# Inside event horizon, space and time are reversed. How, precisely ?

1. Jun 28, 2014

In "The Hidden Reality", Brian Greene mentions almost off-hand that inside the event horizon of a black hole, space and time are reversed. But no details are given. What, precisely, does this mean? Does it mean that in one's equations, if one is using a (-,-,-,+) signature, then everything switches to a (+,+,+,-) signature? Does it mean that time splits into three dimensions and space into one?? Does it mean that one can see both past and future? That space flows like time did? Many of these possibilities would give lie to the idea in general relativity that an observer traveling towards the singularity of a black hole in free fall would not notice any difference when entering the event horizon (putting aside for the moment the enigma of the possible firewall). I would be grateful to anyone who clears up this statement with some details.

2. Jun 28, 2014

### Staff: Mentor

Ah, another misconception peddled by Brian Greene.

What he's talking about is the fact that, in a certain set of coordinates (Schwarzschild coordinates), the coordinate labeled $t$, which is timelike outside the horizon, becomes spacelike inside the horizon, and the coordinate labeled $r$, which is spacelike outside the horizon, becomes timelike inside. But that's an artifact of that particular coordinate system. There are other coordinate systems in which no such "swap" occurs.

There is one aspect of spacetime inside the horizon which could be thought of as "space" taking on an aspect of "time". The coordinate usually called $r$ can be given a coordinate-independent meaning by observing that the spacetime of a black hole is spherically symmetric, and any 2-sphere in this spacetime can be labeled by a number $r$ such that the area of the 2-sphere is $4 \pi r^2$. Outside the horizon, as one moves into the future in time, one can move in either direction relative to $r$--i.e., one can move in the direction of decreasing $r$ (2-spheres decreasing in area) or increasing $r$ (2-spheres increasing in area), or one can just stay at the same $r$.

But inside the horizon, one no longer has this freedom: moving into the future in time *requires* moving in the direction of decreasing $r$. There is no way to move in the direction of increasing $r$, or even to stay at the same $r$, if you are moving into the future in time. (Another way of saying this is that, inside the horizon, you would have to move faster than light to avoid decreasing your $r$.)

Yes, they certainly would (and kudos to you for spotting that, many people miss it). Which is why it's a good thing none of them are true.

3. Jun 28, 2014

Thanks very much, PeterDonis, for clearing that up!

4. Jun 30, 2014

### Vorpal

The standard Schwarzschild coordinates are actually two disjoint coordinate charts. So even in saying that the space and time flip in Schwarzschild coordinates, we have to be careful to interpret that sensibly, because there are two, not related by any transition map. But there are other considerations.

However, the Schwarzschild geometry has a certain Killing vector field (an infinitesimal symmetry generator--intuitively, the geometry 'remains the same' in the direction of this vector field) that's timelike outside the horizon, null at the horizon, and spacelike inside the horizon. The external Schwarzschild coordinate chart actually uses it for its time coordinate, but both the existence of the Killing vector field and its signature-flipping behavior across the horizon is a a completely coordinate-independent notion.

What the external Schwarzschild chart does is use this Killing vector field for the time coordinate, which is why the metric is independent of Schwarzschild time. If we insist on the same labels both for both $r > 2m$ and $r < 2m$:
$$\mathrm{d}s^2 = -\left(1-\frac{2m}{r}\right)\mathrm{d}t^2 + \left(1-\frac{2m}{r}\right)^{-1}\mathrm{d}r^2 + r^2\,\mathrm{d}\Omega^2\text{,}$$
then $\partial_t$ is the Killing vector field in both the exterior and the interior chart. Thus, despite the two charts being disconnected, we can sensibly identify the exterior timelike coordinate $t$ with the interior spacelike coordinate $t$, justifying the usage of the same label. So we can say that time and space flip in the Schwarzschild coordinates after all.

However, what actually flips is not space and time, but the symmetry of the Schwarzschild spacetime: we go from having a symmetry in a timelike direction to having a symmetry in a spacelike direction. Again, the behavior of the Killing vector field is independent of coordinates; the Schwarzschild coordinates just simplify its expression because they match it.

5. Jun 30, 2014

### Staff: Mentor

Yes, good point; people often talk about "the" Schwarzschild chart, but that's sloppy wording, there are actually two, covering the exterior and interior regions respectively, and they are disconnected because both are singular at the horizon.

Yes.

Yes, although many people (myself included) would say that this causes more problems than it solves. But for better or worse, using the same labeling for both charts appears to be standard practice.

But even if the "time changing to space" aspect is coordinate-independent, because it is connected with the signature of the KVF, the "space changing to time" aspect is still specific to the Schwarzschild charts, because the $r$ coordinate is not timelike inside the horizon in other charts. So if we are trying to describe the *spacetime*, which is what Brian Greene appeared to be trying to do, we can't say that "time and space flip"; instead we should say basically what you say at one point in your post, that the spacetime goes from having a time translation symmetry outside the horizon to having a space translation symmetry inside the horizon.

6. Jun 30, 2014

### Vorpal

Oh, I agree with your position that space/time reversal is "an artifact of that particular coordinate system". My primary concern was that even a statement like that is prima facie suspect and might not be meaningful, because there are two disconnected coordinate charts. The alternative would that "space and time are reversed" is simply gobbledygook, wrong objectively and wrong with respect to any particular coordinates. So basically the dilemma was between
1) Greene's statement is very misleading for reasons of coordinate dependence, as you say.
2) Greene's statement is even more wrong than you imply.
My conclusion was that we can indeed interpret as an artifact of coordinates after all.

Sorry, I guess the "but there are other considerations" in the first paragraph made it look like I was switching topics, but I intended to give a sensible interpretation to the statement that "space and time flip in Schwarzschild coordinates". To do so, one would need a reason to consider them the 'same' coordinates beyond "we choose to call them by the same labels". Using KVFs can provide some justification for that (we'd actually need at least two more, but those exist on the nested 2-spheres)--and if that approach rejected, then we can't even sensibly consider the reversal to be a coordinate artifact. It'd be just be plain wrong.

7. Jul 17, 2014

### Staff: Mentor

PeterDonis: I have the idea that any radius interior to a BH must also be an EH, and therefore light can never move radially outward. That sounds similar to what you said in the above quote; correct?

So PeterDonis, are you saying that a person falling in would our would not notice any difference?

8. Jul 17, 2014

### Staff: Mentor

No. An event horizon is not a surface at which light can't move radially outward; it is a surface at which outgoing light stays at the same radius. Only the EH itself has that property. Inside the EH, outgoing light *decreases* in radius with time; i.e., even "outgoing" light actually moves inward, with respect to the radial coordinate.

Would not.

9. Jul 17, 2014

### Staff: Mentor

Thanks PeterDonis, my concept of the definition of EH was incorrect.

But if "Inside the EH, outgoing light *decreases* in radius with time; i.e., even "outgoing" light actually moves inward, with respect to the radial coordinate." then the following seems like it should be true. The person inside the BH holding his hand out toward the center would not see his hand because light from the hand would always move inward faster than the eye moves inward.

Similarly, a person could hover outside the EH with his hand inside the EH. Wouldn't he notice that?

I fear that my questions must repeat similar questions asked endlessly on PF. Sorry to belabor, but some concepts are darn tough to internalize.

10. Jul 17, 2014

### Staff: Mentor

No. Look carefully at what I wrote: I said outgoing light moves inward *with respect to the radial coordinate*. That is *not* the same as moving "inward" with respect to someone free-falling inward inside the BH. If you are free-falling and you hold your hand out in front of you, the light from your hand moves normally to your eye; to you, the light is moving "outward" (assuming that you are facing "inward"). Motion is relative.

If you insist on trying to visualize what is happening in this instance with respect to the radial coordinate, you can think of the light traveling from the person's hand to the person's eye as "falling" more slowly than the hand and eye: so the light leaves the hand (because the light is falling more slowly than the hand), and the eye catches up to it (because the eye is falling faster than the light). But this viewpoint has serious limitations: it requires you to think of the radial coordinate as designating a "place in space"--but something that was "at rest" at this "place in space" would have to be traveling faster than light (because even light is "falling inward" with respect to the radial coordinate inside the BH).

No, he couldn't. His hand would have to move faster than light to stay connected to his body. Since that's not possible, what would happen, assuming that the person's body stayed hovering, is that the hand would be severed from the body and would fall inward. From the person's point of view, since he has to be accelerating very hard to remain hovering, what would happen is that his rocket engine (which is keeping him hovering) would accelerate him away from his hand so quickly that his hand could not "keep up" and would be severed.