Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 4 / 12B (Dynamics and Relativity)

Two-body energy-momentum conservation in the pion rest frame gives $E_e=(m_{\pi }^2+m_e^2)c^2/(2m_{\pi })$ and $p_e=(m_{\pi }^2-m_e^2)c/(2m_{\pi })$. Thus $v_e=c(m_{\pi }^2-m_e^2)/(m_{\pi }^2+m_e^2)$.

Let the incident positron total energy be $E$. For perpendicular photon momenta, real positive photon energies require
$(E+m_e{c}^2{)}^2-4m_e{c}^2(E+m_e{c}^2)\ge 0$.
Thus the threshold is $E=3m_e{c}^2$.

Squaring $p_H=p_1+p_2$ gives $m^2{c}^4=2E_1{E}_2(1-\cos \theta )=4E_1{E}_2{\sin }^2(\theta /2)$, proving the formula. If the rest-frame photons are along $\pm y$, an $x$-boost gives equal energies and $\sin (\theta /2)=1/\gamma$, hence $v=c\cos (\theta /2)$.