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I had an interesting chat the other day about what sort of information an arbitrary observer needs to recover the comoving time in the FLRW universe.

Suppose one performs the following procedure:

-Look out at the surrounding dust. It is, for general observers, relativistically beamed and thus anisotropic.

-Map the anisotropic dust field to an isotropic one. Correct the observed redshift distribution to be isotropic as well. This should allow you to measure your speed relative to the expansion at any given moment.

-Measure the Hubble constant as a function of your proper time from the "corrected" isotropic dust. Since we also know our speed, we can recover the map from proper to comoving time as well.

Therefore, in the approximation that Universe is exactly FLRW, the comoving frame appears to be accessible to all observers from local measurements. Of course this is equally true of any frame, but it's still kind of interesting. Would this procedure work?

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  • $\begingroup$ What dust do we see that is cosmological with local measurements? Do the same but observe what you can, stars, galaxies, and the CMB. They are not local. There is no local cosmological measurement you can do, it's all affected more strongly by local gravity (earth/sun, solar system, Galaxy, cluster) that the cosmological averages. $\endgroup$ – Bob Bee Nov 6 '16 at 7:54
  • $\begingroup$ I'm not talking about the actual universe, I'm talking about the FLRW solution. $\endgroup$ – AGML Nov 6 '16 at 15:57
  • $\begingroup$ Dust is not everywhere, it is a cosmological approximation. This is physics, not mathematical FLRW, it was never meant to be taken that rho (mostly dust density, ignoring dark energy) could be measured locally. Your question is a math question then. As such, mathematically yes. It means nothing. It's like other things young physics students get confused about: like, how can a force be non-conservative when all (ignore Einsteins gravity) elementary forces (EM, weak, strong, Newtonian gravity) are conservative? The answer is statistical mechanics and thermodynamics. Cosmological rho is not local $\endgroup$ – Bob Bee Nov 6 '16 at 19:06
  • $\begingroup$ How exactly do the local inhomogeneities prevent us from doing the above? Suppose we were inside a ~MPc sized spacecraft, if it makes you feel better. $\endgroup$ – AGML Nov 6 '16 at 19:25
  • $\begingroup$ It's like if I asked "does a simple harmonic oscillator oscillate periodically" and you responded "well no because real pendulums have friction". a) That is not the question I asked; and b) There are perfectly physical situations in which you can ignore the friction. $\endgroup$ – AGML Nov 6 '16 at 19:27
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The definition of "local" could mean confined to an arbitrarily small region of space, or confined to an arbitrarily small region of space and time. Either way, I think the question ends up being somewhat trivial.

If we're only confined to an arbitrarily small region of space, but we can observe for long times, then, e.g., we could locally observe a clock starting at the big bang, and we would know the comoving time (possibly after correcting for motion relative to the Hubble flow).

If we're confined to an arbitrarily small time span as well, then we can't even tell that cosmological expansion is occurring or that there was ever a big bang. All we see is a perfect fluid at a certain temperature and pressure.

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It is possible using only local measurements to determine one's speed relative to the Hubble flow, and hence the current cosmological time (up to an additive constant), even in a universe that's only FLRW at large scales, like ours. I know this because we've managed to do it using only experiments performed in the vicinity of Earth's orbital path over the last century or so (which is pretty small by cosmological standards) and there's nothing special about Earth's location in the universe.

These experiments involve what we call "looking at" objects that are billions of light years away "through" telescopes, but they're really local experiments because we aren't traveling out there to collect data, we're just collecting photons and neutrinos and gravitons in our immediate vicinity and inferring the existence and properties of distant objects from that.

With what we now know about the universe, the process can be streamlined a lot. You can find your velocity relative to the Hubble flow very precisely by measuring the CMBR dipole, and if you know the FLRW scale factor at all times and the initial temperature of the CMBR, you can calculate the time since the CMBR was emitted from its current temperature (after correcting for the dipole). This gives you an absolute cosmological time, albeit with its zero point at the CMBR emission rather than the big bang.

In principle you can get a more accurate answer from the neutrino and graviton backgrounds, since the universe is more transparent to them than it is to light, but the same properties that make them better in principle also make them very hard to detect.

Edit: Maybe the intent of your question is that you're in a windowless room that blocks the CMBR, and don't have precise enough equipment to detect neutrinos or gravitational waves. Can you detect the large-scale shape of the universe just from the distribution of matter in the room? In this case the answer is yes if the room is large enough that the matter in it is FLRW distributed, and no if it's smaller. If the universe is FLRW at all scales then you can detect it at all scales, but technically neither you nor the room could exist in that case since they would break the symmetry.

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