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S06. Redshifts - SOLUTIONS

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

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