# Static spacetimes

1. Nov 8, 2004

### big bob

Why do static spacetimes (i.e. with zero initial expansion, twist, and shear), admit timelike Killing vectors? Any explanation(s) would be much appreciated :)

2. Nov 8, 2004

### Stingray

What are you claiming has zero expansion, twist, and shear? These are properties of congruences in a spacetime, not the spacetime itself.

A static spacetime is usually defined by imposing an extra condition on a stationary spacetime (eg Wald p. 119). Stationary spacetimes are in turn defined as those which admit timelike Killing vectors. So there's really no explanation to give. It's a definition.

3. Nov 8, 2004

### big bob

Sorry, I meant spacetimes with geodesic congruences that have zero expansion, shear, and twist.

Is there any way to see how timelike Killing vectors arise by looking at the Raychaudhuri equation for a static spacetime?

4. Nov 8, 2004

### Stingray

I wasn't aware that this was a definition of a static spacetime. In Schwarzchild, neither the timelike nor axial killing fields have these properties, so I'm skeptical that your congruence exists.

Anyways, if you do have such a congruence, then it is possible to write an arbitrary covariant derivative of it in terms of its acceleration (zero since the field is geodesic), and the expansion, twist, and shear. When you do that, you see that your conditions imply that all derivatives of this field are zero. It is therefore a Killing vector.

5. Nov 8, 2004

### big bob

Thanks. I have a related question:

How can you show that a spacetime with a geodesic congruence that is converging admits a timelike Killing vector?

Thanks again.

6. Nov 8, 2004

### Stingray

I'm not sure. I'm also skeptical about this statement though. If a geodesic congruence is converging in some neighborhood, that certainly does not imply the existence of a timelike Killing vector there. Are you saying the congruence is converging everywhere?

If the statement were true in some sense (with more conditions), then the congruence is not itself the Killing field. All Killing vectors have zero expansion.

7. Nov 8, 2004

### big bob

Hmmm...

I'm trying to show, from Raychaudhuri's equation, that if a fluid is flowing on geodesics with zero shear and zero expansion, then spacetime will have a timelike Killing vector.

I've read that static spacetimes (i.e. with congruences with zero shear, twist and expansion) admit timelike Killing vectors. But I'm confused as to how to make a connection to my problem.

Using a few assumptions I derived that the congruence is initially converging, but I'm not even sure if I'm heading in the right direction. Do you have any ideas?

8. Nov 8, 2004

### Stingray

If the expansion is zero then the fluid is by assumption not converging or expanding. (I'm assuming that you're talking about the entire 4D spacetime, and not initial data on hypersurface).

Is this a vacuum spacetime (with non-gravitating fluid)? If so, then Raychaudhuri's equation implies that the twist equals zero. Then the covariant derivatives of the congruence are all zero. It therefore defines a timelike Killing vector field.

Am I using too many assumptions?

9. Nov 8, 2004

### big bob

Well, the question I'm trying to answer doesn't specify whether to consider zero-expansion/shear over the entire spacetime or just intially.

I'm not sure what you mean by "non-gravitating". My text describes the fluid as a perfect fluid, pressureless, flowing on timelike geodesics.

I'm also not sure what you mean by "covariant derivatives of the congruence". What terms are actually being covariantly differentiated?

Sorry for all the questions. I really appreciate your help.

10. Nov 8, 2004

### Stingray

Label the fluid's 4-velocity by $$u^{a}$$. Then you can show that

$$\nabla_{a} u_{b} = \frac{1}{3} \theta h_{ab} + \sigma_{ab} + \omega_{ab} = \omega_{ab}$$

where the last equality is from the assumptions.

Looking at things again, you don't have to use Raychaudhuri's equation at all.

$$\nabla_{(a} u_{b)}=\omega_{(ab)} = 0$$

since the twist is antisymmetric. It follows that $$u^{a}$$ is a timelike Killing vector.

By the way, which text is this?

11. Nov 8, 2004

### big bob

That's exactly how I originally solved the problem; however, I was told that we can't consider the expansion/shear to be always zero (the text doesn't indicate what to assume). I've been trying for four days to figure how to solve this question in a general way.

The book is Spacetime and Geometry (S. Carrol). The question is 3.15.

Does the problem make sense to you if the expansion and shear are not always zero?

12. Nov 9, 2004

### Stingray

I don't have that book, so I can't look it up. Here's my generalization, though.

Theorem: If there exists a Killing vector field $$\xi^{a}$$ such that $$\xi^{a}=u^{a}$$ and $$\nabla_{a} \xi_{b} =\nabla_{a} u_{b}$$ on some initial spacelike hypersurface, then that Killing vector is equal to $$u^{a}$$ everywhere (for a timelike geodesic congruence).

Use the Killing transport equations to show this. Given the Killing vector and its derivatives are equal to $$u^{a}$$ at a point, its values at other points along a fluid particle's worldline are found through the equations,
$$u^{a} \nabla_{a} \xi_{b} = D\xi_{b}/d\tau = u^{a} L_{ab}$$
$$u^{a} \nabla_{a} L_{bc} = DL_{bc}/d\tau = -R_{bca}^{d} \xi_{d} u^{a}$$

Initially, it is clear that $$D\xi_{b}/d\tau = u^{a} \nabla_{a} \xi_{b} = u^{a} \nabla_{a} u_{b} = 0$$. Also, $$DL_{bc}/d\tau = 0$$ due to Riemann symmetries. It follows that $$u^{a} = \xi^{a}$$ is the only solution to these equations, which proves the theorem.

Now, the argument before about showing that $$u^{a}$$ is Killing holds wherever the expansion and shear are zero. I'm assuming this is on a spacelike hypersurface. Then the assumption of the theorem is true.

So if there exists a global Killing vector, then the congruence represents that vector everywhere. I think that the argument can be generalized to explicitly show existence, but I'm too tired to think of how to do it right now.

13. Nov 9, 2004

### Stingray

And here's a much simpler intuitive idea you can probably work into a proof:

Since there exists a timelike Killing vector on a hypersurface, the metric immediately "above" that hypersurface is identical. It follows that a new congruence may be chosen infinitesimaly higher than the first which satisfies the same properties on the new hypersurface. Then there is a timelike Killing vector here too. Repeat ad infinitum.