# BAO : Relation between redshift, Hubble constant and radial

• I
From this link https://en.wikipedia.org/wiki/Baryon_acoustic_oscillations#Measured_observables_of_dark_energy , I can't get this relation :

##c\Delta z = H(z)\Delta \chi\quad\quad(1)##

with ##z## redshift, ##H(z)## Hubble constant at redshift = ##z## and ##\chi## radial coordinates.

One starts from angle of object ##\Delta \theta## which is equal to the ration :

##\Delta\theta = \dfrac{\Delta\chi}{\text{d}_{a}(z)}\quad\quad(2)##

with ##\text{d}_{a}(z)## the angular diameter distance at redshift=##z##.

It is indicated also on this page the relation for angular diameter distance ##\text{d}_a(z)##:

##\text{d}_a(z)\propto \int_{0}^{z}\dfrac{\text{d}z'}{H(z')}\quad\quad(3)##

Actually, I know that ##\text{d}_a(z)## is expressed as a function of cosmological horizon ##\text{d}_{h}(z)## and redshift ##z## like this :

##\text{d}_{a}=\dfrac{\text{d}_{h}(z)}{1+z}\quad\quad(4)##

with ##\text{d}_{h}(z)=c\int_{0}^{z}\dfrac{\text{d}z}{H(z)}\quad\quad(5)##

So from ##eq(5)##, what I can only write is (by considering a little ##\Delta## and a curvature parameter ##\Omega_{k}=0##) :

##c\Delta z=\text{d}_{h}(z)H(z)\quad\quad(6)##

Now, taking the expression of ##\text{d}_{h}(z)## into ##eq(6)## :

##c\Delta z=\text{d}_{a}(z)(1+z)H(z)\quad\quad(7)##

Then :

##c\Delta z=\dfrac{\Delta\chi}{\Delta\theta}(1+z)H(z)\quad\quad(8)##

As you can see in ##eq(8)##, this is not the same form as in ##eq(1)##.

How can I make disappear the factor ##(1+z) /\Delta\theta## in order to have simply for the right member : ##H(z)\Delta \chi## instead of ##\dfrac{\Delta\chi}{\Delta\theta}(1+z)H(z)## into ##eq(8)## ?

##\Delta\chi## represents for me the variation ##\Delta## of radial coordinate, doesn't it ?

Any help is welcome.

## Answers and Replies

Can anybody give some help to find the demonstration of (1) :

##c\Delta z = H(z)\Delta \chi\quad\quad(1)## ??