# Black holes suck

In the last post, we introduced the Schwarzschild spacetime. This spacetime models the gravity outside of a spherically-symmetric, unchanging star.

In spherical spacetime coordinates, the metric for this spacetime is $ds^2 = -\left(1-\dfrac{2M}{r}\right)dt^2 + \left(1-\dfrac{2M}{r}\right)^{-1} dr^2 + r^2 (d\phi^2 + \sin^2(\phi)d\theta^2)$.

The letter $M$ represents the mass of the star in the middle, in appropriate “geometric” units1. For our sun, for instance, this mass is about 1.5 kilometers.

This metric works just fine… until you get down to $r=2M$. There, you divide by zero!

Usually this isn’t a problem. The spacetime is only valid outside the star, and the sun’s radius is a bit bigger than 3 kilometers. But we can imagine something so compact that all the mass fits within that radius of $2M$, the Schwarzschild radius.

What happens then? Is spacetime ripping apart?

Certainly, as $r$ approaches $2M$, the metric blows up (i.e., goes to infinity). That’s bad. But does it mean the spacetime and metric completely break down there?

For a long time, physicists and mathematicians would have agreed with that. They thought the spacetime metric simply broke down at $r=2M$, and there was no sense in looking at the $0 part of the spacetime. Some claimed that nothing could collapse to be so dense that the radius of the star could be less than $2M$ anyway.

But this turns out to be wrong.

If the metric were irredeemable at the Schwarzschild radius, you would expect something horrible to happen. For instance, you might expect gravity to be so strong that you are ripped into pieces or mashed into a tiny ball or stretched into spaghetti.

In relativity, gravity is represented by the bending of spacetime, or, in other words, by the curvature.2 So, if one of these horrible things happens, some part of the curvature needs to go to infinity.

A convenient way to measure total curvature is the Kretschmann scalar, $K$. The Kretschmann scalar is a positive (or zero) number that essentially adds up all of the possible types of curvature. If the scalar is exactly zero at a point of a spacetime, that means the spacetime is perfectly flat there, with no curvature at all. If the Kretschmann scalar goes to infinity, some kind of curvature is also going to infinity.

A curvature singularity in a spacetime is somewhere where the Kretschmann scalar blows up. In other words, curvature singularities are where gravity grows infinitely strong.

If the Schwarzschild metric is irredeemably broken at the Schwarzschild radius, $r=2M$, we would expect there to be a curvature singularity there.

Fortunately, the Kretschmann scalar is fairly straightforward to calculate. For the Schwarzschild metric (using geometric units), the Kretschmann scalar is $K = \dfrac{48 M^2}{r^6}$.

The only place that $K$ blows up is at the center of the spacetime, $r=0$!

In other words, something horrible would happen if you ever got close to $r=0$. The near-infinite gravity would tear you apart in one way or another.

However, there is no singularity at $r =2M$! While something interesting is going on at the Schwarzschild radius, there isn’t infinite gravity or anything unpleasant like that going on there.

So what is going on at $r=2M$? Why does the metric seem to blow up there?

To see what’s happening there, let’s consider a ray of light.

If we shoot a ray of light at the sun, the light should hit the sun some amount of time later. If the radius of the sun was close to the Schwarzschild radius, $r=2M$, it might take a bit longer (farther to travel), but it shouldn’t be that much longer.

Rays of light, as we talked about earlier, travel along null geodesics. While generally it’s difficult to calculate the exact path of a null geodesic, for the Schwarzschild spacetime it’s much easier.

While it’s not hard, I’ll leave the technical explanation to a footnote.3 If we solve the correct equations, a ray of light shot directly toward the sun must travel along a path of the form $(r,t) = (r, C-r-2M \ln(r-2M))$, where $C$ is some number depending on how far away from the sun you started. (We can ignore the $\phi$ and $\theta$ coordinates because the light is traveling directly towards the center of the spacetime, so these coordinates are not changing.)

Something weird happens when $r$ gets close to $2M$. The $t$ coordinate for our light ray goes to infinity!4 But you’d expect light to eventually get to whatever’s in the middle!

Whatever the $t$ coordinate represents, it is not the time it takes to fall into the star!

(The $t$ coordinate actually does represent something important. If you were to push Hitler out the airlock, his $t$ coordinate roughly represents where you would see him at your time $t$. [It’s not exactly this, but you get the idea.] In other words, you’d never actually see Hitler get to $r=2M$, even though he would get there in a finite amount of (his) time. I’ll probably talk more about this in the next post. Spoilers!)

It appears that whats happening is that the time coordinate somehow gets stretched infinitely much near the Schwarzschild radius. The basic idea to fix the metric is to compress the time coordinate back down.

To do this, we will do a change of coordinates. Remember, a change in coordinates does not change the spacetime; it simply changes its presentation.

Since the time coordinate is going to infinity near the Schwarzschild radius, we want to create a new time coordinate, with that pesky, infinite $\ln(r-2M)$ term canceling out.

So, let’s define a new time coordinate “t bar”: $\bar t = t + 2M\ln\left|\frac{r}{2M} -1\right|$. If we use this coordinate change to rewrite the Schwarzschild metric, the metric can be written as $-d\bar t^2 + dr^2 + \dfrac{2M}{r}(d\bar t + dr)^2+ r^2 (d\phi^2 + \sin^2(\phi)d\theta^2)$.5

Notice, in these coordinates, the metric doesn’t go to infinity near $r=2M$. The Schwarschild spacetime is not really singular in any way at $r=2M$; it was simply an artifact of an unfortunate choice of coordinates!

While this isn’t too surprising, since the curvature was so well behaved there, it is nice to confirm that the metric and spacetime are nice everywhere except at the curvature singularity at $r=0$.

In these coordinates, what do the paths of light look like?

We can use a similar trick to the one we used before to find the paths of light.6 If we shoot one ray of light directly towards and one directly away from the center of the spacetime, the paths look like this.

Note that the outgoing light doesn’t travel at 45 degrees, since the spacetime is warped by gravity. But, otherwise, the light is traveling as we expect. The light shooting toward the center reaches the Schwarzschild radius, $r=2M$, and even the center, $r=0$, in finite time. The light traveling away from the center escapes to infinitely far away from the center.

Of course, if we started our light closer or farther away from the center, the paths would look different. So let’s draw a few more of these paths.

For light that starts outside the Schwarzschild radius, everything behaves like normal. Except, as you get closer to $r=2M$, the outgoing light ray seems to just barely escape to infinitely far way.

Right at the Schwarzschild radius is where the weird things start happen.

The ingoing light ray takes a perfectly normal path. But the “outgoing” light ray stays right on the Schwarzschild radius! And for any light starting from within the Schwarzschild radius, both the ingoing and the outgoing light rays fall into the singularity at $r=0$!

Even light, the fastest of all things7, cannot escape from the center of the spacetime! Spacetime is so bent inside the Schwarzschild radius that everything eventually gets sucked into the singularity at $r=0$.

Once a star becomes so compressed that all of its mass sits within its Schwarzschild radius, no light can ever escape!

That is why we call something this compact a black hole.

Another term you might have heard of is event horizon. A horizon in every day life is the edge of the part of the world you can see. In other words, there is more world beyond the horizon, you just can’t see it without going there.

For a (Schwarzschild) black hole, we call the sphere at the Schwarzschild radius $r=2m$ an event horizon. If you stay outside the event horizon (i.e., outside the black hole), you can never see any object or event that occurs inside the event horizon. Any messages someone tried to send from inside the horizon would just be bent back and fall into the singularity a bit after that person fell in.

What goes on in a black hole, stays in a black hole.

And then gets eaten by the singularity.

1. To remind you, geometric units means that both the speed of light $c$ and the gravitational constant $G$ are set to be 1 (unitless). This ends up meaning that length, time and mass are all measured in units of meters.
2. You can read our introduction to general relativity here (part 1) and here (part 2)
3. A null path is one for which the tangent vector is null, i.e., for which $ds^2 = 0$. If we say that the path is only traveling in the $(r,t)$ directions, we can assume that $d\phi^2$ and $d\theta^2$ are zero. If we rearrange the terms of the metric, we find that $\left(\frac{dt}{dr}\right)^2 = (1-2M/r)^{-2}$. (Technically, you can’t really “divide by $dr$.” If you are just a bit more careful, though, you’ll get the same equation more rigorously.) That differential equation is easy to solve, giving $t(r) = C\pm(r+2M\ln|r-2M|)$
4. As a bit of review, the natural log, $\ln(x)$, goes to $-\infty$ when $x$ gets close to 0. In our equation, we have minus natural log, so it goes to infinity.
5. These coordinates are a variation of the Eddington-Finkelstein coordinates
6. This time, we look at $-d\bar t^2 + dr^2 + \dfrac{2M}{r}(d\bar t + dr)^2 = 0$. The easiest solution is if $\bar t = - r + r_0$, for any constant $r_0$. These represent the ingoing rays. The outgoing rays are a bit trickier in these coordinates, but you can check that $(r, \bar t) = (r, C+r+4M \ln|r-2M|)$ works.
7. Hypothetical tachyons aside.