# Difference between revisions of "Composite Models"

(→Problem 3: no solution) |
|||

Line 61: | Line 61: | ||

=== Problem 3 === | === Problem 3 === | ||

When are $L_p$ and $L_e$ equal? It is interesting to know whether both horizons might have or not the same values, and if so, how often this could happen. | When are $L_p$ and $L_e$ equal? It is interesting to know whether both horizons might have or not the same values, and if so, how often this could happen. | ||

− | |||

<div id="Horizon37"></div> | <div id="Horizon37"></div> | ||

+ | |||

=== Problem 4 === | === Problem 4 === | ||

Find the particle and event horizons for any redshift $z$ in the standard cosmological model -- $\Lambda$CDM. | Find the particle and event horizons for any redshift $z$ in the standard cosmological model -- $\Lambda$CDM. |

## Revision as of 16:32, 9 October 2013

### Problem 1

Consider a flat universe with several components $\rho=\sum_i \rho_i$, each with density $\rho_i$ and partial pressure $p_i$ being related by the linear state equation $p_i =w_i \rho_i$. Find the particle horizon at present time $t_0$.

The particle horizon $L_p$ at the current moment $t_0$ is (see \ref{LpDp}) \[L_p =\lim\limits_{t_e \to 0} D_p (t_e ,t_0)\] The proper distance can be written as \[D_p = a_0\int_t^{t_0} \frac{dt'}{a(t')} =- \int_{z(t)}^{z(t_0)} \frac{dz'}{H(z')} =\int_0^z \frac{dz'}{H(z')}. \] The $i$-th energy density in terms of redshift $z$ is \[\rho_i =\rho_{0i}(1+z)^{3(1+w_i)},\] and \[H(z)=H_0 \sqrt{\sum \Omega_{0i}(1+z)^{3(1+w_i)}},\] so \begin{equation} L_p (t_0)=H_0^{-1}\int \limits_{0}^{\infty} \frac{dz}{\sqrt{\sum \Omega_{0i}(1+z)^{3(1+w_i)}}} \end{equation}

### Problem 2

Suppose we know the current material composition of the Universe $\Omega_{i0}$, $w_i$ and its expansion rate as function of redshift $H(z)$. Find the particle horizon $L_p (z)$ and the event horizon $L_p (z)$ (i.e the distances to the respective surfaces along the surface $t=const$) at the time that corresponds to current observations with redshift $z$.

For the particle horizon we rewrite the previous result in terms of redshifts \begin{equation} L_p (z)=H^{-1} (z)\int \limits_{0}^{\infty} \frac{dz'}{\sqrt{\sum \Omega_{i}(z)(1+z')^{3(1+w_i)}}}, \end{equation} and take into account how the partial densities $\Omega_i$ depend on time: they are defined to satisfy \begin{equation} H^2 (z)=H_0^2 \sum \Omega_{i0} (1+z)^{3(1+w_i)}. \end{equation} at any $z$ (or, equivalently, $t$), thus $\Omega_i (z)$ by definition is the ratio of the $i$th term of the sum to the whole sum at any moment of time: \begin{equation} \Omega_i (z) = \Omega _{i0} \frac{H_0^2 }{H^2 (z)}(1 + z)^{3(1+w_i)}. \end{equation} Then for the particle horizon (and for event horizon in the same way) we obtain \begin{align} L_p (z) &= \frac{1}{H(z)}\int_0^\infty \frac{dz'}{\sqrt {\sum\limits_i \Omega_i (z)(1 + z')^{3(1 +w_i )}}}; \label{LpZ}\\ L_e (z) &= \frac{1}{H(z)}\int_{-1}^{0} \frac{dz'}{\sqrt {\sum\limits_i \Omega_i (z)(1 + z')^{3(1 +w_i )}}} \label{LeZ}. \end{align}

### Problem 3

When are $L_p$ and $L_e$ equal? It is interesting to know whether both horizons might have or not the same values, and if so, how often this could happen.

### Problem 4

Find the particle and event horizons for any redshift $z$ in the standard cosmological model -- $\Lambda$CDM.

Using the general formulae (\ref{LpZ}--\ref{LeZ}), one obtains \begin{align} L_p(z) &= \frac{1}{H(z)}\int_0^\infty \frac{dz'}{\sqrt {\Omega_m (z) (1 + z')^3 + \Omega_\Lambda (z)}},\\ L_e(z) &= \frac{1}{H(z)}\int_{-1}^{0} \frac{dz'}{\sqrt {\Omega_m (z) (1 + z')^3 + \Omega_\Lambda (z)}},\\ &\Omega_m (z)=\Omega _{m0}\frac{H_0^2}{H(z)^2} (1 + z)^3 ,\\ &\Omega_\Lambda (z) = \Omega_{\Lambda 0} \frac{H_0^2}{H(z)^2}. \end{align}

### Problem 5

Express the particle $L_p (z)$ and event $L_e (z)$ horizons in $\Lambda$CDM through the hyper-geometric function.

\begin{align} L_p (z) &= \frac{2\sqrt {A(z)}}{H_0 \sqrt {\Omega _{\Lambda 0}}} F\left(\frac{1}{2},\frac{1}{6},\frac{7}{6}; - A(z) \right),\\ L_e (z) &= \frac{1}{H_0 \sqrt {\Omega _{\Lambda 0}}} F\left(\frac{1}{2},\frac{1}{3},\frac{4}{3}; - \frac{1}{A(z)} \right),\\ &A(z) = \frac{\Omega_{\Lambda 0}}{\Omega _{m0}} (1 + z)^{-3}. \end{align}