Difference between revisions of "Solutions of Friedman equations in the Big Bang model"

a) Radiation: $p=\frac{1}{3}\rho$, and from energy conservation \[\dot{\rho}+4\frac{\dot{a}}{a}\rho=0, \quad\Rightarrow\quad \rho\sim \frac{1}{a^{4}}.\] On substitution of this into the first Friedman equation we obtain the equation for $a(t)$ (we use normalization $a_{0}=1$) \[\frac{\dot{a}^{2}}{a^{2}} =\frac{8\pi G}{3}\cdot\frac{\rho_{r0}}{a^4}.\] After integration \[a(t) = \sqrt[4]{\tfrac{32}{3} \pi G{\rho _{r0}}}\;t^{1/2} \quad\mbox{and}\quad \rho_r (t)= \frac 1{\frac {32}{3}\pi G{t^2}}.\] The main result is: \begin{equation}\label{dyn-radiation} \rho_{r}\sim \frac{1}{a^{4}}, \quad a(t) \sim t^{1/2}, \quad \rho_{r}(t)\sim t^{-2}. \end{equation} b) For non-relativistic matter $p=0$, so the energy conservation equation is \[\dot{\rho}+3\frac{\dot{a}}{a}\rho=0, \quad\Rightarrow\quad \rho\sim \frac{1}{a^{3}}.\] In the same way as for the case of radiation domination we get \[a(t) = \sqrt[3]{{6\pi G{\rho _{m0}}}}\;t^{\frac{2}{3}} \quad\mbox{and}\quad \rho_m (t) = \frac{1}{6\pi Gt^2}.\] Then \begin{equation}\label{dyn-matter} \rho_{m}\sim\frac{1}{a^3}, \quad a(t) \sim t^{\frac{2}{3}}, \quad \rho_m (t) \sim t^{-2 }. \end{equation} Note that the dependence $\rho(t)$ is the same in both cases.

The temporal dependence of energy density in a Universe filled with only one component with equation of state $p=w\rho$ (where $w=const$) can be written as \[\rho(t)=\rho_{0}\frac{t_0^2}{t^2},\] where $\rho_0$ is the current density and \[t_0=\frac{1}{(w+1)\sqrt{6\pi G\rho_0}}\] is the current age of the Universe. When both Universes are twice older, density will be four times smaller in each of them.

• 

  Asymptotic behavior of $\rho(t)$ for the case of background interaction of matter and radiation.

If matter and radiation do not interact, as is assumed, we can write down energy conservation laws for them separately. Then for radiation $\rho_{r}\sim a^{-4}$ and for matter $\rho_{m}\sim a^{-3}$, so energy density of radiation in an expanding Universe always decreases faster than that of non-relativistic matter. Thus the dynamics of the Universe can be divided into two stages: the first one when the dominating component is radiation, and the second one when the dominating component is non-relativistic matter.
The first stage: radiation domination. The first Friedman equation is \[\left(\frac{\dot a}{a}\right)^2 = \frac{8\pi G}{3}(\rho_r + \rho_m)\] When $\rho_r \gg \rho_m$, it takes the form \[\left( \frac{\dot a}{a} \right)^2 \approx\frac{8\pi G}{3}\rho_r.\] Then $a(t)\sim t^{1/2}$ and $\rho_r \sim t^{-2}.$
The dominating component determines the behavior of scale factor as function of time. Therefore we have to plug $a(t)$ for the Universe with dominating radiation into $\rho_m = \rho_m(a)$. Then $\rho_m \sim a^{-3} \sim t^{3/2}$. The relative density of matter increases with time, so the stage of radiation domination is succeeded by the stage of matter domination.
Stage 2: matter domination. Following the same train of thought we get \[ a \sim t^{2/3};\quad\rho _m \sim \frac{1}{t^2}; \quad\rho _r \sim \frac{1}{a^4} \sim \frac{1}{t^{8/3}}.\]

a) matter domination $H \equiv\dot a/a = 2/(3t).$
b) radiation domination $H =1/(2t).$

• 

  Scale factor $a(t)$ for a flat Universe filled with dust and radiation.

• 

  Scale factor $a(t)$ for a flat Universe filled with dust and radiation.

Let us present the total energy density as the sum $\rho=\rho_m+\rho _r$ and plug it into the first Friedman equation \[\left( \frac{\dot a}{a} \right)^2 = \frac{8\pi G}{3}(\rho _r + \rho _m).\] The components' densities $\rho _m$ and $\rho _r$, according to problem, are \[\rho_m=\rho_{m0}\left(\frac{a_0}{a}\right)^{3},\quad \rho _r =\rho_{r0}\left(\frac{a_0}{a}\right)^{4}.\] By definition, $\Omega _i = \rho _i/{\rho _{cr}}$, where \[\rho _{cr} \equiv \frac{3H_0^2}{8\pi G}.\] Then \[\frac{\dot a}{a}= H_0\sqrt{\Omega_{m0}\left(\frac{a_0}{a} \right)^{3} +\Omega _{r0}\left(\frac{a_0}{a} \right)^{4}},\] therefore \begin{equation}\label{dyn-MRgenSol} t = \frac{1}{H_0}\int\limits_{0}^{a/a_0} \frac{dx}{x\sqrt{\Omega_{m0}x^{-3} +\Omega_{r0}x^{ - 4}}} =\frac{1}{H_0}\int\limits_{0}^{a/a_0} \frac{x\,dx}{\sqrt{\Omega_{m0}x +\Omega_{r0}}}. \end{equation}

The condition which determines the time $t_{md}$ when radiation and matter densities became equal is, obviously, $\rho_m=\rho_r$. As $\rho_{m}\sim a^{-3}$ and $\rho_{r}\sim a^{-4}$, then (see problem) \[a_{md} = \frac{\Omega _{r0}}{\Omega _{m0}}a_0.\] Using the results and notation of problem, we get \[t_{md} = \frac{1}{H_0}\int\limits_0^{a_{md}/a_{0}} \frac{x\,dx}{\sqrt {\Omega_{m0}x+ \Omega_{r0}}} =H_{0}^{-1} \frac{\Omega_{r0}^{3/2}}{\Omega_{m0}^{2}} \int\limits_{0}^{1}\frac{y\,dy}{\sqrt{y+1}}=\frac{2(2-\sqrt{2})}{3H_0} \frac{\Omega _{r0}^{3/2}}{\Omega _{m0}^2}.\]

For the case of matter domination $a\sim t^{2/3}$ (see problem and problem), so $H=\dot{a}/a=2/(3t)$ and \[t_{0}=\frac{2}{3H_{0}}.\] For the case of radiation domination $a\sim t^{1/2}$, $H=\dot{a}/a=1/(2t)$ and \[t_{0}=\frac{1}{2H_{0}}\] Using the observed value $H_0 \approx 7.15\cdot 10^{ - 11}\mbox{yr}^{-1}$, we get $t_0 \approx 9.3 \cdot 10^{9}\mbox{yr}$ for matter and $t_0 \approx 7 \cdot 10^{9}\mbox{yr}$ for radiation.

The energy conservation equation is \[\dot \rho +3 \frac{\dot a}a(1+w)\rho=0.\] On separation of variables and integration with the initial condition $\rho(1)=\rho_0$ (assuming $a_0=1$), we get \begin{equation}\label{dyn-rho(a-w)} \rho(a)=\rho_0a^{-3(1+w)}. \end{equation} Plugging the obtained $\rho(a)$ into the first Friedman equation, we have \[\left( \frac{\dot a}{a} \right)^2 = \frac{8}{3}\pi G\rho_0a^{ - 3(1 + w)}\] and therefore \[a(t) = \left[ \frac{2t}{3(1 + w)} \sqrt {\frac{8}{3}\pi G{\rho _0}} \right]^{\frac{2}{3(1 + w)}},\] thus \begin{equation}\label{dyn-a(t-w)} a(t)=A_{0} t^{\frac{2}{3(1+w)}}. \end{equation}

Using the result of the previous problem (\ref{dyn-a(t-w)}) and the definition of the Hubble parameter, we get \[H = \frac{\dot a}{a}= \frac{2}{3\left( {1 + w} \right)t}.\]

The dynamics of the Universe filled with different components with energy densities $\rho_i$ and equations of state $p_i=w_i\rho_i$ is described by equations \begin{align*} \left( \frac{\dot a}{a} \right)^2 &= \frac{8\pi G}{3} \sum\limits_i \rho _i - \frac{k}{a^2 },\\ \frac{\ddot a}{a} &= - \frac{4\pi G}{3} \sum\limits_i \left(\rho _i + 3p_i \right). \end{align*} Let us use dimensionless variables \[x \equiv \frac{a}{a_0}; \quad \tau \equiv t\left| {H_0 } \right|; \quad \Omega _{i0} \equiv \frac{\rho _{i0}}{\rho _{cr0}}; \quad \rho _{cr0} \equiv \frac{3H_0^2}{8\pi G}; \quad \Omega _{k0}\equiv-\frac{k}{a_0^2 H_0^2}.\] It is assumed that $H_0 \neq 0$. The Friedman equations in terms of new variables are (the dot now denotes derivative by $\tau$) \begin{align*} \frac{1}{2}\dot{x}^2 &=+\frac{1}{2}\sum\limits_i \Omega _{i0} x^{-(1+3w_i)}+\frac{1}{2}\Omega _{k0};\\ \ddot x &=-\frac{1}{2}\sum\limits_i \Omega _{i0} (1 + 3w_i )x^{ - (2 + 3w_i )}. \end{align*} They can be rewritten as \begin{equation}\label{BBpotential1} \frac{1}{2}\dot x^2 + V(x) = E;\qquad \ddot x = - \frac{dV}{dx}, \end{equation} where \begin{equation}\label{BBpotential2} E = \frac{1}{2}\Omega _{k0} ;\qquad V(x) = - \frac{1}{2}\sum\limits_i \Omega _{i0} x^{ - (1 + 3w_i )}. \end{equation}

For matter $w_{m}=0$, for radiation $w_{r}=1/3$: \begin{equation}\label{BBpotential3} V(x) = - \frac{\Omega _{m0}}{2x} - \frac{\Omega _{r0}}{2x^2}. \end{equation} We see that $dV/dx>0$ for all $x$, therefore $\ddot{x}=-dV/dx<0$.

Following the consideration of the dynamics of the Universe in terms of an effective potential for variable $a=a/a_{0}$, for the Universe filled with dust ($w=0$) and radiation ($w=1/3$) we use the formulas (\ref{BBpotential1}),(\ref{BBpotential2}), and (\ref{BBpotential3}). Then the solution in quadratures is \[\tau=\int\limits_{0}^{x}\frac{x\,dx} {\sqrt{\Omega_{k0}x^{2}+\Omega_{m0}x+\Omega_{r0}}},\] which can be rewritten as \[t=\sqrt{\frac{a_0}{H_0}}\;\int\limits_{0}^{a/a_0} \frac{x\,dx}{\sqrt{-kx^2+\omega_{m}x+\omega_{r}}},\] where $\omega_{i}=8\pi G a_{0}^{2}\rho_{i0}$.

Plugging the first and second Friedman equations into the numerator and denominator correspondingly, we obtain \begin{equation}\label{dyn-q(w)} q = - \frac{\ddot{a}/a}{H^2} = \frac{\frac{4\pi G}{3}\rho(1+3w)} {\frac{8\pi G}{3}\rho}= \frac{1 + 3w}{2}. \end{equation}

For a spatially flat one-component Universe with equation of state $p=w\rho$, where $w=const$, the scale factor depends on time as (\ref{dyn-a(t-w)}) \[ a(t)\sim t^{2/[3(1+w)]}.\] Then \begin{align*} H=\frac{\dot{a}}{a}&=\frac{2}{3(1+w)}\;\frac{1}{t},\\ \frac{\ddot{a}}{a}&=\frac{2}{3(1+w)} \Big(\frac{2}{3(1+w)}-1\Big)\frac{1}{t^{2}}, \end{align*} and we are immediately led to \[q=\frac{1+3w}{2}=const; \quad t_{0}=\frac{2H_{0}^{-1}}{3(1+w)},\] thus $q$ is constant and \[t_{0}=\frac{1}{3H_{0}q}.\]

Excluding the combination $8\pi G$ from the Friedman equations, we get \[\frac{\ddot{a}}{a}= -\frac{1}{2}\left(H^{2}+\frac{k}{a^2}\right)(1+3w),\] therefore \[q = \frac{1+3w}{2} \left(1 + \frac{k}{a^{2}H^{2}}\right).\] We can derive another generalization by simply dividing one of the Friedman equations by the other: \[q=-\frac{\ddot{a}/a}{H^{2}}=\frac{1+3w}{2} \left(1-\frac{k}{a^{2}H^{2}} \frac{\rho_{cr}}{\rho}\right)^{-1}.\]

The energy conservation law can be rewritten as \[a\,d\rho+3da(\rho+p)=0\quad\Leftrightarrow\quad \frac{a}{\rho}\frac{d\rho}{da}=-3(1+w).\] Then plugging $(1+w)$ into the formula for $q$ (\ref{dyn-q(w)}), we obtain \[q=-1-\frac{1}{2}\,\frac{d\ln\rho}{d\ln a}.\]

Let us write the second Friedman equation for the multi-component model as \[\frac{\ddot a}{a} = \frac{4\pi G}{3}\sum\limits_i ( \rho _i + 3p_i).\] Plugging it into the definition of the deceleration parameter and using $H^2 = 8\pi G\rho_{cr}/3$, we get \[q = - \frac{\ddot aa}{\dot a^2} = \frac{\ddot{a}/a}{H^2} =\frac{\frac{4\pi G}{3}\sum\limits_i ( \rho _i + 3p_i)}{\frac{8\pi G}{3}\rho _{cr}}.\] Then \[q = \frac{\Omega }{2} + \frac{3}{2}\sum\limits_i w_i\Omega _i.\]

As $\dot{H}=-{H}^{2}(1+q)$, taking into account (\ref{dyn-q(w)}), we obtain \[\dot{H}<0\quad\Leftrightarrow\quad q=\frac{1+3w}{2}<-1\quad\Leftrightarrow\quad w<-1.\]

Let's use normalization of scale factor by the current value $a_{0}=1$. Taking into account that for matter $\rho\sim a^{-3}$, the first Friedman equation in terms of conformal time $d\eta=dt/a(t)$ is rewritten as \[\frac 1{a^4}\left(\frac{da}{d\eta} \right)^2 =\frac{2a_{\star}}{a^3}-\frac{1}{a^2},\] where \begin{equation}\label{aStar} a_{\star}=\frac{4\pi G \rho_{0}}{3}. \end{equation} After separation of variables and integration, one obtains \[\eta = \int\limits_{0}^{a/a_{\star}} \frac{dx}{\sqrt{2x-x^2}} =-\arccos \Big( 1 - \frac{a}{a_{\star}}\Big).\] Then \begin{align} \label{dyn-closed} a(\eta)&=a_{\star}(1-\cos\eta),\\ t(\eta)=\int a d\eta&=a_{\star}(\eta-\sin\eta), \qquad \eta\in(0,2\pi).\nonumber \end{align} Restoring $a_{0}$ by dimensionality, we get \begin{equation}\label{dyn-closed-dim} \frac{a}{a_0}=\alpha_{\star} (1-\cos\eta),\qquad \frac{t}{a_0}=\alpha_{\star} (\eta-\sin\eta),\qquad \alpha_{\star}=a_{0}^{2}a_{\star} =\frac{4\pi G}{3}\rho_{0}a_{0}^{2}. \end{equation}

We make use of the exact solution in the form (\ref{dyn-closed}). As the range of values of $\eta$ is $[0,2\pi]$, the total lifetime is $T=2\pi a_{\star}$, while the maximal size is reached at $\eta=\pi$: $a_{\max}=2a_{\star}$. Then \[T=\pi a_{\max}.\]

The current value of $a_0$ can be expressed through the observed quantities using the definition of the deceleration parameter $q$: \begin{equation}\label{dyn-a0} q=-\frac{\ddot{a}/a}{H^2} =\frac{\frac{4\pi G}{3}\rho}{H^2} =\frac{1}{2}\Big(1+\frac{1}{a^2 H^2}\Big), \quad\Rightarrow\quad a_0^2 H_0^2=\frac{1}{2q_0 - 1}. \end{equation} Then \[\alpha_{\star}=\frac{4\pi G }{3} \cdot \rho_0 a_{0}^{2} =\frac{a_{0}^{2}}{2} \Big(H_{0}^{2}+\frac{1}{a_{0}^{2}}\Big) =\frac{a_{0}^{2}H_{0}^{2}+1}{2} =\frac{q_0}{2q_0 -1}.\] The scale factor reaches its maximum at $\eta=\pi$ and \begin{equation}\label{dyn-alphaStar} \frac{a_{\max}}{a_0} =2\alpha_{\star}=\frac{1}{1-1/2q_{0}}. \end{equation} Then the maximal value of the size of the Universe can be written as \begin{equation}\label{dyn-Amax} a_{\max}=H_{0}^{-1}\frac{2q_{0}}{(2q_0 -1)^{3/2}}. \end{equation}

From the general solution (\ref{dyn-closed-dim}) \[t_{0}=a_{0}\alpha_{\star}(\eta_0 - \sin\eta_0),\] where, taking into account (\ref{dyn-alphaStar}), \[\eta_{0}=-\arccos(1-\alpha_\star^{-1}) =-\arccos(q_{0}^{-1}-1).\] On substitution of $a_0$ from (\ref{dyn-a0}), we get \[t_{0}=\frac{H_{0}^{-1}}{2q_0 -1}\Big( 1+\frac{q_0}{\sqrt{2q_0 -1}} \arccos(q_{0}^{-1}-1)\Big).\]

First of all, note, that in a stationary closed Universe a photon (as any other particle) that is emitted from a given point in arbitrary direction, travels "around the world" in finite time and returns to the initial point from the opposite direction, very much like a particle moving in a "straight line" on a two-dimensional sphere does. The trigonometric radial coordinate $\chi$, such that $d{\chi ^2} = d{r^2}/(1 - {r^2})$ (see problem of chapter 2), changes during the round-trip from $0$ to $\pi$ (on the opposite pole) and back to zero. In terms of conformal time $d\eta / a(t)$ and $\chi$ the interval for a radially propagating photon $(d\theta = d\varphi = 0)$ is \[d{s^2} = a{(t)^2}(d{\eta ^2} - d{\chi ^2})=0,\] We know (see problem) that the lifetime of the Universe, i.e. the interval between two zeros of $a(\eta)$ corresponds to the interval $\Delta\eta=2\pi$. During this time a photon will travel distance $\Delta\chi=2\pi$, thus making exactly one round-trip around the Universe.

a) The volume of a closed Universe with scale factor $a_0$ was found in problem: $V_{0}=2\pi^2 a_{0}^{3}$. Using the expression for $a_{0}$ through the measured quantities $H_0$ and $q_0$ (\ref{dyn-a0}), we get then \[V_0 = 2\pi^2a_0^3 = 2\pi^2 H_0^{-3}(2q_0-1)^{-3/2}.\] b) The spatial metric of the closed Universe is \[dl^{2}=a^{2}\big(d\chi^2 +\sin^{2}\!\chi\; d\Omega^{2}\big), \quad d\Omega^{2}=d\theta^2 +\sin^{2}\theta d\varphi^2,\] so the volume element between spherical layers of "radii" $\chi$ and $\chi+d\chi$ is \[dV_{\chi}=4\pi\cdot a^{2}\sin^{2}\chi \cdot a\,d\chi =4\pi a^{3}\sin^{2}\chi d\chi.\] At present we have received light emitted with the radial comoving coordinate $\chi$ equal to at most the conformal age of the Universe \[\eta_{0}=-\arccos(1-\alpha_{\star}^{-1}) =-\arccos(q_{0}^{-1}-1),\] where in the second equality sign we used (\ref{dyn-alphaStar}) . Then \begin{align*} V_{b}=4\pi a_{0}^{3} \int\limits_{0}^{\eta_0}d\eta \sin^{2}\eta \end{align*} Assuming here $\eta_{0}=\pi$, which corresponds to the moment of maximal expansion, we obtain the total volume of the Universe calculated in problem at this same moment. At later times $\eta>\pi$ we can see all of the Universe, moreover some points from different directions, so the formula is not applicable. c) The volume that we are observing directly differs from the calculated above as we see the distances between objects as they were at the moment of emission, not as they are "now", so one $a(t)$ shoulf be left inside the integral. Otherwise following the same steps, we arrive to \[V_{c}=4\pi (a_{0}\alpha_{\star})^{3} \int\limits_{0}^{\eta_{0}}d\eta\, (1-\cos\eta)^{3}\sin^{2}\eta.\] In more rigorous terms: consider the $4$-volume $\Omega$ filled with worldlines of all particles which we are currently observing (i.e. the ones the light from which took less time to arrive to us than the age of the Universe). The previous volume $V_{b}$ was the spatial volume that the observed matter occupies "at present", i.e. it is the volume of the section $t=t_{0}$ of $\Omega$. The volume $V_{c}$ is the 3-volume of the section of $\Omega$ by the light "cone" directed to the past with the vertex at the observer (it is actually a cone only in Minkowski spacetime, and is quite deformed for Friedmannian spaces).

The solution differs from the one for $k=+1$ (problem) only by the sign of $k$. Likewise the first Friedman equation is rewritten as \[\frac 1{a^4}\left(\frac{da}{d\eta} \right)^2 =\frac{2a_{\star}}{a^3}+\frac{1}{a^2}, \quad\text{where}\quad a_{\star}=\frac{8\pi G \rho_{0}}{3},\] and on integration we obtain \[ \eta=\int\limits_{0}^{a/a_{\star}} \frac{dx}{\sqrt{2x+x^{2}}} =\text{arcosh}\Big(1+\frac{a}{a^{\star}}\Big),\] so trigonometric functions are replaced by the hyperbolic ones and \begin{align}\label{dyn-open} a(\eta)&=a^{\star}(\cosh\eta-1),\\ t(\eta)=\int a d\eta&=a^{\star}(\sinh\eta-\eta), \qquad \eta\in(0,\infty).\nonumber \end{align}

Let $\rho(t) = b a(t)^{-n}$, then expressing $a$ and $\dot{a}$ in the first Friedman equation through $\rho$ and $\dot{\rho}$, we are led in result to equation \[\frac{d\rho}{dt}=n\sqrt{ \frac{8\pi G}{3}}\rho^{3/2},\] and after integration to \[ \Delta t =\frac{1}{n} \sqrt{\frac{3}{8\pi G}} \int\limits_{\rho_1}^{\rho_2}\!d\rho\;\rho^{-3/2} =\frac{2}{n}\sqrt{ \frac{3}{8\pi G}}\;\; \big(\sqrt{\rho_1}-\sqrt{\rho_2}\big).\]

As \[\frac{d}{dt}\frac{1}{H}=\frac{d}{dt}\frac{a}{\dot{a}} =1-\frac{a\ddot{a}}{a^2}=1+q,\] from \[H=\frac{2}{3(1+w)t}\] we obtain the formula already derived above (\ref{dyn-q(w)}) \[q=\frac{3(1+w)}{2}-1=\frac{1+3w}{2}.\] Then a) for radiation $q=\frac{1}{2}$; b) for matter $q=1$.

By definition \[p_{tot}=w_{tot}\rho_{tot},\] so \begin{equation}\label{w_tot_i} w_{tot}=\frac{\sum\limits_i w_i \rho_i}{\sum\limits_i \rho_i}=\frac{\sum\limits_i w_i \Omega_i}{\sum\limits_i \Omega_i}. \end{equation} For a flat Universe $\sum\Omega_{i}=\Omega=1$, and in case of two-component medium \begin{equation}\label{w_tot_d_m} w_{tot}=\Omega_{m}w_{m}+\Omega_{d}w_{d}. \end{equation}

First let us note that from the conservation laws for each of the components separately and both of them together \begin{align*} \dot{\rho}_{i}&=-3H(1+w_{i})\rho_{i};\\ \dot{\rho}_{tot}&=-3H(1+w_{tot})\rho_{tot}. \end{align*} On substitution of $\dot{\rho}_{i}$ into the time derivative of $\Omega_{i}=\rho_{i}/\rho_{cr}$ we get \[\frac{1}{H}\dot{\Omega}_{i}=\frac{1}{H} \frac{\dot{\rho}_{i}\rho_{tot}- \rho_{i}\dot{\rho}_{tot}}{\rho_{tot}^{2}} =3\Omega_{i}(w_{tot}-w_{i}).\] For a two-component flat Universe we can use the result (\ref{w_tot_d_m}) for $w_{tot}$, thus \[\dot{\Omega}_{m}=-\dot{\Omega}_{r} =3H\Omega_{m}\Omega_{r}(w_{r}-w_{m}).\]

Let us rewrite the conservation law in the Friedmannian Universe \[\dot{\rho}+3H(\rho+p)=0\] (here $c=1$) in the form \[\dot{\rho}+\frac{\dot{V}}{V}(\rho+p)=0,\] and passing to differentiation by pressure $p$ instead of time $t$, as \[\frac{d\rho}{dp}+\frac{1}{V}\frac{dV}{dp} (\rho+p)=0.\] For adiabatic expansion \[pV^\gamma = const\quad \Rightarrow\quad V\sim p^{-1/\gamma}\quad \Rightarrow\quad \frac{1}{V}\frac{dV}{dp}=-\frac{1}{\gamma p}.\] Plugging this into the conservation law, we get a differential equation for $\rho(p)$: \[\gamma p\cdot \rho'=\rho+p.\] Its solution is %\[\rho=\rho _m + \frac{p}{\gamma-1}.\] \[\rho(p)=\frac{p}{\gamma-1}+C\,p^{1/\gamma}.\] Using the initial conditions then \[\rho(p)=\frac{p}{\gamma-1} +\Big(\rho_{0}-\frac{p_{0}}{\gamma-1}\Big) \left(\frac{p}{p_{0}}\right)^{1/\gamma}.\]

Let us consider the spherical region of radius $R$ with density $\rho_{cr}$. Putting down the escape velocity as $\dot R$, we obtain \[\dot R^2 = 2\frac{G\rho _{cr}\cdot\frac{4}{3}\pi R^3}{R^2}R.\] Then taking into account the definition of $H = \dot R/R$, \[\rho_{cr}=\frac{3H^2}{8\pi G}.\]

From $r=\rho_{m}/\rho_{X}$ \[\dot r = \frac{{\dot \rho}_m}{\rho _X} - {\dot \rho }_X{\rho_m}{\rho _X^2}.\] Then on substitution of ${\rho _m}$ and ${\rho _X}$ from the corresponding conservation laws, we get \[\dot r = 3Hwr.\]

Similarly to problem, on substitution of the right hand parts of the Friedman equations into the numerator and denominator, we get \[q = - \frac{\ddot{a}/a}{H^2} = \frac{\frac{4\pi G}{3}(\rho + 3p)}{\frac{8\pi G}{3}\rho } = \frac{1}{2}\frac{\rho_m + \rho_X + 3w\rho _X}{\rho_m +\rho _X} =\frac{1}{2} + \frac{3w}{2(1+r)}.\] Another way: \[q = - 1 + \frac{d}{dt}\left( \frac{1}{H} \right) = - 1 - \frac{\dot H}{H^2};\] \[\dot H = - 4\pi G(\rho + p), \quad {H^2} = \frac{8\pi G}{3}\rho.\]

The Friedman equations can be written as \[3{H^2} = 8\pi \rho - 3\frac{k}{a^2},\] \[\dot H = - 4\pi G\left( {\rho + p} \right) + \frac{k}{a^2}.\] Then \[2\frac{{\dot{H}}}{H} =-3H\left( 1+\frac{p}{\rho } \right) -\frac{k}{{{a}^{2}}H}\left( 1+3\frac{p}{\rho } \right).\] Combining this with the conservation equation for $\rho_X$, we get \[\frac{\dot{\rho}_X}{\rho_X} -2\frac{\dot{H}}{H} =-\frac{3Hwr}{1+r}+\frac{k}{{{a}^{2}}H} \left( 1+3\frac{w}{1+r} \right)\] In the considered case (see problem) $\dot{r}=3Hwr$, so we arrive to \[\dot{r}=-\left( 1+r \right)\left[ \frac{{{{\dot{\rho }}}_{X}}}{{{\rho }_{X}}}-2\frac{{\dot{H}}}{H}-\frac{k}{{{a}^{2}}H}\left( 1+3\frac{w}{1+r} \right) \right].\] If ${{\rho }_{X}}\propto {{H}^{2}}$, then \[\dot{r}=(1+r) \frac{k}{{{a}^{2}}H}\left( 1+3\frac{w}{1+r} \right).\] Thus we obtain immediately that for a flat Universe $\dot{r}=0$.

Combining \[\dot{r}=(1+r)\frac{k}{{{a}^{2}}H}\left( 1+3\frac{w}{1+r} \right)\] with \[q=\frac{1}{2}\left( 1+\frac{k}{{{a}^{2}}{{H}^{2}}} \right)\left( 1+3\frac{w}{1+r} \right)\] we find \[\dot{r}=2\left( 1+r \right)\frac{k}{{{a}^{2}}H}\frac{q}{1+\frac{k}{{{a}^{2}}{{H}^{2}}}}\] Using the relative densities \[{{\Omega }_{m}}=\frac{8\pi G}{3{{H}^{2}}}{{\rho }_{m}},\quad {{\Omega }_{X}}=\frac{8\pi G}{3{{H}^{2}}}{{\rho }_{X}},\quad {{\Omega }_{curv}}=-\frac{k}{{{a}^{2}}{{H}^{2}}}\] we can transform the combination $1+r$ to \[1+r=1+\frac{{{\rho }_{m}}}{{{\rho }_{X}}}=\frac{1-{{\Omega }_{curv}}}{{{\Omega }_{X}}}=\left( 1+\frac{k}{{{a}^{2}}{{H}^{2}}} \right)\frac{1}{{{\Omega }_{X}}}\] Finally \[\dot{r}=-2H\frac{{{\Omega }_{curv}}}{{{\Omega }_{X}}}q\] Yet again we confirm that for the considered model in the flat case $\dot{r}=0$.

The result immediately follows from the ralation obtained in the previous problem \[\dot{r}=-2H\frac{{{\Omega }_{curv}}} {{{\Omega }_{X}}}q\] and definition \[{\Omega }_{curv}=-\frac{k}{{{a}^{2}}{{H}^{2}}}.\]

As $H=\dot{a}/a=\alpha/t$, using the definition for the deceleration parameter, we get \[q=-\frac{\ddot{a}/a}{H^2}=\frac{1-\alpha}{\alpha},\] so \[\alpha=\frac{1}{1+q},\qquad H=\frac{1}{(1+q)\;t}.\]

In the general case, using the relation $(1+z)=1/a$, we can relate the age of the Universe to the expansion history as \[t=\int dt=\int \frac{da/a}{H} =\int\limits_{z}^{\infty} \frac{dz}{(1+z)H}.\] Then using the result of the previous problem), on integration we find \[t(z)=\frac{\alpha/H_{0}}{(1+z)^{1/\alpha}}.\]

The luminosity distance is \[{D_L}(z) = (1 + z)\int_0^z \frac{dz'}{H(z')} = \frac{1 + z}{H_0}\frac{\alpha }{1-\alpha} (1 + z)^{\frac{\alpha-1}{\alpha}}\Big|_{0}^{z} =\frac{H_{0}^{-1}}{q} (1+z)\big[1-(1+z)^{1-q}\big].\]