Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 48 / 1 / a

Assume positive fluid density, ${\rho }_0>0$, and write $m=1+3w>0$. A perfect fluid with constant equation of state obeys the cosmological perfect-fluid continuity equation, givingIn the stated units the Friedmann equation is $H^2=\rho /3-k/a^2+\Lambda /3$. Multiplication by $a^2/2$ yields the Friedmann effective potential for a constant-equation-of-state fluid,The allowed region has $V\le 0$. The Friedmann acceleration equation is equivalently $\ddot{a}=-V^{\prime }(a)$, including turning points by continuity. Thus a zero-energy mechanical trajectory reproduces the cosmological evolution, with $a>0$.

For $k=0$, $\Lambda <0$, the potential increases strictly from negative infinity to positive infinity. Its unique zero givesExpansion from the Big Bang stops there, with negative acceleration, then reverses into a Big Crunch. Both the turning point and the final singularity occur in finite proper time: $dt=da/\sqrt{-2V}$ is integrable near a simple turning point and behaves as a constant times $a^{m/2}da$ near zero.

For $\Lambda =0$, $k=+1$, $V$ increases from negative infinity to $1/2$ and crosses zero atThis is again expansion followed by finite-time recollapse. For $k=-1$, the same increasing curve has asymptote $-1/2$ and never meets zero. Expansion continues without a finite maximum; asymptotically $\dot{a}\to 1$ and $a\sim t$, as spatial curvature dominates the diluted fluid.

For $k=0$, $\Lambda >0$, $V$ tends to negative infinity at both ends. It has a maximum atThere is no turning point. Expansion initially decelerates, then accelerates once $a>a_{\ast }$, and approaches de Sitter spacetime expansion $a\propto e^{\sqrt{\Lambda /3}t}$. Thus positive flat dark-energy expansion is unbounded; negative flat dark energy and positive curvature without dark energy recollapse.