Continuation of Coordinates Across Black Hole Horizons

In summary, the conversation discusses the regularization of coordinate singularities in the context of general relativity. The first case involves a coordinate transformation in Schwarzschild geometry, which allows for an extension of the solution below the horizon. While the new coordinates are not equivalent to the original ones, the solution itself remains equivalent as it describes the same geometric objects. The second case involves a local Lorentz transformation in a tetrad basis, which can suppress divergences but does not remove the singularity in the same way as a coordinate transformation. However, this is not surprising as the solution is still equivalent as it describes the same geometric objects. The thread is marked as A-level, indicating that technical explanations are accepted.
  • #1
Antarres
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Studying and tinkering with some solutions, I've come to some realizations and questions regarding the regularization of coordinate singularities, so I'd like to see if my conclusions are good, and I guess I have some questions as well. There are two questions/conclusions, but since they contrast each other in a way, I will post them in the same thread instead of creating separate threads.

Take, for simplicity, Schwarzschild geometry, which is given in a standard way as:
$$ds^2 = \left(1-\frac{2M}{r}\right)dt^2 - \frac{dr^2}{\left(1-\frac{2M}{r}\right)} - r^2 d\Omega_2$$
where ##d\Omega_2## is the metric of a 2-sphere.
Now, we apply the standard regularization through Eddington-Finkelstein coordinates:
$$dr^* = \frac{dr}{\left(1-\frac{2M}{r}\right)} \qquad t^* = t - r^*$$
We obtain the metric:
$$ds^2 = \left(1-\frac{2M}{r}\right)dt^{*2} + 2dt^*dr - r^2d\Omega_2$$

In the first case, if we say that our coordinates are valid for ##r>2M##, we find after the coordinate transformation that our new coordinates can be extended below ##r=2M##, as well. Therefore, we redefine the solution to include this new region. In this sense, this new solution is not equivalent to the first one? The first solution is the same as the extended one if we restrict the extension to the ##r>2M##, but otherwise, the extension is a bigger solution. Maybe saying they're not equivalent is not a good way to phrase it, but correct me if I'm wrong. At the same time, we can argue that the first coordinates are also defined for ##r<2M##(but on that patch the timelike coordinate is now ##r## instead of ##t##, I guess), and then our extension allowed us to merge these two regions. But these two regions don't overlap, so I would say the previous remark about 'inequivalence' still holds(since on a manifold, the patches should have smooth overlaps). Maybe I'm being pedantic, but I want to contrast it to another situation, which I describe next.

Let's say now, that we have some other solution, but this time, given in a tetrad basis instead of a coordinate basis. If we denote the tetrad field by ##e_i^{\hphantom{i}\mu}##, with Latin indices being Lorentz indices, let's say that we have curvature components denoted by ##R^{i}_{\hphantom{i}jkl}##. Assume that some of these components are divergent at the horizon, while the curvature scalars do not diverge. Such situations exist in literature(there is a review paper by Ellis named Singular spaces from 1977, which classifies various types of singularities), but my focus here is general, so I feel it's not of use to provide concrete examples. Now, in order to regulate these divergences, we would apply some local Lorentz transformation.
But when you think about it, this Lorentz transformation, if it is supposed to regulate the curvature components, will be singular at the horizon. Moreover, this Lorentz transformation will essentially suppress the divergent factor into the new tetrads that are obtained after the transformation. Therefore, we essentially don't remove this singularity in the same way we do with coordinate transformation, if I'm correct.
This doesn't seem like a big surprise to me, since even in the case of coordinate transformations, the regulating transformation is singular at the horizon and the divergent factors are suppressed in new coordinates, but what we do afterwards is that we redefine coordinates, taking these new coordinates as the new definition, which we're(I think), not able to do with local Lorentz basis.

Would like to hear thoughts on what I explained, to see if my conclusions are sound.

P.S. I labeled the thread as A-level, but since I didn't post for long, I forgot if the level of the thread should reflect my own level of knowledge, or the level of the question itself. In any case, I accept explanations/comments that can be technical at A-level. Otherwise, the thread can be moderated to I-level, if that's more appropriate, I'm not sure.
 
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  • #2
Antarres said:
In this sense, this new solution is not equivalent to the first one?
They are not separate solutions. The Einstein field equations relate the stress energy tensor to the metric tensor. Together with the topology, the metric tensor defines the spacetime manifold.

It is important to understand that the tensors and the manifold are geometric objects. They can be described using different coordinates, but the solution to the equations is the underlying geometric object, not the coordinates.

So the coordinates are indeed unequivalent because they cover different open regions of the manifold. But the solution is equivalent because it is describing the same geometric objects.

Antarres said:
I accept explanations/comments that can be technical at A-level.
Then it should remain A. The label is guidance to respondents regarding what level you would like answers to target.
 
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  • #3
Yeah, what you wrote makes sense. After all, on a manifold I can have maximal atlas, and so both of these coordinates can be present, I can just pick the ones I need. So basically the first form of metric tensor doesn't define the manifold but just how this tensor looks on a patch of it.

I was coming to this conclusion thinking about differences between dealing with singularities in tetrad components vs coordinate components, since I've been working in tetrad formalism for a while now, so I was thinking of a way to explain these differences to people unfamiliar with their existence(assuming of course, that I'm correct about those differences existing, which I think I am), hence the contrast I made in the OP.
 
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  • #4
Antarres said:
So basically the first form of metric tensor doesn't define the manifold but just how this tensor looks on a patch of it.
Exactly.

Antarres said:
since I've been working in tetrad formalism for a while now, so I was thinking of a way to explain these differences to people unfamiliar with their existence(assuming of course, that I'm correct about those differences existing, which I think I am), hence the contrast I made in the OP.
I think that was a good contrast.

One thing that I think helps is to recognize that a coordinate chart is a mapping (smooth and invertible) between a region of the manifold and a region of ##\mathbb{R}^n##.

So a coordinate transform is not an operation directly between two coordinate charts. Instead it goes from one chart, to the manifold, and then to the other chart. So it isn’t problematic that if one chart doesn’t cover some region of the manifold, then the transform between charts fails in that region. That is by design.
 
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  • #5
Antarres said:
In this sense, this new solution is not equivalent to the first one?
You are using the term "solution" ambiguously. The usual description of what you are doing would be as follows:

You start with a particular form of the metric which is valid on an open region, namely ##r > 2M##. However, if you investigate closely, you will find that this open region is geodesically incomplete: there are geodesics that only go to a finite value of their affine parameters in the limit ##r \to 2M##, and if you check, you find that all scalar invariants are finite in that limit. That suggests that the coordinate chart you used to derive this particular form of the metric only covers a portion of the full manifold.

The transformation to Eddington-Finkelstein coordinates allows you to extend the coverage of your coordinate chart all the way down to the limit ##r \to 0##. So indeed the suggestion above was correct: your original coordinate chart did only cover a portion of the full manifold. In the limit ##r \to 0##, there are still geodesics that only reach a finite value of their affine parameter, so the extended manifold that your new coordinates cover is still geodesically incomplete, but you can also check that scalar invariants increase without bound in the limit ##r \to 0##, indicating that there is a genuine curvature singularity there, not just a coordinate singularity.

Note, btw, that even Eddington Finkelstein coordinates do not cover the full maximal extension of the manifold in question (Schwarzshild spacetime, considered as a geometric object). There is another limit you can take in those coordinates that will indicate another coordinate singularity and a possible further extension of the manifold. (If you are familiar with Kruskal coordinates, you have already seen the maximal extension, so you just need to ask yourself what regions of that maximal extension are covered by the coordinates you've looked at so far.)
 
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  • #6
@PeterDonis

Thank you for your comment. Yes, I'm aware of the Kruskal coordinates, I was just trying to keep it simple by picking Eddington-Finkelstein. My thoughts were coming from some analysis I did on Kerr solution, and the use of local Lorentz transformations(which are used to regulate tetrads), their effect on curvature divergences(in tetrad components) etc. But I wanted to revert back and pick Schwarzschild and these coordinates as the simplest example, that came to my mind, in order to brush up on the basics, so that I don't say anything that can be misinterpreted to some people that are not aware of deeper things in GR, but have looked on it superficially.

Either way, I agree with what you said, when I say solution, I usually don't mean what I wrote in OP(a specific form of the metric, that is), so now that the issue looks clarified to me, your comment looks obvious. But still thank you, the purpose of the post was to verify/clarify what I was trying to say.
 
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