= Solution
{c}
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 define
$$
H_\Lambda=\sqrt{\frac{8\pi G\varrho_0}{3}}>0.
$$
The <Friedmann equation> and the original <Friedmann acceleration equation> become
$$
\dot a^2+k=H_\Lambda^2a^2,\qquad \ddot a=H_\Lambda^2a.
$$
Solve the second equation as $a=Ae^{H_\Lambda t}+Be^{-H_\Lambda t}$. The first equation then requires $4H_\Lambda^2AB=k$. Choosing the time origin separately in each solution gives all positive branches:
$$
\boxed{
\begin{array}{c|c|c}
k&a(t)&\text{domain and finite past zero on the expanding branch}\\ \hline
0&A_0e^{\pm H_\Lambda(t-t_c)},\ A_0>0&-\infty<t<\infty;\ \text{none}\\
+1&H_\Lambda^{-1}\cosh[H_\Lambda(t-t_b)]&-\infty<t<\infty;\ \text{none}\\
-1&H_\Lambda^{-1}\sinh[H_\Lambda(t-t_B)]&t>t_B;\ a\to0\text{ at }t_B
\end{array}}
$$
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 $\ddot a=H_\Lambda^2a$ 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 $T>0$.
\Image[/past-exam-of-the-general-relativity-ii-course-of-the-university-of-oxford/2015-q3-vacuum-scale-factors.png]
{title=Scale factors for positive vacuum energy: exponential flat expansion, a closed bounce, and open expansion from a finite-time coordinate boundary}
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.
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