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.