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Under the slow-roll approximation,For $V=m^2{\varphi }^2/2$ and the branch $\varphi >0$, setThen $H\simeq Am\varphi$ andWith $\varphi (0)={\varphi }_0$ and $a(0)=a_0$,and integrating $\dot{a}/a=H$ gives

The slow-roll approximation assumesThe first condition makes the energy density potential dominated, and the second neglects inertial acceleration relative to Hubble friction. The scalar and Friedmann equations reduce to

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

From $3H\dot{\varphi }=-c^2{V}^{\prime }$ and $H^2=8\pi GV/(3c^2)$,The slow-roll approximation fails when this kinetic contribution is comparable with $V$, giving

The number of e-folds isUsing the estimate above,For $n=6$ and the roughly $60$ e-folds needed for the flatness problem,This is the monomial slow-roll e-fold count.

The potential slow-roll parameter isFrom the slow-roll approximation,ThereforeThus $ϵ\ll 1$ implies ${\dot{\varphi }}^2\ll V$, consistently with potential-energy domination.

The same equations giveOn the branch $V^{\prime }/V>0$, the definition of $ϵ$ turns this into

The homogeneous inflaton obeysUnder the slow-roll approximation, the acceleration and kinetic-energy corrections may be neglected, so the field equation and Friedmann equation becomeThus a positive field rolls away from the hilltop while $H$ is approximately constant.

For a canonical inflaton under the slow-roll approximation, $\dot{\varphi }/H\simeq -M_{\mathrm{Pl}}^2{V}^{\prime }/V$ and $ϵ\simeq {ϵ}_V=(M_{\mathrm{Pl}}^2/2)(V^{\prime }/V{)}^2$. Differentiating $\log {ϵ}_V$ along this trajectory givesHere $\eta =d\log ϵ/d\log a$, not the alternate convention $-\ddot{\varphi }/(H\dot{\varphi })$. The logarithmic definition requires $ϵ>0$; exact de Sitter expansion is treated as a limit.

For a quadratic inflaton potential ending at the leading estimate ${ϵ}_V=1$, $\Phi /M_{\mathrm{Pl}}=\sqrt{4N+2}$. Cosmological horizon exit gives $d\log k/dN=-1+{ϵ}_H$. A narrow logarithmic interval therefore probes a field span $2\Delta \log k/(\sqrt{4N+2}(1-{ϵ}_H))$, to the accuracy of the slow-roll approximation. At sixty e-folds, five logarithmic wavenumber units probe about $0.64M_{\mathrm{Pl}}$ around a background field near $15.6M_{\mathrm{Pl}}$.