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For an expanding pressureless matter universe with positive curvature, zero cosmological constant and present scale factor one, set $u_0=\arccos (2/{\Omega }_0-1)\in (0,\pi )$. The closed matter-dominated Friedmann solution then gives the displayed age of an FLRW universe. Its flat limit is $2/(3H_0)$, despite the individually vanishing numerator and denominator. To see this, write $\delta ={\Omega }_0-1$; then $u_0=2\arctan \sqrt{\delta }$ and $u_0-\sin u_0=(4/3){\delta }^{3/2}+O({\delta }^{5/2})$.

With positive curvature, zero cosmological constant and radiation density ${\rho }_0{a}^{-4}$ in units $8\pi G=1$, $A=\sqrt{{\rho }_0/3}$. The scale factor is positive for $0<\tau <\pi$, reaches its maximum at $\pi /2$, and has a Big Crunch at conformal time $\pi$ or proper time $2A$. This complements the closed matter-dominated Friedmann solution.

The cycloidal closed matter-dominated Friedmann solution reaches its maximum scale factor at $u=\pi$ and the Big Crunch at $u=2\pi$. Thus its total lifetime is twice its time of maximum expansion. The displayed formula uses a presently expanding normalization with $H_0>0$ and $a_0=1$; the remaining time is this lifetime minus the age of a closed dust universe.

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 $2{\pi }^2{a}^3$.

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.

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.

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 constant equation-of-state parameter $w$, the cosmological continuity equation integrates to $\rho ={\rho }_0{a}^{-3(1+w)}$ using today's normalization. In conformal time, $\mathcal{H}=a^{\prime }/a=\dot{a}$, so the Friedmann equation becomesSince ${\mathcal{H}}^{\prime }=a\ddot{a}$, the Friedmann acceleration equation givesFor $k{c}^2=1$, put $q=(1+3w)/2$. When $q\ne 0$, integration gives $\mathcal{H}=\cot (q(\tau -{\tau }_0))$. Integrating $a^{\prime }/a=\mathcal{H}$ then yields the closed constant-equation-of-state Friedmann scale factorChoose a connected interval with positive sine; the origin of conformal time can absorb ${\tau }_0$. For $w>-1/3$ this is the expansion and recollapse branch between consecutive zeros of the sine. Negative $q$ instead gives the corresponding minimum-radius branch on its positive-sine interval. The printed power formula excludes $w=-1/3$: in that exceptional case $\mathcal{H}=H_0$ is constant and $a=A{e}^{H_0\tau }$, with $H_0^2+1=8\pi G{\rho }_0/3$. A real solution requires the latter constant to be at least one; equality permits a static solution.

For pressure-free matter, $w=0$, so $\beta =1/2$. Set the big-bang origins to zero. The closed matter-dominated Friedmann solution follows by ${\sin }^2(\beta \tau )=(1-\cos 2\beta \tau )/2$ and integrating $dt=ad\tau$:The entire cycle is $0\le 2\beta \tau \le 2\pi$; maximum expansion has $a=\alpha$ at $2\beta \tau =\pi$, and the big crunch occurs at $t=\alpha \pi /(2\beta )$.

Use conformal time, $dt=ad\tau$, and let primes mean $d/d\tau$. Then $a^{\prime }=a\dot{a}$ and $a^{\prime \prime }=a{\dot{a}}^2+a^2\ddot{a}$. For $k=1$, $\Lambda =0$, the Friedmann equation and acceleration equation giveSubstitution yieldsUse the supplied solution without deriving it. Its first maximum has sine equal to one, so its amplitude must equal the physical turnaround scale in part (a):Choose the expanding branch with $B=0$ and the first zero at $\tau =0$.

For pressureless matter, $w=0$, the closed matter-dominated Friedmann solution isThe Big Crunch is at ${\tau }_c=2\pi$, or proper time $t_c=\pi A=\pi {\rho }_0/3$, measured from the Big Bang.

For radiation, $w=1/3$, the closed radiation-dominated Friedmann solution isThe Big Crunch is at ${\tau }_c=\pi$, or $t_c=2A$.

A radial null geodesic in FLRW spacetime obeys $d\chi /d\tau =\pm 1$. The spatial slice is a unit three-sphere times $a$, so a great circle has comoving circumference $2\pi$. The dust lifetime supplies conformal distance $2\pi$: one circumference, with the return occurring only in the crunch limit. Strictly before the singular endpoint no full return is completed. The radiation lifetime supplies conformal distance $\pi$: the photon reaches the antipode, half a great circle, in the crunch limit. These are limiting null rays from near the initial singularity, not photons at a regular event on the singular surface.

For pressureless matter, $w=0$ and part (a) gives $\rho =3\mu /(8\pi a^3)$. With $\Lambda =0$, the Friedmann equation becomesorThis is the $\Lambda =0$ case of the Friedmann effective potential for pressureless matter. The potential rises monotonically from $-\infty$ at $a=0$ to zero from below as $a\to \infty$.

For $k=+1$, $C=-1$ intersects the potential at $a_{\max }=\mu$. An initially expanding closed matter-dominated Friedmann solution reaches this turning point and recollapses to a Big Crunch. For $k=0$, $C=0$ gives expansion for all time with $\dot{a}\to 0$ and $a\propto t^{2/3}$. For $k=-1$, $C=1$ gives eternal expansion with $\dot{a}\to 1$ in these units. Thus positive curvature recollapses, whereas zero and negative curvature expand forever.

For pressureless matter, ${\rho }_m{a}^3={\rho }_{m,0}$ when $a_0=1$. Setthen ${\dot{a}}^2=D/a-k$. Substitution of $a=A(1-\cos \theta )$ makes this identity hold whenIndeed $dt/d\theta =B(1-\cos \theta )=Ba/A$, and henceafter setting $t=0$ at $\theta =0$. This is the closed matter-dominated Friedmann solution.