Dark photons and the Universal ground-state energy

 In a previous post we introduced the idea that our current Universe has boundary conditions.   

Classical 1+1 dS space-time visualized as basketball hoop, with the up/down slam-dunk direction being the time dimension, and the hoop circumference the space dimension.   

We also showed a diagram similar to Figure 1 below. Except here, we are once again thinking about the future dS state.

Figure 1. For an observer at O inside the cosmic event horizon (CEH) with radius $l_{\Lambda}$, the universe can be divided into two sub-vacuums, $(A)$ inside the CEH, and $(B)$, outside. The horizon surface $\Sigma$ has entanglement entropy $S_{dS}$ and rest energy $E_H$

 

Figure 2. The maximum entropy of the Universe (credit: Lineweaver).  

Now, a comoving volume partition of the Universe can be treated as a closed system for which $dS \geq0$. The maximum entropy of a closed system, in this case  (Figure 2) with $L=2 \pi l_{\Lambda}$, the circumference of a circle with radius  $l_{\Lambda}$,  is obtained when all the energy $E_{H}$ within the system is worked, i.e. degraded into the smallest bits possible. All energy is converted into minimal energy dark photons with wavelengths as large as the system $L$. From the Compton wavelength relation, the minimum quanta of energy $E_{0}$ of this system is then:

$$\begin{equation} \notag E_{0}=\frac{hc}{\lambda_{max}}=\frac{\hslash c}{l_{\Lambda}}=m_{0}c^2 \end{equation}$$

A long time ago, Wesson conjectured that in a Universe with a positive cosmological constant, there must be a quantum-scale rest mass $m_p$. In fact, it is fairly simple to show that $m_p=m_0=m_{GR}$ (Wesson used the Planck constant, as he was only utilizing dimensional analysis), and $m_{GR}$ was discussed in this post.  In 2014, Barrow and Gibbons confirmed this lower-bound analysis, and also pointed out that the classical upper bound $m_{PE}$ for any mass in a Universe with a positive cosmological constant is (here $m_{CEH}$ is the mass of the cosmic event horizon of our Figure 1 system):  

$$\begin{equation}m_{PE} = \frac{c^{2}}{G}\sqrt{\frac{3}{\Lambda }} =\frac{c^2 l_{\Lambda}}{G}=2 \ m_{CEH} = 4 \ m_{0} \end{equation}$$ As we know from cosmological observations that $m_{CEH}$ is the observed (effective) mass of the CEH, I posit that $m_{PE}$ and also $m_{o}$ are unphysical (aka unobservable), and therefore are "bare" masses. 

Now, quantum field theory can be formulated in terms of harmonic oscillators. So, if we consider our Figure 1 system as a quantum harmonic oscillator,  with $E_0 = m_0 c^2$ as the ground state (zero-point) energy and  $\omega_{0}$ being the ground state angular frequency.    

$$\begin{equation} \notag E_{0}=\hslash \frac{\omega_{0}}{2} \end{equation}$$   Now, the de Broglie relation is then:
$$\begin{equation} \notag E_{0}=\hslash \omega_{B} \end{equation}$$ As  $\omega_{0}=2\omega_{B}$ we might consider $\omega_{0}$ as the zitter frequency of vacuum. Things get even more interesting when we realise that  $\omega_{B} = c/l_{\Lambda} =H$ implying that the Hubble Constant can be considered as a `de Broglie'  (or effective) frequency of vacuum; or equivalently, that time is a geometric property of space.   

Wait, I hear some of you saying, $H_0$ is usually presented as (km/s/Mpa) rather than ($s^{-1}$),  and you are right, but that simply due to cosmologists confusing things, using more convenient units. 

Now, Rugh and Zinkernagel's  paper from 2000 works out the QFT vacuum energy density from a quantum harmonic oscillator in a box, where $\rho^{QFT}_{vac} \ c^2 = E/V$: 

 

This zero-point vacuum energy density is a "bare" result, i.e. before interactions are considered. If we consider the QFT framework is valid up to the Planck energy scale, the standard approach is to insert $E_{Planck}=\hbar \ w_{max}$  into the QFT calculation above, which gives:

$$\begin{equation} \rho^{QFT}_{vac} \ c^2 = \frac{c^7}{8 \pi^2 \hbar G^2} \end{equation}$$ The expression on the RHS is the famous cosmological constant problem, i.e. this result is 120 orders of magnitude greater than the observed vacuum energy density! The observed vacuum density $\rho^{obs}_{vac}$ is the CEH rest mass-energy divided by the volume of a sphere with radius $l_{\Lambda}$ i.e. our Figure 1 system:

$$\begin{equation} \rho^{obs}_{vac} \ c^2 = \frac{3 \ m_{CEH} \ c^2}{4\pi l_{\Lambda}^3} =  \frac{c^4 \Lambda} {8 \pi G} \end{equation}$$ We have previously associated  $l_{\Lambda}=c/H$ as the de Sitter characteristic length scale, associated with the future de Sitter Hubble constant $H$. The other de-Sitter like epoch of our Universe was just before inflation, i.e. Planck energy scale, where $l_{\Lambda}=2L_{Planck}$.  In this post we showed how the reduced Planck constant is the angular momentum of our quantum scale rest mass $m_0$:

$$\begin{equation} \hbar = \frac{l_\Lambda^2 c^3}{4G} = m_0 \ l_\Lambda \ c \end{equation}$$ If we associate $l_\Lambda$ with $\Lambda$ and substitute for $\hbar$ into the QFT vacuum energy result we get:

$$\begin{equation} \rho^{QFT}_{vac} \ c^2 = \frac{m_{CEH} \ c^2}{l_{\Lambda}^3 \pi^2} \end{equation}$$ With $l_{\Lambda}=c/H \approx 1.5\times 10^{26}$m we now have the same order of magnitude from the QFT result and the observed vacuum energy density. This was also pointed out by Ali et al in their paper, although they didn't realise the de Sitter connection.  Of course, the above QFT equation needs a "correction factor" to give the observed vacuum density exactly:  

$$\begin{equation} \rho_{vac} \ c^2 = \frac{ m_{CEH} \ c^2}{l_{\Lambda}^3 \pi^2} \frac{3\pi}{4}  = \frac{c^4 \Lambda} {8 \pi G} \end{equation}$$

Presto, no more cosmological constant problem. The vacuum energy density being contingent on the future Hubble horizon (or varying Hubble horizon with some extra tweaks), is also how the holographic dark energy model solves the cosmological constant problem.  For more, you can also read this post.