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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.

Thanks in advance.

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!

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  • $\begingroup$ The greater than symbol should be a small than. you are estimating a negative quantity $\endgroup$ – lalala Apr 9 at 20:01
  • $\begingroup$ You're right, how emberassing haha. Gonna probably the post since it's now totally irrelevant. $\endgroup$ – GSofer Apr 9 at 20:05
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Energy considerations in Čerenkov is not a fruitful path, as the phenomenon is driven by the phase velocity of light in the material.

For example: in air, the electron threshold is 21 MeV and there are about 30 photons per meter emitted, so that's about 1 millionth the total energy. Moreover, ionization losses are much greater, one the order of an MeV per gram per square-centimeter.

The spectrum is given by the Frank-Tamm formula:

$$ \frac{d^2E}{dxd\omega}=\frac{e^2}{4\pi}\mu(\omega)\,\omega\left[1-\frac{c^2}{v^2n^2(\omega)}\right]$$

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.

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I think what your equation is saying is that ν is greater than zero as you would expect.

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