Past exam of the astrophysics course of the University of Oxford / 2021 / 3 / b / Solution

For fixed $\beta$, differentiate the apparent speed with respect to angle:The maximum lies at $\cos \theta =\beta$ and has value ${\beta }_{\mathrm{app},\max }=\beta /\sqrt{1-{\beta }^2}=\gamma \beta$. Hence strict apparent superluminal motion requiresThe threshold is an infimum: equality yields exactly ${\beta }_{\mathrm{app}}=1$ at $\theta =\pi /4=45^{\circ }$, rather than a strictly superluminal value. For a chosen larger $\beta$, the optimal inclination is $\theta =\arccos \beta$, and the full superluminal interval satisfies $\sin \theta +\cos \theta >1/\beta$.