- #1

Antarres

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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.