Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 53 / 1

Set $c=1$ and let $H=\dot{a}/a$ denote the proper-time Hubble parameter. The Friedmann equation is the expansion constraint: density drives expansion, while spatial curvature and the cosmological constant contribute separately. The Friedmann acceleration equation shows that the active gravitational density is $\rho +3P$; positive pressure gravitates, while positive vacuum energy accelerates expansion.

Subtracting the constraint from the acceleration equation gives $\dot{H}=-4\pi G(\rho +P)+k/a^2$. Differentiating the constraint gives $2H\dot{H}=(8\pi G/3)\dot{\rho }+2kH/a^2$. Eliminate $\dot{H}$ to obtainThis cosmological continuity equation is local energy conservation. For a comoving volume $V\propto a^3$, it is $d(\rho V)=-PdV$, the adiabatic first law: expansion does work against pressure.

Define the deceleration parameter $q=-a\ddot{a}/{\dot{a}}^2$, the critical density ${\rho }_c=3H^2/(8\pi G)$, and the cosmological density parameters ${\Omega }_m={\rho }_m/{\rho }_c$ and ${\Omega }_{\Lambda }=\Lambda /(3H^2)$. The dust-plus-vacuum acceleration equation then gives $q={\Omega }_m/2-{\Omega }_{\Lambda }$. Dust conservation gives ${\rho }_m\propto a^{-3}$, while ${\rho }_{\Lambda }=\Lambda /(8\pi G)$ is constant. At the acceleration transition, ${\rho }_m=2{\rho }_{\Lambda }$, so

For spatial flatness and $a_0=1$, the constraint today sets ${\Omega }_{\Lambda 0}=1-{\Omega }_{m0}$. Thus $H^2=H_0^2[{\Omega }_{m0}{a}^{-3}+1-{\Omega }_{m0}]$. Put $M={\Omega }_{m0}$, $L=1-M>0$ and $y=a^{3/2}$. Then $\dot{y}=(3H_0/2)\sqrt{M+L{y}^2}$. Integrating from $y(0)=0$ gives $\mathrm{arsinh}(y\sqrt{L/M})=(3H_0/2)\sqrt{L}t$, hence the flat dust-and-vacuum expansionFor small $t$, $a\simeq [(3/2)H_0\sqrt{M}t{]}^{2/3}$, the matter-dominated law. At late times, $a\simeq [M/(4L){]}^{1/3}{e}^{H_0\sqrt{L}t}$, approaching de Sitter spacetime. Negative spatial curvature adds a positive $a^{-2}$ contribution to $H^2$. This is subdominant to dust's $a^{-3}$ at small $a$ and to the constant vacuum term at large $a$, so it does not change either leading limit.

The model describes the transition from matter domination to the accelerated late-time expansion indicated by cosmological distance observations, with nearly flat geometry and vacuum-like dark energy. Its formal $t\to 0$ dust limit is not the actual earliest universe: radiation, neglected here, dominates then and gives $a\propto t^{1/2}$.

For an expanding universe, cosmic inflation requires $\ddot{a}>0$. Including any vacuum contribution in the total density and pressure, the acceleration equation gives $\boxed{\rho +3P<0}$, or $w=P/\rho <-1/3$ for positive density. Equivalently the comoving Hubble radius $(aH{)}^{-1}={\dot{a}}^{-1}$ decreases. Since $|{\Omega }_{\mathrm{total}}-1|=|k|/(aH{)}^2$, prolonged acceleration suppresses curvature and addresses the flatness problem. A sufficiently long inflationary phase also stretches a region that was initially causally connected to encompass the observed region, addressing the horizon problem. The condition on acceleration must last long enough; an arbitrarily short interval is insufficient.

For a homogeneous canonical scalar field $\Phi$ with potential $V(\Phi )$,Thus $V>{\dot{\Phi }}^2$ is sufficient for inflation. A suitably flat positive potential gives slow-roll inflation: Hubble damping in $\ddot{\Phi }+3H\dot{\Phi }+V_{,\Phi }=0$ keeps the kinetic energy small compared with $V$, so $P_{\Phi }\simeq -{\rho }_{\Phi }$.

A physical positive scale factor here requires $t>1/C$. In that domain $\dot{a}=1/t>0$ but $\ddot{a}=-1/t^2<0$. There is no inflationary expansion. The range $0<t<1/C$ has a negative printed scale factor and cannot supply a positive expanding branch of this model.

For $t>0$, $\dot{a}=n{t}^{n-1}>0$ and $\ddot{a}=n(n-1)t^{n-2}$. Consequently inflation occurs exactly when $\boxed{n>1}$. The case $n=1$ is coasting, and $0<n<1$ is decelerating. The Hubble slow-roll parameter is ${ϵ}_H=-\dot{H}/H^2=1/n$, giving the same acceleration criterion ${ϵ}_H<1$.

The Hubble parameter is $H=An{t}^{n-1}>0$, andFor $n=1$ this is positive at every $t>0$, so expansion is exponential and inflationary throughout. For $0<n<1$, acceleration begins only afterThe interval $0<t<t_{\ast }$ is decelerating, while $t=t_{\ast }$ is the transition. Thus the model admits inflation, but not arbitrarily near its specified time origin when $n<1$; choosing parameters can place the transition early. Equivalently ${ϵ}_H=(1-n)/(An{t}^n)$ becomes smaller than one at the same time.