another relaxed pondeing of Jz asym REVIEWED

Another Relaxed Pondering of Appendix D Jz Asymmetry PhL 8.12.14 Reviewing this doc on 9.10.14 1. The High ω explanation of why Jz tracks n(θ) 1 2. So OK, what happens for small ω to this same argument? 2 3. Do it again keeping Eθ stuff. 3 4. A Little Review of Appendix D's Verification of things. [ 8/13/14] 4 4. What happens to am and Km as ω→ 0 ? 6 5. What happens to the Bi as ω→ 0 with G = 0 ? 7 6. Try to find better expressions for the Bi fields. 10 7. Show again that Bz = constant 15 8. Show again that Br and Bθ diverge at ω = 0 16 1. The High ω explanation of why Jz tracks n(θ) My most basic explanation linking Jz to n(θ) appears in Section 6.5 (d). Question: why do I replace ∂z by βd0 here? As shown in (1.5.1b), this is just βd in the special case that G = 0 which means σd = 0. Fine. And βd02 = ω2μdεd But I think really I should have ∂z → k, not → βd0 since the ansatz z dependence is k. jk = a + jb = = so then k = -j if G = 0, this is NOT the same as βd0 = ω/vd. I am making no assumptions about ω here, so I guess I better fix things up: For r = a we know from (3.7.0) that Eθ(a,θ) = 0 and ∂θEθ(r,θ) = 0, so for r just below the surface we expect ∂r (r Er(r,θ,z)) + r ∂zEz(r,θ,z) ≈ 0 // near r = a ∂r (r Er(r,θ)) - jk r Ez(r,θ) ≈ 0 // using ∂z → -jk, see (D.1.16), then cancel ejkz factors Ez(r,θ) ≈ (1/jk) (1/r) ∂r (r Er(r,θ)) . // near r = a (6.5.15) BUT, I have at the start specified the "skin effect regime", so in that case yes k = βd0 as shown in (Q.2). So that rescues things at least for large ω. OK, I read the section and I agree that it shows that Ez(r,θ) "tracks" n(θ) in the high ω limit. [ I think the above section is OK ] 2. So OK, what happens for small ω to this same argument? I think the argument begins the same way and we get as shown above Ez(r,θ) ≈ (1/jk) (1/r) ∂r (r Er(r,θ)) and then Ez(r,m) ≈ (1/jk) (1/r) ∂r (r Er(r,m)) = (1/jk) (1/r) ∂r (r [(j/4) ηm I Rdc (ak) gm]) = (1/jk) (j/4) ηm I Rdc (ak) (1/r) ∂r (r gm(r)) = (1/4) ηm I Rdc (a) (1/r) ∂r (r gm(r)) Now comes the asymmetry ratio Ez(r,m)/ Ez(r,0) = ηm ∂r (r gm(r)) / ∂r (r g0(r)) I could compute this stuff for general ω, but instead lets go to small ω where gm = (r/a)m+1 + (r/a)m-1 g0 = 2 (r/a) Let x = r/a for the moment. Then ∂r = ∂/∂r = ∂/∂x* ∂x/∂r = ∂x * (1/a) = (1/a)∂x ∂r(r F(x)) = F(x) + r ∂r F(x) = F(x) + r (1/a)∂x F(x) = F(x) + x ∂xF(x) Then we have ∂r(r gm(x)) = gm(x) + x ∂x gm(x) = [xm+1 + xm-1] + x∂x[xm+1 + xm-1] = [xm+1 + xm-1] + x[ (m+1)xm + (m-1)xm-2] = [xm+1 + xm-1] + [ (m+1)xm+1 + (m-1)xm-1] = [ (m+2) xm+1 + (m) xm-1] Next, ∂r(r g0(x)) = g0(x) + x ∂x g0(x) = (2x) + x ∂x(2x) = 2x + 2x*1 = 4x The asym ratio is then Ez(r,m)/ Ez(r,0) = ηm ∂r (r gm(r)) / ∂r (r g0(r)) = [ (m+2) xm+1 + (m) xm-1] / 4x = (1/4) [(m+2) xm + (m) xm-2] = (1/4) [(m+2) (r/a)m + (m) (r/a)m-2] For small ω, then, this ratio does not vanish and thus we have asym Jz. For r ≈ a which is where the path above is justified, we then get = (1/4) [(m+2) + (m)] = (1/4)(2m+2 ) so the argument again leads to Jz asym. [ I agree, my theory does lead to Jz being asym at low ω ] 3. Do it again keeping Eθ stuff. Start with ∂r (r Er(r,θ,z)) + ∂θEθ(r,θ,z) + r ∂zEz(r,θ,z) = 0 Go to partial waves and use the ∂z form as above, ∂r (r Er(r,m,z)) + jmEθ(r,θ,z) - rjk Ez(r,m,z) = 0 Cancel the e-jkz factors to get ∂r (r Er(r,m)) + jmEθ(r,m) - rjk Ez(r,m) = 0 Solve for Ez rjk Ez(r,m) = ∂r (r Er(r,m)) + jmEθ(r,m) Install the general solutions to get rjk Ez(r,m) = ∂r (r (j/4) ηm I Rdc (ak) gm) + jm(1/4) ηm I Rdc (ak) hm = ηm I Rdc (ak) (j/4) ∂r (r gm) + jm(1/4) ηm I Rdc (ak) hm = (j/4) ηm I Rdc (ak) [ ∂r (r gm) + m hm ] Thus [ here I divide both sides by k and my limit of interest is k→ 0 ] r Ez(r,m) = (1/4) ηm I Rdc (a) [ ∂r (r gm) + m hm ] The ratio now is a little more complicated than before Ez(r,m)/ Ez(r,0) = ηm [ ∂r (r gm) + m hm ] / [ ∂r (r g0) + m h0 ] Above I already computed these two items ∂r(r gm(x)) = [ (m+2) xm+1 + (m) xm-1] ∂r(r g0(x)) = 4x Meanwhile, we know that hm = (r/a)m+1 - (r/a)m-1 h0 = 0 So our ratio is then Ez(r,m)/ Ez(r,0) = ηm [ ∂r (r gm) + m hm ] / [ ∂r (r g0) ] = ηm [ (m+2) xm+1 + (m) xm-1 + m ((r/a)m+1 - (r/a)m-1)] / [4x ] = ηm [ (m+2) xm+1 + (m) xm-1 + m (x)m+1 - mxm-1)] / [4x ] = ηm [ (2m+2) xm+1 + 0] / [4x ] = ηm [ (2m+2) xm ] Thus, we still have the asym keeping both terms. And for x ≈ 1, this replicates the ratio above. Remember that I want this ratio to → 0 as ω → 0. [ Above I show that keeping Eθ in the "divE = 0" argument does not fix Jz asym at ω→ 0. ] 4. A Little Review of Appendix D's Verification of things. [ 8/13/14] I start off in Appendix D by solving a set of four equations: three Helmholtz for Ei and div E = 0. The solutions for Ei are stated in (D.2.21) in terms of TBD Km and am constants. Are these the most general possible solutions, or have I omitted possible solutions? I start with the homogeneous Ez equation and reject the Ym solution since require finite at r = 0 in the wire, so (D.2.4) has only Ez(r,m) = Czm Jm(β'r). Then for Er I get a particular solution and to that I add a homo solution again only with J and not Y. Thus we have (D.2.15). The Eθ solution first comes from solving div E = 0 and that solution is (D.2.19) and there is no freedom in this solution, no Y that gets thrown out. I then summarize the three solutions in (D.2.21). At this point, I use Maple to verify that the (D.2.21) solutions do in fact solve the four starting equations. I just updated this code replacing βd by k yesterday. The verification shows that I have at least found "a" solution set for the problem. I later comment on the Eθ Helmholtz equation to say that it may not be independent of the other 3 equations. I then just note that Maple showed that my obtained solution for Eθ did cause all four equations to be solved. Probably I could have started off with the θ equation in place of the r equation and obtained the same solution, always rejecting any Y type solution. Note that even Y0 blows up at r = 0. So at this point I have (D.2.21) and have shown that they solve the 4 equations for any choice of the constants am and Km. The next topic is the two boundary conditions, and I reviewed the CPBC yesterday and found it to be good for G = 0. One would think the Eθ = 0 condition would be very good for ω = 0! I then do some very messy algebra to get expressions for am and Km as in (D.2.28). These coefficient solutions involve moments Nm since Nm appears in the CPBC (D.2.25). I then assume ohm's law and integrate Jz to get I, and this results in N0 = (k/2πωa) I as in (D.2.31). After more messy algebra, I end up with the "famous" solutions shown in (D.2.33) which involve the fm type functions. Comparing to (D.2.21) you see that the am x-1 Jm(x) + Jm+1(x) combination for Er now appears as a linear combination of Jm+1(x) and Jm-1(x). My next task is to verify that this new solution set solves the four equations and I do that in line. While there, I also verify that the two BC's are met. This code was updated a little bit today. So at this point I know the following: (D.2.33) for Ei solve the three Helmholtz equations, the div E = 0 equation, and satisfy the two boundary conditions. I think that this is the most general solution that works. Could I add constants to these solutions? That would be OK for div E = 0, but would not be OK for the three Helm equations! Have I omitted some other solutions? I don't think so. My next big step is to compute the B fields using the curl E equation. I am changing the code here and this will require repagination of App D after (D.5.4), a price I can pay. Then staying in the same Maple program, I really do compute the "other three" Maxwell equations and show they are all satisfied by the solution set (D.4.9). I directly verify each of these three Maxwell equations, no question about it!! At this point, I now know that my solution set for the E and B fields has this property: They satisfy: divE = divB = 0, curl E = -jωB and curl B = j (β2/ω) E (2+β2) E = 0, (2+β2) B = 0 , + the two boundary conditions I did not verify that (2+β2) B = 0 but I can prove that right here curl B = j (β2/ω) E curl curl B = j (β2/ω) curl E = j (β2/ω) [-jωB] = β2[B] (B) - 2B = β2[B] - 2B = β2[B] => (2+β2) B = 0 I might at this statement into App D at some point: ******* " The solution set (D.4.9) satisfies the 4 Maxwell equations, the 6 Helmholtz equations, and the 2 boundary conditions. " This concludes my little review of Appendix D's computation of the E and B fields, and while doing it I updated the code with βd replaced by k. [ the above discussion seems correct on 9.10.14 ] 4. What happens to am and Km as ω→ 0 ? I have never really studied what the B fields do. In (D.4.9) there are many pieces to worry about. I will do it all right here. First, the coefficients: am = (j/4) (ak) ηm I Rdc * 2m = (j/4) (ak) ηm I Rdc * [ – ] (+ ) = (j/4) (ak) ηm I Rdc * [ + ] . For ω→0 I expect small xa and then from Section D.11: First for m > 0, and then for m = 0: [ omitting lead factors for the moment ] = [ - ] = [ - ] = (m+1)! (xa/2)-m-1 - (m-1)! (xa/2)-m+1 = (m-1)! (xa/2)-m-1 [ (m+1) m - (xa/2)2 ] ≈ (m-1)! (xa/2)-m-1 (m+1) m = (m+1)! (xa/2)-m-1 = blows up! (2nd term nada) = 2/J1(xa) = 2 / [ xa/2] = 4/xa = blows up First am for m > 0, and then for m = 0: [ omitting lead factors for the moment ] am = 2m / Jm-1(xa) = 2m / [(xa/2)m-1 / (m-1)! ] = 2m (xa/2)-m+1 (m-1)! a0 = 2m / [-xa/2] = -2m (xa/2)-1 This wants to blow up, but m = 0 seems top stop it and we get a0 = 0 as elsewhere. First (+ ) for m > 0, and then for m = 0: [ omitting lead factors for the moment ] This is the same as for except there is a minus sign between first and second terms, at least for m>0. Thus (+ ) = = (m+1)! (xa/2)-m-1 = blows up! What about for m = 0 ? I claim above (D.2.28) that ( + ) = 0 Summary: am = (j/4) (ak) ηm I Rdc 2m (xa/2)-m+1 (m-1)! m > 0 a0 = 0 m = 0 = (j/4) (ak) ηm I Rdc (m+1)! (xa/2)-m-1 m > 0 = (j/4) (ak) ηm I Rdc 4/xa m = 0 (+ ) = (j/4) (ak) ηm I Rdc (m+1)! (xa/2)-m-1 m > 0 ( + ) = 0 m = 0 So these things are generally blowing up in our ω = 0 limit! [ Yes, things blow up as ω→0. This was the first sign of this problem, more on it below. ] 5. What happens to the Bi as ω→ 0 with G = 0 ? What then about the B fields?? I have generally neglected these, but I wonder if something bad happens? Bz(r,m) = (β'/ω) ( + )Jm(x) Br(r,m) = j(β'/ω){ + ( m - am) x-1Jm(x) + ( + ) Jm+1(x) } Bθ(r,m) = (β'/ω){ - ( m - am) x-1Jm(x) + ( + ) Jm+1(x) } , (D.4.8) First off, I assume G = 0 and from Section D.11 I have β2 = -jωμσ (1.5.1d) for good conductor k ≈ ω1/2 (1-j) = ω1/2 e-jπ/4 (Q.4) (highly damped) k2 = RCω(-j) = -jωRC β'2 = β2 - k2 ≈ -jωμσ +jωRC = -jω(μσ - RC) (D.11.2) This means that (β'/ω) = / ω = (1/) = blows up (β'/k) = / [ω1/2 e-jπ/4 ] = / [ e-jπ/4 ] = constant (k/β') = inverse of the above constant Meanwhile, again for small ω, Jm(x) = (x/2)m / m! m > 0 J0(x) = 1 m = 0 Jm+1(x) = (x/2)m+1 / (m+1)! J-1(x) = - (x/2) Now assemble the pieces: Bz(r,m) = (β'/ω) ( + )Jm(x) = (1/) [(j/4) (ak) ηm I Rdc (m+1)! (xa/2)-m-1] (x/2)m / m! = (1/) [(j/4) (aω1/2 e-jπ/4) ηm I Rdc (m+1) (xa/2)-m-1] (x/2)m = [(j/4) (a e-jπ/4) ηm I Rdc (m+1) (xa/2)-m-1] (x/2)m = [(j/4) (a e-jπ/4) ηm I Rdc (m+1) (xa/2)-1] (x/xa)m = [(j/4) (a e-jπ/4) ηm I Rdc (m+1) (xa/2)-1] (r/a)m Bang! We have a problem Houston! Due to the xa-1 factor, this field is blowing up at ω = 0! On the other hand, we know that I = 2πω (a/k) N0 and we can then consider I /xa = I/(β'a) = 2πωa N0 / (β'k) = 2πωa N0 / [ (ω1/2 e-jπ/4)] = 2πa N0 / [(e-jπ/4)] = constant Then we end up with Bz(r,m) = a constant in this limit. The constant is this [(j/4) (a e-jπ/4) ηm Rdc (m+1) 2] (r/a)m 2πa N0 / [(e-jπ/4)] = [(j/4) (a ηm (1/σπa2) (m+1) 2] (r/a)m 2πa N0 = [(j) ( ηm (1/σ) (m+1) ] (r/a)m N0 = j Nm (1/σ) (m+1) ] (r/a)m Since σ is very large, this Bz field is very small, but it approaches this small constant value! Bz(r,0) = (β'/ω) ( + )Jm(x) = 0 so the Bz field exists only in the moments for m > 0. So far, Bz did not blow up! What about the other components? It is a real mess, maybe I need to do this all in Maple. I want to convince myself that B does not blow up. Another approach would be to start with the small ω fields for Ei Ez(r,m) = (1/2) ηm I Rdc (r/a)|m| (|m|+1) Er(r,m) = (j/4) ηm I Rdc (ak) [(r/a)|m|+1 + (r/a)|m|-1] (D.11.7) Eθ(r,m) = (1/4) ηm I Rdc (ak) [(r/a)|m|+1 - (r/a)|m|-1] Ez(r,0) = I Rdc Er(r,0) = (j/2) I Rdc (ak) (r/a) Eθ(r,0) = 0 // low ω E fields and compute from these B fields Br(r,m) = (j/ω) [curl E]r = (j/ω) [r-1jmEz +jkEθ] Bθ(r,m) = (j/ω) [curl E]θ = (j/ω)[-jkEr - ∂rEz] Bz(r,m) = (j/ω) [curl E]z = (j/ω) [r-1∂r(rEθ) - r-1jmEr] . (D.4.7) Recall I = 2πω (a/k) N0 so we can look at the singular nature of each B field. Here for m > 0 Br(r,m) ~ (1/ω) [ I + I k2] = (1/ω) I (A + Bk2) = (1/ω) 2πω (a/k) (A + Bk2) = 2π (a/k) (A + Bk2) = 2πaA/k + 2πaBk = 2πaA/k = blows up! Ouch! Even using the I protection, Br is blowing up at ω = 0 !! Non-physical, something is wrong./ Conclusion: I need to do more study of the B fields in Appendix D and compute their behavior in various limits as well! It seems that they blow up as ω→ 0 and that would seem to be a problem with the entire Appendix if true! [ Above is where I first realized that Br blows up as ω→ 0. Later this was called the curl E problem and I wrote a separate doc about it. ] 6. Try to find better expressions for the Bi fields. [ The work of this section below is too messy to be useful, and besides, in the curlE doc I find a very simple and direct way to show that Br and Bθ blow up at DC, so details below are not needed. ] I would like to get something like (D.2.33) for the B fields. Here is where I left things off Bz(r,m) = (β'/ω) ( + )Jm(x) Br(r,m) = j(β'/ω){ + ( m - am) x-1Jm(x) + ( + ) Jm+1(x) } Bθ(r,m) = (β'/ω){ - ( m - am) x-1Jm(x) + ( + ) Jm+1(x) } , (D.4.8) am = (j/4) (ak) ηm I Rdc * 2m = (j/4) (ak) ηm I Rdc * [ – ] (+ ) = (j/4) (ak) ηm I Rdc * [ + ] . Let's define the following symbol Q = (j/4) (ak) ηm I Rdc Then we can write am = Q 2m = Q [ – ] (+ ) = Q [ + ] . and then to recap: Bz(r,m) = (β'/ω) ( + )Jm(x) Br(r,m) = j(β'/ω){ + ( m - am) x-1Jm(x) + ( + ) Jm+1(x) } Bθ(r,m) = (β'/ω){ - ( m - am) x-1Jm(x) + ( + ) Jm+1(x) } , (D.4.8) Now install coefficients to get Bz(r,m) = (β'/ω) Q [ + ]Jm(x) Br(r,m) = j(β'/ω){ + ( m Q [ – ] - Q 2m ) x-1Jm(x) + Q [ + ] Jm+1(x) } Bθ(r,m) = (β'/ω){ - ( m Q [ – ] - Q 2m ) x-1Jm(x) + Q [ – ] ( + ) Jm+1(x) } , (D.4.8) Now define S = (β'/ω) Q so we then have Bz(r,m) = S [ + ]Jm(x) Br(r,m) = jS{ + ( m [ – ] - 2m ) x-1Jm(x) + [ + ] Jm+1(x) } ok Bθ(r,m) = S{ - ( m [ – ] - 2m ) x-1Jm(x) + [ – ] ( + ) Jm+1(x) } , (D.4.8) ok Pick one and play: Br(r,m)/jS = ( m [ – ] - 2m ) x-1Jm(x) + [ + ] Jm+1(x) ok = ( [ – ] - 2 ) m x-1Jm(x) + [ + ] Jm+1(x) ok = [ – ] m x-1Jm(x) ok 1 - 2 m x-1Jm(x) ok 2 + [ + ] Jm+1(x) ok 3 = { m x-1Jm(x) + Jm+1(x) } ok 3 1 2 + { Jm+1(x) - m x-1Jm(x) - 2 m x-1Jm(x) } = { m x-1Jm(x) + Jm+1(x) } ok + { Jm+1(x) - m x-1Jm(x) ( - 2 ) } = { Jm+1(x) + m x-1Jm(x) } ok + { Jm+1(x) - m x-1Jm(x) ( - ) } I think the above is as simple as I can make things. I could rearrange in this way = Jm+1(x) [ + ] + m x-1Jm(x)[ - ( - ) ] = Jm+1(x) [ + ] + m x-1Jm(x) [ - (1 - ) ] Going another route we get, = { Jm+1(x) + m x-1Jm(x) } ok + { Jm+1(x) - 2m x-1Jm(x) ( - ) } Is there any way to simplify this result? I do know that (2m/x)Jm(x) = Jm-1(x) + Jm+1(x) NIST (10.6.1) so the above can be rewritten = { Jm+1(x) + 2 m x-1Jm(x) + } ok + { Jm+1(x) - [ Jm-1(x) + Jm+1(x)] ( - ) } = { Jm+1(x) + [ Jm-1(x) + Jm+1(x)] } ok + { Jm+1(x) - Jm-1(x) ( - ) - Jm+1(x) ( - ) } = { Jm+1(x) ( + ) + Jm-1(x) } ok + { Jm+1(x) ( -) - Jm-1(x) ( - ) } which does not seem very simple. Next, try Bθ(r,m)/S = - ( m [ – ] - 2m ) x-1Jm(x) + [ – ] ( + ) Jm+1(x) // copy from above ok = ( - m [ – ] + 2m ) x-1Jm(x) + [ – ] ( + ) Jm+1(x) // sign on first line ok = - m [ – ] x-1Jm(x) + 2m x-1Jm(x) + [ – ] ( + ) Jm+1(x) // remove () first line ok = { -m x-1Jm(x) + ( + ) Jm+1(x) } + { +m x-1Jm(x) + 2m x-1Jm(x) - ( + ) Jm+1(x) } ok = { ( + ) Jm+1(x) - m x-1Jm(x) } + { - ( + ) Jm+1(x) + m x-1Jm(x) + 2m x-1Jm(x) } = { ( + ) Jm+1(x) - m x-1Jm(x) } + { - ( + ) Jm+1(x) + m x-1Jm(x)( + ) } = { Jm+1(x) ( + ) - m x-1Jm(x) } + { - Jm+1(x) ( + ) + m x-1Jm(x)( + ) } Here are my results (need Maple checking for sure! ) Bz(r,m)/S = [ + ]Jm(x) Br(r,m)/jS = { Jm+1(x) + m x-1Jm(x) } ok + { Jm+1(x) - m x-1Jm(x) ( - ) } Bθ(r,m)/S = { Jm+1(x) ( + ) - m x-1Jm(x) } + { - Jm+1(x) ( + ) + m x-1Jm(x)( + ) } S = (β'/ω) Q = (β'/ω) (j/4) (ak) ηm I Rdc = (j/4) a ηm I Rdc (kβ') (1/ω) Now notice that (kβ') = k2 (kβ') = β'2 (kβ') ( - ) = β'2 - 2k2 (kβ') ( + ) = β'2 + 2k2 Now off to the side, let's take a sneak peak at the low-ω limit of things for G = 0. 7. Show again that Bz = constant For low ω and G= 0, we use instead these expressions β2 = -jωμσ (1.5.1d) for good conductor k ≈ ω1/2 (1-j) = ω1/2 e-jπ/4 (Q.4) (highly damped) k2 = RCω(-j) = -jωRC β'2 = β2 - k2 ≈ -jωμσ +jωRC = -jω(μσ - RC) (D.11.2) Note that β'/k = / [ ω1/2 e-jπ/4 ] = / [ e-jπ/4] = a constant at low ω where both R and C are constants. Start with Bz(r,m) = S [ + ] Jm(x) Jn(x) = (x/2)n / n! n ≥ 0 Keeping only most divergent term we get S = (j/4) a ηm I Rdc (kβ') (1/ω) Bz(r,m) = S [] Jm(x) = S (xa/2)-m-1 (m+1)! * (x/2)m / m! = S (xa/2)-m (xa/2)-1 (x/2)m (m+1) = S (x/xa)m (m+1) (2/xa) = S (2/a) (x/xa)m (m+1) (1/β') = (j/4) a ηm I Rdc (kβ') (1/ω) * (2/a) (x/xa)m (m+1) (1/β') = (j/2) ηm I Rdc (k) (1/ω) (r/a)m (m+1) = (j/2) ηm[ 2πω (a/k) N0] Rdc (k) (1/ω) (r/a)m (m+1) = (j/2) ηm[ 2π (a) N0] Rdc (r/a)m (m+1) = (j/2) ηm[ 2π (a) N0] (1/σπa2) (r/a)m (m+1) = j ηm N0 (1/σa) (r/a)m (m+1) = j Nm (1/σa) (r/a)m (m+1) The main point: this Bz field is FINITE as it well should be, and it is a constant value once we have taken the ω→0 limit. This was something I wanted to get, is Bz finite. 8. Show again that Br and Bθ diverge at ω = 0 How about the other components? 1 2 Br(r,m)/jS = { Jm+1(x) + m x-1Jm(x) } ok + { Jm+1(x) - m x-1Jm(x) ( - ) } 3 4 Now all the k and β' ratios are constant. The most divergent term has largest index if in bottom and smallest index if on top. The first line should then dominate as ω→0 so write Br(r,m)/jS = { Jm+1(x) + m x-1Jm(x) } Jn(x) = (x/2)n / n! n ≥ 0 The first term has ratio (x/xa)m+1 = (r/a)m+1. We then have Br(r,m)/jS = (r/a)m+1 + m x-1 Jm(x)/ Jm+1(xa) Looking above, the last ratio is Jm(x)/ Jm+1(xa) = (2/a) (r/a)m (m+1) (1/β') so we then have Br(r,m)/jS = (r/a)m+1 + m x-1 (2/a) (r/a)m (m+1) (1/β') = (r/a)m+1 + m (1/rβ') (2/a) (r/a)m (m+1) (1/β') = (r/a)m+1 + m (1/r) (2/a) (r/a)m (m+1) = m (1/r) (2/a) (r/a)m (m+1) where I keep only the divergent second term. Then we get Br(r,m) = jS m (1/r) (2/a) (r/a)m (m+1) = j[(j/4) a ηm I Rdc (kβ') (1/ω)] m (1/r) (2/a) (r/a)m (m+1) = -(1/2) ηm I Rdc (1/ω) m (1/r) (r/a)m (m+1) = -(1/2) ηm [2πω (a/k) N0] Rdc (1/ω) m (1/r) (r/a)m (m+1) = - (1/2) ηm [2π (a/k) N0] Rdc m (1/r) (r/a)m (m+1) and now we have a problem! The factor 1/k causes Br(r,m) → ∞ as ω → 0. These conclusions about Bz and Br agree with those found earlier by much simpler means, which I summarize right here: Ez(r,m) = (1/2) ηm I Rdc (r/a)|m| (|m|+1) Er(r,m) = (j/4) ηm I Rdc (ak) [(r/a)|m|+1 + (r/a)|m|-1] (D.11.7) Eθ(r,m) = (1/4) ηm I Rdc (ak) [(r/a)|m|+1 - (r/a)|m|-1] Br(r,m) = (j/ω) [curl E]r = (j/ω) [r-1jmEz +jkEθ] Br(r,m) ~ (1/ω) [ I + I k2] = (1/ω) I (A + Bk2) = (1/ω) 2πω (a/k) (A + Bk2) = 2π (a/k) (A + Bk2) = 2πaA/k + 2πaBk = 2πaA/k = blows up! Similarly, Bθ(r,m) = (j/ω) [curl E]θ = (j/ω)[-jkEr - ∂rEz] Br(r,m) ~ (1/ω) [ k kI - I ] = (1/ω) I [ k2 - 1 ] = (1/ω) I (A + Bk2) = blows up as above Question: For the infinite transmission line, as ω → 0, how do we interpret the fact that both magnetic field components Br and Bθ become infinite? I saw this problem arise also in my reflection scenario. This would seem to cast a lot of doubt on the theory since if flunks this ω→0 reasonableness test. This problem first appears in (D.4.7) of Appendix D where we compute the B fields. Here is a very concise statement of the problem: Br(r,m) = (j/ω) [curl E]r = (j/ω) [r-1jmEz +jkEθ] ≈ (j/ω) [r-1jmEz] ≈ (j/ω) [r-1jm(1/2) ηm I Rdc (r/a)|m| (|m|+1)] ~ (1/ω) I ~ 1/k → ∞ So now I have two problems with Appendix D. (1) As ω → 0, Jz remains asymmetric (2) As ω → 0, both Br and Bθ diverge. To fix this second problem, we really need to have Ez ~ ω but we seem to have instead Ez ~ I ~ (ω/k) ~ One repair would somehow be to have k → constant. This is the case when G ≠ 0. 9. Just compute curl E from (D.2.33) fields. [curl E]r = [r-1jmEz +jkEθ] Ez(r,m) = (1/2) ηm I Rdc (r/a)|m| (|m|+1) Er(r,m) = (j/4) ηm I Rdc (ak) [(r/a)|m|+1 + (r/a)|m|-1] (D.11.7) Eθ(r,m) = (1/4) ηm I Rdc (ak) [(r/a)|m|+1 - (r/a)|m|-1] assumes β' → 0 [curl E]r = [r-1jmEz +jkEθ] = (1/4) ηm I Rdc (r/a)|m|2 (|m|+1) + jk (1/4) ηm I Rdc (ak) [(r/a)|m|+1 - (r/a)|m|-1] = (1/4) ηm I Rdc { 2 (|m|+1) (r/a)|m| + j a k2 [(r/a)|m|+1 - (r/a)|m|-1] } Even if k → constant, since I → 0 we do get [curl E]r = 0. Here are all three curl components [curl E]r = [r-1jmEz +jkEθ] = I + k2I [curl E]θ = [-jkEr - ∂rEz] = k2I + I [curl E]z = [r-1∂r(rEθ) - r-1jmEr] . = k I + kI and they all go to zero as ω→0 assuming this means β' → 0. Therefore, my solutions (D.2.3) satisfy both div E = 0 and curl E = 0 at DC. The problem is that we really are getting for the r and θ curl components that curl E ~ and then (1/jω) curl E blows up. [ agreed, it is a problem ] 10. Whiteboard drawing This shows various paths to bugs. 11. Problem with k → 0 ? In Appendix Q we find that Fact 4: In the low frequency limit with G = 0 , (Q.4) Re(k) ≈ ω1/2 + ω3/2 Im(k) ≈ - ω1/2 + ω3/2 ω << R/L This implies for a 108 meter long transmission line at as you go to DC, there will be no loss! After all, you are assuming that things go as e-jkz and Im(k) is the attenuation constant. The fact that the above limit vanishes as is quite obvious from the general form of k : k ≡ -j → -j But I always explain this by saying that I ~ in this limit so there is no loss at ω = 0 since there is no current to make a loss. For very small ω there IS a loss which should be dV/dz = -IR. But I cannot make this simple DC limit work. For example, -jkV(0)e-jkz = -I R e-jkz -j k V(0) = -I R -j V(0) / I = -R / k = -j Z0 This last equation is R/k = j Z0 R = j Z0k = j (1/) 1/ (1-j) * ω1/2 (1-j) = j (1/)2 R 1/ (1-j) * ω1/2 (1-j) = j (1/)2 R (1-j) * (1-j) = j R e-jπ/2 = j (-j) R = R so OK. [ So I have discovered the curl E problem in addition to the Jz asym problem. The question remains as to where I should put the blame for these anomalies! ]