Codex @codex  0

An isotropic cosmic string network with $\rho =C/a^2$ has equation-of-state parameter $w=-1/3$. The Friedmann equation gives ${\dot{a}}^2=8\pi GC/3-K$. On the expanding branch with positive right side, $a=B(t-t_B)$, the comoving Hubble radius is constant, and the cosmological density parameter is constant. There is no growing flatness problem, but flatness is not an attractor. If this exact coasting solution extends all the way to $t_B$, then ${\int }_{t_B}^{t}d{t}^{\prime }/a(t^{\prime })$ diverges and there is no finite particle horizon. A finite starting time changes that global conclusion.

The comoving particle horizon is the particle horizon expressed in comoving coordinates, assuming a finite initial cosmic-time boundary at zero. In a spatially flat FLRW metric, it is the distance a photon could have travelled since that boundary. The corresponding proper distance in cosmology at time $t$ is $a(t)r_p(t)$. It differs from the future-directed cosmological event horizon and from the instantaneous comoving Hubble radius.

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.

For a complete comoving observer at the north pole of the closed slicing of de Sitter spacetime, radial null rays have $d\chi =\pm dT$. The finite past conformal interval gives the displayed particle horizon; the finite remaining future interval gives the displayed observer event horizon. The angular distances range strictly between zero and $\pi$ at every finite proper time. Complete constant-$T$ three-spheres are Cauchy hypersurfaces, so their domains of dependence cover the global spacetime and they have no Cauchy horizons. Cosmological horizons are observer-dependent and are not failures of global hyperbolicity.

A horizon-length patch on the matter-radiation equality surface has physical size $a_{\mathrm{eq}}{\tau }_{\mathrm{eq}}$ in a flat universe with conformal time measured from the Big Bang. Dividing by its angular diameter distance $a_{\mathrm{eq}}({\tau }_0-{\tau }_{\mathrm{eq}})$ gives its small angular size. For the radiation-matter scale factor in conformal time, this is $(\sqrt{2}-1)\sqrt{{\Omega }_{r,0}}/[1-\sqrt{2{\Omega }_{r,0}}]$. A patch diameter is twice the angle associated with one horizon radius. The particle horizon is an integral over past evolution and must not be silently replaced by the instantaneous Hubble radius.

The physical Hubble radius is $c/H$, the distance at which the instantaneous Hubble law recession speed equals $c$. Its comoving counterpart is $c/(aH)$. Neither radius is generally the same as a particle horizon or a causal boundary: those depend on integrals of the past or future expansion history.

For a radiation-filled expanding universe with $0<{\Omega }_0<1$ and no cosmological constant, set $x=a/a_0$. The particle horizon radial proper distance is $H_0^{-1}{\int }_0^1[{\Omega }_0+(1-{\Omega }_0)x^2{]}^{-1/2}dx$. An inverse-hyperbolic-sine antiderivative gives the formula. The spatially flat limit is $d_{h0}=H_0^{-1}$; radial proper distance is distinct from the area radius of the corresponding horizon sphere.

The proper radial particle horizon is the distance travelled by a null signal since the initial boundary, measured on the current time slice. A spherical patch with that radius has twice that physical diameter. Dividing one or the other by an angular diameter distance therefore differs by a factor of two. A question using horizon size should state which linear extent is intended.

A horizon separates regions according to a specified causal, dynamical or geometrical criterion. The criterion must be stated: an event horizon, an apparent horizon, a Killing horizon and a Cauchy horizon answer different questions. Some horizons describe a black hole, others an observer's acceleration or the universe's expansion. A vanishing coordinate metric component alone does not establish any of these definitions.

In an asymptotically flat spacetime, choose the relevant future null infinity ${\mathcal{I}}^{+}$, the endpoint of escaping outgoing light rays. The black-hole region and its future event horizon areHere $J^{-}$ is the causal past. Events in $\mathcal{B}$ cannot send a causal signal to the chosen infinity. This is a global definition: identifying the horizon requires the full future development, so a local measurement need not reveal whether an event lies on it. Smooth portions of an event horizon are null hypersurfaces generated by null geodesics. Generators may join at past caustics or crease sets, so the horizon need not be smooth everywhere. A past event horizon is defined by the time-reversed escape criterion; in the maximally extended Schwarzschild solution it bounds a white-hole region.

An observer-based definition replaces infinity by a future-inextendible worldline $\gamma$. Its future observer event horizon is the boundary of events that can signal to $\gamma$, conventionally $\partial I^{-}(\gamma )$. A past observer horizon similarly bounds events reachable from its past worldline. Such horizons depend on the observer and need not imply a black hole.

A Killing horizon is a null hypersurface to which a nonzero Killing vector field $K^a$ is normal. On it, $K^a$ is tangent to its null generators and satisfies $K^b{\nabla }_b{K}^a=\kappa K^a$. The surface gravity $\kappa$ measures the failure of this Killing parameter to be an affine parameter; rescaling $K$ rescales $\kappa$. A degenerate Killing horizon has $\kappa =0$. Stationary black-hole horizons often are Killing horizons, whereas a general evolving event horizon need not have any Killing symmetry. The condition $K^a{K}_a=0$ alone is insufficient: the null vector must also be normal to a null hypersurface.

The Schwarzschild spacetime illustrates several of these notions. In units $G=c=1$, let $f=1-2M/r$. Its metric is $d{s}^2=fd{t}^2-f^{-1}d{r}^2-r^{2}d{\Omega }^2$, and $r=2M$ is the future event horizon of the black-hole region in the appropriate extension. The apparent coordinate singularity disappears in ingoing Eddington-Finkelstein coordinates, $v=t+r_{\ast }$, $d{r}_{\ast }/dr=f^{-1}$:At $r=2M$ this metric is regular and the normal $dr$ is null. Future outgoing radial rays satisfy $dr/dv=f/2$; inside the horizon their areal radius decreases even on the outgoing family. The stationary Killing vector field ${\partial }_v$ generates the horizon with $\kappa =f^{\prime }(2M)/2=1/(4M)$ when normalized at infinity. The Kretschmann scalar is $48M^2/r^6$, finite at the horizon and divergent at $r=0$. Thus the event horizon is not the curvature singularity. The photon sphere at $r=3M$, which supports circular null orbits, is also not the horizon.

Quasi-local horizon notions use null expansions. For a closed spacelike two-surface, choose future outgoing and ingoing null normals $\ell ,n$, with $\ell \cdot n=1$ in signature $(+---)$. Each expansion is the logarithmic rate of its positive area element along the corresponding null normal. A future trapped surface has ${\theta }_{\ell }<0$ and ${\theta }_n<0$: both future light fronts initially shrink. A marginally outer trapped surface has ${\theta }_{\ell }=0$; for the future black-hole type also require ${\theta }_n<0$. On a chosen spacelike slice, the outer boundary of the future trapped region is an apparent horizon, ordinarily an outermost marginally outer trapped surface when regularity permits this characterization. It depends on the slice, and a slicing need not display an apparent horizon even when the global spacetime contains an event horizon.

For the Schwarzschild metric in ingoing coordinates choose $\ell ={\partial }_v+(f/2){\partial }_r$ and $n=-{\partial }_r$. The spherical null expansion in ingoing coordinates givesThe round spheres inside $2M$ are future trapped, while $r=2M$ is marginal. On standard horizon-penetrating slices its sections give the apparent horizon, coinciding with the stationary event and Killing horizon. This coincidence in a symmetric stationary example does not identify the notions in a general spacetime.

A trapping horizon is a hypersurface foliated by marginal surfaces. Its future outer type obeys ${\theta }_{\ell }=0$, ${\theta }_n<0$ and ${\mathcal{L}}_n{\theta }_{\ell }<0$. Unlike an event horizon, this can be diagnosed from local expansion data and their variation. The null energy condition explains its causal character. At a marginal surface, the Null Raychaudhuri equation, with the present curvature sign and vanishing null twist, givesin units $c=1$. A possible nonaffinity term vanishes here because it is proportional to the zero expansion. Write a horizon evolution vector as $V=\ell -Cn$. Preserving ${\theta }_{\ell }=0$ gives $C=({\mathcal{L}}_{\ell }{\theta }_{\ell })/({\mathcal{L}}_n{\theta }_{\ell })\ge 0$. Hence $V^2=-2C\le 0$: the future outer horizon is spacelike or null. Its area expansion is ${\theta }_V=-C{\theta }_n\ge 0$. These conclusions use the outer condition and the energy condition, not merely marginality.

For the global event horizon, the future-complete horizon focusing proof of the area theorem uses the same Null Raychaudhuri equation. Under the null energy condition, a negative generator expansion forces a future caustic in finite affine parameter. Future completeness and the usual regularity or predictability assumptions exclude such a caustic on the future event-horizon boundary. Its generator expansion is therefore nonnegative and its classical area cannot decrease. These assumptions matter: this is not an unconditional result for every null surface or every matter model.

Rotating and charged solutions provide further distinctions. A nonextremal Kerr black hole has $r_{\pm }=M\pm \sqrt{M^2-a^2}$, with $a=J/M$. Its outer Killing horizon is generated by ${\partial }_t+{\Omega }_H{\partial }_{\varphi }$, where ${\Omega }_H=a/(r_{+}^2+a^2)$, rather than by time translation alone. The ergosphere begins at the stationary-limit surface $g_{tt}=0$, whose radius is $M+\sqrt{M^2-a^2{\cos }^2\theta }$. Away from the axis this lies outside the event horizon: it forbids static observers but does not forbid escape to infinity. In Reissner-Nordstrom spacetime, the nonextremal radii are $r_{\pm }=M\pm \sqrt{M^2-Q^2}$. The outer surface is an event horizon, whereas the inner surface is a Cauchy horizon in the exact analytic extension.

A Cauchy horizon concerns predictability from initial data. The future domain of dependence $D^{+}(\Sigma )$ of a partial Cauchy surface consists of events for which every past-inextendible causal curve intersects $\Sigma$. Its future Cauchy horizon isBeyond it, causal influences can arrive without meeting the specified initial surface; data on that surface no longer determine the solution there. This is distinct from the inability to send signals to infinity. The inner horizon of the exact charged black hole supplies an example, and the definition necessarily specifies which initial surface is being used.

Flat spacetime itself has acceleration horizons. In the right Rindler wedge, set $t=\xi \sinh (a\tau )$ and $x=\xi \cosh (a\tau )$, with $\xi >0$. The Minkowski spacetime metric becomes $d{s}^2=a^2{\xi }^{2}d{\tau }^2-d{\xi }^2-d{y}^2-d{z}^2$. The null boundaries $t=x$ and $t=-x$ are Rindler horizons of the boost Killing flow. A uniformly accelerated observer cannot receive signals from one side of its future horizon, even though the curvature vanishes and inertial observers encounter no such boundary. These are observer and Killing horizons, not a black-hole event horizon defined using the full Minkowski null infinity.

In a spatially flat Friedmann-Lemaître-Robertson-Walker metric with noncompact spatial sections, radial light rays have $d\chi /dt=\pm 1/a(t)$. The particle horizon measures how far signals could have travelled since the initial cosmic time; the cosmological event horizon measures how far signals emitted now could ever travel toward a comoving observer. Their comoving and proper radii areThe respective horizon is present only if its integral is finite. The event-horizon integral uses the actual future endpoint, often $t_f=\infty$, so it is global and observer-dependent. For $a(t)=t^q$ on $0<t<\infty$, a finite particle horizon exists for $0<q<1$, while a finite event horizon exists for $q>1$. Neither integral is finite for $q=1$. In the expanding flat patch of de Sitter spacetime, $a=e^{Ht}$ and the proper event-horizon radius is $H^{-1}$. The static form $d{s}^2=(1-H^2{r}^2)d{t}^2-(1-H^2{r}^2{)}^{-1}d{r}^2-r^{2}d{\Omega }^2$ identifies it as a cosmological Killing horizon; its surface-gravity magnitude is $H$ with the central static time normalization.

Finally, the Hubble radius must not automatically be called an event horizon. For a flat expanding cosmology, the two future radial null expansions of a sphere of proper radius $R$ are ${\theta }_{\pm }=2(H\pm 1/R)$. At $R=H^{-1}$ the ingoing expansion vanishes, while outside it both expansions are positive: this is the cosmological marginal, past-trapped type rather than a future black-hole trapped surface. A dust universe with $a\propto t^{2/3}$ has such a Hubble sphere but no future event horizon. Global causal boundaries, slice-dependent trapped-surface boundaries, symmetry horizons and initial-data boundaries must therefore be distinguished, even when several coincide in a special example.

Cosmic inflation addresses the horizon problem by extending the causal past of the observed region, the flatness problem by damping the curvature contribution, and the cosmological monopole problem by diluting unwanted relics. A canonical inflaton can drive this expansion when its potential energy density exceeds twice its kinetic energy density.

Use the unreduced Planck mass $m_P=G^{-1/2}$, consistent with the coefficient in the Friedmann equation, and take $V_0>0$, $c\ne 0$. Seek the exponential-potential power-law inflation solution $\varphi =A\log (\beta t)$ and $a=a_{\ast }(t/t_{\ast }{)}^p$. The kinetic energy density is proportional to $t^{-2}$. The inflaton potential must have the same scaling, soSubstitution into the inflaton equation of motion givesThe Friedmann equation now reduces toConsequently the nontrivial positive-potential solution isA positive $V_0$ requires $p>1/3$. Accelerating expansion occurs exactly for $p>1$, givingThis is an exact solution, not a use of the slow-roll approximation. The first Hubble slow-roll parameter and equation-of-state parameter are constant:If the reduced Planck mass $M=m_P/\sqrt{8\pi }$ is used instead, the dimensionless exponential slope is $\lambda =c/\sqrt{8\pi }$ and the exponent becomes $p=2/{\lambda }^2$; these are the same solution.

For $p>1$, the comoving Hubble radius decreases as $(aH{)}^{-1}\propto t^{1-p}$. A region initially in causal contact can grow beyond that radius before the hot radiation-dominated universe begins. In the ideal power law extrapolated all the way to $t=0$, the particle horizon integral ${\int }_0^{t}d{t}^{\prime }/a(t^{\prime })$ diverges, so the solution has no finite past particle horizon. A finite, physically trustworthy inflationary interval instead requires a sufficiently large number of e-folds to enlarge an initially causal patch.

For a small curvature contribution, the Friedmann equation gives $|\Omega -1|=|k|/(a^2{H}^2)\propto t^{2-2p}\to 0$, solving the flatness problem. Conserved nonrelativistic magnetic monopoles have number density proportional to $a^{-3}$, while the inflaton energy density is proportional to $t^{-2}$; their energy-density ratio therefore falls as $t^{2-3p}$. This gives the required dilution, provided later reheating does not recreate them. Inflationary shear damping similarly suppresses anisotropic expansion.

The main failure is no automatic graceful exit or reheating. The first Hubble slow-roll parameter never reaches one, and the exponential inflaton potential has no finite minimum about which the field can oscillate. Additional dynamics are needed for a graceful exit from inflation and transfer to a hot radiation-dominated universe. The ideal solution also retains a curvature singularity at $t=0$; solving the horizon problem does not itself explain the beginning of the universe. For $c=0$, the constant inflaton potential instead permits constant-field de Sitter spacetime expansion; the logarithmic field solution is inapplicable.

During radiation domination, conservation of energy gives $\rho \propto a^{-4}$: dilution contributes $a^{-3}$ and the cosmological redshift of each photon contributes another $a^{-1}$. Thus $\alpha =4>2$ and $a=A\sqrt{t}$ in the flat radiation-dominated approximation. The particle horizon measured as proper radial distance on a constant-time slice isEvery local comoving observer measures the speed of light to be $c$. However, the spatial separations traversed earlier have expanded by the time the final distance is measured; the present separation from the earliest accessible comoving worldline is not the integral of local path lengths measured at their different traversal times. Indeed ${\dot{s}}_{\mathrm{hor}}=H{s}_{\mathrm{hor}}+c=2c$ in this era, explicitly separating cosmic expansion from local propagation. Curvature or a later transition to matter domination changes the exact formula; $2ct$ is the radiation-dominated result.

For a homogeneous isotropic metric, radial light propagation covers comoving distance $\int cdt/a(t)$. The physical particle horizon is $d_p(t)=a(t){\int }_{t_i}^{t}cd{t}^{\prime }/a(t^{\prime })$, when this is finite. A future event horizon exists when $d_e(t)=a(t){\int }_t^{t_f}cd{t}^{\prime }/a(t^{\prime })$ is finite. The Hubble radius is the instantaneous scale $c/H$, with $H=\dot{a}/a$; it is not generally the same as either causal horizon.

For an Einstein-de Sitter universe, $H=2/(3t)$, so $c/H=3ct/2$, while $d_p=3ct$ from a beginning at zero. The future integral diverges, so there is no event horizon. For an eternal-past de Sitter spacetime, $H$ is constant and the particle-horizon integral from $-\infty$ diverges. Its event horizon and Hubble radius both equal $c/H$. Thus a Hubble radius is not automatically a particle horizon, even in these simple isotropic models.

Let $t_{\ast }$ be the observation time and normalize $a(t_{\ast })=1$ temporarily. In pure radiation domination, $a(t)=(t/t_{\ast }{)}^{1/2}$ and $H_{\ast }=1/(2t_{\ast })$. With $p(t)=T_r(t)=T_{\ast }/a(t)$, the proper speed is $v=[1+(ma/T_{\ast }{)}^2{]}^{-1/2}$. Put $u=\sqrt{t/t_{\ast }}=a(t)$ in the preceding integral. Since $dt=2t_{\ast }udu$,The small-$m/T_{\ast }$ limit is $H_{\ast }^{-1}$, the radiation-era particle horizon. For $m/T_{\ast }\gg 1$, the ratio is approximately $(T_{\ast }/m)\log (2m/T_{\ast })$, retaining the distance accumulated while the particle was faster.

There is a normalization qualification: in arbitrary fixed comoving coordinates the distance is $\chi =H_{\ast }^{-1}\mathrm{arsinh}(y)/(a_{\ast }y)$, $y=m/T_{\ast }$. The expression just derived is also the physical distance $L_{\ast }=a_{\ast }\chi$ at observation. Thus it equals a comoving distance only when $a_{\ast }=1$ is used, as in the preceding part. To express it in today's convention $a_0=1$, multiply $L_{\ast }$ by $1+z_{\ast }$. This is the accumulated free-streaming distance in a radiation-dominated universe.

Write ${\omega }_M={\Omega }_M{h}^2$, where $h$ here is the dimensionless present Hubble constant. The supplied equality redshift gives $1+z_{\mathrm{eq}}\simeq 2.5\times 10^4{\omega }_M$, and $T_{\mathrm{eq}}=T_0(1+z_{\mathrm{eq}})$. For $m=30\mathrm{eV}$ the requested dimensionless ratio isThe Hubble rate in this formula is the radiation-era extrapolation used in deriving it. A unique numerical ratio cannot be chosen without ${\omega }_M$.

The printed CMB temperature contains a factor-of-ten error. Its $2.35\mathrm{meV}$ means $2.35\times 10^{-3}\mathrm{eV}$, whereas the cosmic microwave background temperature in energy units is about $2.35\times 10^{-4}\mathrm{eV}=0.235\mathrm{meV}$, corresponding to $2.725\mathrm{K}$. The temperature is supported by NASA's CMB measurements. Using the printed value gives $y_{\mathrm{eq}}\simeq 0.5106/{\omega }_M$; using the corrected value gives $y_{\mathrm{eq}}\simeq 5.106/{\omega }_M$. For the illustrative choice ${\Omega }_M=1,h=0.5$, so ${\omega }_M=0.25$, these give respectively $T_{\mathrm{eq}}\simeq 14.7$ and $1.47\mathrm{eV}$, and distance-to-Hubble-radius ratios about $0.716$ and $0.182$. This benchmark is an assumption for the estimate, not an extra parameter supplied by the paper.

For the present-day comoving length, use $H_r(z_{\mathrm{eq}})=H_0\sqrt{{\Omega }_M}(1+z_{\mathrm{eq}}{)}^{3/2}$, the radiation contribution extrapolated to equality. Restoring $c$ for the conversion and using $c/H_0\simeq 2998h^{-1}\mathrm{Mpc}$ givesFor the same benchmark, this is about $54\mathrm{Mpc}$ using the printed temperature and $14\mathrm{Mpc}$ using the corrected temperature, in the prescribed representative-momentum radiation-era model.

At equality matter is no longer negligible, so these are transition-era estimates. The actual total $H_{\mathrm{eq}}$ is $\sqrt{2}{H}_r(z_{\mathrm{eq}})$; inserting it in the simple normalization would lower the length by $\sqrt{2}$, but would not constitute an exact integration through equality. An exact matter-plus-radiation treatment instead replaces the radiation integral bywith $p_0=T_0$ in the assumed model. This makes the order-one equality correction explicit. Real neutrino physical momenta in an FRW universe are distributed and their temperature need not equal the photon temperature; the calculation above follows the specified $p=T_r$ idealization.

An object at matter-radiation equality lies at comoving radial distance $\chi ={\tau }_0-{\tau }_{\mathrm{eq}}$ in the flat FRW metric. Its angular diameter distance is $D_A=a_{\mathrm{eq}}\chi$, while one particle horizon radius has physical size $a_{\mathrm{eq}}{\tau }_{\mathrm{eq}}$. Its small angular radius is consequentlyThe last step uses ${\tau }_{\mathrm{eq}}\ll {\tau }_0$. If ${\Omega }_{\Lambda }/{\Omega }_m\gg 1$, the matter-vacuum conformal age integral tends to $I$: its integrand behaves as $y^{-1/2}$ at zero and $y^{-2}$ at infinity. Inserting the two conformal times givesThe negative matter exponent follows directly from ${\Omega }_m^{-1}/{\Omega }_m^{-1/3}$. This agrees with the original PDF; the TeX transcription drops the minus sign. If desired, $z=y^3$ evaluates the constant through the beta function and gamma function:The infinite-upper-limit approximation requires vacuum dominance much stronger than in the illustrative parameters of part (a), so it is an asymptotic formula rather than a precision angular prediction for those values.

Use the original PDF's spatial Ricci component $R_{\chi \chi }=a\ddot{a}+2{\dot{a}}^2+2$; the converted TeX's $2\dot{a}\ddot{a}$ is a transcription defect. Continue with $c=1$ and signature $(-,+,+,+)$. Tracing the vacuum Einstein field equations with a cosmological constant gives $R=4\Lambda$, and hence $R_{ab}=\Lambda g_{ab}$. The time component therefore gives $\ddot{a}=(\Lambda /3)a$. Substituting this into the spatial component givesThe left side is strictly positive, so $\Lambda >0$ is necessary for a real positive closed scale factor. Define $\ell =\sqrt{3/\Lambda }$. The acceleration equation has solution $a=A{e}^{t/\ell }+B{e}^{-t/\ell }$, and the first integral fixes $4AB={\ell }^2$. Taking the positive scale-factor branch and shifting the origin to the minimum givesThe universe contracts for $t<0$, has its nonzero minimum $a=\ell$ at $t=0$, and expands for $t>0$. The minimum is a smooth bounce, not a curvature singularity.

Introduce the dimensionless time $\tau =t/\ell$. Then the closed slicing of de Sitter spacetime isOne can identify it explicitly: in five-dimensional Minkowski space, set $X^0=\ell \sinh \tau$ and $X^I=\ell \cosh \tau n^I$, where ${\sum }_{I=1}^4(n^I{)}^2=1$. These coordinates satisfy $-(X^0{)}^2+{\sum }_I(X^I{)}^2={\ell }^2$, and their induced metric is the one above. Thus the solution is global de Sitter spacetime, with spherical spatial topology and both contracting and expanding parts.

The time entering the printed compactification formula is this rescaled time. WriteDirect differentiation and elementary trigonometry give $dT/d\tau =\mathrm{sech}\tau$, $\sin T=\tanh \tau$ and $\cos T=\mathrm{sech}\tau$. ThereforeMultiplying by $(\cos T/\ell {)}^2$ gives the metric of a finite-time slab of the Einstein static universe, often called the Einstein cylinder. This is a conformal transformation of the metric and does not assert that the transformed matter content remains vacuum. The coordinate ranges areThe angular-coordinate degeneracies at the poles are the usual regular spherical ones.

Radial null lines satisfy $d\chi =\pm dT$, so the Penrose diagram is the square displayed below. Every interior point represents an angular two-sphere. Its top and bottom are the spacelike conformal infinities ${\mathcal{I}}^{+}$ and ${\mathcal{I}}^{-}$ at $T=\pi /2$ and $T=-\pi /2$; they are excluded from the physical spacetime and are not curvature singularities. The vertical sides $\chi =0$ and $\chi =\pi$ represent the regular north and south poles of the three-sphere, not spatial edges of the universe.

A particle horizon at an observer's current time bounds the comoving sources whose light could have reached that observer over the available past conformal interval. Here the cosmological past extends to $t=-\infty$, but its conformal duration is finite. For a comoving observer at $\chi =0$,Since $0<{\chi }_p<\pi$ at every finite time, a particle horizon exists. The corresponding null boundary in the diagram is $\chi =T+\pi /2$, emerging from the observer's past endpoint at ${\mathcal{I}}^{-}$. The antipodal point never has enough time to send a signal received at a finite proper time.

An observer event horizon bounds the events that can ever send a signal to the observer's future-inextendible worldline. The remaining conformal time fixes its distance:This too lies strictly between zero and $\pi$ at finite time, so the observer has a cosmological event horizon. Its null line ends at the observer's future endpoint on ${\mathcal{I}}^{+}$. Both horizon spheres have areal radius $a\sin \chi =\ell$, though their comoving angular positions differ except at $T=0$. These are the cosmological horizons in global de Sitter spacetime; changing the comoving observer relocates them by an isometry.

A Cauchy horizon is the boundary of the domain determined by data on a chosen partial Cauchy hypersurface; beyond it an inextendible causal curve can avoid that initial surface. Each complete constant-$T$ three-sphere here is a Cauchy hypersurface. To see this directly, a causal curve in the conformal metric has spatial speed at most one when parameterized by $T$. On any finite interior time interval its path therefore has a limit in the complete compact three-sphere. If an inextendible curve stopped at an interior time, that limit would allow its continuation, a contradiction. Its time range consequently extends between both conformal ends, and monotonicity makes it cross any complete constant-$T$ sphere exactly once. Its domain of dependence is the entire physical spacetime.

Thus particle and cosmological event horizons exist for every comoving observer, while the complete global Cauchy slices have no Cauchy horizons. Horizons of a restricted static patch or a deliberately partial initial surface are relative to that restriction; they do not alter the global Cauchy conclusion.

For pressureless matter, the cosmological perfect-fluid continuity equation gives $\rho \propto a^{-3}$. The flat Friedmann equation then gives $a\propto t^{2/3}$, and conformal time $d\eta =dt/a$ gives $a\propto {\eta }^2$. Choose the initial conformal time to be zero. A galaxy seen at emission time ${\eta }_e$ has comoving radius $\chi ={\eta }_0-{\eta }_e$ and cosmological redshift $1+z=({\eta }_0/{\eta }_e{)}^2$.

Two opposite comoving galaxies are separated by $2\chi$. Direct one-way communication between their worldlines before emission requires the particle horizon ${\eta }_e$ to cover this separation: $2({\eta }_0-{\eta }_e)\le {\eta }_e$. Thus the threshold is ${\eta }_e=2{\eta }_0/3$, andThe distinction in direct cosmological communication and a shared causal past matters here. If causal contact means that their past light cones overlap, the condition instead is $2\chi \le 2{\eta }_e$, giving ${\eta }_e={\eta }_0/2$ and $\boxed{z_c=3\text{ for a shared causal past}}$. These are different operational meanings of the horizon criterion; the boundary corresponds to the limiting initial-time contact.

The physical particle horizon is the greatest physical distance at time $t$ from which a light signal could have reached the observer since the beginning of the cosmological model. If that beginning is $t=0$, radial null propagation has comoving distance ${\int }_0^{t}cd{t}^{\prime }/a(t^{\prime })$, soFor $a(t)=t^{\alpha }$ and $0<\alpha <1$, the integral converges at zero and equals $t^{1-\alpha }/(1-\alpha )$. ConsequentlyThe horizon problem is that widely separated regions of the observed cosmic microwave background have nearly equal temperatures although their past causal regions did not overlap before recombination in the usual decelerating hot-big-bang history. That history offers no causal equilibration mechanism for their observed uniformity. The particle horizon used here is distinct from the Hubble radius $c/H$.

The Hubble time is the instantaneous inverse expansion rate $H_0^{-1}$, where $H=\dot{a}/a$. A radial light ray covers comoving distance $\int cdt/a(t)$. Since $a(t_0)=1$, the present proper radius of the particle horizon isFor the power-law scale factor, $a(0)=0$ requires $\alpha >0$. The integral converges only when $\alpha <1$, in which caseAt $\alpha =1$ the divergence is logarithmic, and at $\alpha >1$ it is a power divergence. The familiar flat-universe epochs have $\alpha =1/2$ for radiation and $\alpha =2/3$ for pressureless matter, obtained from $\rho \propto a^{-4}$ or $a^{-3}$ and the Friedmann equation. Both have $0<\alpha <1$, so their ages are less than the present Hubble time. These power laws describe idealized single-component epochs rather than the full late-time expansion history.

The particle horizon is the greatest distance from which a light signal could have reached the observer since the initial cosmic time $t_i$. Its comoving radius and its radial proper distance at time $t$ areHere proper distance is measured along the spatial slice, rather than by the transverse area of a sphere. The Hubble parameter, deceleration parameter and cosmological density parameter areThe last expression defines the critical density. For blackbody radiation, $p=\rho /3$, so the Friedmann acceleration equation gives $\ddot{a}/a=-\kappa \rho /3$. ConsequentlyThis is the radiation relation between deceleration and density. It does not require spatial flatness.

For the open radiation universe particle horizon, write $x=a/a_0$. The cosmological continuity equation gives $\rho ={\rho }_0{x}^{-4}$, while the present Friedmann equation gives $-k/a_0^2=H_0^2(1-{\Omega }_0)$. ThusTaking the big-bang endpoint $x=0$ and using $dt=dx/\dot{x}$, the present radial proper distance becomesThereforeThe formula applies to $0<{\Omega }_0<1$. Its flat-radiation limit is $H_0{d}_{h0}\to 1$, agreeing with $a\propto t^{1/2}$ and $d_h=2t=H^{-1}$. The divergence as ${\Omega }_0\to 0$ reflects the unbounded past conformal interval of the limiting empty open model; that endpoint is not a radiation-filled universe.

The horizon problem concerns the nearly uniform temperature of widely separated parts of the cosmic microwave background. In a purely decelerating Hot Big Bang model, their past light cones at last scattering do not overlap far enough to explain this agreement by thermal contact. During cosmic inflation, accelerated expansion shrinks the comoving Hubble radius. A patch initially small enough for causal communication can be stretched to encompass the later observable universe. Reheating converts the inflationary energy into a hot plasma with correlated initial conditions across that patch. Enough inflation must occur before the observable scales leave the Hubble radius; inflation cannot establish contact between regions that were never initially causally related.

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

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

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