The Hubble radius is the proper distance $R_H (t)=c H^{-1}(t)$. The sphere of this radius is called the Hubble sphere. From definition, the Hubble recession "velocity" of a comoving observer on the Hubble sphere is $v(R_H)=H R_H =c$ and equal to the speed of light $c$. This is true, of course, at the same moment of time $t$, for which $H (t)$ is taken.
16.6.1Problem 16.6.1
All galaxies inside the Hubble sphere recede subluminally (slower than light) and all galaxies outside recede superluminally (faster than light). This is why the Hubble sphere is sometimes called the "photon horizon". Does this mean that galaxies and their events outside the photon horizon are permanently hidden from the observer's view? If that were so, the photon horizon would also be an event horizon. Is this correct?
solution
At first glance, observation of galaxies beyond the Hubble sphere appears to be an unsolvable problem. As Eddington in 1933 wrote: ``Light is like a runner on an expanding track with the winning-post receding faster than he can run. But never give up!
In most models of the Universe the Hubble parameter $H$ is not constant. In a decelerating universe the Hubble radius $R_H$ increases with time. Moreover, the Hubble sphere expands faster than the Universe, so that the edge of the Hubble sphere -- the photon horizon -- overtakes the receding galaxies. Light rays outside the Hubble sphere moving toward us may therefore eventually be overtaken by the photon horizon. They will then be inside the Hubble sphere and will at last start approaching us. Eddington's runner sees the winning-post receding, but he must keep running and not give up; the expanding track is slowing down and eventually the winning-post will be reached.
16.6.2Problem 16.6.2
Show, by the example of static universe, that the Hubble sphere does not coincide with the boundary of the observable Universe.
solution
The observable Universe is limited by the particle horizon, and if the Hubble sphere and the observable Universe were the same, the latter in a static Universe would be infinitely large ($H_0 =0$, so $R_H =\infty$). But static Universes of finite age have particle horizons at finite distance. So, the Hubble sphere cannot be the boundary of the observable Universe.
16.6.3Problem 16.6.3
Estimate the ratio of the volume enclosed by the Hubble sphere to the full volume of a closed Universe.
16.6.4Problem 16.6.4
Show that in a spatially flat Universe ($k=0$), in which radiation is dominating, the particle horizon coincides with the Hubble radius.
16.6.5Problem 16.6.5
Find the dependence of comoving Hubble radius $R_H /a$ on scale factor in a flat Universe filled with one component with the state equation $\rho=w p$.
16.6.6Problem 16.6.6
Express the comoving particle horizon through the comoving Hubble radius for the case of domination of a matter component with state parameter $w$.
16.6.7Problem 16.6.7
Show that
\begin{equation}
\frac{dR_H }{dt}=c(1+q),\label{dRHdt}
\end{equation}
where
\begin{equation}
q=-\frac{\ddot{a}/a}{H^{2}}
\end{equation}
is the deceleration parameter.
solution
As $R_H =H^{-1}=a/\dot{a}$, the answer is obtained after one differentiation.
16.6.8Problem 16.6.8
Show that
\begin{equation}
\frac{dL_p}{dt}=1+\frac{L_p}{R_H}.
\end{equation}
solution
Using the definition $L_p =a \int^t dt a^{-1}$, we obtain this after differentiation.
16.6.9Problem 16.6.9
Show that in the Einstein-de Sitter Universe the relative velocity of the Hubble sphere and galaxies on it is equal to $c/2$.
16.6.10Problem 16.6.10
Find $a(t)$ in a universe with constant positive deceleration parameter $q$.
solution
Using
\begin{equation}
\dot{H}=\frac{\ddot{a}}{a}-\frac{\dot{a}^2}{a^2},
\end{equation}
and the condition $q=const$, we get
\begin{equation}
\frac{d}{dt}H^{-1}=1+q ,
\end{equation}
thus $H^{-1}=(1+q)t$ and $a\sim t^{n}$ with
\begin{equation}
n=\frac{1}{1+q}.
\end{equation}
Also $q>0$ means $n<1$.
16.6.11Problem 16.6.11
Show that in universes of constant positive deceleration $q$, the the ratio of distances to the particle and photon horizons is $1/q$.
16.6.12Problem 16.6.12
Show that the Hubble sphere becomes degenerate with the particle horizon at $q=1$ and with the event horizon at $q=-1$.
16.6.13Problem 16.6.13
Show that if $q$ is not constant, comoving bodies can be inside and outside of the Hubble sphere at different times. But not so for the observable universe; once inside, always inside.
solution
From (\ref{dRHdt}) we see that recession velocity of the Hubble sphere can be lesser or greater than $c$, depending on the sign of $q$. As recession velocity of matter on it is exactly $c$, be definition, this means that matter can cross the Hubble sphere in both directions at different times.
"Horizons are like membranes; the photon horizon acts as a two-way membrane (comoving bodies can cross in both directions depending on the value of $q$), and the particle horizon acts like a one-way membrane (comoving bodies always move in and never out)." E. Harrison, Science of the Universe.