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To obtain the Schwarzschild metric from Einstein equations of general relativity, we suppose that the energy density is a distribution :

$$ \rho (\vec{r}) = M \delta(\vec{r})$$

The Schwarzschild radius, corresponding to the horizon, is $R_s = 2 M$, in units $G=c=1$.

A physical interpretation of this, is that you cannot put more energy than $M$ in a sphere of radius $R_s$.

The way I understand this, is because the total energy of the black hole, sum of its mass energy (positive) and its auto-gravitational energy (negative), should remain non-negative.

However, for the Schwarzschild black hole, for each radius $r$ between $0$ and $R_s$, the total mass in the sphere of radius r, is : $$ m(r) = \int_0^r \rho(\vec{u}) d^3u =\int_0^r \delta(\vec{u}) d^3u = M$$

So it seems to violate the principles cited above, and so the Schwarzschild black hole should be unphysical.

If it makes sense, a physical black hole (static, with spherical symmetry) should have a mass/energy density $\rho (r)$, such that, whatever the value of $r$ is, we have the inequality (in units $G=c=1$):

$$m(r) = \int_0^r \rho (u) 4\pi u^2du \le \frac{r}{2}$$

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    $\begingroup$ Nope, your distribution is incorrect, and therefore everything else is incorrect, too. The singularity inside the Schwarzschild is actually spacelike, not timelike: it is a moment of time, not a point in space. Moreover, the curvature around this singularity or any other BH singularity is so big that you can't just write a trivial delta-function for the density: it strongly depends on the coordinates you choose around the singularity and there are no "canonical coordinates" near a singularity that are better than all other coordinates. $\endgroup$
    – Luboš Motl
    Oct 26, 2012 at 10:21
  • $\begingroup$ OK, but what happens if we approach this singularity model by a serie of models, which each has a standard energy density, so that the limit of these models is the model with the singularity ? $\endgroup$
    – Trimok
    Oct 26, 2012 at 10:45
  • $\begingroup$ More precisely, We could take a model with a collapsing spherical cell, where the horizon is progressively created. $\endgroup$
    – Trimok
    Oct 26, 2012 at 11:04
  • $\begingroup$ The way I understand this, is because the total energy of the black hole, sum of its mass energy (positive) and its auto-gravitational energy (negative), should remain non-negative. The distinction between these two types of energy isn't a valid one in GR. There are ways of defining the total mass-energy of an isolated object in GR (e.g., Komar mass), but such a definition does not break down into the kind of sum you have in mind. The idea of summing up two terms like this is only valid in Newtonian gravity. $\endgroup$
    – user4552
    Aug 17, 2013 at 19:30
  • $\begingroup$ @LubošMotl: The singularity inside the Schwarzschild is actually spacelike, not timelike But this is a case where two wrongs make a statement that's technically right, since inside the event horizon, the Schwarzschild $r$ coordinate is timelike rather than spacelike. So inside the horizon, a surface of constant $r$ is spacelike. The rest of your comment hits the nail on the head, however. $\endgroup$
    – user4552
    Aug 17, 2013 at 20:39

3 Answers 3

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I can't improve on Luboš' comment, but I would add that a Schwarzschild black hole is indeed unphysical because it is time independant. A Schwarzschild black hole has existed for an infinite time and this is obviously unphysical. However we expect the Schwarzschild metric to be an excellent approximation to a real black hole.

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  • $\begingroup$ Interesting remark. $\endgroup$
    – Trimok
    Oct 26, 2012 at 11:00
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    $\begingroup$ But its unlikely for a real black hole that it has zero angular momentum. $\endgroup$
    – jjcale
    Oct 26, 2012 at 19:49
  • $\begingroup$ The Schwarzschild solution assumes a static, spherically symmetric spacetime. Clearly, a static spacetime is unrealistic. Remarkably, the static assumption only holds in the exterior region of the maximally extended solution. In the regions within the horizon, the spacetime geometry is time dependent. $\endgroup$
    – Alfred Centauri
    Oct 27, 2012 at 2:06
  • $\begingroup$ @Alfred Centauri : We have to distinguish between the outside observer point of view, and the inside observer point of view. $\endgroup$
    – Trimok
    Oct 27, 2012 at 9:32
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The horizon is not "progressively created" in the collapsing spherical shell (Oppenheimer-Snyder model for black hole formation), it just appears to an exterior observer once the shell falls through. The source of the Schwarzschild metric is not the singularity, but can be thought of as distributed on the horizon.

The unphysical thing about Schwarzschild is that it is unstable to perturbations. if you set it rotating, or give it a bit of charge, the interior changes completely, opening a second Cauchy horizon, and a bridge to another exterior. This property is the main issue with Schwarzschild. It has too much symmetry, so it's singularity is spacelike, it is an instant of time where all observers end.

The timelike singularity in a rotating/charged black hole repels massive matter, and only serves as an end/begin point for null geodesics, the paths of light rays. There is no way to get rid of the null-singularity by the singularity theorem--- the null rays must change from focusing to defocusing at some point, and this must be at a singular place.

The question of whether the singularity inside a generic black hole is spacelike or timelike is open, with Penrose saying "spacelike" and everyone else aping him, and me saying "timelike" and I'm pretty confident. The idea was that the Cauchy horizon would turn into a singularity under a perturbation, but it doesn't in any clear way in simulations, and the AdS/CFT understanding gives more insight into the expected behavior of charged black holes, and suggest they should emit cold matter that falls into them.

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    $\begingroup$ If you look fig 1.12, page 21, of Susskind/Lindesay book (An introduction to Black Holes, information and the String Theory Revolution : The holographic universe), the horizon is described as been progressively created. $\endgroup$
    – Trimok
    Oct 27, 2012 at 9:23
  • $\begingroup$ I think that the main problem is complementarity, the world seen by a outside observer, and the world seen by a inside observer are widely incompatible. For a outside observer, the singularity should be timelike, but for a inside observer, the singularity is maybe spacelike. $\endgroup$
    – Trimok
    Oct 27, 2012 at 9:26
  • $\begingroup$ Your remarks about stability are interesting. $\endgroup$
    – Trimok
    Oct 27, 2012 at 9:28
  • $\begingroup$ @Trimok: It's progressively created in a certain way of speaking, in the exterior view, as the matter gets closer to the classical horizon, and this is fine, but it's semantics--- there is no argument that the black hole will settle down to a Schwarzschild like horizon quickly from the exterior observer's vantage point, and (firewalls aside) there is no doubt that from the infalling point of view, the infalling matter just crosses the horizon at a finite time, corresponding roughly to the time when its strings spread out over a fraction of the black hole from the exterior point of view. $\endgroup$
    – Ron Maimon
    Oct 29, 2012 at 5:29
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My take is that the Schwarzschild black hole is indeed physical in the following sense. The S-metric has a timelike Killing vector, which indicates that the solution is symmetric with respect to time, hence, the "static" solution.

In GR, I generally interpret physically real objects to be scalars, that is, quantities that cannot be transformed away by any coordinate transformations, and indeed the $r=0$ singularity is a feature of all S-metrics.

The generic feature to determine if the black hole is physically plausible or not is the calculation of the Kretschmann scalar, $K = R^{abcd} R_{abcd} = constant/r^6$, and because this is an invariant quantity, one must conclude that indeed the r=0 singularity is physically significant.

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    $\begingroup$ Welcome to physics.SE! Sorry to have to use a negative comment as a greeting, but -- As Alfred Centauri pointed out in a comment on John Rennie's answer, the Schwarzschild metric is not static on its interior (although it is asymptotically static). I also don't really see how this relates to the rest of your answer. The fact that it's asymptotically static suggests that it's actually not physically realizable, in the sense that it can't have formed from gravitational collapse. $\endgroup$
    – user4552
    Aug 17, 2013 at 19:49

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