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In accordance with the above diagram, a photon with energy $E$ is reflected off of a moving mirror with speed $v$. I think I am supposed to use the conservation of four momentum to find the reflected photon's energy, $E^*$. But this makes me think $E^* = E$. Could someone explain the answer (with an emphasis on the basics, since I'm new to this topic)?

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I believe I have solved it:

The photon's four momentum before the collision is given by:

$p^0 = p^1 = \frac{E}{c}$

Then using the lorentz boost transformation:

$(p^0)^{'} = \gamma (p^0 - \frac{vp^1}{c}) = \frac{E}{\sqrt{1-\frac{v^2}{c^2}}}(\frac{1}{c}-\frac{v}{c^2})$

Then we have $$E^* = c(p^0)^{'} = E\frac{1-\frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2}}} = E\sqrt{\frac{c-v}{c+v}}$$

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