# BAO : Relation between redshift, Hubble constant and radial

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## Main Question or Discussion Point

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.

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