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Use units $c=\hslash =k_B=1$, and write $H=\dot{a}/a$ for the Hubble parameter, with dots denoting cosmic time. Differentiating the Friedmann equation givesOn an interval where $H\ne 0$, use the Hubble parameter identity $\dot{H}=\ddot{a}/a-H^2$ and the Friedmann acceleration equation to findSubstitution cancels the curvature term and gives the continuity equationThe cosmological perfect-fluid continuity equation extends by continuity through a regular isolated turning point. Physically it states that the change of energy in a comoving volume is the negative of the pressure work: $d(\rho a^3)=-Pd(a^3)$.

For separately conserved cosmological fluids, each component obeys this equation individually. A constant equation-of-state parameter $w_i=P_i/{\rho }_i$ therefore giveswhere the present scale factor is normalized to $a_0=1$. Thus the component density laws areThe extra factor for radiation in cosmology is the cosmological redshift of each photon's energy; pressureless matter has only number dilution, while the specified dark energy is a cosmological constant.

The critical density at a given expansion rate is the total density that makes the spatial curvature vanish:These cosmological density parameters use the critical density at that same time. Put $E(a)=H(a)/H_0$ and ${\Omega }_{K,0}=-K/H_0^2$. The Friedmann equation yieldsConsequently the fractional-density evolution, rather than just the component-density evolution, isIn particular ${\sum }_i{\Omega }_i=1+K/(a^2{H}^2)$. The simple powers of $a$ alone apply to ${\rho }_i$, or to ${\rho }_i/{\rho }_{\mathrm{crit},0}$, not to the instantaneous cosmological density parameters. At a recollapse turning point $H=0$, those instantaneous ratios are undefined even though the component densities remain finite.

Use units $c=1$ and cosmic time $t$. Define the Hubble parameter $H=\dot{a}/a$, the equation-of-state parameter $\omega =P/\rho$, and the cosmological density parameterHere ${\rho }_{\mathrm{crit}}$ is the critical density. The definition requires $H\ne 0$; assume positive energy density and an expanding branch so that $\log a$ is a valid time coordinate. The Friedmann equation gives $\Omega -1=K/(a^2{H}^2)$. The Friedmann acceleration equation and Hubble parameter identity giveDifferentiating the Friedmann equation and combining with this identity yields the cosmological perfect-fluid continuity equation $\dot{\rho }=-3H(\rho +P)$. ConsequentlyThus the cosmological density parameter flow isThis identity allows time-dependent $\omega$; constancy is needed only for the simple power-law integration in the next part.

Differentiate the Friedmann equation and use energy conservation to obtain $\dot{H}=-4\pi G(\rho +P/c^2)$ on the expanding branch, where $H\ne 0$. Since $\ddot{a}/a=\dot{H}+H^2$,For pressureless matter, conservation gives ${\rho }_M={\rho }_{M0}{a}^{-3}$. In terms of the specified cosmological density parameters, flatness implies ${\Omega }_M+{\Omega }_{\Lambda }=1$, andSet $b=a^{3/2}$ and take ${\Omega }_M>0$. Then $\dot{b}=\frac{3}{2}{H}_0\sqrt{{\Omega }_M+{\Omega }_{\Lambda }{b}^2}$. Integrating with the Big Bang at $t=0$ givesFor another time origin replace $t$ by $t-t_B$. The cosmological constant has effective density ${\rho }_{\Lambda }=3H_0^2{\Omega }_{\Lambda }/(8\pi G)$. Equality with the matter density means $a^3={\Omega }_M/{\Omega }_{\Lambda }$, soIf ${\Omega }_M=0$, the expanding solution is pure exponential de Sitter scale factor in flat slicing and there is no finite matter-density equality time or dust Big Bang; the stated hyperbolic-sine formula presumes a nonzero dust component.

Choose a reference epoch with $H_0=\dot{a}(t_0)/a_0\ne 0$ and let $x=a/a_0$. Define the critical density and each cosmological density parameter byFor a closed universe ${\Omega }_{K,0}<0$. Take the dark-energy component to be a cosmological constant, as indicated by its $\Lambda$ label. The constant-equation-of-state density scaling givesSubstitution into the Friedmann equation givesAt the reference epoch, ${\Omega }_{r,0}+{\Omega }_{m,0}+{\Omega }_{s,0}+{\Omega }_{w,0}+{\Omega }_{\Lambda ,0}+{\Omega }_{K,0}=1$. The Friedmann acceleration equation isStrings have ${\rho }_s+3P_s=0$ and make no direct acceleration contribution, although they affect the Friedmann constraint. The normalized density parameters are undefined at a static epoch, where $H_0=0$; the static solution must therefore be described directly with dimensional densities.

In the decelerating hot Big Bang model, the particle horizon at recombination encloses only a finite comoving region. Widely separated parts of the Cosmic microwave background therefore had no overlapping past light cones in that model, although they have almost the same temperature. This is the Horizon problem. During cosmic inflation, accelerated expansion makes the comoving Hubble radius $(aH{)}^{-1}$ shrink. A small region that was initially in causal contact can then expand to contain the whole presently observable universe.

The Flatness problem follows fromDuring ordinary radiation- or matter-dominated decelerating expansion this quantity grows, so the observed closeness of the cosmological density parameter $\Omega$ to one requires extreme tuning at early times. Inflation makes $aH$ grow nearly exponentially and therefore drives $|\Omega -1|$ rapidly toward zero. A sufficiently large number of e-folds makes spatial flatness a dynamical outcome rather than a finely tuned initial condition.

For a component with constant barotropic equation of state $P=w\rho$, the cosmological perfect-fluid continuity equation givesHence ${\rho }_i={\rho }_{i,0}{a}^{-3(1+w_i)}$. Choose the present scale factor to be $a_0=1$, so the cosmological redshift obeys $1+z=a^{-1}$, and define the present cosmological density parameter bySubstitution in the spatially flat Friedmann equation then yields

The cosmological perfect-fluid continuity equation isFor constant equation-of-state parameter $w=p/\rho$, integration giveswhere $a_0=1$. Define the present cosmological density parameterThe Friedmann equation then givesSpatial curvature may be included as an effective $w=-1/3$ component. Once the parameters are specified, the scale factor follows from the first-order equation $\dot{a}=aH(a)$, or equivalently

For a component with the barotropic equation of state $P=w\rho$, the cosmological perfect-fluid continuity equation isUsing $H=\dot{a}/a$ makes this a separable differential equation:Taking $a_0=1$ gives the constant-equation-of-state density scalingThe spatially flat Friedmann equation and the cosmological density parameters therefore giveAlong a radial light ray, $d\chi =dt/a$. Since the redshift-time relation is $dt=-dz/[(1+z)H]$ and $a=(1+z{)}^{-1}$,Hence the comoving radial distance is

The Flatness problem is that the observed cosmological density parameter is close to $\Omega =1$, although in an ordinary decelerating universe any departure from one grows with time. Extrapolation backward therefore requires implausibly precise early cancellation of the curvature term.

The cosmological perfect-fluid continuity equation isConsequentlyFor an expanding universe $H>0$. Ifthen $\rho a^2$ increases. The Friedmann equation impliesIt therefore decreases during this period, driving $\Omega$ toward one. By the Friedmann acceleration equation, the same condition gives $\ddot{a}>0$ when the cosmological constant is included in the effective fluid. This is the inflationary solution of the flatness problem.

For the scalar field, the slow-roll approximation meansThe equations reduce toTaking the positive-field branch gives $H\simeq m\varphi /\sqrt{3}$, and thereforeWith $\varphi (0)={\varphi }_i$,Now integrate $H=\dot{a}/a\simeq m\varphi /\sqrt{3}$:Thus the quadratic-potential slow-roll solution isFinally,so equivalently

For a component with $P=w\rho$ and constant $w$, the cosmological perfect-fluid continuity equation becomesIntegration gives the constant-equation-of-state density scalingDefine the present cosmological density parameter bySubstitution into the spatially flat Friedmann equation gives the Hubble parameter for constant-equation-of-state componentsSpatial flatness is what allows the expression to contain only the listed density components, with ${\sum }_i{\Omega }_{i,0}=1$.

For radiation in cosmology with zero cosmological constant, $p=\rho /3$. The Friedmann acceleration equation gives $\ddot{a}/a=-8\pi G\rho /3$, so the deceleration parameter equals the cosmological density parameter, regardless of spatial curvature. This is a relation for a pure radiation fluid, not a general identity for a mixture.

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.