general relativity - Why is a black hole black?

In general relativity (ignoring Hawking radiation), why is a black hole black? Why nothing, not even light, can escape from inside a black hole? To make the question simpler, say, why is a Schwarzschild black hole black?

Answer

It's surprisingly hard to explain in simple terms why nothing, not even light, can escape from a black hole once it has passed the event horizon. I'll try and explain with the minimum of maths, but it will be hard going.

The first point to make is that nothing can travel faster than light, so if light can't escape then nothing can. So far so good. Now, we normally describe the spacetime around a black hole using the Schwarzschild metric:

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

but the trouble is that the Schwarzschild time, $t$, isn't a good coordinate to use at the event horizon because there is infinite time dilation. You might want to look at my recent post Why is matter drawn into a black hole not condensed into a single point within the singularity? for some background on this.

Now, we're free to express the metric in any coordinates we want, because it's coordinate independent, and it turns out the best (well, simplest anyway!) coordinates to use for this problem are the Gullstrand–Painlevé coordinates. In these coordinates $r$ is still the good old radial distance, but $t$ is now the time measured by an observer falling towards the black hole from infinity. This free falling coordinate system is known as the "rainfall" coordinates and we call the time $t_r$ to distinguish it from the Schwarzschild time.

Anyhow, I'm going to gloss over how we convert the Schwarzschild metric to Gullstrand–Painlevé coordinates and just quote the result:

$$ds^2 = \left(1-\frac{2M}{r}\right)dt_r^2 - 2\sqrt{\frac{2M}{r}}dt_rdr - dr^2 -r^2d\theta^2 - r^2sin^2\theta d\phi^2$$

This looks utterly hideous, but we can simplify it a lot. We're going to consider the motion of light rays, and we know that for light rays $ds^2$ is always zero. Also we're only going to consider light moving radially outwards so $d\theta$ and $d\phi$ are zero. So we're left with a much simpler equation:

$$0 = \left(1-\frac{2M}{r}\right)dt_r^2 - 2\sqrt{\frac{2M}{r}}dt_rdr - dr^2$$

You may think this is a funny definition of simple, but actually the equation is just a quadratic. I can make this clear by dividing through by $dt_r^2$ and rearranging slightly to give:

$$ - \left(\frac{dr}{dt_r}\right)^2 - 2\sqrt{\frac{2M}{r}}\frac{dr}{dt_r} + \left(1-\frac{2M}{r}\right) = 0$$

and just using the equation for solving a quadratic gives:

$$ \frac{dr}{dt_r} = -\sqrt{\frac{2M}{r}} \pm 1 $$

And we're there! The quantity $dr/dt_r$ is the radial velocity (in these slightly odd coordinates). There's a $\pm$ in the equation, as there is for all quadratics, and the -1 gives us the velocity of the inbound light beam while the +1 gives us the outbound velocity. If we're at the event horizon $r = 2M$, so just substituting this into the equation above for the outbound light beam gives us:

$$ \frac{dr}{dt_r} = 0 $$

Tada! At the event horizon the velocity of the outbound light beam is zero so light can't escape from the black hole. In fact for $r < 2M$ the outbound velocity is negative, so not only can light not escape but the best it can do is move towards the singularity.