# Minimal frequency for Cherenkov radiation

I'm trying to figure out the possible range of frequencies of the photons emitted in the Cherenkov effect, and came across something strange:

If I assume that the emitting electron's initial energy is $$E$$, and I want to find the energy of the emitted photon $$E_\gamma$$, then from conservation of energy I get:

\begin{align} E&=E_\gamma +\sqrt{m^2c^4+p^2c^2}\\[1ex] E_\gamma &=E-\sqrt{m^2c^4+p^2c^2}\\[1ex] \nu &=\frac{E-\sqrt{m^2c^4+p^2c^2}}{h}\geq \frac{E-mc^2}{h} \end{align}

Meaning - I get that there is a minimal possible frequency for the emitted photons. While I expect there to be a maximal frequency for a given energy $$E$$, I did not expect there to be a minimal one - I thought that the frequency of the emitted photon can be arbitrarily small, depending on the momentum of the electron after emission.

Could anyone explain why that is? Is there a reason for this "leap"? Or are my calculations somehow wrong? I'm not very familiar with this effect, sorry if there is some obvious answer.

edit - seems that I made a foolish mistake with the greater than sign - it should be smaller than. The equation indeed provides a maximal frequency, and not a minimal one. Thanks for pointing out my mistake!

• The greater than symbol should be a small than. you are estimating a negative quantity – lalala Apr 9 at 20:01
• You're right, how emberassing haha. Gonna probably the post since it's now totally irrelevant. – GSofer Apr 9 at 20:05

$$\frac{d^2E}{dxd\omega}=\frac{e^2}{4\pi}\mu(\omega)\,\omega\left[1-\frac{c^2}{v^2n^2(\omega)}\right]$$
so that the high energy cut-off occurs because anomalous dispersion where $$n(\omega) < 1$$. Most photons are ultraviolet, many are blue, and almost none are red. Lower frequencies just aren't considered.