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[iceball3]

Well moving over and object is the same as "infalling" really. It will still take infinity to hit the event horizon from any external observer. so the stuff should smear over the event horizon surface.

Observationally will just appear splattered on the surface of the black hole in question? Interesting.
And by this i mean in the less obvious manner, of course someone hit by a black hole wouldn't look anything like someone hit by a Shinkasen.

[Reelya]

As for center of gravity. A hollow Earth would have the same center of gravity as a solid one, if the mass is the same. And the total gravitational field strength would be identical.

No different for a black hole.

In other words, a black hole with the mass outside it's event horizon should be gravitationally identical to the same black hole with the mass inside the event horizon, in the same way as a hollow planet would be. If you're inside a hollow planet, you're not "stuck" to the outside: you're drawn to the center of gravity.

[Sergarr]

The events must be the same for all observers. From the outside perspective, you don't fall into the black hole, therefore, you don't fall there from any perspective, including your own. Sounds logical for me.

No, of course not. Simultainety is relative. An event that already happened in one frame may not have yet happened in another.

Anyway, I won't pretend to understand GR - I only have a shallow, layman's understanding gained from evesdropping on other people's discussions. However, from what I read about it, the singularity at the event horizon exists only in the so-called Shwarztshild coordinates, which can't be used for the infalling observer.
Nowhere will you find the inability to cross the horizon as an obstacle to BHs' existence. That's why the aforementioned paper(and earlier papers on Planck stars, e.g.: http://arxiv.org/abs/1401.6562 ) start with the information paradox as the problem to be solved.

The idea is that no atoms actually go through the event horizon from ANY point of view, because there exists at least one outside observer which observes them as not in black hole. And all events must be the same from ALL points of view. The position and timing of these events may be vastly different, but the event which happened from one point of view happens in ALL of them, and visa versa.

[Darvi]

I assume Icebro meant that the matter "stuck" to the EH would be unevenly distributed, which would indeed cause a shift in the centre of gravity.

[iceball3]

I assume Icebro meant that the matter "stuck" to the EH would be unevenly distributed, which would indeed cause a shift in the centre of gravity.

Yeah, this essentially. Also this is a wild guess, but doesn't the hollowness of mass distribution also affect tidal forces or is this the same too?

[Sergarr]

As for center of gravity. A hollow Earth would have the same center of gravity as a solid one, if the mass is the same. And the total gravitational field strength would be identical.

No different for a black hole.

In other words, a black hole with the mass outside it's event horizon should be gravitationally identical to the same black hole with the mass inside the event horizon, in the same way as a hollow planet would be. If you're inside a hollow planet, you're not "stuck" to the outside: you're drawn to the center of gravity.

Actually, you can prove that inside a hollow planet the gravitational field would be zero.

RE:
http://en.wikipedia.org/wiki/Shell_theorem
"If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell."
Proven by Isaac Newton.

RE2:
The center of gravity approach works only when you're far away from the object. It stops working if you're inside of it.

[Reelya]

Isn't that only when you hit the center? The gravitational field right at the center of the Earth will also be zero. Equal pull from all directions.

[iceball3]

Isn't that only when you hit the center? The gravitational field right at the center of the Earth will also be zero. Equal pull from all directions.

I thiiiink it could me more that when standing underneath the crust you would be up against the ceiling. I THINK. I dunno actually.

[Reelya]

You're pulled towards center of gravity of any body. You definitely don't "stick" to the inner surface. You'd have 1g force towards the center, if the mass of the planet was the same and the distance to the center was the same as Earth's radius.

[Darvi]

No, you would stick against the crust, because if we had a hollow Earth everybody woulda moved inside and developped gravity generators so they would never have to look at that scary black thing that appears regularly up above.

[Reelya]

Apparently, you are near weightless inside a hollow sphere, my bad.

But, the fact that a hollow sphere of the same mass should have the same external gravitational force / center of gravity gives credence to the idea of a black hole with all the matter outside the event horizon: from further out it should not be gravitationally any different.

[Il Palazzo]

The idea is that no atoms actually go through the event horizon from ANY point of view, because there exists at least one outside observer which observes them as not in black hole. And all events must be the same from ALL points of view. The position and timing of these events may be vastly different, but the event which happened from one point of view happens in ALL of them, and visa versa.

Yes, I heard you. I'm saying this is incorrect, not to mention internally contradictory. For example, I choose the infalling object's coordinates. I pass the horizon in finite time. Using you logic, this would mean I pass the horizon in all coordinates(by which I understand you mean that it al all happens, not that the time of occurence is the same). Now I choose outside observer. The object never crosses the horizon. Again, this should mean it never crosses it in all coordinate systems. All you're doing is showing preference for one coordinates over the other when you make that definitive statement.

[miauw62]

http://science-beta.slashdot.org/story/14/07/24/172221/black-holes-not-black-after-all-theorize-physicists

Spoiler (click to show/hide)

Black holes are singularities in spacetime formed by stars that have collapsed at the end of their lives. But while black holes are one of the best known ideas in cosmology, physicists have never been entirely comfortable with the idea that regions of the universe can become infinitely density. Indeed, they only accept this because they can't think of any reason why it shouldn't happen. But in the last few months, just such a reason has emerged as a result of intense debate about one of cosmology's greatest problems — the information paradox. This is the fundamental tenet in quantum mechanics that all the information about a system is encoded in its wave function and this always evolves in a way that conserves information. The paradox arises when this system falls into a black hole causing the information to devolve into a single state. So information must be lost.

Earlier this year, Stephen Hawking proposed a solution. His idea is that gravitational collapse can never continue beyond the so-called event horizon of a black hole beyond which information is lost. Gravitational collapse would approach the boundary but never go beyond it. That solves the information paradox but raises another question instead: if not a black hole, then what? Now one physicist has worked out the answer. His conclusion is that the collapsed star should end up about twice the radius of a conventional black hole but would not be dense enough to trap light forever and therefore would not be black. Indeed, to all intents and purposes, it would look like a large neutron star.

Interesting stuff. It fits with the relativistic calculations: time dilation should hit infinity as you approach an event horizon. Therefore nothing can actually go beyond the event horizon until infinite external time has passed. What this means is that rather than an empty "hole" you fall through, in practice you get this relativistic sludge which builds up outside the black hole, and resembles a neutron star. At least that's the new theory.

This reminds me of Revelation Space :V
Neutron stars that are actually black holes?

E:
More science-y, does this have any consequences for how small we can measure things? My layman's terms can't explain it without looking it up on Wikipedia.

[Sergarr]

The idea is that no atoms actually go through the event horizon from ANY point of view, because there exists at least one outside observer which observes them as not in black hole. And all events must be the same from ALL points of view. The position and timing of these events may be vastly different, but the event which happened from one point of view happens in ALL of them, and visa versa.

Yes, I heard you. I'm saying this is incorrect, not to mention internally contradictory. For example, I choose the infalling object's coordinates. I pass the horizon in finite time. Using you logic, this would mean I pass the horizon in all coordinates(by which I understand you mean that it al all happens, not that the time of occurence is the same). Now I choose outside observer. The object never crosses the horizon. Again, this should mean it never crosses it in all coordinate systems. All you're doing is showing preference for one coordinates over the other when you make that definitive statement.

Ah-ha, but you forget that the black holes evaporate due to Hawking radiation! Therefore, "I pass the horizon in finite time." might never actually happen, because due to time dilation the black hope will cease to exist after a period of time. If this happens, then there's no inside observer and universe becomes consistent.

The consistency-breaking and information-conservation-breaking of something going inside the black hole is precisely why Hawking decided to basically ban black holes.

[Il Palazzo]

If you don't allow falling through the horizon in finite time, then there can be no Hawking radiation, as one of the virtual particle pair can never be trapped inside.