Difference between revisions of "Realization of interaction in the dark sector"

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(Problem 17)
(Problem 15)
 
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Solution of the equation (\ref{EulerDE}) for $x(y)$, where $x(\tau)$, is easy to obtain.
 
Solution of the equation (\ref{EulerDE}) for $x(y)$, where $x(\tau)$, is easy to obtain.
 
The scale factor and the Hubble parameter can be found using the relations (\ref{xTOa}) and (\ref{xDefn}), resulting in the following
 
The scale factor and the Hubble parameter can be found using the relations (\ref{xTOa}) and (\ref{xDefn}), resulting in the following
\begin{eqnarray}
+
\begin{equation}
a(\tau) & = & \tau^{1/3\gamma} \, \left( c_1 \tau^{m_0} + c_2
+
a(\tau) = \tau^{1/3\gamma} \, \left( c_1 \tau^{m_0} + c_2
 
                                   \tau^{-m_0} \right)^{2/3\gamma}
 
                                   \tau^{-m_0} \right)^{2/3\gamma}
\label{aSoln-2a-2} \\
+
\label{aSoln-2a-2}  
H(\tau) & = & \frac{2H_0}{3\gamma} \, \left[ \frac
+
\end{equation}
 +
\begin{equation}
 +
H(\tau) = \frac{2H_0}{3\gamma} \, \left[ \frac
 
               {m_1 c_1 \tau^{m_0} + m_2 c_2 \tau^{m_0}}
 
               {m_1 c_1 \tau^{m_0} + m_2 c_2 \tau^{m_0}}
 
               {\tau \, ( c_1 \tau^{m_0} + c_2 \tau^{-m_0})} \right] ,
 
               {\tau \, ( c_1 \tau^{m_0} + c_2 \tau^{-m_0})} \right] ,
 
\label{HSoln-2a-2}
 
\label{HSoln-2a-2}
\end{eqnarray}
+
\end{equation}
 
where  $m_0 \equiv \frac{1}{2} \sqrt{1+(3\gamma\tau_0)^2 \lambda_0}$.
 
where  $m_0 \equiv \frac{1}{2} \sqrt{1+(3\gamma\tau_0)^2 \lambda_0}$.
 
It is easy to see that $a(\tau)$ diverges at $\tau=0$ for $\lambda_0>0$ if
 
It is easy to see that $a(\tau)$ diverges at $\tau=0$ for $\lambda_0>0$ if
 
$c_2 \ne 0$. Thus the expressions (\ref{aSoln-2a-2}) and (\ref{HSoln-2a-2}) take on the simplified form
 
$c_2 \ne 0$. Thus the expressions (\ref{aSoln-2a-2}) and (\ref{HSoln-2a-2}) take on the simplified form
\begin{eqnarray}
+
\begin{equation}
a(\tau) & = & \left( \frac{\tau}{\tau_0} \right)^{2m_1/3\gamma}
+
a(\tau) = \left( \frac{\tau}{\tau_0} \right)^{2m_1/3\gamma}
\label{aSoln-2a-3} \\
+
\label{aSoln-2a-3}  
H(\tau) & = & \frac{2H_0}{3\gamma} \, \left[ \frac{m_1}{\tau} \right] ,
+
\end{equation}
 +
 
 +
\begin{equation}
 +
H(\tau) = \frac{2H_0}{3\gamma} \, \left[ \frac{m_1}{\tau} \right] ,
 
\label{HSoln-2a-3}
 
\label{HSoln-2a-3}
\end{eqnarray}
+
\end{equation}
 
where $m_1 \equiv 1/2+m_0,$ à $\tau_0 = 2m_1/3\gamma.$
 
where $m_1 \equiv 1/2+m_0,$ à $\tau_0 = 2m_1/3\gamma.$
 
<br/>
 
<br/>
 
'''b)''' $\lambda_0 = -1/(3\gamma\tau_0)^2.$
 
'''b)''' $\lambda_0 = -1/(3\gamma\tau_0)^2.$
 
Proceed in analogous way to obtain
 
Proceed in analogous way to obtain
\begin{eqnarray}
+
\begin{equation}
a(\tau) & = & \tau^{1/3\gamma} \, (c_3 + c_4 \ln \tau)^{2/3\gamma}
+
a(\tau) = \tau^{1/3\gamma} \, (c_3 + c_4 \ln \tau)^{2/3\gamma}
\label{aSoln-2b-1} \\
+
\label{aSoln-2b-1}  
H(\tau) & = & \frac{2H_0}{3\gamma} \, \left[ \frac
+
\end{equation}
 +
\begin{equation}
 +
H(\tau) = \frac{2H_0}{3\gamma} \, \left[ \frac
 
               {(c_3+2c_4) + c_4 \ln\tau}
 
               {(c_3+2c_4) + c_4 \ln\tau}
 
               {2\tau ( c_3 + c_4 \ln\tau)} \right] ,
 
               {2\tau ( c_3 + c_4 \ln\tau)} \right] ,
 
\label{HSoln-2b-1}
 
\label{HSoln-2b-1}
\end{eqnarray}
+
\end{equation}
 
where $c_3,c_4$ are arbitrary constants.
 
where $c_3,c_4$ are arbitrary constants.
 
In order to make $a(\tau)$ finite at $\tau=0$ set $c_4=0$, then
 
In order to make $a(\tau)$ finite at $\tau=0$ set $c_4=0$, then
\begin{eqnarray}
+
\begin{equation}
a(\tau) & = & (c_3 \sqrt{\tau})^{2/3\gamma}
+
a(\tau) = (c_3 \sqrt{\tau})^{2/3\gamma}
\label{aSoln-2b-2} \\
+
\label{aSoln-2b-2}  
H(\tau) & = & \frac{H_0}{3\gamma\tau} .
+
\end{equation}
 +
\begin{equation}
 +
H(\tau) = \frac{H_0}{3\gamma\tau} .
 
\label{HSoln-2b-2}
 
\label{HSoln-2b-2}
\end{eqnarray}
+
\end{equation}
 
Substitute $H(\tau_0)=H_0$ into (\ref{HSoln-2b-2}) to find age of the Universe in the considered model
 
Substitute $H(\tau_0)=H_0$ into (\ref{HSoln-2b-2}) to find age of the Universe in the considered model
 
\begin{equation}
 
\begin{equation}
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\label{sec:deSitter}
 
\label{sec:deSitter}
 
The dependencies $a(\tau)$ and $H(\tau)$ now take on the following form
 
The dependencies $a(\tau)$ and $H(\tau)$ now take on the following form
\begin{eqnarray}
+
\begin{equation}
a(\tau) & = & \tau^{1/3\gamma} \, \left[ c_5 \sin (m_3 \ln\tau) +
+
a(\tau) = \tau^{1/3\gamma} \, \left[ c_5 \sin (m_3 \ln\tau) +
 
               c_6 \cos (m_3 \ln\tau) \right]^{2/3\gamma}
 
               c_6 \cos (m_3 \ln\tau) \right]^{2/3\gamma}
\label{aSoln-2c-1} \\
+
\label{aSoln-2c-1}  
H(\tau) & = & \frac{2H_0}{3\gamma} \, \left\{ \frac
+
\end{equation}
 +
\begin{equation}
 +
H(\tau) = \frac{2H_0}{3\gamma} \, \left\{ \frac
 
               {(c_5-2m_3c_6)\sin(m_3\ln\tau)+(c_6+2m_3c_5)\cos(m_3\ln\tau)}
 
               {(c_5-2m_3c_6)\sin(m_3\ln\tau)+(c_6+2m_3c_5)\cos(m_3\ln\tau)}
 
               {2\tau[c_5\sin(m_3\ln\tau)+c_6\cos(m_3\ln\tau)]} \right\} ,
 
               {2\tau[c_5\sin(m_3\ln\tau)+c_6\cos(m_3\ln\tau)]} \right\} ,
 
\label{HSoln-2c-1}
 
\label{HSoln-2c-1}
\end{eqnarray}
+
\end{equation}
 
where $m_3 \equiv \frac{1}{2} \sqrt{-(3\gamma\tau_0)^2 \lambda_0 - 1}$
 
where $m_3 \equiv \frac{1}{2} \sqrt{-(3\gamma\tau_0)^2 \lambda_0 - 1}$
 
and $c_5,c_6$ are arbitrary constants. Proceeding as usual, express $c_5$ and $c_6$ in terms of $\tau_0$:
 
and $c_5,c_6$ are arbitrary constants. Proceeding as usual, express $c_5$ and $c_6$ in terms of $\tau_0$:
\begin{eqnarray}
+
\begin{equation}\label{c5defn-2}
c_5 & = & \frac{1}{\sqrt{\tau_0}} \sin(m_3\ln\tau_0) + \frac{1}{m_3}
+
c_5 = \frac{1}{\sqrt{\tau_0}} \sin(m_3\ln\tau_0) + \frac{1}{m_3}
 
           \left[ \left( \frac{3\gamma}{2} \right) \sqrt{\tau_0} -
 
           \left[ \left( \frac{3\gamma}{2} \right) \sqrt{\tau_0} -
           \frac{1}{2\sqrt{\tau_0}} \right] \cos(m_3\ln\tau_0) \\
+
           \frac{1}{2\sqrt{\tau_0}} \right] \cos(m_3\ln\tau_0)  
\label{c5defn-2}
+
\end{equation}
c_6 & = & \frac{1}{\sqrt{\tau_0}} \cos(m_3\ln\tau_0) - \frac{1}{m_3}
+
\begin{equation}
 +
c_6 = \frac{1}{\sqrt{\tau_0}} \cos(m_3\ln\tau_0) - \frac{1}{m_3}
 
           \left[ \left( \frac{3\gamma}{2} \right) \sqrt{\tau_0} -
 
           \left[ \left( \frac{3\gamma}{2} \right) \sqrt{\tau_0} -
 
           \frac{1}{2\sqrt{\tau_0}} \right] \sin(m_3\ln\tau_0) .
 
           \frac{1}{2\sqrt{\tau_0}} \right] \sin(m_3\ln\tau_0) .
 
\label{c6defn-2}
 
\label{c6defn-2}
\end{eqnarray}
+
\end{equation}
 
Unlike the previous cases, it is now impossible to keep finite values $a(\tau)$ at $\tau=0$ turning one of the constants $c_5,c_6$ to zero.</p>
 
Unlike the previous cases, it is now impossible to keep finite values $a(\tau)$ at $\tau=0$ turning one of the constants $c_5,c_6$ to zero.</p>
 
   </div>
 
   </div>
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\end{equation}
 
\end{equation}
 
where as usual $a(t_0)=a_0=1,\rho(a_0)=\rho_0$, and the function $f(a)$ takes on the form
 
where as usual $a(t_0)=a_0=1,\rho(a_0)=\rho_0$, and the function $f(a)$ takes on the form
\begin{eqnarray}
+
\begin{equation}
f(a) & \equiv & 1 + \kappa_0 \times \left\{ \begin{array}{ll}
+
f(a) \equiv 1 + \kappa_0 \times \left\{ \begin{array}{ll}
 
                     \frac{ {\textstyle m(a^{3\gamma-m}-1)} }
 
                     \frac{ {\textstyle m(a^{3\gamma-m}-1)} }
 
                         { {\textstyle 3\gamma-m} }
 
                         { {\textstyle 3\gamma-m} }
Line 523: Line 535:
 
                           & \mbox{ if } m=3\gamma
 
                           & \mbox{ if } m=3\gamma
 
                     \end{array} \right.
 
                     \end{array} \right.
                     \label{fDefn} \\
+
                     \label{fDefn}  
\kappa_0 & \equiv & {\cal B}/8\pi G \rho_0 .
+
\end{equation}
 +
\begin{equation}
 +
\kappa_0 \equiv {\cal B}/8\pi G \rho_0 .
 
                     \label{kappaDef1}
 
                     \label{kappaDef1}
\end{eqnarray}
+
\end{equation}
 
If $m=0$ then $f(a)=1$ and the equation (\ref{NewRho}) recovers the usual result $\rho\sim a^{-3}$ for non-relativistic matter ($\gamma=1$) and $a^{-4}$ for radiation ($\gamma=4/3$). The new parameter $\kappa_0$ can be determined by the decay law (\ref{Bam}) and astronomical observations implying ${\cal B} = \Lambda_0 = 3H_0^2\lambda_0$. Substitute this result into (\ref{kappaDef1}) to find:
 
If $m=0$ then $f(a)=1$ and the equation (\ref{NewRho}) recovers the usual result $\rho\sim a^{-3}$ for non-relativistic matter ($\gamma=1$) and $a^{-4}$ for radiation ($\gamma=4/3$). The new parameter $\kappa_0$ can be determined by the decay law (\ref{Bam}) and astronomical observations implying ${\cal B} = \Lambda_0 = 3H_0^2\lambda_0$. Substitute this result into (\ref{kappaDef1}) to find:
 
\begin{equation}
 
\begin{equation}
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<div id="IDE_87_8"></div>
 
<div id="IDE_87_8"></div>
 
<div style="border: 1px solid #AAA; padding:5px;">
 
<div style="border: 1px solid #AAA; padding:5px;">
 +
 
=== Problem 17 ===
 
=== Problem 17 ===
 
Find dependence of deceleration parameter on the scale factor for the model of previous problem.
 
Find dependence of deceleration parameter on the scale factor for the model of previous problem.

Latest revision as of 02:24, 11 November 2013





Vacuum Decay into Cold Dark Matter

Let us consider the Einstein field equations \[R_{\mu\nu}-frac12Rg_{\mu\nu}=8\pi G\left(T_{\mu\nu}+\frac\Lambda{8\pi G}g_{\mu\nu}\right).\] According to the Bianchi identities, (i) vacuum decay is possible only from a previous existence of some sort of non-vanishing matter and/or radiation, and (ii) the presence of a time-varying cosmological term results in a coupling between $T_{\mu\nu}$ and $\Lambda$. We will assume (unless stated otherwise) coupling only between vacuum and CDM particles, so that \[u_\mu,T_{;\nu}^{(CDM)\mu\nu}=-u_\mu\left(\frac{\Lambda g^{\mu\nu}}{8\pi G}\right)_{;\nu}= -u_\mu\left(\rho_\Lambda g^{\mu\nu}\right)_{;\nu}\] where $T_{\mu\nu}^{(CDM)}=\rho_{dm}u^\mu u^\nu$ is the energy-momentum tensor of the CDM matter and $\rho_\Lambda$ is the vacuum energy density. It immediately follows that \[\dot\rho_{dm}+3H\rho_{dm}=-\dot\rho_\Lambda.\] Note that although the vacuum is decaying, $w_\Lambda=-1$ is still constant, the physical equation of state (EoS) of the vacuum $w_\Lambda\equiv=p_\Lambda/\rho_\Lambda$ is still equal to constant $-1$, which follows from the definition of the cosmological constant.

(see Can vacuum decay in our Universe?, Interpreting Cosmological Vacuum Decay)


Problem 1

Since vacuum energy is constantly decaying into CDM, CDM will dilute in a smaller rate compared with the standard relation $\rho_{dm}\propto a^{-3}$. Thus we assume that $\rho_{dm}=\rho_{dm0}a^{-3+\varepsilon}$, where $\varepsilon$ is a small positive small constant. Find the dependence $\rho_\Lambda(a)$ in this model.


Problem 2

Solve the previous problem for the case when vacuum energy is constantly decaying into radiation.


Problem 3

Show that existence of a radiation dominated stage is always guaranteed in scenarios, considered in the previous problem.


Problem 4

Find how the new temperature law scales with redshift in the case of vacuum energy decaying into radiation.


Vacuum decay into CDM particles

Since the energy density of the cold dark matter is $\rho_{dm}=nm$, there are two possibilities for storage of the energy received from the vacuum decay process:

(i) the equation describing concentration, $n$, has a source term while the proper mass of CDM particles remains constant;

(ii) the mass $m$ of the CDM particles is itself a time-dependent quantity, while the total number of CDM particles, $N=na^3$, remains constant.

Let us consider both the possibilities.



Problem 5

Find dependence of total particle number on the scale factor in the model considered in problem #IDE_78.


Problem 6

Find time dependence of CDM particle mass in the case when there is no creation of CDM particles in the model considered in problem #IDE_78.


Problem 7

Consider a model where the cosmological constant $\Lambda$ depends on time as $\Lambda=\sigma H$. Let a flat Universe be filled by the time-dependent cosmological constant and a component with the state equation $p_\gamma= (\gamma-1)\rho_\gamma$. Find solutions of Friedman equations for this system [1].


Problem 8

Show that the model considered in the previous problem correctly reproduces the scale factor evolution both in the radiation-dominated and non-relativistic matter (dust) dominated cases.


Problem 9

Find the dependencies $\rho_\gamma(a)$ and $\Lambda(a)$ both in the radiation-dominated and non-relativistic matter dominated cases in the model considered in problem #IDE_84.


Problem 10

Show that for the $\Lambda(t)$ models \[T\frac{dS}{dt}=-\dot\rho_\Lambda a^3.\]



Time-dependent cosmological "constant"

Problem 11

Consider a two-component Universe filled by matter with the state equation $p=w\rho$ and cosmological constant and rewrite the second Friedman equation in the following form

     \begin{equation}\label{mainDE}
\frac{\ddot{a}}{a} = \frac{1}{2}\left( 1 + 3w\right)
                     \left( \frac{\dot{a}^2}{a^2} + \frac{k}{a^2} \right)
                   + \frac{1-3w}{6} \Lambda .
\end{equation}


Problem 12

Consider a two-component Universe filled by matter with the state equation $p=w\rho$ and cosmological constant with quadratic time dependence $\Lambda(\tau)=\mathcal{A}\tau^2$ and find the time dependence for of the scale factor.


Problem 13

Consider a flat two-component Universe filled by matter with the state equation $p=w\rho$ and cosmological constant with quadratic time dependence $\Lambda(\tau)=\mathcal{A}\tau^{\ell}$. Obtain the differential equation for Hubble parameter in this model and classify it.


Problem 14

Find solution of the equation obtained in the previous problem in the case ${\ell =1}$. Analyze the obtained solution.


Problem 15

Solve the equation obtained in the problem #tauell for ${\ell =2}.$ Consider the following cases
a) $\lambda_0 > -1/(3\gamma\tau_0)^2,$
b) $\lambda_0 = -1/(3\gamma\tau_0)^2,$
c) $\lambda_0 < -1/(3\gamma\tau_0)^2$
(see the previous problem). Analyze the obtained solution.


Problem 16

Consider a flat two-component Universe filled by matter with the state equation $p=w\rho$ and cosmological constant with the following scale factor dependence \begin{equation} \Lambda = {\cal B} \, a^{-m}. \label{Bam} \end{equation} Find dependence of energy density of matter on the scale factor in this model.


Problem 17

Find dependence of deceleration parameter on the scale factor for the model of previous problem.