Past exam of the short options course of the University of Oxford / 2021 / 8 / c

Let $V>0$ be the observer's speed, so its laboratory velocity is $-V{\mathbf{e}}_x$. The particle's laboratory speed along $+y$ follows from the energy–momentum relation:Here $E>m{c}^2$ because the massive particle is moving. Boosting to the observer's frame with velocity $-V$ gives, by the relativistic velocity-addition formula,Equivalently its four-momentum has $p_x^{\prime }=\gamma VE/c^2$, $p_y^{\prime }=p_y$. Thus $\tan \varphi =u/(\gamma V)$. Solving this equation, rather than using the Galilean transverse velocity, yieldsFor a strictly moving observer, $0<\varphi <\pi /2$. At $\varphi =\pi /2$ the limiting observer speed is zero; as $\varphi \downarrow 0$ it tends to $c$. The direction of the observer's velocity remains negative $x$; the displayed $V$ is its positive magnitude.

For a photon, $m=0$ and $u=c$, so the result from (i) reduces toThis is relativistic aberration for a photon initially transverse to the boost. It also follows directly from $p_x^{\prime }=\gamma VE/c^2$ and $E^{\prime }=\gamma E$: the observed direction cosine is $c{p}_x^{\prime }/E^{\prime }=V/c$. The transverse observed speed is $c/\gamma =c\sin \varphi$, and the total photon speed is still $c$. If desired the observed photon energy is $E^{\prime }=E/\sin \varphi$ for $0<\varphi \le \pi /2$.