= Maximum deflection in soft-photon capture
{title2=$\phi_{\max}\simeq3\sqrt3 e/(8E_0),\quad\theta_{\max}\simeq120^\circ$}
When $e\ll m\ll E_0$, the incoming massive-particle <energy> is $E\simeq2E_0/(1-\cos\theta)$. <Momentum> addition then gives $\tan\phi\simeq[e/(2E_0)]\sin\theta(1-\cos\theta)$. 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 $e/E_0$.
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