Paper 1, Section II, E

Cosmology
Part II, 2011

A homogeneous and isotropic universe, with scale factor aa, curvature parameter kk, energy density ρ\rho and pressure PP, satisfies the Friedmann and energy conservation equations

H2+kc2a2=8πG3ρρ˙+3H(ρ+P/c2)=0\begin{aligned} &H^{2}+\frac{k c^{2}}{a^{2}}=\frac{8 \pi G}{3} \rho \\ &\dot{\rho}+3 H\left(\rho+P / c^{2}\right)=0 \end{aligned}

where H=a˙/aH=\dot{a} / a, and the dot indicates a derivative with respect to cosmological time tt.

(i) Derive the acceleration equation

a¨a=4πG3(ρ+3P/c2)\frac{\ddot{a}}{a}=-\frac{4 \pi G}{3}\left(\rho+3 P / c^{2}\right)

Given that the strong energy condition ρc2+3P0\rho c^{2}+3 P \geqslant 0 is satisfied, show that (aH)2(a H)^{2} is a decreasing function of tt in an expanding universe. Show also that the density parameter Ω=8πGρ/(3H2)\Omega=8 \pi G \rho /\left(3 H^{2}\right) satisfies

Ω1=kc2a2H2\Omega-1=\frac{k c^{2}}{a^{2} H^{2}}

Hence explain, briefly, the flatness problem of standard big bang cosmology.

(ii) A flat (k=0)(k=0) homogeneous and isotropic universe is filled with a radiation fluid (wR=1/3)\left(w_{R}=1 / 3\right) and a dark energy fluid (wΛ=1)\left(w_{\Lambda}=-1\right), each with an equation of state of the form Pi=wiρic2P_{i}=w_{i} \rho_{i} c^{2} and density parameters today equal to ΩR0\Omega_{R 0} and ΩΛ0\Omega_{\Lambda 0} respectively. Given that each fluid independently obeys the energy conservation equation, show that the total energy density (ρR+ρΛ)c2\left(\rho_{R}+\rho_{\Lambda}\right) c^{2} equals ρc2\rho c^{2}, where

ρ(t)=3H028πGΩR0a4(1+1ΩR0ΩR0a4)\rho(t)=\frac{3 H_{0}^{2}}{8 \pi G} \frac{\Omega_{R 0}}{a^{4}}\left(1+\frac{1-\Omega_{R 0}}{\Omega_{R 0}} a^{4}\right)

with H0H_{0} being the value of the Hubble parameter today. Hence solve the Friedmann equation to get

a(t)=α(sinhβt)1/2a(t)=\alpha(\sinh \beta t)^{1 / 2}

where α\alpha and β\beta should be expressed in terms ΩR0\Omega_{R 0} and ΩΛ0\Omega_{\Lambda 0}. Show that this result agrees with the expected asymptotic solutions at both early (t0)(t \rightarrow 0) and late (t)(t \rightarrow \infty) times.

[Hint: dx/x2+1=arcsinhx\int d x / \sqrt{x^{2}+1}=\operatorname{arcsinh} x.]