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In minimal $\Lambda\mathrm{CDM}$, there is a parameter labeled $\Omega_\Lambda$, and current fits place it at around $\left( \Omega_\Lambda \sim 0.73\right)$. Meanwhile, $\Lambda$ enters the Einstein field equations as an offset to the Ricci scalar. The smallness of $\Lambda$ is a much-decried problem in physics today. My question is, what is the relation between $\Lambda$ and $\Omega_\Lambda$ in $\Lambda\mathrm{CDM}$? Are they independent?

We are currently in a cosmological constant-dominated era (since $\Omega_\Lambda$ is large). In the FRLW cosmology, we passed through epochs of matter domination, radiation domination, etc. so clearly $\Omega_\Lambda$ was changing through those epochs. How did $\Lambda$ change during that time?

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They are proportional so essentially the same, but $\Omega_\Lambda$ is a convenient dimensionless number. Straight out of Weinberg's newer cosmology book:

$$ \Lambda = 8 \pi G \rho_V, $$

where $\rho_V$ is the vacuum energy density, and

$$ \rho_{V0} = \frac{3 H_0^2 \Omega_\Lambda}{8\pi G}. $$

Putting them together $\Lambda = 3 H_0^2 \Omega_{\Lambda0}$. Note the subscript $0$ represents the present day value. $H_0$ is the present day Hubble rate. Since $\Lambda$ is a constant you can infer $\Omega_\Lambda \sim H^{-2}$. $\Omega_\Lambda$ asymptotically approaches one as the vacuum dominates ever more and $H$ goes to the de Sitter value $H_{dS} = \sqrt{\Lambda / 3}$. In an earlier stage you have to work out $H$ using the Friedmann equation.

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  • $\begingroup$ +1, although it might be better to write the more explicit formulae $\Lambda=8\pi G\rho_V/c^2$ and $\Lambda=3H_0^2\Omega_{\Lambda 0}/c^2$. Relativists like to set $c=1$, but in this case it's not obvious that $\Lambda$ has the dimensions of [length]${}^{-2}$. $\endgroup$
    – Pulsar
    Aug 1 '13 at 15:12
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If $\Omega +\Lambda =1$ then it will be proved that universe is flat.

If $\Omega<1$ then universe will fall.

if $\Omega>1$ then universe is contractimg.

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    $\begingroup$ I edited this for formatting and latex, although I have severe concerns about this answer. Please try adding more detail and explanation. Also, what does "The universe will fall" even mean? $\endgroup$
    – Sean
    Mar 20 '15 at 12:55
  • $\begingroup$ Every line in this answer is wrong. You can't add $\Omega$ to $\Lambda$. $\Omega$ must equal $1$, unless you mean $\Omega_\Lambda$, which is less than $1$. The universe can't "fall". Nothing about the value of $\Omega$ indicates whether or not the universe is contracting $\endgroup$
    – Jim
    Mar 20 '15 at 14:38

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