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My understanding is that dark energy, or equivalently a positive cosmological constant, is accelerating the expansion of the universe and I have read that this gives empty space-time positive curvature, ie de Sitter geometry. I also understand that parallel geodesics converge when curvature is positive and diverge when it is negative. I would expect accelerating expansion of space to make parallel space-time geodesics diverge and thus make curvature negative. Is there a nice visual explanation why dark energy actually produces positive curvature?

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  • $\begingroup$ WMAP found that space appears to be "flat" and that there is no evidence of any positive curvature. IMHO a correct understanding of GR means this should be no surprise. See where Einstein described a gravitational field as space that was neither homogeneous nor isotropic then look at the FLRW metric which "starts with the assumption of homogeneity and isotropy of space". $\endgroup$ – John Duffield Sep 7 '15 at 19:26
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    $\begingroup$ Christoph's answer seems fine, but I would home in on what seems to be the fundamental conceptual error that led to the OP's confusion. GR describes curvature of spacetime, not just curvature of space. If you slice along an xt plane, then initially parallel lines will later diverge, and this is due to the cosmological constant. Spatial curvature would be a slice along a plane such as xy. In this type of plane, the current evidence is that the geometry is nearly Euclidean. $\endgroup$ – Ben Crowell Apr 26 '18 at 21:29
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A positive cosmological constant leads to positive scalar curvature by definition. Just trace over the Einstein equation and you end up with $$ R = 4\Lambda - 8\pi T $$ which is just $$ R = 4\Lambda > 0 $$ in vacuum.

The implicit, but more interesting questions are probably the following ones:

Why can we interpret the cosmological constant as dark energy?

Modelling matter as a fluid in equilibrium, ie $$ T_{\mu\nu} = (\rho + p) u_\mu u_\nu + p g_{\mu\nu} $$ the Einstein equation reads $$ R_{\mu\nu} - \frac 12 R g_{\mu\nu} = 8\pi (\rho + p) u_\mu u_\nu + (8\pi p - \Lambda) g_{\mu\nu} $$ Now, if we want to fold the $\Lambda$ term into the matter terms, we require $$ \rho_\Lambda + p_\Lambda = 0 \\ 8\pi p_\Lambda = -\Lambda $$ which is $$ \rho_\Lambda = -p_\Lambda = \frac\Lambda{8\pi} $$ a positive energy density with negative pressure.

Take note that this pressure is not directly responsible for any acceleration or deceleration of the cosmological expansion: It is uniform across space and stays constant in time, and lacking a gradient, does not induce any forces. Its effect is purely gravitational in nature - after all, this is just the cosmological constant in disguise.

Does positive spacetime curvature actually lead to parallel geodesics converging?

Not necessarily due to the Lorentzian signature of the metric. Take 1+1 de Sitter space, which can be realized as a hyperboloid in Minkowski space and would look like this (picture taken from Wikimedia Commons):

hyperboloid

We get geodesics from the intersections of planes through the origin of the ambient Minkowski space with the hyperboloid, and time-like ones from those which are angled less than 45° towards the time axis.

The vertical lines thus correspond to time-like geodesics and clearly do not converge.

This is where slicing into space-like hypersurfaces comes in: In FLRW cosmology, there's a preferred slicing where the galactic fluid is homogeneous. In de Sitter space, there's no matter and thus no preferred slicing, but we can nevertheless use it to illustrate various features of the cosmological standard model.

The horizontal circles, which we obtain by intersecting a parallel family of planes in the ambient space with the hyperboloid, correspond to a spatially closed universe. Choosing appropriate coordinates yields the metric $$ ds^2 = -dt^2 + \alpha^2\cosh^2\left(\frac t\alpha\right)d\Omega^2 $$ where $d\Omega$ is the metric of the Euclidean sphere and $\alpha=\sqrt{3/\Lambda}$.

By tilting our planes, we can also create flat slicings with corresponding metric $$ ds^2 = -dt^2 + e^{2t/\alpha}dy^2 $$ and open slicings with metric $$ ds^2 = -dt^2 + \alpha^2\sinh^2\left(\frac t\alpha\right)dH^2 $$ where $dH$ is the metric of the Euclidean hyperbolic space.

While the light-like geodesics shown above - corresponding to particles at rest in case of the closed slicing - diverge, the spatial curvature will determine what happens to particles in parallel motion through space. However, this is not something that can be shown in our picture of a 1+1 spacetime.

How does this result in an accelerated expansion of the universe?

Looking at the spatial part of the metrics, all three slicings ultimately lead to an exponential expansion of space, which, in case of a de Sitter universe, is just a matter of geometry. In the closed case however, accelerated expansion happens only after a decelerating collapse to some minimal size determined by the value of the cosmological constant.

In Friedmann models, as long as the cosmological constant dominates over the matter content, we'll eventually approach de Sitter geometry and thus also exponential expansion.

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  • $\begingroup$ Thanks, @Christoph, this is looking great! The second question is the one that was puzzling me. If I read your answer correctly, then the constant time circles on the hyperboloid are not geodesics since they are not formed by planes passing through the origin. Correct? $\endgroup$ – Daniel Mahler Feb 8 '14 at 20:06
  • $\begingroup$ Also I am not famimilar with the equation $T_{\mu\nu} = (\rho + p) u_\mu u_\nu + p g_{\mu\nu}$. Could you define $\rho$, $p$ and $u$. $\endgroup$ – Daniel Mahler Feb 8 '14 at 20:08
  • $\begingroup$ Actually $\rho$ and $p$ are clear from the later equations. Is $u$ the fluid velocity? I guess this is a tensor form of Bernoulli's equation. $\endgroup$ – Daniel Mahler Feb 8 '14 at 20:29
  • $\begingroup$ @DanielMahler: your guess for $u$ is correct, and I added a link to the corresponding Wikipedia page; you're also right that the horizontal circles aren't geodesisc, but we can use them to model a closed universe that's contracting in the lower half and expanding in the upper one $\endgroup$ – Christoph Feb 8 '14 at 20:36
  • $\begingroup$ Great. Thanks. One more bit of confusion is the minus sign in $\rho_\Lambda = -p_\Lambda = \frac\Lambda{8\pi}$. This is saying that accelarating expansion is caused by negative pressure, which seems counterintuitive. $\endgroup$ – Daniel Mahler Feb 8 '14 at 21:37
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This is really a comment, but it got a bit long for the comment box. It's a comment because I kept meaning to go off and research this properly but have failed to find the time (and probably never will). So I'll post my initial thoughts, but treat this as suggestions for things to look at rather than a definitive answer.

When you say I also understand that parallel geodesics converge when curvature is positive I bet you have a mental image of a 2-sphere (apologies if I'm libelling you, but this is definitely my immediate mental image of positive curvature). The thing is that the 2-sphere is a Riemannian manifold i.e. the metric is positive definite. By contrast the manifolds we use in relativity are pseudo Riemannian i.e. the metric is not positive definite and indeed has the signature (-+++) or (+---) depending on your preferred convention. This matters because the scalar curvature is:

$$ R = g^{\alpha\beta} R_{\alpha\beta} $$

So the positive scalar curvature of de Sitter space doesn't mean it's like a sphere. It would if the metric is positive definite, but it isn't.

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  • $\begingroup$ Yes, my main misunderstanding was that I was not aware that the relationship between scalar curvature & geodesics did not apply to indefinite metrics. I had once seen an explanation of curvature in terms geodesics in a book and thought it was meant to be universal. $\endgroup$ – Daniel Mahler Feb 10 '14 at 3:01

protected by Qmechanic Aug 1 '15 at 13:38

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