app A 6 REVIEWED

Appendix A is now stable, so I don't think this doc is needed long term. Perhaps the doc title refers to appendix A section 6 which maybe is now section 7. 7. The Lorentz Gauge and QED: This section is certainly off the transmission-lines beaten path, but the author thought the reader might find it interesting. It is true that the nature of a transmission line results from photons "jumping back and forth" between the conductors. Unlike elsewhere in this document, everything is not fully explained in the following quick outline. In relativistic notation one uses 4-vectors which have one time component and three spatial components such as xμ = (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 4-vector and the corresponding covariant 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.5.1) which in relativistic notation is A'i = Ai + ∂iΛ = Ai - ∂iΛ i = 1,2,3 φ' = φ - ∂tΛ = φ - c ∂0Λ (A.7.1) The components of a classical vector like A, normally written as Ai, are in fact the contravariant components Ai in relativistic notation. If we now 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.7.2) Suppose we want ∂μA'μ = 0 (implicit sum on μ = 0,1,2,3). If we could find a potential A'μ with this property, that would be very convenient for the following reason: In general aμbμ (= aμbμ = a b) is the same in all frames of reference related by Lorentz Transformations. If ∂μA'μ = 0 in one frame, it is 0 in all frames, and that makes computational life simple. 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.7.3) 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 Lorentz Gauge. In this gauge (now dropping the prime on A), we have ∂μAμ = 0. As just noted, this gauge is so named because the resulting equation ∂μAμ = 0 is "covariant" under all Lorentz transformations in that both sides of this equation transform as the same tensor object, in this case a "scalar". That means the equation has the same form in all frames of references which are related by Lorentz transformations ( which include rotations, velocity transformations, and combinations of same). One can interpret ∂μAμ = 0 as ∂A = 0 where ∂ is a 4-divergence operator. Thus, in the Lorentz gauge, the 4-divergence of Aμ is always exactly 0 at every point in spacetime. In relativistic quantum field theory (aka quantum electrodynamics, or QED), the potential Aμ is interpreted as the quantum field of a 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.7.4) 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. Because of gauge invariance (A.7.2), a gauge transformation 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 a 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.7.2). More generally, symmetries (invariances) of the action always result in conserved quantities. Since 1949, unusual names have been given to similar conversed quantities: isospin, strangeness, color, charm, etc. The association of a conserved quantity with a differential symmetry of the action is known as Noether's Theorem, in honor of Emmy Noether who first showed this connection in 1915. __________________________________________________ 4. Fact 4: If div B = 0 and curl E = - ∂B/∂t , then there exist both A' and φ' such that B = curl A' and E = - grad φ'-∂A'/∂t, and the quantity div A' may be set to any function f. Proof: We know from Fact 2 that A' exists such that B = curl A' and such that div A' equals any arbitrary function f. 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.2.3) 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, this says that - 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 (A.4.1) where 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, so there are an infinite number of gauge transformations which leave the electric and magnetic fields invariant. We are free to choose a gauge such that div A' = f for any f we like.