Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 67 / 2

In a spatially flat FLRW metric, a radial null geodesic satisfies $d\chi =cdt/a(t)$. Normalize $a_0=1$. The comoving particle horizon at time $t$ is therefore $r_h=c{\int }_0^{t}d{t}^{\prime }/a(t^{\prime })$, using the initial singularity as the time origin. The cosmological redshift relation $1+z=1/a$ givesFor matter and a cosmological constant in a spatially flat model, the Friedmann equation gives $H(z)=H_0[{\Omega }_{m,0}(1+z{)}^3+{\Omega }_{\Lambda ,0}{]}^{1/2}$. The initial singularity corresponds to $z\to \infty$, so reversing the integration limits yieldsThis is a comoving particle horizon, not a distance to the source currently observed at redshift $z$, and not a future-directed cosmological event horizon. The corresponding proper horizon radius at that epoch is $r_h(z)/(1+z)$.

For an Einstein-de Sitter universe, the Hubble parameter is $H(z)=H_0(1+z{)}^{3/2}$. The comoving particle horizon becomesThusIt tends to zero towards the initial singularity and increases without bound as $z\to -1$ in the infinite future. At the present time it is $2c/H_0=3c{t}_0$, since the Einstein-de Sitter universe has age $t_0=2/(3H_0)$.

Use the stipulated common normalization $A=H_0{\Omega }_{m,0}$ and plot $R=A{r}_h/c$. The Einstein-de Sitter universe has $H_{0,E}=A$, whereas the $({\Omega }_{m,0},{\Omega }_{\Lambda ,0})=(0.3,0.7)$ model has $H_{0,\Lambda }=A/0.3$. Their comoving particle horizons are thereforeBoth decrease with increasing $z$, or increase with cosmic time. The Einstein-de Sitter universe curve diverges at $z=-1$ and has $R_E(0)=2$, $R_E(2)=2/\sqrt{3}$. The matter-plus-vacuum curve remains finite at $z=-1$: the integrand approaches a constant there, and its high-redshift tail is integrable. Moreover it lies below the Einstein-de Sitter curve for this particular normalization, since, with $u=1+z^{\prime }>0$,The high-redshift ratio of the two curves tends to $\sqrt{0.3}$, not one. Holding $H_0\sqrt{{\Omega }_{m,0}}$ fixed would give a different comparison; it is $H_0{\Omega }_{m,0}$ that the original question specifies.

A positive cosmological constant eventually makes the scale factor grow exponentially. The future integral ${\int }_t^{\infty }cd{t}^{\prime }/a(t^{\prime })$ then converges, so the remaining conformal time is finite. The comoving particle horizon approaches a finite limiting visibility radius; it does not shrink. The remaining distance to that limit is the comoving radius of the cosmological event horizon. Thus a signal emitted sufficiently far away today can never reach us, even though the proper particle-horizon radius continues to grow. This is the total conformal lifetime of a flat matter-Lambda universe.

Normalize the Einstein-de Sitter universe by $a(t_0)=1$, giving $a(t)=(t/t_0{)}^{2/3}$. A photon emitted inward today at the current comoving particle horizon starts at ${\chi }_0=3c{t}_0$. By time $t$ it has traversed the comoving distanceAt arrival at Earth this equals ${\chi }_0$, hence $(t/t_0{)}^{1/3}=2$ andThe emitting point is at the horizon location today, not a photon already arriving today from the initial singularity. The calculation of photon arrival from an Einstein-de Sitter particle horizon follows the future inward null geodesic from that event.