[trhway]

Ok. To start she addresses only "Gravitationally Collapsing Star"s. Interesting result, may be many star lifecycles don't end in a black hole. The "before-black-hole" state she describes still would involve significant time dilation and thus what we see as Hawking radiation from black hole may just be Hawking radiation from "before-black-hole" star. Looks the same :)

The result doesn't say that black hole is impossible. It says that a star's collapse loses mass faster than acceptable for the collapse to end in black hole.

Some misconceptions in the web-post:

>They are the ultimate unknown – the blackest and most dense objects in the universe that do not even let light escape.

a black hole isn't necessary most dense. After all a black hole with a mass of visible Universe would be only 125 times denser than current average density of the visible Universe (i.e. 45B light years radius of visible Universe have 10B light years Schwarzschild radius)

> So the story went, an invisible membrane known as the event horizon surrounds the singularity and crossing this horizon means that you could never cross back. It’s the point where a black hole’s gravitational pull is so strong that nothing can escape it.

"membrane" is very bad illustration. It is constantly changing solution to the gravitational equations. One moment it is here, another moment - it has moved because gravitation of black hole changes and a lot of oscillations/perturbations happen. One moment you're inside, another - outside.

"gravitational pull is so strong that nothing can escape it" - that's bad wording too. It is actually "gravitational well is so deep that nothing can escape it". Feel the difference :) A huge black hole may have pretty weak gravitation at its horizon.

Thus no-escape is valid only in the sense that escaping object just would never reach "a point at infinity" from the black hole and the light would get red-shifted into full oblivion.

[lutusp]

> A huge black hole may have pretty weak gravitation at its horizon.

Not really. The amount of spacetime curvature at the event horizon is fixed in the theory -- it's always the same. It's how the event horizon is defined. Near the event horizon, light orbits endlessly (in principle), and (again in principle) if you were located at an event horizon and there was sufficient illumination, anywhere you turned you would see the back of your own head.

> Thus no-escape is valid only in the sense that escaping object just would never reach "a point at infinity" from the black hole and the light would get red-shifted into full oblivion.

From a more distant frame of reference, yes, but not at the event horizon itself.

[dragonwriter]

> The amount of spacetime curvature at the event horizon is fixed in the theory -- it's always the same.

No, its greater the smaller the horizon is (so, equivalently, less the bigger the horizon is).

> Near the event horizon, light orbits endlessly (in principle)

Sure, at the event horizon, light orbits endlessly. But the amount of curvature needed to do that is less the further across the horizon is.

[lutusp]

>> The amount of spacetime curvature at the event horizon is fixed in the theory -- it's always the same.

> No, its greater the smaller the horizon is (so, equivalently, less the bigger the horizon is).

We're using different meanings of "curvature". For all black holes regardless of their properties, the event horizon has the same spacetime curvature -- that required to produce orbiting photons that cannot escape. Outside the event horizon, photons can escape. Inside, they cannot (and those photons don't orbit either, but cross the horizon). All the same.

> But the amount of curvature needed to do that is less the further across the horizon is.

No, it's the same. Same curvature, different large-scale geometry. For a sufficiently small zone near the event horizon, the conditions are identical.

Again, this is about the meaning of "curvature". The circumference of the horizon is greater for a large mass than a small one, but the spacetime curvature at the horizon is the same -- it's the specific value that causes photons to orbit perpetually, and interestingly from a local perspective, the entire horizon surface appears as a plane of infinite extent, sort of like two facing mirrors but with more dimensions.

From the perspective of a hypothetical observer at the horizon, he wouldn't be able to judge the size of the black hole using local observations -- there would be a plane of infinite extent for any black hole regardless of size.

[darkmighty]

You seem to be knowledgeable, so is there a plausible bridge between the impossibility of this kind of collapse and complete impossibility of black holes?

I mean, what if could construct a collapse to your desire, e.g. smashing dense cold chunks, would those necessarily radiate enough energy to prevent collapse too?

[trhway]

>is there a plausible bridge between the impossibility of this kind of collapse and complete impossibility of black holes?

the shown impossibility of star collapse to BH is somewhat questionable for massive stars because - as i cited wikipedia in another comment - the Hawking radiation carries away less mass/energy than cosmic background brings in - the net is positive for the collapsing star.

>smashing dense cold chunks, would those necessarily radiate enough energy to prevent collapse too?

we associate "collapse/smashing" with BH i think mostly because of collapsing stars. One can scale this process out to a lot of small chunks of cold matter as you suggest flying simultaneously toward common center. Imagine the sphere with radius 1000 times radius of Sun filled with material of density of water - this is already a black hole as its Schwarzschild radius is a bit larger than those 1000 radii of Sun. Now lets say we scale the sphere radius additional 10 times - the mass scales 1000 times and the Schwarzschild radius increase 1000 times too - thus all this water may be separated by a lot of space and still be a black hole. Now imagine that all this water is outside the Schwarzschild radius and spread on some imaginary sphere and simultaneously released to fly toward the center - once all these drops cross the schwarzschild radius, even without touching each other, and with each drop feeling only pretty small gravitation from the rest of the water, they nevertheless become a BH. What kind of miracles would happen in the moment these drops cross the imaginary sphere of Schwarzschild radius? Nothing. The only change here is that any light radiated by the drops close to crossing would reach far observers as a very redshifted and any light radiated at the moment of crossing and after would never be seen by the far observers. The article suggests that the loss of mass due to Hawking radiation would be fast enough to shrink the Schwarzschild radius faster than the drops can fall into. All this cold drops flying in empty cold space would supposedly cause very intense radiation close to the schwarzschild radius border? Nope. As mentioned above in the wikipedia citation Hawking radiation is pretty cold, low intense, for that size of black hole, and in particular it is much colder than CMB, ie. the losses are less than what is brought in by CMB. Though may be in the absence of CMB if the math in the article is right the losses (given the time slowing) would be enough to shrink the Schwarzschild radius fast enough. Yet not in our real Universe with CMB.