Difference between revisions of "Cosmological horizons"

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 \[L_{p}=t\;\Big(1+\frac{2}{1+3w}\Big).\]

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

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 \[H = \frac{\dot a}{a} = \frac{1}{2t},\quad \text{and}\quad R_H = H^{-1}=2t = {L_{p}}.\]

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

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

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.

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.

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.

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

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*}

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.

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 $t_0$".

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.