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For an expanding pressureless matter universe with zero cosmological constant, fixed current Hubble parameter $H_0$ and current matter cosmological density parameter $\Omega >0$, the age of a dust universe without a cosmological constant is $H_0{t}_0={\int }_0^1\sqrt{x}/\sqrt{\Omega +(1-\Omega )x}dx$. For every $0<x<1$ the integrand decreases strictly as $\Omega$ increases. Thus open models with $0<\Omega <1$ are older than the flat model, and closed expanding models with $\Omega >1$ are younger. The flat value is $2/3$ and the empty Milne universe limit is $1$. Specifying what is held fixed matters: at fixed present scale factor and energy density, the integral $t_0={\int }_0^{a_0}da/\sqrt{C/a-k}$ instead increases with $k$, provided all three models reach $a_0$ on their expanding branches.

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.

The stated real trigonometric substitution applies to the closed case $q_0>1/2$. Set $A=2q_0-1$, so $x=q_0(1-\cos \theta )/A$, $dx=q_0\sin \theta d\theta /A$, andThe upper limit obeys $1-\cos {\theta }_{\ast }=A/q_0$, with $0<{\theta }_{\ast }<\pi$ on the expanding branch. Substitution into the age of a dust universe without a cosmological constant yieldsOn this interval, the integrand equals $1-\cos \theta$, so it can also be evaluated:For $q_0=1/2$, the original integral instead gives $H_0{t}_0=2/3$. For $0<q_0<1/2$, use a hyperbolic substitution to obtain $H_0{t}_0=q_0(\sinh {\eta }_{\ast }-{\eta }_{\ast })/(1-2q_0{)}^{3/2}$, where $\cosh {\eta }_{\ast }=(1-q_0)/q_0$. A real $\theta$ cannot implement the printed substitution in that open case.

Normalize the scale factor by $x=a/a_0$. Constant-equation-of-state density scaling for pressureless matter gives $\rho ={\rho }_0{x}^{-3}$. Multiplying the Friedmann equation by $x^2$ and substituting the expressions from part (i),Take the expanding branch $\dot{x}>0$ and set the origin of cosmic time at the Big Bang, $x=0$. The age of a dust universe without a cosmological constant is consequently