Cosmological Horizons

A cosmological horizon is a boundary that a finite-age, finite-light-speed universe draws around every observer, separating what can in principle have touched you from what cannot. Two horizons matter. The particle horizon is the largest comoving distance from which a signal could already have reached you — the edge of everything your past can contain. The event horizon is the largest comoving distance to which a signal you send now can ever arrive — the edge of everything your future can touch. Both are pure kinematics of null rays in the Friedmann–Lemaître–Robertson–Walker (FLRW) metric, fixed entirely by the expansion history $a(t)$ , hence by the matter–energy content through the Friedmann equations. The whole subject collapses to a single change of clock. In conformal time $\eta=\int\mathrm{d}t/a$ , radial light rays satisfy $\mathrm{d}\chi=\pm c\,\mathrm{d}\eta$ , so photons travel on straight $45^\circ$ lines. The comoving particle horizon is then just the conformal time elapsed since the Big Bang, and the comoving event horizon the conformal time remaining until the end. Whether a horizon exists becomes a question of whether a conformal-time integral converges — at the lower limit for the particle horizon, at the upper for the event horizon. One distinction organizes everything and trips up nearly everyone: the horizon is not the Hubble radius $R_H=c/H$ . The Hubble radius is local and instantaneous — the proper distance at which recession reaches $c$ right now — while the particle horizon integrates the entire past. In a decelerating universe the comoving Hubble radius grows and scales enter it; during inflation or dark-energy domination it shrinks and scales exit it. In concordance $\Lambda$ CDM the particle horizon sits at $\approx14\ \mathrm{Gpc}$ comoving ( $\approx46$ Gly), a cosmic event horizon exists at $\approx16$ Gly, and the far future closes slowly around the Local Group.

Key takeaways

In an expanding universe light covers a finite comoving distance in finite time, so only a bounded region can have influenced you — the particle horizon — and only a bounded region can ever be reached — the event horizon.
In conformal time radial null rays obey $\mathrm{d}\chi=\pm c\,\mathrm{d}\eta$ , so the comoving particle horizon is $c\,\eta$ since $t=0$ and the comoving event horizon is $c\,(\eta_{\max}-\eta)$ remaining until the end.
A particle horizon is finite whenever early expansion decelerates enough ( $a\propto t^n$ , $n<1$ ); an event horizon exists only when late expansion accelerates ( $w<-\tfrac13$ ), so that $\int^{\infty}\mathrm{d}t/a$ converges.
The Hubble radius $c/H$ is a distinct, local scale often confused with the horizon: its comoving version can shrink while the particle horizon always grows, and objects can re-enter it.
The finite particle horizon at recombination is the horizon problem motivating inflation; the $\Lambda$ CDM event horizon means distant galaxies will freeze, redshift, and fade from view.
Key idea: radial null rays obey $\mathrm{d}\chi=c\,\mathrm{d}\eta$ , so the particle horizon is the conformal time since $t=0$ and the event horizon is the conformal time left until $t_{\max}$ — and each horizon exists only if its conformal-time integral converges.

The core insight

Radial null rays obey $\mathrm{d}\chi=c\,\mathrm{d}\eta$ in conformal time, so the particle horizon is the accumulated conformal time since $t=0$ and the event horizon is the conformal time remaining until $t_{\max}$ ; a horizon exists only if the corresponding conformal-time integral converges.

Where this lives today

Cosmological horizons set the frontiers of working astronomy. The particle horizon is the radius of the observable universe, and the surface of last scattering just inside it is mapped in detail by CMB experiments — COBE, WMAP, and above all Planck, whose acoustic peaks both quantify the horizon problem and confirm the inflationary initial conditions that solve it. The James Webb Space Telescope now images galaxies at $z>13$ , pushing observation toward that horizon; nothing electromagnetic can be seen beyond it, though relic gravitational waves and neutrinos in principle probe still earlier. At the other end of cosmic time, the $\Lambda$ CDM event horizon governs the far future: galaxies beyond $\approx16$ billion light-years will never send us light emitted after today, and over the next $10^{11}$ years everything outside the bound Local Group will redshift, dim, and cross it — a fate quantified by Krauss and Starkman. The de Sitter horizon also carries a genuine thermodynamics: the Gibbons–Hawking temperature $T=\hbar H/2\pi k_B$ and its entropy tie cosmology to black-hole physics and holography. DESI, Euclid, and the Vera Rubin Observatory are now measuring the dark-energy equation of state $w(z)$ precisely, because whether $w<-\tfrac13$ persists is exactly what decides the fate of our event horizon.

The physical picture

The cleanest way to see a horizon is to redraw the expanding universe in coordinates where light moves on straight diagonal lines. Comoving distance $\chi$ removes the stretching of space; conformal time $\eta=\int\mathrm{d}t/a$ reparametrizes the clock so the FLRW line element for a radial ray becomes $\mathrm{d}s^2=a^2(-c^2\mathrm{d}\eta^2+\mathrm{d}\chi^2)$ , and a null ray has $\mathrm{d}\chi=\pm c\,\mathrm{d}\eta$ — a perfect $45^\circ$ line, as in flat-space special relativity. The two horizons are then simply the light cones of the here-and-now: the past cone dropped to the Big Bang, the future cone raised to the end of time.

Figure

A conformal spacetime diagram in comoving distance χ and conformal time η where radial light rays run at 45° obeying dχ = ±c dη. The past light cone of the here-and-now reaches the Big Bang at the comoving particle horizon χ_p = c η₀, while the future light cone reaches conformal infinity at the event horizon χ_e = c(η_max − η₀). A source beyond χ_e can never be seen emitting after today, though sources out to χ_p are already visible.

Now separate the horizon from the Hubble radius, because conflating them is the single most common error in the topic. The Hubble radius $c/H$ is a snapshot — "where things recede at $c$ right now"; the particle horizon is a biography — the whole integrated past. In a decelerating universe the comoving Hubble radius grows, so the Hubble sphere sweeps outward and you keep seeing new galaxies. Under accelerating expansion it shrinks, the door slowly closes, and galaxies you can see today will one day be beyond reach — yet nothing ever re-enters the particle horizon, which only grows. Picture light as a runner on a stretching sheet: decelerating stretch lets it reach ever-farther markers; accelerating stretch means beyond some marker it never arrives. The second figure makes the three scales concrete for our own universe. Because $\Lambda$ CDM accelerates, all three are finite today and nest: the Hubble sphere smallest, the event horizon just outside it, the particle horizon far beyond both. The shell between the Hubble sphere and the particle horizon is exactly the region of galaxies that recede faster than light yet stay visible — the fact that dissolves the "nothing travels faster than light" paradox once horizon and Hubble radius are kept distinct.

Figure

The nested cosmological horizons of ΛCDM as proper distances measured today, with the observer at the centre. The Hubble sphere sits at 14.4 Gly (where recession speed equals c), the cosmic event horizon at about 16 Gly, and the particle horizon at 46 Gly which equals 14 Gpc comoving and bounds the observable universe. Galaxies in the shaded shell between the Hubble sphere and the particle horizon recede faster than light yet are still seen, because observability is set by the particle horizon.

The core equations

Everything begins with the FLRW line element, the metric of a homogeneous, isotropic, expanding universe — the arena in which all horizon integrals live, with the scale factor $a(t)$ as the only dynamical input.

\mathrm{d}s^2=-c^2\mathrm{d}t^2+a^2(t)\left[\mathrm{d}\chi^2+S_k^2(\chi)\,\mathrm{d}\Omega^2\right] \qquad (1)

where: $a(t)$ — scale factor, dimensionless, normalized $a(t_0)=1$ ; $\chi$ — comoving radial coordinate (length); $S_k(\chi)$ — curvature function equal to $\chi$ (flat), $\sin(\sqrt{k}\chi)/\sqrt{k}$ (closed), or $\sinh(\sqrt{|k|}\chi)/\sqrt{|k|}$ (open); $c$ — speed of light. Reading: proper distances stretch as $a(t)$ grows while comoving labels $\chi$ ride along unchanged. For horizons we need only radial light rays, so set $\mathrm{d}s^2=0$ and $\mathrm{d}\Omega=0$ . The metric hands us the master relation, and conformal time $\eta$ turns it into the $45^\circ$ rule.

c\,\mathrm{d}t=a(t)\,\mathrm{d}\chi \quad\Longrightarrow\quad \mathrm{d}\chi=\frac{c\,\mathrm{d}t}{a(t)}=c\,\mathrm{d}\eta \qquad (2)

where: $\eta$ — conformal time, defined by $\mathrm{d}\eta=\mathrm{d}t/a$ . Reading: in $\chi$ – $\eta$ coordinates light advances $c$ units of comoving distance per unit of conformal time whatever the expansion does — causal contact is built into $\eta$ . The comoving particle horizon is the total comoving distance light has covered since the beginning — simply $c$ times the conformal age.

\chi_p(t)=\int_0^{t}\frac{c\,\mathrm{d}t'}{a(t')}=c\,[\eta(t)-\eta(0)] \qquad (3)

where: $\chi_p$ — maximum comoving distance from which light could have reached the observer by time $t$ ; $t'=0$ — Big Bang, or the start of the relevant epoch. Reading: a source could have influenced you iff its comoving distance is below $\chi_p$ ; the proper size of the observable universe is $a(t)\,\chi_p$ . The comoving event horizon is the mirror image — the comoving distance light will cover between now and the end, i.e. $c$ times the conformal time that remains.

\chi_e(t)=\int_t^{t_{\max}}\frac{c\,\mathrm{d}t'}{a(t')}=c\,[\eta(t_{\max})-\eta(t)] \qquad (4)

where: $\chi_e$ — comoving distance of the farthest source a signal emitted at $t$ can ever reach; $t_{\max}$ — end of the universe ( $\infty$ if it expands forever, or the Big Crunch time). Reading: a signal you send now reaches a source only if it lies within $\chi_e$ ; if the integral diverges there is no event horizon and every source is reachable. To evaluate either integral in a real cosmology, change variable from time to scale factor via $\mathrm{d}a=aH\,\mathrm{d}t$ and insert the Friedmann rate — the form you actually compute with.

\chi_p=\int_0^{a}\frac{c\,\mathrm{d}a'}{a'^2 H(a')},\qquad H(a)=H_0\sqrt{\Omega_r a^{-4}+\Omega_m a^{-3}+\Omega_k a^{-2}+\Omega_\Lambda} \qquad (5)

where: $H(a)$ — Hubble parameter from the Friedmann equation; $\Omega_r,\Omega_m,\Omega_k,\Omega_\Lambda$ — present density parameters for radiation, matter, curvature, dark energy. Reading: the horizon is a weighted integral over the whole expansion history, so it "knows" every epoch the universe has passed through. Finally, the Hubble radius — the quantity most often mistaken for the horizon — a local scale set by the instantaneous expansion rate, not an integral over the past.

R_H=\frac{c}{H},\qquad R_H^{\text{com}}=\frac{c}{aH}=\frac{c}{\dot a} \qquad (6)

where: $R_H$ — proper Hubble radius, the proper distance where recession speed $v=Hd$ equals $c$ ; $R_H^{\text{com}}$ — comoving Hubble radius. Reading: $R_H^{\text{com}}$ grows while $\ddot a<0$ (scales enter) and shrinks while $\ddot a>0$ (scales exit) — the behaviour that separates it from the ever-growing particle horizon.

Derivation AParticle horizon in a single-component power-law universe

The goal is to compute the particle horizon exactly for the workhorse case of a flat universe dominated by one perfect fluid, where $a\propto t^n$ , and to extract the compact result that the proper horizon is $d_p=ct/(1-n)$ , a fixed multiple of both $ct$ and the Hubble radius. Along the way we see why the integral converges only for $n<1$ , why the prefactor is not one, and how the horizon compares to the naive guess $ct$ — the calculation that fixes the familiar " $2ct$ " for radiation and " $3ct$ " for matter.

d_p(t)=\frac{c\,t}{1-n}=\frac{n}{1-n}\,\frac{c}{H} \qquad\text{(power-law horizon)}

Step 1. Model the epoch by $a(t)=(t/t_0)^{n}$ with $0<n<1$ , so that $H=\dot a/a=n/t$ and $R_H=c/H=ct/n$ . The comoving horizon is $\chi_p=\int_0^t c\,\mathrm{d}t'/a=c\,t_0^{\,n}\int_0^t t'^{-n}\,\mathrm{d}t'$ . Why: reducing the cosmology to one power law makes the null-ray integral elementary while capturing every decelerating single-fluid universe.

Step 2. Do the integral: $\int_0^t t'^{-n}\,\mathrm{d}t'=t^{1-n}/(1-n)$ , finite at the lower limit precisely when $n<1$ . Hence $\chi_p=c\,t_0^{\,n}\,t^{1-n}/(1-n)$ . Why: convergence at $t'=0$ is what gives a decelerating universe a finite past light cone.

Step 3. Convert to proper distance by multiplying by $a(t)$ : the $t_0^{\,n}$ factors cancel and $d_p=a\chi_p=ct/(1-n)$ . Why: the cancellation shows the horizon is independent of the arbitrary normalization time $t_0$ .

Step 4. Form the ratio to the Hubble radius: $d_p/(c/H)=[ct/(1-n)]/(ct/n)=n/(1-n)$ . For radiation $n=\tfrac12$ this gives $d_p=2ct=c/H$ — one Hubble radius. For matter $n=\tfrac23$ it gives $d_p=3ct=2c/H$ — two Hubble radii. Why: the horizon is the same order as the Hubble radius but never equal in general; the integer coincidences are special to these two equations of state, and getting the factor right is where careless treatments slip. *Direct integration gives $d_p=2ct=c/H$ (radiation) and $d_p=3ct=2c/H$ (matter), with comoving forms $\chi_p=c/(aH)$ and $\chi_p=2c/(aH)$ respectively.*

Go deeper · Derivation B · 1 minThe event horizon of a de Sitter universe

Derivation BThe event horizon of a de Sitter universe

The goal is to show that exponential expansion produces a genuine event horizon by making the future conformal-time integral converge, and to derive the striking result that the proper event-horizon distance is constant in time and exactly equal to the Hubble radius $c/H$ . This is the late-time attractor of $\Lambda$ CDM that our own universe is settling into, and the calculation reveals both why a horizon appears and why the comoving version keeps shrinking while the proper version stands still.

d_e=a(t)\,\chi_e=\frac{c}{H}=\text{constant (proper)} \qquad\text{(de Sitter event horizon)}

Step 1. Take $a(t)=e^{Ht}$ with constant $H=\sqrt{\Lambda c^2/3}$ , the de Sitter solution for a pure cosmological constant. The comoving event horizon is $\chi_e=\int_t^{\infty}c\,\mathrm{d}t'/a=c\int_t^{\infty}e^{-Ht'}\,\mathrm{d}t'$ . Why: with $\Lambda$ dominant the scale factor grows without bound, and the $e^{-Ht'}$ in the integrand is what could make the future light cone terminate at a finite comoving distance.

Step 2. The integral converges because the integrand decays: $\chi_e=c\left[-e^{-Ht'}/H\right]_t^{\infty}=(c/H)\,e^{-Ht}=c/(aH)$ , so the comoving event horizon is finite and shrinks as $a^{-1}$ . Why: convergence at the upper limit is the signature of an event horizon — a comoving distance no future signal of yours can cross.

Step 3. Convert to proper distance: $d_e=a\,\chi_e=c/H$ , independent of time. Why: the shrinking comoving horizon and the growing scale factor cancel exactly, leaving a fixed proper sphere — special to constant $H$ , and the reason de Sitter space carries a thermodynamic Gibbons–Hawking horizon like a black hole.

Step 4. Read the physics: any source now beyond $\chi_e=c/(aH)$ can never receive a photon you emit from now on, and light it emits toward you redshifts without bound as it asymptotes to the horizon, fading rather than vanishing. Why: this is the "door slowly closing" of the accelerating universe stated quantitatively — the boundary is fixed in proper size, so galaxies drift across it outward one by one. *The result $d_e=c/H$ is the standard de Sitter event horizon, consistent with the Gibbons–Hawking horizon and the $\Lambda$ CDM late-time limit computed numerically below.*

Go deeper · Derivation C · 1 minWhen does an event horizon exist

Derivation CWhen does an event horizon exist

The goal is to turn the existence of an event horizon into a sharp condition on the equation of state, showing that one exists if and only if the late-time expansion accelerates, $w<-\tfrac13$ . The whole question reduces to whether $\int^{\infty}\mathrm{d}t/a$ converges at its upper limit, and that is a one-line power-law $p$ -test once the late-time scale factor is written $a\propto t^n$ . The result draws the exact line between a universe where everything is eventually reachable and one where the door closes.

\text{event horizon exists}\iff \int^{\infty}\frac{\mathrm{d}t}{a(t)}<\infty \iff n>1 \qquad\text{(existence condition)}

Step 1. For a fluid of constant $w>-1$ the Friedmann equation gives $a\propto t^{\,n}$ with $n=2/[3(1+w)]$ at late times. Convergence of the future integral is set by the tail: $\int^{\infty} t'^{-n}\,\mathrm{d}t'$ is finite iff $n>1$ . Why: only the behaviour as $t'\to\infty$ can make the event-horizon integral diverge, so a $p$ -test on the power-law tail settles existence completely.

Step 2. Translate $n>1$ into the equation of state: $n=2/[3(1+w)]>1\iff w<-\tfrac13$ . Why: $w<-\tfrac13$ is exactly the condition $\ddot a>0$ for acceleration, so "the universe accelerates" and "an event horizon exists" are the same statement.

Step 3. Read off the cases. Radiation ( $w=\tfrac13$ , $n=\tfrac12$ ) and matter ( $w=0$ , $n=\tfrac23$ ) both have $n<1$ : the integral diverges, no event horizon, every source eventually observable. A cosmological constant ( $w=-1$ ) gives convergence and an event horizon. Why: a decelerating Big Bang universe has no event horizon while our dark-energy universe does — a change of causal structure driven purely by the sign of $\ddot a$ . *A cosmological event horizon exists precisely when the expansion accelerates ( $w<-\tfrac13$ ), consistent with the de Sitter limit $w=-1$ derived above.*

Limiting cases

Horizon behaviour is dictated by the dominant equation of state $w$ through $a\propto t^{2/[3(1+w)]}$ (for $w>-1$ ). The table collects the standard single-fluid universes, with the correct factors relating $\chi_p$ to the comoving Hubble radius $c/(aH)$ .

Case	Result	Meaning
Radiation ( $w=\tfrac13$ )	$a\propto t^{1/2},\ \ d_p=2ct=c/H,\ \ \chi_p=c/(aH)$	Particle horizon finite and growing, equal to one Hubble radius. No event horizon: $\int^{\infty}\mathrm{d}t/a$ diverges.
Matter ( $w=0$ )	$a\propto t^{2/3},\ \ d_p=3ct=2c/H,\ \ \chi_p=2c/(aH)$	Particle horizon is two Hubble radii; comoving $\chi_p\propto t^{1/3}$ keeps growing. No event horizon.
de Sitter ( $w=-1$ )	$a\propto e^{Ht},\ \ d_e=c/H,\ \ \chi_e=c/(aH)$	Event horizon exists with constant proper radius $c/H$ ; comoving event horizon shrinks as $a^{-1}$ .
Empty / Milne ( $a\propto t$ )	$n=1,\ \ \chi_p=c\!\int t'^{-1}\mathrm{d}t'$ diverges at $t'=0$	Marginal case: the particle-horizon integral diverges logarithmically at the Big Bang, so there is no finite particle horizon. No event horizon.

Common trapThe particle horizon is the Hubble radius

It is not. They agree only in order of magnitude and only for power-law expansion — radiation gives $d_p=c/H$ but matter gives $d_p=2c/H$ , so even the factor is not universal. Far more importantly, the two behave oppositely under acceleration: the comoving particle horizon $\chi_p$ grows monotonically forever, whereas the comoving Hubble radius $c/(aH)$ shrinks whenever $\ddot a>0$ (inflation, dark energy). A comoving scale can therefore leave the Hubble sphere and later re-enter it — the entire language of "modes exiting and re-entering the horizon" in inflationary perturbation theory — but nothing ever re-enters the particle horizon. When a cosmologist says "the horizon" in structure formation they usually mean the Hubble radius; for causal contact since the Big Bang they mean the particle horizon. Keep the two apart and the superluminal-recession paradoxes evaporate.

Worked examples

1 · Horizon versus Hubble radius in radiation and matter

In a flat single-fluid universe with $a\propto t^n$ , show explicitly how the proper particle horizon compares to $ct$ and to the Hubble radius for radiation and for matter, and interpret the difference.

Step 1. Use the boxed power-law result $d_p=ct/(1-n)$ . Radiation ( $n=\tfrac12$ ): $d_p=\mathbf{2ct}$ . Matter ( $n=\tfrac23$ ): $d_p=\mathbf{3ct}$ . Both exceed the naive $ct$ because space expanded while the light was in transit, carrying the emission point farther away than the light-travel time alone suggests.

Step 2. Compare to $R_H=c/H=ct/n$ . Radiation: $R_H=2ct$ , so $d_p/R_H=\mathbf{1}$ — one Hubble radius. Matter: $R_H=1.5ct$ , so $d_p/R_H=\mathbf{2}$ — two Hubble radii.

Step 3. The ratio $d_p/R_H=n/(1-n)$ rises as $n\to1$ : the more slowly the universe decelerates, the more the horizon outruns the Hubble radius. Quoting "the horizon is $ct$ " is wrong by a factor of 2–3, and "horizon equals Hubble radius" is right only for radiation. Radiation: $d_p=2ct=c/H$ . Matter: $d_p=3ct=2c/H$ .

2 · The three horizon distances of our universe (fully numerical)

Take the concordance values $H_0=67.7\ \mathrm{km\,s^{-1}\,Mpc^{-1}}$ , $\Omega_m=0.31$ , $\Omega_\Lambda=0.69$ . Compute the present Hubble distance, the asymptotic de Sitter event horizon, and compare with the tabulated particle and event horizons.

Step 1. Hubble distance now: $R_{H,0}=c/H_0=(2.998\times10^5)/67.7=4428\ \mathrm{Mpc}=4.43\ \mathrm{Gpc}$ . With $1\ \mathrm{Mpc}=3.262\ \mathrm{Mly}$ this is $\approx\mathbf{14.4\ Gly}$ — the proper distance at which galaxies recede at exactly $c$ today.

Step 2. Asymptotic event horizon. As dark energy takes over, $H\to H_\Lambda=H_0\sqrt{\Omega_\Lambda}=67.7\sqrt{0.69}=56.2\ \mathrm{km\,s^{-1}\,Mpc^{-1}}$ , so $c/H_\Lambda=5335\ \mathrm{Mpc}\approx\mathbf{17.4\ Gly}$ . The universe is not yet fully $\Lambda$ -dominated, so the present cosmic event horizon is a little smaller, $\approx\mathbf{16\ Gly}$ proper ( $\approx5\ \mathrm{Gpc}$ comoving), growing toward $17.4$ Gly as matter dilutes.

Step 3. Particle horizon. Numerically integrating $\chi_p=\int_0^1 c\,\mathrm{d}a/(a^2 H(a))$ gives $\chi_p\approx14.0\ \mathrm{Gpc}$ comoving; since $a_0=1$ the proper distance today is $\approx\mathbf{46\ Gly}$ — far larger than $c\,t_0\approx13.8$ Gly because space expanded during the journey.

Step 4. The three proper distances today nest as $R_{H,0}=14.4<d_e\approx16<d_p\approx46$ Gly. The event horizon lies just outside the Hubble sphere, the particle horizon more than three times larger — which is why so many observable galaxies recede faster than light. Hubble 14.4 Gly, event horizon ≈ 16 Gly, particle horizon ≈ 46 Gly.

3 · The horizon problem at recombination

Estimate how large the causally connected patch was at recombination ( $z\approx1100$ ) as seen on today's sky, and state the puzzle.

Step 1. At recombination the universe is matter-dominated, so the comoving particle horizon there, $\chi_p(t_{\mathrm{rec}})\approx0.25\ \mathrm{Gpc}$ , is a small fraction of the comoving distance to last scattering, $\chi_{\mathrm{ls}}\approx14\ \mathrm{Gpc}$ .

Step 2. The angle it subtends is $\theta\approx\chi_p(t_{\mathrm{rec}})/\chi_{\mathrm{ls}}\approx0.25/14\approx0.018\ \mathrm{rad}\approx\mathbf{1^\circ\text{–}2^\circ}$ , so the CMB sky divides into tens of thousands of patches never in causal contact before they emitted.

Step 3. Yet those disconnected patches share the same temperature to one part in $10^5$ . Decelerating expansion offers no mechanism to equilibrate them — this is the horizon problem, resolved by an early inflationary phase ( $\ddot a>0$ ) in which the comoving Hubble radius shrank so the whole sky was in causal contact before being stretched apart. The recombination horizon spans only $1^\circ$ – $2^\circ$ , yet the CMB is uniform across the whole sky.

Physics of the portrait

The modern vocabulary of cosmological horizons — the distinction between "what could have influenced me" and "what I could ever influence" — was set down in one remarkably clear 1956 paper.

Origins · 1 minWolfgang Rindler and how this was discovered

Wolfgang Rindler

1924–2019 · relativistic cosmology and the causal structure of expanding world-models

Wolfgang Rindler was an Austrian-born physicist who fled to England as a child in 1938 and spent most of his career at the University of Texas at Dallas, one of the great expositors of relativity. His 1956 paper "Visual horizons in world-models" (MNRAS 116, 662) is the source of the now-universal terminology: he separated the particle horizon, bounding the particles that could already have been seen, from the event horizon, bounding the events an observer could ever come to see, and showed that each is governed by the convergence of an integral over the expansion history. Before Rindler these ideas floated informally in the work of de Sitter, Lemaître, and Eddington; after him they had crisp definitions every cosmology textbook still uses. His name also attaches to the "Rindler coordinates" and "Rindler horizon" of a uniformly accelerated observer in flat spacetime — the special-relativistic cousin of the cosmological event horizon, and a keystone of the later Unruh-effect and black-hole-thermodynamics literature.

History

1917

— de Sitter finds the exponentially expanding vacuum solution, the first model with a cosmological event horizon

1929

— Hubble establishes the expansion of the universe, making horizon distances physically real

1956

— Rindler defines the particle horizon and the event horizon of world-models

1981

— Guth proposes inflation, resolving the horizon problem with an early accelerating phase

1998

— supernova surveys reveal cosmic acceleration, giving ΛCDM a genuine event horizon

Going further

Texts. Ryden, Introduction to Cosmology, 2nd ed., ch. 5 (proper distance and horizons) · Dodelson & Schmidt, Modern Cosmology, 2nd ed., ch. 2–3 (conformal time, the horizon problem) · Mo, van den Bosch & White, Galaxy Formation and Evolution, §3.2 · Carroll & Ostlie, An Introduction to Modern Astrophysics, 2nd ed., ch. 29 · Hobson, Efstathiou & Lasenby, General Relativity: An Introduction for Physicists, ch. 15 · Rindler, "Visual horizons in world-models," MNRAS 116, 662 (1956) — the classic paper defining both horizons. All results follow from the FLRW metric and the Friedmann equation. Key results to keep in mind: (1) the power-law horizon $d_p=ct/(1-n)$ , giving $d_p=2ct=c/H$ for radiation and $d_p=3ct=2c/H$ for matter — note the radiation case is one Hubble radius, correcting a common " $2c/H$ " slip; (2) the de Sitter event horizon $d_e=c/H$ ; (3) the existence condition $w<-\tfrac13$ ; (4) the numerical ladder for Planck $\Lambda$ CDM ( $H_0=67.7$ , $\Omega_m=0.31$ , $\Omega_\Lambda=0.69$ ): Hubble distance $14.4$ Gly, present event horizon $\approx16$ Gly ( $\approx5$ Gpc comoving), asymptotic de Sitter horizon $17.4$ Gly, particle horizon $46$ Gly ( $14$ Gpc comoving).

Prerequisites. The FLRW Metric · The Expansion of the Universe and the Scale Factor · Comoving and Proper Coordinates · The Friedmann Equations · Cosmological Redshift. Leads to. Equation of State in Cosmology · The Fluid and Acceleration Equations · Spatial Curvature of the Universe.

Ask the tutor: Ask me to draw the conformal diagram for any expansion history, or to work through why an accelerating universe closes an event horizon while a decelerating one never does.

What is the difference between the particle horizon and the observable universe?

They are essentially the same thing viewed slightly differently. The particle horizon is the maximum comoving distance from which light could have reached us by now; the sphere of that radius, centred on us, is the observable universe. Its proper radius today is about 46 billion light-years, and its outer edge is the surface of last scattering we see as the cosmic microwave background — the oldest light there is. "Observable" refers to electromagnetic signals; relic gravitational waves or neutrinos in principle originate even closer to the Big Bang, but the particle horizon remains the hard limit for any causal signal since $t=0$ .

How can galaxies be 46 billion light-years away if the universe is only 13.8 billion years old?

Because the distance and the age measure different things. The light from the most distant matter took about 13.8 billion years to arrive, but during that journey space itself expanded, carrying the source ever farther away. By the time the light lands, the galaxy that emitted it is about 46 billion light-years away in proper distance. Nothing travelled through space faster than light; the space in between simply stretched. This is why the proper particle horizon, 46 Gly, is more than three times the naive light-travel figure $ct_0\approx13.8$ Gly.

If a galaxy recedes faster than light, why can we still see it?

Recession velocity $v=Hd$ exceeds $c$ for everything beyond the Hubble sphere at 14.4 Gly, and most galaxies we observe are past that. Superluminal recession is a coordinate effect — space expanding, not objects moving through space — so it violates no law of relativity. Visibility is governed not by the Hubble sphere but by the larger particle horizon: a photon emitted long ago passed through regions where local expansion was slower, made net progress toward us, and eventually arrived. As long as the emitter lies inside the particle horizon, its ancient light reaches us even while the galaxy now recedes faster than light.

Does every expanding universe have an event horizon?

No — an event horizon requires accelerating expansion. It exists only if the future integral $\int^{\infty}\mathrm{d}t/a$ converges, and for a fluid of equation of state $w$ that happens exactly when $w<-\tfrac13$ , i.e. when $\ddot a>0$ . A radiation- or matter-dominated universe that expands forever has no event horizon: given enough time, a signal reaches any comoving destination. Our universe acquired a real event horizon only because dark energy ( $w\approx-1$ ) came to dominate around six billion years ago.

Why is the de Sitter event horizon equal to the Hubble radius, and is that always true?

In de Sitter space $H$ is constant, so the comoving event horizon $\chi_e=c/(aH)$ shrinks as $a^{-1}$ while the scale factor grows as $a$ , and the two cancel to give a constant proper distance $d_e=c/H$ . The equality is therefore special to exponential expansion, not a general rule. In $\Lambda$ CDM today the two differ — the event horizon (≈16 Gly) is a bit larger than the Hubble radius (14.4 Gly) — because the universe is not yet purely de Sitter. Only in the infinite future, as matter dilutes, will they converge near 17 Gly.

What is the horizon problem, and how does inflation solve it?

At recombination the particle horizon subtended only about one to two degrees on today's sky, so the CMB divides into tens of thousands of patches that had never been in causal contact — yet they share the same temperature to one part in $10^5$ . Decelerating expansion offers no way to have equilibrated them, which is the horizon problem. Inflation solves it with a very early epoch of accelerated expansion during which the comoving Hubble radius shrank dramatically: the whole observable universe was in causal contact before inflation stretched it far beyond the Hubble sphere, so its uniformity was established early and merely stretched smooth. Inflation also drives spatial curvature toward zero, matching the observed flatness.