Where does the interior Schwarzschild metric come from? How is it derived and why does it have NOT a singularity? Would it mean that the singularity is only apparent and for those out of the black hole (who are ruled by the usual, exterior metric)?
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$\begingroup$ There is a somewhat analogous situation in just newtons law of gravitation, the force no longer behaves as an inverse square inside an extended massive body. Usually you have an issue because point particles would experience forces that tend to arbitrarily large values. But inside an extended constant density body for example it linearly goes to zero at the center. $\endgroup$– TriatticusCommented Jul 6, 2023 at 0:35
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$\begingroup$ Schwarzschild interior solution is just an interior part of the full solution of Einstein field equations (EFE) for the case of perfect fluid sphere with constant energy density. You can calculate it using differential equation in physics.stackexchange.com/a/679431 with zero pressure boundary condition ($p_{1}=0$) there. Historically, Schwarzschild solved EFE first for the part of spacetime without matter (exterior) and later for the part with matter (interior). $\endgroup$– JanGCommented Jul 6, 2023 at 18:43
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$\begingroup$ It has "singularity" at center if $r_{S}/R$ is equal $8/9$. The energy density is finite there whereas the pressure diverges as $p\sim 1/r^2$. $\endgroup$– JanGCommented Jul 6, 2023 at 18:48
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2 Answers
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Why would it have a singularity if it only applies to homogenous spheres at least 9/8 larger than a black hole, the earth also doesn't have a singularity at its center. The original work with the derivation can be found here (it is in german though, but the equations are self explaining and for the text you can use a translator).
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In this context, "interior" does not mean inside the event horizon of a black hole. Rather, one imagines computing the metric for a spacetime which features the presence of a star, modeled as a spherically-symmetric fluid. The exterior solution corresponds to the (vacuum) region outside of the star, and the interior solution corresponds to the (non-vacuum) region within the star itself.
If the entire mass of the star is compressed to a radius less than the Schwarzschild radius $R_s = 2GM/c^2$, then the star undergoes gravitational collapse and a singularity forms. However, this radius is far smaller than the radius of a typical star (the Schwarzschild radius of the sun is $R_s \sim 10$ km while its actual radius is $R_{\odot }\sim 10^5$ km), so no collapse occurs and no singularity exists.
As a non-relativistic analogy, note that if you compute the gravitational field strength of a spherical mass distribution with uniform density, total mass $M$, and radius $R$ in a Newtonian framework, you obtain
$$\mathbf g(r) = \begin{cases} \frac{GM}{R^2} \frac{r}{R} \hat r & r< R \\ \frac{GM}{r^2}\hat r & r\geq R\end{cases}$$
Outside of the star, the gravitational field strength goes like $\sim 1/r^2$ and therefore increases with decreasing $r$. However, once you get inside the star, the gravitational field strength decreases smoothly back to $0$ at the origin and there is no singularity.
