Cosmological horizons

§ 1

Particle horizon and the Hubble sphere

Problems 3.6.1 – 3.6.8

3.6.1

Problem 3.6.1: Particle horizon in one-component Universe

Calculate the particle horizon for a Universe with dominating

a) radiation,

b) dust,

c) matter with state equation $p=w\rho$.

solution

Using the definition \[L_{p}\left( t \right) = a\left( t \right)\int_0^t \frac{dt'}{a\left(t' \right)},\] so if $a(t)\sim t^{n}$, we get \[L_{p}=\frac{t}{1-n}.\] Then in each specific case
a) $a\sim t^{1/2}\;\Rightarrow\;L_{p}( t) = 2t$,
b) $a\sim t^{2/3}\;\Rightarrow\;L_{p}( t) = 3t$,
c) From the energy conservation law in the form $\rho a^{3(1+w)}=const$ and the first Friedman equation we get: \[a(t)\sim t^{\frac{2}{3}(w+1)^{-1}}.\] Then \[L_{p}=t\;\Big(1+\frac{2}{1+3w}\Big).\]

3.6.2

Problem 3.6.2: Particle horizon and the age of the Universe

Show the the comoving particle horizon equals to the age of the Universe in conformal time.

solution

The comoving horizon is \[L_{p}^{(comov)}(t)\equiv\frac{L_{p}(t)}{a(t)} = \int_0^t \frac{dt'}{a(t')} = \int_0^\eta d\eta '= \eta.\]

3.6.3

Problem 3.6.3: Radiation domination

Show that, if ultrarelativistic matter is dominating in the matter content of a spatially flat Universe ($k=0$), its particle horizon coincides with the Hubble radius.

solution

Using the expression for particle horizon in one-component Universe, we see that for radiation $L_{p}=2t$. On the other hand, from $a\sim t^{1/2}$ we have \[H = \frac{\dot a}{a} = \frac{1}{2t},\quad \text{and}\quad R_H = H^{-1}=2t = {L_{p}}.\]

3.6.4

Problem 3.6.4: Comoving Hubble

Find the comoving Hubble radius $R_{H}/a$ as function of the scale factor for a spatially flat Universe that consists of one component with equation of state $p=w\rho$.

solution

Again using $a(t)=A_{0} t^{\frac{2}{3(1+w)}}$, we get \[\frac{R_H}{a} = (aH)^{-1} = H_0^{-1}a^{\frac{1}{2}(1 +3w)}.\]

3.6.5

Problem 3.6.5: Hubble radius and particle horizon

Express the comoving particle horizon $L_{p}/a$ through the comoving Hubble radius $R_{H}/a$ for the case of domination of a substance with state parameter $w$.

solution

Starting from the definition, \[L_{p}^{(comov)}\equiv\frac{L_{p}}{a} = \int\limits_{0}^{t} \frac{dt'}{a\left( t' \right)} =\int\limits_{0}^{a}\frac{da'}{H(a')^2} = \int\limits_{0}^{a} R_Hd(\ln a').\]

3.6.6

Problem 3.6.6: Observed Universe

Show that in an open Universe filled with dust the number of observed galaxies increases with time.

solution

The velocity of Hubble sphere's recession is $V = c(1 + q)$. In an open Universe filled with dust $q>0$ so the Hubble sphere's velocity exceeds $c$ by $cq$ and overtakes the galaxies that are situated on it at the moment. Thus the galaxies originally outside of the Hubble sphere enter within and the number of observed galaxies increases.

3.6.7

Problem 3.6.7: Particle horizon and curvature scale

Show that even in early Universe the scale of particle horizon is much less than the curvature radius, and thus curvature does not play significant role within the horizon.

solution

At present $L_{p}\sim t_{0}\sim H_{0}^{-1}$ and using the expression $a_{0}=\frac{H_{0}^{-1}}{\sqrt{|\Omega_0 -1|}}$ for $a$ through the observables \[ a_{0}=\frac{H_{0}^{-1}}{\sqrt{|\Omega_{curv}|}},\] we get that \[\frac{L_{p}}{R_{curv}}\bigg|_{present} \sim \sqrt{|\Omega_{curv}|}.\] Since the observational data dictate $|\Omega_{curv}|\leq 0.02$, at present we have $L_{p}/R_{curv}\ll 1$. Taking into account that for a power-law evolution of $a(t)$ we have $L_{p}\sim t\sim H^{-1}$, for the time evolution of this ratio we obtain \[\frac{L_{p}}{R_{curv}}\sim \frac{t}{a(t)}.\] Both for a Universe dominated by matter and by radiation the power exponent in $a(t)\sim t^{n}$ is less then unity, so $L_{p}/R_{curv}$ increases with time. In the present epoch $n=2/3$ so $L_{p}/R_{curv}\sim t^{1/3}$. In the early Universe, filled with radiation, $n=1/2$ and \[\frac{L_{p}}{R_{curv}}\sim t^{1/2}.\] Thus if curvature is negligible in some sence now, then it was all the more insignificant at any moment in the past.

3.6.8

Problem 3.6.8: The observable portion of the Universe

Estimate the ratio of the volume enclosed by the Hubble sphere to the total volume of the closed Universe.

solution

We use again the expression for the radius of the $3$-sphere ($ a_{0}=\frac{H_{0}^{-1}}{\sqrt{|\Omega_0 -1|}}$) through observables in form $a_{0}=(H_{0}\sqrt{|\Omega_{curv}|})^{-1}$, and the one for its volume $V=2\pi^{2}a^{3}$ (problem of chapter 2). Then neglecting curvature on the scales of the Hubble sphere (see problem), we obtain \[\frac{V_H}{V_U} \approx\frac{(4\pi/3)R_{H}^3}{2\pi^2 a_{0}^{3}} =\frac{2}{3\pi}|\Omega_{curv}|^{3/2}.\] Taking into account that observational data provide $|\Omega_{curv}|<0.02$, we get $V_H/V_U<0.06\%$, so the considered Universe contains more than 1 600 Hubbe spheres. More realistic estimate, with particle horizon instead of the Hubble sphere in the Standard cosmological model, is made in chapter 11.

§ 2

FLRW in terms of proper distances

Problems 3.6.9 – 3.6.13

The following five problems are based on work by F. Melia (arXiv:0711.4181, arXiv:0907.5394).

Standard cosmology is based on the FLRW metric for a spatially homogeneous and isotropic three-dimensional space, expanding or contracting with time. In the coordinates used for this metric, $t$ is the cosmic time, measured by a comoving observer (and is the same everywhere), $a(t)$ is the expansion factor, and $r$ is an appropriately scaled radial coordinate in the comoving frame.

F.Melia demonstrated the usefulness of expressing the FRLW metric in terms of an observer-dependent coordinate $R=a(t)r$, which explicitly reveals the dependence of the observed intervals of distance, $dR$, and time on the curvature induced by the mass-energy content between the observer and $R$; in the metric, this effect is represented by the proximity of the physical radius $R$ to the cosmic horizon $R_{h}$, defined by the relation \[R_{h}=2G\,M(R_h).\] In this expression, $M(R_h)$ is the mass enclosed within $R_h$ (which terns out to be the Hubble sphere). This is the radius at which a sphere encloses sufficient mass-energy to create divergent time dilation for an observer at the surface relative to the origin of the coordinates.

3.6.9

Problem 3.6.9: Cosmic horizon in a flat Universe

Show that in a flat Universe $R_{h}=H^{-1}(t)$.

solution

By definition \[R_{h}=\Big(\frac{3}{8\pi G \rho}\Big)^{1/2} =\frac{1}{H(t)}.\]

3.6.10

Problem 3.6.10: FLRW in terms of proper distances

Represent the FLRW metric in terms of the observer-dependent coordinate $R=a(t)r$.

solution

It is convenient to recast FLRW metric using a new function $f(t)$ \[a(t)=e^{f(t)}.\] In that case \[d{s^2} = {c^2}d{t^2} - {e^{2f(t)}}\left( {d{r^2} + {r^2}d{\Omega ^2}} \right).\] Making the coordinate transformation to the radial coordinate $r=R\,e^{-t}$, we obtain \[ds^2 = \Phi \Big[dt +\big(R\dot f\big){\Phi^{ - 1}}dR\Big]^2 -\Phi^{- 1} dR^2 - R^2 d\Omega^2,\] where for convenience we have defined the function \[\Phi=1-\big(R\dot f\big)^{2}.\] It is easy to see, that the radius of the cosmic horizon for the observer at the origin is \[R_{h}=1/\dot{f}=\frac{a}{\dot{a}}=\frac{1}{H}\] and \[\Phi=1-\frac{R}{R_h}.\] Finally, we obtain \begin{align*} ds^{2}&=\Phi\Big[dt+\frac{R}{R_h}\Phi^{-1}dR\Big]^{2} -\Phi^{-1}dR^2 -R^{2}d\Omega^2 =\\ &=\Big(1-\frac{R}{R_h}\Big) \Big[dt+\frac{R/R_h}{1-R/R_h}dR\Big]^{2} -\frac{dR^2}{1-R/R_h}-R^2 d\Omega^2. \end{align*}

3.6.11

Problem 3.6.11: Divergences and the horizon

Show, that if we were to make a measurement at a fixed distance $R$ away from us, the time interval $dt$ corresponding to any measurable (non-zero) value of $ds$ must go to infinity as $r\to R_h$.

solution

Let us examine the behavior of the interval $ds$ connecting any arbitrary pair of spacetime events at $R$. For an interval produced at $R$ by the advancement of time only $dR=d\Omega=0$, metric obtained in previous problem gives \[ds^{2}=\Phi dt^{2}.\] Function $\Phi\to 0$ as $R\to R_h$, thus for any measurable (non-zero) value of $ds$ the interval $dt$ must go to infinity as $R\to R_h$. In the context of black-hole physics (see Chapter 4), we recognize this effect as the divergent gravitational redshift measured by a static observer outside of the event horizon.

3.6.12

Problem 3.6.12: Horizon only increases

Show that $R_h$ is an increasing function of cosmic time $t$ for any cosmology with $w>-1$.

solution

It is straightforward to demonstrate from Friedman equations that \[\dot{R}_{h}=\frac{3}{2}(1+w).\] Consequently, $\dot{R}_{h}>0$ for $w>-1$. "$R_h$ is fixed only for de Sitter, in which $\rho$ is a cosmological constant and $w=-1$. In addition, there is clearly a demarcation at $w=-1/3$. When $w<-1/3$, $R_h$ increases more slowly than lightspeed ($c=1$ here), and therefore our universe would be delimited by this horizon because light would have traveled a distance $t_0$ greater than $R_{h}(t_0)$ since the big bang. On the other hand, $R_h$ is always greater than $t$ when $w>-1/3$, and our observational limit would then simply be set by the light travel distance $t_0$".

3.6.13

Problem 3.6.13: Simple examples

Using FLRW metric in terms of the observer-dependent coordinate $R=a(t)r$, consider specific cosmologies:

a) the De Sitter Universe;

b) a cosmology with $R_h =t$, ($w=-1/3$);

c) radiation dominated Universe ($w=1/3$);

d) matter dominated Universe ($w=0$).

solution

a) De Sitter \[H=H_{0}=const,\quad a(t)=e^{H_{0}t},\quad f=\ln a(t)=H_{0}t.\] In this case $\dot{R}_{h}=0$ and therefore $R_h$ is fixed \[R_{h}=\frac{1}{H_0}.\] b) the equation of state $w=-1/3$ is the only one for which the current age, $t_0$, of the Universe can equal the light-crossing time, $t_{h}=R_h$. In this case \begin{align} d{s^2}& = \Phi {\Big[ {cdt + \big(\frac{R}{t}\big) {\Phi ^{ - 1}}dR} \Big]^2} - \Phi^{-1}d{R^2} - R^{2}d{\Omega ^2},\\ \Phi& = 1 - \left(\frac{R}{ct}\right)^2 \end{align} The cosmic time $dt$ diverges for a measurable line element as $R\to R_h =t$.
c) In the case of radiation domination \begin{align*} a(t)& = {\left( {2{H_0}t} \right)^{1/2}}, \quad f(t) = \frac{1}{2}\ln \left( {2{H_0}t} \right), \quad \dot f = \frac{1}{2t}\\ d{s^2} &= \Phi {\Big[ {dt + \left( {\frac{R}{2t}} \right){\Phi ^{ - 1}}dR} \Big]^2} - {\Phi ^{ - 1}}d{R^2} - {R^2}d{\Omega ^2},\\ \Phi & = 1 - \left( {\frac{R}{2t}} \right)^2. \end{align*} Thus, measurements made at a fixed $R$ and $t$ still produce a gravitationally-induced dilation of $dt$ as $R$ increases, but this effect never becomes divergent within that portion of the Universe (i.e., within $t_0$) that remains observable since the Big Bang.
d) Matter domination: \begin{align*} a(t) &= {\left( {3/2{H_0}t} \right)^{2/3}},\quad f(t) = \frac{2}{3}\ln \left( {3/2{H_0}t} \right),\quad \dot f = \frac{2}{3t}\\ d{s^2} &= \Phi {\Big[ {dt + \left( {\frac{R}{3t/2}} \right){\Phi ^{ - 1}}dR} \Big]^2} - {\Phi ^{ - 1}}d{R^2} - {R^2}d{\Omega ^2},\\ \Phi &= 1 - {\left( {\frac{R}{3t/2}} \right)^2} \end{align*} The situation is similar to that for a radiation dominated universe, in that $R_h$ always recedes from us faster than lightspeed. Although dilation is evident with increasing $R$, curvature alone does not produce a divergent redshift.