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

[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.

 

[WannabeNewton]

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

 

[Mentz114]

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.

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.

 

[WannabeNewton]

This sounds almost tautologous.

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

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.

 

[kevinferreira]

Well the question tells you

... the metric can be static only if the fluid four-velocity...

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 [itex]A,~g_S[/itex] and its (space) derivatives. Then put everything in Einstein's equations and see what constraints appear. Hopefully something nice, with a bit of work.

 

[WannabeNewton]

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

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

 

[Mentz114]

...
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.)
...

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.

 

[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.

 

[WannabeNewton]

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.

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.

 

[WannabeNewton]

tangent velocities of dust GR 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).

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.

 

[TrickyDicky]

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, as you said using the energy momentum tensor of dust, 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.

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.

 

[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).

 

[TrickyDicky]

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).

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

 

[WannabeNewton]

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

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.

 

[TrickyDicky]

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.

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.

 

[PeterDonis]

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.

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 [itex]\partial_t[/itex], and the metric is independent of [itex]t[/itex], the fluid 4-velocity must be of the form [itex](1 / \sqrt{g_{tt}}) \partial_t[/itex]. (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 [itex]u \left( \partial_t + \omega \partial_{\phi} \right)[/itex], where [itex]\phi[/itex] 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 [itex]dt d\phi[/itex] "cross term" in order for the Einstein Field Equation to be satisfied.

 

[PeterDonis]

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.

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.

 

[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.

 

[PeterDonis]

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.

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 [itex]f[/itex] (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 [itex]\partial_t[/itex]) such that [itex]f u^a[/itex] satisfies Killing's equation! This would show that [itex]f u^a[/itex] itself is a timelike Killing vector field. But a non-flat spacetime can have at most one timelike KVF.

 

[PeterDonis]

The problem is in showing that ##u^{a} = \alpha \xi^{a} ## is actually true.

One other thing to note here: since we already know it's a contradiction to have [itex]u^a[/itex] 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} ##.

 

[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.

 

[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## ?

 

[PeterDonis]

I'm having trouble understanding what you mean by a static foliation and a static fluid.

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.

 

[WannabeNewton]

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## ?

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)##.

 

[Mentz114]

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)##.

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.

 

[WannabeNewton]

Another way to approach this, which is kind of suggested by the above, would be to show that one can find some scalar function [itex]f[/itex] (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 [itex]\partial_t[/itex]) such that [itex]f u^a[/itex] satisfies Killing's equation! This would show that [itex]f u^a[/itex] itself is a timelike Killing vector field. But a non-flat spacetime can have at most one timelike KVF.

Well if the goal is to show that ##fu^{a} = \xi^{a}## by use of the uniqueness of the time-like killing vector field then we know that ##f## necessarily has to be ##(-\xi_{a}\xi^{a})^{1/2}## so it comes down to showing that ##\nabla_{(a}(-\xi_{a}\xi^{a})^{1/2}u_{b)} = 0## but I am having trouble showing this in both a coordinate free manner and even in the coordinates adapted to the time-like KVF.

One interesting thing is that since ##T_{ab} = \rho u_{a}u_{b}## has to be time-independent (otherwise the metric tensor generated by this ##T_{ab}## will be time dependent) we know that ##u^{a}## has to be time-independent. If we compute the lie derivative ##\mathcal{L}_{\xi}u^{b} = \xi^{a}\partial_{a}u^{b} - u^{a}\partial_{a}\xi^{b}## in the coordinates adapted to the time-like KVF we find that ##\mathcal{L}_{\xi}u^{\nu} = \delta^{\mu}_{t}\partial_{\mu}u^{\nu} - u^{\mu}\partial_{\mu}\delta^{\nu}_{t} = \partial_{t}u^{\nu} = 0## implying ##\mathcal{L}_{\xi}u^{b} = 0## for any coordinate system although I cannot really think of anything relating to vanishing lie derivatives at the moment that would imply the two vector fields would have to be linearly dependent. Maybe if I can somehow connect this to the fact that ##\xi^{a}## is hypersurface-orthogonal...

The other thing is that all we know beforehand is ##u^{a}## is the 4-velocity field of dust and that the metric tensor generated by the dust is static i.e. there exists a time-like killing vector field ##\xi^{a}## such that in a preferred set of coordinates ##\xi^{a} = \beta \nabla^{a}t## (the t = const. hypersurfaces necessarily all have the same geometry since ##g_{\mu\nu}(t_1) = g_{\mu\nu}(t_2)## for any ##t_{1},t_{2}##). How would we gather from just this that ##u^{a} = \alpha \nabla^{a}t## because this is what we want to ultimately show? Any thoughts or hints?

 

[PeterDonis]

If we compute the lie derivative ##\mathcal{L}_{\xi}u^{b} = \xi^{a}\partial_{a}u^{b} - u^{a}\partial_{a}\xi^{b}## in the coordinates adapted to the time-like KVF we find that ##\mathcal{L}_{\xi}u^{\nu} = \delta^{\mu}_{t}\partial_{\mu}u^{\nu} - u^{\mu}\partial_{\mu}\delta^{\nu}_{t} = \partial_{t}u^{\nu} = 0## implying ##\mathcal{L}_{\xi}u^{b} = 0## for any coordinate system

You realize that "vanishing Lie derivative" is the definition of a Killing vector field, right?

 

[WannabeNewton]

You realize that "vanishing Lie derivative" is the definition of a Killing vector field, right?

The definition of a killing vector field is that ##\mathcal{L}_{\xi}g_{ab} = 0## i.e. the metric tensor is constant along the flow of the KVF. There is nothing that says ##\mathcal{L}_{\xi}u^{a} = 0## a priori.

 

[PeterDonis]

The definition of a killing vector field is that ##\mathcal{L}_{\xi}g_{ab} = 0## i.e. the metric tensor is constant along the flow of the KVF. There is nothing that says ##\mathcal{L}_{\xi}u^{a} = 0## a priori.

No, but when you translate the Lie derivative of the metric, [itex]\mathcal{L}_u g_{ab}[/itex], into component form, you get:

[tex]\nabla_a u_b + \nabla_b u_a = 0[/tex]

where [itex]\nabla_a[/itex] is the covariant derivative with respect to the metric. (Note that you want the Lie derivative with respect to [itex]u[/itex], not [itex]\xi[/itex], because you want to show that [itex]u[/itex] is a KVF.)

This should be equivalent to what you wrote in your previous post; I know the covariant derivative includes terms in the connection coefficients, but I think they will end up canceling in this case.

 

[WannabeNewton]

No, but when you translate the Lie derivative of the metric, [itex]\mathcal{L}_u g_{ab}[/itex], into component form, you get:

[tex]\nabla_a u_b + \nabla_b u_a = 0[/tex]

where [itex]\nabla_a[/itex] is the covariant derivative with respect to the metric. (Note that you want the Lie derivative with respect to [itex]u[/itex], not [itex]\xi[/itex], because you want to show that [itex]u[/itex] is a KVF.)

I was not trying to show that ##u^{a}## was a killing vector field in that calculation. I was trying to show that it was lie transported along ##\xi^{a}## and seeing if I can link this to ##\xi^{a}## being hypersurface orthogonal to see if I can conclude something about the linear dependence of the two vector fields. ##\mathcal{L}_{\xi}u^{a} = 0## has nothing to do with the definition of a killing vector field as something that lie transports the metric tensor.

Also, ##\nabla_{(a}u_{b)} = 0## is not even true for this case if ##u^{a} = \alpha \xi^{a}## because ##\frac{1}{2}\nabla_{(a}\alpha\xi_{b)} = \xi_{a}\nabla_{b}\alpha + \xi_{b}\nabla_{a}\alpha = \alpha^{3}(\xi_{a}\xi^{c}\nabla_{b}\xi_{c} + \xi_{b}\xi^{c}\nabla_{a}\xi_{c})## and in the coordinates adapted to ##\xi^{a}## this turns out to be ##\frac{1}{2}\nabla_{(\mu}\alpha\xi_{\nu)} = \alpha^{3}(\delta ^{t}_{\mu}\Gamma ^{t}_{t\nu} + \delta ^{t}_{\nu}\Gamma ^{t}_{t\mu}) = \alpha^{3}(\frac{1}{2}\delta ^{t}_{\mu}g^{tt}\partial_{\nu}g_{tt} + \frac{1}{2}\delta ^{t}_{\nu}g^{tt}\partial_{\mu}g_{tt})## (after using the fact that ##g_{ti} = 0## for any spatial index ##i## and that time derivatives of ##g_{\mu\nu}## vanish). This doesn't vanish in general.

 

[PeterDonis]

Also, ##\nabla_{(a}u_{b)} = 0## is not even true for this case if ##u^{a} = \alpha \xi^{a}##

Hm, yes, I had forgotten that if ##u^{a} = \alpha \xi^{a}##, ##\alpha## can't be a constant; it has to depend in general on the spatial coordinates. A non-constant multiple of a KVF is not itself a KVF. So that trick won't work.

 

[WannabeNewton]

Hm, yes, I had forgotten that if ##u^{a} = \alpha \xi^{a}##, ##\alpha## can't be a constant; it has to depend in general on the spatial coordinates. A non-constant multiple of a KVF is not itself a KVF. So that trick won't work.

Yeah. Either we are missing something very obvious or this problem was made to make me wrack my head in fury lol. Either way I don't think I'm getting much sleep tonight :p

 

[TrickyDicky]

I think the best way to proceed here is to see the implications of what it would mean to have a static dust perfect fluid(ignoring for a moment that such a solution for the EFE doesn't exist). You have to find a rest frame for the fluid, and in order to do this a logical way would be to choose a congruence of timelike geodesics for which one can say that the fluid as a whole is in a rest frame, the natural choice seems to be a congruence that is vorticity-free, and this would have a tangent vector hypersurface orthogonal and therefore parallel to the timelike killing vector.

 

[WannabeNewton]

Well here's a rather simple way to argue it. First go to the coordinates ##(t,x^1,x^2,x^3)## adapted to the time-like and hypersurface orthogonal killing vector field. Since the dust generates a static space-time solution to the EFEs, we have that ##R_{ti} - \frac{1}{2}g_{ti}R = 8\pi T_{ti}## where the ##i##'s run over the spatial indices. Now we know that ##g_{ti} = 0## so we are left with ##R_{ti} = R^{j}{}{}_{t}{}{}_{j}{}{}_{i} = 8\pi T_{ti}##. We then have that ##R^{j}{}{}_{t}{}{}_{j}{}{}_{i} = \partial _{j}\Gamma ^{j}_{it} - \partial _{i}\Gamma ^{j}_{jt} + \Gamma ^{j}_{jk}\Gamma ^{k}_{it} - \Gamma ^{j}_{ik}\Gamma ^{k}_{jt}##. Now, for any two spatial indices ##i,j##, we find that ##\Gamma ^{j}_{it} = \frac{1}{2}g^{jk}(\partial _{i}g_{kt} + \partial _{t}g_{ik} - \partial _{k}g_{it}) = 0## hence ##R_{ti} = R^{j}{}{}_{t}{}{}_{j}{}{}_{i} = 0## thus ##T_{ti} = \rho u_{t}u_{i} = 0## identically which means that either ##u_{t} = 0## or ##u_{i} = 0## for all ##i## but ##u_{t} = g_{\mu t}u^{\mu} = g_{tt}u^{t} ## and since the 4-velocity field is time-like we know that ##u^{t}## cannot vanish everywhere and we also know that ##g_{tt}## cannot vanish everywhere (if it did then ##g_{\mu t} = 0## for all ##\mu## but this would make ##g_{\mu\nu}## singular which cannot happen) implying ##u_{t}## cannot vanish everywhere so we conclude that ##u_{i} = 0## for all ##i##.

We also know that in these coordinates, ##\xi^{\mu} = \beta \nabla^{\mu}t## and that for any ##v^{\mu}\in \Sigma_{t}##, ##v^{\mu}\nabla_{\mu}t = v^{t} = 0##. Thus, ##v^{\mu}u_{\mu} = v^{t}u_{t} + v^{i}u_{i} = 0## for any ##v^{\mu}\in \Sigma_{t}## therefore ##u_{\mu} = \zeta \nabla_{\mu}t\Rightarrow u^{\mu} = \zeta \nabla^{\mu}t ## so since both ##\xi^{\mu}## and ##u^{\mu}## are parallel to ##\nabla^{\mu}t##, we can finally conclude that ##u^{\mu} = \alpha \xi^{\mu}## and since this is covariant we can claim ##u^{a} = \alpha \xi^{a}## for any coordinate system.

 

[TrickyDicky]

It's great how you can translate hand-waving into math.

Well here's a rather simple way to argue it. First go to the coordinates ##(t,x^1,x^2,x^3)## adapted to the time-like and hypersurface orthogonal killing vector field. Since the dust generates a static space-time solution to the EFEs, we have that ##R_{ti} - \frac{1}{2}g_{ti}R = 8\pi T_{ti}## where the ##i##'s run over the spatial indices. Now we know that ##g_{ti} = 0## so we are left with ##R_{ti} = R^{j}{}{}_{t}{}{}_{j}{}{}_{i} = 8\pi T_{ti}##. We then have that ##R^{j}{}{}_{t}{}{}_{j}{}{}_{i} = \partial _{j}\Gamma ^{j}_{it} - \partial _{i}\Gamma ^{j}_{jt} + \Gamma ^{j}_{jk}\Gamma ^{k}_{it} - \Gamma ^{j}_{ik}\Gamma ^{k}_{jt}##. Now, for any two spatial indices ##i,j##, we find that ##\Gamma ^{j}_{it} = \frac{1}{2}g^{jk}(\partial _{i}g_{kt} + \partial _{t}g_{ik} - \partial _{k}g_{it}) = 0## hence ##R_{ti} = R^{j}{}{}_{t}{}{}_{j}{}{}_{i} = 0## thus ##T_{ti} = \rho u_{t}u_{i} = 0## identically which means that either ##u_{t} = 0## or ##u_{i} = 0## for all ##i## but ##u_{t} = g_{\mu t}u^{\mu} = g_{tt}u^{t} ## and since the 4-velocity field is time-like we know that ##u^{t}## cannot vanish everywhere and we also know that ##g_{tt}## cannot vanish everywhere (if it did then ##g_{\mu t} = 0## for all ##\mu## but this would make ##g_{\mu\nu}## singular which cannot happen) implying ##u_{t}## cannot vanish everywhere so we conclude that ##u_{i} = 0## for all ##i##.

Couldn't this part be saved by saying the 4-velocity u in the rest frame has components (1,0,0,0) ?

We also know that in these coordinates, ##\xi^{\mu} = \beta \nabla^{\mu}t## and that for any ##v^{\mu}\in \Sigma_{t}##, ##v^{\mu}\nabla_{\mu}t = v^{t} = 0##. Thus, ##v^{\mu}u_{\mu} = v^{t}u_{t} + v^{i}u_{i} = 0## for any ##v^{\mu}\in \Sigma_{t}##

Could you put this bit into words? I think I get the drift but just to make sure.