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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 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