Does GR explain cosmological time dilation?

Question

There is plenty of evidence for a cosmological time dilation effect. For instance a supernova that takes 20 days to decay will appear to take 40 days to decay when observed at redshift $z=1$ (see Ned Wright's cosmology FAQ, Ned Wright on Tired Light).

I understand how time dilation comes about when considering a stationary clock near a gravitating body. By using the Schwarzschild metric with $dr=d\theta=d\phi=0$ we find that the proper time interval $d\tau$ experienced by the clock at radial coordinate $r$ is related to an interval of coordinate time $dt$ of an observer by: $$d\tau=d t\sqrt{1-\frac{2GM}{rc^2}}.\tag{1}$$ But contrast this with a stationary clock in the expanding universe. By using the FRW metric with $dr=d\theta=d\phi=0$ we find that for a stationary clock its interval of proper time $d\tau$ is simply related to an interval of coordinate time $dt$ by $$d\tau=dt.\tag{2}$$ Where in Eqn.$(2)$ is there any explanation for a time dilation effect?

Cosmological redshift can be understood in GR as photon wavelength stretching with expanding space but as I understand it cosmological time dilation is a different phenomenon.

Answer 1

I think your main confusion is thinking that time dilation is $dt/dτ$. That definition is meaningless, because coordinate systems are arbitrary and meaningless. There may not even be a coordinate named $t$.

The reason $dt/dτ$ is useful in Schwarzschild coordinates is that Schwarzschild coordinates have a translational symmetry $t\mapsto t+δt$. You can therefore make the following argument: suppose you emit two light pulses at $(t_e, \mathbf x_e)$ and $(t_e{+}δt_e, \mathbf x_e)$, and they are received at $(t_r, \mathbf x_r)$ and $(t_r{+}δt_r, \mathbf x_r)$. The path followed by the light may be complicated, but regardless of the details, it follows from symmetry that $δt_e = δt_r$. Therefore, the redshift/time dilation factor, $δτ_r/δτ_e$, is equal to $(δt_e/δτ_e)/(δt_r/δτ_r)$.

In FLRW coordinates, you can make a similar argument. Suppressing two spatial dimensions, the metric is $dt^2-a(t)^2 dx^2$. This has a translational symmetry $x\mapsto x+δx$. Therefore, if you emit two beams from $(t_e,x_e)$ and $(t_e,x_e{+}δx_e)$ and they are received at $(t_r,x_r{+}δx_r)$, by symmetry $δx_e=δx_r$ and the ratio of received and emitted wavelengths is $(δx_e/δχ_e)/(δx_r/δχ_r) = a(t_r)/a(t_e)$ (where $χ=a(t)\,x$). If $δx$ is small enough, then the $(t,χ)$ coordinates are essentially Minkowskian and $δt_r/δt_e = a(t_r)/a(t_e)$ as well.

These two special cases are given the names "gravitational redshift" and "cosmological redshift", but they're distinguished from general redshifts only by the fact that they're easy to calculate by particular symmetry arguments. There's only one kind of redshift in general relativity, and in principle you can always calculate it by working out the paths of light beams emitted at slightly different spacetime positions.

Answer 2

Time dilation and redshift are tightly connected to each other.

A thought experiment

Imagine an alien in a distant galaxy pointing two lasers towards earth. One is very long wavelength and the other is very short. The alien has arranged things so that at every cycle of the long wavelength laser, they send a short pulse of short wavelength. You can imagine the short pulse being emitted at the same time of one of the crests of the long wavelength laser.

If we call $\lambda_L$ the long wavelength (as measured by the alien), then the other laser emits pulses at intervals of $\Delta t =\lambda_L / c$ (as measured by the alien).

What do you see from Earth? Well, because of redshift, you get a wavelength $$\lambda_L' = (1+z)\lambda_L.$$ At what intervals do the pulses of the other laser come? Well, in vacuum, the group velocity of light is the same as the phase velocity, if the short pulse is emitted at the same time as one of the crests, it will arrive to you at the same time as the crest. So it will arrive at intervals of $$\Delta t' = \lambda_L'/c = (1+z)\lambda_L / c = (1+z)\Delta t.$$ This is time dilation. The alien can use $\Delta t$ as a unit of time, and measure all its activities using that unit. If they celebrate new year's eve every $10^{10}\Delta t$, you will see the fireworks every $(1+z)10^{10}\Delta t$.

(We ignored the fact that the laser pulse will spread. You think of the modulation of the maxima of intensity of the short laser, or sending the pulses every $n$ crests, so that the spread does not cause overlap. Anyhow, GR allows for massless point particles, so you can just think of that.)

Confusion alert

Note that when I said `see' above, I meant it. It is a description of observations of EM phenomena on Earth. There is a lot of confusion out there when talking about time-dilation in relativity.

Look at this example. Every bit of the discussion I did above applies equally well for observers that are receding from you in Minkowski spacetime, where $z>0$. It leads you to conclude that when an alien is moving away from you in a spaceship in flat space you see them in slow motion.

But you could run the same discussion with the alien coming towards you in Minkowski spacetime, with $z<0$. In that case, you would conclude that you see the alien sped up. But we both know that that can't be right, can it? Didn't we learn that time goes slower when a system is moving relative to us?

Different meanings of time dilation

To clear up this confusion, let us be a little more formal. Let's start in an inertial frame $(t,x)$ in Minkowski space (we only need 1+1 dimensions). Have an observer trace a worldline given by $x = f(t)$. Then time-dilation is the statement that $$\int_{t_0}^{t_1} dt \sqrt{\eta_{\mu\nu}{\dot x(t)}{\dot x(t)}} < t_1 - t_0,$$ or, in words, that the proper time elapsed along the trajectory between events $(t_0,x(t_0))$ and $(t_1,x(t_1))$ is less than the difference in the coordinate times $t_1-t_0$.

Ok, but remember that coordinates don't mean much in relativity, and you should be careful when interpreting them. If you don't connect them to some coordinate independent thing, you risk getting in trouble.

For example, in the twin paradox, you have Bob at sitting at rest at the origin of the coordinates, and Alice in her rocket that starts at the origin and comes back to Bob. Now we have two worldlines that intersect twice, and we can ask: how do the proper times along Alice's and Bob's worldline compare? Bob is sitting at the origin of the the inertial coordinates, so his proper time is just the difference in the coordinate's time, i.e. the right-hand side of the equation above. So now we can give an operational meaning to time-dilation, without invoking coordinates: when Alice and Bob meet again, Alice's clock has ticked fewer times than Bob's.

But does that mean that Bob sees Alice moving in slow motion? No. The statement above is about the comparison between two event that happen when Alice's and Bob's worldlines meet, while Bob sees is a question about events along Bob's worldline only. More precisely, it's a question about events where photons leaving Alice intersect Bob's worldline. So to answer what Bob sees, we looks at the discussion above with redshift. Bob sees Alice redshifted for a while (and thus in slow motion) and blueshifted towards the end of the journey. See my answer here for more detail.

Cosmological time

So we have two different concepts of time dilation. Which one is better? It depends on the application. Note that the one about comparing clocks does not apply to the receding galaxies.

Note that your derivation that $d\tau = dt$ for all comoving observers is absolutely right. In the FLRW metric, there exists a coordinate system so that the time coordinate equals the proper time of most galaxies. This is also the coordinate in which most galaxies are at rest. However, to do astrophysics, you need to translate these facts into things you can observe from Earth. Well, the distance between galaxies (as measured by photon time-of flight) increases, and the light from galaxies is redshifted and distant astrophysical phenomena are seen running in slow motion.

Answer 3

Cosmological redshift can be understood in GR as photon wavelength stretching with expanding space but as I understand it cosmological time dilation is a different phenomenon

From the quoted material it appears that Ned Wright is using gravitational time dilation as another term for gravitational redshift. At most he appears to consider one as different ways of saying the same thing. From the first quote

time dilation is a consequence of the standard interpretation of the redshift: a supernova that takes 20 days to decay will appear to take 40 days to decay when observed at redshift z=1

Where he says that the time dilation is an interpretation of the redshift.

The tired light model does not predict the observed time dilation of high redshift supernova light curves.

Which he supports with a plot with the redshift (z) as the horizontal axis.

In the Schwarzschild metric there is a gravitational potential and the time dilation can be considered a function of the potential. But the FLRW metric does not have a potential. So by that analogy, as you pointed out, there is no gravitational time dilation in the cosmological spacetime. You can, of course, change the meaning of the term between spacetimes, as it appears Wright is doing. But by doing so the same rules don’t apply.