2.6: S06. Redshifts - SOLUTIONS

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Nov 8, 2022
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- 1.37: Redshifts
- Back Matter

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Exercise 6.1.1

Answer

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

Answer

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

Answer

The universe has expanded by a factor

$\begin{equation*} \begin{aligned} a(t_r)/a(t_e) = 1 + z = 1 + 8.2 = 9.2. \end{aligned} \end{equation*}$

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Exercise 6.1.1

Exercise 6.2.1

Exercise 6.3.1

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