Cosmology Lecture 02

The Friedmann Equations, continued

Recall we had the following equations:

${\displaystyle \left({{\dot {a}} \over a}\right)^{2}={8\pi \over 3}G\rho -{k \over a^{2}}\,\!}$
${\displaystyle {\dot {\rho }}=-3{{\dot {a}} \over a}(\rho +P)\,\!}$
${\displaystyle {{\ddot {a}} \over a}=-{4\pi \over 3}G(\rho +3P)\,\!}$

To close the equations, we need to relate ${\displaystyle P}$ and ${\displaystyle \rho }$ with an equation of state:

${\displaystyle {P=w\rho (c^{2})}\,\!}$

Note that we will generally set ${\displaystyle c=1}$ in this class. Combined with (2), this gives us:

{\displaystyle {\begin{aligned}{\dot {\rho }}&=-3{{\dot {a}} \over a}(1+w)\rho \\{{\dot {\rho }} \over \rho }&=-3(1+w){{\dot {a}} \over a}\\\rho &\propto a^{-3(1+w)}\\\end{aligned}}\,\!}

Note that we’ve assumed ${\displaystyle {\dot {w}}=0}$, which is okay most of the time. Some special cases of interest are:

• Pressure-less “dust” ${\displaystyle P=0}$, ${\displaystyle w=0\Rightarrow \rho \propto a^{-3}}$ because volume goes as ${\displaystyle V\propto {1 \over a^{3}}}$.
• Relativistic particles (photons, bosons): ${\displaystyle w={1 \over 3}}$, ${\displaystyle P={\rho \over 3}\Rightarrow \rho \propto a^{-4}}$ because ${\displaystyle VV\propto {1 \over a^{3}}}$, and energy is given by ${\displaystyle E\propto {1 \over a}}$.
• (${\displaystyle \Lambda }$)/Dark Energy: ${\displaystyle w=-1}$, ${\displaystyle P=-\rho \Rightarrow \rho =}$ constant in time.

To get density (${\displaystyle \rho }$) as a function of time, want to solve for ${\displaystyle w}$.

Critical Density, ${\displaystyle \rho _{crit}}$

We define ${\displaystyle \rho _{crit}}$ to be the critical density at which ${\displaystyle k=0}$ (and ${\displaystyle E=0}$):

${\displaystyle {\rho _{crit}}={3H^{2} \over 8\pi G}\,\!}$

Today, we measure: ${\displaystyle \rho _{crit,0}={3H_{0}^{2} \over 8\pi G}={3 \over 8\pi }{\left(100h{km \over sMpc}\right)^{2} \over G}=2.78\cdot 10^{11}h^{2}{M_{\odot } \over Mpc^{3}}=1.88\cdot 10^{-29}h^{2}{g \over cm^{3}}}$. Note that the mass of the sun is ${\displaystyle M_{\odot }=2\cdot 10^{33}g}$, and the mass of the proton is ${\displaystyle M_{p}=1.67\cdot 10^{-24}g}$.

Density Parameter, ${\displaystyle \Omega (t)}$

${\displaystyle \Omega }$ measures the ratio of the density of the universe to the critical density:

${\displaystyle \Omega (t)\equiv {\rho (t) \over \rho _{c}(t)}={8\pi G\rho (t) \over 3H^{2}(t)}\,\!}$
${\displaystyle \Omega {\begin{cases}\leq 1&\Rightarrow open\ (k\leq 0)\\=1&\Rightarrow flat\ (k=0)\\\geq 1&\Rightarrow closed\ (k\geq 0)\end{cases}}\,\!}$

In general, ${\displaystyle \Omega }$ can consist of multiple components: ${\displaystyle \Omega =\sum _{i}{\Omega _{i}}}$ e.g.

${\displaystyle \Omega {\begin{cases}r&=radiation\\m&=matter\ (dark\ and\ luminous)\\b&=baryons\ (dark\ and\ luminous)\\\nu &=neutrinos\\\Lambda &=dark\ energy\end{cases}}\,\!}$

${\displaystyle \Omega =1}$ is an unstable equilibrium; any perturbation from ${\displaystyle \Omega =1}$ in the early universe ensures ${\displaystyle \Omega }$ is far from 1 today. That we measure ${\displaystyle \Omega _{m}\approx 0.3}$ today implies that the early universe must have been extremely finely tuned.

Evolution of ${\displaystyle H(t)}$

Today (at ${\displaystyle a=1}$): ${\displaystyle k={8\pi \over 3}G\rho _{0}-H_{0}^{2}=H_{0}^{2}(\Omega _{0}-1)}$. Using the ${\displaystyle 1^{st}}$ Friedmann equation (1), we have:

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

(THE Friedmann Equation) where ${\displaystyle H_{0}\equiv {\sqrt {8\pi G\rho \over 3{\rho _{crit}}}}}$, and it is understood that ${\displaystyle \Omega _{0}=\sum _{i}{\Omega _{i,0}}}$. Each ${\displaystyle \Omega _{i}}$ has it’s own ${\displaystyle w_{i}}$, so really ${\displaystyle {\Omega _{0} \over a^{3+3w}}=\sum _{i}{\Omega _{i,0} \over a^{3+3w_{i}}}}$.

Evolution of ${\displaystyle \Omega (a)}$

We’ll show in PS#1 that for any single component:

${\displaystyle {1-\Omega (a) \over \Omega (a)}={1-\Omega _{0} \over \Omega _{0}}a^{1+3w}\,\!}$

Plotting ${\displaystyle \Omega (a)}$, we will find that for early ${\displaystyle a}$, ${\displaystyle \Omega }$ is extremely close to 1.

Evolution of ${\displaystyle a(t)}$: Solving the Friedmann Equation

The evolution of the expansion of the universe is governed by:

${\displaystyle \left({{\dot {a}} \over a}\right)^{2}={8\pi \over 3}G\rho -{k \over a^{2}}\,\!}$

We can apply this to several models of the universe:

• The Einstein-deSitter (flat) Model: ${\displaystyle k=0}$, ${\displaystyle \Omega (a)=\Omega _{0}=1}$. Using that ${\displaystyle \rho \propto a^{-3(1+w)}}$, we have:
{\displaystyle {\begin{aligned}\left({{\dot {a}} \over a}\right)^{2}&\propto a^{-3(1+w)}\\a^{-1}a^{{3 \over 2}(1+w)}da&\propto dt\\a^{-{3 \over 2}(1+w)}&\propto t\\\end{aligned}}\,\!}
${\displaystyle {a(t)\propto t^{2 \over 3(1+w)}}\,\!}$

Thus, the rate of expansion of the universe depends on ${\displaystyle w}$: (a) The matter-dominated era: ${\displaystyle \Omega \approx \Omega _{m}\Rightarrow w=0,P=0,a\propto t^{2 \over 3}}$. (b) The radiation-dominated era: ${\displaystyle \Omega \approx \Omega _{r}\Rightarrow w={1 \over 3}\Rightarrow a(t)\propto t^{\frac {1}{2}}}$. (c) The ${\displaystyle \Lambda }$-dominated era: ${\displaystyle \Omega \approx \Omega _{\Lambda }\Rightarrow w=-1,P=-\rho ,\rho }$ constant in time ${\displaystyle \Rightarrow a(t)\propto e^{Ht}}$, where H is now actually a constant. This is exponential inflation. We used to think that this only happened early on (like ${\displaystyle 10^{-34}}$ seconds), but now we think that this has also been happening recently. Next time, we will do the harder two cases: open and closed.