Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 1 / 15D (Cosmology)

Multiply the Friedmann equation by $a^2$ and differentiate:Insert the cosmological continuity equation to obtainOn a nonstatic branch divide by $2a\dot{a}$; continuity extends the result across isolated turning points. The resulting Friedmann acceleration equation isAn everywhere static branch needs the independent spatial Einstein field equations: the two differentiated constraints alone cannot determine the pressure when $\dot{a}\equiv 0$. This is the qualification to deriving the acceleration equation solely by division.

For constant equation-of-state parameter $w$, the cosmological continuity equation integrates to $\rho ={\rho }_0{a}^{-3(1+w)}$ using today's normalization. In conformal time, $\mathcal{H}=a^{\prime }/a=\dot{a}$, so the Friedmann equation becomesSince ${\mathcal{H}}^{\prime }=a\ddot{a}$, the Friedmann acceleration equation givesFor $k{c}^2=1$, put $q=(1+3w)/2$. When $q\ne 0$, integration gives $\mathcal{H}=\cot (q(\tau -{\tau }_0))$. Integrating $a^{\prime }/a=\mathcal{H}$ then yields the closed constant-equation-of-state Friedmann scale factorChoose a connected interval with positive sine; the origin of conformal time can absorb ${\tau }_0$. For $w>-1/3$ this is the expansion and recollapse branch between consecutive zeros of the sine. Negative $q$ instead gives the corresponding minimum-radius branch on its positive-sine interval. The printed power formula excludes $w=-1/3$: in that exceptional case $\mathcal{H}=H_0$ is constant and $a=A{e}^{H_0\tau }$, with $H_0^2+1=8\pi G{\rho }_0/3$. A real solution requires the latter constant to be at least one; equality permits a static solution.

For pressure-free matter, $w=0$, so $\beta =1/2$. Set the big-bang origins to zero. The closed matter-dominated Friedmann solution follows by ${\sin }^2(\beta \tau )=(1-\cos 2\beta \tau )/2$ and integrating $dt=ad\tau$:The entire cycle is $0\le 2\beta \tau \le 2\pi$; maximum expansion has $a=\alpha$ at $2\beta \tau =\pi$, and the big crunch occurs at $t=\alpha \pi /(2\beta )$.