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From Section 9.1, in General Relativity by Woodhouse:

For a normal star, the Schwarzschild radius is well inside the star itself. As it is not in the vacuum region of space-time, the Ricci tensor does not vanish at $r=2m$, and so the Schwarzschild solution is not valid there. Inread the metric is that of an 'interior' Schwarzschild solution, found by solving Einstein's equations for a static spherically symmetric metric, with the energy-momentum tensor of an appropriate form of matter on the right hand side. In such metrics, generally nothing exceptional happens at the Schwarzschild radius. But in the extreme case, all of the body lies within its Schwarzschild radius and the vacuum solution extends down to $r=2m$. In this case, we have a spherical black hole.

1) What is the connection to the Ricci tensor here and why does this mean that the Schwarzschild solution is not valid there?

2) What is an $\textbf{interior}$ Schwarzschild solution?

3) Why does nothing exceptional happen here?

4) Why do we care that all of the body needs to be inside $2m$? I would have thought that the Schwartzchild solution just gives a problem at $r=2m$ where you are dividing by zero effectively. See $$ds^2 = \bigg(1-\frac{2m}{r}\bigg)dt^2-\frac{dr^2}{1-2m/r}-r^2(d\theta^2+\sin^2 \theta d\phi^2) $$

This is probably more a mathematical problem as far as I can see the only problem with the Schwartzchild solution is there.

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    $\begingroup$ Dear OP, note the name is Schwarzschild (schwarz+schild = black+shield). $\endgroup$ – Pedro Tamaroff Dec 25 '18 at 13:39
  • $\begingroup$ The Schwarzschild metric is a vacuum-solution (i.e. there's no matter flowing), this means that Stress-energy tensor vanishes, and then the ricci tensor vanishes too. for the interior of a star (which is like a perfect-fluid) the stress-energy tensor is of the form: en.wikipedia.org/wiki/Perfect_fluid. $\endgroup$ – Zober Dec 25 '18 at 13:48
  • $\begingroup$ The coordinate field $\partial_t$ is timelike for $r>2m$. Inside the black hole, $r <2m$ and $\partial_t$ is spacelike, so that even though the geometries inside and outside the black hole can be studied simultaneously (the metric tensor is formally the same), you cannot use geometry to make physical conclusions. Unless, of course, you extend the whole thing using Kruskal-Szekeres coordinates, but this is more complicated. $\endgroup$ – Ivo Terek Dec 25 '18 at 13:59
  • $\begingroup$ @PedroTamaroff Thanks for the insight. My level of German is nonexistent $\endgroup$ – Permian Dec 25 '18 at 14:33
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    $\begingroup$ Crossposted from physics.stackexchange.com/q/450294/2451 $\endgroup$ – Qmechanic Dec 26 '18 at 16:43
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1) The Schwarzschild solution has $R=0$, so is an invalid approximation of the star's interior.

2) A slightly different result that accounts for the star's mass not being compressed to within $r=r_s<2Gm/c^2$, but instead spreading out to $r=r_g>r_s$. The result is $$ds^2=\frac{1}{4}\left(3\sqrt{1-\frac{r_s}{r_g}}-\sqrt{1-\frac{r^2r_s}{r_g^3}}\right)dt^2-\frac{dr^2}{1-r^2r_s/r_g^2}-r^2(d\theta^2+\sin^2\theta d\phi^2).$$You may wish to work out what happens at $r_g=r_s$.

3) Smearing the mass removes the physical singularity.

4) Matter outside $r=2m$ doesn't contribute to gravity at $r=2m$, so it's no longer an event horizon.

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  1. The Schwarszchild solution is a solution of the vacuum field equations $R_{ij}=0$. In other words it is a very good approximation of the field generated by a static spherical mass in vacuum, i.e. outside the spherical mass. However, it is possible to determine a solution describing the behavior inside the spherical mass.

  2. The metric inside a static spherical mass can be computed considering a non Schwarszchild metric having a spherical symmetry and using this as a solution of the field equations. For computing that you need the energy-momentum tensor. The computation is quite long. You should get \begin{equation} ds^2= -\left(\frac{3}{2}(1-K R^2)^{1/2}-\frac{1}{2}(1-K r^2)^{1/2} \right)c^2 dt^2+ \frac{1}{1-K r^2}dr^2+r^2 d \Omega^2 \end{equation} where $K=8 \pi G \rho/ 3c^2$, $\rho$ is the matter density. For $r=R$ this metric matches the vacuum Schwarzschild one.

  3. At the Schwarzschild radius nothing interesting happens because the singularity of the Schwarzschild metric is not a physical singularity. As a matter of fact, it can be easily removed by changing the coordinates. In other words,the singularity $r = 2m$ is a coordinate singularity and not an actual singularity in the geometry of the specetime.

  4. Only matter inside the star contributes to the geometry of the spacetime and that independently of the mass distribution inside the star.

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