Past exam of the symmetry and relativity course of the University of Oxford / 2017 / 3

Three relevant observable effects are the relativistic Doppler effect in spectral-line shifts, including the transverse time-dilation contribution; relativistic aberration, observable as a change of apparent light direction with observer motion; and relativistic beaming of moving emitters, associated with strongly asymmetric jet brightness. Apparent superluminal motion supplies a related light-travel-time test: $v_{\mathrm{app}}/c=\beta \sin \vartheta /(1-\beta \cos \vartheta )$ can exceed one without a material speed exceeding $c$. These interpretations require an emission and source-motion model; a cosmological or gravitational redshift by itself is not a direct test of special-relativistic kinematics.

For Compton scattering from an initially stationary electron, let incoming and outgoing photon energies be $\epsilon$ and ${\epsilon }^{\prime }$ and their mutual angle be $\psi$. Four-momentum conservation gives $P+Q=P^{\prime }+Q^{\prime }$. Squaring $P^{\prime }=P+Q-Q^{\prime }$ and using both electron mass shells and both photon null norms givesConsequently the outgoing angular frequency and shift areEquivalently, the Compton wavelength shift is ${\lambda }^{\prime }-\lambda =h(1-\cos \psi )/(m_{e}c)$. The electron recoil is essential; treating its energy as unchanged would suppress the effect.

A boost into a frame moving with velocity $\mathbf{V}$ transforms the temporal component of the four-wavevector as $K^{\prime 0}=\Gamma (K^0-\mathbf{V}\cdot \mathbf{k}/c)$. Therefore the general relativistic Doppler effect iswhere $\vartheta$ is the laboratory angle between the boost and propagation directions. This is also the frequency measured from the phase rate along the observer's world line, ${\omega }^{\prime }=K\cdot U_{\mathrm{obs}}$. The frequency is an observer-dependent contraction of the invariant wave phase.

Take the wave along positive $x$, so $k_x=\omega /c$. A longitudinal boost $\mathbf{V}=(v_x,0,0)$ givesA receiver moving in the propagation direction measures a redshift. The same signed expression gives a blueshift when $v_x<0$. If the original wave propagates along negative $x$ instead, retain $k_x=-\omega /c$ in the general formula.

For $\mathbf{V}=(0,v_y,0)$, the laboratory wavevector is perpendicular to the boost. The dot product vanishes, soThis transverse-observer transformation is a blueshift. It must not be confused with the inverse factor often quoted when transverse refers to a ray perpendicular to the source velocity in the receiver's frame. The angle condition is specified in different frames in those two statements; relativistic aberration relates them.

The asserted necessity of equal masses is false. Equal unsigned outgoing angles imply equal longitudinal momenta when the transverse momenta cancel; they do not imply equal final speeds or energies for unequal masses.

An explicit counterexample uses $c=1$, masses $M_1=m$ and $M_2=2m$, initial projectile energy $E_1=2m$, and a stationary target. The initial total energy is $4m$ and momentum is $(\sqrt{3}m,0,0)$. Choose final four-momentaThey sum to the incoming four-momentum. Their squared norms are respectively $m^2$ and $4m^2$, so the collision is elastic. Both trajectories make the same acute angle with positive $x$:Thus no proof of the printed claim is possible.

The general centre-of-mass frame construction shows why. Let $p^{\ast }$ be the outgoing momentum magnitude there, ${\theta }^{\ast }$ the projectile scattering angle, and $E_1^{\ast },E_2^{\ast }$ the two energies. After a boost ${\beta }_{\mathrm{CM}}$ along $x$, the equal-angle condition for two forward nonzero trajectories isWith $E_1$ the initial projectile energy, its solution isprovided the right side lies between minus and plus one and both final longitudinal momenta are positive. Equal masses are one allowed case, giving ${\theta }^{\ast }=\pi /2$; they are not necessary. For the counterexample the right side is $3/16$. Even equal masses do not force equal angles for every scattering direction: the symmetric outgoing configuration is an additional choice.

Take the stated static vector potential with zero scalar potential. In the laboratory,Let $S^{\prime }$ move along $z$ at the electron's reference velocity $v$. The inverse transformation is $z=\gamma (z^{\prime }+v{t}^{\prime })$. The Lorentz transformation of electromagnetic fields giveswith all other components zero. The sign follows from $\mathbf{v}\times \mathbf{B}=v{B}_x{\mathbf{e}}_y$.

Although an observer fixed in $S^{\prime }$ sees an oscillating field, this is not an exact source-free propagating electromagnetic wave. The phase covector has ${\omega }^{\prime }=-\gamma v{k}_u$, $k_z^{\prime }=\gamma k_u$, so its squared norm isIt is the boost of a static spatial grating, not a null vacuum wave. Equivalently, $E^{\prime }/B^{\prime }=v<c$ and $B^{\prime 2}-E^{\prime 2}/c^2=B^2$ is nonzero away from nodes; a vacuum plane wave would have vanishing electromagnetic field invariants. In the ultrarelativistic limit it can be a useful nearly wave-like approximation along the electron's path, but the exact distinction remains.

A charged electron cannot actually maintain the given constant velocity through this nonzero transverse field without an external constraint: the Lorentz force deflects it. The calculation therefore uses the prescribed straight reference trajectory, a weak-deflection approximation, or an instantaneously comoving inertial frame. A globally fixed electron rest frame is not available for its exact deflected motion.