 Solution

The four-momentum is $(E/c,p_x,p_y,p_z)$. It transforms by the same Lorentz boost as other four-vectors. For an observer moving at $v$ in the positive $x$ direction, its time and longitudinal components transform asSubstituting the given momentum components proves all three displayed transformation relations.

A photon has zero rest mass, so the relativistic energy-momentum relation gives $E=cp$ on the positive-energy branch. Lorentz invariance preserves $E^2-c^2{p}^2=0$, and $E^{\prime }=\gamma E(1-(v/c)\cos \theta )>0$ for $|v|<c$. Therefore $E^{\prime }=c{p}^{\prime }$. Dividing the transformed longitudinal momentum by $p^{\prime }=E^{\prime }/c$ gives the relativistic aberration formulaAs $v\to c^{-}$, for every fixed $\theta \ne 0$ the ratio tends to $-1$: photon directions are concentrated along the negative $x$ direction in the observer's frame. The exactly forward photon is exceptional: if $\theta =0$, then $\cos {\theta }^{\prime }=1$ for every $v<c$, and its energy tends to zero. For other fixed angles $E^{\prime }$ diverges. The limiting angular behavior is not uniform near $\theta =0$, and $v=c$ itself is not an admissible inertial observer velocity.