How would light's energy increase after bouncing off a moving mirror? Suppose we have a mirror moving with some speed through space. Both the mirror's velocity and mass could potentially be very large. What would happen, if a laser beam would be fired at its front face (the face facing in the direction of its movement) perpendicular to it? I am not sure if this is right, but I think that the mirror would slow down, and the light beam would increase its energy. It would happen like that because of conservation of momentum - mirrors momentum (speed) would decrease, while light's momentum (frequency) would increase. Am I right?
I am interested in calculating the amount of energy the light beam would gain after bouncing off (it would be in Joules per second if the beam would be constantly shining). I don't really know how to do it myself, as I have limited knowledge from what I read somewhere.
 A: Light beam's length: L
Light beam's speed: c
Mirror's speed: v
Light beam's power: P
duration of beam - mirror collision: L / (c + v)
light beam's length after collision: c * duration of collision =
c * (L / (c + v)) = L * c / (c+v)
The beam becomes shorter by factor c / (c + v).
That means the wavelength becomes shorter by that factor, and frequency changes by factor (c + v ) / c. Same for energy.
Finally the increased energy is spread on shortened length, so the power of the reflected beam is:   $$  P *  ((c + v ) / c) ^2 $$
A: It is better to solve the problem in the mirror's frame, where its own initial momentum is zero and energy $E_{m0} = M$, where $M$ in the mass of the mirror and for units where $c = 1$.
The initial energy of the beam is $E_{b0} = p_{b0}$ where $p_{b0}$ is its momentum.
After reflecting, total energy is conserved:
$M + p_{b0} = E_{m1} - p_{b1}$ (beam moment after collision is negative)
Total momentum is also conserved:
$p_{b0} = p_{m1} + p_{b1}$
From this 2 equations and from the expression for the energy of a massive object ($E^2 = M^2 + p^2)$ we can get:
$$p_{b1} = \frac{-Mp_{b0}}{M + 2p_{b0}}$$
Until now, it was used only classical field theory. But with the QM relation between light momentum and wavelength: $|p| = \frac{h}{\lambda}$
$$\lambda_1 = \frac{\lambda_0 M + 2h}{M}$$
Finally, we can move to the frame where the mirror has a velocity $v$. The wavelengths change according to the relativistic Doppler effect:
$$\lambda' = \lambda \frac{1+v}{1-v}$$
For the incident beam:
$$\lambda_0' = \lambda_0 \frac{1-v}{1+v}$$
For the reflected beam:
$$\lambda_1' = \lambda_1 \frac{1+v}{1-v}$$
Naming $\alpha = \frac{1+v}{1-v}$
$$\lambda_1' = \frac{M\lambda_0' + 2h \alpha}{M\alpha ^2} $$
For the energy $E = \frac{h}{\lambda}$
$$E_1' = \frac{M\alpha ^2}{\frac{M}{E_0'}+2\alpha}$$
Here a positive value for $v$ means the mirror going away from the observer.
