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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?

JanG
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  • "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). – Andrew Steane Dec 11 '21 at 09:36
  • 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. – JanG Dec 11 '21 at 11:20
  • The interpretation of focusing geodesics as effect of positive scalar curvature, as described in https://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? – JanG Dec 11 '21 at 16:48
  • 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 $R^{i}_{klm}$. – JanG Dec 12 '21 at 09:26
  • 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) – KP99 Dec 16 '21 at 07:27
  • @KP99, thanks for your comment. If I understand it right then in static spherically symmetric spacetime null geodesics are identical with radial falling or outgoing rays, they focusing or defocusing dependent on their orientation. Thus, how to interpret geometrically the sign change mentioned in the question? – JanG Dec 16 '21 at 11:49
  • Sorry for the late response. Sign of Ricci tensor is not a true signature for convergence or divergence of timelike Or null congruence, there are other geometrical conditions. You should read about Raychoudhuri's equations – KP99 Dec 18 '21 at 13:15
  • @KP99, no problem. Raychoudhuri's equation contains the term $\varepsilon +3p$. If I am not wrong, it is not the same as $\varepsilon -3p$, i.e. the trace of Ricci tenors. – JanG Dec 18 '21 at 15:17
  • The difference is Raychoudhuri's equation contains scalar $R_{ik} t^{i} t^{k}$, whereas I ask about scalar $R_{ik} g^{ik}$. – JanG Dec 18 '21 at 15:31
  • I was checking the EFE in NP formalism. If $l$ and $n$ represents two transverse null direction in orthonormal frame ($l,n,m,\bar{m}$) and say $\rho$ and $\rho'$ represents the associated divergence along these null directions, then $l^a\nabla_a\rho'=\cdots -R/12$ and $n^a\nabla_a\rho=\cdots -R/12$, i.e. Ricci scalar has some negative impact on , say, evolution of divergence of longitudinal vector along transverse direction. But this is not the complete picture, you have to solve the complete Spin coefficient equations to arrive at a complete conclusion – KP99 Dec 18 '21 at 18:29
  • Thanks for your comment. I look forward to your complete conclusion. – JanG Dec 18 '21 at 18:32
  • @KP99, regarding my question I have possibly found a solution after reading Sayan Karand, Soumitra SenGupta paper entitled "The Raychaudhuri equations: A brief review" https://arxiv.org/abs/gr-qc/0611123 . Because radial geodesics in interior solution are focusing ($\varepsilon+p>0$), the Raychaudhuri equation has to remain unchanged despite of the sign change in the trace part of Ricci tensor $\Theta$. This is possible only if geodesics reverse their direction, too ($\lambda \rightarrow -\lambda$). What do you think? – JanG Dec 22 '21 at 15:41
  • I just realized, isn't $\Theta$ just the divergence of velocity field? However, if we use ideal fluids to model interior solution then by definition , $\Theta=\sigma=\omega =0$ locally. But these inhomogeneities in null rays is still non-zero, for which we can use NP formalism as mentioned in my previous comment. See chapter 6 & 7 of "Exact solutions of Einstein's field equations" by Stephani, Kramer, MacCallum, Herlt – KP99 Dec 25 '21 at 19:24
  • @PP99, thanks for your ideas. I have already realized too, that $\Theta$ is not the trace of Ricci tensor. However, I think that the use of NP formalism for the physical or geometrical interpretation of the sign switch in $R$ is an 'overkill'. Figuratively speaking, the energy density warps space and the pressure time. It has to be something about which warping becomes dominating within the change of compactness parameter $\alpha$. – JanG Dec 26 '21 at 17:02

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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.

user321746
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  • @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. – JanG Dec 12 '21 at 18:50