How to interpret the sign change of scalar Ricci curvature in Schwarzschild interior solution?

Question

I ponder about interpretation of scalar curvature in Schwarzschild interior solution. It reads: \begin{equation}\label{scalarcurvature} -S=\varepsilon-3p \tag{1} \end{equation} where dimensionless scalar curvature is defined as $S\equiv R_{l}^{l}R^{2}$, energy density $\varepsilon\equiv {\varepsilon}~{\kappa}~c^2R^2$ and pressure $p\equiv {p}~{\kappa}~R^2$, with $\kappa$ Einstein's gravitational constant $8\pi G/c^4$ and $R$ curvature radius of a perfect fluid sphere (for more details see https://physics.stackexchange.com/a/679431/281096). The minus sign on the left side of equation \eqref{scalarcurvature} is due to metric signature definition (+,-,-,-). For interior Schwarzschild solution it applies \begin{equation} \label{energyandpressure} \varepsilon=3\alpha,~~~p=3\alpha~\frac{\sqrt{1-\alpha u}-\sqrt{1-\alpha}}{3\sqrt{1-\alpha}-\sqrt{1-\alpha u}}.~~~~~~~~~u\equiv r^{2}/R^{2}.\tag{2} \end{equation}

It means that scalar curvature can have both signs. Especially, at the origin ($u=0$), the equation \eqref{scalarcurvature} results in \begin{equation}\label{origincurvature} -S(0,\alpha)=6\alpha~\frac{3\sqrt{1-\alpha}-2}{3\sqrt{1-\alpha}-1}. \tag{3} \end{equation} Due to equation \eqref{origincurvature} the central scalar curvature ($-S_{0}$) is positive for $\alpha<5/9$, zero for $\alpha=5/9$, and negative for $\alpha>5/9$. The radial geodesics are converging in all cases. However, converging geodesics are normally interpreted as a feature of a positive scalar curvature. How is to understand the zero and the negative scalar curvature there?

Answer 1

The scalar curvature is equal to $\epsilon-3p$. For a small body, such as the earth, this expression is dominated by the mass density, so it's positive. For a black hole, the pressure near the center is going to blow up to positive infinity, which indicates that the static solution is no longer self-consistent. Since we arbitrarily (and not very realistically) pretend that $\epsilon$ is always constant (the fluid is perfectly incompressible), the scalar curvature near the center must be negative for large enough bodies.