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:

  1. 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?

  • $\begingroup$ What has the density of dark energy got to do with the epoch of reionization? $\endgroup$
    – ProfRob
    Commented Mar 28, 2022 at 19:04
  • $\begingroup$ ProfRob It just coincides with the time when Lambda was equal to Radiation. $\endgroup$
    – Vick
    Commented Mar 28, 2022 at 19:36
  • $\begingroup$ What is the meaning of what you've added? What is the new maths (it isn't an equation)? Where is it from? You have an answer for when the dark energy and radiation densities are similar. Optical depth to scattering is not directly connected to this an involves star and galaxy formation. $\endgroup$
    – ProfRob
    Commented Mar 28, 2022 at 19:59

1 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.


a = (Ω_R/Ω_L)^(1/4)


z = (Ω_L/Ω_R)^(1/4) - 1.

  • $\begingroup$ This is good. But I was hoping to find a formula that includes the observed and independent variable called reionization optical depth ($\tau$). Do you have any idea about how to do so? $\tau = 0.0561$ as per Planck 2018 @ arxiv.org/pdf/1807.06209.pdf $\endgroup$
    – Vick
    Commented Mar 28, 2022 at 15:07
  • $\begingroup$ I found several places in the Planck paper by searching for " r " I can not understand any of the discussion about this variable, In particular I am unable to understand any equation explaining what this "r" variable is about. I am sorry I cannot help you about this. $\endgroup$
    – Buzz
    Commented Mar 28, 2022 at 18:28
  • $\begingroup$ Buzz that's not "r" but Greek letter "Tau".... $\endgroup$
    – Vick
    Commented Mar 28, 2022 at 18:58
  • $\begingroup$ Ah! That makes all the difference. I am completely unable to do any search for Greek letters. $\endgroup$
    – Buzz
    Commented Mar 29, 2022 at 14:08

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