Difference between revisions of "Transient acceleration"

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(Problem 5)
(Problem 6)
 
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\end{eqnarray}
 
\end{eqnarray}
 
Than using result of the previous problem we can find
 
Than using result of the previous problem we can find
\begin{eqnarray}
+
\begin{equation}
% \nonumber to remove numbering (before each equation)
+
q \equiv -\frac{\ddot a}{a H^2} = \frac{1}{2}\sum_i \Omega_i (1 + 3w_i) = \frac{1}{2}(1+\Omega_r+3 w_Q\Omega_Q)\label{q1},
q &\equiv& -\frac{\ddot a}{a H^2} = \frac{1}{2}\sum_i \Omega_i (1 + 3w_i) = \frac{1}{2}(1+\Omega_r+3 w_Q\Omega_Q)\label{q1}~,\\
+
\end{equation}
D_L &\equiv& H_0 d_L = (1+z)\int_0^z~dz' \frac{H_0}{H(z')} = e^{-N}\int_N^{0}~dN' e^{-N'}\frac{H_0}{H}~,\label{DL}\\
+
\begin{equation}
H_0 t_0 &=& \int_{0}^{+\infty}~dz \frac{H_0}{(1+z)H(z)} = \int_{-\infty}^{0}~dN \frac{H_0}{H}~,\label{t0}
+
D_L \equiv H_0 d_L = (1+z)\int_0^z~dz' \frac{H_0}{H(z')} = e^{-N}\int_N^{0}~dN' e^{-N'}\frac{H_0}{H}~,\label{DL}
\end{eqnarray}
+
\end{equation}
 +
\begin{equation}
 +
H_0 t_0 = \int_{0}^{+\infty}~dz \frac{H_0}{(1+z)H(z)} = \int_{-\infty}^{0}~dN \frac{H_0}{H}~,\label{t0}
 +
\end{equation}
 
with
 
with
  

Latest revision as of 04:44, 11 November 2013



Problem 1

Consider a simple model of transient acceleration with decaying cosmological constant \begin{equation} \label{ec} \dot{\rho}_{m} + 3\frac{\dot{a}}{a}\rho_{m} = - \dot{\rho}_\Lambda\;, \end{equation} where $\rho_{m}$ and $\rho_\Lambda$ energy density DE and cosmological constant $\Lambda$. At the early stages of the expansion of the Universe, when $\rho_\Lambda$ is quite small, such a decay does not influence cosmological evolution in any way. At later stages, as the DE contribution increases, its decay has an ever increasing effect on the standard dependence of the DM energy density $\rho_{m} \propto a^{-3}$ on the scale factor $a$. We consider the deviation to be described by a function of the scale factor - $\epsilon(a)$. \begin{equation} \label{dm} \rho_{m} = \rho_{m, 0}a^{-3 + \epsilon(a)}\;, \end{equation} where $a_0 = 1$ in the present epoch. Other fields of matter (radiation, baryons) evolve independently and are conserved. Hence, the DE density has the form \begin{equation}\label{decayv} \rho_{\Lambda} = \rho_{m0} \int\limits_{a}^{1}\frac{\epsilon(\tilde{a}) + \tilde{a}\epsilon' \ln(\tilde{a})}{\tilde{a}^{4 - \epsilon(\tilde a)}} d\tilde{a} + {\rm{X}}\;, %\tilde\rho_{\Lambda 0} + \end{equation} where the prime denotes the derivative with respect to the scale factor, and ${\rm{X}}$ is the integration constant. If radiation is neglected, the first Friedmann equation takes the form \begin{equation} \label{friedmann} {H}= H_0\left[\Omega_{b,0}{a}^{-3} + \Omega_{m0}\varphi(a) + {\Omega}_{\rm{X,0}}\right]^{1/2}, \end{equation} Using the assumption that the function $\epsilon (a)$ has the following simple form \begin{equation} \label{Parametrization_a} \epsilon(a) = \epsilon_0a^\xi\ = \epsilon_0(1+z)^{-\xi}, \end{equation} where $\epsilon_0$ and $\xi$ can take both positive and negative values, find function $\varphi(a)$ and relative energy density $\Omega_b(a)$, $\Omega_{m}(a)$ and $\Omega_{\Lambda}(a)$.


Problem 2

Using the results of the previous problem, find the deceleration parameter for this model is $q(a)$. Draw the graph deceleration parameter as a function of $\log(a)$ for various values of $\epsilon_0$ and $\xi$: $\xi = 1.0$ and $\epsilon_0 = 0.1$, $\xi = -1.0$ and $\epsilon_0 = 0.1$, $\xi = 0.8$ and $\epsilon_0 = 0.5$, $\xi = -0.5$ and $\epsilon_0 = -0.1.$.


Problem 3

Consider the possibility of an accelerating transient regime within the interacting scalar field model using potential of the form \begin{equation} \label{V_fransient} V(\phi)=\rho_{\phi\,0}[1-\frac{\lambda}{6}(1+\alpha \sqrt{\sigma}\phi)^2)] \exp{[-\lambda\sqrt{\sigma}(\phi+ \frac{\alpha \sqrt{\sigma}}{2}\phi^2)]}, \end{equation} where $\rho_{\phi\,0}$ is a constant energy density, $\sigma=8\pi G/\lambda$, and $\alpha$ and $\lambda$ are two dimensionless, positive parameters of the model, that the deceleration parameter is non-monotonically dependent on the scale factor. Plot the deceleration parameter as a function of the scale factor.


Problem 4

Consider a simple parameterization: \begin{equation} \label{Qsimple} Q=3\beta(a)H \rho_{de} \end{equation} with a simple power-law ansatz for $\beta(a)$, namely: \begin{equation} \label{cas32} \beta(a)=\beta_0 a^\xi. \end{equation} Substituting this interaction form into conservation equations for DM and DE: \begin{eqnarray}\label{eom1} \dot{\rho}_{dm}+3H\rho_{dm}=Q, \end{eqnarray} \begin{eqnarray}\label{eom2} \dot{\rho}_{de}+3H(\rho_{de}+p_{de})=-Q, \end{eqnarray} we get \begin{equation} \label{rhophi2} \rho_{de}=\rho_{de0}\, a^{-3(1+w_0)}\cdot \exp{\left[\frac{3\beta_0(1-a^\xi)}{\xi}\right]}, \end{equation} where the integration constant $\rho_{de0}$ is value of the dark energy at present, and the dark energy EoS parameter $w\equiv p_{de}/\rho_{de}$ is a constant-$w_0$. Substituting Eq. (\ref{rhophi2}) into Eq. (\ref{eom2}), we get the dark matter energy density, \begin{equation}\label{rhom2} \rho_{dm}=f(a)\rho_{dm0}, \end{equation} where \begin{equation} \label{f1} f(a)\equiv \frac{1}{a^3}\left\{1-\frac{\Omega_{de0}}{\Omega_{dm0}}\frac{3\beta_0 a^{-3w_0}e^{\frac{3\beta_0}{\xi}}}{\xi}\cdot\left[a^\xi E_{\frac{3w_0}{\xi}}\left(\frac{3\beta_0 a^\xi}{\xi}\right)-a^{3w_0} E_{\frac{3w_0}{\xi}} \left(\frac{3\beta_0}{\xi}\right)\right]\right\}, \end{equation} where $\rho_{dm0}$ is dark matter density at present day, and $E_n(z)=\int_1^\infty t^{-n}e^{-xt}dt$ the usual exponential integral function. Note however that Eq. (\ref{rhom2}) is an analytical expression, while in the corresponding expressions were left as integrals and were calculated numerically. Obviously, in the case of non-interaction (that is, for $\beta_0=0$), Eq. (\ref{rhom2}) recovers the standard result $\rho_{dm}=\rho_{dm0}/a^3$. For the special case $\xi=0$ find dimensionless Hubble parameter $E^2(z)\equiv \frac{H^{2}}{H^{2}_0}$, the evolution of the density parameters $\Omega_b(a)$, $\Omega_{dm}(a)$ and $\Omega_{de}(a)$ and $q(a)$. For what values of $\beta_0$ the cosmic acceleration is transient?


Problem 5

Consider the flat FLRW cosmology with two coupled homogeneous scalar fields $\Phi$ and $\Psi$: \begin{eqnarray} \dot{\rho_b} &=& -3 H \gamma_b \rho_b \\ \ddot{\Phi} &=& -3H\dot{\Phi} - \partial_\Phi V \\ \ddot{\Psi} &=& -3H\dot{\Psi} - \partial_\Psi V \\ \dot{H} &=& -4\pi G(\gamma_m\rho_m+\gamma_r\rho_r+\gamma_Q\rho_Q)~, \end{eqnarray} \begin{equation} H^2 = \frac{8\pi G}{3}(\rho_m+\rho_r+\rho_Q)-\frac{k}{a^2}~. \end{equation} Here a dot denotes a derivative with respect to the cosmic time $t$, the subscript $b$ refers to the dominant background quantity, either dust (m) or radiation (r) while $Q$ refers to the Dark Energy sector, here the two quintessence scalar fields.

The quintessence fields with potential $V$ have the following energy density and pressure: \begin{eqnarray} \rho_Q =\frac{1}{2}\dot{\Phi}^2 + \frac{1}{2}\dot{\Psi}^2 + V(\Phi,\Psi)\\ p_Q =\frac{1}{2}\dot{\Phi}^2 + \frac{1}{2}\dot{\Psi}^2 - V(\Phi,\Psi) \end{eqnarray} with $p_Q = (\gamma_Q-1)\rho_Q$. It is convenient to define the following new variables: \begin{equation} X_{\Phi}=\sqrt{\frac{8\pi G}{3H^2}}~\frac{\dot{\Phi}}{\sqrt{2}}, X_{\Psi}=\sqrt{\frac{8\pi G}{3H^2}}~\frac{\dot{\Psi}}{\sqrt{2}}, X_V=\sqrt{\frac{8\pi G}{3H^2}}~\sqrt{V}. \end{equation} Find expressions for the $\Phi'$, $\Psi'$, $X_{\Phi}'$, $X_{\Psi}'$, $X_V'$ where a prime denotes a derivative with respect to the quantity $N$, the number of e-folds with respect to the present time, \begin{equation} N\equiv {\rm ln} \frac{a}{a_0}~,\label{N} \end{equation} and we have also $H=\dot{N}$.


Problem 6

Using result from the previous problem find the relative energy density for matter, radiation and quintessence, $\Omega_m$, $\Omega_r$ and $\Omega_Q$, the deceleration parameter $q$, the Hubble-parameter-free luminosity distance $D_L$ and the age of the Universe $t_0$.