4
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Very helpful answer. My thanks to Kyle Oman. $\endgroup$ Commented May 3, 2019 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
    Commented May 3, 2019 at 12:46

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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
    Commented May 3, 2019 at 12:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.