Past exam of the physics 1 paper of the University of Oxford / 2017 / 9 / c

Let the incident photon energies be $E$ and $\epsilon$, meeting at angle $\psi$ in the laboratory frame. Add their four-momenta; because each is null, the squared total invariant energy isFor photon-photon pair production, its threshold value is $(2m_e{c}^2{)}^2$: the electron and Positron have zero relative momentum in their centre-of-momentum frame. The minimum possible $E$ occurs for a head-on collision, $\psi =\pi$, soWith $m_e{c}^2\simeq 0.511\mathrm{MeV}$ and $\epsilon =10^{-3}\mathrm{eV}$,The quoted background photon energy is used directly, without substituting a different characteristic value of the cosmic microwave background. A less opposed collision has a higher threshold; parallel photons cannot produce this massive pair.

At the threshold found in i, both products are stationary in the centre-of-momentum frame, so they have the same velocity in the laboratory frame. Take the energetic incident photon along $+x$. The total energy is $E+\epsilon$ and total momentum is $(E-\epsilon )/c$, so the centre-of-momentum frame moves atFor the numerical threshold,This is extremely close to, but strictly below, the speed of light. Each product has energy $(E+\epsilon )/2$ in the laboratory frame. Above threshold the same ratio of total momentum to total energy determines the motion of the pair's centre-of-momentum frame, but the individual electron and Positron need not have a common velocity.