 Solution

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.