Current observations show that \begin{equation} \rho \gg p \end{equation} (in appropriate units!); the universe is presently “matter dominated”. We therefore consider models with \begin{equation} p = 0 \end{equation} which are called Friedmann-Robertson-Walker models, or just Friedmann models.

With $p=0$, the spatial component of Einstein's equation becomes \begin{equation} 2a\ddot{a}+\dot{a}^2+k = \Lambda a^2 \end{equation} Remarkably, this equation can be integrated exactly, which is most easily done after first multiplying both sides by $\dot{a}$. After integration, we obtain \begin{equation} a\dot{a}^2 + ka = \frac{\Lambda a^3}{3} + C \label{fried0} \end{equation} where $C$ is an integration constant. Rewriting this equation slightly yields Friedmann's equation, namely \begin{equation} \dot{a}^2 = \frac{C}{a} + \frac{\Lambda a^2}{3} - k \end{equation} But the time component of Einstein's equation tells us that \begin{equation} a (\dot{a}^2+k) = \frac{8\pi\rho+\Lambda}{3}\>a^3 \end{equation} and comparison with ($\ref{fried0}$) implies that \begin{equation} C =\frac{8\pi\rho a^3}{3} \end{equation} Thus, the energy density $\rho$ goes like $1/a^3$, which seems reasonable, since $a^3$ has the dimensions of volume.