Difference between revisions of "New oleg"
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\end{equation} | \end{equation} | ||
will deviate from the standard case ($\rho_{m0}$ is the present value of $\rho_m$). | 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 $q, j, l, s$ | ||
<div class="NavFrame collapsed"> | <div class="NavFrame collapsed"> | ||
<div class="NavHead">solution</div> | <div class="NavHead">solution</div> | ||
<div style="width:100%;" class="NavContent"> | <div style="width:100%;" class="NavContent"> | ||
| − | <p style="text-align: left;"></p> | + | <p style="text-align: left;">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 previous section 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} | ||
| + | </p> | ||
</div> | </div> | ||
</div></div> | </div></div> | ||
Revision as of 21:49, 9 April 2015
Problem 1
problem id:
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 previous section 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
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Problem 3
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Problem 4
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Problem 5
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Problem 6
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