Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 15B (Cosmology)

Differentiate the Friedmann equation:The scalar-field equation gives $\ddot{\varphi }+c^2{V}^{\prime }=-3H\dot{\varphi }$, henceBy continuity this also holds through isolated zeros of $H$, so the Raychaudhuri equation is

Differentiating the ansatz givesThe Raychaudhuri equation therefore becomesSince $d[\coth (2\lambda t)]/dt=-2\lambda {\mathrm{csch}}^2(2\lambda t)$, the requested specific solution takes the integration constant to be zero:Now $\dot{a}/a=H$, soand

Put $u=\tanh (\lambda t)=e^{\varphi /{\varphi }_0}$. The hyperbolic cotangent double-angle identity givessoAlsoSolving the Friedmann equation for the scalar potential,therefore gives

Write $r={\varphi }_0^2/M_{\mathrm{Pl}}^2$. As $t\to 0$,so $\varphi \to -\infty$ andBecause $2/3<r<2$, the power $r/2$ lies in $(1/3,1)$; thus $\ddot{a}<0$ and the early expansion decelerates. As $t\to \infty$, $\varphi \to 0$ and $\sinh (2\lambda t)\sim e^{2\lambda t}/2$, givingso the late universe has accelerating expansion.

The transition satisfies $\ddot{a}/a=H^2+\dot{H}=0$. Substitution givesor $r{\cosh }^2(2\lambda t)=2$. ThereforeThis is the transition described by the exact hyperbolic-sine scalar-field cosmology.