# Condition for close timelike curve?

1. Jun 27, 2013

### ngkamsengpeter

How to test if a metric contain close timelike curve?

I read somewhere that if the space coordinate change from positive to negative then it contain close timelike curve. For example, a metric gmn=-Adt2+Bdr2+Cdθ2+Ddz2, if C is negative, then it contain close timelike curve. Is that correct? Is there any other conditions?

Beside that, what is the meaning if A change from positive to negative? For example A=1-r. When r=1, A=0. What is meaning of A=0? Time stop? For r>1, A<0, what is the meaning for A<0? Is this physically possible?

Thanks.

2. Jun 27, 2013

### WannabeNewton

No that is incorrect. Such a change happens in the Schwarzschild coordinate representation of the Schwarzschild metric but that space-time has no CTCs. Coordinates can contain physically meaningless peculiarities whereas CTCs are an intrinsic property of the global causal structure of a space-time.

Consider first the strongly causal condition for space-times: let $(M,g_{ab})$ be a space-time and let $p \in M$. This space-time is strongly causal if for every neighborhood $U\subseteq M$ of $p$, there exists a neighborhood $V\subseteq U$ of $p$ such that no causal curve intersects $V$ more than once.

If a space-time fails to be strongly causal then it can contain CTCs but it doesn't necessarily have to because it can also fail to be strongly causal if it contains causal curves which come arbitrarily close to intersecting themselves. For this reason, we also define a stably causal space-time as follows: let $p\in M$ and $t^{a} \in T_p M$ be time-like and define a new metric by $\tilde{g}_{ab} = g_{ab} -t_{a}t_{b}$. A space-time $(M,g_{ab})$ is stably causal if there exists a continuous time-like vector field $t^{a}$ such that the space-time $(M, \tilde{g}_{ab})$ contains no CTCs, with $\tilde{g}_{ab}$ defined in the way above using the continuous time-like vector field $t^{a}$.

EDIT: Let me add that $(M,g_{ab})\text{ is stably causal}\Leftrightarrow \exists f\in C^{1}(M) \text{ such that} \nabla^{a}f \text{ is a past directed time-like vector field}$. For a proof of this see Wald p. 198.

Last edited: Jun 27, 2013
3. Jun 27, 2013

### ngkamsengpeter

I am not quite understand. Is there any meaning for the change of sign? Mallett spacetime contain close timelike curve, can you show me how to prove that Mallett spacetime contain close timelike?

4. Jun 27, 2013

### WannabeNewton

Peculiarities in coordinates that can be transformed away with a coordinate transformation have no physical meaning. I edited my post #2 because I made a huge typo and I also added a new condition for the intuitive notion of non-existence of CTCs for general space-times. The argument for why the Mallet space-time contains CTCs is shown in the original paper: http://www.phys.uconn.edu/~mallett/Mallett2003.pdf and is shown explicitly using the exterior metric tensor.

5. Jun 27, 2013

### ngkamsengpeter

The paper said that if l<0, then it contain CTC. That's why it give me an impression that a metric contain CTC if it change sign. Is he use the condition you gave in post #2 to reduce the metric to ds^2=-l dθ^2 and then argue that the metric contain CTC if l<0?

6. Jun 27, 2013

### WannabeNewton

You have to be more specific. In your original post you said if the coefficient in front of any space-like coordinate changes sign. As I mentioned this happens in the Schwarzschild case but there are no CTCs in such a space-time. On the other hand, the Mallet space-time is axisymmetric so it contains an axial killing vector field $(\frac{\partial }{\partial \varphi})^{a}$ whose integral curves are by definition closed. We then choose a coordinate system adapted to the axial killing vector field using an axial coordinate $\varphi$. When $g_{\varphi\varphi}$ changes sign it means $(\frac{\partial }{\partial \varphi})^{a}$ goes from being space-like to being time-like and since the integral curves of $(\frac{\partial }{\partial \varphi})^{a}$ are closed as mentioned above, we now have CTCs. This is also what happens in the Kerr space-time.

The reason it works here is because we have a prescribed vector field whose integral curves form a one-parameter family of closed space-like curves with parameter $\varphi$ which we can adapt a coordinate system to, as mentioned above, and detect when the curves become time-like by checking to see if $g_{\varphi\varphi}$ changes sign. Such a prescription won't exist for arbitrary space-times which is why we have things like strongly causal and stably causal to characterize the non-existence of CTCs in a general setting.

7. Jun 28, 2013

### ngkamsengpeter

Can I say that for any axissymmetry metric of the form gmn=-A(r)dt2+2B(r)dθdt+C(r)dθ^2+D(r)(dr2+dz2 contain a closed timelike curve if C(r)<0?

8. Jun 28, 2013

### WannabeNewton

Yes sir. This is because the coordinate $\theta$ is adapted to a killing vector field $(\frac{\partial }{\partial \theta})^{a}$ whose integral curves are by definition closed. Note that $g_{\mu\nu}(\frac{\partial }{\partial \theta})^{\mu}(\frac{\partial }{\partial \theta})^{\nu} = g_{\theta\theta} = C(r)$ so $C(r)$ is nothing more than the norm of $(\frac{\partial }{\partial \theta})^{a}$. If $C(r) > 0$ then, using your sign convention, $(\frac{\partial }{\partial \theta})^{a}$ will be space-like but if $C(r) < 0$ then $(\frac{\partial }{\partial \theta})^{a}$ will become time-like. Since the integral curves of $(\frac{\partial }{\partial \theta})^{a}$ are closed by definition, when $C(r) < 0$ we will get closed time-like curves. The reason this is physically meaningful even though we used coordinates to get the result is that the norm of a vector field is independent of coordinates i.e. it is a purely geometric quantity. If you want another example of this, check out the Kerr metric. I hope I have been of help :)

9. Jun 28, 2013

### ngkamsengpeter

How to know if a metric is axially symmetry? Why is gmn=-A(r)dt2+2B(r)dθdt+C(r)dθ^2+D(r)(dr2+dz2 is axially symmetry? Because it only depend on r? Is gmn=-A(r)dt2+2B(r)dzdt+C(r)dθ^2+D(r)dr2+E(r)dz2 axially symmetry?
Yes. You help a lot. Thanks.

10. Jun 28, 2013

### WannabeNewton

In your example, the metric can be put into a form that doesn't depend on $\theta$.

11. Jun 28, 2013

### ngkamsengpeter

For this metric gmn=-A(r)dt2+2B(r)dzdt+C(r)dθ^2+D(r)dr2+E(r)dz2, can I say that it contain closed timelike if C<0?

The curve are closed if the killing vector field exist?

Thanks.

12. Jun 28, 2013

### WannabeNewton

Unless I'm missing something completely obvious or something very subtle, yes if $C < 0$ in some open subset of the space-time described by the metric you gave, there should be CTCs (I'm assuming $\theta$ is the coordinate corresponding to an axial killing vector field).

13. Jun 28, 2013

### ngkamsengpeter

What is the meaning of coordinate corresponding to an axial killing vector field? I am new to this, can you provide an example or reference on how to calculate this axial killing vector field?
Thanks.

14. Jun 28, 2013

### WannabeNewton

Forget the stuff about killing vector fields for now. What I'm saying is that since $\theta\in [0,2\pi]$ is a periodic coordinate, the integral curves of the vector field $(\frac{\partial }{\partial \theta})^{a}$ will be closed and if $C(r) < 0$, $(\frac{\partial }{\partial \theta})^{a}$ will also be time-like so the integral curves will correspond to closed time-like curves.

15. Jun 28, 2013

### ngkamsengpeter

Can i say that for any periodic coordinate which change to timelike corresponding to the existence of closed timelike curve?

16. Jun 28, 2013

### WannabeNewton

Yeah if you have a periodic coordinate $\varphi$, the coordinate curves obtained by keeping all coordinates except $\varphi$ constant will be closed. If $\varphi$ becomes time-like in the sense described above then you will have CTCs. I hope that helps. Is this for a project or something or are you asking for your own interest?

17. Jun 28, 2013

### ngkamsengpeter

Ok. Although I still dont understand much on the killing vector part, You help a lot. Thanks.

18. Jun 28, 2013

### WannabeNewton

Don't worry about the killing vector stuff at all. I just mentioned it because that's where the axial coordinate is coming from in the coordinate system in which the metric is independent of said coordinate. The important thing is that the coordinate is periodic so the coordinate curves associated with said coordinate are closed; then if you find a region where said coordinate turns time-like you can say you have closed time-like curves. Best of luck!

Last edited: Jun 28, 2013