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A graceful exit from inflation ends sustained accelerating expansion and connects it to a hot radiation-dominated universe. Crossing ${ϵ}_H=1$ ends acceleration in a homogeneous expanding model; reheating requires an additional mechanism to transfer inflaton energy into other species. Pure exponential-potential power-law inflation keeps ${ϵ}_H$ constant and its inflaton potential has no finite minimum, so neither an automatic endpoint nor oscillatory reheating follows from that potential alone.

Write $V=V_0{e}^{-A\varphi /m_{\mathrm{pl}}}$, where the stated model has $V_0=m_{\mathrm{pl}}^4$, and take $A>0$. Slow roll givesThus $d[e^{A\varphi /(2m_{\mathrm{pl}})}]/dt=A^2\sqrt{V_0}/(\sqrt{96\pi }{m}_{\mathrm{pl}})$. Absorb the integration constant into $b$ to obtainIt follows that $H=16\pi /[A^2(t+b)]$, soThe exponential potential has constant ${ϵ}_V=A^2/(16\pi )$ and ${\eta }_V=A^2/(8\pi )$. Inflation requires $p>1$, whereas reliable slow-roll approximation requires $A^2/(8\pi )\ll 1$. This is the slow-roll normalization for exponential inflation.

The stated field normalization is a slow-roll result, not an exact solution of the original full equations. Keeping the acceleration and kinetic terms yields the same $p$ but $C^2=A^4{V}_0/[2m_{\mathrm{pl}}^2(48\pi -A^2)]$, as follows by substituting $\dot{\varphi }=2m_{\mathrm{pl}}/[A(t+b)]$ into both full equations. It agrees to leading order for small slope and is the exponential-potential power-law inflation solution.

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.

Adding and subtracting the canonical scalar field density and pressure givesA real scalar with positive kinetic energy consequently needs $w\ge -1$. On the expanding scalar-dominated branch, divide $\dot{\varphi }$ by $H$ and use the Friedmann equation:Choose the positive rolling sign. Integration and elimination of $a$ using ${\rho }_{\varphi }={\rho }_{\varphi }^0(a/a_0{)}^{-3(1+w)}$ give the scalar-field reconstruction for a constant equation of state:Changing the rolling sign reverses the exponential slope. At $w=-1$, $\varphi$ is constant and $V={\rho }_{\varphi }^0$; elimination by an invertible $\varphi (a)$ then degenerates. The cosmological continuity equation gives $\ddot{\varphi }+3H\dot{\varphi }+V_{,\varphi }=0$ whenever $\dot{\varphi }\ne 0$, so the reconstruction satisfies the field equation as well as the Friedmann equation.

Compare the reconstructed exponent with $V_0{e}^{-\lambda \sqrt{8\pi G}\varphi }$. In this constant-$w$, scalar-dominated scaling solution,Strict ${\rho }_{\varphi }+P_{\varphi }>0$ requires $w>-1$, and acceleration requires $w<-1/3$. HenceThe accelerating branch has $V_0>0$. The zero-slope limit is a constant potential with $\rho +P=0$, so it does not meet the strict inequality. This result characterizes the exponential-potential power-law inflation solution; an arbitrary trajectory with the same potential need not have constant $w$ throughout its evolution.