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# S06. Redshifts - SOLUTIONS

[ "article:topic", "Box Solutions", "authorname:knoxl" ]

Box $$\PageIndex{1}$$

Exercise 6.1.1: Show that if $$\Delta t_e$$ is the time interval between emitted pulses (as measured by a stationary observer located where the emission is happening) and if $$\Delta t_r$$ is the time interval between reception of first pulse and second pulse (as measured by a stationary observer located where the reception is occurring) then $$\Delta t_r/\Delta t_e = a(t_r)/a(t_e)$$. To do so, use conformal time defined by $$d\tau = dt/a(t)$$ and draw the pulse trajectories on an $$x$$ vs. $$\tau$$ diagram. By stationary observer we mean one that is at a constant value of $$x$$; i.e., is at rest in the cosmic rest frame. You can assume that $$\Delta t_r$$ and $$\Delta t_e$$ are very short time scales compared to the time scale over which $$a(t)$$ changes appreciably. Practical applications of this result often have $$a(t)$$ changing on billion-year time scales and the $$\Delta t$$s shorter than nanoseconds so this assumption is well justified!

Solution $$\PageIndex{1.1}$$

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 $$\Delta t_r/\Delta t_e = a(t_r)/a(t_e)$$.

Box $$\PageIndex{2}$$

Exercise 6.2.1: Imagine propagation of electromagnetic waves. Use the above result to show that the wavelength of the waves emitted at time $$t_r$$ and observed at time $$t_e$$ is stretched so that:

\begin{equation*} \begin{aligned} z \equiv (\lambda_{\rm received} - \lambda_{\rm emitted})/\lambda_{\rm emitted} = a(t_r)/a(t_e) - 1 \end{aligned} \end{equation*}

Solution $$\PageIndex{2.1}$$

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*}

Box $$\PageIndex{3}$$

Exercise 6.3.1: The most distant object for which a redshift has been measured is called a gamma-ray burst and it has a redshift of $$z=8.2$$. By what factor has the universe expanded since light left that object?

Solution $$\PageIndex{3.1}$$

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*}