Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 9D / b / Solution

In a decelerating universe the comoving Hubble radius grows, so regions now in causal contact extrapolate to many disconnected early Hubble patches. Inflation shrinks the comoving Hubble radius and stretches one causal patch beyond the present horizon. Radiation scaling gives a present-horizon physical size at $10^{-35}$ s of roughly $c{t}_0(a/a_0)\sim c{t}_0(t/t_0{)}^{1/2}\sim 1$ m. Comparing this with the inflationary Hubble length (about $ct\sim 10^{-27}$ m at that epoch) requires $e^N\sim 10^{27}$, or $N\simeq 62$ (order 60).