2.6: S06. Redshifts - SOLUTIONS
( \newcommand{\kernel}{\mathrm{null}\,}\)
Exercise 6.1.1
- Answer
-
It's clear from the graphic that Δτe=Δτr. We can integrate up dτ=dt/a(t) by approximating a(t) as constant over these short time intervals to get Δτ=Δt/a(t). Then we can easily see that Δtr/Δte=a(tr)/a(te).
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 Te, on to the two pulses you considered in the previous exercise. The time between the pulses upon emission is Δte=Te. The time between the reception of the first pulse and reception of the second pulse is the period of the wave upon reception Tr=Δtr=(a(tr)/a(te))Δte=(a(tr)/a(te))Te. Wavelength is proportional to period so we also have λr=(a(tr)/a(te))λe or
λrλe=a(tr)a(te).
We already know that z≡(λreceived−λemitted)/λemitted, so we can rewrite it as
z≡λrλe−1=a(tr)a(te)−1
Exercise 6.3.1
- Answer
-
The universe has expanded by a factor
a(tr)/a(te)=1+z=1+8.2=9.2.