Past exam of the physics 1 paper of the University of Oxford / 2017 / 9

For a massive free particle with rest mass $m$ and velocity $\mathbf{v}$,The relativistic energy-momentum relation follows directly:This is the squared Minkowski norm of the four-momentum multiplied by $c^2$. The Lorentz transformation preserves that norm, so it has the same value in every inertial frame. In the particle's rest frame, the momentum is zero and the total energy is its rest energy $m{c}^2$.

Choose particle 1's incoming velocity positive and particle 2's negative. The final particle is stationary, so four-momentum conservation requires zero total incoming momentum:It follows that ${\gamma }_1{\beta }_1=1/\sqrt{2}$. Squaring gives ${\beta }_1^2/(1-{\beta }_1^2)=1/2$, henceThe initial total energy becomes the final rest energy. ThereforeThe extra invariant mass is the incoming kinetic energy retained as internal energy of the merged particle: $(M-3m)c^2$. It does not violate conservation of energy or four-momentum conservation; the sum of the separate rest masses is not a conserved quantity in an inelastic collision.

Let $S^{\prime }$ move at $+v$ along $x$ relative to $S$. The Lorentz transformation of energy and momentum, from primed to unprimed components, isThe inverse equations have $v$ replaced by $-v$ and the primes interchanged. A photon making angle ${\theta }^{\prime }$ with $+x$ in $S^{\prime }$ has $p_x^{\prime }=(E^{\prime }/c)\cos {\theta }^{\prime }$. Thus its general Doppler factor isThe requested blueshifted expression assumes propagation parallel to the boost, ${\theta }^{\prime }=0$:The propagation-direction assumption matters: for an oppositely travelling photon, $D=\gamma (1-\beta )$, the reciprocal of the displayed factor. Relative frame speed alone does not select the longitudinal relativistic Doppler effect.

For the forward ray of i, the ultrarelativistic longitudinal Doppler factor follows from the exact expression in terms of the Lorentz factor:Indeed, $\gamma \beta =\sqrt{{\gamma }^2-1}$ for a positive boost. If $\beta =1-ϵ$, the exact formulas and their consistent asymptotic expansions areEquivalently,The printed equality is not exact. For a concrete ultrarelativistic example, $\gamma =10$ gives the correct $D=10+\sqrt{99}\simeq 19.949874$, whereas $\sqrt{4{\gamma }^2-1}=\sqrt{399}\simeq 19.974984$. The latter expands as $2\gamma -1/(4\gamma )+O({\gamma }^{-3})$, with the wrong first correction.

One can obtain the printed expression by using $D^2=2/ϵ-1$ but replacing ${\gamma }^2$ by only its leading approximation $1/(2ϵ)$. This retains a constant term in the first equation while discarding a same-order term in the second. Thus $D\approx \sqrt{4{\gamma }^2-1}$ is admissible only as a leading relative-accuracy estimate, of the same order as $D\sim 2\gamma$; it is not the stated identity or a correct next-order approximation.

Let the incident photon energies be $E$ and $\epsilon$, meeting at angle $\psi$ in the laboratory frame. Add their four-momenta; because each is null, the squared total invariant energy isFor photon-photon pair production, its threshold value is $(2m_e{c}^2{)}^2$: the electron and Positron have zero relative momentum in their centre-of-momentum frame. The minimum possible $E$ occurs for a head-on collision, $\psi =\pi$, soWith $m_e{c}^2\simeq 0.511\mathrm{MeV}$ and $\epsilon =10^{-3}\mathrm{eV}$,The quoted background photon energy is used directly, without substituting a different characteristic value of the cosmic microwave background. A less opposed collision has a higher threshold; parallel photons cannot produce this massive pair.

At the threshold found in i, both products are stationary in the centre-of-momentum frame, so they have the same velocity in the laboratory frame. Take the energetic incident photon along $+x$. The total energy is $E+\epsilon$ and total momentum is $(E-\epsilon )/c$, so the centre-of-momentum frame moves atFor the numerical threshold,This is extremely close to, but strictly below, the speed of light. Each product has energy $(E+\epsilon )/2$ in the laboratory frame. Above threshold the same ratio of total momentum to total energy determines the motion of the pair's centre-of-momentum frame, but the individual electron and Positron need not have a common velocity.