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A hyperbolic change of integration variable uses identities such as $1+{\sinh }^{2}u={\cosh }^{2}u$ to simplify square roots. For an open matter-dominated Friedmann solution, $a=b{\sinh }^2(\eta /2)$ makes the cosmic-time integral elementary.

For an open matter-dominated Friedmann solution, put $u=H_0\sqrt{1-{\Omega }_0}\tau$, $a={\Omega }_0(\cosh u-1)/[2(1-{\Omega }_0)]$ and $\cosh u_0=2/{\Omega }_0-1$. The cosmological bolometric background from conserved sources has $J(u)=\sinh (2u)/4-2\sinh u+3u/2$, since $J^{\prime }=(\cosh u-1{)}^2$ and $dt=adu/[H_0\sqrt{1-{\Omega }_0}]$. At fixed $H_0$, the limit ${\Omega }_0\to 0$ is $cL{n}_0/(2H_0)$, whereas the Einstein-de Sitter universe gives $2cL{n}_0/(5H_0)$. Keeping the cosmic age fixed instead reverses this ordering: the two values are respectively $cL{n}_0{t}_0/2$ and $3cL{n}_0{t}_0/5$.

Introduce the positive dimensionless quantity $x=(1-\Omega )a/\Omega$ and the time scale $A=\Omega /[H_0(1-\Omega {)}^{3/2}]$. The Friedmann equation gives the separable relationFor $x=(\cosh \theta -1)/2={\sinh }^2(\theta /2)$, $\theta \ge 0$, the hyperbolic function identities giveChoose the constant of integration so that $t=0$ at $a=0$. The resulting parametric equation for the open matter-dominated Friedmann solution isFor $\theta \ll 1$, the Taylor series giveEliminating $\theta$ yields $a^{3/2}\simeq (3/2)H_0\sqrt{\Omega }t$, exactly the early matter-dominated universe result. For $\theta \gg 1$, $\sinh \theta \sim \cosh \theta \sim e^{\theta }/2$, whereas $\theta /e^{\theta }\to 0$. The ratio of the two parametric expressions tends to $a/t=H_0\sqrt{1-\Omega }$, recovering the curvature-dominated universe result including its coefficient.

Use the same conserved-source interpretation, comoving number density $n_0$ and normalization $a_0=1$ as in the preceding part. DefineThe open matter-dominated Friedmann solution then has $a=C(\cosh u-1)$, $t=(C/\kappa )(\sinh u-u)$ and $dt=(C/\kappa )(\cosh u-1)du$. The present parameter satisfiesThe cosmological bolometric background from conserved sources is thereforeSince ${\cosh }^{2}u=(1+\cosh 2u)/2$, the open dust bolometric background integral isThis leaves the answer in the requested parameter form and makes no use of a radial-distance approximation. For arbitrary proper number density instead, the same calculation gives

For the conserved population, compare models with the same present $n_0,L,H_0$. As ${\Omega }_0\to 0$, $u_0\to \infty$ and $\cosh u_0\sim 2/{\Omega }_0$. The leading term in the square bracket is $\sinh (2u_0)/4\sim 2/{\Omega }_0^2$; the terms linear in $\sinh u_0$ and $u_0$ vanish after multiplication by ${\Omega }_0^2$. HenceThis is also immediate in the Milne model, for which $a=t/t_0$ and $t_0=H_0^{-1}$. If one compares at fixed cosmic age instead, the results are $L{n}_0{t}_0/2$ and $3L{n}_0{t}_0/5$, so the empty open model gives $5/6$ of the flat-model flux. Specifying what is held fixed is necessary because the two expansion histories have different ages at the same Hubble constant.

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.

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.

For the open matter-dominated Friedmann solution, write $\Omega ={\Omega }_{m,0}$, so $0<\Omega <1$. The present Friedmann equation gives $-K=H_0^2(1-\Omega )$. With $A=H_0^2\Omega$ and $B=H_0^2(1-\Omega )>0$, the expanding equation becomesChoose the hyperbolic substitutionThen $da=(A/(2B))\sinh \eta d\eta$ and $\sqrt{a/(a+A/B)}=\tanh (\eta /2)$. Their product givesTake the Big Bang to be $t=0$, $\eta =0$, and integrate. The parametric solution isThe parameter is determined explicitly byThus $\eta$ is conformal time multiplied by $\sqrt{-K}$, with its origin at the Big Bang; it is not equal to unscaled cosmic time. In particular today ${\eta }_0=\mathrm{arcosh}(2/\Omega -1)$. Small $\eta$ gives $a\propto {\eta }^2$ and $t\propto {\eta }^3$, hence $a\propto t^{2/3}$. Large $\eta$ gives $a/t\to H_0\sqrt{1-\Omega }=\sqrt{-K}$, in agreement with the curvature-dominated universe limit.