A1.16

Statistical Physics and Cosmology
Part II, 2004

(i) Consider a homogeneous and isotropic universe with mass density ρ(t)\rho(t), pressure P(t)P(t) and scale factor a(t)a(t). As the universe expands its energy EE decreases according to the thermodynamic relation dE=PdVd E=-P d V where VV is the volume. Deduce the fluid conservation law

ρ˙=3a˙a(ρ+Pc2).\dot{\rho}=-3 \frac{\dot{a}}{a}\left(\rho+\frac{P}{c^{2}}\right) .

Apply the conservation of total energy (kinetic plus gravitational potential) to a test particle on the edge of a spherical region in this universe to obtain the Friedmann equation

(a˙a)2=8πG3ρkc2a2\left(\frac{\dot{a}}{a}\right)^{2}=\frac{8 \pi G}{3} \rho-\frac{k c^{2}}{a^{2}}

where kk is a constant. State clearly any assumptions you have made.

(ii) Our universe is believed to be flat (k=0)(k=0) and filled with two major components: pressure-free matter (PM=0)\left(P_{\mathrm{M}}=0\right) and dark energy with equation of state PQ=ρQc2P_{\mathrm{Q}}=-\rho_{\mathrm{Q}} c^{2} where the mass densities today (t=t0)\left(t=t_{0}\right) are given respectively by ρM0\rho_{\mathrm{M} 0} and ρQ0\rho_{\mathrm{Q} 0}. Assume that each component independently satisfies the fluid conservation equation to show that the total mass density can be expressed as

ρ(t)=ρM0a3+ρQ0,\rho(t)=\frac{\rho_{\mathrm{M} 0}}{a^{3}}+\rho_{\mathrm{Q} 0},

where we have set a(t0)=1a\left(t_{0}\right)=1.

Now consider the substitution b=a3/2b=a^{3 / 2} in the Friedmann equation to show that the solution for the scale factor can be written in the form

a(t)=α(sinhβt)2/3a(t)=\alpha(\sinh \beta t)^{2 / 3}

where α\alpha and β\beta are constants. Setting a(t0)=1a\left(t_{0}\right)=1, specify α\alpha and β\beta in terms of ρM0,ρQ0\rho_{\mathrm{M} 0}, \rho_{\mathrm{Q} 0} and t0t_{0}. Show that the scale factor a(t)a(t) has the expected behaviour for an Einstein-de Sitter universe at early times (t0)(t \rightarrow 0) and that the universe accelerates at late times (t)(t \rightarrow \infty).

[Hint: Recall that dx/x2+1=sinh1x\int d x / \sqrt{x^{2}+1}=\sinh ^{-1} x.]