*Please excuse my lack of understanding of dark energy, the GR courses that I have taken so far haven't covered it

How would we relate the cosmological constant to a dark energy density? Say we move the $\Lambda g_{\mu\nu}$ term to the right side of the equation, how would we translate the metric to something in the form of a stress tensor not involving the metric? or does the metric not have to disappear to be on the matter side of the equation? i.e.: Would we just say $G_{\mu\nu} = 8\pi T_{\mu\nu} - 8\pi\rho_{DarkEnergy}g_{\mu\nu}$ then?


1 Answer 1


The easy way to see the relationship between the cosmological constant and energy density is to look at the second Friedmann equation:

$$ \frac{\ddot{a}}{a} = -\frac{4\pi G}{3} \left(\rho + \frac{3p}{c^2}\right) + \frac{\Lambda c^2}{3} $$

where $\rho$ is the matter density. If we bring $\Lambda$ inside the bracket we get:

$$ \frac{\ddot{a}}{a} = -\frac{4\pi G}{3} \left(\rho + \frac{3p}{c^2} - \frac{\Lambda c^2}{4\pi G}\right) $$

So the energy density associated with $\Lambda$ is:

$$ \rho_\Lambda = \frac{\Lambda c^2}{8\pi G} $$

  • $\begingroup$ oh I see, so then the $\rho_\Lambda$ you stated would be the $\rho_{DarkEnergy}$ in the question? $\endgroup$
    – B K
    Dec 5, 2018 at 9:44

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