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It's evident and well known that light traveling across an expanding FLRW universe is redshifted via an equation: $$\frac{\lambda_{arriving}}{\lambda_{emitted}}=\frac{a_{now}}{a_{then}}$$ Where $a$ is the cosmological scale factor when the light is emitted and observed (denoted then and now respectively).

Let's say the light was traveling through a waveguide over that same distance. Calculations shouldn't be effected, and the redshift would follow the same equation.

If we now take that same waveguide and make it a large circle of the same total length, would that effect the redshift equation? I don't see how, but maybe someone here knows better.

If light is still redshifted the same it seems we can shrink the size of the waveguide arbitrarily down to a small local system. Does cosmological redshift happen locally? I've found arguments that energy isn't lost to bound systems lacking.

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  • $\begingroup$ a waveguide large enough for cosmological redshift to become relevant within it would also be large enough to significantly affect the metric around it causing all sorts of distortions and bad stuff to happen. No matter how you construct it. $\endgroup$
    – Jim
    Aug 9, 2016 at 12:09

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The cosmological expansion can be seen only with very large structures. Its effective "force" is so weak that even galaxies are not affected, gravity keeps them bound and invariant .

Thus, the Andromeda galaxy, which is bound to the Milky Way galaxy, is actually falling towards us and is not expanding away. Within the Local Group, the gravitational interactions have changed the inertial patterns of objects such that there is no cosmological expansion taking place. Once one goes beyond the Local Group, the inertial expansion is measurable, though systematic gravitational effects imply that larger and larger parts of space will eventually fall out of the "Hubble Flow" and end up as bound, non-expanding objects up to the scales of superclusters of galaxies.

Structures bound by the stronger interactions like electromagnetic and strong are of course not affected. The raisin bread analogue helps understand this:

raisinbred

Animation of an expanding raisin bread model. As the bread doubles in width (depth and length), the distances between raisins also double.

the dough is expanding, but the raisins are stable in size because the electromagnetic bindings are not affected by the yeast in the dough.

The waveguide you are envisaging, is bound together with the electromagnetic force, and any interactions with electromagnetic waves will be within the "raisin".

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  • $\begingroup$ are you saying light won't be redshifted in a waveguide connecting here to a distant star? Is there a more calculational way of showing this? The notion that you can't write a conservation of mass equation in a nonstationary spacetime I was thinking indicates a local loss of energy to bound systems. Still learning though (: $\endgroup$
    – R. Rankin
    Aug 9, 2016 at 6:12
  • $\begingroup$ If that's true, then the distance between the distant star and us would have to remain constant (assuming we were initially at rest with respect to one another) during the light transit time. I understand bound systems wouldn't expand, but wouldn't they still lose energy? If we use springs to connect the distant stars, expansion would effectively do work, where is the line drawn, does it scale all the way down $\endgroup$
    – R. Rankin
    Aug 9, 2016 at 6:28
  • $\begingroup$ Read the link, bound states even with gravity at short distances are not affected by the expansion. The four forces, at distances smaller than galactic clusters define a stable space time because the expansion is so very very weak $\endgroup$
    – anna v
    Aug 9, 2016 at 6:32
  • $\begingroup$ the em waves, the waveguide etc are all within this frame $\endgroup$
    – anna v
    Aug 9, 2016 at 6:33
  • $\begingroup$ I understand that expansion wouldn't take place locally, but from an energy perspective, a decrease in energy would take place? I wrote another question which perhaps is clearer: physics.stackexchange.com/questions/273383/… $\endgroup$
    – R. Rankin
    Aug 9, 2016 at 6:39
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Let's simplify: Instead of using a (possibly dielectric) waveguide (the picture your description summoned in my mind was a ring of optical fibre), just have a photon bounce between mirrors in a vacuum.

At first glance, you should indeed get the same effect from bouncing the photon a few times between mirrors that are far apart and bouncing the photon many times between mirrors close together: In Friedmann cosmology, the total redshift the photon 'accumulates' during its travel through curved spacetime will depend only on the time of emission and absorption.

But you also have to take into account what happens at the mirrors, and ask yourself if it will make a difference if the mirrors are, say, both comoving with the Hubble flow (eg mounted in different galaxies), or if they are kept at constant proper distance and thus decreasing comoving distance (eg through use of a rigid frame).

When deriving cosmological redshift, your starting point are comoving source and observer. Relative to this situation, a mirror at fixed proper distance will be moving towards the observer, and the photon should pick up some energy on its bounce. I suspect this might compensate the redshift, though I have yet to verify this by calculation (or even better, by coming up with a convincing argument).

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  • $\begingroup$ Thanks for that. The heart of what I'm trying to get to is that some locally bound system, to maintain the same size, must effectively do work against expansion, thus losing energy. As an alternative route, the fact that mass in an expanding universe can't be written as a conserved quantity, seems to indicate such systems lose energy. This should be calculable, though I'm not sure how $\endgroup$
    – R. Rankin
    Aug 9, 2016 at 20:20
  • $\begingroup$ @Christoff One can formulate this question another way, as per my comments to Anna above $\endgroup$
    – R. Rankin
    Aug 11, 2016 at 20:12

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