Size of the Hubble volume in 4 billion years What will the size of the observable universe be in 4 billion years?
At present day, light would only reach us if it came from a galaxy within 13.5 billion1 light years, with anything outside that receding at a velocity higher than the speed of light (so never reaching us), what I understand as the Hubble volume.
In exactly 4 billion years, from the Milky Way-Andromeda galaxy, how far would an object be before it recedes at beyond the speed of light?
I imagine it's terribly simple to work out, but from what I understood, the Hubble constant changes (decreasing, it seems) so I'd need to work out what it is in 4 billion years time to accurately work out the size of the Hubble volume at said point in time.
I'm writing a hard sci-fi book about a human empire limited to the speed of light, so it's important (for me) that I work out the maximum size of the empire before it begins to lose contact with its periphery.
I checked out some of the duplicate/similar threads but they don't seem to answer the question. I understand if it can't be answered, but could anyone give a guesstimate? We won't be alive anyway when the time comes and it turns out to be wrong :)
 A: As I stated in my comment, our observable universe is much larger than the Hubble radius: we can observe galaxies that are receding from us faster than the speed of light. I refer to this post of mine and links therein for more info:
https://physics.stackexchange.com/a/63780/24142
Also, in the standard cosmological model, where the density of dark energy is treated as a cosmological constant, our observable universe keeps growing. That is, distant galaxies never disappear from our view, despite the fact that their recession velocities increase. That being said, their redshifts also increase and their brightness decreases, so they do become harder to observe. I hope this doesn't mess up your sci-fi story.
With this in mind, I have made my own cosmological calculator, so I can give you the numbers. Of course, the results depend on the values of the cosmological parameters, so take the precision with a grain of salt; I've used the values listed in the post that I linked to. The distances are proper distances (again see the link).
At the current cosmic age:
$$
\begin{align}
\text{time:}&&  t_0 &= 13.8\;\text{billion years}\\
\text{scale factor:}&& a(t_0) &= 1\\
\text{Hubble parameter:}&& H(t_0) &= 67.3\;\text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1}\\
\text{Hubble radius:}&& D_H(t_0) &= 14.5\;\text{billion lightyears}\\
\text{radius observable universe:}&&D_\text{ph}(t_0) &= 46.2\;\text{billion lightyears}\\
&&&
\end{align}
$$
In 4 billion years:
$$
\begin{align}
\text{time:}&&  t_1 &= 17.8\;\text{billion years}\\
\text{scale factor:}&& a(t_1) &= 1.3\\
\text{Hubble parameter:}&& H(t_1) &= 61.3\;\text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1}\\
\text{Hubble radius:}&& D_H(t_1) &= 16.0\;\text{billion lightyears}\\
\text{radius observable universe:}&&D_\text{ph}(t_1) &= 64.6\;\text{billion lightyears}\\
&&&
\end{align}
$$
A: I get a different figure for the radius observable universe:
The equation is:
$$
DC = {1 \over\sqrt{(Ω_R(1+z)^4)+(Ω_M(1+z)^3)+(Ω_K*(1+z)^2)+Ω_Λ*(1+z)^{3*(1+w)}}}, [z, inf]
$$
where:
DC = comoving distance
$Ω_R$ = Radiation (neutrino + photons) parameter
$Ω_M$ = Matter (dark + baryonic) parameter
$Ω_K$ = Curvature parameter
$Ω_Λ$ = Dark energy parameter
$z$ = Redshift (in this case the scale factor of $1.3$ comes to about $z=-0.23$)
and $w$ = equation of state
The final result is $DC*D_H(t_0)$ which is about 49.8455134 Gly
