Chapter 11 · Section 3 of 11
Find the scale factor dependence for the dark matter density assuming that the interaction between the dark matter and the dark energy equals $Q=\delta(a) H\rho_{dm}$.
The equation for the evolution of the DM energy density reads \[\rho_{dm} + 3H\rho_{dm} - \delta(a) H \rho_{dm} = 0.\] The scale factor variation of DM energy density is \[\rho_{dm}(a)= \rho_{dm0}a^{-3}\exp\left(-\int\limits_a^1\delta(a')d\ln a'\right).\]
Obtain the equation for the evolution of the DE energy density for $Q=\delta(a) H\rho_{dm}$.
The conservation equation for DE density is \[\dot\rho_{de}+3H(1+w_{de})\rho_{de} + \delta(a) H\rho_{dm}=0\] It is more useful to work with the variable $u=\ln a$. Substitute the solution for $\rho_{dm}$, obtained in previous problem \[\rho_{dm}(u)=\rho_{dm0}e^{-3u}\exp\left(-\int\limits_u^0\delta(u')du'\right)\] to finally obtain \[{\rho'}_{de}+3(1+w_{de}(u))\rho_{de}(u) + \delta(u)\rho_{dm0}e^{-3u}\exp\left(-\int\limits_u^0\delta(u')du'\right)=0.\] where the prime denotes derivation with respect to $u$.
Find $\rho_{dm}$ and $\rho_{de}$ in the case $Q=\delta H\rho_{dm}$, $\delta=const$, $w_{de}=const$.
In the considered case \[\rho_{dm}(a) = \rho_{dm0}a^{-3+\delta},\] \[\rho_{de}(a) = \rho_{de0}a^{-3(1+w_{de})} + \frac{\delta}{\delta+w_{de}}\rho_{dm0}\left(a^{-3(1+w_{de})}-a^{-3+\delta}\right).\] The first term of the solution describes the usual evolution of DE without the coupling to DM. From this solution it is easy to see that one must require a positive value of the coupling $\delta>0$ in order to have a consistent positive value of $\rho_{de}$ for earlier epochs of the Universe.
As was shown above, interaction between dark matter and dark energy leads to non--conservation of matter, or equivalently, to scale dependence for the mass of particles that constitute the dark matter. Show that, within the framework of the model of previous problem ($Q=\delta H\rho_{dm}$, $\delta=const$, $w_{de}=const$) the relative change of particles mass per Hubble time equals to the interaction constant.
Using the result of the previous problem \[\rho_{dm}(a) = \rho_{dm0}a^{-3+\delta},\] one finds \[m(a) = m_0 a^\delta.\] Then \[\frac1m\frac{dm}{dt}\frac1H=\delta.\]
Find $\rho_{dm}$ and $\rho_{de}$ in the case $Q=\delta H\rho_{de}$, $\delta=const$, $w_{de}=const$.
\[\rho_{de}(a) = \rho_{de0}a^{-3(1+w_{de})+\delta},\] \[\rho_{dm}(a) = \frac{\delta\rho_{de0}}{3w_{de}+\delta}a^{-3(1+w_{de})+\delta} + \left(\rho_{dm0}\frac{\delta\rho_{de0}}{3w_{de}+\delta}\right)a^{-3}.\]
Find $\rho_{dm}$ and $\rho_{de}$ in the case $Q=\delta(a) H\rho_{de}$, $\delta(a)=\beta_0a^\xi$, $w_{de}=const$.
(after [1])
Substituting the interaction $Q$ into conservation equation for $\rho_{de}$, one obtains \[\rho_{de}(a) = \rho_{de0}a^{-3(1+w_{de})}\exp\left[\frac{3\beta_0(1-a^\xi)}{\xi}\right].\] As for the DM energy density, combine $\rho_{de}$ and the first Friedmann equation to obtain \[\rho_{dm}(a) = \rho_{dm0}a^{-3}\left\{1 - \frac{\Omega_{de0}}{\Omega_{dm0}} \frac{3\beta_0a^{-3w_{de}}e^{\frac{3\beta_0}{\xi}}}{\xi}\left[ a^\xi E_{\frac{3w_{de}}{\xi}} \left(\frac{3\beta_0a^\xi}{\xi}\right) - a^{-3w_{de}} E_{\frac{3w_{de}}{\xi}} \left(\frac{3\beta_0}{\xi}\right)\right]\right\},\] where \[E_n(z)=\int\limits_1^\infty t^{-n} e^{-zt}dt\] is the usual exponential integral function.
Let's look at a more general linear model for the expansion of a Universe that contains two interacting fluids with the equations of state \[p_1 = (\gamma_1-1)\rho_1,\] \[p_2 = (\gamma_2-1)\rho_2,\] and energy exchange \[\dot\rho_1+3H\gamma_1\rho_1 = -\beta H\rho_1 + \alpha H\rho_2,\] \[\dot\rho_2+3H\gamma_2\rho_2 = \beta H\rho_1 - \alpha H\rho_2.\] Here $\alpha$ and $\beta$ are constants describing the energy exchanges between the two fluids. Obtain the equation for $H(t)$ and find its solutions (After [2], [3]).
Using energy exchange equations and the first Friedmann equation one can eliminate the densities to obtain the equation for $H(t)$ \[\ddot H+H\dot H(\alpha+\beta+3\gamma_1+3\gamma_2) +\frac32H^3(\alpha\gamma_1 + \beta\gamma_2 + 3\gamma_1\gamma_2) = \ddot H+ AH\dot H +B H^3=0,\] \[A = \alpha+\beta+3\gamma_1+3\gamma_2, \quad B = \frac32(\alpha\gamma_1 + \beta\gamma_2 + 3\gamma_1\gamma_2).\] The equation has a simple solution \[H=\frac h t, \quad h\ne0,\] as long as the following demand holds \[Bh^2 - Ah +2 =0.\] Since the solution of this equation is \[h_\pm=\frac{A\pm\sqrt{A^2-8B}}{2B},\] real power-law solutions for $H(t)$ exist if $A^2>8B$.
Show that the energy balance equations (modified conservation equations) for $Q\propto H$ do not depend on H.
The required statement can be easily proven if in the energy balance equations one makes transformation of independent variable from time to $N=\ln a$: \[\frac{d}{dt}\to H\frac{d}{dN}.\] Then for a system of two interacting fluids with $Q\propto H$ the phase space is two-dimensional.
The Hubble parameter is present in the first Friedmann equation quadratically. This gives rise to a useful symmetry within a class of FLRW models. Because of this quadratic dependence, Friedmann's equation remains invariant under a transformation $H\to-H$ for the spatially flat case. This means it describes both expanding and contracting solutions. The transformation $H\to-H$ can be seen as a consequence of the change $a\to1/a$ of the scale factor of the FLRW metric. If, instead of just the first Friedmann equation, we want to make the whole system of Universe-describing equations invariant relative to this transformation, we must expand the set of values that undergo symmetry transformations. Then, when we refer to a duality transformation, we have in mind the following set of transformations \[H\to\bar H=-H,\quad \rho\to\bar\rho=\rho,\quad p\to\bar p=-2\rho-p,\quad \gamma\equiv\frac{\rho+p}{\rho}\to\bar\gamma\equiv\frac{\bar\rho+\bar p}{\bar\rho}=-\gamma.\]
Generalize the duality transformation to the case of interacting components.(after [4].)
For the interacting components $\rho_1$ and $\rho_2$ \[3H^2=\rho_1+\rho_2\] \[\dot\rho_1+3H\gamma_1\rho_1=-3H\Pi\] \[\dot\rho_2+3H\gamma_2\rho_2=3H\Pi\] Automatically, the above dual symmetry gets restricted to the following transformation: \[H\to-H,\quad \rho_i\to\rho_i,\quad \gamma_i\to-\gamma_i,\quad \Pi\to-\Pi\] with the overall barotropic index transforming as $\gamma\to-\gamma$. As a consequence, superaccelerated expansion (for example, phantom) can be obtained from decelerated ones and vice versa without affecting the field equations also in the case of interacting DM and DE.