BAO : Relation between redshift, Hubble constant and radial

Question

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 ?

CORRECTION 1 :

As suggested by Michelle Grosso, the eq(3) is not the definition of for angular diameter distance $\text{d}_{a}(z)$ but the definition of radial coordinate :

$$\chi = \int_{0}^{z}\dfrac{c\text{d}t}{R(t)}$$

Taking $\text{d}t=\dfrac{\text{d}t}{\text{d}z} dz$ with $1+z=\dfrac{R_{0}}{R(t)}$, so we get :

$\dfrac{\text{d}z}{\text{d}t} = -\dfrac{R_{0}\,\dot{R(t)}}{R(t)^2} = -H(z)(1+z)$

$$\Longrightarrow\quad \dfrac{\text{d}t}{\text{d}z} = -\dfrac{1}{(H(z)(1+z)}$$

$$\Longrightarrow\quad \chi = \int_{0}^{z}\dfrac{c\text{d}z}{(1+z)\,R(t)\,H(z)}$$

$$\chi = \int_{0}^{z}\dfrac{c\,\text{d}z}{R_{0}\,H(z)}$$

we have finally :

$$c\Delta z = H(z)\,R_{0}\,\Delta\chi\quad\quad(1)$$

Is is a better calculus ?

Answer 1

The angular diameter distance is defined as $d_A = \Delta S / \Delta \theta$ where $\Delta S$ is the proper transverse size of an object at redshift $z$ and $\Delta \theta$ is its observed angular diameter.

Eq(2) is not correct as you posted as proper transverse size $\Delta \chi$, which is instead the variation of the radial coordinate.

Eq(1) can be worked out via the RW (Robertson-Walker) metric written as
$ds^2 = -dt^2 +a^2(t) R_0^2 [d\chi^2 + S_k^2(\chi) d\Omega^2]$
where:
$c = G = 1$ natural units
$a(t)$ scale factor (dimensionless)
$R_0$ radius of the universe as today $(t = t_0)$
$\chi$ radial coordinate
$S_k(\chi) =$
$sin(\chi), k = 1$ positive curvature (closed universe)
$\chi, k = 0$ no curvature (flat universe)
$sinh(\chi), k = -1$ negative curvature (open universe)
$d\Omega^2 = d\theta^2 + sin^2\theta d\phi^2$ metric on the two-sphere
Today $a(t) = a(t_0) = a_0 = 1$

On a null geodesic (photon), chosen radial for convenience we have
$0 = ds^2 = -dt^2 + a^2 R_0^2 d\chi^2$
$d\chi = R_0^{-1} \frac{dt}{a} = R_0^{-1} \frac{da}{a^2 H(a)}$
where:
$H = \dot a / a = (da/dt) / a$ Hubble parameter

Converting the scale factor to redshift via $a = 1 / (1 + z)$ we have
$d\chi = R_0^{-1} \frac{dz}{H(z)}$
This is your eq(1). Just note that I used a dimensionless radial coordinate $\chi$.

Note: Eq(3) is not the angular diameter distance. It is the radial coordinate distance.