Physics Asked on December 8, 2020
As we observe a remote galaxy, we see it with a redshift. The most distant galaxy discovered to date is GN-z11 visible with the redshift of $z=11.09$. For simplicity, let’s assume no gravitational redshift.
In Special Relativity, the Doppler effect has two components, the Doppler component $1+beta$ and the time dilation component, which is simply $gamma$. The combined relativistic effect is $z+1=(1+beta)gamma$.
In case of the expanding universe, the Doppler effect would seem to have similar components, the Doppler component due to the galaxies recession speed and the time dilation component due to the space expansion. Some argue that there is no time dilation in this case, based on the grounds of comoving time. However, this argument holds neither logically, because the relative observed time is different from the cosmological time, nor practically, because without the time dilation component the maximum observed redshit would be $z=1$ for $betaapprox 1$ near the particle horizon.
Could someone please clarify if there is a relative time dilation in the expanding universe? Do we observe time of remote galaxies moving slower? Otherwise, if there is no such a time dilation, then what additional factors make the Doppler effect redshift so significant for distant objects?
Here is a derivation of the cosmological redshift (redshift due to cosmological expansion of space): https://en.wikipedia.org/wiki/Redshift#Expansion_of_space
All the $beta$ and $gamma$ terms are coming from Lorentz transformations between inertial frames in the framework of Special Relativity. However, an expanding universe cannot be described by Special Relativity, you need the tools of General Relativity, more specifically, one of GR's solutions used in cosmology, the FLRW metric. If you have no knowledge of GR, I'd recommend you the book "Physical foundations of cosmology" by V. Mukhanov.
Correct answer by Photon on December 8, 2020
General relativity doesn't have a general definition of gravitational time dilation that applies to all spacetimes. This only works in a static spacetime. In a static spacetime, the metric is derivable from a potential $Phi$, and a gravitational time dilation is of the form $e^{DeltaPhi}$. Cosmological spacetimes aren't static.
A similar example is a Schwarzschild black hole, which is static outside the event horizon but not inside it. This is why we can't define a gravitational time dilation between a location inside the event horizon and a location outside it.
There is no reason to expect cosmological Doppler shifts to be analyzable into factors like the ones you used for your argument for the longitudinal Doppler shifts in SR. Actually GR doesn't have a way to define the relative velocities of distant objects, so there would be no way to define a $beta$. When people talk about cosmological expansion in terms of the velocities of distant objects relative to us, that's just a popularized explanation.
Even in the case of SR, I don't think your derivation of the Doppler shift really works. The SR Doppler shift isn't a nonrelativistic Doppler shift multiplied by a correction factor. If it were, then we would have something like $[(1+beta_o)/(1+beta_s)]gamma$, where $beta_o$ is the velocity of the observer relative to the medium and $beta_s$ is the source relative to the medium. But this is not in fact the form of the relativistic Doppler shift, in which there is only one velocity (of o relative to s).
Answered by user4552 on December 8, 2020
Great Question..
Because objects move slower through time as they move faster through space; Earth Time Dilation from its entropy (velocity) is approximately .9999979972 seconds per Second. That is if one stacks the orbital speeds of Hubble, Earth around sun, Sun through the Galaxy, and the Milky Way through the universe (approximately 600 Million mps).
That seems like a very small number but it's very large when one considers the affect over a distance from 30 to 300 megaparsecs. Put another way... the photons we collect with the Hubble Telescope live 100 Million Years (earth years that is) before being observed in the form of light from candles at approximately 30 parsecs away(Quadrillions of seconds). So Looking at .99999979972 seconds per second factors out to a huge amount of time (approximately 221 years or a 7 billion second differential).
While we observe candles moving forward through space very fast.. they are also moving forward through time (As seen from Earth); and the closer they are to earth.. The smaller the impact of Entropic Time Dilation.
Put another way; from a Cosmological Perspective, we are not looking at present time. We are always looking at the past; and when we see a candle move further away than we think it should, we are looking into that object's past and taking measurements of it movement forward through both time and space.. not merely space alone.
"If you eliminate the impossible, whatever remains, however improbable, must be the solution"
Answered by Richard Bradford on December 8, 2020
the Doppler component $1+beta$ and the time dilation component [...] $gamma$
That decomposition really makes no sense from a special or general relativistic perspective. The only reason you might write the Doppler shift that way is if you're trying to make a connection with nonrelativistic Doppler shift.
A manifestly covariant formula for relativistic redshift is $$1{+}z = frac{p_text{light} cdot v_text{detector}}{p_text{light} cdot v_text{emitter}}$$
where the $v$s are four-velocities and the $p$ is the four-momentum of the light, or any nonzero scalar multiple of it (i.e., any four-vector pointing along the path of the light).
This formula works in general relativity too, in all cases, if you parallel transport the vectors to a common location along the path of the light before taking the dot products. All light frequency shifts in general relativity are aspects of the same underlying phenomenon. The decompositions into "cosmological redshift", "gravitational redshift" and so on are human inventions.
Some argue that there is no time dilation in this case
What John Rennie is saying there is just that the situation is symmetrical, like two people in Minkowski space who are moving apart. They see each other redshifted forever, but they're not mutually aging more slowly than each other. In some sense, they are aging "at the same rate" if you consider the symmetry. In another sense there's no way to compare the rates. The "universal time coordinate" (FLRW time) that he mentions is the proper time of the receding people, or the proper time of clocks moving with the Hubble flow.
Doppler component due to the galaxies recession speed and the time dilation component due to the space expansion
There is a redshift from the recession speed of the galaxies, and there's a redshift from the expansion of space, but they don't combine with each other, they're just equal to each other, because "recession speed" and "expansion of space" are different names for the same thing.
In the special case of linear expansion, $a(τ) = τ/τ_0$, spacetime is flat and you can actually put a global Minkowski coordinate system on it. The Minkowski coordinates $(x,t)$ are related to the FLRW coordinates $(χ,τ)$ by $$begin{eqnarray} t &=& τ,cosh,(χ/τ_0) x &=& τ,sinh,(χ/τ_0) end{eqnarray}$$
and the cosmological recession is just SR relative motion and the cosmological redshift is SR redshift given by the SR formula. (As I said, you can always use the SR formula if you're willing to parallel transport the vectors, but in this case you don't have to transport them if you use Minkowski coordinates.)
You shouldn't attach too much physical significance to this coordinate system (if any at all), but it illustrates that there's no real difference between redshifts that are usually attributed to different physical mechanisms.
Answered by benrg on December 8, 2020
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