# How exactly does spacetime change inside a black hole? [closed]

There are a few coordinate systems to describe the inside of the BH:

1. Schwarzschild, behaves oddly inside the BH

2. Kruskal-Szekeres, space and time do not swap

3. Gullstrand-Painlevé, $$t$$ is the time measured on a clock held by an observer falling freely into the black hole

4. Eddington-Finkelstein, incomplete

I have read a lot of questions and answers on this site about the interior of the black hole and spacetime there. There are very different descriptions, some say time and space swap, some say they do not, some say spacetime bends extremely, and basically that the only future is towards the singularity, $$r$$ becomes $$t$$.

It is not possible to list all questions. But none of them answers my question, since I am just looking for some clarification about the different statements in different answers.

Black holes: when to use which metric

Schwarzschild equation physical meaning

Is space becoming time-like inside an event horizon a consequence of our coordinate system?

Kruskal Solution to Black hole

Future light cones inside black hole

where dannygoldstein says:

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

Falling into a black hole

where Nanashi No Gombe says:

What you are missing is the observation that as you cross the event horizon, the lightcone gets so squashed that space and time swap their dimensions.

Time paradox inside a black hole

where John Rennie says:

Time and space don't swap places inside a black hole. The point of all this is that the coordinates are not spacetime - they are just labels we attach to spacetime.

These are very different descriptions and I would like some clarification on this. What exactly does swap and what does not in theory? The point of the question is only clarification, there is no other answer on this site that would answer my question because it is specifically, the amount of information and the different statements that are confusing.

Question:

1. How exactly does spacetime change inside a black hole?
• For which of the above statements is the mathematics unclear to you? And exactly what conflict do you see between the precise mathematical content of these various posts, as opposed to some superficial apparent conflict between the necessarily imprecise English words that different authors have chosen to use? – WillO Jan 24 at 3:22
• @WillO space and time swap and they do not, these are opposing, not imprecise? which one? I am just looking for clarification. physically I am asking what happens to spacetime. Not the metric. What is our current theory of the physical spacetime inside the BH? To what extent do the spatial and temporal dimensions change or change at all? Do they change role at all in any way? – Árpád Szendrei Jan 24 at 3:37
• I believe the answers you've quoted are quite clear. Unless you pinpoint exactly what's unclear to you, I'm not sure how to help you. Surely there's no point in my just retyping those answers. – WillO Jan 24 at 3:41
• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Jan 25 at 13:31
• @Ben Crowell can you please reopen? I got a good answer. "Any “swapping” of anything at the event horizon is an artifact of a particular coordinate system, with no physical significance. Coordinates are assigned by humans, not by nature. Inside and outside the horizon, there are three spatial dimensions and one temporal dimension, and you fall on a smooth geodesic through them until you reach the singularity." – Árpád Szendrei Jan 25 at 18:28

## 4 Answers

You fall right across the event horizon without even knowing it is there unless you are paying attention.

In classical General Relativity, spacetime at the event horizon is locally Minskowskian, just like it is everywhere else in the universe where there is no curvature singularity. Thinking that something locally-bizarre happens at the event horizon — like space and time ”swapping”, whatever that means — is like thinking something bizarre happens at the North Pole. The Earth is just as locally-flat there, to first order in Riemann normal coordinates, as it is everywhere else.

Any “swapping” of anything at the event horizon is an artifact of a particular coordinate system, with no physical significance. Coordinates are assigned by humans, not by nature.

Inside and outside the horizon, there are three spatial dimensions and one temporal dimension, and you fall on a smooth geodesic through them until you reach the singularity.

As for whether quantum mechanics modifies this picture in a dramatic way (horizon firewall, anyone?), I don’t know, but I doubt it.

As for people who point out that outside observers don’t see you fall through the horizon, they're right, and they’re missing the point. What they see is completely irrelevant to what you experience.

• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Jan 25 at 13:31

A basic way to put this is that the points of a fall can define the distance. If you fall through a cloud then you might fall for 3 minutes along a distance of 60,000 feet with the underside of the cloud being point A and the ground being point B. Point A remains where it is. Therefore there is a spatial distance. Once inside the Event Horizon everything is falling together so you don't fall along a static distance anymore. You cannot define a distance because space itself is moving. Point A (space and anything else) inside the horizon is falling at the same rate as yourself. Normally we could plot your fall in terms of feet per second but inside the Event Horizon we could only plot your fall in terms of seconds because the place you fell from is falling too. Therefore even though you fell for three minutes it would only be in time.

• I think if you delete this answer you will lose the rep you gained from it, but now you can comment. Anyway I think this is a valid answer. – Renan Jan 24 at 14:58

As can be seen from the Schwarzschild metric for an r-coordinate r < 2M the term (1-2M/r) becomes negativ. This means that r- and t-coordinate interchange which does not mean that space and time interchange. This has two consequences:

1. The spacetime within the horizon is not stationary because r has the character of time, r decreases inevitably. The r-coordinate of a photon emitted upwards decreases. Hovering at constant r is not possible (time doesn't stand still).

2. r = 0 means that the singularity is a point in time (not in space).

There are very different descriptions, some say time and space swap, some say they do not

Space and time do not swap, what swaps is the sign of the $$g_{rr}$$ and $$g^{tt}$$ components of the metric tensor, but that is only the case in Droste coordinates, not in Raindrop or Finkelstein coordinates.

some say spacetime bends extremely

For the curvature you have to calculate the Kretschmann scalar, as it is done in the links above. For a Schwarzschild black hole the curvature invariant is $$K=R_{\rm abcd} \ R^{\rm abcd} = 48/r^6$$ (in natural units). That is finite and smooth at the horizon and blows up at the singularity. For rotating black holes though you can get hard curvature gradients that switch abruptely from strong positive to strong negative curvature even before the singularity (for an example see here), but there is no special curvature at the horizon even then.