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How can the redshift be found when $\rho_m = \rho_\Lambda$ and when the cosmological constant begins to dominate from matter? Given that $\Omega_M \approx 0.3$ and $\Omega_\Lambda \approx0.7$ .

I've derived the equation below, however I'm not sure how to deduce the value of $\Omega_{CMB}$ when $\Lambda$ starts to dominate.

$\rho(z)=\rho_c (t_0)[\Omega_M (1+z)^{3}+\Omega_{CMB}(1+z)^{4}+\Omega_\Lambda]$

(Assuming expansion due to non-zero cosmological constant and that the universe content is matter and radiation.)

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The radiation density has been negligable for most of the lifetime of the universe, so we can consider just the matter and dark energy densities. Looking at your equation, the expansion will be matter dominated when $\Omega_M (1+z)^{3} \gg \Omega_\Lambda$ and dark energy dominated when $\Omega_\Lambda \gg \Omega_M (1+z)^{3}$, so the crossover will be when $\Omega_M (1+z)^{3} = \Omega_\Lambda$ giving us:

$$ (1+z)^{3} = \frac{\Omega_\Lambda}{\Omega_M} $$

Feeding in the value from Planck, $\Omega_M = 0.315, \Omega_\Lambda = 0.685$ I get $z = 0.3$.

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I think what you are missing is that $\Omega_{CMB}$ is quite negligible ouside of the very early universe. If you treat the CMB as a black body radiation field with total energy density given by $$ u = \frac{4\sigma}{c} T^4$$ With $T=2.73$ K, this gives an energy density of $4 \times 10^{-14}$ J/m$^{3}$.

The present-day critical density is $8\times 10^{-10}$ J/m$^{-3}$, so $\Omega_{CMB} \sim 5\times10^{-5}$ and even though the energy density of the CMB increases faster than the matter density as we move to smaller scale factors, it is still pretty small even for redshifts out to $\sim 100$.

Thus to find when the energy densities of matter and dark energy were the same $$\Omega_{M} (1+z)^3 = \Omega_{\Lambda}$$ $$ z = \left( \frac{\Omega_{\Lambda}}{\Omega_M}\right)^{1/3} -1$$

For the parameters you suggest, this occurs at $z= 0.326$.

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  • $\begingroup$ @IamThankful but not thankful enough for an upvote... $\endgroup$ – Rob Jeffries Mar 15 '16 at 21:04
  • $\begingroup$ Haha, I did but do not currently have sufficient Reputation points so it won't be displayed. $\endgroup$ – IamThankful Mar 18 '16 at 15:38

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