Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 4 / 9A (Dynamics and Relativity)

Use momentum conservation and conservation of energy in the laboratory inertial frame. The photon has momentum magnitude $E/c$, while the outgoing Electron satisfies the relativistic energy-momentum relation. Its total energy and momentum obeySubstitute these into $E_e^2=m^2{c}^4+c^2|{\mathbf{p}}_e{|}^2$. Cancelling $m^2{c}^4$ and $(E_1-E_2{)}^2$ gives $m{c}^2(E_1-E_2)=E_1{E}_2(1-\cos \theta )$. Both photon energies are positive, so division yieldsThis is Compton scattering. The energies include the Electron's rest energy; using only its kinetic energy in the mass-shell equation would give the wrong result. The outgoing photon has unchanged energy at $\theta =0$ and smaller energy for a nonzero scattering angle.

Let ${\mathbf{p}}_1$ and ${\mathbf{p}}_2$ be the incoming and proton momenta. The other incoming atom is at rest, so its total energy is $m_1{c}^2$. By four-momentum conservation, the tritium atom hasThe specified perpendicular emission gives ${\mathbf{p}}_1\cdot {\mathbf{p}}_2=0$. Therefore the relativistic energy-momentum relation impliesDividing by $c^4$ givesHere $E_1,E_2$ are total energies, not just kinetic energies. This identity is conditional on the collision admitting the stated emission: the calculated energies must satisfy the positive-energy mass-shell conditions, in particular $E_2\ge m_2{c}^2$ and $E_3\ge m_3{c}^2$. No nonrelativistic approximation has been used.