In static spherically symmetric perfect fluid solutions of Einstein field equations the central pressure diverges when the compactness parameter $\alpha~(\equiv r_{S}/R)$ reaches some critical value $\alpha_{c}\leq 8/9$. This fact is commonly understood as the birth of central gravitational singularity and event horizon. What is first, the event horizon or the singularity? How to understand the initial event horizon at the star center, which have a zero surface area?
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What is first, the event horizon or the singularity?
The horizon forms first. Every event on the horizon has the singularity in its causal future. No event anywhere in spacetime has any part of the singularity in its causal past. So unambiguously the horizon is first.
How to understand the initial event horizon at the star center, which have a zero surface area?
It is not a particularly different interpretation from the rest of the horizon. It is simply the first event at which light does not escape to future null infinity.
Dale
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1Thanks for your quick response. I agree with you of course but would you try to explain me how the initial horizon, geometrically a non-dimensional point, transforms into a two-dimensional sphere? – JanG Feb 19 '22 at 20:10
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@JanGogolin the same thing happens with any flash of light. The flash event is the apex of the light cone. The horizon is a lightlike surface, and that point is simply the apex of the light cone – Dale Feb 19 '22 at 20:15
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@JanGogolin The initial surface never coincides with a point. The event horizon comes into being for every distribution of fluid mass, however small. – MatterGauge Feb 19 '22 at 20:29
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@Dale, if the horizon forms first, what is also my opinion, that how do we know that something like "singularity" will arise behind it? – JanG Feb 20 '22 at 15:52
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2@JanGogolin we don’t “know”. We just predict it based on our best current theories. We assume that our current theories hold everywhere, and if they do then the singularity follows. It is likely that assumption breaks down at some point, but we don’t currently know what conditions are required for our current theories to break down nor what physics is like under those conditions. It is possible that those unknown breakdown conditions occur somewhere between the horizon and the singularity – Dale Feb 20 '22 at 16:06
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@Dale, it is my strong believe that it is not Einstein's GR that breaks down there but our interpretation of the corresponding equations. The most feasible interpretation for me is a black hole with no interior spacetime, see Aharon Davidson, Ilya Gurwich, https://arxiv.org/abs/1007.1170 . That a stable static solutions above the critical compactness value exists showed Pawel O. Mazur and Emil Mottola, https://arxiv.org/abs/1501.03806 . However, their theory fails at the Occam's razor principle introducing a more complex black hole model (gravastar). – JanG Feb 20 '22 at 17:09
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1@JanGogolin we currently have no evidence supporting those over GR. In the absence of such evidence I think a “strong belief” is unwarranted, but certainly you are free to “hypothesize” that such alternative theories are correct if you like. I choose to remain agnostic, but recognize that others are allowed to choose differently – Dale Feb 20 '22 at 18:01
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1@Dale, you are right - "believe" is not a physical category. I should say, outgoing from working assumption that Einstein's theory does not break down, I have found some mathematical evidence supporting my these. I have already tried to publish it, however, so far without success. – JanG Feb 20 '22 at 18:46
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It depends. For a faraway observer a black hole takes an infinite time to form. The innermost particles collapse to a small black hole first. The mass of a sphere increases with $R^3$, while the Schwarzschild radius increases linearly with $M$, the mass of the hole. So if the whole sphere lies still outside its associated Schwarzschild radius, every sphere inside lies outside its SR also. So only when the whole sphere lies within the associated SR, the black hole will have formed.
The density of the sphere of (non-rotating) dust is not constant throughout the sphere and the center will first develop an associated SR. As the compactness parameter shows. The radius will then grow to the radius associated with the total mass. So the radius grows from the inside to the outside. The matter collapses towards one point (as assumed in GR, but in reality particles are not pointlike, but a soft and smoothened, continuous singularity can still develop), and at the moment 8/9 of the mass finds itself behind the SR it still needs some time to collapse to a point. So, event horizon first, and then the singularity.
Seen rom the outside the 1/9 left outside the horizon will never reach the hole and seems to be frozen in time above the horizon.
I have read that some doubt black holes actually never develop at all. I can't find it though and I'm not sure what they mean.
This holds for all continuous non-rotating spherical masses. Up to the Planck mass. There are no processes in nature though that develop a black hole when their mass is smaller than about 1.4 solar masses, but in principle a small spherical, non-rotating distribution of mass (say without charges, which normally stop collapse) can collapse to a BH. With a minimum though of the Planck mass.
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thanks for your answer. It confirms my understanding of that process, especially the growth of event horizon from center outwards. – JanG Feb 19 '22 at 20:04
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2"the 1/9 left outside the horizon will never reach the hole and seems to be frozen in time above the horizon." This is a common misconception that isn't actually true. All but the infinitesimal outer surface of the dust will fall inside the black hole in finite time. See e.g. https://arxiv.org/pdf/1805.04056.pdf – TLW Feb 20 '22 at 08:05
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1(The issue is effectively: the standard approximation assumes that the dust has a negligible effect on the metric, but because the dust has nonzero mass and it falls infinitesimally close to the horizon, any nonzero effect on the metric eventually becomes non-negligible.) – TLW Feb 20 '22 at 08:10
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And in the presence of a nonzero amount of matter between the observer and the black hole, even the infinitesimal outer surface falls inside the event horizon in finite time. – TLW Feb 20 '22 at 08:52
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@TLW Only for an outside observer the last 1/9 will never fall in. The question mentions this 8/9 I had never heard of but it sounds very reasonable. Of course it all falls in in the BH frame. – MatterGauge Feb 20 '22 at 09:08
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@Felicia can we please concentrate of the interior Schwarzschild solution? The dust solution is somewhat special because pressure there does not play any role. – JanG Feb 20 '22 at 11:48
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@JanGogolin Don't you write in your question that the SR of a sphere of dust is there when the cloud of dust has collapsed such that 1/9 of the radius is still outside the event horizon? No matter what the mass of the sphere? So if a very small amount of mass could fall freely towards a center, without rotation, the small hole will form at a very tiny radius of the whole sphere, with some dust still outside. Seen from faraway this will never enter the hole, but in the freely falling frame, all will still have to fall to the center. This happens fast in this frame, but takes an eternity for us. – MatterGauge Feb 20 '22 at 12:23
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@Felicia, a perfect fluid is not a dust. Schwarzschild interior solution describes constant energy density (or constant density) star in equilibrium, thus static. By collapse one understand usually a time dependent solution. – JanG Feb 20 '22 at 15:42
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@TLW, I have read your interesting paper. However, I try to understand the emergence of event horizon and singularity on the example of Schwarzschild interior solution, the simplest of the SSSPF (Static Spherically Symmetric Perfect Fluid) solutions. – JanG Feb 20 '22 at 15:49
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Great discussion on this question.
It seems to me, though, that the event horizon can never form, in principle.
If we take as a working hypothesis that an event horizon does form, it will have to form initially around some point. This point will have to be the density maximum of the immediate vicinity.
The initial event horizon around this initial point would have to have an extremely tiny initial radius.
One of two things would then happen (given the assumption that it formed at all). Either (1) it would evaporate so quickly due to Hawking radiation that it could not grow virtually at all in that short time, or (2) it would "consume" whatever was in its immediate vicinity, but after that, anything else (like the next quark over, say) would be so far away from its tiny radius that such growth would appear to take infinite time to happen, so effectively not happen.
