Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 15E / b

With $P=-2\rho /3$, the continuity equation becomeshence $\rho a$ is constant. Since $a(t_0)=1$, one has $\rho ={\rho }_0/a$. The Friedmann equation is thereforewhere

The expanding branch obeysIntegrating from $a=0$ at $t=0$ givesSolving for $a$,

Setting $a(t_0)=1$ givesAt the present epoch,soFor $\beta \gg \gamma$, this tends to $1$, giving $t_0\sim H_0^{-1}$. For $\gamma \gg \beta$, it tends to $2$, giving $t_0\sim 2H_0^{-1}$.

From the explicit solution, $\ddot{a}=\gamma /2$. Since $a(t_0)=1$ and $\dot{a}(t_0{)}^2=\beta +\gamma$,Because an open universe has $\beta >0$, this is strictly greater than $-1/2$. It approaches $-1/2$ only in the zero-curvature limit, so this model cannot attain $q_0\le -0.5$ for the permitted parameters and therefore cannot reproduce $q_0\approx -0.55$.