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For the quadratic-potential slow-roll solution, $V^{\prime }/V=2/\varphi$, so $ϵ=\eta =2M_{\mathrm{pl}}^2/{\varphi }^2$. Inflation ends when $ϵ\simeq 1$, giving ${\varphi }_{\mathrm{end}}^2=2M_{\mathrm{pl}}^2$. The slow-roll e-fold count remaining before that endpoint isWith $N_{\ast }=50$, this gives ${\varphi }_{\ast }^2=202M_{\mathrm{pl}}^2$ and ${ϵ}_{\ast }={\eta }_{\ast }=1/101$. The quadratic-inflation amplitude calibration isMatching the supplied amplitude therefore givesTaking $M_{\mathrm{pl}}\simeq 2.4\times 10^{18}\mathrm{GeV}$, this is $m\simeq 1.7\times 10^{13}\mathrm{GeV}$. Retaining ${\varphi }_{\mathrm{end}}$ produces the small $+1$ correction; dropping it gives almost the same leading slow-roll estimate.

The scalar spectral index isIt agrees with the supplied central value and lies well inside the stated uncertainty interval. This conclusion concerns the scalar tilt; the tensor prediction is a separate test of the same model.

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