I'm not asking about how we worked backward from an expanding universe to the age of the big bang, but rather what is the meaning of time in a near infinitely dense point in the context of general relativity? Wouldn't time flow infinitely slowly for a theoretical (though physically impossible) observer?

  • $\begingroup$ Related: physics.stackexchange.com/q/104153/2451 $\endgroup$
    – Qmechanic
    Mar 21, 2014 at 20:56
  • $\begingroup$ You have gravitational time dilation backwards. The closer a clock is to a source of gravitation, the more slowly time passes. So instead we might expect time to flow infinitely slowly for your impossible observer. $\endgroup$
    – David H
    Mar 21, 2014 at 20:56

1 Answer 1


No. The standard metric for cosmology is given by:

$$ds^{2} = - dt^{2} + a(t)^{2}\left(d^{3}{\vec x}^{2}\right)$$

where the term inside the parenthees represents the 3-metric of a homogenous three space. As you can see, there is no difficulty with evaluating the age of the universe:

$$ T = \int\sqrt{-g}\,\,x^{a}y^{a}z^{a}\epsilon_{abcd} = \int dt$$

where the integral is evaluated from the time when $a = 0$ to now.

  • $\begingroup$ I'll take your word for it but I was hoping for more of an explanation of how the notion of time holds up to extreme time dilation effects $\endgroup$
    – schmop
    Mar 21, 2014 at 22:24
  • $\begingroup$ @schmop: the short answer is that the time dilation effects don't happen at the big bang singularity for cosmological solutions, they happen at the cosmological horizon. The picture is more complicated than the simple intuition of "time runs slower near a large agglomeration of mass" $\endgroup$ Mar 21, 2014 at 22:43
  • $\begingroup$ But the real answer is that you can consistently compute the age of the universe through the simple computation above. It's really just a matter of solving for $a(t)$ $\endgroup$ Mar 21, 2014 at 22:45

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