# How does time become space inside a black hole? What is the space the old time variable "becomes"?

The Schwarzschild metric is

$$ds^2 = - (1 - r_s/r) dt^2 + (1 - r_s/r)^{-1}dr^2 + r^2 d\Theta^2,$$ where $$d \Theta^2 = \sin^2 \varphi\, d \theta + d\varphi^2$$ is the metric on the sphere and $$r_s$$ is the Schwarzschild radius. For $$r > r_s$$, all the coordinates have meanings that an observer can in principle compute and hence locate their position in space-time:

1. $$r=$$ areal radius (it gives the correct area for fixed $$t,r$$ of a sphere of radius $$r$$,

2. $$\theta,\varphi$$ has the standard geometric meanings, and

3. $$t$$ is the time of a distant observer.

For $$0 < r < r_s$$, the metric is $$ds^2 = - (r_s/r - 1)^{-1} dr^2 + (r_s/r - 1) dt^2 + r^2 d\Theta^2,$$ where $$r_s/r - 1 > 0$$. Since the coefficient of $$dr^2$$ is negative and that of $$dt^2$$ is positive, we see that $$r$$ is now a "time-like" variable and $$t$$ is "space-like" and in this sense, "time becomes space" (and the space variable $$r$$ becomes time) inside a black hole.

Since a person that fell through the event horizon is headed toward $$r = 0$$ and can do nothing about it (and cannot avoid it just like you can't avoid heading toward tomorrow) it's clear that $$r$$ behaves just like time.

If I'm inside the event horizon and I asked myself, "what are my coordinates," I can say what my time "$$r$$" is conceptually and I can say what $$\theta,\varphi$$ are conceptually, but how do I conceptually determine the space "$$t$$" coordinate?

To make my question clearer: For $$r < r_s$$, I know

1. $$r$$ measures "time". And, I understand conceptually what $$r$$ is in reference to time --- for example if I'm falling freely, knowing $$r$$ is like knowing "when" (a time concept) I hit the singularity.

2. $$\theta,\varphi$$ have the standard geometric meanings as points on a sphere, so I understand how to find $$\theta,\varphi$$.

3. Question: $$t$$ is a space-like variable, but what space concept does $$t$$ represent? In other words, if you're inside the event horizon and someone asks "what are your coordinates" is there some geometric and conceptual way to get my $$t$$ spacial coordinate?

I'm looking for something like what happens outside the event horizon, where the space variable "$$r$$" represented areal radius, does the space variable "$$t$$" inside the event horizon have some other geometric concept. Of course, one can go to other coordinates, like Kruskal-Szekeres coordinates, then convert back to interpret $$t$$, but I'm looking for a simpler answer if one is possible. Also, $$t$$ is "just an abstract mathematical variable" that is space-like, which is true, but I hope to see some concrete meaning about what it is.

• You seem to know what you are talking about, but just wanted to check. You should be aware that your wristwatch is not suddenly going to start measuring distance inside the horizon. You are describing a coordinate-dependent effect. Oct 26, 2021 at 12:13
• @m4r35n357 Thanks, no I certainly don't and never have believed wristwatches (which measure proper time) magically turn into rulers. I hope I never gave that impression. Oct 26, 2021 at 16:19
• OK please pardon my intrusion! Oct 26, 2021 at 16:54
• @safesphere Thank you for your comment. Would you mind making your comment into answer with more details? I would like to see more thoroughly exactly what you are saying. Oct 27, 2021 at 20:24

The coordinate basis vector $$\partial_t$$ points radially outward. It is spatially perpendicular to the $$\partial_\theta$$ and $$\partial_\phi$$ vectors.

Although $$r$$ is proportional to the area of the spherical surface $$\partial_r$$ is timelike (as you mentioned) so it doesn’t point in the radial direction.

There is no geometrical significance to the value of $$t$$ itself. The components of the metric are unchanged under the transformation $$t \rightarrow t+a$$, so you can set $$t$$ to any value you like by choosing $$a$$ without changing anything else.

• But, it's obvious (since the metric has no cross terms) that $\partial_t$ is spacially perpendicular to $\partial_\theta$ and $\partial_\varphi$. This has nothing to do with my question. Oct 26, 2021 at 2:56
• It has everything to do with it. You want to know what spatial direction t represents. There is only one spatial direction that is perpendicular to both latitude and longitude. That is altitude. That is as concrete a meaning for t as you can get. Pointing out the obvious fact is intended to help you recognize that the spatial direction of t is obvious. You asked for a “concrete meaning” a “simpler answer” and a “geometric concept”. What can fulfill those requests better than pointing out that the only direction perpendicular to both latitude and longitude is altitude?
– Dale
Oct 26, 2021 at 4:06
• If I asked "what are my (x,y) coordinates in the plane", and you say $\partial_x$ is perpendicular to $\partial_y$, it doesn't help me. Similarly, my question asks if someone asks you "what are your coordinates" is there some geometric and conceptual way to get my t spacial coordinate? Simply saying "$\partial_t$ is orthogonal to $\partial_\theta$ and $\partial_\varphi$ does not answer to the question about the $t$ coordinate, it is a description about orthogonality of axes and not about actual coordinates. Oct 26, 2021 at 4:23
• In this case you already know $\theta$ and $\phi$. So using them as a reference is clear. In your current example, if you already understood $x$ then identifying $y$ as the coordinate such that $\partial_y$ was perpendicular to $\partial_x$ would be a clear geometric answer. In both cases there is only one coordinate foliation that meets that geometric constraint.
– Dale
Oct 26, 2021 at 5:07
• But it's not clear: how do you know $t$ is positive or negative or large or small or zero? Simply knowing $\partial_t$ is orthogonal to $\partial_\theta$ and $\partial_\varphi$ tells you absolutely nothing about the value of the coordinate $t$. Please note that my question is not about axes, it is about determining the value of $t$ conceptually/geometrically. Oct 26, 2021 at 5:17

While not immediately obvious from the form of the Schwarzschild coordinates, the fixed $$r$$ surfaces in side the black hole horizon are in fact homogeneous. (The key thing to realize here is that $$\partial_t$$ is a Killing vector.) At fixed $$r$$ (inside the horizon), ($$t$$, $$\theta$$, $$\phi$$) are just coordinates on $$\mathbb{R}\times S^2$$. The coordinate $$t$$ simply indicates where we are along $$\mathbb{R}$$, and has no further geometric meaning.