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2.6: S06. Redshifts - SOLUTIONS

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Exercise 6.1.1

Answer

spacetime6_1.png

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(λreceivedλemitted)/λemitted, so we can rewrite it as

zλrλe1=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.


This page titled 2.6: S06. Redshifts - SOLUTIONS is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Lloyd Knox.

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