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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:

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

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

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

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

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

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

Therefore the spectrum of produced photons is still a power law, with index $ \frac{a+1}{2}$
next up previous contents
Next: Modeling the Gamma-Ray Emission Up: The physical processes producing Previous: Electron Bremsstrahlung   Contents
Andrea Giuliani 2005-01-21