0
$\begingroup$

Is the cosmological time grosso modo isochrone? by analogy with space isotropy. Or else do we have possibly great differences by analogy with great voids in the space.

We know that it's not strictly exact if it is. Special relativity teaches us that travelling clocks may show different times where they come back to their starting reference point.

Events in the sky may occur on objects having speeds in our rest frame on Earth. All the objects we observe are somehow travelling clocks.

We know that black holes may generate great differences in their neighborhood.

Let's take galaxies moving at 3000 km/s in the Earth frame. It's a big speed. Their clocks will have a relative offset of $10^{-4}$ with the ours, hence an absolute offset about 1.4 million years.

But, I wonder if there are big areas in our universe where clocks which had started with the Big Bang , one day will have hundreds of million years of difference ( and more ) with our clocks ( the day when these clocks will meet again ). Do astronomers observe galaxies with relative speeds of $\frac c {10}$ and more?


Edit : Note that it is a variant of the twins case : I insisted with clocks *starting at the BB** and showing differences when they meet again. Just to avoid useless critics , though it's conventional in relativity to suppose it without assumptions on the knowledge of the poster. the "for all practical purposes" of the comment is the computation of cosmological constants, like the Hubble one.

$\endgroup$
6
  • $\begingroup$ Astronomers have been observing objects with redshifts as high as 1100 (the CMB). The CMB is, of course, light, i.e. it connects the present with the time just 300,000 years after the big bang with zero proper distance. IF you want to find baryonic matter moving at relativistic velocities, there are plenty of jets and cosmic rays to chose from. Cosmological time is therefor just a special choice representing a hypothetically completely homogeneous universe. $\endgroup$
    – CuriousOne
    Commented Feb 13, 2016 at 6:27
  • 5
    $\begingroup$ I don't understand what you mean by grosso modo isochrone and googling doesn't help. Can you clarify what you are asking? $\endgroup$ Commented Feb 13, 2016 at 6:32
  • 1
    $\begingroup$ @JohnRennie: The way I read it the OP is trying to ask if there is a global cosmological time coordinate "for all practical purposes". Are there relativistic objects relative to the CMB? Yes, there are... but not on the galactic scale if I am not mistaken... although some jets contain quite a bit of matter. $\endgroup$
    – CuriousOne
    Commented Feb 13, 2016 at 6:56
  • $\begingroup$ igael, are you asking whether any regions of the universe differ significantly from the comoving frame? That is, are there regions of the universe with significant peculiar velocities? $\endgroup$ Commented Feb 13, 2016 at 7:41
  • $\begingroup$ @JohnRennie : yes. Iso chrone = same time. Relativistic speeds or anything else which makes the clocks different at big Crunch time. I can except that this difference will introduce different datas when computing ie the Hubble constant $\endgroup$
    – user46925
    Commented Feb 13, 2016 at 14:11

1 Answer 1

2
$\begingroup$

When we talk about the expanding universe we normally mean the FLRW metric (or some minor perturbation to it) and in the FLRW metric one of the assumptions is that the distribution of matter is completely homogeneous. In this situation the gravitational potential of all observers is the same, and the velocities of all observers are exactly described by the Hubble law. Put another way, the peculiar velocity of all observers everywhere is zero.

in the FLRW universe the proper time measured by all observers since the Big Bang is the same, or more colloquially their clocks all run at the same rate.

But of course the distribution of matter is not homogenous, and hasn't been since at least the time the CMB was emitted. Wherever the density is slightly greater than average the density increases with time, and wherever the density is slightly lower than average the density decreases with time. These density differences cause a relative time dilation between different observers, so in practice different observers measure a different proper time since the Big Bang. There are two reasons for this, both of which can be related to gravitational potential energy.

An observer in high density region will have a greater (more negative) gravitational potential than the average, and vice versa for an observer in a low density region. Let's take the average gravitational potential to be zero, then we can write the potential energy (PE per unit mass) of an observer as $U$, where $U$ is negative in a high density region and positive in a low density region. In the weak field limit (The weak field limit applies when $U/c^2 \ll 1$) this potential is related to the time dilation by:

$$ \frac{d\tau}{d\tau_\text{av}} = \sqrt{1 + \frac{2U}{c^2}} \tag{1} $$

The other effect is that the virial theorem tells us that in a gravitationally bound system the kinetic and potential energy are related by:

$$ T = -\frac{U}{2} $$

For a unit mass $T = \tfrac{1}{2}v^2$, so we get:

$$ v^2 = -U $$

and the time dilation due to this velocity is:

$$ \frac{d\tau}{d\tau_\text{av}} = \sqrt{1 - \frac{v^2}{c^2}} = \sqrt{1 + \frac{U}{c^2}} \tag{2} $$

In the weak field limit we can just multiply together the gravitational time dilation given by (1) and the time dilation due to motion given by (2) to get:

$$ \frac{d\tau}{d\tau_\text{av}} = \sqrt{1 + \frac{3U}{c^2}} \tag{3} $$

where we have discarded terms in $(U/c^2)^2$ on the grounds that they are negligably small.

And equation (3) is the one we need to answer your question, at least in principle. If we measure proper time relative to the average proper time then the elapsed time relative to the average is just given by integrating equation (3):

$$ \tau = \int_0^\tau \, \sqrt{1 + \frac{3U(\tau')}{c^2}} d\tau' $$

The problem is that $U(\tau)$ is a poorly known function. We known that $U$ was effectively zero at the time the CMB was emitted because the CMB is so smooth. We can get an idea of $U$ at the moment. For example in large clusters like the Virgo cluster the peculiar velocities can be up to 1600 km/s implying that $U$ can be as high as $2.5 \times 10^{12}$ J, though that still only makes $U/c^2 \approx 2.8 \times 10^{-5}$. To calculate how much the proper time differes from the average we'd need to know how $U$ has changed with time in between. Possibly this is known from modelling, but I have to admit that I don't know what the results are.

So I guess I'm admitting that I don't know the answer to your question, though as discussed above I do know how to calculate it if you can find suitable modelling data. To your specific question whether galaxies exist with peculiar velocities comparable to $c/10$, no they do not. The highest peculiar velocities we have found are in the range 1000 to 2000 km/s.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.