Chap 5 scratch work REVIEWED

Chapter 5 scratch work PhL 1.8.14 This is obs. The φ Section 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) which we write as φ(x,y,z) = q(z) φt(x,y) (5.2.6) φt(x,y) = !Syntax Error, Idz'{ !Syntax Error, Idx1' dy1' α1(x1',y1') – !Syntax Error, Idx2' dy2' α2(x2',y2') } 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) 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) 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) Thus (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) 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 // original (5.2.7_ 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. For reasons that will be clear later, we write this constant as kφ2. Then we get = kφ2 [ + 2π ] = - β2 - kφ2 (5.2.9) We can rewrite these as [ t2 + (β2 +kφ2)] φt(x,y) = -2πρt(x,y) (5.2.10) [ ∂z2 - kφ2 ] q(z) = 0 (5.2.11) The Az Section 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) which we write as Az(x,y,z) = i(z) Azt(x,y) 5.2.6) Azt(x,y) = !Syntax Error, Idz'{ !Syntax Error, Idx1' dy1' b1(x1',y1') – !Syntax Error, Idx2' dy2' b2(x2',y2') } We also had Jz1(x,y,z) = b1(x,y) i1(z) . (4.7.3) Jz2(x,y,z) = b2(x,y) i2(z) . (4.1.2) Jz (x,y,z) = Jz1 (x,y,z) + Jz2(x,y,z) = [ b1(x,y) - b2(x,y)] i(z) ≡ Jzt(x,y) i(z) The Helmholtz equation for Az in the dielectric region is given by (1.5.4), (2 + β2)Az(x,y,z) = 0 // (1.5.4) Thus (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 or + = -β2 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. For reasons that will be clear later, we write this constant as kA2. Then we get = kA2 = - β2 - kA2 (5.2.9) We can rewrite these as [ t2 + (β2 +kA2)] Azt(x,y) = 0 (5.2.10) [ ∂z2 - kA2 ] i(z) = 0 (5.2.11) which agrees with original (5.2.16) . That which comes next Gathering from above we have [ ∂z2 - kφ2 ] q(z) = 0 [ ∂z2 - kA2 ] i(z) = 0 But, φ(x,y,z) = q(z) φt(x,y) (5.2.6) Az(x,y,z) = i(z) Azt(x,y) 5.2.6) And therefore [ ∂z2 - kφ2 ] φ(x,y,z) = 0 [ ∂z2 - kA2 ] Az(x,y,z) = 0 Recall that V(z) = φ( x1, y1, z ) - φ( x2, y2, z ) W(z) = Az( x1, y1, z ) - Az( x2, y2, z ) (5.2.18) where x1 and x2 are points on the conductors at the same z. Applying (***to these differences gives, [ ∂z2 - kφ2 ] V(z) = 0 [ ∂z2 - kA2 ] W(z) = 0 Now, application of ∂z to the transmission line equations ∂zV(z) = - z i(z) ∂zi(z) = - y V(z) gives ∂2zV(z) = - z∂z i(z) ∂z2i(z) = - y ∂zV(z) or combining the last two pairs of equations, ∂2zV(z) - zy V(z) = 0 ∂2zW(z) - zy W(z) = 0 huh?? where did W(z) come from? or [ ∂z2 - zy ] V(z) = 0 [ ∂z2 - zy ] W(z) = 0 Comparison *** shows that kA2 = kφ2 = zy ≡ k2 (5.2.22) The resulting transverse equations are then [ t2 + (β2 +k2)] Azt(x,y) = 0 (5.2.10) [ t2 + (β2 +k2)] φt(x,y) = -2πρt(x,y) (5.2.10) If I show that k2 ≈ -βd2 then I get [ t2 + (β2 - βd2)] Azt(x,y) = 0 (5.2.10) [ t2 + (β2 - βd2)] φt(x,y) = -2πρt(x,y) (5.2.10) If I wander off now to Appendix D, I see there this typical equation [r2∂r2 + r ∂r - m2 +r2( β2- βd2)] Ez(r,m) = 0 . (D.1.15) This is of course in a partial wave expansion business for a round wire, but you see (β2 - βd2) appearing there just as it is now appearing in Chapter 5. What do I know about problems like this: [ 2D2 + (β2 +k2)] Azt(x,y) = 0 Azt(boundary) = known (5.2.10) [ 2D2 + (β2 +k2)] φt(x,y) = -2πρt(x,y) (5.2.10)