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# 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 $$cz=H_0 d_{\rm lum}$$.