Difference between revisions of "Interactions in the Dark Sector"

A general coupling between a dark energy (with density $\rho _{_{DE}}$) and dark matter (with density $\rho _{_{DM}}$) may be described in the background by the modified conservation equation \[ \begin{array}{l} \dot \rho _{_{DM}} + 3H\rho _{_{DM}} = Q, \\ \dot \rho _{_{DE}} + 3H\left( {\rho _{_{DE}} + p_{_{DE}} } \right) = - Q. \end{array} \] Here $Q$ is the rate of energy density exchange in the dark sector. Sign of $Q$ determines the direction of energy transfer: if $Q < 0$, then dark matter continuously decays into the dark energy, if $Q > 0$ - the opposite process takes place.
A satisfactory model requires that $Q$ should be expressed in terms of the energy densities and other covariant quantities.

We can define effective equation of state parameters as \[ \begin{array}{l} \dot \rho _{_{DM}} + 3H(1+w_{DM,eff})\rho _{_{DM}} = 0 \\ \dot \rho _{_{DE}} + 3H(1+w_{DE,eff})\rho _{_{DE}} = 0 \end{array} \] This definition implies that \[ w_{DM,eff}=-\frac{Q}{3H\rho _{_{DM}}},\ w_{DE,eff}=w_{DE}+\frac{Q}{3H\rho _{_{DE}}} \]

\[ {\it if}\ Q<0 \Rightarrow\left\{ \begin{array}{ll} w_{DM,eff}>0 & {\it dark\ matter\ redshifts\ faster\ than}\ a^{-3}\\ w_{DE,eff}<w_{DE} & {\it dark\ energy\ has\ more\ accelerating\ power} \end{array} \right. \] \[ {\it if}\ Q>0 \Rightarrow\left\{ \begin{array}{ll} w_{DM,eff}<0 & {\it dark\ matter\ redshifts\ slower\ than}\ a^{-3}\\ w_{DE,eff}>w_{DE} & {\it dark\ energy\ has\ less\ accelerating\ power} \end{array} \right. \]

Use the result of the previous problem. When $Q<0$ it is possible that $w_{DE,eff}<-1$. This means that the coupling quintessence behaves like a phantom uncoupled model, but without any negative kinetic energy. (Dynamics of dark energy with a coupling to dark matter, Christian G. Bohmer, Gabriela Caldera-Cabral, Ruth Lazkoz, and Roy Maartens, 0801.1565)

Substitution of the energy density and pressure of the scalar field into the conservation equation with the source $Q$ gives the following: \[\dot{\varphi}\ddot{\varphi}+3H\dot{\varphi}^2+\frac{dV}{d\varphi}\dot{\varphi}=Q\] and the required result follows immediately.

It is convenient to introduce the ratio $r\rho_{DM}/\rho_{DE}$ of the energy densities, which is characterized by the dynamics \[\dot r=r\left[\frac{\dot\rho_{DM}}{\rho_{DM}}-\frac{\dot\rho_{DE}}{\rho_{DE}}\right]\] and to replace the derivatives with respect to the cosmic time by derivatives with respect to $\ln a^3$ , denoted by a prime, $\dot\rho_i=3\rho'_iH$. Then the dynamics of the two-component system is given by \[\frac{\rho'_{DM}}{\rho_{DM}}=-1-\frac{\Pi}{\rho_{DM}}\] \[\frac{\rho'_{DE}}{\rho_{DE}}=-(1+w)+\frac{\Pi}{\rho_{DE}},\ w=\frac{p_{DE}}{\rho_{DE}}\] Or, alternatively, by \[\rho'=-\left(1+\frac{w}{1+r}\right)\] \[r'=\left[w-\frac{(1+r)^2}{r\rho}\Pi\right]r,\ \rho=\rho_{DE} +\rho_{DM}\] (F.Arevalo, A.Bacalhau and W. Zimdahl, arXiv: 1112.5095)

\[\rho_\varphi=\frac{3H^2}{\kappa^2}(x^2+y^2),\ p_\varphi=\frac{3H^2}{\kappa^2}(x^2-y^2)\] \[w_\varphi=\frac{p_\varphi}{\rho_\varphi}=\frac{x^2-y^2}{x^2+y^2},\ w_{tot}=\frac{p_\varphi}{\rho_{tot}}=\frac{x^2-y^2}{\Omega_{tot}}=x^2-y^2\] \[q=-\frac{\ddot a}{aH^2}=\frac12+\frac32(x^2-y^2)\] Relation $\Omega_{tot}=1$ implies that $x^2+y^2\le1$.

The required statement can be easily proved by the replacement $\dot\rho\rightarrow\rho'H$ ($\dot\varphi\rightarrow\varphi'H$) in the energy balance, where prime denotes $d/dN$.

The autonomous system of equation in this case is \[x'=-3x+\lambda\frac{\sqrt6}{2}y^2+\frac32x(1+x^2-y^2)+\alpha\frac{1-x^2-y^2}{2x},\] \[y'=\lambda\frac{\sqrt6}{2}xy+\frac32y(1+x^2-y^2),\] where a prime denotes $d/dN$.

\[ \rho _{_{DM}} = \rho _{_{DM0}} a^{ - 3} \exp \left( {\int {\delta d\alpha } } \right);\quad \alpha \equiv \ln a. \]

\[ \begin{gathered} \rho _{_{DE}} = \rho _{_{DE0}} a^{ - \left[ {3(1 + w_{_{DE}} ) + \delta } \right]} ; \\ \rho _{_{DM}} = \frac{{ - \delta \rho _{_{DE0}} }} {{3w_{_{DE}} + \delta }}a^{ - \left[ {3(1 + w_{_{DE}} ) + \delta } \right]} + \left( {\rho _{_{DM0}} + \frac{{\delta \rho _{_{DE0}} }} {{3w_{_{DE}} + \delta }}} \right)a^{ - 3}. \end{gathered} \]

In the considered model $$ \dot{\rho }_{_{DM}} +3\frac{\dot{a}}{a} \rho _{_{DM}} =-\dot{\rho }_{_{DE}} -3\frac{\dot{a}}{a} \left(\rho _{_{DE}} +p_{_{DE}} \right). $$ Inserting $\rho_{_{DM}} =\rho _{_{DM,0}} a^{-3+\delta } $ into the latter equation and integrating the obtained differential equation, one obtains $$ \rho _{_{DE}} (a)=\rho _{_{DE0}} a^{-3(1+w)} +\frac{\delta \rho _{_{DM0}} }{3\left|w\right|-\delta } a^{-3+\delta } $$ It is evident that in absence of interaction between the components the latter equation reproduces the usual scale factor dependence for quintessence (or cosmological constant for the case $w=-1$).

\begin{equation} \frac{\rho _{_{DM}}} {\rho _{_{DE}}} \propto Aa^{ - \xi } \label{44_1} \end{equation} \begin{equation} \begin{gathered} \dot \rho _{_{DM}} + 3H\rho _{_{DM}} = Q; \\ \dot \rho _{_{DE}} + 3H\rho _{_{DE}} (1 + w_{_{DE}} ) = - Q;\label{44_2} \\ \end{gathered} \end{equation} Let us show that appropriate choice of the interaction $Q$ allows to reproduce the required behavior of the density ratio$\frac{\rho _{_{DM}}}{\rho _{_{DE}}}.$ From the equation (\ref{44_1}) \begin{equation} \rho _{_{DE}} = \frac{\rho _{_{DM}} } {A(1 + z)^\xi}\label{44_3} \end{equation} it follows that \begin{equation} \begin{gathered} \rho _{_{DM}} = \rho _{_{DE}} A(1 + z)^\xi \\ \dot \rho _{_{DM}} = \dot \rho _{_{DE}} A(1 + z)^\xi - \xi \rho _{_{DE}} A(1 + z)^\xi H \label{44_4} \\ \end{gathered} \end{equation} Substitution of equations (\ref{44_2}) and (\ref{44_3}) into (\ref{44_4}) gives the following: $$ Q - 3H\rho _{_{DM}} = - (Q + 3H\rho _{_{DE}} (1 + w_{_{DE}} ))b(1 + z)^\xi - \xi \frac{\rho _{_{DM}}} {b(1 + z)^\xi}b(1 + z)^\xi H.$$ And one finally obtains: $$ Q = - 3H\frac{\frac{\xi } {3} + w_{_{DE}} } {1 + b(1 + z)^\xi }\rho _{_{DM}}.$$

\[ r = 1 + \frac{9}{2}\frac{w}{1 + R}\left[ {1 + w - \frac{\alpha }{\rho _{_{DE} }} - \frac{\dot w}{3wH}} \right]; \] \[ s = 1 + w - \frac{\alpha }{\rho _{_{DE}}} - \frac{\dot w}{3wH}; \] \[ R = \frac{\rho _{_{DM}} }{\rho _{_{DE} }};\quad p_{_{DE}} = w\rho _{_{DE}}.\]

The conservation equation for the Chaplygin gas component reads \[ \dot \rho_{ch} + 3H\left( { - \frac{A}{\rho _{ch}} + 1} \right)\rho _{ch} = - Q = - 3\Gamma H\rho _{ch} .\] The above equation has the following solution \[ \rho _{ch} = \sqrt {\frac{A}[[:Template:1 + \Gamma]] + Ca^{ - 6(\Gamma + 1)} }, \] where $C > 0$ is the integration constant. For large $a$ \[ \lim\limits_{a \to \infty } w_{ch} = - (\Gamma + 1) < - 1. \]

et $\rho _{_{DM}} = m(a)n_{_{DM}}$. Using the result of problem \[ \rho _{_{DM}} = \rho _{_{DM0}} a^{ - 3} \exp \left( {\int {\delta d\alpha } } \right);\quad \alpha \equiv \ln a, \] one obtains \[ \frac{1}{m}\frac{dm}{d\ln a} = \frac{1}{m}\frac{dm}{Hdt} = \delta. \]

Use the standard definition for the dark matter density \begin{equation}\label{11_51_1} \rho _{_{DM}} = m_{_{DM}} (\varphi )n_{_{DM}}. \end{equation} The number density satisfies the equation \begin{equation}\label{11_51_2} \dot n_{_{DM}} + 3Hn_{_{DM}} = 0. \end{equation} Taking derivative over time in (\ref{11_51_1}) and using (\ref{11_51_2}), one obtains \[ \dot \rho _{_{DM}} + 3H\rho _{_{DM}} = \frac{1}{m_{_{DM}} (\varphi)}\frac{dm_{_{DM}} (\varphi )}{d\varphi }\dot \varphi \rho _{_{DM}}. \] In the case of mass independent from $\varphi $ it reproduces the usual equation for the energy density of dark matter particles \[ \dot \rho _{_{DM}} + 3H\rho _{_{DM}} = 0.\] As general covariance lead to conservation law for total energy of dark matter and the scalar field, one finally gets \[ \dot \rho _{_{DE}} + 3H(\rho _{_{DE}} + p_{_{DE}} ) = - \frac{1}{m_{_{DM}} (\varphi )}\frac{dm_{_{DM}}(\varphi )}{d\varphi }\dot \varphi \rho _{_{DM}}.\]

Substitution of standard definitions of energy density and pressure of the scalar field into the conservation equation gives \[ \ddot \varphi + 3H\dot \varphi + \frac[[:Template:\partial V(\varphi )]][[:Template:\partial \varphi]] = - \frac{1}{{m_{_{DM}} (\varphi )}}\frac{{dm_{_{DM}} (\varphi )}}[[:Template:D\varphi]]\rho _{_{DM}}.\]

\begin{eqnarray} &&\rho_m=\rho_{_{DE}}f(a),\label{eq66}\\ &&\rho_{_{DE}}=\frac{\rho_m}{f(a)},\label{eq67}\\ &&\dot{\rho}_m=\dot{\rho}_{_{DE}}f+\rho_{_{DE}}f^\prime \dot{a},\label{eq68}\\ &&\dot{\rho}_{_{DE}}=\frac{\dot{\rho}_m}{f}- \frac{\rho_m\dot{a}f^\prime}{f^2},\label{eq69} \end{eqnarray} where the prime denotes derivative with respect to the scale factor. Substitution of the expressions \eqref{eq68} and \eqref{eq66} into the conservation equation for the dark matter gives the following: \begin{equation}\label{eq700} \dot{\rho}_{_{DE}}f+\rho_{_{DE}}f^\prime\dot{a}+3H\rho_{_{DE}}f=Q, \end{equation} Inserting into the latter the expression for $\dot{\rho}_{_{DE}}$, obtained from the conservation equation for the dark energy, one gets the following: \begin{equation}\label{eq70} Q=\frac{H\rho_{_{DE}}f}{1+f}\left(\frac{f^\prime a}{f}-3w_{_{DE}}\right). \end{equation} The first Friedman equation takes the form: \begin{equation}\label{eq701} \frac{3H^2}{8\pi G}=\rho_{_{DE}}+\rho_m=\rho_{cr}, \end{equation} where $\rho_{cr}$ is the critical density. Therefore one can rewrite that \begin{eqnarray} &&\Omega_{_{DE}}=\frac{\rho_{_{DE}}}{\rho_{cr}}=\frac{1}{1+f},\label{eq72}\\ &&\Omega_m=\frac{\rho_m}{\rho_{cr}}=\frac{f}{1+f}.\label{eq730} \end{eqnarray} And finally we obtain the expression for $Q$, by substitution of \eqref{eq730} into \eqref{eq70}: \begin{equation}\label{eq71} Q=H\rho_{_{DE}}\Omega_m\left(\frac{f^\prime a}{f}-3w_{_{DE}}\right) =H\rho_m\Omega_{_{DE}}\left(\frac{f^\prime a}{f}-3w_{_{DE}}\right). \end{equation}

\begin{equation}\label{eq79} f(a)=f_0\,a^\xi, \end{equation} where $\xi$ is constant; $f_0=f(a=1)=\frac{\rho_{m0}}{\rho_{_{DE0}}} =\frac{\Omega_{m0}}{1-\Omega_{m0}}= \frac{1}{\Omega_{_{DE0}}}-1.$ From the conservation equation for both the densities one finds that \begin{equation}\label{eq702} \frac{d \ln\rho_{tot}}{d\ln a}=-3(1+w_{eff}), \end{equation} where \begin{equation}\label{eq73} w_{eff}=\frac{p_{tot}}{\rho_{tot}} =\frac{\rho_{_{DE}}w_{_{DE}}}{\rho_{_{DE}}+\rho_m} =\frac{w_{_{DE}}}{1+f}=\Omega_{_{DE}}w_{_{DE}}, \end{equation} and therefore \begin{equation}\label{eq74} \frac{d \ln\rho_{tot}}{d\ln a}=-3(1+\Omega_{_{DE}}w_{_{DE}}). \end{equation} from the latter equation it follows that \begin{equation}\label{eq75} \rho_{tot}=Ca^{-3}\exp\left(-\int 3\Omega_{_{DE}}w_{_{DE}}\,d\ln a \right) \end{equation} where $C$ is the integration constant, which can be determined from the condition that $\rho_{tot}(a=1)=\rho_{tot,0}=3H_0^2/(8\pi G)$.
Taking into account the equations (\ref{eq79}) and (\ref{eq75}), and requiring that $\rho_{tot}(a=1)=\rho_{tot,0}$, one determines the integration constant and finds that \begin{equation}\label{eq80} \rho_{tot}=\rho_{tot,0}\,a^{-3}\left[\,\Omega_{m0}+ \left(1-\Omega_{m0}\right)a^\xi\,\right]^{-3w_{_X}/\,\xi}. \end{equation} Substitution of the latter expression into the Friedman equation gives the following: \begin{equation}\label{eq81} \begin{gathered} E^2=\frac{H^2}{H_0^2}=a^{-3}\left[\,\Omega_{m0}+\left(1-\Omega_{m0}\right)a^\xi\, \right]^{-3w_{_X}/\,\xi}\\ =(1+z)^3 \left[\,\Omega_{m0}+\left(1-\Omega_{m0}\right) (1+z)^{-\xi}\,\right]^{-3w_{_X}/\,\xi}. \end{gathered} \end{equation}

The distance module is defined as \begin{equation}\label{test1} \mu_{th}(z_i)\equiv 5\log_{10}D_L(z_i)+\mu_0\,, \end{equation} where $\mu_0\equiv 42.38-5\log_{10}h$, and $h$ is the Hubble constant $H_0$ in units of $100 \,\mbox{km}/\mbox{s}/\mbox{Mpc}$. The photometric distance is defined by \begin{equation}\label{test2} D_L(z)=(1+z)\int_0^z \frac{d\tilde{z}}{E(\tilde{z})}\,, \end{equation} where $E\equiv H/H_0$. In the case $\frac{\rho_{m}}{\rho_{_{DE}}}=f_0a^\xi,$ one has the following \begin{equation} E^2=\frac{H^2}{H_0^2}=(1+z)^3 \left[\,\Omega_{m0}+\left(1-\Omega_{m0}\right) (1+z)^{-\xi}\,\right]^{-3w_{_X}/\,\xi}, \end{equation} and therefore $E\equiv H/H_0$ equals to \begin{equation} E=\frac{H}{H_0}=\sqrt{(1+z)^3 \left[\,\Omega_{m0}+\left(1-\Omega_{m0}\right) (1+z)^{-\xi}\,\right]^{-3w_{_X}/\,\xi}} \end{equation} Inserting the latter expression into \eqref{test2}, one recovers \eqref{test1}. It is useful to compare the obtained results with the observational data for the distance module [arXiv:0901.4804]. Figure presents the dependence of the distance module $\mu$ on the redshift $z$, solid line give theoretical predictions obtained in frame of the considered model, points give the results of the type Ia supernovae observations arXiv:0901.4804.

• 

  Dependence of the distance module on the redshift theoretically calculated in frame of the model at $h=0,65$ $\,\mbox{km}/\mbox{s}/\mbox{Mpc}$ and $\xi=2.9$ compared to one found from observational data for the type Ia supernovae.

\begin{equation} \label{eq10} {Q=(\alpha+3(1+w_1))HC_1a^\alpha+(\beta+3(1+w_1))HC_2a^\beta}. \end{equation}

In the considered case $\rho_1=\rho_{m}$, $\rho_2=\rho_{_{DE}}$, and therefore $w_1=0$, so $ Q=\gamma H \rho_{1}$ can be rewritten as $Q=\gamma H\rho_{m}$ and the expressions for the energy densities take the form: \begin{equation}\label{eq24} \rho_{m}=\rho_{m0}a^{\gamma-3}, \end{equation} \begin{equation}\label{eq25} \rho_{_{DE}}=\rho_{_{DE0}}a^{-3(1+w_2)}-\frac{\gamma\rho_{m0}}{\gamma+3w_2}a^{\gamma-3}, \end{equation} where $\rho_{_{DE0}}$ and $\rho_{m0}$ are constant. For $-1\leq w_2<-1/3$ the dark energy becomes the quintessence, and for $w_2<-1$ it is phantom energy. The case $\gamma=0$ corresponds to absence of interaction, which means that the dark matter and dark energy evolve independently.
As the dark matter density must decay during the expansion of the Universe, it follows from the equation \eqref{eq24} that $\gamma-3<0$. On the other hand the energy densities of dark matter and dark energy must be positive during all the evolution of Universe, thus $\frac{\gamma}{\gamma+3w_2}<0$, $\rho_{_{DE0}}-\frac{\gamma}{\gamma+3w_2}>0$. The condition $\frac{\gamma}{\gamma+3w_2}<0$ implies that $\gamma>0$ and $\gamma+3w_2<0$ or vice versa $\gamma<0$ and $\gamma+3w_2>0$. All other combinations of $\gamma$ and $w_2$ inevitably lead to the result that the energy densities become negative at some value of the scale factor in the process of the Universe evolution. it is worth noting also that if $\gamma<0$ and $\gamma+3w_2>0$, then one obtains $w_2>0$, which is impossible by definition. And finally the allowed range for variation of $\gamma$ is the following: \[0<\gamma<3.\]

On early stages of Universe evolution the dark energy density grows behaves as $\rho_{_{DE}}\simeq-\frac{\gamma \rho_{m0}}{\gamma+3w_2}a^{-3+\gamma}$ and therefore $r=\rho_{m}/\rho_{_{DE}}\simeq-(\gamma+3w_2/\gamma)$. On later stages of the evolution of Universe $r\rightarrow 0$, and it corresponds to the fact that the dark energy dominates over the dark matter. Thus the coincidence problem is partly solved in the considered model, because the densities ratio is constant on early stages of evolution of Universe (see Figure).

• 

  Dependence of $r=\rho_{m}/\rho_{_{DE}}$ on the scale factor for the model with $Q=\gamma H\rho_{m}$.

• 

  Dependence of $r=\rho_{m}/\rho_{_{DE}}$ on the scale factor for the SCM.

For the model under interest the equilibrium $r(a_{eq})=1$ between the dark matter and dark energy is possible under the following condition: \begin{equation}\label{eq26} a_{eq}=\left(\frac{\gamma+3w_2}{2\gamma+3w_2}\frac{\rho_{_{DE0}}}{\rho_{m0}}\right)^{1/(\gamma+3w_2)} \end{equation} Let us now find the expression for the scale factor at the moment of time when the accelerated expansion of Universe started. For that purpose one should set $\ddot{a}=0$ in the second friedman equation to obtain the following: \begin{equation}\label{eq27} a_{ac}=\left(\frac{(1+3w_2)(\gamma+3w_2)}{3w_2(\gamma-1)}\frac{\rho_{_{DE0}}}{\rho_{m0}}\right)^{1/(\gamma+3w_2)} \end{equation} It is easy to see that the two latter expressions do not coincide. One can see also that for $w_2<-2/3$ one gets $a_{ac}<a_{eq}$. It implies that the expansion of Universe already entered the accelerated phase, though the dark matter still dominates. For $-2/3<w_2<-1/3$ one has $a_{ac}>a_{eq}$, so the acceleration started when the dark energy already dominated. Figure show the dependencies of relative densities and deceleration parameter on the scale factor respectively. As can be seen from Figure, dynamics of Universe on late stages is indistinguishable from the SCM.

• 

  The scale factor dependencies for relative densities of all components for the model with $Q=\gamma H\rho_{m}$ at $\gamma=0.5$. Here $\Omega_{DE},\,\Omega_{DM},\,\Omega_{b},\,\Omega_{\gamma}$ are densities of dark energy, dark matter and baryons respectively.

• 

  Dependence of the deceleration parameter on the scale factor. Solid line - for the model with $Q=\gamma H\rho_{m}$ at $\gamma=0.5$, dash line - for SCM.

In such a case equation for the ratio $r=\rho_{DM}/\rho_{DE}$ reduces to \[r'=r(w+\gamma)\] where $w=p_{DE}/\rho_{DE}$, the prime denotes the derivative with respect to $\ln a^3$. The solution of this equation is \[r=r_0a^{3(w+\gamma)}.\] Substitution of the latter solution into \[\rho'=-\left(1+\frac{w}{1+r}\right)\rho\] gives the total energy density \[\rho=\rho_0a^{-3(1+w)}\left[\frac{1+r_0a^{3(w+\gamma)}}{1+r_0}\right]^\frac{w}{w+\gamma}.\] The densities of the components are \[\rho_{DM}=\rho_{DM0}a^{-3(1-\gamma)}\left[\frac{1+r_0a^{3(w+\gamma)}}{1+r_0}\right]^{-\frac{\gamma}{w+\gamma}},\] \[\rho_{DE}=\rho_{DE0}a^{-3(1+w)}\left[\frac{1+r_0a^{3(w+\gamma)}}{1+r_0}\right]^{-\frac{\gamma}{w+\gamma}},\] \[\rho_{DM0}=\frac{r_0}{1+r_0}\rho_0,\ \rho_{DM0}=\frac{1}{1+r_0}\rho_0.\]

Proceeding like in the previous problem, one finds: \[r=r_0\frac{w}{(w-\gamma r_0)a^{-3w}-\gamma r_0},\] \[\rho=\rho_0a^{-3\left(1-\frac{w\gamma}{w-\gamma}\right)}\left[\frac{(w+\gamma r_0)a^{-3w}+r_0(w-\gamma)}{w(1+r_0)}\right]^\frac{w}{w-\gamma}.\]

For $w<0$ the solution reads: \[r=\left(r_0-\frac{\gamma}{|w|}\right)a^{-3|w|}+\frac{\gamma}{|w|},\] \[\rho=\rho_0a^{-3\left(1-\frac{|w|^2}{|w|+\gamma}\right)}\left[\frac{|w|+\gamma (|w|r_0-\gamma)a^{-3|w|}}{|w|(1+r_0)}\right]^\frac{|w|}{|w|+\gamma}.\]

Substitute the Chaplygin gas state equation into the conservation equation for the expanding Universe and separate the variables to obtain the following: $$\frac[[:Template:D\rho]]{{\rho - \frac{A}{\rho }}} = - 3\frac[[:Template:Da]]{a}.$$ Integrate the latter equation with the initial conditions $\rho \left( {a_0 } \right) = \rho '$ and introduce the notion $\rho '^2 - A = \rho _0^2 $ to get the require dependence: \[ \rho \left( a \right) = \sqrt {A + \rho _0^2 \left( {\frac[[:Template:A 0]]{a}} \right)^6 } \]

Use the result of the previous problem \[ \rho \left( a \right) = \sqrt {A + \rho _0^2 \left( {\frac[[:Template:A 0]]{a}} \right)^6 } \] For early Universe $a_0/a\gg 1$ and $$ A \ll \rho _0^2 \left( \frac{a_0 }{a} \right)^6 ,$$ then $\rho \left( a \right) \approx \rho _0 \left( \frac{a_0 }{a} \right)^3$, which corresponds to matter with zero pressure. In later stages of evolution of Universe $a_0 /a \ll 1$ and $A \ll \rho _0^2 \left(\frac{a_0 }{a} \right)^6$, so that $\rho \left( a \right) \approx A = const $. The latter equation of state corresponds to the cosmological constant.

The condition for the accelerated expansion reads $$ \rho + 3p < 0,\;\rho > 0; $$ $$ \rho + 3p = \rho _m + \rho _{ch} + 3p_{ch} = \rho _m + \rho _{ch} - \frac{3A}{\rho _{ch} }; $$ $$ 0 < \rho _{ch} < \frac{ - \rho _m + \sqrt {\rho _m^2 + 12A} } 2 $$

he sound speed in the Chaplygin gas (see the problems \ref{DE84}, \ref{DE85}) equals to \[c_{s}^{2}=\frac{dp}{d\rho }=\frac{A}{{{\rho }^{2}}}=\frac{A{{a}^{6}}}{B+A{{a}^{6}}}=\left\{ \begin{aligned} & 0\quad a\ll 1 \\ & 1\quad a\sim 1 \\ \end{aligned} \right.\] High sound speeds $\left( {{c}_{s}}\sim c \right)$ in late Universe $\left( a\sim 1 \right)$ represent the main defeat of the model. This problem was partially fixed in the so-called generalized Chaplygin gas models (see the problem \ref{DE90}).

$c_s = \sqrt {\frac{dp_{ch}}{d\rho _{ch} }} = \frac{\sqrt A }{\rho _{ch} } \propto t^2$

Let the scalar field energy density equals to that of the Chaplygin gas \begin{equation} \label{DE78n_1} \rho_{\varphi}=\frac{1}{2}{{\dot{\varphi }}^{2}}+V(\varphi )=\sqrt{A+\frac{B}{{{a}^{6}}}}, \end{equation} then for ${{p}_{\varphi }}=-A/{{\rho }_{\varphi }}$ one finds \begin{equation} \label{DE78n_2} p_{\varphi }=\frac{1}{2}\dot{\varphi}^2-V(\varphi )=-\frac{A}{\sqrt{A+\frac{B}{a^6}}}. \end{equation} It follows from (\ref{DE78n_1}), (\ref{DE78n_2}) that \begin{eqnarray} \label{DE78n_3} {{\dot{\varphi }}^{2}}=\frac{B}{{{a}^{6}}\sqrt{A+\frac{B}{{{a}^{6}}}}}\\ V(\varphi)=\frac{2{{a}^{6}}\left( A+\frac{B}{{{a}^{6}}} \right)-B}{2{{a}^{6}}\sqrt{A+\frac{B}{{{a}^{6}}}}}. \label{DE78n_4} \end{eqnarray} Using the first Friedmann equation, the relation (\ref{DE78n_3}) (for the flat case in the units $8\pi G=1$) can be presented in the following form: \begin{equation} \label{DE78n_5} \frac{d\varphi}{da}=\frac{\sqrt{B}}{a{{\left( A{{a}^{6}}+B \right)}^{1/2}}}. \end{equation} The latter relation can be easily integrated to give the following result: \begin{equation} \label{DE78n_6} {{a}^{6}}=\frac{4B\exp (6\varphi )}{A{{\left( 1-\exp (6\varphi ) \right)}^{2}}} \end{equation} Substitution of (\ref{DE78n_6}) into (\ref{DE78n_4}) gives the following simple expression for the scalar field potential: \begin{equation} \label{DE78n_7} V(\varphi )=\frac{1}{2}\sqrt{A}\left[ \cosh (3\varphi )+\frac{1}{\cosh (3\varphi )}. \right] \end{equation} The latter potential does not depend on the integration constant $B$, it is determined only by the constant $A$, which enters the state equation of the Chaplygin gas.

$$ \rho _{ch}^{(gen)} = \left[ {A + Ba^{ - 3(1 + \alpha )} } \right]^{\frac 1 {1 + \alpha }} $$ where $B > 0$ is the integration constant.

As is known, $w = \frac p{\rho }$. Transform the explicit solution $\rho (a)$, obtained in the previous problem $$ \rho = \left[ {A + Ba^{ - 3(1 + \alpha )} } \right] = \rho _0 \left[ {A^* + (1 - A^* )a^{ - 3(1 + \alpha )} } \right]^{\frac 1{1 + \alpha }} $$ where $A^* \equiv A\rho _0^{ - (1 + \alpha )} ;\;a_0 = 1$. Use the following $p = - \frac A {\rho ^\alpha}\quad \left( {A > 0,\alpha> 0} \right), $ to obtain $$ w = - \left[ {1 + \frac{1 - A^* }{A^* }a^{ - 3(1 + \alpha )} } \right]^{ - 1}. $$ One makes sure again that in early Universe (small values of the scale factor) the generalized Chaplygin gas mimics the non-relativistic matter $w \simeq 0$, and at later epoch---as the cosmological constant $w \simeq - 1$.

$$ c_s^2 = \frac{dp}{d\rho } = - \alpha w $$ For the values $\alpha > 1$ on later stages of the evolution of Universe this quantity can exceed the lightspeed, because $w \simeq - 1$ in that case.

$$ w = {\frac{p_{ch}^{(gen)} }{\rho _{ch}^{(gen)} }} = {\frac{p_{DE} } {\rho _{DE} + \rho _{DM} }} = {\frac{w_{DE} \rho _{DE} }{\rho _{DE} + \rho _{DM}}} $$ Using the explicit expression for $\rho _{ch}^{(gen)} $, obtained in the previous problem, and the model definition $p_{ch}^{(gen)} $ $$ \rho _{ch}^{(gen)} = \left[ {A + B(1 + z)^{3(1 + \alpha )} } \right]^{{\frac 1{1 + \alpha }}} ; $$ $$ p_{ch}^{(gen)} = \frac A {\left[ {A + B(1 + z)^{3(1 + \alpha )} } \right]^{{\frac{\alpha}{1 + \alpha }} }} $$ one finds $$ \rho _{DE}^{(gen)} = {\frac{\rho _{DM}^{(gen)} }{1 + w_{DE} \left[ {1 + {\frac B A}(1 + z)^{3(1 + \alpha )} } \right]}} $$ The requirement $\rho _{DE}^{(gen)} > 0$ leads to the condition $w_{DE} \le 0$ for early Universe$\left( {z \gg 1} \right)$ and $w_{DE} \le - 1$ for future $\left( {z = - 1} \right)$. Therefore the condition $w_{DE} \le - 1$ garantees the possibility to describe whole history of universe in frames of the chosen model.

$$ \rho = \left[ {{\frac A {\gamma} } + {\frac C {a^{3\gamma (1 + \alpha )} }}} \right]^{{\frac 1{1 + \alpha }}} ;\quad $$ where $C$ is the integration constant. The latter solution can be rewritten in the following form $$\rho = \rho _0 \left[ {A^* + {\frac{1 - A^* } {a^{3\gamma (1 + \alpha )} }}} \right]^{{\frac 1 {1 + \alpha }}};\quad A^* = {\frac A {\gamma} }\rho _0^{ - (\alpha + 1)} ;\quad a_0 = 1.$$ For $\gamma = 1$ it reproduces the dependence $\rho (a)$ for the generalized model of the Chaplygin gas (see the problem of the present Chapter), and for $\gamma = 1,\;\alpha = 1$---for the usual Chaplygin gas model (see the problem of the present Chapter).

Equation of state for the Chaplygin gas is given by \[p=-\frac A\rho,\ A>0.\] As shown earlier \[\rho=\sqrt{A+Ca^{-6}},\] where $C$ is a real positive constant. Consequently, \[w=\frac p\rho=-\frac{A}{\rho^2}=-\frac{A}{A+Ca^{-6}}.\] We have $-1<w<0$ for $a\in(0,\infty)$, hence $w=-1$ cannot be crossed in this model.

Equation of state of the generalized Chaplygin gas is given by \[p=-\frac{A}{\rho^\alpha},\ A>0,\ \alpha\ge0.\] Using the conservation equation, find density in terms of the scale factor \[\rho=\left[A_s+\frac{1-A_s}{a^{3(\alpha+1)}}\right]^{\frac{1}{1+\alpha}},\ A_s\equiv\frac{A}{\rho_0^{1+\alpha}}.\] The observations favor $0<A_s<1$. Then the state equation parameter of the generalized Chaplygin gas model is \[w=-A_s\left(\frac{\rho_0}{\rho}\right)^{1+\alpha}=-\frac{A_s a^{3(1+\alpha)}}{1-A_s+A_s a^{3(1+\alpha)}}.\] It is easy to see that with the restriction $A_s>0$ one obtains \[-1<w<0.\] Therefore such model excludes phantom region $w<-1$.

The background matter is assumed to be described by a perfect fluid with barotropic equation of state \begin{equation}\label{DAMA_72_1} \rho'_m+3(1+w_m)\rho_m=0, \end{equation} where $\rho_m$ is the energy density of the background matter, a prime denotes the derivative with respect to $N=\ln a$, and $-1<w_m<1$. In particular, $w_m=0$ and $w_m=1/3$ correspond to dust matter and radiation, respectively. And for the dark energy component \begin{equation}\label{DAMA_72_2} \rho'_{DE}+3(1+w_{DE})\rho_{DE}=0, \end{equation} where $\rho_{DE}$ is the energy density of the dark energy for which $w_{DE}=p_{DE}/\rho_{DE}$. Friedmann equations for the whole system read \begin{equation}\label{DAMA_72_3} \begin{array}{l} H^2=\frac{\rho_{tot}}{3m^2_{Pl}}=\frac{1}{3m^2_{Pl}}(\rho_m+\rho_{DE}),\\ \dot H=-\frac{1}{2m^2_{Pl}}(\rho_{tot}+p_{tot})=-\frac{(1+w_{tot})\rho_{tot}}{2m^2_{Pl}},\\ w_{tot}=\frac{\frac{\Omega_m}{\Omega_{DE}}w_m+w_{DE}}{1+\frac{\Omega_m}{\Omega_{DE}}},\\ \Omega_i\equiv\frac{\rho_i}{3m^2_{Pl}H^2}. \end{array} \end{equation} Combining (\ref{DAMA_72_1})-(\ref{DAMA_72_3}) we can get the equations of motion for dynamical variables $\Omega_i$ \[\Omega'_m=3f_{ni}\Omega_m\Omega_{DE}\] \[\Omega'_{DE}=-3f_{ni}\Omega_m\Omega_{DE}\] where $f_{ni}=w_{DE}-w_m$. The index $ni$ means there is no interaction between the background matter and the dark energy.

The fixed points are $(\Omega_m,\Omega_{DE})=\{(1,0),(0,1)\}$. The stability matrix reads: \[\hat M=3f_{ni}\left( \begin{array}{lr} \Omega_{DE} & \Omega_m\\ -\Omega_{DE} & -\Omega_m \end{array} \right)\] Let the dark energy is realized in form of the cosmological constant, and matter---in form of dust $w_m=0$ or radiation $w_m=1/3$. In both cases $f_m<0$. For the critical point $(\Omega_m,\Omega_{DE})=(1,0)$ one obtains $\lambda_\pm=\{(0,1)\}$ and the considered critical point is unstable. Addition to the Universe of some portion of dark energy in form of the cosmological constant makes it evolve from the state $(\Omega_m,\Omega_{DE})=(1,0)$ into the state $(\Omega_m,\Omega_{DE})=(0,1)$. For the critical point $(\Omega_m,\Omega_{DE})=(0,1)$ one has $\lambda_\pm=\{(0,-1)\}$ and therefore the considered critical point is stable: if the Universe, filled by the cosmological constant, is perturbed by some portion of matter, then its relative density will drop to zero due to the expansion of Universe.

The considered dynamical system is described by the equations: \[\rho'=-\left(1+\frac{w}{1+r}\right)\rho,\ r\equiv\frac{\rho_{DM}}{\rho_{DE}},\] \[r'=\left[w-\frac{(1+r)^2}{r\rho}\Pi\right]r,\ \rho\equiv\rho_{DE}+\rho_{DM}.\] The condition \(\rho'=0\) provides the critical point \[r_c=-1-w.\] (We disregard the critical points with $r_c=\rho_c=0$ and the unphysical $r_c=-1,\,\rho_c=0$.) Consequently, for positive values of $r$, the existence of a critical point requires $w<-1$, i.e. dark energy of phantom type. This conclusion does not depend on the interaction. The condition $r'=0$ provides \[\rho_c=-\frac{w}{1+w}\Pi_c.\]

Since $w<-1$, a positive stationary energy density $\rho_c$ requires $\Pi_c<0$, equivalent to $Q_c>0$.

Recall that in the case $Q\propto H$ the energy balance equation is independent from $H$, thus phase space of such coupling model is two-dimensional. Then the dynamics of the two-component system is given by \[\rho'=-\left(1+\frac{w}{1+r}\right)\rho,\] \[r'=-\left[w-\frac{(1+r)^2}{r\rho}\Pi\right]r,\ \rho=\rho_{DE}+\rho_{DM}.\] Here the prime denotes derivative with respect to $\ln a^3$. Solution of the equation $\det|\hat M-\lambda\hat I|=0$ for the eigenvalues $\lambda$ of the stability matrix leads to the following equation: \[\lambda^2+\left[2+w-w(1+w)\frac{\partial_r \Pi}{\Pi}\right]\lambda+(1+w+w\partial_\rho\Pi)=0.\] This equation has the following solutions \[\lambda_\pm=\frac12\left\{\left[w(1+w)\frac{\partial_r \Pi}{\Pi}-(2+w)\right] \right.\] \[\left.\pm\sqrt{\left(2+w-w(1+w)\frac{\partial_r \Pi}{\Pi}\right)^2-4(1+w+w\partial_\rho\Pi)}\right\}.\]

In that case the eigenvalues equal to the following: \[\lambda_\pm=-\frac12\left\{[2=s+(1+n+s)w]\mp\sqrt{2+s+(1+n+s)^2+4(m-1)(1+w)}\right\}.\]

For $m\ne1$ the general classification provides us with the following set of critical points:
1. Stable node for $m>1$ and \[s<\frac{2+(1+n)w}{1+w}-2\sqrt{\frac{1-m}{1+w}};\]
2. Unstable node for $m>1$ and \[s>\frac{2+(1+n)w}{1+w}+2\sqrt{\frac{1-m}{1+w}};\]
3. Saddle for $m<1$, arbitrary $n$ and $s$;
4. Center for $m>1$ and $2+s+(1+n+s)w=0$ for $n>1$;
5. Stable focus for $m>1$ and $2+s+(1+n+s)w>0$;
6. Unstable focus for $m>1$ and $2+s+(1+n+s)w<0$.