Difference between revisions of "New oleg"
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<p style="text-align: left;">For model with interaction term $Q=3H\delta\rho_{de}$ the equivalent Friedmann equation expressed in cosmographic parameters has the form: | <p style="text-align: left;">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} | \begin{equation}\label{fridman_Q_rde} | ||
| − | 1+3(w+\delta)=-2(j-3q(1+w+\delta)) | + | 1+3(w+\delta)=-2(j-3q(1+w+\delta)). |
\end{equation} | \end{equation} | ||
| − | |||
And coupling constant $\delta$ | And coupling constant $\delta$ | ||
\begin{equation}\label{delta-rde} | \begin{equation}\label{delta-rde} | ||
Revision as of 21:53, 9 April 2015
Problem 1
problem id: G-interact-1
Let us consider a very simple model with the following conservation equations: \begin{equation}\label{conserv_Qm} \dot{\rho_{dm}}+3H\rho_{dm}=Q, \end{equation} \begin{equation}\label{conserv_Qe} \dot{\rho_{de}}+3H\rho_{de}(1+w)=-Q, \end{equation} where $Q=3H\delta\rho_{dm}$ is the source of interaction and $w$ is the state equation parameter for dark energy. The positive small coupling constant $\delta$ will eventually characterize how evolution of the matter energy density \begin{equation}\label{rho-m1} \rho_{dm}= \rho_{dm0}a^{-3(1-\delta)}, \end{equation} will deviate from the standard case ($\rho_{m0}$ is the present value of $\rho_m$).
For this model find dependence of the dark energy density on the scale factor and using the method described in the problem New_problems#1301_38 express Friedmann equation in terms of the cosmographic parameters $q, j, l, s$
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}
Problem 2
problem id: G-interact-1
Make the same calculations as in the previous problem for model with interaction term $Q=3H\delta\rho_{de}$.
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} 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}
Problem 3
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Problem 4
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Problem 5
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Problem 6
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