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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.

For an expanding Friedmann-Lemaître-Robertson-Walker metric with positive energy density, put $\Omega =8\pi G\rho /(3H^2)$ and $w=P/\rho$, in units $c=1$. The Friedmann equation, Friedmann acceleration equation and cosmological perfect-fluid continuity equation giveFor constant $w$, linearizing at $\Omega =1$ gives $\Omega -1\propto a^{1+3w}$. For variable $w$, replace the power by $\exp \int (1+3w(a))d\log a$. This describes the flatness problem for decelerating expansion, away from epochs at which $H=0$.

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.

Physical Review D 23, 347–356. The original inflation proposal discusses the horizon and flatness problems, unwanted relics, and the difficulty of completing its original phase-transition scenario.

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 $|\Omega -1|=|k|/(aH{)}^2$.

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 ${\Omega }_M=1$.

The numerical fractions describe physically very different components. Use $c=1$ so that pressure and energy density have the same units. The stress-energy conservation equation for a homogeneous component, together with its equation of state $p_i=w_i{\rho }_i$, givesNonrelativistic baryons and cold dark matter have $w\simeq 0$ and dilute as $a^{-3}$; radiation has $w=1/3$ and dilutes as $a^{-4}$; vacuum energy has $w=-1$ and remains constant. The curvature term scales as $a^{-2}$ in the Friedmann equation but is a geometric contribution, not an additional material fluid.

There is consequently a coincidence in the present mixture. Normalize $a_0=1$. ThenAt early times vacuum is negligible, while at late times it dominates overwhelmingly. Observing order-unity matter and vacuum fractions during their comparatively recent crossover motivates the cosmic coincidence problem: why should this crossover occur near our observing epoch?

The absolute scale gives a second puzzle. Restoring SI units, $H_0\simeq 2.27\times 10^{-18}{\mathrm{s}}^{-1}$ yieldsIts energy density is ${\rho }_{\Lambda ,0}{c}^2\simeq 5.8\times 10^{-10}\mathrm{J}{\mathrm{m}}^{-3}$. Quantum vacuum contributions associated with ordinary particle-physics scales can be far larger than this observed value. Explaining the tiny net value, including cancellations among contributions, is the cosmological constant problem. The required negative pressure also differs qualitatively from ordinary dilute matter.

The matter budget supplies another explanatory gap: only $0.045$ is baryonic, while $0.255$ must be nonbaryonic matter in this benchmark. Together with $0.7$ in vacuum energy, most of the cosmic density is in components whose microscopic explanation is not supplied by fitting their fractions. Finally, near-zero curvature raises the flatness problem. For nonzero $k$,During radiation domination $H\propto a^{-2}$, so $|{\Omega }_k|\propto a^2$; during matter domination $H\propto a^{-3/2}$, so $|{\Omega }_k|\propto a$. Near-flatness today therefore requires extremely small early curvature in an unmodified decelerating history. Cosmic inflation offers a mechanism that reduces this curvature fraction through accelerated expansion.

The apparent contrivances are the recent matter-vacuum coincidence, the tiny vacuum-energy scale, the near-flat initial geometry and the unexplained microscopic dark components. Their observationally inferred values impose concrete requirements on a deeper physical account.

With a prime denoting conformal-time differentiation, $a^{\prime }=a\dot{a}$ and $\mathcal{H}=a^{\prime }/a=\dot{a}$. Multiplying the Friedmann equation by $a^2$ gives ${\mathcal{H}}^2+k{c}^2=8\pi G\rho a^2/3$, and henceThe Friedmann acceleration equation and $P=w\rho c^2$ giveDifferentiate $\Omega -1=k{c}^2/{\mathcal{H}}^2$ and substitute this expression:For $k=0$, $\Omega =1$ identically and the same equation holds directly. The relations assume $\mathcal{H}\ne 0$, as is true on the expanding branch in the question.

For pressureless matter, $w=0$, the cosmological density parameter flow points away from $\Omega =1$: initial $\Omega >1$ increases, while $0<\Omega <1$ decreases. In the closed case the expanding branch can end at a turnaround, where $\mathcal{H}=0$ and this density parameter diverges. For vacuum domination, $w=-1$, the sign reverses: values above one decrease and positive values below one increase toward one. The plotted curves are representative conformal-time solutions with $c=1$ and separately chosen origins; inflationary future infinity occurs at a finite conformal time.

In a decelerating matter or radiation era, small deviations from flatness grow: near one, $d\log |\Omega -1|/d\log a\simeq 1+3w>0$. Observed near-flatness therefore requires extremely small earlier deviations, the flatness problem. During accelerated expansion with $w<-1/3$, this exponent is negative and curvature is driven toward negligible relative size. For $w=-1$, $|\Omega -1|$ decays approximately as $a^{-2}$. Sufficiently long cosmic inflation supplies the inflationary solution of the flatness problem before ordinary decelerating evolution resumes.

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 $P_{\Phi }\simeq -{\rho }_{\Phi }$.

Differentiate the Friedmann equation and use cosmological energy conservation:For an expanding universe $H>0$, division by $2H$ and substitution of the Friedmann equation give the Friedmann acceleration equationThe strong energy condition makes $\ddot{a}\le 0$. Since $aH=\dot{a}$ and $\dot{a}>0$,Thus the quantity is nonincreasing, with strict decrease when the energy condition is strict. Dividing the Friedmann equation by $H^2$ yieldsFor fixed nonzero curvature, the magnitude of $\Omega -1$ grows as expansion decelerates. The observed closeness to one therefore requires much tighter early-time closeness to one in the standard decelerating model: this is the flatness problem. Accelerated expansion during inflation reverses this evolution and suppresses the curvature contribution.

Take a fixed comoving volume, whose physical volume is $V=V_{\ast }{a}^3$. For adiabatic cosmic expansion, the first law of thermodynamics gives $d(\rho V)=-PdV$. ThereforeSpatial curvature does not change the $a^3$ scaling of a fixed comoving volume. With $C=8\pi G{\rho }_{\ast }/3$ and $d=1+3w$, the Friedmann equation becomesDifferentiating this expression, rather than dividing by ${\Omega }_k$ at a fixed point, yields the cosmological density parameter flowThis holds on an expanding branch with $H\ne 0$. The spatially flat solution ${\Omega }_k=0$ is a fixed point. For $|{\Omega }_k|\ll 1$, ${\Omega }_k\propto a^{1+3w}$. Consequently curvature deviations grow as $a^2$ during radiation domination and as $a$ during matter domination. A small present curvature therefore requires a much smaller initial deviation: this is the flatness problem of a decelerating Big Bang. The issue applies to either sign of curvature, not just an open universe. Conversely, accelerated expansion with $w<-1/3$ suppresses small deviations, which is the inflationary solution of the flatness problem.

The flatness problem is the extreme closeness to spatial flatness required in the early universe if ordinary decelerating expansion is to produce the observed near-flat universe. From the Friedmann equation, with $c=1$ and curvature constant $k$,During decelerating expansion $|\Omega -1|$ grows. The Friedmann acceleration equation gives $\ddot{a}/a=-4\pi G(\rho +3P)/3$. Thus $\rho +3P<0$ makes an expanding universe accelerate, increases $\dot{a}$ and suppresses $|\Omega -1|$: this is the inflationary solution of the flatness problem.

The slow-roll approximation neglects ${\dot{\varphi }}^2/2$ compared with $V$ and $\ddot{\varphi }$ compared with $3H\dot{\varphi }$. For the positive-field branch, taking $m>0$,Integrating the field equation and $\dot{a}/a=H$ gives the slow-roll solutionThe potential slow-roll parameters are ${ϵ}_V={\eta }_V=2/{\varphi }^2$. They cease to be small when $\varphi$ is of order $1$ in these units; the customary endpoint ${ϵ}_V=1$ is $\boxed{{\varphi }_{\mathrm{end}}\simeq \sqrt{2}}$. Equating the approximate kinetic and potential energies instead gives $\sqrt{2/3}$, another order-one indicator, not a precise endpoint of an approximation that is already failing.

Put $\Omega =1+\delta$ with $|\delta |\ll 1$. Linearizing the cosmological density parameter flow gives $d\delta /d\log a=(1+3\omega )\delta +O({\delta }^2)$. For constant equation-of-state parameter,For variable $\omega$, the correct replacement is $\delta (a)=\delta (a_i)\exp [{\int }_{a_i}^a(1+3\omega (\tilde{a}))d\log \tilde{a}]$. The printed power assumes constant $\omega$.

For pressureless matter, the departure from flatness grows as $a$; for radiation in cosmology, it grows as $a^2$. Hence in a conventional decelerating expanding history, maintaining $|\Omega -1|\ll 1$ at a late epoch requires extremely small departures at early times. This is the flatness problem, not an assertion that exact flatness is inconsistent: $\Omega =1$ is an exact solution, but it is unstable to small departures in spatial curvature of an FLRW universe during these eras. With $\omega <-1/3$ the perturbation instead decays, which supplies the inflationary solution of the flatness problem.

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.

Let $\mathcal{H}=a^{\prime }/a=\dot{a}$, where a prime denotes differentiation with respect to conformal time $d\tau =dt/a$. Multiplying the Friedmann equation by $a^2$ givesThe critical density is the value obtained when $k=0$, so

Since ${\mathcal{H}}^{\prime }=a\ddot{a}$, the pressure-free Raychaudhuri equation givesorDifferentiating $\Omega =1+k{c}^2/{\mathcal{H}}^2$ and substituting this equation yields

For an expanding universe $\mathcal{H}>0$. The Einstein-de Sitter universe $\Omega =1$ is an unstable constant solution: curves starting with $0<\Omega <1$ decrease towards $0$, while curves starting with $\Omega >1$ rise away from $1$ and diverge as the closed model reaches maximum expansion. This is the flatness problem.

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.

The flatness problem is that the observed cosmological density parameter is close to $\Omega =1$, although in an ordinary decelerating universe any departure from one grows with time. Extrapolation backward therefore requires implausibly precise early cancellation of the curvature term.

The cosmological perfect-fluid continuity equation isConsequentlyFor an expanding universe $H>0$. Ifthen $\rho a^2$ increases. The Friedmann equation impliesIt therefore decreases during this period, driving $\Omega$ toward one. By the Friedmann acceleration equation, the same condition gives $\ddot{a}>0$ when the cosmological constant is included in the effective fluid. This is the inflationary solution of the flatness problem.

For the scalar field, the slow-roll approximation meansThe equations reduce toTaking the positive-field branch gives $H\simeq m\varphi /\sqrt{3}$, and thereforeWith $\varphi (0)={\varphi }_i$,Now integrate $H=\dot{a}/a\simeq m\varphi /\sqrt{3}$:Thus the quadratic-potential slow-roll solution isFinally,so equivalently

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.