# How to prove that something is a non-curvature singularity?

1. Oct 18, 2015

### bcrowell

Staff Emeritus
In general, what methods are there for proving that something is a non-curvature singularity?

As a specific example, take the 1+1-dimensional version of the NUT metric, $ds^2=t^{-1}dt^2-t d\theta^2$, which is discussed in Hawking and Ellis in two places. H&E's treatment is long and meandering, and I have a hard time figuring out what they're getting at.

What I can figure out is that the null geodesics are of the form $\theta=(\text{const})\pm\ln(\pm t)$, and that $t$ itself qualifies as an affine parameter. This shows that the geodesic specified by this equation doesn't continue to arbitrarily large affine parameters. If the opposite had happened, then clearly we would have had geodesic completeness. But the converse does not seem to be a reliable criterion, because, if I'm understanding H&E correctly, they're able to do a change of coordinates such that a particular null geodesic at $t>0$ is glued end-to-end with another null geodesic at $t<0$, making it complete. However, they aren't able to find coordinates that accomplish this for all null geodesics at once.

So to prove that a space is geodesically complete, when the singularity is not a curvature singularity, do we have to prove that no coordinate system exists in which all geodesics are complete? That seems like a tall order. How would you prove such a nonexistence result?

2. Oct 19, 2015

### martinbn

I am probably confused but this doesn't seem enough, nor necessary. You can easily have a complete manifold with no coordinate system in which all geodesics are complete. For example a sphere (for the Riemannian case, which of course is quite different).

3. Oct 19, 2015

### bcrowell

Staff Emeritus
I'm confused myself, so that makes two of us :-)

They happen to be using a single chart to cover the whole spacetime, but I don't think that really matters much. (Well, technically they need at least two charts to handle the periodic behavior of the $\theta$ coordinate, but they don't bother doing that explicitly, because it's obvious how it could be carried out.) I guess you could replace "coordinate system" with "atlas" throughout this discussion.

Although coordinates aren't intrinsic, I don't think it's unusual to define intrinsic properties in terms of the existence or nonexistence of coordinate systems having certain properties. For example, a stationary spacetime can be defined as one for which coordinates exist such that the metric has a certain form. However, in that example we have alternative characterizations of stationarity, so that to prove something is not stationary, you don't have to cook up a direct proof in which you somehow consider all possible coordinate systems and prove that none exists with the right property.

In the example we're talking about here, the coordinate transformation is $\theta' = \theta \pm \ln t$, which is not a diffeomorphism because it has a singularity at $t=0$. I think that means that you can get nontrivial changes in the physics by doing it, as opposed to a diffeomorphism, which never has any physical consequences. I think it's equivalent to doing a cutting and gluing operation on the manifold.

4. Oct 19, 2015

### martinbn

That is ok, I didn't mean to say that it must be coordinate-less. What I meant was that the criteria is probably not right. For example for the sphere there is no coordinate system that covers all of it, it must miss at least one point. Then in any coordinate system there will be geodesics that are not complete, namely those that pass through points not covered by the system. But the sphere is complete.

5. Oct 19, 2015

### Demystifier

6. Oct 19, 2015

### bcrowell

Staff Emeritus
That's what I was addressing in my #3 with the discussion of the atlas versus the coordinate system.

7. Oct 19, 2015

### martinbn

I missed that part completely. So, what is the statement in terms of the atlas instead of coordinate system?

8. Oct 19, 2015

### bcrowell

Staff Emeritus
I think it's just the usual thing you do when you have an angular coordinate that runs around in a circle. Instead of letting the angle $\theta$ go from 0 to 2 pi, where there's a discontinuity, you split it into two charts that overlap. I don't think this issue has any impact on the analysis of the singularity at t=0.

9. Oct 19, 2015

### andrewkirk

I don't know how difficult the calculations would be for this, but if we could show that the Ricci scalar (or some other scalar measure of curvature) has a finite limit at the singularity, would that prove that the singularity is non-curvature? One would not need to consider geodesics in testing for a limit.

10. Oct 19, 2015

### martinbn

I think I understand now. You are talking about extensions. Since the singularity is non-curvature one would hope that the manifold can be extended through the singular points to obtain a complete manifold. But if you can prove that there is no extension in which all incomplete geodesics (of the original manifold) are complete then it is indeed singular.

11. Oct 19, 2015

### bcrowell

Staff Emeritus
I think that only works the other way around. If you can show that a geodesic reaches, within a finite affine parameter, a point at which a particular curvature scalar diverges, then you have a curvature singularity. But that's a sufficient condition, not a necessary one.

What I'm asking about is how to distinguish (1) the lack of a singularity from (2) a non-curvature singularity.

12. Oct 19, 2015

### bcrowell

Staff Emeritus
Exactly. I'm asking about techniques for proving the nonexistence of such an extension.

13. Oct 19, 2015

### SlowThinker

This is obviously well beyond my level, but... what other kinds of singularity are there besides
- coordinate (false) singularity, and
- curvature (true) singularity?
The curvature singularity can be found by looking at the Riemann curvature tensor, if it is infinite somewhere. If it's finite everywhere, there is no true singularity.

This is pretty basic stuff, right? Why does that not work for you?
In any case, if you want to prove something about all coordinate systems, you'll certainly need the Riemann curvature tensor somewhere along the way.

14. Oct 19, 2015

### bcrowell

Staff Emeritus
A good example would be a conical singularity.

The paper by Geroch linked to from #5 discusses this point starting at the bottom of p. 528.

15. Oct 19, 2015

### bcrowell

Staff Emeritus
After a little more thought and digging around I've come to the conclusion that this must be an open problem. There simply is no test that allows us to get a yes/no answer in such cases. The reason I think this is the following paper:

Clement et al., "Rehabilitating space-times with NUTs," http://arxiv.org/abs/1508.07622

What this paper claims is remarkable if true -- and it ended up published in Phys Lett B, so I assume it's not just obviously wrong. (The details are complicated, and I haven't made an attempt to check them myself.) They say that the Taub-NUT spacetime, which has been known since the 1950s, is actually geodesically complete, even though people had believed for the last 50 years that it was geodesically incomplete. The method by which they prove this appears to be a bespoke calculation, not some general test that could be applied to any such spacetime.

16. Oct 19, 2015

### andrewkirk

That's not quite the same test as I was suggesting.
I am wondering whether the following may be a necessary and sufficient condition for the point $p$ to be a curvature singularity:

(1) There exists a curve $\gamma:\mathbb{R}_+\to M$ and $a>0$ such that $\lim_{\lambda\to a}S(\gamma(\lambda))$ does not exist.

(2) There exists a geodesic $\gamma:\mathbb{R}_+\to M$ and $a>0$ such that $\lim_{\lambda\to a}S(\gamma(\lambda))$ does not exist.

I have a feeling that, while (2) is not a necessary condition for $p$ to be a curvature singularity (since there may be points that are not reached by any geodesics, as in the H&E example), (1) probably is.

Strictly speaking, $p$ is not a point in the manifold. But what we can say is that, if a manifold has curves (not necessarily geodesics) whose domain is bounded above (ie the curve disappears off the manifold when a certain parameter value is reached) but for which curvature does not diverge as the disappearance point is approached, then we can divide all such curves into equivalence classes based on the following equivalence relation:

If $\gamma_1,\gamma_2$ are curves (not necessarily geodesics) in $M$ then $\gamma_1\sim\gamma_2$ iff $\exists a_1,a_2>0$ such that:
1. for $i=1,2,\ \gamma_i$ maps $[0,a_i)$ to $M$;
2. for $i=1,2,$ the domain of $\gamma_i$ is $[0,a_i)$;
3. for $i=1,2,$, $\lim_{\lambda_i\to a_i}S(\gamma_i(\lambda_i))$ exists and $\lim_{\lambda_1\to a_1}S(\gamma_1(\lambda_1))=\lim_{\lambda_2\to a_2}S(\gamma_2(\lambda_2))$
4. $$lim_{\lambda_1\to a_1,\lambda_2\to a_2}d(\gamma_1(\lambda_1),\gamma_2(\lambda_2))=0$$

Each such equivalence class would correspond to a missing point in the manifold, where the curvature did not diverge. Since it is not a curvature singularity, perhaps we can extend the manifold by adding the missing point, without breaking any necessary criteria for the Riemannian/pseudo-Riemannian manifold.

17. Oct 19, 2015

### bcrowell

Staff Emeritus
@andrewkirk: In your #16, is lambda an affine parameter? Affine parameters are only defined along geodesics, not other curves.

Also, what you're proposing is an alternative definition, not a test, and it's an alternative definition of a curvature singularity, but I'm talking about non-curvature singularities.

18. Oct 19, 2015

### andrewkirk

@bcrowell: Not necessarily affine. I'm just using it as an ordinary curve parameter.

19. Oct 19, 2015

### bcrowell

Staff Emeritus
See the Geroch paper, p. 529, at "One possible way ..."

20. Oct 20, 2015

### andrewkirk

Golly, that's a great paper! I had no idea there were such subtle difficulties in this area. He presents them very engagingly too.