 Compton recoil kinematics

Choose the incident photon direction as $+x$ and the plane of the collision as the $xy$ plane, with the outgoing photon above the $x$ axis. The Electron initially has zero momentum. Conservation of energy and momentum conservation, using the relativistic energy–momentum relation, giveSquaring the Electron energy and the momentum magnitude in these two equations yieldsCancel common terms to obtain $m_e{c}^2(E-E^{\prime })=E{E}^{\prime }(1-\cos \theta )$. Division by $E{E}^{\prime }>0$ gives $1/E^{\prime }-1/E=(1-\cos \theta )/(m_e{c}^2)$. Substitution of $\lambda =hc/E$ proves both Compton scattering formulas. Using the nonrelativistic approximation $T_e=|{\mathbf{p}}_e{|}^2/(2m_e)$ here would lose the exact relativistic result.

Let $\varphi$ be the nonnegative Electron recoil angle below the incident direction. For $0<\theta <\pi$, the vector components above implyThis assumes exactly the same initially stationary free Electron and single outgoing photon. At $\theta =\pi$, the Electron moves forward with $\varphi =0$; at $\theta =0$ it stays at rest, so its recoil direction is undefined even though the limiting angle as $\theta \downarrow 0$ is $\pi /2$. At $\theta =90^{\circ }$ in the chosen example, $c{\mathbf{p}}_e=(70,-61.566)\mathrm{keV}$ and $\varphi \simeq 41.33^{\circ }$.