Appendix S INSTALLED

This was installed on 9/24/14. Appendix S: Details of the Chapter 4 Averaging Procedure Here we redo the averaging process outlined in Section 4.11 (b) in more detail, and we maintain the transverse derivatives of the vector potential which were neglected in that Section. Since the process is explained there, here we just show what happens to the various equations. Corresponding equation numbers from Chapter 4 are shown in italics. The function arguments suppressed in Chapter 4 for points x1 and x2 are shown in red as (x1,x2). New terms arising from the previously neglected transverse vector potential derivatives are shown in blue. The potential difference between two conductors in the transmission line limit is given by V(x1,x2) ≡ φ12(x1) - φ12(x2) = q(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' α1(x1',y1') – !Syntax Error, Idx2' dy2' α2(x2',y2') } – q(z) !Syntax Error, Idz' { !Syntax Error, Idx1' dy1' α1(x1',y1') – !Syntax Error, Idx2' dy2' α2(x2',y2') } (4.4.1) (S.1) and later V(x1,x2) = q(z) {!Syntax Error, Idx1' dy1' α1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' α2(x2',y2') ln(s222/s122) } s212 = (x2-x1')2 + (y2-y1')2 s222 = (x2-x2')2 + (y2-y2')2 (4.4.6) s112 = (x1-x1')2 + (y1-y1')2 s122 = (x1-x2')2 + (y1-y2')2 . (S.2) Then = (S.3) = {!Syntax Error, Idx1' dy1' α1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' α2(x2',y2') ln(s222/s122) } = K(x1,x2) and V(x1,x2) = q(z) K(x1,x2) (4.4.7) where K(x1,x2) = {!Syntax Error, Idx1' dy1' α1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' α2(x2',y2') ln(s222/s122) } (S.4) Continuing along, Le(x1,x2) = (μd/4π)K(x1,x2) . (4.4.11) (S.5) Z0(x1,x2) = (1/4π) K(x1,x2) (4.4.14) (S.6) We them move from the V section to the W section which is Section 4.10 : W(x1,x2) = i(z) {!Syntax Error, Idx1' dy1' b1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' b2(x2',y2') ln(s222/s122) } (4.10.4) (S.7) W(x1,x2) = Le(x1,x2) i(z) . (4.10.7) (S.8) Le(x1,x2) = = KL(x1,x2) (4.10.8) (S.9) where KL is a dimensionless function we can compare to K, KL(x1,x2) ≡ !Syntax Error, Idx1' dy1' b1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' b2(x2',y2') ln(s222/s122) . (4.10.9) (S.10) K(x1,x2) ≡ !Syntax Error, Idx1' dy1' α1(x1',y1') ln(s212/s112) -!Syntax Error, Idx2' dy2' α2(x2',y2') ln(s222/s122) . (4.4.8) Moving along to Section 4.12 (a) we find (evaluated at some point x ) E = - grad φ - ∂tA (1.3.1) div A = - μdεd ∂tφ - μσφ . // the King gauge (1.3.18) (S.11) Both the above equations are exact and can be rewritten as: Ez(x) = - ∂zφ(x) - jωAz(x) ∂zAz(x) + (∂xAx + ∂yAy) = - j (βd2/ωφ(x) . (4.12.3) (S.12) where the blue term (∂xAx + ∂yAy) was assumed to vanish in Chapter 4, but now we maintain it in what follows, continuing through the development of Section 4.12. Taking into account both conductors gives Ez(x) = - ∂zφ12(x) - jωAz12(x) (4.12.4a) ∂zAz12(x) + (∂xAx12(x) + ∂yAy12(x)) = - j (βd2/ωφ12(x) . (4.12.4b) (S.13) Then evaluate at x1 and x2 and subtract to get Ez(x1) - Ez(x2) = -∂z[φ12(x1) - φ12(x2)] - jω[Az12(x1) - Az12(x2)] (4.12.5a) ∂z[Az12(x1) - Az12(x2)] + (∂xAx12(x1) - ∂xAx12(x2)) + (∂yAy12(x1) - ∂yAy12(x2)) = - j (β2/ω[φ12(x1) - φ12(x2)] . (4.12.5b) (S.14) We then define V(x1,x2) ≡ φ12(x1) - φ12(x2) (4.4.1) W(x1,x2) ≡ Az12(x1) - Az12(x2) (4.10.1) (S.15) and rewrite the previous equation pair as Ez(x1) - Ez(x2) = - ∂zV(x1,x2) - jωW(x1,x2) (4.12.6a) (S.16) ∂zW(x1,x2) = (∂xAx12(x1) - ∂xAx12(x2)) + (∂yAy12(x1) - ∂yAy12(x2)) - j (βd2/ωV(x1,x2) (4.12.6b) Then define function T to represent the Transverse derivatives. T(x1,x2) ≡ (∂xAx12(x1) - ∂xAx12(x2)) + (∂yAy12(x1) - ∂yAy12(x2)) . // dim(T) = tesla (S.17) We can then "double average" V,W and T as demonstrated in (4.12.13) to obtain from (S.16), ∂zW(z) = T(z) - j (βd2/ωV(z) (S.18) where W(z) ≡ <W(x1,x2)>C1,C2 (4.12.7) V(z) ≡ <V(x1,x2)>C1,C2 (4.12.7) T(z) ≡ <T(x1,x2)>C1,C2 . (S.19) Equation (4.12.6a) from (S.16) is then double-averaged to give [Ez1(z) - Ez2(z)] = - ∂zV(z) - jω W(z) (4.12.6a) (S.20) where for example, Ez1(z) ≡ <Ez1(x1)>C1,C2 = (1/P1) ∫C1 ds1 (1/P2)∫C2 ds2 Ez1(x1) = (1/P1) ∫C1 ds1 Ez1(x1) = <Ez1(x1)>C1 . (S.21) Then Ez1(x1) = Zs1(x1) i1(z) gets double-averaged in the same way to define Zs1. At this point we have [Zs1 + Zs2] i(z) = - ∂z V(z) - jω W(z) ∂zW(z) = - j (βd2/ωV(z) + T(z) (4.12.11) (S.22) where T(z) was not present in (4.12.11). From (S.9) we write the Le equation then double-average it, Le(x1,x2) = → Le = . (4.10.8) (S.23) Since then W(z) = Le i(z), the equation pair (S.22) maybe be rewritten, [Zs1 + Zs2] i(z) = - ∂z V(z) - jω W(z) Le∂zi(z) = - j (βd2/ωV(z) + T(z) (S.24) or ∂zV(z) = - [ Zs1+ Zs2+ jωLe] i(z) ∂z i(z) = - [ jβd2/(ωLe)] V(z) + T(z)/Le (4.12.14) (S.25) These are the classical transmission line equations, usually written as = - z i(z) = - y V(z) + T(z)/Le (4.12.15) (S.26) where z = Zs1+ Zs2 + jωLe = R + jωL // z and R are ohms/m (S.27) y = jβd2/(ωLe) = G +jωC = jωC' . // y and G are mhos/m (4.12.16) (S.28) Applying ∂z to the transmission line equations (S.26) then re-using them results in the following second order transmission line equations: - zy V(z) = (-z/Le)T(z) - zy i(z) = (1/Le) ∂zT(z) . (4.12.17) (S.29) which are now inhomogeneous differential equations due to T(z) ≠ 0. In the z equation (S.27) both R and L can be represented as double averages, R = < Re(z)>C1,C2 = <Re[Zs1(x1) + Zs2(x2) + jωLe(x1,x2)] >C1,C2 jωL = < Im(z)>C1,C2 = <Im[Zs1(x1) + Zs2(x2) + jωLe(x1,x2)] >C1,C2 . (S.30) However, in the y equation (S.28) this cannot be done because the number Le ≡ <Le(x1,x2)>C1,C2 is in the denominator instead of the numerator. Thus we must regard the numbers G,C and C' in (S.28) as being defined by the quantity jβd2/(ωLe). However, we can invert both sides of the y equation to get 1/y = ωLe/(jβd2) = (1/C') /(jω) (S.31) which can be interpreted as ω/(jβd2) * < Le(x1,x2) >C1,C2 = (1/jω) * < >C1,C2 (S.32) or ωLe/(jβd2) = < > /(jω) (S.33) or Le< >-1 = (βd2/ω2) = μdξd . // using (1.5.1a) (4.12.19) (S.34) Comparing (S.33) with (S.31) shows that the number C' appearing in (S.31) and (S.28) can be interpreted in this manner, C' = < >-1 . (S.35) Going back a bit, we had = V(x1,x2)/q(z) = K(x1,x2) (4.4.7) (S.3) Le(x1,x2) = = KL(x1,x2) (4.10.8) (S.9) which can be double averaged to get < > = V(z)/q(z) = K (S.36) Le = = KL . (S.37) Inserting these last expressions into (S.34) gives, Le< >-1 = μdξd or [ KL ] [4πξd/K] = μdξd or KL = K . (4.12.20) (S.38) This equality has thus survived the double averaging procedure.