Light Cones in Black Holes

1. Jun 6, 2006

discjockey

What exactly does it mean in practical terms when light cones narrow and tip over while they are approaching the event horizon of a black hole? Does this mean that the object has no choice but to eventually travel at the speed of light? Or am I totally wrong?

2. Jun 6, 2006

coalquay404

Actually, the fact that light cones "close up" as they approach the event horizon is evident only in certain coordinate systems. If you take standard Schwarzschild coordinates, the metric for the Schwarzschild black hole can be written as

$$ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \left(1-\frac{2M}{r}\right)^{-1} dr^2 + r^2 d\Omega^2$$

where $$d\Omega^2$$ is the standard metric on $$S^2$$. In these coordinates, showing that the light cones "close up" is easy. All that you need to do is to consider radial null geodesics (that's a fancy way of saying that we're looking at the path of light rays aimed directly at the centre of the black hole), for which

$$ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \left(1-\frac{2M}{r}\right)^{-1} dr^2 = 0$$.

You can fiddle around with this to show that the slope of the light cones close up as $$1/r$$ as $$r\to 2M$$. The interesting thing is that if you consider a different set of coordinates, such as one where you define a new radial coordinate $$r^*$$ by

$$r^* \equiv r + 2M\log|r-2M|$$

then you see that the light cones never appear to close up.

Physically, what it means is that signals sent from an infalling body will take longer and longer to reach an observer a very large distance away from the black hole. The infalling body never travels at the speed of light; it will simply continue past the event horizon and travel onwards to the singularity. However, an observer at a very large distance won't be able to see this happen since he will see the body appear to freeze just outside the horizon (it appears to freeze because it takes longer and longer from light to reach the observer). There's a more mathematical reason for this too: the event horizon(s) of a black hole are something called null surfaces. Have a look, for example, at Carroll's book or his online lecture notes for a pretty nice discussion of this.

Once the body goes past the event horizon its light cone does indeed "tip over." In practical terms, this means that there's no way for it to escape hitting the singularity at $$r=0$$ since the singularity lies in all future-directed paths. This is what is meant by saying that there's no escape from a black hole.

It's also useful to note that the situation I've described above isn't common to all black holes. The Reissner-Nordstrom solution, for example, contains a scenario where the event horizon around the singularity doesn't exist. Technically, one could approach arbitrarily close to the singularity without ever entering it. (This isn't really of any physical use since everyone believes that the cosmic censorship hypothesis is true and hence that the $$M<|Q|$$ version of the Reissner-Nordstrom solution I've mentioned here can't actually exist.)

Last edited: Jun 6, 2006
3. Jun 6, 2006

pervect

Staff Emeritus
It means that objects inside the event horizon can't sit still. I wouldn't say that objects have to go "at the speed of light", because light will always travel on the lightcone, and objects will always have to travel inside the light cone. Thus physical objects will never be able to quite travel on the light-cone, though they can get arbitrarily close by accelerating.

Note that in most coordinate systems, the slope of the light cone in/near a black hole is not constant, therfore the "speed" of light in most coordinate terms is not constant. The speed of light according to local observers falling into the black hole is still constant, of course, even though the coordinate speed of light varies.

Note that both outgoing and ingoing light rays get closer and closer to the singularity (r decreases) as time goes on when the light rays start inside the event horizon.