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I always think that it is not possible to escape from within event horizon. However, some one recently told me with deep conviction that it is possible with sustained energy output. I countered with calculations in Schwarzschild metric showing that any objects below event horizon in the simplistic black hole will hit the singularity within a finite proper time, and was met with the response that Schwarzschild metric was a bad choice for coordinate system.

Now I understand that the singularity at $r=\frac{2GM}{c^2}$ in Schwarzschild metric is an artifact of the coordinate choice, and thus it seems that he could have a point. So my question is, can one prove that it is (or not) possible to escape from within the event horizon of a simple static black hole using a better metric?

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4 Answers 4

up vote 4 down vote accepted

See Why is a black hole black?

It's quite true that the Schwarzschild coordinates misbehave near the horizon, but there are plenty of other coordinate systems we can use. My answer to the question I've linked above uses Gullstrand-Painlevé coordinates, but you can also use Eddington-Finkelstein or Kruskal-Szekeres coordinates. The conclusion is the same in all coordinate systems - once you have crossed the event horizon there is no way back.

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The other person is likely thinking about the concept of escape velocity. There are people who claim no man-made spaceship could ever have gone into space because they cannot reach the escape velocity for the earth. What they fail to acknowledge is that the escape velocity for the earth is not a constant, it depends on how high above the surface you are.

So what you do to get into space is to reach the escape velocity at the height above the surface you are at. This is in practice obtained by maintaining an energy output until $v(t) = v_{\text{escape}}(h)$ where $h = y(t)$ is the vertical distance of the spacecraft to the earth's surface at time $t$. (see also this answer) When that condition is fulfilled, no additional thrust is needed and the spacecraft will escape the earth.

However, things are different in the case of a black hole. Within the event horizon the (timelike) geodesics of Schwarzschild spacetime are closed curves leading inevitably to the singularity at $r=0$. A clear picture is painted by the Kruskal spacetime diagram:

enter image description here

Here, the event horizon $r=2GM$ is a diagonal line bisecting the right angle between the Kruskal coordinate axes $X$ and $T$ and the singularity $r = 0$ is contained within the blue region marked $II$. The benefit of using Kruskal coordinates to describe Schwarzschild's spacetime is that (radial) light cones are defined by $X = \pm T$ and therefore the causal structure of spacetime is very clear.

Indeed, any observer inside the blue region of spacetime is doomed. Their light cones are completely contained within this region and the singularity is unavoidable. Another interesting way to see this is by simply looking at the Schwarzschild metric:

$$ds^2 = -\left(1-\frac{2GM}{r}\right) \text{d}t^2 + \left(1-\frac{2GM}{r}\right)^{-1} \text{d}r^2 + r^2 d\Omega_2^2$$

where $d\Omega_2^2$ is the metric on a unit two-sphere (don't worry about this part, it contains angular coordinates only and we're interested in radial curves, i.e. curves without angular dependence).

Shamelessly ignoring the coordinate singularity at $r=2GM$, let's see what happens on either side of the event horizon. We notice that for $r>2GM$ this metric has signature (-+++). But when $r<2GM$, the factor in front of $\text{d}t$ becomes positive while that in front of $\text{d}r$ becomes negative. So for $r<2GM$ the time-coordinate becomes spacelike and the space-coordinate becomes timelike! Thus you can no more stop yourself from moving toward the singularity than you can stop yourself from getting older.

Outside the event horizon (region $I$ in the above diagram), a (Schwarzschild) black hole behaves like any other celestial body and you can perfectly well have stable orbits and talk about escape velocities in this region of spacetime. But inside the event horizon there isn't even a notion of escape velocity, since it is defined as the velocity needed to reach infinity without additional forces and everything inside the event horizon is completely cut off from spacelike infinity.

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The causal structure of space time, here that of the Schwarzschild-metric, does not depend on the choice of coordinates. What you say is in fact tautologically true. The event horizon is defined as the boundary of the region from within which you cannot escape.

I recommend you look at the Penrose-Carter diagram of the Schwarzschild-metric, which depicts its causal structure. You see there very clearly that once you've crossed the horizon there is no timelike or even lightlike curve that will get you anywhere except into the singularity. To avoid hitting the singularity, you would have to move on a spacelike curve, which isn't possible regardless of how much energy you use. Again, these statements are entirely independent of the choice of coordinate system.

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I'm thinking that they're looking at an extended Kerr diagram and stating that every past horizon is paired with a future horizon, where it is possible to escape back out to a "black hole exterior":

http://jila.colorado.edu/~ajsh/insidebh/penrose_kerr.gif

It should be stated, however, that this exterior is a different one than the one from which you entered, and the boundary conditions that led to the formation of the black hole, specifically, the fact that the hole formed out of collapsing matter and will eventually decay into hawking radiation, will kill the magic, and you won't even get this structure of escaping back out into something that has a conformal infinity again (other than just escaping the black hole after it has evaporated).

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