3
$\begingroup$

Wikipedia says,

When $n=4$ (3 space dimensions plus time), it is (the de Sitter space) a cosmological model for the physical universe; see de Sitter universe.

It appears to me that the statement means that our Universe described by the FRW metric is really a de Sitter Universe. I'm not sure that I correctly interpret the Wikipedia statement because it isn't obvious$^1$ to me how the FRW spacetime is related to the de Sitter spacetime. Moreover,the de Sitter spacetime arises as the maximally symmetric vacuum solution (i.e., for $T_{\mu\nu}=0$) of Einstein's field equations with a positive cosmological constant $\Lambda>0$ while for our Universe is $T_{\mu\nu}\neq 0$. Maybe this comment refers only to a very early universe dominated by $\Lambda$ that looked like de Sitter. I'm not quite sure what it means to convey.

$^1$ A de Sitter space is a submanifold of the Minkowski space described by the hyperboloid of one sheet $-x_0^2+\textbf{x}^2=\alpha^2$ where $\alpha$ is some nonzero constant having the dimension of length.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ Could you link to where Wikipedia says this? $\endgroup$
    – TimRias
    May 14, 2018 at 13:26
  • $\begingroup$ Your citation is from de Sitter space article, right? Look at sections below about flat, closed and open slicings and see FRW metric. It is obviously not our universe, however think what will happen in $\Lambda$-CDM in asymptotic future $\endgroup$
    – OON
    May 14, 2018 at 13:34
  • $\begingroup$ Why the downvote? @OON: I think you're confused. Our universe is currently quite well approximated by de Sitter space. $\endgroup$
    – user4552
    May 14, 2018 at 13:39
  • $\begingroup$ @BenCrowell I downvoted because this question shows minimal research for the level shown by SRS in other questions. Right now, currently you can say that it is well approximated by de Sitter space as $\Lambda$ amounts to $70%$. However that's still quite far from $100%$. And our current life and what we study in cosmology is largely determined by the earlier epochs when it was not de Sitter $\endgroup$
    – OON
    May 14, 2018 at 13:50
  • $\begingroup$ @BenCrowell I've added a footnote to make it clearer what the confusion is. $\endgroup$
    – SRS
    May 14, 2018 at 13:50

2 Answers 2

Reset to default
4
$\begingroup$

it isn't obvious to me how the FRW spacetime is related to the de Sitter spacetime.

De Sitter space is the FRW solution in which there is no baryonic or dark matter, only dark energy.

Maybe this comment refers only to a very early universe dominated by Λ that looked like de Sitter.

You have this the wrong way around. In a cosmology with nonzero $\Lambda$, i.e., dark energy, dark energy always dominates at late times. This is because the contribution of dark energy to the stress energy stays the same as expansion continues, whereas contributions from other matter fields fall off like some negative power of the scale factor $a$. The early universe was radiation-dominated, because radiation has an exponent of $-4$, which is the biggest.

Our universe is currently quite well approximated by de Sitter space.

Improve this answer
$\endgroup$
2
  • $\begingroup$ A de Sitter space is a submanifold of the Minkowski space described by the hyperboloid of one sheet $-x_0^2+\textbf{x}^2=\alpha^2$ where $\alpha$ is some nonzero constant having the dimension of length. How is FRW same as this? @BenCrowell $\endgroup$
    – SRS
    May 14, 2018 at 13:47
  • $\begingroup$ @SRS As I said in the comments to the question, take a look at the page you quote en.wikipedia.org/wiki/De_Sitter_space there is metric in different coordinates - flat, open and closed. If you don't recognize that you obtain FRW-type metric you really should take a few steps back in your learning $\endgroup$
    – OON
    May 14, 2018 at 14:01
2
$\begingroup$

If the only contribution to density is dark energy, solving the Friedmann equations obtains $a\propto\exp Ht$ with $H:=\sqrt{\Lambda/3}$. This scale factor corresponds to one of several coordinate representations of a de Sitter space.

Improve this answer
$\endgroup$

Your Answer

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

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