# Two Robertson-Walker observers, at what time will a light signal be received?

Here is a question I have that is inspired by this question here.

The spacetime metric of a radiation-filled, spatially flat ($k = 0$) Robertson-Walker universe is given by$$ds^2 = - dT^2 + T[dx^2 + dy^2 + dz^2].$$Consider two "Robertson-Walker observers" [i.e., observers with $4$-velocity $(\partial/\partial T)^a$], the first one of which has spatial coordinates $(0, 0, 0)$ and the second of which has spatial coordinates $(x, 0, 0)$. At time $T = T_1$, the first observer sends a light signal to the second. At what time, $T_2$, will this signal be received?

Although the radiation-dominated (RD) era is long in comparison to the matter-dominated (MD) and $\Lambda$-dominated ($\Lambda$D) eras, it is nice to have an answer which can be adapted easily for any cosmological era. If we assume that the Universe is permeated by a perfect fluid we may use the equation of state

$$w = \frac{P}{\rho},$$

where $P$ is the pressure and $\rho$ the energy density. The two Friedmann equations (or the conservation of the stress-energy tensor $\nabla_{\mu} T^{\mu \nu}=0$) give us

$$\dot{\rho} + 3H\rho(1+3w) = 0, \label{F3}$$

which can be solved for $\rho$ in terms of the scale factor $a(t)$ as

$$\rho = \rho_0 \Big(\frac{a}{a_0}\Big)^{-3(1+w)},$$

where $a_0$ is the scale factor today (and we will set $a_0 = 1$ from now on) and $\rho_0$ is the total energy density of the Universe today. Plugging this back into the first Friedmann equation then gives $a(t)$ as

$$a(t) = At^{\frac{2}{3(1+w)}},$$

where $A$ is a horrible constant which I have calculated to be $A = (8 \pi G/3)^{1/3(1+w)}$ and I have assumed that at some 'initial time' $t_i = 0$, the value of the scale factor was $a_i = 0$.

Now we consider the photon. As in my answer here, the photon follows a radial geodesic in FRW spacetime. In a flat Universe as you specified, we have

$$ds^2 = -dt^2 + dr^2 = 0$$

for a photon. We can define an origin $(0,0,0)$ and from there send a photon to any point at radial distance $r = x$. You specified $(x,0,0)$ but as the Universe is homogeneous and isotropic, any such point will yield the same answer. Using the photon line element, this is

$$\int^x_0 dr = \int^{T_2}_{T_1} \frac{dt}{a(t)}$$

Now we use the expression for $a(t)$ in terms of $t$ and $w$ which we derived earlier, evaluate the integrals in terms of $x$ and $t$, and set $T_1 = 0$ so we can see what is going on, arriving at

$$x = \frac{3(1+w)}{A(3w+1)}T_2^{1-\frac{2}{3(1+w)}},$$

with the previously-stated value of $A$.

RD era ($w = 1/3$): Substituting this value of $w$ we find

$$x \propto \sqrt{T_2} \implies T_2 \propto x^2,$$

so given an infinite amount of time, we can send a signal out to infinity.

MD era ($w = 0$): Substituting this value of $w$ we find

$$x \propto T_2^{1/3} \implies T_2 \propto x^3,$$

so again we can send a signal to infinity but it will take $O(x)$ longer than in the RD era. This is because the Universe is expanding faster, as $a(t) \propto t^{2/3}$ rather than $a(t) \propto t^{1/2}$ in the RD era.

$\Lambda$D era ($w \rightarrow -1$): For this case it is better to write $x$ in terms of $a$, in which case we obtain

$$x \propto a^{-1}.$$

This means that as the Universe grows in the dark-energy regime, its expansion accelerates such that the region over which we can communicate shrinks! If we send out a photon now, the maximum distance it can reach in an infinite time will be greater than if we send out a photon tomorrow. This is the phenomenon of the shrinking Hubble sphere and it means that if the Universe continues to be dominated by dark energy as it is now, then the observable Universe will shrink until we can see only the closest astronomical objects.

In the geometrical optics approximation light ray is represented by a null geodesic. Therefore you only need to find a null geodesic connecting points $(t_0,0,0,0)$ and $(t_1,x,0,0)$ for some $t_1$ (and this condition will determine $t_1$ uniquely). This is probably quite easy to do directly in this case, but in general for investigation of null curves in FLRW spacetime conformal time defined by $\mathrm d \eta = \frac{\mathrm d T}{a(T)}$ (with $a(T)=T^{1/2}$ in your case) is especially convenient.

Therefore you need to calculate the future light cone

$$LC_{proper}=\int_{a(t_0)}^{a(t_1)} \frac{c\cdot a(t_1)}{\alpha^2\cdot H(\alpha )} \, \text{d}\alpha$$

In comoving coordinates you divide that by the scale factor of the time at absorption

$$LC_{comoving}=\frac{LC_{proper}}{a(t_1)}$$

with H as the Hubble parameter

$$H(a)=H_0\cdot \sqrt{\frac{\Omega_R}{a^4}+\frac{\Omega_M}{a^3}+\frac{\Omega_K}{a^2}+\Omega_{\Lambda} }$$

and c the speed of light, t0 the time of emission and t1 the time of absorption.

If you neglect the radiation density you can use

$$a(t)=\sqrt[\frac{3}{2}]{\sqrt{\frac{\Omega_M}{\Omega_{\Lambda} }} \cdot \sinh(\frac{3\cdot H_0\cdot \sqrt{\Omega_{\Lambda} }\cdot t}{2} )}$$

which simplifies the equation a little bit and gives good approximations if you don't go too far back in time , but the light cone will still be an integral with no explicit inverse function.

This leads to a numerical computation with no analytical solution. The exact computation is a little long if one takes matter, dark energy and radiation into account, so forgive that I won't translate everything into Latex.

If I take the cosmological parameters from the Planck mission and calculate for example how long it will take a photon to travel to a distance which is now located 1 Gigalightyear away, light will take 1.036 Gigayears to get there. If the distance is 10 Gigalightyears away today, light will take 15.736 Gigayears until it gets to that comoving coordinate:

Since the Hubble parameter is evolving with time it not only depends on the distance but also on the time when the photon is emitted.

Unfortunately this all has to be solved numerically, so I can't give you an explicit solution for $t_1(LC_{comoving})$, but at least I can show you how to solve for that solution with a computer.

I hope this helps nevertheless, if something is unclear with the code feel free to ask. Maybe it also helps to see the spacetime diagrams here and here.