# Dust generated static space-time implications on fluid 4-velocity

1. Apr 6, 2013

### WannabeNewton

Imagine we have a perfect fluid with zero pressure (dust), which generates a solution to Einstein's equations. Show that the metric can be static only if the fluid four-velocity is parallel to the time-like (and hypersurface orthogonal) Killing vector characterizing the static metric.

So, I know the following things:
Energy momentum tensor of dust with 4-velocity field $u^{a}$: $T_{ab} = \rho u_{a}u_{b}$
Hypersurface orthogonality condition of time-like killing vector $\xi ^{a}$: $\xi_{[a}\nabla_{b}\xi_{c]} = 0$

I haven't been able to come up with much. We know that $\nabla_{a}T^{ab} = 0$ and $u^{a}u_{a} = -1$. This tells us that $\nabla_{a}(\rho u^{a}u^{b}) = \rho u^{a}\nabla_{a}u^{b} + u^{b}\nabla_{a}(\rho u^{a}) = 0$ and $u_{a}\nabla_{b}u^{a} = 0$. Hence, $\rho u^{a}u_{b}\nabla_{a}u^{b} + u_{b}u^{b}\nabla_{a}(\rho u^{a}) = 0 \Rightarrow \nabla_{a}(\rho u^{a}) = 0$. This further implies that $\rho u^{a}\nabla_{a}u^{b} = 0$ identically thus $u^{a}\nabla_{a}u^{b} = 0$ so the dust travel on geodesics. However I am having a hard time figuring out how any of this will be useful (if at all) in connecting it to $\xi ^{a}$ and the fact that $\xi_{[a}\nabla_{b}\xi_{c]} = 0$. I highly doubt any of the above will be useful though, it doesn't seem like it would be.

So the goal is to show that if $\xi^{a}$ is in fact a time-like and hypersurface orthogonal killing vector field then $u^{a} = \alpha \xi^{a}$ where $\alpha$ is just the normalization factor. It seems one possible route would be as follows: if I am in the coordinate system $(t,x^{1},x^{2},x^{3})$ adapted to the time-like KVF, i.e. the one where the time derivatives of the metric vanish and $\xi ^{a} = (\frac{\partial }{\partial t})^{a}$ (which in the coordinate basis is just $\xi ^{\mu} = \delta^{\mu}_{t}$), then this amounts to showing that $u^{i} = 0$ (where the $i$'s run over the spatial indices) because then $u^{\mu} = \alpha \delta ^{\mu}_{t} = \alpha \xi ^{\mu}$ and this expression is covariant so it would then have to be true for all coordinate systems which is what we want. Apparently this has to somehow use the fact that $u^{a}$ is the 4-velocity field of dust. However I am unsure of how to go about showing that $u^{i} = 0$ or even if this approach is practical / doable.

Last edited by a moderator: Apr 6, 2013
2. Apr 6, 2013

### WannabeNewton

By the way this is problem 5 of chapter 4 in Carroll's text.

3. Apr 6, 2013

### Mentz114

This sounds almost tautologous.

static spacetime -> time-like KV -> conservation of energy on geodesic -> K.u = E -> parallel u and KV ?

As usual, I probably missed the point, or got it back-to-front.

4. Apr 6, 2013

### WannabeNewton

Well the fact that the dust travel on geodesics comes out of local conservation of the energy momentum tensor of the dust and the normalization of the 4-velocity field (as shown above) $\nabla^{a}T_{ab} = 0,u^{a}u_{a} = -1\Rightarrow u^{a}\nabla_{a}u^{b} = 0$ and the existence of the time-like KVF $\xi ^{a}$ implies $u^{a}\nabla_{a}(\xi_{b}u^{b}) = u^{b}u^{a}\nabla_{a}\xi_{b} + \xi_{b}u^{a}\nabla_{a}u^{b} = 0$ so indeed $\xi_{a}u^{a} = \text{const. along worldlines of dust}$ but why would this then imply $u^{a} = \alpha \xi^{a}$? We also never made use of the fact that $\xi^{a}$ is hypersurface orthogonal i.e. (in the coordinates adapted to the time-like KVF) $\xi^{a} = \beta \nabla^{a}t$ which implies $\xi_{[a}\nabla_{b}\xi_{c]} =0$, which is a covariant expression.

Last edited: Apr 6, 2013
5. Apr 6, 2013

### kevinferreira

Well the question tells you
so why not do the following: assume a general static spacetime metric in a given local coordinate basis
$$g=-A(x)dt^2+g_S(x)$$
Compute the Ricci tensor in function of $A,~g_S$ and its (space) derivatives. Then put everything in Einstein's equations and see what constraints appear. Hopefully something nice, with a bit of work.

6. Apr 7, 2013

### WannabeNewton

On a second read, I am actually not sure what you really intended to say. It also seems you might have meant to say that if $\xi^{a}$ is parallel transported along $u^{a}$ then the two vector fields must be parallel but unfortunately that also won't be true (you can easily parallel transport a vector field along another even if the two vector fields are pointwise orthogonal everywhere - parallel transport will only guarantee that the transported vector field remains constant / unchanged from the perspective of the other vector field). Sadly even if, in the coordinates adapted to the time-like KVF, $u^{\mu} = \alpha \xi^{\mu} = \alpha \delta^{\mu}_{t}$, $u^{\mu}\nabla_{\mu}\xi^{\nu} = \alpha\nabla_{t}\delta^{\nu}_{t} = \alpha\Gamma ^{\nu}_{tt} = \frac{1}{2}\alpha g^{\nu\beta}\partial _{\beta}g_{tt}$ is not equal to zero identically, in general.

Kevin I tried what you said but unfortunately I didn't get anything that would seem to be of use (of course I could have easily made computational errors). The problem is I'm having trouble connecting the fact that $\xi^{a}$ is orthogonal to the space-like foliations and a killing vector field, to the fact that $u^{a}$ is the 4-velocity of dust. All I can deduce from the latter is that the dust travel on geodesics so I can't see an immediate relation between this and the fact that the time-like KVF is hypersurface orthogonal. The only connection I can see between them is that, as mentioned above by Mentz, $\xi_{a}u^{a}$ is constant along $u^{a}$ (which is of course related to the conserved energy of the orbit). I can't find any other connections no matter how hard I try I can't even remotely see how the fact that $\xi_{[a}\nabla_{b}\xi_{c]} = 0$ would somehow relate to the fact that the dust travel on geodesics or even more generally to the fact that the metric of this KVF was generated by the energy momentum tensor of the dust through Einstein's equations (the only reason I mention this hypersurface orthogonality condition is because Carroll explicitly makes you derive it in the previous part of the problem).; a part of me also doesn't want to slog through the calculations involved in finding constraints via the EFEs by plugging in the static metric again - it is so painful T_T lol

7. Apr 7, 2013

### Mentz114

I also can't see any connection with the irrotionality of the KVF. Also a static spacetime may have an irrotational timelike KVF but geodesics may have some vorticity, so I don't think it is a necessary condition.

8. Apr 7, 2013

### TrickyDicky

I'm not sure if this could be of any help but just in case: Both the tangent velocities of dust GR solutions and timelike killing vectors of GR static solutions share the spatial hypersurface orthogonality property, that makes them parallel. In the case of the velocity 4-vector of the dust this comes about due to its stress-energy being a perfect fluid (vorticity-free).
There are of course no static dusts that are solutions of the EFE. This is part c) of Carroll's exercise.

9. Apr 7, 2013

### WannabeNewton

Indeed in general even dust particles (equivalently a time-like geodesic congruence defined by their 4-velocity field) can have non-vanishing vorticity (e.g. Godel space-time) so maybe it might suffice to show that for the special case of a static space-time the time-like geodesic congruence represented by the dust particles must have vanishing vorticity? This would tell us that the 4-velocity field would be hypersurface orthogonal and it is time-like so it should be (not entirely sure just using intuition here) orthogonal to the space-like foliations implying proportionality to the time-like and hypersurface orthogonal KVF but I'll have to see how it goes, thanks.

10. Apr 7, 2013

### WannabeNewton

Hi Tricky Buns (sorry couldn't resist lol)! This is very similar to my most recent post above which I typed out before I saw your response. There are space-time solutions generated by dust in which the 4-velocity field of the dust does not have vanishing vorticity (like the Godel metric) so they don't have to be hypersurface orthogonal in general, unless I am misreading something. However maybe it would be feasible to show that for the case of a static space-time the dust do have vanishing vorticity in which case they must be hypersurface orthogonal and since they are time-like this should (again not too sure) imply they are hypersurface orthogonal to the space-like foliations.

11. Apr 7, 2013

### TrickyDicky

There are no static dusts solutions of the GR eq., remember, it is a kind of trick question I guess.
Perhaps the exercise is only considering the EFE without cosmological constant so Godel solution wouldn't qualify.

12. Apr 7, 2013

### WannabeNewton

How about this space-time: http://en.wikipedia.org/wiki/Van_Stockum_dust because here the dust particles again have non-zero vorticity. I'm not seeing why the 4-velocity field of dust has to necessarily have vanishing vorticity for even a stationary space-time (linked example).

13. Apr 7, 2013

### TrickyDicky

Stationary spacetimes' timelike killing vectors don't have to be hypersurface orthogonal, only static ones have that requirement.

14. Apr 7, 2013

### WannabeNewton

Sure but what does that have to do with the fluid 4-velocity? In your other post you said, if I read you right, that the 4-velocity field of dust is always hypersurface orthogonal i.e. it always has vanishing vorticity which I can't see as being true as exemplified in that link.

15. Apr 7, 2013

### TrickyDicky

You are right, I was only thinking of dusts that are homogeneous and isotropic. And that is not imposed in the exercise, sorry about the red herring.

16. Apr 7, 2013

### Staff: Mentor

As TrickyDicky already pointed out, this seems like a trick question, because I don't see how it's possible to have a static solution with dust; dust has zero pressure, and a perfect fluid can only be held static against its own gravity by positive pressure. In order for there to be a solution with zero pressure but nonzero energy density, the fluid has to be rotating, so it can't be static; it can be stationary, but not static.

I suspect that the real point behind the question may be to help you realize *why* what I just said above is true. [Edit: I see from TrickyDicky's post #8 that indeed it is.] Here's how the argument goes: for the metric to be static, in a coordinate chart in which the timelike KVF is $\partial_t$, and the metric is independent of $t$, the fluid 4-velocity must be of the form $(1 / \sqrt{g_{tt}}) \partial_t$. (This is basically what the problem wants you to prove.) But the proper acceleration of such a 4-velocity is nonzero, as is easily proved (we've done it in other threads). This contradicts the assumption that the fluid has zero pressure.

However, suppose that instead the fluid's 4-velocity is $u \left( \partial_t + \omega \partial_{\phi} \right)$, where $\phi$ is the usual "longitude" angular coordinate. Now a solution is possible with the fluid proper acceleration being zero; you already linked to one such solution (the Van Stockum dust--I don't know if that's the most general solution possible for these conditions). But such a spacetime will not be static; if you work out the metric you will see that there must be, at the very least, a $dt d\phi$ "cross term" in order for the Einstein Field Equation to be satisfied.

17. Apr 7, 2013

### Staff: Mentor

You're right that a general dust does not have to have vanishing vorticity (as you point out, the Van Stockum dust does not). But a *static* dust does; its 4-velocity does have to be hypersurface orthogonal, even though a general dust's 4-velocity does not.

18. Apr 7, 2013

### WannabeNewton

Indeed if we are dealing with dust, which has zero pressure, it is shown (post #1) that the 4-velocity of the fluid satisfies the geodesic equation i.e. $u^{a}\nabla_{a}u^{b} = 0$. However if $u^{a} = \alpha \xi^{a}$ then $u^{a}\nabla_{a}u^{b} = \alpha^{2} \xi^{a}\nabla_{a}\xi^{b} + \alpha^{4} \xi^{a}\xi^{b}\xi^{c}\nabla_{a}\xi_{c} = \alpha^{2} \xi^{a}\nabla_{a}\xi^{b}$ which is not zero in general (as can be seen in the coordinates adapted to $\xi^{a}$). This is of course a contradiction because the 4-velocity field of dust must satisfy the geodesic equation (Carroll says to use Raychaudhuri's equation to find a contradiction but the idea is the same because in the end we are talking about a vanishing vorticity time-like congruence that happens to be geodesic which is impossible because there must be forces on the congruence in order to keep it from having vanishing vorticity).

This is all fine, I am comfortable with the intuition. The problem is in showing that $u^{a} = \alpha \xi^{a}$ is actually true. My plan for this was to show that $u^{a}$ must have vanishing vorticity IF the space-time is static (as I said in a previous post) which must then imply that it is orthogonal to some space-like foliations since $u^{a}$ is time-like but I see no reason why this would imply $u^{a}$ and $\xi^{a}$ must be orthogonal to the same space-like foliations so being parallel is still eluding me.

Last edited: Apr 7, 2013
19. Apr 7, 2013

### Staff: Mentor

It's not just that $u^a$ is orthogonal to *some* spacelike foliation; it has to be orthogonal to a *static* one, in which every spacelike hypersurface is identical. Otherwise the fluid wouldn't be static. If the spacetime is not flat, I believe there is a theorem to the effect that there can be at most one such foliation, which is present only if the spacetime is static (and if so, it is obviously the foliation that's orthogonal to the timelike KVF).

Another way to approach this, which is kind of suggested by the above, would be to show that one can find some scalar function $f$ (which can depend on the space coordinates, but not the time coordinate, i.e., we pick a chart in which the fluid's 4-velocity is a multiple of $\partial_t$) such that $f u^a$ satisfies Killing's equation! This would show that $f u^a$ itself is a timelike Killing vector field. But a non-flat spacetime can have at most one timelike KVF.

20. Apr 7, 2013

### Staff: Mentor

One other thing to note here: since we already know it's a contradiction to have $u^a$ both a geodesic and static, you may have to relax the geodesic assumption (which amounts to relaxing the dust assumption) to prove that $u^{a} = \alpha \xi^{a}$. The intent of the problem may have been to bring in the geodesic assumption only later on, to derive the contradiction, not to derive $u^{a} = \alpha \xi^{a}$.

21. Apr 7, 2013

### WannabeNewton

Peter I'm having trouble understanding what you mean by a static foliation and a static fluid. I've only seen the word static used for metrics / space-times myself so I'm having trouble extrapolating from your comments on how the "static fluid" must have vanishing vorticity and how the "static space-like foliations" must be identical etc.

22. Apr 7, 2013

### Mentz114

Just a thought. Can a geodesic congruence $u^\mu$ be written as a combination of Killing vector fields ?
So that $u^\mu = \alpha^0{\xi_t}^\mu + \alpha^i{\xi_i}^\mu$. If this is the case, then if there is no rotation doesn't the second term disappear ( no conserved momenta ) and leave $u^\mu = \alpha^0{\xi_t}^\mu$ ?

23. Apr 7, 2013

### Staff: Mentor

I don't know if those are actually standard terms, but they seem to me to be obvious usages of the term "static" given its meaning for a spacetime.

A static foliation is a set of hypersurfaces filling the spacetime that (a) are orthogonal to the timelike KVF, and (b) all have identical spatial geometry.

A static fluid is a fluid with similar properties: the fluid can be foliated by spacelike slices that (a) are everywhere orthogonal to the fluid's 4-velocity, and (b) all have identical spatial geometry.

24. Apr 7, 2013

### WannabeNewton

Well first note that you can't always write a set of killing vector fields $\xi ^{(i)}$ as $\frac{\partial }{\partial x^{i}}$ in case you were trying to use that to say they could span $T_{p}M$ and be linearly independent. There are conditions on the vanishing of the commutators of the killing vector fields. Secondly, a set of killing vector fields won't in general form a basis for $T_{p}M$ so there is no a priori reason why $u^{a}\in T_{p}M$ could be written as a linear combination of only the killing vectors. For example in the case of a stationary axisymmetric space-time you only have in general two killing vector fields: the axial and time-like killing vector fields $\xi^{a}, \psi^{a}$ and since $[\xi,\psi] = 0$ you can indeed claim there exist coordinates where $\xi^{a} = (\frac{\partial }{\partial t})^{a}, \psi^{a} = (\frac{\partial }{\partial \phi})^{a}$ but you still need the other two coordinate vector fields $(\frac{\partial }{\partial x^{2}})^{a}, (\frac{\partial }{\partial x^{3}})^{a}$ in order to form a basis for $T_p(M)$.

25. Apr 7, 2013

### Mentz114

Good answer. But I only require that the KVFs form a basis for geodesics, not all $T_{p}M$. I'm also assuming that the KVFs are orthogonal.