1 Answers
In pair production, a photon creates an electron positron pair. In the process of photons scattering in air , the most important interaction is the scattering of photons at the nuclei of atoms or molecules. The full quantum mechanical process of pair production can be described by the quadruply differential cross section given here:
d 4 σ = Z 2 α fine 3 c 2 2 ℏ | p + | | p − | d E + ω 3 d Ω + d Ω − d Φ | q | 4 × × 2 − p + 2 sin 2 Θ + 2 + 2 ℏ 2 ω 2 p + 2 sin 2 Θ + + p − 2 sin 2 Θ − + 2 | p + | | p − | sin Θ + sin Θ − cos Φ ] . {\displaystyle {\begin{aligned}d^{4}\sigma &={\frac {Z^{2}\alpha _{\textrm {fine}}^{3}c^{2}}{^{2}\hbar }}|\mathbf {p} _{+}||\mathbf {p} _{-}|{\frac {dE_{+}}{\omega ^{3}}}{\frac {d\Omega _{+}d\Omega _{-}d\Phi }{|\mathbf {q} |^{4}}}\times \\&\times \left^{2}}}\left\right.\\&-{\frac {\mathbf {p} _{+}^{2}\sin ^{2}\Theta _{+}}{^{2}}}\left\\&+2\hbar ^{2}\omega ^{2}{\frac {\mathbf {p} _{+}^{2}\sin ^{2}\Theta _{+}+\mathbf {p} _{-}^{2}\sin ^{2}\Theta _{-}}{}}\\&+2\left.{\frac {|\mathbf {p} _{+}||\mathbf {p} _{-}|\sin \Theta _{+}\sin \Theta _{-}\cos \Phi }{}}\left\right].\\\end{aligned}}}
with
d Ω + = sin Θ + d Θ + , d Ω − = sin Θ − d Θ − . {\displaystyle {\begin{aligned}d\Omega _{+}&=\sin \Theta _{+}\ d\Theta _{+},\\d\Omega _{-}&=\sin \Theta _{-}\ d\Theta _{-}.\end{aligned}}}