0
$\begingroup$

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?

$\endgroup$
14
  • $\begingroup$ "radial geodesics are converging in all cases" Are you sure? (the issue is not whether they converge linearly, but whether they turn towards or away from one another as they go). $\endgroup$ Dec 11, 2021 at 9:36
  • $\begingroup$ I am not sure, I have assumed that. I would say, they all turned towards from one another as they go. I suppose, it is the only way for them under spherical symmetry condition. $\endgroup$ Dec 11, 2021 at 11:20
  • $\begingroup$ The interpretation of focusing geodesics as effect of positive scalar curvature, as described in en.wikipedia.org/wiki/Scalar_curvature , is possibly valid only in Riemannian geometry. But in GR we deal with a pseudo-Riemannian manifold. Is that maybe the difference? $\endgroup$ Dec 11, 2021 at 16:48
  • $\begingroup$ My second to the last sentence about converging geodesics due to positive scalar curvature is wrong. The focusing or defocusing of geodesics is determined by the Riemann tensor $R^{i}_{klm}$. $\endgroup$ Dec 12, 2021 at 9:26
  • $\begingroup$ Indeed. $R_{abcd}$ appears in the jacobi equation, but it's effect is pretty broad. It's the Ricci scalar functions $\Phi_{ab}$ with $a,b =0,1,2$ which are related to convergence of null geodesics (in particular, $\Phi_{00}>0$ implies positive focusing) while Weyl scalars $\Psi_a$ contributes to astigmatic focusing. These Weyl scalars introduces shear in null rays (for details, look for Newman Penrose formalism) $\endgroup$
    – KP99
    Dec 16, 2021 at 7:27

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ @Christian, you confirm my finding but do not really answer the question on the meaning of the sign change. I think all the time about situation before blow up of pressure. The condition of energy density constancy is not a problem. The same situation arises in every static spherically symmetric perfect fluid solution. $\endgroup$ Dec 12, 2021 at 18:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.