2.8: S08. The Distance-Redshift Relation - SOLUTIONS

Exercise 8.1.1

Answer

$$\begin{equation*} \begin{aligned} c\int_{a_e}^{1} \frac{da}{a^2H} = \int_{0}^{d} dr = d \end{aligned} \end{equation*}$$

$H = \frac{\dot a}{a}$ and $\dot a =$ constant, so we have $H = \frac{H_0}{a}$. Now,

$$\begin{equation*} \begin{aligned} d = \frac{c}{H_0}\int_{a_e}^{1} \frac{da}{a} = \frac{c}{H_0}\Big[\ln(1) - \ln(a_e)\Big] \end{aligned} \end{equation*}$$

Exercise 8.2.1

Answer

$$\begin{equation*} \begin{aligned} \frac{c}{H_0}&\Big[-\ln(a_e) \Big] = d \\ \\ d &= \frac{c}{H_0}\ln\Big(\frac{1}{a_e}\Big) \\ \\ d &= \frac{c}{H_0}\ln(1 + z) \\ \\ d &\approx \frac{c}{H_0}z \end{aligned} \end{equation*}$$

Now multiply both sides by $(1+z)$,

$$\begin{equation*} \begin{aligned} d \times (1 + z) &= \frac{c}{H_0}z(1 + z) \\ \\ d_{\rm lum} &= \frac{c}{H_0}z(1 + z) \end{aligned} \end{equation*}$$

Since $z <<1$ we can simplify our result to $cz=H_0 d_{\rm lum}$.