# 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: $$$$\tag{1} \rho_{\Lambda} = \frac{\Lambda c^4}{8 \pi G} \sim 5 \times 10^{-10}~\mathrm{J/m^3}.$$$$ The Planck density is defined as this: $$$$\tag{2} \rho_{\text{P}} = \frac{M_{\text{P}} \, c^2}{L_{\text{P}}^3} = \frac{c^7}{\hbar G^2} \approx 5 \times 10^{113}~\mathrm{J/m^3}.$$$$ So the ratio of (2) to (1) is $$$$\tag{3} \frac{\rho_{\text{P}}}{\rho_{\Lambda}} = \frac{8 \pi c^3}{\hbar G \Lambda} \sim 10^{123},$$$$ which is interpreted as the "$$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 $$k_{\text{B}}$$): $$$$\tag{4} S_{\Lambda} = \frac{A}{4 L_{\text{P}}^2},$$$$ where $$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$$). Whatever its status, it gives $$$$\tag{5} S_{\Lambda} = \frac{3 \pi c^3}{\hbar G \Lambda} \approx 4 \times 10^{122}.$$$$ This is almost exactly the same as (3) (except for the numerical factors $$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. Oct 10, 2019 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, 2019 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. Oct 10, 2019 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, 2019 at 16:02

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 $$\Lambda\sim H^2$$ 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 $$\Lambda\sim H^2$$, 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.

As to the 'why' - from the holographic dark energy conjecture, the CCP and the BH entropy are in fact related. Eqn (1) and (2) can also be expressed $$kg/m^3$$ as:

$$$$\tag{1} \rho_{\Lambda} = \frac{\Lambda c^2}{8 \pi G} \$$$$

$$$$\tag{2} \rho_{\text{P}} = \frac{c^5}{\hbar G^2}$$$$

In flat de Sitter space the Hubble radius $$R_h$$ is equivalent to the cosmic event horizon radius. Also: $$$$\Lambda =\frac{3}{R_h^2}$$$$

The cosmological constant problem (CCP) can be expressed by observing that the number of degrees of freedom of dark energy in QFT is much too large to explain the observational data.

That is, the QFT approach relates degrees of freedom to a sphere volume $$V$$. We can see this easily if we can write (3) with $$R_h=2L_{Planck}$$ (the past cosmic event horizon); as opposed to using the future cosmic event horizon $$R_h\approx 16.1Gly$$ as per the OP:

$$$$\tag{3} \frac{\rho_{\text{P}}}{\rho_{\Lambda}} = \frac{4}{3}\pi 2^3 = \frac{32\pi} {3}=V$$$$

The CCP can be overcome by considering the holographic principle which equates the actual number of degrees of freedom $$N_s$$ of a region to its area not its volume. This is equivalent to stating that the quantum vacuum energy in a region cannot be larger than a black-hole mass of the same size. Then the BH entropy relation comes in via:

$$$$N_s=4S_{\Lambda}=16\pi$$$$

So we can write (3) as:

$$$$\frac{\rho_{\text{P}}}{\rho_{\Lambda}} = \frac{2}{3}N_s = \frac{8}{3}S_{\Lambda}=\frac{32\pi} {3}$$$$

TLDR: The OP's numerical correspondence comes from the holographic-inspired solution to the CCP, which uses the BH (de Sitter) entropy.