I have a simple question about the relation between the Ricci scalar curvature and the $k$ constant in the Friedmann–Lemaître–Robertson–Walker solution. Assuming $k=0$, such that the space can be considered flat, the metric has the form (in natural units): $$ ds^2=-dt^2+a^2(t)\left(dx^2+dy^2+dz^2\right). $$

After some algebra I found the Ricci tensor $$R_{00}=-\frac{3\ddot{a}}{a}$$ $$R_{ij}=a^2\delta_{ij}\left[2\frac{\dot{a}^2}{a^2}+\frac{\ddot{a}}{a} \right]$$ and the term $R_{0i}=0$. Then I evaluated the Ricci scalar, finding $$R=6\left[\frac{\ddot{a}}{a}+\frac{\dot{a^2}}{a^2}\right].$$ So my question is: How can I have a flat spacetime if the Ricci scalar curvature is non zero? Moreover, which is the relation between the $k$ and the $R$ scalar?

  • $\begingroup$ To answer your question we need to consider Einstein's equations as well. The spatial curvature index $k$ and spacetime curvature defined by Ricci tensor related with density and pressure of matter due to Einstein's equations. $\endgroup$ Jan 18, 2022 at 4:27

1 Answer 1


$k$ is not a measure of the curvature of spacetime, but rather of the spatial sections. $k = 0$ doesn't mean that spacetime is flat (your example isn't: since $a(t) \neq a_0$, it can't be Minkowski spacetime), but rather that the spatial sections are just the Euclidean $\mathbb{R}^3$ space rather than, for example, a $3$-sphere.

It is also possible to get it the other way around: for $R = 0$, one can get a FLRW solution with $k \neq 0$. The Milne model is such an example (it is just the future light-cone of the origin in Minkowski spacetime).

As one can see from explicit computation for the general $k$ FLRW metric (or by looking up on Wikipedia, as I'm doing right now), the full expression for $R$ including the contribution from $k$ is $$R = 6 \left(\frac{\ddot{a}}{a} + \frac{\dot{a}^2}{a^2} + \frac{k}{a^2}\right).$$

You can, though, compute the Ricci scalar for the spatial section. Using the metric $$\text{d}s^2 = - \text{d}t^2 + a(t)^2 \left(\frac{\text{d}r^2}{1 - k r^2} + r^2 \text{d}\Omega^2\right),$$ which has spatial metric $$\text{d}l^2 = \frac{a(t)^2 \text{d}r^2}{1 - k r^2} + a(t)^2 r^2 \text{d}\Omega^2,$$ I got (with the aid of the Mathematica package OGRE) the expression $${}^{(3)}R = \frac{6k}{a^2(t)}.$$

Hence, you can understand $k$ as being proportional to the Ricci scalar of the spatial section.

  • $\begingroup$ Oh thank you for your explanation. So the $k$ is only a measure of the space curvature, while the $R$ scalar is the curvature of the spacetime? How can this be physically understood? $\endgroup$
    – Stefano98
    Jan 17, 2022 at 23:06
  • $\begingroup$ @StefanoManzoni precisely. Some spacetimes can be written with different choices of $k$ (the Milne model is an example in Minkowski spacetime, and de Sitter spacetime admits all three values). Each value of $k$ leads to different notions of time and different observers. For example, inertial observers in Minkowski spacetime experience $k=0$, while observers following different geodesics through a piece of spacetime can experience $k=-1$ (Milne observers). $\endgroup$ Jan 18, 2022 at 2:13
  • $\begingroup$ In other words, $R$ is a property of the spacetime itself. $k$ is a property of how you are decomposing the spacetime into space + time. $\endgroup$ Jan 18, 2022 at 2:14

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