# Why is the Schwarzschild radius the radius of an event horizon?

I've been searching the web and many references without much success. My question is how do we know that, in the Schwarzschild black hole solution, the surface with coordinate $r=2M$ (in the geometric unit system) defines an event horizon, in the sense that every timelike or null curve that crosses that surface - to the interior - will end up in the singularity at $r=0$?

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Remember that the definition of an event horizon is "the past boundary of future null infinity". What this means, in common language is -- take all of the light rays that escape to infinity. Then, find the one that just barely doesn't make it back out to infinity. The surface formed by these light rays is the event horizon.

Now, look at a the Kruskal diagram of the Schwarzschild spacetime (this is the Schwarzschild spacetime written in coordinates that don't go singular at the horizon). The solid black diagonal lines instersecting the middle correspond to $r = 2M$. They are obviously the asymptotes of the curves at the top and the bottom representing $r=0$, so $r=2M$ is the event horizon of Schwarzschild spacetime. (well, technically, the upper triangle formed by the $r=2M$ surfaces is the "future" event horizon of the spacetime, and the bottom triangle is the "past" event horizon of the white hole, but you probably don't care about pedantry like that.)

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Also, for the second part of your answer, note that this coordinate system is defined in such a way that future pointing null lines always point in either 45${}^{\circ}$ to the left or right of the vertical, which means that timelike lines are always constrained to be at a smaller angle with the vertical. From this, it should be obvious that once you're inside the horizon, you need to be spacelike to leave the interior of the horizon. – Jerry Schirmer Aug 13 '14 at 17:49

In Schwarzschild coordinates, if you look at the $g_{tt}$ and $g_{rr}$ parts fo the metric, they flip signs at $r=2M$. Therefore for $r<2M$ the $r$ direction is timelike and the $t$ direction is spacelike. The future-timelike light cone of any event inside the horizon points toward smaller values of $r$.

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In Schwarzschild coordinates, $r=2M$ is a coordinate singularity, not the event horizon. There are no geodesics that cross it, and it isn't really clear that the interior region has anything to do with the exterior region. – benrg Aug 13 '14 at 17:26
@benrg: the coordinates blow up, but you have a finite travel time to the horizon, as measured in proper time (just not as measured in coordinate time): academic.reed.edu/physics/courses/Physics411/html/411/page2/… – Jerry Schirmer Aug 13 '14 at 17:30
@JerrySchirmer, I know, but as an intuitive explanation this doesn't really work. It would make more sense in Eddington-Finkelstein coordinates. Even then it's not clear how to show finite travel time, but the question didn't ask for that. – benrg Aug 13 '14 at 17:35
@benrg: the question says "geometric coordinate system", which gives us an interesting question of whether they're referring to Schwarzschild or Kruskal coordinates. I'm definitely happier answering this question in Kruskal coordinates, certainly, since the answer is obvious in them. – Jerry Schirmer Aug 13 '14 at 17:44

I'll attempt an answer from a different perspective from the rest.

The Schwarzschild solution is the vacuum solution for a static, spherically symmetric spacetime.

The Schwarzschild coordinates are 'nice' for the solution for at least two reasons:

(1) the line element far away from the spatial origin, in these coordinates, approaches the line element of flat spacetime in spherical coordinates

(2) the surface area of a sphere at radial coordinate $r$ is $4\pi r^2$

However, it turns out that the coordinates do not cover the entire spacetime since the requirement that the spacetime is static cannot be satisfied for the entire spacetime.

This is essentially the reason that there is a coordinate singularity at $r = 2M$, the boundary between the exterior region, where the geometry is static, and interior region where the geometry is dynamic.

There is a coordinate transformation from the Schwarzschild $r, t$ coordinates to the Kruskal–Szekeres $U, V$ coordinates as follows:

For the exterior region,

$$V = \left(\frac{r}{2M} - 1\right)^{1/2}e^{r/4M}\sinh\left(\frac{t}{4M}\right)$$

$$U = \left(\frac{r}{2M} - 1\right)^{1/2}e^{r/4M}\cosh\left(\frac{t}{4M}\right)$$

and, for the interior region,

$$V = \left(1 - \frac{r}{2M}\right)^{1/2}e^{r/4M}\cosh\left(\frac{t}{4M}\right)$$

$$U = \left(1 - \frac{r}{2M}\right)^{1/2}e^{r/4M}\sinh\left(\frac{t}{4M}\right)$$

In these coordinates, the line element is

$$ds^{2} = \frac{32M^3}{r}e^{-r/2M}(-dV^2 + dU^2)$$

where $r$ is implicitly defined by

$$V^2 - U^2 = \left(1-\frac{r}{2M}\right)e^{r/2M}$$

Thus, in these coordinates, there is no coordinate singularity at $r=2M$. Indeed, we have

$$U^2 = V^2$$

for $$r = 2M$$

But, from the line element, we see that when

$$dU = \pm dV$$

the interval is null (light-like) thus, world lines with $U^2 = V^2$ are light-like; the surface $r=2M$ is a null surface, i.e., lies on a lightcone. Only massless entities can remain at $r=2M$.

There is a true spacetime singularity for $r=0$. In the $U,V$ diagram, $r=0$ corresponds to the hyperbola

$$V^2 - U^2 = 1$$

The asymptotes are the $r=2M$ light-cone. It immediately follows that the singularity is in the future of any world-line in the interior region.

Below is an image, from MTW's "Gravitation", of a comparison of various geodesics of the Schwarzschild spacetime in Schwarzschild coordinates and Kruskal–Szekeres coordinates

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I believe that your answer is by far the most complete so I'm inclined to set it as the correct answer. Nevertheless, I'm having an hard time figuring out how do you know that the coordinate $U$ is spacelike for $r>2M$ and timelike otherwise. Do you conclude that from the expressions for $U$? Because I can't see it. Another problem am having is figuring out why does the fact that $U$ is timelike in the interior region implies that the light cone always point to the singularity. – PML Aug 13 '14 at 23:39
Btw, just to make sure I understand what you're saying: when you say that, for example, $U$ is spacelike you're saying that if I consider curve, say, $c$, parameterized by a parameter $\lambda$, given by the parametric equations $c(\lambda)=(c^v,c^u,c^\theta,c^\phi)$ whose tangent vector is $\dot{c}(\lambda)=(0,\dot{c^u},0,0)$ then $c^uc_u>1$? – PML Aug 13 '14 at 23:45
@PML, I fixed a typo - I want to expand and clarify this answer a bit but I'm in the middle of grilling supper right now so allow for some time. – Alfred Centauri Aug 14 '14 at 0:10
I laughed so much. Thank you for your time. I'll check Physics SE later, then. – PML Aug 14 '14 at 0:16

This what Penrose and Hawking proved with the Penrose-Hawking singularity theorems. Specifically, and I quote from the linked article:

Penrose concluded that whenever there is a sphere where all the outgoing (and ingoing) light rays are initially converging, the boundary of the future of that region will end after a finite extension, because all the null geodesics will converge. This is significant, because the outgoing light rays for any sphere inside the horizon of a black hole solution are all converging

This proves any world line crossing the event horizon must end at the singularity.

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