# Curvature singularity with well-behaved Kretschmann scalar

1. Nov 11, 2015

### bcrowell

Staff Emeritus
Does anyone know of an example, preferably a simple one, that can be used to demonstrate that we can have a curvature singularity without a singularity in the Kretschmann scalar?

2. Nov 11, 2015

### PAllen

I found the following reference:

which classifies curvature singularities in various ways. The Kretschmann scalar blowing up makes it an instance of a scalar polynomial singularity. On page 116, they analyze a family of Tolman-Bondi-Lemaitre models. At the bottom, they conclude that for some of the parameter range of these solutions, the curvature singularity is not singular per the Kretschmann invariant.

3. Nov 11, 2015

### pervect

Staff Emeritus
The "wave of death" has a "strong nonscalar (null) curvature singularity". See for instance https://www.physicsforums.com/threads/wave-of-death.93654/ or https://en.wikipedia.org/w/index.php?title=Pp-wave_spacetime&oldid=666287121 (near the end, search for wave of death). I'm not sure how simple a good discussion of it can be made, but it would be amusing in a talk for a general audience.

I believe after some disucssion on PF I/we came to the conclusion that "nonscalar" in this context meant that the Kretschmann scalar didn't "blow up". Hopefully it's a correct conclusion :-).

My understanding is that the Aichelberg-sexl ultraboost https://en.wikipedia.org/w/index.php?title=Aichelburg–Sexl_ultraboost&oldid=686381232 would be an example of a "wave of death", but I don't have a solid reference for that. Certainly it's got a dirac delta function in the metric definition from Wiki. My somewhat intuitive understanding of this is that if you had a baseball floating in free space, and a Aichelberg-sexl p-p wave comes by, when you take the appropriate limit the baseball will be accelerated basically instantaneously to a non-zero velocity by the passage of the wave. This is rather similar to how a field of test articles is accelerated by the passage of an ultra-reltavisitic flyby in Guarino, R. C. (1985). "Measuring the active gravitational mass of a moving object". The total proper time required to accelerate the baseball will be on the order of the proper length of the baseball divided by the speed of light. This process will distort the baseball a bit, but probably less so than being hit by a baseball bat.

4. Nov 11, 2015

### bcrowell

Staff Emeritus
I tried fiddling around with Chris Hillman's wave of death example from https://www.physicsforums.com/threads/wave-of-death.93654/#post-1176988 . Here is my Maxima code:

Code (Text):
ct_coords:[t,x,y,z];
u:bessel_j(0,exp(t-z))^2;
lg:matrix([1,0,0,0],
[0,-u*exp(-2*exp(t-z)),0,0],
[0,0,-u*exp(2*exp(t-z)),0],
[0,0,0,-1]);
cmetric();
ricci(true);
lriemann(false);
uriemann(false);
exptdispflag:false;
scurvature();/* scalar curvature */
rinvariant(); /* Kretchmann */
Both the scalar curvature and the Kretchmann scalar vanish identically. The Ricci tensor comes out grotty looking, but maybe it does simplify to zero; I think CH's standards for "can easily verify" are a little different from mine. CH only defines the metric for $t-z<c$ "for a certain positive real constant c." I assume this c is the determined by the criterion that the metric becomes degenerate at the first zero of the Bessel function. Presumably this is also where geodesics terminate? It's not obvious to me how you can verify that this spacetime is singular at all.

It also seems to me that the metric becomes degenerate at every zero of the Bessel function

It seems to me that it ought to be relatively easy to come up with a simple, artificial example along these lines if you don't care whether it's a vacuum solution. For that matter, it would probably be equally useful as an educational example to have a Riemannian example. What would be nice would be to find an example where the Kretchmann scalar doesn't diverge, but some other curvature scalar does diverge at some finite affine parameter. Then you could demonstrate that there is actually a singularity.

Last edited: Nov 11, 2015
5. Nov 12, 2015

### bcrowell

Staff Emeritus
Wald has a nice example on p. 216 that is a lot easier to think about than CH's version of the wave of death. I came up with the following example, which is similar to Wald's but simpler and more explicit.

Let $g=\Omega^2\eta$, where $\eta$ is the Minkowski metric in 1+1 dimensions, and let $\Omega^2=1/(1+e^t)$. I used the following Maxima code to analyze its behavior.

Code (Text):
dim:2;
ct_coords:[t,x];
u:1/(1+exp(t));
lg:matrix([u,0],
[0,-u]);
cmetric();
ricci(true);
lriemann(true);
uriemann(true);
exptdispflag:false;
scurvature();/* scalar curvature */
rinvariant(); /* Kretchmann */
From the output, you can tell pretty easily that both the scalar curvature and the Kretchmann invariant are finite everywhere. (The upper- and lower-index forms of the Riemann tensor are also finite everywhere in this coordinate system. The only nonvanishing component of the upper-index one is $R^{xxtt}=-(1/2)e^t(1+e^t)$.)

By symmetry, the line $x=0$ is a geodesic. However, if you integrate the proper time along this geodesic, it approaches a finite limit as $t\rightarrow\infty$. This seems to show that the spacetime is geodesically incomplete.

Does this seem right? Obviously it's not as physically well motivated as CH's example, but it seems to work.

Last edited: Nov 12, 2015
6. Nov 12, 2015

### PAllen

Is this really a curvature singularity or just a case of coordinates describing a manifold that can be readily treated as a submanifold of one that has no geodesic inompleteness?

7. Nov 12, 2015

### bcrowell

Staff Emeritus
Good question, I don't know. What I didn't understand about Wald's example was that he carefully constructed it to be a flat spacetime except within a certain timelike tube. I didn't understand why he did that, thought maybe he just felt like making it asymptotically flat. But maybe this prevents it from being embedded in a geodesically complete manifold? I don't know how to determine that.

Another thing that hadn't occurred to me earlier was that I think it's necessary to show that the curvature scalars are finite not just at all finite t but also as t approaches infinity. After all, an observer in this spacetime sees $t=\infty$ not as an idealized point at infinity but as an actual event that occurs at a finite time. But it does happen to be true that both scalars are finite as $t\rightarrow\infty$:

$R=(1+e^{-t})^{-1}$

$K=(1+e^{-t})^{-2}$

(The expression for K is my own simplification by hand of Maxima's output, checked numerically for equality.)

8. Nov 12, 2015

### PAllen

Well, for starters, while the upper index curvature tensor blows up as t→∞, I'm thinking (in my head, no full calculation), that the lower index does not blow up at t approaching infinite. This suggests t infinite is perfectly regular and only geodesically incomplete by chopping.

9. Nov 12, 2015

### jinawee

Hawking and Ellis mentions that Taub-NUT space has finite scalar polynomials, but is b-incomplete. Although it doesn't seem very physical.

10. Nov 12, 2015

### bcrowell

Staff Emeritus
I don't think Taub-NUT is going to work as a pedagogical example, since it took 50 years to establish whether or not it was geodesically incomplete: http://arxiv.org/abs/1508.07622 .

It occurred to me why Wald confined the curvature to a certain spatial region. It's because his purpose in this example is to show that you can have a spacetime that has incomplete timelike geodesics but not incomplete spacelike or lightlike geodesics.

11. Nov 12, 2015

### bcrowell

Staff Emeritus
It seems like the same considerations apply to Wald's example, within the region where it has curvature.

Last edited: Nov 12, 2015
12. Nov 13, 2015

### PAllen

It seems to me that the geodesic incompleteness definition of a singularity has little merit without methodology for ruling out removal by continuation. Consider the manifold defined by the exterior SC coordinates. It is geodesically incomplete, and in standard SC coordinates several of the curvature tensor components blow up a the horizon. None of the curvature scalars blow up. How is this different from the manifold bcrowell has proposed (or Wald, though I don't have that book to check)? Yet the SC horizon is not considered a singularity.

The reference I posted earlier proposes that if you have a strong curvature singulariy of the type they define, then you can sidestep the (difficult) question possibility of continuation because it won't remove the singularity in a physical sense even if there is some mathematical possibility. For the Tolman-Bondi-Lemaitre models they consider, the parameter range where the Kretschmann scalar does not blow up is also not a strong curvature singularity.

An open question (for me) then, is whether there is any case of a strong curvature singularity (which is not defined in terms of scalars) that is not also a scalar polynomial singularity? That reference also makes the claim that for a 'broad range of manifolds', if you have singularity that cannot be removed by a continuous extension, then it is a strong curvature singularity.

At the moment, I am suspicious that the at least the weak singularities mentioned in this thread are removable by continuous extension.

13. Nov 13, 2015

### bcrowell

Staff Emeritus
@PAllen: What does SC stand for?

BTW, google books won't let me access the link you posted in #2.

What is a strong curvature singularity? Does it mean the same thing as a curvature singularity? Is it defined in the link from #2?

14. Nov 13, 2015

### PWiz

I'm guessing Schwarzschild.

15. Nov 13, 2015

### PAllen

SC stands for Schwarzschild as PWiz guessed.

No, a strong curvature singularity is not the same as a curvature singularity. The reference I posted distinguishes weak and strong curvature singularities. A strong curvature singularity has the feature that a bundle of timelike world lines corresponding to a small volume some 'distance from the singularity' is forced by compression/tension to have zero volume on on approach to the singularity. They define this in a rigorous way. They claim that for most any manifold except type D that is not continuously exensible, if there is any curvature singularity, it is strong. They suggest that many weak curvature singularities may be removable by continuation (especially if one does not require analytic continuation).

16. Nov 13, 2015

### PWiz

This would indicate that the tidal forces experienced by the object approach infinity as the object reaches the end of its geodesic (i.e., as it approaches the point at which the object's worldline will be maximally extended). I can't see how the Riemann tensor/contraction of the Riemann tensor doesn't blow up here.

I don't get this. A singularity isn't a point on the manifold - rather, a manifold is singular if it has one or more geodesics of finite affine length. You can't "deal away" with a singularity if a manifold is maximally extended.

Last edited: Nov 13, 2015
17. Nov 13, 2015

### PWiz

Also, wouldn't it be a good idea to define singularities only for maximally extended spacetimes? This way, you avoid cases like the Schwarzchild one for the EH, and you don't have to concern yourself with continuation issues.

18. Nov 13, 2015

### PAllen

Determining whether a manifold is maximally extended wrestles with the same question as the incompleteness definition of a singularity. Things become much harder if one does not demand global analyticity. For example, there are many non-analytic, smooth extensions of the exterior SC manifold that are topologically distinct from the Kruskal manifold (which is only unique as a globally analytic maximal extension). Some years ago, I posted papers (whose main subject was what is the correct formal statement of Birkhoff's theorem) describing 'odd' smooth manifolds formed by gluing patches of Kruskal together in unusual ways to end up with everywhere spherically symmetric manifolds that are topologically distinct from Kruskal.

Last edited: Nov 13, 2015
19. Nov 13, 2015

### bcrowell

Staff Emeritus
Geroch 1968 has a nice discussion on p. 530 of the role of curvature scalars. He discusses the Kundt type I and II invariants, type I being curvature polynomials and type II being something else that I confess I don't understand very well. (I happen to own a copy of the book in which the Kundt paper appeared. I looked at the paper, and I can't make heads or tails of it.) Geroch says, "For example, in the plane wave solutions all the type I invariants vanish, yet the Riemann tensor is not zero." I don't understand why this would be, but assuming it's true, it seems like a pretty strong motivation to consider criteria other than curvature scalars.

BTW, there is a Stanford Encyclopedia of Philosophy article http://plato.stanford.edu/entries/spacetime-singularities that gives a nice nonmathematical overview of the subject.

Geroch, "What is a singularity in general relativity?," Ann Phys 48 (1968) 526

20. Nov 15, 2015

### bcrowell

Staff Emeritus
I did a little more fiddling around with the example from #5 to see if I can understand it better. Summarizing, with a slight change of notation, I have $ds^2=A(dt^2-dx^2)$, with $A=1/(1+e^t)$, which is conformally related to 1+1-dimensional Minkowski space. For convenience, let $B=1/(1+e^{-t})$, i.e., the time-reversed version of the function $A$. The upper-index Riemann tensor is $R^{xxtt}=-B/2$, the scalar curvature is $R=B$, the Kretchmann invariant is $K=B^2$, and the Christoffel symbols are $\Gamma^t_{tt}=\Gamma^x_{tx}=\Gamma^x_{xt}=\Gamma^t_{xx}=-B/2$. The timelike geodesic $x=0$ is incomplete.

I wanted to see if lightlike geodesics were also complete. For the lightlike geodesic $x=t$, the geodesic equation is $\ddot{t}=B\dot{t}^2$, and similarly for $x$. I wasn't able to solve this differential equation explicitly, but its asymptotic behavior is easy to figure out. For large negative $t$, we have $x=t=\lambda$, while for large positive $t$ we have $x=t=e^\lambda$. So this is kind of a funky situation where at least some of the timelike geodesics are incomplete, but the lightlike geodesics are complete. Since $\lambda$ only goes to infinity logarithmically with $t$ for large $t$, I'm thinking that this example happens to be just on the edge of showing this behavior. Probably with a different choice of the function $A(t)$ one could make the lightlike geodesics incomplete. I don't know whether all the timelike geodesics are incomplete, or only the ones with $x=\text{const}$.

This leaves me wondering how it would be possible to extend the spacetime so as to remove the geodesic incompleteness. Since it's conformal to Minkowski space, I'm visualizing it using the diamond that is the Penrose diagram of Minkowski space. The first thing that occurs to me to try is to tile the plane with diamonds, but several things seem to go wrong here. If we join them along their J+ and J- surfaces, it doesn't seem to make sense, because the lightlike geodesics are already complete, so we have copies of Minkowski space that can't communicate and might as well not be connected. Also, if we're connecting the i+ of this spacetime with the i- of another one in the future, in order to complete the timelike geodesics, there doesn't seem to be any sensible way to define what the metric should be on the second copy, or how to make a coordinate chart that would allow us to make a transition from one to the next.

21. Nov 15, 2015

### Staff: Mentor

22. Nov 15, 2015

### Staff: Mentor

Thread reopened; off-topic posts and responses have been removed.

23. Nov 25, 2015

### bcrowell

Staff Emeritus
I calculated the full set of Carminati-McLenaghan invariants for the example in #5 using the (not very well tested) code described here: https://www.physicsforums.com/threads/fun-with-carminati-mclenaghan-invariants.844556/ . The results are as follows:

Code (Text):

t  t  - 1  2 t  2 t  t  - 1
R= %e  (%e  + 1)  R1= %e  (16 %e  + 32 %e  + 16)  R2= 0  R3=
4 t  4 t  3 t  2 t  t  - 1
%e  (1024 %e  + 4096 %e  + 6144 %e  + 4096 %e  + 1024)  M3=
4 t  4 t  3 t  2 t  t  - 1
%e  (576 %e  + 2304 %e  + 3456 %e  + 2304 %e  + 576)  M4= 0
2 t  2 t  t  - 1
W1= %e  (24 %e  + 48 %e  + 24)  W2=
3 t  3 t  2 t  t  - 1
%e  (288 %e  + 864 %e  + 864 %e  + 288)  M1=
3 t  3 t  2 t  t  - 1
%e  (96 %e  + 288 %e  + 288 %e  + 96)  M2=
4 t  4 t  3 t  2 t  t  - 1
%e  (576 %e  + 2304 %e  + 3456 %e  + 2304 %e  + 576)  M5=
5 t
(%i + 1) %e
5 t  4 t  3 t  2 t  t  - 1
(3456 %e  + 17280 %e  + 34560 %e  + 34560 %e  + 17280 %e  + 3456)
test_end.mac

Unfortunately this Maxima output isn't readable in a proportional font, so you'd probably have to cut and paste into a text editor to see it in a monospaced font. Looking at the curvature scalars that are nonvanishing, they all seem to have finite limits as $t\rightarrow\infty$. However, the CM invariants are just the algebraically complete set of invariants of order 0, i.e., those that don't depend on derivatives of the Riemann tensor. It's possible that there are higher-order curvature scalars that do blow up.