How does one derive the Schwarzschild Metric as a solution of squashing matter together?

Question

So we have $2$ statements of a blackhole.

If we postulate a spherically symmetric Lorentzian metric then the solution is the Schwarzschild Metric. (Birkhoff's theorem)
If I squash enough matter together a blackhole is formed.

How can I show statement $1$ is a solution of statement $2$? Or what initial conditions give rise to singularity of the type seen in the Schwarzschild metric (without starting with one)?

It is not quite clear what you are asking. Are you perhaps looking for Oppenheimer-Snyder collapse, other simple metrics for stellar collapse, or Penrose-Hawking singularity theorems? — Void, Feb 05 '20 at 09:50
I was asking what initial conditions give rise to singularity of the type seen in the Schwarzschild metric (without starting with one)? — More Anonymous, Feb 05 '20 at 09:57
In any coordinates, the hirizon is lightlike. This means that a tangent vector to a radial timelike geodesic at the horizon is lightlike. This is prohibited in differential geometry where timelike geodesic cannot becomes lightlike under any circumstances. So all timelike geodesic end at the horizon. Extending them by hand is mathematically forbidden. With no matter crossing the horizon, there is no singularity, so your question cannot be answered. — safesphere, Feb 05 '20 at 16:36

Frederic Thomas · Accepted Answer · 2020-02-05T15:39:40.500

In the question a couple of misconceptions are mentioned.

First of all, Birkhoff's theorem says the following:

"The external gravitational field of a spherical-symmetric mass distribution is equivalent to a field of a pointlike assembly of the total mass in the center."

The Schwarzschild metric can be applied to the outer empty space around all types of non-rotating spherical mass distributions. So it applies not only to a non-rotating black hole, but also to the metric around the sun for example.

From this it is clear that from point 2) is not a prerequisite of point 1) respectively 1) can be concluded from 2). A black hole is not necessary for the establishment of a metric which equals the Schwarzschild metric.

Actually the primary "singularity" of the Schwarzschild metric is at $r=\frac{2GM}{c^2}\equiv r_{schwarzschild}$. In order to see it clearly, the metric is given here:

$$ds^2 = (1-\frac{2GM}{c^2 r})dt^2 - \frac{1}{1-\frac{2GM}{c^2 r}}dr^2 -r^2 d\theta^2 -r^2 sin^2(\theta) d\phi^2$$

At $r=\frac{2GM}{c^2}$ the metric component $g_{11}$ behaves $g_{11}\rightarrow \infty$.

So actually the Schwarzschild metric in its original form is only valid for $r>r_{schwarzschild}$. So a priori the Schwarzschild metric does tell us nothing what happens inside a black hole, respectively its singularity at $r=0$ (but it tells us the metric outside of a non-rotating black hole without charge). However, the "singularity" at $r=\frac{2GM}{c^2}$ is only a problem due to (unfortunate) choice of coordinates. By a change of coordinates it can be extended to another metric called Kruskal-metric which also extends up to $r=0$.

By the the way I would consider a (in cartesian, polar or spherical coordinates written) Lorentzian metric as flat (here in spherical coordinates) :

$$ds^2_{Lorentzian} = dt^2 -dr^2 -r^2(d\theta^2 + \sin^2(\theta)d\phi^2)$$

@safesphere When I speak of Schwarzschild metric in its original form, I mean the metric before it is written in Kruskal coordinates. 2) According to Sexl/Urbantke Gravitation & Cosmology the Kruskal coordinates cover the whole space-time apart from $r=0$ of course, but they seem to be piecelike defined and simplified speaking the borders of the pieces (apart from $r=0$) meet at the Schwarzschild radius. According to the source mentioned it does not seem to happen something mathematically very severe there or at least the border can be approached from both sides as close as possible. — Frederic Thomas, Feb 06 '20 at 12:29

How does one derive the Schwarzschild Metric as a solution of squashing matter together?

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