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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?

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  • $\begingroup$ Welcome to astronomy SE! I suggested some minor edits... $\endgroup$
    – B--rian
    Jul 9, 2021 at 6:43

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$\epsilon_\Lambda={\rm constant}$ 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 $\Omega_\Lambda > \Omega_m$.

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

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