Why does vacuum energy density cause repulsion in Friedmann equations?

[Helpsearcher]

Hi there,



I really hope someone can help me with my stupid but urgent problem of understanding something crucial about the Friedman equations.



So; one of them looks like this (forget about the constants; it is about the principles):



change of the scale factor with time - density - cosmol. constant = -k (curvature term)


Then this is sometimes rewritten in terms of densities, which gives:



change of the scale factor with time - (density of matter + vacuum energy density) = -k (curvature term)




Now; here is what I do not get.



Generally the density of the vacuum (or equivalently the cosmol. constant) are treated just like the density of matter; so they have the same effect on the curvature, which somehow should be understandable as energy=matter and so both curve the spacetime.

But then, it is usually stated that the cosmol. constant, and so the vacuum energy density, are working against gravitation (repulsive).

However; I do not understand, how to see this in the equations above. I mean; both seem to have the same effect: energy=matter -> attraction (simplified).



Where is my error of thinking?

I really hope that someone here can enlighten me.



Thx

 

[George Jones]

Cosmological gravitational repulsion means that $\dot{a}$, the rate of change of the scale factor, is increasing. i.e., that $\ddot{a}$ is positive. The Friedmann equation about which you wrote is

$$\dot{a}^2 = \frac{8\pi}{3} \rho a^2.$$

Differentiating this equation with respect to time gives

$$2\dot{a}\ddot{a} = \frac{8\pi}{3} \left( \dot{\rho} a^2 +2\rho a \dot{a} \right).$$

Consequently, there is cosmological repulsion when $0 < \dot{\rho} a^2 +2\rho a \dot{a}$.

It is somewhat difficult to see what is going on from this, but a couple of things can be noted:

1) in an expanding universe, there is no gravitational repulsion only when $\dot{\rho} a^2$ is sufficiently negative (as it is for normal matter);

2) for an expanding universe that consists solely of vacuum energy, which has $\dot{\rho} = 0$, there is repulsion, since then $\ddot{a} > 0$.

Things become a little clearer when the other Friedmann equation is considered,

$$\ddot{a} = -\frac{4\pi}{3} \left( \rho + 3p \right) R.$$

Clearly, $\ddot{a} > 0$ when $w = p / \rho$ is less than -1/3. Vacuum energy/cosmological constant has w = -1.

Roughly, (for non-exotic matter that has positive energy density) repulsion happens when pressure is sufficiently negative.

 

[Helpsearcher]

Hi;

thank you for the answers.
I totally missed (or have overseen) the second more important equation in this context (the second derivative), which shows the dependencies on the pressure terms.

Will have a deeper look into now that the examination is over. :) Was a little confused the day before.

Thx