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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 numerical prefactor in a horizon-crossing time depends on the chosen horizon scale. Let $a(t_0)=1$, $a_{\mathrm{eq}}=(1+z_{\mathrm{eq}}{)}^{-1}$, $t_{\mathrm{eq}}/t_0=a_{\mathrm{eq}}^{3/2}$ and $k_0=2\pi /(c{t}_0)$. In a sharp radiation-to-matter transition model,Using the order-of-magnitude horizon size $ct$ and setting the physical wavelength $2\pi a/k$ equal to it givesThese familiar powers describe radiation-era and matter-era scaling. For the particle horizon convention in cosmological horizon crossing, use $d_p=ac\tau$ with $\tau ={\int }_0^{t}d{t}^{\prime }/a(t^{\prime })$. The same piecewise model gives $\tau /t_0=2a_{\mathrm{eq}}^{-1/4}(t/t_0{)}^{1/2}$ before equality and $\tau /t_0=3(t/t_0{)}^{1/3}-a_{\mathrm{eq}}^{1/2}$ afterwards. Therefore the corresponding crossing times areBoth branches agree at equality. Explicitly retaining the particle horizon integral fixes the order-unity factors that the $ct$ estimate omits.

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 particle horizon at cosmic time $t$ is the greatest comoving distance from which a light signal could have reached an observer since the initial time:Regions separated by more than twice this distance cannot have exchanged signals by time $t$.

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 a physical diameter twice the radius. 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.

The cosmic microwave background consists of relic photons from the early hot, nearly thermal photon-baryon fluid. Before cosmological recombination, abundant free electrons repeatedly scatter photons by Thomson scattering. Their mean free path ${\lambda }_{\gamma }=(n_e{\sigma }_T{)}^{-1}$ is short and their Thomson scattering rate $n_e{\sigma }_{T}c$ is much greater than the Hubble parameter. An earlier photon therefore cannot stream directly to us: subsequent scattering repeatedly changes its direction.

As cosmic expansion cools the plasma to about $T\simeq 3000\mathrm{K}$, electrons and protons combine into neutral hydrogen. The free-electron abundance drops rapidly, the scattering optical depth to the observer becomes of order one, and photon decoupling occurs. Since the cosmic microwave background temperature scales as $T=T_0(1+z)$, with $T_0\simeq 2.725\mathrm{K}$, this occurs at $z\simeq 1100$. The small baryon-to-photon ratio delays neutralization to well below $13.6\mathrm{eV}/k_B$: the high-energy tail contains enough photons to ionize hydrogen until the temperature is much lower than the Hydrogen binding energy expressed as a temperature. The observed cosmic microwave background is primarily this pre-existing radiation undergoing its last scattering; its nearly blackbody spectrum is then preserved by cosmological redshift.

For the requested estimate take a flat matter-dominated universe, $a(t)=(t/t_0{)}^{2/3}$. Restore $c$ in distances. The comoving particle horizon at cosmological recombination and the comoving radial distance to the Cosmic microwave background last-scattering surface areThe proper particle horizon is $a_{\ast }{r}_{\mathrm{hor},\ast }=3c{t}_{\ast }$ and the angular diameter distance is $D_A=a_{\ast }{\chi }_{\ast }$. Therefore the angular particle-horizon size in a power-law universe, taking one horizon radius as the transverse length, isThe angular diameter of a patch of that radius is about $3.56^{\circ }$. The factor of two is a radius-versus-diameter convention; either gives a horizon scale of only a few degrees. This is a particle horizon, rather than the smaller sound horizon. Including early radiation in cosmology changes the numerical estimate but preserves its order of magnitude.

The horizon problem is that the cosmic microwave background has almost the same temperature across the whole sky, including pairs of regions whose past light cones do not overlap in this noninflating model. Local thermal equilibration cannot explain that observation within this history. A sufficiently long period of cosmic inflation makes the comoving Hubble radius decrease: a region once small enough to share a causal past can expand to encompass our entire observed Cosmic microwave background last-scattering surface. After reheating, it can be homogeneous across many later particle horizon patches. The matter-only estimate integrates from a decelerating Big Bang; extending the earlier history by cosmic inflation changes that causal conclusion.

The official notes, PDF p. 24, handwritten p. 17 integrate the physical particle horizon and derive its early-time redshift dependence, explicitly stating the present matter-dominated approximation used in the printed expression.

The official notes, PDF p. 27, handwritten p. 20 derive the early-time photon count, its vanishing limit, and the failure of thermal equilibrium when the count is of order one. Two details need supplementing.

First, their constant $N_{\gamma 0}$ is the normalization obtained by extrapolating the early-time particle horizon formula to $z=0$. It is not the count within today's actual particle horizon. Let $N_0=N_{\gamma }(t_0)$ and let $r_{h0}$ be that horizon's physical radius. Using the paper's early-time approximation givesFor a present matter-dominated approximation, $r_{h0}\simeq 2c/H_0$. Alternatively, retaining both components in the flat model exactly, ${\Omega }_R=1/(2+z_{\mathrm{eq}})$, and integrating the particle horizon givesThe second boxed formula uses the exact two-component normalization and the radiation-dominated early-time limit. Its large-$z_{\mathrm{eq}}$ leading term agrees with the first expression and $r_{h0}\simeq 2c/H_0$.

Second, a particle horizon containing only order-one photons cannot support a smooth, locally representative radiation density: discrete-particle fluctuations are of the same order as the mean, and a single photon's direction is strongly anisotropic. Thus the fluid description assuming spatial homogeneity and cosmological isotropy is not self-consistent on that causal scale. This does not by itself disprove an ensemble-averaged cosmological symmetry; it limits the extrapolation of the smooth thermal equilibrium description.

The scaling solution has $\rho =3M^2{p}^2/t^2$ and $\rho +P={\dot{\phi }}^2=4M^2/({\lambda }^2{t}^2)$. ThereforeMore generally any real canonical scalar field with positive total energy density obeys $\rho +P={\dot{\phi }}^2\ge 0$, so it cannot have $w<-1$. A negative potential does not reverse this conclusion.

The proper particle horizon is $a(t){\int }_0^{t}d{t}^{\prime }/a(t^{\prime })$. For $a\propto t^p$ the lower-limit integral diverges exactly when $p\ge 1$. Thus, assuming the same power law extends to the initial boundary, there is no finite particle horizon for $\boxed{0<{\lambda }^2\le 2}$. For the strict inequality, $p>1$, the proper cosmological event horizon and present age areAt ${\lambda }^2=2$, $a\propto t$ has an infinite particle-horizon integral but also an infinite future event-horizon integral, so no finite event horizon exists. The age remains $H_0^{-1}$. The $\lambda =0$ constant-potential case is separate: its flat de Sitter patch has event-horizon radius $H_0^{-1}$ and no finite-age power-law Big Bang.

As the temperature falls, electrons and protons form neutral hydrogen through cosmological recombination. The free electron fraction drops sharply, and so does the scattering rate ${\Gamma }_{\gamma }=n_e{\sigma }_{T}c$. When this rate becomes comparable to or smaller than the Hubble parameter, the plasma becomes transparent and photon decoupling occurs. The Thomson scattering cross-section itself does not abruptly vanish. The rapid change is in the abundance of free targets.

The Hydrogen binding energy is $13.6\mathrm{eV}$, but recombination occurs at a much smaller thermal energy, of order $0.3\mathrm{eV}$, or a few thousand kelvin. There are roughly a billion photons per baryon, so the high-energy tail can keep hydrogen ionized long after $k_{B}T$ drops below the binding energy. The falling exponential ionizing tail then makes the transition comparatively sharp. Last scattering follows when sufficiently few electrons remain free.^[ref]

To estimate a horizon, specify an expansion model as well as the redshift. The proper particle horizon at emission isAs a transparent analytic estimate, approximate a spatially flat matter-dominated universe by $a(t)=(t/t_{\mathrm{now}}{)}^{2/3}$ throughout. ThenTaking the illustrative present value $H_0=70\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$, this model has $t_{\mathrm{now}}=2/(3H_0)\simeq 9.3\times 10^9\mathrm{yr}$. At $z_{\ast }=10^3$,This is a proper horizon radius at emission; its corresponding present comoving length is about $270\mathrm{Mpc}$.

For the same model the comoving distance to the Cosmic microwave background last-scattering surface and its angular diameter distance areTherefore the angular extent associated with one radial horizon length isThis is an order-one-degree estimate. Including the earlier radiation domination and late dark energy changes the coefficient; it does not make the horizon span the sky. A horizon patch diameter has twice this angle, whereas the acoustic scale uses a sound horizon and differs again. The redshift by itself, without an expansion model or normalization, is insufficient for a unique horizon distance.

The horizon problem is that the observed cosmic microwave background has nearly the same mean temperature across angular separations vastly greater than this scale. In the purely decelerating Hot Big Bang model, sufficiently separated last-scattering regions had neither time to exchange signals nor a common post-Big-Bang causal region that could establish their observed thermal uniformity. The puzzle concerns the missing causal explanation of those initial correlations; it is not an inconsistency of specifying uniform initial conditions mathematically. An earlier period of cosmic inflation can enlarge a formerly causally connected region beyond the later observable horizon.

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.