# Critical Density

1. May 29, 2010

### CaptainMarvel

Using the FRW:

$$\left( \frac {\dot{a}} {a} \right)^2 = \frac {8 \pi G \rho} {3} - \frac {k c^2} {a^2}$$

We define critical density by setting k = 0 and rearranging to get:

$$\rho_c = \frac {3 H^2} {8 \pi G}$$

Where:

$$H = \left( \frac {\dot{a}} {a} \right)$$

My question is does $$\rho$$ include the density contribution for Cosmological Constant (dark energy) $$\Lambda$$ or is this derivation only for a Universe with no cosmological constant?

How does one then actually measure the density of Universe?

I know that the density has been measured to be slightly less than the critical density, but I thought we are meant to live in a flat Universe? Is this due to the cosmological constant and how is this reconciled with $$\rho$$ not being exactly $$\rho_c$$?

Finally, I am right in saying that a Universe with $$\rho_c$$ will stop expanding after infinite time, one with $$\rho > \rho_c$$ will collapse back on itself and one with $$\rho < \rho_c$$ will expand forever?

Many thanks.

Last edited: May 29, 2010
2. May 29, 2010

### nicksauce

Yes, $\rho$ includes contribution from the cosmological constant. In other words, we can write the density as a function of scale factor as
$$\rho = \rho_c\left(\Omega_Ma^{-3} + \Omega_Ra^{-4} + \Omega_{\Lambda}\right)$$

This is basically correct, although I don't think, "stop expanding after infinite time" is a well-defined notion.

Fit supernova data and/or CMB data to different models and see what works best.

The measurement of $$\rho$$ is within error of being less than, equal to, or greater than the critical density. People say we live in a "flat universe", because the measured value is very close to the critical density.

Last edited: May 29, 2010
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