Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 53 / 3 / iii

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.