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

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