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According to Schwarzschild metric:

$$ c^2~\mathrm d\tau^2 = \left(1-\frac{r_s}{r}\right)c^2 ~\mathrm dt^2 - \left(1-\frac{r_s}{r}\right)^{-1}~\mathrm dr^2 - r^2\left(\mathrm d\theta^2 + \sin^2\theta~\mathrm d\varphi^2 \right) $$

,where $r_s$ is Schwarzschild radius: $ r_s = \frac{2GM}{c^2} $

the density element $$ \rho(r) = \frac{\mathrm dm}{\mathrm dV} $$ becomes infinite on the event horizon if there is mass $$ \mathrm dm(r) > 0, r \rightarrow r_s, r>r_s $$very near at the outside of event horizon.

The density is then:

$$ \delta(r)_\infty = \frac{\mathrm dm_{r,\infty}}{\mathrm dV_{r,\infty}} = \frac{\mathrm dm_{r,\infty}}{(1-\frac{r_s}{r})^{1/2}(r~\mathrm dr~\mathrm d\theta ~\sin\theta~\mathrm d\varphi)} = \frac{\mathrm dm_{r,\infty}}{~\mathrm dV_{\infty,\infty} }\left(\sqrt{\frac{r}{r-r_s}}\right) $$

and at the limit $r \rightarrow r_s$: $$ \delta(r)_{\infty}\to \infty , \text{when}~ (r \rightarrow r_s)~\text{and}~ (\mathrm dm_{r,\infty} > 0 ) $$

Note that this equation does not yet describe the density distribution. The mass element $$ \mathrm dm_{r,\infty} $$ is the mass inside volume element $$\mathrm dV_{r,\infty} $$. In this equation, this can be anywhere between 0 and

$$ \text{max}(\mathrm dm_{r,\infty}) = \frac{M}{A_{BH}}= \frac{M}{4 \pi r_s^2} = \frac{c^4}{16\pi G^2M} $$

,where the maximum is in the situation when all of the mass of BH is concentrated into very thin layer above the event horizon. minimum 0 is the situation that all of the mass is inside the event horizon and none outside of it.

The Schwarzschild metric describes only the metric of spacetime outside of the Black Hole event horizon, not inside.

Birkhoff's theorem states that spacetime in the outerior of non-rotating thin spherical shell has Schwarzchild metric. And the interior of thin spherical shell has flat Minkowsky metric

$$ds^2 = c^2dt^2 - (dx^2 + dy^2 + dz^2)$$

And according to this theorem the interior metric of spherical mass distribution depends only on the spherical mass distribution that is inside the radius r. For example Spherical mass distribution with constant density has following metric:

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

I don't know at the moment what is the general solution for interior metric of spherical mass distribution, but i have a guess that it depends largely on the equation of state of the matter, that determines what kind of density and pressure distribution the matter will eventually have when it is in dynamical and thermical equilibrium.

My question is, what do we know about the mass distribution of density distribution of the black hole if it is observed by distant observer? In other words, what are $$ \delta(r) , m(r) $$ of the ideal Schwarzschild Black Hole, if it is observed by observer that is infinitely far away from BH?

I see here two different kind of possiblities for BH mass distribution:

  1. All or some of the mass is concentrated into very thin layer above the event horizon, when the density distribution is (more or less): $$ \rho (r) \rightarrow \delta (r_s) = \left \{ +\infty ,when r = r_s,= 0, when r > r_s \right \} $$
  2. All of the mass is inside the event horizon,when the density distribution is: $$ \rho (r) = 0 , when r \geqslant r_s $$ One possibility could be that all of the BH mass is concentrated into a very thin layer above the event horizon, and that there is no mass inside the event horizon. The reason why I think that it is possible is that the time dilation of BH becomes near infinity just above the horizon relative to the clock that is infinitely far away from the black hole and the matter stops falling down for this reason, if it is observed by observer that is far away from the black hole. Also the falling objects have nearly infinite radial length contraction relative to the metric infinitely far away from BH.

$$ \left(\frac{\tau_r }{\tau_\infty }\right) = \sqrt{1-\frac{r_s}{r}} \to 0, ~\text{when}~ (r \to r_s ) $$

$$ \left ( \frac{dr_{r}}{dr_{\infty}}\right ) = \sqrt{1 - \frac{r_s}{r}}\rightarrow 0, when (r\rightarrow r_s) $$

I am not sure what happens to the matter of BH in the birth of the black hole. Does the matter enter inside the event horizon or does it stay outside of the event horizon of the newborn black hole? IF the matter stays outside event horizon at the birth of BH, then it may be possible that all of the mass of BH is outside event horizon and the black hole is hollow inside in this sense.

On the other hand I am not sure about this, but the quantum tunneling effect and Heisenberg uncertainty principle may make possible that some of the particles can enter inside the event horizon.

So the question is, how the mass is distributed in ideal Schwarzchild Black hole in the viewpoint of distant observer?

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    $\begingroup$ How can the mass be in a shell above the event horizon? What would hold it there? What's the difference between inside and outside? The free falling observer can't tell such a difference. He will always be in an inertial system (with ever increasing tidal forces). $\endgroup$ – CuriousOne Jul 20 '16 at 7:36
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    $\begingroup$ @CuriousOne "what would hold it there?" - this is answered in the third-to-last paragraph of the question. From the perspective of a distant observer, time dilation prevents matter from ever crossing the horizon, so there is a sense in which the matter is concentrated in a thin shell above the horizon. Of course this is not the case for an inertial observer falling into the hole. I imagine it's not very meaningful to define a function $\delta(r)$, but a good answer could explain why. $\endgroup$ – Nathaniel Jul 20 '16 at 7:54
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    $\begingroup$ It seems that OP is conflating an ideal Schwarzschild black hole with, say, a black hole with infalling matter. $\endgroup$ – Qmechanic Jul 20 '16 at 10:20
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    $\begingroup$ Three comments: (1) The Schwarzchild metric is in fact a solution for both the exterior and interior of a black hole (at least in standard vacuum GR.) (2) The radial coordinate $r$ is not terribly well-behaved as $r \to r_s$, and I suspect that this ill-behaviour may be the source of your infinities. If you really want to see what's happening at the horizon, a better choice would be Kruskal-Szekeres coordinates. ... $\endgroup$ – Michael Seifert Jul 20 '16 at 14:48
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    $\begingroup$ (3) If you don't have a shell of infalling matter that you're treating as a "test mass", then the only meaningful way you can talk about the mass density in the spacetime is via the stress-energy tensor; but Schwarzchild is a vacuum solution and so $T_{\mu \nu} = 0$ everywhere. $\endgroup$ – Michael Seifert Jul 20 '16 at 14:48

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