Special Relativity | Part IB, 2004

For a particle with energy EE and momentum (pcosθ,psinθ,0)(p \cos \theta, p \sin \theta, 0), explain why an observer moving in the xx-direction with velocity vv would find

E=γ(Epcosθv),pcosθ=γ(pcosθEvc2),psinθ=psinθ,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 γ=(1v2/c2)12\gamma=\left(1-v^{2} / c^{2}\right)^{-\frac{1}{2}}. What is the relation between EE and pp for a photon? Show that the same relation holds for EE^{\prime} and pp^{\prime} and that

cosθ=cosθvc1vccosθ\cos \theta^{\prime}=\frac{\cos \theta-\frac{v}{c}}{1-\frac{v}{c} \cos \theta}

What happens for vcv \rightarrow c ?

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