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At time $t_0$, the accelerated Proton has momentum $p=qE{t}_0$ and, by the relativistic energy-momentum relation, energyThe incident Cosmic microwave background photon adds energy $E_{\gamma }$. Since the final state contains only the Delta-plus baryon, conservation of relativistic energy gives its laboratory-frame Lorentz factorIts rest-frame decay time $t_{\Delta }$ is a proper time, so time dilation gives the laboratory decay time

In the parent's rest frame, momentum conservation makes the massive daughter and photon momenta equal and opposite, of magnitude $p$. The relativistic two-body decay formulas giveSince $v=p{c}^2/E_R$ and $r=m/M$,The daughter's Lorentz factor isIf it decays after proper time $\tau$, time dilation gives $t=\gamma \tau$, while uniform motion gives $x=v\gamma \tau$. Therefore the event coordinates in the parent frame are

Since $r>3M$,Relative to the distant observer, the static clock $O_2$ suffers gravitational time dilation. The orbiting clock $O_1$ suffers the same gravitational effect together with the kinematic time dilation caused by its orbital motion, so it accumulates the least proper time.