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Match flat Friedmann equations histories at scale factor $R_1$ and Hubble parameter $H_1$, and give them identical subsequent evolution. Below $R_1$, constant vacuum energy gives $H=H_1$, while a radiation-dominated universe gives $H=H_1(R_1/R{)}^2$. From $ϵR_1$ to $R_1$ the former takes longer by $H_1^{-1}{\int }_{ϵ}^1(1/x-x)dx>0$. Its duration diverges as $ϵ\to 0$; the radiation duration tends to $1/(2H_1)$. A finite period of cosmic inflation requires a finite initial state and a specified matching before such age comparisons are meaningful.

The cosmological monopole problem is the excessive relic density of stable magnetic monopoles produced in some early-universe symmetry-breaking scenarios. Cosmic inflation dilutes their number density as $a^{-3}$. Successful later reheating must also avoid recreating an excessive abundance.

Use natural units, the unreduced Planck mass $m_P=G^{-1/2}$, $V_0>0$ and $c\ne 0$. In a flat universe dominated by a canonical inflaton, $\varphi =(2m_P/c)\log (\beta t)$ and $a\propto t^p$ solve the inflaton equation of motion and Friedmann equation exactly when $p=16\pi /c^2$ and ${\beta }^2=V_0{c}^2/[2m_P^2(3p-1)]$. The positive-potential scaling solution requires $p>1/3$, while cosmic inflation requires $p>1$. Its constant equation-of-state parameter is $w=-1+2/(3p)$. In terms of the reduced Planck mass $M=m_P/\sqrt{8\pi }$, the exponential slope is $\lambda =c/\sqrt{8\pi }$ and the same exponent is $p=2/{\lambda }^2$.

For positive slope parameter $A$, the exponential-potential field increases during cosmic inflation. The quantity $aH$ grows by $e^{(1-{ϵ}_H)N}$ with constant ${ϵ}_H=A^2/(16\pi )$ and $N=8\pi ({\varphi }_R-{\varphi }_i)/(A{m}_{\mathrm{pl}})$. Starting from order-one curvature and demanding order-one or smaller present curvature gives the displayed upper bound on the initial field. A specified tighter final curvature tolerance adds one-half its logarithm to the numerator. Thus sufficient cosmic inflation sets an upper, not a lower, initial-field bound for this sign of the potential slope. Instantaneous reheating and subsequent entropy conservation determine $a_R{H}_R/a_0{H}_0$.

The traceless mixed extrinsic curvature of a spatial hypersurface can have leading solution ${\tilde{K}}^i{}_j=C^i{}_j(\mathbf{x})a_{\mathrm{loc}}^{-3}$. Cosmic inflation suppresses this anisotropic expansion. An anisotropic spatial shape can nevertheless freeze to a nonzero constant, so shear damping does not prove the absence of tensor cosmological perturbations.

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.

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.

Write the equation of state as $P=w\rho c^2$. Integrating the cosmological perfect-fluid continuity equation for constant $w$ gives $\rho \propto R^{-3(1+w)}$. With spatial flatness and the geometric cosmological constant set to zero, the Friedmann equation is $H^2=8\pi G\rho /3$.

In S1, $w=-1$ describes vacuum energy: $T_{\mu \nu }=-\rho c^2{g}_{\mu \nu }$ and $\rho$ is constant. Its negative pressure drives accelerated expansion, $\ddot{R}/R=8\pi G\rho /3=H_1^2$, and the expanding solution is $R\propto e^{H_{1}t}$. This is the ideal de Sitter spacetime approximation to cosmic inflation. An exactly constant vacuum component cannot end its phase unaided; a phase transition or reheating mechanism is required for the assumed change to S2.

In S2, $w=1/3$ describes an isotropic bath of ultrarelativistic particles, such as photons, with negligible rest mass relative to their energy. The number per physical volume falls as $R^{-3}$, while the energy per particle redshifts as $R^{-1}$, so $\rho \propto R^{-4}$. This radiation-dominated universe has $R\propto (t-t_{\ast }{)}^{1/2}$ on its expanding branch.

In S3, $w=0$ describes pressureless matter, or relativistic dust, with negligible random kinetic energy. Conserved particle number and approximately fixed rest energy give $\rho \propto R^{-3}$. A flat universe containing only this component is the Einstein-de Sitter universe, with $R\propto (t-t_{\ast }{)}^{2/3}$. The constants $t_{\ast }$ are integration offsets in the individual phases; they need not represent one common start time when phases are matched.

For monotonically expanding scale factor $R$, $dt=dR/(RH)$, so the elapsed cosmic time isThe second expression uses cosmological redshift $1+z=R_0/R$; the age of the universe is the limit to its earliest specified state, if that limit is finite.

An age comparison for an initial vacuum-dominated phase needs a common normalization. Match the two histories at the end of S1, with $R=R_1$ and $H=H_1>0$, and give them identical subsequent S2 and S3 evolution, including transition radii and today's state. Below $R_1$, the history with S1 has constant $H=H_1$. The comparison history extends S2 backwards, so $\rho \propto R^{-4}$ and $H=H_1(R_1/R{)}^2$. Starting both at $R_i=ϵR_1$, where $0<ϵ<1$, their early durations areThe difference is positive becauseThe common later evolution adds the same time to both histories. Thus S1 makes this matched universe older. More strongly, as $R_i\to 0$, the exact S1 duration diverges logarithmically, whereas a radiation beginning has finite duration $1/(2H_1)$ to the match. The exact de Sitter spacetime expanding patch reaches $R=0$ only as $t\to -\infty$ for its comoving observers; a finite “age since $R=0$” does not exist in this ideal model. This does not establish past geodesic completeness of the patch.

The matching assumption is essential for a finite cutoff or realistic finite period of cosmic inflation: arbitrary changes of the present Hubble parameter or transition densities prevent a universal age comparison. The literal zero-size limit in this question is instead the infinite-duration idealization just derived.

The actual large-scale expansion is well described by the Lambda-CDM model: radiation in cosmology, ordinary baryonic matter, cold dark matter and late-time dark energy coexist. A component becomes dominant through continuous changes in their density ratios, rather than every previous component disappearing at an instantaneous phase boundary. Matter pressure is small in its nonrelativistic regime, but radiation persists, and massive neutrinos interpolate between relativistic and nonrelativistic behavior.

There is also a late accelerated era, absent from a model ending permanently in S3. It is consistent with nearly constant vacuum energy, whose negative pressure eventually dominates because pressureless matter dilutes as $R^{-3}$. The scale factor then ceases to follow the pure $t^{2/3}$ Einstein-de Sitter universe law.

An early phase of cosmic inflation is a well-motivated model rather than a directly observed exact $w=-1$ phase. Successful cosmic inflation must end and transfer its energy to matter and radiation through reheating; realistic models generally have slowly varying $H$ and $w$ close to, rather than exactly equal to, $-1$. Extrapolating a strictly constant $H$ to arbitrarily small $R$ is not a measurement of the universe's past duration. Finally, spatial homogeneity and cosmological isotropy apply statistically on large scales: galaxies, gravitational waves and other perturbations are absent from the exact FLRW metric, and small spatial curvature remains an observational possibility.

For $\Lambda =0$, use the instantaneous cosmological density parameterInsert the Friedmann equation from the preceding part and multiply numerator and denominator by $a^4$:This holds on an expanding or contracting branch with $H\ne 0$; in a recollapsing model the density parameter diverges at turnaround.

During radiation domination, while curvature is small, $|\Omega -1|\simeq |{\Omega }_0-1|a^2/{\Omega }_{R0}$. During matter domination the corresponding behavior is $|\Omega -1|\simeq |{\Omega }_0-1|a/{\Omega }_{M0}$. Thus the exactly flat solution $\Omega =1$ is unstable against relative curvature during decelerating expansion: a small departure grows rather than being dynamically driven to zero. The flatness problem is that a universe with modest present curvature must have had an extremely tiny initial curvature-to-density ratio at an early radiation epoch. The fact that every such solution tends to $\Omega =1$ when extrapolated to $a\to 0$ does not supply a mechanism selecting that small departure at a specified finite initial epoch. An early period of cosmic inflation can instead shrink the comoving Hubble radius and suppress $|\Omega -1|=|k|/(aH{)}^2$.

For a homogeneous canonical scalar, ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and $P_{\varphi }={\dot{\varphi }}^2/2-V$. The acceleration equation requires $V>{\dot{\varphi }}^2$ for cosmic inflation, equivalently ${ϵ}_H=-\dot{H}/H^2<1$ in the flat scalar-dominated limit. Inflation need not be exactly de Sitter.

A standard sufficient regime for a long period is positive potential energy dominating the kinetic energy, physical gradient energy and curvature. Homogeneity over an expanding approximately Hubble-sized patch allows the gradient term to redshift away. In comoving coordinates the gradient energy is $|{\nabla }_{\mathrm{com}}\varphi {|}^2/(2a^2)$; the gradient in the displayed energy formula must be interpreted physically if no $a^{-2}$ is written there. RequireThe slow-roll equations are then $3H\dot{\varphi }\simeq -V^{\prime }$ and $H^2\simeq 8\pi V/(3m_{\mathrm{pl}}^2)$. Sufficient potential conditions areThe potential must remain sufficiently flat over a large enough field interval to produce the required expansion. These slow-roll conditions are stronger than the bare necessary condition for acceleration.

There is no minimum about which this scalar can undergo the damped oscillations normally used for reheating. Its slow-roll parameters are constant, so an inflating exponential solution has no automatic exit. An independent stopping and energy-transfer mechanism is needed.

At the assumed exit, $V_R=m_{\mathrm{pl}}^4{e}^{-30}$. Equating this to the thermal radiation density givesfor $g_{\ast }=10^3$. Corrections from the small kinetic fraction do not change this rough estimate.

Curvature obeys $|\Omega -1|\propto (aH{)}^{-2}$. For constant ${ϵ}_H=A^2/(16\pi )$ and $N=\log (a_R/a_i)$, its suppression during cosmic inflation is $e^{-2(1-{ϵ}_H)N}$. Starting from order-one curvature and requiring present curvature no larger than order one givesAfter instantaneous reheating, entropy conservation implies $a_R/a_0=(T_0/T_R)(g_{\ast s,0}/g_{\ast s,R}{)}^{1/3}$, with $H_R\simeq 1.66\sqrt{g_{\ast }}{T}_R^2/m_{\mathrm{pl}}$. Present scales make the numerator roughly $60$; using the literal approximate units from question2 gives about $63$. A tighter present curvature tolerance adds its corresponding half-logarithm. In slow-roll approximation this is the conventional requirement of roughly sixty e-folds.

Because the field increases toward ${\varphi }_R$, enough expansion requiresThis is the flatness bound on the initial field in exponential inflation. It is an upper bound on the initial field, not a nonzero minimum: starting earlier at a smaller field produces more expansion. If one additionally restricts the initial potential below $m_{\mathrm{pl}}^4$, then ${\varphi }_i\ge 0$ supplies a lower bound, but that is a separate restriction. Without initial-curvature and final-tolerance choices there is no unique numerical initial-field value.

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.

Use units $c=1$ and take the initially expanding branch with ordinary positive matter or radiation density. In the Friedmann-Lemaître-Robertson-Walker metric, $k=+1,0,-1$ respectively describe positive, zero and negative spatial curvature. Separately conserved pressureless matter has ${\rho }_M\propto a^{-3}$ and radiation in cosmology has ${\rho }_R\propto a^{-4}$, so either dominates the curvature and vacuum terms sufficiently far back on a hot Big Bang branch.

With negligible cosmological constant, a spatially flat radiation universe has $a\propto t^{1/2}$ and a flat matter universe has $a\propto t^{2/3}$; both expand forever while decelerating. An open universe also expands forever and eventually becomes approximately curvature dominated, with $a\propto t$. A closed universe containing only ordinary matter/radiation reaches a maximum size and recollapses. For example, dust gives ${\dot{a}}^2=C/a-k$ with $C>0$, so only $k=+1$ admits the positive turning point $a=C$.

Curvature alone does not determine the fate when the cosmological constant is retained. Positive $\Lambda$ can cause late accelerated expansion and prevent even a closed universe from recollapsing; whether a closed expanding branch turns around depends on the relative matter and vacuum terms. Negative $\Lambda$ can cause recollapse even for flat or open spatial sections. These statements follow by inspecting ${\dot{a}}^2=C/a-k+\Lambda a^2/3$ for dust; the curvature sign is a geometrical classification, not by itself a universal destiny criterion.

Define the critical density ${\rho }_c=3H^2/(8\pi G)$ and cosmological density parameters ${\Omega }_M={\rho }_M/{\rho }_c$, ${\Omega }_R={\rho }_R/{\rho }_c$, ${\Omega }_{\Lambda }=\Lambda /(3H^2)$ and ${\Omega }_k=-k/(a^2{H}^2)$. The Friedmann equation gives ${\Omega }_M+{\Omega }_R+{\Omega }_{\Lambda }+{\Omega }_k=1$. Dividing the Friedmann acceleration equation by $-H^2$ givesConsequentlyThe deceleration parameter measures the sign and size of the acceleration relative to $a{H}^2$: $q>0$ means deceleration, $q<0$ accelerated expansion and $q=0$ instantaneously zero acceleration. In particular a falling Hubble parameter alone does not imply deceleration, since $q=-1-\dot{H}/H^2$.

For conformal time $d\eta =dt/a$, the conformal Hubble parameter is $\mathcal{H}=aH$. Using matter conservation and differentiating ${\Omega }_M\propto {\rho }_M/H^2$,This is the matter density flow with a cosmological constant. Neglecting ${\Omega }_{\Lambda }$ during matter domination reduces it to the requested relationThat last equation is not exact with nonzero $\Lambda$. For example, a flat dust-plus-vacuum state with $({\Omega }_M,{\Omega }_{\Lambda })=(0.9,0.1)$ gives ${\Omega }_M^{\prime }/\mathcal{H}=-0.27$, whereas the matter-only expression gives $-0.09$.

For the flatness problem, let ${\Omega }_{\mathrm{tot}}={\Omega }_M+{\Omega }_R+{\Omega }_{\Lambda }$. Curvature satisfies ${\Omega }_{\mathrm{tot}}-1=k/(aH{)}^2$, soIn decelerating near-flat radiation and matter eras, deviations therefore grow as $a^2$ and $a$ respectively. Maintaining small curvature at a late epoch requires very small early deviations unless a prior mechanism suppresses them. Cosmic inflation, with $q\simeq -1$, drives the deviation down approximately as $a^{-2}$. If vacuum energy is present, flatness means ${\Omega }_{\mathrm{tot}}=1$, not ${\Omega }_M=1$.

Use natural units and the unreduced Planck mass $M_P=G^{-1/2}$; it is $\sqrt{8\pi }$ times the reduced Planck mass. A homogeneous canonical inflaton has ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and $P_{\varphi }={\dot{\varphi }}^2/2-V$. Vacuum domination by a scalar field requires ${\dot{\varphi }}_i^2/2\ll V({\varphi }_i)$, with other matter and the curvature term negligible in the Friedmann equation. The equation-of-state parameter is then near $-1$, and the Friedmann acceleration equation gives accelerating expansion.

For sustained slow-roll inflation, Hubble friction must dominate the inflaton acceleration: $|\ddot{\varphi }|\ll 3H|\dot{\varphi }|$, with the scalar potential changing only slightly in one Hubble time. In terms of the potential slow-roll parameter and second potential slow-roll parameter, this requiresFor the quartic scalar potential,Thus $|{\varphi }_i|\gg M_P$ supplies a broad slow-roll range when $\lambda \ll 1$. At an initial density of order the Planck density, potential domination givesso both potential slow-roll parameters are small for small $\lambda$. Small initial kinetic energy is a separate condition; the height of the scalar potential alone does not enforce it. These classical equations describe the post-Planck evolution under the model assumptions, rather than resolving quantum gravity at the initial endpoint.

Take ${\varphi }_i>0$; the negative branch follows by the symmetry $\varphi \mapsto -\varphi$. Dropping kinetic energy from the Friedmann equation and acceleration from the inflaton equation of motion givesWriting $b=M_P\sqrt{\lambda /(6\pi )}$, the quartic-potential slow-roll solution isIndeed, $\dot{\varphi }\simeq -b\varphi$, and integrating $H=\dot{a}/a$ gives $\log (a/a_i)=\pi ({\varphi }_i^2-{\varphi }^2)/M_P^2$. These expressions apply during slow-roll inflation; their formal late-time saturation must not be used beyond the end of that approximation.

The exact end of accelerating expansion is ${ϵ}_H=-\dot{H}/H^2=1$, equivalently ${\dot{\varphi }}^2=V$ for the scalar-dominated flat model. The quartic-inflation endpoint estimate uses the leading potential condition ${ϵ}_V\simeq 1$, givingThe numerical coefficient is an endpoint estimate: the slow-roll approximation breaks down there. Imposing $|{\eta }_V|=1$ estimates when slow roll loses accuracy and gives a different order-one coefficient, not a different exact acceleration condition.

The slow-roll e-fold count isThe quartic-inflation expansion from Planck density combines this initial estimate with the conventional endpoint to giveFor small $\lambda$ the dominant term is enormous; the order-one endpoint uncertainty does not affect that conclusion.

To estimate the reheating temperature, assume prompt, efficient reheating converts the end-of-inflation energy into a thermal radiation bath. With the same endpoint convention, $V_e\simeq \lambda M_P^4/(4{\pi }^2)$, and equating this to the radiation energy density givesThe quartic-inflation prompt-reheating estimate fixes the scaling; the endpoint kinetic energy changes the coefficient by order one. Without an inflaton coupling and its thermalization history, $V$ alone cannot specify the actual reheating temperature.

Finally, during cosmic inflation the comoving Hubble radius decreases:An initially causally connected patch is enlarged enough that the present observable region can lie within it. Its portions may subsequently be far outside one another's Hubble radii while retaining their common earlier thermal history. This solves the horizon problem if the number of e-folds is sufficiently large, conventionally of order sixty for the relevant thermal history; $N\sim 2\pi /\sqrt{\lambda }$ can readily exceed that. The necessary count depends on reheating and later expansion. The causal mechanism and the conversion back to a hot universe are discussed in David Tong's account of inflation.

For an expanding decelerating power law $a\propto t^{\alpha }$ with $0<\alpha <1$, the particle horizon is $ct/(1-\alpha )$, and the comoving Hubble radius is $c/(aH)\propto t^{1-\alpha }$. It increases with time, so regions that are mutually visible today can have been out of causal contact at early times. The nearly uniform early thermal conditions inferred across such regions give the horizon problem.

During cosmic inflation, accelerated expansion makes the comoving Hubble radius decrease. A patch initially within one causal communication scale can expand into the region now observed, after which its separated parts are temporarily outside the Hubble radius and later re-enter during deceleration. Sufficient duration before the hot radiation era permits a common causal origin; accelerated expansion is not itself a claim that signals propagate faster than light locally.

Use $T={\xi }^0$ and ${\xi }^i={\partial }^i\lambda$, and define the physical background Hubble parameter by $H=\dot{a}/(\bar{N}a)$. At first order, the spacetime metric components are $\delta g_{00}=-2{\bar{N}}^2\Phi$, $\delta g_{0i}=a^2{B}_{,i}$ and $\delta g_{ij}=-2a^2(\Psi {\delta }_{ij}+E_{,ij})$. Apply the given passive cosmological gauge transformation component by component. The temporal component changes by $2\bar{N}\dot{\bar{N}}T+2{\bar{N}}^2\dot{T}$; the mixed component changes by ${\bar{N}}^2{T}_{,i}-a^2{\dot{\lambda }}_{,i}$; and the spatial component changes by $-2a\dot{a}T{\delta }_{ij}-2a^2{\lambda }_{,ij}$. HenceThese are the scalar gauge transformations with a background lapse in the paper's sign convention.

Starting in synchronous gauge, preserving both $\Phi =0$ and $B=0$ requiresThe first equation integrates to $\bar{N}T=C(\mathbf{x})$; substituting in the second and integrating givesThe arbitrary lower integration limit is absorbed into $D$. Thus specifying synchronous lapse and shift does not fix the origins of the freely falling clocks or all spatial coordinate labels. This is the residual gauge freedom in synchronous gauge.

During cosmic inflation, choosing these clocks remains arbitrary. For example, the $C$ mode shifts the spatial potential by $HC$ and the density perturbation by $3HC(\bar{\rho }+\bar{P})$. In the de Sitter approximation, $H$ varies little, so a nearly constant metric perturbation can contain a pure coordinate contribution. An inflaton fluctuation or density contrast in synchronous gauge therefore cannot be declared physical solely from its time behavior. Use a uniform-density curvature perturbation or an appropriate field-based gauge-invariant variable, or fix a physical initial slicing. The exact de Sitter case has no evolving homogeneous density clock, which makes density-defined slicing degenerate.

During the standard hot Big Bang, a cold-matter rest frame supplies a convenient additional condition: imposing zero cold-matter velocity removes the nontrivial spatially varying time-shift mode. Before that condition is imposed, the pure-gauge density contrast for constant $w$ is $3H(1+w)C$, which decays as $t^{-1}$ in either radiation or matter domination. In conformal time this is proportional to ${\tau }^{-2}$ during radiation domination and ${\tau }^{-3}$ during matter domination. It can be confused with an independent decaying physical mode unless gauge freedom is fixed. The $D$ mode is a time-independent spatial relabeling and must likewise be fixed by a coordinate or initial-metric convention.

Finally, a scalar density transforms by evaluating the homogeneous scalar at the shifted time:To reach Newtonian gauge from synchronous quantities, impose $\tilde{E}=\tilde{B}=0$. For the nonhomogeneous scalar modes, chooseUsing background stress-energy conservation, $\dot{\bar{\rho }}=-3\bar{N}H(\bar{\rho }+\bar{P})$, gives the synchronous-to-Newtonian density transformation with a lapse:The expression includes the lapse factors because dots refer to the arbitrary time coordinate, not automatically to proper time.

With spatial flatness, the scalar as the only source, and $M_{\mathrm{Pl}}=(8\pi G{)}^{-1/2}$, the Friedmann equation isThe scalar equation gives $\dot{\rho }=\dot{\varphi }(\ddot{\varphi }+V_{,\varphi })=-3H{\dot{\varphi }}^2$, which is the energy-conservation equation $\dot{\rho }+3H(\rho +P)=0$. Differentiating the Friedmann equation on the expanding branch yields $6M_{\mathrm{Pl}}^{2}H\dot{H}=-3H{\dot{\varphi }}^2$, so $\dot{H}=-{\dot{\varphi }}^2/(2M_{\mathrm{Pl}}^2)$. Combining this with $\ddot{a}/a=H^2+\dot{H}$ gives the Friedmann acceleration equation:Thus cosmic inflation occurs exactly when $V(\varphi )>{\dot{\varphi }}^2$ in this canonical scalar model. Equivalently $P<-\rho /3$, or the first Hubble slow-roll parameter ${ϵ}_H=-\dot{H}/H^2$ is smaller than one.

Assume $V>0$ and that its derivatives do not change substantially over the field interval traversed in the estimate. With initially negligible kinetic energy, $H^2\simeq V/(3M_{\mathrm{Pl}}^2)$. The force $-V_{,\varphi }$ changes the field velocity on the Hubble friction time. Keeping $H$ and $V_{,\varphi }$ approximately constant for that local estimate, and starting at rest, solves the field equation explicitly:After one Hubble time,Writing $A=M_{\mathrm{Pl}}^2{V}_{,\varphi }^2/V^2$, the condition $A<1$ keeps this kinetic-to-potential ratio below about $0.31$, comfortably inside the inflationary bound. Integrating once more gives the fractional potential change estimateover the same interval. Thus the characteristic time for an order-one potential change is $H^{-1}/A$, up to constants of order one. The requested local Hubble-timescale criterion isUsing a strict inequality with one as the order-one threshold is conventional; its coefficient is not an exact universal end-of-inflation bound. Much smaller $A$ gives the controlled slow-roll regime and many Hubble times of cosmic inflation if the bounds persist.

After the transient in relaxation to the slow-roll attractor, $3H\dot{\varphi }\simeq -V_{,\varphi }$. Differentiating this approximate relation givesTherefore the standard sufficient conditions for negligible acceleration throughout slow roll are ${ϵ}_V\ll 1$ andThe absolute value matters: large negative curvature also drives rapid evolution. The more precise local acceleration criterion is $|{\eta }_V-{ϵ}_V|\ll 1$, with the usual separate smallness conditions avoiding a tuned cancellation. If a field starts exactly at rest on a nonzero slope, its initial acceleration is not negligible; the displayed exponential transient describes its approach to the attractor.

These are local timescale and slow-roll conditions, rather than a global duration theorem from derivatives at a single point. A smooth potential can have a narrow flat shoulder followed by a steep drop, ending cosmic inflation soon despite small starting derivatives. The conditions must hold over the traversed interval. This is the additional smooth-evolution assumption implicit in the estimate.