Cosmology Lecture 04

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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.

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