# What is the energy threshhold to produce Cherenkov radiation?

I am in a nuclear course right now and am getting some misleading information from different sources. I am trying to figure out what the minimum total energy is that a proton must have in order to make Cherenkov radiation while passing through air? I am guessing it is in the TeV range? Furthermore to this question, does this mean the LHC produces Cherenkov radiation since the beam is about 7 TeV?

• In the LHC the beam passes through a vacuum not air Apr 6, 2015 at 10:30
• Ya right, of course. Does this mean there is no Cherenkov radiation in the LHC? since then v = c and the below equations fall apart? Apr 6, 2015 at 19:44
• And if the LHC were in air, the 7TeV would for sure produce Cherenkov radiation? Apr 6, 2015 at 21:11
• If the LHC were running in air it would be blasting the nitrogen and oxygen nuclei to bits so it would generate floods of radiation. In principle there would be Cherenkov radiation present, but it would be lost in the mass of proton-nucleus collisions. Apr 7, 2015 at 5:15

The refractive index of air is about 1.0003, so the speed of light in air is about 0.9997$c$. You can work out the energy of the proton at this speed using:
$$E^2 = p^2c^2 + m^2c^4$$
$$p = \frac{m_p v}{\sqrt{1 - v^2/c^2}}$$
The number of photons emitted by the Cherenkov radiation (per unit length of the particle track) is proportional to $$\left(1-\frac{1}{n^2\beta^2}\right)$$ where $n$ is the refractive index of the medium and $\beta=v/c$, but only when the expression above is positive (emission is zero for the negative case). The threshold is thus reached when this expression is exactly equal to zero, or $$\beta_\text{th}=\frac{1}{n}.$$ Since $E=\gamma mc^2$ and $\gamma=(1-\beta^2)^{-1/2}$ we get the threshold energy as $$E_\text{th}=\frac{mc^2}{\sqrt{1-\tfrac{1}{n^2}}}.$$ For proton with $m=938\,\text{MeV}/c^2$ in air with $n=1.0003$ this gives you a threshold energy of $E_\text{th}=38\,\text{GeV}$.