Introduction to Electrodynamics · Problem 11.20

Electrodynamics and Relativity — David J. Griffiths

Physics · radiation-reaction-mechanism

Problem

Deduce Eq. 11.100 from Eq. 11.99. Here are three methods: (a) Use the Abraham-Lorentz formula to determine the radiation reaction on each end of the dumbbell; add this to the interaction term (Eq. 11.99). (b) Method (a) has the defect that it uses the Abraham-Lorentz formula—the very thing that we were trying to derive. To avoid this, let F(q)F(q) be the total dd-independent part of the self-force on a charge qq. Then F(q)=Fint(q)+2F(q/2),F(q) = F^{\text{int}}(q) + 2F(q/2), where FintF^{\text{int}} is the interaction part (Eq. 11.99), and F(q/2)F(q/2) is the self-force on each end. Now, F(q)F(q) must be proportional to q2q^2, since the field is proportional to qq and the force is qEq\mathbf{E}. So F(q/2)=(1/4)F(q)F(q/2) = (1/4)F(q). Take it from there. (c) Smear out the charge along a strip of length LL oriented perpendicular to the motion (the charge density, then, is λ=q/L\lambda = q/L); find the cumulative interaction force for all pairs of segments, using Eq. 11.99 (with the correspondence q/2→λ dy1q/2 \to \lambda\,dy_1, at one end and q/2→λ dy2q/2 \to \lambda\,dy_2 at the other). Make sure you don't count the same pair twice.

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