There are several cosmological parameters that are more or less of the same order $10^{-60}$ (in Planck unit), namely,

inverse of the age of the universe $$ \frac{1}{t_0}, $$ Hubble constant $$ H_0, $$ square root of the Cosmological Constant $$ \Lambda^{\frac{1}{2}}, $$ square root of the mass density of the universe $$ \rho^{\frac{1}{2}}, $$ and square root of the dark matter density of the universe $$ \rho_{DM}^{\frac{1}{2}}. $$

The last four parameters are related via the Friedmann equation (neglecting the radiation portion, and assuming flat universe/critical total density which is supposedly explained by the inflationary model): $$ H^2 =\frac{8\pi}{3}\rho + \frac{8\pi}{3}\rho_{DM}+\frac{1}{3}\Lambda $$ There is no a priori reason why the individual terms should be of the same order.

Given that these cosmological parameters evolve in a different manner with time, how do we make sense of these cosmological coincidences? Do we happen to be observing a special epoch of the universe, thus tossing out the Copernican principle?

Does multiverse or anthropic principle help in this case?

Any comments on the following two takes on the "why now" problem below?

"Is the Cosmological Coincidence a Problem?" (https://arxiv.org/abs/1203.4197)

"The $R_h=ct$ universe" (https://academic.oup.com/mnras/article/419/3/2579/1069736)

  • $\begingroup$ can you add the actual values of these numbers $\endgroup$ – michael Jun 21 '18 at 20:38
  • 7
    $\begingroup$ A rough estimate of $1/t_0$ is $H_0$ actually, so no great mystery here, is there? Then the solution of Friedman equations in a flat universe also give $H^2 \propto \Lambda$. Etc. $\endgroup$ – frapadingue Jun 21 '18 at 20:45
  • $\begingroup$ That's the answer, @frapadingue. You should write it up. $\endgroup$ – pela Jun 21 '18 at 20:47
  • $\begingroup$ The Friemann equation requires (assuming flat universe): $H^2=\frac{8\pi}{3}\rho + \frac{8\pi}{3}\rho_{DM} + \frac{1}{3}\Lambda$. There is no a priori reason why any of the individual term on the right side should be of the order of the left side. $\endgroup$ – MadMax Jun 21 '18 at 20:59
  • $\begingroup$ It's exactly what it seems: a coincidence. $\endgroup$ – astronat Jun 21 '18 at 21:22

You almost answer this yourself, in the question.

$$H^2 = \frac{8\pi}{3} \sum_i \rho_i,$$

where you sum over all of the different stuff in the universe. It's true, the energy density of baryonic matter, dark matter, and dark energy (cosmological constant) are all nearly of the same order of magnitude. The energy density of EM radiation and gravitational waves are much, much less, so you ignored them. You could just as easily ask why radiation isn't of that order.

What you observe is a coincidental consequence of when we are.

In the early universe radiation was the dominate energy. In the late universe dark energy will dominate. When studying cosmology it's easy to do calculations for a radiation, matter, or dark energy dominated universe. In each case only one $\rho_i$ matters. These assumptions apply to different epochs of the universe's evolution.

These days we're in a universe where matter (dark and baryonic) and dark energy both contribute.


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.