If a laser is bounced off two mirrors repeatedly for several years, will it g et red-shifted?

Question

I understand that red shift is what shows how far away the star light came from, or how far back in time the light was emitted? If so, should we not see red-shift if a laser is bounced repeatedly off two mirrors back and forth for a long time (years). Has this experiment been carried out?

Answer 1

No, the red-shifting of star light is not caused by the distance or amount of time that the light have traveled, so a light going back and forth between two stationary mirrors will not get red-shifted. Rather, red-shift can be caused by other effects:

1. Relative motion of the light source and receiver, relativistic doppler effect - this is similar to how the frequency of an ambulance siren changes as it drives past you (as it initially comes towards you and then away from you, its siren's frequency appears to shift).
2. Cosmological redshift - caused by the expansion of the universe, which causes faraway galaxies to appear to speed up relative to each other. Arguably the circumstances and proof of cosmological redshift are not identical to doppler redshift, but they are often thought to be related.
3. Gravitational redshift - caused by photons escaping a star with large gravity losing energy, and therefore being redshifted.

None of these causes for redshift apply to your stationary mirror setup, and the light would not get redshifted. Of course, if the mirrors are not perfect, some of the photons might be absorbed by the mirrors, but in an ideal world with perfect mirrors - the light will be reflected indefinitely and not be redshifted.

Answer 2

Assuming that the distance between the two mirrors changes slowly over many reflection cycles, the frequency of the light will scale in inverse proportion with that distance. This follows, for example, from conservation of adiabatic invariants.

Specifically, in the limit that the fractional change in the mirror separation $L$ during each cycle is much smaller than 1, the action $J=\oint p \mathrm{d}q$ is conserved, where $p$ is the photon momentum and the integral is carried out over a cycle of the photon's path. But $p=hf$, where $h$ is the Planck constant and $f$ is the photon frequency, so $J=2Lhf$. Since this quantity is conserved, $f\propto 1/L$.

If the mirrors are maintained at a fixed distance, there will be no redshift or blueshift.