Cherenkov radiation

This happens every time charged particles pass through matter, but we don't see Cherenkov radiation normally. So what? The reason is that these waves cause destructive interference with each other and cancel out shortly after emission, leaving no real radiation as a whole. The difference occurs when the particle's speed exceed the speed of light in the medium:

In this case, the particle moves faster than the wave can propagate, which causes them to sum coherently and not cancel out, thus letting the radiation reach far distances.

The angular opening $\theta_{C}$ of the conical wavefront is called the Cherenkov emission angle. It can be found through purely geometrical considerations.

$\beta ct$ is the hypotenuse and $ct/n$ is its cosine so

Rearrange

There's a few considerations to make:

1. The velocity must be $\beta\geq1/n$, otherwise $1/n\beta>1$ which is impossible for a cosine. In other words, this is true only if the particle moves faster than the speed of light in the medium, as we claimed before.
2. Since $n$ is assumed to be known, measuring $\theta_{C}$ is equivalent to measuring $\beta$. Measuring $\theta_{C}$ is easier than measuring $\beta$ directly.
3. The maximum angle occurs when $\beta=1$, in which $\theta_{C}=\arccos(1/n)$. For regular glass ($n=1.52$), this is about $\theta_{C,\text{max}}=48.9°$.
4. There's no requirement on what $n$ needs to be. Any transparent material can be exhibit Cherenkov radiation.

Cherenkov radiation is also inversely proportional to the wavelength (not proven here, search "Frank-Tamm formula" if you're curious) and it emits a continuous spectrum straddling ultraviolet and visible light, with a peak at 420 nm. For this reason, Cherenkov radiation is visible to the naked eye and appears blue/violet. It's a common sight in nuclear reactor pools, which is why blue is often used in nuclear energy logos and symbols.

This phenomenon is geometrically identical to a sonic boom in sound or the wake formed by a boat moving faster than sea waves.

Energy loss#

The energy loss due to Cherenkov radiation depends on the number of photons emitted per unit path length. The number $N$ of these in turn depends on $\theta_{C}$ (indeed the number emitted per unit length is $dN/dx\propto \sin^{2}\theta_{C}$). It turns out that the energy lost per unit path is

usually measured in keV/cm, a thousand times smaller than collisional loss. For this reason, it can be neglected.

Detectors#

It's sometimes useful to express the threshold speed as a gamma:

Particle speeds can be quite intense, so using $\gamma$ is often advised to avoid having to deal with minuscule differences between $\beta$ and $c$. The threshold $\gamma$ increases as $n\to 1$, meaning as the materials stops refracting and just lets light through. As such, barely-refracting materials can be used to make very high speed Cherenkov thresholds, useful to fine-tune the value for experimental use in detectors. For instance, pressurized gases can be used to manipulate $n$ to be very near one. This is particularly useful at very high speeds where the Bethe-Bloch formula stops providing clear distinctions between particles¹.

The emission angle can be measured using a photosensitive screen or ring on which the radiation cone impacts. From the radius of the circle measured on the screen, we can determine the emission angle.