# Cosmology Lecture 04

## Contents

### Time-Redshift Relations and the Age of the Universe

Last time we found the age of a flat universe. in a flat (Einstein-deSitter) universe: k = 0, Ωm = 1, ΩΛ = 0, so:

$t_0={1\over H_0}\int _0^\infty {{dz\over (1+z)^{5\over 2}}} ={2\over 3H_0}\approx 6.7h^{-1}Gyr \,\!$

Alternatively, recall that for a matter-dominated era, $a(t) = ({t\over t_0})^{2\over 3}.$ Thus, $H = {\dot a \over a}= {2\over 3t} \Rightarrow t_0={2\over 3H_0}$.

If we have $\Lambda \ne 0$: k = 0, Ω0,M + Ω0,Λ = 1, then:

$t_0={1\over H_0}\int _0^\infty {dz\over (1+z)\sqrt {\Omega _{0,M}(1+z)^3+\Omega _{0,\Lambda }}} \,\!$

Assuming $0.1\le \Omega _{0,M}\le 1$, this integral is solvable:

$t_0={2\over 3H_0}{1\over \sqrt {1-\Omega _{0,M}}}\ln \left({1+\sqrt {1-\Omega _{0,M}}\over \sqrt {\Omega _{0,M}}}\right)\approx {2\over 3H_0}(0.7\Omega _{0,M}+0.3-0.3\Omega _{0,\Lambda })^{-0.3} \approx {2\over 3H_0}\Omega _{0,M}^{-0.3} \,\!$

Generally, in a flat universe, $t_0\propto {1\over H_0}$. If $\Omega _0\le 1$, it will be longer.

In an open universe: k < 0, Ω0 < 1(Ω0,Λ = 0). Recall:

$\begin{matrix} t(\theta )={\Omega _0\over 2H_0(1-\Omega _0)^{3\over 2}}(\sinh \theta -\theta ), & a(\theta )={\Omega _0\over 2(1-\Omega _0)}(\cosh \theta -1) \end{matrix} \,\!$

So today:

\begin{align} a(\theta _0)& ={\Omega _0\over 2(1-\Omega _0)}(\cosh \theta _0-1)=1\\ \cosh \theta _0& ={2(1-\Omega _0)\over \Omega _0}+1={2-\Omega _0\over \Omega _0}\\ \sinh \theta _0& =\sqrt {\cosh ^2\theta _0-1}={2\over \Omega _0}\sqrt {1-\Omega _0}\\ t_0& =t(\theta _0)-{\Omega _0\over 2H_0(1-\Omega _0)^{3\over 2}}\left[{2\over \Omega _0} \sqrt {1-\Omega _0}-\cosh ^{-1}\left({2-\Omega _0\over \Omega _0}\right)\right]\\ \end{align} \,\!
${t_0={1\over H_0}\left[(1-\Omega _0)^{-1}-\frac12(1-\Omega _0)^{-{3\over 2}} \cosh ^{-1}\left({2-\Omega _0\over \Omega _0}\right)\right]} \,\!$

Thus $t_0>{2\over 3H_0}$ for $\Omega _0\le 1$, and $t_0={1\over H_0}$ for Ω0 = 0 (an empty universe).

In a closed universe: k > 0, Ω0 > 1, 0,Λ = 0). Recall:

\begin{align} t(\theta )& ={\Omega _0\over 2H_0(\Omega _0-1)^{3\over 2}}(\theta -sin\theta )\\ a(\theta )& ={\Omega _0\over 2(\Omega _0-1)}(1-\cos \theta )\\ \end{align} \,\!

Thus, today:

$a(\theta _0)={\Omega _0\over 2(\Omega _0-1)}(1-\cos \theta )=1 \,\!$
${t_0={1\over H_0}\left[(1-\Omega _0)^{-1}+\frac12\Omega _0(\Omega _0-1)^{-{3\over 2}} \cos ^{-1}\left({2-\Omega _0\over \Omega _0}\right)\right]} \,\!$

### The Robertson-Walker Metric

Lorentz invariance dictates that two inertial frame (x,y,z,t) and $(x^\prime , y^\prime , z^\prime , t^\prime )$, with one moving with respect to the other at velocity $\hat v=v\hat x$, are related by:

$\begin{matrix} x^\prime =\gamma (x-vt), & y^\prime =y, & z^\prime =z, & t^\prime =\gamma \left(t-{v\over c^2}x \right) \end{matrix} \,\!$

where $\gamma \equiv {1\over \sqrt {1-{v^2\over c^2}}}$. Note, to give a taste of tensor forms, this all may be written as ${x^\prime }^\alpha = \Lambda ^\alpha _\beta x^\beta +I_0^\alpha$.

Remember the Lorentz invariant interval, which is conserved between frames:

$ds^2=c^2dt^2-(dx^2+dy^2+dz^2) \,\!$

Light travels a ds2 = 0 path. In tensor form, this equation looks like:

$ds^2=g_{\alpha \beta }dx^\alpha dx^\beta \,\!$

where gαβ, the metric tensor, is given by:

$g_{\alpha \beta }\equiv \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{pmatrix} \,\!$

Look at Weinberg, Ch. 13 for full proof, but for a homogeneous, γ-isotropic space, the metric looks like:

$ds^2=c^2dt^2-a^2(t)\left[{dr^2\over 1-kr^2}+r^2d\Omega \right] \,\!$

where r is a radial direction (in comoving coordinates), and dΩ = dθ2 + sin2θd2φ is the differential angle seperation of two points in space. As usual, k is the measure of curvature.

The k = 0 Model:

$dr^2+r^2(d\theta ^2+\sin ^2\theta d\phi ^2)=dx^2+dy^2+dz^2 \,\!$

so we recover the Minkowski metric for flat space, using comoving coordinates.

The k > 0 (closed) Model:

We get a coordinate singularity at $r={1\over \sqrt {k}}$, so this universe has a finite volume. For k > 0, we need to define “Polar Coordinates” in 4-D (to describe a 3-sphere embedded in 4-D). Here is a comparison of how we define polar coordinates for a 3-sphere in 4-D versus for a 2-sphere in 3-D:

$\begin{matrix} 3-sphere & 2-sphere \\ (x,y,z,w)\Leftrightarrow (R,\alpha ,\beta ,\gamma ) & (x,y,z)\Leftrightarrow (R,\theta ,\phi ) \\ w=R\cos \alpha & z=R\cos \theta \\ z=R\sin \alpha \cos \beta & y=R\sin \theta \cos \phi \\ y=R\sin \alpha \sin \beta \cos \gamma & x=R\sin \theta \sin \phi \\ x=R\sin \alpha \sin \beta \sin \gamma & \\ x^2+y^2+z^2+w^2=R^2 & x^2+y^2+z^2=R^2 \\ \end{matrix} \,\!$

Take a line element on a 2-sphere:

$d\gamma =R^2(d\theta ^2+\sin ^2\theta d\phi ^2) \,\!$

Changing variables for $v\equiv \sin \theta$:

$dv=\cos \theta d\theta =\sqrt {1-v^2}d\theta \,\!$

Then $d\theta ^2={dv^2\over 1-v^2}$, so rewriting our line element, we get:

$d\gamma ^2=R^2\left({dv^2\over 1-v^2}+v^2d\phi ^2\right) \,\!$

For a 3-sphere,

$d\gamma ^2=R^2(d\alpha ^2+\sin ^2\alpha d\Omega ^2) \,\!$

where $d\Omega ^2\equiv d\beta ^2+\sin ^2\beta d\gamma ^2$. Again, using a change of variables so that $v\equiv \sin \alpha$, $d\alpha ={dv\over \sqrt {1-v^2}}$, we get that:

${d\gamma ^2=R^2\left({dv^2\over 1-v^2}+v^2d\Omega ^2\right)} \,\!$

This is what Robertson-Walker showed.