 Solution

Use energy units $m=m_p{c}^2$, $M=m_A{c}^2$, $e=E_{\gamma }$ and $P=pc=\sqrt{E^2-m^2}$. The soft-photon resonant capture threshold follows from four-momentum conservation:For fixed proton energy the left side is largest at head-on incidence. Its threshold is $E+P=2E_0$. Combining that with $(E-P)(E+P)=m^2$ givesEquality is allowed for a head-on photon. Here $e=k_{B}T=2.3482\times 10^{-4}\mathrm{eV}$, soA very soft photon needs an enormous proton Lorentz factor to reach the specified invariant mass. A true thermal photon distribution and a finite-width resonance would smear this monoenergetic threshold.

For $\theta =0$, $E-P=m^2/(E+P)\le m$, whereas $2E_0\gg m$, so production is impossible. For each nonzero angle, sufficiently large energy is possible, and $P\simeq E$ givesTo find the maximum deflection in soft-photon capture, note that the final particle's transverse momentum comes entirely from the photon:Writing $x=\cos \theta$, maximization gives $(1-x)(2x+1)=0$ and the interior maximum $x=-1/2$. Therefore, to the same extraordinarily accurate ultrarelativistic order,The neglected relative corrections are of order $m^2/E_0^2$ and $e/E_0$, both around $10^{-24}$. The final particle is effectively collinear with the proton despite a large photon incidence angle.