Inverse Compton scattering

Remarks

Example: let $\gamma \gg 1$ and let the photons be isotropic, in the Thomson regime $\gamma \varepsilon \ll m_e c^2$. Then the mean energy after scattering is about $\frac{4}{3} \gamma^2 \varepsilon$, so $\varepsilon_1 > \varepsilon$ on average. The net power transferred to the photons is $P = \frac{4}{3} \sigma_T c \gamma^2 \beta^2 U_{\text{rad}}$. Here $\beta = v/c$ with $v$ the speed of the electron, $\sigma_T$ is the Thomson cross-section and $U_{\text{rad}}$ is the energy density of the radiation. Non-example: an X-ray photon scattering off an electron at rest loses energy; this is ordinary Compton scattering. The loss is given by the Compton formula $\lambda_1 - \lambda = \frac{h}{m_e c}(1 - \cos\theta)$. Here $\lambda, \lambda_1$ are the wavelengths before and after, $h$ is the Planck constant and $\theta$ the scattering angle. The ratio of inverse Compton to synchrotron power of the same electron is $U_{\text{rad}} / U_B$, where $U_B$ is the energy density of the magnetic field.