# Cosmology Lecture 04

### 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: ${\displaystyle k=0}$, ${\displaystyle \Omega _{m}=1}$, ${\displaystyle \Omega _{\Lambda }=0}$, so:

${\displaystyle t_{0}={1 \over H_{0}}\int _{0}^{\infty }{dz \over (1+z)^{5 \over 2}}={2 \over 3H_{0}}\approx 6.7h^{-1}Gyr\,\!}$

Alternatively, recall that for a matter-dominated era, ${\displaystyle a(t)=({t \over t_{0}})^{2 \over 3}.}$ Thus, ${\displaystyle H={{\dot {a}} \over a}={2 \over 3t}\Rightarrow t_{0}={2 \over 3H_{0}}}$.

If we have ${\displaystyle \Lambda \neq 0}$: ${\displaystyle k=0}$, ${\displaystyle \Omega _{0,M}+\Omega _{0,\Lambda }=1}$, then:

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

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

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

Generally, in a flat universe, ${\displaystyle t_{0}\propto {1 \over H_{0}}}$. If ${\displaystyle \Omega _{0}\leq 1}$, it will be longer.

In an open universe: ${\displaystyle k<0}$, ${\displaystyle \Omega _{0}<1(\Omega _{0,\Lambda }=0)}$. Recall:

${\displaystyle {\begin{matrix}t(\theta )={\Omega _{0} \over 2H_{0}(1-\Omega _{0})^{3 \over 2}}(\sinh \theta -\theta ),&a(\theta )={\Omega _{0} \over 2(1-\Omega _{0})}(\cosh \theta -1)\end{matrix}}\,\!}$

So today:

{\displaystyle {\begin{aligned}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{aligned}}\,\!}
${\displaystyle {t_{0}={1 \over H_{0}}\left[(1-\Omega _{0})^{-1}-{\frac {1}{2}}(1-\Omega _{0})^{-{3 \over 2}}\cosh ^{-1}\left({2-\Omega _{0} \over \Omega _{0}}\right)\right]}\,\!}$

Thus ${\displaystyle t_{0}>{2 \over 3H_{0}}}$ for ${\displaystyle \Omega _{0}\leq 1}$, and ${\displaystyle t_{0}={1 \over H_{0}}}$ for ${\displaystyle \Omega _{0}=0}$ (an empty universe).

In a closed universe: ${\displaystyle k>0}$, ${\displaystyle \Omega _{0}>1}$, ${\displaystyle (\Omega _{0,\Lambda }=0)}$. Recall:

{\displaystyle {\begin{aligned}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{aligned}}\,\!}

Thus, today:

${\displaystyle a(\theta _{0})={\Omega _{0} \over 2(\Omega _{0}-1)}(1-\cos \theta )=1\,\!}$
${\displaystyle {t_{0}={1 \over H_{0}}\left[(1-\Omega _{0})^{-1}+{\frac {1}{2}}\Omega _{0}(\Omega _{0}-1)^{-{3 \over 2}}\cos ^{-1}\left({2-\Omega _{0} \over \Omega _{0}}\right)\right]}\,\!}$

### The Robertson-Walker Metric

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

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

where ${\displaystyle \gamma \equiv {1 \over {\sqrt {1-{v^{2} \over c^{2}}}}}}$. Note, to give a taste of tensor forms, this all may be written as ${\displaystyle {x^{\prime }}^{\alpha }=\Lambda _{\beta }^{\alpha }x^{\beta }+I_{0}^{\alpha }}$.

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

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

Light travels a ${\displaystyle ds^{2}=0}$ path. In tensor form, this equation looks like:

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

where ${\displaystyle g_{\alpha \beta }}$, the metric tensor, is given by:

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

Look at Weinberg, Ch. 13 for full proof, but for a homogeneous, ${\displaystyle \gamma }$-isotropic space, the metric looks like:

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

where ${\displaystyle r}$ is a radial direction (in comoving coordinates), and ${\displaystyle d\Omega =d\theta ^{2}+\sin ^{2}\theta d^{2}\phi }$ is the differential angle seperation of two points in space. As usual, ${\displaystyle k}$ is the measure of curvature.

The ${\displaystyle k=0}$ Model:

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

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

The ${\displaystyle k>0}$ (closed) Model:

We get a coordinate singularity at ${\displaystyle r={1 \over {\sqrt {k}}}}$, so this universe has a finite volume. For ${\displaystyle k>0}$, 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:

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

Take a line element on a 2-sphere:

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

Changing variables for ${\displaystyle v\equiv \sin \theta }$:

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

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

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

For a 3-sphere,

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

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

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

This is what Robertson-Walker showed.