$$\require{cancel}$$

# Exercise 13.1.1

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

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 $$\ddot a < 0$$.

\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

\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, $$\ddot a = H_0\dot a > 0$$.