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The two scalings give $\rho ={\rho }_{r0}{a}^{-4}+{\rho }_{s0}{a}^{-2}$. Flatness and the Friedmann equation give $H^2=8\pi G\rho /3$, with $H_0^2=(8\pi G/3)({\rho }_{r0}+{\rho }_{s0})$. ThusOn the expanding branch, putting the Big Bang at zero, separation givesIts positive square root is the required solution. If $\beta >0$, its early behaviour is $a\propto t^{1/2}$, as for radiation. The values $a=1$ and $a^2=\beta$ giveFor $\beta =0$ there is only the string component, with linear expansion and no radiation era.

As ${\Omega }_M\downarrow 0$, use $\mathrm{arsinh}z=\log (z+\sqrt{1+z^2})\sim \log (2z)$ to obtainAt fixed Hubble constant, the age of a flat matter-Lambda universe diverges logarithmically. The limiting purely vacuum-dominated de Sitter spacetime has an exponential scale factor and no Big Bang at a finite value of cosmic time in this flat slicing.

For $0<{\Omega }_M<1$, set $x=a^{3/2}$ and choose the expanding branch of the Friedmann equation. ThenSeparating variables and choosing cosmic time $t=0$ at the Big Bang givesThus the flat dust-and-vacuum expansion isSetting $a(t_0)=1$ gives the age of a flat matter-Lambda universe,The two matter-vacuum age endpoint limits follow below.

During radiation domination, the cosmological continuity equation gives ${\rho }_R\propto a^{-4}$. The flat Friedmann equation in conformal time is $(a^{\prime }/a{)}^2=8\pi G{a}^2{\rho }_R/3$. Hence $a^{\prime 2}$ is constant, and choosing the Big Bang at $\tau =0$ yieldsFor the CDM density equation in a matter-radiation universe, neglect ${\rho }_C{\delta }_C$ while retaining ${\delta }_C$ itself. The adiabatic initial conditions give ${\delta }_R=4{\delta }_C/3$, soThis Euler-Cauchy equation has powers ${\tau }^p$ with $p(p-1)+p-4=0$, namely $p=2,-2$. Selecting the growing superhorizon adiabatic CDM mode in radiation domination gives

Differentiate the Friedmann equation and use the cosmological continuity equation to obtain $\dot{H}+H^2=-(4\pi G/3)(\rho +3P/c^2)$. The relation extends to turning points by continuity. Radiation obeys ${\rho }_R={\rho }_{R0}{a}^{-4}$, while dark energy with $P_{\Lambda }=-{\rho }_{\Lambda }{c}^2$ has constant density. Substitution at $a(t_0)=1$ gives $k{c}^2=\beta H_0^2$, and at arbitrary $a$ givesPut $x=a^2$. For $\beta >4$, the upward-opening polynomial $P(x)=x^2-\beta x+\beta$ has roots $x_{\pm }=(\beta \pm \sqrt{\beta (\beta -4)})/2$, with $1<x_{-}<2<x_{+}$. The expanding Big Bang branch reaches $x_{-}$ in finite time. There $H=0$ but $\dot{H}<0$, so expansion turns to contraction; it returns to $a=0$ in finite time, a Big Crunch. A sketch of $a^4{H}^2/H_0^2=P(x)$ has the forbidden negative interval between the roots, while $a^4\dot{H}/H_0^2=\beta (x-2)$ is a line crossing zero at two.

For $\beta =4$, $P(x)=(x-2{)}^2$. The branch starting at zero has $\dot{x}=2H_0(2-x)$, givingAt small $t$, $a\sim 2\sqrt{H_{0}t}$, the radiation-dominated scale factor. At large $t$, $a=\sqrt{2}[1-\frac{1}{2}{e}^{-2H_{0}t}+O(e^{-4H_{0}t})]$. It approaches an Einstein static universe with equal radiation and dark-energy densities, rather than recollapsing or entering unbounded expansion.

On the expanding branch of the Friedmann equation, dropping the vacuum term gives the conformal timeThe lower limit sets $\tau =0$ at the Big Bang. The approximation requires ${\Omega }_{\Lambda }{a}^4\ll {\Omega }_{m}a$, and remains safe earlier when radiation in cosmology dominates. At matter-radiation equality, ${\rho }_{m,0}{a}^{-3}={\rho }_{r,0}{a}^{-4}$, so $a_{\mathrm{eq}}={\Omega }_r/{\Omega }_m$. Substitution yieldsIn conformal coordinates the radial FRW metric is $d{s}^2=a^2(-d{\tau }^2+d{r}^2)$. A radial light ray therefore travels a comoving distance $dr=d\tau$. The maximum distance traversed since the Big Bang is ${\int }_0^{t_{\mathrm{eq}}}dt/a={\tau }_{\mathrm{eq}}$: this is the comoving particle horizon, rather than the instantaneous comoving Hubble radius or the acoustic horizon. Restoring $c$ in the numerical conversion givesFor ${\Omega }_m=0.3$ and $h=0.7$, the comoving particle horizon at matter-radiation equality is about $110\mathrm{Mpc}$. The corresponding scale factor is $a_{\mathrm{eq}}\simeq 2.9\times 10^{-4}$.

With zero scalar potential and a homogeneous scalar field, the preceding perfect fluid identification givesThis equation of state is that of a stiff fluid. The field equation integrates to $a^3\dot{\varphi }=K$, since ${\partial }_t(a^3\dot{\varphi })=0$. Hence $\rho =K^2/(2a^6)$, agreeing with the cosmological continuity equation for $w=1$. If this nonzero component dominates a flat universe, the Friedmann equation giveson the expanding branch. Integrating once more, $a^3=3\sqrt{4\pi G/3}|K|(t-t_B)$. Choosing the Big Bang at $t_B=0$ yieldsThe homogeneous free scalar as a stiff fluid construction requires $K\ne 0$; a constant zero-potential field has zero energy density and cannot dominate this expansion.

At $z=0$, spatial flatness gives $\tan {\theta }_0=\sqrt{(1-{\Omega }_{\Lambda ,0})/{\Omega }_{\Lambda ,0}}$, hence $\cos {\theta }_0=\sqrt{{\Omega }_{\Lambda ,0}}$ and $\sin {\theta }_0=\sqrt{1-{\Omega }_{\Lambda ,0}}$. ThereforeFor ${\Omega }_{\Lambda ,0}\to 0$, $\mathrm{arsinh}u\sim u$ in the previous integral gives $t_0\to 2/(3H_0)$, the Einstein-de Sitter universe result. For ${\Omega }_{m,0}=0$, the positive-vacuum flat slicing has no finite-age Big Bang, explaining the other endpoint divergence.

The age of an FLRW universe at an epoch is the elapsed cosmic time since the Big Bang, not the lookback time from today. For positive matter and positive cosmological constant, integrating $dt=da/\dot{a}$ givesPut $u=\sqrt{{\Omega }_{\Lambda ,0}/{\Omega }_{m,0}}{a}^{\prime 3/2}$; then $du=(3/2)\sqrt{{\Omega }_{\Lambda ,0}/{\Omega }_{m,0}}\sqrt{a^{\prime }}d{a}^{\prime }$. The integral becomes the age of a flat matter-Lambda universe:Choose $0<\theta <\pi /2$ such that $\tan \theta =\sqrt{{\Omega }_{m,0}/{\Omega }_{\Lambda ,0}}{a}^{-3/2}$. The upper-limit value of $u$ is $\cot \theta$, andConsequentlyThe integration constant is fixed by $t\to 0$ as $z\to \infty$, where $\theta \to \pi /2$. Equivalently, differentiating $u=\cot \theta$ changes the last integral to $-\int d\theta /\sin \theta$, recovering the supplied trigonometric identity. The formula assumes positive ${\Omega }_{\Lambda ,0}$ and ${\Omega }_{m,0}$; its zero-vacuum limit is regular even though this parametrization is singular.

The agreement of the main light-element abundances with a hot early nuclear-burning phase, and the independent baryon-density consistency between nucleosynthesis and the microwave background, gives strong support to the hot Big Bang framework. These measurements connect the first minutes to recombination, despite their very different temperatures, densities and observables. In particular, the large primordial helium mass fraction and residual deuterium are natural outputs of Big Bang nucleosynthesis, rather than arbitrary products of subsequent star formation.

The cosmological lithium problem prevents claiming that the simplest abundance calculation explains every observation without qualification. One must distinguish nuclear-rate uncertainties, stellar depletion, atmosphere and abundance systematics, and possible new particle physics. Thus the broad hot Big Bang framework is strongly supported, while lithium tests the completeness of its simplest nucleosynthesis and stellar-abundance interpretation. An unresolved lithium discrepancy alone does not erase the independent successes of deuterium, helium and the cosmic microwave background.

Separate fluid conservation gives ${\dot{\rho }}_M=-3H{\rho }_M$, hence ${\rho }_M={\rho }_{M0}{a}^{-3}$ after normalization at today. For $P_{\Lambda }=-{\rho }_{\Lambda }{c}^2$, the conservation equation gives ${\dot{\rho }}_{\Lambda }=0$. ThusAssume both densities are positive and set $A=8\pi G{\rho }_{M0}/3$, $B=8\pi G{\rho }_{\Lambda 0}/3$. On the expanding flat branch, ${\dot{a}}^2=A/a+B{a}^2$. With $b=a^{3/2}$ this becomes $\dot{b}=(3/2)\sqrt{A+B{b}^2}$. Choosing the Big Bang at $t=0$ and integrating givesThe normalization $a(t_0)=1$ requires $\sinh (\beta t_0)=\sqrt{{\rho }_{\Lambda 0}/{\rho }_{M0}}$; equivalently $\alpha ={\sinh }^{-2/3}(\beta t_0)$. Thus the age $t_0$ and the two positive densities are not independent once this normalization and bang time are fixed.

As $t\downarrow 0$, $a\sim \alpha (\beta t{)}^{2/3}$, the matter-dominated law. As $t\to \infty$, $a\sim \alpha 2^{-2/3}{e}^{\sqrt{B}t}$ and $\ddot{a}/a\to B>0$, giving accelerated dark-energy expansion. Direct differentiation gives $\ddot{a}/a=B-A/(2a^3)$, so acceleration begins once $a^3>A/(2B)$. If dark energy vanishes, the limiting matter-only solution must be taken instead; negative dark-energy density would not have the stated late-time accelerating form.

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

The comoving particle horizon is the maximum radial distance a photon could have travelled since the initial Big Bang, expressed in comoving coordinates. It describes possible past causal contact, whereas a cosmological event horizon concerns photons that can reach an observer in the future.

For a radial null geodesic in the FLRW metric, $dr/\sqrt{1-k{r}^2}=cdt/a$. Spatial flatness sets $k=0$, so with $a_0=1$,The given Hubble parameter then supplies $a^{2}H=H_0\sqrt{{\Omega }_{r,0}+{\Omega }_{m,0}a+{\Omega }_{\Lambda ,0}{a}^4}$. Since cosmological redshift obeys $a=(1+z{)}^{-1}$,The corresponding proper horizon at that epoch is $a(z)r_p(z)$.

Assume positive fluid density, ${\rho }_0>0$, and write $m=1+3w>0$. A perfect fluid with constant equation of state obeys the cosmological perfect-fluid continuity equation, givingIn the stated units the Friedmann equation is $H^2=\rho /3-k/a^2+\Lambda /3$. Multiplication by $a^2/2$ yields the Friedmann effective potential for a constant-equation-of-state fluid,The allowed region has $V\le 0$. The Friedmann acceleration equation is equivalently $\ddot{a}=-V^{\prime }(a)$, including turning points by continuity. Thus a zero-energy mechanical trajectory reproduces the cosmological evolution, with $a>0$.

For $k=0$, $\Lambda <0$, the potential increases strictly from negative infinity to positive infinity. Its unique zero givesExpansion from the Big Bang stops there, with negative acceleration, then reverses into a Big Crunch. Both the turning point and the final singularity occur in finite proper time: $dt=da/\sqrt{-2V}$ is integrable near a simple turning point and behaves as a constant times $a^{m/2}da$ near zero.

For $\Lambda =0$, $k=+1$, $V$ increases from negative infinity to $1/2$ and crosses zero atThis is again expansion followed by finite-time recollapse. For $k=-1$, the same increasing curve has asymptote $-1/2$ and never meets zero. Expansion continues without a finite maximum; asymptotically $\dot{a}\to 1$ and $a\sim t$, as spatial curvature dominates the diluted fluid.

For $k=0$, $\Lambda >0$, $V$ tends to negative infinity at both ends. It has a maximum atThere is no turning point. Expansion initially decelerates, then accelerates once $a>a_{\ast }$, and approaches de Sitter spacetime expansion $a\propto e^{\sqrt{\Lambda /3}t}$. Thus positive flat dark-energy expansion is unbounded; negative flat dark energy and positive curvature without dark energy recollapse.

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.

The cosmological perfect-fluid continuity equation gives ${\rho }_r={\rho }_{r,0}{a}^{-4}$ and ${\rho }_m={\rho }_{m,0}{a}^{-3}$. In conformal time, $\dot{a}=a^{\prime }/a$ and $H=a^{\prime }/a^2$. Thus the flat Friedmann equation isChoose the expanding branch and put the Big Bang at $\tau =0$. Integration givesInverting it supplies the radiation-matter scale factor in conformal time:The early linear term gives radiation domination, while the late quadratic term gives matter domination. The solution presumes the stated radiation-plus-matter model, without a cosmological constant.

At matter-radiation equality, $a_{\mathrm{eq}}={\Omega }_{r,0}/{\Omega }_{m,0}$. Since ${\Omega }_{r,0}+{\Omega }_{m,0}=1$,In units $c=1$, the comoving particle horizon at equality is ${\tau }_{\mathrm{eq}}$; its physical size is $a_{\mathrm{eq}}{\tau }_{\mathrm{eq}}$. The angular diameter distance to the equality surface is $a_{\mathrm{eq}}({\tau }_0-{\tau }_{\mathrm{eq}})$. Therefore the horizon angle at matter-radiation equality corresponding to one horizon length isFor ${\Omega }_{r,0}=10^{-4}$, this is $4.20\times 10^{-3}$ radians, or $0.241^{\circ }$, about $14.4$ arcminutes. If quoting the full diameter of a horizon-sized patch, the result is twice this, $0.481^{\circ }$. The small-angle approximation is excellent. Expanding in $\sqrt{{\Omega }_{r,0}}\ll 1$ would give $4.14\times 10^{-3}$ radians; retaining both cosmic components near equality avoids an inaccurate pure-matter extrapolation there. The scale factor cancels between physical size and angular diameter distance, and $H_0$ cancels from the ratio.

If cosmological recombination occurs only a modest expansion after equality, its causal patch still subtends a small part of the sky. Widely separated regions of the observed cosmic microwave background have nearly the same temperature even though their past particle horizons did not overlap in the ordinary decelerating history. Local thermalization after the Big Bang therefore cannot explain this large-angle uniformity. This is the horizon problem; inflation supplies an earlier connected region that can grow to encompass the observed sky.

Take a fixed comoving volume, whose physical volume is $V=V_{\ast }{a}^3$. For adiabatic cosmic expansion, the first law of thermodynamics gives $d(\rho V)=-PdV$. ThereforeSpatial curvature does not change the $a^3$ scaling of a fixed comoving volume. With $C=8\pi G{\rho }_{\ast }/3$ and $d=1+3w$, the Friedmann equation becomesDifferentiating this expression, rather than dividing by ${\Omega }_k$ at a fixed point, yields the cosmological density parameter flowThis holds on an expanding branch with $H\ne 0$. The spatially flat solution ${\Omega }_k=0$ is a fixed point. For $|{\Omega }_k|\ll 1$, ${\Omega }_k\propto a^{1+3w}$. Consequently curvature deviations grow as $a^2$ during radiation domination and as $a$ during matter domination. A small present curvature therefore requires a much smaller initial deviation: this is the flatness problem of a decelerating Big Bang. The issue applies to either sign of curvature, not just an open universe. Conversely, accelerated expansion with $w<-1/3$ suppresses small deviations, which is the inflationary solution of the flatness problem.

For $0<\Omega <1$, choose the Big Bang to occur at $\tau =t=0$ and retain the expanding solution. The first integral found above isUse $a=(\Omega /b){\sinh }^{2}u$. Substitution gives $u^{\prime }=H_0\sqrt{b}/2=\alpha /2$, so $u=\alpha \tau /2$. This gives the flat matter-coasting-fluid Friedmann solution:Integrate $dt/d\tau =a$, with the same zero of time:As a check, $a^{\prime }=(\Omega \alpha /2b)\sinh (\alpha \tau )$, and the identity ${\sinh }^{2}v=(\cosh v-1)(\cosh v+1)$ verifies $a^{\prime 2}=H_0^2(\Omega a+b{a}^2)$. Also $t^{\prime }=a$ exactly. At early times $a\simeq \Omega H_0^2{\tau }^2/4$ and $t\simeq \Omega H_0^2{\tau }^3/12$, reproducing the matter-dominated relation $a\propto t^{2/3}$.

The endpoint $\Omega =1$ is obtained by taking the smooth limit: $a=H_0^2{\tau }^2/4$ and $t=H_0^2{\tau }^3/12$. At $\Omega =0$ the coasting cosmic-string universe instead has $a=H_{0}t$ and $\mathcal{H}=H_0$. Its Big Bang is at $\tau =-\infty$, so a finite conformal-time origin at the bang is no longer available; choosing $a=1$ at $\tau =0$ gives $a=e^{H_0\tau }$ and $t=H_0^{-1}{e}^{H_0\tau }$. The physical-age limit remains regular.

Insert the component-density laws into the Friedmann equation, using ${\rho }_{i,0}=3H_0^2{\Omega }_{i,0}/(8\pi G)$ and $a_0=1$. Multiplying by $a^2$ givesThe radiation in cosmology and pressureless matter contributions add; the printed first formula has lost the plus sign between them. In the Friedmann acceleration equation, $\rho +3P=2{\rho }_r+{\rho }_m-2{\rho }_{\Lambda }$, soEvaluating the first equation today gives $K=H_0^2({\Omega }_{r,0}+{\Omega }_{m,0}+{\Omega }_{\Lambda ,0}-1)$.

Now remove dark energy and assume a positive pressureless matter or radiation in cosmology density, starting on an expanding branch. Let $f(a)=H_0^2({\Omega }_{r,0}/a^2+{\Omega }_{m,0}/a)$. It decreases from infinity to zero, and ${\dot{a}}^2=f(a)-K$, while $\ddot{a}<0$. This immediately distinguishes all three cases of spatial curvature of an FLRW universe.

The early small-$a$ solution has $a\propto t^{1/2}$ if radiation in cosmology is present, and $a\propto t^{2/3}$ for pressureless matter alone. The figure isolates the effect of curvature in three matter-only examples, rather than comparing models with the same present density parameters.

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.