retired Section 7_4 ARCHIVE

Archive this rejected section of Chapter 7 retired on 9/21/14 7.4 A Second Sign of Trouble: Infinite B fields as ω→0 In Appendix D we compute the E fields inside a round wire as outlined above, then the corresponding B fields are computed from the Maxwell curl E equation, curl E = -jωB => B = (j/ω) curl E In the partial wave m space, and in cylindrical coordinates, this Maxwell equation reads, B = (j/ω) { [ r-1jmEz+ jkEθ] + [-jkEr - ∂rEz] + [ r-1∂r(rEθ) - r-1jmEr ] } or Br(r,m) = (j/ω) [ r-1jmEz(r,m)+ jkEθ(r,m)] Bθ(r,m) = (j/ω) [-jkEr(r,m) - ∂rEz(r,m)] Bz(r,m) = (j/ω) [r-1∂r(rEθ(r,m)) - r-1jmEr(r,m)] where as usual we have replaced ∂θ → jm and ∂z→ -jk. We are interested in examining these B fields as ω→0. Near this limit, we know the behavior of k from Appendix Q. We first assume that Gdc = 0 (vacuum dielectric) which means ωd = 0 in Appendix Q. We then find -- assuming that the equation k = k(ω) ≡ -j is valid in this limit -- that Fact 4: The small ω limit for k(ω), assuming ωd = 0, is given by (Q.4.9) Re(k) ≈ + ( 1 - tanL/2) + O(ω3/2) Im(k) ≈ - ( 1 + tanL/2) + O(ω3/2) where Rdc is the total resistance per length of both transmission line conductors. Ignoring the small loss tangent, this says k(ω) = (1 - j) = e-jπ/4 k2 = -j RdcC ω . Meanwhile, we know from ** that β2 = -jμσω. It then follows that β'2 ≡ β2 - k2 → -jμσω + j RdcC ω = j [-μσ + RdcC] ω As ω→ 0, β' becomes very small, and xa ≡ β'a and x ≡ β'r (for r ≤ a) become << 1. We can therefore use the E field expressions given in (D.11.7) , Ez(r,m) = (1/2) ηm I R (r/a)|m| (|m|+1) Er(r,m) = (j/4) ηm I R (ak) [(r/a)|m|+1 + (r/a)|m|-1] (D.11.7) Eθ(r,m) = (1/4) ηm I R (ak) [(r/a)|m|+1 - (r/a)|m|-1] Ez(r,0) = I R R = 1/(σπa2) Er(r,0) = (j/2) I R (ak) (r/a) Eθ(r,0) = 0 // low ω E fields Whereas Rdc is the DC resistance of both conductors, R is the DC resistance just of the round wire. In these expressions, we may write, using ** and **, I = 2πa(ω/k)N0 = 2πa(ω/k){q/2πa} = q (ω/k) = CV (ω/k) But for small ω, (ω/k) = ω / [ e-jπ/4 ] = ejπ/4 so I = CV (ω/k) = { ejπ/4 } V This implies that Z0-1 = I/V = { ejπ/4 }, something we can quickly verify (as ω→ 0) : Z0 = = => Z0-1 = = ejπ/4 . Notice that I → constant * and Z0 → constant / . In the final ω→0 limit, I → 0 and Z0 → ∞, but we are interested in what happens very close to that limit. We can rewrite our E fields for small ω using I = CV (ω/k). On 3 lines we replace (ω/k) (ak) = a, while on two lines we use C (ω/k) = ejπ/4 . The resulting small ω fields are Ez(r,m) = (1/2) ηm V ejπ/4 R (r/a)|m| (|m|+1) Er(r,m) = (j/4) ηm CV (ω) R (a) [(r/a)|m|+1 + (r/a)|m|-1] (D.11.7) Eθ(r,m) = (1/4) ηm CV (ω) R (a) [(r/a)|m|+1 - (r/a)|m|-1] Ez(r,0) = V ejπ/4 R Er(r,0) = (j/2) CV (ω) R (a) (r/a) Eθ(r,0) = 0 // low ω E fields In preparation for evaluating the B field, we compute these quantities ∂rEz(r,0) = 0 m=0 ∂rEz(r,m) = (1/2) ηm V ejπ/4 R |m| (r/a)|m|-1 (1/a) (|m|+1) m>0 r ∂r(rEθ(r,0)) = 0 r ∂r(rEθ(r,m)) = r ∂r [(1/4) ηm CV (ω) R (a) [(r/a)|m|+2 - (r/a)|m|]] m=0 = r (1/4) ηm CV (ω) R (a) { (|m|+2) (r/a)|m|+1 - (|m|) (r/a)|m|-1 } (1/a) = (1/4) ηm CV (ω) R (a) { (|m|+2) (r/a)|m|+1 - (|m|) (r/a)|m|-1 } (r/a) = (1/4) ηm CV (ω) R (a) { (|m|+2) (r/a)|m|+2 - (|m|) (r/a)|m| } m > 0 As a space-saving measure, we redisplay the above fields showing only the ω dependence, making up dummy names for the combination of factors. Ez(r,m) = Am Er(r,m) = Bm ω (D.11.7) Eθ(r,m) = Cm ω Ez(r,0) = A0 Er(r,0) = B0 ω Eθ(r,0) = 0 // low ω E fields which we simplify to read Ez(r,m) = Am Er(r,m) = Bm ω // for all m, and C0 = 0 Eθ(r,m) = Cm ω We may now insert these expressions for the small-ω fields into our B = (j/ω) curl E equations above, Br(r,m) = (j/ω) [ r-1jmEz(r,m)+ jkEθ(r,m)] Bθ(r,m) = (j/ω) [-jkEr(r,m) - ∂rEz(r,m)] Bz(r,m) = (j/ω) [r-1∂r(rEθ(r,m)) - r-1jmEr(r,m)] For Br Br(r,m) = (j/ω) [r-1jm (1/2) ηm V ejπ/4 R (r/a)|m| (|m|+1) The E fields computed in Appendix D may be written as 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 = [ - ] This is too messy, I want another way to show this problem.