transmission lines pre mu epsilon change REVIEWED

In the original doc my symbols ε and μ were "relative", but now they are "absolute". For example, previously μ = 1 in vacuo, but now μ = μ0 in vacuo. So here I have saved the entire doc before making this change. I guess most changes were of the form μμ0 → μ and εε0 → ε Top Chapter 1: Basic Equations. 2 1.1 Maxwell's equations in a medium. 2 1.2 The Field Wave Equations. 4 1.3 The Potential Wave Equations. 5 1.4 The Retarded Potentials: a path not taken. 6 1.5 The Potential Wave Equations in the Frequency Domain 7 Chapter 2: The Round Wire and Skin Effect 10 2.1 Derivation of E(r), B(r) and J(r) for a round wire. 10 2.2 A study of the solution of a round wire. 15 2.3 The Surface Impedance of a Round Wire 17 2.4 Surface Impedance for a Transmission Line 19 Chapter 3: Preliminaries 22 3.1 Why is there no charge inside a conductor? 22 3.2 How thick is the surface charge layer on a conductor? 22 3.3 How does loss tangent affect dielectric conductivity? 23 3.4 Normal (radial) current conservation at the surface of a conductor. 24 3.5 The TEM mode fields and currents for an IDEAL transmission line . 26 3.6 The TEM mode fields and currents for a REAL transmission line . 28 3.7 The general shape of fields, charges, and currents on a transmission line. 32 3.8 Transmission Line Preliminaries 37 Chapter 4: Transmission Line Equations 40 4.1 Computation of φ due to one conductor of a transmission line. 40 4.2 Computation of V(z) . 44 4.3 Computation of z due to one conductor of a transmission line. 46 4.4 Computation of W(z) . 47 4.5 The Classic Transmission Line Equations. 48 4.6 Example: the wide-spaced, two-wire transmission line. 51 4.7 Example: the coaxial cable. 55 Chapter 5: The Transverse Problem 57 5.1 Philosophy 57 5.2 The Helmholtz Equations and Separation of Variables 59 5.3 A Formal Solution to the Transverse Problem 63 5.4 . An approximate solution to the transverse problem. 67 Chapter 6: An Example 70 6.1 Why logarithms? 71 6.2 The equipotentials of ln[s2/s1] are circles. 72 6.3 Aligning the circles. 73 6.4 Reduction to special cases 76 Appendix A 1.1: Gauge Invariance 81 Appendix B 1.2: The Complex Dielectric Constant. 90 1. A conducting parallel plate capacitor. 90 2. Continuity and free charge in a medium 91 3. Complex ε and the loss tangent of a dielectric. 92 Appendix C 2.1: DC Properties of a Wire 93 1. The DC resistance of a wire. 93 2. The DC surface impedance of a wire. 93 3. The DC inductance of a round wire. 93 3.1 Internal inductance 93 3.2 External inductance 94 4. The DC inductance of a wire of arbitrary cross section 95 4.1 Statement of a Plan of Attack 95 4.2 The divergence problem 96 4.3 Avoiding the divergence problem. 97 Appendix D 2.2: Electric Field in a Round Wire 99 1. The General Method and Solution for Ez 99 2. The Solutions for Er and Eφ 101 3. Statement of the Results 104 4. The Low Frequency Limit 106 5. What about φ, A and B ? 108 Appendix E 3.1: Surface Charge 110 Appendix F 3.2: Waveguides 113 1. A waveguide solution. 113 2. A waveguide interpretation. 115 Appendix G 4.1: Chapter 4 Support 117 1. Evaluation of the integral in 4.1 (9). 117 2. Examination of higher order terms in 4.2 (2). 118 Chapter 1: Basic Equations. In this chapter we state or derive all basic equations that will be be used later in the calculation of transmission line parameters. Appendices 1.1 and 1.2 contain information that is relevant to this chapter. 1.1 Maxwell's equations in a medium. All equations in this document are expressed in SI units, formerly known as rationalized mks units. The first four equations below are Maxwell's equations, though we are supposed to now call them "the Maxwell equations" in a manner similar to "the Bessel functions". These equations may be found for example in Jackson page 2 (1.1a). Our equation of continuity (1.1.7) appears in Jackson as page 2 (1.2). Jackson's symbols J and ρ correspond to our JT and ρT as explained below (T = Total). We always assume isotropic media. We have assumed the validity of Ohm's Law J = σE for the conduction current J. This law appears in Jackson p 219 (5.159) and this law is included in our (1.1.1) where JT = J + Ja = σE + Ja. Whereas Jackson writes D = εE and B = μH at the top of page 296 , we write D = εε0E and B = μμ0H so that our ε and μ symbols are dimensionless. curl H = ∂D/∂t + JT = ∂D/∂t + [ σE + Ja ] (1.1.1) curl E = - ∂B/∂t (1.1.2) div D = ρT = [ ρ + ρa ] (1.1.3) div B = 0 (1.1.4) B = μμ0H (1.1.5) D = εε0E (1.1.6) Notes: (a) The equations above are all expressed in SI units, formerly known as rationalized mks units (a) JT refers to the total current density. Anticipating our first application, we have decomposed JT into a sum of two terms. The term J = σE is the "conduction current". Wherever there is an E field in a conducting medium, there is such a current. This is Ohm's Law, so we are assuming we are operating in a regime where Ohm's law is applicable. The term Ja is meant to represent some sort of "externally applied" current, such as the current in an antenna the generates radiation, or the current in the conductor of a transmission line. (b) Similarly, ρT is the total charge density. We can make a corresponding decomposition of the charge density into a part which corresponds to the conduction current, and an applied part, although this is not particularly useful. Again, the charge density ρa represents an "externally applied" charge density such as might lie on the surface of a radiating antenna, or on the conductors of a transmission line. (c) Inside a medium such as a dielectric or a conductor, and at frequencies of interest to us, there can exist no net charge densities, so ρ = ρT = ρa = 0. In a dielectric there are no available charges. In a conductor, any departure from neutrality would be instantly restored. All charge densities for our application reside on the surfaces of conductors only. If we were interested in the behavior of a transmission line embedded in an electron plasma, things would be different. (d) The notion of "applied" sources like Ja and ρa is useful in approximate solutions to problems where the external sources are regarded as being somewhat removed from the control of Maxwell's equations. This is discussed more in Section 5.1. (e) The "equation of continuity" expresses the fact that charge cannot be created or destroyed. It has the following differential form: div JT = - ∂ρT/∂t (1.1.7) In regions away from the conductors which are regarded as supplying Ja and ρa, we have div(J) = div(σE) = - ∂ρ/∂t (f) We will be deriving many equations based on the above set. In any such derived equation, we can at any time choose to set σ = 0 and replace Ja with JT . This fact follows from (1.1.1). (g) In general, the quantities μ and ε can be functions of space and time. In this entire document, we shall assume the use of "isotropic media", which means that μ and ε are only functions of time. When we transform to the frequency domain, this means we will have μ(ω) and ε(ω) depending on frequency. Since μ and ε are assumed to be constant in space, these quantities pass through operators like div, grad and curl just as ordinary numerical constants. (h) The above equations are written in Système Internationale (SI) units. In this system, formerly known as "rationalized m.k.s.", the speed of light is concealed in the symbols μ0 and ε0. Here are the usual historical names given to the symbols appearing in these equations, along with one expression of the SI units for each symbol: E = electric field (volts/m) H = magnetic field (amps/m) D = electric displacement (coulomb/m2) B = magnetic field (tesla = amp-henry/m2) J = current density (amps/m2) ρ = charge density (coulombs/m3) σ = conductivity of the medium ( mho/m) μ = relative magnetic permeability of the medium (dimensionless) ε = relative dielectric constant (dimensionless) μ0 = permeability of free space = 4π x 10-7 henry/m ε0 = permittivity of free space = 8.8541877 x 10-12 farad/m Here are some related facts, henry-farad = sec2 farad = sec/ohm henry / farad = ohm2 henry = ohm-sec c = 1/= 2.9979246 x 108 m/sec = speed of light Z0 = = 376.73032 ohms = impedance of free space 1.2 The Field Wave Equations. If one applies the usual vector identity curl curl F = grad divF - 2F twice to Maxwell's equations above, one quickly ends up with the damped wave equations for E and H as follows: ( 2 - - σ μμ0) H = -curl Ja (1.2.1) ( 2 - - μμ0) E = (1/εε0) grad ρT + μμ0 (1.2.2) If one assumes exp(jωt) time dependence, the operators on the left become ( 2 + β2 ) as described below. Clearly the applied current Ja drives the magnetic field H, while a combination of the total charge density ρT and Ja drives the electric field E. Away from boundaries, the above equations are damped wave equations for the E and H fields, and the right sides vanish. One usually solves one of these equations for its field, then uses Maxwell's equations to get the other field, and one ends up with the usual notion of an open-medium wave propagation which has an implied relative orientation between the two field vectors and the direction of propagation. Here we have the added the complication of a lossy medium due to σ 1.3 The Potential Wave Equations. Although the fields in (1.2.1) and (1.2.2) are decoupled into separate equations, one for H and one for E, things are still quite complicated due the right hand sides. For this reason, it is sometimes better to start all over from Maxwell's equations above, replacing the B and E fields with the potentials A and φ according to the following rules: [ reference for second equation below? ] B = curl A E = - grad φ - ∂A/∂t (1.3.1) When this is done, we end up with the following two equations: 2φ + = - ρT/εε0 (1.3.2) 2A - - σ μμ0 - grad [ div A + σ μμ0 φ+ ] = - μμ0 Ja (1.3.3) At first glance, this looks like a horrific mess. Although the right hand sides are simpler than in (1.2.1) and (1.2.2), we have both potentials tangled together in both equations. At this point we pause to comment on the results of Appendix 1.1 on gauge invariance. The first thing shown there is that it is permissible to replace fields E and B with the potentials A and φ, according to (1.3.1) above. It is not really obvious that this is possible. Second, it is shown that the potentials A and φ can be selected such that div A equals any function one wants. This freedom in choosing the potentials is loosely called gauge invariance. Looking at (1.3.3) above, we can kill the entire term in square brackets by making this gauge choice for div A: div A = - σ μμ0 φ - (1.3.4) This is seen then to have the added benefit of removing A from (1.3.2) and replacing it with something that gives us two completely symmetrical decoupled equations, to wit, ( 2 - - σ μμ0) A = - μμ0 Ja (1.3.5) ( 2 - - σ μμ0) φ = - (1/εε0)ρT (1.3.6) If σ = 0, the gauge choice (4) matches the Lorentz gauge mentioned in Appendix 1.1. Thus, we shall refer to (4) as a "modified Lorentz gauge". [ except we have μ and ε here and not in the appendix.] Special Relativity Note. At first encounter, one is amazed at how similar these two equations appear. Here we shall show that it can be no other way. As noted in comment 1.1(f), we can set σ=0 and replace Ja by JT in the above equations without changing anything to get, ( 2 - ) A = - μμ0JT (1.3.7) ( 2 - ) φ = - (1/εε0)ρT (1.3.8) If we define the speed of light in the medium to be v, then (1/v2) = μμ0εε0 . We can then write the second equation as ( 2 - )[φv] = - μμ0 [ v ρT ] (1.3.8)' If you made it to the end of Appendix 1.1, you saw that A0 = [φv] and JT0 = [vρT] are the first components of Lorentz 4-vectors Aμ and JTμ. Moreover, the operator in (8) can be written in terms of the 4-gradient ∂μ such that ( 2 - ) = ∂μ∂μ ≡ ∂2. Thus, (1.3.7) and (1.3.8)' are both parts of the same equation, ∂2Aμ = - μμ0 JTμ (1.3.9) Both sides of (1.3.9) transform as Lorentz 4-vectors, since ∂2 is a Lorentz scalar. Even in 3-space, you are never allowed to have an equation in which different components of a vector transform differently. For example, in F = ma, both sides transform as rotational 3-vectors, and mass m is a rotational scalar. Eq.(1.3.9) is like Fμ = maμ. Equations (1.3.7) and (1.3.8)have the same form because they are really components of the same equation when viewed in the light of special relativity. [ needs clean up ] 1.4 The Retarded Potentials: a path not taken. In simple situations it is possible to solve the above wave equations for φ and A directly in the time-domain. For example, if σ = 0 ( so Ja = JT) and μ and ε are not time dependent (eg, μ = ε = 1 as in free space) one can define a Green's Function G by ( 2 - ) G(x,t;x',t') = - 4πδ(3)(x-x') δ(t-t') (1.4.1) [ In light of our previous note, this is ∂2G(xμ; x'μ) = - 4π δ(4)(xμ-x'μ).] The solution of (1.4.1) is G(x,t;x',t') = (1/R) δ( t' + (R/v) - t ) (1.4.2) where R = |x-x'|, and v = c/= the phase velocity in the medium . This lets one write down the solution to either potential equation as follows: φx,t = (1.4.3) A(x,t) = (1.4.4) These expressions for the potentials are typically used in radiation problems to find the fields of accelerating particles or antennas. Notice that the time in the sources is "retarded" by R/v. With the medium σ loss term present, and with frequency dependent (hence time dependent) μ and ε, it is better to work in the frequency domain instead of the time domain. Thus, although they are interesting, we shall not use these retarded potentials. 1.5 The Potential Wave Equations in the Frequency Domain We now assume exp(+jωt) time dependence on all currents and fields, and we thereby project the whole problem into the ω domain. The wave equations (1.3.5) and (1.3.6) then turn into time-independent "Helmholtz" equations: ( 2 + β2 ) A = - μμ0Ja (1.5.1) ( 2 + β2 ) φ = - (1/εε0)ρT (1.5.2) where β ≡ ω (1.5.3) with ξ ≡ [ εε0 + σ/(jω) ] . (1.5.4) The quantity ξ is the complex dielectric constant discussed at length in Appendix 1.2. Quantity β is then the complex wavenumber, more on β later. Meanwhile, the "modified Lorentz gauge" condition (1.3.4) above now appears as div A = - j (β2/ωφ (1.5.5) The next step is to solve the above equations for φ and A. As before, we first define the Green's Function, (2 + β2)G(x,x') = - 4πδ(3)(x-x'). (1.5.6) The solution is G(x,x') = R = |x - x'| (1.5.7) and so the solutions for the potentials are given by: A(x) = (1.5.8) φ(x) = (1.5.9) where R = |x - x'|. For transmission line problems involving a dielectric driven by an applied charge and current density, we can set ρT = ρa in (9), since ρ = 0 . See note 1.1(c) above. See Appendix 1.1 for an explanation of Green's Functions and why the above integral expressions solve (1.5.1) and (1.5.2). Finally, it should be kept in mind in (1.5.8) and (1.5.9) that time t has been replaced with frequency ω and everything is implicitly a function of ω. To emphasize this fact, we rewrite the above one more time as: A(x,ω) = φ(x,ω) = King uses these last two integrals extensively to compute the parameters of various transmission line geometries. The basic idea is simple. One assumes some distribution of charges and currents on the conductors, then one carries out the above integrations to compute the potentials. From these one gets the E and B fields using (1.3.1). Then one knows everything there is to know. In practice, various assumptions must be made, and even then the integrals are no cakewalk. We shall carry out this program in Chapter 4, after the Preliminaries of Chapter 3. Chapter 2: The Round Wire and Skin Effect Chapter 1 dealt with the generalities of electromagnetic theory. Maxwell's equations were stated ex machina, as it were, and wave equations for the fields and potentials were then derived. Formal integral solutions of the potential wave equations were also derived using the Green's Function method. It was noted that the potentials φ and A are parts of the same Lorentz 4-vector. Whereas the approach of Chapter 1 was very general and abstract, the discussion of this chapter is highly specific. The goal here is to learn about the behavior of a very simple device -- a uniform round piece of wire. If very little of Chapter 1 made sense to the practical reader, we think this chapter will be somewhat different. Although transmission lines are not always made out of round wires, there is a wealth of useful practical information that arises from the study of this simple example which applies to arbitrary geometries. The major issue here is called the "skin effect". At high frequencies, current is forced away from the central regions of a conductor and concentrates at the surface in a thin layer that has a characteristic depth called δ, the skin depth. In this chapter it will be shown exactly why this occurs. The significance of the effect is that the resistance (or impedance) of a wire increases drastically at high frequency. In the context of a transmission line, this effect is "felt" through a property of the wire called its surface impedance.. Our development is an extension of the excellent discussion of Matick Chapter 4. It is fastest to solve the round wire problem starting with the general wave equation (1.2.2) for the E field in the absence of "applied" sources. Instead, we have chosen to start from the basic Maxwell curl equations and use simple "loops" to derive the basic equations. One sees better in this way the underlying "cause" of the exclusion of current from a conductor. Moreover, the general technique of putting "loops" in opportune places is extremely useful in analyzing the more complicated situation which arises in a transmission line. 2.1 Derivation of E(r), B(r) and J(r) for a round wire. Here are Maxwell's curl equations along with corresponding integral form statements. The usual exp(+jωt) time dependence has been assumed: curl B = μμ0 (jωεε0 E + J ) ∫B•ds = μμ0 ∫(jωεε0E + J )•dA (2.1.1) curl E = jωB ∫E•ds = -jω∫B•dA (2.1.2) The ds integration is a line integral around some open loop, and the dA integration is over a surface which spans the loop. The two terms on the right side of (2.1.1) have names: jωεε0 E = displacement current (2.1.3) J = σE = conduction current The sum of both currents may be written as ( jωεε0 + σ) E . (2.1.4) For any metal conductor such as copper, the displacement term is completely negligible as long as ωεε0 << σ. Copper has ε = 1 and σ = 5.81 x 107 mho/m, ε0 = 8.85 x 10-12 F/m, so the condition is f << (σ/2πεε0 ) = 1.04 x 1018 Hz ≈ one billion GHz Therefore, the displacement current is always ignored inside a conductor for any transmission line application. At this point, we make the assumption that (a) our round wire is perfectly symmetric, and also that (b) the fields and currents are also symmetric. By "symmetric" is meant that the current flow is longitudinal in the wire and has no azimuthal variation. The electric field has the same shape, since J = σE. The B field goes around in circles centered at the wire axis. The relation between the direction of B and the current flow J is given by the right hand rule. Here is a another way to state assumption (b). We search for a symmetric solution of Maxwell's equations for the round wire, and if we find one, we accept it as a possible way fields and currents could exist in the wire. In Appendix 2.2 we attack the much more formidable problem of finding the general solution for the E field and current in a round wire, subject to transmission line boundary conditions. Such solutions allow the possibility of azimuthal dependence. Consider now the loop shown in Fig 1: // picture is damaged d According to (2.1.1) with J = σE and no displacement current, ∫B•ds = (μμ0σ ∫E•dA (2.1.5) The two sides of this equation can be easily evaluated for the loop of Fig. 1: [ B(r+dr) (r+dr) - B(r) r] θ = (μμ0) E(r) [ rθ dr ] (2.1.6) STOP. What is the meaning of B(r) and E(r) here? Unclear. Perhaps B(r) = Bφ(r) ? Yes, because earlier I said that B goes in circles around the wire. What about E(r)? This must be Ez(r). I need to make these facts clear! which simplifies to = (μμ0σr E(r) ] (2.1.7) Now consider the loop shown in Fig. 2: According to (2.1.2), ∫E•ds = -jω∫B•dA The two sides of this equation can be easily evaluated for the loop of Fig. 2: [ - E(r+dr) + E(r)] s = -jωB(r) [ s dr ] (2.1.8) which simplifies to = jωB(r) (2.1.9) This equation is worth interpreting. It says that the longitudinal electric field in the conductor must vary with radius, unless ω = 0. Since J(r) = σE(r), this means that the current must change as a function of radius. From (2.1.9) alone one cannot know which way it changes, but it must be changing. This fact is what is going to lead to the skin effect! Now solve (2.1.9) for B(r) and put this into (2.1.7) to get = (jωμμ0σE(r) (2.1.10) The operator on the left is 2 in cylindrical coordinates for a function that does not depend on φ or z. Thus, (2.1.10) is really a special case of the following [ 2 - (jωμμ0σE(r) (2.1.11) This in turn is a special case of the damped wave equation (1.2.2), which we can write as [ 2 + β2 E = 0 (2.1.12) where β2 = ω2 μμ0ξ= ω2 μμ0 [ εε0 + σ/jω] (2.1.13) and we have neglected the first displacement current term in β2. β = 2π/λ is the wavenumber and has dimensions of m-1. It is "radians of wave per meter". We could have started out with (2.1.12) and skipped all the above analysis of loops, but this method of using loops is extremely useful for understanding more complex situations. The next step is to expand (2.1.10) as follows: + + β2E(r) = 0 (2.1.14) Change variables to x = βr, which is dimensionless. Then E(r) = E(x/β) , so that ∂E/∂r = β∂E/∂x . Thus (2.1.14) becomes, x2 + x + x2f(x) = 0 (2.1.15) where f(x) = E(x/β). Now (2.1.15) happens to be Bessel's Equation with n=0, and the solution must therefore be, f(x) = C J0(x) + DY0(x) (2.1.16) So far, we still do not know which way ∂E/∂r in (2.1.9) is changing, but we are about to find out. Since f(x) represents the current and the electric field, we know f(0) cannot be infinite. But Y0(x) blows up at x=0. Therefore constant D = 0. We now have an exact solution for the electric field in the wire: E(r) = f(x) = C J0( βr) (2.1.17) where from (2.1.11) and (2.2.12), β = (2.1.18) Notice that β is at +135° in the complex plane. The following definition is usually made δ ≡ (2.1.19) so that β = (/ δ ) e[ (3/4)πj ] = (/ δ ) β2 = -2j/ δ2 (2.1.20) It is convenient to divide (2.1.17) by itself evaluated at r=a which we shall assume is the radius of our round wire, so E(r) = E(a) [J0(βr) / J0(βa)] (2.1.21) This equation tells us how E(r) decreases as we move away from the outer surface of the wire. Again, since J(r) = σE(r), this is a description of the skin effect. From (2.1.9) and (2.1.21) we can easily find the B field, B(r) = (β/jω) E(a) [J0'(βr) / J0(βa)] (2.1.22) since ∂J0(βr)/∂r = β∂J0(x)/∂x = βJ0'(x). An alternate form results from dividing (2.1.22) by itself evaluated at the wire surface, so B(r) = B(a) [J0'(βr) / J0'(βa)] (2.1.23) It happens that J0'( x) = -J1(x), so the above results can be restated as B(r) = - (β/jω) E(a) J1(βr) / J0(βa) = + B(a) J1(βr) / J1(βa) (2.1.24) Let us gather up all the main results obtained so far and put them in a box: (maybe a plot here? ) Solution of a Round Wire = (μμ0σr E(r) ] = jωB(r) E(r) = E(a) [J0(βr) / J0(βa)] J(r) = J(a) [J0(βr) / J0(βa)] B(r) = - (β/jω) E(a) [J1(βr) / J0(βa)] = + B(a) [J1(βr) / J1(βa)] β = / δ ) e[ (3/4)πj ] = (/ δ ) δ ≡ β2 = -2j/ δ2 (2.1.25) 2.2 A study of the solution of a round wire. The real and imaginary parts of a Bessel function having an argument with phase (3/4)π have the following historic names (Bessel real and Bessel imaginary) called Kelvin functions: Jn( z) = Jn( e[ (3/4)πj ] z) = Bern(z) + j Bein(z) (2.2.1) Thus, the above solution E(r) may be written as, E(r) = E(a) z = (r/δ) (2.2.2) Since J(r) = σ E(r), we could replace E with J on both sides of (2.2.2). Since these Bessel forms occur a lot, there are standard functions for the magnitude and phase, Jn( z) = Mn(z) ejθ(z) (2.2.3) Of particular interest is the magnitude of E(r). Applying (2.2.3) to the boxed E(r) gives |E(r)| = |E(a)| (2.2.4) A log plot of M(z) = M0(z) appears in Abramowitz and Segun, here is a rough sketch: (Maple plot!) Figure 1. Plot of ln M0(z) versus z. For argument larger than about 3, the log plot is linear. In this range, one can use the following large z formulas (AS 9.10.21 and 9.10.22) Mn(z) ≈ [ 1 - + O(1/z2) ] θn(z) ≈ (z/) + (π/2) [ n - 1/4 ] + + O(1/z2) (2.2.5) which implies that ln Mn(z) ≈ (1/) z = .707 z z > 3 (2.2.6) so the slope of the above plot in the linear region is .707. In general, then, we can use the exponential formula (2.2.5) except within a few skin depths of r=0, and in that range the function tapers off smoothly to the very small value of 1, since J0(0) = 1. This value is miniscule compared to the value that the growing exponential has a few skin depths away from r=0. So inserting (2.2.5) into (2.2.4) gives, = e(r-a)/δ r/δ > 3/= 2.1 (2.2.7) This is the famous skin depth result as it appears for a round wire. This fraction is 1 at the surface, and then drops off exponentially with characteristic distance δ. One sees now why the was included in the definition of δ. The factor has negligible variation compared to the exponential. Equation (2.2.7) is valid down to within 2 skin depths of the center line of the wire. In general, one can assume the field E(r) is zero for all practical purposes a few skin depths in from the surface. Here are some skin depth values in copper based on δ, σ = 5.81 x 107mho/m, and μ = 1, μ0 = 4π x 10-7 H/m. Selecting a reference point of 1 GHz, we have, δ = = 2.09 μ x (2.2.8) The radius of the center conductor of Belden 8281 coax is 15.5 mil = 394 μ. Here is a little table then of copper skin depths, f δ  f δ 10 GHz 0.66μ 1 KHz 0.21 cm 1 GHz 2.09μ 100 Hz 0.66 cm 100MHz 6.61μ 10 Hz 2.09 cm 10 MHz 20.9μ 1 Hz 6.61 cm 1 MHz 66.1μ 100 KHz 209μ 10 KHz 661μ As we get into the lower frequencies, the exponential decay no longer applies for Belden 8281 . For very low frequencies, we can use the small z limit of J0(z) to see how the distortion begins at low frequency, J0(z) = 1 - z2/4 z << 1 (2.2.9) Using the expressions for E(r) and β2 appearing in the above box, we get = = (2.2.10) Here we see the very early phase of the skin effect happening at low frequencies. This would apply for example in Belden 8281 at 1 KHz and below. There is a very slight dip in the E(r) and J(r) distribution at r=0 compared to r=a. The shape of the magnitude is a quartic in r. (Plot?) 2.3 The Surface Impedance of a Round Wire A piece of round wire can be thought of as a resistor. Consider Fig. 1: Here our piece of "imperfect" wire is attached to a pair of "perfect" contacts having σ = ∞. The total impedance of the wire is then determined by Z = V/I. Alternatively, one could probe the wire along its surface as shown by the two arrows separated by ∆z. There is some voltage between the probes due to the field Ez(a) at the surface of the wire. By definition, the surface impedance per unit length is Zs ≡ (- ∆V/∆z)/I = Ez / I ohms/m (2.3.1) Since the fields and currents derived under the assumptions of Section 2.1 vary only with r, for our resistor we get Z = ZsL (2.3.2) In a more complex situation, where J = J(r,φ,z), the impedance measured by the two probes spaced a small distance dz apart would be a function of φ and z, giving Zs(φ,z) as the surface impedance function of a round wire. The total wire impedance Z is still V/I and is a constant and cannot depend on φ and z. The surface impedance is a more local concept, and one that plays a role in transmission line attenuation. The general relation between Z and Zs follows from (2.3.1) which says that dV(z) = - Zs(φ,z)I(z) dz. Integration gives, Z = V/I = (1/I) !Syntax Error, Idz [ - Zs(z,φ)/ I(z)] (2.3.3) If the two endplates are perfect conductors, then they are equipotential surfaces, so then V cannot be a function of φ. One can select an arbitrary φ to make the above computation. I(z) is the total current in the wire at position z which in general can be non-constant. To compute the surface impedance of our round wire, we have to make a connection to the total current I in the wire. This time, our "loop" is a circular belt lying on the wire surface. We apply (2.1.1) on this loop to get: 2πaB(a) = μμ0 I (2.3.4) Thus, from our Zs definition (2.3.1), Zs = E(a)/I = ( μμ0 /2πa) E(a)/B(a) = (1/2πa) E(a)/H(a) (2.3.5) Looking at the B(r) equation in the box (2.1.25), we may write B(a) = - (β/jω) E(a) [J1(βa) / J0(βa)] (2.3.6) From (2.3.6) we have the ratio E(a)/B(a) needed for Zs, so Zs(ω) = (-jωμμ0/ 2πaβ) [J0(βa)/ J1(βa)] (2.3.7) where we emphasize the fact that Zs is a function of frequency. Equation (2.3.7) is, as expected, rather complex. We certainly expect Zs(ω) to increase with frequency since the current density is forced out to the surface by the skin effect. As usual, we can look at the small ω and large ω limits of (2.3.7), and expect the two limits to be smoothly joined together by the actual function. (Plot?) To get the low frequency limit long wavelength limit, λ is large and β is small, so we expand both Bessel functions for small argument: J0(x) ≈ 1 - x2/ 4 J1(x) ≈ (x/2)(1 - x2/8) 1/J1(x) ≈ (2/x) (1 + x2/8) (2.3.8) Using these in (2.3.7) along with the fact that ωμμ0 = 2/(σδ2), we find Zs(ω) = (1/σπa2) + jω (μμ0 / 8π) (2.3.9) The first term is the DC resistance of the wire, normally written ρ/A, see (C.2.3) of Appendix 2.1. The second term is jω times the DC internal inductance Li = (μμ0/8π) H/m, as derived in 3 (5) of Appendix 2.1. Recall that this is exactly 50 nH/m, quite small, and independent of radius. To get the high frequency limit of (2.3.7) we apply (2.2.3) to get Zs(ω) = (-jωμμ0/2πaβ) [ M0(a/δ) / M1(a/δ) ] exp[ j{θ0(a/δ) - θ1(a/δ)}] (2.3.10) According to (2.2.5) we find that M0(z) / M1(z) = [ 1 + + O(1/z2) ] θ0(z) - θ1 (z) = - [ (π/2) + + O(1/z2) ] (2.3.11) Insertion of these large argument formulas into (2.3.10) gives Zs(ω) = [ 1 + δ/(16a) ] e -j(π/2) e-j(δ/16a) (2.3.12) Combining all powers of j (except the last factor) results in ej(π/4) = [ 1 + j ] /  . Thus, Zs(ω) = { [ 1 + δ/(16a) ] e-j(δ/16a) } (2.3.13) The factor in { } brackets is the first order correction term. For δ << 16a, { } = 1 and we get the famous result, Zs(ω) = (2.3.14) Writing this as the sum of a resistive and inductive part, Zs(ω) = Ri(ω) + jω Li(ω) (2.3.15) we find Ri(ω) = = ω Li(ω) = XL(ω) (2.3.16) Li(ω) = but this is a different Li ? (2.3.17) The resistance has a simple interpretation. It is R = 1/σA where area A = (2πa)δ . This is the area of a thin washer at the periphery of the wire of thickness δ. The inductance is harder to understand. Its origin is best traced back to (2.3.5) above which shows that the phase of Zs is equal to the phase of the ratio E(a)/B(a). It is a result of Maxwell's curl equations that this phase is 45° at the surface of a conductor in the skin effect limit. The inductive reactance is the same as the resistance, but the inductance itself increases as frequency decreases, behaving as L ~ 1/. 2.4 Surface Impedance for a Transmission Line What is the surface impedance of an arbitrary conductor? As we have seen, a large amount of work is needed to obtain the exact result even for the simple geometry of a round wire. Once can repeat this calculation for other geometries, such as a stripline. The general nature of the result is always the same, when δ is much smaller than the depth of the conductor. That result is this: Zs(ω) = (2.4.1) where D is the effective distance around the cross sectional surface of a conductor where significant current flows. For the round wire this was D = 2πa, the circumference. For a thick stripline of width w, D = w. Consider these two possible transmission line cross sections: Figure 1 : stripline Figure 2: twin lead In both cases we assume a frequency ω such that skin depth δ is small compared to the thickness of the conductors. Although the total cross sectional perimeter of one of the stripline strips is 2w + 2t, it seems clear that the length of the "active surface" is only w, and one sets D = w in the surface impedance formula. For the twin lead case, assumed far apart, both conductors are immersed in roughly uniform active fields, so the full 2πa is applicable. As the two round wires are brought very closer together, certainly there will develop an asymmetry so that the currents are largest on the parts of the wires closest to the other wires. In this case, one must make an estimate of the "effective distance" . Here is a picture, Here we have indicated a graphical estimate of the "active region" of current flow. King (p 30) quotes an approximate surface impedance result for the case of Figure 3. The effective distance is equal to, D = 2π a (2.4.2) If the gap between the conductors is (1/6)a, a rough estimate for Fig 3, then the radical in this formula becomes .38, so the dark lines shown should cover 38% of the circumference. If the conductors almost touch, then D becomes extremely small. King makes the interesting remark in his 1955 (updated 1965) book that "accurate formulas for the internal (i.e., surface) impedance of one cylindrical conductor in the presence of another with different radius are not available." We think that such formulas can be derived with moderate effort using the Zs formula given in Appendix 2.2 equation 3 (8) together with the potential φ given in Section 6.3, box (13) . In general, the high frequency skin current will be large where the E and B fields are large. These fields are large where the electric field would be large in a capacitor whose "plates" are the two conductors in cross section. There is an interesting transmission line "paradigm shift" which occurs as one moves from the low frequency domain to that of high frequency. For small ω, one thinks of the current in the two conductors of a transmission line as being there because they are "applied" by some external agency. The current then creates a B field around each wire. In the high frequency skin-effect limit, it is easier to think of the currents in the conductor surfaces as being generated by the field activity near the surfaces. The E and B fields just outside the conductors force themselves slightly into the surface. The resulting E field in the surface layer is then what creates the current. Chapter 3: Preliminaries 3.1 Why is there no charge inside a conductor? There can be charge inside a conductor, but only if the conductor is excited at an extremely high frequency. A charge density ρ inside the conductor implies an electric field E according to div E = ρ/εε0. This E field then causes a current J = σE which attempts to drain the charge off to the surface. Here is a simple way to estimate the time constant for this process. Imagine a one dimensional conductor with some charge density ρ inside. Then from div E = ρ/ε0 we get ∂Ex/∂x = ρ/ε0 (3.1.1) The electric field so generated causes a current J = σ E, so ∂Jx/∂x = σ ∂Ex/∂x (3.1.2) But this drains the charge away according to div J = -∂ρ/∂t, which is Jx ??? ∂Jx/∂x = -∂ρ/∂t (3.1.3) Combining these three equations we get ∂ρ/∂t = - (σ/ε0) ρ (3.1.4) This implies that ρ decays exponentially with time constant T = ε0 / σ (3.1.5) For copper, σ = 5.81 x 107 mho/m, and ε0 = 8.85 x 10-12 F/m, so T = 1.52 x 10-19 sec. Thus, any process which occurs inside a conductor at a frequency much less than 1019 Hz always allows plenty of time for any interior charge to move to the surface. This is 1010 GHz, far beyond the operating frequency of a transmission line. We may therefore conclude that: Fact 1: In a transmission line, charge exists only on the surface of conductors. 3.2 How thick is the surface charge layer on a conductor? This is a fascinating subject and the interested reader will find an analysis in Appendix 3.1. It turns out that the charge density decays exponentially away from the surface into the conductor and drops to 1/e of its surface value at a distance called the Debye length. This is the distance over which any charge is "screened" inside a conductor. For copper, this distance is roughly 0.55A (Angstroms) , which is 5.5 x 10-11 m. The crystal spacing for copper is 3.6A, and the copper atom radius is about 1A. Thus, Fact 2: The thickness of the surface charge density on the surface of a conductor is incredibly small. For copper, it is less than the radius of one copper atom, and the general result applies to any metal. In Section 2.2, we noted that the skin depth δ for copper at 100 GHz is about 0.2 microns which is 2x10-7 m = 2000A. Even at this huge frequency, the skin depth is still about 4000 times larger than the thickness of the surface charge layer. At 1 GHz this ratio is 40,000. Fact 3: Whereas current can exist "deep" under the surface of a conductor, even when the skin effect is dominant, the surface charge can always be thought of as being exactly on the surface. 3.3 How does loss tangent affect dielectric conductivity? The total current in a dielectric may be written, Jtot = [jωεε0 E + σE ] (3.3.1) The first term is the displacement current, and the second term is the conduction current. At high frequencies (say 1 GHz) , the dielectric constant ε acquires a small imaginary part due to the presence of absorption resonances at much higher infrared frequencies. One can write , ε = ε' - jε" = ε' [ 1 - j (ε"/ε' ] = ε' [ 1 - j tanL ] (3.3.2) The loss tangent is the ratio of the small imaginary part to the dominant real part of ε. It is sometimes referred to as the dissipation factor. When this expression is inserted into (3.3.1) the result is, Jtot = [jωε'ε0 E + ( σωε' ε0 tanL) E ] (3.3.3) In effect, the dielectric has now acquired an effective conductivity, σeff = ( σωε' ε0 tanL) (3.3.4) Because the DC conductivity of a good dielectric is so small, the loss tangent contribution to σeff dominates even at quite low frequencies. For polyethylene, for example, we have these numbers, σ ≈ 10-15 mho/m ε' = 2.26 tanL ≈ 2x10-4 (3.3.5) Even at 1 Hz, the loss tangent contribution dominates. In fact, tanL is a function of frequency and is smaller than the above number at such a low frequency. Nevertheless, 2 x 10-4 is a good ball park upper bound for frequencies up to 100 GHz. Using the above figure for tanL, here are a few values of σeff versus frequency: f (GHz) σeff(mho/m) 0 ~1 x 10-15 0.1 2.5 x 10-6 1 2.5 x 10-5 100 2.5 x 10-3 3.4 Normal (radial) current conservation at the surface of a conductor. The normal component of total current is conserved at the boundary between a dielectric and conductor. The reason that this is true follows from the integral form of Maxwell (1.1.1) , ∫H•ds = ∫(jωεε0E + σE)•dA (3.4.1) Imagine a "loop" which lies parallel to the surface. The integral of H around this loop does not change as the loop is moved through the surface, because the parallel component of H is continuous at a surface. [ This latter fact follows from (3.4.1) applied to a thin loop which bisects the surface. ] Thus, the total current, represented as the integrand of the right side of (3.4.1) must be continuous at the surface. We can then write, (jωεdε0 + σdEd = (jωε0 + σcEc (3.4.2) where d = dielectric and c = conductor. On the conductor side, total current is dominated by the conduction current since σc is so large. In fact, for frequencies below 1018 Hz, we can completely neglect jωε0 relative to σc. On the dielectric side, exactly the opposite is true. In Section 3.3 we showed that the DC conductivity of the dielectric is completely negligible, and the loss tangent contribution is on the order of 10-4 times the displacement current. Thus, the displacement current is the dominant term in the dielectric. The ratio of normal Ed to normal Ec at a boundary between conductor and dielectric is therefore given by ≈ (3.4.3) Again, we can look at some typical numbers. σc = 5.81 x 107 mho/m (copper) εd = 2.26 (polyethylene) (3.4.4) At a "high" frequency of perhaps 500GHz this ratio is approximately 106, and at lower frequencies the ratio increases, since the displacement current drops off. Thus, we arrive at these useful facts: Fact 1: The total current in a dielectric is dominated by displacement current, while that in a conductor is dominated by conduction current. Fact 2: At a boundary between dielectric and conductor, the normal E field is at least 1 million times larger in the dielectric than it is in the conductor for frequencies under 500 GHz. 3.5 The TEM mode fields and currents for an IDEAL transmission line . By "ideal" we mean that the conductors have infinite conductivity and the dielectric has zero conductivity. Let us consider a cross sectional view of one conductor of a transmission line having arbitrarily shaped conductors (the shape is however uniform in the z direction). At some point on the surface, we define a little local coordinate system where r = radial direction = the normal outward from the surface (local x) φ = azimuthal direction = tangential to the surface in the cross section plane (local y) z = tangential to the surface along the transmission line (local and global z) pick some other surface spot! The following table shows the sizes of various components of E,B and J (conduction current) at the surface of a transmission line conductor. Several regions of space are of interest: 1. In the dielectric, just outside the super-thin surface charge layer. 2. In the conductor, just under the surface charge layer, and in the current layer. 3. In the conductor, under the surface charge layer and under any current layer. The purpose of defining region 2 is to make the connection to a real transmission line. Table 1: E,B,J for an ideal transmission line 1. In the dielectric, just outside the super-thin surface charge layer. // how justify? Er = large Br = 0 Jr = 0 Eφ = 0 Bφ = large Jφ = 0 Ez = 0 Bz = 0 Jz = 0 2. In the conductor, just under the surface charge layer, and in the current layer. Er = 0 Br = 0 Jr = small Eφ = 0 Bφ = large Jφ = 0 Ez = 0 Bz = 0 Jz = very large 3. In the conductor, under the surface charge layer and under any current layer. Er = 0 Br = 0 Jr = 0 Eφ = 0 Bφ = 0 Jφ = 0 Ez = 0 Bz = 0 Jz = 0 Region 3: In the absence of "applied" charges and currents, we know that E and B satisfy Helmholtz equations of the form (1.5.1). Due to the powerful exponential effect of this equation at the boundary, we know that E and B fields cannot exist deep inside the conductor, and can exist only in the skin depth region. A "perfect conductor" has σ extremely large, and δ ≈ 0, since δ = . Thus, E and B fields can only exist right at the surface. In region 3 of the above table, we show these fields as being 0 underneath the surface. Region 2: We know that Jz = σEz . Since Ez = 0 in the perfect conductor interior, we presume that Jz = 0 there as well. Thus, all current is in a very thin sheath at the surface of thickness δ. Since the thickness is tiny, the current density there is "very large" as indicated in the table. In this thin layer, there is some radial pumping of charge to the surface to "feed" the surface charge which is always changing in time, so we indicate a small Jr term. A ballpark estimate for this current is given later in this section. Since B = 0 inside, magnetic field lines run in a bundle inside the current layer, parallel to the surface, and we have Bφ. This field is 0 on the inside edge of the skin depth region, and builds up to a large value at the surface. Region 1: The Hφ field is continuous through the surface charge layer, and also continues into the space just outside the surface. This follows from Maxwell (1.1.1) since there are no infinite electric fields at the surface. If the conductor and the dielectric have different μ, then Bφ outside the surface will differ somewhat from Bφ inside. In any event, Bφ is still "large" outside the surface. This change in Bφ would only be dramatic if the conductor were made of a ferromagnetic material, in which case Bφ is larger inside. For copper and normal dielectrics, μ = 1 so this is a non-issue. More significantly, since we are now outside the surface charge layer, we now have a large radial electric field Er which is supported by this charge density. There are no tangential E field components since charges would run on the conductor and cancel them out infinitely quickly. See formula (3.1.5) for this time scale when σ = ∞ (ie, T = 0). Notice that just outside the surface, the E and B fields are perpendicular. They are both in the transverse plane, hence this is a TEM mode of the transmission line. Their cross product is the Poynting vector E x B which is in the +z direction, see the above figure. This is the direction of power flow down the transmission line. Finally, since we are now in the dielectric, and since it has zero conductivity, all current density components are zero. Estimate of Jr under the surface. The current Jr is the radial conduction current in the conductor which "feeds" the displacement current in the dielectric, as discussed in Section 3.4 above. This displacement current emits from a relatively large area of the conductor. We can estimate this area as the product of λ/2 in the longitudinal direction, and perimeter p in the transverse plane. As shown in Figure 2 below, over a distance of λ/2, we expect the total displacement current to be 2I, where I is the peak current. Since we are doing ballpark estimates, we omit factors of 2 and make this estimate for the average radial current, <Jr> ~ I/(λ p) ~ (I f) / (pv) (3.5.1) where we have replaced λ = v/f where v is the speed of light in the dielectric, and f the frequency. We expect this dependence on f since the displacement current "fed" by this current is proportional to f. As a rough but conservative estimate of the size of Jr , let v ≈ c, p ≈ 2.5 mm, I ≈ 100 mA, f = 1 GHz. This perimeter applies to a small #20 wire as found in the center of Belden 8281 cable. Then Jr ≈ 1.33 x 102 amps/m2. As we shall see in later examples, this is a relatively "small" current density, so we have indicated this in the above table. For larger conductor size and lower frequency, Jr is even less. 3.6 The TEM mode fields and currents for a REAL transmission line . Now let us "turn on" the imperfections of the transmission line. As soon as σ in the conductor becomes large but finite, the infinitely thin current sheath spreads out over some reasonable skin depth δ. At very low frequencies, the current Jz is spread throughout the entire conductor. [ Note, by the way, that Jz will be non-uniform if the surface charge is non-uniform. This non-uniformity of Jz is therefore most noticeable in fat, closely spaced conductors, and it causes the surface impedance to vary somewhat with point of contact. ] We are no longer interested in "region 3" in this case, but we are still interested in region 2 which is just inside the surface charge layer. Recall from Section 3.2 above that the surface charge layer remains nearly infinitely thin even for a non-perfect conductor. So here is our new table. Table 2: E,B,J for a real transmission line 1. In the dielectric, just outside the super-thin surface charge layer. Er = large Br = 0 Jr = leakage (d) Eφ = 0 Bφ = large Jφ = 0 Ez = small (a) Bz = leakage (b) Jz = leakage (b) 2. In the conductor, just under the surface charge layer, and in the current layer. Er = small (c) Br = 0 Jr = small (d) Eφ = 0 Bφ = large Jφ = 0 Ez = small (a) Bz = leakage (b) Jz = large (a) Items which are significantly different from the previous section are indicated by superscripts which refer to the section below in which each is discussed. In order to make magnitude estimates, we assume that the transmission line is 75 ohms, is properly terminated, and is driven by a voltage of amplitude 7.5 volts, so the current is 100 mA. (a) Ez and Jz in the conductor Inside the conductor, a non-zero Ez arises due to the current flow in the z direction and the now-finite conductivity of the conductor. As a ball park estimate, assume that the conductor is a wire of diameter 1 mm, and is operating at 1 GHz with a skin depth δ of 2 microns. The cross sectional area for current flow is then about 2πrδ = 2π x 10-9 m2. If 100 mA flows through this wire, then Jz = 1.6 x 107 amps/m2. Then Ez = Jz/σ = 1.6 x 107 / 5.81 x 107 = 0.3 volts/meter. [ Such a conductor is dissipating P = I2R = ( 0.1 amp )2 /(σA) ≈ 27 mW/meter, so we are not trying to do something impossible. ] At lower frequencies where skin depth is larger, Ez is less, so we have a worst case here for HDTV frequencies. By contrast, if the conductor separation is 0.5 cm, the transverse field Er is 7.5 volts/ 5 x 10-3 m = 1500 volts/m. This is what we mean by "large" in the above tables. So: Er ~ 1500 V/m Ez ≤ 0.3 V/m ratio (Ez/ Er) ≤ 2 x 10-4 Since there now exists an Ez inside the conductor at the surface, there must also exist an Ez in the dielectric of the same size just outside the surface. This is because tangential E fields are continuous through a boundary. [ This in turn follows from Maxwell (1.1.2) since there are no infinite B fields at a boundary. ] Is Ez continuous thru boundary? (b) Leakage: Bz in the dielectric By "leakage" is meant conduction through the dielectric. As shown in Section 3.3, the effective conductivity in the dielectric is given by σeff = ( σωε' ε0 tanL) (3.6.1) which has a value of about 2.5 x 10-5 mho/m at 1 GHz. This is still 12 orders of magnitude smaller than the σ of copper, but it is 10 orders of magnitude larger than the DC conductivity of the dielectric. To estimate the significance of this leakage at high frequencies, we can compare the leakage current to the displacement current in the dielectric. = = tanL ≈ 2 x 10-4 (3.6.2) Thus, even at high frequencies, the effect of leakage on the current flowing through the dielectric is quite small, when compared to the displacement current. Since this current has a resistive phase, we included it in Table 2 for Jr in region 1. If the transmission line carries 100 mA of current, then the displacement current flowing across one half wavelength worth of transmission line is 100 mA, and the leakage current is 20 µA over this same distance. This leakage current must be fed by a small radial current Jr within the conductor surface of the same size. Since there is a leakage current flowing across the transmission line, there will be a small Bz field resulting. We can estimate the size of this B field relative to the main Bφ field as follows: loop around one conductor in transverse plane: p <Hφ> ~ I, p = perimeter loop around one half wave in z direction: 2 (λ/2) <Hz> = Ileak Thus, Bz /Bφ ≈ Hz/ Hφ = (λ/p) (Ileak/I) = (λ/p) tanL ≤ 2 x 10-5 (3.6.3) Here we assume , consistent with the transmission line limit, that λp ≥ 10. Since the dielectric has conductivity σeff , and since there is an Ez field at the conductor surfaces, there will be some leakage current parallel to the conductors, Jz, in the dielectric. Due to the huge discrepancy between σ in the conductor (108) and in the dielectric (10-4 at 10 GHz), the magnitude of this effect is on the order of 10-12 , so we certainly shall not worry much about it. (c) Er inside the conductor We have already estimated that Er inside the conductor surface is less than 10-6 what it is outside the surface, see Section 3.4. Thus, if Er outside is 1500 volts/m as in our section (a) example, Er inside is less than 1.5 mV/m at 500 GHz, and is proportionally less than this at lower frequencies. (d) Jz and Jr in the conductor. We shall now make an estimate of the relative size of Jr to Jz in the conductor. We shall assume a frequency that is high enough that the skin effect is operative and controls the cross sectional area in the conductor through which conduction current can flow. For a round wire conductor of radius r, we can write, Jz = I/(2πrδ) (3.6.4) where δ is the skin depth. That is, the current must flow through a thin shell near the surface of the conductor. The displacement current, by contrast, flows through the much larger area between the two conductors of a transmission line. From 3.5 (1) we have, <Jr> ~ I/(λ p) = I/(2πrλ) (3.6.5) The ratio is therefore Jr / Jz ≈ δλ) = (3.6.6) where we have used λ = 2πv/ω and δ = . As expected, this ratio increases with frequency f, since the displacement current increases faster than the skin depth decreases. Assume v ~ c, the speed of light. μ= 1 for the conductor, and for copper we use σ = 5.81 x 107 mho/m. Thus, Jr / Jz ≈ (1/2πc) (3.6.7) If we use f = 10 GHz = 1010 Hz, we find that Jr / Jz ≈ 2 x 10-5. Thus, we can estimate that Jr (conductor) ≈ 2 x 10-5 Jz at 10 GHz, and is smaller at lower frequencies. At the surface, Jr is converted to displacement current in the dielectric, except that a fraction tanL of this current remains resistive, as noted in (3.6.2) above. Thus, Jr(dielectric) = tanL Jr (conductor). At low frequencies, (3.6.4) is no longer valid, and we have to use Jz = I/A, where A is the cross sectional area of our conductor. In this case, Jr/Jz ≈ A/(λp). However, since we are low frequency, λ is now very large, so we expect this ratio to be small. To check this claim, assume again that the conductor is round, so A = πr2 and p = 2πr. Then In this case we may write Jr/Jz ≈ (1/2)(r/λ) (3.6.8) Now, by "low frequencies" we mean frequencies where δ ≥ r. This inequality means that ω≤ 2/ (μ0σ r2) (3.6.9) Since λ = 2πv/ω, the this becomes, (λ/r) ≥ (πvμ0σr) (3.6.10) Using v ≈ c, σ = 5.81 x 107mho/m for copper, and r = 1 mm, we find (πvμ0σr) ≈ 107. In this case, we get that Jr/Jz ≤ 5 x 10-8. Therefore, our high frequency result of 2 x 10-5 represents a much worse case, so we will stick with that number. An argument for why Br = 0 is given in the next section [ see Fact 5 ]. We now display once again the table for a real transmission line, showing relative estimates of the sizes of things . A few primes have been added to remove ambiguities. [ huh? ] Table 3: E, B and J for a real transmission line 1. In the dielectric, just outside the super-thin surface charge layer. Er = large [ 1500 v/m] Br = 0 J'r ≤ 2x10-4 Jr (d) Eφ = 0 Bφ = large Jφ = 0 Ez ≤ 2x10-4 Er (a) Bz ≤ 2 x 10-5 Bφ (b) Jz ' ≈ 10-12 Jz (b) 2. In the conductor, just under the surface charge layer, and in the current layer. Er' ≤ 10-6 Er (c) Br = 0 Jr ≤ 2 x 10-5 Jz (d) Eφ = 0 Bφ = large Jφ = 0 Ez ≤ 2x10-4 Er(a) Bz ≤ 2 x 10-5 Bφ (b) Jz = large [ ~107 A/m2 ] 3.7 The general shape of fields, charges, and currents on a transmission line. Based on the information in Table 3 above, we are in a position to draw the general field structure for a real transmission line. Some general rules are now apparent: Fact 1: In a cross sectional sketch of a transmission line, the E field lines land on the conductors at right angles to the conductor surface. This is exactly true for the TEM mode, and applies to all points on the conductor surfaces. Proof: There are no transverse surface currents in the TEM mode, so Eφ = 0 exactly. Even if this were not true, we would expect the right angle rule to be very nearly exact, since transverse E fields must in any event by miniscule. See Fact 2 below. Fact 2: In a longitudinal sketch of a transmission line, the E fields still land on the conductors at very close to right angles. The deviation from angle π/2 is less than 2x10-4 radians, and the deviation is in the direction of current flow at each conductor. Proof: This follows from the Table 3 claim that Ez ≤ 2x10-4 Er . The Ez field is really caused by current flowing just inside the conductor boundary. Fact 3: Apart from an overall scale factor, the cross sectional field shape of a TEM wave on a transmission line is independent of position z along the transmission line, and is independent of time t. The shape is also independent of ω. Discussion: For sinusoidal excitation, the TEM mode on a transmission line is a wave pattern that moves down the line in z. At any instant in time, as one scans in z one sees the overall amplitude of the structure rise and fall for each half? wavelength λ of transmission line. Since there is some loss in a real transmission line, the scale of the cross sectional shape may gradually reduce as z increases. Apart from the scale factor, the shape of the field structure is a constant in z. Later we will see that this shape can be found by solving a certain Helmholtz equation, and we find that the shape is determined entirely by the boundaries of the conductors. Similarly, if one sits at a particular value of z and watches the wave go by in time, the scale of the cross sectional field shape rises and falls with each half? period T, but the shape itself does not change. As ω is altered, the speed of the rise and fall is altered, but again the shape is constant. One should not confuse the constancy of the cross sectional field shape with the possible dispersion of the longitudinal shape of a pulse going down the line in z. Fact 4: In a cross sectional sketch of a transmission line, the E and B field lines are perpendicular at every point in the dielectric. Proof: Consider Maxwell (1.1.2) which we can write in the notation of Section 1.5 as, curl B = j(β2/ω) E ≡ kE To show that the cross sectional E and B fields are perpendicular, we will show that E•B = 0 in the transverse plane. We have: kE•B = (kEx)Bx + (kEy)By = ( ∂yBz - ∂zBy)Bx + ( ∂zBx - ∂xBz)By Since Bz ≈ 0 ( see estimate in previous section), we are left with only two terms kE•B = By(∂zBx) - Bx(∂zBy) = By2 ∂z (Bx/By) However, we argued in the previous Fact that the shape of fields does not vary with z, only their total magnitude. Thus, the ratio of two components like Bx/By cannot vary. Thus, E•B = 0 in the transverse direction, so the E and B lines are perpendicular everywhere in the dielectric. Fact 5: In a cross sectional sketch of a transmission line, the B field lines just outside the conductor surfaces are exactly parallel to the surface. Thus, Br = 0 at the surface. Proof: We have shown in Fact 4 that B lines must be perpendicular to the E lines everywhere in the dielectric, and this includes just outside the conductor. But from Fact 1 we know that E = Er at the surface. Therefore B = Bφ and Br = 0. This then is the justification for maintaining the condition Br = 0 in Table 3 where we have a non-perfect conductor. Comment: Fact 5 is a non-trivial and non-obvious fact, and applies to arbitrarily shaped conductor cross sections, as do all our facts here. The result seems obvious for round wires, but is in fact non-obvious even in that case. If a transmission line is made of two fat round conductors closely spaced, one imagines that the B lines due to current in one conductor are perfectly circular about the center of that conductor. In fact this must be false, because we are claiming in Fact 4 that the vector sum of both conductor's B fields (ie, the total or actual B field) has a round contour line at the surface of each conductor. Thus, the B field contours due to one conductor alone must not be circular. In fact, the current density inside each conductor is non-uniform by just the right amount to make this work out. For widely spaced round conductors, the effect is not very noticeable because the effect of one conductor's B field at the surface of the other conductor is so small. We are now in a position to draw some sketches of fields on a transmission line. Let's start with the transverse or cross section picture: Although this figure is drawn for two round conductors, its general features apply to any conductors. The figure is a snapshot at one instant in time. The • and indicate current flow in the conductors. Positive charge exists on the surface of the left conductor, and is strongest on the face of that conductor which is closest to the other conductor. Negative surface charge lies on the other conductor. The electric fields are as shown and are strongest in the region between the conductors. The magnetic fields derive from the right hand rule relative to the current in each conductor. The B fields of the two conductors are additive between the conductors. The magnitude of the E field is completely determined by the potential difference between the conductors and the geometry. It is independent of frequency. Similarly, the magnitude of the B field is determined by the size of the current in either conductor and is also independent of frequency. Consider a 75Ωtransmission line that is properly terminated and is driven by a 7.5 volt amplitude sine wave oscillator. Regardless of frequency ω, the magnitude of the current in this transmission line is 100 mA, and the magnitude of the potential difference is 7.5 volts. Of course both these quantities have sinusoidal time dependence. At some instant in time, the fields and currents are as in Fig 1. We have just argued then that not much happens in the transverse directions x and y as frequency sweeps from very low to very high. Of course the rate at which the pattern oscillates back and forth increases, but the shape of things does not change. At the peak of each cycle, things look like Fig 1 regardless of ω. This may seem contradictory. In general, one is used to ω affecting things due to equation like curl E = -jωB. The resolution is that all the variation happens in the longitudinal direction. Here then is a top view of the same transmission line: Fig 2: Top view of transmission line. The vertical "cut" is a reminder that this figure is highly distorted. One half wavelength of a longitudinal wave is represented, but the transmission line limit (to be discussed later) requires that λ be much larger than the transverse dimensions of the transmission line. In this figure the magnetic field is perpendicular to the plane of paper. On the left B is coming up out of paper, on the right it is going into paper. At all places ExB points to the right, so we have a wave propagating to the right (+z). The sine curve plotted in the upper conductor represents the manner in which all quantities behave in the z direction: charge, E, B, current J. The shape of the current at this instant in time is shown in the lower conductor. Since the whole field pattern is travelling to the right at the speed of light in the dielectric, positive charge must now be accumulating on the lower conductor near the center of the figure. Both current arrows contribute to this cause. Notice that these current arrows are also consistent with the direction of the magnetic field on the left and on the right. Similar current of course flows in the upper conductor. As discussed in Section 2.1, at high frequency the current will be constrained to flow within a skin depth δ of the surface of the conductors. Where the current is pumping directly into the surface from the conductor side, and where the charge is building up (center of Fig 2), a corresponding displacement current is being fed from the surface into the dielectric. This displacement current arrives at the other conductor and is "converted" back to real current. One can see in the figure that there can be some "radial" component of the current density within the conductors. When averaged over half a wavelength in z, this radial current is the small Jr noted in Table 3 of Section 3.6. We are now ready to apply the curl equations using the two loops shown. Loop 1 is positioned to pick up magnetic flux, so we use Maxwell (2.1.2) which says, ∫E•ds = -jω∫B•dA (3.7.1) Notice the ω sitting on the right side. We argued in the last section that the amplitude of the B field does not change as ω changes. Thus, the right side of (3.7.1) is proportional to ω. As ω increases, the line integral of the E field around loop 1 must increase. Thus, there must be a rate of change of E in the z direction! In other words, as ω increases, the whole pattern of Fig 2 contracts in the z direction, which causes all z derivatives to increase. A similar argument applies to loop 2. This loop is appears end-on in Fig 2. It is set up to sense the electric field flux. The appropriate curl equation is Maxwell (2.1.1) which says, ∫B•ds = jωμμ0εε0 ] ∫B•dA (3.7.2) Since we are now in the dielectric, we have ignored the small leakage conduction current, and have kept the dominant displacement current. This is just the opposite of what we do in a conductor. Again there is a factor of ω on the right side, arising from a time derivative. As ω increases, the line integral of the B field must increase. Thus, the B field must change in the z direction. Again, the curl equation (3.7.2) is satisfied by having the entire pattern contract in the z dimension. If the frequency ω doubles, the wavelength λ goes to half. This of course is no surprise, since ω and λ are related by the speed of light in the medium ν, β = 2π/λ = ω/ν (3.7.3) The main point of the above discussion is to show how the curl equations force the field pattern to contract in the z direction as ω increases. In the transverse direction, the field pattern shape stays constant. The whole pattern moves to the right on the transmission line. Understanding the low frequency limit. Fig 2 provides a way to understand the extreme low frequency limit of a transmission line. At very low ω, the length L of the entire transmission line becomes much less than λ. In this case, one can think of the entire transmission line as being say 1 mm long in Fig 2. As the field pattern moves over such a transmission line, the line is so short that everything is a constant at all z on the line. Instead of thinking of the transmission line as being fixed and the E and B fields moving to the right in Figure 2 at the speed of light, it is perhaps easier to think of the field pattern as fixed, and the little 1 mm long transmission line is moving to the right in Fig 2. For example, if 7.5 volts DC is applied to a terminated 75 ohm transmission line, 100 mA flows, and it is as if the entire line is located at position z=0 in Figure 2 [ needs to be marked more clearly! ] . The current is completely longitudinal, and the E and B fields are very large. If this DC is raised to a very low frequency ω, then one can think of the 1 mm long transmission line moving to the right in Figure 2. When it reaches the point z = λ/4 , the longitudinal current is zero and the fields are also 0. At this point, there is a small transverse current in the transmission line which feeds the very small displacement current in the dielectric. 3.8 Transmission Line Preliminaries A transmission line has two conductors. The cross sectional shape of these conductors is assumed constant in the direction z along the transmission line. The transverse directions are x and y. A wave propagates down a transmission line in what is called the TEM mode. TEM means that the electric and magnetic fields of a wave traveling down the guide are transverse, as in Fig 3.7 (1) and (2). What this really means is that an electromagnetic wave goes straight down the conductors as guides with no surface reflections, unlike what happens in a waveguide, see Appendix 3.2. Apart from a small drag on the wave due to losses in the conductors, the wave proceeds with wavenumber β and velocity ν as it would in an open medium. The conductors shape the E and B fields, so the wave is not a "plane wave". Nevertheless, at each point in the dielectric, E and B are perpendicular and E x B points down the transmission line. We now summarize a set of basic facts about this TEM mode: Fact 1: The major current for the TEM mode is the longitudinal current Jz. In Table 3 above it was shown that there can be small radial transverse currents Jr in the conductors. It was shown that such radial currents are on average smaller than Jz by a factor of at least 2 x 10-5 . There are no azimuthal transverse currents Jφ at the conductor surface, but there can be very small such currents inside the conductors of a magnitude similar to that of Jr. Fact 2: There is no cutoff frequency one has to operate above. The TEM mode works all the way down to DC (although at low frequencies, the attenuation per wavelength may become large). See Appendix 3.2 for why this is not true in a waveguide. Corollary 2: If one operates a transmission line below the cutoff of the lowest waveguide mode, the TEM mode is the only possible way of moving energy down the line. Fact 3: The simplest expression of the boundary conditions are in terms of potentials, not fields, so the potential wave equations are used to solve problems. For example, a boundary condition might be that the electric potential between the two conductors is 0.8 volts. Fact 4: The transverse components of the vector potential A can be neglected, so Az is the only component of A we have to worry about. Proof: Consider equation (1.5.8): A(x) = (3.8.1) Here, Ja represents some prescribed currents in the conductors and the volume integration is over both conductors in x,y and z, and R = |x - x'|. There is clearly going to be a strong Az component since the predominant conductor currents are in the longitudinal direction. According to Fact 1 above, transverse currents are very small. With respect to a local coordinate system at a surface point on one conductor, it was shown in Table 3 that Jr = small and Jφ = 0. However, inside the conductors, we can have some Jφ the same order as Jr. When we compute A in the above integral, we can still decompose A into Az, Ar and Aφ . These components are, however, with respect to some fixed coordinate system located perhaps on some approximate center line between the two conductors. Thus, each potential of the pair Ar and Aφ will feel the effect of both Jr and Jφ , but these are both very small. Moreover, there is considerable cancellation which takes place as pieces of Jr and Jφ add together, but just how effective this is is hard to tell. We rely on the fact that Jr and Jφ are very small to conclude that Ar and Aφ may be safely neglected. This is very different from what happens with Az. In the region of one conductor, the summation is additive for all nearby pieces of current Jz in that conductor, assuming that the wavelength λ of longitudinal propagation is much larger than any transverse dimension. The only place Az is small is on a longitudinal line between the conductors where their contributions cancel. We conclude then that Ar and Aφ can be neglected relative to Az. Fact 5: The potential φ(x) can be identified with the transverse "voltmeter voltage" . Proof: This is not immediately obvious. The thing one measures as "voltmeter voltage" is the line integral of the electric field between two points. If E = - φ, one can identify φ with this voltmeter voltage, but according to Eq. (1.3.1), we have an extra term to worry about, E = - φ - ∂A/∂t . (3.8.2) However, according to Fact 4, the transverse components of A are negligible, so we have Et = -t φ where t ≡ ∂/∂x + ∂/∂y (3.8.3) If one line-integrates Et from one conductor to the other (keeping z fixed), one gets φ1 - φ2 which is a voltage that a voltmeter would really measure. Fact 6: The potential φ is constant over the surface of either conductor at a fixed z. Proof: This follows from the Section 3.6 Table 3 fact that Eφ = 0. Since there is no azimuthal component of electric field at the surface, we get zero doing a line integral of E between any two points on the surface of a conductor in a cross section slice. According to Fact 5, this gives not only the voltmeter voltage between the start and end of such an integration, but it also gives the potential difference ∆φ between these two points. Since the integration must give 0, we conclude that ∆φ = 0 between any two points, so φ must be constant over the surface at fixed z. Fact 7: The potential Az is constant over the surface of either conductor at a fixed z. Proof A: Consider putting thin sensing math loops in Figure 1 perpendicular to the plane of paper. When such a loop is parallel to the B lines, there is no threading B flux. Since B = curlA, this means that Az is then constant on both sides of such a loop. In other words, since B = curlA, the B field lines represent contours of constant Az. According to Fact 5 of Section 3.7, the B field lines just above the surface of a conductor run parallel to the surface, regardless of the cross sectional shape. These B lines cannot "dip" into the surface. Thus, since B field lines are surfaces of constant Az, we conclude that Az must be constant on either conductor surface in a slice at fixed z. Fact 7 is true regardless of that fact that there can be considerable non-uniformity of the current density Jz inside a conductor, see the Comment after Section 3.7 Fact 5. Proof B: Since Fact 7 is so important, and is fairly non-obvious, we give here another proof. According to (1.5.5), we have a gauge condition divA = -j(β2/ω)φ which relates A and φ. We have argued in Fact 4 that, at least in the dielectric region between the conductors, we can neglect all components of A except Az. In this case, the gauge condition reads ∂zAz = -j(β2/ω)φ. Now, as we will soon be doing in Chapter 5, assume a separation of variables so that Az(x,y,z) = (μμ0/2π) i(z) Azt(x,y) (3.8.4) This separation is justified in the "transmission line limit" to be discussed in Chapter 4. If we insert this separated form into ∂zAz = -j(β2/ω)φ, we end up with, [∂zi(z) /i(z)] Az = -j(β2/ω) φ (3.8.5) Thus, since φ is constant over a cross section conductor surface according to Fact 6, we conclude that Az must also be constant over the same surface, since there are no other functions of (x,y) in the above equation. This concludes our Proof B. Proof C: Here is one more proof, a variant of Proof A. We know that B = curl A. Thus, Bx = ∂yAz - ∂zAx (3.8.6) We have argued in Fact 4 that Ax must be very small because transverse currents are so small, and due to cancellations in the potential integration. Moreover, in the transmission line limit, things change very slowly in the z direction, so ∂zAx is then extremely small. We neglect it to get, Bx = ∂yAz (3.8.7) Looking at the local coordinate system shown in the figure of Section 3.5, we can identify x=r and y=φ. Thus, the above states, Br = ∂φAz (3.8.8) But, according to Section 3.7 Fact 5, Br = 0. Thus, ∂φAz = 0. This says that Az does not change as one moves along the cross section surface of a conductor, since this is locally always the φ direction. Thus, Az is constant over the surface at fixed z. Chapter 4: Transmission Line Equations In this Chapter we use the potential integral expressions derived in Chapter 1 to derive the classic transmission line equations. We learn the all "external" transmission line parameters are determined by a single geometric integral K. The approximations assumed are clearly stated. 4.1 Computation of φ due to one conductor of a transmission line. Our starting point is (1.5.9) which expresses the potential φ at some arbitrary point x in space due to conductor C1 as an integral over the charge density on the surface of conductor C1: φ1(x) = (4.1.1) where R is the distance between x and x'. Consider now this charge density ρ1(x). Following a standard methodology, we will assume that its functional form may be factored in the following manner: ρ1(x,y,z) = a1(x,y) q1(z) (4.1.2) Moreover, without any loss of generality we select the relative scale of these two factors such that the integral of a1(x,y) over a slice of conductor C1 at any z is unity, !Syntax Error, Idx dy a1(x,y) = 1 (4.1.3) Therefore, we can interpret q1(z) as the total charge per unit length on C1 , since !Syntax Error, Idx dy ρ1(x,y,z) = q1(z) !Syntax Error, Idx dy a1(x,y) = q1(z) • 1 = q1(z) . Assume that q2(z) is the charge on the other conductor C2. If q1(z) + q2(z) ≠ 0, then we have a net charge per unit length and the transmission line is acting as a radiating antenna as well as a transmission line. From now on, we ignore this superposed problem and assume that at each value of z, the net charge on both conductors is 0. This means that q2(z) = - q1(z) ≡ -q(z) (4.1.4) To simplify notation, we now dispense with the subscript and denote q1(z) = q(z). However, we maintain the subscript on a1(x,y) to emphasize that the two conductors can have completely different cross sectional shapes. The shape of the transverse distribution of charge on C1 is determined by a1(x,y), but the total charge is q(z) per unit length. How can we justify assumption (4.1.2)? This is called "separation of variables". The idea is that we assume it without any justification, and then we try to find a solution to our problem which is consistent with the assumption. All we really want is to find a solution to our basic differential equations with their boundary conditions, and any assumptions we make can be justified in the end once we have found a solution. On the other hand, if an assumption like( 4.1.2) does not lead to a solution, then it must have been a bad assumption. Now insert (4.1.2) into (4.1.1) to get, φ1(x,y,z) = !Syntax Error, Idx dy a1(x',y') !Syntax Error, Idz' q(z') (4.1.5) The next move is to write a power series expansion for q(z') about the point z: q(z') = q(z) + q'(z) (z'-z) + ... = (4.1.6) where we mean by q(n)(z) the nth derivative with respect to z. If we stick this expansion into φ1 we get, φ1(x,y,z) = !Syntax Error, I(1/n!) q(n)(z) !Syntax Error, Idx' dy' a1(x',y') !Syntax Error, Idz' (z'-z)n (4.1.7) In Appendix 4.1 we investigate the dz' integral on the right. If we define the following symbol, s = (4.1.8) then the integral has the following functional form: !Syntax Error, Idz' (z'-z)n = sn fn (βs) (4.1.9) Without even doing the integral, one could arrive at this general form based on the fact that the integral must have dimensions of (distance)n. It happens that the integral is zero for odd n, since the integrand is then odd and the range of integration is symmetric. It is then convenient to change to summation variable m = n/2 , so (4.1.7) now becomes, φ1(x,y,z) = !Syntax Error, I(1/(2m)!) q(2m)(z) !Syntax Error, Idx' dy' a1(x',y') s2m f2m(βs) (4.1.10) To eliminate any further suspense, we now state the integral from Appendix 4.1: f2m (βs) = 21-m (jβ)-m s-m Km(jβs) (4.1.11) where Km(z) is a certain standard kind of Bessel function. If we stick (4.1.11) into φ1 we get: φ1(x,y,z) = -m q(2m)(z) !Syntax Error, Idx' dy' a1(x',y') sm Km (jβs) (4.1.12) We are now done. The interesting thing is that so far we have made no assumptions whatsoever other than the separation of variables in (4.1.2). If we knew the way the charge was distributed on conductor C1 , as indicated by a1(x',y'), we would have an exact closed form solution for φ at any point in space. Notice that the general form is the following: φ1(x,y,z) = k0 q(z) + k1 q"(z) + k2 q(4)(z) + ... (4.1.13) We can now consider applying our first approximation: The Transmission Line Limit: Assume that λ is large compared to the transverse dimensions of the transmission line as characterized by variable s. Since β = 2π/λ, this means that (βs) is small, so we can then use the small-argument limit of the Bessel function Km(jβs). If we do this, what we find is that the series (4.1.13) is not rapidly convergent and may even diverge. If we make the reasonable assumption that the charge density q(z) behaves as a wave of wavevector β along the line, exp(-jβz), we find that the terms in (4.1.12) drop off on the order of 1/m. This suggests a possible formal logarithmic divergence of (4.1.12). In any event, it is clear that the q"(z) term certainly cannot be "neglected" in (4.1.13). In our transmission line analysis, what we really want is the potential due to both conductors C1 and C2, call this φ12. Moreover, we are interested in the difference V(z) = φ12(x1) - φ12(x2), where x1 is a point on C1, and x2 a point on C2, such that these two points have the same value of z. This difference V(z) is the normal "voltmeter voltage" between the two conductors of the transmission line at z. In the next section we will show that, when this four-term combination V(z) is constructed, the series corresponding to (4.1.12) or (4.1.13) is highly convergent when we assume the transmission line limit. In fact, if we drop terms that are of order β2 and smaller, we will find that V(z) = k0 q(z) and we will then interpret k0 as the inverse capacitance of the line. If we do not assume the transmission line limit, then the result for V(z) is more like (4.1.13) , and we then have some entity that is more complex than a "standard transmission line", since it has some sort of higher "capacitive moments" ki that have to be kept track of. We do not in this case obtain the classic transmission line equations. This is a whole painful world we plan to steer clear of. We are happy to live with the transmission line limit since it also serves to rule out the waveguide modes of the line. The implication of the above discussion is that there is a region of frequency ω where the waveguide modes have not yet been activated, but in which the TEM mode does not really obey the classic transmission line equations. This happens when λ/2 is slightly larger than the transverse dimensions of the transmission line. 4.2 Computation of V(z) . First, as outlined above, we form φ12(x) as the potential of both conductors. We get basically two terms each of the form of (4.1.12). The relative minus sign is due to q2(z) = -q(z): φ12(x,y,z) = -m q(2m)(z) * { !Syntax Error, Idx' dy' a1(x',y') sm Km(jβs) - !Syntax Error, Idx' dy' a2(x',y') sm Km(jβs) } (4.2.1) with distance s still given by (4.1.8). Next, we form V(z) = φ12(x1) - φ12(x2) to get, V(z) = -m q(2m)(z) * { !Syntax Error, Idx' dy' a1(x',y') [ s1m Km (jβs1) - s2m Km (jβs2)] - !Syntax Error, Idx' dy' a2(x',y') [ s1m Km (jβs1) - s2m Km (jβs2)] } (4.2.2) where now si = | xi - x' |. One might wonder why it is that V(z) is independent of the location of the contact points x1 and x2, since this dependence seems to be present on the right side of (4.2.2). If the charge distributions a1 and a2 were prescribed by fiat, V(z) given by (2) would be a function of x1 and x2. However, the charge distributions in fact arrange themselves in such a way as to cause each conductor to be an equipotential surface at any given z. This was Fact 6 of Section 3.8. Since the sliced surfaces are equipotentials, the potential difference cannot possibly depend on where contact is made on each surface, assuming both contacts are in the same z plane. We now make the following claim, and relegate its proof to Appendix 4.1: Fact: In (4.2.2), each term m=1 and higher makes a contribution to the sum which is of order (β2) or smaller. In the small-β limit of the transmission line limit, we can neglect all these terms, so that the only term left is the term with m=0. Here then is the m=0 term in (4.2.2): V(z) = q(z) X { !Syntax Error, Idx' dy' a1(x',y') [ K0(jβs1) - K0(jβs2)] - !Syntax Error, Idx' dy' a2(x',y') [ K0(jβs1) - K0(jβs2)] } (4.2.3) where si = | xi - x' |. Since we have already assumed βsi is small, we should take the small-z limit of the K0(z) Bessel function which is, K0(z) ≈ -ln(z/2) [ 1 + (z/2)2 + order(z4) ] + ψ(1) + (z/2)2 ψ(2) + order(z4 ) (4.2.4) The main item here is -ln(z/2), but we have shown the non-leading terms as well. The ψ things are certain constants relating to the gamma function Γ(z). We can see what happens with these non-leading terms in (4.2.3). A power z2 causes a factor of β2 or β2ln(β) so we can throw these terms out, and the z4 terms are even smaller. The ψ(1) term cancels in each square bracket difference. The bottom line is then this relatively simple result: V(z) = q(z){ !Syntax Error, Idx' dy' a1(x',y') ln(s2/s1) -!Syntax Error, Idx' dy' a2(x',y') ln(s2/s1) } (4.2.5) where si = | xi - x' |. At this point, we invoke the discussion of Appendix 1.2 about the complex dielectric constant to make this replacement: q(z)/εε0 = qeff(z)/ξ (4.2.6) The ratio of qeff(z)/ V(z) is the complex capacitance C' per unit length of the transmission line. This is nothing fancy, we are just saying that if the dielectric has some conductivity σ ≠ 0, then this effective C' is complex, so it includes the parts normally called G and C. C' = C + G/jω (4.2.7) From (4.2.5) we then get, 1/C' = (1/2πξ) {!Syntax Error, Idx' dy' a1(x',y') ln(s2/s1) -!Syntax Error, Idx' dy' a2(x',y') ln(s2/s1) } (4.2.8) This is our first major result. We have used Maxwell's equations and have ended up with an expression for the C and G of a transmission line as a purely geometric transverse integral of the charge densities on the two conductors. Later we will show how these charge distributions can be computed for very general situations. For simple situations, such as twin lead and coaxial cables, we can (and will) use (4.2.8) "as is" to get all the familiar results. 4.3 Computation of z due to one conductor of a transmission line. In this section, we quickly develop a set of expressions for Az which are entirely analogous to those for φ. The similarity is not coincidental but is in fact necessary due to the fact that Az and φ are components of the same Lorentz 4-vector, and we have selected a reasonably covariant gauge. This was discussed in Section 1.3. In Section 3.5 Fact 5 it was noted that Az is the only significant component of A. Our starting point then is the z-component of (1.5.8), which expresses the potential Az at some arbitrary point x in space due to conductor C1 as an integral over the current density Jz on the surface of conductor C1: Az1(x) = (4.3.1) We then make the same assumption of separation of variables to write Jz1(x,y,z) = b1(x,y) i1(z) (4.3.2) where i1 is scaled such that !Syntax Error, Idx dy b1(x,y) = 1 (4.3.3) As before, we can now interpret i1(z) as the total current in C1 at z. Again assuming that there is no net superposed radiating antenna current, we have equal and opposite currents in the two conductors, i2(z) = - i1(z) = -i(z) (4.3.4) We are of course led at once to an analogous version of (4.1.5), Az1(x,y,z) = !Syntax Error, Idx dy b1(x',y') !Syntax Error, Idz' i(z') (4.3.5) This is identical to 4.1 (5) with these replacements: φ1 → Az1 q(z) → i(z) (4.3.6) a1 → b1 (1/εε0) → (μμ0) We can now dispense with duplicating the next several steps and jump right to the bottom line, Az1(x,y,z) = -m i(2m)(z)!Syntax Error, Idx' dy' b1(x',y') sm Km(jβs) (4.3.7) 4.4 Computation of W(z) . Our analogous treatment continues. We first construct Az12 (x) as the potential of both conductors C1 and C2 which gives a result identical to (4.2.1) with changes (4.3.6). We then take the difference of this potential evaluated at the two conductor surfaces, W(z) ≡ Az12(x1) - Az12(x2) (4.4.1) to get, W(z) = -m i(2m)(z) * (4.4.2) { !Syntax Error, Idx' dy' b1(x',y') [ s1m Km(jβs1) - s2m Km(jβs2)] - !Syntax Error, Idx' dy' b2(x',y') [ s1m Km(jβs1) - s2m Km(jβs2)] } where si = | xi - x' |. We go on to assume the transmission line limit so β is small. This leads to a result similar to (4.2.3). We then install the limit (4.2.4) to end up with this final result: W(z) = i(z){ !Syntax Error, Idx' dy' b1(x',y') ln(s2/s1) -!Syntax Error, Idx' dy' b2(x',y') ln(s2/s1) } (4.4.3) where si = | xi - x' |. Recall that functions bi(x,y) describe how the current Jz is distributed in the conductors, and the leading factor of 2 is left over from the 21-m factor in (4.4.2). The "magnetic potential" W(z) is not as familiar to us as the electric potential V(z). You cannot just hook up a magnetic voltmeter between the two conductors and measure W(z). However, looking at the dimensions of the right side of (4.4.3), we can arrive at a certain conclusion. The current densities bi are normalized so that dxdy bi is dimensionless. This is so because the integral of bi across conductor Ci is set to unity. The "dimensions" all went with the i(z) part in (4.3.2). Since the dimensions of μ0 are henries/meter, we may conclude that the integral in (4.4.3) is some kind of inductance, and we thus define Le by W(z) = Le i(z) to get, Le = {!Syntax Error, Idx' dy' b1(x',y') ln(s2/s1) -!Syntax Error, Idx' dy' b2(x',y') ln(s2/s1) } (4.4.4) When we develop the standard transmission line equations below, we will see that Le is the mutual inductance between the two conductors. The subscript e stands for "external" and reminds us that this inductance is due to magnetic field storage between the conductors. We will soon encounter an "internal" inductance as well which arises because there is some B field storage inside the conductors as well. As noted in the V(z) discussion, if we do not assume the transmission line limit, we end up with a transmission line described by several "inductive moments" in addition to (4.4.4). These are the coefficients of the derivatives of i(z), that is, we have a situation similar to (4.1.13). As before, we steer clear of this situation and cling happily to the transmission line limit. Observation : If the dielectric between the conductors has any kind of hysteresis, which is not usual for a dielectric, this would be reflected by μω in (4.4.4) being complex. In this case, Le has a slight imaginary part which we interpret as a resistive loss effect due to the hysteresis. In general, we shall consider Le to be completely real. 4.5 The Classic Transmission Line Equations. The results of the previous sections of this chapter may be succinctly summarized as: qeff(z) = C' V(z) [ q(z) = C V(z) ] W(z) = Le i(z) (4.5.1) where C' is given by (4.2.8) and Le by (4.4.4). Notice that we have made no assumptions whatsoever about the cross-sectional shape of the transmission line. We have only assumed that the transverse dimensions are small compared to the wavelength λ that corresponds to β -- this was the transmission line limit. There are two equations from Chapter 1 which we now wish to press into service: E = - grad φ - ∂A/∂t div A = - σ μμ0 φ - (4.5.2) If we regard B = curl A as the definition of A, then the first equation above can be regarded as the definition of φ. The second equation above is our "modified Lorentz gauge" condition. In the frequency domain these equations become E = - grad φ - jωA div A = - j (β2/ωφ (4.5.3) Since A has only component Az, these equations become, Ez = - ∂φ/∂z - jωAz ∂Az/∂z = - j (β2/ωφ (4.5.4) The potentials in the above equations are those due to both conductors and were denoted as φ12 and Az12 in the previous sections. Looking back out our definitions of V(z) and W(z) as differences, we can rewrite the above as: Ez1 - Ez2 = - ∂V/∂z - jωW ∂W/∂z = - j (β2/ωV (4.5.5) The quantity Ez1 is the longitudinal electric field at the surface of conductor C1. It is related to the conductor's surface current density by Jz = σ Ez. If the conductor were "perfect", we would have σ = ∞ and Ez1 = 0. Real conductors are of course not perfect. As shown in (2.3.5), Ez1 can be related to the total current in the conductor i(z) by a quantity known as the surface impedance, so Ez1 = Zi1 i1(z) Ez2 = Zi2 i2(z) (4.5.6) The surface impedance of a perfect conductor is zero. Since i1(z) = -i2(z) = i(z), we rewrite(4.5.5) as, (Zi1 + Zi2) i(z) = - ∂V/∂z - jωW ∂W/∂z = - j (β2/ωV (4.5.7) The circle now closes when we insert into (4.5.7) the second expression in (4.5.1): = - [ Zi1+ Zi1+ jωLe ] i(z) = - [ jβ2/(ωLe)] V(z) (4.5.8) These are the classic transmission line equations. They are usually written in this form: = - z i(z) = - y V(z) (4.5.9) where z = R + jωL y = G +jωC (4.5.10) Here, z and y are called the transmission line impedance and admittance, and the four numbers R,L,G,C are defined to be the appropriate real and imaginary parts. Looking at (4.5.8), we may therefore conclude that: z = R + jωL = Zi1 + Zi2 + jωLe (4.5.11) y = G + jωC = jβ2/(ωLe) (4.5.12) The expression for z seems quite reasonable, but the one for y seems a bit unusual. This is because we still have more work to do. Note: We have been using bold notation only for vectors, and we now break that guideline by bolding these complex quantities z and y. The purpose of this bolding is to distinguish them from Cartesian coordinates z and y which typically appear in the same problem. Apologies. There is one more equation we have not yet made use of. It is the equation of continuity applied to either conductor. In differential form this is div J = -∂ρ/∂t = -jωρ. When this is applied to a slice of conductor of thickness dz, we conclude that = -∂qeff/∂t = -jω qeff(z) (4.5.13) Here we assume that positive current flows in the z direction in conductor C1. If i(z+dz) is larger than i(z), then the net effective charge qeff(z) within dz must be decreasing. Inserting the first of equations (4.5.1) into (4.5.13) we get, = - [ jωC'] V(z) (4.5.14) Comparison with the second equation (4.5.8) results in the following identity, LeC' = β2/ω2 = μμ0ξ (4.5.15) As shown in Appendix 1.2, Eq. (4), we know that C' = (ξ/εε0) C. And from (1.5.3) we have β2 = ω2 μμ0 ξ. Thus we find, LeC = μεμ0ε0 = μεc2 = 1/v2 (4.5.16) This tells us that that 1/= v = the speed of light in the dielectric medium. Look back now at our expressions for C' in (4.2.8) and Le in (4.4.4), 2πξ /C' = {!Syntax Error, Idx' dy' a1(x',y') ln(s2/s1) -!Syntax Error, Idx' dy' a2(x',y') ln(s2/s1) } 2πLe / μμ0 = {!Syntax Error, Idx' dy' b1(x',y') ln(s2/s1) -!Syntax Error, Idx' dy' b2(x',y') ln(s2/s1) } According to the identity (4.5.15), we conclude that the right hand sides of the two equations above are exactly the same! This seems rather amazing, see further comments in Chapter 5. For now, we simply conclude that the basic parameters of a transmission line, apart from the internal impedances, are entirely determined by the following dimensionless transverse integral, K = {!Syntax Error, Idx' dy' a1(x',y') ln(s2/s1) -!Syntax Error, Idx' dy' a2(x',y') ln(s2/s1) } (4.5.17) In terms of this integral we may write: Le = (μμ0/2π)K (4.5.18) C' = 2πξ/K = C + G/jω Since ξ = εε0 + σ/jω, we can decompose the last into two equations, C = 2πεε0/K (4.5.19) G = 2πσ/K (4.5.20) Thus, the three traditional parameters of a transmission line Le, C and G are all determined by the same geometric constant K. Here then is a summary of the results of this section: Transmission Line Equations = - z i(z) = - y V(z) z = R + jωL = Zi1 + Zi2 + jωLe y = G +jωC Le = (μμ0/2π)K C = 2πεε0/K G = 2πσ/K LeC = μεμ0ε0 = μεc2 = 1/v2 (4.5.21) K = {!Syntax Error, Idx' dy' a1(x',y') ln(s2/s1) -!Syntax Error, Idx' dy' a2(x',y') ln(s2/s1) }, si = | xi - x' | 4.6 Example: the wide-spaced, two-wire transmission line. Here we assume two round wires of radius a1 and a2 whose centers are separated a distance b. In order to evaluate the integral K in the above box, we must assume that b is much larger than both wire radii. In this case, we can assume that the charge distribution on the wire surfaces is azimuthally symmetric. The more general case of arbitrary a1, a2, b will have to wait. The calculation is incredibly easy. For each of the two integrals in K, we assume cylindrical coordinates (r,θ) about the respective conductor. The normalized charge density functions are: a1(r,θ) = δ(r - a1)/(2πa1) a2(r,θ) = -δ(r - a2)/(2πa2) (4.6.1) The normalization factors are such that the transverse integral for each charge density is 1, as required in (4.1.3). We need only do the C1 integral, then the C2 one will be obvious. For the C1 integral we use the following geometry: [ new picture, show θ, show values for s12 and s22 ] Fig 1: Geometry for computing K contribution from C1. Notice that points x1 and x2 have been chosen to lie on the points of closest approach. The C1 integral is: K1 = !Syntax Error, Ir dr !Syntax Error, Idθ [ δ(r-a1)/2πa1] (1/2) ln [ b2/2a12(1 - cosθ) ] = (1/4π) !Syntax Error, Idθ ln [ b2/2a12(1 - cosθ) ] = (1/2) ln [ b2/2a12 ] - (1/4π) { !Syntax Error, Idθ ln [ 1 - cosθ ] } . (4.6.2) The integral in { } is a standard definite integral equaling 2π ln(1/2), so the second term gives -(1/2)ln(1/2) which exactly cancels the 2 of 2a12 in the first term. The result is then K1 = ln(b/a1) . (4.6.3) We evaluate the C2 integral in the same way. The minus sign in the expression for K cancels with the minus sign in the charge density for C2 (equal and opposite charges recall), so the result is then K = K1 + K2 = ln(b/a1) + ln(b/a2) = ln(b2/a1a2) = 2 ln ( b/) (4.6.4) Therefore, as shown in the above box, we obtain these results for the widely-spaced two-wire transmission line: K = 2 ln ( b/) C = πεε0 / ln ( b/) Le = (μμ0 /π ) ln ( b/) G = πσ / ln ( b/) (4.6.5) In these formulas, μ, ε and σ of course refer to the dielectric, not the conductors. Normally μ=1 and σ is extremely small. ε is always in the range 1 to 10. What about the internal inductance and resistance? We have solved this problem exactly in Chapter 2, Section 2.3. The exact answer for either wire (n=1,2) is given by (2.3.7): Rin + jωLin = Zs(ω) = ( -jωμ'μ0/ 2πan) J0(βan) / J1(βan) (4.6.6) where β = (/ δ) δ = . (4.6.7) Here we are using primed symbols to stand for quantities for the conductor. Normally μ' = 1 since one seldom uses ferromagnetic (iron) conductors. In the low frequency limit we get from (2.3.9), combining both wires, Ri(ω=0) = [ + ] (4.6.8) Li(ω=0) = μ'μ0 / 4π (4.6.9) In the high frequency limit, where skin depth δ << a1 and a2, the result is (2.3.15), Ri(ω) = [ + ] = [ + ] Li(ω) = Ri(ω)/ω = [ + ] Since the constant in both expressions is the same in this limit, we can write, Ri(ω) = Xi(ω) = k (4.6.10) Li(ω) = k (1/) k = [ + ] ohm-sec1/2 As the frequency increases, the internal resistance Ri and the internal inductive reactance Xi of the two wires both scale as . Recall that the transmission line parameters are z = R + jωL y = G + jωC In the case that the dielectric constant has a slight imaginary part, we can, according to Appendix 1.2 (9), write εω = Re[εω] [ 1 - j tanL (ω)] (4.6.11) where the imaginary part is represented by the loss tangent of the dielectric. In this case, we modify the above C and G as follows: C = πRe[εε0 / ln ( b/) (4.6.12) G = π σeff / ln( b/) where σeff = σ + ε0 ω Re[ε] tanL We can now combine all these results: Two-wire transmission line, b >> ai , in the skin effect limit L = Le + Li = (μμ0 /π ) ln ( b/) + k (1/) R = k k = [ + ] σ', μ' = conductor C = πRe[εε0 / ln( b/) μσεtanL = dielectric G = π σeff / ln ( b/) σeff = σ + ε0 ω Re[ε] tanL (4.6.13) 4.7 Example: the coaxial cable. This calculation is very similar to the previous example. First, we need a picture: The distances are given by: s1 2 = 2 a12 (1 - cosθ) (4.7.1) s2 2 = a12 + a22 - 2a1a2 cosθ (4.7.2) We use the exact same charge distributions as in the previous example, namely (4.6.1). Thus, the K1 integral becomes, K1 = !Syntax Error, Ir dr !Syntax Error, Idθ [ δ(r-a1)/2πa1] (1/2) ln [ ] = (1/4π) !Syntax Error, Idθ ln [ ] (4.7.3) This is really two integrals of the same form, [ Ref ] !Syntax Error, Idθ ln [ A + B cosθ ] = 2π ln [ ] (4.7.4) Thus, keeping track of the two separate integrals we get K1 = (1/4π) { 2π ln [ ] - 2π ln [ ] } (4.7.5) Since a2 > a1 , the first term becomes 2π ln [ ] , and the overall result is then K1 = (1/2) ln [ a22 / a12] = ln (a2/a1) . (4.7.6) This integral has a fairly striking similarity to (4.6.3). In fact, (4.6.3) is the correct result for the example of Section 4.6 even when b is small, if we could force the charge distributions to be symmetric, and provided we reinterpret b as being the distance from the center of C1 to point x2. We know this is true because, if we distort the C2 circle in 4.6 Fig 1, we get 4.7 Fig 1, and we have just done the exact computation for 4.7 Fig 2 and we got K1 = ln (a2/a1). What about K2? Just looking at Fig 1 above, we can see that K2 = 0. K2 represents the contribution to K from the outer sheath C2. We can get the geometry by just distorting Fig 1 and taking a1↔a2. This contribution to K represents the potential at point x1 due to C2. Since C2 is a cylindrical shell of uniform charge density, we know that the potential at any interior point is 0. To see this, put a Gaussian shell inside. Since no charge is in the shell, the electric field on this shell is zero, and therefore the potential is constant. If all this makes no sense, we will now prove it in one sentence. To get K2, make the change a1↔a2 in (4.7.5) and now the second term cancels the first term, so K2 = 0. We are done! The result for the coaxial cable is K = ln(a2/a1). Since the result in Section 4.6 was K = 2 ln ( b/), and since the conductors are still round wires, we can obtain results for the coax cable by replacing K in the box in Section 4.6 with our new K. The only question one might ask is whether the surface impedance of the outer conductor as a cylindrical shell is the same as for a solid wire of the same radius. In light of the discussion surrounding (2.4.1) , we may conclude that this is indeed precisely the case, provided that the outer sheath has a thickness which is many times the skin depth. In this case, we can regard the sheath as being infinitely thick, and then there are no other dimensions to enter the answer. In a nutshell, we are saying that D = 2πa2 in (2.4.1), which seems quite reasonable. One could solve this problem exactly by selecting the solution Y0(x) in (2.1.16) in place of J0(x). We shall not bother doing this. In the low frequency limit, for a non-infinite sheath C2, the DC resistance and inductance will be different from (4.6.8) and (4.6.9). If the thickness of the sheath is t<<a2, then we can replace (4.7.8) with: Ri(ω=0) = . (4.7.7) Here we have just installed the appropriate cross sectional area, as per Appendix 2.1 2(2). As for the DC inductance of a thin sheath, we can make the following computation based on Appendix 2.1, which is accurate if t << a. Outside the sheath we know that H = I/2πa, and inside H = 0, so we can take the average value in the sheath to be I/4πa. The energy stored in the sheath is U = (1/2)μμ0 H2V where V is the volume of the sheath V = 2πat dz, and H = I/4πa. Setting this equal to (1/2)Li I2 we find this result, Li (thin shell of radius a, thickness t) = (μμ0/8π(t/a) (4.7.8) This is the DC inductance of a wire of any radius times (t/a). As t → 0, Li goes to zero because the field is finite, but the volume goes to 0. So here is our modified DC inductance for the coaxial cable with a thin sheath: Li(ω=0) = ( μ'μ0 / 4π[ 1t/a ] (4.7.9) We conclude with the skin-effect limit results, as transcribed from the box at the end of Section 4.6 replacing K with our new K: Coaxial transmission line, in the skin effect limit L = Le + Li = (μμ0 /2π) ln (a2/a1) + k (1/) R = k k = C = 2πRe[εε0 / ln (a2/a1) μσεtanL = dielectric G = 2π σeff / ln (a2/a1) σeff = σ + ε0 ω Re[ε] tanL (4.7.10) *************** Chapter 5: The Transverse Problem 5.1 Philosophy In Chapter 4 we derived the general structure of the behavior of a TEM mode wave on a transmission line. The situation is summarized in the box at the end of Section 4.5. Under the assumption that the wavelength λ 2πβ is large compared with the transverse dimensions of the transmission line, we ended up with the classic transmission line equations. It was shown that the usual parameters Le, C, and G are all related to each other by the factor K. It is now time to look back at what we really did. It all goes back to Maxwell's equation (1.1.1). There is a current term Ja sitting in this equation which makes its way into (1.5.8) which was really the basis of Chapter 4. Similarly, the charge density ρa in (1.1.3) ended up in (1.5.9). We referred in Section 1.1 to these two sources Ja and ρa as being "externally applied". The approach philosophy was to remove the conductors from Maxwell's world and pretend that currents and charges could be prescribed in and on the conductors. These sources then created potentials φ and A, and these then determined the E and B fields and the problem was seemingly solved. In retrospect, we took this plan to its logical conclusion outlined in the box at the end of Section 4.5. Our big problem is that we have no way to compute the K integral shown in the box, except in certain highly symmetrical situations such as those treated in Sections 4.6 and 4.7. The reason is that we do not know how the sources (charges and currents) are distributed on the conductors in the transverse direction, and this information is what is needed to compute K. We must now change our philosophy. We must think of all space as being Maxwell's province. This means the inside of the conductors as well as the dielectric between the conductors. From this point of view, there is no "externally applied" current Ja, and no "externally applied" charge density ρa. We should set Ja and ρa equal to zero in Maxwell's equations (1.1.1) and (1.1.3). There is of course a current J, and it has already been accounted for in (1.1.1) by the term σE. There will also be a charge density ρ which will arise from the divergence of E, as in (1.1.3). The idea is that we are now looking for a self-consistent TEM wave mode travelling down the conductors. The fields are generated by the current and charge distributions, and the distributions are in turn generated by the fields. For example, the principle current in the conductors Jz is due to the fact that there is some electric field Ez which is creating this current according to Ohm's law J = σ E. At the same time, this created current produces a magnetic field which in turn has an associated electric field according to Maxwell (1.1.2). The entire situation is a self-consistent closed loop. If we can find such a solution, then the solution must exist. Whenever there is a normal component of electric field at the surface of a conductor, there is an associated charge density. This is the ρ appearing in (1.1.3). This charge distribution is associated with the current distribution according to the continuity equation div(σE) = - ∂ρ/∂t. The charge density arises as part of the self-consistent solution to the problem. Like J, ρ is not "externally applied". This situation is not uncommon in electromagnetic problems. For example, an accelerating particle radiates, but then the radiation acts back on the particle in a phenomenon known as radiation damping. An antenna radiates, but the radiation has an effect on the current distribution in the antenna. These all serve as reminders that Maxwell's equations, although simply stated, are incredibly complex, and with very few exceptions, no problems have ever been solved exactly. In Chapter 4, the integral expressions (1.5.8) and (1.5.9) formed the basis for everything! We had prescribed Ja and ρa , and we used these expressions to compute the potentials φ and A. In our new point of view, these integral expressions need to be re-interpreted. Equation (1.5.8) would seem to indicate that A = 0, since Ja = 0. In fact, (1.5.8) is only a "particular" solution of the differential equation (1.5.1). Now we have to find the "homogeneous solutions" of (1.5.1), which is of the classic Helmholtz form (2 + β2)A = 0. Equation (1.5.9) which now contains the unknown charge density ρ can be interpreted as an integral equation which is equivalent to the differential equation (1.5.2). So in our new mindset, the integral formulas are not incorrect, they are just not too useful. The time has now come to face the differential equations directly and find solutions. One might at this point think of Chapter 4 as a great waste, but this is not true. Chapter 4 did provide the overall structure of things, and this will be heeded in solving the differential equations. 5.2 The Helmholtz Equations and Separation of Variables Since Ja and ρa no longer exist, as described above, we set Ja = 0 in (1.5.1) and ρa= 0 in (1.5.2), so we have this pair of differential equations, ( 2 + β2 ) A = 0 (5.2.1) ( 2 + β2 ) φ = - (1/εε0) ρ (5.2.2) where β ≡ ω (5.2.3) with ξ ≡ [ εε0 + σ/(jω) ] (5.2.4) Since φ and A are the potentials arising from the presence of both conductors C1 and C2, the φ and A appearing in (5.2.1) and (5.2.2) correspond to φ12and A12 of Chapter 4. The above equations apply at all points in space, both in the dielectric between the conductors, and inside the conductors as well. A differential equation of the form ( 2 + β2 ) f = 0 is usually referred to as the Helmholtz equation, in honor of Hermann von Helmholtz (1821-1894), an early electromagnetic researcher. Let us first consider the equation (5.2.2) for φ. We shall do a separation of variables for both φ and ρ. For ρ, we perform a separation very similar to (4.1.2) through (4.1.4), ρ(x,y,z) = ρt(x,y) q(z) (5.2.5) where q(z) is the charge per unit length on conductor C1. This defines a transverse charge density ρt(x,y). The integral of ρt(x,y) over the surface of a C1 slice is +1, and over the surface of a C2 slice is -1. For the potential, we make the following separation: φ(x,y,z) = (1/2πεε0) q(z) φt(x,y) (5.2.6) We are of course free to set the scale factor arbitrarily, since changing the scale factor just changes the definition of φt(x,y). Our motivation for the form shown in (5.2.6) is the work of Chapter 4, in particular equation (4.2.5) In light of (4.1.13) and the related discussion, we realize that the separable form (5.2.6) is only possible if we are working in the "transmission line limit" where the wavelength λ along the transmission line is much larger than all transverse dimensions. When (5.2.5) and (5.2.6) are inserted into (5.2.2), the result is [ + 2π ] + = - β2 (5.2.7) which has the general form, [ h(x,y) ] + g(z) = - β2 (5.2.8) The only way this can be true for all x,y,z in a region is if g(z) = some constant. For reasons that will be clear later, we write this constant as kφ2. Then we get = kφ2 [ + 2π ] = - β2 - kφ2 (5.2.9) We can rewrite these as [ t2 + (β2 +kφ2)] φt(x,y) = -2πρt(x,y) (5.2.10) [ z2 - kφ2 ] q(z) = 0 (5.2.11) We now repeat the process for the vector potential component Az, which we know is the only significant component of concern. We make the following separation, similar to (5.2.6), which is motivated by (4.4.3), Az(x,y,z) = (μμ0/2π) i(z) Azt(x,y) (5.2.12) Putting (5.2.12) into (5.2.1) yields, [ ] + = - β2 (5.2.13) Again, this can only work if the second term is some constant, which we here call kA2. = kA2 [ ] = - β2 - kA2 (5.2.14) which leads to, [ t2 + (β2 +kA2 )] Azt(x,y) = 0 (5.2.15) [ z2 - kA2 ] i(z) = 0 (5.2.16) Consider now equations (5.2.11) and (5.2.16), which we now apply to the full potentials by making use of (5.2.6) and (5.2.12), - kA2 Az = 0 - kφ2 φ = 0 (5.2.17) We seek next to find the constants kA and kφ. Consider the differences that played such a major role in Chapter 4, V(z) = φ( x1, y1, z ) - φ( x2, y2, z ) W(z) = Az( x1, y1, z ) - Az( x2, y2, z ) (5.2.18) where x1 and x2 are points on the conductors at the same z. Applying (5.2.17) to these differences gives, - kA2 W = 0 - kφ2 V = 0 (5.2.19) Now, application of ∂/∂z to the transmission line equations (4.5.9) gives, - zy i = 0 - zy V = 0 (5.2.20) where constants z and y are the impedance and admittance of the transmission line. According to (4.5.1), W(z) = Le i(z), so we replace the first equation above with an identical one in W: - zy W = 0 - zy V = 0 (5.2.21) Comparison of (5.2.21) with (5.2.19) gives us the result we seek, kA2 = kφ2 = zy ≡ k2 (5.2.22) In retrospect, we are not surprised that the constants are equal in light of the Lorentz covariance discussion relating to equation (1.3.9). From (4.5.11) and (4.5.12) we know that z = Zi + jωLe, and y = jβd2/(ωLe ), where Zi is the total internal impedance of both conductors, and where βd is β of the dielectric. From (4.5.15) we can also say y = jωC'. Therefore k2 + βd2 = Zi(jω C') = Zi j βd2/ (ωLe) (5.2.23) For perfect conductors, Zi = 0. We might define a "low loss" transmission line as one where the right side of (5.2.23) is much smaller than βd2. Using β2 = ω2μμ0ξ this condition becomes, with μ=1, Zi << μ0f K = 4πK x 10-7 f ≈ K f(MHz) ohms/meter (5.2.24) where K is the dimensionless geometric integral defined in Chapter 4. Typically K is a number in the range 1-10, so the above statement is quite clear as a definition of "low loss". For a typical coax cable, Zi is about 1 ohm/m at 1 GHz, K ≈ 3, so the above condition is well met since 1 << 3000. Fact: For a "low loss" transmission line, as defined above, k2 ≈ - βd2 . There remains one important final connection to be made to our work of Chapter 4. The constants in the variable separations for φ and Az given in (5.2.6) and (5.2.12) were carefully selected to yield the following result, φt(x1) - φt(x2) = Azt(x1) - Azt(x2) = K (5.2.25) where x1 and x2 are any points lying on C1 and C2 in the same z plane. Recall that K determines the three transmission line parameters G, C and Le. The ingredients needed to show that (5.2.25) is true are (5.2.6) and (5.2.12) of this section, and (4.5.1) , (4.5.18) and (4.5.19). We now summarize the key results of this section: Separated Transmission Line Equations Variable separations: Eigenvalue relation: ρ(x,y,z) = ρt(x,y) q(z) k2 = zy φ(x,y,z) = (1/2πεε0) q(z) φt(x,y) (βd2 + k2) = jωZi C' Az(x,y,z) = (μμ0/2π) i(z) Azt(x,y) ≈ 0 "low loss" Transverse equations: Transverse boundary conditions: [ t2 + (β2 + k2)] φt(x,y) = -2πρt(x,y) φt(x1) - φt(x2) = K [ t2 + (β2 + k2)] Azt(x,y) = 0 Azt(x1) - Azt(x2) = K Longitudinal equations: β2 = ω2 μμ0ξ [ z2 - k2 ] q(z) = 0 βd2 ≈ ω2 μμ0εε0 = ω2/ v2 [ z2 - k2 ] i(z) = 0 βc2 ≈ j(2/δ2) , δ = (5.2.26) 5.3 A Formal Solution to the Transverse Problem The eigenvalue problem. In either the dielectric or the conductor, both φ and all Cartesian components of A satisfy the same transverse Helmholtz equation, [ + + (β2+ k2) ] f = 0 (5.3.1) The fact that there exists a surface charge affects the φ equation only at the surface. We know that for a reasonably low loss transmission line, k2 ≈ - βd2 ( k ≈ +jβd) which is a "small" quantity, especially compared to β2 in the conductors. The amount by which k2 differs from - βd2 is what we seek to find. We will have an eigenvalue problem for k2. The longitudinal equation tells us that (wave travels in +z direction) = - kf. (5.3.2) It frequently happens in this kind of problem that things contrive to make solutions f exist only for certain specific values of the eigenvalue k2. In each region, the solution must have a certain type of behavior, and then at the boundaries, various quantities must be continuous. These conditions are so stringent that only specific values of k2 permit any solution at all. The parameter β is vastly different in these two media, as shown in the box (5.2.26). In the dielectric, β2 is mostly real, with a slight imaginary part due to the small conductivity σ of the dielectric. The solution f in this region will therefore tend to be oscillatory, since then second derivatives create negative contributions to cancel the positive β2 contribution in (5.3.1). The solution for the lowest eigenvalue will have an extremely mild oscillatory behavior with no nodes. In fact, the oscillation is so mild that the main fields are approximately constant across the dielectric. Inside the conductor, σ is huge, so β2 becomes large and negative imaginary, β2 = -jωμμ0 σ-j(2/δ2) (5.3.3) In a 1-dimensional problem if one has an equation of the form [∂2/∂x2 - j(2/δ2)]f = 0, the solution must look like, exp( ±x) = exp[ ± ( 1 + j) x/δ ] (5.3.4) If we interpret x as going into the surface of the metal, we see that in addition to a strong oscillation, there is a strong exponential growth or decay. Exponential growth is unphysical, so we must select the minus sign. This is the well-known skin effect which we have already encountered several times. So, we are solving for f = φ and A and we know that in the dielectric f is doing some reasonable pattern, and in the conductors f is decaying exponentially to 0. The boundary conditions can be expressed in several ways, one of which is to say that various components of the E and B fields which derive from φ and A must be continuous at each dielectric-conductor interface. In particular, Ez and Hφ must be continuous at each boundary. In terms of our two main potentials, these continuity requirements are, Ez(x,y,z) = - kφ - jωAz = - k(1/2πεε0) q(z) φt(x,y) - jω(μμ0/2π) i(z) Azt(x,y) = continuous Hφ(x,y,z) = (1/μμ0) Bφ = (1/μμ0) ( - ∂rAz) = ( i(z) /2π) ( - ∂rAzt(x,y) ) = continuous (5.3.5) The first line is satisfied if φt(x,y)/ε and Azt(x,y)/μ are continuous, while the second line requires that ∂rAzt(x,y) be continuous, where ∂r means the normal derivative at the surface. Here we have continued to neglect other components of A, and other components of the E and B fields. Note that ∂rφ is definitely not continuous due to the charge density on the surface. In any event, as noted, when the dust settles, the eigenvalues for k2 emerge. The lowest eigenvalue will have the smoothest behavior of Azt between the conductors with no nodes, and this is the behavior that describes our TEM mode. Higher eigenvalues have increasing numbers of nodes (that is, more waves) in the transverse direction between the conductors. These are the waveguide modes discussed in Appendix 3.2. When this lowest eigenvalue is found for k2 by the above analysis, we then know zy = k2. Since we are in the frequency domain, we expect that k2 = zy will be a function of ω. Knowledge of zy then allows an analysis of longitudinal waves down the transmission line. The interested reader will find in Matick (Sections 4.5 and 4.8) samples of the above eigenvalue method applied to stripline, the simplest of all possible geometries. Since the eigenvalue k2 is directly related to the internal impedance Zi of the transmission line, according to (5.2.23), these discussions are also discussions of the internal impedance problem. Determination of the line parameters . When the eigenvalue problem noted above is solved, we end up with a number for k2 = zy, and a functional form for Azt(x,y) and φt(x,y). By evaluating either of these between the two surfaces, we learn, according to (5.2.25), the magic number K, and this in turn determines the three line parameters according to, Le = (μμ0/2π) K C = 2πεε0/K G = 2πσ/K (5.3.6) where μ, ε and σ refer to properties of the dielectric. Equality of potentials in the dielectric. In the dielectric we notice that Azt(x,y) and φt(x,y) obey the same transverse Helmholtz equation, and they have the same boundary conditions, namely, constant on each conductor surface, and the difference between the two surfaces must be K. Therefore we may conclude that: φt(x,y) = Azt(x,y) in the dielectric (5.3.7) This is true to the extent that Az really is constant on the boundaries. We know from the various "proofs" of Az = constant given in Section 3.8 for Fact 7 that certain approximations are involved, such as the neglect of terms associated with other components of A, or such as the neglect of Bz. In fact, Az deviates a small amount from being constant on the boundary of each conductor cross section. This becomes clear when one evaluates the electric field at the boundary: Ez(x,y,z) = - kφx,y,z - jωAz(x,y,z) x,y on boundary (5.3.8) For a perfect conductor, Ez = 0 on the boundary, and the two terms cancel. For a real conductor, the two terms do not cancel, and we get a small difference Ez(x,y,z). We know that Ez(x,y,z) can have some variation on the conductor boundary because Ez = Jz/σ and Jz has variation due to the non-uniformity of the current distribution in the conductors, most noticeable when they are fat and closely spaced. Since φ = constant on the boundary, the only way we can have Ez(x,y,z) vary on the boundary is if Az(x,y,z) varies. This variation of Ez(x,y,z) corresponds to a variation of the surface impedance of the conductor at different points on the surface. Although the variation of Az(x,y,z) on the boundary is miniscule, we have in (5.3.8) that Ez is the difference between two relatively large quantities which very nearly cancel out. Thus, the very small variation in Az(x,y,z) on a boundary can become a large variation in Ez.. Comparison of potentials inside the conductors. Although φt(x,y) and Azt(x,y) are nearly identical in the dielectric, they are grossly different inside the conductors. The reason is that the surface charge density ρt(x,y) affects φt(x,y) but not Azt(x,y), see the transverse equations in box (5.2.26) above. We have seen already that ∂rAzt(x,y) is continuous through the boundary, whereas ∂rφt(x,y) has a large discontinuity due to the surface charge. As noted in Section 3.2, the surface charge on the conductors is for all practical purposes infinitely thin. Table 3 of Section 3.6 shows that outside this charge layer, the radial electric fields can be very large. Inside the charge layer, there is some very small radial electric field Er which goes with the radial current Jr which supplies the charge layer. The potential φt(x,y) has the same value on both sides of the surface charge layer and is constant over the surface, Section 3.8 Fact 6. Rubber sheeting of fields and potentials inside the conductors Inside the conductors, the general behavior of all potentials and fields depends on the frequency ω, which is to say, the behavior depends on the size of the skin depth δ relative to the cross sectional dimensions of the conductors. For high frequency and small skin depth, all potentials and fields have the same characteristic behavior. They all "rubber-sheet" down to zero as you move away from the boundary toward the interior, and they do so exponentially right in the skin depth layer. The reason for this is that all fields and potentials satisfy the Helmholtz equation with β2 = -j(2/δ2). As one moves out of this limit to lower frequencies, things change and the pattern of fields and potentials is frequency dependent. At very low frequency, the field Ez exists everywhere inside the conductors, although it can be non-uniform. The azimuthal and radial components of A, which were neglected in the dielectric, now become important. This fact becomes clear when one considers what happens to divA = -j(β2/ω)φ as one passes through a boundary. Whereas Az and φ are continuous, β2 undergoes a dramatic change, so the other terms in divA must be significant. Nature of the surface charge density. The radial electric field Er is much larger outside the conductor surface than inside. In fact, in Section 3.6 Table 3 we argued that the exterior Er field is typically 1 million times larger than the interior field. By placing a gaussian box at the surface, using Maxwell (1.1.3), and ignoring the interior Er field, one finds this expression for the surface charge density n: n(x,y,z) = εε0 Er = - εε0 ∂rφ-(1/2πq(z) ∂rφt(x,y) (5.3.9) Thus, the surface charge density is completely determined by the radial electric field at the surface, which in turn is controlled by the radial gradient of the transverse potential at the surface. We expect that the surface charge density so obtained can be highly non-uniform over the conductor surfaces. The charge should be larger where the conductor surfaces have their closest approach, and the E fields are largest. If the conductors are large and very closely spaced, the non-uniformity of n over the surface should be quite dramatic. This leads to the non-uniformity in Jz which has been noted several times. Self Consistent Currents. We have noted above the fact that the potentials penetrate into the surfaces of non-perfect conductors and create Ez(z) and Jz(z) in the conductors. As noted in Section 5.1 above, the currents in the wires are not externally applied, they just fall out as part of the self-consistent solution. If the skin depth is small compared to the conductor thickness (diameter), the current in the conductors only flows in a small region near the surface of the conductor. This result was not obvious from the approach taken in Chapter 4, where we thought of the current as more or less uniformly spread over the cross section of a conductor, since we imagined that the "external driving source" made the current this way. Here we see that current is really due to the potentials and fields, and is thus a surface effect. This, then, is the formal solution of the transverse problem. We shall not attempt this program, but the point should be clear that the problem and the solution are well defined. With a computer, one could solve an arbitrary cross section transmission line in this manner, to any degree of accuracy desired. Caveat on Accuracy We should not let this presumed perfect accuracy make us lose sight of the "transmission line limit" of Chapter 4 which is always operative. As suggested by the series in (4.1.13), there are really higher order terms in the transmission line equations (4.5.9) which we ignore because we assume β is small. By doing a "perfect" solution of the transverse problem, we get "perfect" values for the first order coefficients (line parameters) which appear in (4.5.9). But we still have to assume small β if we want to ignore the higher order terms such as one proportional to i"(z). The small-β corrections ( β = 2π/λ) are of order β2, as claimed in Section 4.2, and as shown in Appendix 4.1. Thus, for example, we expect our overall accuracy to be on the order of (d/λ)2, where d is the largest transverse dimension of our transmission line. For example, if d = 0.5 cm for some coaxial line, and λ = 1 meter, we expect our accuracy to be roughly 1 part in 105, which is pretty good. Note that λ is the wavelength in the dielectric, and so is affected by the dielectric constant. 5.4 . An approximate solution to the transverse problem. In Section 5.3 we described a "formal solution" of a transmission line which required the solution of the transverse Helmholtz potential equations. Once the lowest eigenvalue k2 is found, along with the associated solutions for the potentials, we could compute everything, including the surface impedance of the transmission line. Since this eigenvalue problem is hard to solve, we describe here an approximation method that gives reasonable results. Only the dielectric part of the problem is treated, and the conductors are at first ignored, then later their influence is approximately included through estimates of their surface impedance. Notice that the factor (β2 + k2) appears in the transverse equations for Azt(x,y) and φt(x,y). From 5.2 (23) we know that k2 + β2 = jωZi C' (5.4.1) where Zi is the total surface impedance of both conductors. If we are willing to assume that these conductors are "perfect", then Zi ≈ 0, so (β2 + k2) ≈ 0 as well. The degree of "perfection" can be measured using our "low loss" condition (5.2.24) which says Zi must be less than μ0f K. Then we have the following approximated transverse equations from (5.2.26), t2 φt(x,y) = -2πρt(x,y) (5.4.2) t2 Azt(x,y) = 0 and we know that surface charge ρt(x,y) exists only on the conductor surfaces. In the dielectric, both equations have the same form, the same boundary conditions, and in fact the same solution, as already noted in Section 5.3. Thus, we set, f(x,y) = φt(x,y) = Azt(x,y) / in the dielectric (5.4.3) Then f(x,y) must satisfy the following equation and boundary conditions: t2 f(x,y) = 0 (2D Laplace) (5.4.4) f(x,y) = constant on each conductor surface slice (5.4.5) The plan is then to solve the 2D Laplace equation for f(x,y) subject to the condition of constancy on the two conductor cross sectional surfaces. When this has been done, we can evaluate the geometric factor K from, f(x1) - f(x2) = K (5.4.6) and we then have good estimates for the line parameters Le, C and G which are functions of K. What we do not get by this approximation method is an estimate for the surface impedance, since we assumed it was zero. We must therefore estimate Zi by some independent method. For round wires, a method was presented in Chapter 2. For other standard conductor shapes, see Matick Chapter 4. An estimate of error in the above approximation. Once we have an estimate for Zi, we can go back and see how good (or bad) our approximation was for the potential f(x,y) and hence for K. Plugging this Zi into (5.4.1), we get an estimate for the size of the quantity (β2+k2) which we assumed was zero in our first solution (in the dielectric). Basically this involves doing perturbation theory on the solution of a differential equation. One needs to identify a dimensionless smallness parameter s and expand the solution as a power series of functions weighted with powers of s. The solution is highly convergent if s is small. In fact, one can use s itself as a measure of the size of the correction terms to the basic solution we have found above. In our case, the dimensionless smallness parameter is s = | | = = = πσδ2 |Zi|/K (5.4.7) Beware: since β2 appearing above is in the dielectric, the symbols σ and δ refer to the dielectric. As an example of how small s might be, consider our two-wire transmission line solution given in the box (4.6.13), with a1 = a2 = a. Using the round wire estimates, we found in the skin depth limit that, |Zi| = Ri = / (πσ'δ'a) (5.4.8) K = 2 ln(b/a) where primed quantities refer to parameters of the conductor. Thus we get, s = (σ/σ') (δ/δ')2 (δ'/a)(/ K) ≈ (δ'/a)(/K) (5.4.9) where we assume that μ ≈ μ'. Since we have already assumed the skin depth limit ) << 1, we have s << 1, and our approximation is a good one. We expect any fractional errors to be on the order of s . Next, we consider the low-frequency limit. In this case we have, |Zi| ≈ Ri = (2/πσ'a2) (5.4.10) so that, s = (2/K)(σ/σ')(δ/a)2 ≈ (2/K) (δ'/a)2 = (δ'/a)2 / ln(b/a) (5.4.11) This suggests that at very low frequencies, such that s~ 1, our approximation is no good. One way to say this is that a very low frequencies, the inductive reactance ωLe of the transmission line per unit length is no longer much larger than the resistive losses Ri in the conductors. When we set (β2 + k2) ≈ 0 in the transverse equation for f(x,y), the equation becomes the Laplace equation which has a very smooth solution which just meets the boundary conditions in the smoothest possible fashion. When (β2 + k2) becomes larger, then the transverse solution of **** starts becoming oscillatory in the dielectric. The solution must acquire a higher transverse curvature to cancel out the (β2 + k2)f term. The solutions for the potentials and fields then no longer match our intuitive notion of a simple transmission line, and in fact become more like those of a waveguide. It is a characteristic of the Helmholtz equation that when things become oscillatory in x and y, they become exponential in z. Thus, one is not surprised to find that in the low [high!?] frequency limit, the transmission line starts developing a severe attenuation per wavelength. As shown in (5.2.23), we have: k2 = β2 [ - 1] huh? (5.4.12) and waves propagate down the transmission line as e- kx. When Zi ≈ Ri at low frequencies, and Ri/ωLe is large, we get: k ≈ (2π/λ) / (5.4.13) In this case, waves are attenuated over a distance of λ / . If ~ 1, then our wave goes about one wavelength on the transmission line and dies. Power transmission lines approach this low frequency regime. As shown in Section 2.2, the skin depth in copper is about 0.85 cm at 60 Hz, 1.09 cm for aluminum. For wires of diameter 1 cm, separated a distance of 1 m, (11) becomes s ≈ (δ'/a)2 / ln(b/a) ≈ 4/ ln(200) = 0.75 Thus, we would expect significant attenuation over one wavelength of such a power transmission line. Since a wavelength is about 5000 km at 60 Hz, this is presumably acceptable. A similar result obtains for Belden 8281 coaxial cable in the range of 10-100 KHz. ************************** Chapter 6: An Example In this Chapter we use the method outlined in Section 5.4 to solve for the parameters of a transmission line consisting of two round wires of radius a1 and a2 whose center lines are separated by a distance b. The geometry is as shown in Section 4.6 Fig 1. The big difference here is that we put no restrictions on the size of b relative to the two radii. The methods of Section 4.6 cannot be used in this case, since the charge distribution becomes non-symmetric on each wire. To review, our problem then is to solve this equation, t2 f(x,y) = 0 (2D Laplace) such that f(x,y) = constant on each circular conductor slice When this has been done, we can evaluate the geometric factor K from, f(x1) - f(x2) = K and we then have good estimates for the line parameters Le, C and G which are functions of K. First we will show that logarithmic functions are the natural solutions of the 2D Laplace equation, and then we will find combinations of such logarithms which have circles as equipotentials. The final step will be to line up two of these circles with the boundaries of our round wires. 6.1 Why logarithms? In 3D space, it is well known that the potential of a point charge is (1/4πεε0) q/r. This potential must satisfy the 3D Laplace equation. That it does so is very easily shown in 3D spherical coordinates centered at the point charge. Since the potential has no dependence on θ and φ, only the radial portion of the Laplacian matters. In this case we have, 2 f(r) = (1/r2) ∂r [r2 ∂rf(r) ] With f(r) = 1/r, the inner bracket [..] becomes -1, and ∂r[ -1 ] = 0, so 2(1/r) = 0 for r ≠ 0. In 2D space, a similar thing happens, except in 2D, the potential of a point charge behaves as ln(r) instead of (1/r). In a manner similar to the above, we can prove quickly that ln(r) solves the 2D Laplace equation. This time, we use the transverse Laplacian in cylindrical coordinates, and we ignore the longitudinal z coordinate. Since ln(r) does not depend on θ, only the radial part of the Laplacian survives, so we get t2 f(r) = (1/r) ∂r [ r ∂rf(r) ] With f(r) = ln(r), the inner bracket [..] becomes 1, and ∂r[ 1 ] = 0, so t2 ln(r) = 0 for r ≠ 0. If we now shift the origin of our cylindrical transverse coordinate system by some arbitrary 2D vector x1, then the function that was ln(r) becomes ln |x - x1| . Here is a drawing to support this claim. Things are viewed along the z axis, the vectors lie in the plane of paper. The original coordinate system is on the upper left, the new one is on the lower right. Changing the origin of a coordinate system cannot change the fact that a function satisfies the Laplace equation, so we may conclude without further ado that: Fact: The function f(x,y) = ln s1 where s1 = |x - x1| is a solution of t2f(x,y) = 0. This is true for any 2D vector x1. Of course any linear combination of such functions like A ln s1 + B ln s2 also solves the 2D Laplace equation, where s1 = |x - x1| and s2 = |x - x2|. If A = -B, then we find that the function B ln(s2/s1) is a solution. Comment: The above discussion hopefully gives some insight as to why the ln(s) factors keep appearing in our Chapter 4 equations, such as (4.2.5). The integral expression for V(z) came from the difference of the potential φ between two conductors. As noted in equation (5.2.10), in the transmission line limit, this φ can be written as an integral involving ln(s) , φ12(x,y,z) = q(z) { !Syntax Error, Idx' dy' a1(x',y') ln(s) -!Syntax Error, Idx' dy' a2(x',y') ln(s) } where the dependence on x,y is contained in s = . This potential must satisfy the transverse Laplace equation. Application of the 2D Laplacian to both sides gives, t2 φ12(x,y,z) = q(z) { !Syntax Error, Idx' dy' a1(x',y')t2ln(s) -!Syntax Error, Idx' dy' a2(x',y')t2 ln(s) } = 0 The result is 0 since t2 ln(s) = 0, as noted in the above Fact. 6.2 The equipotentials of ln[s2/s1] are circles. In light of Section 6.1, we know that the following function satisfies the 2D Laplace equation: f(x,y) = ln[s2/s1] (6.2.1) where s1 = |x - x1| and s2 = |x - x2|, and x1 and x2 are arbitrary points in 2D space. Consider now the surface defined by ln[s2/s1] = B, (6.2.2) where B is a constant. This means that s2/s1 = eB or (s2)2 = e2B (s1)2 . Thus, [(x-x2)2 + (y-y2)2 ] = e2B [(x-x1)2 + (y-y1)2 ] (6.2.3) Since this is a quadratic form in which the coefficient of x2 is the same as the coefficient of y2, this must represent a circle. Assuming then the form, (x - xc)2 + (y - yc)2 = r2 (6.2.4) where (xc, yc) is the location of the circle center, and r is the radius, we find these results by multiplying out the terms in (6.2.3) and completing the two squares, xc = (x2 - kx1) / (1-k) (6.2.5) yc = (y2- ky1) / (1-k) (6.2.6) r2 = [k(x12 + y12 ) - (x22 + y22 )]/(1-k) + xc 2 + yc 2  (6.2.7) where we have defined, k = e2B (6.2.8) 6.3 Aligning the circles. Since the equipotential surfaces of f(x,y) = ln[s2/s1] are circles, the problem remains to cause two of these circles to align with the boundaries of our two wires. If we put the centers of the two wires at y=0, we see from (6.2.6) that we should select y1 = y2 = 0. This causes all equipotential circles to be centered at y=0. What values should be used for x1 and x2? Without loss of generality, we write x1 = D + d x2 = D - d (6.3.1) Equations (6.2.5) and (6.2.7) for the circle center and radius then become xc = D + d = D + d coth(B) (6.3.2) r = d = d= d |csch(B)| (6.3.3) The hyperbolic function expressions in terms of B follow directly from (8). Here are some crude plots of these two hyperbolics: Fig 1: Plots of the radius and x-center of equipotential circles versus B. It is now extremely helpful to make a plot of the family of circles that arises as the constant B is varied. We have already shown in Section 6.2 that each B corresponds to a circle, so we will now learn where all the circles lie. Fig 2: Plots of equipotential circles for various values of B ( drawn for D=0). The circles on the right correspond to positive B. As B → ∞, Fig 1 shows that r → 0 and xc → d. Thus, the circles shrink down around the point x=d on the right. For negative B, the same thing happens on the left. For B = 0, the circle is the plane at x=0. Apollonius? In Fig 2 we have indicated in heavy ink where our two wires might lie. The one on the left has radius a2 and that on the right has radius a1. Notice that the centers of the circles move out from x = |d| as the circles get larger. The distance between the centers of our wires is b. As the figure shows, b ≥ 2d. We are now ready to require an alignment of the equipotential circles shown in Fig 2 with the cross sections of our two wires. The constraints are as follows: a1 = d csch(B1) / radius of right circle is a1 (B1 > 0) (6.3.4) a2 = d csch(-B2) / radius of left circle is a2 (B2 < 0) (6.3.5) b = d [ coth(B1) - coth(B2) ] / distance between centers is b (6.3.6) Notice that the constant D which appears in (6.3.1) and (6.3.2) does not appear in the above three equations. It cancels out in (6.3.6), and has absolutely no bearing on the problem. We therefore set D = 0. This corresponds to the x=0 plane being as drawn in Fig 2 above. So, we now have three equations in three unknowns, B1, B2, and d. Inserting (6.3.4) and (6.3.5) into (6.3.6) and squaring twice gives us an expression for d as a function of a1, a2, b: d = (1/2b) (6.3.7) The potentials on the two wire surfaces are given by B1 = f(x1) = csch-1 (a1/ d) = ln [ (d/a1 ) + ] (6.3.8) B2 = f(x2) = - csch-1 (a2/ d) = - ln [ (d/a2 ) + ] (6.3.9) And the potential f(x,y) at all points in space between the conductors is given by, f(x,y) = ln(s2/s1) = ln [ ] (6.3.10) Putting (6.3.4) and (6.3.5) into (6.3.2), we get these useful formulas for the circle center locations: xc1 = d coth(B1) = d xc2 = d coth(B2) = - d (6.3.11) Finally, we arrive at our evaluation for K , K = f(x1) - f(x2) = B1 - B2 = ln [(d/a1) + ] + ln [(d/a2) + ] (6.3.12) If we are interested in a transmission with one conductor contained inside the other, Fig 3: Situation when conductors are concentric. then the above analysis still applies, except in this case B2 is positive, so there is no minus sign in (6.3.9), and there is a minus sign before the second term in (6.2.12). Here then is a summary box for our two wire transmission line. General solution of two-wire transmission line : φ(x,y,z) = (1/2πεε0) q(z) f(x,y) Le = (μμ0/2π)K Az (x,y,z) = (μμ0/2π) i(z) f(x,y) C = 2πεε0/K G = 2πσ/K f(x,y) = ln [ ] K = ln [ (d/a1) + ] ± ln [ (d/a2) + ] + sign for conductors as in Figure 2 (separated) – sign for conductors as in Figure 3 (concentric) d = (1/2b) (6.3.13) 6.4 Reduction to special cases (a) Two wire line: b >> a1, a2 In the limit that b >> a1, a2 we find from the box (6.3.13) that, d ≈ b/2 (6.4.1) K ≈ ln [ b/a1] + ln [ b/a2] = ln [ b2/(a1a2)] = 2 ln [b/] (6.4.2) and we are much relieved to find that we have duplicated the result (4.6.4). If we further assume that a1 = a2 = a, the result becomes, K = 2 ln (b/a) (6.4.3) Two wire line, b>>a1, a2 K = 2 ln [ b/] Le = (μμ0/2π)K C = 2πεε0/K G = 2πσ/K (6.4.4) (b) Twin Lead: a1 = a2 = a In the case that a1 = a2 = a with arbitrary b, we find that d/a = (b/2a) (6.4.5) K = 2 ln [ (d/a) + ] (6.4.6) When (6.4.5) is inserted into (6.4.6), we find K = 2 ln (b'/a) b' = b (6.4.7) Comparison of (6.4.7) with (6.4.3) shows that the large b result applies at arbitrary b if b is replaced by the b' shown. Notice that b' < b. One can interpret this as saying that, when the conductors are close together, the charges and currents are offset in each conductor toward the other conductor, so the effective separation is less than the nominal b. Two wire line, a1 = a2 = a K = 2 ln (b'/a) b' = b Le = (μμ0/2π)K C = 2πεε0/K G = 2πσ/K (6.4.8) (c) Off center coax: b<<a2 Look at Figure 3 in the previous section. Here we are talking about a coaxial cable where a2 refers to the radius of the outer conductor, and b is the offset between the conductor center lines. Normally b=0, but we are interested here to see how a manufacturing irregularity might affect the coaxial line properties. Let us define these dimensionless quantities: α = (b/a2) β = (a2/ a1) > 1 (6.4.9) From our general formula for d given in (6.3.13), we may write d = a1 f (6.4.10) where 2f ≡ (1/αβ) (6.4.11) Then we get (d/a1) = f (d/a2) = (f/β) (6.4.12) Inserting these into the formula (6.3.13) for K, we get K = ln(β) + ln [ ] (6.4.13) Notice that ln(β) = ln(a2/a1) is the usual result for a coaxial cable that is perfectly centered, as we obtained in (4.7.6), so the second term is the result of being off-centered. So far we have been exact, now we make an approximation. We assume that α << 1. In this case, looking at (6.4.11) above, we see that f >> 1. Then we can approximate the second term in (6.4.13) using standard methods, and we get this result K ≈ ln(β) - (6.4.14) Again, from (6.4.11) we keep only the largest term inside the radical to get 2f ≈ ( 1 - β2) / αβ (6.4.15) Putting this into (6.4.14) gives K ≈ ln(β) - = ln(a2/a1) - (6.4.16) Example: Consider Belden 8281 coaxial cable. Here are the basic numbers a2 = 2510μ a1 = 394μ β = a2/a1 = 6.37 lnβ = 1.85 (6.4.17) For these numbers, the denominator in the second term in (16) is essentially 1, so we get K = 1.85 - b/a2)2 (6.4.18) Suppose that the coaxial cable is off center by 20%, which seems a large amount, but an amount that could conceivably occur in very poor manufacturing. In this case α = b/a2 = 0.2, α2 = .04, so we get K = 1.85 - 0.04 = 1.81 Therefore, even such a large 20% defect results in a change in K that is a mere 2%. With a 10% defect we get (1/2)% error in K. We conclude that coaxial cables are fairly immune to eccentricity variations! By way of interpretation, notice that the correction term is negative. As the center wire moves off center, the capacitance of the cable increases, since K decreases. In the limit that the inner conductor almost touches the outer one, we have a1 + b = a2 which says αβ = β - 1. When this is put into (11) for f, we find that f = 0. Looking then at (13), we get the result that K = ln(β) + ln(1/β) = 0, so the capacitance is infinite. Off center coaxial cable b = distance between center lines a2 > a1 K = n(a2/a1) - Le = (μμ0/2π)K C = 2πεε0/K G = 2πσ/K (6.4.19) (d) Wire of radius a a distance h above a ground plane. This limit arises by taking a2 = ∞ in Figure 3 so that circle a2 lines up with the plane x=0. In this case, xc1 = h, the height of the a1 wire center above the ground plane. From 6.3 (11) we have then h = d (20) or h2 = d2 + a12 and (d/a1)2 = (h/a1)2 - 1 (6.4.21) From 6.3 (5), if a2 = ∞, then B2 = 0. Therefore the second logarithm in 6.3 (12) is zero, and we get K = ln [ (d/a1) + ] (6.4.22) Inserting (21) into (22) gives K = ln(h'/a1) , h' = h ( 1 + ) (6.4.23) This result has an easy interpretation. It represents one half of the twin lead situation discussed in (b) above. If we set h = b/2, then the result (23) represents 1/2 of the result shown in (8). Since K is half, the capacitance of a wire over a ground plane is twice that of the corresponding twin lead situation. It should be clear that to the right of x=0, the potential and field lines in both cases are identical. And in the limit of a thin wire, the above becomes K = ln(h/a1). Wire over ground plane h = height of center line over plane a = radius of wire K = ln(h'/a) , h' = h ( 1 + ) Le = (μμ0/2π)K C = 2πεε0/K G = 2πσ/K (6.4.24) Appendix A 1.1: Gauge Invariance Here we show why it is that, in choosing potentials φ and A, one is allowed to set the divergence of the vector potential A equal to an arbitrary function. Roughly speaking, this freedom of setting div A is called gauge invariance. Summary of this Appendix: 0. Fact 0: The Poisson Equation -2φ = ρ/ε0 has a unique solution φ(x) = ∫d3x' . 1. Fact 1: If div B = 0, there exists an A such that B = curl A and div A = 0. 2. Fact 2: If div B = 0, there exists A' such that B = curl A' and div A' = f(x), where f(x) is an arbitrary scalar field which "drops off" in some reasonable (sufficient) manner as x → ∞. 3. Fact 3: If curl E = 0, there exists a φ such that E = - grad φ. . 4. Fact 4: If div B = 0 and curl E = - ∂B/∂t , then there exist both A and φ such that B = curl A and E = - grad φ-∂A/∂t, and the quantity div A may be set to any function f. 5. Fact 5: Assume that potentials A and φ have been obtained as in Fact 4. There exists an infinite set of other potentials A' and φ' which yield the same E and B fields. They are given by A' = A + grad Λ φ' = φ - ∂Λ/∂t where Λ is an arbitrary scalar function of x and t. 6. Definition: Gauge Invariance 7. The Lorentz Gauge and QED 0. Fact 0: The Poisson Equation -2φ = ρ/ε0 has a unique solution as stated below. (a) Imagine some static charge distribution ρ(x) that is constrained to a localized region near the origin within infinite space. The distribution ρ(x) includes all charges in this region. Here are some types of charges which would be included in ρ(x): point charges which are "glued down" to certain points in space. linear continuous charge densities that are glued down along curved filaments in space or which are stable on conducting filaments. surface charge densities that are either glued to certain surfaces, or which are stable because they lie on the surfaces of pieces of conductor (like metal). These two charge types account for "Dirichlet boundary conditions", "Neumann boundary conditions", and "mixed boundary conditions", but these phrases need not concern us here. In the Neumann case, a piece of surface has two charge density layers of opposite sign very closely spaced (but not superposed), which is sometimes called a dipole layer. 3D continuous charge densities that are "glued down" somehow in 3D space so they cannot move. By including all these types of charge in ρ, we are able to avoid the complicating issue of "boundary conditions" in our discussion below, and our only boundary of interest is The Great Sphere which is a sphere of infinite radius surrounding our localized region of interest. From Coulomb's Law (in SI units) we know that the electric potential φ of a point charge q located at point x' is φ(x) = (1/4πε0)q/r where ε0 is a certain constant appropriate to SI units and r = |x-x'| is the distance between charge q at x' and an observation point x. Such a point charge is described by ρ(x) = qδ(x-x'). The general equation which relates φ(x) to ρ(x) is the Poisson Equation, -2φ(x) = ρ(x)/ε0 . (A.0.1) Since this is a linear equation (the operator 2 is linear), we may superpose the potentials of multiple charges to get the potential resulting from a distribution of charges. Thus, we at once obtain this superposed version of Coulomb's law, φ(x) = ∫d3x' . (A.0.2) Here d3x' ρ(x') = dq(x') = a differential chunk of charge located at x' contained in tiny volume d3x'. Thus, (A.0.2) must be a solution of (A.0.1). If we allow the observation point x to move right on top of some point charge in the distribution ρ, we will get φ = ∞, so we generally avoid such points. (b) We would like to explicitly show that (A.0.2) is a solution of (A.0.1) for a general distribution ρ. To this end, we digress to consider the following equation and its solution, -2g(x,x') = δ(x-x')/ε0 => g(x,x') = (A.0.3) The equation on the left is Poisson's Equation where ρ consists of a positive point charge of q = 1 unit sitting at position x'. Recall from above that ρ(x) = qδ(x-x') for a point charge. Coulomb's law gives the solution shown on the right. Therefore it must be true that -2{ } = δ(x-x')/ε0 or -2{ } = 4π δ(x-x') . (A.0.4) This fact can be verified directly using the theory of distributions which says -2(1/r) = 4πδ(r), but we have already shown it is true, given Coulomb's Law and superposition. We can now show that (A.0.2) is a solution of (A.0.1) for an arbitrary distribution ρ as follows: -2φ(x) = ∫d3x' ρ(x') {-2 } = ∫d3x' ρ(x') 4π δ(x-x') = ρ(x)/ε0 QED. The assisting function g(x,x') has various names with respect to (A.0.3): the Green's Function, the fundamental solution, the free-space propagator. It is nothing more than the potential created by a point charge of 1 unit located at x' and viewed from x. (c) We have found the particular solution of (A.0.1) given by (A.0.2). There are many other solutions which can be obtained by adding to the solution (A.0.2) a solution of -2u = 0. This last equation, usually written 2u = 0, is called the Laplace Equation, and it is the "homogeneous" form of the Poisson Equation, that is, the right side of the Poisson Equation is set to 0. Solutions u are called homogeneous solutions. One obvious solution is u = 2, so we could then add 2 to (A.0.2) and get a new solution to (A.0.1). Since we have specified that our charge distribution ρ(x) is localized to some region of space, we expect that as x→ ∞, we must have φ → 0. The solution (A.0.2) meets this requirement, but if we add 2, then our physical requirement is not met, so we must rule out adding a 2. We would also rule out 2x + 3, for example, or 7xy. Recall that 2 = ∂x2+ ∂y2+ ∂z2. It turns out that the only solution of 2u = 0 which meets the requirement u→0 as x→∞ in all directions is the trivial function u(x) = 0. In 2D one intuitively sees this because a massless thin rubber sheet tied down to height u = 0 around a large circular perimeter is going to be a flat rubber sheet with u = 0 everywhere. The solutions to the 3D equation 2u = 0 are called harmonic functions, and it is not hard to show that any harmonic function must take both is max and min values on the boundary, which here is a 3D great sphere. Thus umax = 0 and umin = 0, so the only possibility is that u(x) ≡ 0 everywhere. The implication of the previous paragraph is that (A.0.2) is the only possible solution of (A.0.1) because the only homogeneous solution one is allowed to add to (A.0.2) is u = 0. One can suppose there are two different solutions of -2φ = ρ/ε0 called φ and φ' both of which go to 0 on the great sphere. Then -2(φ-φ') = 0 with (φ-φ') → 0 on the great sphere. But then (φ-φ') = 0 so φ' = φ and there cannot then exist two different physical solutions of (A.0.1). 1. Fact 1: If div B = 0, there exists an A such that B = curl A and div A = 0. The gauge choice div A = 0 is known as the Coulomb gauge or the Transverse gauge. Proof: There are several parts to the proof: (a) If A exists such that B = curl A, then it will certainly be true that div B = 0, since div curl A = 0 for any vector A. The problem is showing that A exists, and moreover, that an A exists with div A = 0. (b) Consider the following differential equation (at this point A is some undefined vector field): -2A = curl B (A.1.1) or, in Cartesian coordinates, -2(Ai) = [curl B]i . This is a Poisson Equation for each individual Cartesian component i. We know that a Poisson equation of the form (A.0.1) has a unique solution of the form (A.0.2), so the solution of (A.1.1) is given by A(x) = . (A.1.2) As with ρ in the previous section, we think of curl B as being localized in some region near the origin and dropping off at large distances. Perhaps B is generated by some currents in this localized region. (c) Take the divergence of both sides of (A.1.2) [ implied sum on i ] div A(x) = ∂iAi(x) = . (A.1.3) We can replace ∂i by - ∂'i acting on 1/|x - x'| . Then we can do parts integration and move ∂'i onto [curl' B(x')]i with a parts sign change. In doing so, we assume that at infinity we pick up no "parts" since curl B is assumed to drop off sufficiently fast. We end up then with: div A(x) = . (A.1.4) But div curl F = 0 for any vector F , so the integral vanishes. Thus, we conclude that div A = 0 . (A.1.5) (d) Next, take the curl of both sides of (A.1.2). Here is the ith component [ implied sums on j and k ] : [curl A(x)]i = εijk∂jAk(x) = + εijk . (A.1.6) As before, we replace ∂j by -∂'j acting on (1/|x - x'|). Then do parts to move ∂'j onto [curl' B(x')]k. As before, there is no "parts contribution". The result can then be put back into full vector notation to give: curl A(x) = + . (A.1.7) Now use the vector identity curl curl B = grad div B - 2 B = - 2 B , since div B = 0. This gives curl A(x) = - . (A.1.8) The next step is to move the operator '2 onto the other integrand factor 1/|x - x'| by doing a double parts, and again for each parts operation there is no parts contribution from the Great Sphere at infinity. We then use the fact (A.0.4) that 2(1/|x - x'|) = - 4π δ(x-x') to get curl A(x) = - . (A.1.9) Thus, assuming div B = 0, we have formally constructed a vector field A such that B = curl A and div A = 0, and this was the claim of Fact 1 stated above. 2. Fact 2: If div B = 0, there exists A' such that B = curl A' and div A' = f(x), where f(x) is an arbitrary scalar field which "drops off" in some reasonable (sufficient) manner as x → ∞. Proof: From Fact 1, we first find A such that B = curl A and div A = 0. We then define A' ≡ A + Λ (A.2.1) where Λ is some so-far arbitrary function (scalar field). It follows that div A' = div A + 2 Λ = 2Λ . (A.2.2) We would like to have div A' = f, so we must find Λ such that 2Λf. But this is once again Poisson's Equation which from Fact 0 has this unique solution, Λ(x) = . (A.2.3) Meanwhile, from (A.2.1) we also conclude that, curl A' = curl A + curl grad Λ = curl A = B . (A.2.4) Thus, assuming div B = 0, we have formally constructed a vector field A' such that B = curl A' and div A' = f(x) where f(x) is any function we like that drops off sufficiently fast as x → ∞, and this is the claim of Fact 2. If f(x) drops off away from the origin, this is like the ρ(x) of Fact 0, and we find that Λ → 0 as x → ∞ in any direction. Then since Λ = 0 on the Great Sphere, we know that there are no homogenous solutions to 2Λ = 0 which could be added to (A.2.3) and so Λ(x) is uniquely determined by our select function f(x). 3. Fact 3: If curl E = 0, then there exists a φ such that E = - grad φ . Proof: This proof is almost identical to that of Fact 1, but a little simpler. (a) If φ exists such that E = - grad φ, then it will certainly be true that curl E = 0, since curl grad φ = 0 for any function φ. The problem is showing that φ exists. (b) Consider the following differential equation (at this point φ is some undefined scalar field): 2φ = - div E . (A.3.1) This is a Poisson equation for φ, we solve it as in (A.0.2) to get, φ(x) = . (A.3.2) As usual, we assume that div E drops off in some sufficient manner away from the origin going to infinity. Perhaps E is generated by a charge distribution in some region near the origin. (c) Next, take the grad of both sides of (A.3.2). Here is the ith component: ∂iφ(x) = . (A.3.3) As usual, we replace ∂j by -∂'j acting on (1/|x - x'|). Then do parts to move ∂'j onto div' E(x') with a second sign change, and also as usual there is no "parts contribution" from the Great Sphere. The result can then be put back into full vector notation to give: grad φ(x) = + . (A.3.4) Now use the vector identity grad div E = curl curl E + 2 E = 2 E , since curl E= 0. This gives grad φ(x) = . (A.3.5) As before, move the operator '2 onto the other term 1/|x - x'| by doing a double parts. We then use the fact (A.0.4) that 2(1/|x - x'|) = - 4π δ(x-x') to get grad φ(x) = (A.3.6) Thus, assuming curl E = 0, we have constructed a function φ such that E = - grad φ, so φ must exist, and this is the claim of Fact 3. 4. Fact 4: If div B = 0 and curl E = - ∂B/∂t , then there exist both A and φ such that B = curl A and E = - grad φ-∂A/∂t, and the quantity div A may be set to any function f. Proof: We know from Fact 2 that A exists such that B = curl A and such that div A equals any arbitrary function f (we dispense with prime on A). Consider the vector field E' = E + ∂A/∂t. Apply curl to find that curl E' = curl E + ∂B/∂t = 0. Now apply Fact 3 to this vector E' to conclude that there exists φ such that E' = -grad φ. But this says E = -grad φ - ∂A/∂t. Thus we have expressly found A and φ which satisfy the requirements of Fact 4. 5. Fact 5: Assume that potentials A and φ have been obtained as in Fact 4. There exists an infinite set of other potentials A' and φ' which yield the same E and B fields. They are given by A' = A + grad Λ φ' = φ - ∂Λ/∂t (A.5.1) where Λ is an arbitrary scalar function. It is by making such transformations that div A' can be set to an arbitrary function f, as discussed in Fact 2. Proof: We already showed in Fact 2 that A' = A + grad Λ does not alter B. The second condition is required so that E = -grad φ - ∂A/∂t also remains unaltered. That is, E' = - grad φ'- ∂A' /∂t = -grad (φ - ∂Λ/∂t) - ∂/∂t (A + grad Λ) = - grad φ - ∂A/∂t = E . 6. Definition: Gauge Invariance. In electromagnetism, the situation of Fact 4 arises for B = magnetic field A = vector potential E = electric field φ = scalar potential The two equations shown in Fact 5 are called a gauge transformation of the potentials. One transforms from A,φ to A',φ' without altering the physical electromagnetic fields E and B. The electromagnetic fields are thus invariant under such a gauge transformation, and one says that the classical theory of electromagnetism is gauge invariant. 7. The Lorentz Gauge and QED: This section is certainly off the transmission-lines beaten path, but the author thought the reader might find it interesting. It is true that the nature of a transmission line results from photons "jumping back and forth" between the conductors. Unlike elsewhere in this document, everything is not fully explained in the following quick outline. In relativistic notation one uses 4-vectors which have one time component and three spatial components such as xμ = (t,x,y,z). This is a contravariant 4-vector and the corresponding covariant 4-vector is xμ = (t,-x,-y,-z). Thus, one has x0 = x0 but xi = -xi. We are assuming here the "Bjorken-Drell metric" gμν = diag(1,-1,-1,-1). The gradient operator ∂i "transforms as" the covariant part of a 4-vector, and one can write ∂i = -∂i just as xi = -xi for i = 1,2,3. A four-vector gradient operator can be written ∂μ = (∂0, ∂i) and ∂μ = (∂0, ∂i) = (∂0, -∂i) where ∂0 = ∂0 = ∂t = ∂/∂t. The Laplacian is 2 = ∂i∂i (implied sum on i) while the corresponding object ≡ ∂μ∂μ = ∂t2 - 2 is the D'Alembertian which appears in wave equations. As is normal in discussions of this type, the speed of light is taken to be c = 1. Consider the gauge transformation (A.5.1) which in relativistic notation is A'i = Ai + ∂iΛ = Ai - ∂iΛ i = 1,2,3 φ' = φ - ∂0Λ (A.7.1) The components of a classical vector like A, normally written as Ai, are in fact the contravariant components Ai in relativistic notation. If we now identify φ ≡ A0 we can combine the two gauge transformation equations into a single equation involving three 4-vectors (one of which is ∂μΛ), A'μ = Aμ - ∂μΛ μ = 0,1,2,3. (A.7.2) Suppose we want ∂μA'μ = 0 (implicit sum on μ = 0,1,2,3). If we could find a potential A'μ with this property, that would be very convenient for the following reason: In general aμbμ (= aμbμ = a b) is the same in all frames of reference related by Lorentz Transformations. If ∂μA'μ = 0 in one frame, it is 0 in all frames, and that makes computational life simple. So, is it possible to have ∂μA'μ = 0 ? Writing this out we get ∂tφ + ∂iA'i = 0 or ∂tφ + div A' = 0 or div A' = -∂tφ. (A.7.3) But we showed in Fact 2 that given any A, we can find an equivalent A' which has div A' = any f(x) we want, so we just select our arbitrary function f(x) to be -∂φ/∂t. By selecting this f(x), we are selecting the Lorentz Gauge. In this gauge (dropping the prime on A), we have ∂μAμ = 0. As just noted, this gauge is so named because the resulting equation ∂μAμ = 0 is "covariant" under all Lorentz transformations. That means the equation has the same form in all frames of references which are related by Lorentz transformations ( which include rotations, velocity transformations, and combinations of same). One can interpret ∂μAμ = 0 as ∂A = 0 where ∂ is a 4-divergence operator. Thus, in the Lorentz gauge, the 4-divergence of Aμ is always exactly 0 at every point in spacetime. In relativistic quantum field theory (aka quantum electrodynamics, or QED), the potential Aμ is interpreted as the quantum field of a vector particle called the photon. In the Lagrangian density which describes the interaction between the photon and electron, the term JμAμ appears, L = ... - JμAμ Jμ = e0 γμ ψ (A.7.4) where Jμ is the electric current, an operator built from the quantum field ψ of the electron. Because of gauge invariance (A.7.2), a gauge transformation creates a new term Jμ ∂μΛ in the Lagrangian density. The physics of QED is determined by S = ∫d4x L = ∫d3x ∫ dt L which is called the action. If we insert the gauge term Jμ∂μΛ into the action and do part integration to move ∂μ from Λ to Jμ, we end up with a gauge term of the form ΔS = ∫d4x (∂μJμ)Λ. But at every point in spacetime, we know that ∂μJμ = 0 (shown in a moment) so we find that ΔS = 0 which means the action S is invariant under a gauge transformation. The reason ∂μJμ = ∂μJμ = 0 is because Jμ = (ρ, Ji) where ρ is charge density and Ji is electric current, and then the statement ∂μJμ = 0 says that ∂tρ + ∂iJi = 0 or div J = -∂ρ/dt. This is the equation of continuity which says that if there is a current flowing out of a tiny volume of space, the charge density in that volume must be correspondingly decreasing. In other words, charge is "conserved". We can reverse our logic to conclude that the reason electric charge is conserved and cannot leak away "into the vacuum" is due to the invariance of the QED action under gauge transformations (A.7.2). More generally, symmetries (invariances) of the action always result in conserved quantities. Since 1949, unusual names have been given to similar conversed quantities: isospin, strangeness, color, charm, etc. The association of a conserved quantity with a differential symmetry of the action is known as Noether's Theorem, in honor of Emmy Noether who first showed this connection in 1915. Appendix B 1.2: The Complex Dielectric Constant. In our transmission line analysis, we follow the general approach used in King (see Refs). One notion that King finds useful is that of a complex dielectric constant ξ which incorporates the conductivity σ of a medium along with the normal dielectric constant ε. Associated with this complex dielectric constant is a so-called "effective" surface charge neff . The term "dielectric constant" is somewhat of a misnomer. As noted below, ε and ξ are not constants but in fact are strong functions of frequency ω. When small ranges of ω are considered, ε appears to be constant if all resonances are far away. Summary of this appendix: 1. A conducting parallel plate capacitor. 2. Continuity and free charge in a medium. 3. Complex ε and the loss tangent of a dielectric. 1. A conducting parallel plate capacitor. A good way to understand ξ and neff is to consider a simple "device": a parallel plate capacitor filled with a conducting dielectric. One can regard this device as a parallel RC circuit. Let us first imagine that the R portion is turned off and we have just the C part. Then, assuming the usual exp(jωt) time dependence, we can say that Q = CV = nA, where A is the area of a plate and n is the surface charge density per unit area. Doing a time derivative then gives I = jωnA, so we have a simple relationship between the current I and the area surface charge density n: I = jωnA If we now turn on the R part of our device, we get an extra contribution to the current, so now: I = jωnA + V/R R = L/(σA) // R = V/IR = EL/JA = (L/A)(E/J) = (L/A)(1/σ) Here L is the spacing between the plates, and σ is the dielectric conductivity. We know that V = EL, where E is the electric field between the plates, and we also know that E = n/εε0. This last fact follows from (1.1.3) that divD = ρ applied to a thin gaussian box containing one of the plate surfaces. Using these two facts, we can rewrite the above as I = jωnA + (nL/εε0 ) (σA/L) = jωAn[ εε0 + σ/(jω) ]/εε0 jωAn(ξ/εε0where ξ ≡[εε0 + σ/(jω)] (B.1) = jωAneff where neff ≡ n (ξ/εε0) (B.2) By defining neff in this manner, we are able to incorporate the effect of the R part of our device into the C part. In solving a problem, we can of course work either with the true surface charge n, or with the effective defined quantity neff: Notice that, neff/ξ = n/εε0. (B.3) The advantage of using neff is its simple relationship to the total current I in a situation where both R and C effects are present. Main Point: For our parallel plate RC device, I = jωneffA, which mimics the result I = jωnA when there is no resistive loss. Before departing this section, it is useful to write down a formula for the admittance Y of our "device", which the reader can easily derive from the above expressions for I and V, Y ≡ I/V = G + jωC where G = 1/R = σA/L C = εε0A/L = jω(ξ/εε0)C] = jω C' where C' = (ξ/εε0)C (B.4) Again, using the definition of ξ, we can absorb the R effect into the C part of our device by replacing C with the complex capacitance C'(ξ/εε0)C. 2. Continuity and free charge in a medium Consider the continuity equation div J = -∂ρ/∂t. This applies to the conduction current J = σE in a medium, and ρ is the free charge. Using div J = σ divE = σρ/εε0) we get: ρσ/εε0) = -∂ρ/∂t . This is a first order differential equation whose solution is ρ(t) = ρ(0) exp(- t/τ) τ = εε0/σ (B.5) This says that if there is free charge out in the middle of a conducting medium, it will run away and park on the boundaries. In our capacitor, this happens with time constant τ despite the fact that the capacitor plates are changing polarity at rate ω. For copper, (1/σ) = 1.72 x 10-8 ohm-m, ε = 1, ε0 = 8.85 x 10-12 F/m, so we find τ = 1.5 x 10-19 sec. For polyethylene, (1/σ) ~ 1015, so  ~ 8850 seconds = 2.5 hours. References! Our reason for bringing this subject up at this point is to show that it is not reasonable to replace ∂ρ/∂t with jωρ out in the middle of a medium. Doing so in div J = -∂ρ/∂t with divJ = ρσ/εε0) results in ρ(σ/εε0) = -jωp or [ jω + σ/εε0 ] ρ = 0. The solution to this equation is ρ = 0. In other words, there is no steady-state sinusoidal free charge density out in the middle of a medium, conducting or otherwise. Even if free charge could somehow exist at t = 0 inside a medium, after t >> τ free charge is only allowed on boundaries. One can of course have a current J in a medium, but the charge is balanced so there is zero net charge density, ρ = 0. 3. Complex ε and the loss tangent of a dielectric. The normal dielectric constant in fact has a real and imaginary part, usually expressed as: ε = ε' - jε" (B.6) Thus, we can write our complex dielectric constant ξ as ξ = [εε0 - jσ/ω] = ε'ε0 - j(σ + ωε"ε0)/ω] The cause of ε" is a time-lag loss mechanism in the dielectric. In polar dielectrics, this is due to friction which keeps the polar molecules from instantly following the E field at high frequencies. In non-polar dielectrics (those used in coax cables) this is mainly due to the classical resonance absorption mechanisms, such as molecular vibrational modes. The standard classical (non-quantum) theory for ε gives this general form for the behavior of ε, [ Reference! ] ε(ω) = 1 + (B.7) This is basically the absorption of a damped harmonic oscillator of effective mass m, charge q, damping Γ with resonance at ω0. N is the density of such absorbers per cubic meter. A real dielectric is a superposition of terms of the above form, with appropriate quantum interpretations of things. The main point of showing the above form is that in general ε(ω) always has an imaginary part, which comes from the jΓ damping term in the denominator. In general ε" is small except in the region of a resonance (in the above, at ω such that ω ~ ω0). We can conveniently incorporate both loss mechanisms (conduction of dielectric and absorption) by defining an effective conductivity of the dielectric, σeff ≡ σ + ωε"ε0 => (B.8) ξ = [ε'ε0 - jσeff/ω] = ε'ε0 [ 1 - jσeff/(ωε'ε0)] = ε'ε0( 1 - j tanL) (B.9) tanL ≡ σeff/(ωε'ε0) = [σ(ω) + ωε"(ω)ε0]/[ωε'(ω)ε0] . (B.10) This last quantity is called the "loss tangent" (or dissipation factor) of the dielectric because it is (minus) the ratio of the imaginary and real parts of ξ, and there is a "loss angle" in the complex ξ plane which is θL = tan-1(tanL) = tan-1[σeff/(ωε'ε0)]. We have written ω in lots of places to emphasize that the loss tangent is a function of frequency. In practice, the main frequency dependence comes from ε"(ω). ok to here Appendix C 2.1: DC Properties of a Wire 1. The DC resistance of a wire. The resistance per unit length R of a differential piece of wire is derived as follows: V = E dz, J = σ E, I = J dA, Rdz = V/I = Edz/JdA = (1/σ) dz/dA R = (1/σ) /dA If the conductivity σ is constant across the wire, then R = 1/(σA), where A is the cross sectional area. Using resistivity ρ = 1/σ we have R = ρ/A (C.1.1) For round wire of radius a, A = πa2, so R = ρ/(πa2). ok to here 2. The DC surface impedance of a wire. Imagine a wire carrying current I. If, at the surface of the wire, we put voltmeter probes at longitudinal spacing dz, we will get some potential difference which is dV = Ezdz. When probed at the surface, the wire appears to have this impedance per unit length (Zsdz) = dV/I The quantity Z is called the surface impedance per unit length and is thus given by Zs = Ez / I (C.2.1) where Ez is the component of electric field at the surface in the direction of the wire. For a wire operating at DC, the current density is uniform across the wire so we have Jz = I/A and Ez = Jz / σ = I/(Aσ). Thus, Zs = 1/(Aσ) = ρ/A (C.2.2) where A is the cross sectional area. For a round wire of radius a, A = πa2, so Zs = ρ/(πa2) (C.2.3) If the wire is a perfect conductor, ρ = 0 and Zs = 0. 3. The DC inductance of a round wire. 3.1 Internal inductance Inductance is a slightly harder problem. We compute it from the fact that the energy density stored in a magnetic field is [ reference? ] dU = (1/2) BH dV = (1/2) μμ0H2 dV, (C.3.1) Since power P = IV and V = L dI/dt, we can write P = d/dt[ (1/2)L I2 ] . Thus, the energy stored in the magnetic field of an inductor is U = (1/2)L I2. If we integrate this over the interior of a conductor to get the total stored energy U, we can solve for the internal inductance per unit length Li by setting U = (1/2) (Li dz) I2 (C.3.2) If we transverse put a circle at radius r, the enclosed current is clearly I(r/a)2, where I is the total current in the wire. Again, we are assuming that J is uniform across the wire. We set this enclosed current equal to 2πrH(r) [ using some law ] to get the magnetic field at r, H(r) = (I/2πa2) r r ≤ a (C.3.3) This field is in the azimuthal direction. In cylindrical coordinates it would be called Hφ. The energy stored in a ring of volume 2πrdrdz is dU(r) = πμμ0H2dz rdr = (μμ0 dz I2 / 4πa4) r3dr Integrate this from r=0 to r=a to get the total stored energy, Ui = (μμ0 dz/ 16π) I2 (C.3.4) Set this equal to (1/2)Lidz I2 to find, Li = μμ0/ 8π (C.3.5) Since μ0 = 4π x 10-7 H/m, we conclude that Li = 50μ nH / m [ external verification? ] Interestingly, this result is independent of the radius of the wire. It had to be because there is no combination of μ0 and radius a which has dimensions H/m. For a given current I, the total field energy stored in the wire is independent of a. For small a, the field is stronger but in a smaller volume. 3.2 External inductance What about the external inductance of a round wire? We can compute it by the same method. If we put a circle at radius r ≥ a, the enclosed current is I. We set this enclosed current equal to 2πrH(r) to get the magnetic field at r, H(r) = (I/2π) r-1 (C.3.6) The energy stored in a ring of volume 2πrdrdz is dU(r) = πμμ0 H2dz rdr = (μμ0 dz I2 / 4π) r-1 dr Integrate this from r=a to some large radius r=R to get Ue = (μμ0 dz/ 4π) ln(R/a) I2 (C.3.7) Set this equal to (1/2)Ledz I2 to find, Le = (μμ0 dz/ 4π) ln(R/a) (C.3.8) If we set R = ∞ to get the total external inductance per unit length of a round wire, the result is logarithmically divergent. The total magnetic energy stored per unit length of an infinitely long round wire in isolation is infinite. It takes an infinite amount of work to build up such a field. Suppose there were two wires with currents flowing in opposite directions. In this case, we could compute the magnetic field H at any point in space as the vector sum of the fields of the two wires, then we could integrate H2 over all space to get the total energy U, and then the external inductance Le. In this case, the ln(R) divergence does not appear. In effect, it cancels between the two wires. Since we will be doing this computation by another means in the text, we do not bother with this calculation here. We really only care about the internal inductance Li of a wire in our transmission line analysis because the external inductance Le is already accounted for by the techniques of Chapter 3. That is, Le is computed by considering the magnetic potential Az ( or W ) between the wires. 4. The DC inductance of a wire of arbitrary cross section The main purposes of this section are to show how divergences can be handled, and to remind ourselves that we really have to attack differential equations directly to get real solutions, except in the simplest cases. 4.1 Statement of a Plan of Attack Here is a possible program for computing the DC inductance of a wire of arbitrary cross section. It is very similar to the above except for the starting point. (a) Compute Az(x,y) for an arbitrary wire at DC using formula (1.5.8) with β = 0. (b) Compute B = curl A. (c) The magnetic energy density is then dU/dV = (1/2)B2/μμ0. Integrate this over the interior or exterior of a slice of the wire to get Ui or Ue. For Ue we know we have to use a cutoff because the result is going to be log divergent as was the case for the round wire. (d) Set U = (1/2) L I2 to extract the appropriate inductance. We are mainly interested in the internal inductance. 4.2 The divergence problem Since we are at DC, we know that the current density will be uniform in the conductor, no matter how weird it's cross sectional shape (assumed uniform in z). Thus we are allowed to remove Jz from the integration, leaving the following form, where R = |x-x'|: [where did this come from? (1.5.8) ω = 0 ?] Az(x,y,0) = (C.4.1) Since the wire is infinitely long, we can try to do the dz' integration. We at once get our first sign of trouble, because this integral diverges. This is a reflection of the same problem noted earlier, that there is something unphysical about an infinite wire. Define, s = (C.4.2) Then install a very large cutoff Λ/2 on the dz integration and write: = 2 = 2 = 2 ln[ z+ ] | = 2ln[Λ/2 + = 2ln(Λ) - 2lns = - 2ln(s/Λ) When the transverse portion of the integration is done, the 2ln(Λ) term is just a large constant. When we take the curl of A to get B, this infinite constant will have no effect, and we will get a finite B field, so we can continue with the program. To keep track of dimensions, it is useful to keep Λ around, so we have, Az(x,y,0) = - ∫ dx'dy' ln(s/Λ) (C.4.3) where the transverse integral is over the shape of the wire. Whatever shape wire we choose, we can in principle do the above integration and carry out the program. At worst, we have to resort to numerical techniques. For a round wire, the above integration becomes Az(r) = - !Syntax Error, Ir' dr'!Syntax Error, Idθ ln (/Λ ) (C.4.4) Although it looks like a mess, these are standard integrals, and the results of the previous section are easily duplicated. One subtlety in the above integration occurs when r is less than a, so we are computing the potential inside the wire. In this case, the dr' integration must be broken up into two pieces, one going from 0 to r, and the other from r to a. The reader may notice a similarity between (3) and results of Chapter 4 such as 4.4 (3) which contain factors ln(s1/s2). This is no coincidence of course since we took the small β limit in Chapter 4, and here we are dealing with the β = 0 limit (DC). 4.3 Avoiding the divergence problem. A way to avoid this divergence business is to apply transverse derivatives to both sides of (C.4.1) right at the start, before doing any integrations. It is really these derivatives of A that we need to compute B in Step (b) of the program outlined above. For example, the x derivative of (C.4.1) becomes = - ∫dx'dy' (x-x') !Syntax Error, Idz' (C.4.5) The z' integration is now a convergent integral equal to, !Syntax Error, Idz = 2/s2 (C.4.6) where s is given in (C.4.2). This leaves a straightforward integral over the conductor cross section, = -∫dx'dy' (C.4.7) The other derivatives can be computed in a similar fashion. These resulting integrations are all well behaved and doable. For strange geometries, numerical methods are needed. In any event, we have outlined a method by which the internal DC inductance of any wire can be calculated. For a round wire we know from (C.3.5) above that Li = μμ0 /8π . For a rectangular conductor of dimensions a and b, the result can be obtained using (C.4.7) and its y counterpart. We do not know the answer, but it is computable, and it must have the form Li = (μμ0 /8πf(a/b) where f(x) is the function one obtains by doing the computation. A square wire would then have f(1). As a square wire is gradually deformed into a round wire, f(1) gradually deforms into 1. The method presented in this section is not very useful for an AC calculation, because then we can no longer assume that the current density is uniform across the wire. This will become clear later when we compute the exact current density for a round wire at arbitrary frequency. Again, this is similar to Chapter 4 results such as (4.4.3). We need some other method to compute the distribution of charge and current on a conductor. Appendix D 2.2: Electric Field in a Round Wire In this Appendix we present a somewhat lengthy discussion of the electric field (and therefore the current) in a round wire without assuming that such fields are symmetrical about the axis. 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, along with the current Jz, is in fact non-uniform across the wire. This result is true even in the DC limit of a transmission line. 4. We suggest a method by which one can compute 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 at the end. 1. The General Method and Solution for Ez The starting point for the calculation is the Helmholtz equation for the E field . [ 2 + β2 ] E(r,φ z,t) = 0 (D.1.1) We use cylindrical coordinates, as appropriate for a 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) = exp[ +jω(t - βd z) ] 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 travelling down a transmission line in the +z direction. We assume that one of the conductors of this transmission line is our round wire. The other conductor is unspecified. Partial Wave Expansion The next step is to do a partial wave (Fourier) expansion of E(r,φ) in terms of "azimuthal harmonics", so that the variable φ is replaced with the partial wave index m: E(r,φ) = (D.1.3) Since the usual complex part has been extracted in (D.1.2), we assume that E(r,φ) is real, which implies that E(r,-m) = E(r,m)* . We can then re-express the above as E(r,φ) = E(r,0) + = E(r,0) + (D.1.4) Below we shall find an exact solution for the partial wave amplitudes E(r,m). One should imagine these solutions sitting in (D.1.4), so that the azimuthal dependence is clearly understood. Now, let n(φ) represent the surface charge density (Coulombs/m2) which resides on the surface r=a of our round wire. We expand it in partial waves exactly as done above to get: n(φ) = = N0 + (D.1.5) The Nm are the "moments" of the surface charge distribution. They may be computed as, [ Ref? ] Nm = !Syntax Error, Idφ n(φ) e-jmφ (D.1.6) The Nm may be complex. This simply keeps track of cos(mφ versus sin(mφ) components. Charge Pumping Boundary Condition The reason we are so interested in n(φ) is that it acts as a driving source of the electric field in the wire through the following boundary condition on the radial current: Jr(r=a,φ) = jω n(φ) (D.1.7) This is a statement of divJ = -jωρ at the surface of the wire. 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.1.8) Since we have parallel partial wave expansions for both sides, we can rewrite (D.1.8) in the m space as Er(r=a,m) = (jω/σ) Nm . (D.1.9) Thus, the radial electric field must have a certain value at the 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. Due to divJ = 0 inside the wire, this radial current is converted into the usual longitudinal current one expects to find inside the conductors of a transmission line. The Ez Solution Here is the Helmholtz equation (D.1.1) for Ez ∂r2 Ez + (1/r) ∂r Ez + (1/r2) ∂φ2 Ez + ∂z2 Ez + β2Ez = 0 . (D.1.10) According to (D.1.2), we make the replacements ∂z = -jβd and ∂φ = +jm, and multiply through by r2. The result is r2∂r2 Ez + r ∂r Ez + (r2 β'2 - m2) Ez = 0 (D.1.11) where β'2 = β2 - βd2 (D.1.12) In a conductor, β is huge compared to the dielectric βd, so we could ignore the distinction between β and β'. Equation (D.1.11) is Bessel's Equation, and its solution -- subject to the condition that Ez not blow up at r=0 -- is this, Ez(r,m) = Czm Jm(β'r) (D.1.13) where Czm is an arbitrary constant for each partial wave m. Equation (D.1.13) is in agreement with (2.1.21). In Section 2.1 we dealt only with the m=0 partial wave, which embodies the symmetrical part of the problem. Also, we implicitly assumed constant z behavior, so the question of β and β' never arose. 2. The Solutions for Er and Eφ The Er Solution It is a characteristic of curvilinear coordinate systems that the Laplacian operator applied to a vector quantity causes a cross coupling between field components which does not occur in Cartesian coordinates. In our case, this affects the Er and Eφ field components, but not Ez. Here is (D.1.1) for Er: 2Er - (2/r2 ) ∂φEφ - (1/r2) Er + β2 Er = 0 (D.2.1) The fact that Eφ is cross-coupled in is a minor inconvenience. We eliminate it by using the condition that divE = 0, which in cylindrical coordinates appears as ∂r (r Er) + ∂φEφ + r ∂zEz = 0 (D.2.2) Expanding the 2 operator in (D.2.1), and eliminating ∂φEφfrom (D.2.2) , we get: ∂r2 Er + (3/r) ∂rEr + (1/r2) Er + (1/r2) ∂φ2 Er + ∂z2 Er + β2Er = (-2/r) ∂z Ez (D.2.3) We got rid of Eφ , but now we are stuck with Ez on the right. Actually, we are happy to have this factor on the right, because this is how the longitudinal current in the wire gets coupled into the radial current which feeds the surface charge. As before, we set ∂φ2 = -m2 and ∂z = -jβd . Then we can insert our solution (D.1.3) on the right: ∂r2 Er + (3/r) ∂rEr + (1/r2) Er + (-m2/r2) Er + β'2Er = (2jβd/r) Czm Jm(β'r) (D.2.4) In order to get the left side into something recognizable, we define Er(r,m) = x-1 fm(x) (D.2.5) where x ≡ β'r . (D.2.6) This x is a dimensionless radial variable which will play a major role in the following. Also, we will use the notation xa = (β'a) for x at the surface. Equation (D.2.4) with (D.2.5) now becomes, x2 fm" + x fm' + (x2 - m2) fm = Km x2 Jm(x) (D.2.7) where we have defined Km = 2j (βd/β') Czm (D.2.8) The left side of (D.2.7) is the normal Bessel operator, 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 the Ym(x) since they blow up at x=0 and thereby cause our 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 particular solution to (D.2.7) = (1/2) Km [ x Jm+1(x) ] . (D.2.9) Therefore, we now have this full solution for Er(r,m) in (D.2.5): Er(r,m) = am x-1 Jm(x) + Jm+1(x) . (D.2.10) For each value of m, there are two as-yet undetermined constants, am and (Km/2). However, looking at (D.2.10), we see that , since at small J0(x) ≈ 1, we must have a0 = 0 (D.2.11) to keep Er finite. The Eφ Solution We convert the divE=0 equation (D.2.2) to variable x and insert Ez from (D.1.13) with (D.2.8) to get jmEφ (r,m) = (Km/2) x Jm(x) - ∂x (x Er(r,m)) (D.2.12) When (D.2.10) is inserted for Er we find that, jEφ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) (D.2.13) We have already noted that a0 = 0. For m=0, (D.2.13) is invalid because we divided (D.2.12) by m. In fact, we really know nothing at all about the field Eφ(r,m=0) from our solution method, because we obtained the Eφ fields from the ∂φ Eφ equation (D.2.2). Since Eφ(r,m=0)ej0φ does not depend on φ, this simply gives 0, but does not tells us what Eφ(r,m=0) might be. Presumably we could find it from the φ Helmholtz equation which we have so far ignored. Conjecture: It seems likely that Eφ(r,m=0) = 0. In the m=0 partial wave where everything is symmetrical in azimuth, it is hard to imagine a constant (in φ) field Eφ circulating around the axis of the wire. Thus, we interpret the quantity (+ ) in (D.2.13) as being 0 when m=0. Application of the Boundary Conditions. We have two boundary conditions to impose: Er(r=a,m) = (jω/σ) Nm (D.2.14) Eφ(r=a,m) = 0 (D.2.15) The first is the charge pumping condition shown in (D.1.9) above, while the second is a condition requiring that there be no tangential E field at the surface of the wire. If there were, these fields would move the charge distribution n(φ) in the azimuthal direction, and then we do not have a self-consistent solution to our problem. Moreover, we like this condition because we are used to having the surface of a transmission line conductor be an equipotential of φ(x) at constant z. The algebra is a little messy, but when the above conditions are applied to the fields shown in (D.2.10) and (D.2.13), we are able to determine the two unknown constants in each partial wave. We find: = (jω/2σ) Nm { - } m≥0 (+ ) = (jω/2σ) Nm { + } m>0 am = m (jω/σ) Nm { } m≥0 (D.2.16) The middle equation of the above set should not be used for m=0, due to the 0/0 condition, but the other two are correct. For K0 , the two inverted factors in the first equation are the same, giving = (jω/σ) N0 { } (D.2.17) We have now completely solved the problem. To review, here are the three fields gathered together in one spot. For Ez we use (D.2.8) to replace Czm . jEφ(r,m) = - am x-1 Jm(x) + (+ ) Jm+1(x) Er(r,m) = am x-1 Jm(x) + Jm+1(x) Ez(r,m) = -j(β'/βd) Jm(x) (D.2.18) 3. 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, rather than the factors Nm as is done above. The first step in doing this is to compute the total current I in the wire, I = !Syntax Error, Idφ !Syntax Error, Ir dr Jz(r,m) ejmφ (D.3.1) It is clear from (D.3.1) that only the m=0 partial wave can make any contribution, so we write I = !Syntax Error, I2πrdr σEz(r,m=0) = 2πσ [-j(β'/βd ) ] !Syntax Error, Ir dr J0(x) = -2πσj {} !Syntax Error, Idx x J0(x) = -2πσj [(jω/σ) N0 { }] xaJ1(xa) = -2πσj N0 a2 (D.3.2) At this point it is convenient to introduce the DC resistance per unit length of our wire, Rdc = (D.3.3) along with a new symbol to indicate the relative surface charge moment, ηm ≡ (D.3.4) Using (D.3.3) and (D.3.4) in (D.3.2) gives (jω/σ) Nm = (j/2) (aβd) ηm I Rdc (D.3.5) We can use (D.3.5) to get our coefficients in (D.2.16) into a form that is then normalized to the total current I. Putting all the pieces together, we get: E fields in a Round Wire, m > 0 jEφ(r,m) = (j/4) (aβd) ηm I Rdc [ - + { + } ] Er(r,m) = (j/4) (aβd) ηm I Rdc [ + + { - } ] Ez(r,m) = (1/4) ηm I Rdc [ - ] (D.3.6) E fields in a Round Wire, m = 0 jEφ(r,m) = 0 Er(r,m) = (j/2) (aβd) I Rdc [ ] Ez(r,m) = (1/2) I Rdc [ ] (D.3.7) Comments about the solution: (1) Notice that 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. (2) 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 and hence Jz is non-uniform across the surface of the wire. 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. (3) Because Ez is non-uniform, we conclude that the surface impedance Zs = Ez/I will also be non-uniform around the boundary of the wire. In fact, here is our explicit formula for the surface impedance of a round wire. At r=a, it is a function only of azimuth angle φ : Zs (φ) = (1/2)Rdc { [ ] + } (D.3.8) 4. The Low Frequency Limit If we assume that ω is small enough that xa = β'a << 1, we can greatly simplify the above results. Another way to state this limit is that (a/δ) << 1, which means the skin depth is much larger than the radius of the wire. All limits come from the leading term of Jn(x) which is Jn(x) ≈ xn/ (2nm!) must be Jn(x) ≈ xn/ (2nn!) (D.4.1) In the formulas for Eφ and Er in the box (D.3.6), the leading term is the first term, the other terms may be neglected. We find that, ≈ (r/a)m-1 (D.4.2) and the first factor in Ez is its leading term, ≈ 2(m+1) (r/a)m (D.4.3) For the m=0 fields we notes these facts, [ ] ≈ (r/a) [ ] ≈ 2 (D.4.4) So here is a complete summary of the low frequency results: E fields in a Round Wire, m > 0, Low Frequency Limit jEφ(r,m) = (j/4) (aβd) ηm I Rdc [ - (r/a)m-1 ] Er(r,m) = (j/4) (aβd) ηm I Rdc [ + (r/a)m-1 ] Ez(r,m) = (1/2) ηm I Rdc [ (m+1) (r/a)m ] (D.4.5) E fields in a Round Wire, m = 0, Low Frequency Limit jEφ(r,0) = 0 Er(r,0) = (j/2) (aβd) I Rdc (r/a) Ez(r,0) = I Rdc (D.4.6) In this very last equation, we recover the fact that a piece of wire really does act like a resistor at low frequencies. However, there is still a small linear radial field which serves to feed the symmetric surface charge in the m=0 partial wave. 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. See Section 3.7, Figure 2 for a drawing of the current we have just computed. 5. What about φ, A and B ? Inside the conductor, we know that φ satisfies (D.1.1), and we know that φ is constant at r=a. 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.5.1) 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.5.2) 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.5.3) 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.5.4) from which we can determine φ0The potential Ar is discontinuous at r=a. Appendix E 3.1: Surface Charge It is often said that surface charges only exist very close to the surface of a conductor. In this section, we will show how extremely true this statement is. Here is a crude sketch of what we expect surface charge distributions might look like at the plates of a capacitor. ( redo this drawing with larger E and ρ) The heavy plot is charge density ρ, and the lighter plot is the electric field magnitude. The figure suggests that the charge distribution might have an exponential decay at each surface, with some characteristic distance which we seek to find. The reader might wonder: is it the skin depth δ? The answer to that question is: most definitely not! We are all used to using Ohm's law J = σE in various forms. Application of this law in the regions of charge density in the above figure leads to a contradiction. In the DC static case, nothing moves, so there can be no J, but there is clearly some E, so how can J = σE ? The reason is that Ohm's law only applies in a neutral medium. When there is a net charge density, the correct Ohm's law is this: (reference? ) J = σE - D grad ρ (E.1) The second term represents the pressure of the non-uniform charge density acting on the mean velocity of charge carriers which make up J. In the static case with no current, the second term balances the first term in a surface charge region, σE = D grad ρ (E.2) As electrons pile up on the boundary, they resist further pileup by their higher density. Basically this is a diffusion effect, and D is a diffusion coefficient. There is another more familiar equation which relates E and ρ, namely div E = ρεε0 (E.3) Since one cannot polarize electrons relative to ions inside a metal, the dielectric constant ε = 1. Taking the divergence of (E.2) and using (E.3) we get this result 2ρ = (σ/Dε0) ρ (E.4) The combination of symbols in (E.4) is the square of something called the Debye length, λD2 = (Dε0/ σ) (E.5) which is associated with charge screening in plasmas (such as the electrons in a metal). Thus, (E.4) becomes 2ρ = (1/λD2) ρ (E.6) In our one-dimensional problem of the above figure above, the solution of this equation is ρ(x) = ρ(0) e-x/λD fix this (E.7) where x is a coordinate going into the surface. This says that the thickness of the charge surface layer is basically λD. If the electron cloud inside the metal is treated as a classical gas of particles of mass m, charge q, temperature T, and density n, one gets formulas for the various coefficients. Here is a full set of expressions: J = nqv v = average drift velocity τ = mean life time between collisions μ(v/E) = (q/m)τ = mobility D = (kTτ/m) = diffusion coefficient (k = Boltzmann constant) σ = (nq2τ/m) = conductivity λD = = Debye length (E.8) This set of equations represents a classical model for the free charge in a metal. One major and one minor adjustment is needed when quantum theory is applied because electrons are fermions. This means that they cannot all park in the same state, so they "pile up" in higher and higher states in something known as the Fermi sphere. Only electrons at the surface of this sphere ( " the Fermi surface") can do anything useful. Due to the pileup, the temperature of the active electrons is very much higher than one might think using classical physics. One finds this temperature by setting kT = EF where this latter is the Fermi energy, EF = (h2/ 8π2m) (3π2n)2/3 = kTF (E.9) The appearance of the Plank constant h is the clue that this is a quantum result. This was the major quantum adjustment. The minor one is that T in the Debye formula gets replaced by (2/3)T. Thus, λD = = Debye length (quantum correct) (E.10) We shall now do some numbers. Here are the basics, n = 8.45 x 1028 electrons/ m3 for Copper k = 1.38 x 10-23 = Boltzmann m = 9.1 x 10-31 kg = electron mass h = 6.63 x 10-34 J sec = Planck Plugging these into (E.9) gives the following effective electron temperature TF = 8.17 x 104 ° K = pretty hot (E.11) We can now compute the Debye length, using ε0 = 8.85 x 10-12 F/m q = 1.60 x 10-19 C The bottom line is, λD = 5.55 x 10-11 m = 0.55 A (Angstroms) (E.12) The crystal spacing in copper is 3.6A, and the copper atomic radius is about .8A. Thus, we come to the dramatic conclusion of this section: Fact: The thickness of the surface charge density on the surface of a conductor is incredibly small. For copper, it is less than the radius of one copper atom, and the general result applies to any metal. Thus, the surface charge decays away right in the very first atomic layer of a metal. The charge is "screened" by the electron cloud over a distance that is the Debye length. From Section 2.2, we noted that the skin depth δ for copper at 100 GHz is about 0.2 microns which is 2x10-7m = 2000A. Even at this large frequency, the skin depth is still about 4000 times larger than the thickness of the surface charge layer. At 1 GHz this ratio is 40,000. Fact: Whereas surface current can exist "deep" into the surface of a conductor, even when the skin effect is dominant, the surface charge can always be thought of as being exactly on the surface. Appendix F 3.2: Waveguides Much insight can be obtained about the assumptions made for a transmission line by considering the very different case of a waveguide. Normally a transmission line has two conductors and a waveguide has only one, but there is no reason why "waveguide action" cannot take place in a transmission line. 1. A waveguide solution. The standard assumptions made about a "perfect conductor" are that the fields vanish inside, and that all action takes place at the surface, which is basically a mirror. At the surface, tangential E fields and normal B fields vanish. For a waveguide, these facts become convenient mathematical boundary conditions, so the field wave equations are always used instead of the potential wave equations. There is no ρ or J inside a waveguide, so we write (1.2.2) converted to the frequency domain as, ( 2 + β2 ) E = 0 . (F.1.1) If we define symbol ν to be the "complex phase velocity" of the medium, ν ≡ c / (F.1.2) we can express β appearing in (F.1.1) in a simple form β ≡ ω = ω/ν (F.1.3) Here, ξ ≡ [ εε0 + σ/(jω) ] is the complex dielectric constant mentioned earlier and discussed in Appendix B. Parameters σ, ε and μ refer to properties of the dielectric which is inside the waveguide (normally air). For σ ≠ 0, ν and β have small imaginary parts. The first step is to try to find a solution to the waveguide problem by separation of variables. If a solution is found, then this separation is justified. Starting back in the time domain, we then try this form of solution E(x,y,z,t) = E(x,y) exp [ j(ωt ± kz)] (F.1.4) where the - sign is for waves traveling in the +z direction. Here ω is the (angular) frequency and k is the (angular) wavenumber. Eq. (F.1.1) then becomes, ( + + γ2 ) E(x,y)= 0 (F.1.5) where γ2 = β2 - k2 = - k2 (F.1.6) As a simple but illustrative example, consider a waveguide consisting of only two plates separated by distance a (in coordinate x). We look for a solution with E(x,y) = Ey(x) so we have ( + γ2 ) Ey(x) = 0 (F.1.7) A candidate solution is Ey(x) = A sin(γx). To meet the boundary conditions that Ey vanish at x=0 and x=a, we are forced to set γ = γm = (mπ/a). One says that (F.1.7) and its boundary conditions comprise an eigenvalue problem, and γm are the eigenvalues. Here is a sketch of the waveguide, viewed looking down the line, for the m = 1 mode: Fig 1: Cross sectional view of a simple waveguide. equation 8 is missing If we solve (F.1.6) for the wavenumber k, k = (1/v) (F.1.9) we see clearly that ω must be larger than ν γm, (the mode cutoff) in order to get a predominantly real k. Below this ω, k is imaginary and there is no propagation as in (F.1.4), only attenuation. If the dielectric conductivity σ appearing in ξ is small but non-zero, then even above cutoff the k in (F.1.9) will have some small imaginary part. This results in a slight exponential decay in (F.1.4) caused by heating of the dielectric as the wave moves down the guide. The other loss mechanism, and a more important one, occurs at the "mirror" surface. Since the walls are not really perfect conductors, the E and B fields do in fact penetrate some distance (the skin depth), and currents are created according to J = σE inside the conductor, so there are heat losses at the walls as well. These same loss mechanisms occur in transmission lines as well, and will be addressed in later chapters. 2. A waveguide interpretation In the previous section, we obtained this approximate solution for the electric field inside the two-plate waveguide, Ey(x,z) = A sin(γmx) e-ikz with k2 = β2 - (γm)2 (F.2.1) The solution was approximate because we ignored skin depth effects. It is convenient to think of (F.2.1) as the superposition of two plane waves of wavenumber β traveling at some skew angle ±θ relative to the z direction, Ey(x,z) = (2A/j) [ exp( -jβ1• r) - exp( -jβ2• r)] (F.2.2) where β1 = βx βz β2 = - βx βz (F.2.3) (β1)2 = (β2)2 = βx2 + βz2 = β2 tanθ = βx / βz (F.2.4) Adding the two terms in (F.2.2) gives Ey(x,z) = Asin( βx x) exp( -j βz z) (F.2.5) Comparing with (F.2.1), we conclude that βx = γm = mπ/a βz = k tanθ = γm/k (F.2.6) The two plane waves have wavenumber β = ω/ν and not k. Their sum is the superposed wave going down the guide in the z direction with wavenumber k shown in (F.2.9). As ω approaches cutoff from above, the angle θ gets closer to 90° and at cutoff the plane waves bounce back and forth sideways and there is no propagation down the guide at all. Here is a picture: Consider what happens during the bounces of the plane waves off the interior surfaces of the waveguide. In the skin depth region, current J flows in the direction of the parallel E field, which is the y direction in our example. This current serves to both cancel the incoming wave, and to create the reflected wave. Thus, there can be significant tangential transverse currents in the walls of a waveguide. The second point to be made involves wavelength. The plane waves have wavelength λ = 2π/ β = 2πν/ω Suppose we try to form a waveguide solution by jamming an integral number of half waves of the plane wave between our two plates at right angles, knowing that in this way we meet the boundary conditions at the two plates: mλ/2) = a Using the above expression for λ gives ω = (νγm), which is the cutoff frequency for mode m. This is exactly the situation described above when θ = 90°. As we move above cutoff in ω, the angle θ decreases from 90, the wavelength λ gets shorter, so the wavenumber β gets larger (number of radians of wave per meter). The boundary conditions are maintained by keeping the product of β and cosθ constant: βx = β cosθ = γm fi cosθ = (γmν)/ω = ωm / ω There are several major points that the above discussion is intended to convey. These concern the waveguide modes TE and TM, and not the TEM mode which is what a transmission line does. One should compare the following facts one for one to the corresponding facts which appear in Section 3.8 for the transmission line TEM mode. Fact 1: In waveguide modes, there can be large tangential transverse currents (Jy). That is, such currents can be large relative to longitudinal currents in the conductor. Fact 2: The lowest cutoff frequency of a parallel plate waveguide can be obtained by setting the distance between the plates equal to half a wavelength, a = λ/ 2. In general, for an arbitrary waveguide, the cutoff occurs when λ/2 exceeds some similar characteristic transverse dimension of the guide. For ω below cutoff, there can be no waveguide propagation. Fact 3: In solving waveguide problems, one uses wave equations for the fields since the boundary conditions are expressed in terms of fields. Appendix G 4.1: Chapter 4 Support This Appendix provides technical support for certain claims made in Chapter 4. 1. Evaluation of the integral in 4.1 (9). This integral appeared as follows: = sn fn(βs) R = . (G.1) where R = | x - x'| and s = R(z = z'). Thus, we have R = . The first step is to define z" = z'-z. Due to the infinite integration range, we get: = sn fn(βs) R = . (G.2) In other words, the integral in (1) is independent of z, so we thing of (2) as (1) with z = 0. Now we define x = sz" and the above becomes, = fn(α) αβs (G.3) For odd integer n, the integral clearly vanishes since the integrand is then an odd function of x. For this reason, we replace n = 2m with m = 1,2,3 ... to list off the non-vanishing integrals. At the same time, we make one more change of variables to y = . The result is then 2 !Syntax Error, Idy (y2-1)m-1/2 e-jay = f2m(α) α = βs (G.4) Finally we have something we can find in the standard tables. Gradshteyn-Ryzhik page 322 shows that the result is: f2m(α) = (2/) (2/jα)m Γ(m+1/2) Km(jα) (G.5) Using the standard properties of the Gamma function(z) we can set Γ(m+1/2) = 2-2m (2m)! / m! (G.6) to get the final result, f2m(α) = 21-m (jα)-m [(2m)! / m!] Km(jα) (G.7) This is the result quoted in 4.1 (11) with α = βs. Km(z) is known as a modified Bessel function of the second kind, and is described in any reference on Bessel functions, such as GR quoted above. For small z, we have these approximations: K0(z) ≈ -ln(z/2) [ 1 + (z/2)2 + O(z4) ] + ψ(1) + (z/2)2 ψ(2) + O(z4) (G.8) Km(z) ≈2m-1(m-1)! z-m - 2m-3 (m-2)! z-m+2 + O( z-m+4) m= 1,2,3... Notice that K0 is fundamentally different from the other Km in this limit. K0(z) is logarithmically divergent at z = 0, whereas Km(z) diverges as a power z-m. 2. Examination of higher order terms in 4.2 (2). The expression 4.2 (2) reads, V(z) = q(2m)(z) X (G.9) { !Syntax Error, Idx' dy' a1(x',y') (jβ)-m [ s1m Km (jβs1) - s2m Km (jβs2)] - !Syntax Error, Idx' dy' a2(x',y') (jβ)-m [ s1m Km (jβs1) - s2m Km (jβs2)] } We wish to show that in the small β limit, all terms with m= 1,2,3... can be neglected. Let us insert the small z expansion (8) for Km(z) into the square bracketed factors in (9). We get: (jβ)-m [ s1m Km (jβs1) - s2m Km (jβs2)] = + 2m-1(m-1)! [ 1 - 1 ] /leading term - 2m-3 (m-2)! (jβ)2 [ s12 - s22 ] / second term, order(β2) + 2m-5 (m-3)! (jβ)4 [ s14 - s24 ] / third term, order(β4) + ... The leading term, which would have been significant, completely cancels. The remaining terms are of order β2 and smaller. For example, the second term has a coefficient which contains β2 and involves finite integrals over the charge density of the following form: !Syntax Error, Idx' dy' a1(x',y') [ s12 - s22 ] Thus, each term in the sum over m in (9) is of order β2 or less. That is, it might happen that the integral above vanishes, so then each term would be of order β4. Since β = 2π/λ is a smallness parameter, we conclude that only the m=0 term is significant. This term was analyzed in the text of Chapter 4. References Ronald W.P. King, Transmission Line Theory, Dover, 1965.