Introduction to Electrodynamics · Problem 11.18

Electrodynamics and Relativity — David J. Griffiths

Physics · radiation-reaction

Problem

A point charge qq, of mass mm, is attached to a spring of constant kk. At time t=0t = 0 it is given a kick, so its initial energy is U0=12mv02U_0 = \frac{1}{2}mv_0^2. Now it oscillates, gradually radiating away this energy. (a) Confirm that the total energy radiated is equal to U0U_0. Assume the radiation damping is small, so you can write the equation of motion as x¨+γx˙+ω02x=0,\ddot{x} + \gamma\dot{x} + \omega_0^2 x = 0, and the solution as x(t)=v0ω0e−γt/2sin⁡(ω0t),x(t) = \frac{v_0}{\omega_0} e^{-\gamma t/2}\sin(\omega_0 t), with ω0≡k/m\omega_0 \equiv \sqrt{k/m}, γ=ω02τ\gamma = \omega_0^2 \tau, and γ≪ω0\gamma \ll \omega_0 (drop γ2\gamma^2 in comparison to ω02\omega_0^2, and when you average over a complete cycle, ignore the change in e−γte^{-\gamma t}). (b) Suppose now we have two such oscillators, and we start them off with identical kicks. Regardless of their relative positions and orientations, the total energy radiated must be 2U02U_0. But what if they are right on top of each other, so it's equivalent to a single oscillator with twice the charge; the Larmor formula says that the power radiated is four times as great, suggesting that the total will be 4U04U_0. Find the error in this reasoning, and show that the total is actually 2U02U_0, as it should be.

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