While writing my notes on cosmology, I found this little puzzle, which isn't described in any of my books on general relativity.
Consider a flat space ($k = 0$) filled with dust-like matter, of energy density $\rho \propto a^{- 3}$, without any cosmological constant ($\Lambda = 0$). It's easy to find the cosmological scale factor that solves the Friedmann-Lemaître equations : $a(t) \propto t^{2/3}$. Now, the particle horizon distance is this : $$\tag{1} \mathcal{D}_P(t_0) = a(t_0) \int_0^{t_0} \frac{1}{a(t)} \; dt = 3 \, t_0. $$ The luminosity distance is defined by $\mathcal{D}_L = \sqrt{L/4\pi F}$, where $F$ is the bolometric flux at the observer's location and $L$ is the absolute luminosity of the light source. It can be expressed exactly as a function of the redshift parameter $z \equiv a(t_0)/a(t_e) - 1$, where $t_e$ is the time of light emission. We can write $t_e = t_0 - \delta t$, where $\delta t$ is the propagation time of the light signal. For the dust universe introduced above, we have $$\tag{2} 1 + z = \frac{a(t_0)}{a(t_e)} \quad \Rightarrow \quad 1 - \frac{\delta t}{t_0} = \frac{1}{(1 + z)^{3/2}}. $$ For $k = 0$, the luminosity distance can be expressed exactly as this : $$\tag{3} \mathcal{D}_L(t_0, z) = (1 + z) \, a(t_0) \int_{t_e}^{t_0} \frac{1}{a(t)} \; dt. $$ For the dust universe, this formula gives this : $$\tag{4} \mathcal{D}_L(t_0, z) = \sqrt{1 + z} \, \big( \sqrt{1 + z} - 1 \big) \, 3 \, t_0 \equiv \sqrt{1 + z} \, \big( \sqrt{1 + z} - 1 \big) \, \mathcal{D}_P. $$ So the question is
Can the luminosity distance be larger than the particle horizon distance ? Does it make sense to have $\mathcal{D}_L > \mathcal{D}_P$ ? If not, what is the maximal value of the redshift parameter $z$ ?
Equation (4) gives $\mathcal{D}_L \le \mathcal{D}_P$ if $z \le \tfrac{1}{2}(1 + \sqrt{5}) \approx 1.618$ (amazingly the Golden ratio !).
I didn't knew there was a maximal value for the redshift parameter!
Equation (2) then gives $\delta t_{\text{max}} = (3 - \sqrt{5}) \, t_0 \approx 0.764 \, t_0$, instead of $\delta t_{\text{max}} = t_0$ for $z \rightarrow \infty$ ($t_0$ is the age of the universe).
If we do the same exercice in the case of pure radiation ($\rho \propto a^{- 4}$, $a(t) \propto t^{1/2}$), we get $z \le 1$ and $\delta t_{\text{max}} = \frac{3}{4} \; t_0 = 0.75 \, t_0$.
Are the calculations shown above making sense ?
What is the interpretation of a redshift $z > z_{\text{max}}$ ?
