App B_6 (c) REVIEWED

Appendix B.6 section (c) PhL 2.5.14 This is my first writing of the proof of the "theorem" as opposed to the "lemma" which comes earlier. In the theorem, I worry about the effect of other conductors on the boundary conditions for Az. This stuff is now all installed. (c) Statement and Proof of The Jm Theorem The Jm Theorem. A transmission line consists of two conductors called C2 and C3, so that the index 1 can refer to the dielectric. The total "particular" vector potential Az(x) due to the conduction currents in these two conductors is, according to (1.5.3), A(x)part = Σi=23∫ μiJci(x') dV' . R = |x-x'| The theorem claims that (1) the correct total potential is given by A(x) = Σi=23∫ [ μiJci(x') + μ0Jmi(x')] dV' . (B.6.27) where Jmi(x') represents the surface current at the surface of conductor Ci, and (2) this correct total potential satisfies the boundary conditions (B.6.0) at the surface of both conductors. We claim the theorem is also true for a transmission line consisting of any number of conductors, but we restrict our interest to two conductors. We shall show for (B.6.27) that (B.6.0) is valid at the surface of conductor C2 and a mirror argument then shows it is also valid at the surface of C3. Since we are operating in the transmission line limit, the actual claim being made is this: Az(x) = Σi=23∫ [ μiJczi(x') + μ0Jmzi(x')] dV' . (B.6.28) To enhance clarity, we shall give each conductor its own custom integration variable. The expression above contains four terms which we now write out: Az(c2)(x) = ∫ [μ2 Jcz2(x2')] dV2' R2 = |x - x2'| Az(c3)(x) = ∫ [μ3 Jcz3(x3')] dV3' R3 = |x - x3'| (B.6.4)' Az(m2)(x) = ∫ [μ0 Jmz2(x2')] dV2' R2 = |x - x2'| Az(m3)(x) = ∫ [μ0 Jmz3(x3')] dV3' R3 = |x - x3'| (B.6.5') Az(x) = Az(c2)(x) + Az(c3)(x) + Az(m2)(x) + Az(m3)(x) . (B.6.6)' Without loss of generality, we shall consider x → s where s is a point on the surface of conductor C2. The "conduction solutions" Az(ci) (that is to say, the particular solutions) are naturally smooth at point s, as described in the text surrounding (B.6.0). Thus, we know that Az(ci)(s+) = Az(ci)(s-) ∂nAz(ci)(s+) = ∂nAz(ci)(s-) . i = 2, 3 (B.6.7)' Since s = s+ = s- for these functions, we may trivially write ∂n[Az(c2)(s+) + Az(c3)(s+)] – ∂n[Az(c2)(s-) + Az(c3)(s-)] = + [ - ] ∂n[Az(c2)(s) + Az(c3)(s)]. (B.6.8)' Since this is non-zero, the term [Az(c2)(s+) + Az(c3)(s+)] on its own does not meet the required slope boundary condition (B.6.0) at an interface between μ1 and μ2, and that is precisely why we need the Az(m) terms. Our goal is to show that ∂n[Az(m2)(s+) + Az(m3)(s+)] – ∂n[Az(m2)(s-) + Az(m3)(s-)] = - [ - ] ∂n[Az(c2)(s) + Az(c3)(s)] . (B.6.9)' so that when we add the four terms of (B.6.6)' we will get ∂nAz(s+) – ∂nAz(s-) = 0 (B.6.10) as required by (1.1.46). The concludes our proof outline, and it remains then to demonstrate (B.6.9)'. At this point, we skip over several equations of the Lemma proof since they are all generalized simply by adding 2 or 3 subscripts in the right places. For example, the surface currents are given by, Kz2 = - ( - ) Hθ2 on C2 Kz3 = - ( - ) Hθ3 on C3 . (B.1.10) We arrive then at (B.6.16)' as follows ( recall that E2 is the 2D Helmholtz propagator ), Az(m2)(x) = C2 ds2' [(μ1-μ2) Hθ2(x2')] E2(x|x2') Az(m3)(x) = C3 ds3' [(μ1-μ3) Hθ3(x3')] E2(x|x3') . (B.6.16)' A major difference appears at the next step (B.6.18)' , ∂nAz(m2)(s±) = C2 ds2' [(μ1-μ2) Hθ2(x')] ∂nE2(s|x2') ∓ [(μ1-μ2) Hθ2(s)]/2 . ∂nAz(m3)(s±) = C3 ds3' [(μ1-μ3) Hθ3(x')] ∂nE2(s|x3') . (B.6.18)' The "extra Stakgold term" only appears when a point s lies on the surface being integrated over since it is this integration which gives rise to the singular situation. Since our s lies on C2 and not on C3, there is no "extra term" in the last equation above. Now since we want to prove (B.6.9)', we first evaluate its left hand side using each equation of (B.6.18)' twice, ∂n[Az(m2)(s+) + Az(m3)(s+)] – ∂n[Az(m2)(s-) + Az(m3)(s-)] = { C2 ds2' [(μ1-μ2) Hθ2(x2')] ∂nE2(s|x2') - [(μ1-μ2) Hθ2(s)]/2 } – { C2 ds2' [(μ1-μ2) Hθ2(x2')] ∂nE2(s|x2') + [(μ1-μ2) Hθ2(s)]/2 } + { C3 ds3' [(μ1-μ3) Hθ3(x3')] ∂nE2(s|x3') } – { C3 ds3' [(μ1-μ3) Hθ3(x3')] ∂nE2(s|x3') } = (μ1-μ2) [ - ] C2 ds2' Hθ2(x2') ∂nE2(s|x2') – [ + ] (μ1-μ2) Hθ2(s)/2 + [ - ] (μ1-μ3) C3 ds3' Hθ3(x3')] ∂nE2(s|x3') (B.6.19)' where the last line was not present in (B.6.19). Our task of showing that (B.6.9)' is true then boils down to showing that the last expression above is equal to - [ - ] ∂n[Az(c2)(s) + Az(c3)(s)] . That is to say, we have to show (μ1-μ2) [ - ] C2 ds2' Hθ2(x2') ∂nE2(s|x2') – [ + ] (μ1-μ2) Hθ2(s)/2 + [ - ] (μ1-μ3) C3 ds3' Hθ3(x3')] ∂nE2(s|x3') = - [ - ] ∂n[Az(c2)(s) + Az(c3)(s)] ? (B.6.20)' As before, a question mark indicates an equation that we want to show is true, but have not yet done so. Cancelling (μ1-μ2) factors gives [ - ] C2 ds2' Hθ2(x2') ∂nE2(s|x2') – [ + ] Hθ2(s)/2 - (μ1-μ3) C3 ds3' Hθ3(x3')] ∂nE2(s|x3') = + ∂n[Az(c2)(s) + Az(c3)(s)] ? or (B.6.21)' (μ2-μ1) C2 ds2' Hθ2(x2') ∂nE2(s|x2') – (μ2+μ1) Hθ2(s)/2 + (μ3-μ1) C3 ds3' Hθ3(x3')] ∂nE2(s|x3') = + ∂n[Az(c2)(s) + Az(c3)(s)] ? (B.6.22)' The integrals in (B.6.22)' can be replaced using (B.6.18)' with the s+ choice, ∂nAz(m2)(s+) = C2 ds2' [(μ1-μ2) Hθ2(x2')] ∂nE2(s|x2') - (1/2)[(μ1-μ2) Hθ2(s) ∂nAz(m3)(s) = C3 ds3' [(μ1-μ3) Hθ3(x2')] ∂nE2(s|x3') (B.6.18)+ so (μ2-μ1) C2 ds2' Hθ2(x2') ∂nE2(s|x2') = – ∂nAz(m2)(s+) - (1/2)[(μ1-μ2) Hθ2(s) (μ3-μ1) C3 ds3' Hθ3(x3')] ∂nE2(s|x3') = – ∂nAz(m3)(s) . (B.6.23)' Equation (B.6.22)' then becomes – ∂nAz(m2)(s+) - (1/2)[(μ1-μ2) Hθ2(s) – (μ2+μ1) Hθ2(s)/2 – ∂nAz(m3)(s) = ∂n[Az(c2)(s) + Az(c3)(s)] ? or – ∂nAz(m2)(s+) – μ1 Hθ2(s) – ∂nAz(m3)(s) = ∂n[Az(c2)(s) + Az(c3)(s)] ? or - μ1Hθ2(s) = ∂n[Az(c2)(s) + Az(c3)(s) + Az(m2)(s+) + ∂nAz(m3)(s) ] ? or - μ1Hθ2(s) = ∂nAz(s+) ? // using (B.6.6)'. (B.6.24)' But this last equation is true as shown in Fig B.6 (with C = C2) and (B.6.26), so we can then go back and erase all the question marks and we have then verified equation (B.6.9)' and our proof is complete. By then taking s to be a point on the surface of C3, we would find - μ1Hθ3(s) = ∂nAz(s+) for (B.6.24)' and then Fig B.6 with C = C3 would verify this result as well.