Difference between revisions of "New from June"

(a) The surface that is expanding is two dimensional; the "center" of the balloon is in the third dimension and is not part of the surface, which has no center.
(b) The balloon is expanded by the pressure difference between the inside and the outside, but the Universe is not being expanded by pressure.
(c) Pressure couples to gravity in the Einstein equation, so the addition of (positive) pressure to the Universe would slow, not increase, the expansion rate. It is the negative pressure which is only capable to generate the accelerated expansion of the Universe.
(d) If the dots on the balloon represent galaxies, they too will expand. But real galaxies do not expand due to general Hubble expansion because they are gravitationally bound objects. We can make a better analogy by gluing solid objects (like 10 cent coins) to the surface of the balloon to represent galaxies, so that they do not expand when the balloon expands.

\[\frac{d\dot a}{da}=\frac{d\dot a}{dt}\frac{dt}{da}=\frac{\ddot a}{\dot a}=\frac{\ddot a}{aH}\equiv-Hq.\]

The energy density has a time dependence determined by the conservation equation, \[\dot\rho=-3H\rho-3Hp.\] The two terms define the behavior of the uniform fluid which contains the energy in a dynamic Universe. The $H$ term provides the "friction", while the density term tracks the reduction in density due to the volume increase during expansion and the pressure term tracks the reduction in pressure energy during expansion.

\begin{align} \nonumber \Omega&=\frac\rho{3H^2},\quad (8\pi G=1),\\ \nonumber N&=\log\frac{a}{a_0},\\ \nonumber \frac d{dN}&=\frac1H\frac{d}{dt},\\ \nonumber \frac{d\Omega}{dN}&=\frac13\frac{\frac{d\rho}{dN}H^2-2\rho H\frac{dH}{dN}}{H^4},\\ \nonumber \frac{d\rho}{dN}&=\frac1H\frac{d\rho}{dt}=-3\rho(1+w),\\ \nonumber \frac{dH}{dN}&=-(1+q)H,\\ \nonumber \frac{d\Omega}{dN}&=-[2q-3w-1]\Omega. \end{align} For a single-component fluid \[q=\frac12(1+3w)\Omega\] and one finally obtains \[\frac{d\Omega}{dN}=-(1+3w)(1-\Omega)\Omega.\]

The evolution equations for an $n$-fluid model use the density parameters $\Omega_1$, $\Omega_2\ldots\Omega_n$, as dynamical variables. Relation \[\frac{d\Omega}{dN}=-[2q-3w-1]\Omega\] can easily be generalized or an $n$-fluid model to give \[\frac{d\Omega_i}{dN}=-[2q-(1+3w_i)]\Omega_i,\quad i=1\ldots n,\] where \[q=\frac12\sum\limits_i(1+3w_i)\Omega_i.\] For the total density parameter \[\Omega_{tot}=\sum\limits_{i=1}^n\Omega_i\] one obtains the evolution equation \[\frac{d\Omega_{tot}}{dN}=-2q(1-\Omega_{tot}).\]

Rewrite the first Friedmann equation in the form \[H^2=\frac{\rho_{eq}}{3}\left[\left(\frac{a_{eq}}{a}\right)^3+\left(\frac{a_{eq}}{a}\right)^4\right],\] where $a_{eq}$ is the scale factor when the densities of matter and radiation are the same, and $\rho_{eq}$ is the density in each component at that time. This equation can be solved exactly using conformal time \[\frac{da}{dt}=\frac1a\frac{da}{d\eta},\quad \left(\frac{da}{d\eta}\right)^2=\frac13\rho_{tot}a^4,\] \[\frac{a(\eta)}{a_{eq}}=2(\sqrt2-1)\left(\frac\eta{\eta_{eq}}\right)+[1-2(\sqrt2-1)]\left(\frac\eta{\eta_{eq}}\right)^2,\quad \eta_{eq}\equiv\frac{2(\sqrt2-1)}{a_{eq}}\sqrt{\frac{3}{\rho_{eq}}}.\] We see a smooth transition between the two characteristic behaviors of radiation domination $\eta\ll\eta_{eq}$, $a\propto\eta$ and matter domination $\eta\gg\eta_{eq}$, $a\propto\eta^2$. There is no good way to rewrite this relation in terms of the cosmic times.

\[Q=\frac{\dot H}{H^2}=\left(\frac{\ddot a}a-H^2\right)\frac1{H^2}=-(q+1);\] \[J=\ddot H \frac{H}{\dot H^2}=\left(\frac{\dddot a}a-\frac{\dot a\ddot a}{a^2}-2H\dot H\right)\frac{H}{\dot H^2}=\frac j{Q^2}+\frac q{Q^2}-\frac2Q=\frac{j+3q+2}{(1+q)^2}.\]

\[\frac{d}{d\ln a}=\frac1H\frac{d}{dt};\] \[\bar Q\equiv\frac{\dot H}{H^2},\quad \bar J\equiv\frac{\ddot H}{H^3}-\frac{\dot H^2}{H^4},\quad \bar S\equiv\frac{\dddot H}{H^4}+3\frac{\dot H^3}{H^6}-4\frac{\dot H\ddot H}{H^5}.\]

\begin{align} \nonumber R&= -6\left[(\bar Q+2)H^2+\frac k{a^2}\right];\\ \nonumber \dot R&= -6H\left\{\left[\bar J+\bar Q(\bar Q+4)\right]H^2-\frac{2k}{a^2}\right\};\\ \nonumber \ddot R&= -6H^2\left\{\left[\bar S+4\bar J(\bar Q+1)+(\bar Q+8)\bar Q^2\right]H^2-2(2-\bar Q)\frac{k}{a^2}\right\}. \end{align}

\[c_S^2\equiv\frac{d p}{d\rho}=w(a)+\rho\frac{dw(a)}{d\rho};\] \[\frac{dw}{d\rho}=\frac{dw}{da}\frac{da}{d\rho};\] \[\dot\rho+3H(\rho+p)\Rightarrow\frac{da}{d\rho}=-\frac13\frac a{\rho(1+w)},\] \[c_S^2=w(a)+\rho\frac{dw}{d a}\left(-\frac13\frac a{\rho(1+w)}\right)=w(a)-\frac13\frac{d\ln(1+w)}{d\ln a}.\]

Take time derivative of the conservation equation to obtain \begin{align}\nonumber \frac{d}{dt}[\dot\rho+3H(\rho+p)]&=\ddot\rho+3\dot H(\rho+p)+3H(\dot\rho+\dot p)=\\ \nonumber &=\ddot\rho+3\dot H(\rho+p)+3H\dot\rho(1+c_S^2)=\\ \nonumber & =\ddot\rho+H(\rho+p)-9H^2(\rho+p)(1+c_S^2)=0. \end{align} It then follows that \[\ddot\rho=-3[\dot H+3H^2(1+c_S^2)](\rho+p).\]

\[\frac{d^2 a^3}{dt^2}=3\left(2H^2+\frac{\ddot a}a\right)a^3=3\left[\frac23\rho-\frac16(\rho+3p)\right]a^3=\frac32(\rho-p)\varphi.\]

From the definitions of redshift $1+z=1/a$ we have \[\frac{dz}{dt}=-\frac{\dot a}{a^2}=-H(z)(1+z)\] or \[dt=-\frac{dz}{H(z)(1+z)}.\] The lookback time is defined as \[t_0-t=\int\limits_t^{t_0}dt=\int\limits_0^z \frac{dz'}{H(z')(1+z')}=\frac1{H_0}\int\limits_0^z \frac{dz'}{E(z')(1+z')}\] where \[E(z)=\sqrt{\Omega_r(1+z)^4+\Omega_m(1+z)^3+\Omega_k(1+z)^2+\Omega_w\exp\left(3\int\limits_0^z dz'\frac{1+w(z')}{1+z'}\right)}.\] From the definition of lookback time it is clear that the cosmological time or the time back to the Big Bang, is given by \[t(z)=\int\limits_z^\infty\frac{dz'}{H(z')(1+z')}.\]

In the case of the power law expansion with $\alpha<1$ (the decelerated expansion) the Hubble radius indeed grows faster than the expanding Universe: $R_H=H^{-1}\propto t$, while $a(t)\propto t^\alpha$. In power law situations the Hubble radius has an expansion velocity \[\frac{d}{dt}\left(\frac1H\right)=\frac1\alpha\] greater than light speed. This behavior is true only for a decelerating Universe composed of matter and radiation. This is not a physical velocity, violating special relativity, but the velocity of expansion of the metric itself.

In an empty (Milne) Universe since $H=\dot a/a=t^{-1}$, the age of the Universe $t_0$ is equal to the Hubble time $t_0=H_0^{-1}$.

A naive argument tells us that as we pile up more and more material of constant density $\rho_0$, the ratio $M/R$ increases: \begin{equation}\label{new2015_1_e1} \frac M R=\frac43\pi R^2\rho_0. \end{equation} This equation would seem to suggest that dark stars could indeed form. However, we must include the binding energy $U$, \begin{equation}\label{new2015_1_e2} U=-\int\frac{GMdM}{r} =-\int\limits_0^R\frac{G}{r}\left(\frac43\pi r^3\rho_0\right)\frac4\pi r^2\rho_0 dr=-\frac{16G\pi^2}{15}\rho_0^2R^5. \end{equation} The total mass $M_T$ of the hypothetical dark star is given by the rest mass $M$ plus the binding energy \begin{equation}\label{new2015_1_e3} \frac{M_T}R=\frac43\pi R^2\rho_0-\frac{16G\pi^2}{15}\rho_0^2R^4=\frac M R\left[1-\frac35\frac G{c^2}\frac M R\right] \le\frac5{12}\frac{c^2}G, \end{equation} where the upper limit is obtained by maximizing the function in the range (\ref{new2015_1_e1}). Thus, the dark star criterion (\ref{new2015_1_e1}) is never satisfied, even for the unrealistic case of constant-density matter. In fact, the endpoint of Newtonian gravitational collapse depends very sensitively on the equation of state, even in spherical symmetry.

The position of quanta inside a black hole can only be known within $\Delta x=2r_{Sh}=4GM/c^2$. Thus $\Delta t =2r_{Sh}/c=4GM/c^3$. According to the uncertainty principle $\Delta E\Delta t\ge\hbar$. Thus \[\Delta T=\frac{\Delta E}k\approx\frac{\bar c^3}{4kGM}\] Up to a numeric factor this relation coincides with the black hole Hawking temperature $T=\hbar c^3/8\pi kGM$.

The cosmological constant do provide the accelerated expansion of Universe needed to realize the inflation: $a(t)\propto e^{Ht}$, $q=-1$. However, in approximation of the cosmological constant, $H$ is constant for all time. Therefore a dynamical mechanism for the limited time of inflation is needed. The physical mechanism for the existence of an approximately constant value of $H$ which lasts for a limited time is given by a scalar field. For a large initial potential energy of scalar field the state equation parameter $w\approx-1$ and the scalar field in process of the "slow roll" imitates the cosmological constant for sufficiently long period of time to solve the flatness and causal problems. Then due to shape of the potential the scalar field exits from the slow-roll regime, oscillates about its' potential minimum decaying into less massive particles insuring that inflation time is finite.

Using definition of the parameter $\varepsilon$ one finds \[\varepsilon=-\frac{\dot H}{H^2}.\] In the slow-roll approximation \[H^2=\frac{8\pi}{3M_P^2V},\] \[3H\dot\varphi=-\frac{dV}{d\varphi}.\] Therefore \[\frac{\ddot a}a=H^2(1-\varepsilon).\] The condition $\varepsilon=1$ is equivalent to $\ddot a=0$. When the value $\varepsilon=1$ is reached due to variation of the potential shape the Universe exits the regime of the accelerated expansion (inflation). Around tend, the inflaton field(s) typically begin oscillating around the minimum of the potential. (see Inflation and the Higgs Scalar, DANIEL GREEN (1412.2107).)

\[N\propto\int\limits_{\varphi_{end}}^{\varphi_{initial}}d\varphi\frac{V}{dV/d\varphi} \propto\int\limits_{\varphi_{end}}^{\varphi_{initial}}d\varphi/\sqrt\varepsilon.\] The number of $e$-folds depends on how fast the field is $f$-decreasing. The number of e-folds, $N$ , is inversely proportional to the square root of the slow roll parameter $\varepsilon$ or proportional to the inverse fractional change of the potential with the field, $V/(dV/d\varphi)$.

\begin{align} \nonumber L_{dS}&=\sqrt{1/\Lambda}\\ \nonumber T_{dS}&=\frac1c\sqrt{1/\Lambda}\\ \nonumber M_{dS}&=\frac\hbar c\sqrt{\Lambda} \end{align}

The Einstein equation with cosmological constant $\Lambda$ is \[R_{\mu\nu}-\frac12g_{\mu\nu}R=8\pi GT_{\mu\nu}+g_{\mu\nu}\Lambda.\] Contracting both sides with $g^{\mu\nu}$, one gets \[R=-8\pi GT-4\Lambda,\] where $T=T_\mu^\mu$ is the trace of the matter (including dark matter) energy-momentum tensor. This can be substituted in the original equation to obtain \[R_{\mu\nu}8\pi G\left(T_{\mu\nu}-\frac12T\right)-g_{\mu\nu}\Lambda.\] In the Newtonian limit, one can decompose the metric tensor as \[g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu},\quad |h_{\mu\nu}|\ll1.\] Specifically we are interested in the $00$-component of the Einstein equation. We parameterize the $00$th-component of the metric tensor as \[g_{00}=1+2\Phi,\] where $\Phi$ is the Newtonian gravitational potential. To leading order, one can show that [S. Weinberg, Gravitation and Cosmology, John Wiley \& Sons, Inc., 1972.] \[R_{00}\approx\frac12\vec\nabla^2g_{00}=\vec\nabla^2\Phi.\] In the inertial frame of a perfect fluid, its $4$-velocity is given by $u_\mu=(1,0)$ and we have \[T_{\mu\nu}=(\rho+p)u_{\mu\nu}-pg_{\mu\nu}=diag(\rho,p).\] where $\rho$ is the energy density and $p$ is the pressure. For a Newtonian (non-relativistic) fluid, the pressure is negligible compared to the energy density, and hence $T\approx T_{00}=\rho$. As a result, in the Newtonian limit, the 00-component of the Einstein equation reduces to \[\vec\nabla^2\Phi=4\pi G\rho-\Lambda,\] which is just the modified Poisson equation for Newtonian gravity, including cosmological constant. This equation can also be derived from the Poisson equation of Newtonian gravity, $\vec\nabla^2\Phi=4\pi G(\rho+3p)$, with source terms from matter and dark energy; $p\approx 0$ for non-relativistic matter, and $p=-\rho$ for a cosmological constant. Assuming spherical symmetry, we have \[\vec\nabla^2\Phi=\frac1{r^2}\frac\partial{\partial r}\left(\frac{\partial\Phi}{\partial r}\right)\] and the Poisson equation is easily solved to obtain \[\Phi=-\frac{GM}{r}-\frac\Lambda6r^2,\] where $M$ is the total mass enclosed by the volume $4/3\pi r^3$. The corresponding gravitational field strength is given by \[\vec{g}=-\vec\nabla\Phi=\left(-\frac{GM}{r^2}+\frac\Lambda3r\right)\frac{\vec{r}}r.\]

From the definition of the energy density and the pressure for the scalar field it follows that \begin{equation}\label{150_023_e1} V(\varphi)=\frac12(\rho_\varphi-p_\varphi). \end{equation} The deceleration parameter by definition is \[q=-\frac{\ddot a}{aH^2}=\frac1{6M_{Pl}^2H^2}(\rho_m+\rho_\varphi+3p_\varphi).\] Substituting $\rho_m+\rho_\varphi=3M_{Pl}^2H^2$, one finds \[p_\varphi=(2q-1)M_{Pl}^2H^2.\] Substituting $\rho_m=3M_{Pl}^2H^2-\rho_\varphi$ and $p_\varphi=(2q-1)M_{Pl}^2H^2$ into (\ref{150_023_e1}), one finds \[V[\varphi(z)]=\rho_{m0}\left[\frac{(2-q)}{3\Omega_{m0}}\frac{H^2}{H_0^2}-\frac12(1+z)^3\right].\]

From the definition of the energy density and the pressure for the scalar field it follows that \begin{equation}\label{150_024_e1} \dot\varphi^2=\rho_\varphi+p_\varphi. \end{equation} Substitute the relations $\rho_m=3M_{Pl}^2H^2-\rho_\varphi$ and $p_\varphi=(2q-1)M_{Pl}^2H^2$ (see the previous problem) into (\ref{150_024_e1}) to obtain \[\left(\frac{d\varphi}{dz}\right)^2=M_{Pl}^2\left[\frac{2(1+q)}{(1+z)^2-3\Omega_{m0}}(1+z)\frac{H_0^2}{H^2}\right].\]

\[c_s^2=\frac{\delta p}{\delta\rho}=\frac{\dot p}{\dot\rho}=-\alpha w+(1+\alpha)B,\quad w=\frac p\rho.\] (place after the problem 81, chapter 9 of the book.)

\begin{align} \nonumber \Omega_m&=\frac{\rho_m}{3H^2}=\frac{\Omega_{m0}\exp(-3N)}{\Omega_{m0}\exp(-3N)+\Omega_{\Lambda0}};\\ \nonumber \Omega_\Lambda&=\frac{\rho_\Lambda}{3H^2}=\frac{\Omega_{\Lambda0}}{\Omega_{m0}\exp(-3N)+\Omega_{\Lambda0}}. \end{align}

\begin{align} \nonumber H&=H_0\sqrt{\Omega_{m0}\exp(-3N)+\Omega_{\Lambda0}},\\ \nonumber q&=-1+\frac32\Omega_m=-1+\frac32\frac{\Omega_{m0}\exp(-3N)}{\Omega_{m0}\exp(-3N)+\Omega_{\Lambda0}},\\ \nonumber j&=1. \end{align}

When the new term is so large that the ordinary first term can be neglected, we find \[a\propto t^{\frac2{3n}}.\] Indeed, assuming that the Universe is filled solely by the non-relativistic matter, one finds \[H\propto\rho^{n/2}\propto a^{-3n/2},\quad a\propto a^{-3n/2+1},\quad a^{-3n/2-1}da\propto dt\Rightarrow a\propto t^{2/(3n)},\] so the expansion is superluminal (accelerated) for $n<2/3$. For example, for $n=2/3$ we have $a\propto t$; for $n=1/3$ we have $a\propto t^2$; and for $n=1/6$ we have $a\propto t^4$. The case of $n=2/3$ produces a term $H^2\propto a^{-2}$ in the FLRW equation; such a term looks similar to a curvature term but it is generated here by matter in a universe with a flat geometry. Note that for $n=1/3$ the acceleration is constant, for $n>1/3$ the acceleration is diminishing in time, while for $n<1/3$ the acceleration is increasing (the cosmic jerk).

Use the result of the previous problem \[a\propto t^{2/(3n)}.\] As \[a\propto t^{\frac{2}{3(w+1)}},\] one finds that \[w=n-1\]in the considered case.

The relation \[\rho_X(z)=\rho_{X0}\exp\left[3\int\limits_0^zdz'\frac{1+w_X(z')}{1+z'}\right]\] holds for arbitrary component with the EoS $p_X=w_X\rho_X$. Consequently \[\frac{d\rho_X}{dz}=3\rho_X\frac{1+w_X(z)}{1+z}.\] The same result can be obtained immediately from the conservation equation using the relation \[\frac d{dt}=\frac{dz}{dt}\frac d{dz}=-(1+z)H.\] Taking into account that $\rho_x=\rho^n$ and \[\rho=(1+z)^{3(w_X+1)}\] for $w_X=const$, one finds \[\rho_X=(1+z)^{3(1+w)n}.\] Substitution of the latter expression into the equation for $\rho_X$ gives \[w_X=n-1.\]

Use the result of the previous problem \[\frac{d\rho_X}{dz}=3\rho_X\frac{1+w_X}{1+z}.\] In the considered case \[\rho_X=\rho^n=\left(\rho_m+\rho_r\right)^n=\left(\rho_{m0}(1+z)^3+\rho_{r0}(1+z)^4\right)^n.\] Substitution of the latter expression into the equation for $\rho_X$ gives \[n\rho^{n-1}\left(\rho_{m0}(1+z)^2+\rho_{r0}(1+z)^3\right)=3\rho^n\frac{1+w_X}{1+z};\] \[w_X=\frac n{3\rho}\left(\rho_{m0}(1+z)^3+\rho_{r0}(1+z)^4\right)-1=\frac{3n\rho+n\rho_{r0}(1+z)^4-3\rho}{3\rho},\] and one finally obtains \[w_X=n-1+\frac n{3\rho}\rho_{r0}(1+z)^4,\] which is very slowly varying function with the redshift. If $\rho_{r0}\ll\rho_{m0}$, at late times, parameter $w_X$ is almost constant and it is identical to a dark energy component with a constant equation of state. But in early times, as one can not ignore the radiation component, one has to take the general EoS parameter $w_X$ which is not constant.

Equivalence of the equations \[H^2=A\rho_m+B(\rho_m)^n\] and \[H^2=A\rho_{m0}a^{-3}+Ba^{-3n},\] taking into account that $\rho_i\propto a^{-3(1+w_i)}$ allows to interpret the Cardassian empirical term in the modified Friedmann equation as the superposition of a quintessential fluid with $w=n-1$ and a background of dust with $w=0$.

The second term starts to dominate at the redshift $z_{eq}$ when $A\rho(z_{eq})=B\rho^n(z_{eq})$, i.e., when \begin{equation}\label{150_cardas6_e1} \frac B A=\rho_0^{1-n}(1+z_{eq})^{3(1-n)}. \end{equation} In the Cardassian model today, we have \begin{equation}\label{150_cardas6_e2} H_0^2=A\rho_0+B\rho_0^n, \end{equation} so that \begin{equation}\label{150_cardas6_e3} A=\frac{H_0^2}{\rho_0}-B\rho_0^{n-1}. \end{equation} From Eqs.(\ref{150_cardas6_e1}) and (\ref{150_cardas6_e3}), we have \[B=\frac{H_0^2(1+z_{eq})^{3(1-n)}}{\rho_0^n\left[1+(1+z_{eq})^{3(1-n)}\right]}.\]

From \[H^2=A\rho+B\rho^n\] and (see the previous problem) \[\frac B A=\rho_0^{1-n}(1+z_{eq})^{3(1-n)}\] we have \[H^2=A\left[\rho+\rho_0^{1-n}(1+z_{eq})^{3(1-n)}\rho^n\right].\] Evaluating this equation today with $A=8\pi/3M_{Pl}^2$, we obtain \[H_0^2=\frac{8\pi}{3M_{Pl}^2}\rho_0\left[1+(1+z_{eq})^{3(1-n)}\right].\] The energy density $\rho_0$ is, by definition, the critical density. Consequently, \[\rho_0=\rho_{cr}=\frac{3H_0^2M_{Pl}^2}{8\pi\left[1+(1+z_{eq})^{3(1-n)}\right]}.\] This result can be presented in the form \[\rho_{cr}=\rho_{crFLRW}\times F(z_{eq},n),\quad \rho_{crFLRW}=\frac{3H_0^2M_{Pl}^2}{8\pi},\quad F(z_{eq},n)\equiv\left[1+(1+z_{eq})^{3(1-n)}\right]^{-1}.\] For example, if $z_{eq}=1$, then $F=\{1/3,1/5,3/20\}$ for $n=\{1/3,2/3,1/6\}$. We see that the value of the critical density can be much lower than in the standard Friedmann cosmology.

\[\rho(z_{eq})=\rho_m(z_{eq})+\rho_r(z_{eq})=\rho_{m0}(1+z_{eq})^3\left[1+\frac{\Omega_{r0}}{\Omega_{m0}}(1+z_{eq})\right].\]

In the case of adiabatic expansion one has \begin{align} \nonumber dE+pdV&=0,\\ \nonumber d(g(\rho)a^3)+p_{tot}d(a^3)&=0,\\ \nonumber a^3dg+3a^2gda+3p_{tot}a^2da&=0,\\ \nonumber \dot g+3Hg&=-3p_{tot}H,\\ \nonumber \frac{dg}{dt}&=\frac{dg}{d\rho}\frac{d\rho}{dt},\\ \nonumber \frac{d\rho}{dt}&=-3H\rho,\\ \nonumber -\frac{dg}{d\rho}3H\rho+3Hg&=-3p_{tot}H,\\ \nonumber p_{tot}&=\rho\frac{dg}{d\rho}-g. \end{align} The same relation is commonly used also in presence of the radiation. This is incorrect, as the relation \[\frac{d\rho}{dt}=-3H\rho\] does not hold with radiation.

The Cardassian model \[H^2=\frac{8\pi}{M_{Pl}^2}\rho_m+B\rho_m^n\] can equivalently be written as \[H^2=\frac{8\pi}{M_{Pl}^2}\rho_m\left[1+\left(\frac{\rho_m}{\rho_{card}}\right)^{n-1}\right]\equiv\frac{8\pi}{M_{Pl}^2}\rho.\] In the previous problem we have shown that \[p_{tot}=\rho\frac{dg}{d\rho}-\rho.\] It allows to represent the speed of sound as \[c^2=\frac{\partial p_{tot}}{\partial\rho}=\frac{\partial p_{tot}/\partial\rho_m}{\partial\rho/\partial\rho_m}=\rho_m\frac{\left(\partial^2\rho/\partial\rho_m^2\right)}{\partial\rho/\partial\rho_m}.\] All the derivatives are calculated at constant entropy. Finally for the sound speed in the model we obtain \[c^2=-\frac{n(1-n)}{n+\left(\frac{\rho_{card}}{\rho_m}\right)^{1-n}}.\] This value is not guaranteed to be positive. So this model should be considered as an effective description at scales where $c^2>0$.

\begin{align} \nonumber q(z)&=\frac12\frac{(1+z)}{E^2(z)}\frac{dE^2(z)}{dz}-1,\\ \nonumber E^2&\equiv\frac{H^2}{H_0^2}=\Omega_{m0}(1+z)^3+(1-\Omega_{m0})(1+z)^{3n},\\ \nonumber q(z)&=\frac12-\frac32(1-n)\frac{\kappa(z)}{1+\kappa(z)},\\ \nonumber \kappa(z)&\equiv\left(\frac{1-\Omega_{m0}}{\Omega_{m0}}\right)(1+z)^{-3(1-n)}. \end{align}

The conservation equation in terms of the scale factor and the first Friedmann equation are \[a\frac{d\rho_m}{da}-3(3H\xi-\rho_m)=0,\] \[H^2=\frac{8\pi G}3\rho_m.\] Here $\rho_m$ is total density of the baryon and dark matter components. Having excluded the Hubble parameter and changed the independent variable from the scale factor $a$ to the redshift $z$, one finds \[(1+z)\frac{d\rho_m}{dz}-3\rho_m+\gamma\rho_m^{1/2}=0,\quad \gamma\equiv9\sqrt{8\pi G/3}\xi.\] The solution of this equation is: \[\rho_m(z)=\left[\frac\gamma3+\left(\rho_{m0}^{1/2}-\frac\gamma3\right)(1+z)^{3/2}\right]^2.\]

Substitute the solution $\rho_m(z)$ obtained in the previous problem into the first Friedmann equation to obtain \[H(z)=H_0\left[\frac{\bar\gamma}3+\left(\Omega_{m0}^{1/2}-\frac{\bar\gamma}3\right)(1+z)^{3/2}\right],\quad \bar\xi\equiv\frac{24\pi G}{H_0}\xi.\] In the considered model the bulk viscous matter is the only component of the flat Universe. Consequently, $\Omega_{m0}=1$ and one finally obtains \[H(z)=\frac13H_0\left[\bar\gamma+\left(3-\bar\gamma\right)(1+z)^{3/2}\right].\] The obtained expression allows to write the scale factor in terms of the cosmic time. Let us transform the expression for the Hubble parameter \[H(a)=\frac13H_0\frac{\bar\xi a^{3/2}+3-\bar\xi}{a^{3/2}}\] to the following form \[H(t-t_0)=3\int\limits_1^a\frac{{a'}^{1/2}}{\bar\xi {a'}^{3/2}+3-\bar\xi}\ da'.\] For $\xi\ne0$ and $\bar\xi a^{3/2}+3-\bar\xi>0$ ($\bar\xi a^{3/2}+3-\bar\xi<0$ implies $H<0$ and contradicts the observations) one finds \[a(t)=\left[\frac{3\exp\left(\frac12\bar\xi H(t-t_0)-3+\bar\xi\right)}{\bar\xi}\right]^{2/3}.\]

(a) $0<\bar\xi<3$
When $t\to\infty$ then the obtained solution reproduces the de Sitter-like Universe, \[a(t)\propto\exp\left[\frac{\bar\xi}3H(t-t_0)\right]\] (b) $\bar\xi=3$
In this case the considered model exactly reproduces the de Sitter-like Universe, \[a(t)=\exp\left[H_0(t-t_0)\right]\] The model predicts an Universe in an eternal accelerated expansion.

(c) $\bar\xi>3$
In this case the Universe expands forever (decelerated expansion epoch is absent).

Using the result of Problem #150_9 \[a(t)=\left[\frac{3\exp\left(\frac12\bar\xi H(t-t_0)-3+\bar\xi\right)}{\bar\xi}\right]^{2/3},\] one finds the time of the Big Bang as $a(t_{BB})=0$: \[t_{BB}=t_0+\frac2{\bar\xi H_0}\ln\left(1-\frac{\bar\xi}3\right).\]

For $\xi\to0$ \[H_0(t_0-t_{BB})=\frac23.\]

Using \[H(a)=\frac13H_0\frac{\bar\xi a^{3/2}+3-\bar\xi}{a^{3/2}},\] one obtains \[\frac{d\dot a}{da}=H+a\frac{dH}{da}=\frac13H_0\left(\bar\xi+\frac{\bar\xi-3}{2a^{3/2}}\right).\] In order to find the scale factor value $a_t$, corresponding to the transition from the decelerated expansion to the accelerated one, we use the fact that $d\dot a/da\equiv -Hq$. Thus the condition $q=0$ is equivalent to the requirement $d\dot a/da=0$, therefore \[a_t=\left(\frac{3-\bar\xi}{2\bar\xi}\right)^{2/3},\quad z_t=\left(\frac{2\bar\xi}{3-\bar\xi}\right)^{2/3}-1.\]

1. For $0<\bar\xi<1$ the transition between the decelerated epoch to the accelerated one takes place in the future $a_t>1$.
2. For $\bar\xi\to0$ the value $a_t\to\infty$ (the distant future).
3. For $\bar\xi=1$ the transition takes place today ($a_t=1$).
4. For $1<\bar\xi<3$ the transition takes place in the past of the Universe ($0<a_t<1$).
5. For $\bar\xi\to3$ the transition is close to the Big-Bang time ($a_t\to0$).

\[q(a)=-\frac{\ddot a}{aH^2}.\] The term $\ddot a/a$ can be calculated from the second Friedmann equation, that for a matter-dominated universe with bulk viscosity reads: \[\frac{\ddot a}a=-\frac{4\pi G}{3}(\rho_m-9\xi H).\] From the definition of \[\bar\xi\equiv\frac{24\pi G}{H_0}\xi,\] using \[\rho_m=\frac3{8\pi G}H^2\] one obtains that \[\frac{\ddot a}a=\frac12(\bar\xi H_0-H)H.\] Using \[H(a)=\frac13H_0\frac{\bar\xi a^{3/2}+3-\bar\xi}{a^{3/2}},\] we find \[q(a,\bar\xi)=\frac12\left[\frac{3-\bar\xi(1+2a^{3/2})}{3-\bar\xi(1-a^{3/2})}\right].\]

(a) For the case $\bar\xi=0$ we have $q=1/2$ that corresponds to a matter-dominated universe with null bulk viscosity.
(b) For the case $\bar\xi=3$ we have $q=-1$ that corresponds to the de Sitter Universe.
(c) For the case $0<\bar\xi<3$ it is always a decreasing function from $q(0)=1/2$ to $q(\infty)=-1$ with a transition from positive to negative values in the value of the scale factor \[a_t=\left[\frac{3-\bar\xi}{2\bar\xi}\right]^{3/2}.\] (d) For the case $\bar\xi>3$ the Universe expands forever, and this expansion is always accelerated (there is no decelerating epoch or acceleration-deceleration transition).
When $a\to a_*\equiv(1-3/\bar\xi)^{2/3}$ $q\to-\infty$ and when $a\to\infty$ then $q\to-1$. It is a negative and increasing function but it never becomes positive, its maximum value is $-1$.
(e) The case $\bar\xi<3$ predicts an eternal decelerated expanding Universe. The universe begins with a Big-Bang and expands forever until to reach its maximum value $a_*=(1-3/\bar\xi)^{2/3}$ when $t\to\infty$. Deceleration parameter is a positive and increasing monotonic function where its minimum value is $1/2$.

From the expression \[a_t=\left[\frac{3-\bar\xi}{2\bar\xi}\right]^{3/2}.\] we obtain the deceleration parameter evaluated today as \[q(a=1,\bar\xi)=\frac{1-\bar\xi}{2}.\] When $\bar\xi=1$ the transition from the decelerated to accelerated epochs of the Universe takes place today ($a=1$). For $\bar\xi<1$ we have a decelerated Universe in the present and for $\bar\xi>1$ we have an accelerated one today.

For a flat Universe \[R=-6\left(\frac{\ddot a}a +H^2\right).\] Using \[\frac{\ddot a}a=\frac12H(\bar\xi H_0-H), \quad H(a)=\frac13H_0\frac{\bar\xi a^{3/2}+3-\bar\xi}{a^{3/2}},\] we find \[R=-3H(a)[\bar\xi+H(a)]\Rightarrow R(a,\bar\xi)=-\frac13H_0^2\left[\frac{(3-\bar\xi)^2}{a^3}+\frac{5\bar\xi(3-\bar\xi)}{a^{3/2}}+4\bar\xi^2\right].\]

The Friedmann equations for the considered model read \[H^2=\frac{8\pi G}{3}\rho;\] \[\dot\rho+3H[(1+w)\rho-3\xi_0\rho^\nu H]=0.\] Substitution of $H$ into the conservation equation gives \[\dot\rho+3H[(1+w)\rho-\xi^*_0\rho^{\nu+1/2}H]=0,\quad \xi_0^*\equiv\sqrt{24\pi G}\xi_0.\] Solution of the latter equation reads \[\rho(a)=\left[\frac{\xi_0^*}{1+w}+\frac C{1+w}a^r\right]\frac{1}{\frac12-\nu},\quad r\equiv 3(\nu-1/2)(1+w).\] Here $C$ is an integration constant.

Use the first law of thermodynamics \[d(\rho a^3)+pd(a^3)=TdS\] and the conservation equation in presence of viscosity \[\dot\rho+3H(\rho+p+\Pi)=0,\quad \Pi=-3H\xi,\] one finds \[\frac{T\dot S}{a^3}=\dot\rho+3H(\rho+p)=-3H\Pi.\] Consequently, \[\Pi=-\frac{T\dot S}{3Ha^3}.\] As $\dot S>0$ in an expanding Universe $(H>0)$ then $\Pi$ is negative. It agrees with the definition $\Pi=-3H\xi$, where $\xi$ is the positive bulk viscosity coefficient.