transmission lines before Ch2 eq num changes REVIEWED

11.27.13 Note: Equation numbers for Chapter 2 need massive adjusting globally in this doc, so I decided to do a full backup of lines doc before doing this. Chapter 1: Basic Equations 4 1.1 Maxwell's Equations in a Conducting Dielectric Medium 4 (a) Notes on Maxwell's Equations 4 (b) Integral Forms of Maxwell's Equations and Continuity 8 (c) Rules for behavior of fields and potentials at a boundary 11 1.2 The Field Wave Equations 16 1.3 The Potential Wave Equations 17 (a) The Potential Wave Equations in the Lorenz gauge 17 (b) Special Relativity Note 18 (c) The Potential Wave Equations in the King and Lorenz Gauges with Conductors 20 1.4 Retarded Solutions in the Lorenz gauge: a path not taken 27 1.5 The Wave Equations in the Frequency Domain verify all equation numbers 28 (a) The Transformed Wave Equations 29 (b) The Helmholtz Integrals in the King Gauge 30 (c) King's leading factor (1/4πξ) and the final Helmholtz Integrals 32 (d) Frequency domain wave equations for fields and potentials in the Lorenz Gauge 38 1.6 Reinterpretation of all equations in terms of complex functions 40 (a) Complex Functions 40 (b) Monochrome time 41 (c) Why complex fields? 41 (d) Monochrome E and B fields 43 (e) A Pitfall to Avoid 43 (f) Maxwell's Equations in ω space 44 Chapter 2: The Round Wire and the Skin Effect 45 2.1 Derivation of E(r), B(r) and J(r) for a round wire 45 2.2 A study of the solution of a round wire 51 (a) Kelvin Functions 51 (b) Plots of |E(r)/E(a)| for various δ values 53 (c) Review of the round wire solution 57 (d) Plots of the round wire solution for Belden 8281 at 5 MHz. 58 2.3 The Surface Impedance Zs(ω) of a Round Wire 60 (a) Expressions for Surface Impedance 61 (b) Low frequency limit of Zs(ω) 63 (c) High frequency limit of Zs(ω) 63 (d) Plots of Zs(ω) versus skin depth δ 65 2.4 Surface Impedance for a Transmission Line 67 Chapter 3: Transmission Line Preliminaries 70 3.1 Why is there no charge inside a conductor? 70 3.2 How thick is the surface charge layer on a conductor? 70 3.3 How does loss tangent affect dielectric conductivity? 71 3.4 Size of E field inside a conductor and conservation of total current at a boundary 72 3.5 The TEM mode fields and currents for an IDEAL transmission line 74 3.6 The TEM mode fields and currents for a REAL transmission line 77 3.7 The general shape of fields, charges, and currents on a transmission line 78 (a) Facts about field structure 78 (b) Drawings of the fields 80 (c) More on the field and current Structure 83 (d) Estimate of the ratio Jr/Jz 84 3.8 Transmission Line Preliminaries 86 Chapter 4: Transmission Line Equations 90 4.1 Computation of φ due to one conductor of a transmission line. 90 4.2 Computation of V(z) 94 4.3 Computation of Az due to one conductor of a transmission line 96 4.4 Computation of W(z) 97 4.5 The Classic Transmission Line Equations 99 4.6 Example: the wide-spaced, two-wire transmission line. 102 4.7 Example: the coaxial cable. 106 Chapter 5: The Transverse Problem 108 5.1 Philosophy 108 5.2 The Helmholtz Equations and Separation of Variables 110 5.3 A Formal Solution to the Transverse Problem 114 5.4 An approximate solution to the transverse problem. 118 Chapter 6: An Example 121 6.1 Why logarithms? 122 6.2 The equipotentials of ln[s2/s1] are circles. 123 6.3 Aligning the circles. 124 6.4 Reduction to special cases 126 Appendix A: Gauge Invariance 131 A.0 The Poisson Equation and its Solution 131 A.1 Existence of A such that B = curl A and div A = 0 133 A.2 Existence of A' such that B = curl A' and div A' = f . 135 A.3 Existence of φ such that E = -grad φ 135 A.4 Existence of A' and φ' such that B = curl A', E = - grad φ'-∂tA', and div A' = f. 136 A.5 Gauge Invariance 137 A.6 The Lorenz Gauge and QED 138 A.7 Finding the gauge function Λ for the Lorentz Gauge 140 Appendix B: Magnetization Surface Currents on a Conductor 143 B.1 Relationship between surface current K and the field H at a conductor boundary 143 B.2 Calculation of H from the current J in a conductor 146 B.3 General Method for computing the surface current Jm on a wire 147 B.4 Surface current on a round wire with uniform J 147 B.5 Surface current on a round wire using the General Method 148 B.6 An Acrobatic Exercise: Compute H from J and Jm via the vector potential A. 152 B.7 Reader Exercise: Repeat the above calculation of A using a 3D analysis 157 Appendix C: DC Properties of a Wire 160 C.1 The DC resistance of a wire 160 C.2 The DC surface impedance of a wire 160 C.3 The DC inductance of a round wire 160 (a) Internal DC inductance of a round wire 161 (b) External DC inductance of a round wire 162 C.4 The DC inductance of a wire of arbitrary cross section 163 (a) Statement of a Plan of Attack 164 (b) The divergence problem and its resolution 164 (c) Avoiding the divergence problem 166 Example: Magnetic Field of a Rectangular Wire 167 Appendix D : The General Electric Field Inside a Round Wire 170 D.1 The General Method and Solution for Ez 170 (a) Partial Wave Expansions 171 (b) The Vector Laplacian and the conversion of equations from φ space to m space 172 (c) The Charge Pumping Boundary Condition 176 (d) The Ez Solution 177 D.2 The Solutions for Er and Eφ 177 (a) The Er Solution 177 (b) The Eφ Solution 179 (c) Application of the Boundary Conditions 181 D.3 What about the Eφ Helmholtz Equation ? 183 D.4 Statement of the Results 185 D.5 The Low Frequency Limit 189 D.6 What about the E fields outside the round wire? 192 D.7 What about φ, A and B ? 193 Appendix E: Surface Charge 195 Appendix F: Waveguides 199 F.1 A waveguide solution 199 F.2 A waveguide interpretation 201 Appendix G 4.1: Chapter 4 Support 204 G.1 Evaluation of the integral in 4.1 (9). 204 G.2 Examination of higher order terms in 4.2 (2). 206 Appendix H : Poisson and Helmholtz Propagators in 3D 208 Appendix I : Poisson and Helmholtz Propagators in 2D 213 References 221 Chapter 1: Basic Equations In this chapter we state the basic equations to be used later in the calculation of transmission line parameters. Appendices 1.1 A and 1.2.B contain information that is relevant to this chapter. The compact notation ∂xf ≡ is used throughout. We write scalar F as div F, and vector x F as curl F. 1.1 Maxwell's Equations in a Conducting Dielectric Medium Our working set of equations is the following: curl H = ∂tD + J Maxwell curl H equation (J = Jc) (1.1.1) curl E = - ∂tB Maxwell curl E equation (1.1.2) div D = ρ Maxwell div D equation (ρ = ρfree) (1.1.3) div B = 0 Maxwell div B equation (1.1.4) B = μH magnetic permeability μ (1.1.5) D = εE electric permeability ε (dielectric constant) (1.1.6) J = σE Ohm's Law (σ = conductivity) (1.1.7) div J = - ∂tρ Equation of Continuity (see item 7 below) (1.1.8) (a) Notes on Maxwell's Equations Although Maxwell's equations ("the Maxwell equations") provide a concise overview of classical electrodynamics, there is lot going on "under the hood" and clarification of the meaning of certain symbols seems useful, hence the following set of notes. 0. It is understood that, in a medium other than the vacuum (that is, a "ponderable" medium), all the mathematical fields shown above like E, D, B, H, J, ρ (and later A and φ) are average fields in the sense discussed by Jackson Sections 4.3 and 6.6. The partial differential equations are meaningful for differential volumes, areas and distances which are very small but still contain enough atoms or molecules (perhaps at least 1000) so that averaging makes sense. We shall refer to the various electric and magnetic fields as "fields" to distinguish them from "potentials" like A and φ, though all these quantities are mathematical fields. 1. The equations above are all expressed in SI units. The connection with cgs/Gaussian units is explained in an Appendix present in all three of the Jackson Classical Electrodynamics editions. The above equations appear in Jackson's third edition at these locations (1.1.1) through (1.14): p 2 (I.1a) Maxwell's Equations (1.1.5) and (1.16): p 296 top permeability constitutive relations (1.1.7) p 219 (5.159) Ohm's Law constitutive relation J = σE (1.1.8) p 3 (I.2) equation of continuity 2. All media (conductors, dielectrics between conductors) are assumed to be homogeneous and isotropic so that the quantities σ, ε and μ are constant scalars in space (not tensors) for a given medium. In the vaccum these constants take the values σ = 0, ε = ε0 and μ = μ0. An implication of σ, μ and ε being constants in space is that they pass through the div, curl, grad and 2 operators just as would any constant like π. One must be a little careful at a boundary between homogeous media since these constants can be different in the two media. In principle, all three quantities can vary in time, and when transformed to the frequency domain, σ(ω), ε(ω) and μ(ω) can (and do) vary with ω. However, we shall assume that for our frequencies of interest, these quantities are constant in ω and are therefore also constant in time so they pass through ∂t. 3. The difference between ε and ε0 is caused by polarization of bound charge in the medium. Equations dealing with polarization are these [ see Jackson pp 153-4 or Panofsky & Phillips pp 28-20 and p 140 on the polarization current ] : PdV = electric dipole moment contained in volume dV of a dielectric (1.1.9) Jpol ≡ ∂tP = polarization current density (1.1.10) ρpol = - div P = polarization charge density (1.1.11) P = ε0χeE // polarization assumed proportional to the polarizing E field (1.1.12) D = ε0E + P = ε0(1 + χe)E = ε E = "the electric displacement " (1.1.13) ε = ε0(1 + χe) // ε = dielectric constant, χe = electric susceptibility (1.1.14) div E = (1/ε0)(div D - div P) = (1/ε0)(ρfree + ρol) "E sees all charges" (1.1.15) The E field causes polarization P either by causing existing tiny dipole objects (e.g., molecules) in the media to "line up", or by causing tiny non-dipole objects (e.g., atoms) to have dipole moments and then those get lined up. See for example Bleaney & Bleaney Chapter 10 " Dielectrics". Comment: Since D = ε0E + P, the D and E fields are scaled differently. It might have been better had the D field been replaced by D = ε0D' in which case D' = E + P/ε0, then one can make clearer statements about D' versus E. For example, in a dielectic capacitor with fixed conductor charges (Q,-Q) there exist both D' and E fields, and D' = (ε/ε0) E > E. The D' field can be interpreted as the E field that would be present were the dielectric replaced by empty space. The dielectric reduces this E and hence V, does not change Q, and, since Q = CV, it increases capacitance C for fixed Q. The increase is by factor (ε/ε0). 4. As noted in (1.1.3), the ρ in div D = ρ is the free charge density ρfree and does not include possible polarization charge density. In contrast, the E field "sees" both free charge ρfree and polarization charge ρpol , as derived above in (1.1.15) from (1.1.13), div E = (1/ε0) ( ρfree + ρpol) = (1/ε) ρfree (1.1.16) In the rightmost expression, the polarisation charge is incorporated into the 1/ε factor. To see how this works, see Appendix Q where we have collected some elementary but useful problems. 5. The J in curl H = ∂tD + J is the conduction current Jc . If polarization current Jpol is present, it is included in the "displacement current" term ∂tD along with the Maxwell "vacuum polarization current" Jvac = ε0∂tE . That is, Jd ≡ ∂tD = ∂t[ε0E + P] = ∂tP + ε0∂tE = Jpol + Jvac ρpol = - div P (1.1.17) The Jvac term ε0∂tE was "added" by Maxwell to the curl H equation (Ampere's Law) to make it self-consistent. Since div curl H = 0, and since curl H = Jd + Jc, one must have div [Jd + Jc] = 0 : div [Jd + Jc] = div [∂tP + ε0∂tE] + div[Jc] = ∂t[div P + ε0 div E] - ∂tρfree = ∂t(-ρpol) + ∂t(ρfree + ρpol) - ∂tρfree = 0 (1.1.18) where we have used continuity div Jc = -∂tρfree, see item 7 below. Comment on "Displacement": In the case of polar molecule polarization, the polarization charge and current can be viewed as being caused by a "displacement of bound charge" as suggested by this very symbolic picture of a parallel plate capacitor Fig 1.1 The applied E field of the plates lines up the polar molecules and thus causes a polarization charge density npol to appear on the side faces of the dielectric, as if it were an "electret" object. One can imagine that, with an AC plate voltage, as the applied E field changes to the other polarity, the polar molecules rotate in place 180 degrees putting the positive bound charge on the opposite plate, and as this happens, there is a polarization current Jpol = ∂tP inside the dielectric. In reality, the molecules are close to randomly oriented and the above effect is obtained for the "average" molecule. In any event, the E field causes surface polarization charge densities npol at the faces of the dielectric, and one then thinks of the normally neutral-everywhere bound charge distribution as being "displaced" such that one face has positive charge and the other negative. It is in this sense that Maxwell started using the word "displacement". Before the Jvac term was added, Maxwell had Jd ≡ ∂tD = Jpol and this associated ∂tD entirely with Jpol and thus with the displacement of the dielectric bound charge, and so Maxwell referred to D as the "electric displacement" and ∂tD as the "displacement current". 6. If there is any magnetization current Jm = curl M, it is absorbed into the distinction between B and H and therefore does not appear on the right side of curl H = ∂tD + Jc . The current Jm is discussed for example in Panofsky & Phillips, Sections 7-12, 7-13 and 8-1. The basic equations are as follows, MdV = magnetic dipole moment contained in volume dV of a medium (1.1.19) Jm = curl M = magnetization current density (1.1.20) M = χm H = magnetization (1.1.21) B = μ0(H+M) = μ0(1+χm)H = μH = "magnetic induction" (informally, magnetic field) (1.1.22) μ = μ0(1+χm) // μ = magnetic permeability, χm = magnetic susceptibility (1.1.23) curl B = μ0(curlH + curlM) = μ0(∂tD + Jc) + μ0Jm = μ0(∂tD + Jc + Jm) " B sees all currents" (1.1.24) 7. The "equation of continuity" (1.1.8) expresses the fact that charge cannot be created or destroyed. Barring ionization of a dielectric, free charge and bound charge (polarization charge) cannot be converted into each other and are therefore separately conserved. Thus we have several different equations of continuity: [ see for example Haus and Melcher, Section 6.2, equations (10) and (13) ] div Jc = -∂tρfree // conservation of free charge (aka true or unpaired charge) (1.1.25) div Jp = -∂tρpol // conservation of polarization charge (aka bound or paired charge) (1.1.26) div [Jc+ Jp] = -∂t[ρfree + ρpol] = -∂tρtot // sum of above two equations (1.1.27) div [Jc+ Jd] = 0 ≠ -∂tρtot // reminder of item 5 above (1.1.12) 8. Ohm's Law J = σE is assumed to be a valid constitutive relation for our media of interest. One should keep in mind that this is an approximation, whereas the Maxwell equations and the continuity equations are not. 9. As will be shown, 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 ρfree = 0 . In a dielectric there are no available free charges, while in a conductor, any departure from neutrality would be instantly restored. All free 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. 10. As noted in item 1, the above equations are expressed 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 our equations, along with one expression of the SI units for each symbol: E = electric field (volts/m) H = magnetic field (amp/m) D = electric displacement (coulomb/m2) B = magnetic field (tesla = amp-henry/m2) tesla = 10,000 gauss J = current density (amps/m2) ρ = charge density (coulombs/m3) σ = conductivity of the medium (mho/m = ohm-1/m) μ/μ0 = relative magnetic permeability of the medium (dimensionless) ε/ε0 = relative electric permittivity = 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 (1.1.28) Here are some unit relations obtainable from Q = CV, V = IR, LC = 1/ω2 , τ = RC = L/R, I = dQ/dt : coulomb = farad-volt volt = ampere-ohm henry-farad = sec2 farad = sec/ohm henry = ohm-sec henry / farad = ohm2 ampere = coulomb/sec mho = ohm-1 mho/F = sec-1 c = 1/= 2.9979246 x 108 m/sec = speed of light Z0 = = 376.73032 ohms = "impedance of free space" (1.1.29) Notice how the names of seven people have become forever embedded into the SI unit system. Comment: Inevitably, any given author will at some point refer to both B and H as "the magnetic field". We shall do that throughout, using the historical symbols B or H to indicate which "kind" of magnetic field we are talking about. Some authors refer to B as the magnetic flux density or the magnetic induction to distinguish B from H. (b) Integral Forms of Maxwell's Equations and Continuity The equations above involving the divergence and curl operators have integral forms thanks to these two fundamental mathematical theorems which have nothing to do with electromagnetism in particular, ∫V div F dV = ∫S F dS // "the divergence theorem" Spiegel 22.59 (1.1.30) ∫S curl F dS = C F ds // "Stokes's theorem" Spiegel 22.60 (1.1.31) Fig 1.2 The first theorem involves a closed boundary surface S which encloses a volume V and says that the volume integral of div F over V equals the surface integral of F over S. The second involves a closed bounding curve C (possibly non-planar) which bounds an arbitrary open surface S (also possibly non-planar) and says that the line integral of F around C equals the surface integral of curl F over S. In the divergence theorem, dS points "out" from the volume, and in Stokes's Theorem, the direction of dS and ds are related by the right-hand rule where fingers fit the boundary curve and then thumb gives the direction of dS. In both theorems the differential vector area patch is dS = dS where is normal to the surface. Both theorems have meanings in n-dimensional space, but of course our interest is for n = 3. Both theorems are not hard to derive and this is done in textbooks usually by breaking up the surface into tiny squares and the volume into tiny cubes. Once one sees these derivations, the theorems become less mysterious. In general terms, the divergence theorem says that div F is somehow a source of the field F and the amount of F flowing out through a closed bounding surface equals the amount of F that is generated inside the volume. When F is the electric field E, the divergence theorem is called Gauss's Law and says that the total electric flux "flowing out" [ that is to say, ∫S EdS ] equals the the total of the source inside the volume [ (1/ε)∫V ρ dV ], usually called "the total charge enclosed". Thus, div D = ρ ∫V ρ dV = ∫S D dS (1.1.32) div E = ρ/ε ∫V ρ dV = ∫S ε E dS (1.1.33) Since the magnetic field has no corresponding charge, one always has ∫S B dS = 0, a theorem which seems to have no name, div B = 0 ∫S B dS = 0 S is any closed surface (1.1.34) The surface integral of an E or B field is often refered to as the total electric or magnetic "flux" passing through the surface, even though nothing is really flowing in a mechanical sense. The divergence operator also occurs in the equation of continuity (1.1.8) so we have div J = - ∂tρ -∂t[∫V ρ dV] = ∫S J dS . (1.1.35) This is the prototype application of the divergence theorem in that it is easily understandable: the total electric current flowing out through some closed surface S must equal the rate at which the total charge inside the surface is decreasing. One can write a similar statement for mass flowing out from a volume in which ρ would be the mass density and J = ρv the mass current, v being the velocity field. The Stokes theorem is a bit more mysterious. Since this theorem is associated with George Stokes, it is called Stokes's theorem, but is sometimes called Stokes' theorem (one would not say Gauss' theorem). The curl of a vector field is associated with the amount of "rotation" the field has at some point in space, and in fact curl is sometimes written Rot. If one considers a tiny patch and finds that the line integral of the field around the boundary of that patch is non-zero, then the vector field has a non-zero curl at that point in the direction normal to the patch. At any point where a fluid has a vortex, the curl is non-zero, for example. When Stokes's theorem is applied to the electric field, one has curl E = - ∂tB C E ds = -∂t[∫S B dS] (1.1.36) This says that the voltage induced around a closed loop (the "electromotive force") is proportional to the rate of change of the magnetic flux through that loop, a principle known as Faraday's Law of Induction. If water power rotates a wire loop in the presence of some magnets, one has an electric generator. On the other hand, when Stokes's theorem is applied to the magnetic field, one gets curl H = ∂tD + J C H ds = ∫S [∂tD+J] dS (1.1.37) curl B = με ∂tE + μJ C B ds = μ ∫S [ε ∂t E + J] dS (1.1.38) μ constant in space, ε constant in time When the situation is static, one has H ds = ∫S dS J which says the line integral of the magnetic field H around some loop equals the total current passing through any open surface whose boundary is that loop, a principle known as Ampere's Law. Later we shall encounter a certain "vector potential A" which is related to the B field by B = curl A. Since we are writing out "integral forms" of differential relationships, we can then add this to the list curl A = B C A ds = ∫S B dS . (1.1.39) If the bounding curve C were a wire carrying a current I which creates both A and B, then both sides of the above integral form will be proportional to I, and the constant of proportionality is by definition the self-inductance L of the loop, C A ds = ∫S B dS = [magnetic flux through surface S] = L I . (1.1.40) There are of course many surfaces S which span a given curve C, and (1.1.39) says that all such surfaces give exactly the same ∫S B dS and thus the same L, so L is really a geometric property of the curve C. Sample L calculations appear elsewhere in this document. We shall be using most of these integral forms in the document below. (c) Rules for behavior of fields and potentials at a boundary Consider the boundary between two different media called 1 and 2. Consider a tiny red "math loop" of width L and height 2s which straddles the media boundary which here is seen edge on, Fig 1.3 For the electric field we have from above (for our loop, dS = dS ) curl E = - ∂tB E ds = - [∫S (∂tB) dS] . (1.1.36) Since the loop is tiny and since the fields are assumed to be non-singular, we can regard E and B as a constant everywhere on the loop (for our purposes here). The line integral around the loop is then (start at lower left corner) E ds = LEx(2) + s Ey(2) + s Ey(1) – LEx(1) - s Ey(1) - s Ey(2) = L [Ex(2)- Ex(1)] . The area integral on the right side of (1.1.36) is -∫S (∂tB) dS = - ∂tBz(1) sL - ∂tBz(2) sL = - sL [∂tBz(1) + ∂tBz(2)] so the integral form in (1.1.36) says [Ex(2)- Ex(1)] L = - s L [∂t Bz(1) + ∂t Bz(2)] . As long as ∂tBz is finite at the surface, as s→0 the right side vanishes and we conclude that [Ex(2)- Ex(1)] = 0 . We then summarize for both the x and z directions by saying (t means tangential to boundary) Et1 = Et2 or (1/ε1)Dt1 = (1/ε2)Dt2 (1.1.41) so the tangential (parallel) components of the electric field is continuous through a boundary. A similar analysis using the curl H equation, curl H = ∂tD + J H ds = ∫S [∂t D+J] dS (1.1.37) leads to [Hx(2)- Hx(1)]L = s L[∂t Dz(1) + Jz(1) + ∂t Dz(2) + Jz(2)] . As long as ∂tDx and Jx are finite (non-singular) at the surface, we conclude from s→0 that Ht1 = Ht2 or (1/μ1)Bt1 = (1/μ2)Bt2 . (1.1.42) However, it is possible to have J be singular at the surface in the form of a surface current K where J = K δ(y) J = amp/m2 K = amp/m (1.1.43) and in this case we find that [Hx(2)- Hx(1)] L = ∫S [J] dS = ∫S Kδ(y) dS = ∫S Kz δ(y) dx dy = ∫S Kz dx ≈ Kz L which we summarize as Ht2 - Ht1 = Kzfree or (1/μ2)Bt2 - (1/μ1)Bt1 = Kzfree (1.1.44) where we imagine = x as the meaning of the z in Kz. Notice that this Kz is a "free" surface current, and not a bound magnetization surface current since such a magnetization current is not "seen" by H. We mention here a result similar to (1.1.44) which applies to a special situation of Fig ** above where we assume a vector potential of the form A = Az(y,z) . This vector potential is constant on the boundary surface in the x direction, and has only an Az component. In this case, B = curl A = (∂yAz - ∂zAy) + (∂zAx - ∂xAz) + (∂xAy - ∂yAx) = (∂yAz) = Bx where Bx = ∂yAz (1.1.45) and then (1.1.44) says (1/μ2) (∂nAz)2 - (1/μ1) (∂nAz)1 = Kz (1.1.46) where ∂n is the derivative of the vector potential Az in a direction normal to the surface (n pointing from medium 2 to medium 1). If μ1= μ2= μ0, we have (∂nAz)2 - (∂nAz)1 = μ0Kz and then Kz is proportional to the normal slope jump in Az at the boundary surface. As earlier, Kz is a "free" surface current. Next, we put a tiny "Gaussian pillbox" straddling the two media. Area A and height 2s are both very small so the fields are approximately the same at all points in the box. Gaussian Pillbox a pill box circa 1830 Fig 1.4 For the electric displacement D we consider div D = ρfree ∫V ρ dV = ∫S D dS (1.1.13) where volume V is of the box shown. The surface integral is ∫S D dS = Dy(1)A - Dy(2)A + contributions from the sides of the box The contributions from the pairs of box sides cancel since the fields are assumed constant and finite over the box on either side of the boundary. Assuming the only charge density is a surface charge density nfree on the boundary between the two media, the volume integral is nfreeA and then the conclusion, generalized to the perpendicular field component, is Dn1 - Dn2 = nfree or [ε1E1n - ε2E2n] = nfree (1.1.47) If the two media are conducting dielectrics with Ohm's law Jc = σE, we can apply continuity (1.1.25) to the Gaussian box to find that div Jc = - ∂tρfree -∂t[∫V ρfree dV] = ∫S Jc dA (1.1.35) so that -∂tnfree = [Jn1- Jn2] = σ1En1 - σ2En2 or for monochrome time dependence (coming soon, along with notation explanation), -jω nfree = σ1En1 - σ2En2 . // frequency domain Recall now from (1.1.47) that nfree = ε1En1- ε2En2 . (1.1.47) Adding the last equation to 1/jω times the previous equation gives 0 = [ε1 + σ1/jω] En1 - [ε2 + σ2/jω]En2 In terms of the complex dielectric constants ξi ≡ εi + σi/jω this says that 0 = ξ1En1 - ξ2En2 so that ξ1En1 = ξ2En2 // frequency domain (1.1.48) In the limit that, say, medium 2 becomes a perfect conductor, ξ2 ≈ σ2/jω → ∞ and En2 → 0, but the product is maintained equal to ξ1En1 . Returning again to the special case in which vector potential A = Az , since E = -φ - ∂tA (as shown later), we have Ey = -∂yφ since Ay = 0. In terms of Fig *** where y is the direction normal to the surface, we have En = -∂nφ and then (1.1.47) may be written ε2(∂nφ)2 - ε1(∂nφ)1 = nfree (1.1.49) which can be compared to (1.1.46). If ε1 = ε2 = ε0, we have (∂nφ)2 - (∂nφ)1 = nfree/ε0 and then nfree is proportional to the normal slope jump in φ at the boundary surface. Finally, the other divergence equation div B = 0 ∫S B dS = 0 S is any closed surface (1.1.34) leads to the conclusion that Bn1 = Bn2 or μ1Hn1 = μ2Hn2 (1.1.49) We now summarize these rules in a box, always assuming that there is no singularity in some quantity to invalidate the claims: Rules for continuity of normal and tangential fields at a boundary: (1.1.50) The fields here are either F(x,t) or F(x,ω) except the last rule which is only for E(x,ω) : t = tangential = parallel = || : Et1 = Et2 or (1/ε1)Dt1 = (1/ε2)Dt2 (1.1.41) Ht2 - Ht1 = Kzfree or (1/μ2)Bt2 - (1/μ1)Bt1 = Kzfree (1.1.44) Special case A = Az(y,z) : (1/μ2) (∂nAz)2 - (1/μ1) (∂nAz)1 = Kzfree (1.1.46) n = normal = perpendicular = : ( symbol n is also used for surface charge density) Bn1 = Bn2 or μ1Hn1 = μ2 Hn2 (1.1.49) Dn1 - Dn2 = nfree or [ε1En1 - ε2En2] = nfree (1.1.47) and for monochrome time dependence: ξ1En1 = ξ2En2 where ξ = ε + σ/jω (1.1.48) Special case A = Az : ε2(∂nφ)2 - ε1(∂nφ)1 = nfree (1.1.49) Tangential and normal boundary conditions can always be written in the following manner, Ft1 = Ft2 n x F1 = n x F2 Fn1 = Fn2 n F1 = n F2 (1.1.51) as can be seen by expanding F = Fn + Ft and noting that x = 0 and = 1. The E and B boundary conditions in the above table appear as follows in King (1945), page 202 (obtained from the University of Utah's robotic automated retrieval center ARC), where 1 = - 2, (n,F) = n F , [n,F] = n x F , and ν = 1/μ. 1.2 The Field Wave Equations In the following, quanties μ and ε are treated as constants, independent of space and time. The E wave equation may be derived using these steps : curl E = - ∂tB // Maxwell (1.1.2) curl curl E = -∂tcurl B = -μ∂t[curl H] = -μ∂t[ ∂tD + J ] // curl both sides and Maxwell (1.1.1) grad divE - 2E = -μ∂t[ ∂t[εE] + J ] // vector identity on left and D = εE (2 - με∂t2)E = μ∂tJ + (1/ε) grad ρ . // div E = ρ/ε The B wave equation uses these steps : curl H = ∂tD + J // Maxwell (1.1.1) curl curl H = curl [∂tD] + curl J // curl both sides grad div H - 2H = ε ∂t(curl E) + curl J // vector identity on left and D = εE (1/μ)grad div B - 2H = εμ ∂t(-∂tH) + curl J // Maxwell (1.1.2) and B = μH twice (2 - με ∂t2)H = - curl J // since div B = 0 (1.1.4) The two results are then (2 - με ∂t2)E = μ∂tJ + (1/ε) grad ρ (1.2.1) (2 - με ∂t2)B = - μ curl J (1.2.2) which agree with Jackson p 246 (6.49) and (6.50). Recall that με = 1/v2 where v is the speed of light in the medium of interest. These two equations are undamped driven wave equations. 1.3 The Potential Wave Equations (a) The Potential Wave Equations in the Lorenz gauge It is possible to work with the scalar and vector potentials φ and A instead of the fields E and B. If φ and A can be determined, then E and B are fully determined by (1.3.1) below. However, in the other direction, if E and B are known, then φ and A are determined only up to a certain "gauge transformation" degree of freedom, a subject discussed in Appendix A. The fields E and B are physically observable quantites while the potentials φ and A in general are not and should be regarded as intermediate "helper" functions. In SI units, the E and B fields are obtained from φ and A in this manner : [ Jackson p 239 (6.7) and (6.9)] B = curl A E = - grad φ - ∂tA . (1.3.1) A = vector potential (tesla-m = amp-henry/m = volt-sec/m) E = volt/m φ = scalar potential (volts) B = tesla . In Appendix A (Fact 4) it is shown that there is a continuum of possible choices (φ,A) all of which give the same physical fields (E,B) according to (1.3.1). It turns out that, along this continuum, div A takes different functional forms. Fact 4 shows that there always exists a choice (φ,A) for which div A = any function you want! Selecting f(x) for div A = f(x) is called "making a gauge choice". Different gauge choices just result in different (φ,A) potentials, but always the same (E,B). In the following derivations of the wave equations for A and φ, we shall be making a certain gauge choice as indicated. The following steps are used to develop the φ wave equation. In the vaccuum one has μ = μ0 and ε = ε0 and με = 1/c2 and these are the parameters one sees in the Jackson equation references below. E = - grad φ - ∂tA // (1.3.1) [= Jackson (6.9)] div E = - div grad φ - ∂t (div A) // take div of both sides 2φ + ∂t[div A] = -ρ/ε // div E = ρ/ε [= Jackson (6.10)] (1.3.2) (2 - με ∂t2)φ = - (1/ε)ρ // apply gauge choice divA = - με ∂tφ . And the following steps are used to develop the A wave equation: curl H = ∂tD + J // Maxwell (1.1.1) (1/μ) curl curl A = με ∂tE + J // H = B/μ , B = curl A from (1.3.1), and D = εE grad divA - 2A = με ∂t[- grad φ - ∂tA] + μJ // vector identity and E = - grad φ - ∂tA (2 - με ∂t2) A = grad [με ∂tφ + divA ] - μJ // [ = Jackson (6.11) ] (1.3.3) (2 - με ∂t2)A = - μJ . // apply same gauge choice divA = - με ∂tφ The results are then (2 - με ∂t2)φ = - (1/ε)ρ [ = Jackson (6.15) ] (1.3.4) (2 - με ∂t2)A = - μJ [ = Jackson (6.16) ] (1.3.5) divA = - με ∂tφ . [ = Jackson (6.14) ] // Lorenz Gauge (1.3.6) As discussed in the comment below, this gauge choice is now called the Lorenz Gauge. One should notice how the gauge choice decouples the two wave equations (1.3.2) and (1.3.3) so one resulting equation only involves φ and ρ, while the other involves only A and J. We end up then with undamped driven wave equations with simple driving terms. Note that ρ = ρfree (does not include polarization charge ρpol) and that J = Jc (does not include magnetization current Jm). Comment 1: For perhaps 100 years pretty much all (non-Danish) papers and textbooks (including Jackson's first two editions in 1962 and 1975 and the initial six printings of his 1998 third edition) referred to the Lorenz gauge as the Lorentz gauge, and it was then convenient to say that the Lorentz gauge condition is Lorentz invariant since it transforms as a scalar equation under Lorentz transformations. Now we have to say that the Lorenz gauge is Lorentz invariant because Lorentz was mistakenly credited for first using this gauge condition, see Jackson's note p 294 added in his 7th printing. Although the Dane Ludvig Lorenz (1829-1891) was 24 years older than the Dutch Hendrick Lorentz (1853 –1928), they were contemporary though independent workers at the time (1867) that Lorenz first published his use of his now-eponymous gauge condition. Lorentz will just have to be content with his transformations, his invariance, his contraction and his force law which says F = q(E + v x B). For more on Lorenz and Lorentz, see Nevels and Shin. Comment 2: We speak of (1.3.6) as "the Lorenz gauge" and divA = 0 as "the Coulomb gauge". These gauges are really conditions on A and do not fully specify A since many vector fields A can have the same divergence. So a gauge specifies a class of possible A fields, not a particular one. (b) Special Relativity Note At first encounter, one is amazed at how similar the two equations (1.3.4) and (1.3.5) appear. Here we shall show why that is. We now assume the medium is the vacuum so με = μ0ε0 = 1/c2. Then the two equations may be written (2 - ∂t2)φ = - (1/ε0)ρ (1.3.7) 2 - ∂t2)A = - μ0J . (1.3.8) As shown in Appendix A.6, one can construct Lorentz 4-vectors Aμ = ( φ, A) and Jμ = (cρ, J) with the identification of A0 ≡ φ/c and J0 ≡ cρ. The above equations can then be written, using proper tensor notation where common vectors are contravariant with an upper index, (2 - ∂t2)[cA0] = - (1/ε0)[J0/c] = - (1/ε0)[J0/c] (c2μ0ε0) = - μ0 cJ0 (1.3.9) 2 - ∂t2)Ai = - μ0Ji (1.3.10) Cancelling the c's in the first equation allows both equations to be written as a single 4-vector equation 2 - ∂t2)Aμ = - μ0 Jμ or Aμ = μ0 Jμ where ≡ ∂μ∂μ = ∂t2 - 2 . (1.3.11) This equation is covariant because both sides transform as a Lorentz 4-vector (the operator transforms as a Lorentz scalar). Special relativity requires that all equations of physics be covariant under Lorentz transformations. This is similar to Newton's Law F = ma being covariant under rotations, where both sides transform as 3-vectors. If we start with the correct law of physics (1.3.11) and work backwards through the equation pairs above, where we add a medium with μ and ε, we end up with our starting point (1.3.4) and (1.3.5) and the similarity of these two equations is then explained as being a requirement of special relativity. Recall the Lorenz gauge choice (1.3.6) which was required to decouple things above, divA = - μ0ε0 ∂tφ = - ∂tφ . (1.3.6) As shown in Appendix A.6, this Lorenz gauge condition can be expressed in covariant form as ∂μAμ = 0 divA = - ∂tφ (1.3.12) while the equation of continuity states ∂μJμ = 0 divJ = -∂tρ . (1.3.13) Both sides of these last two tensor-notation equations transform as a rank-0 tensor (scalar) so the equations are covariant (0 is a scalar). As one changes frames of reference doing Lorentz transformations (rotations and "boosts), the potential wave equation, the gauge condition, and the continuity relation always maintain the same tensor form. In closing this relativity note, we must mention that the four Maxwell equations (with ε = ε0 and μ = μ0 and μ0ε0 = 1/c2) can also be stated in covariant notation. One first defines the following antisymmetric rank-2 tensor (see Appendix A.5 concerning up and down indices etc.) Fμν ≡ ∂μAν - ∂μAν Aμ = ( φ, A) Jμ = (cρ, J) ∂μ = (∂0, ∂i) = (∂0, -∂i) (1.3.14) where obviously Fμν = -Fνμ and Fμμ = 0 for diagonal elements. Then the two Maxwell homogeneous (no sources) equations appear as ∂αFμν + ∂μFνα + ∂νFαβ = 0 // both sides transform as a rank-3 tensor so covariant curl E + ∂tB = 0 and div B = 0 (1.1.2) and (1.1.4) (1.3.15) while the two Maxwell inhomogeneous equations are (implied sum on μ ) ∂μFμν = μ0 Jν // both sides transform as a rank-1 tensor (4-vector) so covariant curl B - μ0ε0∂tE = μ0J and div E = ρ/ε0 (1.1.24) and (1.1.15) (1.3.16) The fields are given by ( ε is the permutation tensor), B1 = -F23 E1 = cF10 or Bi = -(1/2)εijkFjk and Ei = cFi0 B2 = -F31 E2 = cF20 B3 = -F12 E3 = cF30 (1.3.17) The E and B fields are part of the tensor Fμν and so do not transform as four vectors like Aμ. That is to say, there are no 4-vectors of the form Eμ or Bν, so there is no up and down index on a field, so the index is just written down. Jackson states the above facts (but in Gaussian units) in his Section 11.9 along with a description of the notion of covariance. Example: μ0J2 = ∂μFμ2 = ∂0F02 + ∂1F12 + ∂2F22 + ∂3F32 = (1/c)∂t(-1/cE2) + ∂1(-B3) + 0 + ∂3(+B1) = - (1/c2)∂tE2 + [curl B]2 => μ0J = curlB - μ0ε0∂tE in the 2 component (c) The Potential Wave Equations in the King and Lorenz Gauges with Conductors We refer to a certain gauge condition below as "the King gauge" because King (see Refs.) made extensive use of this condition in his books and papers at least as early as 1945. Perhaps this gauge has some official name, but we are not aware of it. We start with this King gauge and treat A and then φ. Then we do the Lorenz gauge case for A and φ, and finally we look at the wave equations for E and B. The motivation for using the King gauge is made clear. Unlike most sources on this subject, we allow for the possibility that the conductors' μi might differ from that of the dielectric. KING GAUGE Wave equation for A Consider the following general cross section of a transmission line which happens to be of coaxial cable type, Fig 1.5 The gray regions 2 and 3 are conductors, while the white region 1 is the (possibly conducting) dielectric. Currents J1, J2 and J3 are conduction currents. We start by selecting the King gauge for region 1 and we apply it to all three regions, div A = - μ1ε1 ∂tφ - μ1σ1φ // King gauge, applied to all of R . (1.3.18) We first obtain the wave equation for A in region 1. Start with (1.3.3) which gives the wave equation for A before any gauge choice is made (2 - μ1ε1 ∂t2) A = grad [μ1ε1 ∂tφ + divA ] - μ1J . // region 1 (1.3.3) Now insert the King gauge (1.3.18) to get (2 - μ1ε1 ∂t2) A = grad [μ1ε1 ∂tφ + (- μ1ε1 ∂tφ - μ1σ1φ) ] - μ1J = - μ1σ1 grad φ - μ1J = - μ1σ1 (-E -∂tA) - μ1(σ1E) . // from (1.3.1) and J = σ1E = - μ1σ1 ( -∂tA) . Thus the wave equation for A in region 1 is (2 - μ1ε1 ∂t2 - μ1σ1∂t) A = 0 // region 1 (1.3.19) This is a damped wave equation with no driving source; the equation is homogeneous. Now we start over with (1.3.3) for region 2: (2 - μ2ε2 ∂t2) A = grad [μ2ε2 ∂tφ + divA ] - μ2J2 // region 2 (1.3.3) As before, we insert the region 1 King gauge expression (1.3.18) for div A, even though we are now working in region 2, (2 - μ2ε2 ∂t2) A = grad [μ2ε2 ∂tφ + (- μ1ε1 ∂tφ - μ1σ1φ) ] - μ2J2 = [μ2ε2 ∂t + (- μ1ε1 ∂t - μ1σ1) ] gradφ - μ2J2 = [ (μ2ε2 - μ1ε1) ∂t - μ1σ1) ] gradφ - μ2J2 = [ (μ2ε2 - μ1ε1) ∂t - μ1σ1) ] (-E-∂tA) - μ2J2 // using (1.3.1) = [ (μ2ε2 - μ1ε1) ∂t - μ1σ1) ] (-J2/σ2-∂tA) - μ2J2 // J2 = σ2E ≈ [ (μ2ε2 - μ1ε1) ∂t - μ1σ1) ] (-∂tA) - μ2J2 // since σ2 is very large in conductor 2 = - [ (μ2ε2 - μ1ε1) ∂t2 - μ1σ1∂t ] A - μ2J2 Notice that we have chosen not to set J2 = σ2E in region 2 for the last term, we just leave it as J2. Moving the first term on the right to the left we get ( 2 - μ2ε2 ∂t2 + [ (μ2ε2 - μ1ε1) ∂t2 - μ1σ1∂t ] ) A = - μ2J2 or ( 2 - μ1ε1 ∂t2 - μ1σ1∂t ) A = - μ2J2 // region 2 (1.3.20) On the left side we see the same region 1 damped wave operator although we are in region 2, and J2 is the conduction current density in region 2. A similar result will be obtained for region 3. Thus we have shown that (2 - μ1ε1 ∂t2 - μ1σ1∂t ) A = 0 region 1 (2 - μ1ε1 ∂t2 - μ1σ1∂t ) A = - μ2J2 region 2 (2 - μ1ε1 ∂t2 - μ1σ1∂t ) A = - μ3J3 region 3 (1.3.21) We can combine these into a single equation which is then valid over all of region R, (2 - μ1ε1 ∂t2 - μ1σ1) A = - μ2J2 - μ3J3 all of region R (1.3.22) with the understanding that the conduction current in region 1 has already been accounted for and Ji represents conduction currents in conductor i .We could generalize this result for a region R containing any number N of conductors labeled i = 2,3...N+1 (2 - μ1ε1 ∂t2 - μ1σ1) A = - Σi=2N+1μiJi all of region R (1.3.23) Wave equation for φ We first obtain the wave equation for φ in region 1. Start with (1.3.2) which gives the wave equation for φ before any gauge choice is made 2φ + ∂t[div A] = -ρ/ε1 . (1.3.2) Now use the same global region R King gauge (1.3.18) for div A. 2φ + ∂t[- μ1ε1 ∂tφ - μ1σ1φ] = - ρ1/ε1 (2 - μ1ε1 ∂t2 - μ1σ1∂t)φ = - (1/ε1)ρ1 // region 1 (1.3.24) Again the same damped region 1 wave operator appears on the left side. Since the King gauge is the same in all three regions, we can write (2 - μ1ε1 ∂t2 - μ1σ1∂t)φ = - (1/ε1)ρ(1) // region 1 (2 - μ1ε1 ∂t2 - μ1σ1∂t)φ = - (1/ε2)ρ(2) // region 2 (2 - μ1ε1 ∂t2 - μ1σ1∂t)φ = - (1/ε3)ρ(3) // region 3 (1.3.25) where ρ always means free charge. The three equations are basically the same because the pre-gauge equation (1.3.2) has no region-specific parameters apart from ε1, in contrast with (1.3.3) quoted above. Now the only actual free charge present is the surface charge on the outside surfaces of the conductors and we shall regard all these charge densities as residing in region 1, the dielectric (just inside the boundaries of region 1). Thus, we write ρ(1) = ρ2 + ρ3 // = Σi=2N+1ρi ρ(2) = 0 ρ(3) = 0 (1.3.26) where ρi is the surface charge density on conductor i. We can then combine the above three equations into a single equation for all of region R (2 - μ1ε1 ∂t2 - μ1σ1∂t)φ = - (1/ε1) Σi=2N+1ρi all of region R (1.3.27) Conclusion for wave equations in the King gauge Here then are the wave equations for φ and A in region R using the region 1 King gauge: Potential Wave Equations in the King Gauge (1.3.28) (2 - μ1ε1 ∂t2 - μ1σ1∂t)φ = - (1/ε1) Σi=2N+1ρi all of region R (1.3.27) (2 - μ1ε1 ∂t2 - μ1σ1∂t)A = - Σi=2N+1 μiJi all of region R (1.3.23) div A = - μ1ε1 ∂tφ - μ1σ1φ King gauge (1.3.18) 1 = dielectric 2,3,4.... N+1= conductors (there are N conductors) ρi = free surface charge density on conductor i Ji = free current density in conductor i (J1 in the dielectric exists but does not appear in ΣiμiJi) To reduce clutter, we now make the change that the dielectric region has no subscript 1, and we renumber the conductors 1 to N instead of 2 to N+1. The above box then becomes Potential Wave Equations in the King Gauge (1.3.29) (2 - με ∂t2 - μσ∂t)φ = - (1/ε) Σiρi all of region R (2 - με ∂t2 - μσ∂t)A = - ΣiμiJi all of region R div A = - με ∂tφ - μσφ King gauge μ,ε,σ = dielectric 1,3,4.... N = conductors Σi = Σi=1N μi = for conductor i ρi = free surface charge density on conductor i Ji = free current density in conductor i (J in the dielectric exists but does not appear in ΣiμiJi) The word "free" us used above to emphasize the fact that possible polarization charge densities and magnetization curent densities are not included in these ρi and Ji. King never writes these wave equations in his transmission-line theory book, so it is difficult to find verification of our logic pathway in his book. However, Panofsky and Phillips do show the equations and we quote the relevant section from p 241 of their book: Their last sentence says that J = σE in the conducting dielectric has been incorporated into the -μσ∂tA term in their first equation, just as we have done above. These authors have assumed that the μ's of the dielectric and the conductors are all the same (normally μ = μ0). In order to obtain the above equations, Panofsky and Phillips use the King gauge (1.3.18) but they refer to this gauge simply as "the Lorentz condition" (illustrating Comments 1and 2 above). From their page 240, LORENZ GAUGE If we carry out exact same program with respect to Fig 1.5 using a global Lorenz gauge for all of R, divA = - μ1ε1∂tφ , (1.3.30) we obtain these results for A, where in region 1 the conduction current is not absorbed into a damping term on the left side, (2 - μ1ε1 ∂t2)A = - μ1J1 region 1 (2 - μ1ε1 ∂t2)A = - μ2J2 region 2 (2 - μ1ε1 ∂t2)A = - μ3J3 . region 3 As before, all three equations have the same wave operator on the left side. Again assuming N conductors, we combine these into a single equation as follows (2 - μ1ε1 ∂t2)A = - Σi=1N+1 μiJi all of region R (1.3.31) Meanwhile, the results for φ are (2 - μ1ε1 ∂t2)φ = -ρ(1)/ε1 region 1 (2 - μ1ε1 ∂t2)φ = -ρ(2)/ε2 region 2 (2 - μ1ε1 ∂t2)φ = -ρ(3)/ε3 region 3 so that with the same comments made earlier in (1.3.26) this becomes (2 - μ1ε1 ∂t2)φ = -(1/ε1) Σi=2N+1 ρi all of region R (1.3.32) We then make the same notational change made above to get these Lorenz-gauge results: (2 - με ∂t2)φ = -(1/ε) Σi=1N ρi all of region R (1.3.33) (2 - με ∂t2)A = - Σi=1N μiJi - μJ all of region R (1.3.34) Notic that no conductivities appear in these equations. COMPARISON We can now do a side-by-side comparison, where Σi is a sum over the conductors i = 1,2..N King Gauge: (2 - με ∂t2 - μσ∂t)φ = - (1/ε) Σiρi all of region R (1.3.29) (2 - με ∂t2 - μσ∂t)A = - ΣiμiJi all of region R (1.3.29) div A = - με ∂tφ - μσφ King gauge (1.3.29) Lorenz Gauge: (2 - με ∂t2)φ = -(1/ε) Σiρi all of region R (1.3.33) (2 - με ∂t2)A = - Σi μiJi - μJ all of region R (1.3.34) divA = - μ1ε1∂tφ Lorenz gauge (1.3.30) In the Lorenz gauge, we get undamped wave operators, but the sum on the right of the A equation includes the current J in the dielectric, whereas this is not the case in the King gauge. In a situation where we have prescribed currents Ji in the conductors, it is inconvenient to have to worry about the dielectric conduction current J which complicates the solution of the problem (this will become more obvious later). In the King gauge, we get damped wave operators but we have to include only the current in the conductors since the current in the dielectric has been incorporated into the damping term. When we transform to the frequency domain and write the Helmholtz equation for A and its Helmholtz Integral solution, we need only integrate over the conductors which makes life easier. This then is the motivation for the King gauge. E AND B WAVE EQUATIONS Meanwhile, the E and B field wave equations of course don't know anything about gauges and from (1.2.1) and (1.2.2) we have, with respect to Fig 1.5, ( ρs = ρ2+ρ3 = Σi=2N+1ρi) (2 - μ1ε1 ∂t2)E = μ1∂tJ1 + (1/ε1) grad ρ(1) = μ1∂tJ1 + (1/ε1) grad ρs // region 1 (2 - μ2ε2 ∂t2)E = μ2∂tJ2 + (1/ε2) grad ρ(2) = μ2∂tJ2 // region 2 (2 - μ3ε3 ∂t2)E = μ3∂tJ3 + (1/ε3) grad ρ(3) = μ3∂tJ3 // region 3 (1.3.35) (2 - μ1ε1 ∂t2)B = - μ1 curl J1 // region 1 (2 - μ2ε2 ∂t2)B = - μ2 curl J2 // region 2 (2 - μ3ε3 ∂t2)B = - μ3 curl J3 // region 3 where Ji = σiE . We cannot unify each group of three equations into a single region R equation as we could in the potential case since the wave operators are different in each region. Using Ji = σiE and curl E = - ∂tB and (1.3.26) (2 - μ1ε1 ∂t2 - μ1σ1∂t)E = (1/ε1) Σi=2N+1 grad ρi // region 1 (2 - μ2ε2 ∂t2 - μ2σ2∂t)E = 0 // region 2 (2 - μ3ε3 ∂t2 - μ3σ3∂t)E = 0 // region 3 (1.3.36) (2 - μ1ε1 ∂t2 - μ1σ1∂t)B = 0 // region 1 (2 - μ2ε2 ∂t2 - μ2σ2∂t)B = 0 // region 2 (2 - μ3ε3 ∂t2 - μ3σ3∂t)B = 0 // region 3 Again the three damped wave operators are different. The solution of these equations requires solving the first equationfor the particular solution in region 1, finding all possible homogenous solutions to all 6 equations in their regions using appropriate harmonic forms with "constants to be determined", then matching these conditions at the two boundaries to evaluate the constants. In contrast, in the potential problem of (1.3.28), (2 - μ1ε1 ∂t2 - μ1σ1∂t)φ = - (1/ε1) Σi=2N+1ρi all of region R (1.3.26) (2 - μ1ε1 ∂t2 - μ1σ1∂t)A = - Σi=2N μiJi all of region R (1.3.23) div A = - μ1ε1 ∂tφ - μ1σ1φ King gauge (1.3.18) (1.3.28) one worries about a single unified region R and there is only one damped wave operator. The method of solution is to find the particular solutions of the φ and A equations, add in homogenous solutions and match boundary conditions. 1.4 Retarded Solutions in the Lorenz gauge: a path not taken In a medium where μ and ε are time-independent, the Lorenz gauge equations (1.3.4) and (1.3.5) apply, (2 - με ∂t2)φ = - (1/ε)ρ (1.3.4) 2 - με ∂t2)A = - μJ . (1.3.5) One approach to solving these equations for A and φ is the method of retarded solutions. Although we shall not use this approach, we outline the method here. We seek to solve an equation of this form (2 - με ∂t2) u = - f , (1.4.1) where for example in (1.3.4) we have u = φ and f = ρ/ε. Since με = 1/v2 where v is the wave velocity in the medium, write (1.4.1) as (∂t2 - v22) u = v2f or u = f where ≡ ∂t2 - 2 . (1.4.2) This last equation is similar to (A.7.2) of Appendix A and we solve it in the same manner. Define a Green's function g as the solution of v2 g(x,t; x',t') = δ(x-x')δ(t-t') with g = 0 when |x-x'|→∞ . (1.4.3) As Appendix A.7 shows, the solution is given by v2 g(x,t; x',t') = (1/4πR)δ(t-t'-R/v) with R = |x-x'| . (1.4.4) Jackson (6.41) and (6.44) uses G(+) = 4πv2g with v = c and refers to the solution as a "retarded Green function". See also Stakgold references in Appendix A. Then the solution to (1.4.1) is u(x,t) = ∫d3x' ∫dt' v2 g(x,t; x',t') f(x',t') as can be verified by applying to both sides and making use of (1.4.3). Inserting (1.4.4) then gives u(x,t) = ∫d3x' ∫dt' (1/4πR)δ(t-t'-R/v) f(x',t') = ∫d3x' (1/4πR) f(x', t-R/v) = ∫d3x' . (1.4.5) Thus, the solutions to (1.3.2) and (1.3.3) are (Lorenz gauge) : φ(x,t) = ∫d3x' (1.4.7) A(x,t) = ∫d3x' . (1.4.6) The potentials at time t are generated by the values the sources had at time t - R/v since the influence of the sources travels at finite velocity v through the medium. Compare (1.4.7) to (A.0.2) which is the solution to the electrostatic Poisson equation. These last equations agree with Jackson p 246 (6.48). Although they are interesting and useful for many problems (such as radiation), we shall not make use of these retarded solutions in our transmission line analysis. 1.5 The Wave Equations in the Frequency Domain verify all equation numbers (a) The Transformed Wave Equations A standard method of solving wave equations involves transforming the equations from the time domain to the frequency ω domain using the Fourier Integral Transform, assuming that the μ, ε and σ are constants (possibly complex). As an example, we start with the φ equation in (1.3.29) and expand φ(x,t) and ρs(x,t) onto their Fourier components [ the overloaded notation is explained in Section 1.6 ] (2 - με ∂t2 - μσ∂t) φ(x,t) = - (1/ε) Σiρi(x,t) (1.3.26) (2 - με ∂t2 - μσ∂t) [(1/2π)!Syntax Error, Idω e+jωt φ(x,ω)] = - (1/ε) [(1/2π)!Syntax Error, Idω e+jωt Σiρi(x,ω) ] !Syntax Error, Idω (2 - με ∂t2 - μσ∂t) e+jωt φ(x,ω) = - (1/ε) !Syntax Error, Idω e+jωt Σiρi(x,ω) !Syntax Error, Idω (2 + μεω2 -jω μσ) e+jωt φ(x,ω) = - (1/ε) !Syntax Error, Idω e+jωt Σiρi(x,ω) !Syntax Error, Idω e+jωt [(2 + μεω2 - jω μσ) φ(x,ω)] = !Syntax Error, Idω e+jωt [- (1/ε) Σiρi(x,ω)] . At this point we invoke the completeness of the set of functions {ejωt} on the interval (-∞,∞) to claim that the integrands must be equal, giving (1.3.26) transformed to the frequency domain, (2 + μεω2 - jω μσ) φ(x,ω) = - (1/ε1) Σiρi(x,ω) . or (2 + β2) φ(x,ω) = - (1/ε) Σiρi(x,ω) where β2 is the following complex "Helmholtz parameter" [of Helmholtz operator (2 + β2) ] β2 = μεω2 - jωμσ = ω2μ ( ε - jσ/ω) = ω2μ ξ ξ ≡ ε - jσ/ω . (1.5.1) Here ξ(ω) is the "complex dielectric constant", nothing more or less than the expression shown. Note: The conventional symbol for the Helmholtz wavenumber parameter is k but we use β since k will have another meaning later, and to be consistent with King p 10 (15a,b,c). King bolds parameters when they are complex, but we do not. Examination of the above transformation shows that any equation can be transformed from the time domain to the frequency domain using these simple rules, ∂t → +jω ∂t2 → -ω2 F(x,t) → F(x,ω) . (1.5.2) where it is understood (Section 1.6) that F(x,t) and F(x,ω) are different functions. Thus, the frequency-domain representations of the King-gauge potential wave equations (1.3.26) and (1.3.23) along with the King gauge (1.3.18) become Potential Wave Equations in the King Gauge (ω domain) (2 + β2)φ = - (1/ε) Σiρi all of region R (1.5.3) (2 + β2)A = - Σi=2N μiJi all of region R (1.5.4) div A = - μεjωφ - μσφ = -jωμ(ε+σ/jω)φ = -jωμξφ = -j(β2/ω) φ King gauge (1.5.5) β2 = μεω2 - jωμσ = ω2μ ( ε - jσ/ω) = ω2μ ξ ξ ≡ ε - jσ/ω parameter β (1.5.1) μ,ε,σ = dielectric 1,3,4.... N = conductors Σi = Σi=1N μi = for conductor i ρi = free surface charge density on conductor i Ji = free current density in conductor i (J in the dielectric exists but does not appear in ΣiμiJi) In these equations, all mathematical fields φ, A, Ji, Jm(i) are functions of x and ω. (b) The Helmholtz Integrals in the King Gauge The next step is to solve the above equations for φ and A. The method was demonstrated in Appendix A.0 and is applied again here. We first define the free-space Green's Function g by this boundary value problem, - (2 + β2)g(x,x') = δ(x-x') where lim|x|→∞ g(x,x') = 0 . (1.5.6) As shown in (H.15), the solution to problem (1.5.6) is g(x,x') = R = |x - x'| (1.5.7) and so the particular solutions for the potentials are given by (dV' = d3x' , N = number of conductors) φ(x,ω) = Σi∫ρi(x',ω) dV' R = |x - x'| (1.5.8) A(x,ω) = Σi∫μiJi(x',ω) dV' R = |x - x'| . (1.5.9) In these equations, β is a function of ω, namely β = ω2μξ as in (1.5.1), and Σi is over the conductors. Since (2 + β2) is the Helmholtz operator, solutions of the form (1.5.8) and (1.5.9) are sometimes called Helmholtz integrals. To verify that the φ of (1.5.8) solves (1.5.3) we write φ(x,ω) = ∫ [Σiρi(x',ω)/ε] g(x,x') dV' so that, - (2 + β2) φ(x,ω) = ∫[ Σiρi(x',ω)/ε] { - (2 + β2)g(x,x') } dV' = ∫[ Σiρi(x',ω)/ε] {δ(x-x')} d3x' = Σiρi(x,ω)/ε . In the limit ω→ 0 we find from (1.5.1) that β(ω) → 0 and then (1.5.9) is the same as (A.0.2) obtained from electrostatics and Poisson's Equation. In (1.5.8) we have the volume density function ρs(x',ω) ≡ Σiρi(x',ω) representing a surface charge density, so it is convenient to represent φ as a surface integral over the corresponding surface charge density ns(x',ω), φ(x,ω) = ∫ns(x',ω)dS' R = |x - x'| (1.5.10) Comments on n, σ and Dirichlet : Usually one uses σ for a surface charge, but σ is already used for conductivity. To further complicate things, in his potential theory discussion of Chapter 6, Stakgold uses σ to represent our surface S enclosing a volume V (his region R) as in our Fig 1.2. Stakgold uses n to indicate a normal derivative, as in this Dirichlet problem solution of the Poisson equation -2φ(x) = q(x), φ(x) = ∫R dx' g(x|x') q(x') – ∫σ dSξ f(ξ) ∂ξng(x|ξ) // Stakgold (6.81) . (1.5.11) Here ∂ξn = ∂/∂nξ where nξ is a local coordinate on the surface σ at point ξ which is normal to the surface. In this equation, q(x) is the Poisson source (think ρ(x)/ε0), g(x|ξ) is the full Green's function, meaning g = 0 on boundary σ, and f(ξ) is the Dirichlet prescribed potential on the enclosing boundary σ. Stakgold also uses n for number of dimensions and his work is always done in n spatial dimensions. In the above equation, the first term is the particular solution, like our Helmholtz integral, while the second term is a homogenous solution to -2φ(x) = 0 which, when added in, makes things work. Our Helmholtz integral, however, uses the free-space Green's Function, so we cannot just add on Stakgold's Dirichlet term to get a solution. The above Poisson Dirichlet solution (1.5.11) seems mysterious at first viewing, but is easily derived using -2g(x|x') = δ(x-x') and the famous Green's 2nd "symmetric" identity, where ∂φ/∂n = φ = the same normal derivative ∂ξn discussed above, ∫V dV ψ 2φ = ∫S dS ψ( ∂φ/∂n) – ∫V dV (ψ φ) Green #1 ∫V dV [ ψ 2φ – φ 2ψ ] = ∫S dS [ ψ(∂φ/∂n) – φ(∂ψ/∂n) ] Green #2 (1.5.12) Here #2 = #1(ψ,φ) + #1(φ,ψ) and #1 is derived from the divergence theorem (1.1.30) with F = ψφ and vector identity (ψφ) = ψ φ + ψ 2φ and dS = dS . Green was a busy man. (c) King's leading factor (1/4πξ) and the final Helmholtz Integrals This is a somewhat subtle point and something that King never discusses much in his transmission-line theory book. The issue is that there are two different entities ns and nc which have units charge/area, and they are related by ns = (ε/ξ) nc where ξ = ε + σ/jω is the complex dielectric constant in the dielectric which incorporates the effect of possible dielectric conductivity. In a transmission line problem, it is nc that is specified by the boundary conditions and not ns (which is the actual surface charge density). For that reason, one replaces (1.5.10) with, φ(x,ω) = ∫nc(x',ω) dS' R = |x - x'| (1.5.13) which explains the leading factor which appears every time King writes down the Helmholtz integral for φ in his books. In the discussion below we describe nc and its relation to ns, and then we show how this relation works in the simple example of a parallel plate capacitor. Consider the situation at a general boundary between dielectric (region 1) and conductor (region 2) where there exists a surface charge density ns : In (1.1.18) it was shown that div [Jd + Jc] = 0 where Jc = σE is the conduction current and Jd the displacement current ∂tD = ε∂tE. The divergence theorem (1.1.30) then says 0 = ∫V div [Jd + Jc] dV = ∫S [Jd + Jc] dA . Applied to the blue pillbox which straddles the boundary in the figure, we find Jd1n + Jc1n = Jd2n + Jc2n where n means normal component. Writing this out, ε1∂tE1n + σ1E1n = ε2∂tE2n + σ2E2n ≈ σ2E2n = Jc2n since σ2 is huge inside the conductor. Therefore, Jc2n = σ1E1n + ε1∂tE1n . (1.5.14) Meanwhile, Gauss's Law (1.1.33) states that div (εE) = ρ ∫V ρ dV = ∫S εE dA . (1.1.18) Applied to the same blue pillbox we find ns = ε1En1 - ε2En2 ≈ ε1En1 since En2 ≈ 0 inside the conductor. Thus, En1 = ns/ε1 and then Jcn1 = σ1En1 = ns(σ1/ε1) (1.5.15) Then (1.5.14) can be written as Jc2n = σ1E1n + ε1∂tE1n = (σ1 + ε1∂t)E1n = (1/ε1)(σ1 + ε1∂t)ns or, writing out the arguments, Jc2n(x,t) = (1/ε1)(σ1 + ε1∂t)ns(x,t) . In the frequency domain with rules (1.5.2) this becomes Jc2n(x,ω) = (1/ε1)(σ1 + ε1jω)ns(x,ω) = (1/ε1) (jω)(ε1 + σ1/jω) ns(x,ω) = (ξ1/ε1) (jω) ns(x,ω) ξ1 ≡ ε1 + σ1/jω = complex dielectric constant . (1.5.16) If we observe the conduction current Jc2n flowing through a unit-area loop (red in figure), we can write Jc2n = ∂tnc where nc is the total amount of conduction charge flowing through that unit-area loop per unit time. Thus we have ∂tnc(x,t) = (1/ε1)(σ1 + ε1∂t)ns(x,t) or jω nc(x,ω) = (ξ1/ε1) (jω) ns(x,ω) or nc(x,ω) = (ξ1/ε1) ns(x,ω) . (1.5.17) where is our result claimed at the start that ns = (ε/ξ) nc. Note that: The quantity ns is the amount of free charge per unit area on the conductor surface. The quantity nc does not represent any kind of surface charge anywhere (free or otherwise). nc is related to the transport of conduction charge carriers through the charge-neutral interior of the conductor just below the surface. There is no unit-area surface which holds nc amount of charge, but both ns and nc have the dimensions of charge/area so both can therefore be called "surface charge". These two areal charge densities are different simply because the dielectric leaks charge off the surface. We are now going to rederive (1.5.17) a different way. We can write, using the blue pillbox and continuity relation (1.1.25), div Jc = - ∂tρfree -∂t[∫V ρfree dV] = ∫S Jc dA . (1.1.25) => - ∂t[∫V ns dA] = ∫S Jc dA => -∂tns = Jcn1 - Jcn2 = σ1(ns/ε1) - ∂tnc // see above: Jcn1 = σ1(ns/ε1), Jcn2 = ∂tnc so ∂tns = ∂tnc - σ1(ns/ε1) // change in ns = flow in - flow out or jωns = jωnc - σ1ns/ε1 => (jω+ σ1/ε1)ns = jωnc => (jωε1+ σ1)ns = jωε1nc => (ε1+σ1/jω)ns = ε1nc => ξ1 ns = ε1 nc => nc = (ξ1/ε1)ns which is the same as (1.5.17). Note that surface charge ns is real, while nc is complex. It is useful at this point to examine the simple case of a parallel plate capacitor to see the meaning of ns and nc. The plate separation s is meant to be very small compared to the transverse dimensions of the plates, so the picture is distorted. We drop the subscript 1 on dielectric properties. First off, a DC analysis of the above device shows that the capacitor has resistance R, R = = = = (s/σA). (1.5.18) Now we assume an AC voltage V. The total current entering the conducting capacitor is I = JcA. If we think of I = ∂tQ then Q is the amount of charge passing through the external wire per unit time. Q is not the total charge on the left plate surface which in fact is Qs = nsA. Since I = JcA we have ∂tQ = (∂tnc) A and therefore Q = ncA. Meanwhile, the voltage V between the plates is V = Es, and we know that E = ns/ε from Gauss's law. Thus V = (s/ε)ns. If we define the (complex) capacitance by Q = C'V, then C' = = = (Aε/s) = (ξ/ε) (Aε/s) = (Aξ/s) . (1.5.19) The capacitance C' is complex because it accounts for both the capacitance and conductance of the dielectric, C' = (Aε/s) = (Aε/s) + (σA/s)/(jω) = C + 1/(jωR) (1.5.20) or jωC' = jωC + 1/R or = + Z = = (1.5.21) which is the rule for computing an impedance Z for a capacitor and resistor in parallel Looking back at this example, it is clear that if one wants to compute the complete impedance of the conducting capacitor, one uses C' = Q/V where Q = Anc. The ratio Qs/V gives only the capacitance C. = = (Aε/s) = C . (1.5.22) In this conducting capacitor problem, the boundary conditions are the voltage V or the total current I. Specification of the current I = ∂t(ncA) is really a specification of nc since in the frequency domain we then have I = jωAnc. In analyzing the problem in full, we are thus interested in working with nc and not ns. So recalling now the King gauge Helmholz integral for φ , φ(x,ω) = ∫ns(x',ω)dS' R = |x - x'| , (1.5.10) since it will more convenient to have nc in the integrand, we use (1.5.17) that ns = (ε/ξ) nc to rewrite the above expression as φ(x,ω) = ∫nc(x',ω) dS' R = |x - x'| (1.5.13) which is just (1.5.13) stated earlier. So here are our final forms of the Helmholtz integrals of interest, where now write nc = Σinci, φ(x,ω) = Σi∫nci(x',ω)dS' R = |x - x'| (1.5.13) A(x,ω) = Σi∫μiJi(x',ω) dV' R = |x - x'| (1.5.9) β2 = μεω2 - jωμσ = ω2μ ( ε - jσ/ω) = ω2μ ξ ξ ≡ ε - jσ/ω (1.5.1) where the sum Σi is over all conductors. If all conductor have the same μi = μc, (1.5.9) simplifies to A(x,ω) = Σi∫Ji(x',ω) dV' R = |x - x'| . (1.5.9)' We now quote directly from King's Transmission-Line Theory book to show how he presents the Helmholtz integrals for φ and A. What we call the King gauge appears as (2b) below. His symbols σ, ε, μ, ξ and β apply to the dielectric. // page 8 // page 9 // page 11 Comments: (1) King's (23) and (24) are for one conductor, while our (1.5.13) and (1.5.9) are for several conductors. (2) Due to time lag effects, ε and σ may be complex, so ε = ε'-jε" and σ = σ'-jσ". In this case ξ ≡ ε - jσ/ω = (ε'-jε") - j(σ'-jσ")/ω = [ε'- σ"/ω] - j [σ' + ωε"]/ω = εeff - jσeff/ω so one would replace ε → εeff and σ → σeff in all equations (see King p 9 footnote). (3) In the same way, time lag effects can cause μ = μ' - jμ" (hysteresis). (4) King uses bold font for vectors and for quantities which are complex. For example his ξ of ξ = ε - jσ/ω is bolded. Similarly, our (1.5.1) that β2 = ω2μξ becomes his equation (10) above, β2 = ω2μξ . (5) King assumes that all conductors and the dielectric have the same μ, something we did not assume. In order to make (23) and (24) look as similar as possible, he defines ν ≡ 1/μ. Since these parameters can both be complex, he writes them as μ and ν. This then explains the factor 1/(4πν) appearing in his (24) which then agrees with our (1.5.9)'. (6) He shows his equation (23) charge density n' in bold, indicating it is complex. His n' is our nc, also complex. He refers to n' as "charge density on the surface" but he really means it to be nc as we have discussed at length above, and this is how he uses it in his calculations. King uses these Helmholtz integrals (23) and (24) for φ and A extensively in his book to compute the parameters of various complicated transmission line geometries and interfaces. We shall pursue this subject more in Chapter 4 for some simple case. We should point out that King makes no attempt to derive his equations (23) and (24) and more or less just pulls them out of his hat. We spent some time perusing several of King's other 11 books looking for some kind of derivation but were unsuccessful. The equations do appear in more or less the same form in his earliest book Electromagnetic Engineering (1945). So in some sense, we have spent the first 30 pages of this document deriving his equations (23) and (24). For that reason, it is worth gathering up the results in a summary box: Potential Solutions for φ and A in the King Gauge (ω space) (1.5.23) φ(x,ω) = Σi∫nci(x',ω)dS' R = |x - x'| (1.5.11) A(x,ω) = Σi∫μiJi(x',ω) dV' R = |x - x'| (1.5.9) μ,σ,ε = dielectric; μi = inside conductor i ; if all μi = μ then A(x,ω) = Σi∫Ji(x',ω)] dV' (1.5.9)' β2 = μεω2 - jωμσ = ω2μ ( ε - jσ/ω) = ω2μ ξ ξ ≡ ε - jσ/ω = ε + σ/jω (1.5.1) div A = - μεjωφ - μσφ = -jωμ(ε+σ/jω)φ = -jωμξφ // King gauge (1.5.5) B = curl A E = - grad φ - ∂tA (1.3.1) The Helmholtz integrals are just "particular solutions" to the potential wave equations. In order to solve a problem, one must add to these particular solutions whatever homogeneous solutions are necessary in order to match all boundary conditions. (d) Frequency domain wave equations for fields, and for potentials in the Lorenz Gauge We use the earlier notation with reference to Fig *** where dielectric = 1 and conductors = 2,3...N+1 for N conductors. The Lorenz gauge is given by (1.3.30) transformed to the ω domain, divA = - μ1ε1jωφ . (1.5.24) Undamped Lorenz-gauge potential wave equations (1.3.32) and (1.3.31) : k12 = ω2μ1ε1 (2+k12)φ = -(1/ε1) Σi=2N+1 ρi all of region R (2+k12)A = - Σi=1N+1 μiJi all of region R (1.5.25) In the Lorenz gauge, the potential wave equations don't have damped operator versions. However, for the field wave equations (which know nothing of gauge) we can write both undamped and damped versions: Undamped field wave equations (1.3.35) : ki2 = ω2μiεi (2+k12)E = μ1jωJ1 + (1/ε1) Σi=2N+1 grad ρi (2+k12)B = - μ1 curl J1 // region 1 (2+k22)E = μ2jωJ2 (2+k22)B = - μ2 curl J2 // region 2 (2+k32)E = μ3jωJ3 (2+k32)B = - μ3 curl J3 // region 3 (1.5.26) Damped field wave equations (1.3.36) : βi2 = ω2μiξi (2+β12)E = (1/ε1) Σi=2N+1grad ρi (2+β12)B = 0 // region 1 (2+β22)E = 0 (2+β22)B = 0 // region 2 (2+β32)E = 0 (2+β32)B = 0 // region 3 (1.5.27) The solution method was outlined earlier: for each inhomogeneous equation compute the particular solution as a Helmholtz integral, then for all equations identify generic homogeneous solutions with unknown constants, and finally determine those constants using boundary conditions from box (1.1.50). The potential approach has the advantage of a single wave operator and only two equations, while the damped field approach has the advantage of not involving any currents, but the disadvantage of having three times more equations and requiring computation of grad ρi. There is a lot more to keep track of. These are all of course vector Helmholtz equations. In a problem having only a single region (having μ,ε,σ) containing current density J and charge density ρ (perhaps inside the region, perhaps just on the surface), the Lorenz-gauge potential wave equations above in (1.5.25) may be written (2+k2)φ = -(1/ε) ρ k2 = ω2με all of region R (2+k2)A = - μJ k2 = ω2με all of region R (1.5.28) These equations may be derived directly from the single-region field wave equations (1.2.1) and (1.2.2) converted to the frequency domain, (2 + k2)E = jωμJ + (1/ε) grad ρ k2 = ω2με (2 + k2)B = - μ curl J (1.5.29) Each of these last four equations has its own Helmholtz integral, φ(x,ω) = ∫ρ(x',ω)dV' R = |x - x'| k2 = ω2με A(x,ω) = ∫J(x',ω) dV' (1.5.30) E(x,ω) = - ∫[ jωμJ(x',ω) + (1/ε) grad ρ(x',ω) ] dV' B(x,ω) = ∫[ curl J(x',ω)] dV' (1.5.31) The A(x,ω) Helmholtz integral (1.5.30) appears on Jackson p 408, Eq. (9.3), with j → -i and μ→ μ0. For this same single-region problem, the damped wave equation (1.5.27) becomes (2+β2)E = (1/ε) grad ρ β2 = ω2μξ (2+β2)B = 0 (1.5.32) where again ρ might be in the volume and/or on the surface of the volume. This follows directly from (1.5.29) using the methods above. 1.6 Reinterpretation of all equations in terms of complex functions It seemed useful to defer the topics of this section to avoid cluttering up the preceding five sections. The Fourier Transform has already been used in the previous two sections, and here we shall discuss it more formally as a motivating factor in changing our point of view from real to complex functions. The general nature of the Fourier transform of complex monochrome (ejωt) fields sets the stage for the analysis of the round wire in Section 2. (a) Complex Functions Up to this point, we have been regarding the following fields as representing real physical quantities, H(x,t) D(x,t) J(x,t) A(x,t) B(x,t) E(x,t) ρ(x,t) φ(x,t) (1.6.1) The fields and sources exist in the real physical world and are related by equations involving real operators like curl and ∂/∂t. We can represent such an equation as Lx,tf(x,t) = g(x,t) where Lx,t is some real differential operator and f and g are real fields ( g does not mean Green's function here). One can extend f and g such that f and g are either both the real or both the imaginary parts of complex functions F and G. Then the equation Lx,tF(x,t) = G(x,t) represents two distinct physical equations which we can write as Lx,tF(x,t) = G(x,t) => Lx,t[f(x,t) + jf'(x,t)] = [g(x,t) + jg'(x,t)] => Lx,t f(x,t) = g(x,t) F(x,t) = f(x,t) + jf'(x,t) Lx,t f'(x,t) = g'(x,t) G(x,t) = g(x,t) + jg'(x,t) . (1.6.2) It is convenient to regard all the mathematical fields listed above in (1.6.1) as complex fields like F and G. For example, we might write the Maxwell curl E equation (1.1.2) in this manner curl E(x,t) = - ∂B(x,t)/∂t E(x,t) = e(x,t) + j e'(x,t) B(x,t) = b(x,t) + j b'(x,t) . (1.6.3) The single left equation of (1.6.3) then represents these two different physical equations with real fields curl e(x,t) = - ∂b(x,t)/∂t curl e'(x,t) = - ∂b'(x,t)/∂t . (1.6.4) (b) Monochrome time The classic application of this idea is the assumption that some complex field is "monochrome" in its time dependence, meaning for example E(x,t) = ej[ωt+φ(x,ω)] e(x,ω1) = ejωt ejφ(x,ω) e(x,ω1) , (1.6.5) where e(x,ω1) is real. All time dependence is in the ejωt factor and all spatial dependence is in the factor [ejφ(x,ω)e(x,ω1)] -- separation of variables. This monochrome field might be regarded as a probe or driver of some system and the solution function e(x,ω1) and phase φ(x,ω1) might depend parametrically on the probe frequency ω1 as well as on position x, hence their explicit second arguments. For (1.6.5) the corresponding physical field assumption is either of these equations, e(x,t) = Re{ E(x,t)} = cos[ω1t + φ(x,ω1)] e(x,ω1) e'(x,t) = Im{ E(x,t)} = sin[ω1t + φ(x,ω1)] e(x,ω1) . (1.6.6) We stress again that the phase φ(x,ω1) might depend on both x and ω1. We shall see this situation arise in the next section when we consider fields inside a conducting round wire. A good prototype example for the ω1 dependence of phase φ(x,ω1) is a damped harmonic oscillator with resonant frequency ω0 which is driven at frequency ω1. The solution is: x(t) = x(0) sin[ω1t + φ(ω1)] tanφ(ω1) = -(ω1/τ)/(ω02- ω12) Of course the solution function x(t) is not a field over R3, so in this case the phase φ has no x dependence. (c) Why complex fields? The reason for using a complex field like E(x,t) instead of the real field e(x,t) has to do with the Fourier Transform (or the Laplace Transform). This transform is almost always needed to solve a non-trivial problem involving Maxwell's equations, and we saw it in action in Section 1.5. In our somewhat sloppy notation, and with the convention that the (1/2π) goes in the expansion formula along with e+jωt, we write this transform as : E(x,ω) = !Syntax Error, Idt E(x,t) e-jωt projection = transform (1.6.7) E(x,t) = (1/2π)!Syntax Error, Idω E(x,ω) e+jωt . expansion = inverse transform = recovery (1.6.8) Here E(x,t) is the original complex field whose real and imaginary parts are physical fields as in (1.6.3) or (1.6.6), while E(x,ω) is the Fourier Transform of E(x,t). Since E(x,ω) is a completely different complex function from E(x,t), one really should use some notation like E(x,ω) or E^(x,ω), but we trust the reader to make the distinction when the ω argument is present or in the general context of some discussion. As (1.6.7) shows, the dimensional units of the Fourier transform of some quantity have an extra sec factor. For example, since dim[E(x,t)] = volt/m, it follows that dim[E(x,ω)] = volt-sec/m. An obvious property of the Fourier Transform is this: ∂tE(x,t) = (1/2π)!Syntax Error, Idω E(x,ω) ∂t e+jωt = (1/2π)!Syntax Error, Idω [jω E(x,ω)] e+jωt which we can write as ( symbol ↔ means "corresponds to") E(x,t) ↔ E(x,ω) ∂tE(x,t) ↔ jω E(x,ω) (1.6.9) which is just another way to state our rule (1.5.2). In the case of assumed monochrome time dependence of the form (1.6.5) ( reflected in (1.6.6) ) one finds that E(x,t) = ej[ωt+φ(x,ω)] e(x,ω1) (1.6.5) E(x,ω) = !Syntax Error, Idt [ejωt ejφ(x,ω)e(x,ω1)] e-jωt = e(x,ω1) ejφ(x,ω) !Syntax Error, Idt ej(ω-ω)t = e(x,ω1) ejφ(x,ω) 2πδ(ω-ω1) . (1.6.10) It is this simple single-δ-function form that motivates the use of complex fields as carriers of the real physical fields. One can of course Fourier-transform the monochrome physical field directly, but the result is clumsy to deal with. For example, e(x,t) = cos[ω1t + φ1(x,ω1)] e(x,ω1) e(x,ω) = !Syntax Error, Idt { cos[ω1t + φ1(x,ω1)] e(x,ω1) }e-jωt = e(x,ω1) (1/2) !Syntax Error, Idt { ej[ωt+φ(x,ω)] + e-j[ωt+φ(x,ω)] } e-jωt = e(x,ω1) [ejφ(x,ω)πδ(ω-ω1) + e-jφ(x,ω)πδ(ω+ω1) ] . (1.6.11) A directly related benefit of using the complex function approach is the fact that math with exponentials is so much simpler than the corresponding math with trig functions, as for example ej(ωt+φ) e-j(ω't+φ') = ej(ω-ω')t ej(φ-φ') // dependence on t isolated to one factor cos(ωt+φ)cos(ω't+φ') = (1/2) { cos[ (ω-ω')t + (φ-φ')] + cos[ (ω+ω')t + (φ+φ')] } . (1.6.12) Another benefit of using the Fourier transform is its close connection with the Laplace Transform. Comment: Using the real cosine form shown as the first line of (1.6.6) along with the Fourier Cosine Transform is not viable because cos[ω1t + φ(x,ω1)] e(x,ω1) is not an even function of t. (d) Monochrome E and B fields One might seek to solve a system using monochrome fields of the form (1.6.5) for both the electric and magnetic fields. Those forms would be (e and b are real) E(x,t) = ej[ωt+φ(x,ω)] e(x,ω1) B(x,t) = ej[ωt+φ(x,ω)] b(x,ω1) (1.6.13) where we assume the same frequency ω1 for both fields, but allow the fields to have different phase functions φe and φb. In this case (1.6.10) becomes E(x,ω) = e(x,ω1) ejφ(x,ω) 2πδ(ω-ω1) B(x,ω) = b(x,ω1) ejφ(x,ω) 2πδ(ω-ω1) . (1.6.14) Suppose the directions of the E and B fields are e and b. Then (1.6.14) says En(x,ω) = en(x,ω1) ejφ(x,ω) 2πδ(ω-ω1) Bn(x,ω) = bn(x,ω1) ejφ(x,ω) 2πδ(ω-ω1) (1.6.15) and one finds that = ej[φ(x,ω)- φ(x,ω)] . (1.6.16) Since e and b are real, the phase of the ratio En/ Bn is determined by the last factor and will in general be a function of both position x and frequency ω1. Again, this type of result will appear in the round wire analysis of the next section. (e) A Pitfall to Avoid Notice that E(x,t) = e(x,t) + j e'(x,t) => E(x,ω) = !Syntax Error, Idt E(x,t) e-jωt = !Syntax Error, Idt [e(x,t) + j e'(x,t)] e-jωt = e(x,ω) + j e'(x,ω) . (1.6.17) Whereas e(x,t) and e'(x,t) are the real and imaginary parts of E(x,t), the functions e(x,ω) and e'(x,ω) are not the real and imaginary parts of E(x,ω) since in general e(x,ω) and e'(x,ω) are both complex functions. In this document we shall never deal with transforms of the type e(x,ω) or e'(x,ω). (f) Maxwell's Equations in ω space In the Maxwell and related equations which include the ∂t operator, if the fields are expanded onto their Fourier transformed components using (1.6.8), then using the rule (1.6.9) one may instantly write the frequency-domain version of these equations, just as in the example of Section 1.5. For example, curl H(x,ω) = jωD(x,ω) + J(x,ω) (1.6.18) curl E(x,ω) = -jωB(x,ω) (1.6.19) div J(x,ω) = -jω ρ(x,ω) . (1.6.20) Other equations in the Section 1.1 list have the same form but in terms of the frequency-domain functions. For example, J(x,ω) = σ(x) E(x,ω) (1.6.21) where we momentarily allow σ(x) to have spatial dependence but not time dependence. Chapter 2: The Round Wire and the 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 properties of a very simple object -- 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 more general 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 (impedance) of a wire increases drastically at high frequency. In the context of a transmission line, this effect is "felt" through a property of a wire called its surface impedance, which is studied below in Section 2.3 and qualitatively in Section 2.4. 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 damped wave equation (1.5.32) which, inside the wire where there is no free charge, says (2 + β2)E = 0 with β2 = ω2μξ where μ and ξ apply to the conductor. Instead, we have chosen to start from the basic Maxwell curl equations and use simple "loops" to derive the basic (first order differential) equations relating E and B fields. The general technique of putting loops in opportune places is extremely useful in analyzing the more complicated situation which arises in a transmission line. This method is carried out in Section 2.1 and the wire's interior solutions are then studied in Section 2.2. In the final section 2.5, we go ahead and do the full "math solution" of the round wire to obtain the E and B fields both inside and outside the wire. The Implicit Wave Context and the Skin Effect In the sections below we don't explicitly consider the notion that a wave is traveling down our round wire, but that is in fact what is happening and this fact deserves a few comments before we delve into the interior solution of the wire. Our stated assumption below is only that we assume the z dependence of E and B fields is weak enough that we can just ignore it. For example, if the z dependence were some e-jkz factor, then any ∂z derivative is proportional to k, and if we just assume k is very small, we throw out such derivatives. In effect then a field E(x,y,z) is replaced by E(x,y) and 2 by 22D. To put this in more context, we can imagine that E really does have a wave form E(x,y,z,t) = ej(ωt-kz) E(x,y) so that the frequency domain field is E(x,y,z,ω) = e-jkz E(x,y,ω) . The E field has to satisfy two damped wave equations, one inside and one outside the wire (2 = 23D), as was shown in (1.5.27) , ( 2 + β2 ) E(x,y,z,ω) = 0 inside wire β = (j - 1) ( 2 + βd2 ) E(x,y,z,ω) = 0 outside wire βd = ω where we use subscript d to indicate dielectric properties. The two β expressions come from the general form (1.5.1) and the first will be explained more below so we accept it for now. If we insert our wave form for E(x,y,z,ω) these two equations become ( 22D + β2 - k2) E(x,y,ω) = 0 inside wire β = (j - 1) = complex ( 22D + βd2 - k2) E(x,y,ω) = 0 outside wire βd = ω = real In the second equation, we then make the ansatz assumption that k = βd which basically says that our wave form ej(ωt-kz) E(x,y) really does describe a wave traveling down the wire with k = βd. This k is then related to the speed of light in the dielectric and is the expected value of k for, say, a radio or light wave travelling through the dielectric with no wire present. In the first equation, since the conductor has such a large σ, |β| is a huge number and |β| >> k (unless ω is very small), so k really plays no role in the first equation for such ω. We then have ( 22D + β2) E(x,y,ω) ≈ 0 inside wire β = (j - 1) = complex 22D E(x,y,ω) = 0 outside wire The second equation says that the E field outside the wire must solve the 2D (vector) Laplace equation. We shall hear more about this in later chapters. It is the first equation for the wire interior that we study below in this section, and hopefully we have now put that equation into the context of a wave travelling down the wire. Although the first equation seems to allow k to be a free small parameter, when we examine where the two solutions meet at the boundary r = a, we conclude that k in the first equation must be βd. If βd has a small imaginary part due to conductivity of the dielectric (see (1.5.1)), the factor e-jβz says that the wave slowly damps out as it travels down the wire due to dielectric ohmic loss, as it well should. On the other hand, β is huge and has equal real and imaginary parts. Due to our axial symmetry, the first equation really says (22D + β2) E(r,ω) = 0 which can be through of as a "wave equation" in the radial direction. Of course it is a damped wave equation of a very extreme sort. As we move in from the surface of the wire toward the center, we claim that over a distance in which the "wave" phase changes by about π/2, the amplitude is already down by a factor 1/e, so one can roughly say that the wave basically damps out before the wave even goes 1/2 wavelength. This is the skin effect described below. To understand this effect, it is useful to consider a 1D version of the situation. Imagine zooming the camera in very close to the left surface of the round wire, so that we see a half space of conductor on the right and a half space of dielectric on the left. Let the radial direction be called x which increases into the conductor with x = 0 at the interface. Then the inside-wire wave equation above says (∂x2 + β2) E(x,ω) = 0 . The solution to this equation is (we select a particular sign for the phase) E(x,ω) = E(0,ω) e+jβx = E(0,ω)exp{ j [(j - 1) ]x} = E(0,ω) exp{- x} exp{ -j x} = E(0,ω) exp{- x/δ} exp{ -j x/δ} δ ≡ = E(0,ω) e-x/δ e-jx/δ . Thus as we move from x=0 to the right, in distance δ the E field amplitude drops to 1/e and the phase has changed by π/2. Quantity δ is called the skin depth, and this is probably the most basic way to understand the notion of the skin effect. It is a result forced by the Helmholtz equation having a complex parameter β of the type shown. 2.1 Derivation of E(r), B(r) and J(r) for a round wire We convert (1.1.38) and (1.1.36) to the ω domain using rules (1.5.2), curl B = μ (jωε E + J ) C B ds = μ∫S [jωεE + J ] dS (2.1.1) curl E = jωB C E ds = -jω∫S B dS . (2.1.2) The two terms on the right side of (2.1.1) have names: jωεE = displacement current (density) // amps/m2 J = σE = conduction current (density) // amps/m2 . The sum of both currents may be written as ( jωε + σ) E(x,ω) . (2.1.3) For any metal conductor such as copper, the displacement term is completely negligible as long as ωε << σ. The value of ε for a metal is not very obvious and is likely in fact to be negative at frequencies below optical frequencies (free electron gas, plasma frequency, Drude model, etc), so we will follow Mattick p 118 and blindly set ε = ε0 for a crude comparison. The condition for negligiable displacement current ωε << σ then becomes f << σ/[2πε0]. Using σ = 5.81 x 107 mho/m, ε0 = 8.85 x 10-12 F/m, one gets f << (σ/2πε ) = 1.04 x 1018 Hz ≈ one billion GHz Therefore, the displacement current is always ignored inside a conductor for any conventional transmission line application. Whatever ε really is, we shall ignore jωε compared to σ. All the current inside a conductor is conduction current. At this point we make a set of assumptions: (a) the round wire is perfectly symmetric and uniform (2.1.4) (b) the current pattern in the wire is axially symmetric (no dependence on azimuth θ; invariant under any rotation of the wire about its center line) (c) the curent is axial (longitudinal), so J = J , so J = Jz (d) the E field is also axial so E = E ( this follows from (c) and J = σE ), so E = Ez (e) The B field lines go 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. If J = Jz > 0, then B = Bθ > 0. (f) The characteristic distance of the variation in the fields E and B in the z direction is much larger than in the r direction, so we treat the fields as constants in z. In practice this means that we assume the wire transverse dimensions (radius a) are much smaller than the wavelength of any wave passing down the wire. Thus, we represent Jz(x,ω) = J(r), Ez(x,ω) = E(r), and Bθ(x,ω) = B(r) -- no dependence on θ or z. The fields E(x,ω), B(x,ω) and J(x,ω) are complex, so E(r), B(r) and J(r) are all complex. They all depend on ω, but we suppress the ω arguments. As noted above, E = E(r) and B = B(r) . Here is a another way to state assumption (b). We search for an axially 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. If the wire were in idealized perfect isolation with an axially symmetric source and load, the invariance of the physical situation with regard to rotation about the wire axis would require (b) to be valid. In Appendix D we attack the much more formidable problem of finding the general solution for the electric field and current in a round wire subject to transmission line boundary conditions. Such solutions allow the possibility of azimuthal variation in the fields. Consider now the thin (width is dr) red loop shown in Fig 2.1: Fig 2.1. End view of wire, current flowing in direction toward viewer According to (2.1.1) with J = σE and no displacement current, C B ds = (μσ) ∫S E dS . (2.1.5) For the CCW loop shown, the "right hand rule" says area dS points out of the plane of paper. The two sides of this equation can be easily evaluated (B = Bθ and E = Ez) [ B(r+dr) (r+dr) - B(r) r] θ = (μσ) E(r) [ rθ dr ] (2.1.6) which simplifies to = (μσ) [r E(r) ] . (2.1.7) Comment: When one says in Fig 2.1 that "J points in the direction", one intepretation might be that the vector J has the form J = Jz and that Jz > 0. That is not the correct interpretation for our pictures. The quoted phrase just means that J = Jz and nothing is implied about the "sign" of Jz. In our case, Jz = J(r) is a complex number which has no "sign". If we said "J points in the - direction" we would just mean that J = Jz(-) = -Jz. Our only interest in clarifying these "directions" is to get the signs right in our application of Stokes's law. The same comment applies to the direction of B in the next figure. By saying that "B points out of the plane of paper", we just mean that B = +B(r) which is consistent with the the fact that J = +Jz according to the right hand rule: thumb in the "direction" of J at the wire axis, curled fingers are in the "direction" of B. Now consider the thin red loop shown in Fig. 2.2, Fig 2.2. Top view of wire's central plane, current flowing in direction (down) According to (2.1.2), C E ds = -jω∫S B dS . (2.1.2) where, for the CCW loop shown, the right hand rule puts dS pointing to the viewer (aligned with B which points to the viewer due to its right hand rule with J ). The two sides of this equation can be easily evaluated (the first term on the left is negative because points down while the red arrow points up) [ - E(r+dr) + E(r)] s = -jωB(r) [ s dr ] (2.1.8) which simplifies to = jωB(r) . (2.1.9) This equation says that E(r) changes with radius as long as ω ≠ 0 and B(r) ≠ 0. Since everything is complex, we cannot really tell from (2.1.9) that |E(r)| increases with radius, but we shall see below that it does, and this fact gives rise to the "skin effect" where current is maximum at the wire surface. Reader exercise: Why can't we take the absolute value of both sides of (2.1.9) and reach the conclusion that ∂r|E(r)| = ω|B(r)| > 0 and conclude that |E(r) increases with r? Answer: ∂x|f| ≠ |∂xf| Now solve (2.1.9) for B(r) and put this into (2.1.7) to get = (jωμσE(r) . (2.1.10) The operator on the left is 2 in cylindrical coordinates for a function that does not depend on θ or z. Thus, (2.1.10) is really a special case of the following [ 2 - (jωμσ)] E(r) = 0 . (2.1.11) This in turn is a special case of the field wave equation (1.5.32) with ρ = 0, which we can write as (2 + β2)E(x,ω) = 0 . // homogeneous vector Helmholtz equation (2.1.12) The Helmholtz parameter β2 is given by (1.5.1) , β2 = μεω2 - jωμσ = ω2μ ( ε - jσ/ω) ≈ -jωμσ . The ε term in β2 has been neglected since σ is very large. Parameter β = 2π/λ is a "wavenumber" and has dimensions of m-1. The complex number -j has two square roots which are ej3π/4 and e-jπ/4, Fig 2.3 and we specify the upper red arrow as the square root in the definition of β, β ≡ ej3π/4 . // β = (j - 1) (2.1.13) We could have started out with (2.1.12) and skipped all the above analysis of loops, but this method of using loops is useful for analyzing more complex situations, as noted earlier. Digression on cause of the skin effect. Before continuing, we can show generically how it is that the Helmholtz equation causes skin effect by considering (2.1.12) in one dimension, say x. Then the equation reads (∂x2+ β2)Ex = 0 which has the solutions exp(±jβx). Since β = (j - 1) these solutions are exp(±j(j - 1) x) = exp(∓x) exp(∓j x) . If we are interested in x ≥ 0 with Ex = 0 at x = +∞, then we must take the - solution so | Ex(x) | = | Ex(0) | exp(-x) = | Ex(0)| exp(-x/δ) δ = . Since Jx = σEx, this means | Jx(x) | = | Jx(0)| exp(-x/δ) so the current density decays as e-x/δ as we move away from x = 0. Thus it is really a combination of the Helmholtz equation, the complex nature of parameter β2, and the boundary condition that causes exponential decay of Ex(x) with characteristic distance δ which will appear below as the "skin depth". In the round wire, the reference point will be r = a, and the boundary condition will be that E be finite at r = 0. Continuing now with the round wire analysis, the next step is to expand (2.1.10) as follows: + + β2 E(r) = 0 . (2.1.14) Change variables to dimensionless x = βr. Then r = x/β x = βr ∂x/∂r = β ∂E/∂r = ∂E/∂x ∂x/∂r = β ∂E/∂x = β ∂xE ∂2E/∂r2 = (∂r [∂E/∂r] ) = ∂x [∂E/∂r] (∂x/∂r) = ∂x [β ∂E/∂x] β = β2 ∂x2E . (2.1.15) Inserting these quantities into (2.1.14) and multiplying through by x2/β2 then gives x2 + x + x2f(x) = 0 (2.1.16) where f(x) = E(x/β). Now (2.1.16) happens to be Bessel's Equation with ν = 0 [NIST 10.2.1], and the solution must therefore be a linear combination of this form, where C and D are constants, f(x) = C J0(x) + DY0(x) . (2.1.17) So far, we still don't 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) β = ej3π/4 (2.1.18) The following definition is usually made (factor of 2 explained later) δ ≡ = skin depth // ωμσ = 2/δ2 (2.1.20) so that β = ej3π/4 (/δ) and β2 = -2j/δ2 . (2.1.21) It is convenient to divide (2.1.18) by itself evaluated at r=a which we shall assume is the radius of our round wire, so (plots coming soon), E(r) = E(a) [J0(βr) / J0(βa)] . (2.1.22) From (2.1.9) and (2.1.22) we can easily find the B field, B(r) = (β/jω) E(a) [J0'(βr) / J0(βa)] (2.1.23) since ∂rJ0(βr) = (∂x/∂r) (∂J0(x)/∂x) = β J0'(x). An alternate form results from dividing (2.1.23) by itself evaluated at the wire surface, so B(r) = B(a) [J0'(βr) / J0'(βa)] . (2.1.24) It happens that J0'(x) = -J1(x) [ NIST 10.6.2 ] , so the above results can be restated as B(r) = - (β/jω) E(a) J1(βr) / J0(βa) = + B(a) J1(βr) / J1(βa) . (2.1.25) Let us gather up all the main results obtained so far and put them in a box: Interior Solution of a Round Wire (2.1.26) = (μσ[r E(r) ] (2.1.7) = jωB(r) (2.1.9) E(r) = E(a) [J0(βr) / J0(βa)] (2.1.22) J(r) = J(a) [J0(βr) / J0(βa)] // above times σ, J = Jz = current density B(r) = - (β/jω) E(a) [J1(βr) / J0(βa)] = + B(a) [J1(βr) / J1(βa)] (2.1.25) β = ej3π/4 (/δ) = ej3π/4 and β2 = -2j/δ2 (2.1.21), (2.1.14) δ ≡ = skin depth (2.1.20) The reader is reminded once again that E(r), B(r) and J(r) are complex functions of r and ω since they are components of the Fourier Integral Transform of the time-domain fields and current density. Since β is complex, the various Jν(βr) are also complex. Thus, the nature of the solutions in the above box is not very obvious at this point. 2.2 A study of the solution of a round wire (a) Kelvin Functions The reader may be aware of the so-called first-kind modified Bessel function defined by Iν(x) ≡ e-jπν/2 Jν(ejπ/2x) , where the Jν function argument has phase π/2. Unfortunately, our Jν(βr) functions have phase 3π/4 so the Iν functions are not particularly useful. 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 [ NIST 10.61.1 ], Jν(ej3π/4z) = berν(z) + j beiν(z) . (2.2.1) In our application ej3π/4z = βr = ej3π/4(/δ) r so that z = (r/δ). Thus, the solution E(r) in (2.1.22) may be written as, E(r) = E(a) . z = (r/δ) (2.2.2) The Kelvin functions are real when the arguments are real and positive, as is the case with almost all special functions (real analytic property). Similar functions kerν and keiν are associated with Kν(ej3π/4z ) where Kν is the second-kind modified Bessel function. Since J(r) = σ E(r), we could replace E with J on both sides of (2.2.2). This equation for J appears in Matick as p 101 (4-18). Note: Lord Kelvin (William Thomson) introduced the ber and bei notation for these functions while considering the same problem we are dealing with here. The functions appear in the Appendix of his 34 page 1889 inaugural address used when he became president of the Institute of Electrical Engineers (see Refs) : We can verify using Maple (which Kelvin would have enjoyed) that these are the ber0 and bei0 functions: Some authors, not liking Kelvin's notation, use Ber, Bei, Ker, Kei for ber, bei, ker, kei. Perhaps the idea is that Be is more obviously Bessel and perhaps Ke is then for Kelvin. Since these Bessel forms occur a lot, there are standard functions for their magnitude and phase [ NIST 10.68.1 ] Jν(ej3π/4z) = Mν(z) ejθ(z) . (2.2.3) Of particular interest is the magnitude of E(r). Applying (2.2.3) to the E(r) in (2.1.22) gives |E(r)| = |E(a)| . (2.2.4) (b) Plots of |E(r)/E(a)| for various δ values Finally we are in a position to make some plots to see how the electric field magnitude varies with radius in a round wire as a function of the skin depth parameter δ ≡ . As ω increases, δ decreases. Our aging Maple V knows about the Kelvin functions but not M, so here is the code and here are plots of |E(r)| / |E(a)| for a = 20 and δ = 1 to 10, The steepest curve is for δ = 1: Fig 2.4 . Plot of |E(r)/E(a)| or |J(r)/J(a)| versus r for a round wire with skin depth taking values δ = 1 (rightmost curve) to δ = 10 (top curve). One sees clearly how the current and electric field magnitude drop off quickly moving in from the edge of the round wire (right edge of graph) toward the wire axis when δ is small relative to radius a. Asymptotic expansions for Mn(z) and θn(z) for large z are given by NIST 10.68.16 and 10.68.18, Mν(z) ≈ [ 1 - + O(1/z2) ] θν(z) ≈ (z/) + (π/2) [ ν - 1/4 ] + + O(1/z2) . (2.2.5) For ν = 0 we find M0(z) ≈ [ 1 + ] θ0(z) ≈ (z/) - (π/8) – . (2.2.6) For z > 3 the correction term in M0(z) is less than .03 so we may ignore it for rough estimates. In this case one gets |E(r)| = |E(a)| = |E(a)| z = (r/δ) za = (a/δ) ≈ |E(a)| = |E(a)| exp([z-za]/) . But [z-za]/ = (r/δ)-(a/δ) = (r-a)/δ and = . Thus we find that = 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 ratio is 1 at the surface and then drops off exponentially with characteristic distance δ as we move inside the wire. One sees now why the was included in the definition of δ: there is then no in equation (2.2.7). Equation (2.2.7) is valid down to within about 2 skin depths of the center axis of the wire. In general, one can assume the field E(r) is zero for all practical purposes perhaps 5 skin depths in from the surface (if a > 5δ). Here are plots of |E(r)|/|E(a)| using the approximate formula (2.2.7) for the same ten δ values as our previous plots, Fig 2.5 . Previous plot using a certain approximation discussed above. and here are the two sets of plots superimposed with some notations added: Fig 2.6 . Superposition of the previous two sets of plots. The wire radius is a = 20, and the curves are for δ = 1 to 10, with δ = 1 being the rightmost and steepest curve. The red (exact) and black (approximate) curves for δ = 1 agree down to r = 2 at least. The δ = 5 red/black pair of curves start to pull apart around r = 10 which is 2 skin depths from the center. The δ = 8 red/black pair of curves start to pull apart around r = 16. We thus verify the claim made above that each red/black pair of curves agree starting at r = a and moving in to about 2 skin depths from the center line (the pull-apart points are marked by dots). One can also see that the electric field is roughly zero about 5 skin depths in from the surface (marked by x's). Here are some skin depth values in copper based on (2.1.20) δ with σ = 5.81 x 107mho/m, and μ = μ0 = 4π x 10-7 H/m. Selecting a reference point of 1 GHz, we have, δ = = = = 2.09 x μ (2.2.8) Here is a little table then of copper skin depths (μ = microns), 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 The radius of the center conductor of Belden 8281 coax is 15.5 mil = 394 μ, so the skin effect restriction occurs for f ≈ 1 MHz and above. At 1 GHz δ is about 1/200th the radius. 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 // Spiegel 24.5 z = (r/δ) . (2.2.9) Using the expression for E(r) and β2 in box (2.1.26) we find = = . (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. For example, with a = 20 and δ = 100 we have z ≤ (a/δ) = (1/5) = .28 for all values of r, so z is "small" in the whole range, and here is the plot. Notice the offset zero so the drop is only 2 parts in 10,000. Fig 2.7 Slight dip in E(r) or J(r) moving from surface to center for a round wire in the low frequency limit. In this case radius a = 20 and skin depth δ = 200. (c) Review of the round wire solution To conclude this section, we state in full notation the solution of the round wire as outlined above, using ω rather than δ as the argument of interest, where recall δ = so z = (r/δ) = r . The following two expressions are (2.1.22) and (2.1.25) with (β/jω) = ejπ/4 / ω as in (2.1.13): E(r,ω) = E(a,ω) E(x,ω) = E(r,ω) (2.2.11) B(r,ω) = - E(a,ω) ejπ/4 B(x,ω) = B(r,ω) (2.2.12) The ratio is then = - ejπ/4 . (2.2.13) If the round wire is driven by an external monochrome (ω1) electric field at r = a (same for all z) , E(a,t) = E0ejωt E(a,t) = E(a,ω) then E(a,ω) = !Syntax Error, Idt e-iωt E(a,t) = E0 2πδ(ω-ω1) . Maybe restate this in terms of having a driving current I which then drives B(a) instead. In this case, it is easy to transform (2.2.11) and (2.2.12) to the time domain using (1.6.8). We do this and then replace ω1 by ω to reduce clutter, E(x,t) = E0 ejωt (2.2.14) B(x,t) = - E0 ejωt ejπ/4 (2.2.15) and of course J(x,t) = σ E(x,t). These field solutions have the promised form (1.6.13) E(x,t) = ej[ωt+φ(r,ω)] e(r,ω) B(x,t) = ej[ωt+φ(r,ω)] b(r,ω) (1.6.13) where e and b are real and ejφ(r,ω) e(r,ω) = E0 = E0 (2.2.16) ejφ(r,ω) b(r,ω) = - E0 ejπ/4 (2.2.17) (d) Plots of the round wire solution for Belden 8281 at 5 MHz. Here is some Maple code to generate various plots for E0 = 1 volt/m, μ = μ0 = 4π x 10-7, σcopper = 5.81*107, ω = 2π [ 5 MHz ], and a = 394 μ -- all as appropriate for the center conductor of Belden 8281 coaxial cable. Notice that the factor = = 10-3 = 3.4 x 10-3 causes B to be small even at the surface r = a. We first set in the parameters just quoted, and then do the plots as follows: e = Magnitude of E φe = Phase of E b = Magnitude of B φb = Phase of B b/e = Magnitude of B/E φb-φe = Phase of B/E The nature of these plots for moderate to large z can be obtained from the large z limit of the Jν functions as noted earlier, Jν(ej3π/4z) = Mν(z) ejθ(z) . z = r (2.2.3) Mν(z) ≈ θν(z) ≈ (z/) + (π/2) [ ν - 1/4 ] (2.2.5) Example: For the electric field in (2.2.16) we have this large z limit, E0 = E0 ≈ E0 ej(r-a) ≈ E0 exp[- (a-r) ] e-j(a-r) which shows both the exponential decay in magnitude and the phase linear in r. φe(r,ω) ≈ -(a-r) . This phase starts at 0 when r = a and then goes negative with slope + , as the graph shows. 2.3 The Surface Impedance Zs(ω) of a Round Wire A piece of round wire can be thought of as a resistor. Consider Fig. 2.8: Fig 2.8: Round wire as a resistor Here a piece of finite-σ wire is attached to a pair of σ = ∞ contacts. The total impedance of the wire is then determined by Z = V/I ohms where V is the voltage applied to the contacts and I is the total current through the wire. Alternatively, one could probe the wire along its surface as shown by the two arrows separated by dz. There is some voltage dV between the probes due to the field Ez(a) ≡ E(a) at the surface of the wire. By definition, the surface impedance per unit length is Zs ≡ (- dV/dz)/I = E(a) / I ohms/m . (2.3.1) Since the fields and currents derived under the assumptions (2.1.4) vary only with r, Zs is independent of z and we get V = V(0) - V(L) = - !Syntax Error, IdV = - !Syntax Error, I(dV/dz) dz = !Syntax Error, I I Zs dz = I Zs L = I Z (2.3.2) so Z = Zs L . (2.3.3) As one might imagine, Zs plays a role in transmission line attenuation. (a) Expressions for Surface Impedance 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 just below the wire surface as shown in red in Fig 2.8. We apply (2.1.1) on this loop (with ε = 0) to get: 2πaB(a) = μI . (2.3.4) Thus, from our Zs definition (2.3.1) and from (1.1.5) that B = μH, Zs = E(a)/I = E(a) μ/[2πaB(a)] = (μ/2πa) E(a)/B(a) . (2.3.5) Looking at the B(r) equation in the box (2.1.26), we may write B(a) = - (β/jω) E(a) [J1(βa) / J0(βa)] => E(a)/B(a) = -(jω/β) [ J0(βa) / J1(βa) ] . (2.3.6) This ratio can then be inserted into (2.3.5) to obtain the surface impedance, Zs = Zs(ω) = - (μ/2πa) (jω/β) [ J0(βa) / J1(βa) ] or Zs(ω) = (2.3.7a) where β = (/δ) ej3π/4 and δ ≡ as in box (2.1.26). Using these last two facts and the fact that ej3π/4 is a square root of -j, the leading factor may be written = = so we have this alternate form for (2.3.7a) in which ω does not explicitly appear, Zs(ω) = β = (/δ) ej3π/4 . (2.3.7b) Below we shall use form (2.3.7b) to plot Zs(ω) as a function of skin depth δ. Equation (2.3.7) is, as expected, rather complex. In terms of the Kelvin functions defined in (2.2.1) we may write (2.3.7a) as Zs(ω) = . (2.3.8) According to (2.2.1) we may write, with α ≡ ej3π/4, berν'(z) + j beiν'(z) = = = αJν'(αz) = ej3π/4 Jν'(ej3π/4z) . (2.3.9) Then since J0'(x) = -J1(x) we find that ber0'(z) + j bei0'(z) = ej3π/4J0'(ej3π/4z) = - ej3π/4 J1(ej3π/4z) = - ej3π/4 [ber1(z) + j bei1(z)] . (2.3.10) Then (2.3.8) may be rewritten as Zs(ω) = (2.3.11) and this form for Zs(ω) appears in Matick p 104 (4-28). (b) Low frequency limit of Zs(ω) Small ω => large δ => small β, so we expand both Bessel functions of (2.3.7) for small argument: [ Spiegel 24.5 and 24.6 ] J0(x) ≈ 1 - x2/4 J1(x) ≈ (x/2)(1 - x2/8) 1/J1(x) ≈ (2/x) (1 + x2/8) => J0(x)/J1(x) ≈ (2/x) (1 + x2/8) (1 - x2/4) ≈ (2/x)(1-x2/8) = 2/x - x/4 => J0(βa)/J1(βa) ≈ 2/(βa) - (βa)/4 . Then from (2.3.7a) Zs(ω) = ≈ [2/(βa) - (βa)/4 ] = + = + // β2 from (2.1.13) or Zs(ω) = + jω = Rs + jωLs // low frequency limit (2.3.12) The first term is the uniform DC resistance of the wire per unit length, normally written ρ/A, see (C.2.3) of Appendix 2.1. The second term is jω times the DC internal inductance Li = (μ/8π) H/m, as derived in 3 (5) of Appendix 2.1. Recall that this is exactly 50 nH/m if μ=μ0, quite small, and independent of radius. (c) High frequency limit of Zs(ω) We first use (2.2.3) to write (2.3.7) as Zs(ω) = (-jωμ/2πaβ) [ M0(a/δ) / M1(a/δ) ] exp[ j{θ0(a/δ) - θ1(a/δ)}] . (2.3.13) Since large ω small δ large arguments for the functions in (2.3.13), we use these large z limits which can easily be obtained from (2.2.5) using ν = 0 and 1, M0(z) / M1(z) = [ 1 + + O(1/z2) ] θ0(z) - θ1(z) = - [ (π/2) + + O(1/z2) ] . (2.3.14) Insertion of these large-argument formulas into (2.3.13) with z = δ/a gives Zs(ω) = (-jωμ/2πaβ) (1 + ) exp(-j [π/2 + ]) = (-jωμ/2πaβ) [ 1 + δ/(16a) ] exp(-j [π/2 + δ/(16a) ]) = [ 1 + δ/(16a) ] e-jπ/2 e-jδ/(16a) . // -j = e-jπ/2 The phasor factors combine to give e-jπ/2 e-j3π/4 e-jπ/2 = e-jπ[1+3/4] = e-jπ[2-1/4] = e-jπ2 ejπ/4 = ejπ/4 = (1+j)/ and then Zs(ω) = (1+j) [ 1 + δ/(16a) ] e-jδ/(16a) . (2.3.15) Then if δ << 16a the last two factors are unity and we have Zs(ω) ≈ (1+j) = (1+j) = (1+j) // using δ2 = 2/ωμσ from (2.1.20) ≈ (1+j) δ << 16a . (2.3.16) Writing this as the sum of a resistive and inductive part, Zs(ω) = Rs(ω) + jω Ls(ω) (2.3.17) we find Rs(ω) = = ω Ls(ω) = XLs(ω) . (2.3.18) The inductance can be written several ways, Ls(ω) = = = . (2.3.19) 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 can be 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 the phase of this ratio as seen in (2.3.16) is π/4 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/ as shown. Quantity Rs(ω) in (2.3.18) is called Rhf by Matick in (4-35), and (2.3.16) appears as (4-36). (d) Plots of Zs(ω) versus skin depth δ From (2.3.7b) we found that Zs(ω) = β = (/δ) ej3π/4 δ ≡ . (2.3.7b) This is in SI units, but we will use a = [a(μ)10-6] m and δ = [δ(μ) 10-6] m and σ = 5.81 x 107mho/m for copper, where a(μ) and δ(μ) means the wire radius and skin depth in microns. Then: Zs(ω) = ohms/m N = 1012 The two limits obtained above were : Zs(ω) ≈ + jω = + j small ω, large δ Real part goes to a constant, imaginary part decays as 1/δ2 Zs(ω) ≈ (1+j) = + j large ω, small δ Real and Imaginary part are the same and blow up as 1/δ In our units above the low frequency constant limit is RLF ≡ = . Here is some Maple code which plots the real (red) and imaginary (black) part of ln Zs(ω) as a function of δ, and also computes the constant limit (gray) just mentioned. The copper wire radius is set to a=1000 μ . Fig 2.9: Plot of surface impedance ln Zs as function of skin depth δ ≈ 40 to 1000 μ for a copper wire of radius 1000 μ. Red is real part, black is imaginary. Here is the same plot for δ = 100 to 1000 without the natural logs (ln = "log" in Maple) : Fig 2.10: Plot of surface impedance Zs as function of skin depth δ = 100 to 1000 μ for a copper wire of radius 1000 μ. Red is real part, black is imaginary. Either plot type realizes the two limits discussed above. For the limited range δ = 1 to 10 μ the above plot has this appearance ( the red and black curves are superposed and the gray constant line at .0055 is indistinguishable from the x axis) : Fig 2.11: Plot of surface impedance Zs as function of skin depth δ = 1 to 10 μ for a copper wire of radius 1000 μ. Red is real part, black is imaginary. 2.4 Surface Impedance for a Transmission Line What is the surface impedance of an arbitrary conductor? As we have seen, a significant amount of work was 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 (with comparison) Zs(ω) ≈ // general case (2.4.1) Zs(ω) ≈ (1+j) δ << 16a . // round wire (2.3.16) 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 2.12 : stripline looking down the channel Figure 2.13: 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 (b >> a), both conductors are immersed in roughly uniform active fields, so the full D = 2πa is applicable. As the two round wires are brought very close together, certainly there will develop an asymmetry so that the currents are largest on the parts of the wires closest to the other wire. 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 Eq (45)] quotes an approximate surface impedance result for the case of Figure 2.14. The effective distance is equal to, D = 2π a . a = wire radius, b = center line separation (2.4.2) If the gap between the conductors is a/6, a rough estimate for Fig 2.14, 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 (p 30) 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) . [ Need to ponder this! ] 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. Matick's Chapter 4 computes the surface impedance for the round wire and for some strip line conductors. Chapter 3: Transmission Line 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/ε. 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 (dimension x) with some charge density ρ inside. Then from div E = ρ/ε0 we get ∂xEx = ρ/ε0 . (3.1.1) The electric field so generated causes a current Jx = σ Ex so apply ∂x to get ∂xJx = σ ∂x Ex . (3.1.2) But this current drains the charge away according to div J = -∂tρ, which says ∂xJx = -∂tρ . (3.1.3) Combining these three equations we get ∂tρ = - (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 E from which we now quote. 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. 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, as noted in (2.1.1), Jtot = (jωε 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) If one plots ε in the complex plane, tanL (called the loss tangent) is the tangent of the small angle θL of the triangle whose perpendicular sides have length ε' and ε" where ε" is normally very small. That is to say, the loses tangent is (minus) 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 E + σE = jωε' [1 - j tanL]E + σE = jωε' E + ( σωε' tanL) E = jωε' E + σeff E . (3.3.3) In effect, the dielectric has now acquired an effective conductivity, σeff = ( σωε' 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 use these numbers, σ ≈ 10-15 mho/m ε' = 2.3 tanL ≈ 2 x 10-4 (3.3.5) taken from http://www.sdplastics.com/polyeth.html San Diego Plastics, Inc. Fig 3.1 Note: This table claims ρ = 1015 ohm-cm = 1013 ohm-m, but most other sources give larger values. We assume ρ ~ 1017 ohm-cm = 1015 ohm-m and therefore σ ~ 10-15 mho/m. It does not matter much! Even at 1 Hz, the loss tangent contribution dominates in (3.3.4). Using the above figure for tanL, here are a few values of σeff versus frequency: (f = .0 is really 100 = 1 Hz) // = 2.6 x 10-3 3.4 Size of E field inside a conductor and conservation of total current at a boundary We know from (1.1.28) that the following E field condition applies at a boundary between two media, where n refers to the normal component, ξ1En1 = ξ2En2 // frequency domain (1.1.28) or (ε1 + σ1/jω) En1 = (ε2 + σ2/jω) En2 . (3.4.1) If 1 = dielectric with very small σ1 and 2 = conductor with very large σ2, we have ε1 En1 ≈ (σ2/jω) En2 => ratio = ≈ 3.4.2) Again, we can look at some typical numbers. σ2 = 5.81 x 107 mho/m (copper) ε1 = 2.3 ε0 (polyethylene) (3.4.3) ε0 = 8.85 x 10-12 farad/m At a high frequency of ~ 500GHz the ratio in (3.4.2) is ~ 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 a good dielectric and a good 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. Fact 3: This large jump in En at the boundary must be supported by a significant surface charge density n on the boundary since, according to (1.1.27), n = ε1En1 - ε2En2 ≈ ε1En1. Imagine now a tiny patch of area (bordered in red) on the surface between a conductor and a dielectric, Fig 3.2 Defining a total current Jtot,n ≡ jωεEn + σEn , we have shown that this total current flows right through the area patch but changes its nature from mostly conduction current on one side to mostly displacement current on the other side. In the next section, we identify the normal direction with the local radial direction. Then the total current passing through a tiny square patch like that in Fig 3.2 can be regarded as being "fed" by the radial current Jr just inside the conductor where Jr = σEr . 3.5 The TEM mode fields and currents for an IDEAL transmission line In this and the next section, we take a crude qualitative look and the various E,B and J components first for an ideal transmission line, then for a real one. An example is repeatedly used in which the conductor of interest is the round center conductor (radius a = 1 mm) of a properly terminated 75 Ω coaxial cable driven by 7.5 volts, and thus having a current of 100 mA. The two tables obtained (one ideal, one real) mainly serve as an exericise in applying the various concepts reviewed in previous sections. By "ideal" we mean that the conductors have near infinite conductivity and the dielectric has zero conductivity. Consider a cross sectional view of one conductor of a transmission line having arbitrarily shaped conductors (the shape is uniform in the z direction). At some point on the surface, define a 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) Fig 3.3 The following table shows the qualitative sizes of various components of E,B and J (conduction current) near the surface of a transmission line conductor. Several regions of space are of interest: 1. In the conductor, under the surface charge layer and under any current layer. 2. In the conductor, just under the surface charge layer, and in the skin current layer. 3. In the dielectric, just outside the super-thin surface charge layer. The reader is warned that the rest of this section and Section 3.5 make very tedious reading because an argument must be made for the general size of every single item in the two large Tables. The reader might consider just perusing the two Tables and then skipping to Section 3.6. First is the Table for the ideal transmission line conductor (comments follow): Table 1: E,B,J for an ideal transmission line Region 1. 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 2. In the conductor, just under the surface charge layer, and in the current layer. Er = small Br = 0 Jr = small Eφ = 0 Bφ = large Jφ = 0 Ez = small Bz = small Jz = very large Region 3. In the dielectric, just outside the super-thin surface charge layer (explanations below): Er = large Br = 0 Jr = 0 Eφ = 0 Bφ = large Jφ = 0 Ez = small Bz = small Jz = 0 Region 1: (the interior) In of interior we know that E must satisfy the Helmholtz equations (2.1.12). Due to the powerful exponential effect of this equation (see digression below (2.1.13) and (2.27) for the round wire), we know that E 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 = extremely small since δ =. Thus, conductor E and B fields can only exist very close to the surface. In region 1 of the above table, we show all fields as being 0 underneath the very thin current sheath. Since E = 0 in the perfect conductor interior, it follows from J = σE that J = 0 there as well (region 1). Thus, all current is confined to the thin current sheath of regions 2. Maxwell (1.1.2) says curl E = -jωB in the ω domain, so if E = 0 in the interior, so also is B. Everything is quiet inside. Region 2: (the current sheath) As just noted, all currents flow in a very thin sheath at the surface of thickness δ. Since the thickness is tiny, the current density Jz there is "very large" as marked in the table. Imagine a total current I flowing down the conductor, but it is restricted to flow only in the thin sheath. 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. As noted in Section 3.4, this same Jr is "feeding" the total current flow through the surface, and the surface converts this total current from conduction current on the inside to displacement current on the outside. An argument will given below for why Jr is small compared with Jz and we duly markl Jr as "small" in region 2. Application of Ampere's Law (1.1.24) to the small red loop in Fig 3.3 (Bφ = 0 on the left long edge) shows that the large Jz sheath current creates a "large" Bφ field in the sheath which grows from 0 on the sheath's inner boundary to some large value at the conductor surface. Ignoring dramatic μ differences, this Bφ then exists just outside the surface as well according to (1.1.26). We thus mark Bφ as "large" in both regions 2 and 3. If I = 100 mA and a = 1 mm for a round conductor, then Bφ = μ0I/(2πa) = 20 μT at the wire surface. (Earth field is 32 μT) . This is a large value for Bφ in our current context. Since E = J/σ, even though Jz is very large, σ is extremely large, so we shall mark Ez as being "small". And since Jr is already marked "small", we mark Er also as "small". The remaining three entries in the region 2 table above (Br, Eφ, Jφ) we leave at 0, though they might have some very tiny values. Region 3: (the dielectric) Since we are now outside the surface charge layer, (1.1.19) says there is a large radial electric field Er which is supported by this charge density (Gauss's Law), so we mark Er as "large" in region 3. The tangential electric fields are continuous through the boundary according to (1.1.25). Therefore, we give Eφ and Ez the same values they had in region 2. We already observed that Bφ continues being "large" just above the surface. It was noted above that there is a radial pumping current Jr inside the conductor. This pumps charge onto the conductor surface, and this Jr is converted to displacement current in the dielectric as discussed above in Section 3.4 (think of a simple parallel plate capacitor where this also happens). This displacement current and Jr are relatively small currents and they create a small Bz field as we now demonstrate. Consider a very tall red loop whose one edge lies parallel to the z direction between the conductors and whose top edge is very distant. Consider Ampere's law (1.1.23) relative to this loop and with respect to the displacement current flowing through the loop between the conductors, H ds = ∫S ∂tD dA . (1.1.18) Integration of the small displacement current ∂tD passing through the loop gives some small non-zero value for the area integral on the right. Meanwhile, the line integral on the left has cancelling contributions from the vertical loop sides, while the loop top is far away so contributes nothing. The result is some small Hz and hence small Bz in the region between the conductors. Since Bz is a tangential field, it will exist also just inside the conductor surface, as indicated by (1.1.26). Both these Bz fields are marked "small" in the table for regions 2 and 3. As a crude estimate, if the loop is λ/2 long with λ = 1 m and if the total displacement current through the loop is I ~ 100 mA, then Bz(λ/2) = μ0 I and Bz = 0.25 μT, which is small compared to our 20μT estimate for Bφ. Since the dielectric has zero conductivity, the conduction current components are all set to zero. In the dielectric, if we ignore the small Ez and Bz field components relative to the large Er and Bφ, we find that (see Fig 3.3) just outside the surface, the E and B fields are perpendicular and are both transverse to the z direction. Hence this is a TEM (Transverse Electric and Magnetic) mode of the transmission line. Their cross product is the Poynting vector (1/μ) E x B which is in the +z direction coming at the viewer in Fig 3.3. This is the direction of power flow along the transmission line. (Jackson 6.109: S = E x H in SI units) 3.6 The TEM mode fields and currents for a REAL transmission line We now "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 across the entire conductor and there is no Region 1. At higher ω there still is a Region 1, but we shall ignore it from now on. We are still interested in region 2 which is just below the surface charge layer. Recall from Section 3.2 that the surface charge layer remains nearly infinitely thin even for a non-perfect conductor. So here is the new table. The superscripts refer to descriptive sections below. Other values are just carried from the previous table. In order to make ballpark 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. Table 2: E,B,J for a real transmission line Region 2. In the conductor, just under the surface charge layer, and in the current layer. Er = small [c] Br = 0 Jr = small Eφ = 0 Bφ = large Jφ = 0 Ez = small [a] Bz = small Jz = large [a] Region 3. In the dielectric, just outside the super-thin surface charge layer. Er = large [a] Br = 0 Jr = small [b] Eφ = 0 Bφ = large Jφ = 0 Ez = small [a] Bz = small Jz = leakage [b] [a] Ez and Jz in the conductor; Ez and Er outside the conductor Inside the conductor, a non-zero Ez exists due to the current flow in the z direction and the finite conductivity of the conductor. As an estimate for a round wire not too close to the other conductor, assume that the wire has diameter 1 mm, and is operating at 1 GHz with a skin depth δ = 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 = 0.1/(2πrδ) = 1.6 x 107 amps/m2, and this Jz is marked "large" for region 2 in the above table. Then Ez = Jz/σ = 1.6 x 107 / 5.81 x 107 = 0.3 volts/meter. This Ez is marked "small" in the region 2 part of the above table. At lower frequencies where skin depth is larger, Ez is less. Since Ez is a tangential (parallel to conductor surface) E field, according to (1.1.25) it has the same value in region 3, so that is also marked "small" above. In contrast, if the conductor separation is 0.5 cm, and if we crudely assume the E field is constant between the conductors, then Er between the conductors is 7.5 volts/ 5 x 10-3 m = 1500 volts/m. This is marked "large" in region 3 above. So in region 3 just outside the conductor, Er ~ 1500 V/m Ez ~ 0.3 V/m ratio (Ez/ Er) ≤ 2 x 10-4 (3.6.1) [b] Leakage: Jr, Ez and Jz in the dielectric By "leakage" is meant conduction through the dielectric. As shown in (3.3.4), the effective conductivity in the dielectric is given by σeff = ( σωε' tanL) . (3.3.4) For polyethylene, σ ~ 10-15 and can be ignored, while ε' ≈ 2.3 ε0 and tanL ≈ 2x10-4 as in (3.3.5). For a frequency of 1 GHZ, we then find σeff ≈ ωε' tanL ≈ 2π 109* [2.3 * 8.85 x 10-12] * 2 x 10-4 ≈ 2.5 x 10-5 (3.6.2) This is 12 orders of magnitude smaller than the σ of copper ~ 107, but it is 10 orders of magnitude larger than the DC conductivity of the dielectric ~ 10-15. To estimate the significance of this leakage at high frequencies, we can compare the ratio of the leakage current to the displacement current in the dielectric (the currents flow through the same area so ratio is Jleak/Jdisp) | | ≈ | | ≈ tanL ≈ 2 x 10-4 . (3.6.3) Thus, even at high frequencies, the effect of leakage on the current flowing through the dielectric is quite small compared to the displacement current. The "radial" current Jr has to support both the leakage current and the more significant displacement current, and we have just seen that the leakage part can be ignored. We carry region 3 "small" Jr from the previous table since the leakage does not alter this fact. Finally, we already noted a small Ez just outside the conductor, and since the dielectric has some very small leakage (σeff), there will be some small Jz in region 3 which we have marked "leakage". [c] Er and Jr 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, so Er in region 2 is marked "small". In the example above we found Ez ≈ .3 V/m inside the conductor. Thus we have Er << Ez inside the conductor which in turn means Jr << Jz . Below we shall provide more support for the idea that Jr << Jz . 3.7 The general shape of fields, charges, and currents on a transmission line (a) Facts about field structure Based on the information in Table 2 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 be 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. Proof: The deviation from π/2 is less than 2x10-4 radians according to (3.6.1), and the deviation is in the direction of current flow at each conductor. 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. Proof: As we shall see below, the TEM form of any field or current is F(x,y,z,t) = ej[ωt-kz+φ(ω)]F(x,y) where F(x,y) is real, so all t and z dependence is in the exponential. We can take the physical field to be the real part as discussed in Section 1.6 so Fphysical(x,y,z,t) = cos[ωt-kz+φF(ω)] F(x,y). Thus, the cross sectional shape of the field is determined by F(x,y) and is the same at all values of z apart from an overall scale factor cos[ωt-kz+φF(ω)]. This scale factor varies between +1 and -1 as one moves down the line in z at some fixed t, or as one observes at some fixed z as time varies. Later we will see that this shape F(x,y) can be found by solving a certain 2D Helmholtz equation, and we find that the shape is determined entirely by the shape of the boundaries of the conductors. Different vector fields (e.g., J and E) might have different phases in this wave motion which we indicate by φF(ω) for F(x,y,z,t). 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: Recall these two equations from earlier sections curl B = μ (jωε E + J ) = μ (jωε E + σE ) = μ(jωε + σ)E (2.1.1) β2 = ω2 μ [ ε + σ/jω] = ω2 μ( jωε + σ) /jω = ωμ( jωε + σ) /j (2.1.13) If follows from these equations that, curl B = j(β2/ω) E ≡ C E . (3.7.1) To show that the E and B fields are perpendicular, we will show that E•B = 0. We have from (3.7.1), C E•B = curl B B = ( ∂xBy - ∂yBx)Bz + ( ∂yBz - ∂zBy)Bx + ( ∂zBx - ∂xBz)By . Since Bz ≈ 0 ( see estimate in previous section), we are left with only two terms C E•B = By(∂zBx) - Bx(∂zBy) = By2 ∂z (Bx/By) . However, we argued in Fact 3 that the shape of fields does not vary with z. Thus, the ratio of two components like Bx/By cannot vary with z. Thus, E•B = 0 so the E and B lines are perpendicular everywhere in the dielectric. ( We also assume Ez ≈ 0 since it is small 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, so B = Bφ . Thus, Br = 0 at the surface. Proof: We have now 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 2 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 5 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. (b) Drawings of the fields 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: Fig 1: Cross section view 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 direction 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 right conductor. The electric fields are as shown and are strongest in the region between the conductors. The magnetic field directions derive from the right hand rule relative to the current in each conductor. The lines of E and B always intersect at right angles. The magnitude of the E field is 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. 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 equations like curl E = -iωB Maxwell curl E equation (1.1.2) The resolution is that all the spatial 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 red E arrows are all of unit length and serve to mark the direction and density of electric field lines lying in the plane containing the center lines of the conductors. The blue B arrows are seen end on and indicate the same for the magnetic field. On the left they come out of the plane of paper and on the right they go into it. Later we shall learn about the "transmission line limit" in which the wavelength λ of the wave propagating down a transmission line is assumed to be much larger than all transverse dimensions of the line. The reader should understand the above picture as being in that limit, but one would have to stretch the picture at least 10X horizontally to make it be reasonable. At all places ExB points to the right, so we have a wave propagating to the right (+z). Let us now apply the Maxwell curl equations using the two loops shown. Loop 1 is positioned to pick up magnetic flux, so we use (1.1.22) which in the frequency domain says curl E = -jωB E ds = -jω[∫S B dA] (3.7.2) 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, the rate of change of E must increase 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, thus increasing E•ds for the same fixed loop 1. Remember that the strength of the E field is indicated in Fig 2 by the density of the red arrows, not by the length of the red arrows. 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 (1.1.24) which says curl B = μεjωE + μJc B ds = μ ∫S [εjωE + Jc] dA ≈ jωμε ∫E•dA . (3.7.3) Since we are now in the dielectric, we have ignored the small leakage conduction current, and have kept the dominant displacement current. 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 faster in the z direction. As ω increases, 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, λ = v/f = 2πv/ω . (3.7.4) The main point of the above discussion is to show how the Maxwell 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. (c) More on the field and current Structure Here we explore in more detail the general distribution of fields and currents in a transmission line. The goal is to establish the phase relationships among the electromagnetic fields and various currents. Once this is done, it is possible to make an estimate of the ratio Jr/Jz and that is done in the following section. Consider the following more elaborate version of Figure 2 : Fig 3: Tilted overhead view of a transmission line The picture is quite complicated and deserves clarifying comments: (1) Unlike in Fig 2, the E and B arrows indicate the E and B vectors, and are not just field direction and field line density indicators. (2) The E and B field vectors are shown along some line which lies in the plane of the center lines of the two conductors and which points in the direction, as do those center lines. (3) The blue B field arrows lie in the blue plane which is meant to be perpendicular to the plane of the conductor center lines, which is the plane of paper. The red E field arrows are in the plane of paper. (4) Looking at E x B, we see that the wave is traveling to the right in the direction. (5) The E field arrows point from positive charge to negative charge, so this is why the + and - signs are distributed as shown. (6) The conductors are fixed to the paper, everything else is moving to the right at velocity v. This includes the E and B arrows and their curves, the charge density and its curve n, and the two current curves drawn on the bottom conductor. (7) At point Q on plane z = zQ, since B is coming out of paper to the viewer, the longitudinal current Jz in the lower conductor must be pointing to the right. This is why Jz is shown positive at this point in the lower conductor, and this calibrates the position of the Jz curve. Maximum Jz occurs with maximum B. (8) There exists a displacement current Jdisp = ∂tD = ε ∂tE in the dielectric whose magnitude is shown as a red curve. For an observer sitting at fixed point P, since the wave is moving to the right, the value of ∂tE is at its instantaneous maximum positive value. This is why the red Jdisp curve has a positive maximum at point P. (9) As discussed in Section 3.4, the displacement current is "fed" by the radial current Jr inside the lower conductor, so the Jr curve also has its maximum positive value at point P. This Jr current is busily radially pumping positive charge to the surface of the lower conductor at point P so that charge will be there when the wave has moved λ/4 to the right. Of course this radial Jr is doing this charge pumping all around the lower conductor, but we only show it in the plane of paper. (10) We have glossed over the fact that the E and B fields track each other in magnitude. For example, they are both maximal at the same longitudinal position zQ. B is maximum there because Jz is maximum, but it is not quite clear why E is maximal at the same point. We know this alignment occurs in a plane wave, but a transmission line is not just a plane wave. Various arguments can be ginned up for the alignment of the E and B maximums. One simple argument involves the green cylindrical Gaussian box drawn inside the lower conductor, and we reverse the discussion above. Note that this box lies entirely inside the conductor and so does not enclose any surface charge. Since there can be no charge inside a conductor, this box has Qenclosed = 0. Thus, the total current flowing into this box must be zero. Since the Jz current flows into both ends of this box (black arrows), the Jr current has to flow out on the cylinder's curved surface. By considering shorter green boxes one can show that Jr is maximal at zP (as drawn). Thus Jdisp is maximum at zP which means ∂tE must be maximum there, which means E = 0 at zP which means E has its maximum in alignment with the maximum of B. (d) Estimate of the ratio Jr/Jz Having drawn and described this elaborate picture, we now consider again the green Gaussian box. At the instant in time for which Fig 3 is drawn, the total current flowing into the endcaps of the box is 2I, where I is the peak longitudinal current -- the magnitude of the longitudinal sine wave. Therefore, the total Jr integrated over the sides of the green cylinder must also be 2I. To obtain a ballpark estimate of the situation, we first assume that the two round conductors are far apart compared to their radii, in which case Jr is roughly symmetric around the conductor surface. Then the total radial current emitted by the curved surface of the green Gaussian cylinder is: radial current total = [ (2/π)Jr ]* 2πa * (λ/2) = 2I Since Jr is a longitudinal sine wave, we have added a factor 2/π to get its value averaged over the length of the Gaussian box. In a more general case, we can replace 2πa with distance p which represents the active portion of the conductor perimeter, as illustrated in Fig ***. Then we have [ (2/π)Jr ]*p * (λ/2) = 2I => Jr = 2πI / (λp) (3.7.5) On the other hand, for a round conductor operating in the skin effect regime where δ < a, Jz ≈ I/(pδ) , (3.7.6) where p is the same active perimeter just mentioned. So Jr/Jz ≈ 2π (δ/λ) . (3.7.7) For δ we had δ ≡ (2.1.20) From (3.7.4) we have λ = v/f = 2πv/ω where v is the wave phase velocity. Then (δ/λ) = = = . (3.7.8) Setting v ≈ c and μ = μ0 = 4π x 10-7 and σ = 5.81 x 107 (copper) and f = 109f(Ghz) we get (δ/λ) ≈ = = = 10-3 = 7 x 10-6 and so Jr/Jz ≈ (2π) (δ/λ) ≈ 4.4 x 10-5 (3.7.9) For f ≤ 10 GHz we then find Jr/Jz ≤ 1.4 x 10-4 . f ≤ 10 GHz skin-effect regime (3.7.10) showing that the radial charge-pumping current density Jr is much smaller than the longitudinal current density Jz in the conductor sheath. What about the low-frequency situation with no skin-effect sheath? For simplicity, we assume now two round conductors of radius a which are widely spaced. No skin effect means roughly δ > a which means > a => ω < 2/(μσa2) or ωa/2 < 1/(μσa) . (3.7.11) In this low frequency regime we must replace (3.7.6) by Jz ≈ I/(πa2) . (3.7.12) Since (3.7.5) is still valid, we find now that Jz ≈ I/(πa2) Jr ≈ 2πI/(pλ) ≈ 2πI/(2πaλ) ≈ I/(aλ) so Jr/Jz ≈ π(a/λ) ≈ (πa)(ω/2πv) ≈ ωa/2v = (ωa/2)(1/v) . (3.7.13) Using (3.7.11) for ωa/2 this says Jr/Jz < 1/(μσav) . (3.7.14) With μ = μ0 = 4π x 10-7, σ = 5.81 x 107 (copper) and v = c = 3 x 108 we find for a wire of radius 1 mm, Jr/Jz < = = 4.6 x 10-8. low frequency (3.7.15) The conclusion is that in general Jr << Jz under 10 GHz and finally we justify entries made in the tables of Sections 3.5 and 3.6. The basic fact is that the green cylinder is long, so the surface area through which Jr flows is much larger than the area through which Jz flows. 3.8 Transmission Line Preliminaries A transmission line normally 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. We just showed in the last section that Jr << Jz. There are no azimuthal tangential currents Jφ. 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 7.5 volts at the driving end. 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. ok to here Proof: Consider equation (1.5.9): A(x,ω) = ∫J(x',ω)dV' (3.8.1) Here, J represents the 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, so the corresponding transverse components of A will also be very small and we shall completely neglect them. 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 are added up in the integration. We rely mainly 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. I DO NOT FOLLOW THIS Proof B: Since Fact 7 is important, and is non-obvious, we give here another proof. According to (1.5.5), we have the King 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 4, assume a separation of variables so that Az(x,y,z) = (μ/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, ∂zAz = (μ/2π) ∂zi(z) Azt(x,y) = ∂zi(z) [Az / i(z) ] so [∂zi(z) / i(z)] Az(x,y,z) = -j(β2/ω) φ(x,y,z) . (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. We can construct a cylindrical coordinate system in Fig 3.3 based on a center line that matches the curvature of the surface at the point shown. Let the surface point be distance r from the coordinate system axis. In this system we find that Br = [curl A]r = (1/r)∂φAz - ∂zAφ ≈ (1/r)∂φAz . (3.8.6) 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) = ∫ ρ1(x') dx'dy'dz' R = |x - x'| (4.1.1) where R is the distance between x and x'. Comment: If we assume that ρ1 has "zero phase", then in general φ1(x) is going to be complex and will have some non-zero phase. So let's assume ρ1 is real with this zero phase. Also, at this point conductors can have any 3D shape you want. Also, ρ1 represents a surface charge in 3D. I think at this point we de-generalize and assume that we have conductors of a transmission line with z = longitudinal 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) C/m3 1/m2 C/m The dimensions of the functions in this factorization are as indicated, so the charge goes with q1. Moreover, without any loss of generality we select the relative scale of the 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 : !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. Sticking this expansion into φ1 gives 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) ok to here and holding In Appendix G we investigate the dz' integral on the right. If we define the following symbol (the transverse part of the distance between x and x' ), s ≡ (4.1.8) then the Appendix shows the integral has the following functional form, !Syntax Error, Idz' (z'-z)n = sn fn (βs) . (4.1.9) One could arrive at this general form based on the fact that the integral must have dimensions of (distance)n. The Appendix shows that the integral is zero for odd n, so it is then convenient to change to summation variable m = n/2. Then (4.1.7) 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) [ wrong ] (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) [wrong] (4.1.12) We are now done. Note that we have made no assumptions whatsoever other than the separation of variables in (4.1.2). If we knew the way the charge was transversely 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(s)q(z) + k1(s)q"(z) + k2(s)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) = κ0 q(z) and we will then interpret κ0 as the inverse capacitance of the line per unit length based on the usual notion that Q = CV. 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) * repair { !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) * { !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 can 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(i) 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 B about the complex dielectric constant to make this replacement shown in (B.6) where q(z) is the integral of the surface charge n(x,y,z) around a band of conductor C1 of width dz, q(z)/ε = qeff(z)/ξ. (4.2.6) As shown in (B.8), qeff(z)/ V(z) = C', the complex capacitance 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 (G = "conductance") C' = C + G/jω . (4.2.7) Since q(z) = (ε/ξ) qeff(z) = (ε/ξ) C' V(z) => 1/C' = (ε/ξ) V(z)/q(z) we find from (4.2.5) that 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 Az 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. [ wrong! The King gauge is not covariant. I am now assuming the Lorenz gauge] 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) = ∫ Jz1(x') dx'dy'dz' R = |x - 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) A/m2 1/m2 A (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/ε) → (μ) 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) repair (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) ≡ A12,z(x1) - A12,z(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)] } repair 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) } repair? (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 Stokes theorem applied to B = curl A says curl A = B A ds = ∫S B dA . (4.4.4) Consider the red loop shown in this top view of the two transmission line conductors. The loop is intended to have a tiny width dz, and the top view obscures the fact that each conductor has an arbitrary cross section. Since we neglect any transverse components of A, the Stokes theorem says [Az1(bottom) - Az2(top) ] dz = [ magnetic flux through red loop] (4.4.5) We can imagine the entire transmission line as forming a long thin horizontal loop of wire as shown in blue. The long loop has some total external inductance which a has this definition: (flux through blue loop) = Ltot,ext I where I is the loop current. This is an external inductance only because it does not account for the stored field energy inside the conductors, as we saw in the case of a round wire in Appendix C.3. The blue loop is a superposition of many red loops, and for the red loop shown we write [magnetic flux through red loop] = (Ledz) i(z) (4.4.6) where recall i(z) is the loop current. Now Le is the external inductance per unit length of the transmission line. Combining the above and cancelling the dz's we find from (4.4.1) that W(z) = Le i(z) (4.4.7) In the context of the red loop bordered by two current carrying "wires", one would refer to Le as a mutual inductance, although it is part of the self-inductance of the entire transmission line. Therefore, from (4.4.3) we have Le = {!Syntax Error, Idx' dy' b1(x',y') ln(s2/s1) -!Syntax Error, Idx' dy' b2(x',y') ln(s2/s1) } (4.4.8) 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 (4.4.8). 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 (4.4.8) 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) or q(z) = C V(z) W(z) = Le i(z) (4.5.1) where C' is given by (4.2.8) and Le by (4.4.8). 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 μ φ - delete red term (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 Lorenz 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) tentatively ok to here and β is the real β !! 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 = μξ (4.5.15) As shown in Appendix 1.2, Eq. (4), we know that C' = (ξ/ε) C. And from (1.5.3) we have β2 = ω2 μ 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 / μ = {!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 = (μ/2π)K (4.5.18) C' = 2πξ/K = C + G/jω Since ξ = ε + σ/jω, we can decompose the last into two equations, C = 2πε/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 = (μ/2π)K C = 2πε/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 = πε / ln ( b/) Le = (μ /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 = (μ /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)μ 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) = (μ/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 = (μ /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/ε) ρ (5.2.2) where β ≡ ω (5.2.3) with ξ ≡ [ ε + 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. Note: 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) = (1/2πε) 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) = (μ/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 = ω2μξ 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) = (1/2πε) q(z) φt(x,y) (βd2 + k2) = jωZi C' Az(x,y,z) = (μ/2π) i(z) Azt(x,y) ≈ 0 "low loss" seems wrong 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 μξ [ z2 - k2 ] q(z) = 0 βd2 ≈ ω2 με = 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 = -jωμ 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(1/2πε) q(z) φt(x,y) - jω(μ/2π) i(z) Azt(x,y) = continuous Hφ(x,y,z) = (1/μ) Bφ = (1/μ) ( - ∂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 = (μ/2π) K C = 2πε/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) = ε Er = - ε ∂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 (1/4πε) 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) = (1/2πε) q(z) f(x,y) Le = (μ/2π)K Az(x,y,z) = (μ/2π) i(z) f(x,y) C = 2πε/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 = (μ/2π)K C = 2πε/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 = (μ/2π)K C = 2πε/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 = (μ/2π)K C = 2πε/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 = (μ/2π)K C = 2πε/K G = 2πσ/K (6.4.24) Appendix A: 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. Our step-by-step approach here is somewhat unconventional and brings in the notion of a Green's function and the particular solution of the Poisson Equation which is quoted in the main text. Some extracurricilar topics are brought up which may or may not interest the reader. A.0 The Poisson Equation and its Solution Fact 0: The Poisson Equation -2φ = ρ/ε0 has a unique solution as stated below. (A.0.0) For electrostatics in an isotropic medium equations (1.1.3) and (1.1.6) indicate that div E = ρ/ε while (1.1.2) says that curl E = 0. Since curl grad f = 0 for any function f, if one lets E = - grad φ, then curl E = 0 for any φ, and div E = ρ/ε becomes - div grad φ = ρ/ε, or 2φ = - ρ/ε, an equation known as the Poisson Equation. The problem of electrostatics ("potential theory") is then to solve 2φ = -ρ/ε for the potential φ, and then E = - grad φ produces the resulting electric field. For a static physical situation (nothing varies with time t), the electrostatic potential φ matches the scalar potential φ appearing in (1.3.1). Here ρ(x) refers to the electric charge density. (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). 3D continuous charge densities that are glued down in 3D space so they cannot move, or which manage to achieve a stable configuration as free charge. surface polarization charge densities not already accounted for by the ε in 2φ = - ρ/ε . 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, and in an isotropic medium of dielectric constant ε) we know that the electric potential φ of a point charge q located at point x' is φ(x) = q/[4πεR] where 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)/ε . (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') is 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 observation 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 (subject to the requirement that for fixed x', g(x,x') = 0 for x on the Great Sphere): -2g(x,x') = δ(x-x') => 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 = ε units 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{ } = 4π δ(x-x') . (A.0.4) This last equation is derived in Appendix H (see H.1.4), but we have already shown it is true, given Coulomb's Law. 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)/ε . QED. The assisting function g(x,x') has various names with respect to (A.0.3): the Green's Function or Green function, the fundamental solution, the free-space propagator, or the kernel. It is nothing more than the potential created by a point charge of 1 unit located at x' and viewed from x. Some authors put a 4π in front of the δ in the left equation of (A.0.3) which causes the 1/(4π) to be absent in the right equation of (A.0.3). (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 taut 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/ε 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). A.1 Existence of A such that B = curl A and div A = 0 Fact 1: If div B = 0, there exists an A such that B = curl A and div A = 0. (A.1.0) The "gauge choice" div A = 0 is known as the Coulomb or Transverse Gauge. More on gauges later. 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 field 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.1.1a) or, in Cartesian coordinates, -2(Ai) = [curl B]i . (A.1.1b) We may regard this as the Poisson equation (A.0.1) where φ → Ai and ρ → ε [curl B]i . We know that a Poisson equation of the form (A.0.1) has a unique physical 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. We take (A.1.2) to be a candidate expression for the vector field A. If we can show that div A = 0 and that B = curl A, then (A.1.2) is a viable expression for A. (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 field F , so the integrand and 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, and εijk is the totally antisymmetric Levi-Cevita permutation tensor used to express curl components ] [curl A(x)]i = εijk∂jAk(x) = + εijk . (A.1.6) As before, 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. A.2 Existence of A' such that B = curl A' and div A' = f . 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| → ∞. (A.2.0) Proof: From Fact 1, we first find A such that B = curl A and div A = 0. We then define A' ≡ A + grad Λ dim(Λ) = volt-sec (A.2.1) where Λ is some so-far arbitrary function (scalar field). As shown below (1.3.1), dim(A) = volt-sec/m, and therefore dim(Λ) = volt-sec. It follows from (A.2.1) 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 . dim(f) = volt-sec/m2 (A.2.3) But this is once again Poisson's Equation (A.0.1) with φ → Λ and ρ → -εf. Translating (A.0.2) we then find that Λ(x) = – . (A.2.4) Meanwhile, from (A.2.1) we also conclude that, since curl grad g = 0 for any function g, curl A' = curl A + curl grad Λ = curl A = B . (A.2.5) 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.2.4) and so Λ(x) is uniquely determined by our selected function f(x). The function Λ(x) is called a gauge function for reasons given below. A.3 Existence of φ such that E = -grad φ Fact 3: If curl E = 0, then there exists a φ such that E = - grad φ . (A.3.0) 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 yet again Poisson's equation (A.0.1) for φ, this time with ρ→ ε div E, 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, 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. A.4 Existence of A' and φ' such that B = curl A', E = - grad φ'-∂tA', and div A' = f. 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. (A.4.0) Proof: We know from Fact 2 that A' exists such that B = curl A' and such that div A' equals any arbitrary function f. If we start with some arbitrary A and φ, the successful A' from (A.2.1) is A' = A + grad Λ where Λ is given by (A.2.4) as an integral over f. What is the corresponding φ' ? Since the E field corresponding to (A,φ) and (A',φ') must be the same, we must have -E = -E' or grad φ+∂tA = grad φ'+ ∂tA' . Since A' = A + grad Λ, then ∂tA' = ∂tA + grad ∂t Λ, so the above reads grad φ = grad φ' + grad ∂t Λ which is satisfied by φ' = φ - ∂tΛ. Thus, the successful potential pair giving div A' = f is this: A' = A + grad Λ φ' = φ - ∂tΛ // dim(Λ) = volt-sec (A.4.1) where from (A.2.4), Λ(x) = – (A.2.3) The pair of equations (A.4.1) is called a gauge transformation and we have just seen in Facts 2 and 4 that a gauge transformation preserves both E and B. Each possible choice f defines a function Λ which then gives the transformation. There are an infinite set of f and corresponding Λ functions, so there are an infinite number of gauge transformations which leave the E and B fields invariant. We are free to choose a gauge such that div A' = f for any f we like. A.5 Gauge Invariance In electromagnetism, the situation of Fact 4 arises for B = magnetic field A = vector potential E = electric field φ = scalar potential Using the gauge transformation (A.4.1), 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. The word "gauge" was first used by Hermann Weyl in the context of general relativity. Gauge invariant there means that a certain "covariant derivative" transforms as a proper tensor object so that things have the same form in different coordinate systems used to measure things. These different coordinate systems were referred to as different "gauges" in the sense that a gauge is a marked-off measuring instrument used to measure something (like the marked-off x-axis of a coordinate system). In general relativity the metric tensor gμν, which defines the meaning of distance in the 4 dimensions of spacetime, is a function gμν(x) of the local location in spacetime x. Weyl considered the effect of rescaling the metric tensor according to gμν(x) → λ(x)gμν(x) where λ(x) was an arbitrary "gauge function" ( like our Λ(x) ). Nowadays, gauge invariance is associated with any continuous degree(s) of freedom of a theory which don't affect physical measurements derived from the theory, such as our gauge transformation (A.4.1). See Quigley. A.6 The Lorenz 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. A more detailed description of the tensor notation used below may be found in the author's Tensor Analysis document. In relativistic notation one uses 4-vectors which have one time component and three spatial components such as xμ = (ct,x,y,z) which denotes a point in "spacetime". The time component t is multiplied by the speed of light c so that all four components have the same units -- distance L. Often people measure distance in light-seconds instead of meters so in such units c = 1, but we shall display the c to keep track of units. This xμ is a "contravariant" (index up) 4-vector and the corresponding "covariant" (index down) 4-vector is xμ = (ct,-x,-y,-z). Thus, one has x0 = x0 (= ct) but xi = -xi. We are assuming here the "Bjorken-Drell metric" gμν = diag(1,-1,-1,-1). The gradient operator ∂i "transforms as" the spatial part of the covariant 4-vector ∂μ, and one can write ∂i = -∂i just as xi = -xi for i = 1,2,3. This four-vector gradient operator can be written ∂μ = (∂0, ∂i) and ∂μ = (∂0, ∂i) = (∂0, -∂i) where ∂0 = ∂0 = ∂t = . The four components of ∂μ all have dimension L-1. 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. Consider then the gauge transformation (A.4.1) which in relativistic tensor notation is A'i = Ai + ∂iΛ = Ai - ∂iΛ i = 1,2,3 φ' = φ - ∂tΛ = φ - c ∂0Λ (A.6.1) The components of a classical vector like A, normally written as Ai, are in fact the contravariant components Ai in tensor notation. If we now define A0 ≡ φ 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.6.2) Suppose we want ∂μA'μ = 0 (implicit sum on μ = 0,1,2,3). This would be a sort of relativistic version of the Coulomb gauge choice that ∂iAi = div A = 0. 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. For example, let S and S" be two frames of reference related by a Lorentz transformation. Then the implication is that ∂μA'μ(xν) = 0 ∂"μA'μ(x"ν) = 0 where ∂μ ≡ ∂/∂xμ and ∂"μ ≡ ∂/∂x"μ frame S observer frame S" observer So, is it possible to have ∂μA'μ = 0 ? Writing this out we get ∂0A'0 + ∂iA'i = 0 => ∂t [φ'] + div A' = 0 => ∂tφ' + div A' = 0 so div A' = - ∂tφ'. (A.6.3) But we showed in Fact 2 that given any A, we can find an E-B-fields-equivalent A' which has div A' = any f(x) we want, so we just select f(x) = -(1/c2) ∂φ'/∂t. By selecting this f(x), we are selecting the Lorenz Gauge. In this gauge (now dropping the prime on A), we have ∂μAμ = 0. Thus, the condition defining the Lorenz Gauge is Lorentz covariant under all Lorentz transformations. The reason is that both sides of ∂μAμ = 0 "transform" as the same kind of tensor object, in this case a scalar object. One can interpret ∂μAμ = 0 as ∂A = 0 where ∂ is a 4-divergence operator. Thus, in the Lorenz gauge, the 4-divergence of Aμ is always exactly 0 at every point in spacetime. [ Lorenz and Lorentz are two different people, see the Comment below equation (1.3.6).] In 3D if we said that div F = ∂iFi = 0 defined something called a gauge condition, it would be clear that F was not uniquely determined by that condition since many vector fields have zero divergence. Just so, the Lorenz gauge condition ∂μAμ = 0 does not uniquely determine Aμ , it is just a condition on Aμ. So in fact there are many pairs (A,φ) which satsify the Lorenz gauge condition, so the term "the Lorenz gauge" is a little misleading, though we shall use it anyway. It is a class of gauges. In relativistic quantum field theory (aka quantum electrodynamics, or QED), the potential Aμ is interpreted as the quantum field of a massless vector particle called the photon. The potentials φ and A are thus promoted from being mere "helper functions" to having their own particle interpretation. In the Lagrangian density for the photon-electron system an interaction term - JμAμ appears, L = ... - JμAμ Jμ = e0 γμ ψ (A.6.4) where Jμ is the electric current, an operator built from the quantum field ψ of the electron. The number e0 is the so-called bare (unrenormalized) charge of the electron. According to (A.6.2), a gauge transformation on Aμ creates a new term - Jμ ∂μΛ in the Lagrangian density. In Lagrangian dynamics, 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 parts integration to move ∂μ from Λ to Jμ, we end up with an action change 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 any gauge transformation. The reason ∂μJμ = ∂μJμ = 0 is because Jμ = (cρ, Ji) where ρ is charge density and Ji is electric current, and then the statement ∂μJμ = 0 says that ∂t(cρ) + ∂iJi = 0 or div J = -∂ρ/dt. This is the equation of continuity (1.1.8) 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.6.2). 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. A.7 Finding the gauge function Λ for the Lorentz Gauge In Fact 4 is was noted that if one already has a potential set (A,φ), it is possible to find a new potential set (A',φ') such that div A' = f for any reasonable f. The method of finding the new set (A',φ') was to find the function Λ from f as shown in (A.2.4) and then use the gauge transformation implied by Λ as shown in (A.4.1) to find the new potentials (A',φ'). In the discussion of the Lorenz Gauge, we thus imagine we have some (A,φ) and we want then to find a potential set (A',φ') such that div A' = - ∂tφ', which is the Lorenz Gauge (A.6.3). We are thus using f = - ∂tφ' where φ' is the partner to A'. One might fairly inquire what this function f actually is in terms of the starting potentials (A,φ), since one does not a priori know what φ' is. In other words, since we don't a priori know what f(x) is, we cannot use (A.2.3) to find the right gauge function Λ to give the right new potentials (A',φ'), so we seem to be in a circular conundrum when we try to fit this Lorentz gauge situation into the framework of our accumulated Facts above. Here is one way to find the right function Λ in terms of (A,φ). We know from (A.2.3) and (A.4.1) that 2Λf = - ∂tφ' = - ∂t [φ - ∂tΛ ] which can be written as (∂t2 - c22)Λ = ∂tφ . (A.7.1) Since we know φ from (A,φ), we can obtain Λ by solving this differential equation. The equation is similar to the Poisson equation (A.0.1) when written this way in terms of the symbol introduced above, c2 Λ = ∂tφ . // Stakgold (5.141) with u→Λ and q→∂tφ (A.7.2) Here and below we include some supporting equation numbers from Stakgold Vol II. The formal solution of (A.7.2) can be found by first defining a Green's Function as we did above in (A.0.3), c2 g(x,t; x',t') = δ(x-x')δ(t-t') // Stakgold (5.142) (A.7.3) The solution Green's function (propagator) is given by ← c2 g(x,t; x',t') = (1/4πR)δ(t-t'-R/c) with R = |x-x'| // Stakgold (5.155) n=3 (A.7.4) We have added an arrow that shows the direction of the propagator: it runs from time t' in the past to time t in the future, in which case t > t'. The propagator vanishes for all t < t' since in that case t-t'-R/c < 0 and the δ function can never get a hit. This g is an example of a "causal" Green's function and it describes an expanding spherical wavefront seen at observation point x at time t propagating at velocity c from a point source at location x' and time t' in the past. Formally one can then express a solution to (A.7.2) in a form similar to (A.0.2), Λ(x,t) = ∫d3x' ∫dt' g(x,t; x',t') ∂t'φ(x',t') . (A.7.5) Application of c2 to both sides of (A.7.5) with use of (A.7.3) reproduces (A.7.2) showing that (A.7.5) is indeed the particular solution of (A.7.2). Inserting the propagator (A.7.4) we find that Λ(x,t) = ∫d3x' R = |x - x'| (A.7.6) Thus we have solved our conundrum in that we have Λ expressed in terms of φ from the set (A,φ). The solution (A.7.6) has the same form as the retarded solutions of Section 1.4. Once we have this Λ, we may use (A.4.1) to find the set (A',φ') given the set (A,φ). Comment: Whereas the Poisson equation with 2 is "elliptic" in nature, the wave equation is "hyperbolic" since the various second derivitives in don't all have the same sign, resulting in a change in the nature of the Green's function solution, the principle fact being that it is a causal function in terms of the time coordinates. For details on the above discussion, see Stakgold Vol II p 61-63 (fundamental solutions) and p 246-256 (Green's functions for the wave equation). Stakgold treats this subject with an arbitrary number of spatial dimensions n. One finds, for example, that for n = 3 the propagator (A.7.4) is an expanding infinitely thin spherical shell with no wake, whereas for n = 2 there is a wake behind the front as in his (5.151) which says g(r,t) = θ(t-r/v) 1/ where v is wave velocity. It is difficult to create a clean unit impulse in water, but here is the rough idea: http://physicsilluminati.blogspot.com/2012/10/wave-optics.html Appendix B: Magnetization Surface Currents on a Conductor Overview When a conductor of magnetic permeability μ2 is embedded in a medium of μ1 with μ1 ≠ μ2, a "bound current" appears on the conductor surface. This so-called magnetization current must be accounted for when computing quantities like the B field or vector potential A. Section B.1 shows how this surface current K is related to the H field at the surface. Section B.2 shows how to compute H from the volume current density J. Section B.3 then outlines a general plan for computing surface current K for an arbitrary conductor. The above sections apply in both AC and DC conditions (ω > 0 and ω = 0). The remaining three sections consider the simple case of a round wire where the volume current Jz is uniform in the conductor, a situation that arises at DC and low frequency where there is no skin effect. Section B.4 computes H and the surface current K for a round wire using symmetry. Section B.5 repeats the calculation using the general method outlined in Section B.3. Section B.6 assumes the K computed above, along with uniform Jz, and computes from these the vector potential A. Then B = curl A and H = B/μ and the simple results of Section B.4 are replicated. The conductors considered here are those of a transmission line in the "transmission line limit" in which it is assumed that the wavelength along the line is much longer than the transverse dimensions of the line. In this case, it is reasonable to use 2D wave equations whose solutions then involve use of the 2D Poisson free-space propagator ln(R/2π) as discussed in Appendix I. B.1 Relationship between surface current K and the field H at a conductor boundary First, consider this blow up of a piece of the boundary between a conductor (medium 2) and a dielectric (medium 1). Both media extend uniformly in the z direction, so we are looking at a piece of the cross section of a transmission line at a particular point on the surface of one of the conductors. Fig B.1 We shall assume that the conduction current is positive in the z direction, so J = Jz with Jz > 0. Since the lower medium is the conductor in the drawing, the B and H field at the boundary are in the - direction, that is to say, they point to the left due to the right hand rule relating J and B or H . At the boundary, we know from (1.1.26) that Hx2 = Hx1 (1/μ1)Bx1 = (1/μ2)Bx2 . (B.1.1) Assuming μ2 ≥ μ1 (which would be the case if μ1 = μ0), the right equation implies |Bx2| ≥ |Bx1| so the B field is larger inside the conductor. But in our picture, both Bx2 and Bx1 are negative, so -Bx2 ≥ - Bx1 which then says Bx2 ≤ Bx1 and finally (Bx2 - Bx1) ≤ 0. Also, Hx2 = Hx1 ≤ 0. For the red loop shown in the figure one then has, B ds ≈ Bx2L - Bx1L = (Bx2 - Bx1)L ≤ 0 . (B.1.2) Now consider (1.1.24) which says ( in the ω domain), B ds = ∫S [ jωεE + μJc + μJm] dA (B.1.3) where dA = dA . Since E dA involves only Ez (parallel to surface), and since by (1.1.25) such Ez is continuous at the boundary, and since Ez ≈ 0 inside the conductor, the εjωE term makes no contribution, giving then B ds = ∫S [μJc + μJm] dA . (B.1.4) Since Jc flows in the + direction, the surface current Jm flows in the - direction, so write Jm = Kz δ(y) (B.1.5) where Kz ≤ 0 is the magnitude of the surface current. The area integral on the right side of (B.1.4) is written as two terms, one for the area above the boundary, and one below: ∫S μ [Jc + Jm] dA = ∫ μ1 [Jc1 + Jm1] dA + ∫ μ2 [Jc2 + Jm2] dA . (B.1.6) The surface current lies on the outside of the conductor surface, so it is region 1. Then ∫S μ [Jc + Jm] dA ≈ μ1Jc1z (s/2)L + μ2Jc2z (s/2)L + μ1 ∫dxdy [Kzδ(y)] dA = [μ1Jc1z + μ2Jc2z ] (s/2)L + μ1Kz L . Then (B.1.4) with (B.1.2) says (Bx2 - Bx1)L = [μ1Jc1z + μ2Jc2z ] (s/2)L + μ1Kz L (B.1.7) Since the volume currents are not infinite at the boundary, as s→ 0 we find μ1Kz = (Bx2 - Bx1) (B.1.8) The signs are consistent with Kz ≤ 0 and (Bx2 - Bx1) ≤ 0 as noted above. Then from (B.1.1) we find μ1Kz = (μ2Hx2 - μ1Hx1) = (μ2-μ1) Hx2 and finally Kz = (μ2/μ1 - 1) Hx2 (B.1.9) As noted earlier, Hx2 < 0 so Kz ≤ 0 is consistent with μ2 ≥ μ1. We now rewrite this result in terms of a different picture: Fig B.2 This shows the cross section of the entire conductor in gray, and Jc is still directed toward the viewer. In this picture a point on the surface is associated with a local coordinate system for which = is normal to the surface and = - is tangent to the surface (so Hθ = -Hx). We are thinking of (r,θ,z) as local cylindrical coordinates at the point shown on the conductor surface, where x = , and the x,y,z directions of the figure match those of the previous figure where as usual x = . Then (B.1.9) says Kz = - (μ2/μ1 - 1) Hθ (B.1.10) where Hθ > 0 and Kz ≤ 0. Conclusion: At the surface of a conductor, if the conduction current is in the + z direction, the surface current is in the - z direction and has a magnitude given by Kz = - (μ2/μ1 - 1) Hθ where Hθ is the tangential H field at the surface of the conductor (either inside or outside since both are the same). If μ1 = μ2 then there is no surface current since there is no imbalance of magnetization current at the boundary. Example: For a round wire of radius a carrying an axially symmetric current distribution, we know that 2πaHθ = I so Hθ = I/(2πa) at the surface. Then Kz = - (μ2/μ1 - 1)[ I/(2πa)] round wire of radius a and μ2, dielectric μ1 (B.1.11) Physical mechanism of the surface current. As a reminder, a surface magnetization current arises at a boundary between media with different μ values just the way surface polarization charge arises at a boundary between media with different ε. In the μ case, here is a suggestive picture : Fig B.3 On the left we look at a round wire end on, while the right shows a top view where the wire has been tilted down. Here μ1= μ0 so there is only vacuum outside the wire. The B field lines up the little magnetic dipoles (or creates them) which we represent schematically as little atoms with orbiting electrons. The atoms in the interior always have cancelling current arrows, but there is an imbalance on the outer surface which is the surface magnetization current. The picture shows why it is that the surface current is directed opposite to the current J which creates it, a sort of magnetic Lenz's Law. The surface current is not seen by H, but it is seen by B. In the case that the outer medium has some μ1 > μ0, both media have surface currents at the boundary, and then when μ1 ≠ μ2 there is a surface current imbalance resulting in a net surface current. If it happens that μ1 < μ2, then the directions shown above are correct, but if μ1 > μ2, the surface current runs in the opposite direction to that shown. B.2 Calculation of H from the current J in a conductor Start with Maxwell's equation (1.1.1), and we are now working inside a conductor so E = 0 and then curl H = jωεE + J = J (B.2.1) where J is the conduction current. Apply curl to both sides and use curl curl = grad div -2 to get grad div H - 2H = curl J . (B.2.2) But in a uniform medium div H = 0 since div B = 0 so 2H = - curl J . (B.2.3) Now let's assume that we have J = Jz(x,y) and assume that the solution H does not depend on z. In that case we have 22D H(x,y) = - curl J J = Jz(x,y) (B.2.4) The particular solution to this PDE is shown in (I.1.8) to be H(x,y) = ∫d2x' [ ln(1/R) ] curl' J(x') R = |x-x'| (B.2.5) or H(x,y) = - ∫d2x' ln(R2) curl' J(x') R = |x-x'| (B.2.6) where ln(1/R) is the Poisson 2D free-space propagator. B.3 General Method for computing the surface current Jm on a wire Here are the steps for a wire of arbitrary cross sectional shape: 1. From the prescribed conduction current J in the wire, compute curl J. As will be shown in the example below, some or all of curl J may lie on the outer surface of the wire. 2. Compute the H field at all points in the wire cross-sectional plane section using (B.2.5) H(x,y) = - ∫d2x' ln(R2) curl' J(x') R = |x-x'| . (B.2.5) 3. Evaluate this H field at xb = (xb,yb) for all points xb on the cross section boundary. 4. Compute the component of H which is tangential to the boundary in the cross sectional plane. Call this component Hθ. 5. The surface current density is then given by (B.1.10), Kz = - (μ2/μ1 - 1) Hθ . (B.1.10) B.4 Surface current on a round wire with uniform J For a round wire of radius a with uniform Jz (as would be the DC case ω = 0) , geometric symmetry makes the calculation of H very easy. One need only apply Ampere's Law separately for a point r outside the wire, and for another point r inside the wire. For the outside case one finds 2πr Hθ(r) = I => Hθ(r) = I/(2πr) r ≥ a . (B.4.1) And then for the inside case the "current enclosed" is determined by a simple area fraction. 2πr Hθ(r) = I (πr2/πa2) => Hθ(r) = I r/(2πa2) r ≤ a . (B.4.2) At the boundary the two expressions agree and we have Hθ = I/(2πa) . (B.4.3) If this wire has magnetic permeability μ2 and is embedded in an infinite medium of μ1, then the surface magnetization current induced on the wire is Kz = - (μ2/μ1 - 1) Hθ = - (μ2/μ1 - 1) I/(2πa) amp/m (B.4.4) Kz = Kz (B.4.5) and this surface current is in the direction opposite J if μ2 > μ1. If μ1 = μ2, the surface current vanishes. This result could be expressed in volume density form as Jm = Kzδ(r-a) amp/m2 . (B.4.6) B.5 Surface current on a round wire using the General Method For a wire of some general cross section, symmetry is not available to allow the simple solution outlined in the previous section. We then have to use the more general method outlined in section 3 above. As a check on the viability of this general method, we shall apply it here to the round wire and attempt to replicate the results found in Section 4 above. The conduction current density in a round wire with uniform Jz is given by Jz(r) = J0θ(a-r) (B.5.1) where θ is the Heaviside step function. Our first step is to compute curl J, and we do this in cylindrical coordinates by just staring at the cylindrical-coordinates curl formula, curl J = [ r-1∂θJz - ∂zJθ] + [∂zJr - ∂rJz] + [ r-1∂r(rJθ) - r-1∂θJr ] (B.5.2) and finding the only non-zero piece which is this (uniform Jz) curl J = [-∂rJz(r)] . (B.5.3) Inserting Jz(r) from above we find ∂r Jz(r) = J0 ∂rθ(a-r) = - J0 δ(r-a) (B.5.4) => curl J(r) = J0 δ(r-a) . (B.5.5) so we have a "ring source of curl J". For use in our integral for H we then have curl' J(r') = ' J0 δ(r'-a) (B.5.6) For a current distribution which tapers off smoothly to 0 at the wire edge one would not have this singular contribution, but for a wire with prescribed uniform current, it is present, and curl J vanishes everywhere but on the boundary. The relevant picture is this: Fig B.4 From (B.2.5) the H field at any point x = (x,y) is then given by H(x,y) = ∫d2x' ln(1/R) ' J0 δ(r'-a) = !Syntax Error, Ir'dr' !Syntax Error, Idθ' (1/2) ln(1/R2) ' δ(r'-a) = !Syntax Error, Idθ' ln(1/R2)|r'=a ' = - !Syntax Error, Idθ' ln(R2)|r'=a ' or H(r,θ) = - !Syntax Error, Idθ' ln [ r2 + a2 - 2ar cos(θ'-θ) ] ' . (B.5.7) The figure shows that ' = cosθ' - sinθ' (B.5.8) so then H(r,θ) = - !Syntax Error, Idθ' ln [ r2 + a2 - 2ar cos(θ'-θ) ] [cosθ' - sinθ' ] . (B.5.9) Next, let x ≡ θ'-θ. Since the ∫dθ' has full range 2π, one can replace !Syntax Error, Idθ' = !Syntax Error, Idx . Then H(r,θ) = - !Syntax Error, Idx ln [ r2 + a2 - 2ar cos(x) ] [cos(x+θ) - sin(x+θ) ] . (B.5.10) Now writing H = Hx + Hy , decompose the above into two equations Hx(r,θ) = + !Syntax Error, Idx ln [ r2 + a2 - 2ar cos(x) ] sin(x+θ) Hy(r,θ) = - !Syntax Error, Idx ln [ r2 + a2 - 2ar cos(x) ] cos(x+θ) (B.5.11) or Hx(r,θ) = + !Syntax Error, Idx ln [ r2 + a2 - 2ar cos(x) ] [ sinxcosθ+cosxsinθ ] Hy(r,θ) = - !Syntax Error, Idx ln [ r2 + a2 - 2ar cos(x) ] [cosxcosθ - sinxsinθ ] . (B.5.12) Since !Syntax Error, Idx is over an even range, throw out odd integrand terms, and then fold the negative range into the positive adding a factor of 2 to get Hx(r,θ) = + sinθ !Syntax Error, Idx ln [ r2 + a2 - 2ar cos(x) ] cosx ≡ sinθ K Hy(r,θ) = - cosθ !Syntax Error, Idx ln [ r2 + a2 - 2ar cos(x) ] cosx = -cosθ K (B.5.13) where K ≡ !Syntax Error, Idx ln [ r2 + a2 - 2ar cos(x) ] cosx . (B.5.14) Before evaluating this integral, we see that H = Hx + Hy = - K [ cosθ -sinθ ] = - K = Hθ . (B.5.15) Thus we find that the resulting H is entirely in the direction and Hθ = - K . (B.5.16) We seek now to evaluate this integral K K ≡ !Syntax Error, Idx ln [ r2 + a2 - 2ar cos(x) ] cosx = !Syntax Error, Idx ln [ {a2}{ (r/a)2 + 1 - 2(r/a) cos(x)} ] cosx = !Syntax Error, Idx { ln (a2) + ln[(r/a)2 + 1 - 2(r/a) cos(x)] } cosx = ln (a2)[!Syntax Error, Idx cosx ] + !Syntax Error, Idx ln[(r/a)2 + 1 - 2(r/a) cos(x)] cosx = ln (a2)[0] + !Syntax Error, Idx ln[α2 + 1 - 2α cos(x)] cosx where α ≡ r/a = !Syntax Error, Idx ln[α2 + 1 - 2α cos(x)] cosx . (B.5.17) This integral is the n=1 special case of the following integral from GR7 page 589, Therefore we find K = (B.5.18) so Hθ = - K = . (B.5.19) Now the total current in the wire is I = J0πa2 so (aJ0/2) = (I/2πa) and then Hθ = = . (B.5.20) Thus, we finally arrive at the same results for H as obtained in (B.4.2) and (B.4.1). We then evaluate Hθ on the boundary Hθ = I/2πa and conclude as in (B.1.11) that the surface current is Kz = - (μ2/μ1 - 1) I/(2πa). B.6 An Acrobatic Exercise: Compute H from J and Jm via the vector potential A. For the round wire with a uniform Jz we are aware of both the volume current JZ and the surface current Kz. In this exercise, we insert both these currents into our formula for the vector potential A, we do the required integrals, and at the end we compute B = curl A and finally use H = B/μ. Hopefully in the end we shall recover once again the results for Hθ shown above. Here again is the picture of interest: Fig B.5 Start with a 2D magnetostatics equation which is (1.5.4) with β2 = 0, 22DA(x) = -μ1Jm(x) -μ2Jc(x) . (B.6.1) This 2D approximation is appropriate for a transmission line conductor in the "transmission line limit". The particular solution from (I.1.8) is A = (1/2π) ∫dV' [μ1Jm(x') + μ2Jc(x')] ln(1/R) R = |x-x'| = (1/2π) { ∫dS' [μ1Km(x')] ln(1/R) + ∫dV'[ μ2Jc(x')] ln(1/R) } so Az = (1/2π) { ∫dS' [μ1Kz(x')] ln(1/R) + ∫dV'[ μ2Jcz(x')] ln(1/R) } = -(1/2π) { ∫dS' [μ1Kz(x')] ln(R) + ∫dV'[ μ2Jcz(x')] ln(R) } = -(1/4π) { μ1∫dS' [Kz(x')] ln(R2) + μ2∫dV'[Jcz(x')] ln(R2) } . (B.6.1) Here ∫dV' represents an integral over the "volume" of the wire slice shown in the figure (which is really a disk) and ∫dS' represents an integral over the "surface" of the wire slice (which is really a circle). We now insert the uniform prescribed current Jcz and our previously computed surface current Kz, Jcz(x') = Jcz = I/(πa2) (B.6.2) Kz(x') = Kz = (1-μ2/μ1) (1/2πa) I (B.6.3) with the result that Az = - (1/4π) { μ1(1-μ2/μ1) (1/2πa) I ∫dS' ln(R2) + μ2 I/(πa2)∫dV' ln(R2) } = - [I/(4π2a2)] { (μ1-μ2) (a/2) ∫dS' ln(R2) + μ2∫dV' ln(R2) } . (B.6.4) The volume integral (per unit length in the z direction, so really an area integral as just noted) is, ∫dV' ln(R2) = !Syntax Error, Ir' dr' !Syntax Error, Idθ' ln [r'2 +r2-2rr' cos(θ-θ')] = 2 !Syntax Error, Ir' dr' !Syntax Error, Idθ' ln [r'2 +r2-2rr' cos(θ-θ')] = 2 !Syntax Error, Ir' dr' !Syntax Error, Idx ln [r'2 +r2-2rr' cosx] . (B.6.5) Define the x integral as Q(r',r) ≡ !Syntax Error, Idx ln [r'2 +r2-2rr' cosx] (B.6.6) so then ∫dV' ln(R2) = 2 !Syntax Error, Ir' dr' Q(r',r) . (B.6.7) Meanwhile, our surface integral of interest (really a line integral) is ∫dS' ln(R2) = !Syntax Error, I[a dθ'] ln [r'2 +r2-2rr' cos(θ-θ')] |r'=a = 2a !Syntax Error, Idx ln [r'2 +r2-2rr' cosx] |r'=a = 2a Q(a,r) (B.6.8) so that both integrals of interest require computation of Q(r',r), which we replicate here, Q(r',r) ≡ !Syntax Error, Idx ln [r'2 +r2-2rr' cosx] . (B.6.6) This integral may be evaluated using GR7 p 531 4.224, with a = r'2 +r2 and b = -2rr' and a2-b2 = (r'2-r2)2 so that = | r'2-r2 | .The condition a > |b| > 0 is met since (r±r')2 > 0 => r2+r'2 > ±2rr' which says a > ±b so a > |b|. Thus, Q(r',r) = !Syntax Error, Idx ln [r'2 +r2-2rr' cosx)] = π ln [ ] = = 2π . (B.6.9) This our two integrals of interest are ∫dS' ln(R2) = 2a Q(a,r) = 4πa (B.6.10) ∫dV' ln(R2) = 2 !Syntax Error, Ir' dr' Q(r',r) = 4π !Syntax Error, Ir' dr' , (B.6.11) where we still have a dr' integral to carry out for the ∫dV' case. For r > a we know that r > r' since r' integrates over the inside of the wire. Then ∫dV' ln(R2) = 4π!Syntax Error, Ir' dr' ln(r) = 4π ln(r) [ a2/2 ] = 2πa2 ln(r) r > a (B.6.12) On the other hand, for r < a we have to break the integral into two parts : ∫dV' ln(R2) = 4π !Syntax Error, Ir' dr' ln(r) + 4π !Syntax Error, Ir' dr' ln(r') = 4πln(r) (r2/2) + 4π [(1/2)x2(lnx-1/2) ]|ar = 2π r2 ln(r) + 4π [(1/2)a2(lna-1/2) - (1/2)r2(lnr-1/2)] = 2π r2 ln(r) + 2π [a2(lna-1/2) - r2(lnr-1/2)] = 2π r2 ln(r) + 2π a2lna - πa2 - 2π r2lnr + πr2 = 2π a2lna + π(r2-a2) . r < a (B.6.13) The result then is the following, where below it we repeat the earlier surface integral result, ∫dV' ln(R2) = (B.6.14) ∫dS' ln(R2) = 4πa . (B.6.15) Now install these integrals into the Az expression (B.6.4) to get Az = - [I/(4π2a2)] {(μ1-μ2) (a/2) ∫dS' ln(R2) + μ2∫dV' ln(R2) } (B.6.4) = - [I/(4π2a2)] {(μ1-μ2) (a/2) 4πa + μ2} (B.6.16) Then write these out for the two separate regions : Az(r>a) = -[I/(4π2a2)] { μ1-μ2) (a/2) 4πa ln(r) + μ2 2πa2 ln(r) } = -[I/(4π2a2)] { μ1-μ2) 2πa2 ln(r) + μ2 2πa2 ln(r) } = -[I/(2π)] { μ1-μ2) ln(r) + μ2 ln(r) } = -[I/(2π)] { μ1ln(r) } (B.6.17) Az(r<a) = -[I/(4π2a2)] { μ1-μ2) (a/2) 4πa ln(a) + μ2( 2πa2ln(a) + π(r2-a2) ) } = -[I/(4πa2)] { μ1-μ2) 2a2 ln(a) + μ22a2ln(a) + μ2(r2-a2) ) } = -[I/(4πa2)] { μ1 2a2 ln(a) + μ2(r2-a2) ) } = -[I/(2π)] { μ1 ln(a) + μ2(r2-a2)/(2a2) ) } (B.6.18) At the boundary r = 0 both expressions give Az(r=a) = -[I/(2π)] { μ1ln(a) } . (B.6.19) The next step is to compute the magnetic field B = curl A given our function Az(r). We find from the usual cylindrical coordinates curl formula (B.5.2), B = curl A = - ∂rAz(r) . (B.6.20) First, for r > a one finds, -∂rAz(r) = ∂r[I/(2π) { μ1ln(r) }] = (I μ1/2π) (1/r) r > a (B.6.21) and then for r < a, -∂rAz(r) = I/(2π) ∂r { μ1 ln(a) + μ2(r2-a2)/(2a2) ) } = (I μ2/2π) μ2 r/a2 (B.6.22) Thus B = (I μ1/2π) (1/r) for r > a which is in the dielectric medium with μ1 B = (I μ2/2π) (r/a2) for r < a which is in the conductor medium with μ2 . (B.6.23) Using B = μH in each region we then get H = (I /2π) (1/r) for r > a => Hθ = (I/2πr) H = (I /2π) (r/a2) for r < a => Hθ = (Ir/2πa2) (B.6.24) Amazingly, these results agree with the elementary calculation results given in (B.4.1) and (B.4.2). The important point here is that these results only come out right when both the volume and surface currents on the round wire are included in the calculation. The elementary Ampere's Law calculation of Hθ is simple because in the Maxwell equation curl H = J (Ampere's Law), the field H does not "see" the surface magnetization current, so one can forget about it. Maple provides plots of Az from (B.6.17,18), Bθ from (B.6.23) and Hθ from (B.6.24) for this round wire situation. Parameters are set to I = 1, a = 2, μ1 = 2, μ2 = 3. Fig B.6 Az wanders down as ~ -ln(r) for large r, Bθ jumps at r = a while Hθ is continuous there. B.7 Reader Exercise: Repeat the above calculation of A using a 3D analysis This is a guided exercise with waypoints. The starting point is this: 2A(x) = -μ1Jm(x) -μ2Jc(x) // 3D wave equation from (1.5.4) with β= 0 A(x) = (1/4π) ∫dV' [μ1Jm(x') +μ2Jc(x')] (1/R) R = |x-x'| . // according to (H.1.8) (a) Insert Jm and Jc from (B.6.2) and (B.6.3) to get Az(x) = (1/4π) (1/2πa) I { (μ1-μ2) ∫dS' (1/ R) + (2/a) μ2∫dV' (1/ R) } where now we have true surface and volume integrals. Notice that we are using currents which do not vay in z, so here we are making the same "transmission line limit" approximation made in the 2D analysis. (b) Show these integrals may be written: ∫dS' (1/R) = ∫(adθ') !Syntax Error, Idz' (1/R) = a !Syntax Error, I dθ' !Syntax Error, Idz' (1/ ) ∫dV' (1/R) = ∫(r'dθ') ∫dr'!Syntax Error, Idz' (1/R) = !Syntax Error, I r'dr' !Syntax Error, I dθ' !Syntax Error, Idz' (1/ ) where s2 = (x-x')2+ (y-y')2 = R2 of the 2D problem. (c) Install a a large cutoff Λ to get !Syntax Error, Idz' (1/ ) → !Syntax Error, Idz' (1/ ) = -2ln(s/Λ) = - ln(R2/Λ2) . (d) Show then that Az(x) = = - (I/4π2a2) { (μ1-μ2) [ a2 Q(a,r) - πa2 ln(Λ2)] + 2 μ2 [ !Syntax Error, Ir' dr' Q(r',r) - ln(Λ2) π a2] } where Q(r',r) is the integral defined in (B.6.6). (e) Finally, show that the Az(x) obtained here is the same as that obtained in the 2D analysis apart from the appearance here of the following extra terms + (I/4π) ln(Λ2) (μ1+ μ2) . (f) Compare the 2D and 3D methods with an interpretion of these extra terms which of course have no effect on B = curl A. Physically why does the infinite term appear in the 3D analysis and not in the 2D analysis? Appendix C: DC Properties of a Wire C.1 The DC resistance of a wire The resistance per unit length R of a differential piece of wire of length dz and area dA 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). C.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, (Zsdz) = dV/I The quantity Zs is 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 // = R of (C.1.1) (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. Of course at DC (ω = 0), Zs = R of (C.1.1). So for DC, we have Zs = R, but for AC this is no longer true. C.3 The DC inductance of a round wire (a) Internal DC inductance of a round wire Inductance is a slightly harder problem. The energy density (joules/m3) stored in an electromagnetic field within a medium of negligible loss is given by u = (ED + BH)/2 [ Jackson p 259 Eq. (6.106) ] . Since μ and ε are assumed to be non-tensor in nature, u = (εE2+μH2)/2. In highly conductive wire, there is some current density J and then E = J/σ. Since σ is very large, E is very small, E2 can be neglected, and then u ≈ μH2/2 as the energy stored in a field per unit volume. Denoting by dU the field energy stored in a volume dV, so that u = dU/dV, one finds dU = (1/2) μH2 dV . (C.3.1) For any inductor of inductance L carrying current I and having potential difference V, we know that V = L dI/dt and power P = IV = L I dI/dt = d/dt[ (1/2)L I2 ]. The total energy U stored in the full field of the inductor must then be U = (1/2)L I2 in order to obtain P = dU/dt. If we consider a piece of uniform round wire of length dz, we can compute the total energy U stored in the field portion that lies inside the wire, and this piece of wire then has "internal inductance" Lidz where then Li is the internal inductance of the wire per unit length. So, Ui = (1/2) (Li dz) I2 . (C.3.2) Looking at (C.3.1), we see that Ui = ∫inside (1/2) μH2dV, so our next task is to determine H inside the wire. Due to the symmetry of the round wire, we know that H = H(r) . We may use the static Maxwell H curl equation (1.1.1) and the static integral form (1.1.23), curl H = J H ds = ∫S [J] dA "Ampere's Law' (1.1.23) Consider a transverse circular loop of radius r centered on the wire's center line. The enclosed current is ∫J•dA = J!Syntax Error, I2πrdr = (I/πa2) (πr2) = I (r/a)2 while H•ds = 2πrH(r). Thus one finds that I (r/a)2 = 2πr H(r) so H(r) = (I/2πa2) r r ≤ a . (C.3.3) From (C.3.1) the energy stored in a ring of volume dV =2πrdrdz is dU(r) = (1/2) μH2 dV = (1/2) μ (Ir/2πa2)22πrdrdz = μ (I2r2/4π2a4) πrdrdz = μ (I2r3/4πa4) drdz = (μdzI2/4πa4) r-3dr . Integrate this from r=0 to r=a to get the total magnetic energy stored inside the wire, Ui = (μdz I2/16π) = (1/2) [μ/8π] dz I2 . (C.3.4) Set this equal to (1/2)Lidz I2 in (C.3.2) to find that Li = μ/8π . (C.3.5) where parameter μ is for the medium of which the wire is composed. If this medium is non-magnetic, then μ = μ0. Since μ0 = 4π x 10-7 henry/m we find that (μ0/8π) = (1/2)x10-7 = 50 x 10-9 so the internal inductance per unit length of a round wire is given by Li = (μ/μ0) * 50 nH/m // nH = nanohenries (C.3.5a) which agrees with Matick p 97 (4-6) . Interestingly, this result is independent of the wire radius a. 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. (b) External DC inductance of a round wire What about the external inductance of a round wire? We can compute it by the same method. If we use a Stokesian loop of 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) From (C.3.1) the energy stored in a ring of volume dV = 2πrdrdz is dU(r) = (1/2) μH2dV = (1/2)μ(I/2πr)2[2πrdrdz] = μ(I2/4π2r2)πrdrdz = μ(I2/4πr)drdz = (μdzI2/4π) r-1 dr . Integrate this from r=a to some large radius r=R to get Ue = (μdz I2/4π) ln(R/a) = (1/2) [ ln ] dz I2 . (C.3.7) Set this equal to (1/2)Ledz I2 in (C.3.2) to find that, with μ = μ0, Le = ln . (C.3.8) Note that μ here is for the medium surrounding the wire, whereas the μ in (C.3.5) for Li is for the medium of the wire itself. 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 even in 1 cm worth of the wire. A "practical wire" is more like a loop of wire than an infinitely long wire. It is difficult to conjure up an experiment to test (C.3.8) even for a very long straight piece of wire without having some return path for the current to return to the driving "battery". For wire and dielectric both having μ0, Jackson shows (p 216-218) that the inductance per unit length of a loop of projected area A of radius-a wire is given by Le+ Li = (μ0/4π) [ ln(ξA/a2) + 1/2], where ξ is a near-unity factor which accounts for messy details of the calculation. The 1/2 term accounts for the internal inductance Li = μ0/8π as in (C.3.5). For a circular loop of radius R, one has A = πR2 and, if R >> a, ξ = 64/(πe4) ≈ .373. So, ln(ξA/a2) = ln(64 πR2/πe4a2) = 2 ln(8R/ae2) = 2 ln(8R/a) + 2 ln(e-2) = 2 ln(8R/a) - 4 and then that the total inductance per unit length is Le + Li = (μ0/4π) [2 ln(8R/a) - 4 + 1/2] = (μ0/2π) [ ln(8R/a) - 2 + 1/4] = (μ0/2π) [ ln(8R/a) -7/4] The total inductance of such a loop is then L = 2πR(μ0/2π) [ ln(8R/a) -7/4] = μ0R [ ln(8R/a) -7/4] in agreement with Jackson Problem 5.32 p 234. If one omits the internal inductance, the last factor is -2 instead of -7/4, and this result is seen in some sources. The point is that this is a finite result, even though the magnetic field of such a loop does extend to infinity. A loop of N turns gets an extra factor N2 because in effect current I → NI in (C.3.4) and (C.3.7), so the total field energy increases by factor N2. Suppose there were two parallel 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 from that the external inductance Le. In this case, the ln(R) divergence does not appear. In effect, the divergence cancels between the two wires, similar to the way opposite short segements of the circular wire cancel to give the finite result quoted above. Since we will be doing this computation by another means in the main text, we do not bother with the parallel wire 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 4. That is, Le is computed by considering the magnetic potential Az ( or W ) between the wires. C.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. (a) 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 first step. The wire is assumed aligned with the z axis. 1. Compute A(x,y) for an arbitrary wire at DC using formula (1.5.9) with β(ω=0) = 0 : A(x) = ∫ dV' [ μ1Jm (x') + μ2Jc(x')] R = |x - x'| ω = 0 (1.5.9) where A(x) means A(x,ω=0) and the same for the J's. Here μ2 is magnetic permeability of the wire, while μ1 is for the dielectric outside the wire. Jm represents the surface magnetization current (expressed as a volume density) that appears on the conductor surface in the case μ1 ≠ μ2. We presume that there is some reasonable prescribed conduction current density Jc inside the wire, and from this one can (with some effort) compute Jm and then one can compute A(x,ω) according to (1.5.9). Since (at DC) Jc is longitudinal along the wire (z direction), Jm is also longitudinal and A = Az . Thus, all equations below should really be written in terms of the z component only, but we continue to use the simpler vector notation. 2. Compute B = curl A. 3. The magnetic energy density is then dU/dV = (1/2)B2/μ. 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. 4. Set U = (1/2) L I2 to extract the appropriate inductance. We shall be mainly interested in the internal inductance. (b) The divergence problem and its resolution Let us look more closely at the integral for A(x,ω) stated above. At DC, both Jc and Jm will be constant in z, so we may write A(x) = ∫dx'dy' [ μ1Jm (x',y') + μ2Jc(x',y')] !Syntax Error, Idz' (C.4.1) where s = and R = At DC Jc will be uniform across the wire so we can write Jc(x',y') = Jc. An attempt to do the dz' integration reveals the 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. So we install a very large cutoff ± Λ/2 on the z integration, as if the wire were of length Λ instead of ∞. Then, !Syntax Error, I → !Syntax Error, I = 2 !Syntax Error, I = 2 ln[ z' + ] | Λ/20 = 2ln[Λ/2 + ≈ 2ln(Λ) - 2lns = -2ln(s/Λ) , where in the last step we assume that the transverse dimensions of the wire are much smaller than the cutoff Λ so that s << Λ for all s in the transverse integration. We are then left with A(x) = ∫dx'dy' [ μ1Jm (x',y') + μ2Jc] { 2ln(Λ) - 2lns } = - ∫dx'dy' [ μ1Jm (x',y') + μ2Jc] { lns - lnΛ} (C.4.2) = - ∫dx'dy' [ μ1Jm (x',y') + μ2Jc] ln[ ] + a constant where the constant is proportional to ln(Λ). When we compute B = curl A , this constant has no effect on B so we just ignore the constant, setting it to 0 for the purpose of computing B. However, in order to keep track of dimensions, we shall keep the constant Λ around, perhaps later setting it to 1 since its value does not matter. So: A(x) = ∫dx'dy' [ μ1Jm (x',y') + μ2Jc] ln(Λ/s) (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. Comment: We could have arrived at (C.4.3) by starting with (1.5.4) treated as a 2D problem since none of the fields depends on z (recall β2 = 0 since we are doing DC treatment ), - (22D )A = [ μ1Jm(i) + μ2J2] . (1.5.4) The solution to this equation is (C.4.3) "by inspection" since the 2D free-space propagator is ln (1/R) where R = s = . This concept is discussed in Appendix I. See in particular (I.1.8). In the case that μ1 = μ2 where Jm = 0, the equation (C.4.3) simplifies to A(x) = - ∫dx'dy' [μ2Jc] ln(s/Λ) = - μ1 Jc∫dx'dy' ln(s/Λ) // μ1= μ2 which involves a purely geometric transverse integral over the wire cross section. For a round wire still with μ1 = μ2 the above integration becomes Az(r) = - !Syntax Error, Ir' dr'!Syntax Error, Idθ' ln (/Λ ) // μ1= μ2 (C.4.4) Although it looks like a mess, these are standard integrals, and the H field results (C.3.3) and (C.3.6) of the previous section are easily duplicated using H = (1/μ1)B = (1/μ1) curl A. The following exercise shows how this works. Need reference to somewhere else on this!! Reader Exercise: In the expression (C.4.4) the -ln(Λ) term creates a constant in Az(r) which we ignore as noted above since we only care about B = curl A, so set Λ = 1. Then use GR7 4.224.9 to show that !Syntax Error, Idθ' ln ( ) = π ln [ ] . The denominator 2 can be ignored since it too just creates a constant. Then show that for r > a, Az(r) = - !Syntax Error, Ir' dr' π ln [ ] = - (μJz/2) a2 ln(r) . Then compute B using the curl in cylindrical coordinates to get B = - ∂rAz(r) = (μJza2/2)r-1 = B(r) where B(r) = (μJza2/2)r-1 . Finally, since I = Jz(πa2), show that one obtains the known result for field outside a round wire, H(r) = (I/2π) r-1 . (C.3.6) For r < a, break up the integral into two parts to obtain the internal results Az(r) = - (μJz/4) r2 => H(r) = (I/2πa2) r r ≤ a (C.3.3) The reader may notice a similarity between (C.4.3) 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). // Remove this paragraph because reader probably has not read Chapter 4 when looking at this appendix. (c) 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 in order to compute B in Step (b) of the program outlined above. For example, in Cartesian coordinates, ∂x R-1 = -R-2∂xR = -R-2∂x = -R-2(1/2)R-1 2(x-x') = - (x-x') R-3 so the x derivative of (C.4.1) becomes ∂x Az(x) = - ∫∫dx'dy'[ μ1Jmz (x',y') + μ2Jcz] (x-x') !Syntax Error, Idz' . (C.4.5) The z' integration is now a convergent integral equal to [ GR7 3.252.7 ], !Syntax Error, Idz' = 2/s2 s = . (C.4.6) so that -By = ∂x Az(x) = - ∫∫dx'dy'[ μ1Jmz (x',y') + μ2Jcz] Bx = ∂y Az(x) = - ∫∫dx'dy'[ μ1Jmz (x',y') + μ2Jcz] (C.4.7) Now the need for a cutoff Λ is completely avoided. In the special case μ1 = μ2 so Jmz = 0 we have -By = ∂x Az(x) = - ∫∫dx'dy' Bx = ∂y Az(x) = - ∫∫dx'dy' // μ1 = μ2 (C.4.8) and the fields are then determined by these purely geometric integrals (see example below) In any event, we have outlined a method by which both the internal and external DC inductance of any wire can be calculated. For a round wire we know from (C.3.5) above that Li = μ /8π . For a rectangular conductor of dimensions a and b, the result can be obtained using (C.4.7). The result must have the form Li = (μ/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. Example: Magnetic Field of a Rectangular Wire As one might expect, the B field for a rectangular conductor is not particularly simple. Here is a Maple calculation of the double integral appearing in the ∂xAz expression of (C.4.7) where we have made a clumsy attempt to minimize the number of terms. We assume μ1 At least the result is expressible in closed form. Using By = -∂xAz as above and Bx = ∂yAz similarly, we can plot the magnetic field lines for a = 1, b = 4 both inside and outside the conductor shown in red: Far away, the field is the same as that of a round wire carrying the same total current. It would take some work to integrate the energy in this field to obtain Li and Le for such a conductor, but it could be done. 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 : The General Electric Field Inside a Round Wire In this Appendix we present a somewhat lengthy calculation of the electric field (and therefore the current) inside a round wire without assuming that such fields are symmetrical about the axis. This wire may be regarded as one conductor of a transmission line down which a wave is propagating. The conclusions one can draw from the analytic solution given below are supportive of the general discussion elsewhere in this monograph: 1. The Er and Eφ field components are very small. In fact, they are smaller than Ez by the factor (a/λ), where a = wire radius, and λ = wavelength of the wave on a transmission line containing the wire. This fraction is always assumed small in any analysis of a transmission line. 2. It is possible to maintain the condition Eφ = 0 on the wire surface. This condition is consistent with the idea that the wire surface is an electrical equipotential at any constant z. 3. We get an explicit formula for the surface impedance. We find that it is non-uniform around the boundary of the wire cross section, and that Jz is non-uniform inside the wire. This result is true even in the DC limit of a transmission line. 4. We suggest a method for computing the E fields in the low frequency limit of a transmission line consisting of round wires. First, solve the "electrostatic" problem illustrated in Chapter 6. Knowing the potential φ, compute the radial electric field at the surface of the conductors. From this compute the surface charge density n(φ) as a function of azimuth around the wire Then, as described below, compute the moments Nm (or ηm) of this charge distribution, and use the formulas below to find the electric field. This is a highly technical appendix. The reader is invited to inspect the boxed results and comments in Section D.4. D.1 The General Method and Solution for Ez The starting point for the calculation is the Helmholtz equation (1.6.2) for the E field inside the wire, ( 2 + β2 ) E(r,φ z, t) = 0 . (D.1.1) Here we use cylindrical coordinates (r,φ,z), as appropriate for a straight round wire. The first step is to expose the assumed t and z dependence, and in doing so, define the field E(r,φ) : E(r,φz,t) = ej(ωt-βz) E(r,φ) . (D.1.2) Here, symbol βd stands for the value that parameter β = ω takes in the dielectric. We reserve the symbol β with no subscript to mean β inside the metal of our wire. The form shown in (D.1.2) is a simple wave traveling down a transmission line in the +z direction. We assume that one of the conductors of this transmission line is our round wire, while the other conductor is (other conductors are) unspecified. One can regard (D.1.2) as an assumed variable-separated form for a solution, an "ansatz". Our second ansatz is that Eφ(r=a,φ) = 0 which means at any z = constant slice, the circular wire perimeter is an equipotential. If a consistent solution to the field equations can be found with these assumptions, they are justified de facto. Convention Comment: Section 1.6 discusses the Fourier Transform (1.6.7) and (1.6.8) where e+jωt appears in the expansion formula (1.6.8). For E(x,ω) = 2πδ(ω-ω1) one gets E(x,t) = e+jωt and then the form of a wave solution is e+j(ωt-kz) with the + sign associated with ωt. In general EE people like to assume time dependence of the form e+jωt (and they like j in place of i for ). The Fourier Transform is of course valid with the other sign choice for the two exponentials, and for that other sign choice one would have E(x,ω) = 2πδ(ω-ω1) => E(x,t) = e-jωt and one would think of a wave as e-j(ωt-kz) = e+j(kz-ωt). This sign convention is common in many physics texts, but in this document we use the e+j(ωt-kz) convention usually used in EE texts. It is just a convention choice and, as in (1.6.6), it just effects the sign of the imaginary physical field under consideration. If one thinks of the physical field under consideration as Re{ E(x,t)}, the sign convention choice makes no difference at all. (a) Partial Wave Expansions The next step is to do a "partial wave expansion" (that is, a complex Fourier series expansion) of E(r,φ) in terms of "azimuthal harmonics" eimφ, so that the variable φ is replaced with the partial wave index m: E(r,φ) =!Syntax Error, I E(r,m) ejmφ E(r,m) = (1/2π) !Syntax Error, Idφ E(r,φ) e-jmφ (D.1.3) Since the usual complex part has been extracted in (D.1.2), we assume that E(r,φ) is real so the right equation above says that E(r,-m) = E(r,m)* . We can then re-express the above as, E(r,φ) = E(r,0) +!Syntax Error, I[ E(r,m)ejmφ + E(r,-m)e-jmφ] = E(r,0) +!Syntax Error, I[ E(r,m)ejmφ + [E(r,m)ejmφ]*] = E(r,0) + 2!Syntax Error, IRe{ E(r,m)ejmφ } = E(r,0) + 2!Syntax Error, I[ Re(E) + j Im(E)] [ cos(mφ) + j sin(mφ)] = E(r,0) + 2!Syntax Error, I[Re(E)cos(mφ) - Im(E) sin(mφ)] + 2j!Syntax Error, I[ Im(E)cos(mφ) + Re(E) sin(mφ)] Since E(r,φ) is assumed real, the second sum must vanish for all r and φ, giving this final result, E(r,φ) = E(r,0) + 2!Syntax Error, I[Re{E(r,m)}cos(mφ) - Im{E(r,m)} sin(mφ)] . (D.1.4) In analogy with (D.1.2) and the discussion of Section 1.6 regarding complex functions, we define a complex surface (r=a) charge density n(φ,z,t) which has the following ansatz variable-separated form, n(φ,z,t) = ej(ωt-βz) n(φ) (D.1.5) where n(φ) is real. We then expand n(φ) as in (D.1.3) and (D.1.4), n(φ) = !Syntax Error, I Nm ejmφ Nm = (1/2π) !Syntax Error, Idφ n(φ) e-jmφ . (D.1.6) n(φ) = N0 + 2!Syntax Error, I[Re{Nm}cos(mφ) - Im{Nm} sin(mφ)] (D.1.7) Since n(φ) is real, Nm = N-m*. These Nm are the "moments" of the surface charge distribution, and their complex nature keeps track of cos(mφ versus sin(mφ) components. (b) The Vector Laplacian and the conversion of equations from φ space to m space Given the following cylindrical-coordinates field components, E(r,φz,t) = Er(r,φz,t) + Eφ(r,φz,t) + Ez(r,φz,t) (D.1.8) we may write out our ansatz wave form (D.1.2) and the Helmholtz equation (D.1.1) in more detail, Er(r,φz,t) = ej(ωt-βz) Er(r,φ) . [2E(r,φ,z,t)]r + β2 Er(r,φ,z,t) = 0 Eφ(r,φz,t) = ej(ωt-βz) Eφ(r,φ) . [2E(r,φ,z,t)]φ + β2 Eφ(r,φ,z,t) = 0 Ez(r,φz,t) = ej(ωt-βz) Ez(r,φ) . [2E(r,φ,z,t)]z + β2 Ez(r,φ,z,t) = 0 . (D.1.9) In Cartesian coordinates, it happens that [2E]i = 2(Ei), but this is not generally true for curvilinear coordinates. In cylindrical coordinates, it is true for the z coordinate only. The operator 2 when applied to a vector field is called the "vector Laplacian" and it is very different from the scalar Laplacian, so much so that some authors replace [2E] by [E] which is defined in this manner [E] ≡ [2E] ≡ grad(div E) – curl (curl E) (D.1.10) whereas 2φ ≡ div(grad φ) = (φ) . (D.1.11) It is the vector Laplacian that appears in our Helmholtz equation (D.1.1). For cylindrical coordinates it turns out that, (2E)r = 2Er - (2/r2) ∂φEφ - (1/r2) Er (2E)φ = 2Eφ + (2/r2) ∂φEr - (1/r2) Eφ (2E)z = 2Ez (D.1.12) where 2 = (1/r)∂r(r∂r) + (1/r2)∂φ2 + ∂z2 = ∂r2 + (1/r)∂r + (1/r2)∂φ2 + ∂z2 . (D.1.13) See for example Morse and Feshbach Vol I p 116, Moon and Spencer p 139, or do a web search on "vector Laplacian"; the author's Tensor Analysis document, Sections 13, 14 and 15, derives these results for arbitrary coordinate systems. Here is a summary of vector differential operators in cylindrical coordinates taken from Morse and Feshbach above, where the last line corresponds to the above discussion: (D.1.14) Using the ansatz form (D.1.2) and partial wave expansions of the form (D.1.3) or (D.1.6), it is a simple matter to convert an equation involving components like Ei(r,φz,t) or n(r,φz,t) to a simpler equation involving components like Ei(r,m) and n(r,m). We shall give an example, then state a simple set of rules that does the conversion automatically. Example 1: The Ez Helmholtz Equation (D.1.9): [2E]z + β2 Ez = 0 Using (D.1.12) and (D.1.13), this may be written, [∂r2 + (1/r) ∂r + (1/r2) ∂φ2 + ∂z2 + β2 ] ej(ωt-βz)Ez(r,φ) = 0 . (1) Inserting the expansion (D.1.3) for Ez(r,φ) and moving the m sum to the left gives !Syntax Error, I [∂r2 + (1/r) ∂r + (1/r2) ∂φ2 + ∂z2 + β2 ] ej(ωt-βz)Ez(r,m) ejmφ = 0 . (2) We can then make the obvious replacements ∂z = -jβd and ∂φ = +jm to get, !Syntax Error, I { [∂r2 + (1/r) ∂r -m2 (1/r2) – βd2 + β2 ] ej(ωt-βz)Ez(r,m) } ejmφ = 0 . (3) Due to the completeness of functions ejmφ on the interval (-π.π), we conclude that { } = 0, or [∂r2 + (1/r) ∂r -m2 (1/r2) – βd2 + β2 ] ej(ωt-βz)Ez(r,m) = 0 . (4) In other words, one applies !Syntax Error, Idφ e-jm'φ to both sides of (3), uses the completeness property !Syntax Error, Idφ ej(m-m')φ = 2π δm,m' , (5) and then change m' to m to get (4). Next, multiply both sides of (4) by r2 e-j(ωt-βz) to get, [r2∂r2 + r ∂r - m2 +r2( β2- βd2)] Ez(r,m) = 0 We may then write these rules for converting equation (1) to equation (6) Conversion Rules: ∂z → -jβd (D.1.15) ∂φ → +jm ∂t→ +jω f(r,φ,z,t ) → f(r,m) We can now practice with these rules to convert various equations of interest . Here a field with unstated arguments has the full arguments (r,φ,z,t). Example 2: The Er Helmholtz Equation (D.1.9): [2E]r + β2 Er = 0 . Using (D.1.12) we find, 2(Er) - (2/r2) ∂φEφ - (1/r2) Er + β2Er = 0 [∂r2 + (1/r)∂r + (1/r2)∂φ2 + ∂z2] Er - (2/r2) ∂φEφ - (1/r2) Er + β2Er = 0 [∂r2 + (1/r)∂r + (1/r2)∂φ2 + ∂z2 - (1/r2) + β2] Er - (2/r2) ∂φEφ = 0 Now apply the conversion rules to get [∂r2 + (1/r)∂r + (1/r2) (-m2) - βd2 - (1/r2) + β2] Er(r,m) - (2/r2) jm Eφ(r,m) = 0 Example 3: The Eφ Helmholtz Equation (D.1.9): [2E]φ + β2 Eφ = 0 . Using (D.1.12) we find, 2(Eφ) + (2/r2) ∂φEr - (1/r2) Eφ + β2Eφ = 0 [∂r2 + (1/r)∂r + (1/r2)∂φ2 + ∂z2] Eφ + (2/r2) ∂φEr - (1/r2) Eφ + β2Eφ = 0 [∂r2 + (1/r)∂r + (1/r2)∂φ2 + ∂z2 - (1/r2) + β2] Eφ + (2/r2) ∂φEr = 0 Now apply the conversion rules to get [∂r2 + (1/r)∂r + (1/r2)(-m2) + (-βd2) - (1/r2) + β2] Eφ(r,m) + (2/r2) jm Er(r,m) = 0 Example 4: The div E = 0 . Using (D.1.14) we write div E = 0 as ∂r (r Er) + ∂φEφ + r ∂zEz = 0 Applying the conversion rules gives ∂r [r Er(r,m)] + jmEφ(r,m) + r (-jβd)Ez(r,m) = 0 [1 + r∂r ] Er(r,m) + jmEφ(r,m) + r (-jβd)Ez(r,m) = 0 Summary of Examples: [2E]z + β2 Ez = 0 : [r2∂r2 + r ∂r - m2 + r2 ( β2- βd2)] Ez(r,m) = 0 (D.1.16) [2E]r + β2 Er = 0 : [r2∂r2 + r∂r - (m2+1) + r2(β2-βd2)] Er(r,m) - 2jm Eφ(r,m) = 0 (D.1.17) [2E]φ + β2 Eφ = 0 : [r2∂r2 + r∂r - (m2+1) + r2(β2-βd2)] Eφ(r,m) + 2jmEr(r,m) = 0 (D.1.18) div E = 0 : ∂r [r Er(r,m)] + jmEφ(r,m) -jβd r Ez(r,m) = 0 (D.1.19) Comment: The scalar Helmholtz equation is fully separable in cylindrical coordinates (see M&S p 15) and the "atomic forms" or "harmonics" are of the generic form ejmφ x ejk'z x Jm(jkr). In solving a problem involving such a scalar Helmholtz equation, one can expand the solution onto these harmonics with some coefficient for each term in the expansion, and then evaluate the expansion on some boundary to obtain those coefficients. We refer to this solution method as "the method of Smythian forms", admittedly an obscure phrase. In this situation, ejmφ is part of the "harmonic" functional form. In contrast, the vector Helmholtz equation is NOT separable in cylindrical coordinates (see M&S p 139), it is not even "R-separable", so there are no associated "harmonics" as there are with the scalar Helmholtz equation. Nevertheless, the functions ejmφ form a complete set for φ in (-π,π) and our expansion of each Ei onto these ejmφ is certainly allowed, even though these ejmφ are not part of any associated harmonics for the vector Helmholtz equation. As seen below, when we do this, we end up with ODE's whose solutions are Bessel functions of a strangely phased argument. (c) The Charge Pumping Boundary Condition The reason we are interested in surface charge n(φ) of (D.1.5) is that it acts as a driving source of the radial electric field in the wire. Recall the equation of continuity (1.1.8) converted to the ω domain divJ = -jωρ ∫S dSJ = -jω ∫V dV ρ . // divergence theorem (D.1.20) When applied to a thin box of radial area dS straddling the wire surface, one finds that ∫S dSJ = -Jr(r=a,φ)dS and ∫V dV ρ = n(φ) dS so that Jr(r=a,φ) = jω n(φ) . (D.1.21) We assume that there is no current outside the wire to get this result. Since J = σE, this is really a boundary condition on the radial electric field, Er(r=a,φ) = (jω/σ) n(φ) . (D.1.22) We convert this to m-space using the conversion rules (D.1.15) to trivially obtain Er(r=a,m) = (jω/σ) Nm . (D.1.23) Thus, the radial electric field must have a certain value at the r=a boundary in each partial wave. And the value it must have is determined by the moment of the charge distribution. By way of interpretation, the surface charge of a transmission line is "pumped" by the radial current in the wire. Since divJ = 0 inside the wire, this radial current is accompanied by the usual longitudinal current one expects to find inside the conductors of a transmission line. ok to here (d) The Ez Solution As shown in (D.1.16), the Helmholtz equation for Ez(r,m) is [r2∂r2 + r ∂r + (r2 β'2 - m2)] Ez(r,m) = 0 (D.1.24) where β'2 = β2 - βd2 . (D.1.25) In a conductor like copper, β is huge compared to the dielectric βd, so we could ignore the distinction between β and β'. Setting x = β'r we find ∂r = β'∂x and then r∂r = x∂x and so on so that (D.1.24) reads [x2∂x2 + x ∂x + (x2 - m2)] Ez(x/β',m) = 0 . x = β'r (D.1.26) This is Bessel's equation [ Spiegel 24.1] and the solution subject to the condition that Ez be finite at r = 0 is Ez(x/β',m) = Czm Jm(x) or Ez(r,m) = Czm Jm(β'r) (D.1.27) where Czm is an arbitrary constant for each partial wave m. Equation (D.1.27) is consistent with (2.1.22). In Section 2.1 we dealt only with the m=0 partial wave, which embodies the symmetrical part of the problem. Also, we assumed constant z behavior so βd = 0 and the question of β versus β' never arose. D.2 The Solutions for Er and Eφ (a) The Er Solution As shown in (D.1.17), the Helmholtz equation for Er(r,m) is [r2∂r2 + r∂r - (m2+1) + r2β'2] Er(r,m) - 2jm Eφ(r,m) = 0 (D.2.1) while the div E = 0 condition was stated in (D.1.19) as [1 + r∂r ] Er(r,m) + jmEφ(r,m) + r (-jβd)Ez(r,m) = 0 or -jmEφ(r,m) = [1 + r∂r ] Er(r,m) - r (jβd)Ez(r,m) . (D.2.2) Inserting this into (D.2.1) gives [r2∂r2 + r∂r - (m2+1) + r2 β'2] Er(r,m) + [2 + 2r∂r ] Er(r,m) - 2r (jβd)Ez(r,m) = 0 or [r2∂r2 + 3r∂r + (1-m2) + r2 β'2)] Er(r,m) = 2r (jβd)Ez(r,m) . (D.2.3) Inserting solution (D.1.27) for Ez(r,m) this becomes [r2∂r2 + 3r∂r + (1-m2) + r2 β'2)] Er(r,m) = 2r (jβd) Czm Jm(β'r) or [r2∂r2 + 3r∂r + (1-m2) + r2 β'2)] Er(r,m) = 2j (βd/β') Czm β' r Jm(β'r) or [r2∂r2 + 3r∂r + (1-m2) + r2 β'2)] Er(r,m) = Km β'r Jm(β'r) (D.2.4) where Km ≡ 2j (βd/β') Czm . (D.2.5) In order to get the left side of (D.2.4) into something recognizable, we define Er(r,m) = x-1 fm(x) (D.2.6) where x is a dimensionless radial variable which will play a major role in the following, x ≡ β'r and xa ≡ β' a . (D.2.7) Then (D.2.4) becomes [x2∂x2 + 3x∂x + (1-m2) + x2] { x-1 fm(x)} = 2j (βd/β') xCzm Jm(x) or x [x2∂x2 + 3x∂x + (1-m2) + x2] { x-1 fm(x)} = K x2 Jm(x) . (D.2.8) Ever eager, Maple expands the left side of (D.2.7), so that (D.2.8) becomes [ x2 ∂x2 + x ∂x + (x2-m2)] fm(x) = Km x2 Jm(x) . (D.2.9) The left side of (D.2.9) is the normal Bessel operator [ Spiegel 24.1] , but the equation is also driven by a power times a Bessel function. The solution to the equation is the homogeneous solution of the Bessel equation plus the particular solution which is the response to the driving function on the right hand side. The homogeneous solution is the usual linear combination of Jm(x) and Ym(x), but we must reject Ym(x) since it blows up at x=0 and thereby causes the field Er to be singular, which it cannot be, smack in the middle of a wire. The particular solution is not very obvious and required some hunting to find. It is this fm(x)particular = (1/2) Km [ x Jm+1(x) ] . (D.2.10) a Maple confirms, continuing the above code, Therefore, we now have this full solution for fm(x) fm(x) = fm(x)particular + fm(x)homogeneous = (1/2) Km [ x Jm+1(x) ] + am Jm(x) and then from (D.2.6) the full solution Er , Er(r,m) = am x-1 Jm(x) + Jm+1(x) . (D.2.11) For each value of m, there are two as-yet undetermined constants, am and (Km/2). However, looking at (D.2.11), we see that, since J0(x) ≈ 1 for small x, we must have a0 = 0 (D.2.12) to keep Er finite at r = 0. We shall obtain expressions for am and (Km/2) below. (b) The Eφ Solution Recall (D.2.2) , jmEφ(r,m) = -∂r[rEr(r,m)] + r (jβd)Ez(r,m) . (D.2.2) We can then insert our known Ez and Er to get Eφ Ez(r,m) = Czm Jm(x) (D.1.27) Er(r,m) = am x-1 Jm(x) + Jm+1(x) . (D.2.11) so (D.22) becomes the following jmEφ(r,m) = -∂r[r{ am x-1 Jm(x) + Jm+1(x)}] + r (jβd) Czm Jm(x) jmEφ(r,m) = -∂x[x{ am x-1 Jm(x) + Jm+1(x)}] + x Jm(x) // Km ≡ 2j (βd/β') Czm jmEφ(r,m) = -∂x[am Jm(x) + x Jm+1(x)] + x Jm(x) jmEφ(r,m) = -am Jm'(x) - Jm+1(x) - x Jm+1'(x) + x Jm(x) jmEφ(r,m) = - am Jm'(x) + [ - Jm+1(x) - x Jm+1'(x) + x Jm(x) ] (D.2.13) At this point we invoke the recurrence relations AS 10.6.2 to write Jm+1' = Jm - (m+1)x-1Jm+1 first relation with ν = m+1 Jm' = -Jm+1 + (m/x)Jm second relation with ν = m (D.2.14) Insert these into (D.2.18) to get jmEφ(r,m) = - am Jm' + [ - x Jm+1' - Jm+1 + xJm] = - am {-Jm+1 + (m/x)Jm } + [ - x { Jm - (m+1)x-1Jm+1} - Jm+1 + xJm] = am Jm+1 - am (m/x)Jm + [ - x Jm + (m+1) Jm+1 - Jm+1 + xJm] = am Jm+1 - am (m/x)Jm + [ m Jm+1] = - am (m/x)Jm + ( m + am ) Jm+1 Dividing by m then gives the final solution: jEφ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) x = β'r . (D.2.15) What about Eφ(r,m=0)? We have already noted that a0 = 0. For m=0, (D.2.15) is invalid because we divided (D.2.13) by m. In fact, we really know nothing at all about the field Eφ(r,m=0) from our solution method because Eφ is not even present in our starting equation (D.2.2) when m = 0, -jmEφ(r,m) = [1 + r∂r ] Er(r,m) - r (jβd)Ez(r,m) . (D.2.2) However, we can use the Eφ Helmholtz equation (D.1.18) to learn more about Eφ(r,m) with m = 0, [r2∂r2 + r∂r - (m2+1) + r2β'2] Eφ(r,m) + 2jmEr(r,m) = 0 (D.1.18) With m = 0 we get [r2∂r2 + r∂r - 1 + r2β'2] Eφ(r,0) = 0 or [x2∂x2 + x∂x - 1 + x2] Eφ(r,0) = 0 x = β' r (D.2.16) Since [..] is the ν = 1 Bessel operator and since Eφ must be finite at r = 0, we find that Eφ(r,0) = Cφ0 J1(β'r) (D.2.17) where Cφ0 is some constant. (c) Application of the Boundary Conditions We have two boundary conditions to impose: Er(r=a,m) = (jω/σ) Nm (D.2.18) Eφ(r=a,m) = 0 (D.2.19) The first is the radial charge pumping condition shown in (D.1.23) above, while the second is our ansatz assumed earlier that any z = constant circle on the wire surface be an equipotential. These two boundary conditions serve to determine the two constants am and Km, though a bit of algebra is required. The first step is to use (D.2.11) for Er and (D.2.15) for Eφ to write out the two boundary conditions as am xa-1 Jm(xa) + Jm+1(xa) = (jω/σ) Nm (1) - am xa-1 Jm(xa) + ( + ) Jm+1(xa) = 0 . (2) Addition and subtraction of these equations gives two new equations, Jm+1(xa) + ( + ) Jm+1(xa) = (jω/σ) Nm (3) 2 am xa-1 Jm(xa) - Jm+1(xa) = (jω/σ) Nm (4) the second of which may be immediately solved for am = (jω/σ) Nm . (5) Using the recursion relation 2m x-1 Jm = [Jm+1 + Jm-1] and using (5) for am , equation (2) may be solved to get ( + ) = (jω/2σ) Nm [ + ] (6) Finally, subtracting (5) from (6) we find = (jω/2σ) Nm [ – ] (7) For m = 0 we use (D.2.17) for Eφ and the two boundary conditions are then J1(xa) = (jω/σ) N0 (D.2.20) Cφ0 J1(xa) = 0 => Cφ0 = 0 (D.2.21) The K0 result is compatible with (7) since J-1= - J1 and the (9) requires that Cφ0 = 0, barring some strange radius which happens to be a zero of the J1(x) function. To summarize, am = 2m(jω/2σ) Nm m≥0 = (jω/2σ) Nm [ – ] m≥ 0 (+ ) = (jω/2σ) Nm { + } m>0 Cφ0 = 0 m = 0 (D.2.22) We have now completely solved the problem. To review, here are the three fields gathered together in one spot: Ez(r,m) = -j(β'/βd) Jm(x) (D.1.27) and (D.2.5) Er(r,m) = am x-1 Jm(x) + Jm+1(x) (D.2.11) (D.2.23) jEφ(r,m) = - am x-1 Jm(x) + (+ ) Jm+1(x) m > 0 (D.2.15) jEφ(r,0) = 0 m = 0 (D.2.17) and Cφ0 = 0 D.3 What about the Eφ Helmholtz Equation ? Except to find Eφ(r,0) we have completely ignored the Eφ Helmholtz equation until now. It is reasonable to wonder whether the solution fields we have found above in fact solve this equation? A related question is whether the three Helmholtz equations and div E = 0 are four independent equations, or is one of the three Helmholtz equations dependent? In Cartesian coordinates suppose we know that (implied sums on repeated indices) (∂j∂j + β2) E1= 0 (∂j∂j + β2) E2= 0 ∂iEi = 0 . (D.3.1) Can we show that (∂j∂j + β2) E3 = 0 so this third Helmholtz equation is dependent? If we apply the operator (∂j∂j + β2) to the last equation above we get (∂j∂j + β2) ∂iEi = 0 or ∂i (∂j∂j + β2) Ei = 0 or ∂1 (∂j∂j + β2) E1 + ∂2 (∂j∂j + β2) E2 + ∂3 (∂j∂j + β2) E3 = 0 or ∂3 [(∂j∂j + β2) E3] = 0 (D.3.2) This does not prove that (∂j∂j + β2) E3 = 0, but it certainly suggests it. Perhaps (∂j∂j + β2) E3 = f(x1,x2) ≠ 0, but other considerations might force f = 0, such as the nature of curvilinear coordinates and boundary conditions. Rather than pursue this question further, we will now directly show that our solutions do in fact satisfy the Eφ Helmholtz equation. The Eφ Helmholtz equation (D.1.18) is this, expressed in terms of x = β'r, [x2∂x2 + x∂x - (m2+1) + x2)] Eφ(r,m) = – 2jmEr(r,m) = 0 (D.1.18) (D.3.3) and we found these solutions for Er and Eφ , Er(r,m) = am x-1 Jm(x) + Jm+1(x) . (D.2.11) jEφ(r,m) = - am x-1 Jm(x) + ( + ) Jm+1(x) x = β'r . (D.2.15) Treating (D.2.22) as LHS = RHS we then have LHS = [x2∂x2 + x∂x - (m2+1) + x2] {- am x-1 Jm(x) + ( + ) Jm+1(x) } RHS = 2m{ am x-1 Jm(x) + Jm+1(x)} . (D.3.4) In order to have LHS = RHS, it must be true for arbitrary am and Km . Thus, we want to show that each of the following equations is valid, [x2∂x2 + x∂x - (m2+1) + x2] ( Jm+1(x) ) = 2m Jm+1(x) [x2∂x2 + x∂x - (m2+1) + x2] {- am x-1 Jm(x) + Jm+1(x) } = 2m am { x-1 Jm(x) } which are the same as [x2∂x2 + x∂x - (m2+1) + x2] (Jm+1(x) ) = 2m Jm+1(x) (D.3.5) [x2∂x2 + x∂x - (m2+1) + x2] {- x-1 Jm(x) + Jm+1(x) } = 2m { x-1 Jm(x) } . (D.3.6) Equation (D.3.5) may be written [x2∂x2 + x∂x - (m2+1) + x2 - 2m] Jm+1(x) = 0 or [x2∂x2 + x∂x + x2- (m+1)2] Jm+1(x) = 0 . But this is Bessel's equation for ν = m+1, thus (D.3.5) really is valid. As for (D.3.6), we leave it to trusty Maple : Thus (D.3.6) is also valid We conclude that the our round wire E field solutions satisfy the Eφ Helmholtz equation as well as the other two Helmholtz equations and the div E = 0 equation. D.4 Statement of the Results Although we have a complete solution to our problem, it is useful to normalize the results to some well defined absolute scale. The first step in doing this is to compute the total current I in the wire using Jz = σEz. The purpose here is relate the current I to the moment N0 of the surface charge distribution. Many steps are required to ferret out this relationship: I = !Syntax Error, Idφ !Syntax Error, Ir dr Jz(r,φ) = !Syntax Error, Idφ !Syntax Error, Ir dr { σ !Syntax Error, I Ez(r,m) ejmφ } // (D.1.3) = σ !Syntax Error, I !Syntax Error, Ir dr Ez(r,m) !Syntax Error, Idφ ejmφ = 2π σ!Syntax Error, Ir dr Ez(r,0) = 2π σ!Syntax Error, Ir dr {-j(β'/βd) J0(x) } // (D.2.21) x = β'r = -j(β'/βd) 2π σ !Syntax Error, Ir dr J0(x) // x = β'r so xdx = β'2 rdr = -j(β'βd)-1 2πσ [!Syntax Error, Idx x J0(x)] = -j(β'βd)-1 2πσ [ xa J1(xa) ] = -j(β'βd)-1 2πσ {(jω/σ) N0 / J1(xa)} [ xa J1(xa) ] // (D.2.20) = (β'βd)-1 2πω N0 xa = (β'βd)-1 2πω N0 β'a = 2πω (a/βd) N0 so that N0 = (βd/2πωa) I . (D.4.1) At this point it is convenient to introduce the DC resistance per unit length of our wire, Rdc = (D.4.2) along with a new symbol to indicate the relative surface charge moment, ηm ≡ (D.4.3) It follows from (D.4.1) that the normalization factor appearing in (D.2.22) may be written as (jω/2σ) Nm = (jω/2σ) N0 = (jω/2σ) ηm [(βd/2πωa) I ] = (j/4) ηm (aβd/σπa2) I = (j/4) (aβd) ηm I Rdc . (D.4.5) We may now construct the final form for our solutions from (D.2.23) and (D.2.22) for m > 0: Ez(r,m) = -j(β'/βd) Jm(x) = -j(β'/βd) (jω/2σ) Nm [ – ] Jm(x) = -j(β'/βd) [(j/4) (aβd) ηm I Rdc] [ – ] Jm(x) = (1/4) ηm I Rdc [- ] Er(r,m) = am x-1 Jm(x) + Jm+1(x) = [(jω/2σ) Nm] { 2m x-1 Jm(x) + [ – ] Jm+1(x) } = (j/4) (aβd) ηm I Rdc { + - } jEφ(r,m) = - am x-1 Jm(x) + (+ ) Jm+1(x) = [(jω/2σ) Nm] { - 2m x-1 + [ + ] Jm+1(x) = (j/4) (aβd) ηm I Rdc { - + [ + ] } We summarize these results in a box. The m=0 results are all obtainable from the general expressions with m = 0 using J-1 = - J1 : E Fields Inside a Round Wire x = β'r β'2 = β2- βd2 ≈ β2 β = ω ξ =[ε + σ/(jω)] ≈ σ/(jω) conductor xa = β'a βd = ω ξd =[εd + σd/(jω)] ≈ εd dielectric E(r,φz,t) = ej(ωt-βz) E(r,φ) (D.1.2) E = Er + Eφ + Ez (for any arguments) E(r,φ) = E(r,0) + 2!Syntax Error, I[Re{E(r,m)}cos(mφ) - Im{E(r,m)} sin(mφ)] = real (D.1.4) where: Ez(r,m) = (1/4) ηm I Rdc [ - ] a = radius ηm ≡ Ez(r,0) = (1/2) I Rdc [] // = (2.1.22) for βd = 0 η0 = 1 Er(r,m) = (j/4) (aβd) ηm I Rdc [ + - ] Er(r,0) = (j/2) (aβd) I Rdc [ ] Eφ(r,m) = (1/4) (aβd) ηm I Rdc [ - + + ] Eφ(r,0) = 0 (D.4.6) Observations about the solution: (1) We looked for a wave solution inside a round wire in which phase fronts propagate down the wire (z direction) with angular frequency ω and wavelength λd = 2π/βd, as indicated by (D.1.2) in the box. We found the solution shown in the box. It satisfies all three components of the Helmholtz equation (D.1.1) as well as the div E = 0 equation (no charge inside conductor). (2) For a good conductor and a good dielectric, one has ξ ≈ σ/(jω) and ξd ≈ εd as shown in the box. These are the complex dielectric "constants". The corresponding wavenumbers are then β = ω ≈ ω = ej3π/4 = ej3π/4 (/δ) (2.1.18) , (2.1.21) βd = ω ≈ ω = ω / vd vd = speed of light in the dielectric (D.4.7) Thus, in our wave solution (D.1.2), the phase fronts propagate down the inside of the wire at vd, the speed of light in the dielectric outside the wire. Although we have been silent about the fields outside the wire, it seems reasonable to presume there is a wave outside also moving down the wire at vd . See below. (3) The fact that the wavenumber β inside the conductor is complex indicates the presence of absorption of energy in the conductor due to σ which one could calculate in detail. (4) In our axially symmetric round wire analysis of Section 2.1 we assumed no z variation of the fields, which means wave number βd = 0 so β' = β. In this case the box above shows Er(r,m) = 0 and Eφ(r,m) = 0 for all m ≥ 0. Our symmetry assumptions in Section 2.1 limited the analysis there to m = 0 and the Ez(r,0) shown in the box above agrees with result (2.1.22). (5) The components Eφ and Er are smaller than Ez by factor (βda). This is the dimensionless smallness parameter which defines the "transmission line limit", see Chapter 4. (6) If there exist moments Nm of the charge distribution on the wire with m > 1, then the corresponding ηm ≠ 0 and it is clear that Ez(r,φ) and hence Jz(r,φ) is non-uniform around the surface of the wire. That is, these fields vary with φ as cos(mφ). The non-uniformity is not "small" but has the full strength of ηm. Of course we only expect to get significant moments of charge density n(φ) when conductors are "fat and close". It seems likely in this case that the largest contribution will come from m=1. (7) The surface impedance from (C.2.1) is just Zs(φ) = Ez(r=a,φ)/I. Using the boxed result for Ez(r,φ) we quickly find that Zs(φ) = (1/2) Rdc { + !Syntax Error, I [ - ] Re[ηmejmφ] } where (D.4.8) Re[ηmejmφ] = Re(ηm)cos(mφ) - Im(ηm) sin(mφ). Thus we see the expected non-uniformity of Zz(φ) around the perimeter of the wire cross section for the m > 0 components. D.5 The Low Frequency Limit If we assume that ω is small enough that |xa| = |β'|a << 1, we can greatly simplify the above results. For a good conductor, β' ≈ β = ej3π/4 (/δ) so |β'|a << 1 => (a/δ) << 1 or (a/δ) << 1, which means the skin depth is much larger than the radius of the wire. In terms of frequency ω this condition is a << 1 => ω << . Using our ongoing example, for the Belden 8281 coax center conductor with a = .394 mm, μ = μ0 = 4π x 10-7 henry/m, and σ = 5.81 x 107 mho/m (copper), the limit requires that ω << = 88 KHz . All limits come from the leading small-x term of Jn(x) which is Jm(x) ≈ xm/ (2mm!) m ≥ 0 // Spiegel (24.2) (D.5.1) To avoid errors, we have Maple compute the limits of the various terms appearing in box ***. The left column is for visual check, the right column then provides the limits. Our limit is xa<<1 and since x ≤ xa we also have x << 1, but the ratio (x/xa) = (r/a) is merely ≤ 1 and not << 1. From the column on the right above we read off the limits: A = ≈ 2(m+1)(x/xa)m B = ≈ (1/2m)(x/xa)m xa2 ≈ 0 since xa << 1 C = ≈ (x/xa)m-1 D = ≈ (x/xa)m+1 E = ≈ (x/xa)m+1 xa2 ≈ 0 F = ≈ 2 G = ≈ (x/xa) (D.5.2) In terms of the letters A through G the E(r,m) fields from box *** are given by Ez(r,m) = (1/4) ηm I Rdc [A - B ] Ez(r,0) = (1/2) I Rdc [F] Er(r,m) = (j/4) (aβd) ηm I Rdc [ C + D - E ] Er(r,0) = (j/2) (aβd) I Rdc [G] Eφ(r,m) = (1/4) (aβd) ηm I Rdc [ - C + D + E ] Eφ(r,0) = 0 (D.5.3) Installing the limits and setting (x/xa) = (r/a) we find E Fields Inside a Round Wire (low frequency limit) Ez(r,m) = (1/2) ηm I Rdc (m+1)(r/a)m Ez(r,0) = I Rdc Er(r,m) = (j/4) (aβd) ηm I Rdc [(r/a)m-1 + (r/a)m+1 ] Er(r,0) = (j/2) (aβd) I Rdc (r/a) Eφ(r,m) = (1/4) (aβd) ηm I Rdc [- (r/a)m-1 + (r/a)m+1 ] Eφ(r,0) = 0 (D.5.4) The equation Ez(r,0) = I Rdc states that in this low frequency limit the wire acts as a resistor having the expected resistance per unit length Rdc = 1/(σπa2) as in (D.4.2). In the transmission line limit aβd << 1 of interest to us, we see that there is always a small radial current which pumps the changing surface charge. In the m = 0 this radial current is linear in r, peaking at the surface. In fact, if you integrate the radial current over the surface of a long piece of wire of length λ/ 2, taking into account the z dependence, you get 2I. [ I think the above has no basis in fact! At DC, we have ω = 0 and Er(a,0) = (j/2) (aβd) I Rdc. But βd = ω ≈ ω = 0 at DC, so the integral of the radial current is 0. For m = 1, the radial current depends on η1 which we don't know, and then Er(a,1) = (j/4) (aβd) ηm I Rdc(2) and we have no conclusion at all. Yes it is proportional to I. I don't know where this strange claim came from! Throwing in the sinφ angle and integrating over half a wavelength is not going to rescue this weird claim. ] See Section 3.7, Figure 2 for a drawing of the current we have just computed. continue here after reading Chapter 3. D.6 What about the E fields outside the round wire? This is an Exercise for the Reader, with some preliminary discussion given below. The Helmholtz equation (D.1.1) outside the wire now contains βd instead of β. The assumed wave solution form is still (D.1.2). In the three component Helmholtz equations this means that β' = βd2 - βd2 = 0 so that: [2E]z + β2 Ez = 0 : [r2∂r2 + r ∂r - m2] Ez(r,m) = 0 (D.1.16)o [2E]r + β2 Er = 0 : [r2∂r2 + r∂r - (m2+1)] Er(r,m) - 2jm Eφ(r,m) = 0 (D.1.17)o [2E]φ + β2 Eφ = 0 : [r2∂r2 + r∂r - (m2+1)] Eφ(r,m) + 2jmEr(r,m) = 0 (D.1.18)o div E = 0 : ∂r [r Er(r,m)] + jmEφ(r,m) -jβd r Ez(r,m) = 0 (D.1.19)o The differential operators appearing in the above equations are no longer Bessel-style operators, they are Euler-style operators. Euler ODE's have the general form [ r2∂r2 + a r ∂r + b] f(r) = 0, and the solutions have this form from p 45 of Polyanin's excellent ODE compendium, For Ez(r,m) the equation (D.1.17)o shown just above is in fact an Euler equation which has a = 1 and b = -m2 so μ = m and the solution forms are these (r ≥a outside the wire), Ez(r,m) = Azmrm + Bzmr-m m > 0 Ez(r,0) = Czln(r) + Dz m = 0 . This Ez Helmholtz equation is the same as the radial equation one obtains when solving the 2D Laplace equation 2u = 0 in separated polar coordinates, see for example Stakgold Vol II p 92 (6.7). It certainly seems reasonable that Azm = 0 to avoid a dramatic blowup at r = ∞, but the m = 0 coefficients are less obvious, since we have already seen logarithmic divergences associated with infinitely long wires. One approach is to mimic the method used for the inside solution, making use of the div E = 0 equation shown above, and thereby coming up with a full set of E field solutions. Along the way one must determine appropriate boundary conditions at r = a so the outside solutions are consistent with the inside solutions. We can at least see that for m > 0 Ez(r,m) ~ (r/a)-m outside the round wire. D.7 What about φ, A and B ? Inside the conductor, we know that potential φ satisfies (D.1.1), and we know that φ at r=a does not depend on angle φ. This means that φ is similar to our Ez solution in the m=0 partial wave, and vanishes for all higher partial waves. Thus we write φ(r,m) = δm,0 φ0 J0(x)/J0(xa) x = β'r (D.7.1) where φ0 is the value of the potential on the surface at r=a. Since we know E(r,m) from Section 3 (6), we can solve for the vector potential A as follows: -jωA = E + grad φ (D.7.2) Converted to partial waves, this says -jωAr(r,m) = Er(r,m) + δm,0 ∂rφ(r,0) -jωAφ(r,m) = Eφ(r,m) - (1/r) jm δm,0 φ(r,0) = Eφ(r,m) -jωAz(r,m) = Ez(r,m) - jβd δm,0 φ(r,0) (D.7.3) Thus, for m≠0 we have -jωA = E. For m = 0 there are extra pieces as shown for Ar(r,0) and Az(r,0). Since A(r,m) is known, A(r,φ,z) is also known, and then so too is B = curl A. This curl can also be performed in each partial wave if desired by replacing ∂φ = -jm and ∂z = -jβd as usual. Since -jωA = E in the higher partial waves, we may conclude that the transverse components of A are small compared to the longitudinal component, just as is the case for E, see (D.3.6). For m=0 we know that Aφ = Eφ = 0, but Ar is a combination of Er and ∂r φ which is not small compared to Az, due to the ∂rφ contribution. Thus, we arrive at the conclusion that, although Ar is negligible in the dielectric, it cannot be ignored inside the conductor. This explains, incidentally, the problem one encounters with the gauge condition (1.5.5) divA = -j(β2/ω)φ. Since φ is continuous at the boundary, and β2 takes a jump of many orders of magnitude (from βd to β'), something on the left side must change violently. But Az is also continuous. It is the m=0 Ar inside the conductor that takes up the slack. In fact, taking the difference inside minus outside we conclude that ∂rAr (r=a-ε) = -j(β'2/ω)φ0 (D.7.4) from which we can determine φ0The potential Ar is discontinuous at r=a. Appendix E: 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. Fig E.1 The red plot is charge density ρ, and the black plot is the electric field magnitude.The charge density is exactly ρ = 0 in the dielectric region between the two plates simply because there are no available charge carriers as there are in a metal ( the electron cloud). Barring a huge E field, electrons cannot just "jump off" the metal surface into the dielectric region because of an energy cost to do so called a work function. The figure suggests that the charge distribution might have an exponential decay going into each metal 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 corrected Ohm's law is this: J = σE - D grad ρ . (E.1) The grad term, associated with Fick's Law, represents a flux of charged particles (a current) created by a gradient of the charge density. The charge flows (diffuses) from a region of high density to one of lower density, hence the minus sign, just as heat flows from a region of higher temperature to one of lower temperature. In a static situation 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ε . (E.3) Inside a metal conductor the dielectric constant ε requires some careful study, but here we shall just set it to ε0 as if there were nothing in the electron cloud of the metal that could be polarized. Taking the divergence of (E.2) and using (E.3) we get this result 2ρ = (σ/Dε0) ρ . (E.4) The inverse 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) may be written, 2ρ = (1/λD2) ρ . (E.6) In our one-dimensional problem of the above figure, the solution of this equation is ρ(x) = ρ(0) e-x/λ (E.7) where x is a coordinate going into the surface. This says that the thickness of the charge surface layer inside the metal 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 are some 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 constant m = 9.1 x 10-31 kg = electron mass h = 6.63 x 10-34 J sec = Planck constant Plugging these into (E.9) gives the following effective electron temperature so TF = 81,702 ° K = pretty hot (E.11) We can now compute the Debye length, using (E.10). ε0 = 8.85 x 10-12 F/m q = 1.60 x 10-19 C so 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: In our simple model, the thickness of the surface charge density below 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. Fact: Then thin layer of negative surface charge on the right plate in Fig E.1 above serves to neutralize the E field which would otherwise be present inside the right conductor due to the positive charge on the surface of the left plate. One says that the E field inside (and to the right of) the right plate is "screened" (killed off) by the negative surface charge layer on the right plate. This is of course the principle behind the ever-popular Faraday Cage (note kids inside): Fig E.2 From Section 2.2, we found 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: 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. F.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. The tangential E field vanishes because the surface is an equipotential, and the normal B field vanishes because the Maxwell curl E equation (1.1.2) says ∂xEy - ∂yEx = -jωBz where we assume a point on a metal surface with z being the normal direction. 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 (1.6.2) says, ( 2 + β2 ) E = 0 . (1.6.2) (F.1.1) Traditional wave motion ej(ωt-kz) has a phase front ωt-kz = constant so k∂tz = ω or kv = ω or k = ω/v where v = ∂tz is the phase front velocity. There is an associated temporal frequency f and period T such that T = 1/f = 1/(2πω) where ω is the angular frequency ω = 2π/T. Similarly there is spatial frequency fs and wavelenght λ such that λ = 1/fs = 1/(2πk) where k is the wavenumber k = 2π/λ : k is the number of radians per wavelength, just as ω is the number of radians per period. In our current context, the wavenumber k is called β and it has the added feature of being complex. Recall, β2 ≡ ω2μ ξ β = complex wavenumber (1.5.3) ξ ≡ ε - jσ/ω . ξ = complex dielectric constant (1.5.4) 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. In analogy with (1.1.10) that speed of light c = 1/ we define a complex phase velocity v ≡ 1/ (F.1.2) and then β = ω/v (F.1.3) just as we had k = ω/v in the traditional wave discussion above. If σ is very small then ξ ≈ ε with a very small imaginary part, and v is also mostly real with a small imaginary part. 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 wave form of solution (wavenumber k), E(x,y,z,t) = E(x,y) ej(ωt±kz) . (F.1.4) The - sign is for waves traveling in the +z direction. 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) . (F.1.8) 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 E field at some z = constant slice for the m = 1 mode: Fig 1: Cross sectional view of a simple waveguide. equation 8 is missing Since the Electric field is in a direction Transverse to the wave motion, this is called a TE mode. If we solve (F.1.6) for the wavenumber k, k = (1/v) λ = 2π/k ωm ≡ v γm = cutoff frequency (F.1.9) we see clearly that ω must be larger than ωm ≡ v γm 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 since this is true of v. 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. F.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) = (Aj/2) [ exp( -jβ1• r) - exp( -jβ2• r)] = sum of two plane waves (F.2.2) where β1 = βx βz = wavevector of first wave β2 = - βx βz = wavevector of second wave (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.1.4). As ω approaches cutoff from above, tanθ → ∞ and θ→π/2 and at cutoff the plane waves just 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 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 = γmv, which is the cutoff frequency for mode m in (F.1.9). This is exactly the situation described above when θ = 90° in (F.2.6). 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: sin(βxa) = 0 => βxa = mπ => (β cosθ)a = mπ => βcosθ = mπ/a = γm => cosθ = (γm/β) = (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. G.1 Evaluation of the integral in 4.1 (9). (a) The first claim is that the integral on the left below has the form shown on the right: In(x) = !Syntax Error, Idz' (z'-z)n = sn fn(βs) (G.1) where R = and s = . Here In(x) = I(x,y,z) is just our name for this integral, please do not confuse this with the modified Bessel function In(x). To show the claim, the first step is to define z" ≡ z'-z to get In(x) = !Syntax Error, Idz" (z")n R = (G.2) which shows that I(x) is independent of z so we can think of (G.2) as (G.1) with z = 0. Next define x ≡ z"/s to get dz" (z")n = sn+1 dx xn R = = s so that In(x) = sn !Syntax Error, I dx xn = sn fn(sβ) which then confirms the claimed form in (G.1). (b) Our task is then to compute this integral fn(α) = !Syntax Error, I dx xn For odd integer n, the integral clearly vanishes since the integrand is then an odd function of x. For even n we replace the integral with double its positive side value and replace n = 2m with m = 1,2,3 ... to list off the non-vanishing integrals. f2m(α) = 2!Syntax Error, I dx x2m m = 0,1,2... Finally, change to y = which says x2 = y2 = 1 and so xdx = ydy. Then dx x2m = (y/x)dy x2m = y dy x2m-1 = y dy (y2-1)m-1/2 and so f2m(α) = 2 !Syntax Error, Idy (y2-1)m-1/2 e-jαy Finally we have something we can find in the standard tables. GR7 3.387 3 with ν = m+1/2 ] shows that the result is: f2m(α) = (2/) (2/jα)m Γ(m+1/2) Km(jα) (G.5) = (2m+1/) Γ(m+1/2) Km(jα) / (jα)m where Km(z) is a modifed Bessel function. For m = 0 one has Γ(m+1/2) = . For m = 1,2,3 one can write Γ(m+1/2) = (2m-1)!! /2m Spiegel 16.6 (G.6) where for example 5!! = 5*3*1. These can be combined into a single formula if one interprets (-1)!! = 1 as well as (0)!! = 1. Then f2m(α) = (2m+1/) [(2m-1)!! /2m] Km(jα) / (jα)m = 2 (2m-1)!! Km(jα) / (jα)m (G.7) and then restoring 2m = n we have fn(α) = 2 (n-1)!! Kn/2(jα) / (jα)n/2 for n even fn(α) = 0 for n odd (G.8) Using then the fact that sn fn(βs) = sn 2 (n-1)!! Kn/2(jβs) / (jβs)n/2 (G.9) we conclude that In(x) = !Syntax Error, Idz' (z'-z)n = sn fn(βs) = (G.10) In particular I0(x) = !Syntax Error, Idz' = 2 K0(jβs) . (G.11) This is the result quoted in 4.1 (11). The modified Bessel function Km(z) is described in any reference on Bessel functions such as GR7 quoted above. For small z, we have these approximations: [ reference ] K0(z) ≈ -ln(z/2) [ 1 + (z/2)2 + O(z4) ] + ψ(1) + (z/2)2 ψ(2) + O(z4) (G.9) Km(z) ≈2m-1(m-1)! z-m - 2m-3 (m-2)! z-m+2 + O( z-m+4) m = 1,2,3... where the ψ(i) are certain constants associated with the gamma function. 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 for m > 0. repair done to this point, I think all the above is correct 10.31.13 G.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. Appendix H : Poisson and Helmholtz Propagators in 3D Note: Appendix I deals with these propagators in 2D rather than 3D. __________________________________________________________________________________ H.1 Overview and Meaning of Free-Space Propagators This appendix proves the following Facts: Fact 1 : -2[1/4πr] = δ(r) (H.2.1) (H.1.1) Fact 2 : -2[h(r)/r] = 4π h(0) δ(r) - h"(r)/ r (H.3.1) (H.1.2) Fact 3 : - (2+k2) (e-jkr/4πr) = δ(r) (H.3.5) (H.1.3) Throughout, 2 is the usual 3D Laplacian operator 2 = ∂x2 + ∂y2 + ∂z2. In the first and last results above, if one replaces r → r-r' (a simple translational shift of origin) ones finds -2[1/4πR] = δ(r-r') R = | r - r' | (H.1.4) - (2+k2) [e-jkR/4πR] = δ(r-r') δ(r-r') = δ(x-x') δ(y-y') δ(z-z') (H.1.5) The quantities in brackets are known as free-space Green's Functions (Green Functions) or propagators, or as "fundamental solutions": 1/4πR = the Poisson 3D free-space propagator (H.1.6) e-jkR/4πR = the Helmholtz 3D free-space propagator (H.1.7) The significance of these propagators is the following: -2 f(x) = s(x) => f(x) = ∫d3x' [1/4πR] s(x') + homogeneous solutions The Poisson Equation (H.1.8) - (2+k2) f(x) = s(x) => f(x) = ∫d3x' [e-jkR/4πR] s(x') + homogeneous solutions The Helmholtz Equation (H.1.9) The equations on the left are inhomogeneous partial differential equations driven by source function s(x). If one is careful to include in s(x) all source contributions (such as those on boundary surfaces), one generally does not have to add any homogeneous solutions on the right. A homogeneous solution refers to -2 fh(x) = 0, for example. The solutions shown on the right above can be instantly verified as follows: f(x) = ∫d3x' [1/4πR] s(x') + fh(x) -2 f(x) = ∫d3x' (-2 [1/4πR] ) s(x') -2 fh(x) = ∫d3x' δ(r-r') s(x') - 0 = s(x) (H.1.10) and similarly for - (2+k2) f = g. A "free space" Green's Function gF in general is a solution of D gF(r, r') = δ(r-r'), gF(r, r') → 0 as r → ∞ (H.1.11) where D is some differential operator. The condition on the right says gF must vanish on the Great Sphere. More generally one can write D g(r, r') = δ(r-r'), g(r, r') = 0 for r on some closed surface enclosing a region of interest (H.1.12) In this second form, the ∫d3x' is over the volume inside that closed surface. We shall not make use of this more general form in this document. George Green (1793-1841), by the way, was an English grain miller. Looking at f(x) = ∫d3x' [1/4πR] s(x') = ∫ gF(x,x') [s(x') d3x'], one can say that the kernel Green's Function gF(x,x') "propagates" a tiny piece of "source" [s(x')d3x'] from location x' to location x so that the solution f(x) is then a sum of all such propagated contributions as the source ranges over the entire volume of interest, which for us is all 3D space where the source is non-vanishing. __________________________________________________________________________________ H.2 Fact 1: -2[1/r] = 4πδ(r) (H.2.1) Proof: Let volume V be all of 3D space. Carve out from V a small spherical cavity of radius a centered at r = 0. If we call this spherical volume Va and then V' = V - Va is the original volume with the spherical cavity carved out: In order to show that some function g(r) = δ(r), one has to show that lima→0 ∫V' dV g(r) = 0 (H.2.2a) lima→0 ∫Va dV g(r) = 1 (H.2.2b) This is basically the definition of δ(r). Since δ(r) has units L-3, g(r) has units L-3. Our candidate function of interest is g(r) = - (1/4π) 2[1/r] . (H.2.3) Using 2 in spherical coordinates acting on a function of r, one finds that, since ∂r(1) = 0, 2[1/r] = (1/r2)∂r(r2∂r) [1/r] = 0 r > 0 (H.2.4) so that g(r) = - (1/4π) 2[1/r] = 0 r > 0 . (H.2.5) Thus, condition (H.2.2a) is trivially satisfied since r > 0 everywhere in volume V'. It remains to verify condition (H.2.2b). Consider the integral appearing in the left side of (H.2.2b) ∫Va dV g(r) = - (1/4π) ∫Va dV 2[1/r] = - (1/4π) ∫Va dV [1/r] . (H.2.6) The divergence theorem says ∫V dV div F = ∫S dS F (H.2.7) where V is any closed volume whose surface is S, and dS points out. Using V = Va and F = [1/r] = ∂r(1/r) = -r-2 we find that LHS (H.2.7) = ∫Va dV div [1/r] = ∫Va dV 2[1/r] = ∫Va dV [-4πg(r)] = -4π ∫Va dV g(r) RHS (H.2.7) = ∫S dS [1/r] = ∫dΩ [a2 ] [1/r]|r=a = ∫dΩ[a2 ] [-a-2] = -4π which tells us that ∫Va dV g(r) = 1 for any a. Thus, lima→0 ∫Va dV g(r) = 1 and we have then verified (H.2.2b). Therefore we conclude that the candidate g(r) of (H.2.3) is in fact the same as δ(r) so - (1/4π)2[1/r] = δ(r) (H.2.8) or 2[1/r] = - 4πδ(r) (H.2.9) which is (H.2.1). QED __________________________________________________________________________________ H.3 Fact 2: 2[h(r)/r] = - 4π h(0) δ(r) + h"(r)/ r (H.3.1) Proof: Start with this vector identity, 2(φψ) = φ2ψ + ψ2φ + 2 φ ψ . (H.3.2) This identity is valid in any number of dimensions (implied sum on i from 1 to N) , ∂i2(φψ)= ∂i[ (∂iφ)ψ + ψ(∂iφ)] = (∂i2φ)ψ + (∂iφ) (∂iψ) + φ(∂i2ψ) + (∂iφ) (∂iψ) . So apply (H.3.2) to the case φ = h and ψ = r-1, 2(h r-1) = h2(r-1) + r-12h + 2 h (r-1) = - h 4π δ(r) + r-12h + 2 [ h' (-r-2) ] // using (H.2.9) = - 4π h(0) δ(r) + r-12h - 2 r-2 h'(r) . (H.3.3) Algebra shows that, using spherical coordinates, 2h = (1/r2)∂r(r2∂r)h(r) = h"(r) + (2/r)h'(r) (H.3.4) so then 2(h r-1) = - 4π h(0) δ(r) + r-1 [h"(r) + (2/r)h'(r) ] - 2 r-2 h'(r) = - 4π h(0) δ(r) + h"(r)/ r which is the claim of (H.3.1). QED Fact 3: - (2+k2) (e-jkr/4πr) = δ(r) (H.3.5) This Fact is just an application of Fact 2 to the case h(r) = e-jkr : h = e-jkr h(0) = 1 h' = -jk e-jkr h" = -k2 e-jkr 2[h(r)/r] = - 4π h(0) δ(r) + h"(r)/ r (H.3.1) so 2(e-jkr/r) = - 4π 1 δ(r) + [-k2 e-jkr ] / r = -4πδ(r) - k2(e-jkr/r) Thus, ( 2+k2) (e-jkr/r) = - 4πδ(r) or - (2+k2) (e-jkr/4πr) = δ(r) as claimed. Appendix I : Poisson and Helmholtz Propagators in 2D Note: Appendix H deals with these propagators in 3D rather than 2D. Sections I.1 and I.2 below are basically "cut, paste and edit" versions of Sections H.1 and H.2, and we have made equation numbers match. However, Section I.3 is something new since it involves a "special function". __________________________________________________________________________________ I.1 Overview and Meaning of Free-Space Propagators This appendix proves two Facts: (H0(1) is a Hankel function ) Fact 1 : -2[ln(1/r)/2π] = δ(r) (I.2.1) (I.1.1) Fact 2 : - (2+k2) [(j/4) H0(1)(kr)] = δ(r) (I.3.1) (I.1.2) Throughout this Appendix, 2 is the usual 2D Laplacian operator, 2 = ∂x2 + ∂y2. and δ(r) = δ(x) δ(y) . (I.1.3) In the two Facts above, if one replaces r → r-r' (a simple translational shift of origin) ones finds -2[ln(1/R)/2π] = δ(r-r') R = | r - r' | (I.1.4) - (2+k2) [(j/4) H0(1)(kR)] = δ(r-r') δ(r-r') = δ(x-x') δ(y-y') (I.1.5) The quantities in brackets are known as free-space Green's Functions (Green Functions) or propagators, or as "fundamental solutions" : ln(1/R) = the Poisson 2D free-space propagator (I.1.6) (j/4) H0(1)(kR) = the Helmholtz 2D free-space propagator (I.1.7) The significance of these propagators is the following: -2 f(x) = s(x) => f(x) = ∫d2x' [ln(1/R)/2π] s(x') + homogeneous solutions The Poisson Equation (I.1.8) - (2+k2) f(x) = s(x) => f(x) = ∫d2x' [(j/4) H0(1)(kR)] s(x') + homogeneous solutions The Helmholtz Equation (I.1.9) The equations on the left are inhomogeneous partial differential equations driven by source function s(x). If one is careful to include in s(x) all source contributions (such as those on boundary curves), one generally does not have to add any homogeneous solutions on the right. A homogeneous solution refers to -2 fh(x) = 0, for example. The solutions shown on the right above can be instantly verified as follows: f(x) = ∫d2x' [ln(1/R)/2π] s(x') + fh(x) -2 f(x) = ∫d2x' (-2 [ln(1/R)/2π] ) s(x') -2 fh(x) = ∫d3x' δ(r-r') s(x') - 0 = s(x) (I.1.10) and similarly for - (2+k2) f = g. A "free space" Green's Function gF in general is a solution of D gF(r, r') = δ(r-r'), gF(r, r') → 0 as r → ∞ (I.1.11) where D is some differential operator. The condition on the right says gF must vanish on the Great Circle. More generally one can write D g(r, r') = δ(r-r'), g(r, r') = 0 for r on some closed curve enclosing a region of interest (I.1.12) In this second form, the ∫d2x' is over the area inside that closed surface. We shall not make use of this more general form in this document. George Green (1793-1841), by the way, was an English grain miller. Looking at f(x) = ∫d2x' [ln(1/R)/2π] s(x') = ∫ gF(x,x') [s(x') d2x'], one can say that the kernel Green's Function gF(x,x') "propagates" a tiny piece of "source" [s(x')d2x'] from location x' to location x so that the solution f(x) is then a sum of all such propagated contributions as the source ranges over the entire volume of interest, which for us is all 2D space where the source is non-vanishing. __________________________________________________________________________________ I.2 Dervivation of Fact 1: 2[ln(1/r)] = - 2πδ(r) (I.2.1) Proof: Let area A be all of 2D space. Cut out from A a small spherical hole of radius a centered at r = 0. If we call this spherical area Aa and then A' = A - Aa is the original area with the spherical hole cut out: In order to show that some function g(r) = δ(r), one has to show that lima→0 ∫A'dA g(r) = 0 (I.2.2a) lima→0 ∫AdA g(r) = 1 (I.2.2b) This is basically the definition of δ(r). Since δ(r) has units L-2, g(r) has units L-2. Comment: When any differential operator like or 2 is applied to ln(r0/r), the result is independent of r0 so we can always take r0 = 1. For example, ∂x [ln(r0/r)] = ∂x [ lnr0 + ln(1/r)] = ∂x ln(1/r). In what follows, ln(r) and ln(1/r) are always acted upon by differential operators, so we can interpret these objects as dimensionless quantities ln(r/r0) and ln(r0/r) for any r0. Then it is clear below that dim [g(r)] = L-2. Our candidate function of interest is g(r) = - (1/2π) 2[ln(1/r)] = +(1/2π) 2 [ ln(r) ] . (I.2.3) Using 2 in polar (cylindrical without the z) coordinates acting on a function of r, one finds that, since ∂r(1) = 0, 2[ln(r)] = (1/r)∂r(r∂r) [ln(r)] = 0 r > 0 (I.2.4) so that g(r) = - (1/2π) 2[ln(1/r)] = 0 r > 0 . (I.2.5) Thus, condition (I.2.2a) is trivially satisfied since r > 0 everywhere in area A' for any a > 0. It remains to verify condition (I.2.2b). Consider the integral appearing in the left side of (I.2.2b) ∫AdA g(r) = - (1/4π) ∫AdA 2[1/r] = - (1/4π) ∫AdA [1/r] . (I.2.6) The divergence Fact in 2D says ∫A dA div F = C ds F (I.2.7) where A is any closed area whose bounding curve is C, and where ds = ds where is normal to C at any given point on C. Notice that this closed area is necessarily planar since everything is 2D here. Using A = Aa = disk of radius a and F = [ln(1/r)] = ∂r(ln(1/r)) = - ∂r(lnr) = [ -r-1] we find that LHS (I.2.7) = ∫AdA div [ln(1/r)] = ∫AdA 2[ln(1/r)] = ∫AdA [-2πg(r)] = -2π ∫AdA dA g(r) RHS (I.2.7) = ∫C ds [ ln(1/r)] = ∫ [adθ ] [ ln(1/r)]|r=a = ∫dθ [a ] [-a-1] = -2π which tells us that ∫Aa dA g(r) = 1 for any a. Thus, lima→0 ∫Aa dA g(r) = 1 and we have then verified (I.2.2b). Therefore we conclude that the candidate g(r) of (I.2.3) is in fact the same as δ(r) so - (1/2π)2[ln(1/r)] = δ(r) (I.2.8) or 2[ln(1/r)] = - 2πδ(r) (I.2.9) which is (I.2.1). QED __________________________________________________________________________________ I.3 Derivation of Fact 2: - (2+k2) [(j/4) H0(1)(kr)] = δ(r) (I.3.1) We seek the solution E(r) of this equation - (2+k2 ) E(r) = δ(r) where E(r→∞) = 0 (I.3.2) which we write as 2E+ k2E = - δ(r). Using polar coordinates this says r-1∂r(r∂rE) + k2E = - δ(r) or E" + r-1E' + k2E = - δ(r) or r2E"(r) + rE'(r) + r2k2E(r) = - δ(r) . (I.3.3) Writing E(r) = F(kr) we get r2k2F"(kr) + rk F'(kr) + r2k2F(kr) = - δ(r) or (rk)2F"(kr) + (rk) F'(kr) +(rk)2F(kr) = - δ(r) or z2F"(z) + z F'(z) + z2F(z) = - δ(r) where z = kr . (I.3.4) Away from r = z = 0, this is Bessel's equation of index 0 (A&S 10.2.1) so solutions are Bessel functions like these F(z) = J0(z), Y0(z), H0(1)(z), H0(2)(z). z = kr (I.3.5) which are Bessel functions of the first, second and third kind. The third kind functions (the H's) are called Hankel Functions. If we assume that k has a tiny positive imaginary part (see Comments later), then of all the functions just listed, only H0(1)(kr) has decaying behavior for large r (A&S 10.2.5). We therefore put forward the following candidate for a delta function g(r) = - (2+k2 ) C H0(1)(kr) . (I.3.6) Recall from Section I.2 that a successful δ(r) candidate must satisfy these two conditions lima→0 ∫A'dA g(r) = 0 (I.2.2a) lima→0 ∫AdA g(r) = 1 (I.2.2b) Our candidate g(r) vanishes within any region A' no matter how small the hole because g(r) = 0 for any r> 0, so the first condition is already met. It remains only to show that the second condition is also met. We must then show that lima→0 ∫A dA {- (2+k2 ) C H0(1)(kr)} = 1 . (I.3.7) Since Aa is a very small disk as we approach the limit, we may use the small argument behavior of our candidate g(r) in studying the situation. We know that H0(1)(kr) ≈ (2j/π) ln(kr) // A&S 10.7.2 (I.3.8) so what we need to show is that lima→0 ∫AdA {- (2+k2) C (2j/π) ln(kr)} = 1 or - C (2j/π) lima→0 ∫AdA { (2+k2) ln(kr)} = 1 or - C (2j/π)2π lima→0 !Syntax Error, Irdr{ (2+k2) ln(kr)} = 1 // ∫dθ = 2π or C (4/j) lima→0 !Syntax Error, Irdr{ (2+k2) ln(kr)} = 1 . (I.3.9) Now consider : lima→0 [!Syntax Error, Irdr ln(kr)] = lima→0 [(1/4)a2{2ln(ka)-1}] = 0 . (I.3.10) Thus, the k2 ln(kr) term in (I.3.9) makes no contribution in the limit, so we then have to show that C (4/j) lima→0 !Syntax Error, Irdr 2 [ ln(kr)] = 1 . (I.3.11) But (I.2.9) says that 2[ln(r)] = 2πδ(r) . (I.2.9) Now δ(r) = δ(x)δ(y) = δ(r)/2πr (I.3.12) since 1 = ∫∫dxdy δ(x)δ(y) = ∫rdr∫dθ δ(r)/2πr = 2π∫rdr δ(r)/2πr = ∫dr δ(r) = 1 . Therefore 2[ln(r)] = δ(r)/r (I.3.13) and then 2[ln(kr)] = 2[ln(k) + ln(r)] = 2[ln(r)] = δ(r)/r . (I.3.14) Inserting this last result into (I.3.11) then gives C (4/j) lima→0 !Syntax Error, Irdr 2 [ ln(kr)] = 1 C (4/j) lima→0 !Syntax Error, Irdr δ(r)/r = 1 C (4/j) lima→0 !Syntax Error, Idr δ(r) = 1 C (4/j) lima→0 1 = 1 C (4/j) = 1 . Thus, we have a solution if we select constant C = (j/4). Therefore, the solution to (I.3.2) is E(r) = C H0(1)(kr) = (j/4) H0(1)(kr) . (I.3.15) Stakgold Vol II page 55 (5.120) confirms this result where = k. Therefore we have shown that - (2+k2) [(j/4) H0(1)(kr)] = δ(r) (I.3.16) which is the Fact stated as (I.3.1). QED On page 54 Stakgold gives the solution to - (2+k2 ) E(r) = δ(r) for n≥2 dimensions as (5.118): Comments: 1. Complex Helmholtz Parameter and Hν(1)(z). Stakgold considers the Helmholtz operator to be λ which is our k2. He regards λ as a complex variable which can lie anywhere in the complex λ plane. If we consider the function k(λ) = λ1/2, we find that it has a branch point at λ = 0. If we take the branch cut to the right, then one of the two Riemann sheets in λ-space for this function maps to the upper half k-plane as shown. This is the branch of λ1/2 that Stakgold selects and that is why we think of k and therefore k2 as having a tiny positive imaginary part when k is "real". The point is that we approach the positive real axis from above, not from below. It is this assumption that causes the large-r-decaying solution to our problem to be H0(1)(kr) instead of H0(2)(kr) . As shown on NIST p 229 10.17.5,6, expansions of the Hankel functions for large argument are, Hν(1)(z) ≈ z-1/2 e+j(z-νπ/2-π/4) Σk=0∞ (+j)k ak(ν) z-k Hν(2)(z) ≈ z-1/2 e-j(z-νπ/2-π/4) Σk=0∞ (-j)k ak(ν) z-k where ak(ν) are some real coeeficiients shown in 10.17.1 which we don't care about right now. The differences are highlighted in red. Here one sees that Hν(1)(z) ~ e+jz = e-Imz ejRez . Thus Hν(1)(kr) ~ e-rImk ejrRek and as long as k is in the upper half plane as shown in the right, Hν(1)(kr) decays exponentially (whereas Hν(2)(kr) blows up). In our round wire application, k = β has phase ej3π/4 which is in the upper half k plane, so Hν(1)(kr) is the appropriate solution function for large r. 2. Helmholtz morphs into Poisson. We have shown that - (2+k2) [(j/4) H0(1)(kr)] = δ(r) . (I.3.16) In the limit that k << 1, we showed above that H0(1)(kr) ≈ (2j/π) ln(kr) // A&S 10.7.2 (I.3.8) In this limit we then have - (2+k2) [(j/4)) (2j/π) ln(kr) ] = δ(r) or - (2) [(1/2π) ln(kr) ] = δ(r) and this is in agreement with the Poisson result (I.1.1). So as the Helmholtz equation morphs into the Poisson equation as k → 0, the Helmholtz propagator morphs into the Poisson propagator. References B.I. Bleaney and B. Bleaney, Electricity and Magnetism, 3rd Ed. (Oxford University Press, London, 1976). That would be Brevis Bleaney and wife Betty Isabelle. Brevis pioneered electron spin resonance independently with Russian Yevgeny Zavoisky in 1944. This book was reissued in 2013 as a two-volume paperback set. J.D. Jackson, Classical Electrodynamics, 3rd Ed. ( Wiley & Sons, New York, 1998). The author was fortunate to have learned his E&M from Dave Jackson in person, circa 1971 (green 1st edition). King, R.W.P., Electromagnetic Engineering, (McGraw-Hill, New York, 1945). This is the first of twelve books that Ronold King wrote or co-authored. His last was an antenna book (his specialty) published in 2002; he died in 2006 at age 100. It happens that I did an "independent study" with Dr. King circa 1969, but regrettably I knew so little that Dr. King could only smile and be encouraging. King, R.W.P., Transmission Line Theory, Dover, 1965. Another of the twelve books. H.A. Haus and J.R. Melcher, Electromagnetic Fields and Energy, (Prentice-Hall, New Jersey, 1989). Though out of print and hard to get, this very detailed and practical book is alive and well on the MIT OpenCourseWare website where all chapters can be read and downloaded. Two of the instructors are the authors. http://ocw.mit.edu/resources/res-6-001-electromagnetic-fields-and-energy-spring-2008/ R. Nevels and C-S Shin, "Lorenz, Lorentz, and the Gauge", IEEE Antennas and Propagation Magazine, Vol 43, No 3, June 2001, pp 70-71. See www.engr.mun.ca/~egill/index_files/7811_w10/lorenz_gauge.pdf and elsewhere. [NIST] F.W.J. Olver, D.W. Lozier, R.F. Boisvert and C.W. Clark, NIST Handbook of Mathematical Functions (Cambridge University Press, 2010). NIST is the U.S. National Institute of Standards and Technology which published the world-famous earlier edition in 1964 with editors Abramowitz and Stegun, known affectionately as "A&S". The greatly expanded 2010 edition (968 p) can be accessed online at dlmf.nist.gov which also has errata. The book (≥ $17) comes with a CD containing a bookmarked PDF file which of course has been bootlegged onto the web. Olver died in 2013. W.K.H. Panofsky and M. Phillips, Classical Electricity and Magnetism, 2nd Ed. (Addison-Wesley, Reading MA, 1962), reissued as a Dover paperback in 2005. Some of the fascinating history of Prof. Wolfgang "Pief" Panofsky appears in Jackson's Jan 2009 Physics Today article "Panofsky agonistes.." which can be found at http://www-theory.lbl.gov/jdj/PT_article.pdf. Quigley, Callum, "On the Origins of Gauge Theory" (2003), www.math.toronto.edu/~colliand/426_03/Papers03/C_Quigley.pdf. S. Lipschutz, M. Spiegel and J. Liu, Schaum's Outlines: Mathematical Handbook of Formulas and Tables (4th Ed.), (McGraw-Hill, 2012). The excellent original 1968 edition by Murray Spiegel has been a dog-eared reliable friend for many years. John Liu was added for the 1999 2nd Ed, and Seymour Lipschutz joined for the 2008 3rd Ed. Not to be confused with a watered-down "Easy Outline" version. This low- cost paperback is an excellent fast reference for well-known mathematical facts. Thomson, W.T. (Lord Kelvin), "Ether, Electricity and Ponderable Matter", The Proceedings of the Institution of Electrical Engineers (founded 1871), Volume 18 (1889), No 77, pp 4-37. The Appendix with ber and bei begins on page 35. Google Books has an unrestricted scan of a Harvard library copy of Vol. 18 which can be downloaded in PDF format: http://books.google.com/books?id=Wy89AAAAYAAJ Add Lucht Tensor Analysis ref used in D.2 Add Polyanin ODE book reference. Add AS2000 reference. Add GR7 reference and use those letters since they are used in the text.