I looked around to see if this precise question was asked before and it appears not to be.

So is it just me or has anyone else noticed that, no matter what consistent set of units you use,

$$ \Lambda \approx H_0^2 \quad ?$$

the values check out (well within an order of magnitude) and the dimensions agree. In Planck units,

$$ H_0 = 1.18 \times 10^{-61} \quad t_\text{P}^{-1} $$


$$ \Lambda = 5.6 \times 10^{-122} \quad t_\text{P}^{-2} \quad . $$

Is there no physical theory that relates the two? Are there any cosmologists studying this apparent coincidence? It can't be just coincidental, can it?

I know that in John Baez's Fundamental Constants Page, he lists $\Lambda$ as one of the 26 dimensionless fundamental constants that define the whole of current physical theory and that cannot be derived from other constants. (The other 25 all have to do with the Standard Model.) So he does not list the Hubble constant as fundamental. But what, then, is the connection between the two?

I don't think I asked this question to sci.physics.research yet, but I may. This coincidence of physical quantity has intrigued me for years.

  • 4
    $\begingroup$ I think the answer is trivial, as the Lamda value is computed from the Omega value using the hubble constant. this answer here has it analytically groups.google.com/forum/#!topic/sci.physics.research/…, which enters as a square. As omega is close to one this is your coincidence $\endgroup$ – anna v Nov 16 '15 at 6:25

Yes there is.

The solution to the Friedmann equation in a flat universe with a cosmological constant is $$H^2 = \frac{8\pi G}{3}\rho + \frac{\Lambda}{3},$$

Thus, as the universe expands and the relative importance of gravitating matter, characterised by density $\rho$, decreases, then $\Lambda \simeq 3H^2$.

We are already (just) in a dark energy dominated universe, so the relationship is already (nearly) true.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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