Actually I have been inspired by the post Why does dark energy produce positive space-time curvature? to ask the following question. In difference to the just cited post I will take the pressure out of the game.

I am wondering about the effect of the cosmological constant $\Lambda$. I will consider 2 cases:

1) an universe with only dust (of matter density $\rho$), to which I can associate an energy density $\epsilon= \rho c^2$, but pressure $p=0$ (similar to the model of Tolman (1934)) and no cosmological constant. The curvature scalar of EFEs is given by the trace of the energy-momentum tensor $T = T^i_i$ (eq. (95,7) in Classical field theory of Landau/Lifshitz, German ed.)

$$R = - 8\pi T$$

assuming $G=c=1$. For this universe the trace of the energy-momentum tensor (based on the energy-momentum tensor of a homogeneous incompressible fluid) is

$$T_i^i = \epsilon- 3p = \epsilon$$

so therefore the curvature scalar has the following form:

$$R = -8\pi \epsilon$$

2) an universe with no matter, but with a positive cosmological constant:

$$R = 4\Lambda$$

Upon association of $\Lambda$ with a positive energy density $\epsilon_\Lambda$ I get an universe of opposite curvature to the one made of Tolman dust. How this is possible ? Under the impression of this observation would it not be more natural to change the sign of $\Lambda$ (change of definition of the cosmological constant) with the effect that an associated energy density $\epsilon_\Lambda$ would have the same effect on the curvature as the energy density of normal matter ? Under the latter assumption, however, a positive energy density $\epsilon_\Lambda$ would lead to a collapsing universe, whereas a negative energy density $\epsilon_\Lambda$ would lead to an accelerating expanding universe.

  • $\begingroup$ I was just about to ask the same in a simpler way. If energy and matter are equivalent, then why is the dark energy repulsive while the dark matter is attractive? $\endgroup$ – safesphere Aug 18 '18 at 15:28
  • $\begingroup$ @safesphere DE in the form of the cosmological constant is distributed homogeneously in spacetime and does not cluster; DM clumps together in the same way as baryonic matter and hence is attractive. Just because E = mc^2 does not mean energy and mass behave equivalently. $\endgroup$ – astronat Aug 19 '18 at 17:46
  • $\begingroup$ @astronat If DM were distributed evenly, it would not be repulsive, would it? Also, mass Is localized energy while the $c^2$ predefined number in the formula is only to convert light-seconds to meters or another unit. So any difference in behavior is the difference between the energy in the rest frame (a.k.a. "mass") and the energy moving with the speed of light. Thus your comment does not explain why DE is repulsive. I am sure that you as a PhD student in cosmology can do a lot better by posting a comprehensive and insightful answer. $\endgroup$ – safesphere Aug 19 '18 at 18:06

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