ISEGORIA / MATH ENCYCLOPEDIA
Cosmology: the expanding universe
The Friedmann equation turned into pictures: how the two density parameters Ω_m and Ω_Λ fix the age, the fate and the acceleration of the universe; why there are three different distances to a galaxy and why the angular-diameter distance turns over; and the conformal diagram in which light travels at 45° and the particle horizon, the event horizon and the Hubble sphere are three different curves.
Before you begin: General relativity and thermodynamics
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. The Friedmann equation
For a homogeneous universe of matter, curvature and a cosmological constant, the Friedmann equation says H² = H₀²(Ω_m a⁻³ + Ω_k a⁻² + Ω_Λ) with Ω_k = 1 − Ω_m − Ω_Λ. Drag the gold point in the (Ω_m, Ω_Λ) plane and the curve a(t) on the right is re-integrated: backwards to the big bang, which fixes the age t₀, and forwards to a big crunch or an endless expansion. The dashed diagonal is the flat universe; the teal line Ω_Λ = Ω_m/2 is where the expansion is neither speeding up nor slowing down today. Push Ω_Λ high enough and the shaded region tells you that the universe never had a big bang at all: a(t) bounces. Push it below the orange boundary and the universe recollapses. Watch the orange dot, where ä changes sign, slide along the curve as you move the point.
Worked example. With \(H_0=67.7\) km/s/Mpc, \(1/H_0=14.44\) Gyr. The Einstein–de Sitter universe \((1,0)\) has \(a\propto t^{2/3}\) and age \(\tfrac23/H_0=9.63\) Gyr; the empty universe \((0,0)\) has \(a\propto t\) and age exactly \(14.44\) Gyr. For ΛCDM, \((0.31,0.69)\), the integral gives \(t_0=13.80\) Gyr and \(q_0=0.155-0.69=-0.535\). Acceleration began when \(a^{3}=\Omega_m/2\Omega_\Lambda=0.2246\), that is \(a=0.608\), \(z=0.645\), when the universe was \(7.63\) Gyr old. The closed matter universe \((3,0)\) is \(7.41\) Gyr old and recollapses \(40.7\) Gyr from now.
Watch out. Radiation is neglected here. It dominated the first fifty thousand years, so the curves are wrong very close to \(a=0\) and the ΛCDM age comes out about \(0.02\) Gyr too high; nothing else changes visibly. The model also treats \(\Lambda\) as a strict constant and ignores the small mass of the neutrinos. The recollapse and no-big-bang boundaries are computed by asking whether \(E^{2}(a)\) has a zero for \(a>1\) or for \(a<1\), and agree with the closed forms of Carroll, Press and Turner to twelve digits.
Curvature appears in the Friedmann equation but not in the acceleration equation. Why does the flat line not affect where the orange dot sits?
Differentiate \(\dot a^{2}=H_0^{2}(\Omega_m a^{-1}+\Omega_k+\Omega_\Lambda a^{2})\) with respect to time: the constant \(\Omega_k\) drops out, leaving \(\ddot a=H_0^{2}(-\tfrac12\Omega_m a^{-2}+\Omega_\Lambda a)\). Curvature is a constant of integration of the motion, like the total energy of a thrown stone: it decides whether the stone escapes, not how hard gravity pulls at any moment. So \(\ddot a=0\) always happens at \(a^{3}=\Omega_m/2\Omega_\Lambda\), whatever \(\Omega_k\) is, and moving the point parallel to the flat line only changes the speed at which the curve passes through that scale factor.
2. Distances and redshift
In an expanding universe there is no single distance to a galaxy. The comoving distance D_C is the integral of c dz/H(z) along the light path; the luminosity distance D_L = (1+z)D_M is what you would infer from how faint a standard candle looks; the angular-diameter distance D_A = D_M/(1+z) is what you would infer from how small a standard ruler looks. Drag the gold point along the redshift axis and read all three. D_A grows, reaches a maximum near z = 1.6, and then falls: the picture on the right shows a galaxy 100 kpc across at that redshift, and beyond the turnover it looks larger again, because the light we receive left when the galaxy was much closer. Change H₀ and the curves rescale; change Ω_m and Ω_Λ and the transverse distance D_M picks up the sinh or sin of an open or closed geometry.
Worked example. For ΛCDM with \(H_0=67.7\) km/s/Mpc, \(c/H_0=4.43\) Gpc. At \(z=1\), \(\int_0^{1}dz/E=0.767\), so \(D_C=3.40\) Gpc, \(D_L=6.79\) Gpc and \(D_A=1.70\) Gpc; the light left \(7.94\) Gyr ago. A galaxy \(100\) kpc across subtends \(\theta=0.1\,\mathrm{Mpc}/1698\,\mathrm{Mpc}=5.9\times10^{-5}\) rad, that is \(12.1\) arcseconds. The largest \(D_A\) is \(1.79\) Gpc at \(z=1.59\), where the same galaxy looks smallest, \(11.5\) arcseconds; at \(z=10\) it is back to \(24\) arcseconds while its flux is down by a further factor \((1+z)^{-2}=1/121\).
Watch out. Radiation is neglected, which matters only at \(z\gtrsim100\), and the redshift is taken as purely cosmological: a real galaxy also has a peculiar velocity of a few hundred km/s, which at \(z<0.01\) dominates. The distances assume a smooth universe; light that threads clumps and voids is lensed, so individual \(D_A\) values scatter by a few per cent. \(D_L\) is defined by bolometric flux, and observing in one band needs the K-correction on top of it. The lookback time is not a distance: light from \(z=10\) has travelled for \(13.3\) Gyr but its source is now \(31\) Gly away.
Why is D_L/D_A = (1+z)² in every Friedmann universe, whatever the curvature and the densities?
Both distances are built on the same transverse comoving distance \(D_M\), the radius of the sphere over which the light is spread today. A ruler of size \(\ell\) at emission subtends \(\ell/(a_e D_M)\) because the sphere was smaller by \(a_e=1/(1+z)\) when the light left, giving \(D_A=D_M/(1+z)\). The flux is \(L/4\pi D_M^{2}\) times two factors of \(1/(1+z)\): each photon arrives with less energy, and photons arrive less often, giving \(D_L=(1+z)D_M\). The ratio is the Etherington relation, and it holds for any metric theory in which photons are conserved, so a measured violation would signal new physics rather than a different cosmology.
3. Horizons and light cones
Change the clock to conformal time, dη = dt/a, and measure distance in comoving coordinates, and the expanding universe becomes a flat spacetime diagram in which every light ray is a 45° line. Galaxies are vertical lines. Our past light cone runs down from today to the big bang at η = 0, and where it meets η = 0 is the particle horizon: the farthest matter whose light has had time to reach us. In ΛCDM the conformal time of the whole future is finite, η_∞, so a second cone comes down from the top: the event horizon, beyond which events can never be seen. The Hubble sphere, where recession equals c, is a third, quite different curve. Drag the gold galaxy along the "now" line and see whether a photon it sends today ever reaches χ = 0; press Play to watch our light cone grow from the big bang.
Worked example. For ΛCDM today, \(\eta_0=46.2\) Gly, so the particle horizon is \(46.2\) Gly away in comoving terms, three times the \(13.8\) Gly the light has travelled. The remaining conformal time is \(\eta_\infty-\eta_0=16.6\) Gly: that is the event horizon. The Hubble radius is \(c/H_0=14.4\) Gly. A galaxy at \(\chi=15\) Gly recedes at \(1.04c\), faster than light, and its light emitted today still arrives, at \(t=54\) Gyr, because the Hubble sphere expands to \(17.4\) Gly and overtakes the photon. A galaxy at \(30\) Gly is inside our particle horizon, so we see its past, but any light it emits today never arrives.
Watch out. This lab includes radiation, \(\Omega_r=9.1\times10^{-5}\), because it changes the particle horizon by nearly one Gly; the ΛCDM age then comes out \(13.78\) Gyr. The diagram is one-dimensional in space, so the two sides are the same direction and its mirror. The de Sitter-like model keeps \(a=1\) today, which puts today at \(t_0=40\) Gyr, deep in the Λ era, and its particle horizon far off the frame. Comoving galaxies are drawn as exactly vertical: real galaxies have peculiar motions, and a bound structure such as our Local Group does not expand at all.
The Hubble sphere and the event horizon both sit near 15 Gly today. Why is the Hubble sphere not a horizon?
The Hubble sphere is where the recession speed equals \(c\) now; it says nothing about the future of a photon. A photon just outside it moves toward us at \(c\) through space that recedes slightly faster than \(c\), so it loses ground at first, but in a decelerating or slowly accelerating universe the Hubble sphere \(c/(aH)\) grows, sweeps past the photon, and from then on the photon gains. Whether it ever arrives depends on the whole future integral \(\int_{t}^{\infty}c\,dt/a=c(\eta_\infty-\eta)\), which is the event horizon. In ΛCDM the Hubble sphere approaches \(c/(H_0\sqrt{\Omega_\Lambda})=17.4\) Gly and the event horizon shrinks to meet it, so only asymptotically do the two coincide; in Einstein–de Sitter the Hubble sphere grows without bound and there is no event horizon at all.