Past exam of the physics 1 paper of the University of Oxford / 2014 / 10

The incident particle obeys the energy–momentum relation $E^2=p^2{c}^2+m_1^2{c}^4$. The stationary target contributes energy $m_2{c}^2$ and zero momentum. A single product inherits the total four-momentum, so its invariant mass satisfiesHenceHere $E$ is the incident total energy, including rest energy. In a one-particle final state, exact production of a specified mass fixes the incident energy; there is no adjustable final relative motion to absorb excess energy. A resonance with a finite decay width relaxes that ideal exact equality over its line shape.

For fixed-target electron-Positron formation, set $m_1=m_2=m_e$ in a and solve for the incident total energy:The incident kinetic energy is smaller by another $m_e{c}^2$, a negligible relative change here. Most of the fixed-target energy becomes motion of the final rho meson in the laboratory.

With equal and opposite incoming momenta, the centre-of-momentum frame is the laboratory. Each incoming particle has the same total energy, and there is no final translational kinetic energy at formation:This large contrast illustrates the invariant-energy advantage of colliding beams. The given $768\mathrm{MeV}/c^2$ resonance mass is used rather than replacing it with a modern tabulated value.

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.