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

[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.

 

[WannabeNewton]

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

What part exactly?

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

Sure. So we know that in this preferred set of coordinates, $\xi^{\mu}$ is parallel to $\nabla^{\mu}t$. Now the space-like hypersurfaces that $\nabla^{\mu}t$ will be orthogonal to will just be the family of surfaces $(\Sigma_{t})_{t \in \mathbb{R}}$ with each member of the family defined by setting the time coordinate $t$ equal to some constant. Take any vector from any tangent space to any member of the family $v^{\mu}\in T_{p}\Sigma_{t}$ then we know that $\nabla^{\mu}t$ will of course be orthogonal to $v^{\mu}$ i.e. $v^{\mu}\nabla_{\mu}t = v^{\mu}\delta ^{t}_{\mu} = v^{t} = 0$. We also know from the above that $u_{i} = 0$ for all spatial indices $i$ thus $v^{\mu}u_{\mu} = v^{t}u_{t} + v^{i}u_{i} = 0$ and since this was an arbitrary vector from an arbitrary tangent space to an arbitrary space-like hypersurface that $\nabla^{\mu}t$ is orthogonal to, this means that the 4-velocity field of the dust $u^{\mu}$ must itself be orthogonal to the space-like hypersurfaces hence parallel to $\nabla^{\mu}t$ which means the 4-velocity is also parallel to $\xi^{\mu}$ since they are mutually parallel to $\nabla^{\mu}t$.

 

[WannabeNewton]

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

I see what you are trying to say now. If we go to a locally inertial frame of a dust particle in the congruence, with locally inertial coordinates $(x'^0,x'^1,x'^2,x'^3)$, then yes in these coordinates $u^{\mu'} = \delta ^{\mu'}_{x'_{0}}$ but this is not covariant (you can see it easily because one of the coordinates explicitly enters the equation). It is only true, in general, in the locally inertial coordinates; the form need not hold in general when you transform to another set of coordinates.

We want to use the preferred coordinates $(t,x^{1},x^{2},x^{3})$ adapted to the time-like and hypersurface orthogonal killing vector field of the static metric because we KNOW that in these coordinates $\xi^{\mu} = \beta \nabla^{\mu}t$ and that for any $v^{\mu}\in T_{p}\Sigma_{t}$, $v^{\mu}\nabla_{\mu}t = v^{t} = 0$; this was the result that was pivotal in proving the final result we wanted. There is no reason why this should hold in the locally inertial coordinates nor even have the same meaning.

If you could always use the locally inertial frame of a particle to claim $u^{i} = 0$ in any coordinate system of any space-time just because it is ubiquitously true in the locally inertial coordinates then test particle trajectories in GR would be rather banal

 

[TrickyDicky]

Thanks, mate.

 

[WannabeNewton]

So anyways, we can now prove a static solution to the EFEs generated by dust cannot exist for positive energy density by explicitly using Raychaudhuri's equation as the problem asks.

First off, we have from above that $u^{\mu} = \zeta \nabla^{\mu}t$ in the preferred coordinate system which implies the covariant equation $u_{[a}\nabla_{b}u_{c]} = 0$. The vorticity tensor is defined as $\omega_{ab} = \nabla_{[b}u_{a]}$. Now, $u_{[a}\nabla_{b}u_{c]} = u_{a}\nabla_{[b}u_{c]} + u_{c}\nabla_{[a}u_{b]} + u_{b}\nabla_{[c}u_{a]} = u_{a}\omega_{cb} + u_{c}\omega_{ba} + u_{b}\omega_{ac}$. Since $\omega_{ab}$ is a purely spatial tensor field and $u^{a}$ is time-like vector field, $u^{a}\omega_{ab} = 0$ so $u_{a}u^{a}\omega_{cb} - u_{c}u^{a}\omega_{ab} + u_{b}u^{a}\omega_{ac} = -\omega_{cb} = 0$ hence the vorticity vanishes identically.

Furthermore, using the fact that $u^{a} = \alpha \xi^{a}$, the expansion $\theta = \nabla_{a}u^{a} = \alpha \nabla_{a}\xi^{a} + \alpha^{3}\xi^{a}\xi^{c}\nabla_{a}\xi_{c} = 0$. Consequently, $\frac{\mathrm{d} \theta}{\mathrm{d} \tau} = u^{b}\nabla_{b}(\nabla_{a}u^{a}) = 0$ thus Raychaudhuri's equation reduces to $\sigma_{ab}\sigma^{ab} + R_{ab}u^{a}u^{b} = 0$ where $\sigma_{ab}$ is the shear. But the strong energy condition tells us that for a positive energy density, $R_{ab}u^{a}u^{b} > 0$ and we know that the shear is also a purely spatial tensor so $\sigma_{ab}\sigma^{ab} \geq 0$ thus $\sigma_{ab}\sigma^{ab} + R_{ab}u^{a}u^{b} > 0$ which is a contradiction.

 

[TrickyDicky]

I see what you are trying to say now. If we go to a locally inertial frame of a dust particle in the congruence, with locally inertial coordinates $(x'^0,x'^1,x'^2,x'^3)$, then yes in these coordinates $u^{\mu'} = \delta ^{\mu'}_{x'_{0}}$ but this is not covariant (you can see it easily because one of the coordinates explicitly enters the equation). It is only true, in general, in the locally inertial coordinates; the form need not hold in general when you transform to another set of coordinates.

We want to use the preferred coordinates $(t,x^{1},x^{2},x^{3})$ adapted to the time-like and hypersurface orthogonal killing vector field of the static metric because we KNOW that in these coordinates $\xi^{\mu} = \beta \nabla^{\mu}t$ and that for any $v^{\mu}\in T_{p}\Sigma_{t}$, $v^{\mu}\nabla_{\mu}t = v^{t} = 0$; this was the result that was pivotal in proving the final result we wanted. There is no reason why this should hold in the locally inertial coordinates nor even have the same meaning.

If you could always use the locally inertial frame of a particle to claim $u^{i} = 0$ in any coordinate system of any space-time just because it is ubiquitously true in the locally inertial coordinates then test particle trajectories in GR would be rather banal

Wait, I wasn't thinking of what you write, rather I was considering why for instance in the usual derivations of the Newtonian limit where a static distribution of matter is used it is stated that this means u0=1, ui=0.

 

[TrickyDicky]

So anyways, we can now prove a static solution to the EFEs generated by dust cannot exist for positive energy density by explicitly using Raychaudhuri's equation as the problem asks.

First off, we have from above that $u^{\mu} = \zeta \nabla^{\mu}t$ in the preferred coordinate system which implies the covariant equation $u_{[a}\nabla_{b}u_{c]} = 0$. The vorticity tensor is defined as $\omega_{ab} = \nabla_{[b}u_{a]}$. Now, $u_{[a}\nabla_{b}u_{c]} = u_{a}\nabla_{[b}u_{c]} + u_{c}\nabla_{[a}u_{b]} + u_{b}\nabla_{[c}u_{a]} = u_{a}\omega_{cb} + u_{c}\omega_{ba} + u_{b}\omega_{ac}$. Since $\omega_{ab}$ is a purely spatial tensor field and $u^{a}$ is time-like vector field, $u^{a}\omega_{ab} = 0$ so $u_{a}u^{a}\omega_{cb} - u_{c}u^{a}\omega_{ab} + u_{b}u^{a}\omega_{ac} = -\omega_{cb} = 0$ hence the vorticity vanishes identically.

Furthermore, using the fact that $u^{a} = \alpha \xi^{a}$, the expansion $\theta = \nabla_{a}u^{a} = \alpha \nabla_{a}\xi^{a} + \alpha^{3}\xi^{a}\xi^{c}\nabla_{a}\xi_{c} = 0$. Consequently, $\frac{\mathrm{d} \theta}{\mathrm{d} \tau} = u^{b}\nabla_{b}(\nabla_{a}u^{a}) = 0$ thus Raychaudhuri's equation reduces to $\sigma_{ab}\sigma^{ab} + R_{ab}u^{a}u^{b} = 0$ where $\sigma_{ab}$ is the shear. But the strong energy condition tells us that for a positive energy density, $R_{ab}u^{a}u^{b} > 0$ and we know that the shear is also a purely spatial tensor so $\sigma_{ab}\sigma^{ab} \geq 0$ thus $\sigma_{ab}\sigma^{ab} + R_{ab}u^{a}u^{b} > 0$ which is a contradiction.

The shear would also be zero for a perfect fluid so Raychaudhuri reduces to $R_{ab}u^{a}u^{b} = 0$ , wouldn't it?

 

[WannabeNewton]

The shear would also be zero for a perfect fluid so Raychaudhuri reduces to $R_{ab}u^{a}u^{b} = 0$ , wouldn't it?

I can't see the shear necessarily vanishing. If it did then it would imply $\nabla_{(a}u_{b)} = 0$ but as I showed in a prior reply to Peter, if the 4-velocity field is parallel to the time-like killing vector field then it won't be a killing vector field itself in general.

 

[WannabeNewton]

Wait, I wasn't thinking of what you write, rather I was considering why for instance in the usual derivations of the Newtonian limit where a static distribution of matter is used it is stated that this means u0=1, ui=0.

The Newtonian limit is a slow motion approximation for the sources on the background flat minkowski space-time so we say beforehand, to first order in velocities, $u^{0} >> u^{i}$ so that $u^{0}\approx 1 + O(v^{2}),u^{i}\approx 0 + O(v^{2})$. In our case we don't know a priori that the dust follows an orbit of the time-like KVF. That is what we are trying to prove assuming only that the dust must generate a static solution to the EFEs. We of course don't want to make any approximations since we are given an arbitrary such static solution.

 

[TrickyDicky]

I can't see the shear necessarily vanishing. If it did then it would imply $\nabla_{(a}u_{b)} = 0$ but as I showed in a prior reply to Peter, if the 4-velocity field is parallel to the time-like killing vector field then it won't be a killing vector field itself in general.

Hmm, then I guess the definition of perfect fluid (and a dust is a pressureless p.f.) that I quote from Wikipedia "Specifically, perfect fluids have no shear stresses", doesn't hold in this particular case? But, why?

The Newtonian limit is a slow motion approximation for the sources on the background flat minkowski space-time so we say beforehand, to first order in velocities, $u^{0} >> u^{i}$ so that $u^{0}\approx 1 + O(v^{2}),u^{i}\approx 0 + O(v^{2})$. In our case we don't know a priori that the dust follows an orbit of the time-like KVF. That is what we are trying to prove assuming only that the dust must generate a static solution to the EFEs. We of course don't want to make any approximations since we are given an arbitrary such static solution.

Yeah, I was not referring to the approximation part, but its true that in the exercise we are not given the static property as starting point , it has to be proved that it would be static only in a specific case.

 

[WannabeNewton]

Hmm, then I guess the definition of perfect fluid (and a dust is a pressureless p.f.) that I quote from Wikipedia "Specifically, perfect fluids have no shear stresses", doesn't hold in this particular case? But, why?

When I said it won't vanish in general I meant for the dust that are generating arbitrary static solutions to the EFEs because they will necessarily follow orbits of the timelike KVF as shown. However as we know such solutions to Einstein's equation can't actually exist so this is where the apparent discrepancy comes from.

 

[WannabeNewton]

Yeah, I was not referring to the approximation part

What were u referring to then?

 

[TrickyDicky]

When I said it won't vanish in general I meant for the dust that are generating arbitrary static solutions to the EFEs because they will necessarily follow orbits of the timelike KVF as shown. However as we know such solutions to Einstein's equation can't actually exist so this is where the apparent discrepancy comes from.

Ok.
I hate these "Imagine...show that.." exercises lol

 

[TrickyDicky]

What were u referring to then?

The text I read said specifically that a static distribution of matter implies ui=0. But one cannot use this in the exercise. I was starting with the assumption that was needed to show, simply that.

 

[WannabeNewton]

The text I read said specifically that a static distribution of matter implies ui=0.

Ok but it's probably good to be careful because the notion of a stationary 4-velocity field only works properly if you're in a stationary space-time where you can define being stationary as following an orbit of the time-like killing vector field. In such a case, $u^{a} = \alpha \xi^{a}$ for any coordinate system we choose. If we choose the coordinate system adapted to $\xi^{a}$ then we see that $u^{\mu} = \alpha \xi^{\mu} = \alpha \delta ^{\mu}_{t}$ hence in this coordinate system $u^{i} = 0$. This of course need not be true for some other coordinate system we choose (it will always be true in a locally inertial frame of course but I already explained why that won't be of much use for this particular exercise in one of the previous posts; it is more advantageous to just choose the coordinate system adapted to the time-like KVF). So all this is in no way restricted to the Newtonian limit.