Composite Models

16.3.1

Problem 16.3.1

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 $t_0$.

solution

The particle horizon $L_p$ at the current moment $t_0$ is (see this relation) \[L_p =\lim\limits_{t_e \to 0} D_p (t_e ,t_0)\] 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}

16.3.2

Problem 16.3.2

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 $z$.

solution

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}

16.3.3

Problem 16.3.3

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.

16.3.4

Problem 16.3.4

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

solution

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}

16.3.5

Problem 16.3.5

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

solution

\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}