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I am reading through these slides about the transition radiation in the optical range. It says that the spectrum of the intesity of the photons as a function of the frequency is given by:

$$\dfrac{dI}{d\omega}=\dfrac{e^2}{6\pi c} \left( \dfrac{\gamma \omega_p}{\omega} \right)^4$$

where $\omega_p$ is the plasma frequency and $\gamma$ is the relativistic gamma factor.

How do I calculate from this formula the number of the photons emitted in a given range of frequencies?

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  • $\begingroup$ Please explain what this question is about. $\endgroup$
    – my2cts
    Commented Apr 25, 2020 at 10:44
  • $\begingroup$ @my2cts I think the question is clear: I have that formula for the intensity of the transition radiation emitted by a particle with a gamma factor $\gamma$ and from that I want to calculate the number of photons emitted instead of the intensity. $\endgroup$
    – mattiav27
    Commented Apr 25, 2020 at 10:49
  • $\begingroup$ Fine but I did not read the slides. It would help if your question were self contained. $\endgroup$
    – my2cts
    Commented Apr 25, 2020 at 12:04

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Considering the intensity is in $W/m^2$ you will need to consider the area over which you want to calculate the number of photons say $A$. You can then multiply the $dI$ by the area to get the infinitesimal power $dP$

$$dP=dIA=\frac{A\alpha}{\omega^4}d\omega$$ where $\alpha=\frac{e^2}{6\pi c}\left(\gamma\omega_p\right)^4$, considering the energy of a photon at a frequency $\omega$ is $E=\hbar\omega$ the number of photons at $\omega$ emmited per unit time for $dI$ is then

$$dN=\frac{A\alpha}{\hbar\omega^5}d\omega$$

So to get the number of photons per unit time emitted over a range of frequencies $\omega_1,\omega_2$ is

$$N=\int_{\omega_1}^{\omega_2}\frac{A\alpha}{\hbar}\omega^{-5}d\omega=\frac{A\alpha}{4\hbar}\left(\omega_1^{-4}-\omega_2^{-4}\right)$$

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  • $\begingroup$ Thank you @jamie1989 $\endgroup$
    – mattiav27
    Commented Apr 25, 2020 at 11:12

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