Realization of interaction in the dark sector - Revision history

Cosmo All: /* Problem 15 */

2013-11-11T02:24:03Z

‎Problem 15

Cosmo All: /* Problem 15 */

2013-11-11T02:22:22Z

‎Problem 15

Cosmo All: /* Problem 16 */

2013-11-11T02:17:37Z

‎Problem 16

Cosmo All: /* Problem 17 */

2013-11-11T02:14:12Z

‎Problem 17

Cosmo All: /* Problem 15 */

2013-11-11T02:12:27Z

‎Problem 15

Cosmo All: /* Problem 7 */

2013-11-11T02:11:12Z

‎Problem 7

Cosmo All: Created page with "7 NOTOC = Vacuum Decay into Cold Dark Matter = Let us consider the Einstein field equations \[R_{\mu\nu}-frac12Rg_{\mu..."

2013-11-11T01:56:41Z

Created page with "7 __NOTOC__ = Vacuum Decay into Cold Dark Matter = Let us consider the Einstein field equations \[R_{\mu\nu}-frac12Rg_{\mu..."

New page

[[Category:Interactions in the Dark Sector|7]]

__NOTOC__

= 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 [http://arxiv.org/abs/astro-ph/0408495 Can vacuum decay in our Universe?], [http://arxiv.org/abs/astro-ph/0507372 Interpreting Cosmological Vacuum Decay])

<div id="IDE_78"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== 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.
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<p style="text-align: left;">Conservation equation for CDM yields
\[\rho_\Lambda=\bar\rho_{\Lambda0}+\frac{\varepsilon\rho_{dm0}}{3-\varepsilon}a^{-3+\varepsilon}.\]</p>
</div>
</div></div>

<div id="IDE_79"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 2 ===
Solve the previous problem for the case when vacuum energy is constantly decaying into radiation.
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<div class="NavHead">solution</div>
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<p style="text-align: left;">By considering that radiation will dilute more slowly compared to its standard evolution, $\rho_r\propto a^{-4}$, and that such a deviation is characterized by a positive constant $\alpha$ one finds
\[\rho_r=\rho_{r0}a^{-4+\alpha}.\]
By inserting this expression into radiation conservation equation
\[\dot\rho_r+4H\rho_r=-\dot\rho_\Lambda,\]
one obtains
\[\rho_\Lambda=\bar\rho_{\Lambda0}+\frac{\alpha\rho_{r0}}{4-\alpha}a^{-4+\alpha}.\] </p>
</div>
</div></div>

<div id="IDE_80"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 3 ===
Show that existence of a radiation dominated stage is always guaranteed in scenarios, considered in the previous problem.
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<div class="NavHead">solution</div>
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<p style="text-align: left;">The ratio between the vacuum and radiation energy densities reads
\[\frac{\rho_\Lambda}{\rho_r}=\frac{\bar\rho_{\Lambda0}}{\rho_{r0}}a^{4-\alpha}+\frac\alpha{4-\alpha}.\]
The first term is asymptotically vanishing at early times whereas the second one is smaller than unity. Therefore, a radiation dominated stage is always guaranteed in this kind of scenarios.</p>
</div>
</div></div>

<div id="IDE_81"></div>
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=== Problem 4 ===
Find how the new temperature law scales with redshift in the case of vacuum energy decaying into radiation.
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<div class="NavHead">solution</div>
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<p style="text-align: left;">For an adiabatic vacuum decay the equilibrium relations are preserved, as happens with the Stefan law, $\rho_r\propto T^4$. As a consequence, one may check that the $Ta^{1-\alpha/4}$. This implies that the new temperature law scales with redshift as
\[T=T_0(1+z)^{1-\alpha/4}.\]</p>
</div>
</div></div>

== 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.

<div id="IDE_82"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 5 ===
Find dependence of total particle number on the scale factor in the model considered in problem [[#IDE_78]].
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<div class="NavHead">solution</div>
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<p style="text-align: left;">Since $\rho_{dm}=nm$, we find, using $\rho_{dm}=\rho_{dm0}a^{-3+\varepsilon}$, that
\[n=n_0a^{-3+\varepsilon}.\]
In the considered case, there is necessarily a source term in the current of CDM particles, that is, $N^\alpha_{;\alpha}$, where $N^\alpha=nu^\alpha$. In terms of the concentration it can be written as
\[\dot n +3Hn=\psi.\]
Inserting $n=n_0a^{-3+\varepsilon}$, one obtains
\[\psi=\varepsilon\frac{\dot a}an\equiv n\Gamma.\]
We have written particle source $\psi$ in terms of a decay rate, $\Gamma\equiv\dot N/N$.</p>
</div>
</div></div>

<div id="IDE_83"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== 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]].
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<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">In this case the concentration of CDM particles satisfies the equation
\[\dot n +3Hn=0,\]
whose solution is $n=n_0a^{-3}$, which implies that $N(t)=const$. Naturally, if CDM particles are not being created, the unique possibility is an increasing in the proper mass of CDM particles. Actually, since $\rho_{dm0}=nm$, expressions \[\rho_{dm}=\rho_{dm0}a^{-3+\varepsilon}\] and \[n=n_0a^{-3}\] imply that the mass of the CDM particles scales as
\[m(t)=m_0a(t)^\varepsilon,\]
where $m_0$ is the present day mass of CDM particles.</p>
</div>
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<div id="IDE_84"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== 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 [http://arxiv.org/abs/0711.2686 Observational constraints on late-time Lambda(t) cosmology].
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">System of equations describing dynamics of the model reads
\begin{align}
\nonumber
\rho_\gamma+\Lambda & = 3H^2,\\
\nonumber
\dot\rho_\gamma + 3H\gamma\rho_\gamma & =-\dot\Lambda,\\
\nonumber
\Lambda & =\sigma H.
\end{align}
Combine this equations to obtain
\[2\dot H+3\gamma H^2-\sigma\gamma H=0.\]
For $H>0$ (expanding Universe) the equation has the following solution
\[a(t)=C\left[\exp\left(\frac{\sigma\gamma t}2\right)-1\right]^{3\gamma/2}.\]
Here $C$ is one of the two integration constants (the equation for scale factor is the second-order one). The second integration constant is determined from the condition $a(t=0)=0$. Using the obtained solution, one finds
\begin{align}
\nonumber
\rho_\gamma & =\frac{\sigma^2}3\left[1+\left(\frac C a\right)^{3\gamma/2}\right]\left(\frac C a\right)^{3\gamma/2},\\
\nonumber
\Lambda & =\frac{\sigma^2}3\left[1+\left(\frac C a\right)^{3\gamma/2}\right].
\end{align}</p>
</div>
</div></div>

<div id="IDE_85"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== 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.
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<div class="NavHead">solution</div>
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<p style="text-align: left;">In the radiation-dominated epoch $\gamma=w+1=4/3$. Therefore
\[a(t)=C\left[\exp\left(\frac{2\sigma t}3\right)-1\right]^{1/2}.\]
For small times $\sigma t\ll1$ the formula reproduces correct dependence $a(t)\propto t^{1/2}$. In the matter dominated case $\gamma=w+1=1$ and
\[a(t)=C\left[\exp\left(\frac{2\sigma t}3\right)-1\right]^{2/3}.\]
At $\sigma t\ll1$ we recover the standard law of non-relativistic matter evolution $a(t)\propto t^{2/3}$. Note that in the limit of long times $\sigma t\gg1$ the scale factor grows exponentially for all values of the parameter $\gamma$.</p>
</div>
</div></div>

<div id="IDE_86"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== 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]].
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Using the expressions for $\rho_\gamma(a)$ and $\Lambda(a)$ obtained in the problem [[#IDE_84]], one finds that in the radiation-dominated epoch $(\gamma=4/3)$
\begin{align}
\nonumber
\rho_\gamma & =\frac{\sigma^2C^4}3\frac1{a^4}+\frac{\sigma^2C^2}3\frac1{a^2},\\
\nonumber
\Lambda & =\frac{\sigma^2}3+\frac{\sigma^2C^2}3\frac1{a^2}.
\end{align}
In the limit $a\to0$ ($t\to0$)
\begin{align}
\nonumber
\rho_\gamma & =\frac{\sigma^2C^4}3\frac1{a^4}=\frac{3}{4t^2},\\
\nonumber
\Lambda & =\frac{\sigma^2C^2}3\frac1{a^2}=\frac\sigma{2t}.
\end{align}
At the matter-dominated epoch ($\gamma=1$)
\begin{align}
\nonumber
\rho_\gamma & =\frac{\sigma^2C^3}3\frac1{a^3}+\frac{\sigma^2C^{3/2}}3\frac1{a^{3/2}},\\
\nonumber
\Lambda & =\frac{\sigma^2}3+\frac{\sigma^2C^{3/2}}3\frac1{a^{3/2}}.
\end{align}</p>
</div>
</div></div>

<div id="IDE_87"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 10 ===
Show that for the $\Lambda(t)$ models \[T\frac{dS}{dt}=-\dot\rho_\Lambda a^3.\]
</div>

= Time-dependent cosmological "constant" =

<div id="IDE_87_2"></div>
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=== 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}
</div>

<div id="IDE_87_3"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== 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.
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<div class="NavHead">solution</div>
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<p style="text-align: left;">Substitute the dependence $\Lambda(t)=Bt^2$ into the equation (\ref{mainDE}) and multiply both sides by $a^2$ to obtain
\begin{equation}
\ddot{a} a = \frac 12 (1+3w)\left(\dot{a}^2 + k\right) +\frac{1-3w}{6} a^2\mathcal{A}t^2\Lambda
\label{mainDE1}
\end{equation}
Seek the solution in the form $a(\tau)=a_0\tau^\beta.$ Substitute this dependence into (\ref{mainDE1}) to show that the equality takes place only under the following condition
$$\mathcal{A} = \frac{6\beta}{1-3w}\left(1 -\frac 1\beta +\frac 32 w \right)$$</p>
</div>
</div></div>

<div id="tauell"></div>
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=== 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.
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<div class="NavHead">solution</div>
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<p style="text-align: left;">Recall that the definition of the Hubble parameter reads
\begin{equation}
H \equiv \frac{\dot{a}}{a} = H_0 \left( \frac{da}{a \, d\tau} \right) ,
\label{Hdefn}
\end{equation}
and use \eqref{mainDE} to obtain:
\begin{equation}
\frac{dH}{d\tau} = \left( \frac{-3\gamma}{2H_0} \right) H^2 +
\left( \frac{\gamma}{2H_0} \right) \Lambda +
\left( 1-\frac{3\gamma}{2} \right) \frac{k}{H_0 \, a^2},
\label{1stDE}
\end{equation}
where $\gamma = 1+w.$ This equation can presented in the following form
\begin{equation}
\frac{dH}{d\tau} = {\mathcal P}(\tau) \, H^2 + {\mathcal Q}(\tau) \, H +
{\mathcal R}(\tau) ,
\label{RiccatiDE}
\end{equation}
where ${\mathcal P} \equiv - 3\gamma/2$, ${\mathcal Q} \equiv 0$ and
${\mathcal R}(\tau) \equiv (\gamma/2) \Lambda(\tau).$

The obtained equation \eqref{RiccatiDE} is of Riccati type. For further convenience we transit to dimensionless variables $H \to x$:
\begin{equation}
H \equiv -\frac{1}{{\mathcal P} x} \, \frac{dx}{d\tau} =
\left( \frac{2H_0}{3\gamma} \right) \frac{dx}{x \, d\tau} ,
\label{xDefn}
\end{equation}
According to the condition of the problem one should consider the spatially flat Universe $k=0,$ so substitute into the equation \eqref{1stDE} the following dependence $\Lambda(\tau)=A\tau^{\ell}:$
\begin{equation}
\tau^{\ell} \; \frac{d\,^2x}{d\tau^2} - \alpha \, x = 0 ,
\label{mainDE-t}
\end{equation}
where
\begin{equation}
\alpha \equiv \frac{3\gamma^2 {\mathcal A}}{4} .
\label{alphaDef1}
\end{equation}
It is easy now to show that the scale factor $a(\tau)$ and the function $x(\tau)$ are linked by the relation
\begin{equation}
a(\tau) = [ x(\tau) ]^{2/3\gamma} .
\label{xTOa}
\end{equation}
The constant $\alpha$ is determined by the relation \eqref{alphaDef1}.</p>
</div>
</div></div>

<div id="IDE_87_5"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 14 ===
Find solution of the equation obtained in the previous problem in the case ${\ell =1}$. Analyze the obtained solution.
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<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">In the considered case the equation \eqref{mainDE-t} takes the form
\begin{equation}
\tau \; \frac{d\,^2x}{d\tau^2} - \alpha \, x = 0 ,
\label{mainDE-t-1}
\end{equation}
The constant $\alpha$ is determined by the relation \eqref{alphaDef1}:
\begin{equation}
\alpha = (3\gamma/2)^2 \lambda_0 \tau_0 .
\label{alphaDefn-1}
\end{equation}
Change the independent variable from $\tau$ to $z \equiv 2\sqrt{-\alpha\tau}$ to obtain
\begin{equation}
z^2 \, \frac{d\,^2x}{dz^2} - z \, \frac{dx}{dz} + z^2 \, x = 0 .
\label{mainDE-t-1b}
\end{equation}
This equation can be reduced to the Bessel one
$x(z) = c_1 z J_1(z) + c_2 z Y_1 (z),$
where $J_1(z)$ and $Y_1(z)$ are respectively Bessel and Neumann functions of first order.

Use the relation (\ref{xTOa}) to obtain the explicit dependence for the scale factor:
\begin{equation}
a(\tau) = \tau^{1/3\gamma} \left[ c_1 J_1(z) +
c_2 Y_1 (z) \right]^{\; 2/3\gamma} ,
\label{aSoln-1}
\end{equation}
where the factor $2\sqrt{-\alpha}$ enters the coefficients $c_1,c_2$.
The Hubble parameter can be determined by substitution $x(z)$ into the expression \eqref{xDefn}:
\begin{equation}
H(\tau) = H_0 \, \sqrt{-\lambda_0 \left( \frac{\tau_0}{\tau} \right)} \,
\left[ \frac{c_1 J_0(z) + c_2 Y_0(z)}
{c_1 J_1(z) + c_2 Y_1(z)} \right] ,
\label{HSoln-1}
\end{equation}
where $J_0(z)$ and $Y_0(z)$ are respectively Bessel and Neumann functions of zero order.
Note that from the definition (\ref{alphaDefn-1}) it follows that $z(\tau),$
$a(\tau)$ and $H(\tau)$, can be real-valued (for $\tau >0 $) if
$\lambda_0 \leq 0.$
Though such possibility exists in theory, it contradicts the observational data.
Thus the case $\ell=1$ is likely to be impossible in real Universe.</p>
</div>
</div></div>

<div id="IDE_87_6"></div>
<div style="border: 1px solid #AAA; padding:5px;">
=== Problem 15 ===
Solve the equation obtained in the problem \ref{tauell} for ${\ell =2}.$ Consider the following cases
<br/>
a) $\lambda_0 > -1/(3\gamma\tau_0)^2,$<br/>
b) $\lambda_0 = -1/(3\gamma\tau_0)^2,$<br/>
c) $\lambda_0 < -1/(3\gamma\tau_0)^2$
<br/>
(see the previous problem). Analyze the obtained solution.
<div class="NavFrame collapsed">
<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">In the case ${\ell =2}$ the equation (\ref{mainDE-t}) takes on the form
\begin{equation}
\tau^2 \; \frac{d\,^2x}{d\tau^2} - \alpha \, x = 0 ,
\label{mainDE-t-2}
\end{equation}
where $\alpha$ is determined by the equation (\ref{alphaDefn}):
\begin{equation}
\alpha = (3\gamma/2)^2 \lambda_0 \tau_0^2 .
\label{alphaDefn-2}
\end{equation}
This equation belongs to Euler type. Make the transformation of variables $y \equiv \ln\tau$ in the equation (\ref{mainDE-t-2}) to obtain
\begin{equation}
\frac{d\,^2x}{dy^2} - \frac{dx}{dy} - \alpha \, x = 0 .
\label{EulerDE}
\end{equation}

'''a)''' $\lambda_0 > -1/(3\gamma\tau_0)^2.$
As the astronomical observations give evidence in favor of positive value of $\lambda_0,$ thais model makes certain interest.
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
\begin{eqnarray}
a(\tau) & = & \tau^{1/3\gamma} \, \left( c_1 \tau^{m_0} + c_2
\tau^{-m_0} \right)^{2/3\gamma}
\label{aSoln-2a-2} \\
H(\tau) & = & \frac{2H_0}{3\gamma} \, \left[ \frac
{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] ,
\label{HSoln-2a-2}
\end{eqnarray}
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
$c_2 \ne 0$. Thus the expressions (\ref{aSoln-2a-2}) and (\ref{HSoln-2a-2}) take on the simplified form
\begin{eqnarray}
a(\tau) & = & \left( \frac{\tau}{\tau_0} \right)^{2m_1/3\gamma}
\label{aSoln-2a-3} \\
H(\tau) & = & \frac{2H_0}{3\gamma} \, \left[ \frac{m_1}{\tau} \right] ,
\label{HSoln-2a-3}
\end{eqnarray}
where $m_1 \equiv 1/2+m_0,$ à $\tau_0 = 2m_1/3\gamma.$
<br/>
'''b)''' $\lambda_0 = -1/(3\gamma\tau_0)^2.$
Proceed in analogous way to obtain
\begin{eqnarray}
a(\tau) & = & \tau^{1/3\gamma} \, (c_3 + c_4 \ln \tau)^{2/3\gamma}
\label{aSoln-2b-1} \\
H(\tau) & = & \frac{2H_0}{3\gamma} \, \left[ \frac
{(c_3+2c_4) + c_4 \ln\tau}
{2\tau ( c_3 + c_4 \ln\tau)} \right] ,
\label{HSoln-2b-1}
\end{eqnarray}
where $c_3,c_4$ are arbitrary constants.
In order to make $a(\tau)$ finite at $\tau=0$ set $c_4=0$, then
\begin{eqnarray}
a(\tau) & = & (c_3 \sqrt{\tau})^{2/3\gamma}
\label{aSoln-2b-2} \\
H(\tau) & = & \frac{H_0}{3\gamma\tau} .
\label{HSoln-2b-2}
\end{eqnarray}
Substitute $H(\tau_0)=H_0$ into (\ref{HSoln-2b-2}) to find age of the Universe in the considered model
\begin{equation}
\tau_0 = 1/3\gamma.
\label{Age-2b}
\end{equation}

Substitute the expression (\ref{Age-2b}) into (\ref{aSoln-2b-2})
and $a(\tau_0)=1$ to find $c_3 = \sqrt{3\gamma}$. Then
\begin{equation}
a(\tau) = \left( \frac{\tau}{\tau_0} \right)^{1/3\gamma}.
\end{equation}
<br/>
'''c)''' $\lambda_0 < -1/(3\gamma\tau_0)^2$
\label{sec:deSitter}
The dependencies $a(\tau)$ and $H(\tau)$ now take on the following form
\begin{eqnarray}
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}
\label{aSoln-2c-1} \\
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)}
{2\tau[c_5\sin(m_3\ln\tau)+c_6\cos(m_3\ln\tau)]} \right\} ,
\label{HSoln-2c-1}
\end{eqnarray}
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 $\tau_0$:
\begin{eqnarray}
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} -
\frac{1}{2\sqrt{\tau_0}} \right] \cos(m_3\ln\tau_0) \\
\label{c5defn-2}
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} -
\frac{1}{2\sqrt{\tau_0}} \right] \sin(m_3\ln\tau_0) .
\label{c6defn-2}
\end{eqnarray}
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></div>

<div id="IDE_87_7"></div>
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=== 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.
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<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Use the conservation equation and the first Friedman equation to obtain
\begin{equation}
\frac{d}{da} \left( \rho a^{3\gamma} \right) = \left( \frac{m{\cal B}}{8\pi G}
\right) a^{3\gamma-(m+1)} .
\end{equation}
Its integration leads to the required dependence of energy density of matter on the scale factor
\begin{equation}
\rho (a) = \rho_0 a^{-3\gamma} f(a) ,
\label{NewRho}
\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
\begin{eqnarray}
f(a) & \equiv & 1 + \kappa_0 \times \left\{ \begin{array}{ll}
\frac{ {\textstyle m(a^{3\gamma-m}-1)} }
{ {\textstyle 3\gamma-m} }
& \mbox{ if } m \neq 3\gamma \\
3\gamma \ln(a)
& \mbox{ if } m=3\gamma
\end{array} \right.
\label{fDefn} \\
\kappa_0 & \equiv & {\cal B}/8\pi G \rho_0 .
\label{kappaDef1}
\end{eqnarray}
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}
\kappa_0 = \lambda_0/\Omega_0.
\label{kappaDefn}
\end{equation}</p>
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<div id="IDE_87_8"></div>
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=== Problem 17 ===
Find dependence of deceleration parameter on the scale factor for the model of previous problem.
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<div class="NavHead">solution</div>
<div style="width:100%;" class="NavContent">
<p style="text-align: left;">Substitute (\ref{Bam}) and (\ref{NewRho}) into the Friedman equation to obtain
\begin{equation}
\frac{da}{d\tau} = a \left[ \Omega_0 a^{-3\gamma} f(a) -
(\Omega_0+\lambda_0-1) a^{-2} +
\lambda_0 a^{-m} \right]^{1/2} ,
\label{Expansion}
\end{equation}
where the function $f(a)$ is determined in the previous problem (see the formula \eqref{fDefn}).
The second Friedman equation can be rewritten in the form
\begin{equation}
\frac{d^{\, 2}a}{d\tau^2} = \left( 1 - \frac{3\gamma}{2} \right)
\Omega_0 a^{1-3\gamma} f(a) +
\lambda_0 a^{1-m} .
\label{Deceleration}
\end{equation}
Substitute this derivatives into the definition of the deceleration parameter to obtain the required dependence
\begin{equation}
q\equiv -\frac{\ddot{a}a}{\dot{a}^2}=-\frac{\left( 1 - \frac{3\gamma}{2} \right)
\Omega_0 a^{1-3\gamma} f(a) +
\lambda_0 a^{1-m}}{a\left[ \Omega_0 a^{-3\gamma} f(a) -
(\Omega_0+\lambda_0-1) a^{-2} +
\lambda_0 a^{-m} \right]}</p>
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@@ Line 439: / Line 439: @@
 Proceed in analogous way to obtain
 \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}
 \end{equation}
@@ Line 474: / Line 474: @@
 The dependencies $a(\tau)$ and $H(\tau)$ now take on the following form
 \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}
 \label{aSoln-2c-1}

@@ Line 411: / Line 411: @@
 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
-\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}
-\label{aSoln-2a-2} \\
+\label{aSoln-2a-2}
-H(\tau) & = & \frac{2H_0}{3\gamma} \, \left[ \frac
+\end{equation}
                {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] ,
 \label{HSoln-2a-2}
-\end{eqnarray}
+\end{equation}
 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
 $c_2 \ne 0$. 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}
 \label{HSoln-2a-3}
-\end{eqnarray}
+\end{equation}
 where $m_1 \equiv 1/2+m_0,$ à $\tau_0 = 2m_1/3\gamma.$
 <br/>
 '''b)''' $\lambda_0 = -1/(3\gamma\tau_0)^2.$
 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}
-\label{aSoln-2b-1} \\
+\label{aSoln-2b-1}
-H(\tau) & = & \frac{2H_0}{3\gamma} \, \left[ \frac
+\end{equation}
                {(c_3+2c_4) + c_4 \ln\tau}
                {2\tau ( c_3 + c_4 \ln\tau)} \right] ,
 \label{HSoln-2b-1}
-\end{eqnarray}
+\end{equation}
 where $c_3,c_4$ are arbitrary constants.
 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}
 \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
 \begin{equation}
@@ Line 464: / Line 473: @@
 \label{sec:deSitter}
 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) +
                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}
                {(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\} ,
 \label{HSoln-2c-1}
-\end{eqnarray}
+\end{equation}
 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 $\tau_0$:
-\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} -
-           \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}
            \left[ \left( \frac{3\gamma}{2} \right) \sqrt{\tau_0} -
            \frac{1}{2\sqrt{\tau_0}} \right] \sin(m_3\ln\tau_0) .
 \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>
    </div>

@@ Line 515: / Line 515: @@
 \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
-\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)} }
                           { {\textstyle 3\gamma-m} }
@@ Line 523: / Line 523: @@
                            & \mbox{ if } m=3\gamma
                      \end{array} \right.
-                     \label{fDefn} \\
+                     \label{fDefn}
-\kappa_0 & \equiv & {\cal B}/8\pi G \rho_0 .
+\end{equation}
                      \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:
 \begin{equation}
@@ Line 539: / Line 541: @@
 <div id="IDE_87_8"></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === Problem 17 ===
 Find dependence of deceleration parameter on the scale factor for the model of previous problem.

@@ Line 565: / Line 565: @@
                              \lambda_0 a^{1-m}}{a\left[ \Omega_0 a^{-3\gamma} f(a) -
                     (\Omega_0+\lambda_0-1) a^{-2} +
-                    \lambda_0 a^{-m} \right]}</p>
+                    \lambda_0 a^{-m} \right]}\end{equation}</p>
    </div>
 </div></div>

@@ Line 380: / Line 380: @@
 <div style="border: 1px solid #AAA; padding:5px;">
 === Problem 15 ===
-Solve the equation obtained in the problem \ref{tauell} for ${\ell =2}.$ Consider the following cases
+Solve the equation obtained in the problem [[#tauell]] for ${\ell =2}.$ Consider the following cases
 <br/>
 a) $\lambda_0 > -1/(3\gamma\tau_0)^2,$<br/>
@@ Line 493: / Line 493: @@
 <div id="IDE_87_7"></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === 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

@@ Line 135: / Line 135: @@
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 === 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 [http://arxiv.org/abs/0711.2686 Observational constraints on late-time Lambda(t) cosmology].
+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 [http://arxiv.org/abs/0711.2686].
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    <div class="NavHead">solution</div>
@@ Line 166: / Line 166: @@
 <div id="IDE_85"></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === 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.

Realization of interaction in the dark sector - Revision history

Cosmo All: /* Problem 15 */

Cosmo All: /* Problem 15 */

Cosmo All: /* Problem 16 */

Cosmo All: /* Problem 17 */

Cosmo All: /* Problem 15 */

Cosmo All: /* Problem 7 */

Cosmo All: Created page with "7 __NOTOC__ = Vacuum Decay into Cold Dark Matter = Let us consider the Einstein field equations \[R_{\mu\nu}-frac12Rg_{\mu..."

Cosmo All: Created page with "7 NOTOC = Vacuum Decay into Cold Dark Matter = Let us consider the Einstein field equations \[R_{\mu\nu}-frac12Rg_{\mu..."