Past exam of the general relativity and cosmology course of the University of Oxford / 2013 / 4 / c / Solution

The scaling solution has $\rho =3M^2{p}^2/t^2$ and $\rho +P={\dot{\phi }}^2=4M^2/({\lambda }^2{t}^2)$. ThereforeMore generally any real canonical scalar field with positive total energy density obeys $\rho +P={\dot{\phi }}^2\ge 0$, so it cannot have $w<-1$. A negative potential does not reverse this conclusion.

The proper particle horizon is $a(t){\int }_0^{t}d{t}^{\prime }/a(t^{\prime })$. For $a\propto t^p$ the lower-limit integral diverges exactly when $p\ge 1$. Thus, assuming the same power law extends to the initial boundary, there is no finite particle horizon for $\boxed{0<{\lambda }^2\le 2}$. For the strict inequality, $p>1$, the proper cosmological event horizon and present age areAt ${\lambda }^2=2$, $a\propto t$ has an infinite particle-horizon integral but also an infinite future event-horizon integral, so no finite event horizon exists. The age remains $H_0^{-1}$. The $\lambda =0$ constant-potential case is separate: its flat de Sitter patch has event-horizon radius $H_0^{-1}$ and no finite-age power-law Big Bang.