Cosmology Lecture 03

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<latex> \documentstyle[11pt]{article} \def\hf{\frac12} \def\imply{\Rightarrow} \def\inv#1Template:1\over \def\aaTemplate:\dot a \over a \def\addaTemplate:\ddot a \over a \def\thnot{\theta_0} \def\etot{\Omega_0} \def\econs{\Omega_{0,\Lambda}} \def\emat{\Omega_{0,M}} \def\econs{\Omega_{0,\Lambda}} \def\p{^\prime} \def\iff{\Leftrightarrow} \def\xvTemplate:\vec x\def\pvTemplate:\vec p \def\vvTemplate:\vec v \def\pptTemplate:\partial\over\partial t \def\ddt{{\frac{d}{dt}}} \def\epotTemplate:8\pi \over 3 \def\atow{a^{-3(1+w)}} \def\atowi{a^{-3(1+w_i)}} \def\athow{a^{-{3 \over 2}(1+w)}} \def\attw{a^{3+3w}} \def\cc{Cosmological constant} \def\addaTemplate:\ddot a \over a \def\jumble{{\Omega _0 \over \attw} + {1 - \Omega _0 \over a^2}} \def\jimble{\sqrt{\Omega _{0,m}(1+z)^3+\Omega _{0,\Lambda}+(1-\Omega _{0,m}-\Omega _{0,\Lambda})(1+z)^2}} \def\paap{\left(\aa\right)} \def\rcr{{\rho_{crit}}}

\usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document}

\subsection*{ Evolution of $a(t)$: Solving the Friedmann Equation, continued }

Last time we did (1), the flat universe:

Einstein-deSitter, $k=0$ (flat), $\Omega =1,\imply$

$a\propto t^{2\over 3(1+w)}$. In general, we consider the evolution of

the universe to be {\it matter dominated} if $a\propto t^{2\over 3}$, {\it radiation dominated} if $a \propto t^{\hf}$, and {\it $\Lambda$ dominated} if $a \propto e^{1+t}$ (that is, $H$ is constant).

Open, $k<0$, $\etot<1$, $\econs=0$, $\emat<1\imply$ our Friedmann Equation $\paap^2=\epot G\rho-{k\over a^2}$ becomes: $$\dot a^2=H_0^2\left({\etot \over a^{1+3w}}+1-\etot\right)$$ Since we are matter dominated, $w=0$. Solving for $a(t)$: $$\begin{align} \int_0^{a_f}{da\over\sqrt{{\etot\over a}+1-\etot}} &=\int_0^{t_f}{H_0\cdot dt}\\ \int_0^{a_f}{a\cdot da\over\sqrt{1-\etot}\sqrt{a^2+{\etot\over1-\etot}a}} &=H_0t_f\\ \end{align}$$ Let's define $2\alpha\equiv{\etot\over1-\etot}$. Then: $$\int_0^{a_f}{a\cdot da\over\sqrt{1-\etot}\sqrt{(a+\alpha)^2-\alpha^2}} =H_0t_f$$ Defining $a^\prime=a+\alpha$, we get: $$\begin{align} \int_0^{a_f+\alpha}{(a^\prime-\alpha)da^\prime\over \sqrt{1-\etot}\sqrt{a^{\prime 2}-\alpha^2}}&=H_0t_f\\ \int_0^{a_f+\alpha}{(a-\alpha)da\over \sqrt{1-\etot}\sqrt{a^2-\alpha^2}} &=H_0t_f\\ \end{align}$$ Now we define $\alpha\cosh\theta\equiv a^\prime$, $\alpha\sinh\theta\,d\theta=da$, so that $\sqrt{a^{\prime 2}-\alpha^2}=\alpha\sinh\theta$. We'll be lazy and just say that $a_f^\prime\to{\etot \over 2(1-\etot)(\cosh\theta -1)}$ because there was nothing special about time $a_f^\prime$. Then we have: $$\begin{align} \int_0^{\theta_f}{\alpha^2(\cosh\theta-1)\sinh\theta\,d\theta\over \sqrt{1-\etot}\alpha\sinh\theta}&=H_0t_f\\ {\alpha\over1-\etot}(\sinh\theta-\theta)\bigg|_0^{\theta_f}&=H_0t_f\\ \end{align}$$

Plugging $\alpha$ back in: $$\begin{align} H_0t_f&={\etot\over2(1-\etot)^{3\over 2}}(\sinh\theta-\theta)\\ t(\theta)&={\etot\over2H_0(1-\etot)^{3\over2}}(\sinh\theta-\theta)\\ \end{align}$$ Thus we have a parametric relationship between $a$ and $t$ with a dummy variable $\theta$: $$\boxed{\begin{align}a(\theta)&={\etot\over 2(1-\etot)}(\cosh\theta-1)\\ t(\theta)&={\etot\over 2\,H_0(1-\etot)^{3\over 2}}(\sinh\theta-\theta)\\ \end{align}}$$ Note that $a$ has no maximum and expands forever.

Closed, $k>0$, $\etot>1$, $\Omega_\Lambda=0$. As usual, we start with the Friedmann Equation: $$(\aa)^2 = \underbrace{\overbrace{\epot G\rho}^{positive} - \overbraceTemplate:K\over a^2^{positive}}_{can\ be\ 0}$$ Therefore, in this model it is possible for $\dot a=0$. We'll again solve for a matter-dominated era, $w=0$. The solution is similar, to the open model, but ($1-\etot<0$), so $\theta\to i\,\theta$. Thus our solution: $$\boxed{\begin{align}a(\theta)&={\etot\over 2(1-\etot)}(\cos\theta-1)\\ t(\theta)&={\etot\over2H_0(1-\etot)^{3\over 2}}(\sin\theta-\theta)\\ \end{align}}$$ is a cycloid solution. That is, $a$ oscillates, with a maximum at $\theta=\pi$.

\subsection*{ Deceleration Parameter $q$}

$$q\equiv-{\ddot aa\over \dot a^2}$$ Note that $q$ is {\it dimensionless}, and has a negative sign. The minus is there because historically, the universe was thought to be decelerating. Thus, $q<0\imply acceleration$, and $q>0\imply deceleration$. We can find out what $q$ does from the $2^{nd}$ Friedmann Equation: $$\adda=-{4\pi\over3}G(\rho+3P)$$ Recall that $P=w\rho$, $\Omega={\rho\over\rcr}$, $\rcr={3H^2\over8\pi G}$, $H^2\Omega=\epot G\rho$. Thus, the above equation becomes: $$\adda=-{4\pi\over3}G\rho(1+3w)=-{H^2\Omega\over2}(1+3w)$$ Substituting in our friend, $q$: $$q={\Omega\over2}(1+3w)=\sum_i{{\Omega_i\over2}(1+3w_i)}$$ $$\boxed{q = {\Omega _m \over 2} - \Omega _\Lambda +\Omega _r}.$$ Note that today, $\Omega_r\ll1$. That we are accelerating $\imply q<0 \imply\econs\ne0$.

\subsection*{ Redshift }

$${\lambda_{obs} \over \lambda_{emit}} \equiv 1 + z$$ Or alternately, $$\boxed{1 + z = {1\over a}}$$ $z$ is what we call the {\it redshift}. Reionization happened somewhere around $z=17\pm6$, and recombination around $z=1091$.

\subsection*{ Time-Redshift Relations and the Age of the Universe }

We seek a relation between $t$ and $z$. Beginning with the Friedmann Equation: $$\begin{align} H^2&=H_0^2\left[\jumble\right]\\ \inv{a}{da\over dt}&=H_0\sqrt{\jumble}\\ dt&=\inv{H_0}\left(\inv{a}\right){da\over\sqrt{\jumble}}\\ dt&=-\inv{H_0}{dz \over (1+z)\jimble}\\ \end{align}$$ We can calculate the time since the Big Bang at redshift $z_1$ by solving the integral: $$t_1=-\inv{H_0}\int_{z_1}^\infty{{dz\over 1+z}\inv{\jimble}}$$ For the Age of the Universe, set $z_1=0$. In the special case of a flat (Einstein-deSitter) universe: $$t_0=\inv{H_0}\int_0^\infty{dz\over(1+z)^{5\over 2}}={2\over3H_0}$$ If $H_0^{-1}\approx10Gyrs$, then $t_0\approx6.7h^{-1}Gyrs$, so the Hubble constant needs to be pretty small to get reasonable estimates of the age of the universe.