# What is critical density

1. Jul 23, 2014

### Greg Bernhardt

Definition/Summary

The critical density is defined to be the density necessary to asymptotically halt the expansion of the universe (i.e. flat or euclidean), slightly less and the universe is 'open' (hyperbolic or saddle shaped), slightly greater and the universe is 'closed' (spherical).

Equations

Einstein field equations (EFE)-

$$G_{\mu\nu}+g_{\mu\nu}\Lambda=\frac{8\pi G}{c^4} T_{\mu\nu}$$

$G_{\mu\nu}$ is the Einstein tensor of curvature (spacetime), $g_{\mu\nu}$ is the metric tensor, $\Lambda$ is the cosmological constant, $8\pi$ is the concentration factor and $T_{\mu\nu}$ is the energy tensor of matter (matter energy).

Friedmann equation-

$$H^2=\frac{8 \pi G}{3} \rho - \frac{k c^2}{a^2}+\frac{\Lambda c^2}{3}$$

where H is a function of time (in this case, the inverse of Hubble time), G is the gravitaional constant, $\rho$ is density in $kg/m^3$, k is the spatial curvature parameter (-1 to +1, -k is hyperbolic, 0 is flat and +k is hyperspherical), a is the time-scale factor (0 to 1, now being 1) ($k/a^2$ being the spatial curvature in any time-slice of the universe) and $\Lambda$ is the cosmological constant. (G, $\Lambda$ and c are universal constants, k is a constant throughout a solution and H, $\rho$, and a are a function of time. a is established by $a=1/(1+z)$ where z is the redshift.

Critical density equation-

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

where $\rho_c$ is the critical density.

Extended explanation

Einstein field equations (EFE)-

$$G_{\mu\nu}+g_{\mu\nu}\Lambda=\frac{8\pi G}{c^4} T_{\mu\nu}$$

c and G are introduced to convert the quantity (which is expressed in physical units) to geometric units (G/c4 is used to convert units of energy into geometric units while G/c2 is used to convert units of mass, when mass is used instead of energy, the c4 is replaced with c2).

Using the EFE to establish Λ (presuming that gμν=1)

$$\Lambda =\frac{8\pi G}{c^4}u_{\Lambda}\ \equiv\ \ \frac{8 \pi G}{c^2}\rho_{\Lambda}$$

where uΛ is vacuum energy and ρΛ is vacuum density (basically, uΛ=ρΛ·c2)

The equation is more commonly known as-

$$\rho_{vac}=\frac{\Lambda c^2}{8\pi G}$$

where ρvac is essentially ρΛ

Critical density (ρc)-

The critical density is derived from the Friedmann equation-

$$H^2=\frac{8 \pi G}{3} \rho - \frac{k c^2}{a^2}+\frac{\Lambda c^2}{3}$$

substituting for Λ, the equation can be rewritten-

\begin{align*} H^2&=\frac{8 \pi G}{3} \rho - \frac{kc^2}{a^2}+\frac{8 \pi G}{3}\rho_{\Lambda}\\ &=\frac{8 \pi G}{3} (\rho+\rho_{\Lambda}) - \frac{kc^2}{a^2} \end{align*}

where $\rho_c=(\rho+\rho_{\Lambda})$, ρ representing baryonic and dark matter, ρΛ representing dark energy.

if we considered a flat universe, then k=0 and the equation can be reduced to-

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

rewriting the equation relative to ρc, the critical density for a flat universe is-

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

Based on a Hubble constant of ~70 (km/s)/Mpc, a critical density of 0.918x10-26 kg/m3 is equivalent to 0.825 joules per km3.

Another short hand derivation based on exact escape velocity (i.e. kinetic energy cancelling out gravitational potential) is-

$$E=0=\frac{1}{2}v^2-\frac{Gm}{r}$$

$$\Rightarrow v_{esc}=\sqrt{\frac{2Gm}{r}}$$

if m is rewritten as-

$$m=V\rho$$

where V is volume of a sphere $V=(4/3)\pi r^3$ and substitute-

$$v_{esc}=\sqrt{\frac{8\pi G}{3}\rho r^2}$$

substituting for Hr=v (multiplying the inverse of Hubble time by the radius of the observable universe equals c) which would imply ρ=ρc-

$$Hr=\sqrt{\frac{8\pi G}{3}\rho_c r^2}$$

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

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

rearrange relative to ρc-

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

* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!