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

Question

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.

Mr Anderson · Answer

Since (Gibbons and Hawking, 1977) we know (in shorthand Planck units) $$S_ds  ≤~1/Λ$$ Written out fully $S_ds  ≤(3πc^3)/(ℏGΛ)$  i.e. equation (5) of the OP.  Now, as noted in the comments, the de Sitter entropy is the same magnitude as the vacuum energy discrepancy because you can also write the vacuum energy discrepancy as $~ 1/Λ$.  Sure, but why?
First - the magnitude of the de Sitter entropy is the maximum possible universal entropy.  Typically, we think of de Sitter entropy in thermodynamical terms, i.e. the amount of energy which is unavailable to do work. Now, entropy can be also formulated as a measure of unavailable information (i.e. entropy is a measure of potential information). These are the same entropies.
Second, let us make a prediction of what the vacuum energy is, i.e. the Planck density, Equation (2) of the OP. However, once we measure the vacuum energy, we get Equation (1)! Turns out our prediction was a terrible fit to the data. In fact, mathematically, in terms of the maximum number of available degrees of freedom in the universe, it is the worst fit it could possibly be, i.e. Equation (3). In other words, our vacuum energy density prediction result was also the maximum possible universal unavailable information – aka entropy.
So that is ‘why’ the vacuum energy discrepancy and the de Sitter entropy are the same magnitude. It is because they are both measures of universal maximum entropy.

Eric David Kramer · Answer

For my own convenience, I will use units where $hbar=c=1$, and will ignore order 1 constants like 2 and $pi$.
The entropy has to be a dimensionless combination of $Lambdasim H^2$ and $M_{rm pl}$ (but we know it scales with the area of the horizon, so it's $M_{rm pl}^2/H^2$.)
The cosmological constant problem can be expressed in any many forms, including  $M_{rm pl}/H$, $M_{rm pl}^2/H^2$, etc. Since the quantity in the Einstein equations is $Lambdasim H^2$, that is the conventional way to express the cosmological constant problem.
So I think the answer is that they are both $H^2$ in units of $M_{rm pl}$. G. Smith wrote the same thing in a comment above.

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

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