 Solution

Let $p$ be the magnitude of the incident particle's momentum. The relativistic energy-momentum relation givesThe stationary particle contributes rest energy $m{c}^2$ and zero momentum. Conservation of four-momentum therefore gives equal photon energies $\epsilon =(E+m{c}^2)/2$. A photon has momentum magnitude $\epsilon /c$, so if the two momentum vectors meet at angle $\theta$,Substitution gives the angular relation for two-photon electron-positron annihilation, whose derivation depends only on the stipulated equal rest masses and equal photon energies:All energies and the angle here are in the laboratory inertial frame.

For perpendicular photon trajectories, $\cos \theta =0$ and hence $E=3m{c}^2$. Since $E=\gamma m{c}^2$ with Lorentz factor $\gamma =(1-v^2/c^2{)}^{-1/2}$,For the small-speed limit, set $\beta =v/c\ge 0$ and $\delta =\pi -\theta$. The angle formula impliesUsing $\arcsin z=z+O(z^3)$ therefore gives $\delta =\beta +O({\beta }^3)$, soAt zero incident speed the momenta are equal and opposite, so $\theta =\pi$, consistent with the expansion.