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

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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, \(\ddot a = H_0\dot a > 0\).