I Why open manifolds in Relativity

1. Apr 13, 2017

davidge

Why manifolds in General (and Special) Relativity have to be open? Would this be because an open manifold have a continous interval? (i.e. an interval with no interruptions)

2. Apr 13, 2017

Demystifier

Because you don't want to mess with boundary.

3. Apr 13, 2017

davidge

What a boundary would mean? I was reading a book that I found on the university library and that book says that a open region $D$ in $\mathbb{R}^n$ means you have an open sphere (that is, the sphere consist of all points $x < r$) centered at some point $x_o$ of $D$ and contained in $D$ for some $r$. Of course, such space have to have a metric so that we can define $d(x,x_o)$, the distance between $x$ and $x_o$.

4. Apr 13, 2017

Staff: Mentor

A lot of the proofs in GR rely on the fact that you can pick any point in the manifold and construct a neighborhood around that point which looks like R4 locally. That is true for any point in an open manifold, but not true for points on the boundary of a closed manifold.

5. Apr 13, 2017

martinbn

What do you mean by open? Open as a topological space? Then the manifold is always open and closed. Or open as in no boundary? The second questions is hard to understand, I cannot guess what you are trying to say.

6. Apr 13, 2017

davidge

And what is the definition of a open manifold?
Yes, I mean as a topological space. Why is it always open and closed? What I said in my post #2 is valid to define a open space?

7. Apr 13, 2017

Staff: Mentor

See https://arxiv.org/abs/gr-qc/9712019 chapter 2.

Shouldn't you have started with this question? Your question in the OP presupposes that the questioner understood this terminology already. Frankly it is rather irritating to have this follow on question given the original question.

8. Apr 13, 2017

davidge

Yes, I actually want to know if the definition that I mentioned in my post #2 is correct for the case of Relativity.

9. Apr 13, 2017

Staff: Mentor

See chapter 2 in Carrolls notes

10. Apr 13, 2017

martinbn

Then you need to learn what a topology and a topological space are, manifolds later.

11. Apr 13, 2017

davidge

Ok. He explains what a open set mean, but he does not explain why Relativity requires open sets

12. Apr 13, 2017

Staff: Mentor

It doesn't - at least in general. The tools continuity, differentiability, atlas and so on do. A boundary means, you will have to do everything twice: for inner points and for points on the boundary which you can approach from only one side. So you double the work, probably more as boundaries can be more difficult, without gaining anything. This doesn't mean that boundaries don't play a role at all. They do if you apply Stoke's theorem, consider cohomologies and so on. But for the analytic investigations what's going on on a manifold, you simply don't need them, at least at the start.

13. Apr 13, 2017

davidge

Ah, ok. Are there cases in Relativity where we cannot consider the manifold to be open? Maybe inside a Black Hole or so.

14. Apr 13, 2017

Staff: Mentor

The concept of manifolds comes into play as one investigates differential equations like EFE, mainly because we want to know their geometric qualities and most important: without having them embedded in some surrounding space where they could have boundaries, or are boundaries of some set itself, like spheres are. As such they are a tool to describe solutions and their behavior. It has nothing to do with physical places, as they are a mathematical instrument. In addition it is a local concept, i.e. what happens at certain points. E.g. do the solutions of a system of differential equations build an attractor or repeller somewhere, what happens if we follow a flow and such. Therefore the consideration of a nicely behaving neighborhood of such a point is needed. And nicely means open, because otherwise you run into terribly many special cases without gaining something from it. And as in the example of a sphere: the manifold itself might be closed in a surrounding space, but the concept of a manifold means: we are on the sphere and there is no outside. So if there is no outside, only the surface of a sphere and nothing else, then it is open. And given the fact, that even an event horizon of a black hole isn't a physical place, it's hard to imagine what a physical boundary should look like.

15. Apr 13, 2017

Staff: Mentor

The Schwarzschild spacetime has no pathologies. The central singularity isn't a point in the manifold, and there's no difficulty finding an open ball around any point "near" it. Likewise, there's no problem finding an open ball around any point on the event horizon; Schwarzschild coordinates refuse to map these points, but that's a problem with the coordinates not the manifold.

16. Apr 13, 2017

davidge

If it isn't a point, what it is then? Please can you say wheter I'm wrong in the following reasoning:

Take a region $C$ and map it to $\mathbb{R}^4$ through a map, say $\varphi$. The image of this map, i.e. all $\varphi (C) \in \mathbb{R}^4$ is the manifold (of course, obeying the other conditions, like having a one-to-one map, being $C^{\infty}$, etc).
So by this, shouldn't the central singularity of a BH be a point?

@fresh_42 I got it. Thanks.

17. Apr 13, 2017

Staff: Mentor

Nothing. The singularity is not part of the spacetime manifold at all. See below.

No, for two reasons:

(1) The singularity is not "central"; it's not a "place in space", it's a "moment of time".

(2) The "moment of time" we refer to as the singularity isn't actually part of the manifold; it's a limit point which is never actually reached. Heuristically, it can be thought of as something like the boundary of the closure of the manifold, i.e., what we would have to "add" to the open manifold of spacetime, topologically, to give it a boundary (or at least a boundary on the "side" that is to the future of the interior of the hole).

18. Apr 13, 2017