For a particle with energy $E$ and momentum $(p \cos \theta, p \sin \theta, 0)$, explain why an observer moving in the $x$-direction with velocity $v$ would find

$$E^{\prime}=\gamma(E-p \cos \theta v), \quad p^{\prime} \cos \theta^{\prime}=\gamma\left(p \cos \theta-E \frac{v}{c^{2}}\right), \quad p^{\prime} \sin \theta^{\prime}=p \sin \theta,$$

where $\gamma=\left(1-v^{2} / c^{2}\right)^{-\frac{1}{2}}$. What is the relation between $E$ and $p$ for a photon? Show that the same relation holds for $E^{\prime}$ and $p^{\prime}$ and that

$$\cos \theta^{\prime}=\frac{\cos \theta-\frac{v}{c}}{1-\frac{v}{c} \cos \theta}$$

What happens for $v \rightarrow c$ ?