# What is the energy of a photon reflected off of a moving mirror?

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)?

## 1 Answer

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