Difference between revisions of "Dynamics of the Universe in terms of redshift and conformal time"

Using the relation $a_{0}/a=z+1$ with normalization $a_{0}=1$, we can rewrite the energy conservation law for every non-interacting component as \[\Omega_{i}=\Omega_{i0}a^{-3(1+w_{i})} =\Omega_{i0}(1+z)^{3(1+w_{i})},\] so for the Universe filled with matter, radiation and curvature the first Friedman equation becomes \[H^{2}(z) = H_0^2\left[ \Omega _{r0}(1+z)^4 + \Omega_{m0}(1+z)^3+ \Omega_{k0}(1+z)^2\right].\] At large $z$ (early Universe) the term with the highest power becomes dominating -- that is, the one with radiation. As the term with curvature is very small at present (see problem), it was all the more negligible before. On the other hand, it should become dominating in the future ($1+z\to 0$), but in the standard cosmological model, which will be discussed in the corresponding chapter later, the cosmological constant enters the play before that, and the curvature term does not ever have the chance to shine.

As \[\eta = \int \frac{dt}{a(t)},\] for the Universe with the radiation component dominating $a(t) \sim t^{1/2}$, so $\eta \sim a$. For the non-relativistic matter $a(t) \sim t^{2/3}$, so $\eta\sim a^{1/2}.$

Using $H(z)$, we can express the first Friedman equation in the form \[\left(\frac{\dot z}{1 + z} \right)^2 =H_0^2 \sum\limits_i{\Omega_i},\] where $\Omega_i$ are the relative densities of all components filling the Universe (including curvature) with state parameters $w_i$. On separating variables, we arrive to \[t_0(z) = \frac{1}{H_0}\int_0^z\frac{dz'} {(1 + z')\sqrt{\sum \limits_i \Omega _i(1 + z')^{3(1 +w_i)}}}.\] In case $\Omega _{m0} = 1,\;\Omega _{r0} =\Omega _{curv} = 0$ then \[t_0(z) = \frac{2/3}{H_0\sqrt{\Omega _{m0}}} \left(1-\frac{1}{(1 + z)^{3/2}}\right).\]

The first Friedman equation takes the form \[\frac{1}{a^4}\left(\frac{da}{d\eta}\right)^2 = H_0^2\frac{\Omega _{r0}}{a^4},\] and its solution is \[a - a_0 = H_0\sqrt{\Omega _{r0}}(\eta-\eta_0 ).\]

Using the result of the previous problem with $a_0 =0$, we have \[t = \int a(\eta )d\eta =\frac{1}{2}H_0\sqrt{\Omega _{r0}}\;\eta^2.\]

In this case the first Friedman equation is \[\frac{1}{a^4}\left(\frac{da}{d\eta}\right)^2 = \frac{H_0^2\Omega _{m0}}{a^3},\] and we obtain \[a(\eta)-a_{0} = \frac{H_0^2\Omega _{m0}}{4}(\eta-\eta_0)^2.\]

The first Friedman equation is \[\frac{1}{a^4}\left(\frac{da}{d\eta}\right)^2 = H_0^2\left(\frac{\Omega _{m0}}{a^3} + \frac{\Omega _{r0}}{a^4} \right),\]so \[\frac{H_{0}\Omega_{m0}}{2}(\eta-\eta_{0}) =\sqrt{\Omega_{r0}+\Omega_{m0}a} -\sqrt{\Omega_{r0}+\Omega_{m0}a_{0}}.\] Let $\eta_{0}=a_{0}=0$. Then \[a=\frac{H_{0}^{2}\Omega_{m0}}{4}\,\eta^{2} -H_{0}\sqrt{\Omega_{m0}}\;\eta\]

As $w_i =w_{i}(t)$ and $t = t(z)$, we have $w_i = w_{i}(z)$. The conservation equation then is \[\dot\rho_{i}+3\frac{\dot z}{1 + z}[1+w(z)]\rho_{i}= 0.\] Separating the variables, one gets \[\rho_{i}(z) =\rho_{i0}\exp\left\{ -3\int_0^z \big[1 + w_{i}(z')\big] \frac{dz'}{(1 + z')}\right\}.\]

From the first Friedman equation \[H^2 = H_0^2\Omega _{m0}(1 + z)^3 \quad\Rightarrow\quad H(z)=H_0\sqrt {\Omega _{m0}} (1 + z)^{3/2}.\]

The equation of null curve (and radial null geodesics) is $ds^{2}=a^{2}(d\eta^{2}-d\chi^{2})=0$, and thus $d\eta=\pm d\chi$. Then the light signal, which comes from a source at comoving distance $\chi$, and which we observe at conformal time $\eta_{0}$, was emitted at conformal time $\eta_{e}=\eta_{0}-\chi$. The observed redshift is thus a function of the time of observation $\eta_0$ and is equal to \[z\left(\eta _0 \right) = \frac{a_0}{a_e} - 1 = \frac{a\left(\eta_0\right)} { a\left(\eta _0 - \chi \right)}-1.\] As $\chi$ does not change (we a looking at the same object), recalling the definition of conformal time $d\eta = dt/a(t)$, we get \begin{align*} \dot z \equiv \frac{dz}{dt_0} &= \frac{1}{a\left( \eta _0\right)} \frac{\partial z}{\partial\eta _0} = \frac{1}{a\left( \eta _0\right)} \frac{\partial }{\partial\eta _0} \left(\frac{a\left(\eta _0 \right)} {a\left( \eta _0- \chi \right)} -1 \right)=\\ &= \frac{1}{a(\eta_0 - \chi)\,a(\eta _0)} \frac{\partial a(\eta _0)}{\partial \eta _0} - \frac{1}{a(\eta _0 - \chi)^2} \frac{\partial a(\eta _0 - \chi)} {\partial(\eta _0- \chi)}=\\ &=\frac{\dot a_0}{a_e} - \frac{\dot a_e}{a_e} = (1+z)\frac{\dot a_0}{a_0} - \frac{\dot a_e}{a_e} = (1 + z)H_0 - H(z) \end{align*} For the Universe with dominating non-relativistic matter then \[\dot z = \left(1 + z\right)H_0 \left(1 - \sqrt{\Omega _{m0}}(1 +z)^{1/2}\right).\]

The comoving distance $\chi$ to the surface we are observing is determined from the condition $ds^2 = dt^2 - a(t)d\chi^2 = 0$. Expressing $\chi$ thought its redshift, we get \[\chi = \int_0^z \frac{dz'}{H(z')}.\] For a model with dominating matter \[\chi = \int_0^z \frac{dz'}{H_0\sqrt{\Omega _{m0} (1 + z')^3}} =\frac{2}{H_0\sqrt{\Omega _{m0}}} \left( 1 - \sqrt {\frac{1}{1 + z}}\right).\] Using $\Omega _{m0} \approx 1$ and the value of redshift at the moment of recombination $z_r \approx 1100$, we arrive to $\chi\approx 7.8$ GPc. Then the particle horizon is \[ L_{p} = \int_0^\infty \frac{dz}{H_0\sqrt {\Omega _{m0}(1 + z)^3}}= \frac{2}{H_0\sqrt{\Omega _{m0}}} \approx 8.02\mbox{ GPc},\] and therefore \[\frac{L_{p}}{\chi} = \Big(1 - \sqrt {\frac{1}{1 + z}}\Big)^{-1} \approx 1+\frac{1}{\sqrt{z}} \approx 1.031.\]

Analogously to the problem, for the model with dominating of radiation and matter we obtain \[L_{p} =\int\limits_{0}^{z} \frac{dz'}{H_0\sqrt{ \Omega _{m0}(1 + z')^3 +\Omega_{r0}(1 + z')^4}} =\frac{2}{\Omega _{m0}H_0} \left(1 - \sqrt{\frac{1+\Omega_{r0}z}{1 + z}}\right).\] We have taken into account here that $\Omega _{m0} + \Omega _{r0} = 1$.