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For an expanding pressureless matter universe with zero cosmological constant, normalize $x=a/a_0$. Dust relation between deceleration and density and mass conservation give ${\dot{x}}^2=H_0^2(1-2q_0+2q_0/x)$. Integrating from the Big Bang yields the age integral above.

For $q_0>1/2$, set $x=q_0(1-\cos \theta )/(2q_0-1)$ to obtain $H_0{t}_0=q_0({\theta }_{\ast }-\sin {\theta }_{\ast })/(2q_0-1{)}^{3/2}$, with $\cos {\theta }_{\ast }=(1-q_0)/q_0$ on the expanding branch. For $0<q_0<1/2$, the corresponding hyperbolic expression is $q_0(\sinh {\eta }_{\ast }-{\eta }_{\ast })/(1-2q_0{)}^{3/2}$ with $\cosh {\eta }_{\ast }=(1-q_0)/q_0$. The flat-dust and empty Milne universe limits are respectively $2/3$ and $1$.

The flat barotropic matter-vacuum solution has $H_{\Lambda }/H_0=\sqrt{{\Omega }_{\Lambda 0}}$ and therefore the displayed age, measured from its Big Bang. At fixed $H_0$ and density fractions it is $1/(1+w)$ times the dust-plus-vacuum age. Its $w=0$ case is the age of a flat matter-Lambda universe.

A positively curved pressureless matter universe with zero cosmological constant has the cycloid parametrization displayed above. In natural units, if the Friedmann equation is ${\dot{a}}^2=C/a-k$ with $C,k>0$, then $A=C/(2k)$ and $u=\sqrt{k}\tau$ for conformal time measured from the Big Bang. Substituting in $a^{\prime 2}=Ca-k{a}^2$ verifies the scale factor; integrating $dt=ad\tau$ gives cosmic time. The universe reaches maximum size at $u=\pi$ and recollapses into a Big Crunch at $u=2\pi$.

For positive constant cosmological constant, zero cosmological spatial curvature and a separate positive-density fluid with constant equation-of-state parameter $w>-1$, set $H_{\Lambda }=\sqrt{\Lambda /3}$ in the time-unit convention. Eliminating the fluid density gives $\dot{H}=-3(1+w)(H^2-H_{\Lambda }^2)/2$. On the expanding Big Bang branch,The origin is the Big Bang. The fluid's early-time power-law expansion approaches exponential vacuum domination at late times.

For pressureless matter, put $C=8\pi G{\rho }_m{a}^3/3>0$ and use units $c=1$. The flat Friedmann equation is ${\dot{a}}^2=C/a+\Lambda a^2/3$. Substitution $u=a^{3/2}$ gives ${\dot{u}}^2=(9/4)(C+\Lambda u^2/3)$. With a Big Bang at $t=0$, positive cosmological constant gives $a=(3C/\Lambda {)}^{1/3}{\sinh }^{2/3}(\sqrt{3\Lambda }t/2)$. Negative cosmological constant replaces the hyperbolic sine by $\sin (\sqrt{3|\Lambda |}t/2)$ and produces recollapse at $2\pi /\sqrt{3|\Lambda |}$. Zero cosmological constant gives $a=(9C/4{)}^{1/3}{t}^{2/3}$.

A flat universe with separately conserved pressureless matter and a coasting fluid satisfies $a^{\prime 2}=H_0^2[{\Omega }_{m}a+(1-{\Omega }_m)a^2]$. For both densities positive, choosing the Big Bang at zero conformal time gives $a={\Omega }_m[\cosh (\alpha \tau )-1]/[2(1-{\Omega }_m)]$, with $\alpha =H_0\sqrt{1-{\Omega }_m}$. Physical time follows by integrating $dt=ad\tau$. The hyperbolic parametrization is formally like the open dust solution, but the spatial geometry here is flat.

A horizon-length patch on the matter-radiation equality surface has physical size $a_{\mathrm{eq}}{\tau }_{\mathrm{eq}}$ in a flat universe with conformal time measured from the Big Bang. Dividing by its angular diameter distance $a_{\mathrm{eq}}({\tau }_0-{\tau }_{\mathrm{eq}})$ gives its small angular size. For the radiation-matter scale factor in conformal time, this is $(\sqrt{2}-1)\sqrt{{\Omega }_{r,0}}/[1-\sqrt{2{\Omega }_{r,0}}]$. A patch diameter is twice the angle associated with one horizon radius. The particle horizon is an integral over past evolution and must not be silently replaced by the instantaneous Hubble radius.

The age of a flat matter-Lambda universe is $H_0{t}_0=2\mathrm{arsinh}\sqrt{(1-{\Omega }_m)/{\Omega }_m}/(3\sqrt{1-{\Omega }_m})$. The large-argument expansion $\mathrm{arsinh}z=\log (2z)+O(z^{-2})$ gives the first limit. Its small-argument expansion $\mathrm{arsinh}z=z+O(z^3)$ gives the second. The divergence at the pure-vacuum endpoint reflects the absence of a finite Big Bang time in the expanding flat de Sitter spacetime patch.

A negatively curved universe with pressureless matter and no dark energy expands forever. With $a_0=1$ and $0<{\Omega }_{m,0}<1$, its scale factor is $a={\Omega }_{m,0}(\cosh \eta -1)/[2(1-{\Omega }_{m,0})]$ and its cosmic time is $t={\Omega }_{m,0}(\sinh \eta -\eta )/[2H_0(1-{\Omega }_{m,0}{)}^{3/2}]$. The parameter is conformal time scaled by $\sqrt{-K}$ and measured from the Big Bang.

The radial null integral from the Big Bang is ${\chi }_p=c{\int }_0^{t}d{t}^{\prime }/R(t^{\prime })=c{t}^{1-p}/[A(1-p)]$. Multiplication by the scale factor gives the proper radial horizon distance. It is $3ct$ for a flat matter-only model and $2ct$ for a flat radiation-only model. A finite initial time different from zero or an earlier expansion era changes the result.

Use $c=1$ and signature $+---$. The positively curved FLRW metric in the usual radial chart isSet $r=\sin \chi$ and introduce conformal time by $dt=ad\eta$. Then $d{r}^2/(1-r^2)=d{\chi }^2$ in the chart, and the spherical polar coordinate $\chi \in [0,\pi ]$ extends the spatial description to the full three-sphere. ThusThe $r$ coordinate itself is not one-to-one across the equator, which is why the new polar coordinate is useful.

For $\Lambda =0$ and $k=1$, the Friedmann equation is ${\dot{a}}^2=(8\pi G/3)\rho a^2-1$. Since $a^{\prime }=a\dot{a}$, it becomesFor pressureless matter, the background continuity equation gives $\rho a^3={\rho }_0{a}_0^3$. Define the conserved normalization $M=(4\pi /3){\rho }_0{a}_0^3$. It makes the equation $(a^{\prime }{)}^2=2GMa-a^2$. The closed matter-dominated Friedmann solution is the cycloidIndeed $a^{\prime }=GM\sin \eta$ and $2GMa-a^2=(GM{)}^2{\sin }^2\eta$. Integrating $dt/d\eta =a$ gives the second expression, with the Big Bang at $t=\eta =0$. It expands to $a=2GM$ at $\eta =\pi$ and recollapses at $\eta =2\pi$. Here $M$ is the constant appearing in the Friedmann equation; it is not the total mass of the entire closed spatial slice, whose volume is $2{\pi }^2{a}^3$.

Choose the Big Bang at $\tau =t=0$. For $0<\Omega <1$, the expanding singular solution of the Riccati equation is $\mathcal{H}=\beta \coth (\beta \tau /2)$. Integrating $a^{\prime }/a=\mathcal{H}$ gives $a=C{\sinh }^2(\beta \tau /2)$. The Friedmann constraint fixes $C=\Omega /(1-\Omega )$, rather than leaving an independent amplitude. ThusNow $dt=ad\tau$. Integrating from the singular origin givesThis derives the flat matter-coasting-fluid Friedmann solution. For small $\tau$, $a\propto {\tau }^2$ and $t\propto {\tau }^3$, hence $a\propto t^{2/3}$. At late time $a\propto t$, as expected when the coasting fluid dominates.

Use natural units and conformal time $dt=ad\tau$. The conformal Hubble parameter is $\mathcal{H}=a^{\prime }/a=aH$. For the barotropic equation of state $P=(\gamma -1)\rho$, the Friedmann acceleration equation givesThe Friedmann equation gives $8\pi G{a}^2\rho /3={\mathcal{H}}^2+k$. Hence the constant-equation-of-state conformal Riccati equation isFor $\alpha \ne 0$, separation and integration of this Riccati equation yield, after choosing the origin of conformal time,The open positive-energy density branch uses the hyperbolic cotangent, rather than the cotangent. Integrating $a^{\prime }/a=\mathcal{H}$ giveswhere the positive constants are fixed by the energy density normalization. These formulas apply on branches where the bases are positive. For $\alpha >0$, take the Big Bang at $\tau =0$. The cosmic time is thenThe integrals also give the general answer after selecting a suitable reference time when a finite Big Bang endpoint does not exist. The exceptional coasting fluid $\gamma =2/3$ has $\alpha =0$, so $\mathcal{H}=h$ is constant: $a=A{e}^{h\tau }$ and $t=(A/h)e^{h\tau }+t_{\ast }$. If $h=0$, instead $a=A$ and $t=A\tau +t_{\ast }$. A flat fluid with $\gamma =0$ is the constant-density de Sitter spacetime case, for which the flat time integral is logarithmic.

During matter domination, $\gamma =1$ and $\alpha =1/2$. Absorbing numerical factors into separate positive constants gives the closed matter-dominated Friedmann solution, the flat Einstein-de Sitter universe and the open matter-dominated Friedmann solution:To compare the age of a dust universe without a cosmological constant, specify the normalization. At fixed present Hubble parameter $H_0$, with $x=a/a_0$ and present matter cosmological density parameter ${\Omega }_m>0$, the Friedmann equation givesThe integrand decreases strictly with ${\Omega }_m$ for $0<x<1$. Therefore age ordering of dust universes at fixed Hubble parameter gives, on the expanding branches,The final bound is approached by the empty Milne universe. This compares models with the same $H_0$; fixing instead the present scale factor and energy density gives a different comparison.

Take $c=1$ and normalize the Big Bang to $t=0$. The cosmological perfect-fluid continuity equation for pressureless matter makes $C=8\pi G{\rho }_m{a}^3/3$ a positive constant. The flat Friedmann equation consequently readsPutting $u=a^{3/2}$ gives ${\dot{u}}^2=\frac{9}{4}(C+\Lambda u^2/3)$, which is directly integrable. The expanding flat dust universe with a cosmological constant hasFor negative cosmological constant, the last expression describes expansion and contraction on $0<t<2\pi /\sqrt{3|\Lambda |}$. Its maximum is $a_m=(3C/|\Lambda |{)}^{1/3}$ at $t_m=\pi /\sqrt{3|\Lambda |}$; it reaches zero again at $2t_m$. Positive or zero cosmological constant gives unbounded expansion and no finite maximum. All three have infimum zero at their initial singularity; negative cosmological constant also gives zero at the final singularity, with no positive minimum.

For positive cosmological constant, expanding the hyperbolic sine shows $a\sim (9C/4{)}^{1/3}{t}^{2/3}$ at early times: pressureless matter dominates. At late times,Thus the expansion changes from matter-dominated deceleration to exponential acceleration. Indeed $\ddot{a}/a=-C/(2a^3)+\Lambda /3$ changes sign at $a^3=3C/(2\Lambda )$.

Normalize the region's reference scale factor to one. Pressureless matter has $\rho ={\rho }_0{a}^{-3}$, so the closed Friedmann equation isThe expanding branch can reach only $a\le a_m$, where $a_m={\Omega }_m/({\Omega }_m-1)$. Substitute $a=a_m(1-\cos \theta )/2$. Since $da/d\theta =a_m\sin \theta /2$ and $\dot{a}=H_0\sqrt{({\Omega }_m-1)(a_m/a-1)}$, cancellation on $0<\theta <\pi$ givesIntegration from the Big Bang establishesThe same parametrization continues through spherical-collapse turnaround at $\theta =\pi$ onto the contracting branch up to $\theta =2\pi$. Its maximum-expansion time is $t_m=\pi {\Omega }_m/[2H_0({\Omega }_m-1{)}^{3/2}]$.

Use natural units with $c=1$, write $H=\dot{a}/a$, and measure cosmic time from the Big Bang. The cosmological continuity equation follows by differentiating the Friedmann equation and using the Friedmann acceleration equation; for pressureless matter it gives $\dot{\rho }+3H\rho =0$, hence $\rho ={\rho }_0{a}^{-3}$ when the present scale factor is $a_0=1$. The present critical density and cosmological density parameter giveConsequently, with $C=H_0^2{\Omega }_0$, the Friedmann equation becomes ${\dot{a}}^2=C/a-k$.

Introduce conformal time through $dt=ad\tau$. A prime denotes a derivative with respect to $\tau$, so $a^{\prime }=a\dot{a}$ andSet $u=\sqrt{k}\tau$, with $u=0$ at the Big Bang. On the expanding branch the substitution $a=C(1-\cos u)/(2k)$ satisfies this equation. Integrating $dt=adu/\sqrt{k}$, with $t(0)=0$, gives the closed matter-dominated Friedmann solution:The parametrization continues smoothly through the maximum scale factor, reached at $u=\pi$, onto the contracting branch. Its curve in the $(t,a)$ plane is a rescaled cycloid.

For a presently expanding universe, $H_0>0$ selects $0<u_0<\pi$. The normalization $a(u_0)=1$ givesSubstitution yields the age of a closed dust universe:The first later zero of the scale factor occurs at $u=2\pi$, so the lifetime of a closed dust universe gives the Big Crunch timeThis is the total lifetime measured from the Big Bang; the remaining cosmic time is $t_{\mathrm{BC}}-t_0$. As a check, the limit ${\Omega }_0\downarrow 1$ gives $H_0{t}_0\to 2/3$, while the Big Crunch recedes to infinite time, as expected for a flat matter-dominated universe with zero cosmological constant.

For ${\Omega }_0>1$, introduce conformal time by $dt/d\tau =a$. Since $a^{\prime }=a\dot{a}$, the constraint becomesOn the initial expanding branch, separate variables and choose $a=t=\tau =0$ at the Big Bang. With $\alpha =H_0\sqrt{{\Omega }_0-1}$,The second formula follows by integrating $dt/d\tau =a$. The model reaches maximum size at $\alpha \tau =\pi /2$ and recollapses at $\alpha \tau =\pi$.

Today $a=1$ and $H_0>0$ selects the expanding branch, soSubstitution yields the requested ageIn particular the numerator is $\sqrt{{\Omega }_0}-1$, not $\sqrt{{\Omega }_0-1}$; the PDF displays the former. The flat-model limit is $1/(2H_0)$.

Put $K=2q_0-1>0$ and parametrize the scale factor byOn the expanding branch $0<\theta <\pi$, the equation from part (ii) givesConsequently,Choose the origin of cosmic time at the Big Bang, $y=0$, $\theta =0$. Integration proves the closed matter-dominated Friedmann solution:The same parametrization continues through maximum expansion at $\theta =\pi$ to collapse at $\theta =2\pi$: $dt/d\theta$ remains nonnegative while $dy/d\theta$ changes sign. The maximum scale factor is $y_{\max }=2q_0/K$, reached at $H_0{t}_{\max }=\pi q_0/K^{3/2}$; the total lifetime is twice this time. The squared equation alone does not prescribe the sign of $\dot{y}$, so this continuation supplies the collapsing branch explicitly.

For $q_0=1/2$, the expanding Friedmann equation reduces to $\dot{y}=H_0{y}^{-1/2}$. Separating variables and placing the Big Bang at $t=0$ givesThus the flat pressureless Einstein-de Sitter universe hasIn particular its present age is $2/(3H_0)$.

Put $B=1-2q_0>0$ and introduce $\psi >0$ throughThe expanding equation becomesDividing these expressions yieldsIntegrating from the Big Bang at $\psi =0$ proves the open matter-dominated Friedmann solution:The scale factor grows indefinitely. At early times $y\propto t^{2/3}$, as in the matter-dominated flat case, whereas at late times $\dot{y}\to H_0\sqrt{B}$ and $y\propto t$ as curvature becomes dominant.