EDIT: I stated the current understanding wrong, thanks Koschi for the comment.

In the $\Lambda$CDM model, it's said that the vacuum energy pretty much "balances" the baryonic matter, such that the overall curvature of space is flat.

But if you consider the spatial Ricci curvature, both dark energy and baryonic matter have positive curvature! To see this, just plug the dark energy equation of state, $p = -\rho$, into the Einstein equation, and choose an orthonormal frame:

$R_{\mu\nu} = T_{\mu\nu} - \frac12 T^{\alpha}_{\alpha}g_{\mu\nu} \\ = -\rho\eta_{\mu\nu} - \frac12 (-\rho - 3\rho)\eta_{\mu\nu} \\ = \rho\eta_{\mu\nu}$

with $\eta_{\mu\nu}$ the (-+++) Minkowski metric. So the spatial curvature is positive.

The difference is that dark energy has negative curvature along timelike directions (hence accelerating expansion), whereas ordinary matter has positive timelike curvature (hence gravity). But if we're talking about spatial curvature, it's all positive, so how can they cancel each other out? Wouldn't space have net positive curvature, such that, if it doesn't abruptly end somewhere, it has to be closed into something like a 3-sphere?

The difference in timelike curvature should just determine whether we keep expanding or undergo the big crunch, so that part I get.

Or does the distinction between spacelike and timelike curvature somehow break down on cosmic scales, kind of like space and time get "mixed up" in a black hole? If so, how?

  • $\begingroup$ I don't think, that the dominating component alone decides about spatial closure or not... is it not in general important, if the total energy density is above, the same as, or lower than the critical density? Today, the $\Lambda$CDM-model assumes dark energy domination, but still a, more or less, flat, not open, universe. $\endgroup$
    – Koschi
    Jul 23, 2021 at 18:03
  • $\begingroup$ @Koschi Thanks for that clarification, I will update the question -- I should have said flat instead of open (flat space could either be open or closed, because that's a question of topology as opposed to curvature). But I still don't get it -- how could space be flat on average, if it has positive curvature everywhere? $\endgroup$ Jul 23, 2021 at 19:51
  • $\begingroup$ I've tended to assume (perhaps mistakenly, so I think your question may be good) that, making the usual assumptions of homogeneity & isotropy (either or both of which may be wrong), the volume of observable space currently appears to be more flat than curved, sort of like something more black than gray would be called black. If you'd be judging the mass rather than the volume, the answer would be different, but, to judge by such long-observed changes as the fragmentation of our star's original 4th planet, as well as ordinary language, spacetime (not space alone) would be under consideration. $\endgroup$
    – Edouard
    Jul 24, 2021 at 16:22
  • $\begingroup$ Thinking it over, I feel the question's good: The disintegration of the 4th planet, if it would've occurred earlier than the estimated time, would've increased purely spatial curvature, as more of the resulting particles (asteroids) would've originally been molten, & consequently rounded (thereby increasing the curvature of adjacent space); whereas, if it would've occurred later, more of them would've been solid, & consequently jagged (thereby tending to leave more of the area of the adjacent spatial surfaces flat). $\endgroup$
    – Edouard
    Jul 24, 2021 at 17:01
  • 1
    $\begingroup$ @Edouard Right on, I'll check it out, thanks for the feedback $\endgroup$ Jul 26, 2021 at 3:53

1 Answer 1


I think it is worth keeping in mind that the intrinsic spatial curvature is not the same thing as the spatial components of the spacetime Ricci tensor. Instead, the intrinsic Ricci tensor $\mathcal{R}_{ij}$ and the spacetime Ricci tensor $R_{\mu\nu}$ are related by a somewhat complicated equation involving the extrinsic curvature $K_{ij}$ of the spatial surface and the acceleration $a_i$ of the unit normal: $$ \mathcal{R}_{ij} = e^\mu_i e^\nu_jR_{\mu\nu} + 2K_i{}^k K_{kj} - K K_{ij} - \mathcal{L}_uK_{ij} + D_i a_j + a_i a_j $$ where $e^\mu_i$ is a projector onto the tangent directions of the spatial hypersurfaces, and $\mathcal{L}_u$ is the Lie derivative with respect to the unit normal $u^\alpha$. In an FRW spacetime, isotropy requires the acceleration $a_i$ to vanish, but the extrinsic curvature $K_{ij}$ is nonvanishing; rather, it is proportional to the induced spatial metric (again by isotropy). Roughly, the extrinsic curvature is the time derivative of the spatial metric, and hence is nonzero due to the time-dependence in the scale factor. For spatially flat FRW spacetimes, the intrinsic geometry really is flat, so $\mathcal{R}_{ij}=0$. However, this doesn't require the spatial projections of $R_{\mu\nu}$ to vanish; instead, they just need to cancel against the extrinsic curvature terms in the above equation. So there is no inconsistency with having the spatial components of $R_{\mu\nu}$ be positive while simultaneously having the intrinsic geometry be flat.

  • $\begingroup$ Thank you so much, this is undoubtedly what I was missing! Now, is there an intuitive geometric interpretation of the meaning of the intrinsic Ricci tensor? At first glance I'm not finding anything helpful in my searches, and as you say, it would be a bit of a nightmare to decode those 5 terms. And why is that the one by which we measure the spatial curvature of the universe? $\endgroup$ Jul 26, 2021 at 21:24
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    $\begingroup$ In FRW spacetimes, we assume the spatial part of the metric is a scale factor times a maximally symmetric space (this follows from the assumptions of homogeneity and isotropy). In this case, the spatial Ricci tensor has all the information about the spatial curvature (in fact, the Ricci scalar has all the information, since $\mathcal{R}_{ij}$ is proportional to the spatial metric). So it is just one of several different quantities that detects whether the spatial geometry is flat, positively curved, or negatively curved. $\endgroup$
    – asperanz
    Jul 27, 2021 at 18:14
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    $\begingroup$ Also in FRW, the equation above simplifies a lot, since the extrinsic curvature is proportional to the spatial metric, $K_{ij} = H h_{ij}$, where $H$ is the Hubble rate. So you would end up with an equation involving $H$ and $\dot{H}$, and using the Einstein equation, the pressure and energy density. This would just be one of the usual differential equations for the Hubble rate that you obtain when finding cosmological FRW solutions. $\endgroup$
    – asperanz
    Jul 27, 2021 at 18:18
  • $\begingroup$ Awesome, thanks for the clarification $\endgroup$ Jul 27, 2021 at 22:54

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