1
$\begingroup$

Did clocks run more slowly in the early universe due to gravitational time dilation? Or, alternatively, do they appear to be running more slowly when observed from today, if that is not the same thing?

$\endgroup$
1
  • $\begingroup$ I think gravitational time dilation is noticeable (and matters) only in high gravitational field. Hence, only during very early stages of universe when radius was the order of $r_c$~$\frac{2GM}{c^2}$. Moreover, since we are in a (almost) Minkowski geometry, we will not experience anything weird (due to high density early universe) in the time except observing gravitational redshift which probably cannot be observed for those early universe because of the dark era. $\endgroup$ Jun 21 at 18:28
1
$\begingroup$

Surprisingly, there is no time dilation in the standard FLRW metric that is usually used to model a homogeneous and isotropic universe: $$ds^2 = dt^2 + a(t)d\Omega^2$$ where $t$ is the time, $a(t)$ the scale factor and $d\Omega^2$ the spatial metric. So clocks that are not moving in co-moving coordinated ($d\Omega^2=0$) will not experience gravitational time dilation.

$\endgroup$
1
  • $\begingroup$ This is not a physical property of FLRW metrics, it's just a convention. See my answer. (Also $a(t)$ should be squared and $+$ should be $-$.) $\endgroup$
    – benrg
    Jun 22 at 3:03
-1
$\begingroup$

Time dilation is meaningless unless you say what it's relative to.

The usual convention is to define the cosmological time $t$ to be the time measured by comoving clocks (clocks that have zero peculiar velocity). Those clocks then don't "dilate" relative to the cosmological time, in any era, simply by definition.

You could instead say that the reason that light from the early universe appears redshifted to us is that clocks ran slower in the early universe. You can define a cosmological time with the property that the ratio of the rate of local comoving clocks to the cosmological time is equal to the Doppler shift ratio. This gets you the so-called conformal time $η$.

Both of these descriptions are equally valid. They're the same thing in different coordinates.

In terms of the usual cosmological time, the metric is $ds^2=dt^2-a(t)^2dΩ^2$, as seen in Anders Sandberg's answer. It looks like there is no time dilation because there is no factor in front of $dt$. In terms of conformal time, the metric is $ds^2=a(η)^2(dη^2-dΩ^2)$, and it looks like there is time dilation because there is a factor of $a(η)$ in front of $dη$.

$\endgroup$
-5
$\begingroup$

In glonal coordinates time ran slower. So seen from present day, time ran slower.At the beginning there was no globality yet. Only localty. Seen locally, time ran with normal (proper) pace.

So seen from our pespective it looks as if the big bang happened an infinire time ago and we will never be able to get access to that moment. Likewise, it seems as if time stands still at the horizon of a black hole.The difference being that we can approah and even cross the horizon in finite time. We can fallfreely to the hole but we can't fall freely to the big bang but if we could it would exactly take the age of the universe (about 14 billion years).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.