Chap 5 rewrite 2 REVIEWED

Chapter 5 rewrite #2 PhL 1.8.14 Chapter 5: The Transverse Problem 5.1 Separation of φ Let φ ≡ φ12(x) of Section 4.2. Then in the transmission line limit we found φ(x) = q(z) !Syntax Error, Idz'{ !Syntax Error, Idx1' dy1' α1(x1',y1') – !Syntax Error, Idx2' dy2' α2(x2',y2') } . (4.3.10) We now rewrite this as φ(x,y,z) = q(z) φt(x,y) (5.1.1) φt(x,y) = !Syntax Error, Idz'{ !Syntax Error, Idx1' dy1' α1(x1',y1') – !Syntax Error, Idx2' dy2' α2(x2',y2') } (5.1.2) where R1 = | x - x1'|, R2 = | x - x2'|, and x is a point in the dielectric. We thus identify φt as a certain "transverse potential" associated with the full potential φ. We also had ρ1(x,y,z) = α1(x,y) q1(z) . (4.1.2) ρ2(x,y,z) = α2(x,y) q2(z) . (4.1.2) so ρ(x,y,z) = ρ1(x,y,z) + ρ2(x,y,z) = [ α1(x,y) - α2(x,y)] q(z) ≡ ρt(x,y) q(z) so that ρ(x,y,z) = ρt(x,y) q(z) ρt ≡ α1(x,y) - α2(x,y) (5.1.3) where ρt is a "transverse charge distribution" associated with ρ. The Helmholtz equation for φ in the dielectric region is given by (1.5.3), (2 + β2)φ(x,y,z) = - (1/ε) ρ(x,y,z) (1.5.3) (5.1.4) where ρ(x,y,z) exists on the boundary of the dielectric region (ie, on the conductor surfaces). Combining equations (5.1.1,3,4), (2 + β2) q(z) φt(x,y) = - (1/ε) ρt(x,y) q(z) or (t2 + ∂z2 + β2) q(z) φt(x,y)= - (1/ε) ρt(x,y) q(z) // t2 = 2D2 = 2 - ∂z2 or t2φt(x,y) q(z) + φt(x,y) ∂z2q(z) + β2 φt(x,y) q(z) = - (4πξ/ε) ρt(x,y) q(z) . Now divide through by φt(x,y) q(z) to get + + β2 = - (4πξ/ε) or [ + 4π(ξ/ε) ] + = - β2 (5.1.5) which has the general form, [ h(x,y) ] + g(z) = - β2 . The only way this can be true for all x,y,z in a region is if g(z) = some constant, which call kφ2. Then, = kφ2 [ + 2π ] = - β2 - kφ2 . (5.1.6) We can rewrite these equations as [ t2 + (β2 +kφ2)] φt(x,y) = -2πρt(x,y) (5.1.7) [ ∂z2 - kφ2 ] q(z) = 0 . (5.1.8) The second equation has the following solution q(z) = q(0) e-jkz => q(z,t) = q(0) ej(ωt-kz) (5.1.9) and we find that q(z) has the form of a wave traveling down the transmission line with wavenumber kφ. The reader of Chapter 2 or of Appendix D will recognize this as the form assumed for the electric field in (2.1.1) or (D.1.1) where it was assumed as an ansatz without much a priori justification. 5.2 Separation of Az Let Az ≡ Az12(x) of Section 4.9. Then in the transmission line limit we found Az(x) = i(z) !Syntax Error, Idz'{ !Syntax Error, Idx1' dy1' b1(x1',y1') – !Syntax Error, Idx2' dy2' b2(x2',y2') } . (4.9.1) We now rewrite this as Az(x,y,z) = i(z) Azt(x,y) (5.2.1) Azt(x,y) = !Syntax Error, Idz'{ !Syntax Error, Idx1' dy1' b1(x1',y1') – !Syntax Error, Idx2' dy2' b2(x2',y2') } (5.2.2) where R1 = | x - x1'|, R2 = | x - x2'|, and x is a point in the dielectric. We thus identify Azt as a certain "transverse vector potential" associated with the full vector potential Az. The Helmholtz equation for Az is given by (1.5.4) for a region including dielectric and conductors, (2 + β2)Az(x,y,z) = - Σi=2N μiJi . (1.5.4) (5.2.3) These Ji are currents inside the conductors. Although there is small conduction current in the dielectric, it has been absorbed into β2 as shown in (1.3.21) in the time domain with the use of the King gauge. If we take our region of interest to be the dielectric alone, we then have (2 + β2)Az(x,y,z) = 0 . (5.2.4) Note: there are no free surface currents on the conductor surfaces which would correspond to the free surface charge appearing in (5.1.4). Inserting (5.2.1) into (5.2.4) yields, (2 + β2) i(z) Azt(x,y) = 0 or (t2 + ∂z2 + β2) i(z) Azt(x,y) = 0 or t2Azt(x,y) i(z) + Azt(x,y)∂z2i(z) + β2 Azt(x,y) i(z) = 0 . Now divide through by Azt(x,y) i(z) to get + + β2 = 0 (5.2.5) which has the general form, [ h(x,y) ] + g(z) = - β2 . The only way this can be true for all x,y,z in a region is if g(z) = some constant, which call kA2. Then, = kA2 = - β2 - kA2 . (5.2.6) We can rewrite these equations as [ t2 + (β2 +kA2)] Azt(x,y) = 0 (5.2.7) [ ∂z2 - kA2 ] i(z) = 0 . (5.2.8) The second equation has the following solution i(z) = i(0) e-jkz => i(z,t) = i(0) ej(ωt-kz) (5.2.9) and we find that i(z) has the form of a wave traveling down the transmission line with wavenumber kA. Comparing (5.2.9) with (5.1.9), it would certainly seem odd if q(z) and i(z) had the form of travelling waves with different wavenumbers kφ ≠ kA. We will formally show in the next section that kφ = kA. 5.3 Statement of the transverse problem Gathering facts from above we have [ ∂z2 - kφ2 ] q(z) = 0 (5.1.8) [ ∂z2 - kA2 ] i(z) = 0 . (5.2.8) But, φ(x,y,z) = q(z) φt(x,y) (5.1.1) Az(x,y,z) = i(z) Azt(x,y) . (5.2.1) Therefore, [ ∂z2 - kφ2 ] φ(x,y,z) = 0 [ ∂z2 - kA2 ] Az(x,y,z) = 0 . (5.3.1) Recall that x1 and x2 are points on the surfaces of conductors C1 and C2. If we write equations (5.3.1) first at x1 and then at x2 and then subtract, we get [ ∂z2 - kφ2 ] V(z) = 0 // V(z) = φ(x1) - φ(x2) [ ∂z2 - kA2 ] W(z) = 0 // W(z) = Az(x1) - Az(x2) (5.3.2) where we have used the definitions V(z) and W(z) from (4.4.1) and (4.10.1). Recall now the transmission line equations (4.11.11), ∂zV(z) = - z i(z) ∂zi(z) = - yV(z) . (4.11.11) Apply ∂z to each equation and move the right side to the left side, ∂z2V(z) + z ∂z i(z) = 0 ∂z2i(z) + y ∂zV(z) = 0 . Now reuse (4.11.11) to replace ∂z i(z) = - yV(z) and ∂zV(z) = - z i(z) [∂z2 - zy ]V(z) = 0 [∂z2 - zy ] i(z) = 0 Finally, use (4.10.8) to replace i(z) = W(z)/Le and then our equations become [∂z2 - zy ]V(z) = 0 [∂z2 - zy ]W(z) = 0 . (5.3.3) Comparison of (5.3.3) with (5.3.2) shows that kφ2 = kA2 = zy ≡ k2 (5.3.4) which fulfills the claim made earlier that kφ = kA . The resulting transverse equations (5.1.7) and (5.2.7) are then, [ t2 + (β2 +k2)] φt(x,y) = -2πρt(x,y) (5.3.5) [ t2 + (β2 +k2)] Azt(x,y) = 0 (5.3.6) k2 = z y . ************************************************************ The resulting transverse equations (5.1.7) and (5.2.7) are then, [ t2 + (β2 +k2)] φt(x,y) = 0 (5.3.5) [ t2 + (β2 +k2)] Azt(x,y) = 0 (5.3.6) k2 = z y .