Understanding Electromagnetic Waves: Retarded Potentials & Radiation, Study notes of Astronomy

Download Understanding Electromagnetic Waves: Retarded Potentials & Radiation and more Study notes Astronomy in PDF only on Docsity! Retarded Potentials and Radiation No, this isn’t about potentials that were held back a grade :). Retarded potentials are needed because at a given location in space, a particle “feels” the fields or potentials of other charges, not where those charges are now, but where they were a light travel time ago. A few lectures ago we talked about how electromagnetism can be phrased in terms of potentials rather than fields. Ask class: if you have a single charge e a distance r away from a given point, what is the electrostatic potential there, ignoring light travel times? It’s just φ = e/r. Now, generalizing, suppose one had lots of (static!) charges at different locations. Ask class: how then would you find the potential? It’s additive, so the total potential would be φ = ∑ i ei/ri , (1) where each source has charge ei and is a distance ri from the point in question. Now, suppose that you had a bunch of moving charges ei. If you pick a particular time t and you know that for each particle the distance was rreti (t) a light travel time ago, Ask class: what will the potential be then? φ = ∑ i ei/r ret i (t) . (2) Note that more distant charges will have had a longer light travel time than nearer charges, so we can no longer evaluate all the particles simultaneously. If we take this formula and write it for a continuous charge density ρ, then the scalar potential at a time t and location r is then φ(r, t) = ∫ [ρ]d3r′ |r − r′| (3) where the brackets mean to evaluate the quantity at the retarded time: [Q] = Q ( r′, t − 1 c |r − r′| ) . (4) Note that here we’ve shifted things slightly: instead of following individual charges, we’re evaluating the potential from fixed points in space, but allowing the charge density to change. It amounts to the same thing. One can go through an identical procedure for the vector potential: A(r, t) = 1 c ∫ [j]d3r′ |r − r′| . (5) Remember that the potentials have gauge freedom, so in writing φ and A this way we’ve chosen a particular gauge, in this case the Lorentz gauge, in which ∇ · A + 1 c ∂φ/∂t = 0 . (6) That’s no big deal in most circumstances, but it’s good to be clear. A specific charge “distribution” that one can imagine is that of a single charge(!). If it has charge q and moves along a trajectory r0(t), so that its velocity is u(t) = ṙ(t), then we can congratulate ourselves on being cool by writing the charge and current densities in terms of Dirac delta functions: ρ(r, t) = qδ(r − r0(t)) , j(r, t) = qu(t)δ(r − r0(t)) . (7) Lovely. If we crank a bit (see §3.1 of Rybicki and Lightman) we get the Liénard-Wiechart potentials φ = [ q κR ] A = [ qu cκR ] . (8) Here the brackets mean an evaluation at the retarded time, and κ(t′) ≡ 1 − n(t′) · u(t′)/c, where n = R/R is the unit vector in the direction to the charge. That κ factor is mighty important. If the charge isn’t moving, u = 0, the whole thing is just 1 and you get the potentials you expect from electrostatics (A = 0, φ = q/R). On the other hand, for speeds close to the speed of light this factor produces strong intensity in the direction of motion of the particle, i.e., a beaming effect. Let us ponder what this does for us. You may remember that the Poynting flux carried by an electromagnetic field is S = c 4π E × H. If you forget, remember that (aside from the 1/4π factor) you can get this just from units, if you remember that B2 and E2 have units of energy density. Anyway, think about a static charge distribution. Ask class: what are φ and A? φ is just the sum of qi/Ri evaluated at the charges’ current position (since they aren’t moving), and A = 0 because the charges are fixed (no current). Ask class: so, what is S in that case? Zero, of course. A static charge distribution has no flux. It is therefore clear that radiation requires motion. What is less clear is that it’s the retarded time that is the key. One can differentiate the potentials to get the fields (RL say that this is “straightforward but lengthy”, always a warning that you may be in for an algebraic challenge). One then ends up (§3.2 in RL) with an electric field that can be written as the sum of two terms: one that does not depend on the acceleration β̇ (where β = u/c), and one that does. Ask class: which of these terms do they expect will produce radiation? From the arguments we used a few classes ago, it must be the one proportional to the acceleration. A static charge distribution emits no radiation, so a charge at constant velocity can’t either (since you could transform to a frame in which it is static). Now an interesting thing is to look at the radius dependences of the two terms for the electric field (and the corresponding magnetic field, which is just B(r, t) = [n × E(r, t)]). The “velocity field” term is proportional to 1/R2, whereas the “radiation field” term is proportional to 1/R. If we consider just the velocity field then the flux S ∝ EB ∝ 1/R4, as d ≡ ∑ i qiri. In an analogous way to the Larmor formula for a single charge’s power, the radiation power is P = 2d̈2/3c3 in this approximation. You may be familiar from other cases with dipole or multipole expansions. For example, in Newtonian gravity you can estimate the potential around a mass distribution by expanding it in powers of the distance, where the first term is that of a point mass at the center of mass, and higher order terms come in as well. It’s similar here, with the caveat that we have to think of nonrelativistic motion to be strictly correct. The book goes into a little more detail about general multipolar expansions. Our last topic for this lecture will be radiation reaction. Since an accelerated particle radiates, it carries away energy, linear momentum, and angular momentum. This means that the motion of the particle itself must be modified. We can get an approximate idea of what this modification does by treating it as an extra force, the force of radiation reaction. First, though, let’s figure out under what circumstances the force can be treated as a perturbation. Suppose the particle has a speed u, so its kinetic energy is ∼ mu2. Then from the Larmor formula the time in which the kinetic energy is changed substantially is T ∼ mu2/P ∼ (3mc3/2e2)(u/u̇)2 . (16) Let’s estimate a typical orbital time for the particle of tp ∼ u/u̇. Then for the energy lost in an orbital period to be small, T/tp ≫ 1, or tp ≫ τ ≡ 2e 2/3mc3 ≈ 10−23 s(!). That’s a mighty small time. It is about the time necessary for light to travel a distance equal to the classical electron radius 2.8 × 10−13 cm. Now, to figure out the force our first inclination would be to set the force times the velocity equal to the power radiated, Frad · u = −2e 2u̇2/3c3. The problem is that (1) Frad can’t depend on u since that would imply a preferred frame, so (2) one side of this equation depends on u while the other doesn’t! Oops. Instead, we can see if this can be satisfied in a time-averaged sense. − ∫ t2 t1 Frad · udt = (2e 2/3c3) ∫ t2 t1 u̇ · u̇dt = (2e2/3c3) [ u̇ · u ∣ ∣ ∣ ∣ t2 t1 − ∫ t2 t1 ü · udt ] . (17) In the second step we integrated by parts. If we assume that the motion or periodic, or at least that u̇ · u is the same at t2 as at t1, then the first term on the right hand side vanishes and we find that Frad = mτ ü in a time-averaged sense. That’s all very well, but as always we want to know the limits of this expression. Ask class: what does this say about a particle with constant linear acceleration? Then ü = 0, so it would imply no radiation reaction force, even though the particle does radiate (since it is accelerated). The problem is that then the expression at the limits doesn’t vanish, so our approximation is not valid. For most cases, though, it is if you average over a long enough time and the motion is bounded. If you put this into a grand equation of motion it reads m(u̇ − τ ü) = F (18) assuming some applied force F. This is a little weird. One normally doesn’t encounter third time derivatives like this. A problem one can encounter in such cases is spurious solutions. For example, suppose F = 0. Then u=constant is obviously a solution, but so is u = u0 exp(t/τ), which is a runaway solution that becomes large quickly. In such cases one must use physical or mathematical considerations to eliminate this solution. The mathematical reason is that u̇ · u(t1) 6= u̇ · u(t2), so in fact the radiation reaction force would have a different form. You have to be careful in cases like this.