How to calculate the redshift of reionization?

Question

I am trying to calculate equalities given Omega parameters.

For example given, $\Omega_L = 0.6889083, \Omega_M = 0.311, \Omega_R = 9.17$ x $10^{-5}$ and $\Omega_K = 0$
$H_0 = 67.7$ km/sec/Mpc

1. The redshift when Matter was equal to Radiation can be calculated as follows: $$ Z_{eq} = \Omega_M/\Omega_R - 1= 0.311/0.0000917 - 1 = 3390.49 $$ $(1)$

2. The redshift when Dark energy and Matter were equal can be calculated as follows: $$ \Omega_L = \Omega_M \rightarrow (\Omega_L/\Omega_M)^{1/3} - 1 = (0.6889083/0.311)^{1/3} - 1 = 0.3036$$ $(2)$

But how do you calculate the following:

3. The redshift when Dark Energy was equal to Radiation (that is at Reionization)?

As per first answer below the answer in part is as follows: $$z = (\Omega_L / \Omega_R)^{1/4} - 1 $$ $(3)$

However, I was hoping for a different answer that includes the Reionization Optical Depth $(\tau)$ as such:

$$z = 92 *(0.03 * (H_0 /100)* \tau / \Omega_bh^{2})^{2/3} * \Omega_M^{1/3} $$ $(4)$

where $\Omega_bh^{2}$ is the physical baryon density parameter as referenced in the link in the comment below.

The problem is the "92" and "0.03" figures. The result from that formula is close to the real one as obtained at (3).

How are these two figures derived?

Answer 1

Dark energy had equal density to radiation when

Ω_L = Ω_R/a^4.

The red shift would be

z = (1/a) - 1.

Therefore

a = (Ω_R/Ω_L)^(1/4)

and

z = (Ω_L/Ω_R)^(1/4) - 1.