Phenomenology of interacting models

11.2.1

Problem 11.2.1

Find the effective state parameters $w_{(de)eff}$ and $w_{(dm)eff}$ that would allow one to treat the interacting dark components as non-interacting.

solution

The system which describes the interacting dark components can be transformed into the standard form that corresponds to non-interacting components by redefining the parameters $w_{de}$ and $w_{dm}$ \[\dot\rho_i+3H(1+w_{(i)eff})\rho_i=0,\quad i=\{de,dm\}.\] where in the case of cold DM ($w_{dm}=0$) \[w_{(de)eff}=w_{de}+\frac{Q}{3H\rho_{de}},\quad w_{(dm)eff} = -\frac{Q}{3H\rho_{dm}}.\] If one "turns off" the interaction ($Q=0$), the original parameters are recovered: \[w_{(de)eff}=w_{de}, \quad w_{(dm)eff}=0.\]

11.2.2

Problem 11.2.2

Using the effective state parameters obtained in the previous problem, analyze dynamics of dark matter and dark energy depending on sign of the rate of energy density exchange in the dark sector.

solution

If $Q>0$: dark energy $\to$ dark matter, $w_{(dm)eff}<0$ and dark matter redshifts slower than $a^{-3}$, $w_{(de)eff}>w_{de}$ and dark energy has less accelerating power.
If $Q<0$: dark matter $\to$ dark energy, $w_{(dm)eff}>0$ and dark matter redshifts faster than $a^{-3}$, $w_{(de)eff}>w_{de}$ and dark energy has more accelerating power.

11.2.3

Problem 11.2.3

Find the effective state parameters $w_{(de)eff}$ and $w_{(dm)eff}$ for the case of the warm dark matter ($w_{dm}\ne0$) and analyze the features of dynamics in this case.

solution

For the warm dark matter \[w_{(de)eff}=w_{de}+\frac{Q}{3H\rho_{de}},\quad w_{(dm)eff} = w_{dm}-\frac{Q}{3H\rho_{dm}}.\] Note that the behavior of $w_{(dm)eff}$ can be quite different if $w_{dm}\ne0$. With the usual assumption of cold DM ($w_{dm}=0$, $Q>0$), which implies that energy is transferred from DE to DM, some kind of exotic dark matter with a negative EoS parameter is driven, assuming, of course, that we are in an expanding Universe ($H>0$). For warm dark matter we would have, depending on the type of interaction considered and the strength of the coupling constant $Q$ appearing, a possibility to change the sign of the effective parameter during the cosmic evolution.
For $Q>0$ the expression for $w_{(de)eff}$ indicates, even for $w_{de}=-1$ (the cosmological constant), that the DE fluid will behave as a quintessence field. Thus an effective phantom behavior can only be obtained if $w_{de}<-1$.

11.2.4

Problem 11.2.4

Show that the quintessence coupled to DM with certain sign of the coupling constant behaves like a phantom uncoupled model, but without negative kinetic energy.

solution

Use the result of the previous problem. When $Q<0$ it is possible that $w_{(de)eff}<w_{de}$. This means that the coupled quintessence behaves like a phantom uncoupled model, but without negative kinetic energy.

11.2.5

Problem 11.2.5

In order to compare dynamics of a model with observational results it is useful to analyze all dynamic variables as functions of redshift rather than time. Obtain the corresponding transformation for the system of interacting dark components.

solution

Use the identity \[\frac{d}{dt} = \frac{d}{dz}\frac{dz}{da}\frac{da}{dt} = -(1+z)H(z)\frac{d}{dz}\] and transform the basic equations \[\dot\rho_{dm} + 3H\rho_{dm} = Q,\] \[\dot\rho_{de} + 3H\rho_{de}(1+w_{de}) = -Q\] to the form \[\frac{d\rho_{dm}}{dz}-\frac{3}{1+z}\rho_{dm} = -\frac{Q(z)}{(1+z)H(z)},\] \[\frac{d\rho_{de}}{dz}-\frac{3}{1+z}(1+w_{de})\rho_{de} = \frac{Q(z)}{(1+z)H(z)}.\]

11.2.6

Problem 11.2.6

Show that energy exchange between dark components leads to time-dependent effective potential energy term in the first Friedman equation. (after [1]).

solution

We assume for simplicity some two-component fluid with effective pressure and energy \[p=p_{de},\quad \rho=\rho_{de}+\rho_{dm}.\] The conservation condition can be rewritten to the form \begin{equation}\label{IDE_9_0_1} \frac1{a^3}\frac d{dt} (\rho_{dm}a^3)+\frac1{a^{3(1+w_{de})}}\frac d{dt}\left(\rho_{de}a^{3(1+w_{de})}\right)=0. \end{equation} Relation (\ref{IDE_9_0_1}) can be written as \begin{equation}\label{IDE_9_0_2} \frac1{a^3}\frac d{dt} (\rho_{dm}a^3) = \gamma(t),\quad \frac1{a^{3(1+w_{de})}}\frac d{dt}\left(\rho_{de}a^{3(1+w_{de})}\right)=-\gamma(t). \end{equation} The function $\gamma(t)$ describes the interaction between two dark components. Integration of (\ref{IDE_9_0_2}) gives \begin{align} \nonumber \rho_{dm}a^3= & \rho_{dm0}a_0^3 + \int\limits_{t_0}^t\gamma(t)a^3dt,\\ \label{IDE_9_0_3} \rho_{de}a^{3(1+w_{de})}= & \rho_{de0}a_0^{3(1+w_{de})} + \int\limits_{t_0}^t\gamma(t)a^{3(1+w_{de})}dt. \end{align} Let us represent the first Friedman equation in the form of energy conservation law for a particle moving in the one-dimensional potential \begin{equation}\label{IDE_9_0_4} \frac{\dot a^2}2 +V(a)=\frac k2,\quad V(a) = -\frac{a^2}6(\rho_{dm}+\rho_{de}). \end{equation} In presence of the interaction the potential function is explicitly time dependent and now takes the following form \begin{equation}\label{IDE_9_0_5} V(a) = -\frac12\left[\frac{A(t)} a+ \frac{B(t)}{a^1+3w_{de}}\right], \end{equation} where \begin{equation}\label{IDE_9_0_6} A(t)=\frac{\rho_{dm}}3,\quad B(t)=\frac{\rho_{de}a^{3(1+w_{de})}}3. \end{equation} In the SCM model both matter and the cosmological constant are treated separately without the interaction, so both functions $A(t)$ and $B(t)$ are constant. The presence of the interaction manifests in the model by appearing the time dependence of the potential function (\ref{IDE_9_0_5}).

11.2.7

Problem 11.2.7

Show that the system of interacting components can be treated as the uncoupled one due to introduction of the partial effective pressure of the dark components

    \[\Pi_{de}\equiv\frac{Q}{3H},\quad \Pi_{dm}\equiv-\frac{Q}{3H}.\]

solution

\[\dot\rho_{dm} + 3H(\rho_{dm} + \Pi_{dm} = 0,\] \[\dot\rho_{de} + 3H(\rho_{dm} + p_{de} + \Pi_{de} = 0,\] In this case, the conservation equations formally look as those for two independent fluids. A coupling between them has been mapped into the relation $\Pi_{de}=-\Pi_{dm}$.

11.2.8

Problem 11.2.8

Assume that the mass $m_{dm}$ of dark matter particles depends on a scalar field $\varphi$. Construct the model of interacting dark energy and dark matter in this case.

solution

Use the standard definition for the dark matter density \begin{equation}\label{IDE_10_1} \rho_{dm} = m_{dm}(\varphi) n_{dm} \end{equation} The number density $n_{dm}$ satisfies the equation \begin{equation}\label{IDE_10_2} \dot n_{dm} + 3Hn_{dm} = 0 \end{equation} Taking time derivatives of (\ref{IDE_10_1}) and using (\ref{IDE_10_2}), one obtains \[\dot\rho_{dm} + 3H\rho_{dm} = \frac{1}{m_{dm}(\varphi)}\frac{d m_{dm}(\varphi)}{d\varphi}\dot\varphi\rho_{dm}.\] In the case of mass independent from $\varphi$ one recovers 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 \[\dot\rho_{de} + 3H(\rho_{de} + p_{de}) = -\frac{1}{m_{dm}(\varphi)}\frac{d m_{dm}(\varphi)}{d\varphi}\dot\varphi\rho_{dm}.\]

11.2.9

Problem 11.2.9

Assume that the mass $m_{dm}$ of DM particles depends exponentially on the DE scalar field $m=m_*e^{-\lambda\varphi}$. Find the interaction term $Q$ in this case.

solution

Using the result of the previous problem one finds \[Q = \frac{1}{m_{dm}(\varphi)}\frac{d m_{dm}(\varphi)}{d\varphi}\dot\varphi\rho_{dm}= -\lambda\dot\varphi\rho_{dm}.\]

11.2.10

Problem 11.2.10

Find the equation of motion for the scalar field interacting with dark matter if its particles' mass depends on the scalar field.

solution

Substitute of standard definitions of energy density and pressure of the scalar field into the conservation equation to obtain \[\ddot\varphi + 3H\dot\varphi + \frac{dV}{d\varphi} = -\frac{1}{m_{dm}(\varphi)}\frac{d m_{dm}(\varphi)}{d\varphi}\rho_{dm}.\]

11.2.11

Problem 11.2.11

Make the transformation from the variables $(\rho_{de}, \rho_{dm})$ to \[\left(r=\frac{\rho_{dm}}{\rho_{de}}, \rho = \rho_{dm} + \rho_{de}\right)\] for the system of interacting dark components.

solution

\[\rho' = -\left(1+\frac{w_{de}}{1+r}\right)\rho,\] \[r' = r\left[w_{de} -\frac{(1+r)^2}{r\rho}\Pi\right],\] where \[\Pi\equiv-\frac{Q}{3H}\] is effective pressure and the prime denote the derivatives with respect to $\ln a^3$.

11.2.12

Problem 11.2.12

Generalize the result of previous problem to the case of warm dark matter.

solution

\[\rho' = -\left(1+\frac{w_{dm} r + w_{de}}{1+r}\right)\rho,\] \[r' = r\left[w_{de} -w_{dm} -\frac{(1+r)^2}{r\rho}\Pi\right].\]

11.2.13

Problem 11.2.13

Calculate the derivatives $dr/dt$ and $dr/dH$ for the case of flat universe with the interaction $Q$.

solution

Using the result of problem 9 and the relation \[\frac d{du}=\frac1{3H}\frac d{dt},\] where $u\equiv\ln a^3,$ one obtains \[\dot r=3Hr\left[w_{de}+\frac Q{9H^3}\frac{(r+1)^2}{r}\right].\] To calculate $dr/dH$ use the relations \[\dot r=\dot H\frac{dr}{dH},\quad \dot H=-\frac12(\rho_{dm}+\rho_{de}+w_{de}\rho_{de}).\] Substitute $\rho_{de}+\rho_{de}= 3H^2$ and \[\rho_{de}=\frac{3H^2}{1+r}\] to obtain \[\dot H=-\frac32\frac{1+w_{de}+r}{1+r}H^2.\] Ultimately the derivative $dr/dH$ takes the form \[\frac{dr}{dH}=\frac I H,\quad I\equiv-2r\frac{1+r}{1+w_{de}+r}\left[w_{de}+\frac Q{9H^3}\frac{(r+1)^2}{r}\right].\]

11.2.14

Problem 11.2.14

It was shown in the previous problem that \[\dot r=r\left(\frac{\dot\rho_{dm}}{\rho_{dm}}-\frac{\dot\rho_{de}}{\rho_{de}}\right) = 3Hr \left(w_{de} +\frac{1+r}{\rho_{dm}}\frac Q{3H}\right)=(1+r)\left[3Hw_{de}\frac{r}{1+r}+\Gamma\right],\quad \Gamma\equiv\frac Q {\rho_{de}}.\] Exclude the interaction $Q$ and reformulate the equation in terms of $\rho_{de}$, $H$ and its derivatives.

solution

A combination of field equations for a spatially flat homogeneous and isotropic Universe \begin{align} \nonumber H^2 & =\frac13\rho_{tot},\\ \nonumber \dot H & =-\frac12(\rho_{tot}+p_{tot}), \end{align} yields \[2\frac{\dot H}{H}=-3H\left(1+\frac{p_{tot}}{\rho_{tot}}\right).\] Using \[\frac{\dot\rho_{de}}{\rho_{de}}=-3H(1+w_{de})-\Gamma,\] we obtain \[\frac{\dot\rho_{de}}{\rho_{de}} - 2\frac{\dot H}{H}=-\left(3Hw_{de}\frac r{1+r}+\Gamma\right).\] Comparing this equation with \[\dot r=(1+r)\left[3Hw_{de}\frac{r}{1+r}+\Gamma\right],\] we find that the dynamics of the ratio $r$ is governed by \[\dot r =-(1+r)\left(\frac{\dot\rho_{de}}{\rho_{de}} - 2\frac{\dot H}{H}\right).\]

11.2.15

Problem 11.2.15

Generalize the result, obtained in the previous problem, for the case of non-flat Universe [2]

solution

A combination of field equations for a spatially flat homogeneous and isotropic Universe \begin{align} \nonumber H^2 & =\frac13\rho_{tot} -\frac k{a^2},\\ \nonumber \dot H & =-\frac12(\rho_{tot}+p_{tot}) + \frac k{a^2}, \end{align} yields \[2\frac{\dot H}{H}=-3H\left(1+\frac{p_{tot}}{\rho_{tot}}\right) -\frac{k}{a^2H}\left(1+3\frac{p_{tot}}{\rho_{tot}}\right).\] In this case \[\frac{\dot\rho_{de}}{\rho_{de}} - 2\frac{\dot H}{H}=-\left(3Hw_{de}\frac r{1+r}+\Gamma\right) + \frac{k}{a^2H}\left(1+\frac{3w}{1+r}\right).\] Comparing this equation with \[\dot r=(1+r)\left[3Hw_{de}\frac{r}{1+r}+\Gamma\right],\] we find that the dynamics of the ratio $r$ is governed by \[\dot r =-(1+r)\left[\frac{\dot\rho_{de}}{\rho_{de}} - 2\frac{\dot H}{H}-\frac{k}{a^2H}\left(1+\frac{3w}{1+r}\right)\right].\]

11.2.16

Problem 11.2.16

Show that critical points in the system of equations obtained in problem 11 exist only for the case of dark energy of the phantom type.

solution

The critical points of the equation \[\rho' = -\left(1+\frac{w_{de}}{1+r}\right)\rho,\] are determined by the condition $\rho'=0$. The relevant critical point is \[r_c=-(1+w_{de}).\] Consequently, for positive values of $r$, the existence of a critical point requires $w_{de}<-1$, i.e., dark energy of the phantom type. This conclusion does not depend on the interaction.

11.2.17

Problem 11.2.17

Show that the result of previous problem holds also for warm dark matter.

solution

The critical points of the equation \[\rho' = -\left(1+\frac{w_{dm} r + w_{de}}{1+r}\right)\rho,\] are determined by the condition $\rho'=0$. The relevant critical point is \[r_c=-\frac{1+w_{de}}{1+w_{dm}}\] If $w_{dm}>0$ (warm dark matter) and since $r$ must be positive, it follows that $w_{de}<-1$, which corresponds to a phantom DE.

11.2.18

Problem 11.2.18

Show that existence of critical points in the system of equations obtained in problem 11 requires a transfer from dark energy to dark matter.

solution

The condition $\rho'=r'=0$ provides \[\rho_c = -\frac{w_{de}}{1+w_{de}}\Pi_c,\quad \Pi_c\equiv \Pi(\rho_c,r_c)\] Since $w_{de}<-1$, a positive stationary energy density $\rho_c$ requires $\Pi_c<0$, which is equivalent to $Q_c>0$. Consequently, the existence of the critical points $\rho_c$ and $r_c$ requires a transfer from dark energy to dark matter. Note that $\rho_c$ remains undetermined for a linear dependence of $\Pi$ on $\rho$.

11.2.19

Problem 11.2.19

Show that the result of previous problem holds also for warm dark matter.

solution

For this case the expression for $\rho_c$ becomes \[\rho_c=\frac{w_{de}-w_{dm}}{(1+w_{de})(1+w_{dm})}\frac{Q}{3H}.\] The condition $w_{de}<-1$ leads in this case to the same result for the sign of $Q$ as in the previous problem. This result also holds for $-1<w_{de}<0.$.

11.2.20

Problem 11.2.20

Assume that the ratio of the interacting dark components equals \[r\equiv\frac{\rho_{dm}}{\rho_{de}}\propto a^{-\xi}, \quad \xi\ge0.\] Analyze how the interaction $Q$ depends on $\xi$.

solution

Consider the time evolution of the ratio $r$, \[\dot r = r\left(\frac{\dot\rho_{dm}}{\rho_{dm}} - \frac{\dot\rho_{de}}{\rho_{de}}\right) = 3Hr \left( w_{de} \frac{1+r}{\rho_{dm}} \frac{Q}{3H}\right).\] For $r\propto a^{-\xi}$ \[\frac{Q}{3H\rho_{dm}}=-\frac{w_{de}+\xi/3}{1+r}.\] The obtained result demonstrates that by choosing a suitable interaction between dark components, one can produce any desired scaling behavior of the energy densities. The uncoupled case, corresponding to $Q=0$, is given by $w_{de} + \xi/3=0$. The SCM model (the special uncoupled case) corresponds to $w_{de}=-1$, $\xi=3$. Generally, interacting models are parameterized by deviations from $\xi=-3w_{de}$.

11.2.21

Problem 11.2.21

Show that the choice \[r\equiv\frac{\rho_{dm}}{\rho_{de}}\propto a^{-\xi}, \quad (\xi\ge0)\] guarantees existence of an early matter-dominated epoch.

solution

For $\xi>0$, the interaction \[Q=-3H\rho_{dm}\frac{w_{de}+xi/3}{1+r}\] (see previous problem) becomes very small for $a\ll1$. Consequently, the interaction is not relevant at high redshifts. This guarantees the existence of an early matter-dominated epoch.

11.2.22

Problem 11.2.22

Find the interaction $Q$ for the Universe with interacting dark energy and dark matter, assuming that ratio of their densities takes the form

   \[r\equiv\frac{\rho_{dm}}{\rho_{de}}=f(a),\] where $f(a)$ is an arbitrary differentiable function of the scale factor.

solution

\[\dot\rho_{dm}=\dot\rho_{de}f+\rho_{de}f'\dot a,\] where the prime denotes derivative with respect to the scale factor. Substitute this expression into the conservation equation for the dark matter to obtain the following: \[\dot\rho_{de}f+\rho_{de}f'\dot a + 3H\rho_{de} f = Q.\] Insert into the latter the expression for $\dot\rho_{de}$, \[\dot\rho_{de}=-Q-3H\rho_{de}(1+w_{de})\] to finally obtain \[Q=-3H\rho_{de}\frac{f}{1+f}\left(w_{de}-\frac13\frac{f'a}{f}\right).\] For $f\propto a^{-\xi}$ this result coincides with the one obtained in the previous problem.

11.2.23

Problem 11.2.23

Let \[Q=\frac{\dot f(t)}{f(t}\rho_{dm}.\] Show that the sign of the deceleration parameter is defined by the ratio \[\frac{\dot f}{fH}.\]

solution

Represent the Friedman equation with the conservation law in the following form \begin{align} \nonumber H^2 &=\frac13\rho_{tot},\\ \nonumber \frac{\dot H}{H^2} &=\frac32\left(1+\frac{p_{tot}}{\rho_{tot}}\right),\\ \nonumber \dot\rho_{dm}+3H\rho_{dm} &=\frac{\dot f}{f}\rho_{dm},\\ \nonumber \dot\rho_{dm}+3H(1+w_{de})\rho_{de} &=-\frac{\dot f}{f}\rho_{dm}. \end{align} From the conservation equations one easily finds \[w_{de}=-\frac{\dot f}{3Hf}\left(1+\frac{\rho_{dm}}{\rho_{de}}\right)= -\frac{\dot f}{3Hf}(1+r).\] The deceleration parameter is \[q=\frac12\left(3\frac{p_{tot}}{\rho_{tot}+1}\right)=\frac12\left(3\frac{w_{de}}{1+r}+1\right)=\frac12\left(1-\frac{\dot f}{Hf}\right).\]

11.2.24

Problem 11.2.24

Show that in the model, considered in the previous problem, the transition from the accelerated expansion to the decelerated one can occur only due to time dependence of the interaction.

solution

Let us analyze the result, obtained in the previous problem. The sign of $q$ is defined by the ratio \[\frac{\dot f}{Hf}.\] For \[\frac{\dot f}{Hf}<1\] we have $q>0$, i.e., decelerated expansion. For \[\frac{\dot f}{Hf}>1\] we have $q<0$---accelerated expansion. If, in particular, $f$ is such that the ratio $\dot f/f$ changes from $1$ to $-1$, this corresponds to a transition from decelerated to accelerated expansion. Consequently, this transition occurs solely due to interaction.