# Dark Matter Density Parameter Variation

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?

• Welcome to astronomy SE! I suggested some minor edits... Jul 9, 2021 at 6:43

$$\epsilon_\Lambda={\rm constant}$$ 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 $$\Omega_\Lambda > \Omega_m$$.
• 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 $H^2$? Jul 9, 2021 at 14:28
• Would this imply a relation like $\Omega_m = \Omega_{m,0} a^{-3}$ is erroneous and instead should be $\Omega_m = \Omega_{m,0} a^{-3} (H_0^2/H^2)$? Jul 9, 2021 at 14:39