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The hot Big Bang model describes an initially hot expanding universe whose standard radiation- and matter-dominated eras are governed by decelerating Friedmann expansion. Without an earlier inflationary era it exhibits the horizon problem and flatness problem.

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.

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 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.

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 $c/H$.

For an expanding universe, cosmic inflation requires $\ddot{a}>0$. Including any vacuum contribution in the total density and pressure, the acceleration equation gives $\boxed{\rho +3P<0}$, or $w=P/\rho <-1/3$ for positive density. Equivalently the comoving Hubble radius $(aH{)}^{-1}={\dot{a}}^{-1}$ decreases. Since $|{\Omega }_{\mathrm{total}}-1|=|k|/(aH{)}^2$, prolonged acceleration suppresses curvature and addresses the flatness problem. A sufficiently long inflationary phase also stretches a region that was initially causally connected to encompass the observed region, addressing the horizon problem. The condition on acceleration must last long enough; an arbitrarily short interval is insufficient.

For a homogeneous canonical scalar field $\Phi$ with potential $V(\Phi )$,Thus $V>{\dot{\Phi }}^2$ is sufficient for inflation. A suitably flat positive potential gives slow-roll inflation: Hubble damping in $\ddot{\Phi }+3H\dot{\Phi }+V_{,\Phi }=0$ keeps the kinetic energy small compared with $V$, so $P_{\Phi }\simeq -{\rho }_{\Phi }$.

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.

The printed $\mathcal{H}$ is the conformal Hubble parameter, $\mathcal{H}=aH=\dot{a}$; ${\mathcal{H}}^{-1}$ is the comoving Hubble radius, rather than the physical radius $H^{-1}$. Since the derivative requested is with respect to cosmic time,Thus the cosmic-time derivative of the comoving Hubble radius hasFor an expanding universe, the same factor $1+3\omega$ determines whether a small departure from flatness grows and whether the comoving Hubble radius grows. Ordinary matter with $\omega >-1/3$ therefore has both signs associated with the conventional flatness problem and horizon problem. Scales whose physical wavelengths now exceed the Hubble radius were even farther outside it, in relative terms, earlier in such an era, rather than being brought inside for causal equilibration. Accelerated expansion with $\omega <-1/3$ reverses both signs.

There is a qualification to the word “always”: an actual horizon problem depends on the complete past history, not just this local sign. In a flat, constant-$\omega$ hot Big Bang model with $\omega >-1/3$, $a\propto (t-t_B{)}^p$ with $p=2/[3(1+\omega )]<1$, so the comoving particle horizon ${\int }_{t_B}^{t}d{t}^{\prime }/a(t^{\prime })$ is finite. The conventional causal problem then accompanies the flatness instability. An earlier accelerated era or a different past boundary can change the causal conclusion even when the present-era comoving Hubble radius is growing. Thus the requested correspondence is valid in that usual expanding hot Big Bang setting, not a universal logical equivalence between global horizons and local stability. At $\omega =-1/3$ both local effects are marginal.

The prescribed cosmic string network scaling gives $\dot{\rho }=-2H\rho$. Comparing with the cosmological perfect-fluid continuity equation givesThe averaged pressure is therefore $P=-\rho /3$, provided the network is treated as the stipulated isotropic homogeneous component. The Friedmann acceleration equation gives $\ddot{a}=0$. On an expanding branch let $\rho =C/a^2$ and $B^2=8\pi GC/3-K>0$; thenThis coasting cosmic-string universe has no growing flatness instability, but it does not drive an initially nonflat universe toward flatness. The comoving Hubble radius remains constant, so there is also no shrinking-radius mechanism of the usual cosmic inflation type.

The global causal issue is more precise than saying that a constant comoving Hubble radius solves every horizon problem. If this ideal coasting solution extends to its zero-size time $t_B$, then ${\int }_{t_B}^{t}d{t}^{\prime }/[B(t^{\prime }-t_B)]$ diverges logarithmically, so it has no finite particle horizon. If string domination starts at a finite earlier time after another cosmological phase, its causal history must instead be evaluated from that earlier phase. The string era alone does not establish that previously disconnected regions had time to equilibrate. A static branch with $B=0$ is a separate possibility; its Hubble-normalized $\Omega$ is undefined and is not covered by the expanding formulas.

The physical particle horizon isFor an Einstein-de Sitter universe, $a(t)\propto t^{2/3}$, soSince $H_0=2/(3t_0)$,The horizon problem is that widely separated regions of the observed universe have nearly identical conditions even though, in a decelerating standard Big Bang history, their past light cones do not overlap early enough to establish causal thermal equilibrium.

In the decelerating hot Big Bang model, the particle horizon at recombination encloses only a finite comoving region. Widely separated parts of the cosmic microwave background therefore had no overlapping past light cones in that model, although they have almost the same temperature. This is the horizon problem. During cosmic inflation, accelerated expansion makes the comoving Hubble radius $(aH{)}^{-1}$ shrink. A small region that was initially in causal contact can then expand to contain the whole presently observable universe.

The flatness problem follows fromDuring ordinary radiation- or matter-dominated decelerating expansion this quantity grows, so the observed closeness of the cosmological density parameter $\Omega$ to one requires extreme tuning at early times. Inflation makes $aH$ grow nearly exponentially and therefore drives $|\Omega -1|$ rapidly toward zero. A sufficiently large number of e-folds makes spatial flatness a dynamical outcome rather than a finely tuned initial condition.

The flatness problem is that $|\Omega -1|$ grows during ordinary decelerating expansion, so its small present value requires the early density to have been extraordinarily close to critical. The horizon problem is the near uniformity of regions whose past light cones do not meet in a noninflationary hot Big Bang model.

Inflation requires accelerated expansion,For a constant equation of state $P=w\rho$, this means $w<-1/3$; representative scale factors are $a\propto t^p$ with $p>1$ and $a\propto e^{Ht}$. Equivalently, the comoving Hubble radius $(aH{)}^{-1}$ decreases. Sincemany e-folds drive the universe toward spatial flatness. A small region that was initially in causal contact is stretched to a size far larger than the later Hubble radius, allowing the observed universe to originate from one equilibrated patch.

cosmic inflation is an expanding epoch with $\ddot{a}>0$, equivalently a shrinking comoving Hubble radius $(aH{)}^{-1}=1/\dot{a}$. In general relativity it requires $\rho +3P/c^2<0$ for the total effective fluid. A scalar field whose potential energy dominates its kinetic energy has $P\simeq -\rho c^2$ and can produce approximately exponential expansion. slow-roll inflation is one realization; rapid expansion alone without acceleration does not have the required effect.

The horizon problem is that widely separated regions of the cosmic microwave background have nearly the same temperature despite lying outside one another's causal domains in a purely decelerating hot Big Bang extrapolation. Sufficient inflation stretches a previously small causally connected region across our observable universe; relevant perturbation wavelengths can begin inside the Hubble radius, leave it during inflation and re-enter later. This does not require local signals to outrun light. Inflation supplies a possible earlier causal history, not a proof that every possible initial state becomes homogeneous.

For the flatness problem, the Friedmann equation gives ${\Omega }_{\mathrm{tot}}-1=k{c}^2/(aH{)}^2$ when $a$ is length-valued. Since $aH$ grows during inflation,Pre-existing unwanted massive relics such as magnetic monopoles are also diluted: conserved number density falls as $a^{-3}$. Enough inflation, followed by reheating that does not recreate the relic problem, can make their abundance negligible. These were central motivations of the original proposal, whose first phase-transition implementation also faced a graceful-exit problem.^[ref]

quantum fluctuations of the inflating field and metric supply seeds for linear cosmological density perturbations. Simple single-field slow-roll models predict a dominant adiabatic mode with nearly Gaussian, nearly scale-invariant scalar cosmological perturbations, consistent with the cosmic microwave background; tensor cosmological perturbations are another possible test. These predictions are model-dependent, and consistency is not unique proof of inflation. reheating must connect the accelerated phase to the hot expanding universe. Inflation addresses causal reach, small cosmological spatial curvature and unwanted relic abundance, while providing a mechanism for primordial structure.

The metric uses signature $(+---)$ and a time-valued interval $ds$: a comoving clock has $ds=dt$. Here $t$ is cosmic proper time, $R(t)$ a scale factor with dimensions of length, $c$ the speed of light, $\chi$ a dimensionless comoving radial coordinate, and $\theta ,\varphi$ the usual polar and azimuthal angles. With curvature normalized to $k=+1,0,-1$, $S(\chi )=\sin \chi ,\chi ,\sinh \chi$, respectively. Homogeneity and isotropy apply to each constant-$t$ spatial slice. A comoving object's fixed coordinates do not imply a constant proper separation: spatial lengths scale with $R$.

Radial null propagation obeys $d\chi =cdt/R(t)$ in magnitude. Two neighbouring wave crests between the same comoving emitter and observer have the same radial integral, giving $\delta t_0/R_0=\delta t_e/R_e$. The wavelength redshift is consequentlyPeculiar Doppler shifts are excluded by the comoving assumption. Define angular diameter distance by $d_{\theta }={\ell }_{\perp }/\delta \theta$, where ${\ell }_{\perp }$ is proper transverse size at emission. The transverse metric gives ${\ell }_{\perp }=R_{e}S(\chi )\delta \theta$, henceThis distance is neither the photon travel time times $c$ nor the object's present proper distance.

The comoving particle horizon at $t_1$, with the initial boundary at the Big Bang, is ${\chi }_p(t_1)={\int }_0^{t_1}cdt/R(t)$. Its proper radial distance is $D_p=R(t_1){\chi }_p$; for a power-law scale factor $R=A{t}^p$, $0<p<1$,Thus a matter-only flat Einstein-de Sitter universe, $p=2/3$, has $D_p=3c{t}_1$, and the flat radiation-only case, $p=1/2$, has $D_p=2c{t}_1$. Strictly, Einstein-de Sitter denotes the pressureless matter model; the radiation calculation uses the flat zero-$\Lambda$ radiation analogue requested here.

For the source seen at time $t_1=t_{\mathrm{rec}}$, the radial coordinate is the integral from recombination to today, not the earlier horizon integral. In these flat models,Taking the linear horizon scale in the question to be $D_p$, its small angular extent isThereforeUsing the supplied times gives ${\theta }_{p,m}=0.02684\mathrm{rad}=1.54^{\circ }$ and ${\theta }_{p,r}=0.004244\mathrm{rad}=0.243^{\circ }$. There is a genuine radius/diameter convention to state: the supplied $3c{t}_1$ is a radial horizon distance. If angular diameter means the full diameter of a spherical horizon patch of radius $D_p$, the two answers are instead $2{\theta }_{p,m}=3.08^{\circ }$ and $2{\theta }_{p,r}=0.486^{\circ }$. Both conventions follow from the same distance calculation; a factor of two should not be hidden in the word horizon.

Matter-only evolution is the better of these two approximations because the universe is already matter dominated around recombination and evolves mostly through a subsequent matter era, rather than remaining radiation dominated to the present. Neither is exact: pre-recombination radiation changes the particle horizon, and late-time dark energy changes the distance to the last-scattering surface. No supplied ages have been replaced by later best-fit values.

The cosmic microwave background originates near last scattering and samples many horizon-sized regions. In an always-decelerating hot-Big-Bang model, widely separated last-scattering patches lack overlapping early causal histories, motivating the horizon problem. The measured acoustic angular scale uses the smaller sound horizon, not automatically this light-travel particle horizon; baryon-photon pressure changes the propagation speed. Inflation supplies an earlier accelerated expansion during which the comoving Hubble radius shrinks, allowing a patch with a common earlier causal history to expand to the size of the observable universe. Quantum perturbations can then seed later acoustic and density fluctuations. This does not require local signals to propagate faster than light, and the Hubble radius must not be equated indiscriminately with the particle horizon. CMB temperature and lensing spectra test the resulting perturbation and expansion model.^[ref]

cosmic inflation is an early interval of accelerated expansion, $\ddot{R}>0$. For expansion this means a shrinking comoving Hubble radius, $d[(RH{)}^{-1}]/dt<0$. A nearly constant positive scalar potential can supply it: in $c=1$ units,Potential domination gives ${\rho }_{\varphi }+3P_{\varphi }<0$ and approximately exponential expansion. Sustaining the interval requires sufficiently slow evolution, not just setting the initial field velocity to zero on a steep potential.

The horizon problem is the near equality of conditions in widely separated microwave-background regions that could not equilibrate in the standard decelerating hot Big Bang history. cosmic inflation permits the currently observed patch to descend from an earlier region small enough to have been causally connected, before that patch is stretched to enormous size. The claim concerns a causal history extended before the conventional hot phase; merely accelerating a collection of already disconnected regions would not explain their common temperature.

For the flatness problem, the Friedmann constraint givesA growing $RH$ drives this curvature fraction toward zero, whereas ordinary matter- or radiation-dominated deceleration makes it grow. cosmic inflation therefore suppresses observable curvature without requiring an equally extreme initial tuning within the later hot evolution.

Pre-existing monopoles and other unwanted relics are also diluted by the volume increase: a conserved particle number density falls as $e^{-3N}$ over $N$ e-folds. The transition to the hot universe must reheat efficiently, and its temperature must not regenerate an excluded relic population. The required number of e-folds is model and reheating dependent; “about sixty” is a useful scale for many scenarios rather than a universal exact requirement.

cosmic inflation also supplies a mechanism for perturbations. quantum fluctuations are stretched beyond the Hubble radius, and nearly scale-invariant comoving curvature perturbations can later seed cosmic structure. Simple single-field slow-roll models commonly predict approximately adiabatic, nearly Gaussian fluctuations, with a small spectral tilt; tensor perturbations would provide an additional test. Acoustic coherence and the observed scalar spectrum are compatible with this account, but they do not single out one inflaton potential. cosmic inflation addresses causal homogeneity, curvature and relic dilution while motivating the initial perturbations; it does not automatically solve the cosmological constant problem or determine reheating.

The angle estimates the largest transverse region at decoupling that could have established causal contact after the big bang in the assumed decelerating model. The cosmic microwave background has nearly the same temperature in directions separated by much larger angles. This is the horizon problem: those regions have no shared causal history in that simple model, yet their thermodynamic conditions are strikingly alike. The particle horizon should be distinguished from the smaller sound horizon, which controls acoustic propagation in the primordial fluid.

An early epoch dominated by a sufficiently large positive effective cosmological constant gives $\ddot{R}>0$ and approximately $R\propto e^{Ht}$. During this cosmic inflation, the comoving Hubble length $c/(RH)$ shrinks. A patch that was initially in causal contact can expand to contain the whole presently observable cosmic microwave background surface, while expansion suppresses spatial curvature. A finite period of cosmic inflation followed by reheating and the usual hot expansion therefore explains the uniformity without superluminal local signals. An eternally large unchanged cosmological constant would prevent the later matter-dominated history; an exit from cosmic inflation and reheating are required. Sufficient duration and a suitable pre-inflationary causal patch are part of this explanation.

Adiabatic photon redshifting gives $T\propto a^{-1}$. Using the stipulated temperatures,In the flat dust approximation of part c, $t=t_0{a}^{3/2}$, so $t_{\mathrm{rec}}\simeq (2/3H_0)1000^{-3/2}\simeq 2.75\times 10^5$ years. This approximation omits the significant radiation contribution near the actual recombination epoch.

The elementary travel-distance estimate is $c{t}_{\mathrm{rec}}\simeq 8.4\times 10^4$ pc, whose corresponding comoving scale today is about $84$ Mpc. In an expanding universe, however, that is not the exact proper particle-horizon radius. Integrating the null geodesic givesfor dust. Thus the proper horizon at this epoch is about $0.25$ Mpc and its present comoving size about $250$ Mpc. The factor three accounts for expansion during light propagation. These scales describe the limited causal domain in the simple decelerating model. The near-uniform CMB temperature across much more widely separated last-scattering regions motivates the horizon problem; the question's rough $ct$ estimate and the exact dust horizon should not be conflated.