Past exam of the physics 1 paper of the University of Oxford / 2013 resit / 9

For a frame moving at velocity $V\hat{\mathbf{x}}$, put $\Gamma =(1-V^2/c^2{)}^{-1/2}$. The Lorentz transformation of the energy-momentum four-vector isExpansion givesThe transverse components are unchanged, hence $\boxed{E^{\prime 2}-c^2|{\mathbf{p}}^{\prime }{|}^2=E^2-c^2|\mathbf{p}{|}^2}$. For a particle of rest mass $m$ this invariant is $m^2{c}^4$; the same algebra also applies to the total four-momentum of a system.

Let $p$ be the incident momentum and $q_1,q_2$ the outgoing magnitudes in the laboratory frame. Transverse conservation of momentum gives $(q_1-q_2)\sin \theta =0$. For the nondegenerate symmetric scattering, $q_1=q_2=q$, and the two outgoing energies are equal, $ϵ=(E+E_0)/2$. ThenCombining these mass shell relations yieldsHere $E$ includes the incident electron's rest energy. As $E\downarrow E_0$ the limiting angle is $45^{\circ }$, consistent with the nonrelativistic equal-mass elastic collision; at large $E/E_0$ the outgoing tracks become forward directed.

For $\theta =30^{\circ }$, the result in b gives $1/3=2E_0/(E+E_0)$, hence $E=5E_0$. Acceleration from rest supplies only the change in kinetic energy:The accelerating electric potential difference is a magnitude; its sign depends on which electrode is taken as the reference, since the electron has negative electric charge.