Blackhole question?

[Gustav32Vasa]

Blackholes are so dense that their gravity wont allow anthing to escape, not even light. But how can gravity, that travels at the speed of light, escape?

[Hawkwings]

Gravity isn't a particle that has speed (at least not in generally accepted physics) so nothing actually escapes from the black hole. The black hole warps space-time because of its mass, causing gravity.

[NoXion]

As far as I know (and since I am not a physicist, that is very little) gravity does not travel at all, but changes in gravity propagate at C.

[TithonusSyndrome]

To quote Kuroneko, "this is a bit like asking if sod can escape a steep hill."

[Starglider]

Blackholes are so dense that their gravity wont allow anthing to escape, not even light.

Even if you're thinking of gravity as a force carried by messenger particles (the graviton model) rather than a distortion of space-time, surely this would just imply that gravitons don't interact with other gravitons. Why would you assume that they did? After all photons don't interact with other photons.

[Kuroneko]

I don't recall saying that, but... you're thinking about this in reverse. It's actually because information can't locally travel faster than light that the gravitational field of a black hole remains the same.

Imagine a spherical mass distribution. Under Newtonian gravity, the shell theorem guarantees that the external field is the same no matter what the radial distribution is, as long as it maintains spherical symmetry: the external field is the same as that of a point-mass. In GTR, there is a similar result (Birkhoff's theorem) that guarantees that the field of such a distribution (nonrotating, uncharged) is always the Schwarzschild solution.

Now imagine a spherically symmetric collapse. At any particular moment, the external field will stay exactly the same as it was, and as the edge of the matter moves inward, it will bend spacetime to match the Schwarzschild solution. When a certain bit of mass crosses the event horizon, that information cannot escape the black hole, and so even when it is crushed out of existence at the singularity, the fact that it no longer exists cannot affect the external field--otherwise, information about the state of the mass bit would escape the black hole. Thus, the impossibility of information escaping a black hole actually requires the external field to stay the same no matter what happens to the matter.

It is also why isolated black hole solutions are vacuum: all the matter meets the singularity and no longer exists, but the fact that it doesn't cannot be communicated past the horizon, so the external field still acts the same. Also because of this, the field still has a well-defined and observable "geometric mass" that can be treated as the black hole's mass.

---

Perhaps I'm speaking a bit too loosely. According to any external static observer, the would-be black hole never finishes collapse, in the sense all of its matter can still be seen (just red-shifted outside of practical limits of observation). For about fifty years after the Schwarzschild geometry was discovered, the Russian literature called black holes "frozen stars" (the English was the slightly less vivid "collapsed stars"); they were seen as the result of a gravitational collapse "frozen in time" by gravitational time dilation. That any particular piece of matter in the collapse crossed the horizon and reaches the singularity in finite proper time wasn't fully appreciated.

An interesting side-effect is that when more matter falls into the black hole, the event horizon expands outwards to meet it: the size of the event horizon depends not only on the matter than has already collapsed, but also on the matter than will collapse in the future.

[Kuroneko]

Even if you're thinking of gravity as a force carried by messenger particles (the graviton model) rather than a distortion of space-time, surely this would just imply that gravitons don't interact with other gravitons. Why would you assume that they did?

I would assume that they did because gravity is known to be non-linear. But if we're talking about quantum forces, they're mediated by virtual particles, which can be superluminal.

[Steel]

Well a further question about black holes.

Having done some GR, when we briefly covered black holes, it seemed that the event horizon was the radius in the swarzschild metric where all geodesics would tend to the origin. Which means that all particles in free fall and all light rays would eventually (in their perspective) hit the singularity.

But what i'm not sure of is if you have something with engines just inside the event horizon, can it escape (disregarding absolutely any practical concerns about durability or power outputs)?

I mean if you take a newtonian argument that inside the event horizon the escape velocity is infinite, and so adding any finite amount of potential energy from the ships fuel is still not going to be much help in lifting you above that. But newtonian thinking isnt too much help for most of GR... Is the situation analagous to how with a pendulum with a rod, there is no starting point you can release it from so that it has a speed greater than zero when it reaches the top, but if you give it a big enough shove then any starting point can go arbitrarily fast when it reaches the top? (ie a free particle cant escape, but something with engines can?)

[Kuroneko]

The short answer is that every massive, non-tachyonic particle must travel in a timelike manner, i.e., it always stays within its local light-cone. That means that if all light signals, which form the edges of the light-cones, inevitably fall inward, then so will every massive particle, no matter how much or in what manner it accelerates.

--

If you look at the Schwarzschild metric ds² = Adt² - [dr²/A + r²(dθ² + sin²θdφ²)], A = 1-2m/r, you'll notice that not only is it singular at r = 2m (Schwarschild radius), but there's a certain kind of transition there: the signs of the coefficients of dt² and dr² switch. Externally, r is a spacelike direction and t is a timelike one, but internally, the opposite is the case. So, just like moving in the time direction is mandatory and unidirectional, one should expect that that inside the horizon moving in the r direction would be likewise.

Slightly more precisely, if one were draw light-cones on a (t,r) half-plane (which have as edges the tangent vectors of intersecting ingoing and outgoing radial light signals at that point), one finds that for r>>2m, they're quite ordinary, with edges of slopes r/t = ±1. However, they 'tip over' the closer one gets to r = 2m, with the lightcones there having one edge on the horizon itself. This means the best one can do in moving outward is to stay right on the horizon (which involves moving at the speed of light); crossing over takes one into a region where all the light-cones point toward the singularity.

[Ariphaos]

Once you are inside the horizon, distance towards the singularity is timelike - it is in your future, you cannot escape it, you cannot go 'backwards'. The only thing turning on your engines does is make you reach the singularity faster, because -every direction- leads to the singularity.

[Ariphaos]

And of course Kuroneko beats me with a better answer -_-

[Surlethe]

This is where the idea that space itself flows into a singularity comes from. If you're in a ship inside the event horizon and you point your engines toward the singularity and blast them as hard as you can, you're still going to fall toward it (faster, in fact, than otherwise). It's like trying to swim against an overpowering current.

It's also where you get that space and time flow out of the Big Bang singularity. Just as a black hole singularity is always in your future when you're inside the event horizon, the Big Bang singularity is always in your past when you're inside its event horizon (which the entire universe is).

[NoXion]

It's also where you get that space and time flow out of the Big Bang singularity. Just as a black hole singularity is always in your future when you're inside the event horizon, the Big Bang singularity is always in your past when you're inside its event horizon (which the entire universe is).

Doesn't that mean that the interior of an event horizon is not causally connected to the rest of the universe, or am I misunderstanding you?

[Enola Straight]

IIRC, photons DO escape from a singularity...albeit infinitely redshifted and thus invisible.

The strength of gravity is apparently independent of the frequency/wavelengths of the gravitic field's component gravitons.

[HRogge]

IIRC, photons DO escape from a singularity...albeit infinitely redshifted and thus invisible.

No, they don't... light cannot escape the event horizon.
(and no, "Hawkins Radiation" is not light coming out of the black hole)

[Wyrm]

It's also where you get that space and time flow out of the Big Bang singularity. Just as a black hole singularity is always in your future when you're inside the event horizon, the Big Bang singularity is always in your past when you're inside its event horizon (which the entire universe is).

Doesn't that mean that the interior of an event horizon is not causally connected to the rest of the universe, or am I misunderstanding you?

If by "causally connected" you are restricted to events in the absolute future of an event within the horizon. The absolute past of the event can still include events outside the horizon — you can still be influenced by events outside the horizon at any point. This is why a black hole can grow.