 Solution

In a spatially flat FLRW metric, a radial null geodesic satisfies $d\chi /dt=\pm c/a(t)$. For a source and observer at fixed comoving separation, $\chi ={\int }_{t_e}^{t_0}cdt/a(t)$. Applying this to successive wave crests gives $d{t}_0/a(t_0)=d{t}_e/a(t_e)$ and therefore the cosmological redshift $1+z=a(t_0)/a(t_e)$.

Each photon's energy is reduced by $1+z$, and the arrival interval is increased by $1+z$. The received power crossing the observation sphere is consequently $L/(1+z{)}^2$. Its physical area is $4\pi a(t_0{)}^2|x{|}^2$, so the radiative flux isThis is the luminosity distance relation with $d_L=(1+z)a(t_0)|x|$. The Euclidean area formula presumes spatial flatness; in a curved FLRW metric the transverse comoving distance replaces $|x|$.