The second Friedmann equation gives the <first Hubble slow-roll parameter>
$$
\boxed{\epsilon=-\frac{\dot H}{H^2}
=\frac{\dot{\bar\phi}^{,2}}{2M_{\rm Pl}^2H^2}}.
$$
Because $\dot{\bar\phi}\propto a^{-3}$,
$$
\frac{\dot\epsilon}{\epsilon H}
=2\frac{\ddot{\bar\phi}}{H\dot{\bar\phi}}
-2\frac{\dot H}{H^2}
=-6+2\epsilon.
$$
Thus
$$
\boxed{\eta_{\rm SR}=-6+2\epsilon}.
$$
As $a\to\infty$, the kinetic energy decays as $a^{-6}$ and the constant $\Lambda$ dominates. Hence $H$ approaches a constant,
$$
\boxed{aH\simeq-\frac1\eta,
\qquad\epsilon\simeq0,
\qquad\eta_{\rm SR}\simeq-6}.
$$
This non-attractor background is <ultra-slow-roll inflation>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 312 / 2 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-312/2/ii/solution