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

AI Thread Summary
Cherenkov radiation appears blue due to high-energy photons traveling at short wavelengths, which correspond to blue and violet light. The Frank-Tamm formula connects the radiation's wavelengths to the refractive index of materials, indicating that higher frequencies yield greater intensity in Cherenkov radiation. Transparent materials typically exhibit a higher refractive index for shorter wavelengths, enhancing the blue appearance. The discussion raises questions about materials that could potentially shift the radiation's color towards red and explores the concept of negative dispersion in relation to absorption bands. Overall, the conversation delves into the complexities of light behavior in different media and its implications for Cherenkov radiation.
.Scott
Science Advisor
Homework Helper
Messages
3,800
Reaction score
1,856
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?
 
  • Like
Likes Drakkith and vanhees71
Physics news on Phys.org
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?
 
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.
 
  • Like
Likes vanhees71 and malawi_glenn
Last edited:
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Thread 'Griffith, Electrodynamics, 4th Edition, Example 4.8. (First part)'
I am reading the Griffith, Electrodynamics book, 4th edition, Example 4.8 and stuck at some statements. It's little bit confused. > Example 4.8. Suppose the entire region below the plane ##z=0## in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility ##\chi_e##. Calculate the force on a point charge ##q## situated a distance ##d## above the origin. Solution : The surface bound charge on the ##xy## plane is of opposite sign to ##q##, so the force will be...
Dear all, in an encounter of an infamous claim by Gerlich and Tscheuschner that the Greenhouse effect is inconsistent with the 2nd law of thermodynamics I came to a simple thought experiment which I wanted to share with you to check my understanding and brush up my knowledge. The thought experiment I tried to calculate through is as follows. I have a sphere (1) with radius ##r##, acting like a black body at a temperature of exactly ##T_1 = 500 K##. With Stefan-Boltzmann you can calculate...
Back
Top