# Why do we say Sun curves space and the Earth moves following the geodesic?

Why do we say sun curves space even if the Ricci Scalar for a Schwarzchild metric, the solution of Einsteins Equations for the Sun, is equal to zero.

The Ricci scalar for the constant time slice is also zero.

Can we define curvature using a single quantity? If yes then what is it?

• Can you explain the relevance of "The Ricci scalar for the constant time slice is also zero." to the rest of the part of your question?
– user87745
Commented Jun 22, 2017 at 21:11
• Also, a few suggestions for the title: "Why do we say the Sun curves the space?" or "Why do we say the Sun curves the space even if the Ricci scalar is zero?". The geodesic part of your current title is irrelevant to the content of your question.
– user87745
Commented Jun 22, 2017 at 21:13
• en.wikipedia.org/wiki/Riemann_curvature_tensor Commented Jun 22, 2017 at 21:14
• I am not 100% sure about all of the details since I come from a condensed-matter background, but I did once see Penrose talk in Leiden and I believe he said that all of the curvature of general relativity can be separated into two parts: (1) the structure of how the light cones at every point of the space have been tilted, and (2) the gravitational time dilation factor at any given point in space. So there is certainly a scalar field which comes out naturally, but it may leave out how light rays are bent... Commented Jun 22, 2017 at 21:18
• Yes, Ricci scalar for Schwarzschild metric is 0. But it does not mean that there is no intrinsic spacetime curvature. Complete information about curvature comes from the Riemann tensor, whose components are not all 0 in Schwarzschild coordinates. If by a single quantity, you mean a scalar, then there are a couple of such invariants. One of them, which uses the Riemann tensor, is the Kretschmann scalar given by $R^{\mu \nu \rho \sigma}R_{\mu \nu \rho \sigma}$ Commented Jun 22, 2017 at 21:26

According to the mathematical definition of the curvature, it represents the commutator of covariant derivative. More precisely, $$[\nabla_\mu,\nabla_\nu]A^\rho=R^\rho_{\kappa\mu\nu}A^\kappa$$
Where $R^\rho_{\kappa\mu\nu}$ is the Riemann curvature tensor. It is the single quantity that defines (or represents) the curvature of spacetime. For a spacetime to be completely flat at a point means all the components of $R^\rho_{\kappa\mu\nu}$ to identically vanish there. But there are some less strict criteria for differently defined flatness, e.g. if $R^\rho_{\kappa\rho\nu}\equiv R_{\kappa\nu}=0$ then it's called Ricci-flatness and if $R_{\mu\nu}g^{\mu\nu}\equiv R=0$ then it's called scalar-flatness. But such kinds of flatness don't imply that the spacetime is not curved. As long as the Riemann curvature tensor has non-zero components the spacetime is curved - which can be verified by making a vector parallel transport along a closed curve and seeing that it doesn't match with what it was originally.
• @A.Ok If $T_{\mu\nu}=\dfrac{\Lambda g_{\mu\nu}}{8\pi}$ then $R\neq 0$ implying $R_{\mu\nu}\neq 0$ - making the space not Ricci flat. For Ricci flat spacetimes, $R_{\mu\nu}$ is simply zero. Tho it follows from Einstein equations that by determining $R_{\mu\nu}$or even $R$ in the vacuum (i.e. when $T_{\mu\nu}=0$), you can calculate the $\Lambda$. The exact relations (that follow from taking trace of Einstein equations) are $R_{\mu\nu}=\dfrac{2\Lambda g_{\mu\nu}}{D-2}$ and $\Lambda=\dfrac{(D-2)R}{2D}$. $D$ is number of spacetime dimensions.
• Isn't $8\pi T_{\mu\nu}=\Lambda g_{\mu\nu}$ just what you get when you set $R_{\mu\nu}=R=0$ in Einstein's field equation? Commented Jun 23, 2017 at 0:05
• @A.Ok Yes, my bad. Actually it would be wrong to check it that way as we don't know a priori that it would be a valid solution. But yes, we can plug in $T_{\mu\nu}$ to be $\dfrac{\Lambda g_{\mu\nu}}{8\pi}$ and get that $R_{\mu\nu}=R=0$. In the previous comment, I was using the results of Einstein equations without a cosmological constant to argue that if $T_{\mu\nu}\neq 0$ then $R \neq 0$ so that's why that reasoning is wrong. Tho the latter part of my previous comment is correct as you might have checked.