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.
