I was trying to calculate the cosmological constant with two different methods proposed on the internet, that are apparently equivalent but they give different results and different dimensional analysis.

We know that: $$\displaystyle \Lambda = \frac{3 H_0^2}{c^2} \, \Omega_{\Lambda_0}\tag1$$ but WolframAlpha and Wikipedia page suggest that $$\displaystyle \Lambda = \frac{8 \pi G}{c^2} \, \rho_{vac}\tag2$$ where $\rho_{vac}$ is the cosmological vacuum energy density.

$\Omega_{\Lambda_0}$ is defined as $\displaystyle\frac{\rho_{vac}}{\rho_{c}}$, where $\rho_{c}$ is by definition $\displaystyle\frac{3 H_0^2}{8 \pi G}\ (3)$; so when we substitute eq. 3 into 1, it's easy to see that we obtain eq.2.

Let's do a dimensional analysis:

$[\Lambda]=m^{-2}$, $[H_0]=s^{-1}$, $[c^{2}]=m^2 s^{-2}$, $[G]=m^3kg^{-1}s^{-2}$, $[\rho_{vac}]=J/m^{3}=kg\space m^{-1}s^{-2}$

for eq. 1 we have: $$\displaystyle \Lambda =\frac{s^{-2}}{m^2s^{-2}}=m^{-2}$$ which is correct. But for eq. 2 we have: $$\displaystyle \Lambda =\frac{m^3kg^{-1}s^{-2}}{m^2s^{-2}}\space kg \space m^{-1}s^{-2}=s^{-2} \neq m^{-2}.$$

How can this be possible? It's the same equation.

In addition WolframAlpha states that $\rho_{vac}\approx 5.36 \times 10^{-10}$, while if taking $\Lambda=1.106 \times 10^{-52}$, $G=6.6743 \times 10^{-11}$ and $c=299\space792\space458$, and inverting eq. 2, we get $\rho_{vac}\approx 5.92585 \times 10^{-27}$, which is a value 17 orders of magnitude smaller!

  • 1
    $\begingroup$ You should add the units to the values you listed at the end. $\endgroup$
    – Cham
    Sep 5, 2019 at 0:53

1 Answer 1


Equation (2) should have $c^4$ in the denominator, and equation (3) should have a $c^2$ in the numerator. Whenever you are off by 17 orders of magnitude, think $c^2$. There is a lot of inconsistency in the cosmological literature over whether $\rho$ means a mass density or an energy density.

  • $\begingroup$ Yes, it should, but it doesn't. So you are suggesting that both $\Omega_{\Lambda_0} = \displaystyle\frac{\rho_{vac}}{\rho_{c}}$ and WolframAlpha are wrong? Actually dimensional analysis of $\rho_{vac}$ gives $kg/m^{3}$ so you are right, but then, how is $\Omega_{\Lambda_0}$ defined? And how could WolframAlpha state something that wrong? $\endgroup$
    – Aleph_0
    Sep 5, 2019 at 0:21
  • $\begingroup$ That expression for $\Omega$ is correct, regardless of whether the $\rho$’s are mass densities or energy densities. I have no idea what you are seeing in Wolfram Alpha, but if it dimensionally incorrect then it is wrong. I would hope that is obvious. $\endgroup$
    – G. Smith
    Sep 5, 2019 at 0:24
  • $\begingroup$ cosmological constant on WA $\endgroup$
    – Aleph_0
    Sep 5, 2019 at 0:25
  • $\begingroup$ It’s wrong. They meant to say mass density. Wolfram isn’t perfect. I’ve found numerous bugs in Mathematica. $\endgroup$
    – G. Smith
    Sep 5, 2019 at 0:27
  • $\begingroup$ I meant that in the ratio for $\Omega$, both numerator and denominator must be an energy density, or both a mass density. $\endgroup$
    – G. Smith
    Sep 5, 2019 at 0:35

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