Past exam of the general relativity and cosmology course of the University of Oxford / 2013 / 4

Throughout this question use natural units with $c=1$. For zero potential and nonzero kinetic energy, the homogeneous canonical scalar field has $P=\rho ={\dot{\phi }}^2/2$, so $\boxed{w=1}$. Its equation of motion becomes $d(a^3\dot{\phi })/dt=0$, giving $\dot{\phi }\propto a^{-3}$ and thereforeThe same scaling follows from cosmological energy conservation with constant equation-of-state parameter $w=1$. If the field is exactly constant, both $P$ and $\rho$ vanish and their ratio is undefined; that is the separate empty solution.

Write $M=M_{\mathrm{Pl}}$ for the reduced Planck mass. For a nontrivial power law with $\lambda \ne 0$, $H=p/t$, $\dot{\phi }=BM/t$, and $V=V_0(t/t_0{)}^{-\lambda B}$. Matching its time dependence to the kinetic and Friedmann equation terms requires $\lambda B=2$. The scalar field equation of motion and the Friedmann equation then reduce toSolving gives the exponential-potential scalar-dominated power-law solutionChoose $t>0$ and the positive square root for $t_0$. The existence conditions for exponential-potential scaling matter: if $V_0>0$, a finite real $t_0$ requires $0<{\lambda }^2<6$, rather than an inclusive bound on signed $\lambda$. At ${\lambda }^2=6$, nonzero $V_0$ is incompatible with this logarithmic ansatz. If $V_0<0$, the corresponding scaling solution instead requires ${\lambda }^2>6$. The sign of $\lambda$ changes the field's rolling direction, not $p$. For $\lambda =0$ the potential is constant and this finite-$p$ ansatz does not apply; with positive $V_0$, a constant field gives de Sitter spacetime expansion.

The scaling solution has $\rho =3M^2{p}^2/t^2$ and $\rho +P={\dot{\phi }}^2=4M^2/({\lambda }^2{t}^2)$. ThereforeMore generally any real canonical scalar field with positive total energy density obeys $\rho +P={\dot{\phi }}^2\ge 0$, so it cannot have $w<-1$. A negative potential does not reverse this conclusion.

The proper particle horizon is $a(t){\int }_0^{t}d{t}^{\prime }/a(t^{\prime })$. For $a\propto t^p$ the lower-limit integral diverges exactly when $p\ge 1$. Thus, assuming the same power law extends to the initial boundary, there is no finite particle horizon for $\boxed{0<{\lambda }^2\le 2}$. For the strict inequality, $p>1$, the proper cosmological event horizon and present age areAt ${\lambda }^2=2$, $a\propto t$ has an infinite particle-horizon integral but also an infinite future event-horizon integral, so no finite event horizon exists. The age remains $H_0^{-1}$. The $\lambda =0$ constant-potential case is separate: its flat de Sitter patch has event-horizon radius $H_0^{-1}$ and no finite-age power-law Big Bang.

For a single constant-$w$ perfect fluid in a flat expanding FLRW metric model, cosmological energy conservation gives $\rho ={\rho }_0{a}^{-3(1+w)}$ and $H(z)=H_0(1+z{)}^{3(1+w)/2}$. With $a_0=1$, the luminosity distance follows by integrating the radial null geodesic:The constant-equation-of-state luminosity distance is consequentlyThis includes $w=-1$ regularly, giving $D_L=z(1+z)/H_0$. Restoring conventional units multiplies the distance formulas by $c$. Expanding about $z=0$ givesThe second result also follows directly from the Friedmann acceleration equation. The expansion agrees with the general coefficient $(1-q_0)/2$ of $z^2$.

For phantom energy, $w<-1$, put $n=3(1+w)/2<0$. Integration of $\dot{a}=H_0{a}^{1-n}$ with $a(t_{\mathrm{now}})=1$ givesBecause $1/n<0$, the scale factor diverges at the Big Rip timeThis is a property of the stipulated constant-$w$ fluid model; it is not realizable by the positive-density canonical scalar of the preceding parts.