What the Universe Is Made Of

Section 3 of 7

The Critical Density

Part 2: The Critical Density

The division above introduced a yardstick: the critical density, the density that makes space exactly flat.

Set the curvature term to zero ( $k = 0$ ) in the Friedmann equation and solve for the density: $\rho_\text{crit} = 3H_0^2/8\pi G$ . A universe denser than this curves closed; less dense curves open; exactly this is flat. Every cosmic density is then quoted as a fraction of it, $\Omega_i = \rho_i/\rho_\text{crit}$ .

Critical density

The density $\rho_\text{crit} = 3H_0^2/8\pi G$ that makes the universe spatially flat. About $9\times10^{-30}~\mathrm{g\,cm^{-3}}$ — roughly five hydrogen atoms per cubic meter. Densities are measured against it as $\Omega_i = \rho_i/\rho_\text{crit}$ .

Worked Example 1Five Atoms per Cubic Meter

Problem

Compute the critical density today from $\rho_\text{crit} = 3H_0^2/8\pi G$ , using $H_0 = 2.27\times10^{-18}~\mathrm{s^{-1}}$ and $G = 6.67\times10^{-8}~\mathrm{cm^3\,g^{-1}\,s^{-2}}$ . Then express it in hydrogen atoms per cubic meter ( $m_H = 1.67\times10^{-24}~\mathrm{g}$ ).

StepPlug in

$\rho_\text{crit} = \dfrac{3(2.27\times10^{-18})^2}{8\pi(6.67\times10^{-8})} = \dfrac{1.55\times10^{-35}}{1.68\times10^{-6}} = 9.2\times10^{-30}~\mathrm{g\,cm^{-3}}$ .

Dimensional check

$\dfrac{(\mathrm{s^{-1}})^2}{\mathrm{cm^3\,g^{-1}\,s^{-2}}} = \dfrac{\mathrm{s^{-2}}}{\mathrm{cm^3\,g^{-1}\,s^{-2}}} = \mathrm{g\,cm^{-3}}$ — a density, as required.

Result

Dividing by $m_H$ : $9.2\times10^{-30}/1.67\times10^{-24} = 5.5\times10^{-6}~\mathrm{atoms\,cm^{-3}}$ , or about 5 hydrogen atoms per cubic meter. The average density of the entire universe is five protons in the volume of a small refrigerator — and that number decides its shape and fate.