Soft-photon resonant capture threshold ($E_{\min }=E_0+m^2/(4E_0)$)

In energy units $m=m_{\mathrm{in}}{c}^2$ and $M=M_A{c}^2$, capture of a photon of energy $e$ by a massive particle into one resonance satisfies $E-P\cos \theta =(M^2-m^2)/(2e)=2E_0$, with $P=pc$. Head-on incidence gives the largest invariant mass at fixed $E$, so the threshold solves $E+P=2E_0$. Using $E^2-P^2=m^2$ gives the title expression. It is the minimum over incident angle; above that energy the exact angle is fixed by the invariant equation, not arbitrary. The soft-photon regime $2E_0\gg m$ excludes exactly co-propagating incidence.

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 $e/E_0$.