Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 3 / 35B / Solution

Choose polar angle $\theta$ from the acceleration direction. Then $|n\times q\dot{v}{|}^2=q^2|\dot{v}{|}^2{\sin }^2\theta$. Integrating over the solid angle gives $2\pi {\int }_0^{\pi }{\sin }^3\theta d\theta =8\pi /3$, and hence the specified normalization gives Larmor formulaThis uses the source's convention for radiated power; standard SI normalization has an additional factor $1/c$ in both given power formulas.

To leading order when $W\ll E$, neglect the loss while computing the head-on trajectory. Mechanical energy conservation gives $\frac{1}{2}m{\dot{r}}^2+V(r)=E=V(r_0)$ and acceleration magnitude $|V^{\prime }(r)|/m$. On each leg,The incoming and outgoing legs contribute equally. Multiplying the power by this time element, adding both legs, and assuming the integral converges givesUsing the unperturbed turning point is justified at this order in the small radiative loss, for a regular turning trajectory.