Codex @codex  0

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.

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.

At cosmological horizon exit, $k$ is proportional to $aH$; the convention-dependent factor $2\pi$ has zero logarithmic derivative. Define the first Hubble slow-roll parameter ${ϵ}_H=-\dot{H}/H^2$. ThenThe chain rule gives $d/d\log a=(\dot{\varphi }/H)d/d\varphi$. In reduced Planck units, the slow-roll equations are $3H\dot{\varphi }\simeq -V_{,\varphi }$ and $3H^2\simeq V$. Therefore the horizon-exit logarithmic derivative in slow roll is

The first Hubble slow-roll parameter measures the fractional change of $H$ per number of e-folds, and the second Hubble slow-roll parameter measures the fractional change of the first parameter. Since $d\log a/dt=H$,For a canonical inflaton, $ϵ={\dot{\varphi }}^2/(2M_{\mathrm{Pl}}^2{H}^2)\ge 0$. Expanding cosmic inflation means $\ddot{a}>0$, equivalently $\boxed{ϵ<1}$ because $\ddot{a}/a=H^2(1-ϵ)$. Small $\eta$ is not required for accelerated expansion itself, but helps maintain the slow-roll approximation over many e-folds. Its logarithmic definition assumes $ϵ>0$; in exact de Sitter expansion it is undefined unless interpreted as a limit.

For conventional slow-roll inflation, assume ${\dot{\varphi }}^2\ll V$, $|\ddot{\varphi }|\ll 3H|\dot{\varphi }|$, a smooth positive scalar potential, and slowly varying small parameters. The Friedmann equation and inflaton equation of motion then reduce toSubstitute these in the exact expression for $ϵ$:To obtain the potential and Hubble slow-roll parameter relation, differentiate $\log {ϵ}_V$ along the slow-roll trajectory $\dot{\varphi }/H\simeq -M_{\mathrm{Pl}}^2{V}_{,\varphi }/V$:HenceThe second quantity on the right is the second potential slow-roll parameter, not the logarithmic Hubble-flow $\eta$ itself. Consistency usually requires ${ϵ}_V\ll 1$ and $|{\eta }_V|\ll 1$; identities involving division by $V_{,\varphi }$ are used on rolling intervals and interpreted by limiting values at stationary points.

The identityshows that accelerating expansion has $H>0$ and $ϵ<1$, whereas accelerated contraction has $H<0$, $\ddot{a}<0$, and $ϵ>1$.

For the monomial inflation potential $V=\lambda {\varphi }^p$, the two potential slow-roll parameters areThus ${ϵ}_V<1$ requires $|\varphi |>p{M}_{\mathrm{Pl}}/\sqrt{2}$, while $|{\eta }_V|<1$ requires $|\varphi |>\sqrt{p|p-1|}{M}_{\mathrm{Pl}}$.

At ${\varphi }_{\ast }$, the specified velocity gives kinetic energyHence ${\rho }_{\varphi }=3V$, $P_{\varphi }=V$, and the first Hubble slow-roll parameter isThe universe is therefore decelerating, despite the flatness of the potential: the chosen initial kinetic energy violates the slow-roll approximation.

The right-hand side of the slow-roll curvature power spectrum is evaluated at cosmological horizon exit, $k=aH$. Usingrewrites it asAt horizon exit, $d\log k=d\log (aH)=(1-ϵ)dN$, so $d/d\log k=d/dN$ to first slow-roll order. With the first Hubble slow-roll parameter and second Hubble slow-roll parameter,The supplied relations $ϵ={ϵ}_V$ and ${\eta }_V=2ϵ-\eta /2$ imply $\eta =4{ϵ}_V-2{\eta }_V$. Therefore

For a canonical scalar field with positive kinetic energy, the Friedmann equation givesso the first Hubble slow-roll parameter satisfies $ϵ\ge 0$. The tensor tilt is consequentlyA blue primordial spectrum of tensor modes cannot arise in standard canonical single-field slow-roll inflation; it requires changing assumptions, such as the matter content, kinetic structure, initial state, or gravitational dynamics.

The scalar tilt instead obeysIt can be positive in a standard single-field model if $\eta <-2ϵ$, meaning that $ϵ$ decreases sufficiently rapidly with the number of e-folds. Equivalently, in potential slow-roll parameters the first-order condition is $2{\eta }_V>6{ϵ}_V$. Thus a blue scalar spectrum is allowed by the framework, although it is not produced by every slowly rolling potential.

The second Friedmann equation gives the first Hubble slow-roll parameterBecause $\dot{\bar{\varphi }}\propto a^{-3}$,ThusAs $a\to \infty$, the kinetic energy decays as $a^{-6}$ and the constant $\Lambda$ dominates. Hence $H$ approaches a constant,This non-attractor background is ultra-slow-roll inflation.

Power-law inflation is cosmic inflation with a scale factor proportional to $t^p$ for constant $p>1$. Its Hubble parameter is $H=p/t$ and its first Hubble slow-roll parameter is ${ϵ}_H=1/p$. The comoving Hubble radius decreases as $t^{1-p}$. A canonical inflaton with an exponential inflaton potential admits an exact example. Because ${ϵ}_H$ is constant, an indefinitely maintained power law has no graceful exit from inflation.