= Equal-energy atmospheric muon-pair survival
{title2=$P_{\rm both}=\exp[-4Hm_\mu c^2/(c\tau_\mu p_{\rm in}c)]$}
A vertically incident <particle> annihilating with a stationary target into two equal-energy <muons> gives each half the incident longitudinal <momentum>, while their transverse <momenta> are opposite. At height $H$ above a locally flat surface, each proper flight time is $2Hm_\mu c^2/[c(p_{\rm in}c)]$, since $\gamma v_z=p_z/m_\mu$. Independent exponential <particle decays> then give the displayed joint survival <probability>. Atmospheric <energy loss>, scattering and curvature are omitted; replacing both trajectories by vertical paths would miss the two-body angular correction.
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