What is the product of a proton interacting with the CMB?

Question

Assuming the CMB is at 2.7 K, if a proton interacts with it, what would be the particles resulting of this collision? I read that at the GZK cut-off (~$10^{21}$ eV), there is photo-pion production, but if the proton has three quarks, how does it produce a pion with two? Does anybody know what would it be the Feynman diagram of the process?

Answer 1

The main reaction is the production of the $\Delta^+$ particle. This has the same uud quark content as the proton but the spins are aligned so it is an excited state with a mass of 1232 MeV. It decays by the strong interaction to $p \pi^0$ or $n \pi^+$,(as $1232>938+140$) or in quark terms a gluon gives a $u \overline u$ or $d \overline d $ pair and the quarks rearrange into a meson and a baryon.