Bug in Section 4.7

Bug in Section 4.7 PhL 3.26.05 Statement of the Problem 4.7 BUG below (4.7.7) talking about μ: Why do I say there is no free surface current on a TL conductor? Later I say that there is in fact a Debye Surface current which is caused by the motion of the "free" surface charge. When I wrote Section 4.7, I was probably unaware of the Debye current. This could be a major problem! Ouch ouch ouch! Time to digress! I start with Az1(x,y,z) = !Syntax Error, Idz' i(z') !Syntax Error, Idx' dy' b1(x',y') . (4.7.6) I claim that there is no "magnetic boundary", so this function and its derivatives are continuous through the conductor boundary. King I think says this as well. But if there is a Debye surface current, then the first derivative is NOT continuous. Both facts cannot be true!! Yeouch! Let's look at the picture Fig D.6 Suppose I regard the "boundary" as being the upper surface. Then on that surface there is no free surface current, so Kz = 0, and then my argument is OK if + and - refer to the two sides of this top surface. I am then putting the Debye surface current inside the conductor along with the regular conduction current. Comments regarding μ This is a subtle subject and is not discussed in King's transmission line theory book. If μ1 = μd, and if there is no surface current on the C1 conductor boundary, then in fact there is no "magnetic boundary" at the conductor surface. The solution (4.7.2) is then smooth at this boundary, and so Az1(x,y,z) "naturally" satisfies these two boundary conditions, Az1(x+) = Az1(x-) (1/μd)∂nAz1(x+) = (1/μ1) ∂nAz1(x-) (4.7.7) where x+ is just outside the conductor surface and x- is just inside. The second equation here is just (1.1.46) in the case there is no free surface current Kzfree. In fact there is a Debye surface current as discussed in Section D.9, but it is totally negligible compared to the bulk conductor current and we therefore just ignore it. Since we have assumed that μd = μ1, this second boundary condition just says ∂nAz1(x+) = ∂nAz1(x+). Since there is no magnetic boundary at the conductor/dielectric interface, the solution (4.7.2) is continuous and all its derivatives are also continuous at the boundary, since nothing special happens at that boundary. Thus, the Helmholtz integral solution provides the whole solution for Az1 since it meets both "boundary conditions" at this pseudo boundary. ***************************************8 Question: Later I say this: φ1(x+) = φ1(x-) [ε1∂nφ1(x+) - ε∂nφ1(x-)] = nfree(x). (4.7.10) where nfree cannot be just neglected. Suppose both C1 and dielectric had ε0. Then this would say [ε0∂nφ1(x+) - ε0∂nφ1(x-)] = nfree(x). (4.7.10) which says ∂nφ1(x+) - ∂nφ1(x-) ≠ 0 But at the same time I argue that the Helm integral provides the complete solution! φ1(x,ω) = ∫ ρ1(x',y',z',ω) dx'dy'dz' . R = |x - x'| (4.1.1) If this is a complete and smooth solution, I would argue that all derivatives match on both sides of the boundary and that includes this one: ∂nφ1(x+) - ∂nφ1(x-) = 0 . Ouch again. This is really a surface layer integral, φ1(x,ω) = ∫C1 ns(x',ω) dS' R = |x - x'| (1.5.13) Stak tells us what happens when we approach this thing from the two sides. u(x) = ∫σ dSξ a(ξ) E(x|ξ) u(s) = ∫σ dSξ a(ξ) E(s|ξ) (B.6.2) ∂νu(x) = ∫σ dSξ a(ξ) ∂νE(x|ξ) ∂νu(s) = [∂nu(x)]x→s± = ∫σ dSξ a(ξ) ∂νE(s|ξ) ∓ a(s)/2 // extra term ! (B.6.3) and this is why ns appears in the jump condition! So the φ Helm integral is NOT so smooth as I was saying. For Az I guess we could consider the integral as two terms, where one is for the bulk current, and the other is for the Debye Surface current. This second contribution would then have a jump. I better then not make any comments about all derivatives being continuous. Comments regarding μ This is a subtle subject and is not discussed in King's transmission line theory book. In this section we regard conductor C1 as having parameters ε1, μ1 while the dielectric outside the conductor has εd and μd. Before dealing with Az and μ, it is useful to start with φ and ε. Our Helmholtz integral solution for φ at a point x in the dielectric, expressed for a single conductor C1, can be written as follows based on (1.5.13), φ(x,ω) = ∫ ns(x',ω) dS' . R = |x - x'| (1.5.13) The potential φ is continuous at the boundary, but ∂nφ is not continuous, having a jump there, though this fact is not obvious since we only have φ stated for x in the dielectric. We do know from (1.1.47) that, [εaEan - εbE2b] = nfree . (1.1.47) Letting a = dielectric and b = conductor, and with pointing out from the conductor, we translate this equation to read, εdEn,d - ε1En,c = ns (4.7.7) where ns is the free surface charge appearing in (1.5.13). According to (4.7.1) we can set transverse Ai components to zero, so then (1.3.1) which says E = - grad φ - jωA tells us that En = -∂nφ. We then have εd (-∂nφ(x+)) - ε1 (-∂nφ(x-)) = ns(x) (4.7.8) where x+ is just outside the conductor surface and x- is just inside. From this result one can compute the discontinuity or jump in ∂nφ at the conductor boundary. If εd = ε1 = ε0, then ∂nφ(x+) - ∂nφ(x- ) = -ns/ε0, for example. Normally, however, En,c ≈ 0 so (4.7.7) reads εdEn,d = ns and then En,d = ns/εd is the normal E field in the dielectric just outside the conductor. And since En,c ≈ 0, we have φ(x,ω) ≈ constant inside the conductor, so we are generally not interested in finding a version of (1.5.13) that is valid inside the conductor. We just evaluate (1.5.13) at the surface for some x+ and that gives φ inside the conductor. Thus, the Helmholtz integral (1.5.13) provides our full solution of interest, and we have no homogeneous adder terms to worry about of the type discussed below. With this as warm up, we now consider the case of Az and μ. Our Helmholtz integral solution for Az at a point in the dielectric, expressed for a single conductor C1, can be written as follows based on (4.7.2), Az1(x,ω) = ∫ Jz1(x',y',z',ω) dx'dy'dz' . R = |x - x'| (4.7.2) where μ1 is for the conductor C1. The potential Az1 is continuous at the boundary but ∂nAz1 is not continuous, having a jump. Appendix D.9 shows that there is a miniscule free surface current KzD which flows on the surface of a transmission line conductor, which we call the Debye surface current. It is ns being moved slightly by Ez at the surface. Equation (D.9.3) shows that the Debye current contribution to the above integral is totally negligible relative to bulk current contribution, so the left side of (4.7.2) is unchanged if we completely ignore this Debye current contribution to Jz. Thus, we are in effect setting KzD = 0 in this well justified approximation. We may then apply (1.1.46) to find that (1/μd) (∂nAz1(x+)) - (1/μ1) (∂nAz1(x-)) = 0 . (1.1.46) Thus we arrive at these boundary conditions on Az1 produced by conductor C1 : Az1(x+) = Az1(x-) (1/μd) ∂nAz1(x+) = (1/μ1) ∂nAz1(x-) (4.7.9) where again x+ is just outside the conductor surface and x- is just inside. If μ1 = μd, there is no "magnetic boundary" at the surface, and (4.7.9) says ∂nAz1(x+) = ∂nAz1(x-), so both the function Az1 and its normal derivative are continuous through the boundary -- nothing special is happening there. Thus, the Helmholtz integral solution (4.7.2) provides the whole solution for Az1 in both the dielectric and conductor since it meets both "boundary conditions" at this pseudo boundary. If on the other hand we have μ1 ≠ μd, then there is a magnetic boundary between conductor and dielectric which we have to worry about. In this case, (4.7.2) applied in both dielectric and conductor cannot possibly satisfy the second boundary condition of (4.7.9) since, as already noted, the Az1 of (4.7.2) satisfies ∂nAz1(x+) = ∂nAz1(x+). Thus, in this case (4.7.2) is not the full solution for Az1. One must add a homogeneous Helmholtz equation solution to (4.7.2) in order to have a proper solution for Az1 that satisfies both equations in (4.7.9). It turns out that the correct total Az1 solution can be generated by adding a certain fictitious surface current term to μ1Jz1 in (4.7.2). Since such a surface current vanishes on both sides of the boundary between μd and μ1, the Helmholtz solution due just to this surface current term is in fact a homogeneous solution to the Helmholtz equation in both the conductor and dielectric regions, away from that boundary. It turns out moreover that the correct fictitious surface current to add is in fact the magnetization surface current Jm which is created at the boundary between μd ≠ μ1. Adding this surface current is just a "trick" in order to generate the correct homogeneous adder solution so that the resulting total Az1 satisfies both boundary conditions in (4.7.9). Formally speaking, the Ji appearing in (1.5.4) and then Jz1 in (4.7.2) should not include such magnetization currents since this J is really the J in Maxwell's equation curl H = ∂tD + J, and this J does not include magnetization currents -- it includes only normal conduction currents. In our current Chapter 4, we want (4.7.2) to represent the complete solution for Az1 and for that reason we must restrict our analysis to the situation where dielectric and all conductors have the same permeability which we shall just call μd. In practice, one normally has μd = μ1 = μ0. In order to handle the more general case of μd ≠ μ1, we have to deal with the inhomogeneous adder solutions or equivalently with the abovementioned fictitious surface current, and this complicates our analysis which is already quite complicated. So, for the moment, we now make the same assumption made by King and other authors: Fact: From now on, conductors and dielectric must have the same permeability μd. (4.7.10) After fully developing this special case, we shall then extend the theory in Section 4.13 to allow for μd ≠ μ1. There are several Appendices which relate to this subject. Appendix G shows for the round wire how the inhomogeneous adder solution is found and how it then causes the boundary conditions (4.7.9) to be met when μ1 ≠ μd. However, in Appendix G the notation is different: 1 = dielectric and 2 = conductor, and μ2 ≠ μ1 is studied. Appendix B shows how the addition of a fictitious surface current term μ0Jm provides an alternate and simpler solution to the same problem of meeting boundary conditions (4.7.9) when μ1 ≠ μd. It then shows exactly how this works in the special case of a round wire (same notation as Appendix G).