Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / Section B / B9

For collinear inelastic capture followed by two photons, each particle has $\mathbf{p}=\gamma m\mathbf{u}$ and $E=\gamma m{c}^2$, with $\gamma =(1-u^2/c^2{)}^{-1/2}$. An isolated collision conserves the vector sum of momentum and total energy, including rest energy, internal excitation and radiation. It need not conserve the sum of individual rest masses. At low speed,recovering the classical kinetic energy after subtracting $m{c}^2$.

Here the incoming Lorentz factors are $5/3$ and $5/4$, soThe composite therefore hasIts rest mass exceeds $7m$ because relative kinetic energy has become internal rest energy.

For the subsequent collinear photon pair, $E_{+}+E_{-}=10m{c}^2$ and $(E_{+}-E_{-})/c=mc$. Thus $E_{+}=11m{c}^2/2$, $E_{-}=9m{c}^2/2$, andThe positive-$x$ photon is the higher-frequency one.