= Solution
In <time> $\Delta t$ the particle advances $\beta c\Delta t$, while a light wavelet travels $c\Delta t/n$. Coherent addition along the direction of emitted <photons> gives the <Cherenkov cone angle>
$$
\boxed{\cos\alpha=\frac1{n\beta},\qquad n\beta>1.}
$$
The angle is measured between <photon> propagation and the particle velocity. It is complementary to the angle between the conical <wavefront> surface and the trajectory; confusing those two angles exchanges sine and cosine. At threshold the <photon> cone closes, $\alpha\to0$.
Back to article page
