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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 $\beta\approx 1$ 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?

safesphere
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  • Here is a derivation of the cosmological redshift (redshift due to cosmological expansion of space): https://en.wikipedia.org/wiki/Redshift#Expansion_of_space – Photon Jan 01 '18 at 09:09
  • @Photon This is no doubt very helpful, thank you! However, the derivation there is not completely convincing. "The subsequent crest is again emitted from $r = R$", - not really, especially if the expansion speed is faster than the speed of light. Wouldn't it be more like $r=R+\beta\lambda,$? where $\beta$ is the expansion speed in light speed units (with possibly $\beta\gt 1$)? – safesphere Jan 02 '18 at 01:33
  • I don't understand, where should the additive term $\beta\lambda$ come from? – Photon Jan 02 '18 at 07:26
  • @Photon Because space expands, the position, from which the second crest is emitted would be farther away. If the relative recession speed is $v$, then it would seem that this position should be shifted by $vt$, where, $t$ is one period of oscillations of light $ct=\lambda$. No? – safesphere Jan 02 '18 at 08:03
  • No, r is given in comoving coordinates, which are not affected by the Hubble flow. – Photon Jan 02 '18 at 08:24
  • @Photon Why not? $r$ is the distance that increases with the expansion of space, no? What am I missing? – safesphere Jan 02 '18 at 08:27
  • No, r is the comoving radial coordinate of the light source. What the distance is, is hard to tell, because there are several notions of distance in cosmology. Here is an overview: https://arxiv.org/abs/astro-ph/9905116v4 – Photon Jan 02 '18 at 08:34
  • @Photon Still $r$ is not constant. It has increased to like 46 Gly in under 14 Gy, so $\beta\gt 3$ in my formula, right? Am I missing something simple here? The derivation makes an assumption that the difference in $r$ is negligibly small compared to $\lambda$, but this assumption is not self evident at all. – safesphere Jan 02 '18 at 08:46
  • You are trying to describe an expanding universe as a Minkowski space-time. This doesn't work. I think, you need to read a bit about general relativity in general and the FLRW space-time in particular. If you have no knowledge of GR, I'd recommend you the book "Physical foundations of cosmology" by V. Mukhanov. – Photon Jan 02 '18 at 08:52
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    @Photon Thanks for the reference. And I see now what you mean. The space increase if all attributed to the scale factor while the comoving distance is presumed constant. Got it, thanks! – safesphere Jan 02 '18 at 09:10
  • Exactly, that's the point! – Photon Jan 02 '18 at 09:20
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    "Could someone please clarify if there is a relative time dilation in the expanding universe?" For what it's worth: A cosmological time-dilation factor of (1 + z) is applied to the type Ia supernova data in The Supernova Cosmology Project (Perlmutter, et al.). The following paper from that project (https://arxiv.org/abs/astro-ph/9602124) claims to show "the first clear observation of cosmological time dilation for macroscopic objects". – D. Halsey Oct 04 '18 at 22:09
  • @D.Halsey Thanks so much for the reference! I appreciate it. – safesphere Oct 04 '18 at 22:12
  • Here's another reference: http://adsabs.harvard.edu/abs/1996ApJ...466L..21L – D. Halsey Oct 04 '18 at 22:18

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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.

There is only one kind of redshift in special and general relativity: an object with a timelike worldline emits two light wavefronts/pulses at times separated by $dτ_e$, they propagate along null geodesics to a second timelike worldline which they intersect at times separated by $dτ_r$, and the redshift is $1{+}z = dτ_r/dτ_e$.

For certain special worldlines and spacetime backgrounds, you can come up with simple formulas for $1{+}z$. Examples are arbitrary worldlines in a Minkowski background (special-relativistic redshift, derived here), stationary worldlines in a static background ("gravitational redshift", derived here), and comoving worldlines in a FLRW background ("cosmological redshift", see previous link). These are just simple, symmetric special cases. In a general spacetime background, there is only general Doppler shift.

Some argue that there is no time dilation in this case

"Time dilation" can refer to a difference in total elapsed proper time along different worldlines that meet twice (the twin effect/paradox), which is a real physical effect. But comoving observers in FLRW cosmology never meet.

It can also mean, in special relativity, $dτ/dt<1$, where $τ$ is the proper time of some worldline and $t$ is the $t$ coordinate of some inertial frame. This is frame-dependent and not really a physical effect, unless all of the clocks that define $t$ really exist. That answer seems to be using this second meaning, but applied to a noninertial frame (FLRW coordinates).

In special relativity, if all of your clocks are at rest relative to your chosen inertial frame, there is no time dilation relative to those coordinates. Likewise, in FLRW coordinates, if all of the clocks are at rest relative to the coordinates, there is no time dilation relative to those coordinates. This doesn't have any physical consequences. It's a near-tautological statement that amounts to saying that $dτ/dt=1$ if $τ=t$. It does mean, though, that if you want to calculate a redshift factor in those coordinates, you shouldn't include a factor for time dilation of comoving observers.

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, which you could even decompose into $1{+}β$ and $γ$ components if you really wanted to.

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.

benrg
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  • You are misinterpreting what John Rennie is saying. In Minkowski space two observers with relative velocity will each see the other aging more slowly, but in the Friedman-Robertson-Walker solution, time is the same for all comoving observers. What is confusing here is that an observer on earth will actually “see” Events occur more slowly because of the red shift, but that is not due to time dilation. In a contracting universe our observer would “see” events in the moving galaxy progress more quickly and there is clearly no analog to that in special relativity – Thomas Tiger Oct 01 '22 at 00:47
  • @ThomasTiger I rewrote some parts of the answer, including that bit. – benrg Oct 01 '22 at 19:38
  • @benrg, "worldline emits two light wavefronts/pulses at times separated by τ, they propagate along null geodesics to a second timelike worldline which they intersect at times separated by τ, and the redshift is 1+=τ/τ.", so exactly why 1+=τ/τ is valid? Thank you. – ABC Oct 24 '22 at 00:12
  • @ABC That's just the definition of redshift, or at least the definition that I'm using. The elapsed proper time between successive wavefronts is the reciprocal of the frequency, so you can also say $1{+}z=f_e/f_r$. – benrg Oct 24 '22 at 18:06
  • @benrg I get that $1+z = a(t_o)/a(t_e) = \lambda_o/\lambda_e = f_e / f_o$, but I'm just having trouble connecting wavelength or frequency to the proper time "between successive wavefronts". For a mechanical water wave or sound wave that follows sinusoidal oscillation, sure the spacing between wavefronts are related to frequency/wavelength. But light emitted from a gas cloud is due to electrons changing energy levels in the atoms, why wouldn't the interval between two photons being emitted be random? Why does the interval relate to wavelength? – ABC Oct 26 '22 at 02:50
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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 $e^{\Delta\Phi}$. 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).

  • Thank you for the answer. The relativistic Doppler effect formula and derivation in my question are straight from Wiki: https://en.m.wikipedia.org/wiki/Relativistic_Doppler_effect#Motion_along_the_line_of_sight - It was just an example anyway, not very important. My question is, what defines the actual observed Doppler effect of distant galaxies? In other words, what is the formula for the redshift in the expanding universe assuming the simplest case of a constant speed expansion with no acceleration and no local gravitational time dilation? Either flat or positively curved space. Thanks! +1 – safesphere Dec 26 '17 at 04:22
  • what is the formula for the redshift in the expanding universe assuming the simplest case of a constant speed expansion with no acceleration and no local gravitational time dilation? You might want to ask this as a separate question. If you want the expansion to be at a constant rate, then you need a universe with no matter in it, and the spacetime is flat. This cosmology is just Minkowski space described in funny coordinates. –  Dec 26 '17 at 14:57
  • It's not a separate question, it's my original question that I am trying to phrase differently to make it as simple as possible to get a specific answer. The acceleration was discovered only recently, ignore it or not, I don't care. No local gravitational time dilation means that light does not come from a neutron star. All I want is the formula. Can you provide it? – safesphere Dec 26 '17 at 16:44
  • @safesphere: The attitude shown in your comment doesn't motivate me to spend any more time trying to help you, but I'll correct your misconception for the benefit of other people who might come across this comment thread. The acceleration was discovered only recently, No, the acceleration was previously believed to be negative, but is now known to be positive. –  Dec 26 '17 at 23:52
  • Sorry if my comment wasn't phrased the best way (I am not a native English speaker). I can assure you that I meant no disrespect. I simply had to ask and rephrase my question several times, because I was unable to get a specific answer. I know you are very knowledgeable and provided great answers before, which I greatly appreciated. So I'm just puzzled why in this case you limited your answer and comments to general statements around and about my question refusing to actually answer it. Whatever the reason is, this is fine Sir, no hard feelings. I am here to learn with no "attitude" to show :) – safesphere Dec 27 '17 at 00:12
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    Perhaps you could add some details to your answer now? Thanks! – safesphere Jan 02 '18 at 01:35
  • As both Safesphere and Ben Crowell have been very helpful to me, I'd like to point out the fact that the 1996 SN1a study under discussion had 25 authors and includes a diagram with a rest frame: It was trying to distinguish between redshift due to "gravitational dissipation of photonic energy" (AKA "tired light") and redshift due to spatial expansion, and included a remark that, "it is clear that we have observed the time dilation of macroscopic clocks at cosmological distances", without ruling out the generally-accepted conclusion that spatial expansion played the main role in redshift. – Edouard Jan 23 '20 at 16:11
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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.

Photon
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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"

Richard Bradford
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    Thanks for the answer. How do you account for the fact that distant galaxies move away from us three times faster than light? – safesphere Aug 16 '20 at 01:44
  • By the time we observe light from an object at 30 megaparsecs the candle is already 220 years further into its future the moment we see it (Earthtime). An object 300 megaparsecs away takes 1000 million years to arrive. By the time we see photons from a candle 300 Mpcs away it moved 2200 years into its future. It did not move through space alone. It moves forward through time. Our time however remains 2,200 earth years behind (It's Time). Relatively speaking.. Nothing travels faster than the speed of light. – Richard Bradford Aug 16 '20 at 17:07
  • When something is.. IS much more important than where something is. – Richard Bradford Aug 16 '20 at 17:11
  • Hi Safesphere, Again thank you for your question. I am recalculating the data I shared with you yesterday in hopes of giving you more precise answer. The general answer remains however: - When we arrive at the conclusion that an object at 300+ megaparsecs moved further through space ; the object does so because it has larger amount of time to get there. It's not because it traveled a further distance. It is traveling for a longer amount of time at its known speed through space. If you have a specific "distant galaxy" you are referring to.. I will be more than happy clarify. – Richard Bradford Aug 17 '20 at 19:42