I found multiple questions where it is stated that dark energy increases as the universe expands. Assuming a big crunch scenario, will this dark energy "go away" again as the size of the universe decreases again, or will there be more energy (=mass) at the Big Crunch than at the Big Bang?


Point A) The same mechanism that causes the amount of dark energy to increase as the universe expands will cause the amount of dark energy to decrease if the universe contracts.

Point B) A universe that includes an increasing amount of dark energy due to the same mechanism that exists in our universe (in theory) makes it nearly impossible for the universe to begin to contract (not completely impossible in general, but it is for the universe we think we live in). Thus, there will $\textit{probably}~^1$ never be a Big Crunch (unless we're talking about the chocolate bar. I'm led to believe that already exists).

$^1$(probably allows for the possibility that the universe we think we live in is not the universe we actually live in)

  • $\begingroup$ unless we're talking about the chocolate bar. I'm led to believe that already exists Sir you have put a smile on my face and I owe you a pound. $\endgroup$ – Matas Vaitkevicius Jul 29 '14 at 14:50
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    $\begingroup$ Point B is wrong. In the $\Omega_M$-$\Omega_\Lambda$ plane, there is a boundary between cosmologies that lead to an eventual recollapse and those that don't. Positive values of $\Omega_\Lambda$ can be compatible with a Big Crunch. $\endgroup$ – user4552 Jul 29 '14 at 15:09
  • $\begingroup$ @BenCrowell Note the word "nearly" before the word "impossible" $\endgroup$ – Jim Jul 29 '14 at 15:42
  • $\begingroup$ and the word "probably" before the word "never" $\endgroup$ – Jim Jul 29 '14 at 15:43
  • $\begingroup$ @Jim: You could edit the answer to make the language more precise, but then I think it would become clear that point B is irrelevant to the question. It's not a matter of probability whether or not we will have a Big Crunch. We know that there will not be. The question asks about a hypothetical cosmology that is not a realistic model of our own universe. $\endgroup$ – user4552 Jul 29 '14 at 15:55

General relativity does not have a conserved, scalar measure of mass-energy that can be defined in all spacetimes. (Such measures do exist for asymptotically flat spacetimes, but cosmological spacetimes aren't asymptotically flat.) Therefore there is no useful way of defining the total mass-energy of the universe, or even the total mass-energy of the observable universe (even for a closed universe, which is spatially finite). Misner, Thorne, and Wheeler has a nice discussion of this sort of thing on p. 457.

So the question as posed does not have an answer. You could ask instead whether, in a recollapsing cosmology, the local density of mass-energy at time $\Delta t$ after the Big Bang is the same as the local density at $-\Delta t$ before the Big Crunch. If recollapsing cosmologies are symmetric under time reversal, then the answer would automatically be yes. I think the answer to the rewritten question is that you can construct cosmologies that are time-reversal symmetric, but they would be descriptions of a universe that had a maximum-entropy Big Bang, so that the universe was at all times in thermal equilibrium, and there was never any 2nd law of thermodynamics. This would be a universe without observers, which would be a little boring.

Therefore if the question is modified in this way to make it answerable within GR, then the answer is probably that there would be a different mass density at the end that at the beginning (at corresponding times). However, this would have no close logical connection with dark energy. (And it is not true as claimed in Jim's answer that you can't have a big crunch in a universe with dark energy.)

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    $\begingroup$ One quibble: some non-asymptotically flat spacetimes do have definiable local energies over things like the cosmological horizon--you just need a null or timelike 3-surface with a null or timelike killing vector, and you can define the energy contained in that surface. This isn't the case for cosmological horizons, though. $\endgroup$ – Jerry Schirmer Jul 29 '14 at 15:42

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