Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 15B (Cosmology)

Write $A=8\pi G/(3c^2)$ and $H=\dot{a}/a$. Differentiating the Friedmann equation givesThe cosmological perfect-fluid continuity equation says $\dot{\rho }=-3H(\rho +P)$, so, first where $H\ne 0$ and then everywhere by continuity,Using $K{c}^2/a^2=A\rho -H^2$ from the original equation givesSince $\dot{H}+H^2=\ddot{a}/a$ for the Hubble parameter, this is the Friedmann acceleration equation.

Separate continuity givesusing the scalings for radiation in cosmology and cosmological constant energy. At $t=t_0$, where $a=1$, the Friedmann equation becomessoAt general $a$,which rearranges to

Part (a), applied to $\rho ={\rho }_r+{\rho }_{\Lambda }$ and $P={\rho }_r/3-{\rho }_{\Lambda }$, givesSubtracting the displayed formula for $H^2$ yieldsThese relations define the closed radiation--cosmological-constant turning polynomial.

Put $x=a^2$. Apart from the positive factor $H_0^2/a^4$,For $\beta >4$, its discriminant $\beta (\beta -4)$ is positive and its two roots areBoth are positive. Moreover $F(0)>0$ and $F(2)=4-\beta <0$, soA universe expanding from $a=0$ follows the allowed region $0\le a^2\le x_{-}$ and reaches $a^2=x_{-}$ with $H=0$. There,so $H$ crosses from positive to negative: expansion reverses into contraction.

As $a\to 0$ on the contracting branch,Hence $ada\sim -H_0\sqrt{\beta }dt$, so zero scale factor is reached in finite time. The universe therefore undergoes the supercritical closed radiation--cosmological-constant recollapse to a Big Crunch.

For $\beta =4$,The branch expanding from $a=0$ hasuntil it approaches $a^2=2$. With $x=a^2$,Solving this linear equation gives the critical closed radiation--cosmological-constant solution

For $H_{0}t\ll 1$,For $H_{0}t\gg 1$,soexponentially from below.