Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 15E (Cosmology)

Let a comoving test particle have physical radius $R=a(t)\chi$, where $\chi$ is fixed. Assume Newtonian gravity, spherical homogeneity, negligible pressure, and no mass flux through this sphere, so its enclosed mass $M=(4\pi /3)\rho R^3$ is constant. The shell theorem gives a conserved specific energySubstitute $R=a\chi$, divide by $R^2/2$, and write $2\mathcal{E}/{\chi }^2=-k{c}^2$ to obtainHomogeneity makes the energy scale as ${\chi }^2$, leaving the same $k$ at every comoving radius. The sign of $k$ encodes the binding-energy sign. This Newtonian derivation applies to pressure-free matter; interpreting $k$ as spatial curvature and extending the equation to relativistic pressure components requires general relativity, rather than conservation of this Newtonian shell energy alone.

Separate fluid conservation gives ${\dot{\rho }}_M=-3H{\rho }_M$, hence ${\rho }_M={\rho }_{M0}{a}^{-3}$ after normalization at today. For $P_{\Lambda }=-{\rho }_{\Lambda }{c}^2$, the conservation equation gives ${\dot{\rho }}_{\Lambda }=0$. ThusAssume both densities are positive and set $A=8\pi G{\rho }_{M0}/3$, $B=8\pi G{\rho }_{\Lambda 0}/3$. On the expanding flat branch, ${\dot{a}}^2=A/a+B{a}^2$. With $b=a^{3/2}$ this becomes $\dot{b}=(3/2)\sqrt{A+B{b}^2}$. Choosing the Big Bang at $t=0$ and integrating givesThe normalization $a(t_0)=1$ requires $\sinh (\beta t_0)=\sqrt{{\rho }_{\Lambda 0}/{\rho }_{M0}}$; equivalently $\alpha ={\sinh }^{-2/3}(\beta t_0)$. Thus the age $t_0$ and the two positive densities are not independent once this normalization and bang time are fixed.

As $t\downarrow 0$, $a\sim \alpha (\beta t{)}^{2/3}$, the matter-dominated law. As $t\to \infty$, $a\sim \alpha 2^{-2/3}{e}^{\sqrt{B}t}$ and $\ddot{a}/a\to B>0$, giving accelerated dark-energy expansion. Direct differentiation gives $\ddot{a}/a=B-A/(2a^3)$, so acceleration begins once $a^3>A/(2B)$. If dark energy vanishes, the limiting matter-only solution must be taken instead; negative dark-energy density would not have the stated late-time accelerating form.