Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 3 / 36C (Electrodynamics)

For a point charge, $p=qx$, so the radiation fields are transverse and satisfy $|E_{\mathrm{Rad}}|=c|B_{\mathrm{Rad}}|$. The radial Poynting vector isChoosing the acceleration as polar axis and integrating ${\sin }^2\theta$ over the sphere gives $\int {\sin }^2\theta d\Omega =8\pi /3$. Hence the Larmor formula is

Ignoring the small radiative back-reaction, conservation of energy gives $\dot{x}=\sqrt{2(E-f(x))/m}$ and Newton's equation gives $\ddot{x}=-f^{\prime }(x)/m$. Integrating the Larmor formula with $dt=dx/\dot{x}$ yields, for $E>V_0$,If $E<V_0$, the particle turns at $x_{\ast }$ defined by $f(x_{\ast })=E$ and traverses the accelerated region twice, so the same formula holds with twice the integral from $0$ to $x_{\ast }$.

For $f(x)=V_{0}x/L$, direct integration givesandThe reflected expression increases up to threshold and the transmitted expression decreases above it. The supremum is therefore approached from below at $E=V_0$ and equals