Past exam of the symmetry and relativity course of the University of Oxford / 2016 / 4 / c

An isolated free electron in vacuum cannot spontaneously emit a nonzero real photon. If $P=P_f+Q$, where $Q$ is the emitted photon four-momentum, $P^2=P_f^2=m^2{c}^2$ and $Q^2=0$, conservation would require $P_f\cdot Q=0$. A future-pointing timelike four-vector has strictly positive inner product with every nonzero future-pointing null four-vector, so this is impossible.

The same conclusion follows from the classical Cherenkov radiation condition $v>c/n$: in vacuum $n=1$, while the electron with $\gamma =100$ has $\beta =\sqrt{1-10^{-4}}<1$. Hence there is no vacuum emission angle for the specified free particle. Acceleration by an external field or scattering from another body would change the physical problem and can permit radiation.

In a transparent uniform medium, constructive phase interference along the trajectory requires $\omega -\mathbf{k}\cdot \mathbf{v}=0$. Since $k=n\omega /c$, the Cherenkov cone angle obeysHere $n\beta =2\sqrt{1-10^{-4}}>1$, so emission is allowed andThe electron is not an isolated vacuum particle: it interacts with the dielectric polarization of the medium. Here the angle describes the photon direction relative to the trajectory; the conical wavefront surface has the complementary angle.

This angle is the classical or negligible-recoil result. At finite photon energy, energy-momentum bookkeeping in a nondispersive medium gives a recoil correction proportional to $\hslash \omega /E$; no photon frequency is supplied for such a correction. In a dispersive medium use $n(\omega )$ separately for each transparent frequency band.