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I have heard the idea that real astronomical black holes cannot be called Schwarzschild BHs, not because of rotation but because Schwarzschild contains no mass and is eternal, due to being a stationary solution. But I would imagine a BH from stellar collapse becomes stationary after reaching a steady state condition? Is it not indistinguishable from an eternal Schwarzschild BH (or maybe it's better to say, the surrounding geometry is well described by the Schwarzschild metric with parameter $M$ corresponding to the mass of the former stellar core?).

Is it splitting hairs to say that non-spinning astronomical BHs (or stars and planets for that matter, outside their surfaces) are not truly Schwarzschild BHs, or is there a deeper difference?

Adding edit to clarify the question without so many double negatives:

When a stellar object collapses and settles into a steady state condition, is there a meaningful difference between it and a Schwartzchild BH? (neglecting rotation)

RC_23
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"But I would imagine a BH from stellar collapse becomes stationary after reaching a steady state condition?" Well, it all depends on how one models stellar collapse. One could use a spherically symmetric collapse of matter, where the metric inside the matter content is either say FRW type or Vaidya metric, but outside the collapsing matter, the metric is Schwarzschild eternally (by virtue of Birkhoff's theorem). The metric is eternal in the sense that if we consider the region outside the last shell of infalling matter, the geometry of this region will be given by Schwarzschild solution for eternity. However, one could also start from a generic Robinson-Trautman type metric, and using Lyapunov-functional argument it can be shown that for any general initial data on outgoing null surface $u=$const (where $u$ is the retarded time), the system will emit gravitational radiation and later settle down to schwarzschild solution (https://link.springer.com/article/10.1007/BF02431882). Likewise, one could come up with other models which is not eternally schwarzschild; it will depend on the symmetry and type of boundary conditions one imposes on the solution for Einstein's field equations. (EDIT: Even if the asymptotic state of solution is given by schwarzschild solution or any member of the Kerr family, it may not necessarily be a BH solution. A BH solution would also mean existence of event horizon and for schwarzschild case, it means that the matter field must be compressed within its schwarzschild radius. This will depend on the internal physics of the stellar matter which requires a separate consideration. For white dwarfs, the Chandrasekhar limit provides an estimate / upper bound of the total mass beyond which the star can collapse into a neutron star or BH, and in this sense the metric describes a schwarzschild BH (assuming spherical symmetry is maintained through out the process))

"Is it splitting hairs to say that non-spinning astronomical BHs (or stars and planets for that matter, outside their surfaces) are not truly Schwartzchild BHs, or is there a deeper difference?" I guess by splitting hairs you are referring to the BH hairs. It was proposed by Hawking that highly energetic collapsing matter (not necessarily spherically symmetric) can introduce super-translations on the generator of event horizon (https://arxiv.org/abs/1509.01147). Since, supertranslation is an infinite dimensional abelian subgroup of the total asymptotic symmetry group (say the BMS - symmetry ), it has been interpreted that these super-translation effects on outer event horizon produces infinitely many soft hairs. One can also artificially implant such soft hairs on Schwarzschild horizon (such as https://link.springer.com/article/10.1007/JHEP05(2017)161). So, if a realistic static non-rotating BH needs to have such infinitely many soft hairs, it is no longer equivalent to the exact Schwarzschild solution.

KP99
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  • These soft hair arguments has been proposed to play a significant role in the resolution of BH information paradox – KP99 Nov 02 '22 at 08:06
  • "Likewise, one could come up with other models which is not eternally schwarzschild; it will depend on the symmetry and type of boundary conditions one imposes on the solution for Einstein's field equations." But in any case the stationary end state has to be a member of the Kerr family. – TimRias Nov 02 '22 at 08:22
  • @TimRias Yes, I meant that one can in principle construct solutions which asymptotically reach schwarzschild or any member of the Kerr family – KP99 Nov 02 '22 at 08:26
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Following comments and discussion to the question I believe to reconcile some views.

The so-called Schwarzschild black hole (SBH) solution is a remnant of its formation process. Although Schwarzschild entitled his paper On the gravitational field of a mass point according to Einstein's theory, in true he had found the first part of it which is valid only outside the spherical body, i.e. in vacuum. The second part of the solution, the so-called interior solution On the gravitational field of a sphere of incompressible fluid according to Einstein's theory, he published short thereafter becoming he second known exact solution of Einstein field equations (EFE). Theoretically, there exists even a cosmological version of that full solution as noted in Boundary condition for gravity on galaxy scale?.

Using full Schwarzschild solution (interior plus exterior) one can study black hole formation in quasi static steps by changing the compactness parameter $\alpha\equiv r_{S}/R$. For the critical $\alpha$, $\alpha_{c}=8/9$, at the star center emerges the initial event horizon that for growing $\alpha$ moves outwards, until $\alpha=1$ is reached ($R=r_{S}$). In this process the interior part of the solution vanishes and only the exterior remains. The mass parameter in SBH metric is only a reminiscence of the formation process. Figuratively speaking, the initial mass fully converts into spacetime curvature. In this sense the SBH solution represents a whole universe without matter and radiation.

Astronomical black hole is in contrary a local black hole in a universe filled by matter, radiation and other black holes. After forming at some places and time they still interact with surroundings together determining the metric of the whole universe. That is the difference I see.

JanG
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  • "Figuratively speaking, the initial mass fully converts into spacetime curvature." - Not just figuratively - it is exactly what actually happens. The total energy of the star becomes the BH gravitational energy, which is the energy of the curved spacetime. The mass of the BH is the energy of its gravitational filed with no other separate contributions, such as from the mass or energy of the collapsing matter - it all has been converted to the energy of an empty curved space +1 – safesphere Nov 04 '22 at 05:31
  • I was conscious imprecise. I did not want to shock people with the idea that behind the event horizon there is no spacetime at all. My paper with the mathematical prove of it has been regrettably not accepted. You have put finger on the right place. How a point transforms into two-sphere with areal radius $r$? They have clearly different topology. – JanG Nov 04 '22 at 06:59
  • About gravitational energy of a BH please look into https://physics.stackexchange.com/a/680710/281096. The reference there supports your view. – JanG Nov 04 '22 at 07:04
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What we know from observation is that there are BH-like "holes" that are deep enough that the redshift at the bottom (if it exists) makes the stuff at the bottom invisible to us. Beyond that, the theorists make models, but most models lack observable consequences. The only way to make this into proper physics is to find models with observable consequences and test them against observations.

John Doty
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