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Take a fixed comoving region, whose physical volume is $V=V_0{a}^3$. Its energy is $E=\rho V$, so the first-law relation $dE=-PdV$ givesDifferentiating with respect to cosmic time and using the Hubble parameter $H=\dot{a}/a$ yields the cosmological perfect-fluid continuity equation

For the barotropic equation of state $P=w\rho$, separation gives the constant-equation-of-state density scalingSubstituting this into the flat Friedmann equation and integrating the expanding branch gives, for $w\ne -1$,For $w=-1$, the density and Hubble parameter are constant and instead $a(t)\propto e^{Ht}$.

Conformal time is defined byAt radiation–string equality, let each component have density ${\rho }_{\mathrm{eq}}$ at $a=a_{\mathrm{eq}}$. Since radiation in cosmology has $w=1/3$ and a cosmic string network has $w=-1/3$,ConsequentlyThusOn the expanding branch, integration givesChoosing the Big Bang to occur at $\tau =0$ gives the radiation--cosmic-string Friedmann solution

The Hubble parameter is the fractional expansion rateHere $\rho$ is the homogeneous energy density, $k\in \{-1,0,1\}$ records the sign of the spatial curvature of an FLRW universe, and $R$ is the fixed curvature length scale used with the dimensionless scale factor $a(t)$.

Energy-momentum conservation in a homogeneous expanding universe gives the continuity equationDifferentiate the Friedmann equationUsing $\dot{a}/a=H$ and the continuity equation givesEliminating the curvature term with the Friedmann equation yieldsSince $\dot{H}+H^2=\ddot{a}/a$, this is the Raychaudhuri equation

Under the strong energy condition $\rho +3P\ge 0$, one has $\ddot{a}\le 0$. Thus $a(t)$ is concave, and for $t<t_0$ it lies below its tangent at the present time:The right side vanishes at $t=t_0-H_0^{-1}$. A positive scale factor therefore cannot extend regularly to that time; it must reach the Big Bang singularity at some $t_{\mathrm{BB}}\ge t_0-H_0^{-1}$. Equivalently,

On long wavelengths the gradient term is negligible. The Friedmann equation impliesDuring matter domination, $a\propto t^{2/3}$ and $H=2/(3t)$, so the linear growth equation becomesThe power-law ansatz $\delta \propto t^s$ gives $s=2/3$ or $s=-1$. Since $a\propto t^{2/3}$, the growing and decaying modes are

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 elementary Newtonian derivation of the Friedmann equation follows a test particle on the boundary of a finite expanding sphere. It therefore introduces a distinguished centre and can misleadingly suggest that all matter expands away from one origin.

The fluid derivation is local: the Euler equation and Poisson equation must be compatible in every comoving neighbourhood. It consequently describes a genuinely homogeneous infinite universe, in which every comoving observer sees the same Hubble flow, without selecting a physical centre. This is the improvement captured by the Newtonian fluid derivation of the Raychaudhuri equation.

Contracting the Ricci tensor gives the Ricci scalarHence the time-time component of the Einstein tensor is $G_{00}=R_{00}-\frac{1}{2}{g}_{00}R=3H^2$. A comoving perfect fluid in general relativity has $T_{00}=\rho$, so the time-time Einstein field equations give the flat Friedmann equation

Separate stress-energy conservation and the constant equation-of-state parameter $w$ give the constant-equation-of-state density scalingBecause $w<-1$, the phantom energy density grows with the scale factor, whereas the pressureless matter density decays as $a^{-3}$. Phantom energy therefore dominates at late times. With $\Lambda =0$, the Friedmann equation becomesIntegrating and absorbing the integration constant into the finite time $t^{\ast }$ givesThe exponent is negative, so $a(t)\to \infty$ as $t\to t^{\ast -}$. This is a Big Rip.

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 $k=-1$, the Friedmann-Lemaitre-Robertson-Walker metric isThe empty first Friedmann equation, with $\Lambda =0$, givesChoosing the expanding branch and shifting the origin of proper time yields

By the shell theorem, only the mass inside the particle's radius contributes:The conserved Newtonian energy per unit particle mass isWriting $r=a(t)r_0$ and using conservation of enclosed mass, so that $\rho a^3$ is constant, givesThis is the first Friedmann equation without a cosmological constant.

The derivation has two central defects. It uses instantaneous Newtonian gravity and omits the relativistic gravitational effect of pressure, so it is not valid for general cosmological matter. It also starts from a finite ball with a preferred centre and boundary, contrary to exact homogeneity and isotropy; the assumption $r\ll L$ only hides that defect locally.

Differentiate the Friedmann equationand use the cosmological perfect-fluid continuity equationAfter dividing by $2H$,Since $\ddot{a}/a=\dot{H}+H^2$, eliminating the curvature term gives the Friedmann acceleration equation

Under the strong energy condition in a Friedmann universe, $\rho +3P\ge 0$, so $\ddot{a}/a\le 0$. Thereforeand wherever $H\ne 0$,

If $H(t_0)>0$, integrating the inequality backward shows that $H^{-1}$ reaches zero no later than $t_0-H(t_0{)}^{-1}$. Thus $H\to +\infty$ and $a\to 0$ at a finite time in the past. If $H(t_0)<0$, the same argument forward in time makes $H\to -\infty$ and $a\to 0$ at a finite future time. The sign of $H$ distinguishes an expanding universe with a past Big Bang from a contracting universe ending in a Big Crunch. Hence a Friedmann universe obeying the strong energy condition cannot remain nonsingular for unlimited proper time in both directions.

The constant $R$ is the curvature radius of the closed spatial slices when $a=1$; at general scale factor their curvature radius is $Ra$ and their scalar curvature is $6/(R^2{a}^2)$.

The cosmological perfect-fluid continuity equation isDuring expansion, $\dot{a}>0$, and thereforeIf $C=\rho (t_0)a(t_0{)}^3$, then throughout the later expanding phaseMultiplying the Friedmann equation by $a^2$ givesThe right-hand side is negative oncewhich is impossible because ${\dot{a}}^2\ge 0$. Thus the scale factor cannot grow without bound.

In fact the expanding phase ends in finite time. Combining the Friedmann and continuity equations yields the Friedmann acceleration equationThe curvature term in the first Friedmann equation also impliesWhile $a$ remains below the finite bound just found, $\ddot{a}$ is bounded above by a strictly negative constant. Hence $\dot{a}$ reaches zero after finite time: the closed universe attains a maximum size and cannot expand forever. This is recollapse of a closed Friedmann universe with nonnegative pressure.

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

Write $A=8\pi G/(3c^2)$ and $H=\dot{a}/a$. Differentiating the Friedmann equation givesThe cosmological perfect-fluid continuity equation says $\dot{\rho }=-3H(\rho +P)$, so, first where $H\ne 0$ and then everywhere by continuity,Using $K{c}^2/a^2=A\rho -H^2$ from the original equation givesSince $\dot{H}+H^2=\ddot{a}/a$ for the Hubble parameter, this is the Friedmann acceleration equation.

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 dynamical dark energy, use $dz/dt=-(1+z)H$ in the continuity equation to obtainTherefore the variable dark-energy equation of state givesCombining this with ${\rho }_m={\rho }_{m,0}(1+z{)}^3$ in the Friedmann equation yieldswhereThis is the Hubble parameter for matter and dynamical dark energy.

For a canonical scalar field with positive kinetic energy, the Friedmann equation givesso the first Hubble slow-roll parameter satisfies $ϵ\ge 0$. The tensor tilt is consequentlyA blue primordial spectrum of tensor modes cannot arise in standard canonical single-field slow-roll inflation; it requires changing assumptions, such as the matter content, kinetic structure, initial state, or gravitational dynamics.

The scalar tilt instead obeysIt can be positive in a standard single-field model if $\eta <-2ϵ$, meaning that $ϵ$ decreases sufficiently rapidly with the number of e-folds. Equivalently, in potential slow-roll parameters the first-order condition is $2{\eta }_V>6{ϵ}_V$. Thus a blue scalar spectrum is allowed by the framework, although it is not produced by every slowly rolling potential.

The cosmological continuity equation and $P_i=w_i{\rho }_i$ giveWith $a_0=1$, $1+z=a^{-1}$ andthe Friedmann equation becomesSpatial curvature may be included as an effective component with $w=-1/3$ and ${\Omega }_{k,0}=-k/H_0^2$.

For the homogeneous background, $X={\dot{\bar{\varphi }}}^{,2}/2$. The second Friedmann equation givesDividing by $H{M}_{\mathrm{Pl}}^2\dot{\bar{\varphi }}$ shows that the coefficient found above is