= Solution
Write $m_e$ for the electron <rest mass> and $\varepsilon=m_ec^2$. The incident positron has total <energy> $E$ and satisfies $p^2c^2=E^2-\varepsilon^2$. Adding the target electron's <four-momentum>, the squared total <centre-of-momentum energy> is
$$
W^2=(E+\varepsilon)^2-p^2c^2=2\varepsilon(E+\varepsilon).
$$
At the <fixed-target production threshold>, all three final particles are at rest relative to one another, so $W=3M_0c^2$. Consequently
$$
\boxed{E_{\min}=\frac{9M_0^2c^4}{2m_ec^2}-m_ec^2.}
$$
This is the positron's total energy, not its <kinetic energy>. It assumes $3M_0\ge2m_e$ and addresses kinematic allowance; other <selection rules> could still forbid a proposed reaction. For $M_0c^2=135\,\mathrm{MeV}$ and $m_ec^2\simeq0.510999\,\mathrm{MeV}$, $E_{\min}\simeq160.494\,\mathrm{GeV}$.
At threshold the three products share the <centre-of-momentum frame>'s laboratory <velocity>. Their common <Lorentz factor> and speed are
$$
\gamma_{\mathrm{CM}}=\frac{E_{\min}+\varepsilon}{3M_0c^2}=\frac{3M_0}{2m_e}\simeq396.283,\qquad\boxed{v=c\sqrt{1-\frac{4m_e^2}{9M_0^2}}\simeq0.999996816\,c.}
$$
For the subsequent <two-photon decay>, each <photon> has rest-frame energy $E_\gamma^*=M_0c^2/2$. A <Lorentz transformation> of a photon emitted at rest-frame angle $\theta^*$ to the boost gives $E_\gamma=\gamma_{\mathrm{CM}}E_\gamma^*(1+\beta_{\mathrm{CM}}\cos\theta^*)$. Therefore the kinematically allowed laboratory range is
$$
\boxed{\frac{\gamma_{\mathrm{CM}}M_0c^2}{2}(1-\beta_{\mathrm{CM}})\le E_\gamma\le\frac{\gamma_{\mathrm{CM}}M_0c^2}{2}(1+\beta_{\mathrm{CM}}),}
$$
or numerically
$$
\boxed{0.08517\,\mathrm{MeV}\ \lesssim E_\gamma\lesssim\ 5.3498\times10^4\,\mathrm{MeV}.}
$$
The two photon energies in a single event are correlated: their sum is the parent's laboratory energy $\gamma_{\mathrm{CM}}M_0c^2$. 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.
Back to article page
