# What are the units of dark energy?

Popular literature seems to equate dark energy with the cosmological constant of the Einstein field equations. We know however that the dimensions of any constituent of the field equations is $${\rm length}^{-2}$$, and that the metric tensor itself is dimensionless. This means that the dimensions of the cosmological constant are $${\rm length}^{-2}$$. These are certainly not the units of energy! Are we to understand that "dark" energy has different dimensions than "normal" energy?

• Very helpful answer. My thanks to Kyle Oman. – Robert Harper May 3 '19 at 12:34
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It is customary to include a term $$\Omega_\Lambda$$, the density of dark energy relative to the critical density for closure, in the Friedmann equations. It is defined:
$$\Omega_\Lambda = \frac{\rho_\Lambda}{\rho_{\rm crit}} = \frac{c^2\Lambda}{3H_0^2}$$
Since the critical density is $$\frac{3H_0^2}{8\pi G}$$, the (mass) density associated with $$\Lambda$$ comes out to be $$\frac{c^2\Lambda}{8\pi G}$$. The energy density would then be $$\frac{c^4\Lambda}{8\pi G}$$.
Indeed, the cosmological constant itself has units of $${\rm length}^{-2}$$, but this is really just convention. One could just as easily define the constant with dimensions of $${\rm energy}$$, then it would have a different numerical value, and a few extra constants would appear in the equations (e.g. $$c$$, $$G$$) to get the dimensions to work out correctly.
• One way to think about it is to move the $\Lambda$ term to the other side of the Einstein equation and then divide the whole equation by $\dfrac{8\pi G}{c^4}$ to change all terms into energy density dimensions. – Cuspy Code May 3 '19 at 12:03