Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 53 / 4 / Solution

For the quadratic potential, the e-fold integral above gives ${\Phi }_N/M_{\mathrm{Pl}}\simeq \sqrt{4N+2}$. At $N\simeq 60$, the field is ${\Phi }_{\ast }/M_{\mathrm{Pl}}\simeq \sqrt{242}\simeq 15.6$. Since $k=aH$ at exit and $N$ counts down,To leading slow-roll order, $|\Delta N|\simeq \Delta \ln k$. The quadratic-inflation field interval from a wavenumber interval is thereforeThe corresponding field endpoints are approximately $\sqrt{232}{M}_{\mathrm{Pl}}$ and $\sqrt{252}{M}_{\mathrm{Pl}}$, or $15.2$ to $15.9$ reduced Planck masses. Longer wavelengths exit earlier at the larger field value. Using $\ln 100=4.605$ instead of the supplied rounded logarithmic interval gives a span of about $0.59M_{\mathrm{Pl}}$; retaining $1-{ϵ}_H$ changes either estimate by less than one percent. The observable interval samples a small part of the large background field value, rather than the entire excursion to the end of inflation.