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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?

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  • $\begingroup$ Very helpful answer. My thanks to Kyle Oman. $\endgroup$ – Robert Harper May 3 '19 at 12:34
  • $\begingroup$ Welcome to the site. If you find an answer helpful, you upvote it(+1), or if you decide it is the best you check it as the best, the v below the votes on the left of the answers to your question $\endgroup$ – anna v May 3 '19 at 12:46
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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.

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  • $\begingroup$ 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. $\endgroup$ – Cuspy Code May 3 '19 at 12:03

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