Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / Section B / B9

For a particle of rest mass $m_r$, $E=\gamma m_r{c}^2$ and $p=\gamma m_{r}v$, with $\gamma =(1-v^2/c^2{)}^{-1/2}$. The low-speed expansion is $E=m_r{c}^2+\frac{1}{2}{m}_r{v}^2+O(m_r{v}^4/c^2)$, separating invariant rest energy from the nonrelativistic kinetic energy term.

Conserve the total laboratory energy and momentum at coalescence: $E_{\mathrm{tot}}=4m{c}^2$ and $p_{\mathrm{tot}}=2.4mc$. Since $V=c^2{p}_{\mathrm{tot}}/E_{\mathrm{tot}}$, the combined particle moves at $0.6c$. Its rest mass is $M=\sqrt{4^2-2.4^2}m=3.2m$. The incident particle is not of mass $m$: its stated energy and momentum imply $m_{\mathrm{incident}}=\sqrt{3^2-2.4^2}m=1.8m$. ThusThe extra $0.4m{c}^2$ is internal rest energy retained by the composite after a perfectly inelastic collision; total energy and four-momentum are conserved, while the sum of individual rest masses need not be.

The composite's Lorentz factor is $1.25$. For a rest-frame lifetime $\tau$, the corresponding laboratory time is $1.25\tau$ and its travel distance isIf $\tau$ denotes a mean radioactive lifetime, this is the mean flight distance, not a fixed distance for every decay. The separate frame diagrams below use the common positive collision axis.

In the laboratory frame the incoming particle and the composite move along $+x$, while the target is initially at rest. In the equal-energy two-photon decay of a moving particle, zero total transverse momentum makes the two photon directions symmetric about $+x$. Each photon has energy $2m{c}^2$ and momentum magnitude $2mc$, so longitudinal momentum conservation gives $2(2mc)\cos \theta =2.4mc$. ThereforeTheir momenta can be written $(1.2,\pm 1.6)mc$. The opening angle is not $180^{\circ }$ in this frame, because the parent carried nonzero total momentum.

Boost along $+x$ at $0.6c$ into the centre-of-momentum frame. The composite is stationary. The incoming particle has $p_x^{\prime }=1.25(2.4-0.6\times 3)mc=0.75mc$, while the target has $p_x^{\prime }=-0.75mc$. After decay the two photons have energies $M{c}^2/2=1.6m{c}^2$ and opposite momenta.

Equal laboratory photon energies require the rest-frame photons to have zero momentum component along the boost: $E_{\mathrm{lab}}=\gamma (E^{\prime }+V{p}_x^{\prime })$ must be equal for two opposite photons. Their directions are therefore $+y$ and $-y$ in the centre-of-momentum diagram, perpendicular to the incoming collision axis. Boosting those transverse photons gives exactly the laboratory momenta in part (a). A stationary composite has no direction-of-travel arrow in this frame.