S08. The Distance-Redshift Relation - SOLUTIONS

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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 \(cz=H_0 d_{\rm lum}\).