Why is Cherenkov radiation blue? And what about refractive index?

[.Scott]

TL;DR Summary

The exact color of Cherenkov radiation is a function of the transparent material used.
For any wavelength, the relative intensity is dependent on refractive index (and more).
Can it be anything other than bright blue? Any neat experiments?

I got this question from my son last night.

If you Google "Why is Cherenkov radiation blue", you get this:

Due to the high energies at play during Cherenkov radiation, the photons travel as waves that have high frequencies and short wavelengths, which are typical of violet and blue colours. The higher the frequencies and the shorter the wavelengths are, the bluer or more violet the light appears to the human eye.

Somewhat more substantial is the Wiki article on the Frank-Tamm formula.
That formula ties the Cherenkov radiation wavelengths to the transmission characteristics at any specified wavelength - and specifically to the refractive index. It also states:

The relative intensity of one frequency is approximately proportional to the frequency. That is, higher frequencies (shorter wavelengths) are more intense in Cherenkov radiation. This is why visible Cherenkov radiation is observed to be brilliant blue.

In general, a transparent material will respond to shorter wavelengths with a higher index of refraction - and thus more Cherenkov. So blue energy is pushed by both the higher energy per photon and the higher refractive index.

My first thought was that perhaps either of the two components of an achromatic lens (flint glass and crown glass) would provide an example of a material where the index of refraction decreases at with shorter wavelengths - but I suspect I am wrong. From what I read, they may create achromatic results even though both would "push blue".

Are there any materials that would "push red" and are there any that would push it enough to make the Cherenkov radiation look something other than blue?

 

[snorkack]

I think that actual negative dispersion is confined to bands of strong absorption. Which means that if a particle travels in a medium which is opaque to blue light, it is possible that blue Cherenkov photons are not just emitted and promptly absorbed but never emitted in the first place.
How wide do absorption bands need to be in order to produce negative dispersion? Can you have a material which is transparent to red, opaque to green and again fairly transparent to blue - but with a smaller refractory index than in red?

 

[Vanadium 50]

It's not blue. It's mostly ultraviolet. Does that help?

In principle, the energy distribution is propoprtional to the frerquency. In practice, this cuts off once the wavelength gets below the atomic scale and "speed of light in the material" ceases to have meaning.

 

[malawi_glenn]

There are ideas to use the moon as a neutrino detector, using the radio-"part" of the Cherenkov spectrum for detection
https://www.sciencedirect.com/science/article/abs/pii/S0168900209004938
https://arxiv.org/abs/1108.2459
https://indico.cern.ch/event/174201/contributions/1432974/attachments/221397/309979/3_YP2012.pdf

Anyway, have you played around with different $\mu(\omega)$ and $n(\omega)$ and see which yields your "desired" outcome?