Past exam of the symmetry and relativity course of the University of Oxford / 2016 / 2

For a Lorentz boost of speed $v$ along $x$, the velocity-addition formula formulas areSubstitute $u_x=u\cos \theta$, $u_y=u\sin \theta$ and take the ratio. The common time-transformation denominator cancels, yieldingTo specify the direction rather than just its tangent, use ${\theta }^{\prime }=atan2(u\sin \theta /{\gamma }_v,u\cos \theta -v)$. This keeps the correct quadrant when the boost reverses the sign of the longitudinal velocity. For physical subluminal speeds the canceled denominator is positive. At $u_x=v$ and nonzero $u_y$, the transformed direction is perpendicular to the boost; if the particle is exactly comoving, its transformed velocity vanishes and the angle is undefined.

The initial momentum magnitude satisfies $p^2/(m^2{c}^2)={\gamma }_i^2-1=288$. Equality of longitudinal and transverse magnitudes therefore givesAn electric field along $z$ changes $p_z$, while the magnetic Lorentz force only rotates the transverse momentum. Its magnitude $p_{\perp }$ remains fixed. At the longitudinal turning point $p_z=0$, henceFor an electron initially moving toward positive $z$, the decelerating field points toward positive $z$ because its charge is $-e$. The electric work over $\Delta z=1\mathrm{m}$ givesThe radius is determined by transverse momentum rather than total energy:so the requested radius plot is horizontal. The cyclotron angular frequency changes as $e|B|/(\gamma m)$ and the helix pitch shrinks to zero. These statements neglect radiation, as stipulated.

Each photon has energy $\epsilon =\hslash \omega$. Their total four-momentum isIts squared invariant is $P^2=2{\epsilon }^2(1-\cos \theta )/c^2$. Therefore the system's rest energy isTwo individually massless particles can thus have positive total invariant mass. Back-to-back photons give rest energy $2\hslash \omega$, whereas exactly parallel photons have zero invariant mass and no massive-system rest frame.

The center-of-momentum frame moves with velocity $\mathbf{V}=c^2{\mathbf{p}}_{\mathrm{tot}}/E_{\mathrm{tot}}$. Using the total four-momentum in solution givesFor $0<\theta \le \pi$, its direction bisects the two photon directions, and its speed decreases from the limiting value $c$ to zero. At $\theta =\pi$, the lab already is the center-of-momentum frame. At $\theta =0$, the formal speed is $c$ and no inertial rest frame exists; this endpoint is a limit, not an allowable boost.

Let $\epsilon =\hslash \omega$ be the incident laser-photon energy and $\gamma =W/(m{c}^2)$. The maximum in the usual soft-laser geometry occurs for a head-on encounter followed by a forward outgoing photon along the electron beam. Boost into the initial electron rest frame: the incident energy becomes ${\epsilon }_{\ast }=\gamma (1+\beta )\epsilon$. Backscattering there gives the Compton scattering recoil relation ${\epsilon }_{\ast ,s}={\epsilon }_{\ast }/[1+2{\epsilon }_{\ast }/(m{c}^2)]$. Boosting this outgoing photon forward to the lab givesThe familiar $4{\gamma }^2\epsilon$ limit additionally requires $\gamma \gg 1$ and ${\epsilon }_{\ast }\ll m{c}^2$. The denominator is the recoil correction, not an optional factor at arbitrary incident energy.

For completeness, if the incidence direction is prescribed rather than optimized, let its unit vector be $\mathbf{n}$ and the outgoing direction ${\mathbf{n}}_s$. Squaring the residual electron four-momentum on its mass shell gives the more general kinematic relationThus an unspecified crossing angle must not silently be identified with head-on incidence; the boxed result is the maximum for the stated head-on soft-photon arrangement.

Numerically $W=2\mathrm{GeV}$ gives $\gamma =3913.89$, and a $1\mathrm{cm}$ photon has energy $1.23984\times 10^{-4}\mathrm{eV}$. The maximum scattered energy is $7.5970\mathrm{keV}$, givingThe rest-frame incident photon is only about one electronvolt, so the small recoil correction is consistent with the amplification estimate.