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We know that some galaxies are moving away from us faster than the speed of light and we know it by measuring the redshift, but how's that possible? If they're moving away say at $2c$, how would the light of the galaxy even reach us? How do we measure "redshift" for something faster than light?

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  • $\begingroup$ I had hoped to use the oft-claimed difference between relative motion and spatial expansion in a simple answer, but, in looking for a reference to Vilenkin (who had mentioned that supposed difference in his pop.-sci. book titled "Many Worlds in One"), I ran across a 2007 argument by Chodorowski, in an essay published by Oxford (titled "A direct consequence of the Expansion of Space", whose web address is a yard long), which disputes EoS, suggesting instead that a conformal difference in travel times between inbound and outbound photons is involved in the effects attributed to it. $\endgroup$
    – Edouard
    Commented Jan 30, 2021 at 18:18

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We know that some galaxies are moving away from us faster than the speed of light and we know it by measuring the redshift, but how's that possible?

The following papers give good explanations:

http://users.etown.edu/s/stuckeym/AJP1992a.pdf

http://arxiv.org/pdf/astro-ph/0011070v2.pdf

In summary, Hubble Law: $v = H(t)D$, where $v$ is recession velocity, $D$ is distance, and $H(t)$ is the Hubble "constant" at a given time, requires that beyond a certain distance velocity is greater than the speed of light. If recession velocity at the location of a traveling photon were greater than the speed of light the entire time the photon from a distance galaxy were traveling, we would never observe the photon. A photon emitted from a galaxy moving away from us faster than light, initially is also receding from us. However, the photon may eventually get to a region of spacetime where recession from us is $<c$. In this case, the photon can reach us. The exact relationship between red shift and velocity depends upon the cosmological model, but according to the above references, galaxies with red shifts greater than ~3 were and are receding from us faster than light.

If they're moving away say at 2c, how would the light of the galaxy? even reach us?

Only if the photons from the galaxy reach a region of spacetime where recession velocity is $<c$.

How do we measure "redshift" for something faster than light?

Red-shift is measured as the change in wavelength of the light, but rather than interpreting the results using special relativity (which would result in $v<c$ for all red shifts), the results are interpreted in the context of a cosmological model and general relativity.

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    $\begingroup$ By the way, it's generally better to edit your existing answer, rather than deleting and posting a new one (unless the new answer is really completely independent of the original one). $\endgroup$
    – David Z
    Commented Apr 15, 2014 at 4:28
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    $\begingroup$ The first answer only considered special relativity. Pulsar pointed out that this was insufficient. I thanked Pulsar in the first line of this answer, but that line was edited out. $\endgroup$
    – DavePhD
    Commented Apr 15, 2014 at 11:58
  • $\begingroup$ I can possibly provide an explanation why a layman like me didn't understood this: I was thinking that the receding speed of a galaxy that is X away, depends on an almost constant factor so it's v =X * a. When galaxy moves to 2X, I was thinking the receding speed will be 2v. But looks like it's not the case. Then I think the other way, can our photo reach the receding galaxy (faster than light)? So I guess I still don't understand how can a photon reach a galaxy that is always receding faster than light $\endgroup$
    – Splash
    Commented Dec 23, 2021 at 4:09
  • $\begingroup$ All galaxies with redshifts greater than ~1.5 have recession velocities v > c (i.e. their distance from us increases by more than 300,000 km/s) and yet we easily see galaxies at much higher redshifts. Moreover, all galaxies with z < ~1.6 (i.e. higher than the 1.5) could emit a light signal right now that, in principle, we might detect some day in the future. $\endgroup$
    – pela
    Commented Oct 17 at 9:14
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Light from beyond the Hubble sphere (the place where recession velocity equals the speed of light) reaches us daily.

I'm not good enough a physicist to come up with a nice layman's explanation for this fact, but it might help to think in comoving coordinates: This is a special coordinate system where the coordinate grid expands with space, ie even though the proper distance between galaxies will increase, their coordinates won't change.

In this coordinate system, light does not get frozen at the Hubble sphere (as one might possibly expect), but steadily moves from emitter to eventual observer, regardless of any change in proper distance.

The moving steadily towards us should actually also hold true for light emitted from beyond the cosmic event horizon (the thing that actually delimits the observational universe) - it just takes the light a longer-than-infinite time to reach us ;)

As to the second part of your question about the redshift: That doesn't depend on recession velocities, but rather on relative velocities as computed by parallel transport along the light path (and should stay below $c$ until you hit the event horizon).

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They simply aren't receding faster than light.

Recession speeds are defined like rapidity in special relativity. If A, B, ..., Z are all moving away from each other in a line, and A and C are receding from B at a speed $v$ that's small enough that the Newtonian relative velocity is a reasonable approximation, and B and D are receding from C at the same speed, and so on, then the relative rapidity of A and Z is $25v$ by definition. If $v=0.05c$ then the relative rapidity is $1.25c$, while the $dx/dt$ relative speed is $0.85c$. Nothing is going faster than light. The speed $c$ has no special meaning when talking about rapidities.

All of this is also true of cosmological recession speeds. It makes no sense to compare them to the speed of light because they're defined in such a way that there is no value representing the speed of light – certainly not $c$.

I think people fall into the trap of thinking that if you have enough galaxies in a line, all receding from each other, you must eventually get to a point where they're really truly moving faster than light, but this simply isn't true. Again, it's untrue even in special relativity. You can add another thousand galaxies to the end of the line of 26 with no problem, all the same distance apart at any given time, as measured by comoving metersticks and clocks that were synchronized when the galaxies were at the same point. There is an unlimited amount of room "just short of $c$". You could say it's because of length contraction that there is unlimited room, but note that this is a completely symmetric situation: you can put any galaxy in the center with a Lorentz boost.

In real cosmology, you can actually think of distant galaxies as being length contracted if you like, at least if the spatial curvature isn't positive. You can conformally embed any negative-curvature FLRW universe in Minkowski space in a way that looks similar to the special-relativistic toy model. In the quasi-Minkowski coordinates, light travels at $|dx/dt|=c$, all of the galaxies have $|dx/dt|<c$, and the ones with speeds close to $c$ are Lorentz contracted.

In ΛCDM cosmology, you have a cosmological horizon, and you'll never receive another signal from a galaxy after it crosses the horizon. This still doesn't mean that it's retreating faster than light in any physically meaningful sense. In conformal embeddings, the horizon arises because the universe ends at a finite conformal time (making it a time reversal of the big bang horizon problem). I'm not suggesting that's the true and correct explanation of the horizon, but it's a correct explanation. It is definitely not the case that a galaxy will ever outrun a light beam, so any explanation along those lines isn't correct.

The accepted answer offers this explanation of the ability of light to outrun expansion:

A photon emitted from a galaxy moving away from us faster than light, initially is also receding from us. However, the photon may eventually get to a region of spacetime where recession from us is $<c$. In this case, the photon can reach us.

This is a correct explanation. It's also a correct explanation of the special relativistic case, if you measure total distance as it's measured in cosmology, as the sum of distances measured by local comoving metersticks at times that are simultaneous according to local comoving clocks. The coordinate system effectively defined in this way is similar to Rindler coordinates with space and time reversed. In Rindler coordinates there is a gravitational redshift even though the spacetime is Minkowski. In these quasi-cosmological coordinates there is "expanding space" and cosmological redshift even though the spacetime is Minkowski.

It's not a wrong explanation but I don't find it very illuminating.

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I am no specialist in gravity or cosmology. Though, I know (without details) that A. Peres proved that the light velocity wasn't the same all along the history of the universe. The reference is

Int. J. Mod. Phys. D, 12, 1751 (2003). DOI: 10.1142/S0218271803004043

International Journal of Modern Physics D (Gravitation; Astrophysics and Cosmology)

Volume 12, Issue 09, October 2003

VARIABILITY OF FUNDAMENTAL CONSTANTS

ASHER PERES, This essay received an "honorable mention" in the 2003 Essay Competition of the Gravity Research Foundation.

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I do not know if the following answer can explain each and every observation, but here goes :

The expansion or moving away of galaxies is dependent on the distance between them, if something is moving away at some rate then previously since it must have been close, it must have moved away at a slower pace.

While making astronomical observations, we are always seeing back in time. 8 minutes back to see the moon and millions of years to see some far of stars. The light of such stars that would be leaving the star when the observation is made will reach us by the time star already dies, it is interesting that the star much before that may have reached some place from where it moves away at speed larger than that of light, and hence it can no more be observed.

Astronomers after observations, calculate the present state of the cellestial bodies and then publish all results, so if they tell you that something is a million light years away and is moving away at 2c, then that is its present position and speed, it was observed due to the light it emitted long tims ago and various observations and calculations allow us to predict its present state.

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    $\begingroup$ actually, recession velocities may well exceed $c$ at time of emission and we may still be able to observe the galaxy; not the Hubble sphere, but the cosmic event horizon places bounds of the observational universe $\endgroup$
    – Christoph
    Commented Apr 10, 2014 at 20:22
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    $\begingroup$ About 8.45 light minutes is the greatest distance from Earth to the Sun, not to the Moon. When the Moon is at its furthest orbital point from Earth, it is about 1.35 light seconds away. $\endgroup$
    – Ralph Dratman
    Commented May 14, 2015 at 6:32

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