Take the 2-minute tour ×

For an interval of this spacetime,

$$ ds^{2} = c^{2}dt^{2} - c^{2}t^{2}(d \psi^{2} + sh^{2}(\psi )(d \theta^{2} + sin^{2}(\theta )d \varphi^{2})), $$

scalar curvature is equal to zero. Also, Ricci tensor is equal to zero. How can it be for curved space-time (the metric tensor isn't represented as diag (1, -1, -1, -1))?


For an explanation odf reason of question creation

share|improve this question
Possible duplicate - physics.stackexchange.com/questions/60437/… –  Kitchi Apr 25 '13 at 14:15
Where did you see duplicate in this question? –  PhysiXxx Apr 25 '13 at 14:27
Hmmm - I thought it was a duplicate too at first, but on closer reading I suppose the questions are logical inverses of one another in some sense. That linked question gives the result "nonzero $R$ implies curved," but here the question is, "Does vanishing $R$ imply flat?" –  Chris White Apr 25 '13 at 15:52
add comment

3 Answers

up vote 4 down vote accepted

You are incorrect to suppose that this spacetime is curved. In fact, up to some conditions on the coordinate ranges, this is simply a piece of Minkowski spacetime. Let me put it in this form: $$ds^2 = dt^2 - t^2(d\psi^2 + \sinh^2\psi\,d\Omega^2)\text{,}$$ where $d\Omega^2 = d\theta^2 + \sin^2\theta\,d\phi^2$ is the metric for a unit $2$-sphere, and we can go from spherical Minkowski form $dT^2 - dR^2 - R^2\,d\Omega^2$ to yours by the substitution $$\begin{eqnarray*}R = t\sinh\psi &\quad\quad& T = t\cosh\psi\end{eqnarray*}$$ Simply, the Ricci tensor is zero because $R_{\mu\nu} = 0$ is a coordinate-invariant condition: if a tensor zero in some coordinate chart (e.g., Minkowski), then it is zero in every coordinate chart (e.g., yours). Naturally, the Riemann tensor is also zero.

Interestingly, this is the also the hyperbolic FLRW ("Big Bang") metric in the limit of zero density, where a galaxies don't interact gravitationally but every galaxy sees itself a center of cosmological expansion. For details, see the Milne universe model.

...the article at the link below says that this interval corresponds to an open flat (Space curvature = -1/R^2 < 0) expanding Universe (look at the picture in the question). Can you explain that?

Sure: spacetime curvature is not the same thing as space curvature, or more generally the curvature of a manifold is not the same thing as the curvature of a submanifold of lower dimensionality. Take Euclidean space $E^3$ in spherical coordinates, $dS^2 = dR^2 + R^2d\Omega^2$, and look at submanifolds of constant radius: $$d\sigma^2 = R^2(d\theta^2 + \sin^2\theta\,d\phi^2)\text{,}\quad R = \text{constant}$$ One can verify by direct computation that the scalar curvature is positive, as one might expect from them being spheres. Obviously, if you 'slice up' $E^3$ by planes, each will have zero curvature instead.

In your case, what's going on is just the same thing, except we're 'slicing up'/'foliating' the Minkowski spacetime $E^{1,3}$ by spacelike hypersurfaces of the form $$d\sigma^2 = t^2\left[d\psi^2 + \sinh^2\psi\,d\Omega^2\right]\text{,}\quad t = \text{constant}$$ In this form, it's not hard to guess that the curvature for each hypersurface is same as above except negative. (Let me know if you need help calculating it explicitly, but for the moment I think intuitively motivating the fact that this makes sense would be more fruitful.)

Since coordinate changes do not change the scalar curvature, we can let $\rho = t\sinh\psi$ to get $d\rho = t\cosh\psi\,d\psi$ and $$t^2\,d\psi^2 = \frac{d\rho^2}{\cosh^2\psi} = \frac{d\rho^2}{1 + \sinh^2\psi} = \frac{d\rho^2}{1+\rho^2/t^2}\text{,}$$ putting it in the same form as (80) except for variable names; the other part of the metric should be more obvious.

share|improve this answer
Thank you! I understood it yesterday, but your response has made the utmost clarity of understanding! But there is a strange thing: the article at the link below says that this interval (before some transformations) corresponds to an open flat (Space curvature = -1/R^2 < 0) expanding Universe (look at the picture in the question). Can you explain that? arxiv.org/pdf/gr-qc/9905046v1.pdf . –  PhysiXxx Apr 26 '13 at 4:00
Also, I added a picture with this words into a body of the question. –  PhysiXxx Apr 26 '13 at 4:33
Sorry, I misread your follow-up question at first. I'll fix it in a bit. –  Stan Liou Apr 26 '13 at 5:07
add comment

Firs of all, the metric depends on the specific coordinate system. If metric is not diagonal, it does not mean that the space is curved -- just take some crazy coordinates in the flat space and you will have a really messy metric. Futhermore, only vanishing Riemann tensor ensures that there are local coordinates in which metric is diag(1,-1,-1,-1), that is, the space is locally flat.

(Pseudo-)Riemann manifolds with vanishing Ricci tensor are called Ricci-flat, and they are the solutions of Einstein equations in empty space-time without cosmological constant. For example, Schwarzschild solution is actually Ricci-flat (except the singularity at the origin).

share|improve this answer
add comment

The point is that saying "spacetime is curved" does not necessarily mean anything specific about the metric or any of the curvature tensors. I think for physics (meaning, for general relativity), "spacetime is curved" should mean that the scalar curvature is not zero, since that's what shows up in the action. Of course, what matters is the variation of the scalar curvature with the metric (which gives the Einstein equations), so even if the scalar curvature is locally zero, it does not need to be zero everywhere. The metric can look complicated if you choose some complicated coordinates, so just looking at the metric is not enough to determine if "spacetime is curved."

For my money, flat spacetimes have $R=0$ everywhere, and Ricci-flat spacetimes have zero Ricci tensor everywhere (and no matter). These are the most common and most precise ways of considering a spacetime to be flat (or curved).

share|improve this answer
I call a spacetime manifold locally flat if it admits an atlas such that the metric components are "flat" (that is, unit-diagonal with appropriate signature) in every coodinate chart of the atlas. I believe that it is the same as saying that the Riemann tensor vanishes. Also, for GR to sensible, it should be coupled to something, and then the metric is the thing that shows up in the action in the first place. –  Peter Kravchuk Apr 26 '13 at 14:10
"For GR to (be) sensible, it should be coupled to something"? Not sure I understand, or if I do understand I don't agree. Vacuum spacetimes are some of the most important models for physics we have, right? –  levitopher Apr 27 '13 at 3:30
I mean, who is doing physics, if there is nothing coupled to gravity?) The main point was that the metric is measurable. –  Peter Kravchuk Apr 27 '13 at 14:43
add comment

Your Answer


