Is the fact that "space behaves like time" inside an event horizon a consequences of our particular coordinate system? Or is it a universal fact?

I am asking based on the statement in this video How Time Becomes Space Inside a Black Hole | Space Time (at 10:44 mark), where its claimed that "There are other coordinate systems where the switch does not happen".

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    $\begingroup$ The separation of two points being timelike or spacelike is a coordinate invariant. $\endgroup$
    – Slereah
    Nov 24, 2017 at 12:31
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    $\begingroup$ @Sleeeah You comment is wrong along with those who upvoted it. The separation of two points is relative. For example, if the points are New York and Chicago and you are at home, then the separation is spacelike, but if you are driving or flying from one point to the other, then in your frame the separation is timelike. This is the difference between points and events. The 4-interval between two events is invariant, but the separation of two points in space or in time is relative (except for some cases inside the event horizon). $\endgroup$
    – safesphere
    Nov 24, 2017 at 19:23
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    $\begingroup$ @safesphere What we physically refer to as events are just points on spacetime, so points and events are indeed the same. In other words, New York and Chicago are both world lines, not points, and it really makes no sense to refer to spacelike or timelike separation of world lines, since depending on what point on the world line you choose it will vary. The point is that although the split into time and space separation obviously is observer dependent (since the notion of space is), the separation of two points being time- or spacelike is not. $\endgroup$ Nov 25, 2017 at 12:14
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    $\begingroup$ @safesphere I believe you're mistaken. The definition of "timelike" and "spacelike" relates to whether it's possible for light from one event to reach the other event. No observer is mentioned in the definition, which makes it entirely independent of observer or choice of coordinates. $\endgroup$ Mar 31, 2018 at 22:31
  • $\begingroup$ @DawoodibnKareem There's nothing here for me to be mistaken about. You simply did not understand my comment, which did not in any way contradict the definition you quoted. $\endgroup$
    – safesphere
    Mar 31, 2018 at 22:39

2 Answers 2


Let's consider the simplest example of a black hole, the Schwarzschild black hole, given by the metric: $$ds^2 = -\left(1-\frac{r_s}{r} \right)dt^2+\left(1-\frac{r_s}{r} \right)^{-1}dr^2 + r^2 d\Omega ^2 $$ where $r_s = 2GM$ is the Schwarzschild radius. We're using units with $c=1$.

The metric signature is $(-,+,+,+)$ (some authors use $(+,-,-,-)$ but it is not relevant for the present discussion). In this sense the first coordinate, $t$, is associated with a negative component of the metric. Notice however that this is valid only for $r>r_s$. If $r<r_s$,that is, inside the event horizon, the component of the metric associated with $t$ becomes positive, while the component associated with $r$ becomes negative. In this sense the $r$ coordinate becomes "time", while the $t$ coordinate becomes part of "space".

However given the metric above becomes singular at $r=r_s$. This is a problem with this coordinate system. We can make a coordinate transformation (for details see these notes, page 182 onward) to bring the metric to the following form (the so called Kruskal coordinates):

$$ds^2 = \frac{4r_s^3}{r} e^{-r/r_s} (-dv^2 + du^2) + r^2 d\Omega^2$$

where $r=r(u,v)$ is the usual $r$, but should be understood in this case as a function of $u$ and $v$.

In this case the role of time (negative component of the metric) is played by $v$, and this is valid for both $r>r_s$ and $r<r_s$. In this sense, in Kruskal coordinates, inside a black hole time remains time and space remains space.

What does this tell us? Essentially that we should be careful in interpreting what the coordinates mean. For example, the time $t$ in the original Schwarzschild coordinates should be understood as the time experienced by an observer infinitely far away from the black hole. However, it is not the time experienced by an observer falling into the black hole: that would be the so-called proper time $\tau$, defined by $dt^2 = -ds^2$. It is known that it will take infinite $t$ for an observer radially falling into the black hole to actually fall, however only finite $\tau$. That is: if you fall into a black hole, you will actually cross an event horizon (finite $\tau$) but your friend far away will never see you cross.

The takeaway is that "timelike" coordinates are not necessarily the time experienced by an observer, they are just a way of describing a spacetime. In fact, $u$ and $v$ above do not have a simple interpretation in terms of time experienced by someone.


As has been discussed in the comments, the separation of two points being timelike or spacelike is observer independent, but this does not really answer the question.

One thing that happens to an observer crossing the event horizon is that once it is crossed the singularity lies in the future. To the outside observer, the black hole (and inside it, the singularity) forms a timelike world-tube, but to the inside observer, the singularity is now a spacelike hypersurface (in your future). In this sense one could say that time becomes space inside a black hole in a coordinate independent manner.

However, what we usually refer to as space is a one-parameter family of spacelike hypersurfaces normal to a timelike vector field (this would define local surfaces of simultaneity for the observers corresponding to the vector field), the integral curves of which defines time. Usually we would introduce coordinates $(t,x^i)$ ($i \in \{1,2,3\}$) such that $t$ is constant on the hypersurfaces, and $x^i$ are constant along the integral curves. In canonical Schwarzschild coordinates the line element takes the form $$ ds^2 = Adt^2 - A^{-1}dr^2 - r^2d\Omega^2, $$ where $d\Omega^2$ is the standard metric of the 2-sphere, and $A = 1 - 2m/r$. Inside the event horizon $2m/r > 1$, and $A < 0$ such that $r$ becomes a timelike coordinate (and $t$ becomes a spacelike coordinate). Thus "space becomes time."

In fact, in the language of differential geometry, canonical coordinates define one chart which covers the exterior of the event horizon, and one separate chart which covers the interior. In this language, that $r$ "becomes" timelike in the interior means simply that we can define coordinates on the interior such that the line element depends on the timelike coordinate similarly to its dependency on the spacelike coordinate $r$ in the exterior region. Since we are dealing with separate charts, covering disjoint regions, one can argue from this that it has got nothing to do with space becoming time (whether coordinate dependent or not), but rather is a statement about the symmetries of the regions (that the timelike Killing vector of the exterior is replaced by one which is spacelike in the interior; in fact these can be joined by a smooth extension which is also Killing, and null on the horizon).


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