# Can the luminosity distance be greater than the particle horizon distance?

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 $z > z_{\text{max}}$ ?

• Please, could you elaborate more ? Why $\mathcal{D}_L$ can't be compared with $\mathcal{D}_P$ ? I know that the former isn't a real physical distance, while the later is a real one. But then how could we give a physical sense to the luminosity distance ? – Cham Oct 11 '16 at 0:49
• A distance is a distance. Both notions have the same units and same interpretation (i.e. a "distance") and reduces to the same thing at small values (i.e. small $z$), so they are comparable. Why they can't be compared ? – Cham Oct 11 '16 at 10:01
• An example : The Hubble radius $\mathcal{D}_{H} = H^{- 1}$ is another notion of "distance" which can be greater than the particle horizon distance (in some models), and it can be compared with the particle horizon. – Cham Oct 11 '16 at 10:08
• I think I've found a partial solution to my puzzle : a particle CAN in fact be at a farther distance than the particle horizon and yet still be visible in the future. So there is no problem having $\mathcal{D}_L > \mathcal{D}_P$. In my query, I have confused the particle horizon with the event horizon. So in the cases of the dust and radiation universes, there's no problem since there is no event horizon. However, the same exercice above could be done for an empty universe with a cosmological constant, without a particle horizon but with an event horizon. Then $D_L > D_P$ is weird. – Cham Oct 11 '16 at 10:33