What is the correct entropy density of the cosmic vacuum? Which of the following two arguments is correct?
(1) The total entropy of the cosmic vacuum should be the same as the entropy of the cosmological horizon (with radius $R$). The horizon entropy $S$ is given by the black hole entropy $A/4$ (in Planck units); the total cosmic vacuum entropy is thus proportional to $R^2$.
The temperature of the horizon, and of the vacuum in its interior that is due to the horizon (and not to other effects), is $T = 1/R$ (again in Plank units).
(2) Alternatively, the total entropy $S$ of the cosmic vacuum is the entropy of the black body radiation from the horizon: $S = V \cdot \sigma \cdot T^3 / c$.  Therefore, the total entropy $S$ does not depend on $R$, because $V$ changes as $R^3$, and $T^3$ changes as $1/R^3$.
Which of the two entropy expressions is correct? Or are both wrong? Why?
 A: Let us evolve the Universe in both arguments to a de Sitter end-point so the 'horizon' is then the future cosmic event horizon in both cases. What is then clearly the same in both arguments is that the entropy is:
\begin{equation}\tag{A}
\ S=S_{Bulk}=S_{dS}\sim \frac{E}{T_{dS}} 
\end{equation}
Also, $T_{dS}\approx 2.4\times10^{-30}K$.
The volume of the space enclosed by the future cosmic event horizon   (radius $l_\Lambda$) is simply:
\begin{equation}\tag{B}
V = \frac{4}{3} \pi{l_\Lambda^3} \sim 1.4 \times 10^{79}~\mathrm{m^3}.
\end{equation}
Or, if you feel this volume relation (B) is only valid in dS space with a horizon-based argument, you can use $V=\pi^2l_\Lambda^3$  in argument (2) below, it doesn't change the outcome.
The entropy density $s$ of the cosmic vacuum is $s=S/V$
What is not the same:
In argument (1) $E=E_H=E_{Bulk}$. This is a holographic relation. Also, $k_BS_{dS}=E_{dS}/T_{dS}$
In argument (2) $E=E_B=aT_{dS}^4V$. This a black body radiation relation. Also, $k_BS_{dS}=4E_{B}/3T_{dS}$
Lets look at (2) first. If this were right, the energy density of the cosmic vacuum could be calculated as $\rho_{\Lambda}=E_B/V=aT_{dS}^4\approx2.5\times10^{-134}\mathrm J/m^3$.
Now, the energy density of the bulk vacuum is constant over time, and from observation is known to be approx.:
\begin{equation}\tag{3}
\rho_{\Lambda} = \frac{\Lambda c^4}{8 \pi G} \sim 6.3 \times 10^{-10}~\mathrm{J/m^3}.
\end{equation}
The ratio of (3) : $\rho_{\Lambda}$ from argument (2) is $\sim10^{124}$. So, what we have with argument (2) is a version of the cosmological constant problem (CCP). In fact, we could expect this, since the CCP arises from relating the degrees of freedom to a sphere volume rather than a sphere surface area, and this is what argument (2) is really doing, trying to relate vacuum energy density to volume. Also, see these answers why this approach is a 'no'.
What about (1)? This is more likely to be correct, since as per this answer, there is a holographic-inspired approach that solves the CCP.
