Statefinder parameters for interacting dark energy and cold dark matter

11.9.1

Problem 11.9.1

(after [1])

Show that in flat Universe both the Hubble parameter and deceleration parameter do not depend on whether or not dark components are interacting. Become convinced the second derivative $\ddot H$ does depend on the interaction between the components.

solution

In flat universe the Hubble parameter is determined solely by the density $H^2\propto\rho_{tot}$. The deceleration parameter \[q=-1-\frac{\dot H}{H^2}=\frac12(1+3w_{de}\Omega_{de}).\] It is obvious, that neither the Hubble parameter nor deceleration parameter depend on whether or not dark components are interacting. Consider the following dimensionless combination \[\frac{\ddot H}{H^3}=\frac92\left(1+\frac{p_{de}}{\rho_{tot}}\right)+\frac92\left[w_{de}(1+w_{de})\frac{p_{de}}{\rho_{tot}} - w_{de}\frac{\Pi}{\rho_{tot}} - \frac{\dot w_{de}}{3H}\frac{p_{de}}{\rho_{tot}}\right].\] Unlike $H$ or $\dot H$, the second derivative $\ddot H$ does depend on the interaction between the components. Consequently, to discriminate between models with different interactions or between interacting and non-interacting models it is desirable to characterize the cosmological dynamics additionally by parameters that depend on $\ddot H$.

11.9.2

Problem 11.9.2

Find statefinder parameters for interacting dark energy and cold dark matter.

solution

Statefinder parameters in the general case can be presented in the following form \[\begin{align} & \frac{d}{dt}\frac{{\ddot{a}}}{a}=\frac{a\dddot{a}-\ddot{a}\dot{a}}{{a}^{2}}=\frac{\dddot{a}}{a}-\frac{\ddot{a}}{a}H, \\ & \frac{\dddot{a}}{a}=\frac{d}{dt}\frac{\ddot{a}}{a}+\frac{\ddot{a}}{a}H \\ \end{align}\] Using second Friedmann equation $\left( \frac{8\pi G}{3}=1 \right)$ find \[\begin{align} & \frac{\dddot{a}}{a}=-\frac{1}{2}\left( \dot{\rho }+3\dot{p} \right)-\frac{1}{2}\left( \rho +3p \right),p \\ & \dot{\rho}=-3H\left( \rho +p \right), \\ & r=\frac{\dddot{a}}{a{{H}^{3}}}=\frac{\rho }{{{H}^{2}}}-\frac{3}{2}\frac{{\dot{p}}}{{{H}^{3}}}, \\ & {H}^{2}= \rho ,\quad H=-\frac{1}{3}\frac{\dot{\rho}}{\left( \rho +p \right)}, \\ & r=1+\frac{9}{2}\frac{\rho +p}{\rho }\frac{{\dot{p}}}{{\dot{\rho }}} \\ \end{align}\] For statefinder $s$ obtain \[\begin{align} & s=\frac{r-1}{3\left( q-1/2 \right)}=\frac{\frac{9}{2}\frac{\rho +p}{\rho }\frac{{\dot{p}}}{{\dot{\rho }}}}{3\left( -\frac{\ddot{a}}{a{{H}^{2}}}-\frac{1}{2} \right)}, \\ & -\frac{{\ddot{a}}}{a{{H}^{2}}}-\frac{1}{2}=\frac{-\frac{\ddot{a}}{a}-\frac{{H}^{2}}{2}}{{H}^{2}}=\frac{3}{2}\frac{p}{\rho }, \\ & s=\frac{\rho +p}{p}\frac{\dot{p}}{\dot{\rho }} \\ \end{align}\] Finally \begin{align} \nonumber r & =1+\frac92\frac{\rho_{tot}+p_{tot}}{\rho_{tot}}\frac{\dot p_{tot}}{\dot\rho_{tot}},\\ \nonumber s & =\frac{\rho_{tot}+p_{tot}}{p_{tot}}\frac{\dot p_{tot}}{\dot\rho_{tot}}. \end{align} Use the relation $\dot\rho_{tot}=-3H(\rho_{tot}+p_{tot})$ and the interaction $Q$ in the form $Q=-3H\Pi$ to obtain \begin{align} \nonumber r & =1+\frac92\frac{w_{de}}{1+\kappa}\left[1+w_{de}- \frac{\Pi}{\rho_{tot}} - \frac{\dot w_{de}}{3w_{de}H}\right], \quad \kappa \equiv\frac{\rho_{dm}}{\rho_{de}},\\ \nonumber s & =1+w_{de}- \frac{\Pi}{\rho_{tot}} - \frac{\dot w_{de}}{3w_{de}H}. \end{align}

11.9.3

Problem 11.9.3

Show that the statefinder parameter $r$ is generally necessary to characterize any variation in the overall equation of state of the cosmic medium.

solution

This becomes obvious from the general relation \[r =1+\frac92\frac{\rho_{tot}+p_{tot}}{\rho_{tot}}\frac{\dot p_{tot}}{\dot\rho_{tot}},\] where $p_{tot}$ in this case of interacting dark energy and cold dark matter reduces to $p_{tot}\approx p_{de}$. Since \[\frac{d}{dt}\left(\frac p \rho\right)=\frac{\dot\rho}\rho\left(\frac{\dot p}{\dot\rho}-\frac p\rho\right)\] it is evident, that an interaction term in $\dot p\approx \dot p_{de}$ according to conservation equation will additionally change the time dependence of the overall equation of state parameter $p/\rho$.

11.9.4

Problem 11.9.4

Find relation between the statefinder parameters in the flat Universe.

solution

Use the relations \begin{align} \nonumber r & =1+\frac92\frac{w_{de}}{1+\kappa}\left[1+w_{de}- \frac{\Pi}{\rho_{tot}} - \frac{\dot w_{de}}{3w_{de}H}\right], \quad \kappa \equiv\frac{\rho_{dm}}{\rho_{de}},\\ \nonumber s & =1+w_{de}- \frac{\Pi}{\rho_{tot}} - \frac{\dot w_{de}}{3w_{de}H}. \end{align} to obtain \[r = 1+\frac92w_{de}\Omega_{de}s,\] Then use \begin{align} \nonumber r & =1+\frac92\frac{\rho_{tot}+p_{tot}}{\rho_{tot}}\frac{\dot p_{tot}}{\dot\rho_{tot}},\\ \nonumber s & =\frac{\rho_{tot}+p_{tot}}{p_{tot}}\frac{\dot p_{tot}}{\dot\rho_{tot}}, \end{align} to obtain \[r=1+\frac92\frac{p_{tot}}{\rho_{tot}}s.\] The first result obviously follows from the second one: \[r=1+\frac92\frac{p_{tot}}{\rho_{tot}}s =1+\frac92\frac{p_{de}}{\rho_{de}+\rho_{dm}}s =1+\frac92\frac{w_{de}\rho_{de}}{\rho_{de}+\rho_{dm}}s=1+\frac92\frac{w_{de}\Omega_{de}}{\Omega_{de}+\Omega_{dm}}s = 1+\frac92w_{de}\Omega_{de}s.\]

11.9.5

Problem 11.9.5

Express the statefinder parameters in terms of effective state parameter $w_{(de)eff}$, for which \[\dot\rho_{de}+3H(1+w_{(de)eff})\rho_{de}=0.\]

solution

Substitute \[w_{(de)eff}=w_{de}-\frac\Pi{\rho_{de}}\] into \[s =1+w_{de}- \frac{\Pi}{\rho_{tot}} - \frac{\dot w_{de}}{3w_{de}H}\] to obtain \[s =1+w_{(de)eff}- \frac{w'_{de}}{3w_{de}},\] where prime denotes derivative with respect to $u=\ln a$. Statefinder parameter $r$ can be found using the relation \[r=1+\frac92w_{de}\Omega_{de}s.\]

11.9.6

Problem 11.9.6

Find the statefinder parameters for $Q=3\delta H\rho_{dm}$, assuming that $w_{de}=const$.

solution

In the considered case \[s=1+w_{de}-\frac\Pi{\rho_{de}}.\] Substitute \[\Pi=-\delta\frac{\rho_{dm}}{\rho_{de}}\] to find \[s=1+w_{de}+\delta(\Omega_{de}^{-1}-1),\quad r=1+\frac92w_{de}\Omega_{de}s.\]

11.9.7

Problem 11.9.7

Find statefinder parameters for the case $\rho_{dm}/\rho_{de}=a^{-\xi}$, where $\xi$ is a constant parameter in the range $[0,3]$ and $w_{de}=const$.

solution

If $w_{de}=const$, the interactions which produce scaling solutions $\rho_{dm}/\rho_{de}=a^{-\xi}$ are given by \[\frac{\Pi}{\rho_{de}}=\left(w_{de}+\frac\xi3\right)\frac{\kappa_0(1+z)^\xi}{1+\kappa_0(1+z)^\xi}\] where $\kappa_0$ denotes the present energy density ratio. Insert this expression into \begin{align} \nonumber r & =1+\frac92\frac{w_{de}}{1+\kappa}\left[1+w_{de}- \frac{\Pi}{\rho_{tot}} - \frac{\dot w_{de}}{3w_{de}H}\right], \quad \kappa \equiv\frac{\rho_{dm}}{\rho_{de}},\\ \nonumber s & =1+w_{de}- \frac{\Pi}{\rho_{tot}} - \frac{\dot w_{de}}{3w_{de}H} \end{align} to get the statefinder parameters \begin{align} \nonumber r & =1+\frac92\frac{w_{de}}{1+\kappa_0(1+z)^\xi}\left[1+w_{de}- \left(w_{de}+\frac\xi3\right)\frac{\kappa_0(1+z)^\xi}{1+\kappa_0(1+z)^\xi}\right],\\ \nonumber s & =1+w_{de}- \left(w_{de}+\frac\xi3\right)\frac{\kappa_0(1+z)^\xi}{1+\kappa_0(1+z)^\xi}. \end{align}

11.9.8

Problem 11.9.8

Show that in the case $\rho_{dm}/\rho_{de}=a^{-\xi}$ the current value of the statefinder parameter $s=s_0$ can be used to measure the deviation of cosmological models from the SCM.

solution

For the scaling models $\rho_{dm}/\rho_{de}=a^{-\xi}$ we have \[r_0 =1+\frac92\frac{w_{de}}{1+\kappa_0}s_0,\quad s_0 =1+w_{de}- \left(w_{de}+\frac\xi3\right)\frac{\kappa_0}{1+\kappa_0}.\] Realizing that \[q_0=\left.q\right|_{t=t_0}= \left.-\frac{\ddot a}{aH^2}\right|_{t=t_0}= \left.-\frac{\frac16(\rho_{tot}+3p_{tot})}{\frac13\rho_{tot}}\right|_{t=t_0} = \frac12\frac{1+\kappa_0+w_{de}}{1+\kappa_0}.\] Introduce \[q_{0SCM}\equiv q_)(w_{de}=-1)=-\frac12\frac{2-\kappa_0}{1+\kappa_0}.\] to present the current value of the statefinder parameter $s_0$ in the following form \[s_0(q_0,\xi)=\frac23\left[q_0-q_{0SCM}+\left(\frac\xi3-1\right)(1+q_{0SCM})\right].\] The first part in the bracket on the right hand side describes the deviation from $w_{de}=-1$, the second part accounts for the deviations from the SCM scaling $\xi=3$. Of course, $\xi=3$ corresponds to the SCM model with $s_0=0$.

11.9.9

Problem 11.9.9

Find how the statefinder parameters enter the expression for the luminosity distance.

solution

Up to second order in the redshift the Hubble parameter is \[H(z)=H_0+\left(\frac{dH}{dz}\right)_{z=0}z+\frac12\left(\frac{d^2H}{dz^2}\right)_{z=0}z^2.\] By virtue of \[\frac{dH}{dz}=\frac{q+1}{z+1}H,\quad \frac{d^2H}{dz^2}=\frac{r-1+2(1+q)-(1+q)^2}{(1+z)^2}H,\] this can be written as \[H(z)=H_0\left\{1+(1+q_0)z+\frac12\left[r_0-1+2(1+q_0)-(1+q_0)^2\right]z^2\right\}.\] The luminosity distance \[d_L=(1+z)\int\frac{dz}H\] becomes \[d_L=\frac z{H_0}\left\{1+\frac12(1-q_0)z+\frac16\left[3(1+q_0)^2-5(1+q_0)+1-r_0\right]z^2\right\}.\] Since the interaction affects $r_0$ but neither $q_0$ nor $H$, it is obvious that the deviation of luminosity distances of different interacting as well as of interacting and non-interacting models manifests itself only in third order in the redshift.