old App D with betad ARCHIVE

Appendix D : The General E and B Fields Inside an Infinite Straight Round Wire This Appendix presents a rather lengthy calculation of the fields and currents inside a round wire without the Chapter 2 assumption that such fields and currents are symmetrical about the axis. This wire is regarded as one conductor of an infinite transmission line down which a wave is propagating. Since this Appendix is quite long, a brief summary is in order (see also Table of Contents) : Section D.1 (a) A longitudinal traveling wave form E(r,θz,t) = ej(ωt-βz) E(r,θ) is assumed inside the round wire and E(r,θz,t) is then shown to satisfy a certain vector Helmholtz equation. (b) The field E(r,θ) and the surface charge n(θ) are both expanded onto "azimuthal partial waves" ejmθ with coefficients E(r,m) and Nm. (c) The Helmholtz equation's vector Laplacian ≡ 2 is stated in cylindrical coordinates. (d) The three Helmholtz component equations and div E = 0 are written out in these coordinates. Section D.2 (a),(b),(c): The z and r Helmholtz equations and the div E = 0 equation are solved for Ez then Er and then Eθ. These solutions are expressed in terms of Bessel J functions of a complex argument and two unknown constants am and Km for each partial wave. (d) a boundary condition relating Er to surface charge density n(θ) is derived (see D.9 below) (e) this and another boundary condition Eθ(a,m) = 0 (see D.8 below) are used to evaluate am and Km and then the solution E field components are stated in box (D.2.33). Section D.3 It is noted that the boxed E field solutions also satisfy the ignored third θ Helmholtz equation. Section D.4 The B fields are computed from the E fields using Maxwell -jωB = curl E, and then box (D.4.9) summarizes both the E and B partial wave fields inside a round wire. Section D.5 These E and B fields are shown to exactly solve the other three Maxwell equations. Section D.6 The m = 0 partial wave results are stated and compared to the results of Chapter 2. Section D.7 The problem of finding an exterior field solution for the round wire is discussed. Section D.8 Arguments supporting the second boundary condition Eθ(a,m) = 0 are presented. Section D.9 The "charge pumping boundary condition" is discussed in relation to surface currents. Section D.10 High frequency limits of the round wire E fields are presented. Section D.11 Low frequency limits of the round wire E fields are presented, including losses. D.1 Partial Wave Expansion Warning: In this appendix, we use the same function name E to represent three different functions, E(r,θz,t) E(r,θ) E(r,m) The functions are distinguished by the arguments shown, and if they are not shown, the general context of the discussion will indicate which function is implied. The symbol E is thus "overloaded". (a) The General Method The starting point for the calculation is the damped wave equation (1.3.36, region 2) for the E field inside the wire. Unsubscripted parameters refer to properties of the wire. (2 - με ∂t2 - μσ∂t)E(r,θz,t) = 0 . (1.3.36) Cylindrical coordinates (r,θ,z) are used, as appropriate for an infinite straight round wire. Recall that the damping term arises when the driving current J on the right of (1.2.1) is replaced by Ohm's Law J = σE. We now make the ansatz that a solution to the above wave equation may be expressed in the following form where the t and z dependence is exposed and where E(r,θ) is a complex function to be determined: E(r,θz,t) = ej(ωt-βz) E(r,θ) . (D.1.1) The idea here is that we take our round wire to be one of two conductors of a transmission line (the other wire may or may not have a round cross section). The form shown in (D.1.1) says that the E field inside our round wire is assumed (an Ansatz!) to be a simple "traveling wave" moving down this transmission line in the +z direction. As this interior wave moves down the line, we expect to have an exterior wave whose E field takes the same general form shown in (D.1.1). If we match the E and B field boundary conditions of the interior and exterior waves, we expect βd to have the same value on both sides of the round wire boundary. For a lossless wave, we expect the conductors to simply deform the exterior fields (for example, causing the E field to be perpendicular to the conductor surfaces), but we expect the exterior wave to travel at the speed of light in the dielectric, with no "drag" from the conductors. In this lossless case, we then expect to have βd = ω = ω/vd where vd is the speed of light in the dielectric. If the dielectric conducts but the conductors are perfect, we have instead βd = ω and then βd has a small negative imaginary part which causes decay along the line, but βd is still a characteristic of the dielectric medium. However, if the conductors are not perfect, then they too contribute to the decay, and in this case we expect that our parameter βd will no longer be a characteristic just of the dielectric, despite the d subscript. For example, we expect it will depend on R, the resistance per unit length of the conductors. We shall continue to use the symbol βd throughout this appendix, but for the reasons just stated, we shall never use the explicit forms just quoted, βd = ω or βd = ω . One should think of βd as a general complex parameter which (hopefully) has a negative imaginary part and whose real part is the wave phase velocity. Only in the special case of a completely lossless line do we have βd = ω/vd. When (D.1.1) is put into the above wave equation (1.3.36), time derivatives can be replaced ∂t→ jω with the result (2 + β2) E(r,θz,t) = 0 (D.1.2a) or [2D2 + (β2-βd2) ] E(r,θ) = 0 2 = 2D2 + ∂z2 (D.1.2b) where β2 = μεω2 - jωμσ = ω2μ (ε - jσ/ω) = ω2μ ξ ξ ≡ ε - jσ/ω . (1.5.1) We could have defined the temporal Fourier Transform of E(r,θz,t), E^(r,θz,ω') ≡ FT{ E(r,θz,t), ω'} = e-jβz E(r,θ) 2πδ(ω-ω') = e-jωt E(r,θz,t) 2πδ(ω-ω') as in (1.6.11) and then (D.1.2a) would be valid as well for E^(r,θz,ω) which would be a more conventional Helmholtz equation, but since E(r,θz,t) is monochromatic, we leave (D.1.2a) as is. One can regard (D.1.1) as an assumed variable-separated form for a solution, an "ansatz". If a consistent solution to the Maxwell equations can be found with this assumption, it is justified de facto. Sign Convention Comment: Section 1.6 discusses the Fourier Transform (1.6.8) where e+jωt appears in the expansion formula. For E^(x,ω) = 2πδ(ω-ω1) one gets E(x,t) = e+jωt and then the form of a wave solution is e+j(ωt-kz) with the + sign associated with ωt. In general, EE people like to assume time dependence of the form e+jωt (and they prefer j in place of i for ). The Fourier Transform is of course valid with the other sign choice for the two exponentials, and for that other sign choice one would have E^(x,ω) = 2πδ(ω-ω1) => E(x,t) = e-jωt and one would think of a wave as e-j(ωt-kz) = e+j(kz-ωt). This sign convention is common in many physics texts [e.g. Jackson (7.8)], but in this document we use the e+j(ωt-kz) convention usually used in EE texts [e.g. Haus-Melcher 13.1 (7)]. Jackson suggests a physics/EE conversion algorithm of i ↔ -j. It is all just a convention choice and, as in (1.6.6), only the sign of the imaginary physical field under consideration is affected. If one thinks of the physical field under consideration as Re{E(x,t)}, the sign convention choice makes no difference at all. (b) Partial Wave Expansions The next step is to do a "partial wave expansion" (that is, a complex Fourier series expansion) of E(r,θ) in terms of "azimuthal harmonics" eimθ, so that the variable θ is replaced with the partial wave index m: E(r,θ) =!Syntax Error, I E(r,m) ejmθ // expansion (D.1.3a) E(r,m) = (1/2π) !Syntax Error, Idθ E(r,θ) e-jmθ . // projection (D.1.3b) In analogy with (D.1.1) we define a surface charge density n(θ,z,t) which has the following ansatz variable-separated form, n(θ,z,t) = ej(ωt-βz) n(θ) . (D.1.4) We then expand n(θ) as in (D.1.3), n(θ) = !Syntax Error, I Nm ejmθ // Coul/m2 (D.1.5a) Nm = (1/2π) !Syntax Error, Idθ n(θ) e-jmθ . // Coul/m2 (D.1.5b) Nm is the "moment" of the surface charge distribution in the mth partial wave. As with E(r,θ), the function n(θ) is also a function of implicit arguments ω and βd. In principle, n(θ) could have a phase which varies with θ. If we momenarily assume this is not the case and assume that n(θ) is real, then (D.1.5b) says N-m = Nm* and then n(θ) = !Syntax Error, I Nm ejmθ = N0 + !Syntax Error, I[ Nm ejmθ + Nm* e-jmθ ] = N0 + 2 !Syntax Error, IRe{ Nm ejmθ} = N0 + 2 !Syntax Error, I{ Re(Nm) cos(mθ) - Im(Nm) sin(mθ) } . (D.1.6) If we furthermore assume that n(θ) is an even function of θ, as symmetry implies for our particular figure below, then (D.1.5b) says the Nm are real and then we have n(θ) = N0 + 2!Syntax Error, INm cos(mθ) . // n(θ) real and even in θ (D.1.7) For a moderately closely spaced twin lead transmission line (we allow for different radii), one might expect the m=0 and m=1 partial waves to be dominant : Fig D.1 Notice that N0 = (1/2π) !Syntax Error, Idθ n(θ) = (1/2π) (1/a)(1/dz) !Syntax Error, I[adθdz] n(θ) = (1/2π) (1/a)(1/dz) Q where Q is the total charge on a thin ribbon (width) dz wrapping the round wire. In (4.3.8) we refer to the quantity Q/dz as q(0), where q(z) = q(0) ejβz = the total charge on the wire per unit length. Thus, N0 = (1/2πa) q(0) (D.1.8) (c) The Vector Laplacian in Cylindrical Coordinates Given the following cylindrical-coordinates field components, E(r,θz,t) = Er(r,θz,t) + Eθ(r,θz,t) + Ez(r,θz,t) we may write out our ansatz wave form (D.1.1) and the Helmholtz equation (D.1.2a) in more detail, Er(r,θz,t) = ej(ωt-βz) Er(r,θ) . [2E(r,θ,z,t)]r + β2 Er(r,θ,z,t) = 0 Eθ(r,θz,t) = ej(ωt-βz) Eθ(r,θ) . [2E(r,θ,z,t)]θ + β2 Eθ(r,θ,z,t) = 0 Ez(r,θz,t) = ej(ωt-βz) Ez(r,θ) . [2E(r,θ,z,t)]z + β2 Ez(r,θ,z,t) = 0 . (D.1.9) where β2 is the Helmholtz parameter of the conductor medium, not to be confused with βd. In Cartesian coordinates, it happens that [2E]i = 2(Ei), but this is not generally true for curvilinear coordinates. In cylindrical coordinates, it is true for the z coordinate only. The operator 2 when applied to a vector field is called "the vector Laplacian" and it is very different from the scalar Laplacian, so much so that some authors (Moon and Spencer) replace [2E] by [E] which is defined in this manner [E] ≡ [2E] ≡ grad(div E) – curl (curl E) = (E) - x ( x E) (D.1.10) whereas 2φ ≡ div(grad φ) = (φ) . (D.1.11) It is the vector Laplacian that appears in our Helmholtz equation (D.1.2). For cylindrical coordinates it turns out that, (2E)r = 2Er - (2/r2) ∂θEθ - (1/r2) Er (2E)θ = 2Eθ + (2/r2) ∂θEr - (1/r2) Eθ (2E)z = 2Ez (D.1.12) where 2 is the scalar Laplacian, given in cylindrical coordinates by 2 = (1/r)∂r(r∂r) + (1/r2)∂θ2 + ∂z2 = ∂r2 + (1/r)∂r + (1/r2)∂θ2 + ∂z2 . (D.1.13) Notice in (D.1.12) that Eθ is mixed into the "r equation" and Er is mixed into the "θ equation". See for example Morse and Feshbach Vol I p 116, Moon and Spencer p 139, or do a web search on "vector Laplacian". The author's Tensor Analysis document, Sections 13, 14 and 15, derives these results for arbitrary coordinate systems. Here is a summary of vector differential operators in cylindrical coordinates taken from Morse and Feshbach, where the last line corresponds to the above discussion: (D.1.14) We use θ for azimuth instead of their φ since φ is our scalar potential. Using the ansatz form (D.1.1) and partial wave expansions of the form (D.1.3) or (D.1.5), it is a simple matter to convert an equation containing the above differential operators and involving components like Ei(r,θz,t) or n(θz,t) to a simpler equation involving components like Ei(r,m) and Nm and this will be done below. (d) The three Helmholtz equations and div E = 0 (in partial waves) 1. The z equation: The Ez Helmholtz Equation from (D.1.9) is [2E]z + β2 Ez = 0. Using (D.1.12) and (D.1.13), the Ez equation may be written, [∂r2 + (1/r) ∂r + (1/r2) ∂θ2 + ∂z2 + β2 ] ej(ωt-βz)Ez(r,θ) = 0 . (1) Inserting the expansion (D.1.3) for Ez(r,θ) and moving the m sum to the left gives !Syntax Error, I [∂r2 + (1/r) ∂r + (1/r2) ∂θ2 + ∂z2 + β2 ] ej(ωt-βz)Ez(r,m) ejmθ = 0 . (2) We can then make the obvious replacements ∂z = -jβd and ∂θ = +jm to get, !Syntax Error, I { [∂r2 + (1/r) ∂r -m2 (1/r2) – βd2 + β2 ] ej(ωt-βz)Ez(r,m) } ejmθ = 0 . (3) Due to the completeness of functions ejmθ on the interval (-π.π), we conclude that { } = 0, or [∂r2 + (1/r) ∂r -m2 (1/r2) – βd2 + β2 ] ej(ωt-βz)Ez(r,m) = 0 . (4) Alternatively one can apply !Syntax Error, Idθ e-jm'θ to both sides of (3), use the orthogonality property !Syntax Error, Idθ ej(m-m')θ = 2π δm,m' , (5) and then change m' to m to get (4). Next, multiply both sides of (4) by r2 e-j(ωt-βz) to get, [r2∂r2 + r ∂r - m2 +r2( β2- βd2)] Ez(r,m) = 0 . (D.1.15) We may then write these rules for converting equation (1) to equation (D.1.15) Conversion Rules: ∂z → -jβd ∂t→ +jω ∂θ → +jm f(r,θ,z,t ) → f(r,m) (D.1.16) We can now practice with these rules to convert various other equations of interest. A field with unstated arguments has the full arguments (r,θ,z,t). 2. The r equation: The Er Helmholtz Equation from (D.1.9) is [2E]r + β2 Er = 0 . Using (D.1.12) we find, 2(Er) - (2/r2) ∂θEθ - (1/r2) Er + β2Er = 0 [∂r2 + (1/r)∂r + (1/r2)∂θ2 + ∂z2] Er - (2/r2) ∂θEθ - (1/r2) Er + β2Er = 0 [∂r2 + (1/r)∂r + (1/r2)∂θ2 + ∂z2 - (1/r2) + β2] Er - (2/r2) ∂θEθ = 0 . Now apply the conversion rules to get [∂r2 + (1/r)∂r + (1/r2) (-m2) - βd2 - (1/r2) + β2] Er(r,m) - (2/r2) jm Eθ(r,m) = 0 . Group like terms and multiply by r2 to get [r2∂r2 + r∂r - (m2+1) + r2(β2-βd2)] Er(r,m) - 2jm Eθ(r,m) = 0 . (D.1.17) 3. The θ equation: The Eθ Helmholtz Equation from (D.1.9) is [2E]θ + β2 Eθ = 0 . Using (D.1.12) we find, 2(Eθ) + (2/r2) ∂θEr - (1/r2) Eθ + β2Eθ = 0 [∂r2 + (1/r)∂r + (1/r2)∂θ2 + ∂z2] Eθ + (2/r2) ∂θEr - (1/r2) Eθ + β2Eθ = 0 [∂r2 + (1/r)∂r + (1/r2)∂θ2 + ∂z2 - (1/r2) + β2] Eθ + (2/r2) ∂θEr = 0 . Now apply the conversion rules to get [∂r2 + (1/r)∂r + (1/r2)(-m2) + (-βd2) - (1/r2) + β2] Eθ(r,m) + (2/r2) jm Er(r,m) = 0 . Group like terms and multiply by r2 to get [r2∂r2 + r∂r - (m2+1) + r2(β2-βd2)] Eθ(r,m) + 2jmEr(r,m) = 0 . (D.1.18) 4. The divE = 0 equation: Using (D.1.14) for div E ( times r) we write div E = 0 as ∂r (r Er) + ∂θEθ + r ∂zEz = 0 . Applying the conversion rules gives ∂r [r Er(r,m)] + jmEθ(r,m) + r (-jβd)Ez(r,m) = 0 or [1 + r∂r ] Er(r,m) + jmEθ(r,m) + r (-jβd)Ez(r,m) = 0 . (D.1.19) Here then is a summary of the above four results: The Three Helmholtz Equations and the div E = 0 equation (in partial waves) (D.1.20) [2E]z + β2 Ez = 0 : [r2∂r2 + r ∂r - m2 + r2 ( β2- βd2)] Ez(r,m) = 0 (D.1.15) [2E]r + β2 Er = 0 : [r2∂r2 + r∂r - (m2+1) + r2(β2-βd2)] Er(r,m) - 2jm Eθ(r,m) = 0 (D.1.17) [2E]θ + β2 Eθ = 0 : [r2∂r2 + r∂r - (m2+1) + r2(β2-βd2)] Eθ(r,m) + 2jmEr(r,m) = 0 (D.1.18) div E = 0 : ∂r [r Er(r,m)] + jmEθ(r,m) -jβd r Ez(r,m) = 0 (D.1.19) Helmholtz Comments: The scalar Helmholtz equation (2+β2)u(r,θz) = 0 is fully separable in cylindrical coordinates and the "harmonics" (we call them atomic forms) are as follows [ Jm(β'r), Ym(β'r)] * [ejmθ, e-jmθ] * [e jβz , e- jβz] (D.1.21) where βd is a free real parameter and where β'2 = β2 - βd2. Here we use parameter names relevant for our particular problem where u = Ez. These atomic forms appear for example in Moon and Spencer p15 with β = κ, m = p, β' = iq, α2 = m2, and -α3 = β'2. Whether a parameter like m or β' is real, imaginary or complex depends on the nature of the problem, and the above is a standard atoms choice for problems of our type. The θ "quantum number" m is quantized to be an integer by the fact that our problem region is the entire range (-π,π) for θ and the solution must be single valued in θ. Our βd is a parameter determined for a lossless line by βd = ω and is thus correlated with the selected frequency ω, whereas our parameter β is always complex as in (1.5.1). In general, in any list of atomic forms like that shown above, two of the three atoms will be oscillatory and the third will be exponential, and in our case Jm(β'r) is the exponential one, hence the skin effect with its exponential damping as shown in (2.3.7). Away from a singular point, any solution to (2 + β2)u(r,θz) = 0 must be writable as a linear combination of the atoms, so u = ∫dβd Σm [Aβ,m Jm(β'r) + Bβ,m Ym(β'r) [Cβ,m ejmθ + Dβ,m e-jmθ ] [Eβ,m e jβz + Fβ,m e- jβz ]. A general solution method is to find a subset of the above most-general form that is appropriate in each "region" of the problem, and then to match boundary conditions between regions. If the problem is well-posed, this will determine all the constants A,B,C,D,E,F. We refer to this solution method as "the method of Smythian forms" (Smythe used this method a lot). Often many of these constants are 0. In contrast, the vector Helmholtz equation is NOT separable in cylindrical coordinates (see Moon and Spencer p 139), it is not even "R-separable", so there are no associated "harmonics" as there are with the scalar Helmholtz equation.. Nevertheless, the functions ejmθ form a complete set for θ in (-π,π) and our expansion of each Ei onto these ejmθ is certainly allowed, even though these ejmθ are not part of any associated harmonics for the vector Helmholtz equation. However, in Cartesian coordinates each Helmholtz component equation is a scalar Helmholtz equation. In cylindrical coordinates z is a Cartesian coordinate, so we should not be surprised when we find below that Ez ~ Jm(β'r) eimθ e- jβz and this fits into the general form noted above. Neither Er nor Eθ will have such a form. D.2 Solutions for Ez,Er and Eθ (a) The Ez Solution As shown in (D.1.15), the Helmholtz equation for Ez(r,m) is [r2∂r2 + r ∂r + (r2 β'2 - m2)] Ez(r,m) = 0 (D.2.1) where β'2 = β2 - βd2 . (D.2.2) In a conductor like copper, |β| is very large compared to the dielectric βd (lossless line), so we could ignore the distinction between β and β', but we won't in order to keep our results exact. Recall from (1.5.1) and (2.2.3) that, for a line with perfect conductors and only slightly conducting dielectric, βd2 = ω2μdξd β2 ≈ - jωμσ => | | ≈ . In scale, μ and μd are about the same, so using numbers from (1.1.28) and (1.1.29), | | ≈ = ≈ = | | ≈ For f = 100 GHz we then find that |β/βd| ≈ 3200, so for f < 100 GHz, |β/βd| > 3200. Nevertheless, we maintain the distinction between β and β'. For a lossy line, we regard βd as some general complex parameter and we don't know a priori the magnitude of βd relative to that of β'. Setting x = β'r one finds ∂r = β'∂x and then r∂r = x∂x and so on so that (D.2.1) reads [x2∂x2 + x ∂x + (x2- m2)] Ez(x/β',m) = 0 . x = β'r (D.2.3) This is Bessel's equation [ Spiegel 24.1] and the solution subject to the condition that Ez be finite at r = 0 is Ez(x/β',m) = Czm Jm(x) or Ez(r,m) = Czm Jm(β'r) (D.2.4) where Czm is an arbitrary constant for each partial wave m. For m = 0, equation (D.2.4) is consistent with (2.2.22) found by other means. In Section 2.1 we dealt only with the m=0 partial wave, which embodies the symmetrical part of the problem. (b) The Er Solution As shown in (D.1.17), the Helmholtz equation for Er(r,m) is, using (D.2.2), [r2∂r2 + r∂r - (m2+1) + r2β'2] Er(r,m) - 2jm Eθ(r,m) = 0 (D.2.5) while the div E = 0 condition was stated in (D.1.19) as [1 + r∂r ] Er(r,m) + jmEθ(r,m) + r (-jβd)Ez(r,m) = 0 or -jmEθ(r,m) = [1 + r∂r ] Er(r,m) - r (jβd)Ez(r,m) . (D.2.6) Inserting this into (D.2.5) gives [r2∂r2 + r∂r - (m2+1) + r2 β'2] Er(r,m) + [2 + 2r∂r ] Er(r,m) - 2r (jβd)Ez(r,m) = 0 or [r2∂r2 + 3r∂r + (1-m2) + r2 β'2] Er(r,m) = 2r (jβd)Ez(r,m) . (D.2.7) Inserting solution (D.2.4) for Ez(r,m) this becomes [r2∂r2 + 3r∂r + (1-m2) + r2 β'2] Er(r,m) = 2r (jβd) Czm Jm(β'r) or [r2∂r2 + 3r∂r + (1-m2) + r2 β'2] Er(r,m) = 2j (βd/β') Czm β' r Jm(β'r) or [r2∂r2 + 3r∂r + (1-m2) + r2 β'2] Er(r,m) = Km β'r Jm(β'r) (D.2.8) where Km ≡ 2j (βd/β') Czm . (D.2.9) In order to get the left side of (D.2.8) into something recognizable, we define Er(r,m) = x-1 Fm(x) (D.2.10) where x is a dimensionless radial variable which will play a major role in the following, x ≡ β'r and xa ≡ β'a . (D.2.11) Then (D.2.8) becomes [x2∂x2 + 3x∂x + (1-m2) + x2] { x-1 Fm(x)} = 2j (βd/β') Czm x Jm(x) or x [x2∂x2 + 3x∂x + (1-m2) + x2] { x-1 Fm(x)} = Km x2 Jm(x) . (D.2.12) Ever eager, Maple expands the left side of (D.2.12), so that (D.2.12) becomes [ x2 ∂x2 + x ∂x + (x2-m2)] Fm(x) = Km x2 Jm(x) . (D.2.13) The left side of (D.2.13) is the normal Bessel operator [ Spiegel 24.1] , but the equation is also driven by a power times a Bessel function. The solution to the equation is the homogeneous solution of the Bessel equation plus the particular solution which is the response to the driving function on the right hand side. The homogeneous solution is the usual linear combination of Jm(x) and Ym(x), but we must reject Ym(x) since it blows up at x=0 and thereby causes the field Er to be singular, which it cannot be, smack in the middle of a wire. The particular solution is not very obvious and required some hunting to find. It is this Fm(x)particular = (1/2) Km [ x Jm+1(x) ] . (D.2.14) as Maple confirms, continuing the above code, Therefore, we now have this full solution for Fm(x) Fm(x) = Fm(x)particular + Fm(x)homogeneous = (1/2) Km [ x Jm+1(x) ] + am Jm(x) and then from (D.2.10) the full solution for Er , Er(r,m) = am x-1 Jm(x) + Jm+1(x) . (D.2.15) For each value of m, there are two as-yet undetermined constants, am and (Km/2). However, looking at (D.2.15), we see that, since J0(x) ≈ 1 for small x, we must have a0 = 0 (D.2.16) to keep Er finite at r = 0. Later we shall obtain expressions for am and (Km/2). (c) The Eθ Solution Recall (D.2.6) in slightly altered form, jmEθ(r,m) = -∂r[rEr(r,m)] + r (jβd)Ez(r,m) . (D.2.6) We can then insert our known Ez and Er to get Eθ : Ez(r,m) = Czm Jm(x) (D.2.4) Er(r,m) = am x-1 Jm(x) + Jm+1(x) . (D.2.15) so (D.2.6) just above becomes the following : jmEθ(r,m) = -∂r[r{ am x-1 Jm(x) + Jm+1(x)}] + r (jβd) Czm Jm(x) jmEθ(r,m) = -∂x[x{ am x-1 Jm(x) + Jm+1(x)}] + x Jm(x) // Km ≡ 2j (βd/β') Czm jmEθ(r,m) = -∂x[am Jm(x) + x Jm+1(x)] + x Jm(x) jmEθ(r,m) = -am Jm'(x) - Jm+1(x) - x Jm+1'(x) + x Jm(x) jmEθ(r,m) = - am Jm'(x) + [ - Jm+1(x) - x Jm+1'(x) + x Jm(x) ] . (D.2.17) At this point we invoke the recurrence relations [ NIST 10.6.2 ], where C is any Bessel function, to write Jm+1' = Jm - (m+1)x-1Jm+1 first relation with ν = m+1 Jm' = -Jm+1 + (m/x)Jm second relation with ν = m . (D.2.18) Insert these into (D.2.17) to get jmEθ(r,m) = - am Jm' + [ - x Jm+1' - Jm+1 + xJm] = - am {-Jm+1 + (m/x)Jm } + [ - x { Jm - (m+1)x-1Jm+1} - Jm+1 + xJm] = am Jm+1 - am (m/x)Jm + [ - x Jm + (m+1) Jm+1 - Jm+1 + xJm] = am Jm+1 - am (m/x)Jm + [ m Jm+1] = - am (m/x)Jm + ( m + am ) Jm+1 . Dividing by m then gives the final solution, jEθ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) x = β'r . (D.2.19) We now gather up the solutions developed above, but first, recall that Km ≡ 2j (βd/β') Czm (D.2.5) which we can solve to get Czm = (1/2j)(β'/βd) Km . (D.2.20) Installing this into (D.2.4), our three E field components are then First summary of the E field solutions (D.2.21) Ez(r,m) = - j (β'/βd) Jm(x) x = β'r (D.1.27) Er(r,m) = am x-1 Jm(x) + Jm+1(x) β'2 = β2 - βd2 (D.2.11) jEθ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) (D.2.15) These solutions were obtained from the z and r Helmholtz equations and from the div E = 0 equation. It is an easy matter to have Maple verify that this solution set solves the div E equation and all three of the Helmholtz equations z, r and θ : In Maple one must be careful with this kind of verification to make sure Maple has not misunderstood something. For example, perhaps it thinks ∂rEz = 0 because it thinks Ez is a constant. This is the purpose of using the "inert" Diff operators (versus diff) and then forcing them to evaluate later with the value() operator. One should always view expressions before simplification to make sure things are kosher. For example, changing the colon to semicolon after value(e1) to force display, one gets Here Maple has duly computed the Bessel function derivatives in expression e1 but does not yet realize that the expression is 0. This is brought out by the simplify(%) command (simplify that last computed expression) and the output of the simplify command is the 0 on the last line. (d) The Charge Pumping Boundary Condition The reason we are interested in the surface charge n(θ) of (D.1.5) is that it acts as a driving source of the radial electric field inside the wire. Recall the equation of continuity (1.1.35) converted to the ω domain div J = - jωρ -jω[∫V ρ dV] = ∫S J dS . (D.2.22) This is meant to be (1.1.25) where J is conduction current and ρ is free charge. When applied to a thin box of radial area dS straddling the wire surface, Fig D.2 one finds that ∫S J dS = -Jr(r=a-ε,θ)dS and ∫V ρ dV = n(θ) dS so that (ε implies just below surface) Jr(r=a-ε,θ) = jω n(θ) . (D.2.23) We assume that there is no conduction current outside the wire to get this result (non-conducting dielectric). Since J = σE, this is really a boundary condition on the radial electric field just below the surface, Er(r=a-ε,θ) = (jω/σ) n(θ) . (D.2.24) We convert this to m-space using the conversion rules (D.1.16) to obtain (dropping the ε) Er(r=a,m) = (jω/σ) Nm . (D.2.25) Thus, the interior radial electric field must have a certain value at the r=a boundary in each partial wave, and this value is determined by the moment of the surface charge distribution. By way of interpretation, the surface charge of a transmission line is "pumped" by the radial current in the wire (in quadrature). This radial current is accompanied by the usual longitudinal current one expects to find inside the conductors of a transmission line. If the dielectric conducts with some σd > 0 but σd << σ, one must make these replacements in (D.2.24) and (D.2.25), n(θ) → (ξd/εd) n(θ) Nm → (ξd/εd) Nm . See (D.9.23) and surrounding discussion. Generally we shall assume σd = 0 in the following work just to avoid having the extra (ξd/εd) factors floating around. (e) Application of the Boundary Conditions Our task here is to derive expressions for the constants am and Km appearing in the above E field component equations. We have two boundary conditions to impose: Er(r=a,m) = (jω/σ) Nm (D.2.26) Eθ(r=a,m) = 0 (D.2.27) The first is the radial charge pumping condition shown in (D.2.25) above. The second boundary condition is an assumption that requires its own discussion in Section D.8 below. It implies that the cross sectional wire surface is an equipotential surface and that therefore Eθ(r=a,θ) = 0. This in turn requires that in each partial wave Eθ(r,m) = 0 since Eθ(r,m) = (1/2π) !Syntax Error, Idθ Eθ(r,θ) e-jmθ (D.1.3b) Eθ(a,m) = (1/2π) !Syntax Error, Idθ Eθ(a,θ) e-jmθ = (1/2π) !Syntax Error, Idθ 0 e-jmθ = 0 . These two boundary conditions serve to determine the two constants am and Km, though a bit of algebra is required. The first step is to use the Er and Eθ expressions shown in summary box (D.2.21) to write out the two boundary conditions as am xa-1 Jm(xa) + Jm+1(xa) = (jω/σ) Nm (1) - am xa-1 Jm(xa) + ( + ) Jm+1(xa) = 0 . (2) Addition and subtraction of these equations gives two new equations, Jm+1(xa) + ( + ) Jm+1(xa) = (jω/σ) Nm (3) 2 am xa-1 Jm(xa) - Jm+1(xa) = (jω/σ) Nm . (4) Using the recursion relation 2m x-1 Jm = [Jm+1 + Jm-1] , the second may be immediately solved for am, = (jω/2σ) 2 Nm . (5) Using this same recursion relation and (5) for am , equation (2) may be solved to get ( + ) = (jω/2σ) Nm [ + ] . (6) Finally, subtracting (5) from (6) we find = (jω/2σ) Nm [ – ] . (7) Notice the following situations for m = 0, a0 = 0 (8) // from (5) = (jω/2σ) 2 N0 = - (jω/2σ) 2 N0 (9) // from (5) = (jω/2σ) N0 [ – ] = ( jω/σ) N0 (10) // from (7) ( + ) = 0 (11) // adding (9) and (10) We summarize the coefficients as follows: am = (jω/2σ) 2m Nm . a0 = 0 (D.2.28) = (jω/2σ) Nm [ – ] = (jω/σ) N0 (+ ) = (jω/2σ) Nm [ + ] ( + ) = 0 The third equation is obvious from adding the first two, and Maple verifies that the first two satisfy (1) and (2). At this point it is convenient to introduce the DC resistance per unit length of the wire (C.1.1), Rdc = (D.2.29) along with a new symbol to indicate the relative surface charge moment, ηm ≡ . (D.2.30) The DC moment N0 can be related to the total current I in the wire as follows: I = !Syntax Error, Idθ !Syntax Error, Ir dr Jz(r,θ) = !Syntax Error, Idθ !Syntax Error, Ir dr { σ !Syntax Error, I Ez(r,m) ejmθ } // (D.1.3a) = σ !Syntax Error, I !Syntax Error, Ir dr Ez(r,m) !Syntax Error, Idθ ejmθ = 2π σ!Syntax Error, Ir dr Ez(r,0) = 2π σ!Syntax Error, Ir dr {-j(β'/βd) J0(x) } // (D.2.21) for Ez(r,0) = -j(β'/βd) 2π σ !Syntax Error, Ir dr J0(x) // x = β'r so xdx = β'2 rdr = -j(β'βd)-1 2πσ [!Syntax Error, Idx x J0(x)] = -j(β'βd)-1 2πσ [ xa J1(xa) ] // GR7 5.52.1 = -j(β'βd)-1 2πσ {(jω/σ) N0 / J1(xa)} [ xa J1(xa) ] // (D.2.28) for = (β'βd)-1 2πω N0 xa = (β'βd)-1 2πω N0 β'a = 2πω (a/βd) N0 so that N0 = (βd/2πωa) I . // I is called i(z=0) in (4.9.2) so I = i(0) (D.2.31) As a check on this last result, recall from (D.1.8) that N0 = (1/2πa) q(0), so (D.2.31) says (1/2πa) q(0) = (βd/2πωa) i(0) or q(0) = (βd/ω) i(0) . Given q(z) = q(0) e-jβz as in (4.3.8), i(z) = i(0) e-jβz as in (4.9.2), and βd = ω/vd as in (2.1.4b) (for a lossless line) we obtain q(z) = i(z) /vd which agrees with (4.11.19a) [ no d subscripts were used in Chapter 4 for the dielectric.] It follows from (D.2.31) that the normalization factor appearing in (D.2.28) may be written as (jω/2σ) Nm = (jω/2σ) N0 = (jω/2σ) ηm [(βd/2πωa) I ] = (j/4) ηm (aβd/σπa2) I = (j/4) (aβd) ηm I Rdc . (D.2.32) We may now construct the final form for our E field solutions in (D.2.21) using the coefficients in (D.2.28) and the replacement (D.2.32) : Ez(r,m) = -j(β'/βd) Jm(x) = -j(β'/βd) (jω/2σ) Nm [ – ] Jm(x) = -j(β'/βd) [(j/4) (aβd) ηm I Rdc] [ – ] Jm(x) = (1/4) ηm I Rdc (aβ') [ - ] Er(r,m) = am x-1 Jm(x) + Jm+1(x) = [(jω/2σ) Nm] { 2m x-1 Jm(x) + [ – ] Jm+1(x) } = (j/4) (aβd) ηm I Rdc { + - } = (j/4) (aβd) ηm I Rdc { + } where in the last line we used the NIST (10.6.1) Bessel identity (2m/x)Jm(x) = Jm-1(x) + Jm+1(x). Next, jEθ(r,m) = - am x-1 Jm(x) + (+ ) Jm+1(x) = [(jω/2σ) Nm] { - 2m x-1 + [ + ] Jm+1(x) = (j/4) (aβd) ηm I Rdc { - + [ + ] } = (j/4) (aβd) ηm I Rdc { - } Gathering up one more time: Second summary of the E field solutions : Rdc = β'2 = β2 - βd2 (D.2.33) Ez(r,m) = (1/4) ηm I Rdc (aβ') fm fm = [ - ] x = β'r Er(r,m) = (j/4) ηm I Rdc (aβd) gm gm = [ + ] xa = β'a Eθ(r,m) = (1/4) ηm I Rdc (aβd) hm hm = [ - ] Maple verification of these solutions is shown below. Observations about the solution: (1) We looked for a traveling wave solution inside a round wire in which phase fronts propagate down the wire (z direction) with angular frequency ω and wavelength λd = 2π/Re(βd). We found the solution shown in the above box. This solution satisfies all three components of the vector Helmholtz equation (D.1.2) as well as the div E = 0 equation. (2) For a lossless line one has ξ ≈ σ/(jω) and ξd ≈ εd. These are the complex dielectric "constants". The corresponding wavenumbers are then β' ≈ β = ω ≈ ω = ej3π/4 = ej3π/4 (/δ) (1.5.1) ,(2.2.19), (2.2.21) βd = ω ≈ ω = ω / vd vd = speed of light in the dielectric (D.2.34) Thus, in our wave solution (D.1.1), the phase fronts propagate down the inside of the wire at vd, the speed of light in the dielectric outside the wire. Although we have been quiet about the fields outside the wire, it seems reasonable to presume there is a wave outside also moving down the wire at vd . See Section D.7. (3) For r near a, where most of the action occurs due do the skin effect, the Bessel function ratios appearing in (D.2.33) are on the general order of unity so we expect the three brackets [...] to be of the same general size. It then follows that the Er and Eθ fields are smaller than Ez by the ratio |βd/β'| which we have shown in the discussion below (D.2.2) is very small at frequencies below 100 GHz (lossless). Since Ez is an electric field inside copper, it is already itself quite small, so the Er and Eθ fields are extremely small. This then justifies their omission from the development of Chapter 2. (4) If there exist moments Nm of the surface charge distribution on the wire with m > 1, then the corresponding ηm ≠ 0 and it is clear that Ez(r,θ) and hence Jz(r,θ) are non-uniform inside the wire. That is, these fields vary with θ as cos(mθ) as well as with r. The non-uniformity is not "small" but has the full strength of ηm. Of course we only expect to get significant moments of charge density n(θ) when conductors are "fat and close". See (6.5.4) for the special case of both conductors being round wires, and then Section 6 (b) for more on this "proximity effect". (5) The surface impedance from (C.2.1) is just Zs(θ) = Ez(r=a,θ)/I. Thus, from (D.1.3a), Zs(θ) = (1/I) !Syntax Error, I Ez(a,m) ejmθ // (D.1.3a) = (1/4) Rdc !Syntax Error, I ηm [ - ] ejmθ // (D.2.33) where, (D.2.35) ηm = Nm/N0 = !Syntax Error, Idθ n(θ) e-jmθ // (D.1.5b) and (D.2.31) Thus we see the expected non-uniformity of Zz(θ) around the perimeter of the wire cross section due to the m ≠ 0 surface charge components. Maple verification of box (D.2.33) We use the same method illustrated below box (D.2.21). The same expressions e1,e2,e3,e4 are entered as the left sides of the four equations whose right sides we expect to be 0. Then: D.3 What about the Eθ Helmholtz Equation ? A review of the above derivation of the three fields Ez, Er and Eθ shows that the Eθ Helmholtz equation has been completely ignored. The Eθ expression was obtained from the div E = 0 equation after the Ez and Er fields were computed. It is reasonable to wonder whether the solution fields we have found above in fact solve this θ Helmholtz equation which mixes the Er and Eθ fields together in a manner similar to the r Helmholtz equation. A related question is whether the three Helmholtz equations and div E = 0 are four independent equations, or is one of the three Helmholtz equations dependent? In Cartesian coordinates suppose we know that (implied sums on repeated indices) (∂j∂j + β2) E1 = 0 (∂j∂j + β2) E2 = 0 ∂iEi = 0 . // div E = 0 (D.3.1) Can we show that (∂j∂j + β2) E3 = 0 so this third Helmholtz equation is dependent? If we apply the operator (∂j∂j + β2) to the last equation above we get (∂j∂j + β2) ∂iEi = 0 or ∂i (∂j∂j + β2) Ei = 0 or ∂1 (∂j∂j + β2) E1 + ∂2 (∂j∂j + β2) E2 + ∂3 (∂j∂j + β2) E3 = 0 or ∂3 [(∂j∂j + β2) E3] = 0 (D.3.2) This does not prove that (∂j∂j + β2) E3 = 0 since (∂j∂j + β2)E3 = f(x1,x2) ≠ 0 also satisfies (D.3.2). Rather than pursue this question further, we simply note that the Maple code below box (D.2.21) verifies that the E field solutions given in that box do indeed satisfy the θ Helmholtz equation (as well as the other two Helmholtz equations and the div E = 0 equation). Reader Exercise: Come up with some reason that this had to be the case. D.4 Computation of the B fields in the round wire The B field components may be computed from the Maxwell curl E equation (1.1.2) - ∂tB = curl E . Maxwell curl E equation (1.1.2) (D.4.1) In cylindrical coordinates one has from (D.1.14), curl E = [ r-1∂θEz - ∂zEθ] + [∂zEr - ∂rEz] + [ r-1∂r(rEθ) - r-1∂θEr ] (D.4.2) where the fields are of the traveling wave form shown in (D.1.1) which we assume also for the B field. Thus, combining (D.1.1) with (D.1.3a), one has E(r,θz,t) = ej(ωt-βz) E(r,θ) = ej(ωt-βz) !Syntax Error, I E(r,m) ejmθ (D.4.3) B(r,θz,t) = ej(ωt-βz) B(r,θ) = ej(ωt-βz) !Syntax Error, I B(r,m) ejmθ . (D.4.4) Inserting the three cylindrical components of the E expansion (D.4.3) into (D.4.2), one finds that these replacements may be made, ∂t → +jω ∂z → -jβd ∂θ → +jm . (D.4.5) Similarly, inserting the B expansion (D.4.4) into -∂tB one may replace ∂t→ +jω. After doing this, both sides of (D.4.1) are expansions having the general form of (D.4.3) and one may then equate terms in the m sum [completeness of the ejmθ on (-π.π)] to find that -jωB(r,m) = [ r-1jmEz +jβdEθ] + [-jβdEr - ∂rEz] + [ r-1∂r(rEθ) - r-1jmEr ] (D.4.6) and this then gives the three components of the B field Br(r,m) = (j/ω) [curl E]r = (j/ω) [r-1jmEz +jβdEθ] Bθ(r,m) = (j/ω) [curl E]θ = (j/ω)[-jβdEr - ∂rEz] Bz(r,m) = (j/ω) [curl E]= (j/ω) [r-1∂r(rEθ) - r-1jmEr] . (D.4.7) It is now a mechanical task to insert our E field components, and such tasks are grist for Maple's mill. We use the E component forms summary box (D.2.21) which have the am and Km constants not yet specified. The alias line "unaliases" I, sets j = in place of the default I, and allows simple reference to the Bessel functions of interest. Diff(Ez,r) represents ∂rEz, but in an "inert" form which is not executed until later after Ez has been specified. The resulting B field expressions are somewhat ugly but can be cleaned up using a few more Maple manipulations. Having seen the results, we extract certain factors as shown in the following commands, which we then translate back into our normal notation, (ω/β')Bz(r,m) = ( + )Jm(x) (ω/jβ')Br(r,m) = + ( m - am) Jm(x) + ( + ) Jm+1(x) (ω/β')Bθ(r,m) = - ( m - am) Jm(x) + ( + ) Jm+1(x) . (D.4.8) The last two equations contain the same term which can be written as (recall x = rβ') ( m - am) = ( m - am) = ( m - am)(1/x) The three equations for the exact B field components in the round wire are then shown in the summary box below which includes the earlier E field results as well: Summary of E and B fields inside a round wire (D.4.9) Ez(r,m) = - j (β'/βd) Jm(x) x = β'r β'2 = β2 - βd2 Er(r,m) = am x-1 Jm(x) + Jm+1(x) . jEθ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) (D.2.21) Bz(r,m) = (β'/ω) ( + )Jm(x) Br(r,m) = j(β'/ω){ + ( m - am) x-1Jm(x) + ( + ) Jm+1(x) } Bθ(r,m) = (β'/ω){ - ( m - am) x-1Jm(x) + ( + ) Jm+1(x) } , (D.4.8) where the constants are given in (D.2.28), which we rewrite using (D.2.32), am = (j/4) (aβd) ηm I Rdc * 2m = (j/4) (aβd) ηm I Rdc * [ – ] (+ ) = (j/4) (aβd) ηm I Rdc * [ + ] . The three constant quantities at the end of the above summary box are roughly the same size in terms of scale. Using this fact, and the fact that for a low loss line |β'| >> βd (so β = ≈ β') we can simplify (D.4.9) to read, Bz(r,m) = (β/ω) ( + )Jm(x) β' ≈ β, x = βr Br(r,m) = j(β/ω){ + ( m ) x-1Jm(x) } // m ≠ 0 Bθ(r,m) = (β/ω){ - ( m ) x-1Jm(x) + ( ) Jm+1(x) } . (D.4.10) The last line of (D.4.10) can be further simplified, Bθ(r,m) = (β/ω){ - ( m ) x-1Jm(x) + ( ) Jm+1(x) } = (β/ω) { - 2mx-1Jm(x) + 2Jm+1(x) } = (β/ω) { - Jm+1(x) - Jm-1(x) + 2Jm+1(x) } // Spiegel 24.17 identity = (β/ω) { Jm+1(x) - Jm-1(x) } . (D.4.11) Therefore, in the limit |β'| >> βd equations (D.4.10) become Bz(r,m) = (β/ω) ( + )Jm(x) β' ≈ β Br(r,m) = j(β/ω) { m x-1Jm(x) } // m ≠ 0 Bθ(r,m) = (β/ω) [ Jm+1(x) - Jm-1(x)] . (D.4.12) For m>0, |Br| and |Bθ| are larger than |Bz| by the large factor | β/βd|. For m = 0, βr ≈ 0 so |βθ| >> |Bz|. It is this large Bθ field which appears in Chapter 2 as Bθ. D.5 Verification that the E and B fields satisfy the Maxwell equations The Maple program discussed above goes on to verify that the exact E and B fields obtained above for the round wire in fact satisfy Maxwell's equations. Since the B equations were obtained from the curl E Maxwell equation, this one is not verified. The other three Maxwell equations are projected into their partial wave versions analogous to (D.4.6) above : div B(r,m) = r-1∂r(rBr) + r-1∂θBθ + ∂zBz = r-1∂r(rBr) +r-1jmBθ -jβd Bz (D.5.1) div E(r,m) = r-1∂r(rEr) + r-1∂θEθ + ∂zEz = r-1∂r(rEr) + r-1jmEθ - jβdEz (D.5.2) curl B(r,m) = [ r-1∂θBz - ∂zBθ] + [∂zBr - ∂rBz] + [ r-1∂r(rBθ) - r-1∂θBr ] = [ r-1jmBz + jβdBθ] + [-jβdBr - ∂rBz] + [ r-1∂r(rBθ) - r-1jmBr ] . (D.5.3) Inside the round wire we expect to find div B = 0 div E = 0 // no free charge curl B = μ J + μ jωεE = μ(σ + jωε) E = μ(jω)( ε - jσ/ω) E = jω μξ E = j (β2/ω) E . // see (1.5.1) Thus, for the divergence equations we just compute the divergence as shown and see if it comes out zero, while for the curl B equation we verify that curl B - j(β2/ω) E = 0 (D.5.4) for each component. The fact (D.2.2) that β'2 = β2 - βd2 is also used. Here then is the Maple code which does the verification of the three Maxwell equations: In Maple % refers to the last quantity computed. Prior to each simplify(%) statement we find a huge mess for the expression at hand, but simplify then shows it is really zero. As an example, here is the execution of the verification that [curl B]z - j(β2/ω) Ez = 0 : No approximations were made in the E fields, the B fields, or in these Maxwell verifications. D.6 The exact E and B fields for the m=0 partial wave The m=0 partial wave is all there is for an axially symmetric problem like that considered in Chapter 2, where the round wire is imagined in isolation, but is operationally the central conductor of a coaxial cable with a very distant return cylinder (outer shield). Here is the reduction of box (D.4.9) for m = 0, making use of the m=0 coefficients noted in box (D.2.28), namely, m = 0 a0 = 0 ( + ) = 0 = (jω/σ) N0 Summary of E and B fields inside a round wire ( m = 0 only ) (D.6.1) Ez(r,0) = - j (β'/βd) J0(x) // large x = β'r β'2 = β2 - βd2 Er(r,0) = J1(x) . // small = (j/2) (aβd) I Rdc jEθ(r,0) = 0 Rdc = Bz(r,0) = 0 Br(r,0) = 0 Bθ(r,0) = (β'/ω) ( + ) J1(x) // large ~ β' (β'/βd) No approximations have been made in these results, but a very good approximation for a low loss line is that |β| >> βd which means β' ≈ β, as discussed below equation (D.2.2). With this approximation, we have commented in the above box on the size of the various field components. The dominant components are Ez(r,0) = - j (β/βd) J0(x) = - j (β/βd) (j/2) (aβd) I Rdc = (1/2) β a I Rdc = (ω/β) (1/2) (aβ2/ω) I Rdc Bθ(r,0) = (β/ω) ( ) (j/2) (aβd) I Rdc = (j/2) (aβ2/ω) I Rdc . As shown in (2.2.3) we can write β2/ω ≈ - jμσ so that (1/2) (aβ2/ω) I Rdc = (1/2) a (- jμσ) I = - j and then the dominant components above become Ez(r,0) = (ω/β) [ - j ] = -j (ω/β) Bθ(r,0) = j [- j ] = . (D.6.2) These results are in agreement with E(r) and B(r) shown in summary box (2.2.30) from the Chapter 2 calculation where we assumed E = E(r) and B = B(r) D.7 What about the E fields outside the round wire? For a lossless wave, the Helmholtz equation (D.1.2) outside the wire contains βd instead of β. The assumed wave solution form is still (D.1.1). In the three component Helmholtz equations this means that β'2 ≡ β2- βd2 = βd2 - βd2 = 0 We can then translate box (D.1.20) by replacing β2- βd2 → 0 and β2 → βd2 to get the following "exterior" versions: (Note that 2 = 22D + ∂z2) [2E]z + βd2 Ez = 0 : // [22DE]z = 0 [r2∂r2 + r ∂r - m2] Ez(r,m) = 0 (D.1.15)ext [2E]r + βd2 Er = 0 : // [22DE]r = 0 [r2∂r2 + r∂r - (m2+1)] Er(r,m) - 2jm Eθ(r,m) = 0 (D.1.17)ext [2E]θ + βd2 Eθ = 0 : // [22DE]θ = 0 [r2∂r2 + r∂r - (m2+1)] Eθ(r,m) + 2jmEr(r,m) = 0 (D.1.18)ext div E = 0 : ∂r [r Er(r,m)] + jmEθ(r,m) -jβd r Ez(r,m) = 0 (D.1.19)ext The differential operators appearing in the above equations are no longer Bessel-style operators, they are Euler-style operators. Euler ODEs have the general form [ r2∂r2 + a r ∂r + b] f(r) = 0, and the solutions have this form ( from p 45 of Polyanin's excellent ODE compendium, or just use Maple), For Ez(r,m) the equation (D.1.15)ext shown just above is in fact an Euler equation which has a = 1 and b = -m2 so μ = m and the solution forms are these (r ≥a outside the wire), Ez(r,m) = Amrm + Bmr-m m > 0 Ez(r,0) = Czln(r) + Dz m = 0 . (D.7.1) Since z is a Cartesian coordinate, [22DE]z = 0 is the same as 22DEz = 0 which is just the 2D Laplace equation. When this equation is solved in polar coordinates (r,θ), one finds Ez = Ez(r,m) ejmθ and the expressions shown above are the standard atomic forms for the radial function. See for example Stakgold Vol II p 92 (6.7) and following discussion. We can mimic our interior solution method presented in Section D.2 above, using the div E = 0 equation to eliminate Eθ, and eventually end up with expressions for the three field components outside the wire. For m > 1 the general form for the exterior solution is found to be, Ez(r,m) = Am rm + Bm r-m Er(r,m) = -(jβd/2) Bm r1-m - 2j βd Am r1+m + Cm rm-1 + Dm r-m-1 jEθ(r,m) = (jβd/2) Bm r1-m + j βd Am r1+m - Cm rm-1 + Dm r-m-1 (D.7.2) where there are now four constants Am, Bm, Cm and Dm to be determined in each partial wave. One could match the three E-field boundary conditions at r = a as per (1.1.50) (subscript d means dielectric) Ez(a,m) = Ezd(a,m) ξ Er(a,m) = ξd Erd(a,m) jEθ(a,m) = jEθd(a,m) (D.7.3) using the interior solutions shown in (D.2.33) where Eθ(a,m) = 0. This gives 3 conditions on the 4 unknown constants so these boundary conditions can be met. The problem with this exterior solution method is that more information is needed to solve the problem. The "Smythian Form" solution (D.7.2) is fine, but it only applies inside a thick cylindrical shell (blue) whose inner diameter is r = a and whose outer diameter is r = b, where b causes this shell to touch the nearest other conductor, as illustrated here, Fig D.3 The reason is that the dielectric E-field wave (Helmholtz) equation is not valid inside the "other conductor", so the form (D.7.2) cannot apply in a region which includes any of this other conductor. Since the blue shell region does not include r = ∞, one cannot rule out coefficients like Am and Cm. One is now stuck worrying about boundary conditions at r = b and the whole problem becomes intractable. But if one could find the complete exact exterior solution, one would find that inside the blue cylindrical shell the solution's partial wave fields would have the form shown in (D.7.2). Reader Exercise: (a) Verify (D.7.2). (b) In Chapter 6 a transmission line with two round conductors is solved "exactly". Convert the solution to a coordinate system like that shown above, compute the Ei(r,m) using (D.1.3b), and verify that these Ei field components fit into the form shown in (D.7.2). D.8 About the boundary condition Eθ(a,m) = 0 We start with a quick review. In earlier sections of this Appendix we examined the electric field inside a round wire (radius a) which was regarded as a conductor in a straight transmission line. The electric field was assumed to have the form of a longitudinal wave traveling down the conductor, E(r,θz,t) = ej(ωt-βz) E(r,θ) , (D.1.1) where βd is the wavenumber parameter of the surrounding dielectric medium. We expanded the function E(r,θ) onto azimuthal partial waves ejmθ and solved the Helmholtz wave equation inside the wire with solutions as shown in box (D.2.21), Ez(r,m) = - j (β'/βd) Jm(x) x = β'r (D.1.27) Er(r,m) = am x-1 Jm(x) + Jm+1(x) . (D.2.11) jEθ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) . (D.2.15) where β'2 = β2-βd2 with β being the (complex) wavenumber parameter of the conductor, and where am and Km are undetermined constants. At this point we applied the two boundary conditions, assuming a non-conducting dielectric, Er(r=a,m) = (jω/σ) Nm (D.2.26) Eθ(r=a,m) = 0 . (D.2.27) where Nm is the mth partial wave moment of the surface charge n(θ) distribution, where n(θ,z,t) = ej(ωt-βz) n(θ) . (D.1.4) These conditions determined the constants am and Km giving the resulting E field inside the wire, Ez(r,m) = (1/4) ηm I Rdc (aβ') [ - ] a = radius ηm ≡ (D.2.33) Er(r,m) = (j/4) ηm I Rdc (aβd) [ + ] x = β'r Eθ(r,m) = (1/4) ηm I Rdc (aβd) [ - ] xa = β'a where Rdc = 1/(πa2σ) is the DC resistance of the wire per unit length, and I is the amplitude of the current in the wire. Everything is an implicit function of frequency ω. It was noted that, for |βd/β'| << 1, the fields Er and Eθ are much smaller than Ez, and this is the case for f ~ 100 GHz or below (but not too small). An implication of the solution is that the E fields inside the wire for each partial wave are described by a single parameter Nm which is the surface charge moment noted above. If the other transmission line conductor(s) were to change their position relative to the round wire and/or to vary their cross sectional shape, the only effect this would have would be to adjust the set of parameters Nm, and the solutions would still be given by (D.2.33) quoted above. Although the set {Nm} is infinite, it seems likely that for reasonable shapes of the other conductor(s), the lowest few Nm partial waves would provide a good approximation to the E fields inside the wire. Since Ohm's Law is assumed to apply inside the wire, one then knows in detail the current densities Jz, Jr and Jz. The magnetic field B inside the wire is then also known and was calculated above. The lowest moment is always N0 = (βd/2πωa) I from (D.2.31). As an example, the following five-conductor transmission line might be expected to have a strong m = 2 quadrupole surface charge moment N2, Fig D.4 A critical ingredient of our solution is the assumption that Eθ(r=a,m) = 0 and that is the subject now addressed. We present two somewhat different arguments as to why Eθ(r=a,m) = 0. It should be noted that King in his Transmission-Line Theory book always assumes that any straight transmission line conductor cross section has an equipotential surface (a ring, see for example middle p 14, top 15, 25 bottom). Due to the presence of small transverse vector potential components, Eθ= 0 and "equipotential" for the scalar potential φ are not the same thing. (a) The Quasi-Static Argument In electrostatics, we are used to metal surfaces being equipotentials. For example, if we put a point charge q near a metal sphere, it induces a surface charge on that sphere. The electric field lines land on the sphere exactly perpendicular to the surface. One argues that if there were even some tiny E field component tangential to the surface, the surface charges would adjust their position to cancel out that tangential field. Since the situation is static, any adjustment has already been made. Since Etan = 0, the sphere's surface is an equipotential surface. If we were to then slowly move the charge q around (perhaps it rotates in a circle around the sphere), the surface charge instantly adjusts at each new position of q, and those E field lines remain perpendicular to the surface, and Etan = 0. While the charges are adjusting position, there is admittedly some very tiny surface current driven by some tiny Etan , but if we move the charge slowly, we are "quasi-static" and the approximation Etan ≈ 0 is very good. One might compare the time constant of the moving sphere (T, the period of q's revolution around the sphere) to the time constant of the surface charge adjustment. For copper the time constant is roughly the mean electron collision time which is on the order of 10-14 sec. The conclusion here is that for frequencies << 1014 Hz, the quasi-static situation prevails and then Etan ≈ 0 is a very good approximation. This then is our first argument for why we claim the boundary condition Eθ = 0 on the surface of the round wire in a transmission line operating at a typical frequency. We note from our solution Eθ(r,m) that if we assume Eθ(a,m) = 0 on the round wire surface, we will still have Eθ(r,m) ≠ 0 inside the wire. This fact is consistent with our argument above since there are no free charges available to adjust themselves inside the wire. However: if Eθ = 0 by this quasi-static argument, then we should expect that Ez = 0 by the same argument, since Ez is also a tangential field at the round wire surface, and since Ez operates at the same frequency ω as Eθ. But we know that Ez ≠ 0 because Jz ≠ 0 just below the wire surface -- there is current flowing there -- and Ez is continuous through the surface by (1.1.50). So the E field lines are not quite perpendicular to the round wire surface in the z direction. This is not too surprising since we expect everything to vary in the z direction as ej(ωt-βz) so we would expect the surface not to be an equipotential in this direction. But what happened to that quasi-static argument we just applied to Eθ ? What happened is that there is external field activity associated with the wave going down the line which forces Ez ≠ 0. One might say the EM wave traveling down the line induces a Jz in the round wire, with its associated Ez ≠ 0. But then perhaps this same thing could somehow happen with Eθ and then our quasi-static argument that Eθ = 0 collapses. We think this could happen in fact, but only if the transmission line is driven by an apparatus which creates a "torsion wave" in the line. For example, the apparatus could drive counter-rotating azimuthal currents onto the round wire surfaces of a twin-lead transmission line as suggested by this picture (which is not meant to imply that other field components vanish), Fig D.5 It seems from our work above that such a wave would satisfy Maxwell's equations and be a viable mode of the transmission line. In this case, Eθ≠ 0 because the EM wave going down the line forces Eθ ≠ 0, just as the normal wave forces Ez ≠ 0. We have not investigated whether this type of torsion wave is really viable. Whether or not it is, we assume in our transmission line discussion that this mode is not activated and that therefore the quasi-static argument for Eθ = 0 is valid at the round wire surface. (b) An Ansatz Argument We make an ansatz that Er,Eθ << Ez in our round wire E field solution, perhaps based on an expectation that most current in the wire will be longitudinal. We assume this is true, and see if this assumption is born out in a final solution of Maxwell's equations. Given that Eθ is then very small, we can make an approximation (another ansatz) that this field Eθ is exactly zero on the surface of the round wire. This may not be exactly true, but again we assume it for our purposes and see where it leads. This is the nature of an "ansatz". When we make this assumption, the cross section of the transmission line may be regarded (Chapter 5) as a two dimensional potential theory problem -- basically a capacitor problem where one conductor has potential V/2 and the other -V/2, say (for a symmetric line, at some fixed value of z). In such a potential problem, one always assumes that the electrostatic potential φ is a constant on the surface of each conductor, and that is precisely what our ansatz says: Eθ = -(φ)θ = 0, φ = constant in the θ direction. Now when we solve the capacitor problem for potential φ, that gives E = - φ in the dielectric between the conductors, and from that we may deduce E at the surface of one of the conductors. For the round wire with a cylindrical coordinate system whose axis is aligned with the wire center, that field is Er. Next, from this surface value of Er (which will be proportional to V) we may compute the surface charge density n(θ) on the round wire using (D.2.24) which says Er(r=a,θ) = (jω/σ) n(θ). For a "fat" twin lead transmission line for example we expect this to have a bulge in n(θ) on the side of the wire facing the other wire (m = 1, dipole), since that is what happens in such a capacitor. In any event, given n(θ) we may compute the moments Nm of the surface charge using (D.1.5b) and this then provides one "boundary condition" on our coefficients am and Km which appear in all the field expressions we found above, Er(r=a,m) = (jω/σ) Nm . (D.2.26) But recall that, in order to carry out this entire process just described, we had to start with the assumption that Eθ = 0 on the conductor cross section surface, so that we could have a capacitor problem in the first place. According to (D.1.3b), if Eθ(r=a,θ) = 0, then Eθ(r=a,m) = 0, so that in fact we must have Er(a,m) being zero in all partial waves m. Thus our assumed ansatz condition is Eθ(r=a,m) = 0 (D.2.27) which is then a second boundary condition on am and Km. Although (D.2.27) might not be exactly true, we know it is very close to being true. More importantly, we know that the above two conditions on am and Km are consistent with each other, even though both boundary conditions might be slightly wrong. We then expect them to give good values for constants am and Km. Using these "perhaps slightly wrong" boundary conditions, we obtain the solutions shown in (D.2.33). It has already been noted above that for copper conductors and normal dielectrics, |βd/β'| << 1 up to at least 100 GHz. The condition |βd/β'| << 1 when applied to the (D.2.33) results shows that in fact our ansatz that Er,Eθ << Ez is born out. There are three footnotes to the above discussion. First, we note that the second boundary condition does not force Eθ(r,m) = 0 for r < a inside the wire. In fact, there will be some small azimuthal "swirling" current inside the wire even if Eθ(r=a,m) = 0, and this is just a result of Maxwell's equations and their solutions above. Second, one might make the argument that the round wire surface is an equipotential since that is the way a line is driven at the source. For example, the center conductor of a coaxial cable plugs into a tiny driving cylinder (jack) in a BNC connector and this drives only the wire surface, and it does so in an azimuthally symmetric way so that one expects to have the wire surface be an equipotential at the driving point; this equipotential surface then moves down the line as the wave progresses. Third, we have the complication that we don't really have a purely electrostatic situation, and the potential is in fact related to E by equation (1.3.1) which says E = - φ - ∂tA . The rescue here comes by claiming that roughly A ≈ A so that the transverse components Ar and Aθ are very small. In this case, we then do get E ≈ -φ so that Eθ = 0 is associated with constant φ on the wire surface. The argument for A ≈ A is that A is driven by J, and J is mostly in the direction, which in turn is related to our starting ansatz (see Appendix M). D.9 About the boundary condition Er(a,θ) = (jω/σ) n(θ) . The "charge pumping boundary condition" appears in (D.2.23) and here we want to examine it more closely. Our concern is that the derivation of (D.2.23) ignores surface currents that we know exist on the surface of a transmission line conductor as the surface charge moves around in response to tangential E fields. The first issue then is to define and quantify the nature of these surface currents. (a) The notion of Debye Surface Currents We continue in the context of our classical treatment of the conductor surface. In Appendix E it is pointed out that the surface charge on a transmission line conductor exists in an incredibly thin surface layer we shall call the Debye layer for want of a better name. For copper the thickness λD of this layer is on the order of one atomic radius. In addition to the normal conduction electrons, this thin layer contains extra free electrons that are piled up just below the surface (negative surface charge) or are depleted from this thin region (positive surface charge), as shown by the red curve in Fig E.1. We want first so show : Fact 1: In a good conductor, the volume density of free electron carriers piled up at a surface (to make up the surface charge) is negligible compared to the volume density of conduction electrons. (D.9.1) Proof: From (E.7) the free charge density in the Debye layer (assume x is the inward surface normal direction) is given by ρ(x) = ρ(0) e-x/λ . The effective free surface charge n is then given by n = !Syntax Error, Idx ρ(x) = ρ(0) !Syntax Error, Idx e-x/λ = ρ(0)λD . The free electron density ne is then ne = ρ(0)/e = n / (eλD) . As a typical example, consider a parallel plate capacitor with close plate spacing s. The E field in the gap is E = V/s and the surface charge density from (1.1.47) is n = εE = εV/s. For V = 10 volts and s = 1 mm we find n = ε0V/s = 8.85 x 10-12 * 10 / 10-3 ≈ 101-12+1+3 = 10-7 Coul/m2 . Then the free electron density is ne = n / (eλD) ≈ 10-7 Cou/m2 / [ 1.6 x 10-19 Coul * 10-10m] ≈ 0.6 * 10-7+19+10 ≈ 1022 electrons/m3 As noted in (N.1.2), in copper the conduction electron density (one electron per atom) is 1029 /m3, QED. Corollary: The conductivity σD inside the Debye later is basically the same as σ outside that layer. (D.9.2) Proof: From (N.1.10) conductivity is σ = (nq2τ/m) where n is the electron density. The Fact above shows that this density is the same in the Debye layer as in the bulk conductor, so σD = σ. (We ignore the possibility that the collision time τ could differ in the Debye layer vs. in the bulk volume. ) QED Consider now this crude drawing which shows a tiny slice of width dx of a piece of a transmission line conductor cross section at its surface. The yellow Debye surface charge layer is greatly exaggerated in thickness and is modeled as if it had a clean lower boundary. Recall from the comment below Fig E.2 that at 100 GHz one has δ ≈ 4000 λD so δ >> λD at all frequencies of transmission line interest. Fig D.6 The Debye layer holds the surface charge, and when this surface charge moves, one has a Debye surface current. We now show : Fact 2: The total current in the Debye layer is negligible compared to that in the skin effect layer. (D.9.3) Proof: The field Ez is transverse to the conductor surface, so we know from (1.1.41) that it is continuous through the boundary at the bottom of the Debye layer. Then the ratio of the currents in the two layers is, = = = ≈ 1 * 1 * = << 1 . QED (b) The role of Debye Surface Currents in the boundary condition Now referring to the Debye surface currents as KzD and KθD we reconsider the derivation of the charge pumping boundary condition of (D.2.24) where we had this figure, Fig D.2 If we include the Debye surface currents in the θ and z direction in our application of continuity, div J = - jωρ -jω[∫V ρ dV] = ∫S J dS , (D.2.22) the result is -jω n(θ,z) = - Jr(r=a-ε, θ, z) + ∂zKz(θ,z) +(1/a) ∂θKθ(θ,z) (D.9.4) where we assume that the dielectric outside the round wire is vacuum with σd = 0. The gaussian box selected here is that shown in red in Fig D.6. The bottom face lies below the Debye layer so Jr(a-ε, θ, z) is the value of Jr in the normal skin effect region close to the surface. The Debye surface currents may be written approximately as KzD = JzD λD = σDEzD λD = σ Ez(r=a,θ) λD // dim(K) = amp/m KθD = JθD λD = σDEθD λD = σ Eθ(r=a,θ) λD = 0 // (D.9.2) and (3.7.0) (D.9.5) where we use the Corollary above that σD = σ. From (3.7.0) we have Eθ = 0 at the surface so KθD = 0 and we have only the Debye current KzD to worry about. Recall that Eθ(a,θ) = 0 is the second boundary condition (D.2.27) used in Section D.2 to evaluate the am and Km coefficients, and that this condition is itself a topic of interest in Section D.8, and we assume it is valid. We then have, -jω n(θ,z) = - Jr(a,θ,z) + ∂zKzD(θ,z) = - Jr(a,θ,z) + ∂z [σ Ez(a,θ,z) λD] (D.9.6) Assuming everything has z dependence ej(ωt-βz) as in (D.1.4), we replace ∂z → -jβd and then suppress the z arguments to get -jω n(θ) = - Jr(a,θ) -jβd [σ Ez(a,θ) λD] = - σEr(a,θ) -jβd [σ Ez(a,θ) λD] = - σEr(a,θ)[ 1 - jβdλD ] . (D.9.7) If we assume that the second term in (D.9.7) can be ignored, we get the charge pumping boundary condition Er(r=a-ε,θ) = (jω/σ) n(θ) (D.2.24) (D.9.8) which in return yields the E fields as stated in (D.2.33) where we see that roughly ~ | | . (D.9.9) Thus, our self-consistent condition for ignoring the second term in (D.9.7) is βd λD * << 1 βd λD * | | << 1 λD |β| << 1 λD | ej3π/4 (/δ)| << 1 (λD/δ) << 1 (δ/λd) >> 1 // ignore But we know from above that (δ/λd) >> 1 for any f < 100GHz, so for such f the second term in (D.9.7) can in fact be ignored. We have just proven: Fact 3: For f < 100 GHz, the Debye surface currents can be ignored in the derivation of the boundary condition Er(a-ε,θ) = (jω/σ) n(θ) . (D.9.10) (c) Where does surface charge n(θ) come from? According to our traveling-wave ansatz (D.1.1), all E field related quantities move down a transmission line at vd as ej(ωt-βz) . For a low-loss line and a vacuum dielectric, vd ≈ c, the speed of light. Therefore, n(θ,z,t) = n(θ,0,0) ej(ωt-βz) βd = (ω/vd) . (D.9.11) One can ponder and then discard a list of hypotheses concerning where n(θ) "comes from" as it increases and decreases over time at some location z on one of the conductors. The first hypothesis might be that the individual electrons which make up n(θ) simply travel at vd in the z direction down the conductor surface, and n(θ) is not fed by any radial currents inside the conductor. In this case one would have KzD(θ) = vd n(θ). But we know this is not what happens. Apart from the massive energy required to achieve relativistic electron velocities, we know from Appendix N.1 that the electrons in the Debye layer in fact drift along at something like ~ 1 mm/sec, just as do the regular conduction electrons in the conductor bulk. The second hypothesis is a variation of the first, where we now allow that the Debye surface current works like any other conduction current, and when one electron moves "to the right" at some point z, a distant electron at z+L moves to the right at nearly the same time, all electrons in a long string moving to the right one position, giving the illusion that a particular electron moved very fast. This does in fact happen, and if it were all that happened, again we would have KzD(θ) = vd n(θ). Comment : Assume some skin depth δ ≤ a/10 so the bulk current is flowing in a sheath of thickness δ just under the conductor surface. The total sheath current is then roughly 2πaδJz(a,θ). We can regard this current flow as due to an effective "full surface current" Kz = Jz δ. Notice that this "surface current" is different from the "Debye surface current". Based on Fact 2 above, we certainly expect Kz >> KzD . A third hypothesis is that somehow n(θ,z,t) is fed by azimuthal Debye surface currents, or some combination of these along with the z-directed KzD(θ). Our condition (3.7.0) that Eθ = 0 puts a stop to the possibility of feeding by azimuthal Debye surface currents. What we have learned from Fact 3 is that none of the above hypotheses explains where n(θ) comes from. The analysis above shows that, although the surface motions of the Debye surface charges do create Debye surface currents, these currents are so small that they play no role in div J = -∂tρ for the Gaussian box shown in red in Fig D.6. The charge n(θ) "comes from" inside the wire and is fed by the radial current density Jr just below the surface according to (D.2.24), Jr(r=a-ε,θ) = jω n(θ) n(θ) = (1/jω) Jr(a-ε,θ) (D.9.12) Here is a suggestive picture, Fig D.7 where the white boxes are little "radial charge pumps" delivering the required Jr needed to feed the changing surface charge n(θ). Apart from the miniscule KzD , charges in n(θ) don't move in the z direction in this picture, they just appear to be doing that due to the choreographed radial pumping in and out at the wire surface. A wave front of the n(θ) wave travels at vd, and this is just a phase velocity. In an analogous situation, in a deep ocean wave the individual particles of water travel in small ellipses and do not travel along with the wave, though there are small scale longitudinal motions due to those ellipses. Comment: In our transmission line theory, the exterior problem in the dielectric is solved using the capacitor method, from which one learns n(θ). The boundary condition Jr(r=a-ε,θ) = jω n(θ) couples this exterior information into the wire interior, allowing one to solve for the fields and currents inside. In Section 6.5 we show how div E = 0 inside the conductor (or div J = 0) forces a relationship between Jr and Jz just below the conductor surface. When that relationship (6.5.18) is combined with the charge pumping boundary condition (D.9.8), one finds that Ez(a,θ) = (-jω/σ) (β/βd) n(θ) (6.5.19) or Jz(a,θ) = (-jω) (β/βd) n(θ) . (D.9.13) This same result is obtained in a different manner as (6.5.13). The "full surface current" Kz was defined in a Comment above as Kz(θ) = δ Jz(a,θ). Thus, Kz(θ) = δ Jz(a,θ) = [δ (-jω) (β/βd)] n(θ) (D.9.14) But = so [δ (-jω) (β/βd)] = δ (-jω) = -j ej3π/4 vd = -j (j-1)/ * vd = (1+j) vd and we end up with Kz(θ) = (1+j) vd n(θ) Re(Kz(θ)) = Im(Kz(θ)) = vd n(θ) (D.9.15) Once again, this last equation gives the illusion that the surface charge density n(θ) moves "to the right" at speed vd to create the real or imaginary part of the full δ-thick surface current Kz. This is the equation that replaces the incorrect equation KzD(θ) = vd n(θ) which assumes there is no radial charge pumping. Reader Exercise: Show using F = ma and F = qE (ignore magnetic fields) that with a time-harmonic E field, a classical electron inside a transmission line conductor traverses a tiny elliptical path and thus never really goes anywhere. That path is traversed once per period T= 2π/ω. Mathematically, show that this amounts to proving that the three equations x = Acos(ωt-a) y = Bcos(ωt-b) z = Ccos(ωt-c) are parametric equations for an ellipse with some orientation in 3D space. As just noted above, this goes-nowhere aspect of the electron is similar to what happens with a droplet of water in an ocean wave. (Hint: first show that the first two equations describe an ellipse in the xy plane and that the semi-major axes in general are not A and B .) (d) Modifications for a Conducting Dielectric Ignoring the Debye surface currents as per section (b) above, if the dielectric has some conductivity σd, the charge pumping boundary condition (D.2.23) becomes Jr(a-α,θ) - Jr(a+α,θ) = jω n(θ) or σ Er(a-α,θ) - σdEr(a+α,θ) = jω n(θ) // this is div J = -jωρ where α > 0 is a tiny distance (ε is already used for dielectric constant). Another boundary condition at the surface is provided by (1.1.47) which says ( points into medium 1 which is the dielectric) [ε1En1 - ε2En2] = nfree // this is continuity of Dn at the surface or [εdEr1 - ε0Er2] = n(θ) or [εd Er(a+α,θ) - ε0 Er(a-α,θ)] = n(θ) . A seeming third boundary condition is (1.1.48), ξ1En1 = ξ2En2 or (εd + σd/jω) Erd = (ε0 + σ/jω) Er ≈ (σ/jω) Er or (εd + σd/jω) Er(a+α,θ) ≈ (σ/jω) Er(a-α,θ) . There seem to be three boundary conditions at the round wire surface, σ Er(a-α,θ) - σdEr(a+α,θ) = jω n(θ) // modified cpbc from div J = -jωρ (D.9.16) εd Er(a+α,θ) - ε0 Er(a-α,θ) = n(θ) // div D = ρ (straddle) (D.9.17) (εd + σd/jω) Er(a+α,θ) ≈ (σ/jω) Er(a-α,θ) // ξ1En1 = ξ2En2 (D.9.18) but only two of these conditions are independent. For example, multiply (D.9.16) by (-1/jω) to get (σd/jω)Er(a+α,θ) - (σ/jω) Er(a-α,θ) = - n(θ) . Adding this to (D.9.17) then gives (εd + σd/jω) Er(a+α,θ) - (σ/jω) Er(a-α,θ) = 0 which is in fact the same as (D.9.18). When we solve the "capacitor problem" as in Section 6.5 (a) to obtain n(θ) on the round conductor surface, we are using (D.9.17) with the assumption that Er(a+α,θ) >> Er(a-α,θ). Typically one just says that in a good conductor Er(a-α,θ) = 0 and then n(θ) = εd Er(a+α,θ). That is fine, but it is not clear what happens to (D.9.16) above. The first term is the product of a large quantity σ times a small quantity Er(a-α,θ) so can be the same size as the other terms in the equation. The resolution is provided by the discussion in Section 1.5 (c) where we encountered the equation (1.5.17) nc(x,ω) = (ξ1/ε1) ns(x,ω) . (1.5.17) which in our current context (1 = dielectric) becomes nc(θ) = (ξd/εd) n(θ) . (D.9.19) In that discussion it is noted that n(θ) is the actual free surface charge density, whereas nc(θ) is a related "transport charge density" having the same dimensions as n(θ). If we multiply (D.9.16) and (D.9.17) by (ξd/εd), our (redundant) triplet of boundary conditions becomes, 1 σ (ξd/εd) Er(a-α,θ) - σd (ξd/εd) Er(a+α,θ) = jω nc(θ) 2 ξd Er(a+α,θ) - (ξd/εd) ε0 Er(a-α,θ) = nc(θ) 3 ξdEr(a+α,θ) ≈ ξ Er(a-α,θ) . (D.9.20) We now use last of these three equations to eliminate Er(a+α,θ) in the first, which then becomes σ (ξd/εd) Er(a-α,θ) - σd (ξ/εd) Er(a-α,θ) = jω nc(θ) or [ σ ξd - σd ξ ]/εd Er(a-α,θ) = jω nc(θ) or [ σ (εd + σd/jω) - σd (ε0 + σ/jω) ]/εd Er(a-α,θ) = jω nc(θ) or [ (σ εd - σdε0)]/εd Er(a-α,θ) = jω nc(θ) // two large terms cancelled or [ (σ - σd(ε0/εd) ] Er(a-α,θ) = jω nc(θ) . Assume now that ε0 (conductor) and εd (dielectric) are the same order of magnitude, and assume that, even though the dielectric conducts, one still has σ >> σd . The last equation then reads Er(a-α,θ) = (jω/σ) nc(θ) = (jω/σ) (ξd/εd) n(θ) (D.9.21) Er(a-α,m) = (jω/σ) (ξd/εd) Nm . // partial waves (D.9.22) The above are the "modified" charge pumping boundary conditions which replaces (D.2.24) and (D.2.25) for a mildly conducting dielectric, Er(r=a,θ) = (jω/σ) n(θ) . (D.2.24) Er(r=a,m) = (jω/σ) Nm . (D.2.25) How then does σd ≠ 0 alter the E field results summarized in box (D.2.33)? If we indicate σd = 0 quantities with no prime and σd > 0 quantities with a prime, then tracing through the development we find that n(θ) → (ξd/εd) n(θ) ≡ n'(θ) Nm → (ξd/εd) Nm ≡ N'm but ηm ≡ Nm /N0 remains the same Km → (ξd/εd) Km ≡ K'm am → (ξd/εd) am ≡ a'm βd = ω → ω ≡ βd' so βd' = βd (D.9.23) But I never used this form for βd anywhere in Appendix D up to this point! The normalization condition (D.2.31) I = 2πω (a/βd) N0 (D.2.31) becomes (where I' is the total current in the wire with σd > 0) I' = 2πω (a/β'd) N0' = 2πω (a/βd) (ξd/εd)N0 = 2πω (a/βd) N0 = I . (D.9.24) Then for example the Ez field becomes [see below (D.2.32)], E'z(r,m) = -j(β'/β'd) Jm(x) = -j(β'/β'd) (jω/2σ) N'm [ – ] Jm(x) = -j(β'/β'd) (jω/2σ) ηm N'0 [ – ] Jm(x) = -j(β'/β'd) [(j/4) (aβ'd) ηm I' Rdc] [ – ] Jm(x) = (1/4) ηm I' Rdc (aβ') [ - ] and for the Er field, E'r(r,m) = a'm x-1 Jm(x) + Jm+1(x) = [(jω/2σ) N'm] { 2m x-1 Jm(x) + [ – ] Jm+1(x) } = (j/4) (aβ'd) ηm I' Rdc { + - } = (j/4) (aβ'd) ηm I' Rdc { + }. The results (D.2.33) then become: Second summary of the E field solutions : Rdc = β'2 = β2 - β'd2 (D.9.25) (conducting dielectric) I' = V/Z0 ξd = εd + σd/jω βd' = βd where βd = ω ( I' = I where I was for σd = 0 ) E'z(r,m) = (1/4) ηm I' Rdc (aβ') fm fm = [ - ] x = β'r E'r(r,m) = (j/4) ηm I' Rdc (aβ'd) gm gm = [ + ] xa = β'a E'θ(r,m) = (1/4) ηm I' Rdc (aβ'd) hm hm = [ - ] Conclusion: Turning on σd causes the total current to "increase" from I to I' = I, which seems intuitive since the total current now has to feed the conductance between the conductors. However, the E field solutions have exactly the same form they had when σd = 0 when expressed in terms of the new larger current I' and the new β'd. This β'd differs from the βd for σd = 0, and this then gets mapped into a slight difference in β' since β'2 = β2 - β'd2 . D.10 Symmetry and the high frequency limit of the round wire E fields Recall from (D2.2.33) that, Summary of the E field solutions : Rdc = β'2 = β2 - βd2 (D.2.33) Ez(r,m) = (1/4) ηm I Rdc (aβ') fm fm = [ - ] x = β'r Er(r,m) = (j/4) ηm I Rdc (aβd) gm gm = [ + ] xa = β'a Eθ(r,m) = (1/4) ηm I Rdc (aβd) hm hm = [ - ] (a) Symmetry of fm, gm and hm and expansions for Ei(r,θ) From NIST (10.4.1) we know that for integer m, J-m(x) = (-1)mJm(x) . (D.10.1) For fm we find that fm = [ - ] f-m = [ - ] = - [ - ] = fm Coefficients gm and hm have the same symmetry, gm = [ + ] g-m = [ + ] = [ + ] = gm hm = [ - ] h-m = [ - ] = [ - ] = - [ - ] = hm Thus we have shown that f-m = fm g-m = gm h-m = hm . (D.10.2) If the surface charge n(θ) happens to be even in θ, we know from (D.1.7) that ηm = Nm/N0 = η-m. In this case, the E field components in (r,θ) can be written as in (D.1.7), Ei(r,θ) = Ei(r,m=0) + 2!Syntax Error, IEi(r,m) cos(mθ) (D.10.3) Then for even n(θ) the E fields are Ez(r,θ) = (1/4) I Rdc (aβ') [ f0 + 2 Σm=1∞ fm ηm cos(mθ) ] Er(r,θ) = (j/4) I Rdc (aβd) [ g0 + 2 Σm=1∞ gm ηm cos(mθ) ] Eθ(r,θ) = (1/4) I Rdc (aβd) [ h0 + 2 Σm=1∞ hm ηm cos(mθ) ] . (D.10.4a) For general n(θ) where ηm and η-m are no longer equal we have instead Ez(r,θ) = (1/4) I Rdc (aβ') [Σm=-∞∞ fm ηm ejmθ ] Er(r,θ) = (j/4) I Rdc (aβd) [ Σm=-∞∞ gm ηm ejmθ ] Eθ(r,θ) = (1/4) I Rdc (aβd) [ Σm=-∞∞ hm ηm ejmθ ] . (D.10.4b) (b) High frequency evaluation of fm, gm and hm and the E fields For large ω we assume β' = β as discussed below (D.2.2). From NIST 10.17.2, keeping a few leading terms in each inverse power expansion, we have this large x behavior for Jm(x) , Jm(x) = (2/πx)1/2 { cos(w) [a0(m) - a2(m)/x2 + O(1/x4)] - sin(w) [a1(m)/x + O(1/x3)] ] } w = x - mπ/2 -π/4 => e-jw = e-j(x-mπ/2-π/4) = e-jx ejπm/2 ejπ/4 a0(m) = 1 a1(m) = ≡ cm a2(m) = ≡ dm . (D.10.5) The expansion is in fact valid for all real and complex values of the parameter m, but we shall only use the expansion for integer m. Using abbreviations cm and dm one gets, Jm(x) = (2/πx)1/2[ cos(w) (1-dm/x2) - sin(w) (cm/x ) ] . (D.10.6) Recall that inside the round wire, δ ≡ = skin depth // ωμσ = 2/δ2 (2.2.20) β = ej3π/4 (/δ) = (j-1)/δ (2.2.21) so x = βr = ej3π/4 (/δ) r = (j-1) (r/δ) xa = βa = ej3π/4 (/δ) a = (j-1) (a/δ) . (D.10.7) Since x has a large positive imaginary part for small δ, so does w. Then cos(w) = [ ejw + e-jw]/2 ≈ (1/2) e-jw sin(w) = [ ejw - e-jw]/2j ≈ -(1/2j) e-jw = (j/2)e-jw . (D.10.8) The large-x expansion above then becomes Jm(x) = (2/πx)1/2 (1/2) [e-jw (1- dm /x2) - j e-jw (cm /x) ] = (1/2πx)1/2 e-jw [ 1 -j cm (1/x) - dm (1/x2) + ... ] = (1/2πx)1/2 e-jx ejπm/2 ejπ/4 [ 1 -j cm (1/x) - dm (1/x2) + ... ] . (D.10.9) It is not hard to show that this agrees with (2.3.5) through order 1/x. Notice that e-jx = e-j(j-1)(r/δ) = e(1+j)(r/δ) giving a convenient hybrid form Jm(x) = (1/2πx)1/2 e(1+j)(r/δ) (j)m ejπ/4 [ 1 -j cm (1/x) - dm (1/x2) + ... ] . (D.10.10) From (D.10.10) we see by inspection that, through O(1/x), = (j)m-n e(1+j)(r-a)/δ ≈ (j)m-n e(1+j)(r-a)/δ [ 1 - jcm/x + jcn/xa ] (D.10.11) and = e(1+j)(r-a)/δ // independent of m (D.10.12) Therefore for large ω, gm = [ + ] = 2 e(1+j)(r-a)/δ hm = [ - ] = 0 fm= [ - ] = (j)-1 e(1+j)(r-a)/δ [ 1 - jcm/x + jcm+1/xa ] - (j)+1 e(1+j)(r-a)/δ [ 1 - jcm/x + jcm-1/xa ] = - j e(1+j)(r-a)/δ [ 2 - 2jcm(1/x) + j(cm+1+cm-1) (1/xa) ≈ - 2j e(1+j)(r-a)/δ The results are then Large ω limits of the E field solutions : Rdc = (D.10.13) Ez(r,m) = (1/4) ηm I Rdc (aβ) fm fm = -2j e(1+j)(r-a)/δ x = β'r Er(r,m) = (j/4) ηm I Rdc (aβd) gm gm = 2 e(1+j)(r-a)/δ xa = β'a Eθ(r,m) = (1/4) ηm I Rdc (aβd) hm hm = 0 As observed earlier, the longitudinal current Jz is much larger than the radial current Jr by factor (β/βd). Notice the standard skin effect behavior both in amplitude and phase for all field components. We saw this earlier in several places: E(x,ω) = E(0,ω) e-x/δ e-jx/δ . x → (a-r) 1D example (2.1.8) = e(r-a)/δ r/δ > 3/= 2.1 . (2.3.7) The configuration space fields from (D.10.4b) are then, Ez(r,θ) = (1/4) I Rdc (aβ') {-2j e(1+j)(r-a)/δ } [ Σm=-∞∞ ηm ejmθ ] Er(r,θ) = (j/4) I Rdc (aβd) { 2 e(1+j)(r-a)/δ } [ Σm=-∞∞ ηm ejmθ ] Eθ(r,θ) = 0 ηm = Nm/N0 . (D.10.14) But the [...] expansions shown here are just n(θ)/N0 from (D.1.5a) where recall that N0 = (βd/2πωa) I . (D.2.31) Since Rdc = (πa2)/σ we find, (1/4) I Rdc(aβd) 2 /N0 = (1/2) I Rdc (aβd) 2πωa/(βdI) = Rdc πωa2 = (ω/σ) so then for large ω (D.10.14) becomes, Ez(r,θ) = - (jω/σ) e(1+j)(r-a)/δ n(θ) (β'/βd) Er(r,θ) = (jω/σ) e(1+j)(r-a)/δ n(θ) Eθ(r,θ) = 0 (D.10.15) Observations on the E fields for large ω In the extreme skin effect (small δ, large ω) regime: 1. There is no azimuthal field Eθ inside or on the surface of the round wire. 2. Both Ez and Er exhibit the standard skin effect form for amplitude and phase 3. The ratio Ez(r,θ)/Er(r,θ) = - (β'/βd) is very large and is constant in r and θ 4. Both Ez and Er track the surface charge density n(θ) for azimuthal dependence 5. If n(θ) ≠ constant, then Jz = σEz ≠ constant in θ and the longitudinal current density is asymmetric across the round wire cross section, which is known as the proximity effect. This implies that the surface impedance Zs is a function of θ. In Section 2.4 the surface impedance was a constant since only the m=0 partial wave was involved. D.11 Low frequency limit of the round wire E fields Our assumed Appendix D wave z dependence is e-jβz from (D.1.1). (a) An Inaccurate Model for Low Frequency Recall from Chapter 4 that for high frequencies we derived the transmission line equations, = - z i(z) = - y V(z) (4.11.14b) and from those we derived the second order uncoupled transmission line equations, - zy V(z) = 0 - zy i(z) = 0 . (4.11.15) These last equations imply the following z-dependence for V(z) and i(z) and then presumably for all other transmission line non-power quantities (like the fields), e-jβ'z where βd' = -j= -j If in Appendix D we replace βd (the wave number of the dielectric) with βd', the field solutions carry through everywhere with the simple replacement βd → βd' and we get, Summary of the E field solutions : Rdc = β'2 = β2 - βd'2 (D.2.33) Ez(r,m) = (1/4) ηm I Rdc (aβ') fm fm = [ - ] x = β'r Er(r,m) = (j/4) ηm I Rdc (aβd') gm gm = [ + ] xa = β'a Eθ(r,m) = (1/4) ηm I Rdc (aβd') hm hm = [ - ] where we have written βd' in three places. Before making further comments, we first review the high frequency situation. For large ω, we show in Appendix Q that For large ω: Re(β'd) = ω Im(β'd) = - ω >> (R/L) and ω >> (G/C) Since the wave z dependence is e-jβ'z = e-jRe(β')z e-[-Im(β')]z we can interpret Re(β'd) as the "wavenumber" of the wave, and we know that this wavenumber = ω/vd where vd is the phase velocity of the wave. Thus, we identify vd = 1/ in terms of which Re(β'd) = ω /vd = βd -Im(β'd) = (RC+GL)(vd/2) . This last quantity is the attenuation parameter sometimes called α, so α = (RC+GL)(vd/2) . Thus we have a reasonable model for high frequency waves on our transmission line, including the attenuation due to the losses caused by R and G. We can go on to evaluate β'2 and take its large ω limit, β'2 = β2 - βd'2 = -jμσω - [ω /vd - jα ]2 → -jμσω - (ω/vd)2 ≈ -jμσω . In the last equality, although it seems the ω2 term should win out at large ω, we have shown elsewhere that for any ω of transmission line interest, σ is large and vd is large so - (ω/vd)2 can be ignored. For example : f = 1000 GHz ω = 2π x 1012 Hz μ0σω = 4π x 5.81 x 2π x 1012 = .46 x 1015 (ω/vd)2 ≈ (ω/c)2 = [2π x 1012/ 3x108]2 = [(2π/3) x 104]2 = .44 x 109 In Section D.10 we went on to take the large β' limit of the fm, gm and hm functions. This led to an interesting asymmetric Jz distribution which we associated with the proximity effect. For small ω things are much more hazy. The main issue is that the transmission line equations quoted above are only justified from Maxwell's equations at high frequencies where the skin effect is at least starting to set in. We show in Appendix K that for the network model of a transmission line, the transmission line equations are valid at all ω. One must keep in mind that this network model is just a model, and we have shown in Chapter 4 that it does not correspond to the real world at low frequencies. Despite this caveat, we can pretend that the transmission line equations are accurate at low ω and see what happens. We quote again from Appendix Q for a transmission line with G = 0 (vacuum dielectric) since this is assumed in our boundary condition (D.2.24), For small ω: Re(β'd) = = α // a new α - Im(β'd) = = α G = 0 and ω << (R/L) This is an extremely lossy wave, since it damps out in a half wave cycle. We then have e-jβ'z = e-jRe(β')z e-[-Im(β')]z = e-jαz e-σz = e-(j+1)αz . This has the exact form shown in (2.1.8) so what we have here is in effect a longitudinal skin effect where the wave is trying to travel down the transmission line but it can't get very far. The longitudinal skin depth is 1/α . We can blindly proceed and compute β'2 = β2 - βd'2 = -jμσω - ωRC/2 → 0 as ω→0 In this case, we want to take the small argument limits of fm, gm and hm. We shall go ahead and do that below just for the sake of "completeness", but it should be realized that the results are not accurate. (b) Low frequency evaluation of fm, gm and hm and the E fields In the following we consider only m ≥ 0 since we know from (D.10.2) that f-m = fm , g-m = gm, h-m = hm. The small x limit for Jm(x) is given by NIST 10.7.3, Jn(x) = (x/2)n / n! . for n = 0,1,2,..... (D.11.14) Since Jm-1 appears in our coefficient expressions and since m = 0 is encountered, we have to deal with m = 0 as a special case since the above limit is not valid for n = -1. To this end we use NIST 10.2.2 which is valid for integer n, J-n(x) = (-1)nJn(x) ≈ (-1)n (x/2)n / n! (D.11.15) so that J-1(x) = - J1(x) ≈ - (x/2). Our small-x forms of interest are then Jn(x) = (x/2)n / n! for n = 0,1,2,..... J-1(x) = - (x/2) for n = -1 . (D.11.16) We now examine the small x limits of fm, gm, and hm . First fm for m > 0, and then for m = 0: fm = [ - ] = [ - ] = [ (m+1) (x/xa)m (2/xa) - (1/m) (x/xa)m(xa/2) ] = (x/xa)m [ (m+1) (2/xa) - (1/m) (xa/2) ] ≈ (x/xa)m (m+1) (2/xa) // as xa→ 0 f0 = [ - ] = [ + ] = 2 = 2 = 4/xa First gm for m > 0, and then for m = 0: gm = [ + ] = [ + ] = (x/xa)m+1 + (x/xa)m-1 g0 = [ + ] = [ + ] = 2 = 2 (x/xa) Results for hm are then obvious since there is only a sign change between the terms in gm, hm = (x/xa)m+1 - (x/xa)m-1 h0 = 0 The results are then ( we replace βd → βd' as discussed in section (a) above ) : Small ω limit of the E field solutions : Rdc = (D.11.17) Ez(r,m) = (1/4) ηm I Rdc (aβ') fm fm = (r/a)m (m+1) (2/β'a) f0 = 4/(aβ') Er(r,m) = (j/4) ηm I Rdc (aβd') gm gm = (r/a)m+1 + (r/a)m-1 g0 = 2 (r/a) Eθ(r,m) = (1/4) ηm I Rdc (aβd') hm hm = (r/a)m+1 - (r/a)m-1 h0 = 0 For m > 0: Ez(r,m) = (1/2) ηm I Rdc (r/a)m (m+1) Er(r,m) = (j/4) ηm I Rdc (aβd') [(r/a)m+1 + (r/a)m-1] Eθ(r,m) = (1/4) ηm I Rdc (aβd') [(r/a)m+1 - (r/a)m-1] For m= 0: Ez(r,0) = I Rdc Er(r,0) = (j/2) I Rdc (aβd') (r/a) Eθ(r,0) = 0 Recall now these field expansions for the case that n(θ) is an even function of θ, Ez(r,θ) = (1/4) I Rdc (aβ') [ f0 + 2 Σm=1∞ fm ηm cos(mθ) ] Er(r,θ) = (j/4) I Rdc (aβd') [ g0 + 2 Σm=1∞ gm ηm cos(mθ) ] Eθ(r,θ) = (1/4) I Rdc (aβd') [ h0 + 2 Σm=1∞ hm ηm cos(mθ) ] . (D.10.4a) For low ω we insert the expressions above to get Ez(r,θ) = I Rdc [ 1+ Σm=1∞ (r/a)m (m+1) ηm cos(mθ) ] Er(r,θ) = (j/2) I Rdc (aβd) [ (r/a) + Σm=1∞ [(r/a)m+1 + (r/a)m-1] ηm cos(mθ) ] Eθ(r,θ) = (1/2) I Rdc (aβd) [ Σm=1∞ [(r/a)m+1 - (r/a)m-1] ηm cos(mθ) ] . (D.11.18) THIS SECTION ABOVE IS ON HOLD Observations on the E fields for small ω 2. The m = 0 term in Ez is just Ez = IRdc which says Jz = σ IRdc = σ I(1/σπa2) = I/(πa2). This is the current one would expect in a wire carrying DC current I. 3. Assuming a non-uniform n(θ) charge density on the wire surface, the ηm moments are non-zero and one concludes that Jz(r,θ) is asymmetric even as ω → 0. This conclusion is incorrect, and the reason is that we have improperly assumed STOP I = V/Z0 and according to (4.11.16), where G = 0 since we assumed a non-conducting dielectric, Z0 ≡ V(z)/i(z) = = . (4.11.16) Thus in the limit ω→0 we get Z0 → ∞. For a finite length transmission line, as one lowers ω, one must increase the size of the termination Z0 to maintain a properly terminated line. Thus, our situation here does not apply to taking the low frequency limit of a transmission line terminated by a fixed 75Ω or 8Ω resistor. See Section 6.5 (e) for more on this subject. The major point here is that the entire model is not valid for low ω and we cannot use it to study the limit ω → 0.