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A black hole is where it's mass is great enough that light can't escape at a radius above the surface of the mass?

I've been told that strange things happen inside the event horizon such as space-time ceases to exist.

Why does the escape velocity of light warrant space-time to not exist or the mass to be confined to a single point of infinite density. I mean what's make the escape velocity being the speed of light so special compared to other speeds?

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3 Answers 3

You've heard a slightly garbled account of the physics inside a black hole event horizon, but what you've heard is not so far from the truth.

The physics of (stationary) black holes is described by the Schwarzschild metric, though unless you're a GR nerd you'll find this a bit opaque. I'll try to describe what is going on in everyday terms, but you need to bear in mind that this is just an analogy and you can't get a full understanding unless you're prepared to learn the necessary maths.

To take your second question first, you're correct that at the event horizon the escape velocity is the speed of light. See this question for more info on this, and this somewhat more involved question for exactly why nothing can escape from the event horizon. Once you move inside the event horizon even light cannot move fast enough to avoid being dragged into the singularity at the centre of the black hole, and because nothing can move faster than light that means anything and everything that passes through the event horizon must end up at the centre of the black hole. That's why we get a point of infinite density. The density can only be finite if whatever has fallen through the event horizon is a non-zero distance away from the centre. But nothing can stay a finite distance away from the centre, so everything must end up exactly at the centre i.e. we have an infinite density.

Now to return to your first question, i.e. whether spacetime ceases to exist inside the black hole, spacetime does not cease to exist inside the event horizon but there is a sense in which it ceases to exist at the singularity at the centre. If you have some object travelling through spacetime, e.g. an apple falling towards Newton's head, then we can use simple physics to predict how the apple moves. That is, if we know where it is at time $t$ we can tell where it will be at a later time $t + \delta t$ for some small time increment $\delta t$. However if the apple is falling into a black hole we can calculate its trajectory right up to the point where it reaches the singularity at the centre of the black hole, but it is impossible to calculate its motion in space or time beyond this point. It is as though both space and time stop at the singularity. This is probably why you've heard that spacetime ceases to exist there. The technical term is geodesic incompleteness: the geodesic is just the trajectory in spacetime traced out by the falling apple.

Incidentally, few of us really believe the density becomes infinite at the singularity and that the geodesic of the apple ends there. Most of us think that some form of quantum gravity will blur out the singularity and keep everything finite. However, at the moment no-one knows what this theory of quantum gravity will turn out to be.

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I'm quite wondered that someone read this answer within a minute. Ok, I've read it now (+1)... ;-) –  Waffle's Crazy Peanut Apr 19 '13 at 16:04
    
The OP asks, "Why does the escape velocity of light warrant space-time to not exist or the mass to be confined to a single point of infinite density." The technical answer to this is that the OP's statement can be refined into the Penrose singularity theorem. The theorem says that if the strong energy condition holds, then the formation of a trapped lightlike surface guarantees the formation of a singularity. –  Ben Crowell Apr 19 '13 at 16:09

First, I'd like to clarify with the title. As it may seem, but blackholes are not at all "special". They attract the crowd only because of their strong curvature of spacetime. Let's consider a Schwarzchild blackhole.

... mass is great enough that light can't escape at a radius above the surface of the mass?

Yes. The mass $m$ is highly compressed to the radius $r_s=\frac{2Gm}{c^2}$ which may seem to imply that light can escape from the event horizon of blackhole. It simply doesn't. Because, the event horizon is the boundary of spacetime where matter and radiation can only fall into. There's no coming out. The light photons can still trace a path very close to the event horizon. But, the light that floating at the horizon (always hover there and so, it can't reach the observer) or that fell into the horizon don't come out. In other words, the paths of spacetime inside the horizon always curve inwards, and light simply travels along the geodesic and finally reach the singularity (as far as we can think)

... inside the event horizon, space-time ceases to exist

That's quite untrue. Spacetime is everywhere. Or, it is assumed to be spread everywhere so that we can learn about the events taking place in space & time. We can still imagine how the spacetime paths would be inside the event horizon and how the light cones would behave inside the horizon. But, there's something true in this phrase (i.e) Spacetime ceases to exist at the singularity where all the physical laws would breakdown.

... speed of light so special compared to other speeds?

That would be a question to SR. More specifically, to its second postulate. BTW, the speed of light is constant only when its measured locally. In GR, especially when you're falling into blackholes (being a prolific physicist), you can prove $c$ to be a variable assuming that you've not crushed...

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I'd say another big thing that make's a black hole "special" is that it may* violate the law of the conservation of energy which state that energy may neither be created nor destroyed.

*Hawking radiation refers to the small amounts of thermal radiation spit out by a black hole. Since the event horizon is defined as the "point of no return", where exactly is this thermal radiation coming from?

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