app D work REVIEWED

New Version of Appendix D PhL 10.4.13 Even though not quite finished, today 10.13.13 I installed this App D into transmission lines doc just because it has been inactive for a long time and I wanted to get Lines to be as complete as possible for making a backup copy today. So now this doc here is just storage. Appendix D : The General Electric Field Inside a Round Wire 1 D.1 The General Method and Solution for Ez 2 (a) Partial Wave Expansions 2 (b) The Vector Laplacian and the conversion of equations from φ space to m space 3 (c) The Charge Pumping Boundary Condition 7 (d) The Ez Solution 8 D.2 The Solutions for Er and Eφ 8 (a) The Er Solution 8 (b) The Eφ Solution 10 (c) Application of the Boundary Conditions 12 D.3 What about the Eφ Helmholtz Equation ? 14 D.4 Statement of the Results 16 D.5 The Low Frequency Limit 19 D.6 What about the E fields outside the round wire? 22 D.7 What about φ, A and B ? 23 Appendix D : The General Electric Field Inside a Round Wire In this Appendix we present a somewhat lengthy calculation of the electric field (and therefore the current) inside a round wire without assuming that such fields are symmetrical about the axis. This wire may be regarded as one conductor of a transmission line down which a wave is propagating. The conclusions one can draw from the analytic solution given below are supportive of the general discussion elsewhere in this monograph: 1. The Er and Eφ field components are very small. In fact, they are smaller than Ez by the factor (a/λ), where a = wire radius, and λ = wavelength of the wave on a transmission line containing the wire. This fraction is always assumed small in any analysis of a transmission line. 2. It is possible to maintain the condition Eφ = 0 on the wire surface. This condition is consistent with the idea that the wire surface is an electrical equipotential at any constant z. 3. We get an explicit formula for the surface impedance. We find that it is non-uniform around the boundary of the wire cross section, and that Jz is non-uniform inside the wire. This result is true even in the DC limit of a transmission line. 4. We suggest a method for computing the E fields in the low frequency limit of a transmission line consisting of round wires. First, solve the "electrostatic" problem illustrated in Chapter 6. Knowing the potential φ, compute the radial electric field at the surface of the conductors. From this compute the surface charge density n(φ) as a function of azimuth around the wire Then, as described below, compute the moments Nm (or ηm) of this charge distribution, and use the formulas below to find the electric field. This is a highly technical appendix. The reader is invited to inspect the boxed results and comments in Section D.4. D.1 The General Method and Solution for Ez The starting point for the calculation is the Helmholtz equation (1.6.2) for the E field inside the wire, ( 2 + β2 ) E(r,φ z, t) = 0 . (D.1.1) Here we use cylindrical coordinates (r,φ,z), as appropriate for a straight round wire. The first step is to expose the assumed t and z dependence, and in doing so, define the field E(r,φ) : E(r,φz,t) = ej(ωt-βz) E(r,φ) . (D.1.2) Here, symbol βd stands for the value that parameter β = ω takes in the dielectric. We reserve the symbol β with no subscript to mean β inside the metal of our wire. The form shown in (D.1.2) is a simple wave traveling down a transmission line in the +z direction. We assume that one of the conductors of this transmission line is our round wire, while the other conductor is (other conductors are) unspecified. One can regard (D.1.2) as an assumed variable-separated form for a solution, an "ansatz". Our second ansatz is that Eφ(r=a,φ) = 0 which means at any z = constant slice, the circular wire perimeter is an equipotential. If a consistent solution to the field equations can be found with these assumptions, they are justified de facto. Convention Comment: Section 1.7 discusses the Fourier Transform (1.7.7) and (1.7.8) where e+jωt appears in the expansion formula (1.7.8). 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 like 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, but in this document we use the e+j(ωt-kz) convention usually used in EE texts. It is just a convention choice and, as in (1.7.6), it just effects the sign of the imaginary physical field under consideration. If one thinks of the physical field under consideration as Re{ E(x,t)}, the sign convention choice makes no difference at all. (a) Partial Wave Expansions The next step is to do a "partial wave expansion" (that is, a complex Fourier series expansion) of E(r,φ) in terms of "azimuthal harmonics" eimφ, so that the variable φ is replaced with the partial wave index m: E(r,φ) =!Syntax Error, I E(r,m) ejmφ E(r,m) = (1/2π) !Syntax Error, Idφ E(r,φ) e-jmφ (D.1.3) Since the usual complex part has been extracted in (D.1.2), we assume that E(r,φ) is real so the right equation above says that E(r,-m) = E(r,m)* . We can then re-express the above as, E(r,φ) = E(r,0) +!Syntax Error, I[ E(r,m)ejmφ + E(r,-m)e-jmφ] = E(r,0) +!Syntax Error, I[ E(r,m)ejmφ + [E(r,m)ejmφ]*] = E(r,0) + 2!Syntax Error, IRe{ E(r,m)ejmφ } = E(r,0) + 2!Syntax Error, I[ Re(E) + j Im(E)] [ cos(mφ) + j sin(mφ)] = E(r,0) + 2!Syntax Error, I[Re(E)cos(mφ) - Im(E) sin(mφ)] + 2j!Syntax Error, I[ Im(E)cos(mφ) + Re(E) sin(mφ)] Since E(r,φ) is assumed real, the second sum must vanish for all r and φ, giving this final result, E(r,φ) = E(r,0) + 2!Syntax Error, I[Re{E(r,m)}cos(mφ) - Im{E(r,m)} sin(mφ)] . (D.1.4) In analogy with (D.1.2) and the discussion of Section 1.7 regarding complex functions, we define a complex surface (r=a) charge density n(φ,z,t) which has the following ansatz variable-separated form, n(φ,z,t) = ej(ωt-βz) n(φ) (D.1.5) where n(φ) is real. We then expand n(φ) as in (D.1.3) and (D.1.4), n(φ) = !Syntax Error, I Nm ejmφ Nm = (1/2π) !Syntax Error, Idφ n(φ) e-jmφ . (D.1.6) n(φ) = N0 + 2!Syntax Error, I[Re{Nm}cos(mφ) - Im{Nm} sin(mφ)] (D.1.7) Since n(φ) is real, Nm = N-m*. These Nm are the "moments" of the surface charge distribution, and their complex nature keeps track of cos(mφ versus sin(mφ) components. (b) The Vector Laplacian and the conversion of equations from φ space to m space Given the following cylindrical-coordinates field components, E(r,φz,t) = Er(r,φz,t) + Eφ(r,φz,t) + Ez(r,φz,t) (D.1.8) we may write out our ansatz wave form (D.1.2) and the Helmholtz equation (D.1.1) in more detail, Er(r,φz,t) = ej(ωt-βz) Er(r,φ) . [2E(r,φ,z,t)]r + β2 Er(r,φ,z,t) = 0 Eφ(r,φz,t) = ej(ωt-βz) Eφ(r,φ) . [2E(r,φ,z,t)]φ + β2 Eφ(r,φ,z,t) = 0 Ez(r,φz,t) = ej(ωt-βz) Ez(r,φ) . [2E(r,φ,z,t)]z + β2 Ez(r,φ,z,t) = 0 . (D.1.9) 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 replace [2E] by [E] which is defined in this manner [E] ≡ [2E] ≡ grad(div E) – curl (curl 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.1.1). For cylindrical coordinates it turns out that, (2E)r = 2Er - (2/r2) ∂φEφ - (1/r2) Er (2E)φ = 2Eφ + (2/r2) ∂φEr - (1/r2) Eφ (2E)z = 2Ez (D.1.12) where 2 = (1/r)∂r(r∂r) + (1/r2)∂φ2 + ∂z2 = ∂r2 + (1/r)∂r + (1/r2)∂φ2 + ∂z2 . (D.1.13) 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 above, where the last line corresponds to the above discussion: (D.1.14) Using the ansatz form (D.1.2) and partial wave expansions of the form (D.1.3) or (D.1.6), it is a simple matter to convert an equation involving components like Ei(r,φz,t) or n(r,φz,t) to a simpler equation involving components like Ei(r,m) and n(r,m). We shall give an example, then state a simple set of rules that does the conversion automatically. Example 1: The Ez Helmholtz Equation (D.1.9): [2E]z + β2 Ez = 0 Using (D.1.12) and (D.1.13), this may be written, [∂r2 + (1/r) ∂r + (1/r2) ∂φ2 + ∂z2 + β2 ] ej(ωt-βz)Ez(r,φ) = 0 . (1) Inserting the expansion (D.1.3) for Ez(r,φ) and moving the m sum to the left gives !Syntax Error, I [∂r2 + (1/r) ∂r + (1/r2) ∂φ2 + ∂z2 + β2 ] ej(ωt-βz)Ez(r,m) ejmφ = 0 . (2) We can then make the obvious replacements ∂z = -jβd and ∂φ = +jm to get, !Syntax Error, I { [∂r2 + (1/r) ∂r -m2 (1/r2) – βd2 + β2 ] ej(ωt-βz)Ez(r,m) } ejmφ = 0 . (3) Due to the completeness of functions ejmφ 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) In other words, one applies !Syntax Error, Idφ e-jm'φ to both sides of (3), uses the completeness property !Syntax Error, Idφ ej(m-m')φ = 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 We may then write these rules for converting equation (1) to equation (6) Conversion Rules: ∂z → -jβd (D.1.15) ∂φ → +jm ∂t→ +jω f(r,φ,z,t ) → f(r,m) We can now practice with these rules to convert various equations of interest . Here a field with unstated arguments has the full arguments (r,φ,z,t). Example 2: The Er Helmholtz Equation (D.1.9): [2E]r + β2 Er = 0 . Using (D.1.12) we find, 2(Er) - (2/r2) ∂φEφ - (1/r2) Er + β2Er = 0 [∂r2 + (1/r)∂r + (1/r2)∂φ2 + ∂z2] Er - (2/r2) ∂φEφ - (1/r2) Er + β2Er = 0 [∂r2 + (1/r)∂r + (1/r2)∂φ2 + ∂z2 - (1/r2) + β2] Er - (2/r2) ∂φEφ = 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 Eφ(r,m) = 0 Example 3: The Eφ Helmholtz Equation (D.1.9): [2E]φ + β2 Eφ = 0 . Using (D.1.12) we find, 2(Eφ) + (2/r2) ∂φEr - (1/r2) Eφ + β2Eφ = 0 [∂r2 + (1/r)∂r + (1/r2)∂φ2 + ∂z2] Eφ + (2/r2) ∂φEr - (1/r2) Eφ + β2Eφ = 0 [∂r2 + (1/r)∂r + (1/r2)∂φ2 + ∂z2 - (1/r2) + β2] Eφ + (2/r2) ∂φEr = 0 Now apply the conversion rules to get [∂r2 + (1/r)∂r + (1/r2)(-m2) + (-βd2) - (1/r2) + β2] Eφ(r,m) + (2/r2) jm Er(r,m) = 0 Example 4: The div E = 0 . Using (D.1.14) we write div E = 0 as ∂r (r Er) + ∂φEφ + r ∂zEz = 0 Applying the conversion rules gives ∂r [r Er(r,m)] + jmEφ(r,m) + r (-jβd)Ez(r,m) = 0 [1 + r∂r ] Er(r,m) + jmEφ(r,m) + r (-jβd)Ez(r,m) = 0 Summary of Examples: [2E]z + β2 Ez = 0 : [r2∂r2 + r ∂r - m2 + r2 ( β2- βd2)] Ez(r,m) = 0 (D.1.16) [2E]r + β2 Er = 0 : [r2∂r2 + r∂r - (m2+1) + r2(β2-βd2)] Er(r,m) - 2jm Eφ(r,m) = 0 (D.1.17) [2E]φ + β2 Eφ = 0 : [r2∂r2 + r∂r - (m2+1) + r2(β2-βd2)] Eφ(r,m) + 2jmEr(r,m) = 0 (D.1.18) div E = 0 : ∂r [r Er(r,m)] + jmEφ(r,m) -jβd r Ez(r,m) = 0 (D.1.19) (c) The Charge Pumping Boundary Condition The reason we are interested in surface charge n(φ) of (D.1.5) is that it acts as a driving source of the radial electric field in the wire. Recall the equation of continuity (1.1.8) converted to the ω domain divJ = -jωρ ∫S dSJ = -jω ∫V dV ρ . // divergence theorem (D.1.20) When applied to a thin box of radial area dS straddling the wire surface, one finds that ∫S dSJ = -Jr(r=a,φ)dS and ∫V dV ρ = n(φ) dS so that Jr(r=a,φ) = jω n(φ) . (D.1.21) We assume that there is no current outside the wire to get this result. Since J = σE, this is really a boundary condition on the radial electric field, Er(r=a,φ) = (jω/σ) n(φ) . (D.1.22) We convert this to m-space using the conversion rules (D.1.15) to trivially obtain Er(r=a,m) = (jω/σ) Nm . (D.1.23) Thus, the radial electric field must have a certain value at the r=a boundary in each partial wave. And the value it must have is determined by the moment of the charge distribution. By way of interpretation, the surface charge of a transmission line is "pumped" by the radial current in the wire. Since divJ = 0 inside the wire, this radial current is accompanied by the usual longitudinal current one expects to find inside the conductors of a transmission line. ok to here (d) The Ez Solution As shown in (D.1.16), the Helmholtz equation for Ez(r,m) is [r2∂r2 + r ∂r + (r2 β'2 - m2)] Ez(r,m) = 0 (D.1.24) where β'2 = β2 - βd2 . (D.1.25) In a conductor like copper, β is huge compared to the dielectric βd, so we could ignore the distinction between β and β'. Setting x = β'r we find ∂r = β'∂x and then r∂r = x∂x and so on so that (D.1.24) reads [x2∂x2 + x ∂x + (x2 - m2)] Ez(x/β',m) = 0 . x = β'r (D.1.26) 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.1.27) where Czm is an arbitrary constant for each partial wave m. Equation (D.1.27) is consistent with (2.1.22). In Section 2.1 we dealt only with the m=0 partial wave, which embodies the symmetrical part of the problem. Also, we assumed constant z behavior so βd = 0 and the question of β versus β' never arose. D.2 The Solutions for Er and Eφ (a) The Er Solution As shown in (D.1.17), the Helmholtz equation for Er(r,m) is [r2∂r2 + r∂r - (m2+1) + r2β'2] Er(r,m) - 2jm Eφ(r,m) = 0 (D.2.1) while the div E = 0 condition was stated in (D.1.19) as [1 + r∂r ] Er(r,m) + jmEφ(r,m) + r (-jβd)Ez(r,m) = 0 or -jmEφ(r,m) = [1 + r∂r ] Er(r,m) - r (jβd)Ez(r,m) . (D.2.2) Inserting this into (D.2.1) 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.2.3) Inserting solution (D.1.27) 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.2.4) where Km ≡ 2j (βd/β') Czm . (D.2.5) In order to get the left side of (D.2.4) into something recognizable, we define Er(r,m) = x-1 fm(x) (D.2.6) where x is a dimensionless radial variable which will play a major role in the following, x ≡ β'r and xa ≡ β' a . (D.2.7) Then (D.2.4) becomes [x2∂x2 + 3x∂x + (1-m2) + x2] { x-1 fm(x)} = 2j (βd/β') xCzm Jm(x) or x [x2∂x2 + 3x∂x + (1-m2) + x2] { x-1 fm(x)} = K x2 Jm(x) . (D.2.8) Ever eager, Maple expands the left side of (D.2.7), so that (D.2.8) becomes [ x2 ∂x2 + x ∂x + (x2-m2)] fm(x) = Km x2 Jm(x) . (D.2.9) The left side of (D.2.9) 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.2.10) a 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.2.6) the full solution Er , Er(r,m) = am x-1 Jm(x) + Jm+1(x) . (D.2.11) For each value of m, there are two as-yet undetermined constants, am and (Km/2). However, looking at (D.2.11), we see that, since J0(x) ≈ 1 for small x, we must have a0 = 0 (D.2.12) to keep Er finite at r = 0. We shall obtain expressions for am and (Km/2) below. (b) The Eφ Solution Recall (D.2.2) , jmEφ(r,m) = -∂r[rEr(r,m)] + r (jβd)Ez(r,m) . (D.2.2) We can then insert our known Ez and Er to get Eφ Ez(r,m) = Czm Jm(x) (D.1.27) Er(r,m) = am x-1 Jm(x) + Jm+1(x) . (D.2.11) so (D.22) becomes the following jmEφ(r,m) = -∂r[r{ am x-1 Jm(x) + Jm+1(x)}] + r (jβd) Czm Jm(x) jmEφ(r,m) = -∂x[x{ am x-1 Jm(x) + Jm+1(x)}] + x Jm(x) // Km ≡ 2j (βd/β') Czm jmEφ(r,m) = -∂x[am Jm(x) + x Jm+1(x)] + x Jm(x) jmEφ(r,m) = -am Jm'(x) - Jm+1(x) - x Jm+1'(x) + x Jm(x) jmEφ(r,m) = - am Jm'(x) + [ - Jm+1(x) - x Jm+1'(x) + x Jm(x) ] (D.2.13) At this point we invoke the recurrence relations AS 10.6.2 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.2.14) Insert these into (D.2.18) to get jmEφ(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: jEφ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) x = β'r . (D.2.15) What about Eφ(r,m=0)? We have already noted that a0 = 0. For m=0, (D.2.15) is invalid because we divided (D.2.13) by m. In fact, we really know nothing at all about the field Eφ(r,m=0) from our solution method because Eφ is not even present in our starting equation (D.2.2) when m = 0, -jmEφ(r,m) = [1 + r∂r ] Er(r,m) - r (jβd)Ez(r,m) . (D.2.2) However, we can use the Eφ Helmholtz equation (D.1.18) to learn more about Eφ(r,m) with m = 0, [r2∂r2 + r∂r - (m2+1) + r2β'2] Eφ(r,m) + 2jmEr(r,m) = 0 (D.1.18) With m = 0 we get [r2∂r2 + r∂r - 1 + r2β'2] Eφ(r,0) = 0 or [x2∂x2 + x∂x - 1 + x2] Eφ(r,0) = 0 x = β' r (D.2.16) Since [..] is the ν = 1 Bessel operator and since Eφ must be finite at r = 0, we find that Eφ(r,0) = Cφ0 J1(β'r) (D.2.17) where Cφ0 is some constant. (c) Application of the Boundary Conditions We have two boundary conditions to impose: Er(r=a,m) = (jω/σ) Nm (D.2.18) Eφ(r=a,m) = 0 (D.2.19) The first is the radial charge pumping condition shown in (D.1.23) above, while the second is our ansatz assumed earlier that any z = constant circle on the wire surface be an equipotential. 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 (D.2.11) for Er and (D.2.15) for Eφ 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) the second of which may be immediately solved for am = (jω/σ) Nm . (5) Using the recursion relation 2m x-1 Jm = [Jm+1 + Jm-1] and using (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) For m = 0 we use (D.2.17) for Eφ and the two boundary conditions are then J1(xa) = (jω/σ) N0 (D.2.20) Cφ0 J1(xa) = 0 => Cφ0 = 0 (D.2.21) The K0 result is compatible with (7) since J-1= - J1 and the (9) requires that Cφ0 = 0, barring some strange radius which happens to be a zero of the J1(x) function. To summarize, am = 2m(jω/2σ) Nm m≥0 = (jω/2σ) Nm [ – ] m≥ 0 (+ ) = (jω/2σ) Nm { + } m>0 Cφ0 = 0 m = 0 (D.2.22) We have now completely solved the problem. To review, here are the three fields gathered together in one spot: Ez(r,m) = -j(β'/βd) Jm(x) (D.1.27) and (D.2.5) Er(r,m) = am x-1 Jm(x) + Jm+1(x) (D.2.11) (D.2.23) jEφ(r,m) = - am x-1 Jm(x) + (+ ) Jm+1(x) m > 0 (D.2.15) jEφ(r,0) = 0 m = 0 (D.2.17) and Cφ0 = 0 D.3 What about the Eφ Helmholtz Equation ? Except to find Eφ(r,0) we have completely ignored the Eφ Helmholtz equation until now. It is reasonable to wonder whether the solution fields we have found above in fact solve this 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 . (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, but it certainly suggests it. Perhaps (∂j∂j + β2) E3 = f(x1,x2) ≠ 0, but other considerations might force f = 0, such as the nature of curvilinear coordinates and boundary conditions. Rather than pursue this question further, we will now directly show that our solutions do in fact satisfy the Eφ Helmholtz equation. The Eφ Helmholtz equation (D.1.18) is this, expressed in terms of x = β'r, [x2∂x2 + x∂x - (m2+1) + x2)] Eφ(r,m) = – 2jmEr(r,m) = 0 (D.1.18) (D.3.3) and we found these solutions for Er and Eφ , Er(r,m) = am x-1 Jm(x) + Jm+1(x) . (D.2.11) jEφ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) x = β'r . (D.2.15) Treating (D.2.22) as LHS = RHS we then have LHS = [x2∂x2 + x∂x - (m2+1) + x2] {- am x-1 Jm(x) + ( + ) Jm+1(x) } RHS = 2m{ am x-1 Jm(x) + Jm+1(x)} . (D.3.4) In order to have LHS = RHS, it must be true for arbitrary am and Km . Thus, we want to show that each of the following equations is valid, [x2∂x2 + x∂x - (m2+1) + x2] ( Jm+1(x) ) = 2m Jm+1(x) [x2∂x2 + x∂x - (m2+1) + x2] {- am x-1 Jm(x) + Jm+1(x) } = 2m am { x-1 Jm(x) } which are the same as [x2∂x2 + x∂x - (m2+1) + x2] (Jm+1(x) ) = 2m Jm+1(x) (D.3.5) [x2∂x2 + x∂x - (m2+1) + x2] {- x-1 Jm(x) + Jm+1(x) } = 2m { x-1 Jm(x) } . (D.3.6) Equation (D.3.5) may be written [x2∂x2 + x∂x - (m2+1) + x2 - 2m] Jm+1(x) = 0 or [x2∂x2 + x∂x + x2- (m+1)2] Jm+1(x) = 0 . But this is Bessel's equation for ν = m+1, thus (D.3.5) really is valid. As for (D.3.6), we leave it to trusty Maple : Thus (D.3.6) is also valid We conclude that the our round wire E field solutions satisfy the Eφ Helmholtz equation as well as the other two Helmholtz equations and the div E = 0 equation. D.4 Statement of the Results Although we have a complete solution to our problem, it is useful to normalize the results to some well defined absolute scale. The first step in doing this is to compute the total current I in the wire using Jz = σEz. The purpose here is relate the current I to the moment N0 of the surface charge distribution. Many steps are required to ferret out this relationship: I = !Syntax Error, Idφ !Syntax Error, Ir dr Jz(r,φ) = !Syntax Error, Idφ !Syntax Error, Ir dr { σ !Syntax Error, I Ez(r,m) ejmφ } // (D.1.3) = σ !Syntax Error, I !Syntax Error, Ir dr Ez(r,m) !Syntax Error, Idφ ejmφ = 2π σ!Syntax Error, Ir dr Ez(r,0) = 2π σ!Syntax Error, Ir dr {-j(β'/βd) J0(x) } // (D.2.21) x = β'r = -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) ] = -j(β'βd)-1 2πσ {(jω/σ) N0 / J1(xa)} [ xa J1(xa) ] // (D.2.20) = (β'βd)-1 2πω N0 xa = (β'βd)-1 2πω N0 β'a = 2πω (a/βd) N0 so that N0 = (βd/2πωa) I . (D.4.1) At this point it is convenient to introduce the DC resistance per unit length of our wire, Rdc = (D.4.2) along with a new symbol to indicate the relative surface charge moment, ηm ≡ (D.4.3) It follows from (D.4.1) that the normalization factor appearing in (D.2.22) 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.4.5) We may now construct the final form for our solutions from (D.2.23) and (D.2.22) for m > 0: 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 [- ] 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 { + - } jEφ(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 { - + [ + ] } We summarize these results in a box. The m=0 results are all obtainable from the general expressions with m = 0 using J-1 = - J1 : E Fields Inside a Round Wire x = β'r β'2 = β2- βd2 ≈ β2 β = ω ξ =[ε + σ/(jω)] ≈ σ/(jω) conductor xa = β'a βd = ω ξd =[εd + σd/(jω)] ≈ εd dielectric E(r,φz,t) = ej(ωt-βz) E(r,φ) (D.1.2) E = Er + Eφ + Ez (for any arguments) E(r,φ) = E(r,0) + 2!Syntax Error, I[Re{E(r,m)}cos(mφ) - Im{E(r,m)} sin(mφ)] = real (D.1.4) where: Ez(r,m) = (1/4) ηm I Rdc [ - ] a = radius ηm ≡ Ez(r,0) = (1/2) I Rdc [] // = (2.1.22) for βd = 0 η0 = 1 Er(r,m) = (j/4) (aβd) ηm I Rdc [ + - ] Er(r,0) = (j/2) (aβd) I Rdc [ ] Eφ(r,m) = (1/4) (aβd) ηm I Rdc [ - + + ] Eφ(r,0) = 0 (D.4.6) Observations about the solution: (1) We looked for a wave solution inside a round wire in which phase fronts propagate down the wire (z direction) with angular frequency ω and wavelength λd = 2π/βd, as indicated by (D.1.2) in the box. We found the solution shown in the box. It satisfies all three components of the Helmholtz equation (D.1.1) as well as the div E = 0 equation (no charge inside conductor). (2) For a good conductor and a good dielectric, one has ξ ≈ σ/(jω) and ξd ≈ εd as shown in the box. These are the complex dielectric "constants" of Appendix B. The corresponding wavenumbers are then β = ω ≈ ω = ej3π/4 = ej3π/4 (/δ) (2.1.18) , (2.1.21) βd = ω ≈ ω = ω / vd vd = speed of light in the dielectric (D.4.7) Thus, in our wave solution (D.1.2), 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 silent about the fields outside the wire, it seems reasonable to presume there is a wave outside also moving down the wire at vd . See below. (3) The fact that the wavenumber β inside the conductor is complex indicates the presence of absorption of energy in the conductor due to σ which one could calculate in detail. (4) In our axially symmetric round wire analysis of Section 2.1 we assumed no z variation of the fields, which means wave number βd = 0 so β' = β. In this case the box above shows Er(r,m) = 0 and Eφ(r,m) = 0 for all m ≥ 0. Our symmetry assumptions in Section 2.1 limited the analysis there to m = 0 and the Ez(r,0) shown in the box above agrees with result (2.1.22). (5) The components Eφ and Er are smaller than Ez by factor (βda). This is the dimensionless smallness parameter which defines the "transmission line limit", see Chapter 4. (6) If there exist moments Nm of the charge distribution on the wire with m > 1, then the corresponding ηm ≠ 0 and it is clear that Ez(r,φ) and hence Jz(r,φ) is non-uniform around the surface of the wire. That is, these fields vary with φ as cos(mφ). 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(φ) when conductors are "fat and close". It seems likely in this case that the largest contribution will come from m=1. (7) The surface impedance from (C.2.1) is just Zs(φ) = Ez(r=a,φ)/I. Using the boxed result for Ez(r,φ) we quickly find that Zs(φ) = (1/2) Rdc { + !Syntax Error, I [ - ] Re[ηmejmφ] } where (D.4.8) Re[ηmejmφ] = Re(ηm)cos(mφ) - Im(ηm) sin(mφ). Thus we see the expected non-uniformity of Zz(φ) around the perimeter of the wire cross section for the m > 0 components. D.5 The Low Frequency Limit If we assume that ω is small enough that |xa| = |β'|a << 1, we can greatly simplify the above results. For a good conductor, β' ≈ β = ej3π/4 (/δ) so |β'|a << 1 => (a/δ) << 1 or (a/δ) << 1, which means the skin depth is much larger than the radius of the wire. In terms of frequency ω this condition is a << 1 => ω << . Using our ongoing example, for the Belden 8281 coax center conductor with a = .394 mm, μ = μ0 = 4π x 10-7 henry/m, and σ = 5.81 x 107 mho/m (copper), the limit requires that ω << = 88 KHz . All limits come from the leading small-x term of Jn(x) which is Jm(x) ≈ xm/ (2mm!) m ≥ 0 // Spiegel (24.2) (D.5.1) To avoid errors, we have Maple compute the limits of the various terms appearing in box ***. The left column is for visual check, the right column then provides the limits. Our limit is xa<<1 and since x ≤ xa we also have x << 1, but the ratio (x/xa) = (r/a) is merely ≤ 1 and not << 1. From the column on the right above we read off the limits: A = ≈ 2(m+1)(x/xa)m B = ≈ (1/2m)(x/xa)m xa2 ≈ 0 since xa << 1 C = ≈ (x/xa)m-1 D = ≈ (x/xa)m+1 E = ≈ (x/xa)m+1 xa2 ≈ 0 F = ≈ 2 G = ≈ (x/xa) (D.5.2) In terms of the letters A through G the E(r,m) fields from box *** are given by Ez(r,m) = (1/4) ηm I Rdc [A - B ] Ez(r,0) = (1/2) I Rdc [F] Er(r,m) = (j/4) (aβd) ηm I Rdc [ C + D - E ] Er(r,0) = (j/2) (aβd) I Rdc [G] Eφ(r,m) = (1/4) (aβd) ηm I Rdc [ - C + D + E ] Eφ(r,0) = 0 (D.5.3) Installing the limits and setting (x/xa) = (r/a) we find E Fields Inside a Round Wire (low frequency limit) Ez(r,m) = (1/2) ηm I Rdc (m+1)(r/a)m Ez(r,0) = I Rdc Er(r,m) = (j/4) (aβd) ηm I Rdc [(r/a)m-1 + (r/a)m+1 ] Er(r,0) = (j/2) (aβd) I Rdc (r/a) Eφ(r,m) = (1/4) (aβd) ηm I Rdc [- (r/a)m-1 + (r/a)m+1 ] Eφ(r,0) = 0 (D.5.4) The equation Ez(r,0) = I Rdc states that in this low frequency limit the wire acts as a resistor having the expected resistance per unit length Rdc = 1/(σπa2) as in (D.4.2). In the transmission line limit aβd << 1 of interest to us, we see that there is always a small radial current which pumps the changing surface charge. In the m = 0 this radial current is linear in r, peaking at the surface. In fact, if you integrate the radial current over the surface of a long piece of wire of length λ/ 2, taking into account the z dependence, you get 2I. [ I think the above has no basis in fact! At DC, we have ω = 0 and Er(a,0) = (j/2) (aβd) I Rdc. But βd = ω ≈ ω = 0 at DC, so the integral of the radial current is 0. For m = 1, the radial current depends on η1 which we don't know, and then Er(a,1) = (j/4) (aβd) ηm I Rdc(2) and we have no conclusion at all. Yes it is proportional to I. I don't know where this strange claim came from! Throwing in the sinφ angle and integrating over half a wavelength is not going to rescue this weird claim. ] See Section 3.7, Figure 2 for a drawing of the current we have just computed. continue here after reading Chapter 3. D.6 What about the E fields outside the round wire? This is an Exercise for the Reader, with some preliminary discussion given below. The Helmholtz equation (D.1.1) outside the wire now contains βd instead of β. The assumed wave solution form is still (D.1.2). In the three component Helmholtz equations this means that β' = βd2 - βd2 = 0 so that: [2E]z + β2 Ez = 0 : [r2∂r2 + r ∂r - m2] Ez(r,m) = 0 (D.1.16)o [2E]r + β2 Er = 0 : [r2∂r2 + r∂r - (m2+1)] Er(r,m) - 2jm Eφ(r,m) = 0 (D.1.17)o [2E]φ + β2 Eφ = 0 : [r2∂r2 + r∂r - (m2+1)] Eφ(r,m) + 2jmEr(r,m) = 0 (D.1.18)o div E = 0 : ∂r [r Er(r,m)] + jmEφ(r,m) -jβd r Ez(r,m) = 0 (D.1.19)o The differential operators appearing in the above equations are no longer Bessel-style operators, they are Euler-style operators. Euler ODE's 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, For Ez(r,m) the equation (D.1.17)o 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) = Azmrm + Bzmr-m m > 0 Ez(r,0) = Czln(r) + Dz m = 0 . This Ez Helmholtz equation is the same as the radial equation one obtains when solving the 2D Laplace equation 2u = 0 in separated polar coordinates, see for example Stakgold Vol II p 92 (6.7). It certainly seems reasonable that Azm = 0 to avoid a dramatic blowup at r = ∞, but the m = 0 coefficients are less obvious, since we have already seen logarithmic divergences associated with infinitely long wires. One approach is to mimic the method used for the inside solution, making use of the div E = 0 equation shown above, and thereby coming up with a full set of E field solutions. Along the way one must determine appropriate boundary conditions at r = a so the outside solutions are consistent with the inside solutions. We can at least see that for m > 0 Ez(r,m) ~ (r/a)-m outside the round wire. D.7 What about φ, A and B ? Inside the conductor, we know that potential φ satisfies (D.1.1), and we know that φ at r=a does not depend on angle φ. This means that φ is similar to our Ez solution in the m=0 partial wave, and vanishes for all higher partial waves. Thus we write φ(r,m) = δm,0 φ0 J0(x)/J0(xa) x = β'r (D.7.1) where φ0 is the value of the potential on the surface at r=a. Since we know E(r,m) from Section 3 (6), we can solve for the vector potential A as follows: -jωA = E + grad φ (D.7.2) Converted to partial waves, this says -jωAr(r,m) = Er(r,m) + δm,0 ∂rφ(r,0) -jωAφ(r,m) = Eφ(r,m) - (1/r) jm δm,0 φ(r,0) = Eφ(r,m) -jωAz(r,m) = Ez(r,m) - jβd δm,0 φ(r,0) (D.7.3) Thus, for m≠0 we have -jωA = E. For m = 0 there are extra pieces as shown for Ar(r,0) and Az(r,0). Since A(r,m) is known, A(r,φ,z) is also known, and then so too is B = curl A. This curl can also be performed in each partial wave if desired by replacing ∂φ = -jm and ∂z = -jβd as usual. Since -jωA = E in the higher partial waves, we may conclude that the transverse components of A are small compared to the longitudinal component, just as is the case for E, see (D.3.6). For m=0 we know that Aφ = Eφ = 0, but Ar is a combination of Er and ∂r φ which is not small compared to Az, due to the ∂rφ contribution. Thus, we arrive at the conclusion that, although Ar is negligible in the dielectric, it cannot be ignored inside the conductor. This explains, incidentally, the problem one encounters with the gauge condition (1.5.5) divA = -j(β2/ω)φ. Since φ is continuous at the boundary, and β2 takes a jump of many orders of magnitude (from βd to β'), something on the left side must change violently. But Az is also continuous. It is the m=0 Ar inside the conductor that takes up the slack. In fact, taking the difference inside minus outside we conclude that ∂rAr (r=a-ε) = -j(β'2/ω)φ0 (D.7.4) from which we can determine φ0The potential Ar is discontinuous at r=a. There is no more to this appendix.