# S13. The Evolution of Mass-Energy Density and a First Glance at the Contents of the Cosmos - SOLUTIONS

- Page ID
- 7786

# Exercise 13.1.1

**Answer**-
If the particles are not being created or destroyed then the number, \(N\) in some comoving volume is fixed. But the volume in the comoving volume is increasing as \(a^3\). The number density \(n = N/V\) is thus proportional to \(a^{-3}\).

# Exercise 13.2.1

**Answer**-
__Matter Case:__\[\begin{equation*}

\begin{aligned}

\dot a^2 = a^2 \frac{8\pi G \rho_0}{3}a^{-3} = \frac{H_0^2}{a}

\end{aligned}

\end{equation*}\]Since a is increasing with time, \(H_0^2/a\) decreases with time, which means \(\dot a^2\) decreases with time.

\(\Longrightarrow\) the magnitude of \(\dot a\) decreases with time, and since \(\dot a > 0\) that means \(\dot a\) decreases with time.

More mathematically:

\[\begin{equation*}

\begin{aligned}

\dot a = \frac{H_0}{a^{1/2}} \quad \Longrightarrow \quad \ddot a = -\frac{H_0}{2a^{3/2}}\dot a

\end{aligned}

\end{equation*}\]Since \(\dot a > 0\), it implies \(\ddot a < 0\).

__Radiation Case:__\[\begin{equation*}

\begin{aligned}

\dot a^2 = a^2 \frac{8\pi G \rho_0}{3}a^{-4}

\end{aligned}

\end{equation*}\]\[\begin{equation*}

\begin{aligned}

\dot a = \frac{H_0}{a}

\end{aligned}

\end{equation*}\]\[\begin{equation*}

\begin{aligned}

\ddot a = -\frac{H_0}{a^2}\dot a

\end{aligned}

\end{equation*}\]Since \(\dot a > 0\) we see \(\ddot a < 0\).

# Exercise 13.2.2

**Answer**-
\[\begin{equation*}

\begin{aligned}

\frac{\dot a^2}{a^2} = \frac{8\pi G \rho_0}{3}a^0 \quad \Longrightarrow \quad \dot a = H_0a

\end{aligned}

\end{equation*}\]Therefore, \(\ddot a = H_0\dot a > 0\).