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can you express the Ricci-scalar in a FLRW-universe in terms of the Hubble function and the deceleration parameter?

Yes.

ReplyDeleteNo; in general, you need one more parameter to express R than just H and q ...

ReplyDeletethe relevant Christoffel-symbols for a flat FLRW-metric are $\Gamma^0_{ii}=a^2H$ and $\Gamma^i_{0i}=H$, from which the Ricci-scalar can be constructed to be $R=6H^2(1-q)$ with the deceleration parameter $q=-a\ddot{a}/\dot{a}^2$. a nonflat FLRW-model with curvature $k$ has an additional term $k\chi_H^2$.

ReplyDeleteCQW would like to add that it's really very nice that the dimensionless acceleration and dimensionless curvature are simply added in the Ricci-scalar. the case of a flat FLRW-model with only radiation would imply a vanishing Ricci-scalar?

Deleteas a matter of fact, yes: there's the relation $2(1+q)=3(1+w)$ between deceleration and equation of state for critical FLRW-models, implying that $q=1$ for $w=1/3$ and therefore $R=0$.

DeleteThe smallness of the Ricci scalar during radiation domination, $R \ll H^2$, has been used in cosmological models of non-minimally coupled scalar field dark energy: a coupling term $f(\varphi) R$ between the scalar field and the Ricci scalar is negligible during radiation domination. This changes when matter starts to dominate and might initiate a change in the evolution of the dark energy. Of course, people wanted to use this to construct models where the late-time accelerated expansion finds some 'natural' explanation.

ReplyDelete