# What is cosmological critical density?

I'm taking a first course in cosmology and I'm struggling to understand exactly what critical density is. I've been told "Critical density is the energy density of the universe that corresponds to the Hubble parameter measured today", but what does that really mean?

Is the critical density of the universe simply the "total" energy density, obtained via

$$\rho_{crit}=\rho_{matter}+\rho_{radiation}+\rho_{darkmatter}+\rho_{dark energy},$$

(omitting the types of energy density not present in the universe in question)?

You can think critical density as a some sort of measure of the curvature.

$$H^2 = \frac{8 \pi G}{3c^2}\sum_i\varepsilon_i - \frac{\kappa c^2}{Ra^2}$$

Let us write the equation in this form,

$$H^2 - \frac{8 \pi G}{3c^2}\sum_i\varepsilon_i = - \frac{\kappa c^2}{Ra^2}~~(1)$$

For $$\kappa = 0$$ we need

$$H^2 - \frac{8 \pi G}{3c^2}\sum_i\varepsilon_i=0$$

Lets call $$\sum_i\varepsilon_i = \varepsilon_{crit}$$

In this case we have

$$\varepsilon_{crit} = \frac{3c^2H^2}{8 \pi G}$$

Let us look Equation (1).

When we have a density parameter $$\varepsilon_x$$ such that $$\varepsilon_x > \varepsilon_{crit}$$ the RHS becomes negative so $$\kappa$$ becomes positive ($$\kappa > 0$$)which corresponds to positive curvature or closed universe.

When we have $$\varepsilon_x < \varepsilon_{crit}$$ the RHS becomes positive which corresponds to negative kappa ($$\kappa < 0$$) which describes negative curvature.