Covectors

dextercioby

Yes. Finally :)

latentcorpse

Now for the final part, it is asking us to show that [latex]u_a[/latex] is parallel to [latex]k_a[/latex]. But we have a static spacetime and so [latex]k_a[/latex] will be hypersurface orthogonal. This means we are asked to show that [latex]u_a[/latex] is also hypersurface orthogonal.

So the question is asking us to show that [latex]u_{[a;b}u_{c]}=0[/latex], correct?

Well we know that this spacetime will have energy momentum tensor [latex]T_{ab}=(\rho+p)u_au_b+pg_{ab}[/latex]

So we have [latex]T_{ac;b}=(\rho+p) u_{a;b}u_c+u_au_{c;b}[/latex] since [latex]g_{ac;b}=0[/latex]

Then I tried writing [latex]u_{[a;b}u_{c]}[/latex] in terms of [latex]T[/latex]'s but it didn't get me anywhere. Can you give me a hint please?

dextercioby

If [itex] p [/itex] and [itex] \rho [/itex] are totally unrelated, then (1) is more that enough to prove your statement.

The hint is to consider a full antisymmetrization on all 3 terms of (1).

latentcorpse

I misstyped - it should be:

[latex]T_{ac;b}=(\rho+p)(u_{a;b}u_c+u_au_{c;b})[/latex]

Anyway, I get a ridiculously long expression:

[latex]T_{ac;b}+T_{cb;a}+T_{ba;c}-T_{ca;b}-T_{ab;c}-T_{bc;a}=(\rho+p)(u_{a;b}u_c+u_{b;c}u_a+u_{c;a}u_b-u_{b;a}u_c-u_{a;c}u_b-U_{c;b}u_a+u_au_{c;b}+u_cu_{b;a}+u_bu_{a;c}-u_cu_{a;b}-u_au_{b;c}-u_bu_{c;a}[/latex]
But all the terms on the RHS cancel and so
[latex]\Rightarrow T_{[ac;b]}=0[/latex]

But if [latex]T_{ac;b}=(\rho+p)(u_{a;b}u_c+u_au_{c;b})[/latex]

then we can conclude

[latex]u_{[a;b}u_{c]}+u_{[a}u_{c;b]}=0[/latex]

which isn't quite what we want is it? We want to show [latex]u_{[a;b}u_{c]}=0[/latex], no?

dextercioby

The energy momentum tensor is the RHS of the Einstein field equations, so it must be a symmetric second order tensor. Antisymmetrizing over the 2 free indices it would give 0. It's no need to give a proof for it. It's a trivial thing.

latentcorpse

Ok. Yes, I should have spotted that. Althoughw e still arrive at the wrong conclusion do we not?

dextercioby

The last point of this problem is really nice. I'll give you a hint: What does it mean for the 4-velocity u_a to be parallel to a covariant Killing vector k_a ? After that, I tell you that the solution of the problem involves the Einstein equations.

latentcorpse

Well if [latex]u_a[/latex] is parallel to [latex]k_a[/latex] then it will also be hypersurface orthogonal and hence it will also satisfy [latex]u_{[a;b}u_{c]}=0[/latex]. Am I correct that this is what we are ultimately trying to prove?

Now the Einstein equations read:

[latex]8 \pi ( \rho + p) u_a u_b + 8 \pi p g_{ab} = R_{ab} - \frac{1}{2} R g_{ab}[/latex]

The only constructive thing I can think to do with this is to antisymmetrise it. This tells us [latex]u_au_b=0[/latex] i.e. the product [latex]u_au_b[/latex] is symmetric but I'm not sure that really helps, does it?

dextercioby

If they're parallel, then they must have the same direction, which mean that they (up to the modulus) should coincide when subject to a parallel transport.

This means that there exists a constant 'p' such that

[tex] k_a = p u_a [/tex]

This means that everything you have proven for [itex]k_a[/itex] applies for [itex]u_a [/itex].

latentcorpse

I understand that argument but I'm not sure that we have actually answered what the question is asking us, have we?

How do we go from the fact that [latex]T_{ab}=(\rho + p) u_au_b + pg_{ab}[/latex] to deducing that [latex]u_a[/latex] and [latex]k_a[/latex] are parallel?

Thanks and sorry for dragging this out!

dextercioby

The hint I'm giving you is to use Einstein's equation, one of the points already proved in the problem and one more fact: you know that in R^3 euclidean geometry, 2 vectors are paralel iff their cross product is 0.

This last fact can be extended to the space-time geometry determined by the existence of that particular T_ab.

latentcorpse

Sorry. I'm just not seeing it. Please don't give me the answer, but could you offer a bit of extra help?

dextercioby

You need to use the following things

1. [tex] E_{ab} = kT_{ab} [/tex]

2. [tex] k_{a} R^{a}_{~[b} k_{c]} = 0 [/tex]

3. [tex] \mbox{a ext b} = 0 \Leftrightarrow \mbox{a} = \lambda \mbox{b} [/tex]

3. contains the 1-forms a and b, lambda is a 0-form.

latentcorpse

Sorry again!

Starting from [latex]R_{ab}-\frac{1}{2}Rg_{ab}=8 \pi \rho u_au_b + 8 \pi p u_au_b + 8 \pi pg_{ab}[/latex], what should I do next?
Antisymmetrise? Or Raise an index with a metric and then contract with [latex]k_a[/latex]? Or something else entirely?

Thanks.

dextercioby

To use 2. you have to enter the Ricci curvature in terms of the 4-velocity. Can you do that ?

latentcorpse

Is it as follows?

[latex]R_{ab}=8 \pi ( \rho + p) u_au_b + ( 8 \pi p + \frac{1}{2} R) g_{ab}[/latex]
[latex]R^a{}_b=8 \pi ( \rho + p) u^a u_b + (8 \pi p + \frac{1}{2} R) \delta^a{}_b[/latex]

dextercioby

The separation is not fully done (there's some explicit and some implicit dependence of the Ricci tensor of u), however the second equation will lead you to a successful use of 1. and 2. in what I wrote above.

So post your final equation.