# Difference between revisions of "New from June"

Line 1: | Line 1: | ||

__TOC__ | __TOC__ | ||

+ | |||

+ | |||

+ | ==To Chapter 2== | ||

<div id="150_0"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 1''' <p style= "color: #999;font-size: 11px">problem id: 150_0</p> | <div id="150_0"></div> <div style="border: 1px solid #AAA; padding:5px;"> '''Problem 1''' <p style= "color: #999;font-size: 11px">problem id: 150_0</p> |

## Revision as of 02:49, 19 June 2015

## Contents

## To Chapter 2

**Problem 1**

problem id: 150_0

Comprehensive explanations of the expanding Universe often use the balloon analogy. Although the balloon analogy is useful, one must guard against misconceptions that it can generate. Point out the misconceptions that appear when using this analogy [see M.O. Farooq, Observational constraints on dark energy cosmological model parameters, arXiv: 1309.3710.]

(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.

**Problem 2**

problem id: 150_1

(into the cosmography and extended deceleration parameter) Show that \[\frac{d\dot a}{da}=-Hq.\]

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

**Problem 3**

problem id: 150_2

Give a physical interpretation of the conservation equation.

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.

**Problem 4**

problem id: 150_04

Find evolution equation for the density parameter $\Omega$ of the single-fluid FLRW models with the linear equation of state $p=w\rho$.

\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 \[\frac{d\Omega}{dN}=-(1+3w)(1-\Omega)\Omega.\]

**Problem 5**

problem id: 150_05

Solve the previous problem for the multi-component case.

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 \[\frac{d\Omega_{tot}}{dN}=-2q(1-\Omega_{tot}).\]

**Problem 6**

problem id: 150_06

Use the conformal time to prove existence of smooth transition from the radiation-dominated era to the matter dominated one.

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.

**Problem 7**

problem id: 150_07

Consider a set of the cosmographic parameters built from the Hubble parameter and its time derivatives [see S. Carloni, A new approach to the analysis of the phase space of f(R)-gravity, arXiv:1505.06015) ] \[Q\equiv\frac{\dot H}{H^2},\quad J\equiv\frac{\ddot H H}{\dot H^2},\quad S\equiv\frac{\dddot H H^2}{\dot H^3},\ldots \] Express them in terms of the *canonic* cosmographic parameters $q,j,s\dots$.

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

**Problem 8**

problem id: 150_08

Consider another set of the cosmographic parameters [see S. Carloni, A new approach to the analysis of the phase space of f(R)-gravity, arXiv:1505.06015) ] \[\bar Q\equiv\frac{H_{,N}}{H},\quad \bar J\equiv\frac{H_{,NN}}{H},\quad \bar S\equiv\frac{H_{,NNN}}{H},\ldots,\] where \[H_{,N}\equiv \frac{dH}{d\ln a}.\] Express them in terms of the Hubble parameter and its time derivatives.

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

**Problem 9**

problem id: 150_09

Express the Ricci scalar and its time derivatives in terms of the $\bar Q$, $\bar J$ and $\bar S$.

\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}

**Problem 10**

problem id: 150_3

Show that for a perfect fluid with EoS $p=w(a)\rho$ the adiabatic sound speed can be represented in the form \[c_S^2=w(a)-\frac13\frac{d\ln(1+w)}{d\ln a}.\]

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

**Problem 11**

problem id: 150_4

Obtain equation for $\ddot\rho(t)$, where $\rho(t)$ is energy density of an ideal fluid participating in the cosmological expansion.

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 \[\ddot\rho=-3[\dot H+3H^2(1+c_S^2)](\rho+p).\]

**Problem 12**

problem id: 150_5

Show that in the case of the flat Friedmann metric, the third power of the scale factor $\varphi\equiv a^3$ satisfies the equation \[\frac{d^2\varphi}{dt^2}=\frac32(\rho-p)\varphi,\quad 8\pi G=1.\] Check validity of this equation for different cosmological components: non-relativistic matter, cosmological constant and a component with EoS $p=w\rho$.

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

**Problem 13**

problem id: 150_6

The lookback time is defined as the difference between the present day age of the Universe and its age at redshift $z$, i.e. the difference between the age of the Universe at observation $t_0$ and the age of the Universe, $t$, when the photons were emitted. Find the lookback time for the Universe filled with non-relativistic matter, radiation and a component with the EoS $p=w(z)\rho$.

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 \[t(z)=\int\limits_z^\infty\frac{dz'}{H(z')(1+z')}.\]

**Problem 14**

problem id: 150_7

Show that the Hubble radius grows faster than the expanding Universe in the case of power law expansion $a(t)\propto t^\alpha$ with $\alpha<1$ (the decelerated expansion).

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.

\paragraph{To chapter 3, section 15, if absent}

**Problem 15**

problem id: 150_015

Show, that in the Milne Universe the age of the Universe is equal to the Hubble time.

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 $t_0=H_0^{-1}$.

## To chapter 4 The black holes

**Problem 16**

problem id: new2015_1

see E. Berti, A Black-Hole Primer: Particles, Waves, Critical Phenomena and Superradiant instabilities (arXiv:1410.4481[gr-qc])

A Newtonian analog of the black hole concept is a so-called "dark star". If we consider light as a corpuscle traveling at speed $c$, light cannot escape to infinity whenever $V_{esc}>c$, where \[V_{esc}^2=\frac{2GN}R.\] Therefore the condition for existence of "dark stars" in Newtonian mechanics is \[\frac{2GN}{c^2R}\ge1.\]

Can this condition be satisfied in the Newtonian mechanics?

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.

**Problem 17**

problem id: 150_017

Derive the relation $T\propto M^{-1}$ from the Heisenberg uncertainty principle and the fact that the size of a black hole is given by the Schwarzschild radius.

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 $T=\hbar c^3/8\pi kGM$.

## To chapter 8

**Problem 18**

problem id: 2501_06

Why the cosmological constant cannot be used as a source for inflation in the inflation model?

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.

**Problem 19**

problem id: 2501_09

Show that inflation ends when the parameter \[\varepsilon\equiv\frac{M_P^2}{16\pi}\left(\frac{dV}{d\varphi}\frac1V\right)^2=1.\]

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).)

**Problem 20**

problem id: 2501_10

How does the number of e-folds $N$ depend on the slow-roll parameter $\varepsilon$?

\[N\propto\int\limits_{\varphi_{end}}^{\varphi_{initial}}d\varphi\frac{V}{dV/d\varphi} \propto\int\limits_{\varphi_{end}}^{\varphi_{initial}}d\varphi/\sqrt\varepsilon.\] 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, $V/(dV/d\varphi)$.

## To chapter 9

**Problem 21**

problem id: 150_021

Using the by dimensional analysis for cosmological constant $\Lambda > 0$, define the set of fundamental "de Sitter units" of length, time and mass.

\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}

**Problem 22**

problem id: 150_022

In the Newtonian approximation, find the force acting on the point unit mass in the Universe filled by non-relativistic matter and cosmological constant. (see Chiu Man Ho and Stephen D. H. Hsu, The Dark Force: Astrophysical Repulsion from Dark Energy, arXiv: 1501.05592)

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 \[\vec{g}=-\vec\nabla\Phi=\left(-\frac{GM}{r^2}+\frac\Lambda3r\right)\frac{\vec{r}}r.\]

**Problem 23**

problem id: 150_023

Consider a spatially flat FLRW Universe, which consists of two components: the non-relativistic matter and the scalar field $\varphi$ in the potential $V(\varphi)$. Find relation between the scalar field potential and the deceleration parameter.

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 \[V[\varphi(z)]=\rho_{m0}\left[\frac{(2-q)}{3\Omega_{m0}}\frac{H^2}{H_0^2}-\frac12(1+z)^3\right].\]

**Problem 24**

problem id: 150_024

Find relation between the deceleration parameter and the derivative $d\varphi/dz$ for the Universe considered in the previous problem.

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 \[\left(\frac{d\varphi}{dz}\right)^2=M_{Pl}^2\left[\frac{2(1+q)}{(1+z)^2-3\Omega_{m0}}(1+z)\frac{H_0^2}{H^2}\right].\]

**Problem 25**

problem id: new_30

Find the sound speed for the modified Chaplygin gas with the state equation \[p=B\rho-\frac A{\rho^\alpha}.\]

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

## A couple of problems for the SCM

**Problem 26**

problem id: 150_026

Let $N=\ln(a/a_0)$, where $a_0=a(t_0)$ and $t_0$ is some chosen reference time. Usually the reference time is the present time and in that case $\tau=-\ln(1+z)$. Find $\Omega_m(N)$ and $\Omega_\Lambda(N)$ for the SCM.

\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}

**Problem 27**

problem id: 150_027

Express the cosmographic parameters $H,q,j$ as functions of $N=\ln a/a_0$ for the SCM.

\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} \end{enumerate}

## Cardassian Model

[K. Freese and M. Lewis, Cardassian Expansion: a Model in which the Universe is Flat, Matter Dominated, and Accelerating, arXiv: 0201229] Cardassian Model is a modification to the Friedmann equation in which the Universe is flat, matter dominated, and accelerating. An additional term, which contains only matter or radiation (no vacuum contribution), becomes the dominant driver of expansion at a late epoch of the universe. During the epoch when the new term dominates, the universe accelerates. The authors named this period of acceleration by the Cardassian era. (The name Cardassian refers to a humanoid race in Star Trek whose goal is to take over the universe, i.e., accelerated expansion. This race looks foreign to us and yet is made entirely of matter.) Pure matter (or radiation) alone can drive an accelerated expansion if the first Friedmann equation is modified by the addition of a new term on the right hand side as follows: \[H^2=A\rho+B\rho^n,\] where the energy density $\rho$ contains only ordinary matter and radiation, and $n<2/3$. In the usual Friedmann equation $B=0$. To be consistent with the usual result, we take \[A=\frac{8\pi}{3M_{Pl}^2},\] where $M_{Pl}^2\equiv1/G$.

**Problem 28**

problem id: 150_cardas1

Show that once the new term dominates the right hand side of the Friedmann equation, we have accelerated expansion.

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).

**Problem 29**

problem id: 150_cardas2

Let us represent the Cardassian model in the form \[H^2\propto \rho+\rho_X,\quad\rho_X=\rho^n.\] Find the parameter $w_X$ of the EoS $p_X=w_X\rho_X$, assuming that the Universe is filled exclusively by the non-relativistic matter.

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.

**Problem 30**

problem id: 150_cardas3

Show that the result obtained in the previous problem takes place for arbitrary one-component fluid with $w_X=const$.

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 \[w_X=n-1.\]

**Problem 31**

problem id: 150_cardas4

Generalize the previous problem for the case of two-component ideal liquid (non-relativistic matter $+$ radiation) with density $\rho=\rho_m+\rho_r$.

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.

**Problem 32**

problem id: 150_cardas5

Show that we can 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.

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 $w=0$.

**Problem 33**

problem id: 150_cardas6

We have two parameters in the original Cardassian model: $B$ and $n$. Make the transition $\{B,n\}\to\{z_{eq},n\}$, where $z_{eq}$ is the redshift value at which the second term $B\rho^n$ starts to dominate.

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 \[B=\frac{H_0^2(1+z_{eq})^{3(1-n)}}{\rho_0^n\left[1+(1+z_{eq})^{3(1-n)}\right]}.\]

**Problem 34**

problem id: 150_cardas7

What is the current energy density of the Universe in the Cardassian model? Show that the corresponding energy density is much smaller than in the standard Friedmann cosmology, so that only matter can be sufficient to provide a flat geometry.

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.

**Problem 35**

problem id: 150_cardas8

Let us represent the basic relation of Cardassian model in the following way \[H^2=A\rho\left[1+\left(\frac\rho{\rho_{car}}\right)^{n-1}\right],\] where $\rho_{car}=\rho(z_{eq})$ is the energy density at which the two terms are equal. Find the function $\rho(z_{eq})$ under assumption that the Universe is filled with non-relativistic matter and radiation.

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

**Problem 36**

problem id: 150_cardas9

Let Friedmann equation is modified to be \[H^2=\frac{8\pi G}{3}g(\rho),\] where $\rho$ consists only of non-relativistic matter. Find the effective total pressure.

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.

**Problem 37**

problem id: 150_cardas10

Find the speed of sound in the Cardassian model. [P.Gandolo, K. Freese, Fluid Interpretation of Cardassian Expansion, 0209322 ]

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 $c^2>0$.

**Problem 38**

problem id: 150_cardas11

Find the deceleration parameter for the canonic Cardassian model.

\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} \end{enumerate}

## Models with Cosmic Viscosity

A Universe filled with a perfect fluid represents quite a simple which seems to be in good agreement with cosmological observations. But, on a more physical and realistic basis we can replace the energy-momentum tensor for the simplest perfect fluid by introducing cosmic viscosity. The energy momentum tensor with bulk viscosity is given by
\[T_{\mu\nu}=(\rho=p-\xi\theta)u_\mu u_\nu+(p-\xi\theta)g_{\mu\nu},\]
where $\xi$ is bulk viscosity, and $\theta\equiv3H$ is the expansion scalar. This modifies the equation of state of the cosmic fluid. The Friedmann equations with inclusion of the bulk viscosity, i.e. using the energy-momentum tensor $T_{\mu\nu}$, read
\begin{align}
\nonumber \frac{\dot a^2}{a^2}&=\frac13\rho,\quad \rho=\rho_m+\rho_\Lambda,\quad 8\pi G=1;\\
\nonumber \frac{\ddot a^2}{a}&=-\frac16(\rho+3p-9\xi H).
\end{align}
*Problems \ref{150_8}-\ref{150_14} are inspired by A. Avelino and U. Nucamendi, Can a matter-dominated model with constant bulk viscosity drive the accelerated expansion of the universe? arXiv:0811.3253*

**Problem 39**

problem id: 150_8

Consider a cosmological model in a flat Universe where the only component is a pressureless fluid with constant bulk viscosity ($\xi=const$). The pressureless fluid represent both the baryon and dark matter components. Find the dependence $\rho_m(z)$ for the considered model.

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: \[\rho_m(z)=\left[\frac\gamma3+\left(\rho_{m0}^{1/2}-\frac\gamma3\right)(1+z)^{3/2}\right]^2.\]

**Problem 40**

problem id: 150_9

Find $H(z)$ and $a(t)$ for the model of Universe considered in the previous problem.

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 \[a(t)=\left[\frac{3\exp\left(\frac12\bar\xi H(t-t_0)-3+\bar\xi\right)}{\bar\xi}\right]^{2/3}.\]

**Problem 41**

problem id: 150_10

Analyze the expression for the scale factor $a(t)$ obtained in the previous problem for different types of the bulk viscosity.

(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).

**Problem 42**

problem id: 150_11

Show that the Universe in the considered model with $\xi=const$ had the Big Bang in the past for all values of the bulk viscosity in the interval $0<\bar\xi<3$ and determine how far in the past (in terms of the cosmic time) it happened.

Using the result of Problem \ref{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 $a(t_{BB})=0$: \[t_{BB}=t_0+\frac2{\bar\xi H_0}\ln\left(1-\frac{\bar\xi}3\right).\]

**Problem 43**

problem id: 150_12

Show that the result of the previous Problem for zero bulk viscosity ($\xi=0$) correctly reproduces the lifetime of the matter-dominated Universe.

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

**Problem 44**

problem id: 150_13

As we have seen in the previous problems, in the interval $0<\bar\xi<3$ the Universe begins with a Big-Bang followed by an eternal expansion and this expansion begins with a decelerated epoch followed by an eternal accelerated one. The transition between the decelerated-accelerated expansion epochs depends on the value of $\bar\xi$. Find the value of the scale factor where the transition happens.

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 \[a_t=\left(\frac{3-\bar\xi}{2\bar\xi}\right)^{2/3},\quad z_t=\left(\frac{2\bar\xi}{3-\bar\xi}\right)^{2/3}-1.\]

**Problem 45**

problem id: 150_14

Analyze the dependence \[a_t=\left(\frac{3-\bar\xi}{2\bar\xi}\right)^{2/3},\] obtained in the previous problem.

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 ($a_t\to0$).

**Problem 46**

problem id: 150_15

Find the deceleration parameter $q(a,\bar\xi)$ for the cosmological model presented in the Problem \ref{150_8}.

\[q(a)=-\frac{\ddot a}{aH^2}.\] 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 \[q(a,\bar\xi)=\frac12\left[\frac{3-\bar\xi(1+2a^{3/2})}{3-\bar\xi(1-a^{3/2})}\right].\]

**Problem 47**

problem id: 150_16

Analyze behavior of the deceleration parameter $q(a,\bar\xi)$ obtained in the previous problem for different values of the bulk viscosity $\bar\xi(\xi)$.

(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 $1/2$.

**Problem 48**

problem id: 150_17

Use result of the problem \ref{150_15} to find the current value of the deceleration parameter and make sure that for $\bar\xi=1$ the transition from the decelerated to accelerated epochs of the Universe takes place today.

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.

**Problem 49**

problem id: 150_18

Find the curvature scalar $R(a,\xi)$ for the cosmological model presented in the Problem \ref{150_8}.

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 \[R=-3H(a)[\bar\xi+H(a)]\Rightarrow R(a,\bar\xi)=-\frac13H_0^2\left[\frac{(3-\bar\xi)^2}{a^3}+\frac{5\bar\xi(3-\bar\xi)}{a^{3/2}}+4\bar\xi^2\right].\]

**Problem 50**

problem id: 150_19

Let us consider a flat homogeneous and isotropic Universe filled by a fluid with bulk viscosity. We shall assume that the EoS for the fluid is $p=w\rho$, $w=const$ and that the viscosity coefficient $\xi(\rho)=\xi_0\rho^\nu$. Find the dependence $\rho(a)$ for the considered model.

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.

**Problem 51**

problem id: 150_0013

Usually the inflationary models of the early Universe contain two distinct phases. During the first phase entropy of the Universe remains constant. The second phase is essentially non-adiabatic, particles are produced through the damping of the coherent oscillations of the inflaton field by coupling to other fields and by its subsequent decay. Find relation between the bulk viscosity and the entropy production [J. A. S. Lima, R. Portugal, I. Waga, Bulk viscosity and deflationary universes, arXiv:0708.3280].

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.