterior BC's are then boundary conditions: Ezd(a,m) = - j (β'/βd) Jm(x) where = (jω/2σ) Nm [ – ] Erd(a,m) = Nm/εd jEφd(a,m) = 0 solutions: Ezd(r,m) = Bzmr-m Erd(r,m) = -(1/2)(m-1)-1 jβd Bzm r1-m + Dmr-m-1 jEφd(r,m) = (1/2) (m-1)-1 jβd Bzm r1-m + Dm r-m-1 The obvious problem is that I have only two constants Bzm and Dm but I have to satisfy three BC's as shown. For example, consider that Erd(r,m) + jEφd(r,m) = 2 Dmr-m-1 Erd(a,m) + jEφd(a,m) = 2 Dma-m-1 Therefore 2 Dma-m-1 = Nm/εd => Dm = (1/2) am+1 Nm/εd Next, consider Erd(r,m) - jEφd(r,m) = - (m-1)-1 jβd Bzm r1-m Erd(a,m) - jEφd(a,m) = - (m-1)-1 jβd Bzm a1-m Therefore - (m-1)-1 jβd Bzm a1-m = Nm/εd => Bzm = j (m-1)am-1(1/εdβd) Nm The first solution must then be Ezd(r,m) = j (m-1)am-1(1/εdβd) Nm r-m dim chck: RHS = L-1 L ε0-1 C/m2 = ε0 C/m2 = m/F *Cm-2 = C F-1m-1 = FV F-1m-1 = V/m and then at r = a this becomes Ezd(a,m) = j (m-1)(1/a)(1/εdβd) Nm But this is supposed to match Ez(a,m) = - j (β'/βd) Jm(xa) where = (jω/2σ) Nm [ – ] which then requires j (m-1)(1/a)(1/εdβd) Nm = - j (β'/βd) (jω/2σ) Nm [ – ] Jm(x) or (m-1)(1/a)(1/εd) = - β' (jω/2σ)[ – ] Jm(xa) For example, for m = 2 this requires that (1/a)(1/εd) = - β' (jω/2σ)[ – ] J2(xa) I can then set β'2 ≈ β2 = ω2μ(-jσ/ω) = -jωμσ to get (1/a)(1/εd) = - (jω/2σ)[ – ] J2(xa) This condition is obviously not met! Question: What is wrong in the above analysis? 12/6/13 1:20 PM I am now going to take a long hard look at Idea #1, it lasts several pages. Idea #1. Suppose I were to assume that the inside solution had k in place of βd where k was a to be determined parameter. How would that change the inside solution? These facts would still be true Er(r=a,m) = (jω/σ) Nm where Nm = ηm N0 = ηm (k/2πωa) I (D.2.26) Eφ(r=a,m) = 0 (D.2.27) Anywhere βd appears, it must be replaced by k. The wave is ej(ωt-kz) now for inside and outside. Thus we have this inside z solution Ez(r,m) = (1/4) ηm I Rdc [ - ] x = β'r β'2 = β2 - k2 Our boundary conditions on the exterior solutions would then be Erd(r=a,m) = (jω/σ) ηm (k/2πωa) I Eφd(r=a,m) = 0 Ezd(r,m) = (1/4) ηm I Rdc [ - ] We then have three conditions for what I suspect will be three constants: k, Am, Bm . If I now go through the processing of Appendix D where I make these changes β'→ β" = βd2 - k2 r → 0 r→∞ as BC I guess my Ez solution would have this form Ez(r,m) = Czm" Hm(1)( β"r) since this would be the well-behaved Bessel function as r → ∞. Then we would get [r2∂r2 + 3r∂r + (1-m2) + r2 β"2)] Erd(r,m) = Km" β" r Hm(1)( β"r) (D.2.8) Km" ≡ 2j (k/β") Czm". (D.2.9) Erd(r,m) = x-1 f"m(x) (D.2.10) Then we end up with [ x2 ∂x2 + x ∂x + (x2-m2)] f"m(x) = K"m x"2 Hm(1)(x") f"m(x)particular = (1/2) Km [ x" Hm+1(1)(x")] fm(x) = fm(x)particular + fm(x)homogeneous = (1/2) Km" [ x" Hm+1(1)(x") ] + a"m Hm(1)(x")] Erd(r,m) = am" x-1 Hm(1)(x") + Hm+1(1)(x") The Bessel recursion is still true and this should lead us to jEφ(r,m) = - am" x"-1 Hm(1)(x") + (+ ) Hm+1(1)(x") So here are my new external solutions: External Ezd(r,m) = - j (β'"/k) Hm(1)(x") x" = β"r β"2 = βd2 - k2 Erd(r,m) = am" x"-1 Hm(1)(x") + Hm+1(1)(x") jEφd(r,m) = - am" x"-1 Hm(1)(x") + (+ ) Hm+1(1)(x") Internal Ez(r,m) = - j (β'/k) Jm(x) x = β'r β'2 = β2 - k2 Er(r,m) = am x-1 Jm(x) + Jm+1(x) . (D.2.11) jEφ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) . (D.2.15) The boundary conditions are now Ezd(a,m) = Ez(a,m) jEφd(a,m) = 0 ξdErd(a,m) = ξ Er(a,m) Installing our forms we get - j (β"/k) Hm(1)(x"a) = - j (β'/k) Jm(xa) // Ez condition - am" x"a-1 Hm(1)(x") + (+ ) Hm+1(1)(x"a) = 0 (2) // Eφ condition am" x"a-1 Hm(1)(x"a) + Hm+1(1)(x"a) = (ξ /ξd) (jω/σ) Nm (1) // Er condition xa = β'a β'2 = β2 - k2 x"a = β"a β"2 = βd2 - k2 Notice that k cancels out in the first equation. We have three unknowns am", Km" and k. We now mimic equations above (D.2.28) to find: Add and subtract (1) and (2) to get (2 + ) Hm+1(1)(x"a) = (ξ /ξd) (jω/σ) Nm (3) 2 am" x"a-1 Hm(1)(x"a) - Hm+1(1)(x"a) = (ξ /ξd) (jω/σ) Nm (4) Solve the second for a"m : am" { 2m x"a-1 Hm(1)(x"a) - Hm+1(1)(x"a)} = (ξ /ξd) (jω/σ) Nm am" { Hm-1(1)(x"a)} = (ξ /ξd) (jω/σ) Nm = (ξ /ξd) (jω/σ) Nm/ Hm-1(1)(x"a) // (5) ( + ) = (ξ /ξd) (jω/2σ) Nm [ + ] = (ξ /ξd) (jω/2σ) Nm [ - ] But now we have to satisfy the Ez condition β" Hm(1)(x"a) = β' Jm(xa) which is this β" (ξ /εd) [ - ] Hm(1)(x"a) = β' [ – ] Jm(xa) The above is the horrible equation which determines k which I guess could be complex. xa = β'a β'2 = β2 - k2 x"a = β"a β"2 = βd2 - k2 β" (ξ /εd) [ - ] Hm(1)( x"a) = [ – ] Jm(a) But certainly β >> k so we can rewrite the above as β" (ξ /εd) [ - ] Hm(1)( x"a) = β [ – ] Jm(βa) and then the RHS is independent of k and seems to be a large number. If k is small relative to βd as well, then perhaps we can use these limits Then in this quantity [ - ] Hm(1)( a) the common factor of -(i/π) cancels and we then have = [ - ] Γ(m)(xa/2)-m = [ - ] Γ(m) = [ - ] Γ(m) then our condition is β" (ξ /εd) [ - ] Γ(m) = β [ – ] Jm(βa) or (x"a/a) (ξ /εd) [ - ] Γ(m) = β [ – ] Jm(βa) or (2/a) (x"a/2) (ξ /εd) [ - ] Γ(m) = β [ – ] Jm(βa) or (2/a) (ξ /εd) [ - ] Γ(m) = β [ – ] Jm(βa) (2ξ /aεd) [ - ] Γ(m) = β [ – ] Jm(βa) [ - ] Γ(m) = (aεd/2ξ) β [ – ] Jm(βa) = + (aεd/2ξ) β [ – ] Jm(βa) = (m-1) + (aεd/2ξ) β [ – ] Jm(βa) Now β2/ξ2 ≈ ω2μ ξ/ξ2 = ω2μ/ξ ≈ ω2μ/(- jσ/ω) = ω2μ(-ω/jσ) = j ω3σ β/ξ ≈ j1/2ω3/2σ1/2 Well, the RHS does not seem particularly small, so this whole approximation with x"a does not seem self consistent. (- jσ/εdω) [ - ] Γ(m) = β [ – ] Jm(βa) xa = β'a β'2 = β2 - k2 x"a = β"a β"2 = βd2 - k2 Now if β" is very small, maybe β"a is also small, so x"a is small, so then we have (- jσ/εdω) [ ] Γ(m) = β [ – ] Jm(βa) (- jσ/εdω) [(x"a/2)-1 ] (m-1) = β [ – ] Jm(βa) (x"a/2) = (- jσ/εdω) (m-1) / { β [ – ] Jm(βa)} Now we have β2 = ω2μ(-jσ/ω) = -jσμω so (x"a/2) = (- jσ/εdω) (m-1) / { [ – ] Jm(βa)} But = β" = x" OK, enough on this path without approximation. Question: is there any reason to expect k to be close to βd ? People seem to assume this for a coaxial cable, I'll bet it is true. Idea #2. Suppose I argue that although "large" by interior standards, Ez at the boundary is really quite small since inside it is Ez inside a good conductor. Then maybe set Bzm = 0 and take the resulting solutions as approximate: We had: Ezd(r,m) = Bzmr-m Erd(r,m) = -(1/2)(m-1)-1 jβd Bzm r1-m + Dmr-m-1 jEφd(r,m) = (1/2) (m-1)-1 jβd Bzm r1-m + Dm r-m-1 And I want to approximate this as Ezd(r,m) ≈ 0 Erd(r,m) ≈ Dmr-m-1 jEφd(r,m) ≈ Dm r-m-1 The problem here is that the Eφd(a,m) BC then sets Dm = 0 !! Idea #3. Maybe my Eφ(r,m) = 0 is wrong! Think of an m = 1 moment. At the ω rate, would not the plus charge flow as shown to the lower side, and would that not constitute some Eφ at the surface? There must be an Eφ field at the surface to make that surface charge move. I guess I was arguing that the charge was controlled by the fixed z pattern flowing down the wire. But regardless of where the charge comes from or what is moving it, there must be some Eφ along that surface because E lines go from + charge to - charge. I argued in my opening section of App D that Eφ = 0 because if not, Eφ ≠ 0 would neutralize the charge. But think electrostatics where the right picture is a metal ball in an external E field. Even at DC you don't get neutralization because some OTHER E field is holding the charge there! And in our coaxial cable, there is an other field caused by the transmission line action. Suppose then I were to forego this boundary condition altogether in my interior discussion. Then I just end up with these fields: First summary of the E field solutions (D.2.21) Ez(r,m) = - j (β'/βd) Jm(x) x = β'r (D.1.27) Er(r,m) = am x-1 Jm(x) + Jm+1(x) . (D.2.11) jEφ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) . (D.2.15) Meanwhile, my external problem would with k = βd (simple case) would have these solutions, Ezd(r,m) = Bzmr-m Erd(r,m) = -(1/2)(m-1)-1 jβd Bzm r1-m + Dmr-m-1 jEφd(r,m) = (1/2) (m-1)-1 jβd Bzm r1-m + Dm r-m-1 Suppose I now impose these BC's : Ezd(a,m) = Ez(a,m) ξdErd(a,m) = ξ Er(a,m) jEφd(a,m) = jEφ(a,m) I still like the idea of tying the "free parameter" to Nm which says Er(r=a,m) = (jω/σ) Nm and where the dielectric is assumed not to conduct to get this little rule as stated. So then I have these various equations: (1) Bzm a-m = - j (β'/βd) Jm(xa) // Ez match (2) εd{- (1/2) (m-1)-1 jβd Bzm r1-m + Dm r-m-1} = ξ {am xa-1 Jm(xa) + Jm+1(xa)} // Er (3) (1/2) (m-1)-1 jβd Bzma1-m + Dm a-m-1 = - am xa-1 Jm(xa) + ( + ) Jm+1(xa) // jEφ match (4) am xa-1 Jm(xa) + Jm+1(xa) = (jω/σ) Nm // tie-in to moment Nm Now if I regard Nm as a given quantity, I have four equations for the four unknowns am Km Bzm Dm Now let's solve these four equations. Rewrite as (1) Bzm = - j am(β'/βd) Jm(xa) // Ez match (2) - (1/2) (m-1)-1 jβd Bzm a1-m + Dm a-m-1 = (ξ/εd) {am xa-1 Jm(xa) + Jm+1(xa)} // Er (3) (1/2) (m-1)-1 jβd Bzma1-m + Dm a-m-1 = - am xa-1 Jm(xa) + ( + ) Jm+1(xa) // jEφ match (4) am xa-1 Jm(xa) + Jm+1(xa) = (jω/σ) Nm // tie-in to moment Nm From (1) we can eliminate Bzm as shown. Add (2)+(3) to get 2 Dm a-m-1 = (ξ/εd) {am xa-1 Jm(xa) + Jm+1(xa)} + {- am xa-1 Jm(xa) + ( + ) Jm+1(xa)} So if we can determine am and Km then we know Bzm and Dm. Now compute (3) - (2) : (m-1)-1 jβd Bzma1-m = - am xa-1 Jm(xa) + ( + ) Jm+1(xa) + (ξ/εd) {-am xa-1 Jm(xa) - Jm+1(xa)} or -(m-1)-1 jβd j am(β'/βd) Jm(xa) a1-m = - am xa-1 Jm(xa) + ( + ) Jm+1(xa) + (ξ/εd) {-am xa-1 Jm(xa) - Jm+1(xa)} or (m-1)-1 a β' Jm(xa) = - am xa-1 Jm(xa) + ( + ) Jm+1(xa) + (ξ/εd) {-am xa-1 Jm(xa) - Jm+1(xa)} or { (m-1)-1 aβ' Jm(xa) - Jm+1(xa) + (ξ/εd) Jm+1(xa) } = am { - xa-1 Jm(xa) + (1/m) Jm+1(xa) - (ξ/εd) xa-1 Jm(xa) } or { (m-1)-1 aβ' Jm(xa) + [(ξ/εd)-1] Jm+1(xa) } = am { (1/m) Jm+1(xa) - [(ξ/εd)+1] xa-1 Jm(xa) } which directly relates Km and am. Now bring in the tie-in equation am xa-1 Jm(xa) + Jm+1(xa) = (jω/σ) Nm am { (1/m) Jm+1(xa) - [(ξ/εd)+1] xa-1 Jm(xa) } - { (m-1)-1 aβ' Jm(xa) [(ξ/εd)-1] + Jm+1(xa) } = 0 We can finally solve for am and Km , but it is extremely messy. Reduce clutter first am xa-1 Jm + Jm+1= (jω/σ) Nm am { (1/m) Jm+1 - [ξ/εd+1] xa-1 Jm } - { (m-1)-1 aβ' Jm + [ξ/εd-1] Jm+1} = 0 or am xa-1 Jm + Jm+1= (jω/σ) Nm am xa-1 Jm { (1/m) xa Jm+1/Jm - [ξ/εd+1] } - { (m-1)-1 aβ' Jm + [ξ/εd-1] Jm+1} = 0 Now replace am xa-1 Jm from the first in the second { (jω/σ) Nm - Jm+1} { (1/m) xa Jm+1/Jm - [ξ/εd+1] } - { (m-1)-1 aβ' Jm + [ξ/εd-1] Jm+1} = 0 [ Jm+1{ (1/m) xa Jm+1/Jm - [ξ/εd+1] } + { (m-1)-1 aβ' Jm + [ξ/εd-1] Jm+1} ] = (jω/σ) Nm { (1/m) xa Jm+1/Jm - [ξ/εd+1] } and finally we have a solution for . I am now ready for approximation! I think I will regard xa and any Jn as of "nominal size". I think we can assume for sure that ξ/εd+1 ≈ ξ/εd since ξ = - jσ/ω and we always assume this is >> εd . So then [ Jm+1{ (1/m) xa Jm+1/Jm - ξ/εd] } + { (m-1)-1 aβ' Jm + [ξ/εd] Jm+1} ] = (jω/σ) Nm { (1/m) xa Jm+1/Jm - [ξ/εd] } Next, we drop terms much smaller than the large number ξ/εd [ Jm+1{ - [ξ/εd] } + { (m-1)-1 aβ' Jm + [ξ/εd] Jm+1} ] = (jω/σ) Nm { - [ξ/εd] } or [ - Jm+1 [ξ/εd] + (m-1)-1 aβ' Jm + [ξ/εd] Jm+1 ] = (jω/σ) Nm { - [ξ/εd] } or [ (m-1)-1 aβ' Jm] = - (jω/σ) Nm (ξ/εd) or = - (m-1)(jω/σ) (1/aβ') (ξ/εd) Nm/Jm Now install β' ≈ β = ej3π/4 ξ ≈ - jσ/ω Then (jω/σ) (1/aβ') (ξ/εd) = (jω/σ)(1/aεd) e-j3π/4 (1/) (- jσ/ω) = (1/aεd) e-j3π/4 (1/) = (1/β'aεd) and then = - (m-1) (1/β'aεd) Nm/Jm Nm = ηm (βd/2πωa) I or = - (m-1) (1/β'aεd) ηm (βd/2πωa) I /Jm or = - (1/2π)(m-1) (βd/β')(1/ωa2εd) ηm I /Jm Meanwhile Bzm a-m = - j (β'/βd) Jm = - j (β'/βd) Jm = + j (β'/βd) Jm (1/2π)(m-1) (βd/β')(1/ωa2εd) ηm I /Jm = + j (1/2π)(m-1) (1/ωa2εd) ηm I dim check: LHS = volt/m RHS = sec m-2(1/ε0) amp = amp secm-2m/f = amp secm-1 /f = amp sec m-1 ohm sec-1 = volt/m Now want approximation for am . Recall { (m-1)-1 aβ' Jm(xa) + [(ξ/εd)-1] Jm+1(xa) } = am { (1/m) Jm+1(xa) - [(ξ/εd)+1] xa-1 Jm(xa) } and now keep only large terms { (m-1)-1 (εd/ξ) aβ' Jm(xa) + Jm+1(xa) } = - am xa-1 Jm(xa) so am xa-1 Jm(xa) = - { (m-1)-1 (εd/ξ) aβ' Jm(xa) + Jm+1(xa) } am xa-1 = - { (m-1)-1 (εd/ξ) aβ' + Jm+1(xa)/ Jm(xa) } am = -xa{ (m-1)-1(εdaβ'/ξ) + Jm+1(xa)/ Jm(xa) } = - (1/2π)(m-1) (βd/β')(1/ωa2εd) ηm I /Jm Bzm a-m = j (1/2π)(m-1) (1/ωa2εd) ηm I Ezd(r,m) = Bzmr-m = j (1/2π)(m-1) (1/ωa2εd) ηm I (a/r)m Status at 7:45PM. As usual, my entire Appendix D has fallen apart again, just plain collapsed. I am once again unhappy with the Eφ = 0 condition on the surface, it seems wrong no. I liked it, then I didn't like it, then I liked it again, now I don't like it again. Poor stability here! I can say this Ez(r,m) = - j (β'/βd) Jm(x) Ez(a,m) = - j (β'/βd) Jm(xa) = r ≤ a βr = ej3π/4 (/δ)r = ej3π/4 z , z = (/δ)r Does this show the skin effect? Yes it does, but you have to do the magnitude to see it Jν(ej3π/4z) = Mν(z) ejθ(z) . (2.3.3) Jν(βr) = Mν(r/δ) ejθ(z) So at least I can get some skin effect without evaluating Km. || = | | = So you could plot this versus r to see about skin effect in higher modes. I would expect to see it still.