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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.

Alan Guth proposed a period of rapid early expansion as a physical explanation of the horizon problem and flatness problem. His 1981 model used a high-energy state whose nearly constant energy density could drive accelerated expansion before ordinary hot Big Bang evolution.

The causal argument is that a region initially small enough to equilibrate can expand to encompass the subsequently observable region. Rapid expansion also suppresses the relative importance of spatial curvature. Ending that stage and obtaining a sufficiently uniform hot state is a separate dynamical requirement; solving the motivating geometric problems does not by itself supply a successful exit.^[ref]

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 $u=2\pi$.

At maximum expansion of a closed matter-dominated Friedmann solution, the Hubble parameter vanishes while the physical mass density stays positive and finite. Its critical density therefore vanishes and the cosmological density parameter diverges. This is a singularity of that ratio rather than a curvature singularity. In the cycloid variable $\theta$, ${\Omega }_M={\sec }^2(\theta /2)$, diverging at $\theta =\pi$ and approaching one at the Big Bang and Big Crunch ends.

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.

For a spatially flat FLRW metric containing conserved pressureless matter and radiation in cosmology, with $a_0=1$ and conformal time zero at the Big Bang, the Friedmann equation gives $a^{\prime 2}=H_0^2({\Omega }_{M,0}a+{\Omega }_{R,0})$. Separating the positive square root gives the displayed scale factor. At matter-radiation equality, $a_{\mathrm{eq}}={\Omega }_{R,0}/{\Omega }_{M,0}$ and ${\eta }_{\mathrm{eq}}=2(\sqrt{2}-1)\sqrt{{\Omega }_{R,0}}/(H_0{\Omega }_{M,0})$. Here the prime denotes differentiation with respect to $\eta$, not $c\eta$.

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.

The cosmological principle permits a Friedmann-Lemaître-Robertson-Walker metric. The Friedmann equation, with vanishing cosmological constant, isFor separately conserved pressureless matter, the cosmological continuity equation gives ${\bar{\rho }}_m={\bar{\rho }}_{m0}{a}^{-3}$: its rest energy is conserved while its physical volume grows as $a^3$. Define the present critical density and matter density parameter byEvaluating the Friedmann equation today gives $-k=H_0^2(1-\Omega )>0$. Thus the model has negative spatial curvature of an FLRW universe; the second contribution to $H^2$ represents that spatial curvature of an FLRW universe, rather than another material fluid. Taking the expanding branch givesHere $0<\Omega <1$. At early times, $a\ll \Omega /(1-\Omega )$, matter domination gives $\dot{a}\simeq H_0\sqrt{\Omega }{a}^{-1/2}$, hence, measuring cosmic time from the Big Bang,At late times the $a^{-2}$ term dominates the Friedmann equation. This curvature-dominated universe has $\dot{a}\to H_0\sqrt{1-\Omega }$ and thereforeThe late expansion approaches the coasting Milne model. The early limit belongs to this idealized matter-only model; including radiation in cosmology would alter the sufficiently early evolution.

The cosmic microwave background consists of relic photons from the early hot, nearly thermal photon-baryon fluid. Before cosmological recombination, abundant free electrons repeatedly scatter photons by Thomson scattering. Their mean free path ${\lambda }_{\gamma }=(n_e{\sigma }_T{)}^{-1}$ is short and their Thomson scattering rate $n_e{\sigma }_{T}c$ is much greater than the Hubble parameter. An earlier photon therefore cannot stream directly to us: subsequent scattering repeatedly changes its direction.

As cosmic expansion cools the plasma to about $T\simeq 3000\mathrm{K}$, electrons and protons combine into neutral hydrogen. The free-electron abundance drops rapidly, the scattering optical depth to the observer becomes of order one, and photon decoupling occurs. Since the cosmic microwave background temperature scales as $T=T_0(1+z)$, with $T_0\simeq 2.725\mathrm{K}$, this occurs at $z\simeq 1100$. The small baryon-to-photon ratio delays neutralization to well below $13.6\mathrm{eV}/k_B$: the high-energy tail contains enough photons to ionize hydrogen until the temperature is much lower than the Hydrogen binding energy expressed as a temperature. The observed cosmic microwave background is primarily this pre-existing radiation undergoing its last scattering; its nearly blackbody spectrum is then preserved by cosmological redshift.

For the requested estimate take a flat matter-dominated universe, $a(t)=(t/t_0{)}^{2/3}$. Restore $c$ in distances. The comoving particle horizon at cosmological recombination and the comoving radial distance to the Cosmic microwave background last-scattering surface areThe proper particle horizon is $a_{\ast }{r}_{\mathrm{hor},\ast }=3c{t}_{\ast }$ and the angular diameter distance is $D_A=a_{\ast }{\chi }_{\ast }$. Therefore the angular particle-horizon size in a power-law universe, taking one horizon radius as the transverse length, isThe angular diameter of a patch of that radius is about $3.56^{\circ }$. The factor of two is a radius-versus-diameter convention; either gives a horizon scale of only a few degrees. This is a particle horizon, rather than the smaller sound horizon. Including early radiation in cosmology changes the numerical estimate but preserves its order of magnitude.

The horizon problem is that the cosmic microwave background has almost the same temperature across the whole sky, including pairs of regions whose past light cones do not overlap in this noninflating model. Local thermal equilibration cannot explain that observation within this history. A sufficiently long period of cosmic inflation makes the comoving Hubble radius decrease: a region once small enough to share a causal past can expand to encompass our entire observed Cosmic microwave background last-scattering surface. After reheating, it can be homogeneous across many later particle horizon patches. The matter-only estimate integrates from a decelerating Big Bang; extending the earlier history by cosmic inflation changes that causal conclusion.

Use mass density units for the $\rho$ appearing in the Friedmann equation; an actual energy density is $\rho c^2$. Conservation, $\dot{\rho }+3H(\rho +P/c^2)=0$, gives $\boxed{{\rho }_M={\rho }_{M,0}{a}^{-3},{\rho }_R={\rho }_{R,0}{a}^{-4}}$ for pressureless matter and radiation in cosmology, respectively, with $a_0=1$. The first scaling counts dilution of particle number, while the second includes the cosmological redshift of each photon's energy.

Define the present cosmological density parameters by ${\Omega }_{X,0}=8\pi G{\rho }_{X,0}/(3H_0^2)$. ThenFor $k=0$, conformal time gives $da/d\eta =a^{2}H$, so the expanding branch obeys $(da/d\eta {)}^2=H_0^2({\Omega }_{M,0}a+{\Omega }_{R,0})$. Set $\eta =0$ at the Big Bang. Integrating givesand hence the flat matter-radiation conformal solutionThe limiting pure-radiation and pure-matter cases follow continuously.

At matter-radiation equality, ${\rho }_{M,0}{a}^{-3}={\rho }_{R,0}{a}^{-4}$, so $a_{\mathrm{eq}}={\Omega }_{R,0}/{\Omega }_{M,0}$. Evaluating the integrated square-root relation givesThis assumes both present densities are positive and uses the same Big Bang origin for conformal time.

For dust alone, write ${\dot{a}}^2=c^2(A/a-k)$, whereSet $\theta =\sqrt{k}c\eta$ and $a=(A/k){\sin }^2(\theta /2)$. Differentiation gives $\dot{a}=c\sqrt{k}\cot (\theta /2)$ once $dt=ad\eta$ is used, verifying the Friedmann equation. Integrating the time relation from the Big Bang yields the closed matter-dominated Friedmann solutionThe Hubble parameter and the cosmological density parameter areThe first decreases from positive infinity through zero at maximum expansion to negative infinity at the Big Crunch. The second approaches one near either singular endpoint and diverges at maximum expansion. This is a density-parameter divergence at a cosmological turnaround: $H=0$ makes the critical density vanish even though the physical dust density there is finite.

The scaling solution has $\rho =3M^2{p}^2/t^2$ and $\rho +P={\dot{\phi }}^2=4M^2/({\lambda }^2{t}^2)$. ThereforeMore generally any real canonical scalar field with positive total energy density obeys $\rho +P={\dot{\phi }}^2\ge 0$, so it cannot have $w<-1$. A negative potential does not reverse this conclusion.

The proper particle horizon is $a(t){\int }_0^{t}d{t}^{\prime }/a(t^{\prime })$. For $a\propto t^p$ the lower-limit integral diverges exactly when $p\ge 1$. Thus, assuming the same power law extends to the initial boundary, there is no finite particle horizon for $\boxed{0<{\lambda }^2\le 2}$. For the strict inequality, $p>1$, the proper cosmological event horizon and present age areAt ${\lambda }^2=2$, $a\propto t$ has an infinite particle-horizon integral but also an infinite future event-horizon integral, so no finite event horizon exists. The age remains $H_0^{-1}$. The $\lambda =0$ constant-potential case is separate: its flat de Sitter patch has event-horizon radius $H_0^{-1}$ and no finite-age power-law Big Bang.

Both positive-density components have nonnegative pressure, so the Friedmann acceleration equation gives $\ddot{a}<0$ whenever $a>0$. Suppose $a(t_1)=a_1>0$ and $\dot{a}(t_1)=v_1>0$. Going backward while the solution is regular, $\dot{a}(t)\ge v_1$. Integrating,and thereforeA positive scale factor cannot persist indefinitely into the past: the right side vanishes at the finite time $t_1-a_1/v_1$.

For completeness, in this model the past endpoint must actually have $a\to 0$, rather than another singularity at positive $a$. From part (d) write ${\rho }_m=C_m/a^3$ and ${\rho }_r=C_r/a^4$ with $C_m,C_r>0$. ThenThe right side strictly decreases with $a$ and is positive at $a_1$, so it remains positive throughout $0<a<a_1$. There is no earlier bounce. It is smooth at every positive $a$, permitting continuation there. The lookback integral isNear $a=0$ its integrand is $a/\sqrt{K_r}$, which is integrable. ThusThe early asymptotic behavior is $a\simeq [2\sqrt{K_r}(t-t_0){]}^{1/2}$. Both densities diverge at this Big Bang boundary. Writing $a(t_0)=0$ denotes the limiting singular endpoint of the classical solution, not an ordinary regular spacetime slice.