1.I.4C

Special Relativity | Part IB, 2008

In an inertial frame SS a photon of energy EE is observed to travel at an angle θ\theta relative to the xx-axis. The inertial frame S′S^{\prime} moves relative to SS at velocity vv in the xx direction and the x′x^{\prime}-axis of S′S^{\prime} is taken parallel to the xx-axis of SS. Observed in S′S^{\prime}, the photon has energy E′E^{\prime} and travels at an angle θ′\theta^{\prime} relative to the x′x^{\prime}-axis. Show that

E′=E(1−βcos⁡θ)1−β2,cos⁡θ′=cos⁡θ−β1−βcos⁡θ,E^{\prime}=\frac{E(1-\beta \cos \theta)}{\sqrt{1-\beta^{2}}}, \quad \cos \theta^{\prime}=\frac{\cos \theta-\beta}{1-\beta \cos \theta},

where β=v/c\beta=v / c.

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