# Inverse Compton Effect

1. Jul 11, 2010

### FunWarrior

Hello !

My question concerns the inverse Compton effect. I would like to determine the energy $$E_{\dif}$$ of a photon after this diffusion. Its initial energy is $$E_{\gamma}$$ and the other particle has an energy $$E$$, a mass $$m_0$$ and a relativistic speed. It is a head-on collision.

$$u$$ is the speed of the particle before the collision, $$v$$ is the speed of the particle after the collision and $$\gamma_u = \frac{1}{\sqrt{1-(\frac{u}{c})^2}}$$ (the same for $$\gamma_v$$).

I used the conservation of the quadrivector energy-impulse and I found the equations $$E_{\gamma} + E = E_{dif} + \gamma_u m_0 c^2$$ and $$E_{\gamma} - E \frac{v}{c} = E_{dif} - \gamma_u m_0 c u$$ where $$E = \gamma_v m_0 c^2$$.

I think that it's not true and I would like to know why.

Thank you