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At the time the Big Bang happened the matter had enormous density. According the GR (I may be wrong here) such density dilates time.

If so, could it be that the time periods just after Big Bang which are usually considered happening in small part of a second (such as the Planck epoch), in reaity took billons of year (or may be, infinity) but due to time dilation appear to us as spanning only microscopic parts of a second? Could it be that the age of the universe is dramatically underestimated?

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  • $\begingroup$ At the time of Big Bang Spacetime was compact too. Don't compare it with Black Holes.. $\endgroup$ – Schrödinger's Cat Jun 16 '14 at 3:45
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If one assumes FRW geometry (homogeneous and isotropic universe) in GR, $$d\tau^2 =dt^2-a^2(t)d\vec{x}^2 $$ the age of the universe is precisely calculated by (derived from equations of motion and assume that $a(t=0)=0$ is the beginning of our universe)

$$t_0 = \int_0^1\frac{dy}{yH_0\sqrt{\Omega_{\Lambda}+\Omega_My^{-3} +\Omega_R y^{-4}}} $$

where $y=\frac{a(t)}{a_0}$,

$H_0$: the present Hubble parameter $\frac{\dot{a}_0}{a_0}$

$\Omega_{\Lambda}$: percentage of dark energy

$\Omega_{M}$: percentage of matter, including known matter and dark matter

$\Omega_{R}$: percentage of radiation

*According to Planck 2013 results. XVI. Cosmological parameters, the lifetime of our universe is around 13.8 billion years

*See Cosmology by Steven Weinberg Ch.1 for references, one can also derive this from Friedmann equations

What you have concerned have been taken into account.

Of course, there are many ways to define the concept of time, such as conformal time, comoving time. However, there is only one physical time in each frame of reference, which is the comoving time defined by the frame of the observer (in FRW metric, which is different from the black hole case), $$ dt = d\tau \quad \mbox{since} \quad d\vec{x}=0 \quad \mbox{for the observer} $$

One can calculate the conformal lifetime of the universe as well but it is confusing the audience since it is different from measurable time. The comoving time is exactly the meaning of "time" by scientists and even ordinary people. Thus it is the most reasonable way to report the lifetime of the universe in comoving time.

There are also other details about your question in cosmology, such as the accuracy of GR at the beginning of the universe, how the beginning of the universe is defined.

(*Note that the speed of light $c$ is taken to be 1 in the above discussion)

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  • $\begingroup$ Welcome to Physics.SE ! Thank you for your answer. Your answer would benefit from references (like books, web links) and more explanations about the models, assumptions and approximations that are used (e.g. here, how to get from the FRW metric to the expression of $t_0$ you have given). $\endgroup$ – Tom-Tom Jun 16 '14 at 8:55
  • $\begingroup$ What is "conformal time"? $\endgroup$ – Anixx Jun 16 '14 at 9:16
  • $\begingroup$ roughly speaking, conformal time is a redefinition of time coordinate such that light cone (or simply the trajectory of light in spacetime coordinate, or null geodesic) is always straight line $$d\tau^2 = a^2(t)(dT^2-d\vec{x}^2) $$ here $T$ is conformal time, or $$ T = \int \frac{dt}{a(t)}$$ $\endgroup$ – mastrok Jun 16 '14 at 10:50
  • $\begingroup$ comoving time is universal in all spacetime. As I mentioned, it is really the time measured by the observer. It applies to flat spacetime, positively/negatively curved spacetime. $\endgroup$ – mastrok Jun 17 '14 at 5:46
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Time is a tricky concept in GR because in general different observers will measure different elapsed times. However there is a well defined property called proper time that is invarient and has the same value for all observers. The proper time is normally represented by the symbol $\tau$, and it's defined as:

$$ d\tau = g_{ab}dx^adx^b $$

where $g$ is the metric of the spacetime. In the particular case of the FLRW spacetime that (approximately) describes our universe this simplifies to the expression given by mastrock:

$$ d\tau^2 =dt^2-a^2(t)d\vec{x}^2 $$

where $dt$ is the movement in time and $d\vec{x}$ is the movement in space (the coordinates in use here are comoving coordinates).

The point of all this is that suppose you choose an observer that is not moving in space, i.e. $d\vec{x} = 0$, then the expression for the proper time becomes:

$$ d\tau^2 =dt^2$$

so the proper time is just the time measured on whatever clock the observer has to hand. So if you are such an observer then the proper time from the Big Bang to the current day is just the time measured on your clock. Furthermore the FLRW metric assumes spacetime is homogeneous, so all stationary (i.e. comoving) observers measure the same time on their clocks - the time since the Big Bang is the same everywhere.

And now we can answer your question. Time wasn't dilated near the Big Bang when the density was higher because (proper) time is by definition what is shown on a comoving observer's clock. Every comoving observer everywhere would have recorded 13.7 billion years since the Big Bang.

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  • $\begingroup$ Well I see it, yet would be the time of the Big Bang if we account for greated density of the time (I see that it will not be the proper time)? $\endgroup$ – Anixx Jun 16 '14 at 12:07
  • $\begingroup$ @Anixx: the proper time to the Big Bang is the only time that makes physical sense. $\endgroup$ – John Rennie Jun 16 '14 at 13:06
  • $\begingroup$ Well can that time be compared to the time passed in an absolutely flat spacetime? $\endgroup$ – Anixx Jun 16 '14 at 13:37
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Is the universe finite or infinite?

If the big bang had infinite mass at the start then time dilation due to gravity would be infinite and time would stand still. That's the problem with trying to conceive of the start of time because it took infinitely long to start.

An infinitesimal length of time under infinite gravity is infinitely long.

If the universe was finite then time dilation would have some maximum value, giving time a starting point.

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  • $\begingroup$ 1) density itself does not cause gravitational effects. An infinitely dense region of zero volume has an undefined mass and there is no reason to assume it must experience time dilation. 2) If the entirety of the universe experiences the same gravitational time dilation, then there exists no frame in which time is not dilated and, thus, no frame to compare with to say "hey, it actually took a long time". In essence, if everybody measures it as a small amount of time, then it was a small amount of time... $\endgroup$ – Jim Oct 21 '16 at 13:45
  • $\begingroup$ And 3) At the time of the big bang, the universe was not an infinitely dense point. It was many causally disconnected points with zero proper distance between each point. The density at any one point was assuredly finite. $\endgroup$ – Jim Oct 21 '16 at 13:46
  • $\begingroup$ I suppose it depends if the total mass in the universe at the big bang finite or infinite. was space empty or full or have any amount at the start? $\endgroup$ – ryanmonk Oct 21 '16 at 13:52
  • $\begingroup$ The total mass in any causally connected regions of the universe was definitely finite. Depending on your definition of the "big bang", it might also be empty. Whether the total mass is infinite across all the causally disconnected regions is irrelevant. Those regions can't influence each other while they are causally disconnected; they behave as if the other regions don't exist. $\endgroup$ – Jim Oct 21 '16 at 14:00
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We don't know if ∞ is the actual age of the Universe or its size. If we presume it is, we have to assume too that Time dilation itself was ∞ (and we wouldn't be here). But we can assume →∞ for each, e.g. a hyperbolic "history" & "size" of the Universe, which make age & size, for all intents & purposes, ∞-1.

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    $\begingroup$ Please elaborate $\endgroup$ – QuIcKmAtHs Jan 13 '18 at 6:12

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