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

The second Friedmann equation gives the first Hubble slow-roll parameterBecause $\dot{\bar{\varphi }}\propto a^{-3}$,ThusAs $a\to \infty$, the kinetic energy decays as $a^{-6}$ and the constant $\Lambda$ dominates. Hence $H$ approaches a constant,This non-attractor background is ultra-slow-roll inflation.