# FRW. Dark energy, negative pressure or density?

1. Mar 3, 2015

### binbagsss

<<Follow-up: futher edited to fix LaTeX tags>>

Context: FRW Metric /universe, perfect comological fluid , dark/vacuum energy

So the equation of state $w \rho=p, w=-1, \rho=-p$
so this clealy implies either $p$ or $\rho$ is negative.
Am I correct in thinking which it is, depends on the cosmological constant?
So $cosmo constant >0, \rho>0, p<0$
$cosmo constant <0, \rho <0, p>0?$

Now I look at a Friedmann equation given by: $\Omega -1 = \frac{k}{H^2a^2}$, where $\Omega = \frac{\rho}{\rho_{c}}$, $H=\frac{\dot{a^{2}}}{a^{2}}$

and solving for $a$ as a function of $t$ for $cosmo constant >0$, my book says that in this case $\Omega <0$ ( which I understand if my above reasoning is correct) and so from the Friedmann equation this is only possible if $k=-1$. So this is fine, I agree , but the denominator needs to be greater than zero.

Anyway, it than solves for $cosmo constant >0$, and says all $k=-1,0,1$ are fine. This is my QUESTION. I dont understand how $k=-1$ can be okay for both $cosmo constant >0, <0$ which in turn say different things above which of $\rho$ and $p$ are negative and positive.

Thanks very much for your help.

Last edited by a moderator: Mar 3, 2015
2. Mar 3, 2015

### Matterwave

You have to fix those Latex tags or else this is pretty unreadable...

I see a note by a mentor saying it's been edited for readability, but this still looks pretty unreadable to me.

3. Mar 3, 2015

### Staff: Mentor

Yes.

I think you mean for $cosmo constant < 0$, correct? That's the case for which $\rho < 0$, which makes $\Omega < 0$.

$k$ describes the spatial curvature; it doesn't describe either the density or the pressure associated with the cosmological constant. All this is saying is that "open" spatial slices (i.e., negative spatial curvature) are compatible with both a positive and negative cosmological constant.