S08. The Distance-Redshift Relation - SOLUTIONS

Box \(\PageIndex{1}\)

Exercise 8.1.1: Plugging \(H = H_0/a\) into Equation 8.4 one can now do the integral on the left-hand side. The right-hand side could not be easier (since we are assuming \(k = 0\)). Check that you find:

\[\begin{equation*}
  \begin{aligned}
        \frac{c}{H_0}\left[\ln(1) - \ln(a_e)\right] = d
  \end{aligned}
\end{equation*}\]

Solution \(\PageIndex{1.1}\)

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

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

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

 

Box \(\PageIndex{2}\)

Exercise 8.2.1: Relate \(a_e\) to the redshift \(z\), and take advantage of \(\ln(1+x) = x\) to first order in \(x\) to derive \(cz = H_0 d\) to first order in \(z\). How is \(d\) here related to luminosity distance? Simplify your result, again assuming \(z <<1\). You should find \(cz=H_0 d_{\rm lum}\). Finally, if \(z\) is replaced with \(z=v/c\) we get Hubble's Law.

Solution \(\PageIndex{2.1}\)

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

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 \(cz=H_0 d_{\rm lum}\).