 Solution

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.