If a spacetime is homogenous and isotropic can I say that $\nabla_\mu R =0$?

I was reading this paper https://arxiv.org/abs/astro-ph/0610483 and, I think that is the justification for the authors setting $\nabla_\mu R =0$ (just below Eq. (3)). (Am I correct here?)

I found some references (such as section 5.1 of Wald) that state a spacelike surface that is homogenous and isotropic will have constant curvature, but what bothers me is that the space like surface of curvature scalar K is not the same curvature scalar for a 3+1 universe with curvature scalar R.

But if this is not the case, I don't understand the author's justification for setting $\nabla_\mu R =0$ in the paper above.

  • $\begingroup$ Or would I be more accurate to say that a spacelike surface that is homogenous and isotropic will have $\nabla_\mu R = 0$? (Since we can't observe the entirety of the spacetime, only (approximate) spheres at various loopback times.) $\endgroup$ – Bob Jan 11 '18 at 20:37
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    $\begingroup$ A good example to keep in mind is that all curvature scalars vanish for a gravitational wave. $\endgroup$ – Ben Crowell Jan 12 '18 at 20:18

The condition $\nabla_\mu R=0$ simply means that scalar curvature is constant. It neither implies homogeneity and isotropy nor follows from it. For example Ricci-flat spacetimes or solutions with cosmologiacal constant (but without matter) would have this condition but they are not necessarily isotropic or homogenous (one example is Kerr-de Sitter solution).

On the other side, spacetimes with homogenous and isotropic $t=const$ slices can have $\nabla_\mu R\ne 0$. Simplest example is FLRW metric with dust-like matter. It has time dependent Ricci scalar $R=-8\pi G \rho(t)$.

As for the paper referenced in question, the condition $\nabla_\mu R=0$ together with $T=0$ was imposed to check that one particular spacetime, the de Sitter vacuum (which has maximal number of spacetime symmetries) is a solution of this model and find the expression for de Sitter Hubble parameter through the model parameter $\mu$.


Indeed, what you say is correct, homogeneous and isotropic spacetimes do not in general have a constant Ricci curvature in the time direction. However, there are exceptions, such as the maximally symmetric spacetimes: flat, de Sitter and anti de Sitter.

If I am not wrong, what they are trying to show in the paper is that the modified gravity theory they are working with admits de Sitter type solutions. So they are trying to show de Sitter is a solution, and de Sitter has constant $R$, so they impose it and see everything is consistent.

  • $\begingroup$ Thanks. Are you saying that (in general), knowing a spacetime to be homogenous and isotropic is not enough information to conclude that $\nabla_\mu R=0$? (I was saying the opposite, that homogenous and isotropic is enough to say $\nabla_\mu R=0$.) Also, I thought flat, dS, and AdS were the only possibilities if we assume homogenous and isotropic spacetimes. Are there other examples of homogenous and isotropic that are not flat, dS, or AdS? $\endgroup$ – Bob Jan 11 '18 at 22:38
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    $\begingroup$ The usual meaning of homogeneous and isotropic refers to the space-like symmetries (that it looks the same in every spatial point and in every spatial direction) but it does not say anything about time. The most general homogeneous and isotropic spacetimes are the FRW spacetimes. Most of them do NOT satisfy $\nabla_\mu R = 0$. However, dS, AdS and flat belong to the FRW family and do satisfy $\nabla_\mu R = 0$. For example, the Einstein Static Universe is a FRW spacetime that solves Einsteins equations (note that for a spacetime to be FRW we are not saying anything about field equations). $\endgroup$ – M. Sasieta Jan 11 '18 at 22:49

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