2.6: S06. Redshifts - SOLUTIONS
Exercise 6.1.1
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It's clear from the graphic that \(\Delta\tau_e = \Delta\tau_r\). We can integrate up \(d\tau = dt/a(t)\) by approximating \(a(t)\) as constant over these short time intervals to get \(\Delta \tau = \Delta t/a(t)\). Then we can easily see that \(\Delta t_r/\Delta t_e = a(t_r)/a(t_e)\).
Exercise 6.2.1
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Imagine successive crests of a single wave. Map one crest, and then the subsequent one, separated in time by the period of the emitted wave \(T_e\), on to the two pulses you considered in the previous exercise. The time between the pulses upon emission is \(\Delta t_e = T_e\). The time between the reception of the first pulse and reception of the second pulse is the period of the wave upon reception \(T_r = \Delta t_r = (a(t_r)/a(t_e)) \Delta t_e = (a(t_r)/a(t_e)) T_e \). Wavelength is proportional to period so we also have \( \lambda_r = (a(t_r)/a(t_e)) \lambda_e\) or
\[\begin{equation*}
\begin{aligned}
\frac{\lambda_r}{\lambda_e} = \frac{a(t_r)}{a(t_e)}.
\end{aligned}
\end{equation*}\]We already know that \(z \equiv (\lambda_{\rm received} - \lambda_{\rm emitted})/\lambda_{\rm emitted}\), so we can rewrite it as
\[\begin{equation*}
\begin{aligned}
z \equiv \frac{\lambda_r}{\lambda_e} - 1 = \frac{a(t_r)}{a(t_e)} - 1
\end{aligned}
\end{equation*}\]
Exercise 6.3.1
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The universe has expanded by a factor
\[\begin{equation*}
\begin{aligned}
a(t_r)/a(t_e) = 1 + z = 1 + 8.2 = 9.2.
\end{aligned}
\end{equation*}\]