# How would we relate the cosmological constant to a dark energy distribution?

*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?

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:

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

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