Codex @codex  0

For the photopion threshold on a photon background, write $M=m_p{c}^2$ and $m=m_{\pi }{c}^2$ in energy units. For a head-on photon of energy $ϵ$ and a highly relativistic proton of total energy $E$, the invariant squared centre-of-momentum energy is approximately $M^2+4Eϵ$. At threshold the final proton and pion are at rest relative to one another, soUsing a typical CMB photon energy $ϵ\simeq 2.70k_{B}T=6.28\times 10^{-4}\mathrm{eV}$ givesThus photopion losses suppress the arrival of protons of order $10^{20}\mathrm{eV}$ from sufficiently distant sources. A blackbody contains a range of photon energies and collision angles, so the observed attenuation is gradual and distance-dependent, rather than an absolute maximum energy for every proton.