Past exam of the physics 1 paper of the University of Oxford / 2019 / 10

For a boost along $+x$ at speed $v$, the energy-momentum four-vector transforms asThe total four-vector is the sum over the particles. Write $E={\sum }_i{E}_i$ and $\mathbf{P}={\sum }_i{\mathbf{p}}_i$. A centre-of-momentum rest frame exists when this total is timelike, with ${\mathbf{V}}_{\mathrm{CM}}=c^2\mathbf{P}/E$. Boost along $\mathbf{P}$ and use $P=E{V}_{\mathrm{CM}}/c^2$:ThereforeThe timelike qualification is needed: exactly parallel photons alone have null total four-momentum and no such rest frame.

Use the squared total energy-momentum four-vector, expressed in energy-squared units. With the proton initially at rest,At the reaction threshold, the outgoing proton and pion have no relative kinetic energy in their centre-of-momentum frame, so $W=(m_p+m_{\pi })c^2$. Solving givesThe numerical value uses the supplied $938$ and $135\mathrm{MeV}$ rest energies. The threshold exceeds the pion's rest energy because the outgoing system must carry the incident photon's nonzero laboratory momentum.

At threshold, the pion is at rest in the final centre-of-momentum frame, which moves in the laboratory withIn the pion's rest frame, its two-photon decay produces opposite photons of equal energy $m_{\pi }{c}^2/2$. The specified forward and backward directions remain collinear under this boost, soAt threshold, $\gamma (1+\beta )=(m_p+m_{\pi })/m_p$ and $\gamma (1-\beta )=m_p/(m_p+m_{\pi })$. HenceThe forward photon is the more energetic one. Their product remains $(m_{\pi }{c}^2/2{)}^2$, as required by the invariant mass of an oppositely directed photon pair, while their sum equals the laboratory pion energy.