Assuming a very high energy photon (energy $E$) crosses the atmosphere and produces an electron-positron pair, I would like to know what is the angle between these to leptons produced. I was trying to calculate it by applying the energy-momentum conservation and realized that in this case the angle could be 0 if the momentum $p$ does not need to be conserved. My question therefore is: does
Question: Does $p$ need to be conserved in the interaction or is it enough that the following relation applies: $E^2=2(p_ec)^2+2(m_ec^2)^2$ ($$ E^2=2\left(p_\text{e}c\right)^2+2\left(m_\text{e}c^2\right)^2 \,,$$where $p_e$$p_\text{e}$ is the momentum of the resulting electron/positron and $m_e$$m_\text{e}$ its mass)?