Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 9D (Cosmology)

Under the slow-roll approximation,For $V=m^2{\varphi }^2/2$ and the branch $\varphi >0$, setThen $H\simeq Am\varphi$ andWith $\varphi (0)={\varphi }_0$ and $a(0)=a_0$,and integrating $\dot{a}/a=H$ gives

For $V=\lambda {\varphi }^4/4$ and $\varphi >0$, putThe slow-roll approximation gives $H\simeq B{\varphi }^2$ andTherefore, with $\gamma =c\sqrt{\lambda /(6\pi G)}$,Since $B/\gamma =2\pi G/c^2$, integration of the Hubble parameter gives