# Entropy of de-Sitter spacetime and the $10^{120}$ vacuum discrepency

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.

• 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. – D. Halsey Oct 10 '19 at 15:16
• @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. – Cham Oct 10 '19 at 15:18
• 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. – G. Smith Oct 10 '19 at 15:58
• @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. – Cham Oct 10 '19 at 16:02

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.