I have to write a paper on Cherenkov detection.

And got a bit of an issue on the relativistic particle/recoil correction of the Cherenkov angle formula.

Normaly $$cos( \theta_c) = \frac{1}{n\beta} $$

If I go to most papers I see the recoil formula as $$cos( \theta_c) = \frac{1}{n\beta} + \frac{\hbar k (1-\frac{1}{n^2})}{2p}$$. But my prof did supply me with a few articles to start me up, and in one of them [http://large.stanford.edu/courses/2014/ph241/alaeian2/] looking at their Cherenkov angle calculation and $cos(\theta_c)=2(p^2 c^2 + m^2 c^4)^{1/2}+\frac{(n^2-1)h\nu}{2 p c n} $ or rewriting it via $p=vm$, $\beta=v/c$, and $hn\nu=\hbar kc$

I'd get $$cos( \theta_c) = \frac{(\beta^2 +1)^{1/2}}{n\beta} + \frac{\hbar k (1-\frac{1}{n^2})}{2p} $$.

And to me this Stanford version makes no sense. Even if we go into the limit where $\hbar k << p $ where the second term falls away as in with the standard recoil version. Then suddenly we have $cos( \theta_c) = \frac{(\beta^2 +1)^{1/2}}{n\beta}$

What thought am I missing here?

EDIT: This would surely mean that for Cherenkov radiation in aerogels where n=1,01 and $\beta$ would be near c that $$cos(\theta_c)=\frac{\sqrt2}{\beta n}$$ somehow I find this hard to swallow without any other paper or book mentioning it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.