The problems of this section need basic understanding of Friedman equations, definitions of proper, comoving and conformal coordinates, the cosmological redshift formula and simple cosmological models (see Chapters 2 and 3).
Let us make our definitions a little more strict.
A particle horizon, for a given observer $A$ and cosmic instant $t_0$ is a surface in the instantaneous three-dimensional section $t=t_0$ of space-time, which divides all comoving particles [1] into two classes: those that have already been observable by $A$ up to time $t_0$ and those that have not.
An event horizon, for a given observer $A$, is a hyper-surface in space-time, which divides all events into two non-empty classes: those that have been, are, or will be observable by $A$, and those that are forever outside of $A$'s possible powers of observation. It follows from definition that event horizon, and its existence, depend crucially on the observer's (and the whole Universe's) future as well as the past: thus it is said to be an essentially global concept. It is formed by null geodesics.
The following notation is used hereafter: $L_p (t_0)$ is the proper distance from observer $A$ to its particle horizon, measured along the slice $t=t_0$. For brevity, we will also call this distance simply "the particle horizon in proper coordinates", or just "particle horizon". The corresponding comoving distance $l_p$ is the particle horizon in comoving coordinates.
Likewise, $L_e$ is the proper distance from an observer to its event horizon (or, rather, its section with the hypersurface $t=t_0$), measured also along the slice $t=t_0$. It is called "space event horizon at time $t_0$", or just the event horizon, for brevity. The respective comoving distance is denoted $l_e$.
16.2.1Problem 16.2.1
The proper distance $D_p (t_0)$ between two comoving observers is the distance measured between them at some given moment of cosmological time $t=t_0$. It is the quantity that would be obtained if all the comoving observers between the given two measure the distances between each other at one moment $t=t_0$ and then sum all of them up. Suppose one observer detects at time $t_{0}$ the light signal that was emitted by the other observer at time $t_e$. Find the proper distance between the two observers at $t_0$ in terms of $a(t)$.
solution
By definition, measuring distance along the slice $t=const$, we have for some given time $t_0$
\begin{equation}
D_p (r, t_0)=\int |ds| = a_0 \int_0^r \frac{dr}{\sqrt {1 - k r^2} },
\quad {a_0} \equiv a (t_0)
\end{equation}
On the other hand, along the worldline of the light ray connecting the two observers $ds=0$, so we have
\[\frac{dt}{a} = \frac{dr}{\sqrt {1 - k r^2 }}.\]
Integrating along the worldline from the time of emission $t_e$ to the time of detection $t_0$, we get
\[\int_{t_e}^{t_0} \frac{dt'}{a(t')}
= \int_0^r \frac{dr}{\sqrt {1 - k r^2}} = \frac{D_p}{a_0},\]
Therefore the proper distance between two observers in an expanding Universe is
\begin{equation}
D_p (t_e, t_0)= a_0\int_{t_e}^{t_0} \frac{dt'}{a(t')}.
\end{equation}
16.2.2Problem 16.2.2
Show, that the proper distance $L_p$ to the particle horizon at time $t_0$ is
\begin{equation}
L_p (t_0)=\lim\limits_{t_e \to 0}D_p (t_e ,t_0). \label{LpDp}
\end{equation}
solution
As define above, particle horizon is the proper distance between the observer that receives the light signal at present and the comoving particle that emitted this light at the very beginning of the Universe (which may correspond to $t\to -\infty$). Thus (\ref{LpDp}).
16.2.3Problem 16.2.3
The past light cone of an observer at some time $t_0$ consists of events, such that light emitted in each of them reaches the selected observer at $t_0$. Find the past light cone's equation in terms of proper distance vs. emission time $D_{plc} (t_e)$. What is its relation to the particle horizon?
solution
\begin{equation}
D_{plc}(t_e ,t_0) =a(t_e) \int\limits_{t_e}^{t_0} \frac{c\,dt}{a(t)}.
\label{Dplc}
\end{equation}
The corresponding comoving distance $d(t_e ,t_0)=D(t_e, t_0)/a(t_e)$ at $t_e \to t_{in}$ ($t_{in}$ is the time of creation singularity or infinite past, whichever is realized) gives us the observer that only now (at $t_0$) just becomes observable, thus determining the particle horizon at $t_0$:
\[L_p (t_0)=a(t_0) d_{plc}(t_{in}, t_0).\]
16.2.4Problem 16.2.4
The simplest cosmological model is the one of Einstein-de Sitter, in which the Universe is spatially flat and filled with only dust, with $a(t)\sim t^{2/3}$. Find the past light cone distance $D_{plc}$ (\ref{Dplc}) for Einstein-de Sitter.
solution
\begin{equation}
D_{plc}(t_e, t_0)=3c (t_e^{2/3}t_0^{1/3}-t_e ).
\end{equation}
16.2.5Problem 16.2.5
Demonstrate that in general $D_{plc}$ can be non-monotonic. For the case of Einstein-de Sitter show that its maximum -- the maximum emission distance -- is equal to $8/27 L_H$, while the corresponding redshift is $z=1.25$.
solution
From $dD_{plc}/dt_e =0$ we see that maximum exists and lies at
\begin{equation}
\frac{t_e}{t_0}=\frac{8}{27},\qquad D_{plc}^{max}=\frac{4}{9}ct_0 .
\end{equation}
The redshift of light signal emitted at maximum emission distance is then
\begin{equation}
z=\frac{a(t_0)}{a(t_e)}-1=\Big(\frac{t_0}{t_e}\Big)^{2/3}-1 =\frac{9}{4}-1=1.25 .
\end{equation}
16.2.6Problem 16.2.6
In a matter dominated Universe we see now, at time $t_0$, some galaxy, which is now on the Hubble sphere. At what time in the past was the photon we are registering emitted?
solution
The emission event is at the intersection of the particle's worldline with the past light cone, so
\begin{equation}
\tfrac{3}{2}ct_0 \;\Big(\frac{t_e}{t_0}\Big)^{2/3}
=3c (t_e^{2/3}t_0^{1/3}-t_e ),
\end{equation}
from which
\[t_e =\frac{t_0}{8}.\]
16.2.7Problem 16.2.7
Show that the particle horizon in the Einstein-de Sitter model recedes at three times the speed of light.
solution
In Einstein-de Sitter $a\sim t^{2/3}$, thus $L_p =3ct$ and $H=2/3t$, so
\begin{equation}
\dot{L}_p =c+ \frac{2}{3t}\cdot 3ct =3c.
\end{equation}
16.2.8Problem 16.2.8
Does the number of observed galaxies in an open Universe filled with dust increase or decrease with time?
16.2.9Problem 16.2.9
Draw the past light cones $D_{plc}(t_e)$ for Einstein-de Sitter and a Universe with dominating radiation on one figure; explain their relative position.
16.2.10Problem 16.2.10
Find the maximum emission distance and the corresponding redshift for power law expansion $a(t)\sim (t/t_0 )^n$.
solution
The comoving distance along a null geodesic is $\int d\eta =\int dt/a(t)$. Then at the time of emission $t_e$ the proper distance between the comoving emitter and detector is
\begin{equation}
L (t_e ,t_0)=a(t_e )\int\limits_{t_e}^{t_0}\frac{dt}{a(t)}=\frac{t_0}{1-n}\big[x^n -1\big],
\end{equation}
where $t_0$ is the time of detection (present), and
\begin{equation}
x=\frac{t_e}{t_0}.
\end{equation}
This is the past light cone of an event at $t_0$ given in terms of proper distance and cosmic time.
The maximum of $L(t_e ,t_0)$ is at
\begin{equation}
x_m=n^{1/(1-n)}, \qquad L_{\max}=L(x_m t_0, t_0)
=\frac{t_0}{n}x_m =\frac{t_0}{n}n^{1/(1-n)}.
\end{equation}
The corresponding redshift is given by the general relation
\begin{equation}
z_{\max}+1 =\frac{a(t_0)}{a(xt_0)}=n^{n/(n-1)}.
\end{equation}
For radiation domination ($n=1/2$) we have
\begin{equation}
L_{\max}=\tfrac{1}{2}ct_0=\tfrac{1}{2}R_H,\qquad z_{\max}=1,
\end{equation}
and for matter domination ($n=2/3$)
\begin{equation}
L_{\max}=\tfrac{4}{9}ct_0 =\tfrac{8}{27}R_H ,\qquad z=1.25.
\end{equation}
16.2.11Problem 16.2.11
Show that the most distant point on the past light cone was exactly at the Hubble sphere at the moment of emission of the light signal that is presently registered.
solution
The recession velocity of the emitter at the time of emission $t_e =x_m t_0$ is
\begin{equation}
v(t_e ,L_{\max})=H(x_m t_0)\cdot L_{\max}=\frac{n}{x_m t_0}\cdot \frac{t_0}{n}x_m =1 ,
\end{equation}
so by definition the emitter (i.e. galaxy) was at that moment exactly on the Hubble sphere.
16.2.12Problem 16.2.12
Show that the comoving particle horizon is the age of the Universe in conformal time
solution
Starting from the definition
\begin{equation}
\frac{L_p (t)}{a(t)} = \int_0^t \frac{dt'}{a(t')} = \int_0^\eta d\eta ' = \eta .
\end{equation}
16.2.13Problem 16.2.13
Show that
\begin{align}
\frac{dL_p}{dt}&=L_p (z)H(z)+1;\\
\frac{dL_e}{dt}&=L_e (z)H(z)-1.
\end{align}
solution
From the definitions (here $c=1$)
\begin{align}
\dot{L_p}&=\frac{d}{dt}\Big(a(t)\int\limits^{t}\frac{dt}{a(t)}\Big)= +1+HL_p , \label{dotLp}\\
\dot{L_e}&=\frac{d}{dt}\Big(a(t)\int\limits_{t}\frac{dt}{a(t)}\Big)= -1+HL_e .\label{dotLe}
\end{align}
16.2.14Problem 16.2.14
Find $\ddot{L_p}$ and $\ddot{L_e}$.
solution
Differentiating $L_p$ and $L_e$ from (\ref{dotLp}-\ref{dotLe}) once again, we get
\begin{align}
\frac{d^2 L_p}{dt^2} &= H(+1-q H L_p),\\
\frac{d^2 L_e}{dt^2} &= H(-1-qH L_e),
\end{align}
where $q=-\frac{\ddot{a}/a}{H^2}$ is the deceleration parameter.
16.2.15Problem 16.2.15
Show that observable part of the Universe expands faster than the Universe itself. In other words, the observed fraction of the Universe always increases.
solution
The particle horizon at distance $L_p$ recedes with velocity $\dot{L}_p$ found in the previous problem, while the galaxies at the particle horizon recede at velocity $HL_p$, hence the horizon overtakes the galaxies with the speed of light $c$.
16.2.16Problem 16.2.16
Show that the Milne Universe has no particle horizon.
solution
In the Milne Universe $a\sim t$, $H=1/t$, $q=0$, the lightcone reaches the beginning of time $t=0$ at an infinite comoving distance and there is no particle horizon. The observable universe fills the entire actual Universe and all galaxies are in principle visible. In other words, all galaxies are visible at some stage in their evolution.
16.2.17Problem 16.2.17
Consider a universe which started with the Big Bang, filled with one matter component. How fast must $\rho(a)$ decrease with $a$ for the particle horizon to exist in this universe?
solution
The comoving particle horizon is $\int_{0}dt a^{-1}(t)$, where $t=0$ is assumed to correspond to the Big Bang singularity. Its existence depends on whether this integral converges at small times or not. Then using the Friedman equation,
\begin{equation}
dt=\frac{da}{\dot{a}}=\frac{da}{\sqrt{\rho a^2}},
\end{equation}
so the particle horizon is
\begin{equation}
l_p = \int\limits_0 \frac{da /a}{\sqrt{\rho a^2}}.
\end{equation}
The integral converges as long as $\rho a^2$ diverges at small $a$. That is, if equation of state is such that $\rho$ changes faster than $\sim 1/a^2$, then light can only propagate a finite distance between Big Bang and now. In particular, the particle horizon exists in models, which are dominated at early times either by radiation $\rho\sim 1/a^4$ or matter $\rho\sim 1/a^3$.
16.2.18Problem 16.2.18
Calculate the particle horizon for a universe with dominating
solution
In radiation-dominated universe $a\sim t^{1/2}$, in matter-dominated $a\sim t^{2/3}$. Then integration of $L =a(t)\int dt/a(t)$ gives
\begin{equation}
L_{p}^{(rad)}=2t,\qquad L_p^{(mat)}=3t.
\end{equation}
16.2.19Problem 16.2.19
Consider a flat universe with one component with state equation $p =w \rho$. Find the particle horizon at present time $t_0$.
solution
The particle horizon $L_p$ at the current moment $t_0$ is (see \ref{LpDp})
\[L_p =\lim\limits_{t_e \to 0} D_p (t_e ,t_0)\]
The proper distance can be written as
\begin{equation}
D_p = a_0\int_t^{t_0} \frac{dt'}{a(t')}
=- \int_{z(t)}^{z(t_0)} \frac{dz'}{H(z')}
=\int_0^z \frac{dz'}{H(z')}.
\end{equation}
The density in terms of redshift $z$ is
\[\rho =\rho_{0}(1+z)^{3(1+w)},\]
and
\[H(z)=H_0 \sqrt{(1+z)^{3(1+w)}},\]
so
\begin{equation}
L_p (t_0)=H_0^{-1}\int \limits_{0}^{\infty}
\frac{dz}{\sqrt{(1+z)^{3(1+w_i)}}}
\end{equation}
16.2.20Problem 16.2.20
Show that in a flat universe in case of domination of one matter component with equation of state $p=w\rho$, $w>-1/3$
\begin{equation}
L_{p} (z)=\frac{2}{H(z) (1+3w)},\qquad \dot{L_p}(z)=\frac{3(1+w)}{(1+3w)}.
\end{equation}
solution
In the considered case
\begin{equation}
L_p (z) = \frac{1}{H(z)}\int_0^\infty \frac{dz'}{\sqrt{(1 + z')^{3(1 + w)}}}
= \frac{2}{H(z)(1 + 3w)}.
\end{equation}
Note that $L_p >0$, i.e. the horizon exists, only if $w>-1/3$.
On differentiating by time and using
\begin{equation}
H^2 = \frac{1}{3}\rho ,\qquad \dot H =- \frac{1}{2}\rho (1 + w),
\end{equation}
we get
\begin{equation}
\frac{dL_p}{dt} = \frac{3(1 + w)}{1 + 3w}.
\end{equation}
16.2.21Problem 16.2.21
Show that in a flat universe in case of domination of one matter component with equation of state $p=w\rho$, $w<-1/3$
\begin{equation}
L_{e} (z)=-\frac{2}{H(z) (1+3w)},\qquad \dot{L_p}(z)=-\frac{3(1+w)}{(1+3w)}.
\end{equation}
solution
In the considered case
\begin{equation}
L_e (z) = \frac{1}{H(z)}\int_{-1}^{0} \frac{dz'}{\sqrt {(1 + z')^{3(1 + w)}}}
= -\frac{2}{H(z)(1 + 3w)},
\end{equation}
and
\begin{equation}
\frac{dL_e}{dt} = -\frac{3(1 + w)}{1 + 3w}.
\end{equation}
Note that $L_e >0$, i.e. the event horizon exists, only if $w<-1/3$. In particular, when we have the cosmological constant, $w=-1$, there is only the event horizon
\[L_e =H .\]
16.2.22Problem 16.2.22
Estimate the particle horizon size at matter-radiation equality.
- ↑ Rindler uses the term "fundamental observers".