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For a constant-$w$ material fluid and positive vacuum energy, the Friedmann acceleration equation gives $\ddot{a}/(a{H}_0^2)={\Omega }_{\Lambda 0}-(1+3w){\Omega }_{m0}(1+z{)}^{3(1+w)}/2$. If $w>-1/3$, its zero determines the displayed transition. A past transition additionally requires the right-hand side of the displayed equation to exceed one. The equation does not imply the same transition for a fluid that already has sufficiently negative pressure.

For a closed Friedmann equation normalized to $k{c}^2=1$, constant equation-of-state parameter $w$, and conformal time $\tau$, let $q=(1+3w)/2$. The cosmological continuity equation and Friedmann acceleration equation give ${\mathcal{H}}^{\prime }=-q(1+{\mathcal{H}}^2)$, where $\mathcal{H}=a^{\prime }/a$. For $q\ne 0$ this integrates to the stated scale factor on any interval where the sine is positive, with ${\alpha }^{2q}=8\pi G{\rho }_0/3$. For $w=-1/3$ the separate solution is $a=A{e}^{H_0\tau }$ with $H_0^2+1=8\pi G{\rho }_0/3$; the sine-power formula is inapplicable.

In an FLRW metric with zero separate cosmological constant, $\varrho +3p\ge 0$ makes the Friedmann acceleration equation imply $\ddot{a}\le 0$. If $a_0=a(t_0)>0$ and $v_0=\dot{a}(t_0)>0$, then the scale factor is a concave function andthroughout its regular past interval. Thus a regular positive scale factor cannot persist earlier than $t_0-a_0/v_0$. If the solution continues regularly whenever $a>0$, its first past zero occurs within $a_0/v_0=H(t_0{)}^{-1}$, where $H$ is the Hubble parameter.

The continuation condition matters: without restrictions on the equation of state, an earlier curvature singularity can occur at positive $a$. For example, in fixed units $a(t)=1+\sqrt{t}$, $t>0$, with $k=0$, defines energy density $\varrho =3(\dot{a}/a{)}^2/(8\pi G)$ and pressure $p=-\varrho /3-\ddot{a}/(4\pi Ga)$. These satisfy the Friedmann equations and $\varrho +3p>0$, yet $a\to 1$ at the singular past boundary. Conversely the Milne model reaches zero without divergent curvature. A finite duration, a zero of the scale factor, and a curvature singularity are distinct conclusions.

For a perfect fluid in general relativity with constant equation-of-state parameter $w$, use conformal time $dt=ad\tau$ and the conformal Hubble parameter $\mathcal{H}=a^{\prime }/a=aH$. The Friedmann acceleration equation gives ${\mathcal{H}}^{\prime }=a\ddot{a}=-4\pi G{a}^2(1+3w)\rho /3$. The Friedmann equation eliminates $\rho$ and gives this Riccati equation. For positive energy density, $\alpha \ne 0$ and appropriately shifted conformal time, its usual branches are $\cot (\alpha \tau )$ for $k=1$, $1/(\alpha \tau )$ for $k=0$, and $\coth (\alpha \tau )$ for $k=-1$. Their scale factors are proportional to ${\sin }^{1/\alpha }(\alpha \tau )$, $(\alpha \tau {)}^{1/\alpha }$ and ${\sinh }^{1/\alpha }(\alpha \tau )$ respectively on intervals where the bases are positive. At $\alpha =0$ the conformal Hubble parameter is constant and the separate solution is $a=A{e}^{h\tau }$.

With $\mathcal{H}=aH=\dot{a}$, the Friedmann acceleration equation givesThe coefficient is positive when $a>0$, $\rho >0$ and $H\ne 0$. Thus growth of the comoving Hubble radius and linear growth of departures from flatness have the same sign criterion in an expanding universe. This local criterion is distinct from a global particle horizon, which depends on the whole past history.

A smooth cosmological bounce is a positive minimum of the scale factor: contraction changes to expansion, with $a(t_b)>0$, $\dot{a}(t_b)=0$ and $\ddot{a}(t_b)>0$ in the nondegenerate case. The closed slicing of de Sitter spacetime gives $a=H_{\Lambda }^{-1}\cosh [H_{\Lambda }(t-t_b)]$, whose minimum is $H_{\Lambda }^{-1}$. The Friedmann acceleration equation with zero separate cosmological constant requires $\varrho +3p<0$ there.

For an expanding Friedmann-Lemaître-Robertson-Walker metric with positive energy density, put $\Omega =8\pi G\rho /(3H^2)$ and $w=P/\rho$, in units $c=1$. The Friedmann equation, Friedmann acceleration equation and cosmological perfect-fluid continuity equation giveFor constant $w$, linearizing at $\Omega =1$ gives $\Omega -1\propto a^{1+3w}$. For variable $w$, replace the power by $\exp \int (1+3w(a))d\log a$. This describes the flatness problem for decelerating expansion, away from epochs at which $H=0$.

Neglect radiation in cosmology and take pressureless matter plus a separately conserved dark energy component with $p_{\mathrm{DE}}=w{\rho }_{\mathrm{DE}}{c}^2$, where $\rho$ here denotes mass-equivalent energy density. The Friedmann acceleration equation givesThus $w<-1/3$ for a positive-density component is necessary for that component to accelerate expansion, but the total must also overcome pressureless matter. For a cosmological constant, $w=-1$ and $q_0={\Omega }_{m,0}/2-{\Omega }_{\Lambda ,0}$, so acceleration requires ${\Omega }_{\Lambda ,0}>{\Omega }_{m,0}/2$. Spatial curvature of an FLRW universe enters the distance relation but has no separate term in this acceleration criterion. A Hubble diagram alone does not identify a microscopic dark energy mechanism.

For pressureless matter with zero cosmological constant, the Friedmann acceleration equation gives $q=4\pi G\rho /(3H^2)={\Omega }_m/2$. Spatial curvature need not vanish. Consequently $k{c}^2/a^2=(2q-1)H^2$; this relation does not hold for a general fluid mixture.

For a perfect fluid in general relativity of constant equation of state $P=w\rho$, conservation gives $\rho ={\rho }_0{a}^{-3(1+w)}$. In units $8\pi G=1$, the Friedmann equation becomes ${\dot{a}}^2/2+V(a)=0$. Allowed scales have $V\le 0$, and the Friedmann acceleration equation is $\ddot{a}=-V^{\prime }(a)$. The curvature term shifts the potential vertically; positive cosmological constant bends it downward and negative cosmological constant bends it upward.

For a homogeneous canonical scalar field, the cosmological perfect-fluid continuity equation reduces to $\dot{\varphi }(\ddot{\varphi }+3H\dot{\varphi }+V^{\prime })=0$. It implies the inflaton equation of motion wherever $\dot{\varphi }\ne 0$, and by continuity at isolated turning points. It cannot by itself exclude a frozen field on a time interval. Indeed, choose $\varphi ={\varphi }_0$ with $V({\varphi }_0)>0$ but $V^{\prime }({\varphi }_0)\ne 0$; a de Sitter scale factor in flat slicing with $H^2=V({\varphi }_0)/(3M_{\mathrm{Pl}}^2)$ satisfies the homogeneous Friedmann equation and Friedmann acceleration equation, but not the scalar equation. Varying the inflaton action in an expanding universe supplies the missing equation also for a frozen field, requiring $V^{\prime }({\varphi }_0)=0$.

For pressureless matter and a positive cosmological constant, the Friedmann equation gives ${\dot{a}}^2=H_0^2[{\Omega }_{m,0}{a}_0^3/a+{\Omega }_{k,0}{a}_0^2+{\Omega }_{\Lambda ,0}{a}^2]$. Differentiating with respect to $a$ gives the displayed unique minimum of this expression for positive matter density. It is a minimum of the actual expanding solution's speed only if that branch reaches it; sufficient positive spatial curvature can cause recollapse first. At the minimum, ${\rho }_m=2{\rho }_{\Lambda }$ and the Friedmann acceleration equation changes sign.

For a FRW metric with positive scale factor, nonnegative dust and radiation, and negative vacuum density, the Friedmann acceleration equation gives $a_{tt}\le -{\omega }^{2}a$, with ${\omega }^2=8\pi G|{\rho }_{\Lambda }|/3$. At maximum expansion, $a(0)=a_{\max }$ and $a_t(0)=0$. The forced-oscillator integral formula gives $a(t)\le a_{\max }\cos (\omega t)$ until the first zero, since its forcing is nonpositive and its sine kernel is nonnegative for $0\le t\le \pi /(2\omega )$. Thus each half of the expansion-recollapse history lasts at most $\pi /(2\omega )$. Nonzero dust or radiation makes the endpoint a physical density singularity; the matter-free limit instead bounds the open anti-de Sitter coordinate patch.

At maximum expansion, $a_{\max }=A/k$. With $a_0=1$, the present Friedmann equation gives $k{c}^2=H_0^2({\Omega }_{M,0}-1)$, so positive curvature requires ${\Omega }_{M,0}>1$. Taking $H_0>0$ for the presently expanding branch givesThe lifetime of a closed dust universe follows by setting $\theta =2\pi$. The deceleration parameter follows from the Friedmann acceleration equation, $\ddot{a}/a=-4\pi G{\rho }_M/3$. Thus this model always decelerates while expanding. Its total lifetime includes the elapsed age; it is not the remaining lifetime.

The supernova discovery of accelerating expansion does not support this dust-only recollapse model as a description of the observed expansion history. Already the 1998 and 1999 Type Ia supernova analyses favoured accelerating expansion, requiring ingredients beyond this zero-cosmological constant, positive-curvature dust model within the usual FLRW metric interpretation.^[ref][ref]

For a single constant-$w$ perfect fluid in a flat expanding FLRW metric model, cosmological energy conservation gives $\rho ={\rho }_0{a}^{-3(1+w)}$ and $H(z)=H_0(1+z{)}^{3(1+w)/2}$. With $a_0=1$, the luminosity distance follows by integrating the radial null geodesic:The constant-equation-of-state luminosity distance is consequentlyThis includes $w=-1$ regularly, giving $D_L=z(1+z)/H_0$. Restoring conventional units multiplies the distance formulas by $c$. Expanding about $z=0$ givesThe second result also follows directly from the Friedmann acceleration equation. The expansion agrees with the general coefficient $(1-q_0)/2$ of $z^2$.

For phantom energy, $w<-1$, put $n=3(1+w)/2<0$. Integration of $\dot{a}=H_0{a}^{1-n}$ with $a(t_{\mathrm{now}})=1$ givesBecause $1/n<0$, the scale factor diverges at the Big Rip timeThis is a property of the stipulated constant-$w$ fluid model; it is not realizable by the positive-density canonical scalar of the preceding parts.

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.

Set $H=\dot{a}/a$, the Hubble parameter. The first supplied component of the Einstein field equations gives the Friedmann acceleration equationSubstitution into the second supplied equation givesTherefore the first Friedmann equation isDifferentiate this equation with respect to cosmic time:Insert the Friedmann acceleration equation on the left and cancel $8\pi Ga/3$, which is nonzero for $a>0$. This givesThis is the cosmological continuity equation, expressing stress-energy conservation for the perfect fluid in general relativity. No division by $\dot{a}$ was made, so the derivation remains valid at a cosmological bounce or an instant of maximum expansion.

For $a>0$, the assumed inequality and the Friedmann acceleration equation imply $\ddot{a}\le 0$. Thus $a$ is a concave function. Let $a_0=a(t_0)>0$ and $v_0=\dot{a}(t_0)>0$. At every earlier time in the same regular expanding interval, $\dot{a}(t)\ge v_0$. Integration gives the concavity bound on past FLRW evolution:The right-hand side reaches zero at $t=t_0-a_0/v_0$. Hence an everywhere regular positive scale factor cannot extend arbitrarily far into the past. With the usual implicit continuation assumption that the cosmological solution remains regular while $a>0$, it reaches $a(t_B)=0$ at a finite past boundary satisfyingPast expansion lasts at most the present inverse Hubble parameter. The inequality is the essential argument, and equality is attained by a linearly growing scale factor.

There is a literal qualification to the claim that $a$ must reach zero: the printed hypothesis alone does not exclude an earlier curvature singularity with positive limiting $a$. In arbitrary fixed units, take $k=0$ and $a(t)=1+\sqrt{t}$ for $t>0$, and define $\varrho ,p$ using the two Friedmann equations and the Friedmann acceleration equation. Then $\dot{a}>0$, $\ddot{a}<0$, and $\varrho +3p=-3\ddot{a}/(4\pi Ga)>0$, but $a\to 1$ while the energy density and pressure diverge as $t\downarrow 0$. Thus the rigorous unconditional conclusion is the finite bound on a regular past interval; reaching zero additionally requires the stated continuation assumption. Nor does a zero of $a$ alone prove a curvature singularity, as the Milne model illustrates.

For the equation of state $p=-\varrho$, the cosmological continuity equation gives $a\dot{\varrho }=0$. Hence $\varrho ={\varrho }_0$ is constant. First take the physically intended positive energy density and defineThe Friedmann equation and the original Friedmann acceleration equation becomeSolve the second equation as $a=A{e}^{H_{\Lambda }t}+B{e}^{-H_{\Lambda }t}$. The first equation then requires $4H_{\Lambda }^{2}AB=k$. Choosing the time origin separately in each solution gives all positive branches:For $k=-1$, there is also the time reverse $a=H_{\Lambda }^{-1}\sinh [H_{\Lambda }(t_B-t)]$ on $t<t_B$, which contracts to a future zero. For $k=+1$, the cosmological bounce joins contraction to expansion at the positive minimum $a(t_b)=H_{\Lambda }^{-1}$. Retaining $\ddot{a}=H_{\Lambda }^{2}a$ at the minimum excludes a spurious permanently static solution that would satisfy the first-order constraint alone.

These are vacuum-energy FLRW solutions: the expanding flat branch, closed slicing of de Sitter spacetime, and open slicing of de Sitter spacetime. To compare them at the same energy density, the figure sets $T=H_{\Lambda }(t-t_c)$, takes $A_0=H_{\Lambda }^{-1}$, and aligns the open boundary and closed minimum at $T=0$. Thus the three plotted functions are $H_{\Lambda }a=e^T$, $\cosh T$, and $\sinh T$, the last only for $T>0$.

Only the expanding $k=-1$ branch has a finite past zero. The expanding $k=0$ branch tends to zero only as $t\to -\infty$, and the $k=+1$ branch is bounded below. All three describe portions of de Sitter spacetime, with Ricci scalar $R=12H_{\Lambda }^2$ and Kretschmann scalar $24H_{\Lambda }^4$. In particular the $k=-1$ zero is a boundary of that slicing, not a curvature singularity. The inequality in part (d) does not apply, because $\varrho +3p=-2{\varrho }_0<0$.

For zero energy density, the remaining solutions are $a=A_0>0$ for $k=0$, and the expanding or contracting Milne model $a=\pm (t-t_B)>0$ for $k=-1$; no positive real solution exists for $k=+1$. If negative energy density is admitted mathematically, write ${\alpha }^2=-8\pi G{\varrho }_0/3>0$. Only $k=-1$ is possible, with $a={\alpha }^{-1}\sin [\alpha (t-t_B)]$ for $0<\alpha (t-t_B)<\pi$, an open slicing of anti-de Sitter spacetime. These limiting and negative-density cases do not alter the positive-density comparison above.

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.