Fluid Tensors and the Cosmological Constant

[Messenger]

\nabla_s \left( g^{ks} \Omega \right) = \Omega \nabla_s g^{ks} = 0

because the covariant derivative of the metric vanishes*. So the divergence of the Lambda vacuum SET is zero.

*See https://www.physicsforums.com/showthread.php?t=479553

A covariant derivative is a derivative from which we've removed any contribution that is purely a result of the choice of coordinates. The first derivatives of the metric are the way they are purely as a result of the choice of coordinates.

Hmmm. Perhaps I would be less confused just reading the forum instead of asking questions. Thanks for the link Mentz.

 

[Mentz114]

Ahhhhh. Perhaps I would be less confused just reading the forum instead of asking questions. Thanks for the link Mentz.

Just to clear things up, this is the metric (b is a constant > 0)
<br /> ds^2={d\phi}^{2}\,{r}^{2}\,{sin\left( \theta\right) }^{2}+{dt}^{2}\,\left( b\,{r}^{2}-1\right) +\frac{{dr}^{2}}{1-b\,{r}^{2}}+{d\theta}^{2}\,{r}^{2}<br />
and the Einstein tensor is -3b\ diag(g_{00},g_{11},g_{22},g_{33}). In any local frame the metric g can be replaced by η, the Minkowski metric.

 

[PeterDonis]

So the metric for the Lambdavacuum solution is not diag(-\Lambda,\Lambda,\Lambda,\Lambda)?

No. The cosmological constant \Lambda *multiplies* the metric in the Einstein Field Equation:

G_{\mu \nu} + \Lambda g_{\mu \nu} = 8 \pi T_{\mu \nu}

To find out what the metric g_{\mu \nu} actually is, you have to *solve* the above equation, given some stress-energy tensor T_{\mu \nu}. The Lambdavacuum solution is the solution for which T_{\mu \nu} = 0; there is no "ordinary" stress-energy present. (Some people prefer to move the \Lambda term to the other side of the equation and call it a form of "stress energy"--"dark energy" or something like that. That's a matter of terminology and doesn't change the physics.)