Why is matter drawn into a black hole not condensed into a single point within the singularity?

Question

When we speak of black holes and their associated singularity, why is matter drawn into a black hole not condensed into a single point within the singularity?

Answer 1

The many comments have covered the main points about the question, but I thought it would be worthwhile explaining how the behaviour is calculated. If we solve the Einstein equation for a point mass we get the Schwarzschild metric:

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

All equations look scary to non-nerds, but don't worry too much about the details. The key points are that the equation involves time, $dt$, and the distance from the centre of the black hole, $dr$, and it calculates a quantity called the line element, $ds$.

The time, $t$, and radial distance, $r$, are the physical quantities measured by an observer outside the black hole, i.e. that's you and me. They are exactly what you would think i.e. the time is what we measure with a stopwatch. The radial distance is what you'd get by measuring the circumference of a circle round the black hole and dividing by $2\pi$ (because the circumference of a circle is $2\pi r$).

The line interval, $ds$, is a bit more abstract but for our purposes $ds$ is the time as measured by someone falling into the black hole. This is called the proper time and usually written as $\tau$. You've probably heard that time slows down as you approach the speed of light, and you get a similar effect here (as dmckee menioned in a comment). That means the time measured by someone falling into the black hole, i.e. the proper time $\tau$, is not the same as the time $t$ that we measure when watching the black hole from the outside.

The point of all this stuff is that you can use the metric to calculate how long it takes to fall from some distance, $R$, outside the black hole to the event horizon. First of all let's calculate this for the person falling into the black hole. This means we have to calculate the proper time, $\tau$. You can find this in any book on GR, or by Googling, and the result is:

$$ \Delta \tau = \frac{2M}{3} \left[ \left( \frac{r}{2M} \right)^{3/2} \right] ^R _{2M} $$

Again, don't worry about the detail. A long as we know the mass of the black hole, $M$, and the starting distance, $R$, we just feed these into a calculator and it gives us $\Delta\tau$ which is the time measured by the person falling into the black hole. The time obviously depends on how far away you start and how big the black hole is, but it's just a number of seconds. In fact if we use $r = 0$ in the expression about we can calculate how long it takes to fall through the event horizon and on to the singularity at the black hole.

So, the point to note down at this stage is that the person falling into the black hole reaches the event horizon and inded the singularity in a finite time.

The next step is to calculate the time measured by you and me sitting outside the black hole. This is a bit more involved, but we end up with an expression:

$$ dt = \frac {-(\epsilon + 2M)^{3/2}d\epsilon} {(2M)^{1/2}\epsilon} $$

where $\epsilon$ is the distance from the event horizon i.e. $\epsilon = r - 2M$. Integrating this to find the time to reach the event horizon is a bit messy, but if restrict ourselves to distances very near the event horizon we find:

$$\Delta t \propto ln(\epsilon)$$

but $\epsilon$ is the distance from the event horizon, so it's zero at the event horizon and $ln(0)$ is infinity. That means that the time you and I measure for our astronaut to reach the event horizon, $\Delta t$, is infinity.

And this is why you get the apparently paradoxical result about falling into black holes. The time measured by the person falling in, $\Delta\tau$, is finite but the time measured by people outside the black hole, $\Delta t$, is infinite.