Cosmology Lecture 05

Finishing the Robertson-Walker Metric

We’ve been following Weinberg’s derivation to show there are discrete metrics. We’ll start with:

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

Recall that $k$ is the curvature constant and $r$ is in comoving coordinates. If we define $U\equiv \sinh \theta$, then:

${\begin{aligned}dU&=\cosh \theta \cdot d\theta \\d\theta &={dU \over {\sqrt {1+U^{2}}}}\\\end{aligned}}\,\!$

The beauty of this metric is that we derived it only using symmetry (no dynamics). There is an alternate way of writing the metric above:

$ds^{2}=c^{2}dt^{2}-a^{2}\left[d\chi ^{2}+S^{2}(\chi )(d\theta ^{2}+\sin ^{2}\theta d\phi ^{2}\right]\,\!$

where:

$S(\chi )\equiv {\begin{cases}\chi &\,(for\ k=0,\ \chi =r)\\k^{-{\frac {1}{2}}}\sin(k^{\frac {1}{2}}\chi )&\,(for\ k>0)\\(-k)^{-{\frac {1}{2}}}\sinh({\sqrt {-k}}\chi )&\,(for\ k<0)\\\end{cases}}\,\!$

Comoving Radial Distance vs. Redshift: the Hubble Diagram

This is the fundamental diagram behind using a standard candle (supernova) to infer the curvature of the universe. What we want here is an algebraic expression relating $r$ and $z$. We’ll look at light propagation ($ds^{2}=0$), and take a radial path ($d\theta =d\phi =0$) to know from the Robertson-Walker metric that:

$c^{2}dt^{2}=a^{2}{dr^{2} \over 1-kr^{2}}\,\!$

Separating out our $r$ dependencies and our $t$ dependencies and integrating, we get:

$\int _{0}^{r_{1}}{dr \over {\sqrt {1-kr^{2}}}}=\int _{t_{1}}^{t_{0}}{c\cdot dt \over a(t)}\,\!$

• For the flat, $k=0$, matter-dominated model ($w=0$, $\Omega _{0,\Lambda }=0$, $\Omega _{0,M}=1$), we’ll start with the Friedmann Equation:

$H={{\dot {a}} \over a}=H_{0}{\sqrt {{\Omega _{0} \over a^{3+3w}}+{1-\Omega _{0} \over a^{2}}}}\,\!$

Recognizing that $a={1 \over 1+z}$, we have:

$H_{0}dt=-{dz \over (1+z){\sqrt {\Omega _{0}(1+z)^{3+3w}+(1-\Omega _{0})(1+z)^{2}}}}\,\!$

These integrals evaluate to (in comoving $r$):

$r_{1}=\int _{t_{1}}^{t_{0}}{c\cdot dt \over a(t)}\,\!$ ${r_{1}={2c \over H_{0}}\left[1-(1+z)^{-{\frac {1}{2}}}\right]}\,\!$

Here, ${2c \over H_{0}}$ is called the Hubble distance and is the definition of how far away we can possibly see–how far light could have traveled since the beginning of time. Notice that as $z\to \infty$, $r={2c \over H_{0}}\approx 10000h^{-1}Mpc$.

• For the open, $k<0$, $\Omega _{0}<1$, $w=0$ model, we’ll substitute ${\sqrt {-k}}r=\sinh \chi$, $dr={\cosh \chi d\chi \over {\sqrt {-k}}}$ for $r$ in the integral above:

${\begin{aligned}r&=\int _{-}^{\chi _{1}}{\cosh \chi d\chi \over {\sqrt {-k}}{\sqrt {1+\sinh ^{2}\chi }}}\\{\chi _{1} \over {\sqrt {-k}}}&={\sinh ^{-1}({\sqrt {-k}}r_{1}) \over {\sqrt {-k}}}\\\end{aligned}}\,\!$

From this, we can use Mattig’s Formula, which states for $w=0$, arbitrary $\Omega _{0,M}=\Omega _{0}$, that:

$r={2c \over H_{0}\Omega _{0}^{2}}{1 \over 1+z}\left[\Omega _{0}z+(\Omega _{0}-2)({\sqrt {1+\Omega _{0}z}}-1)\right]\,\!$

In general, for arbitrary $\Omega _{0,M},\Omega _{0,\Lambda }$ (we’ll derive this in PS#3), one can show that, in comoving $r$:

${r={1 \over {\sqrt {|k|}}}sinn\left({c \over H_{0}}{\sqrt {|k|}}\int _{0}^{z}{dz^{\prime } \over {\sqrt {\Omega _{0,M}(1+z)^{3}+\Omega _{0,\Lambda }+(1-\Omega _{0,M}-\Omega _{0,\Lambda })(1+z)^{2}}}}\right)}\,\!$

where $sinn$ is a funny function:

$sinn\equiv {\begin{cases}\sin &\,for\ k>0\\absent&\,for\ k=0\\\sinh &\,for\ k<0\\\end{cases}}\,\!$

Note that when $w=0$, for arbitrary $\Omega _{0,M}$, we recover Mattig’s Formula.

Angular Diameter Distance

The angular diameter distance is a useful quantity which relates the physical size or separation of objects to the angular size on the sky. For normal, Euclidean geometries, this is trivial trigonometry. For a curved universe, this is not trivial. For example, in some universes, an object pulled far enough away may actually start looking larger (have a larger angular diameter) than a closer object!

This brings us to the end of the smooth universe. We’ve seen $a(t)$, but we have not seen any perturbations off of that. Similarly, we’ve seen $\rho (t)$, but no spatial components of density. We will begin to talk about perturbations off of the Smooth Universe, and we will call this:

The Bright Side of the Universe

Let’s do a quick tour of the the particles out there to give some context to what we’re talking about. Take a look at Review of Particle Physics, which is also at http://pgd.lbl.gov, for some more detailed information.

First, we’ll talk about fermions. Fermions come in two varieties: Leptons and Quarks. Quarks are hadrons and group together to from baryons (made of 3 quarks) and mesons (made of quark-antiquark pairs). Elementary Particles: Fermions (spin ${\frac {1}{2}}$)

${\begin{matrix}Leptons&Charge&Mass&Mean\ Lifetime\\e&-1&0.51099892\pm 0.00000004MeV&\infty \\\nu _{e}&0&<3eV&\infty \\\mu &-1&106MeV&2.2\mu s,\ \mu ^{-}\to e^{-}{\bar {\nu }}_{e}\nu _{p}\\\nu _{\mu }&0&<190keV&\infty \\\tau &-1&1.78GeV&2.9\cdot 10^{-13}\\\nu _{\tau }&0&<18.2MeV&\infty \\\end{matrix}}\,\!$ ${\begin{matrix}Quarks\ (3\ colors\ each)&Charge&Mass\\U&{2 \over 3}&1.5\to 4MeV\\d&-{1 \over 3}&4\to 8MeV\\s&-{1 \over 3}&80\to 130MeV\\c&{2 \over 3}&1.15\to 1.35GeV\\b&-{1 \over 3}&4.1\to 4.4GeV\\t&{2 \over 3}&174.3\pm 5.1GeV\\\end{matrix}}\,\!$