# 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:

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

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

{\displaystyle {\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:

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

where:

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

### 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 ${\displaystyle r}$ and ${\displaystyle z}$. We’ll look at light propagation (${\displaystyle ds^{2}=0}$), and take a radial path (${\displaystyle d\theta =d\phi =0}$) to know from the Robertson-Walker metric that:

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

Separating out our ${\displaystyle r}$ dependencies and our ${\displaystyle t}$ dependencies and integrating, we get:

${\displaystyle \int _{0}^{r_{1}}{dr \over {\sqrt {1-kr^{2}}}}=\int _{t_{1}}^{t_{0}}{c\cdot dt \over a(t)}\,\!}$
• For the flat, ${\displaystyle k=0}$, matter-dominated model (${\displaystyle w=0}$, ${\displaystyle \Omega _{0,\Lambda }=0}$, ${\displaystyle \Omega _{0,M}=1}$), we’ll start with the Friedmann Equation:
${\displaystyle H={{\dot {a}} \over a}=H_{0}{\sqrt {{\Omega _{0} \over a^{3+3w}}+{1-\Omega _{0} \over a^{2}}}}\,\!}$

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

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

These integrals evaluate to (in comoving ${\displaystyle r}$):

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

Here, ${\displaystyle {2c \over H_{0}}}$ 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 ${\displaystyle z\to \infty }$, ${\displaystyle r={2c \over H_{0}}\approx 10000h^{-1}Mpc}$.

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

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

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

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

where ${\displaystyle sinn}$ is a funny function:

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

Note that when ${\displaystyle w=0}$, for arbitrary ${\displaystyle \Omega _{0,M}}$, 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 ${\displaystyle a(t)}$, but we have not seen any perturbations off of that. Similarly, we’ve seen ${\displaystyle \rho (t)}$, 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 ${\displaystyle {\frac {1}{2}}}$)

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