Past exam of the physics 1 paper of the University of Oxford / 2016 / 11

Write $m_e$ for the electron rest mass and $\epsilon =m_e{c}^2$. The incident positron has total energy $E$ and satisfies $p^2{c}^2=E^2-{\epsilon }^2$. Adding the target electron's four-momentum, the squared total centre-of-momentum energy isAt the fixed-target production threshold, all three final particles are at rest relative to one another, so $W=3M_0{c}^2$. ConsequentlyThis is the positron's total energy, not its kinetic energy. It assumes $3M_0\ge 2m_e$ and addresses kinematic allowance; other selection rules could still forbid a proposed reaction. For $M_0{c}^2=135\mathrm{MeV}$ and $m_e{c}^2\simeq 0.510999\mathrm{MeV}$, $E_{\min }\simeq 160.494\mathrm{GeV}$.

At threshold the three products share the centre-of-momentum frame's laboratory velocity. Their common Lorentz factor and speed areFor the subsequent two-photon decay, each photon has rest-frame energy $E_{\gamma }^{\ast }=M_0{c}^2/2$. A Lorentz transformation of a photon emitted at rest-frame angle ${\theta }^{\ast }$ to the boost gives $E_{\gamma }={\gamma }_{\mathrm{CM}}{E}_{\gamma }^{\ast }(1+{\beta }_{\mathrm{CM}}\cos {\theta }^{\ast })$. Therefore the kinematically allowed laboratory range isor numericallyThe two photon energies in a single event are correlated: their sum is the parent's laboratory energy ${\gamma }_{\mathrm{CM}}{M}_0{c}^2$. When one attains the upper endpoint, the other attains the lower endpoint. Angular distributions and detector acceptance can affect which part of this range is observed.