Cosmology Project Log Day 1/2

Consider a universe filled with dust of uniform density $\rho(t)$ and consider a spherical shell of mass $m$ and radius $r(t)$. This shell expands along with the universe, always containing the same dust particles. So, $\rho(t)r(t)^3$ is constant.

The total mechanical energy of the shell is $K + U = E$. As the shell expands, the gravitational pull from the mass inside causes $K$ to decrease while the gravitational P.E. $U$ increases. But, by energy conservation $E$ is constant. We write

(1)
\begin{align} E = -\frac{1}{2}mkc^{2}\bar{\omega}^2 \end{align}

The constant $k$ has dimensions of $(length)^{-2}$ and $\bar{\omega} = r(t_{0})$ is the present radius of the shell. Therefore,

(2)
\begin{align} \frac{1}{2}mv^2 - G\frac{M_{r}m}{r} = -\frac{1}{2}mk^2\bar{\omega}^2 \end{align}

where $M_{r} = \frac{4}{3}\pi r^{3}\rho$ is the mass interior to the shell and $v$ is the recessional speed of the shell. Thus we have

(3)
\begin{align} v^2 - \frac{8}{3}\pi G \rho r^{2} = -kc^2\bar{\omega}^2 \end{align}

The constant $k$ determines the overall geometry of the universe

• $k > 0$: total energy < 0, universe is bounded or closed. The expansion will someday halt and reverse itself.
• $k < 0$: total energy > 0, universe is unbounded or open. The expansion will continue forever.
• $k = 0$: total energy = 0, universe is flat (neither open nor closed). The expansion will continue to slow down coming to a halt only as $t \rightarrow \infty$ and the universe is infinitely dispersed.
page revision: 1, last edited: 10 May 2008 06:39
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