App D actions 2 compute B and check Max REVIEWED

App D Actions 2 PhL 11.30.13 This material is all in Appendix D now. I did a few checks to show new App D agrees with B calculations that were done here. 1. Our first task is to compute the B field 1 2. Our next task is to check the four Maxwell equations 2 3. Here are the B fields read off from the Maple computation 3 I showed in App D Actions 1 that when I computed the corresponding B field, I obtained a pair of fields E and B which satisfies all four of Maxwell's equations, but that was only for m = 0. Here I would like to repeat that same process but for general m. I want to do this without assuming any boundary conditions. My E fields from App D are stated before any BC's are applied as" jEφ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) x = β'r . (D.2.15) Er(r,m) = am x-1 Jm(x) + Jm+1(x) . (D.2.11) Ez(r,m) = Czm Jm(β'r) (D.1.27) Km ≡ 2j (βd/β') Czm (D.2.5) From the last we get Czm = (1/2j)(β'/βd)Km = -j (β'/βd) Thus without any assertion of boundary conditions, our E field App D solutions for general m > 0 are Ez(r,m) = -j (β'/βd) Jm(x) x = β'r Er(r,m) = am x-1 Jm(x) + Jm+1(x) jEφ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) 1. Our first task is to compute the B field Now you have to remember that there is an outside factor of ej(ωt-βz) ejmφ so now we have two rules ∂z → -jβd ∂θ → +jm We identify θ and φ as the same variable. B = (-1/jω) curl E curl E = [ r-1∂θEz - ∂zEθ] + [∂zEr - ∂rEz] + [ r-1∂r(rEθ) - r-1∂θEr ] = [ r-1jmEz+jβdEθ] + [-jβdEr - ∂rEz] + [ r-1∂r(rEθ) - r-1jmEr ] Maple has computed these three components, I won't copy them here right now but they have this form B = Br(r) + Bθ(r) + Bz(r) 2. Our next task is to check the four Maxwell equations (1) div B = 0. div B = r-1∂r(rBr) + r-1∂θBθ + ∂zBz = r-1∂r(rBr) +r-1jmBθ -jβd Bz Maple confirms that div B = 0, I get to use my unapply operator, seems good. (2) div E = 0. div E = r-1∂r(rEr) + r-1∂θEθ + ∂zEz = r-1∂r(rEr) + r-1jmEθ - jβdEz Again Maple confirms. We just have verification within each partial wave since it could be activated all by itself. (3) curl E = -jωB, but this is how we produced B in the first place so OK. (4) curl B = μ(σ+jωε) E This one is the Big Enchilada. I should at the start make this replacement μ(σ+jωε) = (-1/jω) β2 so what I really need to show is this: β'2+ βd2 = β2 curl B = (-1/jω) β2 E = (-1/jω)( β'2+ βd2) E We start then with curl B = [ r-1∂θBz - ∂zBθ] + [∂zBr - ∂rBz] + [ r-1∂r(rBθ) - r-1∂θBr ] = [ r-1jmBz +jβdBθ] + [-jβdBr - ∂rBz] + [ r-1∂r(rBθ) - r-1jmBr ] We want to show that curl B + (1/jω) β2 E = 0 or [curlB]z + (1/jω) β2 Ez = 0 [curlB]r + (1/jω) β2 Er = 0 [curlB]th + (1/jω) β2 Eth = 0 Maple confirms all three of these equations! This is a major happy thing for me right now. 3. Here are the B fields read off from the Maple computation Bz = (β'/ω) ( + )Jm(β'r) Br = (j/ω)( r-1)[ (β'/βd) m - (βd/β')am ] Jm(β'r) + (j/ω)βd (+ ) Jm+1(β'r) Bθ = - (1/ω)( r-1)[ (β'/βd) m - (βd/β')am ] Jm(β'r) + (1/ω) (βd + β'2/βd) Jm+1(β'r) Rewrite to simplify if possible // these three match box (D.4.9) of new App D Bz = (β'/ω) ( + )Jm(β'r) Br = (j/ω)( r-1)[ (β'/βd) m - (βd/β')am ] Jm(β'r) + (j/ω)βd (+ ) Jm+1(β'r) Bθ = - (1/ω)( r-1)[ (β'/βd) m - (βd/β')am ] Jm(β'r) + (1/ω)β' (βd/β' + β'/βd) Jm+1(β'r) or Bz = (1/ω)β' ( + )Jm(x) Br = j (1/ω) βd[ (β'2/βd2) m - am ] Jm(x)/x + j (1/ω)βd (+ ) Jm+1(x) Bθ = - (1/ω) βd[ (β'2/βd2) m - am ] Jm(x)/x + (1/ω) βd (1 + β'2/βd2) Jm+1(x) I don't know how large the constants are, so not sure what terms are large. For m = 0 these say // these match (D.6.1) Bz = (1/ω) β' ( )J0(x) Br = j (1/ω)βd () J1(x) Bθ = (1/ω) βd (1 + β'2/βd2) J1(x) For β >> βd we can write these as Bz = (1/ω) βd (β'/βd) ( )Jm(x) medium Br = j (1/ω)βd () Jm+1(x) small Bθ ≈ (1/ω) βd (β'2/βd2) J1(x) large I know from actions 1 v2 that K0 = jI βd/(πσa) * 1/ J1(β'a) = jI (aβd)/(πσa2) * 1/ J1(β'a) = j (aβd) I Rdc/ J1(β'a) So my m=0 B field components become Bz = (1/2ω) β' (K0 )J0(x) Br = j (1/2ω)βd (K0) J1(x) Bθ = (1/2ω) βd K0 (1 + β'2/βd2) J1(x) or Bz = j (1/2ω) β' (aβd) I RdcJ0(x) / J1(β'a) Br = - (1/2ω) βd (aβd) I Rdc J1(x) / J1(β'a) Bθ = j (1/2ω) βd (aβd) I Rdc (1 + β'2/βd2) J1(x) / J1(β'a) Again in the β >> βd limit these become Bz = j (1/2ω) β' (aβd) I Rdc J0(x) / J1(xa) medium Br = - (1/2ω) βd (aβd) I Rdc J1(x) / J1(xa) small Bθ = j (1/2ω) β'(β'/βd) (aβd) I Rdc (J1(x) / J1(xa) large Now I can rewrite the last as Bθ ≈ j (1/2ω) β2 a I (1/πσa2) (J1(x) / J1(xa) = j (1/2ω) [ω2μσ/jω] I (1/πσa) (J1(x) / J1(xa) = ([μ] I (1/2πa) (J1(x) / J1(xa) and this is the Chapter 2 result. Here then is my general-m summary of all fields Ez(r,m) = -j (β'/βd) Jm(x) x = β'r Er(r,m) = am Jm(x)/x + Jm+1(x) jEφ(r,m) = - am Jm(x)/x + ( + ) Jm+1(x) (ω/β')Bz(r,m) = ( + )Jm(x) (ω/jβd)Br(r,m) = [ (β'2/βd2) m - am ] Jm(x)/x + ( + ) Jm+1(x) (ω/βd)Bθ(r,m) = - [ (β'2/βd2) m - am ] Jm(x)/x + (1 + β'2/βd2) Jm+1(x) And they satisfy the four Maxwell equations exactly. I don't know how to better state these fields.