Question about fluid tensors

Mentz114

Just to clear things up, this is the metric (b is a constant > 0)
[tex]
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}
[/tex]
and the Einstein tensor is [itex]-3b\ diag(g_{00},g_{11},g_{22},g_{33})[/itex]. In any local frame the metric g can be replaced by η, the Minkowski metric.

PeterDonis

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

[tex]G_{\mu \nu} + \Lambda g_{\mu \nu} = 8 \pi T_{\mu \nu}[/tex]

To find out what the metric [itex]g_{\mu \nu}[/itex] actually is, you have to *solve* the above equation, given some stress-energy tensor [itex]T_{\mu \nu}[/itex]. The Lambdavacuum solution is the solution for which [itex]T_{\mu \nu} = 0[/itex]; there is no "ordinary" stress-energy present. (Some people prefer to move the [itex]\Lambda[/itex] 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.)