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I've heard that a cosmological constant can be used to model dark energy (e.g. $\Lambda$-CDM model), and that the constant $\Lambda$ should be positive. But my (quite small) understanding of dark energy is that it acts to expand things, which should correspond to $R_{00} < 0$ rather than $R_{00} > 0$, where here $R_{\mu\nu}$ is the Ricci tensor and $0$ is some fixed timelike direction with, say, $g_{00} = -1$.

But in Einstein's equations with a cosmological constant, $$ R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R + \Lambda g_{\mu\nu} = T_{\mu\nu}, $$ it seems to have the opposite effect, as follows. If $\Lambda > 0$, then putting it on the other side and letting $\mu = \nu = 0$ gives $R_{00} = \cdots + \Lambda$, so $\Lambda$ has the effect of increasing Ricci curvature, i.e. causing the focusing of timelike geodesics, rather than the supposed "spreading out" of things.

Edit 1: This question addresses the positivity of the scalar curvature, but I am interested in how it affects the Ricci curvature, in particular the value of $\text{Ric}(T,T)$ for a timelike vector $T$.

Any help is appreciated!

Qmechanic
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Chris
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    The energy density $\rho$ is positive, but the pressure $p=-\rho$ is negative, and since pressure acts $3\times$ stronger than energy density, the overall effect is $2\times$ negative, see here – Yukterez Mar 05 '24 at 04:03
  • Thanks for your comment. Can you explain what you mean by "pressure acts $3\times$ stronger than energy density"? I'm not sure how this explains the fact that $R_{00}$ is positive given $\Lambda > 0$, leading to spreading-out of geodesics. – Chris Mar 05 '24 at 04:13
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    see here at $\rm \ddot{a}/a = - 4\pi G\left(\rho/3 + p\right)$, if $\rm p=-\rho$ it is $2\times$ the negative of pressureless matter. – Yukterez Mar 05 '24 at 05:17

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