# Is the apparent lack of (Ricci) curvature in the Schwarzschild metric due to a choice of coordinates?

I've been lightly studying GR lately. Something that has been bothering me has been the lack of (Ricci) curvature produced from the Schwarzschild metric in the few lectures I've watched, as well as the few snippets of text book I've been able to read. Why is there no (Ricci) curvature outside this spherically symmetric, non-rotating, uncharged body that still has mass? Shouldn't there always be curvature in the presence of mass or am I missing something? I've read a bit about certain information that is unobtainable when dealing with Schwarzschild coordinates, is the curvature outside the body one of the specific quantities that cannot be defined with these coordinates?

• There are many ways to measure curvature (and more than one type of curvature). The Ricci scalar is one of them, but the Riemann tensor is the one we use to say whether a spacetime is curved or not, and it is nonzero for the Schwarzschild metric, therefore it is curved. Jul 12, 2014 at 13:53
• "Shouldn't there always be curvature in the presence of mass... ?" -- There is no curvature inside a spherically symmetric shell: spacetime is flat there. So it's not true that the presence of mass necessarily guarantees that spacetime is everywhere non-flat. Jul 12, 2014 at 14:10
• In response to the title, please note that the Ricci tensor is a tensor and hence independent of choice of coordinates. Regarding the body of the question, note that in vacuum only the Ricci tensor and scalar need vanish, not the Riemann parts. Dec 20, 2018 at 3:49

I suspect you're getting confused by the fact that the Ricci tensor $R_{\mu\nu} = 0$ and therefore the scalar curvature $g^{\mu\nu}R_{\mu\nu} = 0$. This is always the case in regions of space where the stress-energy tensor is zero. The curvature is certainly not zero in the sense that spacetime is flat. For example the Kretschmann scalar is non-zero:
$$R_{abcd} R^{abcd} = \frac{12 r_s^2}{r^6}$$