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The homogeneous inflaton obeysUnder the slow-roll approximation, the acceleration and kinetic-energy corrections may be neglected, so the field equation and Friedmann equation becomeThus a positive field rolls away from the hilltop while $H$ is approximately constant.

The constant-equation-of-state density scaling givesFor a spatially flat universe dominated by this component, the Friedmann equation gives $H^2\propto a^{-1}$. Integrating $\dot{a}/a\propto a^{-1/2}$ yieldsThis is accelerated expansion because $w=-2/3<-1/3$.

Consider a spatially flat universe containing radiation and a cosmic string network. If their densities are both ${\rho }_{\mathrm{eq}}$ at scale factor $a_{\mathrm{eq}}$, thenIn conformal time $d\tau =dt/a$, the flat Friedmann equation becomesChoosing the Big Bang at $\tau =0$, its expanding solution is

For relativistic species at a common temperature, the mass-equivalent energy density is $\rho ={\pi }^2{g}_{\ast }(k_{B}T{)}^4/(30{\hslash }^3{c}^5)$. Using mass rather than energy density accounts for the powers of $c$ in the Friedmann equation.

In a flat universe containing separately conserved radiation in cosmology and pressureless matter, the Friedmann equation becomes $(a^{\prime }{)}^2=H_0^2({\Omega }_{r,0}+{\Omega }_{m,0}a)$ in conformal time. Setting $a=0$ at $\tau =0$ gives the stated quadratic-plus-linear scale factor. The linear term controls radiation domination and the quadratic term controls matter domination. The formula presumes $a_0=1$, ${\Omega }_{r,0}+{\Omega }_{m,0}=1$ and no cosmological constant.

A noninteracting perfect fluid in general relativity has ${\nabla }_{\mu }{T}_i^{\mu \nu }=0$, so ${\dot{\rho }}_i+3H({\rho }_i+P_i)=0$. If energy is transferred between species, source terms $Q_i$ appear and only their sum vanishes. The Friedmann equation still involves the sum of all component stress tensors because all components share one spacetime metric.

A closed universe containing only a cosmic string network can remain static. Its Friedmann acceleration equation vanishes identically because $P_s=-{\rho }_s/3$, and its Friedmann equation requires ${\rho }_s{a}^2=3K/(8\pi G)$. Once this conserved normalization is tuned, every positive radius gives a static solution. The usual Hubble-normalized cosmological density parameter is undefined at a static epoch.