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While doing some lazy calculations, I came across a curiosity that I'm unable to interpret. It is well known that the cosmological constant $\Lambda \sim 10^{-52}~\mathrm{m^{-2}}$ is usually interpreted as a measure of the vacuum energy: \begin{equation}\tag{1} \rho_{\Lambda} = \frac{\Lambda c^4}{8 \pi G} \sim 5 \times 10^{-10}~\mathrm{J/m^3}. \end{equation} The Planck density is defined as this: \begin{equation}\tag{2} \rho_{\text{P}} = \frac{M_{\text{P}} \, c^2}{L_{\text{P}}^3} = \frac{c^7}{\hbar G^2} \approx 5 \times 10^{113}~\mathrm{J/m^3}. \end{equation} So the ratio of (2) to (1) is \begin{equation}\tag{3} \frac{\rho_{\text{P}}}{\rho_{\Lambda}} = \frac{8 \pi c^3}{\hbar G \Lambda} \sim 10^{123}, \end{equation} which is interpreted as the "$10^{120}$" crisis in fundamental physics (I'm very expeditive on this here).

Now, the entropy of the de-Sitter horizon is defined as this (in units of $k_{\text{B}}$): \begin{equation}\tag{4} S_{\Lambda} = \frac{A}{4 L_{\text{P}}^2}, \end{equation} where $A = 4 \pi \ell_{\Lambda}^2$ is the area of the de-Sitter horizon and $\ell_{\Lambda} = \sqrt{3 / \Lambda}$. The formula (4) is very controversial in the case of the de-Sitter spacetime (with $\Lambda > 0$). Whatever its status, it gives \begin{equation}\tag{5} S_{\Lambda} = \frac{3 \pi c^3}{\hbar G \Lambda} \approx 4 \times 10^{122}. \end{equation} This is almost exactly the same as (3) (except for the numerical factors $8 \Leftrightarrow 3$).

So my question is how should I interpret this "coincidence", i.e. that the ratio of energy density (3) is the same as the horizon entropy (5) ? AFAIK, the entropy has nothing to do with the discrepency in the energy density relative to the Planck density.

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  • $\begingroup$ You are comparing a dimensionless ratio of one thing and a value of something else in one particular set of units. Change the units & see how the coincidences change. $\endgroup$ – D. Halsey Oct 10 '19 at 15:16
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    $\begingroup$ @D.Halsey, it is very natural to express entropy in units of the Boltzman constant. Fundamentaly, entropy (i.e. information) is dimensionless. So there is no problem with units here. $\endgroup$ – Cham Oct 10 '19 at 15:18
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    $\begingroup$ I interpret it as both being $\Lambda^{-1}$ in Planck units. They have to both be some power of the cosmological constant because it is the only parameter of deSitter space. $\endgroup$ – G. Smith Oct 10 '19 at 15:58
  • $\begingroup$ @G.Smith, I agree, but I find it odd that the entropy has the same magnitude as the vacuum density discrepency, while it has nothing to do (apparently) with this problem. I'm probably overlooking something but I don't see what. $\endgroup$ – Cham Oct 10 '19 at 16:02
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Our Universe will to become a de Sitter space, and so the entropy of this 'particle horizon at infinity' is equal in magnitude to the inverse cosmological constant of our Universe.

This is known in the 'lore' of quantum gravity, see accepted answer here Entropy of an empty universe

In natural units, if you divide the de Sitter entropy by $Λ$ the result is $8\pi [L^2]$ i.e. the black hole quantum of area.

As via the holographic principle, a finite-dimensional Hilbert space can describe this ultimate dS space (including the Universe now), as Sean Carroll points out: the cosmological constant problem becomes the question of “Why does Hilbert space have a certain number of dimensions?"

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