Maximum deflection in soft-photon capture (${\varphi }_{\max }\simeq 3\sqrt{3}e/(8E_0),{\theta }_{\max }\simeq 120^{\circ }$)

When $e\ll m\ll E_0$, the incoming massive-particle energy is $E\simeq 2E_0/(1-\cos \theta )$. Momentum addition then gives $\tan \varphi \simeq [e/(2E_0)]\sin \theta (1-\cos \theta )$. Differentiating the angular factor gives its maximum at $\cos \theta =-1/2$. This is an asymptotic result for both the energy-angle constraint and final deflection; corrections are controlled by $m^2/E_0^2$ and $e/E_0$.