Worked Problems — The Robertson–Walker Metric & Friedmann Dynamics

aSignificance of $k$ and $R$

$k$ — the curvature index. A discrete parameter taking values $k \in \{-1,\,0,\,+1\}$ that fixes the sign of the spatial sectional curvature:

$R$ — the (present) radius of curvature. A constant with dimensions of length that sets the scale on which the geometry departs from Euclidean. The physical Gaussian curvature of a $t=\mathrm{const}$ slice is

Today, with the convention $a(t_{0})=1$, the curvature radius of space is just $R$. By rescaling $a$ one could absorb $R$ into the scale factor; it is convenient to keep it explicit so that $x$ has units of length and $a$ remains dimensionless.

bComoving proper distance — the three integrals

At fixed cosmic time and fixed angles, $dt = d\theta = d\phi = 0$, so

Dividing out $a^{2}(t)$ defines the comoving proper distance — the proper distance measured on the slice rescaled to $a = 1$:

(i) Open universe, $k = -1$.

The integrand becomes $(1 + x'^{\,2}/R^{2})^{-1/2}$. Substitute $x' = R\sinh u$, so $dx' = R\cosh u\,du$ and $\sqrt{1 + x'^{\,2}/R^{2}} = \cosh u$:

(ii) Flat universe, $k = 0$.

The integrand reduces to $1$:

(iii) Closed universe, $k = +1$.

Now $(1 - x'^{\,2}/R^{2})^{-1/2}$. Substitute $x' = R\sin u$, $dx' = R\cos u\,du$, $\sqrt{1 - x'^{\,2}/R^{2}} = \cos u$:

This case requires $|x| \le R$; reaching $x = R$ corresponds to $r = \pi R / 2$, one quarter of the way around the 3-sphere. The coordinate $x$ covers only a single chart, while $r$ — the arc length on the sphere — is the geometrically meaningful comoving distance.

cSmall distances: $r \ll R$

Expand each result around $x/R \to 0$ using

With $y = x/R$ this gives

the upper sign for $k = +1$ and the lower for $k = -1$. The fractional correction to the flat result is

negligible for $r \ll R$: all three geometries are locally Euclidean. The universe's curvature is invisible on scales much smaller than its radius of curvature — in exact analogy with how the Earth looks locally flat to a walker. This is also why local laboratory physics is insensitive to the global geometry of the cosmos.

dWhy "open" and "closed"?

The terminology tracks the range of the comoving coordinate $r$ — the genuine arc length on the spatial slice.

Closed, $k = +1$.

Because $r = R\sin^{-1}(x/R)$ and the chart can be continued around the sphere, a complete circumnavigation requires $r \in [0, \pi R]$. The maximum proper distance from any point is $\pi R$, finite. A radial geodesic eventually returns to its starting point. The spatial slice is a compact manifold (topologically $S^{3}$) of finite volume

The geometry is closed in the same sense the surface of a sphere is closed: no boundary, but bounded.

Open, $k = -1$.

$r = R\sinh^{-1}(x/R)$ grows without bound as $x \to \infty$. The slice is the non-compact hyperboloid $H^{3}$ with infinite volume. A radial geodesic recedes forever, never returning. The geometry is open — unbounded in the strong sense.

Flat, $k = 0$.

$r = x \in [0, \infty)$ — also open, but with flat Euclidean topology $\mathbb{E}^{3}$ (modulo global identifications the local metric cannot decide). Conventionally "open" is reserved for $k = -1$, the genuinely hyperbolic case, and "flat" labels $k = 0$ alone.

The terminology therefore tracks a topological / global property — finite versus infinite volume, geodesics returning versus escaping — set by the sign of $k$.

iComoving distance to a source

For a radial null geodesic, $ds^{2} = 0$ with $d\theta = d\phi = 0$:

taking the outgoing-from-source branch with $dx > 0$ as $dt > 0$. Integrate from emission $t_{e}$ to observation $t_{o}$ on the left, and along the corresponding chord $0 \to x$ on the right. The right-hand integral is precisely $(\star)$ — namely $r$:

Two structural remarks are worth emphasizing.

1. The geometric factor $1/\sqrt{1 - kx'^{\,2}/R^{2}}$ has been swept into the definition of $r$. The formula has the same form in all three geometries — which is why the redshift derivation is usually quoted without ever referring to $k$.
2. The integrand $c/a(t)$ is the conformal time element $d\eta \equiv dt/a(t)$ (times $c$). Hence $r = c\,(\eta_{o} - \eta_{e})$: light travels on straight lines in conformal coordinates, exactly as in Minkowski space.

iiFrom this to $\lambda_{o}/\lambda_{e} = a(t_{o})/a(t_{e})$

Consider two successive wave crests emitted from a comoving source at coordinate $x$ (the source is comoving — $x$ does not change between the two emission events). The first crest is emitted at $t_{e}$ and observed at $t_{o}$; the next crest is emitted at $t_{e} + \delta t_{e}$ and observed at $t_{o} + \delta t_{o}$. Both crests satisfy the same null-geodesic relation with the same right-hand side $r$:

Subtracting and reorganizing the limits,

For optical or even radio wavelengths, $\delta t \sim 10^{-15}\!-\!10^{-9}\,\mathrm{s}$ — utterly negligible compared with the Hubble time over which $a(t)$ changes appreciably. So $a(t)$ is effectively constant across each of the two short intervals:

Identifying $\delta t$ with the wave period ($\lambda = c\,\delta t$):

Defining the redshift $1 + z \equiv \lambda_{o}/\lambda_{e}$:

iiiPhysical reading

The wavelength stretches in exactly the same ratio as the spatial slices stretch between emission and observation. Cosmological redshift is not a kinematic Doppler effect from a peculiar velocity — sources here are at rest in comoving coordinates — but a direct consequence of the time-dependence of the spatial metric: the wave propagates through space whose physical scale is growing as $a(t)$, and its wavelength grows in lockstep.

Note what didn't enter: nothing about $k$ or $R$. The redshift formula is universal across open, flat, and closed FRW geometries. Curvature affects the distance–redshift relation (through the form of $x(r)$ once $r = c\!\int dt/a$ is known) and the volume element, but not the kinematics of a single null geodesic's frequency shift.

a · luminosity$\;d_L \equiv \sqrt{L/(4\pi f)}$

(i) Surface area at comoving radius.

Restrict the spatial metric to the 2-surface $\{t = \mathrm{const},\,x = x_k = \mathrm{const}\}$:

a round 2-sphere of physical radius $a(t)\,x_k$. Its area is

At the observer's epoch $a(t_o) = 1$, so the area is $4\pi x_k^{2}$. The operational meaning of $x_k$: photons emitted isotropically by a comoving source spread, at the observer's time, over a sphere of physical area $4\pi x_k^{2}$ — not $4\pi r^{2}$.

(ii) Two factors of $(1 + z)$.

In a static Euclidean universe, $f = L/(4\pi x_k^{2})$. The cosmological expansion modifies this through two effects.

Energy redshift. Each photon arrives with energy $h\nu_{o} = h\nu_{e}/(1+z)$ since $\lambda_{o} = \lambda_{e}(1+z)$. Each photon delivers a factor $1/(1+z)$ less energy than at emission.

Arrival-rate dilation. Photons arrive less frequently by $\delta t_{o}/\delta t_{e} = 1 + z$.

Combining, energy received per unit time is reduced by $(1+z)^{2}$:

(iii) Hence the luminosity distance.

The definition $f = L/(4\pi d_{L}^{2})$ is engineered so $d_L$ is the literal distance in flat static space. Matching:

A high-redshift source therefore appears fainter than its $x_k$ alone would predict — its inferred $d_{L}$ exceeds $x_k$ by the redshift factor.

b · angular$\;d_A \equiv \ell/\theta$

(i) A small perpendicular length.

Orient the polar system so the small ruler of proper length $\ell$ lies along a meridian of constant $\phi$. Then only $\theta$ changes along it:

with $a(t)$ evaluated at the emission time — when the ruler had proper length $\ell$ and emitted the light we now observe. By spherical symmetry the answer cannot depend on orientation.

(ii) Hence the angular diameter distance.

The observer measures $\theta = \Delta\theta$. By definition $\theta = \ell/d_{A}$, so

Using $1 + z = a(t_{o})/a(t_{e}) = 1/a(t_{e})$:

A high-redshift source therefore appears larger in angular size than $x_k$ alone would predict — the light we receive was emitted when the universe (and the ruler at the source's location) occupied a smaller region of space.

c · dualityEtherington's reciprocity

Combining the two distances,

This holds in any metric theory of gravity in which photons travel on null geodesics and photon number is conserved. The factor $(1+z)^{2}$ has a clean accounting: one $(1+z)$ for energy per photon, one $(1+z)$ for photon arrival rate. Departures from this relation are a generic signature of new physics (photon–axion mixing, exotic absorption, modified gravity), and so it is regularly tested with combined SNe Ia, BAO, and cluster data.

aWhen does the choice of distance matter?

The fractional discrepancy between $d_{L}$ and $d_{A}$ at small $z$ is

Setting this equal to a 5% systematic floor:

Above $z \sim 0.025$, the choice of distance indicator pushes the inferred distance by more than the typical systematic. Standard candles (SNe Ia, Cepheids) yield $d_{L}$; standard rulers (BAO, cluster radii, Einstein-radius lensing) yield $d_{A}$.

b · iHubble law and peculiar velocities

The observed recession velocity is the sum of Hubble flow and line-of-sight peculiar velocity:

Inverting via the Hubble law alone gives a fractional distance error $|\Delta d|/d = v_{\rm pec}/(H_{0}\,d)$. Demanding this be $< 10\%$ with $v_{\rm pec} = 10^{3}\,\mathrm{km\,s^{-1}}$:

For $H_{0} \in [50,\,100]\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$:

To be safe across the allowed $H_{0}$ range, work at $d \gtrsim 200\,\mathrm{Mpc}$. This is why classical $H_{0}$-ladder measurements target $\sim 100$–$400\,\mathrm{Mpc}$: close enough that cosmology dependence is mild, far enough that the Virgo-scale peculiar-velocity field is sub-dominant.

b · iiDoes $d_L$ vs $d_A$ matter here?

At $d \sim 100$–$200\,\mathrm{Mpc}$ with canonical $H_{0}$, the relevant redshift is

where the $d_L$–$d_A$ gap is

just below the 10% peculiar-velocity budget but above the 5% systematic floor. At the margin, yes — but barely.

1. At leading order in $z$, $cz = H_{0}d$ is unambiguous — both $d_L$ and $d_A$ Taylor-expand to $cz/H_{0}$.
2. The first-order $\sim 6\%$ correction is sub-dominant but not negligible; in quadrature with 10% it adds $\sim 2\%$ to the total budget.
3. In practice one should quote which distance is meant and use the consistent expansion.

The two equations on the table:

The fluid equation is the local conservation law $\nabla_{\mu} T^{\mu 0} = 0$ for a homogeneous perfect fluid in FRW; it also follows independently from $\nabla_{\mu} G^{\mu\nu} = 0$ applied to Einstein's equations.

Multiply Friedmann by $a^{2}$ and differentiate:

Divide by $2\dot a$ and use the fluid equation, $\dot{\mathcal{E}}\,a^{2}/\dot a = -3 a (\mathcal{E} + P)$:

Therefore

Two readings. First, gravity acts on $\mathcal{E} + 3P$, not on $\mathcal{E}$ alone — positive pressure gravitates and decelerates the expansion in its own right, the pressure-as-source-of-gravity that has no Newtonian analogue. Second, $\ddot a > 0$ requires $P < -\mathcal{E}/3$, i.e. $w < -1/3$: the threshold a fluid must cross to drive cosmic acceleration. A cosmological constant ($w = -1$) qualifies; matter ($w = 0$) and radiation ($w = 1/3$) do not.

a · scaling$\;\mathcal{E} = \mathcal{E}_{0}\,a^{-3(1+w)}$

For $P = w \mathcal{E}$ with $w$ constant, the fluid equation reads

integrating to

with the convention $a_{0} = 1$.

b · componentsThe three substances

The two crossovers that fall out of these scalings — radiation–matter equality at $a_{\rm eq} \approx \Omega_{r,0}/\Omega_{m,0} \sim 3\times 10^{-4}$ and matter–$\Lambda$ equality at $a \approx (\Omega_{m,0}/\Omega_{\Lambda,0})^{1/3} \sim 0.77$ — are the two natural eras of the late universe:

c · FriedmannThe cover-form

Define the critical density as the energy density that would close the universe at the present epoch:

Density parameters: $\Omega_{i,0} \equiv \mathcal{E}_{i,0}/\mathcal{E}_{c}$ for each fluid, and the curvature density parameter

Sign convention: $\Omega_{k} > 0$ for $k = -1$ (open); $\Omega_{k} < 0$ for $k = +1$ (closed). Substituting $\mathcal{E} = \mathcal{E}_{r,0}a^{-4} + \mathcal{E}_{m,0}a^{-3} + \mathcal{E}_{\Lambda,0}$ into Friedmann and dividing by $H_{0}^{2}$:

At $a = 1$ this collapses to the closure constraint $\sum_{i}\Omega_{i,0} = 1$, including $\Omega_{k,0}$.

Vacuum energy ↔ cosmological constant.

Promoting the $\Lambda g_{\mu\nu}$ term to the right-hand side, $G_{\mu\nu} = (8\pi G/c^{4})\,[T_{\mu\nu} + T^{(\Lambda)}_{\mu\nu}]$ with $T^{(\Lambda)}_{\mu\nu} = -(\Lambda c^{4}/8\pi G)\,g_{\mu\nu}$. In the comoving frame this is a perfect fluid with $\mathcal{E}_{\Lambda} = -P_{\Lambda}$, giving

Convert to SI: $H_{0} = 70\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$ becomes

so $H_{0}^{2} = 5.146\times 10^{-36}\,\mathrm{s^{-2}}$, and

The last form is the most physically transparent: it corresponds to $\sim 5.5$ proton rest-energies per cubic metre, or roughly six hydrogen atoms per cubic metre — a useful mental benchmark for what "cosmological mean density" actually means.

From the definition,

so $k = +1$ (closed) and

With $H_{0} = 50\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$, the Hubble distance is $c/H_{0} = 6\,000\,\mathrm{Mpc}$, so

— bigger than the radius of the present observable universe ($\sim 14\,\mathrm{Gpc}$). A universe with this much curvature is still globally finite, but with circumference comfortably exceeding what we can see.

5%Maximum $r$ for which $r \approx x_{k}$

From Problem 1,

The leading fractional correction is

Setting this to $0.05$:

with the corresponding $r = R_{0}\sin^{-1}(0.548) \approx 1.16 \times 10^{4}\,\mathrm{Mpc}$.

The Planck-collaboration value of $H_{0} \approx 50\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$ that "fits" $\Omega_{k} = -0.09$ is therefore not a viable cosmology in isolation — it is one corner of the degenerate ridge in the $(H_{0},\,\Omega_{k})$ plane, ruled out by combining datasets.

Starting from the original equation,

i · time-dependentIn terms of $\Omega_{i}(t)$

Dividing by $H^{2}$ and noticing that each $8\pi G\,\mathcal{E}_{i}/(3 c^{2} H^{2})$ is the present-epoch density parameter $\Omega_{i}(t)$:

Rearranging,

The curvature term has been isolated; everything else is dimensionless. This is the form that most cleanly exposes the link between geometry and total energy budget — and the seed of the next identity.

ii · present-dayIn terms of $\Omega_{i,0}$

Using $\mathcal{E}_{i}(a) = \mathcal{E}_{i,0}\,a^{-3(1+w_{i})}$ for each fluid, dividing Friedmann by $H_{0}^{2}$, and multiplying both sides by $a^{2}$ (since $\dot a^{2} = a^{2} H^{2}$):

where

with $\Omega_{k,0} > 0$ for $k = -1$ (open), $\Omega_{k,0} = 0$ for $k = 0$ (flat), $\Omega_{k,0} < 0$ for $k = +1$ (closed). At $a = 1$ equation (1) reduces to the closure constraint $\sum_{i}\Omega_{i,0} = 1$.

Defining the total density parameter $\Omega \equiv \Omega_{r} + \Omega_{m} + \Omega_{\Lambda}$, the result of 9(i) reads

Evaluated today ($a = 1$, $H = H_{0}$):

Dividing (A) by (B):

The right-hand side is a strictly positive quantity ($H^{2}a^{2} > 0$) multiplied by the constant $(1-\Omega_{0})$. The latter sets the sign for all time; the former is just a rescaling.

a · signSign of $k$ vs. sign of $1-\Omega$

From (A): $H^{2}(1-\Omega) = -kc^{2}/(a^{2}R_{0}^{2})$. Since $H^{2},\,a^{2},\,R_{0}^{2},\,c^{2}$ are all strictly positive,

Equation (2) does more than (A) alone: it tells us this dictionary is time-independent. If $\Omega_{0} = 1$ today, then $\Omega(t) = 1$ for all $t$; if $\Omega_{0} < 1$, then $\Omega(t) < 1$ at every epoch. The Universe never crosses from one geometry to another — once flat, always flat.

b · recollapseClosed, $\Lambda = 0$ universe

With $\Omega_{\Lambda,0} = 0$ and $k = +1$ (so $\Omega_{k,0} < 0$), equation (1) becomes

The right-hand side decreases monotonically with $a$: there exists a unique $a_{\max}$ where $\dot a^{2} = 0$:

For a matter-only closed universe, $a_{\max} = \Omega_{m,0}/(\Omega_{m,0}-1)$. Use the acceleration equation:

For matter ($P = 0$) and radiation ($P = \mathcal{E}/3$), $\mathcal{E} + 3P > 0$ everywhere, so $\ddot a < 0$ at every $a$ — including at $a_{\max}$. The universe arrives at $a_{\max}$ with $\dot a = 0$ and $\ddot a < 0$, and therefore starts contracting, ending in a Big Crunch. Geometrically: the closed-universe analogue of a ball thrown upward that never escapes.

c · accelerationFlat, matter+$\Lambda$ universe

Set $\Omega_{r,0} = 0$ and $\Omega_{k,0} = 0$ in (1):

Differentiating, cancelling $2\dot a$:

The condition $\ddot a > 0$ becomes

For our universe with $\Omega_{m,0} \approx 0.31$, $\Omega_{\Lambda,0} \approx 0.69$: $a_{\rm acc} \approx 0.60$, hence $z_{\rm acc} \approx 0.67$ — the redshift at which the cosmic deceleration parameter changes sign, locked in by Type Ia supernovae circa 1998.

a · limitsRadiation early, $\Lambda$ late

Look again at equation (1):

The four terms scale as $a^{-2},\,a^{-1},\,a^{0},\,a^{2}$:

• As $a \to 0$: $a^{-2}$ wins — radiation dominates regardless of relative coefficient size.
• As $a \to \infty$: $a^{+2}$ wins — $\Lambda$ dominates (provided $\Omega_{\Lambda,0} \neq 0$).

The one exception.

$\Omega_{\Lambda,0} = 0$ — no cosmological constant. Then at late times $a^{2}$ has no coefficient, and the dominant term becomes the curvature term ($a^{0}$, constant) for an open universe (coasting expansion, $\dot a^{2} \to H_{0}^{2}\Omega_{k,0}$), the matter term for a flat universe (Einstein–de Sitter, matter-dominated forever), or no late-time at all for a closed universe (which recollapses).

b · equalityMatter–radiation equality

Set $\Omega_{r}(z) = \Omega_{m}(z)$:

With $\Omega_{m,0} \approx 0.31$ and $\Omega_{r,0} \approx 9.2\times 10^{-5}$, $z_{\rm eq} \approx 3370$, $a_{\rm eq} \approx 3.0\times 10^{-4}$. This is the radiation–matter transition — the epoch before which sub-horizon density perturbations could not grow (the Mészáros effect), and which sets the turnover in the matter power spectrum and the relative heights of CMB acoustic peaks.

c · equalityMatter–$\Lambda$ equality

Set $\Omega_{m}(z) = \Omega_{\Lambda}(z)$:

For our universe, $z_{m\Lambda} = (0.69/0.31)^{1/3} - 1 \approx 0.31$. Compare with the acceleration redshift: $z_{\rm acc} \approx 0.67$. Cosmic acceleration began at $z \approx 0.67$; $\Lambda$ overtook matter only at $z \approx 0.31$. We presently live at $z = 0$ — in the $\Lambda$-dominated, accelerating era, but only just (in $\ln a$ terms).

Three landmark redshifts of $\Lambda$CDM.

Now numerical, with $H_{0} = 74\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$, $\Omega_{m,0} = 0.28$, $T_{0} = 2.7\,\mathrm{K}$. The blackbody energy density is

a · naiveNaive equality redshift

Photon density today:

Critical density today (rescaling Section 7 by $(74/70)^{2} = 1.118$): $\mathcal{E}_{c} \approx 9.25\times 10^{-10}\,\mathrm{J\,m^{-3}}$, hence

So $z_{\rm eq}^{\rm naive} \approx 6\,400$ — roughly twice WMAP's $z_{\rm eq} \approx 3200$.

What we neglected: relativistic neutrinos.

At $z_{\rm eq}$, $T \sim 2.7 \times 3200 \approx 8\,700\,\mathrm{K}$, corresponding to $k_{B}T \sim 0.75\,\mathrm{eV}$ — many orders of magnitude above the neutrino mass scale. So the three Standard Model neutrino species behave as relativistic particles at equality, and they belong in $\mathcal{E}_{r}$ alongside photons. Neutrinos decoupled before $e^{+}e^{-}$ annihilation reheated the photons but not the neutrinos:

Being fermions, each species carries $7/8$ of the bosonic energy density at the same temperature. With $N_{\nu} = 3$:

Corrected:

This sits within ~20% of WMAP's $z_{\rm eq} = 3200$. The residual gap traces to the fact that what actually determines $z_{\rm eq}$ is $\Omega_{m,0} h^{2}$, and the problem's values give $\Omega_{m,0} h^{2} = 0.153$ where observation requires closer to $0.14$. A small further refinement is $N_{\rm eff} = 3.046$ rather than $3$, from incomplete neutrino decoupling during $e^{+}e^{-}$ annihilation — a few-percent effect, well below the discrepancy here.

b · accelerationOnset of acceleration

Flat $\Lambda$CDM with $\Omega_{m,0} = 0.28$, $\Omega_{\Lambda,0} = 0.72$. From Section 11(c),

Cross-check: matter–$\Lambda$ equality is at $z_{m\Lambda} = (0.72/0.28)^{1/3} - 1 \approx 0.37$. Acceleration began at $z \approx 0.73$ — well before $\Lambda$ took over as the dominant component, as expected from the $2\mathcal{E}_{\Lambda}$ vs. $\mathcal{E}_{m}$ asymmetry of the $\mathcal{E}+3P$ source.

For a flat single-fluid universe with equation of state $w$, the cover-form Friedmann equation (1) collapses to $\dot a = H_{0}\,a^{-(1+3w)/2}$. Integrating with $a(0) = 0$ (Big Bang) gives

For radiation, $w = 1/3$:

Evaluating at the present ($a = 1$) fixes the age, and the universal form $a(t) = (t/t_{0})^{1/2}$ follows.

(i) Age.

(ii) Comoving distance to redshift $z$.

With $a_{e} = 1/(1+z)$ and $t_{0} = 1/(2H_{0})$:

(iii) Horizon ($z \to \infty$).

The radiation-universe particle horizon equals exactly the Hubble distance. A useful normalization: $r/r_{\rm hor} = z/(1+z)$, which saturates quickly — half the horizon at $z = 1$, 90% by $z = 9$.

With $\Omega_{m,0} = 1$, $\dot a^{2} = H_{0}^{2}/a$ becomes

so $a^{3/2} = (3/2)\,H_{0}\,t$. At $a = 1$: $t_{0} = 2/(3H_{0})$, hence $a(t) = (t/t_{0})^{2/3}$.

(i) Age.

(ii) Comoving distance.

Since $a_{e} = (t_{e}/t_{0})^{2/3}$, $(t_{e}/t_{0})^{1/3} = a_{e}^{1/2} = 1/\sqrt{1+z}$:

(iii) Horizon.

Twice the Hubble distance. Now $r/r_{\rm hor} = 1 - 1/\sqrt{1+z}$, which saturates more slowly than the radiation case — half-horizon at $z = 3$, 90% at $z = 99$. Light has more time, in comoving terms, to climb out of a matter universe than a radiation one.

With $\Omega_{r,0} = \Omega_{m,0} = 0$, $\Omega_{k,0} = 0$ and $\Omega_{\Lambda,0} = 1$, equation (1) reduces to

so $\dot a/a = H_{0} \equiv H$ — a strictly time-independent Hubble rate. Integrating with $a(t_{0}) = 1$:

The link to $\Lambda$ follows from Friedmann combined with $\mathcal{E}_{\Lambda} = \Lambda c^{4}/(8\pi G)$:

(i) Comoving distance.

Using $e^{H(t_{0}-t_{e})} = 1/a_{e} = 1+z$:

(ii) Horizon.

The de Sitter universe has no particle horizon: any sufficiently distant past event is in causal contact with us.

a · forbiddenWhy $k = +1$ is excluded

From equation (1) with no fluids,

The LHS is non-negative; for $k = +1$ the RHS is strictly negative. No real solution. An empty closed universe cannot exist. A closed universe needs something to curve it.

b · flatMinkowski spacetime

With $k = 0$ and no fluids, $\dot a^{2} = 0$, so $a(t) = \text{const}$. No expansion or contraction. The metric reduces to ordinary flat Minkowski spacetime; cosmology becomes special relativity. Empty flat space is the trivial vacuum solution of GR.

c · MilneOpen empty universe

With $k = -1$ and no fluids, $\Omega_{k,0} = 1$ (closure forces it), so

with $a(0) = 0$ at the Big Bang.

(i) Age.

Exactly the Hubble time — no factor of $2/3$ or $1/2$ in front.

(ii) Comoving distance.

Since $t \propto a$, $t_{0}/t_{e} = 1+z$:

(iii) Horizon.

Like de Sitter, the Milne universe is horizon-free. (Indeed, the Milne metric is just Minkowski spacetime in coordinates adapted to a congruence of inertial observers all emerging from a common origin — an instructive but unphysical limit.) Linear growth $a \propto t$ is sometimes called coasting expansion; it is the dividing case $\ddot a = 0$ between deceleration and acceleration.

The star's age is $13.4 \pm 2.2\,\mathrm{Gyr}$, and the first stars formed $\sim 0.1\,\mathrm{Gyr}$ after the Big Bang. The universe must therefore be at least

Assuming Einstein–de Sitter, $t_{0} = 2/(3H_{0})$, so

Converting via $1\,\mathrm{km\,s^{-1}\,Mpc^{-1}} = 3.241\times 10^{-20}\,\mathrm{s^{-1}}$:

Propagating $t_{0} = 13.5 \pm 2.2\,\mathrm{Gyr}$ gives $H_{0} \approx 48^{+10}_{-7}$ — entirely below the observed range, where every independent determination (local distance ladder, $\sim 73$; Planck CMB, $\sim 67$) sits well above $58$.

For $\Omega_{0} \simeq 1$ with single-fluid domination,

Reading off the three cases:

Feeding these into equation (2) and using $\Omega_{i,0} \simeq 1$ in the dominant component:

Reading the result — the flatness problem and inflation.

Decelerating eras drive $\Omega$ away from $1$. If today $|1 - \Omega_{0}| \lesssim 5\times 10^{-3}$ (Planck), then in the past it must have been spectacularly smaller. Tracing back to the radiation era at the GUT scale $a \sim 10^{-28}$:

Initial conditions of the Big Bang must be fine-tuned to one part in $10^{59}$ for flatness today — the flatness problem, one of the founding motivations for inflation.

Accelerating eras drive $\Omega$ toward $1$. The $\Lambda$ row is qualitatively different: $|1-\Omega|$ shrinks as $a^{-2}$ during de Sitter expansion. An early phase of $\Lambda$-like (quasi–de Sitter) expansion flattens the universe automatically, by exactly the same mechanism that today's $\Lambda$-domination is flattening it now. Inflation invokes a brief, very large-scale-factor period of $\Lambda$-like expansion in the very early universe — typically $a$ grows by $e^{60}$ or more — driving $|1-\Omega|$ down by $e^{120} \sim 10^{52}$ regardless of the pre-inflationary value.

With $\Omega_{r,0} = \Omega_{\Lambda,0} = 0$ and $\Omega_{m,0} < 1$, the cover-form Friedmann reads

Matter is a falling positive contribution, curvature is a positive constant. At early times matter wins (EdS-like); at late times curvature wins (coasting). The dimensionless control parameter is $Q \equiv (1-\Omega_{m,0})/\Omega_{m,0} = \Omega_{k,0}/\Omega_{m,0}$, with $a_{\rm trans} = 1/Q$ marking the matter–curvature transition.

a · integrationBy substitution $\sinh(\theta/2) = \sqrt{Q\,a}$

Rearrange (3) and separate variables:

Hyperbolic substitution is natural because curvature dominates late. From $\sinh(\theta/2) = \sqrt{Q a}$:

so

Substituting and simplifying with $\tanh(\theta/2)\,\sinh\theta = \cosh\theta - 1$:

Integrating with $a = 0$ at $\theta = 0$ ($t = 0$ at the Big Bang):

The function $\sinh\theta - \theta$ is the hyperbolic cousin of $\sin\theta - \theta$ — the cycloid for closed matter universes.

b · verificationBy differentiation

From (4):

The $\Omega_{m,0}/2$ factors cancel in the ratio:

Squaring and eliminating $\theta$ via $\sinh^{2}(\theta/2) = (1-\Omega_{m,0})\,a/\Omega_{m,0}$:

c · limit$\Omega_{m,0} \to 1$ recovers Einstein–de Sitter

The prefactors carry $(1-\Omega_{m,0})$ in the denominator, so for finite $a$ and $t$ in the limit, $\theta$ must be small. Expand:

Substituting and eliminating $\theta$:

d · limit$\Omega_{m,0} \to 0$ recovers Milne

Now $\Omega_{m,0}$ in the prefactors goes to zero, so for finite $a$ we need large $\theta$. Using $\cosh\theta \approx \sinh\theta \approx e^{\theta}/2$:

The parametric solution interpolates smoothly between EdS at the Big Bang and Milne at late times.

Setup: $H_{0} = 74\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$, $\Omega_{m,0} = 0.28$. Hubble time $1/H_{0} = 13.21\,\mathrm{Gyr}$.

Prefactor and $\theta$-recipe:

and (4) inverts to $\cosh\theta = 1 + 2(1-\Omega_{m,0})\,a/\Omega_{m,0} = 1 + 5.143\,a$.

i · age$t_{0}$ (today, $a = 1$)

$\cosh\theta_{0} = 6.143 \Rightarrow \theta_{0} = 2.502$ and $\sinh\theta_{0} = 6.061$.

ii · look-back$t_{0} - t_{e}$ at $z = 3$

$a_{e} = 1/4 \Rightarrow \cosh\theta_{e} = 2.286 \Rightarrow \theta_{e} = 1.468$, $\sinh\theta_{e} = 2.055$, hence $H_{0}\,t_{e} = 0.2292 \times 0.587 = 0.1346$, $t_{e} \approx 1.78\,\mathrm{Gyr}$:

iii · SCDMComparison with the matter-only model

For EdS at the same $H_{0}$, $t = (2/3 H_{0})\,a^{3/2}$:

readingLower density, older universe

The open universe is roughly $2\,\mathrm{Gyr}$ older than the matter-only universe at the same $H_{0}$. Lower mean matter density → weaker gravitational deceleration → the expansion has been more "coasting" than "braking" for cosmic history. At fixed $H_{0}$ today, less deceleration in the past means slower expansion in the past, so the universe took longer to grow to its current $a = 1$.

Open CDM relieves the HE 1523 age tension partially: at $\Omega_{m,0} = 0.28$, $H_{0} = 74$ gives $t_{0} = 10.78\,\mathrm{Gyr}$, still uncomfortably young against $\geq 13.5\,\mathrm{Gyr}$, but headed in the right direction. To reach $13.5\,\mathrm{Gyr}$ with $H_{0} = 74$, open needs $\Omega_{m,0} \lesssim 0.1$ — implausibly low.

For any FRW spacetime, the radial comoving distance is $r = c\!\int dt/a$. Switching from $t$ to $\theta$ via the parametric form of Section 20:

Every $\Omega_{m,0}$, $(\cosh\theta-1)$, and factor of 2 cancels — a small miracle of the substitution. What survives is a constant times $d\theta$:

Integrating from $\theta_{e}$ to $\theta_{0}$:

Comoving distance is linear in the parameter $\theta$. The Hubble length is rescaled by $1/\sqrt{1-\Omega_{m,0}} = 1/\sqrt{\Omega_{k,0}}$ — the same combination that fixed $R_{0}$ in Section 8.

(i) Comoving horizon distance.

Set $\theta_{e} \to 0$. Using $\theta_{0} = 2.502$ and $\sqrt{1-\Omega_{m,0}} = 0.8485$ from Section 21, and $c/H_{0} = 4054\,\mathrm{Mpc}$ for $H_{0} = 74$:

(ii) Comoving distance to $z = 3$.

From Section 21, at $z = 3$: $\theta_{e} = 1.468$. Hence

Comparison with SCDM at the same $H_{0}$.

From Section 15: $r_{\rm hor}^{\rm EdS} = 2c/H_{0} = 8.11\,\mathrm{Gpc}$, $r^{\rm EdS}(z=3) = (2c/H_{0})(1/2) = c/H_{0} = 4.05\,\mathrm{Gpc}$.

a · implied$H_{0}$ from HE 1523–0901 for open $\Omega_{m,0}=0.28$

Take the same age constraint as Section 18: $t_{0} \geq 13.5\,\mathrm{Gyr}$. With $\Omega_{m,0} = 0.28$, Section 21 gave $H_{0}\,t_{0} = 0.816$, so

Propagating $t_{0} = 13.5 \pm 2.2\,\mathrm{Gyr}$: $H_{0} \approx 59^{+12}_{-8}$.

The open universe halves the SCDM tension but does not fully resolve it. A flat $\Lambda$CDM with the same matter density closes the gap to within $1\sigma$.

b · boundaryMaximum $\Omega_{m,0}$ consistent with both

Combining the 1$\sigma$ data ranges:

• HE 1523 + first-star delay: $t_{0} \in [11.3,\,15.7]\,\mathrm{Gyr}$.
• Riess (2009): $H_{0} = 74.2 \pm 3.6 \Rightarrow 1/H_{0} \in [12.57,\,13.85]\,\mathrm{Gyr}$.

The minimum acceptable $H_{0}\,t_{0}$ comes from the low-corner:

Acceptable interval: $H_{0}\,t_{0} \in [0.816,\,1.00]$.

Tabulating via $\theta_{0}$ as intermediate parameter.

From $\cosh\theta_{0} = 1 + 2(1-\Omega_{m,0})/\Omega_{m,0}$:

Reading off the boundary: $\Omega_{m,0} = 0.28$ gives $H_{0}\,t_{0} = 0.816$, exactly the lower limit.

The question's $\Omega_{m,0} = 0.28$ in Section 21 is no coincidence: it sits precisely at the 1$\sigma$ boundary of open-CDM consistency. Anything larger pushes the model into incompatibility.

With $\Omega_{r,0} = \Omega_{\Lambda,0} = 0$ and $\Omega_{m,0} > 1$, the cover-form Friedmann reads

Sign flip from § 2.2: matter is falling positive, curvature is a negative constant. The two terms compete destructively: matter wins early, eventually overtaken by the negative curvature constant, at which point $\dot a^{2} = 0$ and expansion halts. The closed universe recollapses.

Control parameter: $Q \equiv (\Omega_{m,0}-1)/\Omega_{m,0} > 0$, and the substitution is now $\sin(\theta/2) = \sqrt{Q a}$ — trigonometric where the open case was hyperbolic. The closed-universe solution is the celebrated cycloid.

a · integrationBy substitution $\sin(\theta/2) = \sqrt{Q a}$

Rearranging (6):

From the substitution:

so

Substituting and using $\tan(\theta/2)\,\sin\theta = 2\sin^{2}(\theta/2) = 1 - \cos\theta$:

Integrating with $a = 0$ at $\theta = 0$:

The cycloid — trigonometric mirror of the open hyperbolic family.

Maximum scale factor $a_{\max} = \Omega_{m,0}/(\Omega_{m,0}-1)$ confirms the Section 11(b) recollapse argument: at $a_{\max}$, $\dot a = 0$, and the universe contracts symmetrically back to a singularity.

b · verificationBy differentiation

From (7):

The $\Omega_{m,0}/2$ factors cancel in the ratio:

Squaring and eliminating $\theta$ via $\sin^{2}(\theta/2) = (\Omega_{m,0}-1)\,a/\Omega_{m,0}$:

c · limit$\Omega_{m,0} \to 1$ recovers Einstein–de Sitter

The prefactors carry $(\Omega_{m,0}-1)$ in the denominator, so $\theta$ must be small. Using $1 - \cos\theta = \theta^{2}/2 + \cdots$ and $\theta - \sin\theta = \theta^{3}/6 + \cdots$:

EdS is the parabolic limit of the cycloid family.

d · age$t_{0}^{\rm closed} < 2/(3H_{0})$ always

From (6), with $a = 1$ today:

Compare integrands. For $\Omega_{m,0} > 1$, $a \in (0, 1)$:

so the closed integrand is uniformly smaller than EdS:

Sanity check: $\Omega_{m,0} = 2$.

From (7) inverted with $a = 1$: $\cos\theta_{0} = (2 - \Omega_{m,0})/\Omega_{m,0} = 0$, so $\theta_{0} = \pi/2$ and $\sin\theta_{0} = 1$:

comfortably below $2/3 = 0.667$.

The complete age axis — one row per model.

e · distanceComoving distance

The same parametric trick as Section 22:

Integrating from emission to today:

with $\theta_{0} = \arccos\!\bigl((2 - \Omega_{m,0})/\Omega_{m,0}\bigr)$ for the expanding phase. Linear in $\theta$ again, with curvature rescaling $1/\sqrt{|\Omega_{k,0}|}$ — trigonometric counterpart of (5).

Keep $\Lambda$ but drop radiation, with $\Omega_{k,0} = 1 - \Omega_{m,0} - \Omega_{\Lambda,0} < 0$:

Three terms: $a^{-1}$ (matter, falling), $a^{2}$ ($\Lambda$, rising), $a^{0}$ (curvature, negative constant). A contest between a falling positive and a rising positive with a negative constant between.

a · ne fugitWhy $\Lambda$ can save a closed universe

$f(a) = 0$ at $a > 1$ would mean recollapse. Look at the limits:

• $f(a) \to +\infty$ as $a \to 0$ (matter dominates),
• $f(1) = \Omega_{m,0} + \Omega_{\Lambda,0} + \Omega_{k,0} = 1$ (closure),
• $f(a) \to +\infty$ as $a \to \infty$ ($\Lambda$ dominates).

$f$ is positive at both ends. The function dips through a minimum at some intermediate $a_{*}$; only if that minimum is negative does $f$ cross zero. A closed universe with $\Lambda$ avoids recollapse whenever $f(a_{*}) > 0$.

b · $\ddot a = 0$Onset of acceleration is $\Omega_{k,0}$-independent

Differentiating (9): $\ddot a = (H_{0}^{2}/2)(-\Omega_{m,0}/a^{2} + 2\,\Omega_{\Lambda,0}\,a)$.

Setting $\ddot a = 0$:

No $\Omega_{k,0}$. Spatial curvature is a property of constant-$t$ slices, not of the stress-energy sourcing $\ddot a$. Curvature enters $\dot a^{2}$ but not $\ddot a$. Identical to Section 11(c).

c · insufficientWhy small $\Lambda$ doesn't save it

The minimum of $f$ sits at $a_{*} = a_{\rm acc}$. Plugging into $f$:

If $\Omega_{\Lambda,0}$ too small, $f(a_{*}) < 0$: $f$ crosses zero at some $a_{\max} < a_{*}$. Expansion halts before $\Lambda$ takes over.

d · thresholdCubic for $a_{\max}$ and the recollapse relation

At $\dot a = 0$, $f(a_{\max}) = 0$:

The threshold trick. At the boundary, $f$ is tangent to zero: $a_{\max} = a_{\rm acc}$. Substituting $a_{\rm acc}^{3} = \Omega_{m,0}/(2\,\Omega_{\Lambda,0})$:

Cubing and equating to $a_{\rm acc}^{3} = \Omega_{m,0}/(2\,\Omega_{\Lambda,0})$:

Substituting $\Omega_{k,0} = 1 - \Omega_{m,0} - \Omega_{\Lambda,0}$:

The recollapse boundary in the $(\Omega_{m,0},\,\Omega_{\Lambda,0})$ plane.

e · solveIterative solution and the SCP comparison

Setting $\mu = \Omega_{m,0} - 1 > 0$ and $x = \Omega_{\Lambda,0}$:

Leading order:

The 2003 SCP best-fit ($\Omega_{m,0} \approx 0.27$, $\Omega_{\Lambda,0} \approx 0.73$) sits far in the "expands forever" region. Even with $\Omega_{m,0} = 3$, only $\Omega_{\Lambda,0} \approx 0.13$ is enough to escape recollapse.

Einstein's static model requires $\dot a = 0$ and $\ddot a = 0$ both at $a = 1$ — universe motionless in time, with matter and $\Lambda$ in delicate balance.

a · static$\Lambda$ and the curvature radius

Front-cover Friedmann (with $\Lambda$ kept explicit, matter only):

(For $\Lambda$, $P_{\Lambda} = -\mathcal{E}_{\Lambda}$ gives $\mathcal{E}_{\Lambda} + 3 P_{\Lambda} = -2\,\mathcal{E}_{\Lambda}$, equivalent to $+\Lambda c^{2}/3$.)

Condition $\ddot a = 0$.

Dimensions: $[\Lambda] = L^{-2}$, as required.

Condition $\dot a = 0$ at $a = 1$, $k = +1$.

Substituting:

Hence

Radius of curvature equals $\Lambda^{-1/2}$ — the de Sitter length, a single fundamental scale set by $\Lambda$ alone. Two unknowns tied to one observable; no free parameters.

b · unstableInstability under tiny perturbations

Perturb $\mathcal{E}_{m} \to \mathcal{E}_{m,0}(1 + \epsilon)$. Using the unperturbed balance $\mathcal{E}_{m,0} = 2\,\mathcal{E}_{\Lambda}$:

Add a tiny bit of matter ($\epsilon > 0$): the universe begins to contract. Remove a tiny bit: it expands. In either direction the perturbation grows.

Linearized growth rate.

With $\mathcal{E}_{m} \propto a^{-3} \approx \mathcal{E}_{m,0}(1 - 3\,\delta a)$:

Solution: $\delta a \propto \exp(c\sqrt{\Lambda}\,t)$. E-folding timescale:

— the light-crossing time of the universe. No slow, gentle return to equilibrium; the universe runs away on the same timescale that defines it.

The "benchmark" model is flat $\Lambda$CDM: radiation negligible, matter at $\Omega_{m,0}$, vacuum filling the rest. With $\Omega_{m,0} \approx 0.28$ (WMAP-9):

A falling matter term and a rising $\Lambda$ term, no curvature constant. Begins matter-dominated ($a \propto t^{2/3}$), ends $\Lambda$-dominated ($a \propto e^{Ht}$), smooth transition between.

a · integrationBy substitution $\sinh\theta = \sqrt{Q a^{3}}$

Set $Q = (1-\Omega_{m,0})/\Omega_{m,0}$ and $\sinh\theta = \sqrt{Q a^{3}}$, so $1 + Qa^{3} = \cosh^{2}\theta$. Then

Differentiating $a^{3} = \sinh^{2}\theta/Q$: $da/d\theta = 2\sinh\theta\cosh\theta/(3 Q a^{2})$. Combining with $\dot a = (da/d\theta)\,\dot\theta$ and using $a^{3/2} = \sinh\theta/\sqrt{Q}$:

$\theta$ grows linearly with $t$. Integrating from $\theta = 0$ at $t = 0$:

Compact, exact, elegant — the hyperbolic analog of the cycloid.

b · verificationBy differentiation

Let $A = (\Omega_{m,0}/(1-\Omega_{m,0}))^{1/3}$ and $\omega = \tfrac{3}{2}\sqrt{1-\Omega_{m,0}}\,H_{0}$, so $a = A\,\sinh^{2/3}(\omega t)$:

Squaring and using $\sinh^{-2/3}(\omega t) = A/a$, $A^{3} = \Omega_{m,0}/(1-\Omega_{m,0})$, and $\cosh^{2} = 1 + a^{3}/A^{3}$:

c · ageClosed-form age $t_{0}$

Setting $a = 1$ in (11): $\sinh(\omega t_{0}) = \sqrt{(1-\Omega_{m,0})/\Omega_{m,0}}$, so $\cosh(\omega t_{0}) = 1/\sqrt{\Omega_{m,0}}$ and $e^{\omega t_{0}} = (1 + \sqrt{1-\Omega_{m,0}})/\sqrt{\Omega_{m,0}}$:

A logarithm: the signature of the hyperbolic family. For $\Omega_{m,0} \to 0$ this diverges as $\ln(2/\sqrt{\Omega_{m,0}})$ (de Sitter eternity); for $\Omega_{m,0} \to 1$ it tends to $2/(3 H_{0})$ (EdS age).

d · early limitMatter-era parabola

For small $\omega t$, $\sinh^{2/3}(\omega t) \approx (\omega t)^{2/3}$. The $(1-\Omega_{m,0})$ factors cancel between $A$ and $\omega$:

In the limit $\Omega_{m,0} \to 1$ this becomes the EdS solution $a = (t/t_{0})^{2/3}$ with $t_{0} = 2/(3 H_{0})$ exactly.

e · late limitDe Sitter exponential

For large $\omega t$, $\sinh(\omega t) \to \tfrac{1}{2}e^{\omega t}$ and $\sinh^{2/3}(\omega t) \to 2^{-2/3}\,e^{(2/3)\omega t}$:

The exponent must equal the asymptotic Hubble rate because $\dot a^{2} \to H_{0}^{2}(1-\Omega_{m,0})\,a^{2}$ at large $a$, giving $\dot a/a = \sqrt{\Omega_{\Lambda,0}}\,H_{0}$ — the de Sitter rate set by $\Lambda c^{2}/3$.

Fig. 6 makes the structural reading visceral: the matter parabola underestimates $a$ at late times (decelerating gravity would have kept the universe smaller), while the de Sitter exponential overestimates $a$ at early times (it back-extrapolates to $a_{\Lambda}$ rather than zero). The true $\sinh^{2/3}$ interpolates smoothly.

With $H_{0} = 74$ km s$^{-1}$ Mpc$^{-1}$ and $\Omega_{m,0} = 0.28$, Hubble time $H_{0}^{-1} = 13.21$ Gyr.

i · ageFrom the closed-form (12)

ii · look-back to z = 3Inverting (11)

$a_{e} = 1/(1+z) = 1/4$. From (11): $\sinh(\omega t_{e}) = 1.60357 \times 0.125 = 0.20045$, so $\omega t_{e} = 0.19905$ and $t_{e} = 0.15639\, H_{0}^{-1} = 2.07$ Gyr.

The proper distance from a comoving observer to an object of scale factor $a_{e}$ in the benchmark universe is

The substitution $u = \sqrt{a}$ removes the $1/\sqrt{a}$ divergence and produces a polynomial under the radical:

Integrand bounded everywhere on $[0,1]$. Simpson's rule with 8–10 subintervals converges to four digits.

numerical resultsThree universes, $H_{0} = 74$

Comoving distance scale: $c/H_{0} = 4054$ Mpc.

sanity check$\Omega_{m,0} \to 1$ limit

Setting $\Omega_{m,0} = 1$ in (14) collapses the integrand to 1, giving the EdS closed form $d_{p}/D_{H} = 2(1 - \sqrt{a_{e}})$: horizon = 2, $d_{p}(z=3) = 1$. Pushing $\Omega_{m,0} = 0.99$ in the benchmark integrator gives $1.998$ for the horizon — right onto the EdS limit. If your numerics drift from this by more than $\sim 1\%$, the integration has a bug.

physical readingWhy $\Lambda$CDM wins at the horizon

$\Lambda$CDM has the largest comoving distances at every $z$, because vacuum-driven acceleration over the last $\sim 6$ Gyr has stretched comoving separations more than any pure-matter cosmology. EdS has the smallest, since maximal deceleration leaves photons less time to traverse comoving space. The open $\Omega_{m,0} = 0.28$ model lies between. The horizon-distance ratios $13.78 : 11.96 : 8.11$ make the contribution of each ingredient visible at a glance.

For flat $\Lambda$CDM:

Setting $\ddot a = 0$. With $\Omega_{\Lambda,0} = 1 - \Omega_{m,0}$:

numericalFor $\Omega_{m,0} = 0.28$

look-back to z_accFrom (11)

$\sinh(\omega t_{\rm acc}) = 1.60357 \times 0.580^{3/2} = 0.7081$, so $\omega t_{\rm acc} = 0.6594$ and $t_{\rm acc} = 0.5181\, H_{0}^{-1} = 6.84$ Gyr.

What $H_{0}$ would a flat matter-only universe need to reach the same age? EdS age is $t_{0} = 2/(3 H_{0})$, so matching $t_{0} = 12.98$ Gyr:

A matter-only universe needs $H_{0} \approx 50$ — the "low-$H_{0}$" value once championed by Sandage — to age to 13 Gyr without help from $\Lambda$. With $H_{0} = 74$ and no $\Lambda$, the universe would be only $8.81$ Gyr old, younger than the stars in it.

i · look-back times$t_{0} - t_{e}$ for $z \in \{1, 2, 3, 4, 5\}$

Benchmark from (11); matched-EdS from $t_{0}\,[1 - (1+z)^{-3/2}]$.

ii · proper distancesAt the same redshifts

Benchmark from (14); matched-EdS from $d_{p}(z) = 2(c/H_{0}^{\rm EdS})\,[1 - (1+z)^{-1/2}]$ with $c/H_{0}^{\rm EdS} = 5.996$ Gpc. Unmatched-EdS ($H_{0} = 74$) as third baseline.

discussionThe two readings

Look-back times (Fig. 7). Matched-EdS sits uniformly above the benchmark — offset peaks at $z = 1$ ($+1.01$ Gyr), tapers to $+0.25$ Gyr by $z = 5$. Both meet at the horizontal asymptote $t_{0} = 12.98$ Gyr. The difference is in distribution: matched-EdS spends more cosmic time at moderate $a$, less near $a = 1$, because its deceleration packs aging into the late epoch. The benchmark decelerates only for half its history, then accelerates — arriving at $a = 1$ "more rapidly" once $\Lambda$ kicks in.

Proper distances (Fig. 8). At small $z$, $d_{p} \approx c z / H_{0}$ and matched-EdS wins by $74/50 = 1.48$. But the benchmark picks up an integrated boost from acceleration: comoving separations stretched faster between $z_{\rm acc} \approx 0.7$ and today. The boost compounds, and by $z \approx 2.33$ overtakes the matched-EdS Hubble-law advantage. Beyond the crossover the benchmark dominates, asymptoting to $13.78$ Gpc against EdS's $11.99$ Gpc — a $14\%$ horizon excess scaling exactly with the $\sinh^{-1}(\sqrt{(1-\Omega_{m,0})/\Omega_{m,0}})$ factor in (14).