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

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

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

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

Thank you

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# Inverse Compton Effect

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