Vavilov–Cherenkov radiation
An electron moves in a homogeneous, transparent, nonmagnetic medium with refractive index \(n > 1\). If the speed of the electron is \(v > 1/n\), the decay \(e(p) \to e(p') + \gamma(k)\) becomes possible, which gives rise to Vavilov–Cherenkov radiation. The photon energy is \(\omega_k = |\boldsymbol{k}|/n\). Its polarization is \(\varepsilon^\mu_\lambda(\boldsymbol{k}) = (0,\boldsymbol{\varepsilon}_\lambda(\boldsymbol{k}))\).
- Show that the squared matrix element, summed over the spins of the final electron and averaged over the spins of the initial electron, is \[|\mathcal{M}|^2 = 4\pi \alpha \left( 4\left(\boldsymbol{p}\cdot \boldsymbol{\varepsilon}_\lambda(\boldsymbol{k})\right)^2 - k^2 \right).\]
This expression shows that the photon is polarized in the scattering plane spanned by the vectors \(\boldsymbol{p}\) and \(\boldsymbol{k}\).
Show that the spectral and angular distribution of the radiation is \[\frac{d^2\Gamma}{d\omega\, d\Omega_\gamma} = \frac{|\mathcal{M}|^2}{32\pi^2 v E^2}\, \delta\left( \cos\theta - \frac{1}{vn} - \frac{\omega(n^2 - 1)}{2v n E} \right),\] which shows that the polar emission angle of the photon (the \(z\) axis is along the momentum \(\boldsymbol{p}\) of the initial electron) is fixed by the energies of the initial electron and the final photon: \[\cos\theta \equiv \cos\theta_0 = \frac{1}{vn} + \frac{\omega(n^2 - 1)}{2vnE}.\]
Show that the spectral distribution of the radiation is \[\frac{d\Gamma}{d\omega} = \alpha \left[ v \sin^2\theta_0 + \frac{\omega^2 (n^2 - 1)}{2v E^2} \right].\]
The future Nobel laureate Vitaly Lazarevich Ginzburg solved this problem in 1940.
Neglecting terms of order \(\omega/E\), we recover the classical results of two other Nobel laureates, I. E. Tamm and I. M. Frank: \[\frac{d\Gamma}{d\omega} = \alpha v \sin^2 \theta_0, \qquad \cos \theta_0 = \frac{1}{vn}.\]