2
$\begingroup$

Consider the interior Schwarzschild metric were the value $r_g$ of the $r$-coordinate at the body's surface is equal to the Schwarzschild radius $r_s$. In this case, the interior solution simplifies to the following form

$$ds^{2} = \left(1-\frac{r^2}{r_g^2} \right) c^2 dt^2 - \left( 1-\frac{r^2}{r_g^2} \right)^{-1} dr^2 - r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right),$$

where the factor $1/4$ in $g_{00}$ of the original solution has been absorbed into the time-variable (the correponding time-discontinuity across the boundary is not relevant for the following).

Obviously, the above expression is corresponding to a static representation of the de Sitter spacetime with positive and constant curvature $K=1/r_g^2$. From textbooks of the standard model of cosmology we know that such a space has the property to act repulsive in the interior of the horizon. Wouldn't this imply that the equilibrium state of the interior fluid is perfectly homogeneous at the end, instead of collapsing into the center $r=0$?

$\endgroup$
4
  • $\begingroup$ How would your argument go in the case of the (equivalent) Gullstrand–Painlevé coordinates? en.wikipedia.org/wiki/… $\endgroup$ – m4r35n357 Feb 9 '20 at 15:18
  • 1
    $\begingroup$ I can not see a representation of the interior solution on the page you have posted. Actually, my question should apply to every representation which is isometric to the standard form of the interior solution. This should also include the Gullstrand-Painleve representation. $\endgroup$ – user56224 Feb 9 '20 at 15:28
  • 1
    $\begingroup$ The interior Schwarzschild solution is only valid for rg>9rs/8, if you go below that the collapse is unavoidable and the time dilation in the center becomes infinite, therefore static and homogenous solutions with a smaller radius are unphysical. $\endgroup$ – Gendergaga Feb 9 '20 at 18:13
  • $\begingroup$ I agree, to reach the situation $r_g=r_s$ concerning my question, the star has to pass the condition $r_g=9r_s/8$. But the statement that all masses inside will approach the origin is only based on the phenomenological (ad hoc) assumption that the fluid is pressureless ($p=0$) (see Oppenheimer and Snyder, Phys Rev. 56 (1939), p.457, Sec. II). There is no prove of this statement. $\endgroup$ – user56224 Feb 9 '20 at 18:53
2
$\begingroup$

The Schwarzschild interior solution is a non-physical model that that gives the correct qualitative prediction for the interior metric of astrophysical bodies only within a certain parameter range. It is constructed by assuming that the density of the material of the "Schwarzschild star" is constant and the pressure profile is assumed to automatically enforce this constancy of density.

Such a fluid could be imagined to be realized, for instance, by a non-conducting ideal gas with the right temperature profile $T(r)$ such that $P = \rho k_{\rm B}T/m$ where $m$ is the mass of the gas particles and $\rho$ the constant mass density. Omitting the issue of conduction and various convective instabilities, this temperature must be at least finite. In the wikipedia article you link to you can find in the section Pressure and stability that the pressure (or temperature) profile diverges at the center when the ratio of the radius of the "Schwarzschild star" reaches $9/8$ times its Schwarzschild radius. Obviously, there is no physical material that can reach infinite pressures and temperatures, so considering a star radius equal to the Schwarzschild radius makes no sense.

To consider more physical equilibria, you can integrate the Tolman-Oppenheimer-Volkov equations, and consider perturbation theory of such equilibria (treated in detail in Misner, Thorne and Wheeler's Gravitation). Ultimately, the most convincing argument that squeezing a physical material up to the formation of a horizon always leads to a singularity is given by the Hawking-Penrose singularity theorems.

$\endgroup$
3
  • $\begingroup$ Great answer +1 except for the last sentence. The Penrose theorem applies only after the horizon forms, but the horizon never forms. $\endgroup$ – safesphere Feb 10 '20 at 23:01
  • $\begingroup$ At the critical point $9r_s/8$ the body is extremely dense and such a body suffers a gravitational collapse into a black hole. Since this is a time dependent process, the Schwarzschild solution does not hold any longer. The Snyder-Opperheimer collaps is corresponding to a time-dependent solution which is isometric to the FLRW metric (see e.g. G. Ellis). As I already mentioned, the pressure in this approach is defined to be $p=0$ from a phenomenological point of view. Alternatively, one could also chose $p$ such as to get the solution corresponding to the de Sitter spacetime instead. $\endgroup$ – user56224 Feb 12 '20 at 5:36
  • 1
    $\begingroup$ @kaffeeauf Well, in reality you cannot "choose" pressure, it is determined by physical conditions. You can model a real physical situation by choosing a pressure profile that simplifies your computations, but you always have to check whether your choice is at least remotely realistic. A contracting star will never have a deSitter-like stress-energy tensor. Here you should also find that apart from the normal deSitter pressure there is a pathological matter shell at the Schwarzschild radius. You can compute that from the Israel junction conditions (see Eric Poisson's Relativist's toolkit). $\endgroup$ – Void Feb 12 '20 at 10:17
1
$\begingroup$

The interior Schwarzschild metric models a fluid in equilibrium, and it is clearly static. It doesn't represent a collapsing body; for that you need to take a look at, for example, the Oppenheimer-Snyder solution, which models a ball (or was it a shell?) of presureless dust collapsing to a black hole.

$\endgroup$
2
  • $\begingroup$ Isn't it that the OS metric tells us that the event horizon appears at the centre of the collapsing ball and grows outwards until it has grown past the outer edge of the ball, at which point it becomes static? $\endgroup$ – user56224 Feb 9 '20 at 14:20
  • $\begingroup$ Actually, the pressurelessness in the OS approach is phenomenolgically (or ad hoc) introduced. Such that it is obvious that a collaps will occure by defintion. $\endgroup$ – user56224 Feb 9 '20 at 15:33
1
$\begingroup$

If there was a material that could resist collapse so that $r_g=r_s$ were possible it would mean that this material would have a divergent bulk modulus. The bulk modulus $$ B=-V\frac{dP}{dV} $$ for an isentropic process is $PV^\gamma=k$ a constant and so $B=\gamma P$. However, if the collapse stops at the horizon it means there is a large outward acceleration at the horizon. The near horizon spacetime for the Schwarzschild metric defines an acceleration $g=c^2/d$ for d the distance from the event horizon. This means the acceleration required to oppose the gravitational implosion is divergent and the bulk modulus of this material must be similarly divergent. As a result, this harte kugel state means this metric for the interior of a ball of material at the Schwarzschild radius breaks down. However, the idea may not be completely dead as I explain below.

The Schwarzschild metric for the interior of a material ball $$ ds^2=\frac{1}{4}\left(3\sqrt{1-\frac{r_s}{r_g}}-\sqrt{1-\frac{r^2r_s}{r_g^3}}\right)^2dt^2-\left(1-\frac{r^2r_s}{r_g^3}\right)^{-1} -r^2(d\theta^2+sin^2\theta d\phi^2), $$ with the condition $r_g<r<<r_s$ has $$ ds^2=\left[\frac{r_s}{r_g}\left(9-\frac{6r}{r_g}\right)dt^2 – dr^2\right]+\frac{r_s}{r_g}\left(\left(\frac{r}{r_g}\right)^2dr^2~-~\left(\frac{r_g}{r}\right)^2dt^2\right)-r^2(d\theta^2+sin^2\theta d\phi^2) $$ which is the metric for $AdS_2\times\mathbb S^2$ perturbed by a space $(g,{\cal M})$. The metric terms for the square brackets are $(g,{\cal M})$ and terms in the parentheses are for $AdS_2$. It is not hard to show $AdS_2\times\mathbb S^2$ is the near horizon condition for the Kerr or Reisner-Nordstrom metric. Hence the anti-de Sitter spacetime emerges from black hole metrics as well. This condition occurs for a highly accelerated observer positioned above the event horizon. The near singularity condition is what obtains for an observer close to the horizon. This means that an observer very close to a Kerr horizon on an accelerated frame observes a similar spacetime as that near a Schwarzschild singularity. The difference though is that with the Schwarzschild singularity the sphere has a radius that has a dimension given by a timelike dimension. With $AdS$ spacetimes there are also closed timelike curves. In the near horizon Kerr metric by contrast this condition obtains for an accelerated frame, but the $\mathbb S^2$ is parameterized by a spatial radius. There is then some funny relationship between these anti-de Sitter spacetimes and with the de Sitter spacetime..

The $dS$ spacetime above might sound completely fictional. However, there is this oddity with the so-called firewall. This stems from the fact that Hawking radiation emitted by a black hole is entangled with the black hole. Hence this is a bipartite entanglement. As the black hole emits radiation the emitted radiation there is a growing probability this is entangled with previously emitted radiation and with the black hole. Once the black hole emits half its mass, at the Page time of around ${\frac{7}{8}}^{th}s$ the duration of a black hole, this probability approaches unity. This means previously emitted radiation transforms from a bipartite entanglement to a tripartite entanglement. This runs into troubles because the $W$, constructed from bipartite states, and $GHZ$ states are defined on different parts of the $3$-Kirwan polytope for these states. This means there is a $3$-tangle that is a sort of topological obstruction to this sort of unitary transformation. For some more details see V. Coffman, J. Kundu, W. K. Wooters, “Distributed Entanglement,” Phys. Rev. $\bf A~61$:052306 (2000). If one presumes that quantum states or equivalently quantum information is conserved, then the fix is to abandon the equivalence principle. Other people, Unruh and Wald etc., prefer to abandon conservation of quantum information and preserve the equivalence principle. The firewall then might connect with this harte kugel.

Consider the black hole that has reached the Page time and has a firewall. This is a barrier or naked singularity that is for all purposes the material with a divergent bulk modulus. At least on a classical level that is what it appears as. This would then tend to support some idea of the black hole interior as a de Sitter spacetime. The $dS$ and $AdS$ spacetimes are separated from each other in a higher dimension spacetime by a light cone. The $dS$ and $AdS$ spacetimes are connected by their conformal infinite regions or “boundaries.” The $dS$ spacetime is the single hyperboloid outside the cone and there are two $AdS$ spacetimes are inside the opening of the two conical horns. This then may lead to a nest of questions about these relationships, the near horizon condition, near singularity condition and the firewall.

$\endgroup$
4
  • $\begingroup$ This is a very informative post so +1 but it seems to be missing a critical step from the start. The horizon (some type of it) first forms at the point in the center where $r_s=0$. So the rest of your argument seems non physical, at least to me :) $\endgroup$ – safesphere Feb 10 '20 at 23:24
  • $\begingroup$ The formation of the horizon and black hole is a dynamical process that is not easy to understand in an exact form. This page mathpages.com/rr/s7-02/7-02.htm has a diagram of the formation of horizon and deformed spatial surfaces with the implosion of matter. The Schwarzschild metric for bulk material is a static solution, which is a pale approximation to the dynamic process of implosion. $\endgroup$ – Lawrence B. Crowell Feb 11 '20 at 12:31
  • $\begingroup$ Thanks for the link. Kevin Brown of MathPages is my favorite author with original thoughts and clear logic addressing real concerns. Most other sources just repeat the same nonsense after each other. $\endgroup$ – safesphere Feb 11 '20 at 13:49
  • $\begingroup$ See his next to the last paragraph here clarifying that crossing the horizon is a popular assumption, not a logical conclusion from any rigorous argument: mathpages.com/rr/s7-03/7-03.htm $\endgroup$ – safesphere Feb 11 '20 at 14:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy