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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 $2H{m}_{\mu }{c}^2/[c(p_{\mathrm{in}}c)]$, 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.

Treat the atmospheric proton as stationary in the Earth frame, the incident antiproton as vertically downward, and neglect atmospheric energy loss, scattering and the Earth's curvature initially. Take $m_p{c}^2=0.9383\mathrm{GeV}$. Conservation of energy and equal outgoing energies giveEqual energies imply equal momentum magnitudes. Their transverse momenta cancel, so the two muons have equal downward components $p_z=p_{\bar{p}}/2$, whereEach muon therefore makes an angle $\theta$ with the downward vertical satisfyingThe paths are not both vertical: replacing their path lengths by $20\mathrm{km}$ would miss this two-body kinematic correction. Since $v_z=p_z{c}^2/E_{\mu }$, the proper flight time over vertical height $H$ isFor equal-energy atmospheric muon-pair survival, independent exponential particle decays with mean proper muon lifetime ${\tau }_0=2.2\mu \mathrm{s}$, both survive with probabilityEquivalently each travels about $20.55\mathrm{km}$, with proper flight time about $0.404\mu \mathrm{s}$, and survives with probability about $0.832$. The result is decay-only survival to an ideal surface, not a prediction including air interactions or detector efficiency. For a spherical Earth of radius $R_E$, the exact path length is $\ell =(R_E+H)\cos \theta -\sqrt{R_E^2-(R_E+H{)}^2{\sin }^2\theta }$; inserting $P_{\mathrm{both}}=\exp [-2\ell /({\gamma }_{\mu }{v}_{\mu }{\tau }_0)]$ gives a small curvature correction.

The dominant $z^2/{\beta }^2$ factor raises the energy loss as a charged projectile slows, while the logarithm moderates that rise. This produces a Bragg peak close to the end of its particle range. At still lower velocity, changing effective charge, shell effects and nuclear stopping invalidate extrapolation of the elementary Bethe stopping formula to zero energy. A finite initial energy gives a finite continuous-slowing-down range,A higher initial energy increases the particle range; greater electron density increases stopping power per unit length. Absorber $Z/A$ and $I$ also affect the depth. Multiple Coulomb scattering, energy-loss straggling, nuclear reactions and the incident energy spread broaden the measured peak and distal fall-off; nuclear secondaries can produce a small tail. Superposing many incident energies makes a spread-out Bragg peak for an extended target. The schematic curve below illustrates shapes only and is not patient radiation dosimetry.