Difference between revisions of "New oleg"

The dark energy density in the consider model model takes on the form: \begin{equation}\label{rhode} \rho_{de}=\rho_{de0} a^{-3(1+w)}-\frac{a^{-3(1-\delta)}\delta\rho_{dm0}}{w+\delta}, \end{equation} where the integration constant $\rho_{de0}$ equals to the present value of the dark energy density. Substituting the energy density dependence of the scale factor into the Friedmann equation, we get: \begin{equation}\label{habl} H^2=\frac 13\left(\rho_{de0} a^{-3(1+w)}-\frac{a^{-3(1-\delta)}\delta\rho_{dm0}}{w+\delta}+\rho_{dm0}a^{-3(1-\delta)} \right). \end{equation} Note that here we consider a flat ($k=0$) Universe. Using the method described in the problem New_problems#1301_38 we can write the equivalent Friedmann equation expressed in cosmographic parameters: \begin{equation}\label{fridman_Q_rm} (1+3w)(3\delta-1)=2(j-3q(1+w-\delta)). \end{equation} In order to express the coupling constant $\delta$ in terms of the cosmographic parameters we must know the second, the third and the fourth derivative of the scale factor, because we need three independent equations to resolve three independent constants. After the straightforward calculations we obtain \begin{equation}\label{delta-first11} -2q(2+3w)(3\delta-2)+j(-5+3(\delta-w))-qj-s=0, \end{equation} or \begin{equation}\label{delta-first} \delta=\frac{-2q(2+3w)+j(5+3w)+jq+s}{3(j-q(2+3w))}. \end{equation}

For model with interaction term $Q=3H\delta\rho_{de}$ the equivalent Friedmann equation expressed in cosmographic parameters has the form: \begin{equation}\label{fridman_Q_rde} 1+3(w+\delta)=-2(j-3q(1+w+\delta)), \end{equation} for $\delta=0$ and $w=-1$ this equation also take the form (\ref{8}). And coupling constant $\delta$ \begin{equation}\label{delta-rde} -2q(2+3(w+\delta))+j(2+3(w+\delta))+qj+s=0, \end{equation} or \begin{equation}\label{delta-rde} \delta=\frac{2q(2+3w)-j(5+3w)-qj-s}{3(j-2q)}. \end{equation}