In Caroll's Spacetime and Geometry, page 227, he says that from the Schwarzschild metric, you can see than from inside a black hole future events all lead to the singularity. He says you can see this because for $r<2GM$, t becomes spacelike and r becomes timelike. I don't understand his reasoning though. Why does r being timelike mean you can only travel towards the singularity?

  • $\begingroup$ It seems like a glorified way of saying that a geodesic travels faster than the speed of light, therefore, not even light, on the edge of the future cone, can increase its distance from the singularity. $\endgroup$ – Alan Rominger Mar 22 '13 at 18:04
  • $\begingroup$ Can you expand on this? I have a rough idea of what you mean, but am unsure. $\endgroup$ – JLA Mar 22 '13 at 18:11
  • $\begingroup$ See physics.stackexchange.com/questions/28297/… for the restricted case of radial trajectories $\endgroup$ – John Rennie Mar 22 '13 at 18:15
  • $\begingroup$ That's a pretty good answer, though I still don't see how you can see this from the Schwarzschild metric. $\endgroup$ – JLA Mar 22 '13 at 18:37

A good way to understand this is to delve a little more deeply into the meaning of timelike coordinates in flat space, then move up to Schwarzschild space. In flat space, we have the metric

$$ds^2 = -dt^2 + dx^2+dy^2+dz^2. $$

Looking at the metric, we can see that only one of the spacetime coordinate differentials gets a negative sign ($dt^2$). Looking closer, you can see that this term is the only one that will contribute negatively to the metric line element, no matter the values of $dx, dy, dz$, or $dt$. In general, you can think of the timelike coordinate as the one that will contribute negatively to the metric line element.

What makes a coordinate timelike? A good way to think about this question is to think about how time differs from the other spacetime coordinates we are used to in flat space. For us, time can only move forward (as @elfmotat says, you can't avoid getting older), while we can move freely in the other spacetime coordinate directions. So, for particles, if a coordinate contributes negatively to the line element, then you can only move forward in the direction of that coordinate.

To motivate this, I'm just going to appeal to your intuition for lightcones. In Minkowski Space:

enter image description here

Particles are forced to follow timelike geodesics $(ds/d\tau)^2 =-1$ , all of which lie within the future lightcone in the diagram. So the statement that the coordinate that contributes negatively to the metric line element is timelike is kind of a geometrical one. Basically, it confines you to a certain region of spacetime and in so doing only allows forward motion.

Now let's look at the $r, t$ lightcone around a black hole:

enter image description here
(source: asymptotia.com)

Once you've crossed the event horizon, all worldlines move backward in $r$. So we say that $r$ becomes timelike while $t$ becomes spacelike.

In the Schwarzschild metric,

$$ds^2 = -\left(1-\frac{2GM}{r}\right)dt^2 + \left(1-\frac{2GM}{r}\right)^{-1}dr^2+r^2d\Omega ^2, $$ We can see that for $r > 2GM$, the terms in the parenthesis are positive, and therefore the timelike coordinate is $dt$, since it already has a negative sign outside it. But for $r < 2GM$, the terms in the parenthesis are negative, so the time-spacelikeness of $dt, dr$ flip, and $dr$ becomes the only coordinate that contributes negatively to the metric line element. What this means physically is that you can only move forward in $r$, so in that sense $r$ becomes a timelike coordinate. A bit more pondering should reveal that this mathematical treatment of the problem reproduces the well-known black hole behavior exactly. (Can't go back when you've crossed the event horizon).

  • $\begingroup$ I see, however I don't see why the coordinate that gets a negative sign in the metric necessarily implies you can only move in one direction in that coordinate. On the other hand, can't I apply the same reasoning as you did to the time coordinate to infer that inside a black hole you can freely travel back and forth in time? $\endgroup$ – JLA Mar 22 '13 at 18:35
  • $\begingroup$ Actually, yes! The time-like and space-like ness of t and r flip, so you actually gain freedom to "move" in coordinate time. But again, remember that the coordinate time t is just a coordinate, and not what you're conventionally used to thinking of as time. $\endgroup$ – dannygoldstein Mar 22 '13 at 18:57

Objects that follow timelike trajectories always have a unique timelike coordinate associated with each point on the trajectory. This is not true for spacelike trajectories, where a Lorentz transformation can render two points on the path simultaneous.

Hence, an object on a timelike trajectory within the event horizon can only cross a specified radius once along its trajectory. This means that it must either fall all the way inward without stopping or turning back; or it can be completely ejected without stopping or turning back. These two possibilities are mutually exclusive--if one is possible, the other is not (because time only flows in one direction) for a given system. The former describes a black hole; the second a "white" hole.

  • $\begingroup$ To add to this, you can't avoid falling toward the singularity any more than you can avoid growing older. $\endgroup$ – Jold Mar 22 '13 at 18:26
  • $\begingroup$ Since the timelike and spacelike dimensions flip inside a black hole, can you avoid getting older when you cross the event horizon? Even more, can you reverse your aging (just as you can turn around in space)? $\endgroup$ – markovchain Mar 23 '13 at 13:20
  • $\begingroup$ Remember this is just a coordinate singularity, not a physical one. The geodesics are entirely cobntinuous. It is merely the coordinates associated with forward progress along those geodesics that change roles. Even when $r$ is a timelike coordinate instead of $t$, you are still going forward in time. What's happening is more of a failure of the coordinate to fit the geometry. There's still a well-defined notion of time (that, as has been pointed out, does not always align with $t$). $\endgroup$ – Muphrid Mar 23 '13 at 15:52

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