# Lay question about formation of singularity (black hole)?

I have never been able to quite grasp the following- in the collapse of a stellar body that meets the requirements for the creation of a black hole...Upon the initiation of core collapse, spacetime is deformed infinitely, thus according to general relativity light is unable to escape at or beyond event horizon due to warping...This is how it has been explained to me. Wouldn't, from an observational point of view, the collapsing body forever appear to be approaching infinity, rather than be infinitely small, due to the stretching of time (relative) itself approaching zero proportionally with gravitational warping approaching infinite? Wouldn't i only be able to confirm that there is an actual singularity if I were infinitely long lived, or if I entered the black hole? I may misunderstand general relativity (and special relativity, as the speed of collapse would have to increase to c, wherein gen and special relativity meet in regards to falling bodies/gravitation and time dilation), but does a black hole ever actually become a singularity or is it forever approaching it?

• May 26, 2017 at 12:39

You are correct in assuming that an outside observer would never the see collapse all the way through.

Let's look at the Schwarzschild solution, the simplest black hole solution: $$ds^2 = -\left(1-\frac{2M}{r}\right) dt^2 + \left(1 - \frac{2M}{r}\right)^{-1} dr^2 + r^2d\theta^2 + r^2 \sin^2 \theta d\phi^2$$ These coordinates approach flat space as distance goes to $\infty$. Hence, they are adapted to an observer far away from the horizon.

If we want to see how light rays (radially incoming or outgoing) behave in these coordinates, we can solve the equation above with $ds^2 = 0$ and get

$$\frac{dt}{dr} = \pm \left(1- \frac{2M}{r}\right)^{-1}$$

which can be nicely drawn as:

You can see how the light cones, which are bounded by the light rays, begin squeezing, the light ray coming from very near the horizon would therefore to a distant observer be infinitely redshifted.

The solution to understanding the singularity comes in changing the coordinates. Notice that the solution above has a divergence in $r=2M$. This divergence is only due to the choice of coordinates.

We settle this first by doing $$r^*=r + 2M \ln \left(\frac{r}{2M} - 1\right)$$ to get the metric to the following form: $$ds^2 = \left(1 - \frac{2M}{r}\right)\left(-dt^2 + dr^{*2}\right) + [r(r^*)]^2 d\Omega^2$$

Now there is no divergent behavior, and the radial light rays are now described by:

$$\label{Tortoise} t= \pm r^* + \text{const.}$$

The only problem now is that the event horizon is now moved to $r^*(2M)=-\infty$.

Now we sit on the light rays by the following change of coordinates:

$$v=t+r^*\nonumber\\ u= t - r^*$$

A constant $v$ represents light rays going inwards, and a constant $u$ represents light rays going outwards.

If we now choose to use the original coordinate $r$ and instead of the original coordinate $t$ use $v$, then the metric looks like: $$ds^2 =- \left(1 - \frac{2M}{r}\right)dv^2 + (dv dr + dr dv) + r^2 d\Omega$$

These coordinates $(v, r, \theta, \phi)$ are called Eddington-Finklestein coordinates.

The equation for radial light rays is now

$$\frac{dv}{dr}= \begin{cases} 0 \text{ for incoming light ray}\\ 2\left(1-\frac{2M}{r}\right)^{-1} \text{ for outgoing light ray} \end{cases}$$

Which can be nicely drawn as:

Where you can see how after a certain $r$, all light rays, and all geodesics (thus all particles) must hit the singularity.

So if we were the distant observer, the particles we observe going to the horizon will to us appear never to get there. But if follow the path through the horizon (which is here represented by a suitable coordinate change), we will see they will go through and inevitably reach the singularity.