I want to consider a thought experiment. Lets ignore technical problems of actually performing such an experiment. Consider two photons having the same wavelength. We send 1 photon to a distant galaxy (millions of light years away). The photon would hit a mirror there and return to Earth. On Earth (I assume) a cosmological redshift would be detected. At the same time we keep the second photon on Earth in small container with perfect mirrors where it bounces back and forth. Would the photon in the container exhibit the same cosmological redshift as the photon that traveled to the distant galaxy and back? If so, (now back to technical problems) could the redshift be ever measured in the lab or is the effect many orders of magnitude too small to be ever measured in a lab on Earth, not to mention the problem of constructing a perfect mirror to keep the photon?
2 Answers
While your question isn't an exact duplicate of Why does space expansion not expand matter? the underlying principle is the same.
The answer is that the photon in the lab would not be red shifted, or at least that any red/blue shift would be unrelated to the expansion of the universe. This is because at small distances the electromagnetic force completely overwhelms the cosmological expansion so the distance between the mirrors would not change.
However if you put the mirrors a light year apart and far from any other matter (like a galaxy cluster, so this is in deeeeeeeeep space!) then the red shift would be the same.
Let's have a look at your experiment:
The diagram shows your box with the mirrors at each end. The white smiley is you in the rest frame of the box, and the red shifted smiley is an unconstrained observer. Due to the expansion of spacetime the red shifted observer is receding from you at a speed $v$ given by the usual Hubble law:
$$ v = H_0 l $$
where $l$ is the length of the box. However, and this is the key point, the end of your box is not receding from you because the electromagnetic forces that keep it in shape overcome the cosmological expansion.
You emit a photon of light with some well defined frequency $f_0$. When the red shifted observer measures the frequency of the light they find a frequency $f < f_0$, which is exactly what we mean by the cosmological red shift. However from the red shifted observer's point of view the mirror at the end of the box is moving with a velocity $H_0 l$, so when the photon reflects off it the distant observer sees the light is blue shifted. The result of the blue shift is that when you receive the reflected photon its frequency is unchanged i.e. there has been no (net) red shift.
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$\begingroup$ I cannot see any difference between a photon in a box and traveling between galaxies. Sending a photon to a distant galaxy is the same as keeping it in a box. You could in principle construct a 10 000 000 ly long box. The only difference I see is the relative motion of a distant mirror in another galaxy compared to mirrors remaining stationary relative to each other in a box. $\endgroup$– EiverCommented Sep 24, 2013 at 11:05
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$\begingroup$ Well yes. The point is that the mirrors in the box will remain stationary with respect to each other, and the distance between them will not change. The point of putting the mirrors a light year apart is to ensure that any interaction between them doesn't swamp the cosmological expansion. $\endgroup$ Commented Sep 24, 2013 at 11:09
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$\begingroup$ The linked answer tells us why we won't see electron moving away from the nucleus of an atom or why a planet won't move away from the parent star. Thats because the 4 forces are strong enough to keep these objects in place despite space between them is expanding. I think that a photon changing its wavelength is a different thing. $\endgroup$– EiverCommented Sep 24, 2013 at 11:14
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$\begingroup$ @Eiver: I've elaborated on my answer to (hopefully!) make my argument a bit clearer. $\endgroup$ Commented Sep 24, 2013 at 12:19
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$\begingroup$ I want to eliminate any possibility of a Doppler shift. In your picture showing a rigid box with two mirrors do both ends remain stationary relative to each other? You wrote that the red-shifted observer is receding, but at the same time the end of the box is not. Can you clarify? P.S. Thanks for nice illustration. $\endgroup$– EiverCommented Sep 24, 2013 at 12:41
This experiment is essentially being conducted at some level by the Gravitational Wave detectors of LIGO and VIRGO.
Laser light is fed into a Fabry Perot interferometer, so that it effectively bounces up and down the 4km arms about 280 times.
The mirrors are suspended and the gravitational wave has the effect of expanding and contracting the space between them (in the laboratory frame).
If the gravitational wave has a wavelength smaller than $2 \times 280 \times 4$ km, then the response of the interferometer is much reduced because the laser light is "stretched" by a similar amount to the arms and the phase difference engendered by the wave is diminished.
If instead we deal with low frequency, long wavelength gravitational waves, the disturbance of the mirrors lasts much longer than the time spent by the laser light in the instrument arms and the phase difference response is maximised.
Whilst this is not a "cosmological redshift" it is pretty much the same physics. The phase difference accumulated by the laser "in the lab" is caused by the metric perturbation caused by the passing gravitational wave and is similar to the accumulated redshift of a photon travelling across the universe.
