# Does cosmic time (or comoving clock) run at the same pace at all times?

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1. Oct 9, 2014

### Vincentius

My question is if the pace of cosmic time changes over time or not. I think this is identical to asking if the rate of a comoving clock changes over time.

The FRW metric holds at any time, but only the spatial part expands:

ds2=c2dt2-a(t)2dx2

Apparently, there is no time delation in the metric, but that does not mean that a physical clock in the past ran at the same pace as today.

Can anyone explain or give a reference on the subject?
Thanks

2. Oct 9, 2014

### Chalnoth

For a physical clock to have run at a different pace in the past compared to today, the laws of physics would have had to be different (e.g., different physical constants). So far, no such deviation has been found.

3. Oct 9, 2014

### TumblingDice

Time dilation occurs between observers moving relative to each other and in gravitational fields, so relative motion and gravity would be conditions to give a better/best answer. You mentioned a comoving clock, and I'm not expert enough to know if that means CMB frame.

For example, the age of the universe is often calculated using the reference frame of the CMB and away from gravitational influences. Pretty much in 'free fall' with the geometric expansion of the universe. The CMB is the oldest thing we can "see" and we have an excellent idea of what "time" it was created very early in the evolution of the universe. But we're not co-moving with the CMB's rest frame, so attempting to use a clock on Earth to extrapolate an 'age' backward would give a slightly different result.

Here's quote from a post by PeterDonis:

4. Oct 9, 2014

### Vincentius

We know by GR that a clock closer to a gravitational well runs slower, without new laws of physics. How can we be so sure that something similar doesn't happen in a universe of totally different density then today. My question is really what the theoretical reason is for putting just c2t2 in the FRW metric, while they could have chosen any other parametrization, e.g. c2t'2/a(t')2. So what is the theoretical argument to understand that c2t2 is comoving physical clock time?

5. Oct 9, 2014

### Chalnoth

But you're not comparing a clock inside a gravitational well and one outside. You're comparing the same clock at two different points along its own timeline. It doesn't even make sense to compare a clock against itself in this manner.

Another way to understand it is to recognize that we define our time coordinate as that which a particular clock would measure (in this case, the clock would be a co-moving clock, stationary with respect to the CMB).

6. Oct 9, 2014

### Vincentius

What's the difference? Comparing clocks over spatial distance or over time intervals? Above that, looking far into space is looking back in time. And why doesn't it make sense?

7. Oct 9, 2014

### Chalnoth

Because the behavior of the clock is how you define the passage of time.

8. Oct 9, 2014

### TumblingDice

Hope you still mean rate when you speak of comparing clocks (as in your OP):
You've mentioned the possibility of a clock running at a different rate when the density of the universe was different. Have you considered what the maximum rate of change might be based on the required energy density needed in the open space that's used to base universe time on?

9. Oct 10, 2014

### Chronos

This implies a 'universal' clock, which is not known to exist. Under GR, time and space are malleable and have no universal reference frame.

10. Oct 10, 2014

### Torbjorn_L

Maybe I'm too daft to grok this, but the question makes little sense to me. We can't define a "pace" (rate) of time by definition, since dt/dt = 1. It is part of time physics, "the problem of time". What we can do is measure time with by necessity local clocks, so local time.

Then observations of these clocks show:
1) that the clocks experience relativistic effects (time dilation)
2) that the observation of clocks experience relativistic effects (Penrose-Terrell rotation [ http://math.ucr.edu/home/baez/physics/Relativity/SR/penrose.html ] )
3) That poorly chosen clocks and/or their observation experience quantum effects (non-local decoherence, smearing time over the lightcone [ http://en.wikipedia.org/wiki/Quantum_Zeno_effect , for one ])

This means that "cosmic time" would be what we see as invariant time in the CMB rest frame, i.e. cosmological time would observe the cosmological expansion as part of what makes it tick.

Last edited: Oct 10, 2014
11. Oct 10, 2014

### Vincentius

Right, we talk about clocks. And only comoving clocks. And these are supposed to indicate cosmic time t. What's the proof that a clock 'then' ticks at the same pace as a clock 'now'?

Perhaps another way to look at this question: light from far away is redshifted, but this can be attributed to Doppler, gravitational and/or cosmological redshift. Doppler and cosmological redshift are effects of remote observation, but gravitational redshift happens locally, like two clocks at different altitude. Hence, clock rate then would be different from clock rate now if the cosmic potential had changed.

12. Oct 10, 2014

### TumblingDice

Using one second as an example of a 'tick' - whether the "then" or the "now" clock, a tick is a tick and a second is a second. To posit anything different would require introducing another clock in a different reference frame, and since we've already defined universal time as comoving with CMB in open space, another clock would only blow smoke into the picture. (It does not make sense.)

13. Oct 10, 2014

### Staff: Mentor

How would you test, experimentally, for a difference in "tick rate"?

This seems confused to me. If an object right next to me is moving away and emits light that I detect, I will measure it to have a Doppler redshift. Is that a "remote observation"? If so, why wouldn't gravitational redshift I observe in light emitted by an object a little bit below me also be a "remote observation"?

Conversely, if I observe a Doppler shift in light emitted by a galaxy a billion light-years away, why could I not also observe gravitational redshift from light emitted by, say, a supermassive black hole a billion light-years away?

There is no such thing as "cosmic potential". "Gravitational potential" is a concept that only makes sense in stationary spacetimes. The universe is not a stationary spacetime, because it's expanding.

14. Oct 10, 2014

### Vincentius

Comparing clock rates over cosmic distances is probably quite difficult and complicated. This however does not make the question irrelevant or senseless. Just stating a second is a second is not enough and is only true for local observers, like the speed of light is the same for any local observer. But not so the speed of light in the reference system of a remote observer. If for instance one calculates the comoving distance to the particle horizon, then obviously one has to know how fast this horizon propagates, which is dr/dt'=c in proper coordinates (r,t'). The metric tells dr=adx, and dt'=dt. Why would the latter be the case?
So far no one provided a theoretical answer. The metric allows any parametrization of time, but there is only one which matches proper time over cosmic intervals. Why would this be dt'=dt? One would expect formal proof in any text book, but so far I haven't been able to find anything.

15. Oct 10, 2014

### Staff: Mentor

You left out a key qualifier: there is only one which matches proper time for comoving observers over cosmic intervals (or indeed over any intervals).

Because that's how the FRW coordinate chart is defined: coordinate time = proper time for comoving observers.

The real question of physics here is, what is special about comoving observers? The answer is, they are the ones that see the universe as homogeneous and isotropic (for example, they see an isotropic CMBR). No other observers have this property. That's why comoving observers are picked out as the ones whose proper time equals coordinate time in the FRW chart: because it makes the chart itself homogeneous and isotropic (the only change is in the scale factor), which makes it very convenient for calculations.

16. Oct 10, 2014

### Vincentius

One can define any convenient parametrization of time. dt'=dt is surely convenient. But this still does not proof it matches proper time of comoving observers over cosmic intervals. I would appreciate if anyone could provide a good reference on this subject.

17. Oct 10, 2014

### Staff: Mentor

There's nothing to prove; this condition is the definition of coordinate time in the FRW chart. See below.

$dt' = dt$ is not a "parametrization of time". I'm not sure exactly what it is (i.e., what you are trying to convey by writing it down), but it isn't a choice of how to parametrize the time coordinate. That choice is made by choosing $g_{tt}$, i.e., by choosing the coefficient of $dt^2$ in the metric. The FRW chart is defined by the choice $g_{tt} = - 1$; combined with the choice that comoving observers are observers who are at rest in the chart.

Perhaps by "proof" you mean a proof that the above choices are in fact consistent? Showing that they're consistent mathematically is trivial: just write down the line element. If you need physical justification, just observe that the Einstein tensor derived from the FRW metric implies a stress-energy tensor for a homogeneous, isotropic perfect fluid, as described, for example, here:

http://en.wikipedia.org/wiki/Friedmann–Lemaître–Robertson–Walker_metric

18. Oct 10, 2014

### Vincentius

I still don't see any proof. It seems like the subject is ignored in pretty much every text on the subject. Also in the wp article Peter Donis referred to:

From wp:
The FLRW metric starts with the assumption of homogeneity and isotropy of space. It also assumes that the spatial component of the metric can be time-dependent. The generic metric which meets these conditions is

Hence, a possible time dependence of the coefficient gtt is simply not considered. Homogeneity and isotropy are spatial properties, so time can be parametrized in an arbitrary way. There is no condition specified for it. In the Schwarzschild metric it is the gravitational potential which determines gtt. I think it is no different in our universe. The fact that GR does not explicitely deal with a cosmic potential does not mean it is not there. How can any spacetime metric ignore the influence of the huge amounts of cosmic mass around? As if these were irrelevant to the metric. In GR the potential of the cosmic masses appears implicite as -c2, in agreement with Sciama's cosmic potential. How can a cosmic potential not exist if there is mass in the universe? And why wouldn't it change over time with all matter receding (change over time as observed over cosmic distances)?

19. Oct 10, 2014

### Staff: Mentor

Because we have chosen coordinates that match up with the homogeneity and isotropy in a particular way. See below.

Which means they are only true on a certain family of spacelike slices; they are not true on every single spacelike slice you could "cut" in the spacetime. That certain family of spacelike slices automatically gives you a time coordinate with particular properties: see below.

It can be, sure. But it doesn't have to be, in the sense that, if there are a family of spacelike slices that are homogeneous and isotropic, it is always possible to choose a time coordinate such that each such spacelike slice has the same value of the time coordinate everywhere on it. This coordinate choice ensures that $g_{tt}$ is independent of the spatial coordinates. Then it is always possible to choose a parametrization of the time coordinate that makes $g_{tt}$ independent of $t$ as well. Once you have that, choosing $g_{tt} = -1$ is just a matter of scaling; you could rescale the time coordinate so that, say, $g_{tt} = - 1/2$, but what's the point? It's still the same everywhere.

So what? The Schwarzschild metric is a different solution of the Einstein Field Equation, with different properties. One property in particular is that it is stationary; that is what makes it possible to define "gravitational potential" in it. The spacetime describing the universe as a whole is not stationary, so you can't do that.

Why do you think FRW spacetime does this? The Wikipedia page I linked to describes how FRW spacetime is derived as a solution of the Einstein Field Equation with a stress-energy tensor describing the "huge amounts of cosmic mass around". That accounts for it.

No, it doesn't; the $c^2$ in $g_{tt}$ (or $1$ in the units I've been using) has nothing to do with "the potential of the cosmic masses". You really need to review how FRW spacetime is derived.

Reference, please? Sciama did write some papers about a speculative theory of gravity that he thought would be more "Machian", but AFAIK he never claimed it actually applied to cosmology, nor has anyone else.

Because the universe is not stationary, and "gravitational potential" can't be defined in a non-stationary spacetime. I've said this repeatedly.

Because it can't be defined in the first place. The change with time of the distribution of matter in the universe shows up in the dynamics of the scale factor $a(t)$ in the FRW metric; it doesn't need to show up anywhere else.

Last edited: Oct 10, 2014
20. Oct 11, 2014

### Vincentius

I see only definitions. And Sciama was wrong? Please a text which proofs that comoving clocks match a constant gtt. We only agree that GR does indeed account for cosmic masses. But if it is not via c2, where else does it?