Past exam of the general relativity II course of the University of Oxford / 2015 / 3 / e

For the equation of state $p=-\varrho$, the cosmological continuity equation gives $a\dot{\varrho }=0$. Hence $\varrho ={\varrho }_0$ is constant. First take the physically intended positive energy density and defineThe Friedmann equation and the original Friedmann acceleration equation becomeSolve the second equation as $a=A{e}^{H_{\Lambda }t}+B{e}^{-H_{\Lambda }t}$. The first equation then requires $4H_{\Lambda }^{2}AB=k$. Choosing the time origin separately in each solution gives all positive branches:For $k=-1$, there is also the time reverse $a=H_{\Lambda }^{-1}\sinh [H_{\Lambda }(t_B-t)]$ on $t<t_B$, which contracts to a future zero. For $k=+1$, the cosmological bounce joins contraction to expansion at the positive minimum $a(t_b)=H_{\Lambda }^{-1}$. Retaining $\ddot{a}=H_{\Lambda }^{2}a$ at the minimum excludes a spurious permanently static solution that would satisfy the first-order constraint alone.

These are vacuum-energy FLRW solutions: the expanding flat branch, closed slicing of de Sitter spacetime, and open slicing of de Sitter spacetime. To compare them at the same energy density, the figure sets $T=H_{\Lambda }(t-t_c)$, takes $A_0=H_{\Lambda }^{-1}$, and aligns the open boundary and closed minimum at $T=0$. Thus the three plotted functions are $H_{\Lambda }a=e^T$, $\cosh T$, and $\sinh T$, the last only for $T>0$.

Only the expanding $k=-1$ branch has a finite past zero. The expanding $k=0$ branch tends to zero only as $t\to -\infty$, and the $k=+1$ branch is bounded below. All three describe portions of de Sitter spacetime, with Ricci scalar $R=12H_{\Lambda }^2$ and Kretschmann scalar $24H_{\Lambda }^4$. In particular the $k=-1$ zero is a boundary of that slicing, not a curvature singularity. The inequality in part (d) does not apply, because $\varrho +3p=-2{\varrho }_0<0$.

For zero energy density, the remaining solutions are $a=A_0>0$ for $k=0$, and the expanding or contracting Milne model $a=\pm (t-t_B)>0$ for $k=-1$; no positive real solution exists for $k=+1$. If negative energy density is admitted mathematically, write ${\alpha }^2=-8\pi G{\varrho }_0/3>0$. Only $k=-1$ is possible, with $a={\alpha }^{-1}\sin [\alpha (t-t_B)]$ for $0<\alpha (t-t_B)<\pi$, an open slicing of anti-de Sitter spacetime. These limiting and negative-density cases do not alter the positive-density comparison above.

Take the expanding flat branch $a(t)=A{e}^{H_{\Lambda }t}$, with $A>0$. A radial null geodesic in FLRW spacetime satisfies $d\rho /dt=\pm 1/a(t)$. For the incoming past light cone of the spacetime event $(t_0,\rho =0)$, integration giveswith arbitrary angular direction. Equivalently, conformal time $\eta =-1/(H_{\Lambda }a)$ gives $\rho ={\eta }_0-\eta$.

Let the reception time tend to infinity. The largest comoving radial distance from which a signal emitted at time $t$ can reach the central comoving observer isThus the cosmological event horizon has constant proper distance in cosmology:An emitted spacetime event is seen at a finite time if and only if $a(t)\rho <H_{\Lambda }^{-1}$. Equality would require $t_0=\infty$, while an event with $a(t)\rho >H_{\Lambda }^{-1}$ never reaches the observer. This is a bound on events at their emission time; a fixed comoving observer can have earlier visible events and later invisible ones.

The boundary really is a null hypersurface. For $f(t,\rho )=\rho -[H_{\Lambda }a(t){]}^{-1}$, its normal has components ${\partial }_{t}f=1/a$ and ${\partial }_{\rho }f=1$, soThe finite future integral, and hence this observer event horizon, uses the expanding branch. For the contracting flat solution $a=A{e}^{-H_{\Lambda }t}$, the same integral diverges and there is no future observer event horizon of this kind; the static zero-density solution likewise has none.