Transient acceleration - Revision history

Cosmo All: /* Problem 6 */

2013-11-11T04:44:02Z

‎Problem 6

Cosmo All: /* Problem 5 */

2013-11-11T04:42:28Z

‎Problem 5

Cosmo All: /* Problem 4 */

2013-11-11T04:40:53Z

‎Problem 4

Cosmo All: /* Problem 1 */

2013-11-11T04:39:35Z

‎Problem 1

Cosmo All at 04:38, 11 November 2013

2013-11-11T04:38:33Z

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Cosmo All at 11:42, 5 November 2013

2013-11-05T11:42:34Z

Cosmo All at 11:42, 5 November 2013

2013-11-05T11:42:12Z

Cosmo All: Created page with "10 NOTOC"

2013-11-05T11:23:24Z

Created page with "10 __NOTOC__"

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[[Category:Interactions in the Dark Sector|10]]

__NOTOC__

@@ Line 295: / Line 295: @@
 \end{eqnarray}
 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}
 with

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 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}},[[User:Cosmo All|Cosmo All]]
+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}},[[User:Cosmo All|Cosmo All]]
+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}
@@ Line 280: / Line 280: @@
 <div id=""></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === 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 $t_0$.

@@ Line 122: / Line 122: @@
 \end{eqnarray}
 we get
- \begin{equation}
+\begin{equation}
 \label{rhophi2}
 \rho_{de}=\rho_{de0}\, a^{-3(1+w_0)}\cdot
@@ Line 131: / Line 131: @@
 Substituting Eq. (\ref{rhophi2}) into Eq. (\ref{eom2}), we get
 the dark matter energy density,
- \begin{equation}\label{rhom2}
+\begin{equation}\label{rhom2}
 \rho_{dm}=f(a)\rho_{dm0},
 \end{equation}
@@ Line 222: / Line 222: @@
 <div id=""></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === Problem 5 ===
 Consider the flat FLRW cosmology with two coupled

@@ Line 23: / Line 23: @@
 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},
+\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
@@ Line 47: / Line 47: @@
 We note that the case $\epsilon_0 = 0$ corresponds to the SCM, i.e., ${\rm{X}} \equiv {\rho}_{\Lambda}$.
 Using the formulas presented above, it is not difficult to obtain the dependences for the relative densities $\Omega_b(a)$, $\Omega_{m}(a)$ and $\Omega_{\Lambda}(a)$:
 \begin{equation} \label{8a}
 \Omega_{b}(a) = \frac{a^{-3}}{{\rm{A}} + a^{-3} + {\rm{B^{-1}}}\varphi(a)}\;,
@@ Line 57: / Line 56: @@
 \Omega_{{\rm{\Lambda}}}(a) = \frac{{\rm{D}} + \varphi(a) - a^{-3 + \epsilon(a)}}{{\rm{D}} + {\rm{B}}a^{-3}  + \varphi(a)}\;,
 \end{equation}
   where ${\rm{A}} = {\Omega_{{\rm{X}},0}}/{\Omega_{b,0}}$, ${\rm{B}} = {\Omega_{b,0}}/{\Omega_{m0}}$ and  ${\rm{D}} = {\Omega_{{\rm{X}},0}}/{\Omega_{m0}}$.</p>
    </div>
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 <div id=""></div>
 <div style="border: 1px solid #AAA; padding:5px;">
 === 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 $\epsilon_0 = -0.1.$.

← Older revision		Revision as of 11:42, 5 November 2013
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−	[[Category:Interactions in the Dark Sector\|0]]	+	[[Category:Interactions in the Dark Sector\|9]]

	__NOTOC__		__NOTOC__

Transient acceleration - Revision history

Cosmo All: /* Problem 6 */

Cosmo All: /* Problem 5 */

Cosmo All: /* Problem 4 */

Cosmo All: /* Problem 1 */

Cosmo All at 04:38, 11 November 2013

Cosmo All at 11:42, 5 November 2013

Cosmo All at 11:42, 5 November 2013

Cosmo All: Created page with "10 __NOTOC__"

Cosmo All: Created page with "10 NOTOC"