Inverse Compton

The energy of a gamma photon produced via inverse Compton by an electron of energy $E_e$ and a photon of energy $E_{ph}$ is:

$$E_{\gamma} = \frac{4}{3} (\frac{E_{e}}{m_e c^{2}})^2 E_{ph}$$ (1.23)

The cross section of the reaction is the Thompson cross section ($\sigma_t$). The inverse-Compton gamma-ray emissivity of interstellar matter is therefore:

$$g_{IC}(E_{\gamma}) dE_{\gamma}= 4 \pi \sigma_t \frac{U_{ph}}{E_{ph}} f(E_e) dE_e$$ (1.24)

where $U_{ph}$ is the energy density of the radiation field of photons of energy $E_{ph}$. By integrating:

$$g_{IC}(E>E_{\gamma}) = 4 \pi \sigma_t \frac{U_{ph}}{E_{ph}} \int_{E_e}^{\inf} f(E_e^{'}) dE^{'}_e$$ (1.25)

where $E_{\gamma}$, $E_{ph}$ and $E_e$ are related by eq. 1.23. For a distribution of electrons with spectrum $N_e(E)= k E^{-a}$, from eq 1.24 $g_{IC}$ become:

$$g_{IC}(E_{\gamma}) = 4 \pi \sigma_t \frac{U_{ph}}{E_{ph}} f(E_e) \frac{dE_e}{dE_{\gamma}}=$$

$$= \frac{8 \pi \sigma_t}{3} k (m_ec^2)^{1-a} \left( \frac{3}{4 E_{ph}} \right)^{\frac{3-a}{2}} E_{\gamma}^{-\frac{a+1}{2}}$$ (1.26)

Therefore the spectrum of produced photons is still a power law, with index $\frac{a+1}{2}$