Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 66 / 2

For pressureless matter and zero cosmological constant, the Friedmann acceleration equation gives $\ddot{a}/a=-4\pi G\rho /3$. The deceleration parameter therefore satisfies the dust relation between deceleration and density, $q=4\pi G\rho /(3H^2)$. Evaluating today,Substitute this into the present Friedmann equation to obtain $H_0^2+k{c}^2/a_0^2=2q_0{H}_0^2$, orIn particular, $q_0>1/2$, $q_0=1/2$, and $q_0<1/2$ correspond to positive, zero, and negative spatial curvature.

Normalize the scale factor by $x=a/a_0$. Constant-equation-of-state density scaling for pressureless matter gives $\rho ={\rho }_0{x}^{-3}$. Multiplying the Friedmann equation by $x^2$ and substituting the expressions from part (i),Take the expanding branch $\dot{x}>0$ and set the origin of cosmic time at the Big Bang, $x=0$. The age of a dust universe without a cosmological constant is consequently

The stated real trigonometric substitution applies to the closed case $q_0>1/2$. Set $A=2q_0-1$, so $x=q_0(1-\cos \theta )/A$, $dx=q_0\sin \theta d\theta /A$, andThe upper limit obeys $1-\cos {\theta }_{\ast }=A/q_0$, with $0<{\theta }_{\ast }<\pi$ on the expanding branch. Substitution into the age of a dust universe without a cosmological constant yieldsOn this interval, the integrand equals $1-\cos \theta$, so it can also be evaluated:For $q_0=1/2$, the original integral instead gives $H_0{t}_0=2/3$. For $0<q_0<1/2$, use a hyperbolic substitution to obtain $H_0{t}_0=q_0(\sinh {\eta }_{\ast }-{\eta }_{\ast })/(1-2q_0{)}^{3/2}$, where $\cosh {\eta }_{\ast }=(1-q_0)/q_0$. A real $\theta$ cannot implement the printed substitution in that open case.

An empty universe with $\Lambda =0$ has ${\Omega }_{r,0}={\Omega }_{m,0}={\Omega }_{\Lambda ,0}=0$. The present Friedmann equation then requires ${\Omega }_{k,0}=1$, soThe Friedmann acceleration equation gives $\ddot{a}=0$, hence $a\propto t$ after choosing its zero at $t=0$, andThis is the Milne universe. A larger past Hubble parameter does not require negative $\ddot{a}$: even constant $\dot{a}$ produces $H=\dot{a}/a=1/t$. Equivalently, deceleration parameter from the Hubble derivative gives $q=-1-\dot{H}/H^2=-1+1=0$. The age is $t_0=H_0^{-1}$.