Past exam of the short options course of the University of Oxford / 2017 / 8

Use the four-momentum convention $p^{\mu }=(E/c,\mathbf{p})$, with Minkowski signature $(+,-,-,-)$. For a particle of rest mass $m>0$ and three-velocity $\mathbf{v}$,Here $E=\gamma m{c}^2$ and $\mathbf{p}=\gamma m\mathbf{v}$, so $p^{\mu }{p}_{\mu }=m^2{c}^2$.

For a massless particle of energy $E>0$ moving in the unit direction $\mathbf{n}$,Its spatial momentum is $(E/c)\mathbf{n}$, and its four-momentum has Minkowski norm zero. A rest-frame expression $m\gamma (c,\mathbf{v})$ is not used for a massless particle.

The two initial momenta cancel, so the laboratory is the centre-of-momentum frame. Let $\mathcal{E}=2{\gamma }_A{m}_A{c}^2$ be the total initial energy. Four-momentum conservation gives ${\mathbf{p}}_B=-{\mathbf{p}}_{\gamma }$ and $E_B+E_{\gamma }=\mathcal{E}$. The energy–momentum relation then yieldsThusUsing $v_B/c=c|{\mathbf{p}}_B|/E_B=E_{\gamma }/E_B$ givesThis is relativistic production of a massive particle and a photon. A positive-energy photon requires $\mathcal{E}>m_B{c}^2$; equality is the zero-photon-energy threshold.

The velocity vector is ${\mathbf{v}}_B=-v_B{\mathbf{n}}_{\gamma }$. Its direction is opposite the photon, but is not fixed relative to the incoming beam by conservation alone. Kinematics supplies its magnitude; an angular distribution needs the collision dynamics.

The laboratory flight time is $R/v_B$. By time dilation, the accumulated proper time is $\tau =R/({\gamma }_B{v}_B)$. The definition of the rest-frame half-life gives at least one-half survival when $\tau \le T$, or ${\gamma }_B{v}_B\ge R/T$. Equivalently,Solving for the minimum speed givesUnder the usual exponential decay law, the survival probability is exactly $2^{-R/({\gamma }_B{v}_{B}T)}$. This is relativistic decay survival to a detector.

For $R=(3/4)cT$, part (ii) requires ${\gamma }_B{v}_B/c\ge 3/4$, soThe speed formula in part (i) can be written $v_B/c=({\mathcal{E}}^2-m_B^2{c}^4)/({\mathcal{E}}^2+m_B^2{c}^4)$. Requiring it to be at least $3/5$ gives $\mathcal{E}\ge 2m_B{c}^2$, hence ${\gamma }_A{m}_A\ge m_B$.

Since ${\gamma }_A\ge 1$ already, the exact minimum isIn the second case, the rest energy of the incoming pair already suffices for the required outgoing speed. Keeping this mass-ratio case avoids taking the square root of a negative number.