Difference between revisions of "Cosmological models with a change of the direction of energy transfer"
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| − | === Problem | + | === Problem 2 === |
Find deceleration parameter for the case $\alpha=0$ in the model considered in the previous problem. | Find deceleration parameter for the case $\alpha=0$ in the model considered in the previous problem. | ||
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| − | === Problem | + | === Problem 3 === |
Consider the model $Q=q(\alpha\dot\rho_\Lambda+3\beta H\rho_\Lambda)$. Obtain the Hubble parameter $H(a)$ and deceleration parameter for the case $\alpha=0$. | Consider the model $Q=q(\alpha\dot\rho_\Lambda+3\beta H\rho_\Lambda)$. Obtain the Hubble parameter $H(a)$ and deceleration parameter for the case $\alpha=0$. | ||
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Latest revision as of 10:43, 8 November 2013
Let us consider one more type of interaction $Q$, whose sign (i.e., the direction of energy transfer) changes when the mode of decelerated expansion is replaced by the mode of accelerated expansion, and vice versa. The simplest interaction of this type is the one proportional to the deceleration parameter. An example of such interaction is \[Q=q(\alpha\dot\rho+\beta H\rho),\] where $\alpha$ and $\beta$ are dimensionless constants, and $\rho$ can be any of densities $\rho_{de}$, $\rho_{dm}$ or $\rho_{tot}$. In the following problems for simplicity we restrict ourselves to the decaying $\Lambda$ model, for which $\dot\rho_{de}=\dot\rho_\Lambda=-Q$ and $p_{de}=-\rho_{de}$. (after [1])
Problem 1
Construct general procedure to the Hubble parameter and the deceleratio parameter for the case \(Q=q(\alpha\dot\rho_{dm}+\beta H\rho_{dm})\).
Insert $Q$ into the conservation equation for DM, \[\dot\rho_{dm}=\frac{\beta q-1}{1-\alpha q}3H\rho_{dm}.\] Substituting the obtained expression into the definition of $Q$, one gets \[Q=\frac{\beta-\alpha}{1-\alpha q}3qH\rho_{dm}.\] Using \[\dot H = -\frac{\kappa^2}{2}(\rho_{tot}+p_{tot}) = -\frac{\kappa^2}{2}\rho_{dm},\quad \kappa^2 = 8\pi G,\] one obtains \[\rho_{dm}=-\frac2{\kappa^2}\dot H.\] Inserting this into the expression for $\dot\rho_{dm}$ one find that \[\ddot H=\frac{\beta q-1}{1-\alpha q}3H\dot H.\] Transforming from the time derivative to differentiation with respect to the scale factor (denoted by the prime) one obtains \[aH'' +\frac a H H'^2 + H'=\frac{\beta q-1}{1-\alpha q}3H'.\] The deceleration parameter is \[q=-1-\frac{\dot H}{H^2}=-1-\frac a H H'.\] If $\alpha\ne0$, there is no analytical solution for the obtained second-order differential equation.
Problem 2
Find deceleration parameter for the case $\alpha=0$ in the model considered in the previous problem.
With $\alpha=0$ the solution of the equation obtained in the previous problem can be presented in the form \[H(a) = C_1\left[3C_2(1+\beta)-(2+3\beta)a^{-3(1+\beta)}\right]^{1/(2+3\beta)},\] where $C_1$ and $C_2$ are constants of integration. The partial energy density of dark matter is given by \[\Omega_{dm}\equiv \frac{\kappa^2\rho_{dm}}{3H^2}= -\frac{2aH'}{3H} = \frac{2(1+\beta)}{2+3\beta-3C_2(1+\beta)a^{3(1+\beta)}}.\] Requiring $\Omega_{dm}(a=1)=\Omega_{dm0}$ and $H(a=1)=H_0$, one obtains \[C_1 = H_0[3C_2(1+\beta) - (2+3\beta)]^{-1/(2+3\beta)},\] \[C_2 = \frac{\Omega_{dm0}(2+3\beta)-2(1+\beta)}{3\Omega_{dm0}(1+\beta)}.\] Using the relation \[q(z) = -\frac{1+z}{E(z)}\frac{d}{dz}\left(\frac1{E(z)}\right)-1\] \[E\equiv \frac{H}{H_0}=\left\{1-\frac{2+3\beta}{2(1+\beta)}\Omega_{dm0}\left[1-(1+z)^{3(1+\beta))}\right]\right\}^{1/(2+3\beta)}.\] one finally obtains \[q(z) = -1+\frac32\Omega_{dm0}\frac{(1+\beta)^{3(1+\beta)}}{E^{2+3\beta}}.\]
Problem 3
Consider the model $Q=q(\alpha\dot\rho_\Lambda+3\beta H\rho_\Lambda)$. Obtain the Hubble parameter $H(a)$ and deceleration parameter for the case $\alpha=0$.
Following the same procedure as in the two preceding cases, one obtains \[Q=\frac{3\beta qH\rho\Lambda}{1+\alpha q}.\] Using the relation \[\rho_{dm}=-\frac2{\kappa^2}\dot H,\] one finds \[\rho_\Lambda=\frac3{\kappa^2}H^2 -\rho_{dm} = \frac1{\kappa^2}(3H^2+\dot H).\] The equation for the Hubble parameter in terms of the scale factor takes the form: \[aH'' +\frac a H H'^2 +\left(4+\frac{3\beta q}{1+\alpha q}\right)H' + \frac{9\beta q H}{2a(1+\alpha q)}=0.\] The exact solution can be obtained in the case of $Q=3\beta qH\rho_\Lambda$ ($\alpha=0$), \[H(a) =C_1 a^{-3(2-5\beta +\gamma)/[4(2-3\beta)]}\left(a^{3\gamma/2 +C_2}\right)^{1/(2-3\beta)}.\] where $C_1$, $C_2$ are the integration constants and $\gamma\equiv|2-\beta|$. Inserting $H(a)$ into \[\Omega_{dm} = \frac{\kappa^2\rho_{dm}}{3H^2} = \frac{2aH'}{3H},\] one gets \[\Omega_{dm} = \frac1{2(2-3\beta)}\left[2-5\beta +\gamma\left(\frac{2C_2}{a^{3\gamma/2}+C_2}-1\right)\right].\] Requiring $\Omega_{dm}(a=1) = \Omega_{dm0}$ and $H(a=1)=H_0$, one obtains \[C_1 = H_0(1+C_2)^{1/(3\beta-2)}, C_2=-1 + \frac{2\gamma}{2-5\beta +\gamma + 2\Omega_{dm0} (3\beta-2)}.\] Using the relation \[q(z) = -\frac{1+z}{E(z)}\frac{d}{dz}\left(\frac1{E(z)}\right)-1,\] \[E\equiv \frac{H}{H_0} = (1+z)^{3(2-5\beta+\gamma)/[4(2-3\beta)]}\left[\frac{(1+z)^{-3\gamma/2+C_2}}{1+C_2}\right]^{1/(3\beta-2)}.\] one can obtain the deceleration parameter.
