2.8: S08. The Distance-Redshift Relation - SOLUTIONS
( \newcommand{\kernel}{\mathrm{null}\,}\)
Exercise 8.1.1
- Answer
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c∫1aedaa2H=∫d0dr=d
H=˙aa and ˙a= constant, so we have H=H0a. Now,
d=cH0∫1aedaa=cH0[ln(1)−ln(ae)]
Exercise 8.2.1
- Answer
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cH0[−ln(ae)]=dd=cH0ln(1ae)d=cH0ln(1+z)d≈cH0z
Now multiply both sides by (1+z),
d×(1+z)=cH0z(1+z)dlum=cH0z(1+z)
Since z<<1 we can simplify our result to cz=H0dlum.