<p><span class="math-container">$\epsilon_\Lambda={\rm constant}$</span> is the definition of a cosmological constant.</p>
<p><span class="math-container">$\Omega_\Lambda$</span> is <em>not</em> constant. In our flat, or nearly flat, universe, the energy densities of matter and radiation scale as the size of the universe cubed and to the power of 4 respectively. That means that <span class="math-container">$\Omega_m$</span> and <span class="math-container">$\Omega_r$</span> were bigger in the past, yet the sum of <span class="math-container">$\Omega_\Lambda + \Omega_m + \Omega_r \simeq 1$</span>.</p>
<p>In particular:
<span class="math-container">$$\Omega_m =\Omega_{m,0} a^{-3}, \ \ \ \Omega_r = \Omega_{r,0}a^{-4}\ ,{\rm so} $$</span>
<span class="math-container">$$\Omega_\Lambda \simeq 1  - \Omega_{m,0} a^{-3} - \Omega_{r,0}a^{-4}\ . $$</span>
And for the Hubble parameter
<span class="math-container">$$H^2 = H_0^2 \left(  \frac{\Omega_{r,0}}{a^4} + \frac{\Omega_{m,0}}{a^3}  + \Omega_{\Lambda,0}\right) = \frac{8\pi G}{3}\rho + \frac{\epsilon_\Lambda}{3},$$</span></p>
<p>It is only in the last few billion years, as the energy density of matter got smaller, that <span class="math-container">$\Omega_\Lambda &gt; \Omega_m$</span>.</p>

CC BY-SA 4.0
2021-07-09T14:28:15.503Z
Just to clarify, does $(\Omega_{\Lambda} + \Omega_m + \Omega_r) = 1$ mean that $H^2 = H_0^2 (\Omega_{\Lambda} + \Omega_m + \Omega_r)$ is not true, and in fact the $H_0^2$ should be $H^2$?
wheelix
CC BY-SA 4.0
2021-07-09T14:39:15.570Z
Would this imply a relation like $\Omega_m = \Omega_{m,0} a^{-3}$ is erroneous and instead should be $\Omega_m = \Omega_{m,0} a^{-3} (H_0^2/H^2)$?
wheelix
CC BY-SA 4.0
2021-07-09T06:41:12.830Z
2021-07-09T15:07:29.253Z
2021-07-09T15:07:29.253Z
ProfRob
<p>The definition of the dark energy density parameter is <span class="math-container">$$\Omega_{\Lambda} = \frac{\epsilon_{\Lambda}}{\epsilon_c}$$</span></p>
<p>where <span class="math-container">$\epsilon$</span> is the energy density, <span class="math-container">$\Lambda$</span> subscripts represents dark energy and the <span class="math-container">$c$</span> subscript represents the critical density.
I have encountered both</p>
<ol>
<li><span class="math-container">$\epsilon_{\Lambda} = \mathrm{const.}$</span> and</li>
<li><span class="math-container">$\Omega_{\Lambda} = \mathrm{const.}$</span> with respect to <span class="math-container">$z$</span>,</li>
</ol>
<p>but it seems like those statements would be contradictory because <span class="math-container">$$\epsilon_c = \frac{3H^2c^2}{8 \pi G}$$</span></p>
<p>where <span class="math-container">$H$</span> depends on <span class="math-container">$z$</span>. Which of 1 or 2 is correct?</p>

CC BY-SA 4.0
2021-07-09T06:43:01.517Z
Welcome to astronomy SE! I suggested some minor edits...
B--rian
<cosmology><dark-energy>
Dark Matter Density Parameter Variation
wheelix
# Dark Matter Density Parameter Variation

(Asked by: username_0 on 2021-07-09)

The definition of the dark energy density parameter is $$\Omega\_{\Lambda} = \frac{\epsilon\_{\Lambda}}{\epsilon\_c}$$

where $\epsilon$ is the energy density, $\Lambda$ subscripts represents dark energy and the $c$ subscript represents the critical density.
I have encountered both

1. $\epsilon\_{\Lambda} = \mathrm{const.}$ and
2. $\Omega\_{\Lambda} = \mathrm{const.}$ with respect to $z$,

but it seems like those statements would be contradictory because $$\epsilon\_c = \frac{3H^2c^2}{8 \pi G}$$

where $H$ depends on $z$. Which of 1 or 2 is correct?

**Tags:** cosmology, dark-energy
> **Comments:**
> **username_1**: Welcome to astronomy SE! I suggested some minor edits...  

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## 1 answers

### Answer by username_2 on 2021-07-09. Score: 3 | Accepted answer

$\epsilon\_\Lambda={\rm constant}$ is the definition of a cosmological constant.

$\Omega\_\Lambda$ is *not* constant. In our flat, or nearly flat, universe, the energy densities of matter and radiation scale as the size of the universe cubed and to the power of 4 respectively. That means that $\Omega\_m$ and $\Omega\_r$ were bigger in the past, yet the sum of $\Omega\_\Lambda + \Omega\_m + \Omega\_r \simeq 1$.

In particular:
$$\Omega\_m =\Omega\_{m,0} a^{-3}, \ \ \ \Omega\_r = \Omega\_{r,0}a^{-4}\ ,{\rm so} $$
$$\Omega\_\Lambda \simeq 1 - \Omega\_{m,0} a^{-3} - \Omega\_{r,0}a^{-4}\ . $$
And for the Hubble parameter
$$H^2 = H\_0^2 \left( \frac{\Omega\_{r,0}}{a^4} + \frac{\Omega\_{m,0}}{a^3} + \Omega\_{\Lambda,0}\right) = \frac{8\pi G}{3}\rho + \frac{\epsilon\_\Lambda}{3},$$

It is only in the last few billion years, as the energy density of matter got smaller, that $\Omega\_\Lambda > \Omega\_m$.

> **Comments:**
> **username_0**: Just to clarify, does $(\Omega_{\Lambda} + \Omega_m + \Omega_r) = 1$ mean that $H^2 = H_0^2 (\Omega_{\Lambda} + \Omega_m + \Omega_r)$ is not true, and in fact the $H_0^2$ should be $H^2$?  
> **username_0**: Would this imply a relation like $\Omega_m = \Omega_{m,0} a^{-3}$ is erroneous and instead should be $\Omega_m = \Omega_{m,0} a^{-3} (H_0^2/H^2)$?  

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