Infinite B fields directly from D_4_13 REVIEWED

Infinite B fields directly from (D.4.13) PhL 9.30.14 Here I do this twice. The second pass (see red below) is more current, but both worked. Start with (D.4.13) : Summary of E and B fields inside a round wire (D.4.13) Ez(r,m) = (1/4) ηm I Rdc (aβ') fm x = β'r xa = β'a β'2 = β2 - k2 Er(r,m) = (j/4) ηm I Rdc (ak) gm Eθ(r,m) = (1/4) ηm I Rdc (ak) hm Bz(r,m) = (j/4) (a/ω) ηm I Rdc (k β' em ) Br(r,m) = - (1/4) (a/ω) ηm I Rdc ( r-1m β' fm + k2 hm ) Bθ(r,m) = (j/4) (a/ω) ηm I Rdc ( k2 gm - β'2fm [ (m/x) - Jm+1(x)/Jm(x)] ) em = [ + ] gm = [ + ] Rdc = fm = [ - ] hm = [ - ] I argue that both β and k are very small so β'a is very small and I can then use the small argument limits of all these items. I already do the low arg limits below (D.11.5). Here is a new section for em em = [ + ] = [ + ] = [ (m+1) (x/xa)m (2/xa) + (1/m) (x/xa)m(xa/2) ] = (x/xa)m [ (m+1) (2/xa) + (1/m) (xa/2) ] ≈ (x/xa)m (m+1) (2/xa) // as xa→ 0 // only first term matters e0 = [ + ] = [ - ] = 0 So here is a full summary em = (r/a)m (m+1) (2/β'a) e0 = 0 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 But now I also need a limit for this quantity (m/x) - Jm+1(x)/Jm(x) = (m/x) - = (m/x) - = (m/x) - = (m/x) - ≈ m/x = m/(β'r) 2nd term does not matter For m = 0 let's make sure: (0/x) - J1(x)/J0(x) = (0/x) - = (0/x) - = - x/2 = -(β'r)/2 I can now insert these expressions into the above E and B fields for m > 0 only: Summary of E and B fields inside a round wire m > 0 (D.4.13) Ez(r,m) = (1/4) ηm I Rdc (aβ') (r/a)m (m+1) (2/β'a) Er(r,m) = (j/4) ηm I Rdc (ak) [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (1/4) ηm I Rdc (ak) [r/a)m+1 - (r/a)m-1] Bz(r,m) = (j/4) (a/ω) ηm I Rdc (k β' (r/a)m (m+1) (2/β'a) ) Br(r,m) = - (1/4) (a/ω) ηm I Rdc ( r-1m β' (r/a)m (m+1) (2/β'a)+ k2 [(r/a)m+1 - (r/a)m-1] ) Bθ(r,m) = (j/4) (a/ω) ηm I Rdc ( k2 [(r/a)m+1 + (r/a)m-1] - β'2(r/a)m (m+1) (2/β'a) (m/β'r) ) and here I simplify as needed Summary of E and B fields inside a round wire m > 0 (D.4.13) Ez(r,m) = (1/4) ηm I Rdc (r/a)m (m+1) (2) Er(r,m) = (j/4) ηm I Rdc (ak) [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (1/4) ηm I Rdc (ak) [r/a)m+1 - (r/a)m-1] Bz(r,m) = (j/4) (k/ω) ηm I Rdc ((r/a)m (m+1) (2) ) Br(r,m) = - (1/4) (a/ω) ηm I Rdc ( r-1m (r/a)m (m+1) (2/a)+ k2 [(r/a)m+1 - (r/a)m-1] ) Bθ(r,m) = (j/4) (a/ω) ηm I Rdc ( k2 [(r/a)m+1 + (r/a)m-1] - (r/a)m m(m+1) (2/ra) ) Things don't simplify very much as you can see. But next I will install (D.2.31c) I = (ω/k) CV // right here I am assuming G = 0 !! to get Summary of E and B fields inside a round wire m > 0 (D.4.13) Ez(r,m) = (1/4) ηm (ω/k) CV Rdc (r/a)m (m+1) (2) Er(r,m) = (j/4) ηm (ω/k) CV Rdc (ak) [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (1/4) ηm (ω/k) CV Rdc (ak) [r/a)m+1 - (r/a)m-1] Bz(r,m) = (j/4) (k/ω) ηm (ω/k) CV Rdc ((r/a)m (m+1) (2) ) Br(r,m) = - (1/4) (a/ω) ηm (ω/k) CV Rdc ( r-1m (r/a)m (m+1) (2/a)+ k2 [(r/a)m+1 - (r/a)m-1] ) Bθ(r,m) = (j/4) (a/ω) ηm (ω/k) CV Rdc [ k2 [(r/a)m+1 + (r/a)m-1] - (r/a)m m(m+1) (2/ra) ] and then I will try to simplify the above Summary of E and B fields inside a round wire m > 0 (D.4.13) Ez(r,m) = (1/4) ηm CV Rdc (ω/k) (r/a)m (m+1) (2) Er(r,m) = (j/4) ηm CV Rdc (ωa) [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (1/4) ηm CV Rdc (ωa) [r/a)m+1 - (r/a)m-1] Bz(r,m) = (j/4) ηm CV Rdc (r/a)m (m+1) (2) Br(r,m) = - (1/4) ηm CV Rdc ( (2/kr) m (r/a)m (m+1) + ak [(r/a)m+1 - (r/a)m-1] ) Bθ(r,m) = (j/4) ηm CV Rdc [ (ak) [(r/a)m+1 + (r/a)m-1] - (r/a)m m(m+1) (2/rk) ] dim(CV Rdc) = far/m * volts * ohms/m = sec/m2 * volts = volt-sec/m2 = tesla Now finally I reorder the terms in Bθ so Ez(r,m) = (1/4) ηm CV Rdc (ω/k) (r/a)m (m+1) (2) Er(r,m) = (j/4) ηm CV Rdc (ωa) [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (1/4) ηm CV Rdc (ωa) [r/a)m+1 - (r/a)m-1] Bz(r,m) = (j/4) ηm CV Rdc (r/a)m (m+1) (2) Br(r,m) = - (1/4) ηm CV Rdc[ (ak) [(r/a)m+1 - (r/a)m-1] + (2/kr) (r/a)m m(m+1) ] Bθ(r,m) = (j/4) ηm CV Rdc [(ak) [(r/a)m+1 + (r/a)m-1] - (2/rk) (r/a)m m(m+1) ] Now the last two B fields are very similar in appearance. Now use (r/a)m /r = (r/a)m (a/r)(1/a) = (r/a)m-1(1/a) and the last two fields then become Br(r,m) = - (1/4) ηm CV Rdc[ (ak) [(r/a)m+1 - (r/a)m-1] + (2/ak) m(m+1) (r/a)m-1] Bθ(r,m) = (j/4) ηm CV Rdc [(ak) [(r/a)m+1 + (r/a)m-1] - (2/ak) m(m+1) (r/a)m-1] Now replace (r/a)m+1 = (r/a)m-1 (r2/a2) Br(r,m) = - (1/4) ηm CV Rdc[ (ak) [(r/a)m-1 (r2/a2) - (r/a)m-1] + (2/ak) m(m+1) (r/a)m-1] Bθ(r,m) = (j/4) ηm CV Rdc [(ak) [(r/a)m-1 (r2/a2) + (r/a)m-1] - (2/ak) m(m+1) (r/a)m-1] Now factor out (r/a)m-1 to get Br(r,m) = - (1/4) ηm CV Rdc(r/a)m-1 [ (ak) [(r2/a2) - 1] + (2/ak) m(m+1)] Bθ(r,m) = (j/4) ηm CV Rdc (r/a)m-1 [(ak) [(r2/a2) + 1] - (2/ak) m(m+1)] Next, do a little adjusting Br(r,m) = - (1/4) ηm CV Rdc(r/a)m-1 [ (k/a) [r2-a2] + (2/ak) m(m+1)] Bθ(r,m) = (j/4) ηm CV Rdc (r/a)m-1 [(k/a) [r2+a2] - (2/ak) m(m+1)] Then factor out (1/ak) Br(r,m) = - (1/4) ηm CV Rdc(r/a)m-1 (1/ak) [k2 [r2-a2] + (2) m(m+1)] Bθ(r,m) = (j/4) ηm CV Rdc (r/a)m-1 (1/ak) [k2 [r2+a2] - (2)m(m+1)] and final sign fiddle Br(r,m) = (1/4) ηm CV Rdc(r/a)m-1 (1/ak) [- k2 [r2-a2] - 2m(m+1)] Bθ(r,m) = (j/4) ηm CV Rdc (r/a)m-1 (1/ak) [k2 [r2+a2] - 2m(m+1)] or Br(r,m) = (1/4) ηm CV Rdc(r/a)m-1 (1/ak) [- 2m(m+1) - k2 (r2-a2) ] Bθ(r,m) = (j/4) ηm CV Rdc (r/a)m-1 (1/ak) [- 2m(m+1) + k2 (r2+a2)] or Br(r,m) = (1/4) ηm CV Rdc(r/a)m-1 (1/ak) [- 2m(m+1) + k2 (a2-r2) ] Bθ(r,m) = (j/4) ηm CV Rdc (r/a)m-1 (1/ak) [- 2m(m+1) + k2 (a2+r2) ] Gather up all the B field results (only for m > 0) Ez(r,m) = (1/4) ηm CV Rdc (ω/k) (r/a)m (m+1) (2) Er(r,m) = (j/4) ηm CV Rdc (ωa) [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (1/4) ηm CV Rdc (ωa) [r/a)m+1 - (r/a)m-1] Bz(r,m) = (j/2) ηm CV Rdc (r/a)m (m+1) Br(r,m) = (1/4) ηm CV Rdc (r/a)m-1 (1/ak) [- 2m(m+1) + k2 (a2-r2) ] Bθ(r,m) = (j/4) ηm CV Rdc (r/a)m-1 (1/ak) [- 2m(m+1) + k2 (a2+r2) ] This agrees with the m > 0 portion of (7.4.2). You see the problem now that if k → 0, the last two B fields go infinite for m > 0. This is the "infinite B field problem". ************************ Let's do a quick check on the computation of I I = !Syntax Error, Idθ !Syntax Error, Ir dr Jz(r,θ) = !Syntax Error, Idθ !Syntax Error, Ir dr { σ !Syntax Error, I Ez(r,m) ejmθ } // (D.1.3a) = σ !Syntax Error, I !Syntax Error, Ir dr Ez(r,m) !Syntax Error, Idθ ejmθ = 2π σ!Syntax Error, Ir dr Ez(r,0) = 2π σ!Syntax Error, Ir dr {-j(β'/k) J0(x) } // (D.2.21) for Ez(r,0) = -j(β'/k) 2π σ !Syntax Error, Ir dr J0(x) // x = β'r so xdx = β'2 rdr = -j(β'k)-1 2πσ [!Syntax Error, Idx x J0(x)] = -j(β'k)-1 2πσ [ xa J1(xa) ] // GR7 5.52.1 = -j(β'k)-1 2πσ {(jω/σ) N0 / J1(xa)} [ xa J1(xa) ] // (D.2.28) for = (β'k)-1 2πω N0 xa = (β'k)-1 2πω N0 β'a = 2πω (a/k) N0 so that I = 2πω (a/k) N0 (D.2.31a) N0 = (k/2πωa) I . // I is called i(z=0) in (4.9.2) so I = i(0) (D.2.31b) From (D.1.8) we know that N0 = <n(θ)> = (1/2πa) q and since q = CV we get the alternate form, I = 2πa (ω/k) N0 = 2πa (ω/k) (1/2πa) q = (ω/k) CV . (D.2.31c) I cannot find anything wrong with this computation of I. I have checked it many times and just did it again. ************************ 10.4.14. Now restate our results from above Ez(r,m) = (1/4) ηm CV Rdc (ω/k) (r/a)m (m+1) (2) Er(r,m) = (j/4) ηm CV Rdc (ωa) [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (1/4) ηm CV Rdc (ωa) [r/a)m+1 - (r/a)m-1] m > 0 and G = 0 Bz(r,m) = (j/2) ηm CV Rdc (r/a)m (m+1) Br(r,m) = (1/4) ηm CV Rdc (r/a)m-1 (1/ak) [- 2m(m+1) + k2 (a2-r2) ] Bθ(r,m) = (j/4) ηm CV Rdc (r/a)m-1 (1/ak) [- 2m(m+1) + k2 (a2+r2) ] Are these equations consistent with the Maxwell curl B equation? You would think so since I showed this for the general solutions, but let's just check to make sure. Set the coefficients to 1 for simplicity since constant won't affect the result, so Ez(r,m) = (1/2) (ω/k) (r/a)m (m+1) Er(r,m) = (j/4) (ωa) [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (1/4) (ωa) [r/a)m+1 - (r/a)m-1] m > 0 and G = 0 and const = 1 Bz(r,m) = (j/2) (r/a)m (m+1) Br(r,m) = (1/4) (r/a)m-1 (1/ak) [- 2m(m+1) + k2 (a2-r2) ] Bθ(r,m) = (j/4) (r/a)m-1 (1/ak) [- 2m(m+1) + k2 (a2+r2) ] Maybe check ALL of the Maxwell equations at this point! I am suspicious. In the Maple file "B for small omega.mws" (App D) I directly enter the above Ei expressions and that file computes the B expressions. you see above, using B = (j/ω) curl E Right now I will add more code to that mws file in an attempt to verify all of Maxwell's equations. I can steal some of this code from "B field verify 9_20.mws". I am doing this right now only for m > 0. I am able to verify the first three Maxwell equations OK. The fourth one is this: curl B = μ J + μ jωεE = μ(σ + jωε) E = μ(jω)( ε - jσ/ω) E = jω μξ E (D.5.1) = j (β2/ω) E . // see (1.5.1c) which I take simply to be curl B = j (β2/ω) E. However, in my small-x limit in which everything is being done here, we have β' << 1 so to speak, and that means we have β = k so you can replace the above with curl B = j (k2/ω) E In "B for small omega m GT 0.mws" and "B for small omega m EQ 0.mws" I show that my low-x fields for E and B do in fact satisfy all four Maxwell equations, including the curl B one above. I thought there was a problem here at first and wrote " Problem with the curl B Maxwell Equation", but in that doc I realized that there is in fact no problem. 10.8.14. Let's try this all again now that I have a fancier statement of the B fields in D.9. Summary of E and B fields inside a round wire (D.9.39) Ez(r,m) = (1/4) ηm B (ω/k) (aβ') fm x = β'r xa = β'a β'2 = β2 - k2 Er(r,m) = (j/4) ηm B (ωa) gm Eθ(r,m) = (1/4) ηm B (ωa) hm B ≡ (ξd/εd) CV Rdc Bz(r,m) = (j/4) (a) ηm B ( β' em ) (ξd/εd) = 1 + (G/jωC) Br(r,m) = - (1/4) (a) ηm B (1/k) ( r-1m β' fm + k2 hm ) Bθ(r,m) = (j/4) (a) ηm B (1/k) ( k2 gm - β'2fm [ (m/x) - Jm+1(x)/Jm(x)] ) ηm = Nm/N0 em = [ + ] gm = [ + ] Rdc = fm = [ - ] hm = [ - ] G ≥ 0 I know that for small x and xa (and small β') em = (r/a)m (m+1) (2/β'a) e0 = 0 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 Thus the B field expressions for small x are: Bz(r,m) = (j/4) (a) ηm B ( β' em ) Br(r,m) = - (1/4) (a) ηm B (1/k) ( r-1m β' fm + k2 hm ) Bθ(r,m) = (j/4) (a) ηm B (1/k) ( k2 gm - β'2fm [ (m/x) - Jm+1(x)/Jm(x)] ) Bz(r,0) = (j/4) (a) B ( β' e0 ) Br(r,0) = - (1/4) (a) B (1/k) ( r-1m β' f0 + k2 h0 ) Bθ(r,0) = (j/4) (a) B (1/k) ( k2 g0 - β'2f0 [ (m/x) - Jm+1(x)/Jm(x)] ) Now earlier I showed that (m/x) - Jm+1(x)/Jm(x) ≈ m/(β'r) m > 0 (m/x) - Jm+1(x)/Jm(x) ≈ -(β'r)/2 m = 0 So install things to get Bz(r,m) = (j/4) (a) ηm B ( β'(r/a)m (m+1) (2/β'a) ) Br(r,m) = - (1/4) (a) ηm B (1/k) ( r-1m β' (r/a)m (m+1) (2/β'a) + k2 [(r/a)m+1 - (r/a)m-1] ) Bθ(r,m) = (j/4) (a) ηm B (1/k) ( k2 [(r/a)m+1 + (r/a)m-1] - β'2(r/a)m (m+1) (2/β'a) [m/(β'r)] ) Bz(r,0) = (j/4) (a) B ( β' 0 ) Br(r,0) = - (1/4) (a) B (1/k) ( r-1 0 β' 4/(aβ') + k2 0 ) Bθ(r,0) = (j/4) (a) B (1/k) ( k2 2 (r/a) - β'2 4/(aβ') [-(β'r)/2] ) Do a rewrite and simplification Bz(r,m) = (j/4) (a) ηm B ( (r/a)m (m+1) (2/a) ) Br(r,m) = - (1/4) (a) ηm B (1/k) ( r-1m (r/a)m (m+1) (2/a) + k2 [(r/a)m+1 - (r/a)m-1] ) Bθ(r,m) = (j/4) (a) ηm B (1/k) ( k2 [(r/a)m+1 + (r/a)m-1] - (r/a)m (m+1) (2/a) [m/(r)] ) Bz(r,0) = 0 Br(r,0) = 0 Bθ(r,0) = (j/4) (a) B (1/k) ( k2 2 (r/a) + β'2 2r/a ) Now use this (r/a)m /r = (r/a)m (a/r)(1/a) = (r/a)m-1(1/a) to get Br(r,m) = - (1/4) (a) ηm B (1/k) ( r-1m (r/a)m (m+1) (2/a) + k2 [(r/a)m+1 - (r/a)m-1] ) Bθ(r,m) = (j/4) (a) ηm B (1/k) ( k2 [(r/a)m+1 + (r/a)m-1] - (r/a)m (m+1) (2/a) [m/(r)] ) Br(r,m) = - (1/4) (a) ηm B (1/k) ( m (r/a)m-1(1/a) (m+1) (2/a) + k2 [(r/a)m+1 - (r/a)m-1] ) Bθ(r,m) = (j/4) (a) ηm B (1/k) ( k2 [(r/a)m+1 + (r/a)m-1] - (r/a)m-1(1/a) (m+1) (2/a) [m] ) Br(r,m) = - (1/4) (a) ηm B (1/k) ( m(m+1) (r/a)m-1 (2/a2) + k2 [(r/a)m+1 - (r/a)m-1] ) Bθ(r,m) = (j/4) (a) ηm B (1/k) ( k2 [(r/a)m+1 + (r/a)m-1] - (r/a)m-1m (m+1) (2/a2) ) Br(r,m) = - (1/4) (a) ηm B (1/k) ( 2m(m+1) (r/a)m-1 (1/a2) + k2 [(r/a)m+1 - (r/a)m-1] ) Bθ(r,m) = (j/4) (a) ηm B (1/k) ( k2 [(r/a)m+1 + (r/a)m-1] - (r/a)m-12m (m+1) (1/a2) ) Br(r,m) = (1/4) (a) ηm B (1/k) (- 2m(m+1) (r/a)m-1 (1/a2) - k2 [(r/a)m+1 - (r/a)m-1] ) Bθ(r,m) = (j/4) (a) ηm B (1/k) (- 2m(m+1) (r/a)m-1 (1/a2) + k2 [(r/a)m+1 + (r/a)m-1] ) Br(r,m) = (1/4) (a) ηm B (1/k) ((r/a)m-1 {- 2m(m+1) (1/a2) + k2} - k2(r/a)m+1 ) Bθ(r,m) = (j/4) (a) ηm B (1/k) ((r/a)m-1 {- 2m(m+1) (1/a2) + k2} + k2(r/a)m+1 ) Br(r,m) = (1/4) (a) ηm B (1/k) (r/a)m-1 ({- 2m(m+1) (1/a2) + k2} - k2(r/a)2 ) Bθ(r,m) = (j/4) (a) ηm B (1/k) (r/a)m-1 ( {- 2m(m+1) (1/a2) + k2} + k2(r/a)2 ) Br(r,m) = (1/4) (a) ηm B (1/k) (r/a)m-1 (- 2m(m+1) (1/a2) + k2 - k2(r/a)2 ) Bθ(r,m) = (j/4) (a) ηm B (1/k) (r/a)m-1 (- 2m(m+1) (1/a2) + k2 + k2(r/a)2 ) Br(r,m) = (1/4) (a) ηm B (1/k) (r/a)m-1 (1/a2) (- 2m(m+1) + k2a2 - k2r2 ) Bθ(r,m) = (j/4) (a) ηm B (1/k) (r/a)m-1 (1/a2) (- 2m(m+1) + k2a2 + k2r2 ) Br(r,m) = (1/4) (a) ηm B (1/k) (r/a)m-1 (1/a2) (- 2m(m+1) + k2(a2 - r2 ) ) Bθ(r,m) = (j/4) (a) ηm B (1/k) (r/a)m-1 (1/a2) (- 2m(m+1) + k2(a2 + r2 ) ) Br(r,m) = (1/4) ηm B (1/ka) (r/a)m-1 (- 2m(m+1) + k2(a2 - r2 ) ) Bθ(r,m) = (j/4) ηm B (1/ka) (r/a)m-1 (- 2m(m+1) + k2(a2 + r2 ) ) The complete results at this point are then Bz(r,m) = (j/2) ηm B (r/a)m (m+1) Br(r,m) = (1/4) ηm B (1/ka) (r/a)m-1 (- 2m(m+1) + k2(a2 - r2 ) ) Bθ(r,m) = (j/4) ηm B (1/ka) (r/a)m-1 (- 2m(m+1) + k2(a2 + r2 ) ) Bz(r,0) = 0 Br(r,0) = 0 Bθ(r,0) = (j/2) B (kr) And this agrees exactly with (7.4.2), confirmed!