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# The Timeline of Universe Based On the Gravitons Release Due of the Merging of the Primordial Micro-black Holes

Stefan Mehedinteanu

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Stefan Mehedinteanu. The Timeline of Universe Based On the Gravitons Release Due of the Merging of the Primordial Micro-black Holes. 2016. hal-01316347v5

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# The Timeline of Universe Based On the Gravitons Release Due of the Merging of the Primordial Micro-black Holes

Stefan Mehedinteanu¹

¹ (retired) Senior Researcher CITON-Romania; E-Mail: mehedintz@yahoo.com; s.mehedinteanu@gmail.com

**Key words:** LGO, Inflation, reheating, gravitational charges, gravitonmass, quarks, color gluons, micro black holes (μBHs) merging, epochs, Schwinger effect, BH nuclei of galaxies, SM*A*S*H, cosmological constant

## Abstract

Based on the recently LGO discovery, respectively the merging of two black-holes as the base of the space-time deformation, I will advance another scenario for Universe creation and its evolution, respectively, when are created a huge number ≈10⁶² of Micro-black-holes (micro virtual black holes μBHs) as generated due of Quantum fluctuations at Reheating (at 10¹¹ GeV; 10⁻²⁹ s), and which further, during theirs merging its release gravitons which deform the space-time. Thus, at Confinement the μBHs become, firstly, via Quarks Gluons Plasma (QGP), the quarks of nucleons, and later, another's fractions becomes electrons (also via QGP) and others leptons and bosons (W±, Z, H), and very later Black Holes (BH) as the nuclei of galaxies. Also, is proposed a new test for the finally proof of the model as the deduction of the lifetime value of the β⁻-decay of the free neutron. A correspondence it was established with the recent Standard Model Axion Seesaw Higgs portal inflation (SM*A*S*H) which offers a self-contained description of particle physics from the electroweak scale to the Planck scale and of cosmology from inflation until today.

## 1. Introduction

The Universe beginning is when are created a huge number ≈10⁶² of Micro-black-holes (micro-virtual black holes- μBH) as been produced due of Quantum fluctuations [1a] at Electroweak epoch at 10¹¹ GeV; 10⁻²⁹ s, and which during theirs *merging* release gravitons which deform the space-time. At Confinement epoch (V=2GeV) these ex-Micro-black-holes as becoming via Quarks Gluons Plasma (QGP) e⁺e⁻-pairs (quarks) which gives an external electrical field (E) [1a], that produces later at the interface between normal phase and the superconducting phase a superconductor magnetic field (B) generated by the free photons-color gluons (also radiated by μBHs) condensate a Meissner effect.

Scientists first caught wind of this strange phenomenon when they pointed the European Southern Observatory's Very Large Telescope (VLT) at the neutron star RX J1856.5-
3754, which has a magnetic field billions of times stronger than our sun's, reports Popular Mechanics. They noticed that there was a high linear polarization of light waves in the empty space around the star. This is unusual because, according to conventional relativity at least, light should pass freely through a vacuum without being altered. But in this instance it wasn't passing freely at all.

The virtual particles that result from this interaction are not really particles at all, but rather fluctuations in quantum fields. Though they're merely "virtual," the fluctuations nevertheless act like particles in that they can affect light, that it could be our proposal as μBH pairs.

Virtual particles are often popularly described as coming in pairs, a particle and antiparticle, which can be of any kind. These pairs exist for an extremely short time, and then mutually annihilate. In some cases, however, it is possible to boost the pair (like W± bosons) apart using external energy so that they avoid annihilation and become actual particles. In other words the Quantum is a real world.

After the discovery of quark (q) jets in 1975 in $e^{+} - e^{-} \rightarrow q\bar{q}$ at SLAC, detailed studies in understanding the hadronisation process, and hence the energy-momentum profiles of the quark jets, were initiated in 1977 by Feynman and Field. These $e^{+}e^{-}$ pair's collisions generate either as $Y(94.6\,\text{GeV}) \rightarrow 3g$, ($g$ gluons) near the characteristic energy of electroweak symmetry breaking $\approx 100\,\text{GeV}$, or $q\bar{q}g$ ($\approx \text{few GeV}$), a 3-jet like in PETRA, DESY experiments, for more details see [1b]. These $q\bar{q}g$ flux tube is generated by the electrical field E by Schwinger effect or $e^{+} - e^{-}$ collision, or as at SLAC, that it produces at the interface between normal phase and the superconducting phase, a decay of the superconductor magnetic field (B) generated by the color gluons condensate components like the Meissner effect. The inverse value of the penetration length of B could be equivalent with $W^{\pm}$ bosons at high densities, and the effective color gluon mass existing in nucleons at QCD. The first evidence for the existence of quarks came in 1968, in deep inelastic scattering experiments at the Stanford Linear Accelerator Center (SLAC).

Another's fractions of μBHs merging will becomes electrons and, respectively BH's nuclei of galaxies, these later producing further Universe expansion.

## 2. How determine the gravitons production due of μBHs merging the timeline of Universe

First of all, I present some of known data. Thus, in [1a] it is therefore assumed, that "potential energy" caused by gravitation and "kinetic energy" caused by expansion of the Universe are equal to each other (using the relations $R_U = ct$ and $E_U = M_U c^2$):

$$ \frac{GM_U^2}{R} = M_U c^2 \rightarrow G = \frac{c^5 t}{E_U} \rightarrow G = \frac{R_U^5}{t^4 E_U} $$
$R_U = 1.6 \times 10^{26} [m]$, the radius of Universe; $t = 5 \times 10^{17} s$, the age of the Universe;

$M_U = 2.2 \times 10^{53} \text{kg}$, the mass of the Universe. The Planck mass $m_P = (\frac{\hbar c}{G})^{1/2}$; the Planck

length $L_P = (\frac{\hbar G}{c^3})^{1/2}$; the Planck time $t_P = (\frac{\hbar G}{c^5})^{1/2}$.

In fact as already noted [23] from [1b], a Planck mass particle decays via the Bekenstein radiation within a Planck time $10^{-42} s$, see below.

In Inflation models [2], the scale leaving the horizon at a given epoch is directly related to the number $N(\phi)$ of e-folds of slow-roll inflation that occur after the epoch of horizon exit. Indeed, since H -the Hubble length is slowly varying, we

have $d \ln k = d(\ln(aH)) \cong d \ln a = \frac{\dot{a}dt}{a} = Hdt$. From the definition Eq. (38) of [2] this

gives $d \ln k = -dN(\phi)$ as of eq.(46) from [2a], and therefore $\ln(k_{end}/k) = N(\phi)$, or,

$k_{end} = ke^N[m]$ where $k_{end}$ is the scale leaving the horizon at the end of slow-roll inflation, or usually $k^{-1} \ll k_{end}^{-1}[m]$, the correct equation being $k = k_{end}e^N[m^{-1}]$. When the wavelength ($k^{-1}[m]$) is large compared to the Hubble length ($H^{-1}[m]$), the distance that light can travel in a Hubble time becomes small compared to the wavelength, and hence all motion is very slow and the pattern is essentially frozen in.

In this new scenario of Universe evolution are reproduced of the known data.

Also, along the entire Universe dynamics is verified the Einstein formula $E = mc^2$, or when all the created photons transform in mass of the particles.

Since, the FLRW metric of the universe must be of the form $ds^2 = a(t)^2 ds_3^2 - c^2 dt^2$ where $ds_3^2$ is a three-dimensional metric that must be one of (a) flat space, (b) a sphere of constant positive curvature or (c) a hyperbolic space with constant negative curvature, or for small commoving time $dt = \frac{1}{aHc}$, we can consider the distance as $L = ds \cong a = a_{end}$, so the volume is given by:

$$V_{\text{matter}} = (a)^4 \frac{1}{c} [m^3 s]$$

In other words, this model of particle creation by trans-Planckian physics results in a significant part of the present total energy density of matter in the Universe being contained in gravitons with energies $M_P$ that is not compatible with the observed behavior of $a(t)$.

A second example is de Sitter space which contains an event horizon. In this case the temperature $T$ is proportional to the Hubble parameter $H$, i.e. $T \propto H$, such a conclusion being used by author in [2] to calculate the evolution of Universe. To estimate the horizon entry we use some derivations done in [2].

During Universe evolution [2a], the horizon leave is when $a_{leave} = k_{leave}/H_{leave} = 1$, $k_{leave}^{-1} = H_{leave}^{-1} = 10^{-27} [m]$, $t_{leave} = H_{leave}^{-1}/c = 3.3 \times 10^{-36} s$ at the Electroweak epoch
Here, the Hubble constant is defined as resulting from the equations:

$$ \left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G}{3}\rho + \frac{\Lambda c^2}{3} $$

$$ \left(\frac{\dot{a}}{a}\right)^2 = -\frac{8\pi G}{c^2} p + \Lambda c^2 $$

In Newtonian interpretation, the Friedmann equations are equivalent to this pair of equations:

$$ \frac{\dot{a}^2}{2} = \frac{G \frac{4\pi a^3}{3} \rho}{a} \\ \rho[\text{kg/m}^3]; \text{energy density } p = -\rho c^2 $$

If we divide with $a^2$ we obtain for outside the object (BH, planets, stars etc.)

$$ \frac{\dot{a}^2}{2a^2} = H^2 [s^{-2}] = \frac{GM}{a^3} \Rightarrow \frac{GE}{c^2 a^3} \rightarrow \frac{GE}{c^4 a^3} = \frac{G}{c^4 a^3} \cdot n_g \cdot \varepsilon_g [m^{-2}]; \quad (1) $$

where: $n_g = E/\varepsilon_g$; $n_{at-merging} = \frac{n_g}{(\varepsilon_p/\varepsilon_g)}$; $M = \frac{4\pi a^3}{3} \rho$; $E = Mc^2$; $M = n_{\mu BH} m_{\mu BH}$; $R = H^{-1}$.

Or, for inside of these objects:

$$ \frac{\dot{a}^2}{2a^2} = \frac{G \frac{4\pi a^3}{3} \rho}{a^3} [s^{-2}] \rightarrow \frac{4\pi G \rho}{3c^2} = \frac{4\pi G \frac{M_U}{n_{at-merging} l_c^3}}{3c^2} [m^{-2}] \quad (2) $$

During Universe evolution at Electroweak epoch or Reheating due of the quantum fluctuations [2a] a huge number of the micro-black holes as Planck particles $n_p$ are generated, $m_{\mu BH} = m_p$; the graviton energy being at horizon leave;

$\varepsilon_g = \hbar c / a_{end} = 2 \times 10^{-26} \text{ J}$; when $a_{leave} = k_{leave} / H_{leave} = 1$, $M_U = 2.2 \times 10^{53} \text{ kg}$;

$\lambda_C = \hbar c / m_{\mu BH} c^2$; $n_{\mu BH} \approx 1 / H_{-1}^3 \cdot V_{available} \approx n_p \approx 10^{61-62}$,

First step

$n_p = E_U / \varepsilon_p = 10^{70} \text{ J}/10^9 \text{ J} = 10^{61}$ with $\varepsilon_{\mu BH} = 10^{19} / a_{end-ee} = 10^{11} \text{ GeV} \rightarrow 17 \text{ J}$, it results with $k_{leave}^{-1} = H_{leave}^{-1} = 10^{-27} \text{ [m]}$ by iteration for $N = 2.5$; in eq. (2); $k_{end} = 9.2 \times 10^{19} \text{ [m}^{-1}]$; $a_{end-ee} = 8.2 \times 10^7$; $H^{-1} = 10^{-18} \text{ [m]}$; $t = 3.3 \times 10^{-27} \text{ s}$; $R = 6.5 \times 10^{-20} \text{ [m]}$; $l_c = l_p = 1.5 \times 10^{-27} \text{ [m]}$;

Second step
or When immediately, $\varepsilon_{\mu BH} = 10^{11}/a_{EW} = 10^{11} \text{GeV} \rightarrow 17J$, that results with

$k_{leave}^{-1} = H_{leave}^{-1} = 10^{-27} [m]$ by iteration for $N = 16.2$; in eq. (2); $k_{end} = 9.2 \times 10^{19} [m^{-1}]$; $a_{EW} = 0.92$; $H^{-1} = 10^{-20} [m]$; $t = 3.3 \times 10^{-29} s$; $R = 6.5 \times 10^{-20} [m]$;

$l_C = l_P = 1.7 \times 10^{-27} [m]$; $a_{EW} = 0.92$; where $n_g = \frac{\varepsilon_P}{\varepsilon_g} \cdot n_P \rightarrow 10^{96}$ as a “fix” number of

gravitons escaped (an inverse process of a black-hole) from micro-blacks holes (the fraction for matter – 0.25) created in Universe as the Planck particles during theirs totally decaying till the Confinement, see below, and which deforms the spacetime.

In both steps the number of gravitons that has been released following µBHs merging is

only $n_{at-merging} = \frac{n_g}{(\varepsilon_{\mu BH}/10^{-26})} = \frac{10^{96}}{(17/10^{-26})} \cong 10^{68}$, and these generate the curvature radius of the object $R = d_H = 10^{-20}$.

In other words the contribution to the space-time deformation is due of

$n_H^{EW} = 10^{68}/2n_P = 10^6$, for two µBHs merging.

To mention that only this data set match the model.

### Electroweak symmetry breaking-quarks epoch (QGP)

The µBH particles decay to QGP; $m_{QGP} = 1.4 \times 10^{-22} \text{kg} \rightarrow 7.9 \times 10^4 \text{GeV}$, or

$\varepsilon_{\mu BH} = 10^{11}/a_{end-ee} = 7.9 \times 10^4 \text{GeV} \rightarrow 1.26 \times 10^{-5} J$, $k_{end} = k_{leave} e^{-N}$;

$k_{end}^{-1} = 7.9 \times 10^{-14} [m]$, $a_{end\_QGP} = 1.27 \times 10^6$, $\lambda_{C-ee} = 2.3 \times 10^{-21} [m]$, with eq. (2)

$R = 1.2 \times 10^{-7} m$, and $H_{end}^{-1} = 10^{-7} [m]$, $t_{end} = H_{end}^{-1}/c = 3.3 \times 10^{-16} s$,

$H_{leave}^{-1} = k_{leave}^{-1} = H_{EW}^{-1} = 10^{-20} [m]$ we found $N = 15.88$ to match the iterations cycle:

$m_g \rightarrow \hbar\nu \rightarrow k_B T \rightarrow \varepsilon \rightarrow R \rightarrow H_{end}^{-1} \rightarrow a_{end} \rightarrow N$.

The number of gravitons that has been released following µBHs merging is only

$n_{at-merging} = \frac{n_g}{(\varepsilon_{\mu BH}/10^{-26})} = \frac{10^{96}}{(1.26 \times 10^{-5}/10^{-26})} \cong 7.9 \times 10^{74}$, and these generate the

curvature radius of the object $R$.

In other words the contribution to the space-time deformation is due of

$n_H^{QGP} = 7.9 \times 10^{74}/2n_P = 7.9 \times 10^{12}$, for two µBHs merging.

To mention that only this dataset match the model.

### Another proof-the light bending by Earth

The number of graviton in case of Earth is $n_g = U/10^{-26} = 3.2 \times 10^{58}$;

$$U = \frac{GM_{Earth}^2}{R_{Earth}} = 3.2 \times 10^{32} J$$
The number of gravitons that has been released following µBHs (nucleons) merging is only $n_{at-merging} = \frac{n_g}{(\varepsilon_{\mu BH}/10^{-26})} = \frac{3.2 \times 10^{58}}{(3.6 \times 10^{-10}/10^{-26})} \cong 8.8 \times 10^{41}$, and that generate the curvature radius of the object ($\cong r_{Schw}$).

Now, the horizon-entry is when the wave length $k_{end} = k_{leave}e^{-N}$; $k_{end} = 31[m^{-1}]$; the scale factor arrives at $a_{end} = k_{end}/H_{end}$, the frequency is $\nu = c/k_{end} = 9.3 \times 10^9$ Hz, and the Compton length $\lambda_{C-g0} = \hbar/m_{nucleon\_BH}c = 8.2 \times 10^{-17} [m]$, for $\varepsilon_{nucleons} \cong 10^{-10}$ J it results $\varepsilon_p = \varepsilon_{\mu BH} = \varepsilon_{nucleon}/a_{end} = 10^{-10}/a_{end} = 3.6 \times 10^{-10} \rightarrow 2.2$GeV; it results $a_{end\_BH} \cong 1$, and from eq. (2) $H_{end}^{-1} \cong r_{Sch} \cong 8.8 \times 10^{-3} [m]$, $t_{end} = H_{end}^{-1}/c = 2.9 \times 10^{-11}$ s, we found $N = 35.7$ to match the iterations cycle:

$$m_g \rightarrow \hbar\nu \rightarrow k_B T \rightarrow \varepsilon \rightarrow R \rightarrow H_{end}^{-1} \rightarrow a_{end} \rightarrow N.$$

At BHs merging, the curvature radius R from eq. (2) with the number of merging nucleons during which release gravitons as $n_{at-merging}$, $R = H_{end}^{-1} = 5.3 \times 10^{-3} [m] \cong r_{Sch}$. The gravitons release from Earth as a gravitational wave

Therefore, the new horizon leave as gravitational wave (GW) is just when the gravitons escape from the Schwarzschild radius (the Universe is viewed as an inverse big black hole): $a_{leave} = k_{leave}/H_{leave} = 1$, or $k_{leave}^{-1} = H_{leave}^{-1} = r_{Schw} = 7.4 \times 10^{-3} [m]$.

Now, the new horizon-entry is when the wave length $k_{end} = k_{leave}e^{-N}$; $k_{end} = 8 \times 10^{-4} [m^{-1}]$, $k_{end}^{-1} = 1.2 \times 10^{3} [m]$; and the scale arrives factor at $a_{end} = k_{end}/H_{end}$, the Hubble length with Compton length $\lambda_{C-GW} = \hbar/m_{g\_GW}c = 16[m]$; from eq. (1) $H_{end}^{-1} = R = 6.5 \times 10^{6} [m]$; it results $a_{end\_GW} = 5.3 \times 10^{3}$, $t_{end} = H_{end}^{-1}/c = 0.02$s, we found $N = 12$ to match the iterations cycle: $m_g \rightarrow \hbar\nu \rightarrow k_B T \rightarrow \varepsilon \rightarrow R \rightarrow H_{end}^{-1} \rightarrow a_{end} \rightarrow N$. The energy of the graviton becomes at an eventually detector (like LIGO) with the above value at merging $\varepsilon_{g\_GW} = \frac{\varepsilon_{g\_BH}}{a_{end\_GW}} = \frac{10^{-23}}{6.3 \times 10^{17}} = 1.9 \times 10^{-27} [J] \rightarrow 1.1 \times 10^{-17}$ GeV, and the frequency is $\nu = c/k_{end}^{-1} = 2.4 \times 10^{5}$ Hz, the mass is $m_{g\_GW} = 2 \times 10^{-44}$ kg; the number of particles released as the gravitational wave (like the photons of the light wave) remains equally with the above value $n_g = 5 \times 10^{67}$, the total energy initially released is $E_{GW} = E_{mBH} = \varepsilon_g \cdot n_g = M_{Earth}c^2 = 5 \times 10^{41} [J]$, and the curvature radius it results from eq.(1) with the integral (which the starts at the outside of the $r_{Schw}$) graviton energy with $n_g$, and with $a_{end} = a_{end\_GW}$, as $R_{Earth} = 6.5 \times 10^{6} [m]$.

## The strain at Earth surface

Now, based on eq. (1) we can derive for the G-wave effect in the deformation (strain) of the space-time between Earth and a detector site by using the gravitational pressure due of gravity charges on the area of Schwarzschild radius $r_{Schw}$, we have:
$$ \left( \frac{r_{Schw}}{R} \right)^2 = \frac{4\pi}{3} \frac{G \cdot \varepsilon_g \cdot n_g \cdot r_{Schw}^2}{c^4 \cdot a_{end\_GW}^3} = 1.24 \times 10^{-18} $$

Separately,

$$ r_{Schw} = \frac{2G \cdot M_{Earth}}{c^2} = 8.86 \times 10^{-3} [m], \text{ we have} $$

$$ \frac{r_{Schw}^2}{R_{Earth}^2} = 1.34 \times 10^{-18} \rightarrow r_{Schw}/R_{Earth} = 1.1 \times 10^{-9} $$

, so, in both cases the strain is around $\theta = r_{Schw}/R \approx 1.1 \times 10^{-9}$, that is near equally with strain as light bending.

The average magnitude of the electric field (negative charge) in the event horizon of a micro-black-hole is like that of the model electron given in [7], and where the inside “trapped” photon is similar with the “absorbed” photon from thermal energy V in case of μBH particle, or in other words the electron is a decaying μBH particle, see below equation (4).

$$ \langle E \rangle = \sqrt{\frac{6\hbar c}{\pi \varepsilon_0 \lambda^4}} $$

, it results $\langle E \rangle = 2.9 \times 10^{50} [N/C]$, for μBH particle of $\lambda_C = 1.6 \times 10^{-29} [m]$ and where the gravity charge formally corresponds as $\hbar c \leftrightarrow e^2$.

## 2.1 The μBH merging at the base of BH's nuclei of galaxies production

An important contribution to further Universe expansion is given by the BH's nuclei of galaxies which result from another fraction (0.75) of the initially μBHs merging.

Thus, with $\varepsilon_{BH} = 10^{11}/a_{end\_BH} = 8.4 \times 10^{-10} \text{GeV} = 1.35 \times 10^{-19} \text{J}$, and horizon-entry is when $k_{end} = k_{leave}e^{-N}$; $k_{end}^{-1} = 8.4[m]$, $a_{end\_BH} = 1.2 \times 10^{20}$, from eq. (2) $H_{end}^{-1} = 10^{21}[m]$, and the curvature radius results $R = 10^{21}[m]$, $t_{end} = H_{end}^{-1}/c \approx R/c = 3.3 \times 10^{12} s$, with $H_{leave}^{-1} = k_{leave}^{-1} = H_{EW} = 10^{-20}[m]$ we found $N = 48.2$ to match the iterations cycle: $m_g \rightarrow \hbar v \rightarrow k_B T \rightarrow \varepsilon \rightarrow R \rightarrow H_{end}^{-1} \rightarrow a_{end} \rightarrow N$.

$$ T = 9 \times 10^3, \lambda_{C\_BH} = 2.2 \times 10^{-7} [m], \nu = c/k_{end}^{-1} = 10^8 \text{Hz} $$

The number of gravitons that has been released following μBHs merging is only

$$ n_{at-merging} = \frac{n_g}{(\varepsilon_{\mu BH} / 10^{-26})} = \frac{10^{96}}{(1.35 \times 10^{-19} / 10^{-26})} \approx 7.4 \times 10^{88}, \text{ and these generate the curvature radius of the object (Universe) R.} $$

In other words the contribution to the space-time deformation is due of $n_H^{\text{dark}} = 7.4 \times 10^{88}/2n_P = 7.4 \times 10^{26}$, $n_P = 10^{61}$, for two μBHs merging.

Thus, from the overall balance the number of gravitons which are consumed to push the space time is $:= 7.4 \times 10^{26} \triangleleft n_{\mu BH-P} = 10^{35}$.

Therefore, it remain a lot of gravitons embedded in μBHs-BH in a new merging step towards primordial BHs of number $10^{11}$ of masses $10^{14} \div 10^{23} \text{kg}$.
## What is in fact the CMBR?

To note that the time of $3×10^{12} s \cong t_{\text{CMBR}} = 1.13×10^{13} s$, the temperature $\sim 10^4 K$, the wave number of $\sim 8.57$ m (microwave) and the frequency of $10^8 Hz$, all correspond to the situation at cosmic microwave background radiation (CMBR) time. Since, in the all-sky map (pattern) of the CMB appear some voids, these can represent the µBHs merging (clustering) which emit gravitational radiation, and the rest is due of the thermal radiation at Recombination as following of the photon decoupling.

### The formation of the Primordial BHs galaxies nuclei by µBHs merging

Thus, following author model for LGO measurement interpretation [25], with $\epsilon_{\text{mgBH}} = 1.35×10^{-19}/a_{\text{end\_mgBH}} = 1.5×10^{-9} \text{GeV} = 2.4×10^{-19} J$, and horizon-entry is when $k_{\text{end}} = k_{\text{leave}}e^{-N}$; $k_{\text{end}}^{-1} = 176[m]$, $a_{\text{end\_mgBH}} = 0.56$, from eq. (2)

$H_{\text{end}}^{-1} = 106[m] \cong r_{\text{Sch}}$, and the curvature radius results $R=11[m]$, $t_{\text{end}} = H_{\text{end}}^{-1}/c \cong R/c = 3.3×10^{-7} s$, with $H_{\text{leave}}^{-1} = k_{\text{leave}}^{-1} = \lambda_{\text{C\_BH}} = 2.2×10^{-7}[m]$ we found $N=20.5$ to match the iterations cycle:

$m_g \rightarrow \hbar v \rightarrow k_B T \rightarrow \epsilon \rightarrow R \rightarrow H_{\text{end}}^{-1} \rightarrow a_{\text{end}} \rightarrow N$.

$T = 1.7×10^4$, $\lambda_{\text{C\_mgBH}} = 1.26×10^{-7}[m]$, $v = c/k_{\text{end}}^{-1} = 10^6 Hz$

The number of gravitons that has been released following µBHs merging into primordial galaxies is only $n_{\text{at-merging}} = \frac{n_g}{(\epsilon_{\text{µBH}}/10^{-26})} = \frac{10^{60}}{(2.4×10^{-19}/10^{-26})} \cong 4.2×10^{51}$, where the remnant number of gravitons in µBHs is

$n_g = \frac{10^{35}}{7.4×10^{26}} \cdot \frac{1}{n_{\text{galaxies}}} \cdot \frac{10^{60}}{(\epsilon_{\text{mgBH}}/10^{-26})} \cong 4.2×10^{51}$, $n_{\text{galaxies}} \cong 10^{11}$, and these generate the curvature radius of the object (Universe) $R$. To note, that the remnant number of gravitons following this last merging process is only $\frac{10^{62}/10^{11}}{4.2×10^{51}} \cong 4$, or in others words

the gravitational potential which deforms the space time is entirely consumed. The energy released as a gravitational wave which push the space-time is

$E_{\text{GW-mgBHs}} = 4.2×10^{51} \times \epsilon_{\text{mgBH}} = 10^{33} J \rightarrow 10^{16} Kg$ which is in the range of primordial BH galaxies nuclei $10^{14} \div 10^{23} kg$.

### The space-time deformation during µBHs merging into Primordial BHs galaxies nuclei

Thus, following author model for LGO measurement interpretation [25], with $\epsilon_{\text{mgBH\_GW}} = 1.7×10^{-19}/a_{\text{end\_mgBH\_GW}} = 1.04×10^{-22} \text{GeV} = 1.6×10^{-32} J$, and horizon-entry is when $k_{\text{end}} = k_{\text{leave}}e^{-N}$; $k_{\text{end}}^{-1} = 9.7×10^{11}[m]$, $a_{\text{end\_mgBH\_GW}} = 10^{13}$, from eq. (1)

$H_{\text{end}}^{-1} = 10^{25}[m]$, and the curvature radius results $R=10^{25}[m]$,
$t_{end} = H_{end}^{-1}/c \cong R/c = 3.3 \times 10^{16} s$, with $H_{lae}^{-1} = k_{lae}^{-1} = 100[m]$ we found $N = 23$ to match the iterations cycle: $m_g \to \hbar\nu \to k_B T \to \epsilon \to R \to H_{end}^{-1} \to a_{end} \to N$,

$$T = 10^{-9} K, \quad \lambda_{C\_mgBH} = 1.8 \times 10^{6} [m], \quad \nu = c/k_{end}^{-1} = 10^{-3} \text{ Hz}$$

The number of gravitons that has been released during µBHs merging into a primordial galaxies BH nuclei is $n_{mg\_BH\_GW} = \frac{E_{GW-mgBHs}}{10^{-26}} = 10^{59}$, these generate the increase of curvature radius of the object (Universe) to R.

To note, that this process it seems to be equivalent with the use of the cosmological constant in ΛCDM model (but is not necessary to explicit it use-Einstein was right!) .
The strain at LIGO site

Now, based on eq. (1) we can derive for the G-wave effect in the deformation (strain) of the space-time between BH and LIGO site by using the gravitational pressure due of gravity charges on the area of Schwarzschild radius $r_{Schw}$ of BH, respectively with the values calculated in section 5, we have:

$$\left(\frac{r_{Schw}}{R}\right)^2 = \frac{4\pi G \cdot \varepsilon_g \cdot n_g \cdot r_{Schw}^2}{3 c^4 \cdot a_{end\_mgBH\_GW}^3} = 1.6 \times 10^{-72}$$

, where $\varepsilon_g = 2 \times 10^{-26} / a_{end\_BH} = 10^{-26} J$ and $n_g = 10^{59}$

Separately, with the LIGO site curvature radius as $R_{LIGO} = 1.3 \times 10^9 \text{Light\_years} \times c = 1.25 \times 10^{25} m$, and the Schwarzschild radius

$$r_{Schw} = \frac{2G \cdot 3M_{mgBH}}{c^2} = 1.4 \times 10^{-11} [m], \text{ we have}$$

$$\frac{r_{Schw}^2}{R_{LIGO}^2} = \frac{(1.4 \times 10^{-11})^2}{(1.25 \times 10^{25})^2} = 1.5 \times 10^{-72} \to r_{Schw} / R_{LIGO} = 1.2 \times 10^{-36}$$

, so, in both cases at (LIGO laser) the strain is around $\theta = r_{Schw} / R \cong 1.3 \times 10^{-36}$, which is undetectable.

## 2.2 The Confinement into nucleons

In the dual-superconductor ($E \leftrightarrow B$) picture for the QCD vacuum, the squeezing of the - electric flux ($E$) between quarks (as decayed from µBH particles) is realized by the dual Meissner effect, as the result of condensation (as a solenoidal electric current) of **soft photons** as color gluons (bosons) radiated by µBH particles, which is the dual version of the electric charge as the Cooper pair. The order parameter is the vacuum expectation value of the creation operator of gluons (like a condensate under a critical temperature $T_c$!), such as $e\bar{e}$ for Cooper pairing of electrons in superconductors. The color confinement is based on the analogy between the superconductor and the QCD vacuum. In the superconductor, magnetic field is excluded due to the Meissner effect,
which is caused by Cooper-pair condensation. As the result, the magnetic flux is squeezed like the Abrikosov vortex in the type II superconductor [22]. On the other hand, the color-electric flux is excluded in the QCD vacuum, and therefore the squeezed color-flux tube is formed between color sources. Thus, these two systems are quite similar, and can be regarded as the dual version each other. The color-electric field is then excluded in the QCD vacuum through the dual Meissner effect, and is squeezed between color sources to form the hadron flux tube.

Therefore, in case of superconductors an external magnetic flux decreases exponentially into the superconductor, penetrating a distance of the order $\lambda$. This distance is called the London penetration depth. In a dual superconductor the roles of magnetic and electric fields are exchanged and the Meissner effect tries to expel electric field lines. Quarks and antiquarks carry opposite color charges, and for a quark-antiquark pair the 'electric' field lines run from the quark to the antiquark. If the quark-antiquark pair is immersed in a dual superconductor, then the electric field lines get compressed to a flux tube. The energy associated to the tube is proportional to its length, and the potential energy of the quark-antiquark is proportional to their separation.

As resulting from [1a] this Lorenz force appears to be necessary in order to equilibrate the gravity charges (gravitons) embedded in the quarks viewed as ex-micro-black-holes, and these gravitons themselves “deforms” the space-time following Einstein-Friedman equation.

So, in this paper we will continue to follow the development of references [8a; 8b; 23a; 23b], especially, as regarding the single flux-tube solution in the dual Ginzburg-Landau (DGL) theory.

<figure><img src="image_3.png" /><figcaption>Figure 1.a. The mass generation by dual Meissner effect, the $\beta$ decay for a free neutron and, n-normal vacuum, s-superconductor, QW-quantum well.</figcaption></figure>

At Confinement we have for the vortex potential $3 \times q\bar{q}g$ as from [8a],

$$V = \varepsilon = \lambda_c^{-1} = 80 \text{GeV}, \text{ it results with eq. (1a;1.b)}$$

$$E = 1.1 \times 10^{28} [\text{N/C}], \quad B = E/c = 3.7 \times 10^{19} [\text{T}].$$
Hadrons, along with the valence quarks ($q_v$) (white) that contribute to their quantum numbers, contain virtual quark-antiquark ($q\bar{q}$) pairs known as *sea quarks* ($q_s$), see figure 1.b. Sea quarks ($RR'$) form when a gluon of the hadron's color field splits; this process also works in reverse in that the annihilation of two sea quarks produces a gluon. Free particles have a color charge of zero: baryons are composed of three quarks, but the individual quarks can have red (R), green (G), or blue (B) charges, or negatives.

<figure><img src="image_4.png" /><figcaption>Fig.1.b. Fields due to color charges as in sea quarks ($q\bar{q}(RR'$)) of valence quarks ($u$), ($G$ is the gluon field strength tensor). These are "colorless" combinations. **Top:** Color charge has "ternary neutral states" as well as binary neutrality (analogous to electric charge).</figcaption></figure>

The application of DGL theory to the quark-gluon-plasma (QGP) physics in the ultrarelativistic heavy-ion collisions, or in the early universe, where, a new scenario of the QGP formation via the annihilation of color-electric flux tubes based on the attractive force between them is proposed in [22]. The QGP phase is characterized as the deconfinement and the chiral-symmetry restoration.

There, for the same flux tubes with opposite flux direction (e.g. $R-\bar{R}$ and $\bar{R}-R$), one finds $Q_1 = -Q_2$ i.e. $Q_1Q_2 = -e^2/3$, so that these flux tubes are attracted each other. It should be noted that they would be annihilated into dynamical gluons in this case.

For the different flux tubes satisfying $Q_1Q_2 < 0$ (e.g. $R-\bar{R}$ and $B-\bar{B}$), one finds $Q_1Q_2 = -e^2/6$, so that these flux tubes are attractive. In this case, they would be unified into a single flux tube (similar to $G-\bar{G}$ flux tube), [8b].

The effect of the potential $B$ is the same as shifting the *effective mass* $m_*^2 c^4 = m_e^2 c^4 + qB\hbar c^2 (2j+1-\sigma)$ for fermions for each Landau level.

Where, $\lambda_C = \hbar/m_*c = 3.6 \times 10^{-16} [m]$,
the effective mass is
$$m_* = \sqrt{m_e^2 c^4 + qB\hbar c^2} / c^2$$

$$\text{Or } m_q = 9.2 \times 10^{-28} \text{ kg} \rightarrow 0.5 \text{ GeV}$$

which is just the $q\bar{q}$ string tensions.

$$\text{With } \varepsilon_{qq_g} = \varepsilon_{QGP} / a_{end\_qq_g} = 8.9 \times 10^4 / 3.7 \times 10^3 = 2.1 \text{ GeV} \rightarrow 3.4 \times 10^{-10} \text{ J at}$$

Confinement, when $k_{end} = k_{leave}e^{-N}$; $k_{end}^{-1} = 0.4[m]$, $a_{end\_qq_g} = 3.7 \times 10^3$, with eq. (2)

$$H_{end}^{-1} = 1500[m], \quad R = 1447[m]; \quad t_{end} = H_{end}^{-1} / c = 5 \times 10^{-6} s, \text{ and}$$

$$H_{lae}^{-1} = k_{lae}^{-1} = H_{QGP}^{-1} = 10^{-7}[m] \text{ we found } N = 15.2 \text{ to match the iterations cycle:}$$

$$m_g \rightarrow \hbar\nu \rightarrow k_B T \rightarrow \varepsilon \rightarrow R \rightarrow H_{end}^{-1} \rightarrow a_{end} \rightarrow N.$$

The number of gravitons that has been released following µBHs merging is only

$$n_{at-merging} = \frac{n_g}{(\varepsilon_{\mu BH} / 10^{-26})} = \frac{10^{96}}{(3.4 \times 10^{-10} / 10^{-26})} \approx 2.9 \times 10^{79}, \text{ and these generate the}$$

curvature radius of the object R.

In other words the contribution to the space-time deformation is due of

$$n_H^{qq-g} = 2.9 \times 10^{79} / 2n_p = 2.9 \times 10^{17}, \text{ for two µBHs merging.}$$

Thus, from the overall balance the number of gravitons consumed to push the space time is $=1/8 \cdot 7.9 \times 10^{12} \cdot 5.7 \times 10^6 \cdot 2.9 \times 10^{17} \approx 10^{36} \approx n_p = 10^{35}$.

In other words, at Confinement it remains at least 1-2 gravitons of

$$\varepsilon_{nucleon} = 10^{-26} / \lambda_c \approx 10^{-10} \text{ J} \approx 1 \text{ GeV}.$$

To mention that only this data set match the model.

From [2a], we have $H_0$ -an “external” electro-magnetic field of a dipole created by the pair $u\bar{u}$ (the chromoelectrical colors field)

$$H_0 = E_0 = \frac{d \cdot e}{4\pi\varepsilon_0 r^3} = 8.33 \times 10^{24} [N/C]$$

, where $r = 0.05 \text{ fm}$ -is the electrical flux tube radius, $d = 0.48 \text{ fm}$ -the distance between the two quarks charges, this is in fact equilibrated by the gluons field, and respectively, from eq. (2.a;2.b) at a more deep penetration

$$\lambda_{C-qq} = 4.7 \times 10^{-16} > \lambda_{C-gluon} = 8.6 \times 10^{-17} [m], \text{ see below.}$$

Because the magnetic induction of the color magnetic gluons current which is powered by electric field given by a pair of quarks ($H_0$), $B_{gluon} \ge 2H_0 \equiv H_{c2}$, it has the raw flow consequences squeezing this chromoelectrical flux into a vortex line, followed by forcing an organization into a triangular Abrikosov lattice, see figure 1.

The penetration of B is

$$\lambda = \left( \frac{\varepsilon_0 \cdot m \cdot c^2}{n_s \cdot e^2} \right)^{1/2} [m], \text{ m-gluons mass, } n_s \text{-number of gluons per } [m]^3$$

$$\Phi_0 = \pi\hbar c/e \rightarrow \text{usually} \quad \frac{\pi\hbar}{e} = 2.07e-15[Tm^2] \rightarrow Js/C$$

From [2a], we have the lower critical field:
$$B(x) = \frac{2\Phi_0}{2\pi\lambda^2} \log\frac{\lambda}{\xi} = \frac{\pi\hbar c}{\pi\lambda^2 c} \log(\kappa) = 10^{15} \left[ \frac{J}{Am^2} \right], \quad (2.a)$$

where $\xi = 0.1114$, and when

near the axis, for $x = 0.116 = \xi$, when the induction is

$$B(\xi) = 2 \times 10^{15} [T] = H_{c1}, \quad E = cB = 6 \times 10^{23} [N/C]$$

In the case of a homogeneous potential directed along the z-axis [9], the Einstein stress-energy tensor is:

$$T^{00} = T^{11} = T^{22} = -T^{33} = \rho_B = \frac{\varepsilon_0 c^2 B^2}{8\pi}; \quad T^{0i} = 0, \quad \text{where } \rho_B[J/m^3]\text{-the magnetic energy density.}$$

The equivalence between the Lorenz force energy which squeezes the electrical field $E_{e^\pm}$ is $\varepsilon_L = ec\lambda_c B$, and at the interface between normal and superconducting phase we have

$B \cong E/c$, with $e^\pm$ pair giving E as: $k_B T = \hbar v = \varepsilon_L = ec\lambda_c \frac{\hbar}{e\lambda_c^2} = c \frac{\hbar}{\hbar} mc = mc^2$, and accounting that the inverse of the penetration length $\lambda \cong \lambda_c$.

Also, the interaction energy at interface $E-B$ is:

$$\varepsilon = \frac{V_{vol} \varepsilon_0 c^2 B^2}{8\pi} = \rho_B V_{vol} = V[J], \quad (2.b)$$

$V_{vol} = 2\pi\lambda_c\lambda_c(4\lambda_c) \cong 8\pi\lambda_c^3$, at Compton length equally with the penetration length $\lambda_c = \lambda$, that results

$$E^2 = \frac{(V)}{\varepsilon_0 (\lambda_c^{e*})^3} \quad (2.c)$$

With $V = \varepsilon_{gluons}$ as above is obtained $B = E_{q\bar{q}}/c = 1.98 \times 10^{15} [T]$, where $E_{q\bar{q}} = 5.9 \times 10^{23}$ with eq.

(2.a), that are identically with the above values, **indubitable** meaning that this force creates the spacetime curvature and this is equilibrated by the gravity charge, see below.

With equation (4).

$$\langle E \rangle = \sqrt{\frac{6\hbar c}{\pi \varepsilon_0 \lambda^4}}$$

, it results $\langle E \rangle = 3.6 \times 10^{23} [N/C]$, for $q\bar{q}$ particle of $\lambda_c = 4.7 \times 10^{-16} [m]$, that is near the values calculated before by both methods.

## 2.3. The electrons production

Also, the further Universe expansion could be when the $\mu$BH particles via quarks can decay to electrons till $m_{e^+e^-} = 9.2 \times 10^{-31} \text{kg} \rightarrow 0.52 \text{MeV}$, or

$$\varepsilon_{e^+e^-} = \varepsilon_{QGP} = 7.9 \times 10^4 / a_{end-ee} = 5.2 \times 10^{-4} \text{GeV} \rightarrow 8.3 \times 10^{-14} \text{J}, \quad k_{end} = k_{leave} e^{-N};$$

$$k_{end}^{-1} = 66[m], \quad a_{end\_e} = 1.5 \times 10^8, \quad \lambda_{C-ee} = 3.6 \times 10^{-13}[m], \text{ with eq. (2) } R = 2.9 \times 10^9 \text{m, and}$$
$H_{end}^{-1} = 10^{10} [m]$, $t_{end} = H_{end}^{-1}/c = 33s$, $H_{lae}^{-1} = k_{lae}^{-1} = H_{QGP}^{-1} = 10^{-7} [m]$ we found

$N = 20.31$ to match the iterations cycle:

$m_g \rightarrow \hbar v \rightarrow k_B T \rightarrow \epsilon \rightarrow R \rightarrow H_{end}^{-1} \rightarrow a_{end} \rightarrow N$.

The number of gravitons that has been released following µBHs merging is only

$n_{at-merging} = \frac{n_g}{(\epsilon_{\mu BH}/10^{-26})} = \frac{10^{96}}{(8.3 \times 10^{-14}/10^{-26})} \cong 10^{83}$, and these generate the curvature radius of the object R.

In other words the contribution to the space-time deformation is due of

$n_H^{ee} = 10^{83}/2n_p = 1.2 \times 10^{21}$, for two µBHs merging.

Thus, from the overall balance (via quarks-QGP) the number of gravitons which are consumed to push the space time is : $= 7.9 \times 10^{12} \cdot 1.2 \times 10^{21} \cong 10^{34} \cong n_p = 10^{35}$.

Therefore, it remains only a few gravitons embedded in µBHs-quarks (electrons), of mass $10^{-26}/\lambda_{C-ee} \cong 2.7 \times 10^{-14} J = 0.17 \text{MeV}$. To mention that only this kind of dataset matches the model.

### b) A strong prove of the model-the free neutron decay

In the following, we will use some results of section 4.1a, but where $\lambda_c = \hbar/m_*c = 2.3e-18 [m]$, the effective mass is

$m_* = 1.44 \times 10^{-25} \text{kg} \rightarrow V = 81 \text{GeV} \rightarrow E = 1.1 \times 10^{28}$, the critical field

being $E_c = \frac{m_*^2 c^3}{e\hbar} \cong 3.5 \times 10^{28} > E = 1.1 \times 10^{28} [N/C]$; $B = E/c = 3.7 \times 10^{19} [T]$.

From the above section (4.1b), are used the bosons $W^\pm$ pairs generated inside the nucleons as due of one quark $u\bar{u} \leftrightarrow u$ resultant $\times 3$ flux tubes vortex potential, see figure (1.b), respectively $\epsilon = mc^2 \cong 81 \text{GeV}$- which after the release of an electron that it getting the final beta energy as been equally to the out of barrier turning point after the tunneling, and accounting for the valence nucleons interactions (shell-energy levels). The number of assaults of the barrier, like in Gamow theory [20, 21] is $n_a = v_b/R_{inner}$; where the velocity is $v_b \cong (2\epsilon/m)^{1/2} = 2.3 \times 10^8 \text{ m/s}$, where, the inner radius of the barrier is $R_{inner} \cong b = 3.5 \times 10^{-17} [m]$, see below. For only one of the three vortex-flux tubes ($q\bar{q}g$) we have: $\epsilon = \hbar eB/m \cong 4 \times 10^{-9} [J] \rightarrow \cong 25 \text{GeV}$, with the above (B) which is obtained from eq.(1.a) with the resultant potential $V = 81 \text{GeV}$, that corresponds to $m_{q\bar{q}} \cong 29 \text{GeV}$ from 4.1a, the energy of the particle for the first Landau level (as above), and we can see that it results to be equally with $\approx 1/3$ rest mass of the $W^\pm$, that resulting $n_a \cong 7.5 \times 10^{24} \text{ s}^{-1}$.

In case of WKB [20], the transmission coefficient is $T = 2 \frac{\sqrt{2m|V-Q|}}{\hbar} \Delta r$, and the decay constant $\Gamma = n_a e^{-T}$.
For the thick barrier the transmission coefficient is $T = 2\pi \frac{Qb}{\hbar v} = 2\pi \frac{\sqrt{2mQ}}{\hbar} b$ ;
where, the kinetic energy of the particle after the barrier at b

is $Q \equiv \frac{1}{2}mv^2, b = d_b/2\pi = 3.5 \times 10^{-17} [m]$, see below, that results $T = 63$ ; and the decay constant $\Gamma = 3 \times 10^{-3} s^{-1} \rightarrow \approx 324s$

To “materialize” a virtual $e^+ - e^-$ pair in a constant electric field E the separation d
must be sufficiently large $eEd = 2mc^2$

Probability for separation *d* as a quantum fluctuation

$$
P \propto \exp\left(-\frac{d}{\lambda_{\text{Compton}}}\right) = \exp\left(-\frac{2m^2c^3}{e\hbar E}\right) = \exp\left(-\frac{2E_{cr}}{E}\right)
$$

The emission (transmission through barrier) is sufficient for observation when $E \approx E_{cr}$,
with $Q = 1/2mc^2$, results $T = 2\pi \frac{mcb}{\hbar} \approx \frac{2\pi b}{\lambda_c}$, or
$b \approx d_b / 2\pi$.

Now, by using the Schwinger effect as in section 2.1 , the number of W ± pairs produced
inside the nucleon (more inside of the only one resultant flux tube , see figures 1.a; 1.b)
due of the potential resultant uū ↔ 3×vortex( q ¯ q g) of V = 80GeV , results as

$$
R = (E/E_{cr})^2 (c/\lambda^4) (8\pi^3)^{-1} * \exp(-\pi E_{cr}/E), \text{ the rate per unit volume of pair creation } R \\
\text{in a constant and uniform electric field of strength } E, \text{ when this electric field } E \text{ is} \\
\text{induced by quarks pairs as ex-Micro-black-holes, or } E/E_{cr} \ll 1, \text{ positron charge } e, \\
\text{mass } m, \text{ Compton wave-length } \lambda = \hbar/mc \text{ and so-called "critical" electric field} \\
E_{cr} = m^2 c^3/e\hbar, \quad \text{it results}
$$

$R/s = R/V \times V_{vol} = 2.3 \times 10^{18} s^{-1} \approx n_a$, where $R/V = 2 \times 10^{71} [1/m^3 s]$ and the volume is $V_{vol} \approx (\lambda_c)^3 \approx 1.24 \times 10^{-53} [m^3] \ge V_b$, the penetration length being $\lambda_c = 2.3 \times 10^{-18} [m]$, and for a four-volume of $\lambda_c^4/c \approx 9.5 \times 10^{-80} [m^3 s]$, results as a *permanently rate*

$R \approx 10^{-8} W^{\pm}$ pairs. To note, that in the previously version of the work [1a], we have used for $V = v.e.v = 247$ GeV, and since with this value it results $B = 3.3 \times 10^{20} [T]$ and the velocity of W bosons resulted to be $v_b = 7.2 \times 10^8 > c = 2.986 \times 10^8$, greater than c, that it was not acceptable, so it renounced to consider an external Higgs field v.e.v. Otherwise, if this field existed, that means the Universe it remained at about $R = 0.05 [m]$.

Thus, it results a main conclusion of this investigation, namely, that the “interacting”
potentials inside the nucleons are that were already established in [8a],
respectively ≈ 80GeV around the valence quarks (u,d) which it seems to be “locked” at
the electroweak symmetry breaking (≈ 100GeV ); that of the Giant Vortex (see the insert
in fig. 1.a) at the center of the triangle-the Higgs boson H = 125GeV ; and that resulting
from interaction of 2× interpairs of flux tubes as been the neutral boson Z = 90GeV .
Therefore, in other words is proved that all the time inside the nucleon are available
$10^{-8}$W$^{\pm}$ pairs that seems to corresponds to the “weak interaction” coupling constant
$10^{-7}$, which is absorbed or emitted by the quarks, resulting an $e^{+}$, or $e^{-}$ which help
the quarks transformation like ($u \rightarrow d$), respectively ($d \rightarrow u$) for beta-decay. In our
understanding, the created electron takes the energy at the *turning point* out of the
barrier equally with the electron itself for unbounded neutrons, or that of the binding
energy of nucleon in isotope nucleus, when it *passes* the barrier of gluon condensate
characterized by an *quantum tunneling* suppression given as:

$\exp(-\Delta E \tau/\hbar) \approx 7.3 \times 10^{-22}$, where, as the lifetime of $W^{\pm}$ being $\tau \approx 3 \times 10^{-25}$ s. Here, $\Delta E$ corresponds to the height of gluon condensate barrier, due of the *phase slip* with $2\pi - \varphi$ and of a $\Phi_0$ energy release as: $\Delta E = c^2 \Phi_0^2 \epsilon_0 / d_b$; $d_b \approx k\lambda_C = 1.98 \times 10^{-16}$, $k=85$, where the Compton length is just the penetration length for $W^{\pm}$ pair $\lambda_C = 2.3 \times 10^{-18}$ [m], or in other words just the barrier size, and $\Delta E = 1.6 \times 10^{-8}$ [J] $\rightarrow 100$ GeV $\approx 3 \times 25$ GeV as for $\times 3$ sea quarks color flux tubes, see figures 1.a; 1.b. The value of the resulting flux tube it remains as in (4.2.a), respectively of 0.4 GeV as the string strength.

Thus, the probability (rate) to produce $W^{\pm} \rightarrow e^{\pm}$, into a more simple way- without the external interactions of the neutron (free-not bounded), is given as:

$RV \exp(-\Delta E \tau/\hbar) \approx 1.7 \times 10^{-3} s^{-1} \rightarrow \tau_{1/2} \approx 584 \text{ [s]} \approx 612 \text{ s}$, that corresponds for *free neutrons decay* ($\beta^{-}$) by emission of an electron and an electron antineutrino to become a proton $n^0 \rightarrow p^+ + e^- + \bar{\nu}_e$, with half-life of 611s, and $Q_{\beta^{-}} = m_e v^2 = 0.5 \text{ MeV}$.

Now, the energy corresponding to $E_{cr}$ is much higher than $E$, respectively as from eq.

$$
(1.a): \quad \nu = E_{cr}^2 \epsilon_0 (\lambda_{Compton}^3), \quad \lambda_{Compton} = \hbar/mc; \quad \nu = \epsilon = mc^2 \text{ or}
$$

$$
\nu = \frac{m_W^4 c^6}{e^2 \hbar^2} \frac{\hbar^3}{m_W^3 c^3} \frac{4\pi\epsilon_0}{4\pi} = \frac{m_W c^2 \epsilon_0 \hbar c 4\pi}{4\pi\epsilon^2} \approx \frac{M}{4\pi\alpha} \rightarrow \nu \approx 10 \times 80 = 800 \text{GeV},
$$

$$
\alpha = \frac{e^2}{4\pi\epsilon_0\hbar c} = \frac{1}{137}
$$

$$
, \text{ since } (E_c/E)^2 = \left( \frac{3.5 \times 10^{28}}{1.1 \times 10^{28}} \right)^2 \approx 10.
$$

In the classic understanding of $\beta^-$ disintegration $n \rightarrow p+e^-+\bar{\nu}_e$, in ours understanding
this occurs when one of the down quarks (d) in the neutron (udd) transforms into an up
quark (u) due of interacting with the charge of $W^+$ boson of the pair $W^\pm$, transforming
the neutron into a proton (uud). In mean time the other part of this pair $W^-$ boson decays
into an electron and an electron antineutrino $udd \rightarrow uud+e^-+\nu_e$. Probable the claimed
energy of boson $W^-$ is the same as to be the necessarily energy to traverse the gluonic
barrier, when it decays into $e^-$ at the end.
### The free neutron decay

Consequently, for the $\beta^-$ decay process, the energy combines well with the existing one, that releasing an electron which penetrates the barrier:

$$d \rightarrow u+W^{+}+W^{-} \rightarrow u+e^{-}+\bar{v}_{e}$$
$$d(-1/3e)+e^{+}(+3/3e)=u(+2/3e)+e^{-}(-3/3e)$$
, since $W^{-} \rightarrow e^{-}$, and $W^{+} \rightarrow e^{+}$

In case of $\beta^+$ decay, it can only happen inside nuclei when the daughter nucleus has a greater binding energy (and therefore a lower total energy) than the mother nucleus. The difference between these energies goes into the reaction of converting a proton into a neutron, a positron and a neutrino and into the kinetic energy of these particles.

Thus, an opposite process to the above negative beta decay, $\beta^+$ decay of nuclei (only bounded proton) when $p \rightarrow n+e^{+}+v_{e}$, or energy+$uud$+$W^{+}$+$W^{-}$ $\rightarrow$ $udd$+$e^{+}$+$v_{e}$, or, $u(2/3e)+e^{-}(-3/3e)+energy=d(-1/3e)+e^{+}(3/3e)$.

For free proton decay an added energy it seems to be necessarily to reduce the barrier width to $d_{b}=9 \times 10^{-17} [m]$, when the production rate is:

$RV \exp(-\Delta E \tau/\hbar) \approx 7 \times 10^{-29} s^{-1} \rightarrow \tau_{1/2} \approx 10^{28} [s]$, respectively, an increase to $\Delta E = 3.5 \times 10^{-8} [J] \geq 225 \text{GeV}$ from $\Delta E = 1.6 \times 10^{-8} [J] \rightarrow 100 \text{GeV}$, as for the free neutron, or near $\geq \text{v.e.v} = 247 \text{GeV}$, like at LHC when the gluonic “cover” of protons it was “melted (at least 2 gluons)”, and the resulted difference ($\approx 225-100=125 \text{GeV}$) being just that of the Higgs boson (a quanta of energy!) which it was, in this spectacular way “released” [8a] as $2g \rightarrow 2\gamma$.

In the process of electron capture, one of the orbital electrons, usually from K or L electron shell, is captured by a proton in the nucleus, forming a neutron and an electron neutrino.

$$p+e^{-} \rightarrow n+v_{e}$$

About others calculations of beta decay processes of different isotopes, see the author's work [8a].

### The Higgs field

After EW the $\mu$BH becomes later the Higgs field (particle) by capturing two photons which moves inside in opposite directions, thus giving a neutral charge and of spin 0.

The mass is $m_{H_{-g}} = 4.7 \times 10^{-25} \text{kg} \rightarrow 267 \text{GeV}$, or

$\varepsilon_{H_{-g}} = \varepsilon_{\mu BH-EW} / a_{end} = 267 \text{GeV} \rightarrow 4.3 \times 10^{-8} \text{J}$, $\varepsilon_{\mu BH-EW} = 8900 \text{GeV}$; $k_{end} = k_{leave} e^{-N}$;

$k_{end}^{-1} = 3 \times 10^{-4} [m]$, $a_{end} = 33.4$, $\lambda_{C} = 7 \times 10^{-19} [m]$, with eq. (2) $R = 0.01 m$, and

$H_{end}^{-1} = 0.01 [m]$, $t_{end} = H_{end}^{-1} / c = 3.3 \times 10^{-11} s$, $H_{leave}^{-1} = k_{leave}^{-1} = H_{EW}^{-1} = 10^{-5} [m]$ as has been obtained in [1] for hadrons, we found $N = 3.4$ to match the iterations cycle:

$$m_{g} \rightarrow \hbar v \rightarrow k_{B} T \rightarrow \varepsilon \rightarrow R \rightarrow H_{end}^{-1} \rightarrow a_{end} \rightarrow N.$$
The number of gravitons that has been released following µBHs merging is just the primordial one $n_g = 10^{96}$. The number of gravitons that has been released following these

$\mu$BHs merging is only $n_{at-merging} = \frac{n_g}{(\varepsilon_{H-g}/10^{-26})} = \frac{10^{96}}{(4.3 \times 10^{-8}/10^{-26})} \cong 2.3 \times 10^{77}$, and these generate the above curvature radius $R$.

### The Higgs field as GW(alias axion)

The mass is $m_{H-g} = 9.4 \times 10^{-41} \text{kg} \rightarrow 5.2 \times 10^{-14} \text{GeV}$, or

$\varepsilon_{H-g} = \varepsilon_{\mu BH-H} / a_{end} = 5.2 \times 10^{-14} \text{GeV} \rightarrow 8.4 \times 10^{-24} \text{J}$, $\varepsilon_{\mu BH-g} = 10^{-23} \text{J}$; $k_{end} = k_{leave} e^{-N}$; $k_{end}^{-1} = 8.4 \times 10^{24} [m]$, $a_{end} = 1.2$, $\lambda_c = 3.6 \times 10^{-3} [m]$, with eq. (1) $R = 2 \times 10^{27} m$, and $H_{end}^{-1} = 10^{25} [m]$, $t_{end} = H_{end}^{-1} / c = 3.3 \times 10^{16} s$, $H_{leave}^{-1} = k_{leave}^{-1} = H_H^{-1} = 10^{-2} [m]$ as had been obtained above, we found $N = 62$ to match the iterations cycle:

$m_g \rightarrow \hbar v \rightarrow k_B T \rightarrow \varepsilon \rightarrow R \rightarrow H_{end}^{-1} \rightarrow a_{end} \rightarrow N$.

The number of gravitons that has been released following the interaction (merging) of Higgs's $\mu$BHs with any other particle is $n_g = m_H c^2 / 10^{-26} = 4.3 \times 10^{18}$. The number of gravitons that has been released following these $\mu$BHs merging is only

$n_{at-merging} = \frac{n_g}{(\varepsilon_{H-g}/10^{-26})} = \frac{4.3 \times 10^{18}}{(8.4 \times 10^{-24}/10^{-26})} \cong 5 \times 10^{15}$, and these generate the above curvature radius, or this interaction is propagated everywhere till the Universe limit $\cong 10^{26} [m]$.

## 3. A Correspondence with the recently SM*A*S*H model

Moreover, the SM cannot generate the primordial ination needed to solve the horizon and atness problems of the Universe, as well as to explain the statistically isotropic, Gaussian and nearly scale invariant fluctuations of the CMB.

The SM also lacks enough CP violation to explain why the Universe contains a larger fraction of baryonic matter than of antimatter.

Aside from these three problems at the interface between particle physics and cosmology, the SM suffers from a variety of intrinsic naturalness issues. In particular, the neutrino masses are disparagingly smaller than any physics scale in the SM and, similarly, the strong CP problem states that the $\theta$ parameter of quantum chromodynamics (QCD) is constrained from measurements of the neutron electric dipole moment to lie below an unexpectedly small value.

In the new model [26], [27], [29], [30] the solution pointing to a unique new physics scale around a scalar $\sigma = 10^{11} \text{GeV}$ at Reheating, identically with our case where are generated $\mu$BHs by quantum oscillations, that by decaying becomes two Weyl fermions $Q$ and $\bar{Q}$ in the $3$ and $\bar{3}$ representations of SU(3)c and with charges $-1/3$ and $1/3$ under U(1)Y. These charges, $Q$ and $\bar{Q}$ can decay into SM quarks, other fermions etc., as in my model.
Thus, with the graviton in my model, alias axion in SMASH model, that of an energy
$\epsilon_g = \hbar c / a_{end} = 2 \times 10^{-26} \text{ J (when it born !)}$; when $a_{leave} = k_{leave} / H_{leave} = 1$, of mass
$M_U = 10^{-43} \text{ kg}$;

$\lambda_C = \hbar c / m_{\mu BH} c^2$; and with $\epsilon_{\mu BH} = 10^{11} / a_{EW} = 10^{11} \text{ GeV} \rightarrow 17 \text{ J}$ alias $\sigma$ from SMASH, it
results with $k_{leave}^{-1} = H_{leave}^{-1} = 10^{-27} [m]$ by iteration for $N = 16.2$; in eq. (2);

$k_{end} = 9.2 \times 10^{19} [m^{-1}]$; $a_{end-ee} = 0.92$; $H^{-1} = 10^{-20} [m]$; $t = 3.3 \times 10^{-29} s$;

$R = 6.5 \times 10^{-20} [m]$; $l_C = 1.7 \times 10^{-27} [m]$; $a_{EW} = 0.92$; where the remnant number of gravitons into a quark $n_g = 0.9$, and the number of gravitons that has been released following two Weyl fermions of quarks tubes, viewed by ours, as $\mu$BHs merging is only

$n_{at-merging} = \frac{n_g}{(\epsilon_{\mu BH} / 10^{-26})} = \frac{0.8}{(17 / 10^{-26})} \approx 5 \times 10^{-28}$, and these generate the curvature radius inside the nucleon $R = 10^{-20}$, which equilibrate the Coulomb field of quarks, thus diminishing the dipole moment (the strong CP problem), and preventing the collapsing due of the gluons field, as in the following.

Independently, the author in [1c] has obtained the same quantization condition in case of the calculation of nucleons structure.

Thus, the ratio of the forces of gravity between two gravitational charges $Gm_{Planck}^2 = \hbar c$
embedded in quarks together with photons (see the electron structure [1]), and of
electromagnetic force between the vortex and the quark flux tube (electric field), see
figure 1., becomes:

for $\lambda \approx r$

$$
\frac{F_G}{F_{gluon-condensate}} = \frac{\frac{\hbar c}{r^2}}{\frac{ec\pi\hbar c}{\pi ec\lambda^2}} = 1
$$

, where the Lorentz force is $F_{M-\text{condensate}} = ecB$; from above

$$
\begin{gather*}
B|_{x\ll\lambda} = \frac{2\Phi_0}{2\pi\lambda^2 c} \\
\Phi_0 = \pi\hbar c/e \rightarrow \text{usually} \quad \frac{\pi\hbar}{e} = 2.07e-15[Tm^2]
\end{gather*}
$$

If we consider only an electric Coulomb field of the quark dipole in the middle of the gluons condensate, we obtain another very important result, namely, the value of fine structure constant $\alpha_s$:

$$
\frac{F_G}{F_{q-dipole}} = \frac{\frac{\hbar c}{r^2}}{\frac{e^2}{4\pi\epsilon_0 r^2}} = \alpha_s = 137
$$
In other words, it result that the nuclear force, in fact, is due of the gravitational charges embedded in quarks, which by theirs interaction are emitted gravitons continuously at a very small rate (only ≈ 10⁻²⁸, see above) sufficiently to deform the spacetime (≈ 10⁻²⁰ m, see before), finally impeding the electric and the electromagnetically respectively, collapsing.

An experimental prove of ours unifying model is the LGO measurement of gravitational waves, where, to estimate the gravitational waves emitted by the binary BHs is used the Newman-Penrose scalar $\Psi_4$, it can be shown that one of these scalars- $\Psi_4$ in the appropriate frame—encodes the outgoing gravitational radiation of an asymptotically flat system.

The identification of the Weyl scalar $\Psi_4$ with the gravitational radiation content of the spacetime is a result of the peeling theorem, which states that in an appropriate frame the component of the curvature has the slowest fall-off with radius, O(1/r).

To estimate the total energy E emitted in gravitational waves, is used the following formula [28].

$$ \frac{dE}{dt} = \frac{R^2}{4} \int p d\Omega \quad \text{and} \quad p = \int_0^t \Psi_4 dt \cdot \int_0^t \bar{\Psi}_4 dt, $$

where $\bar{\Psi}_4$ is the complex conjugate of $\Psi_4$, and the surface integrated over is a sphere of constant coordinate radius R (in uncompactified coordinates).

## 4. Conclusions

Accounting that the mass of graviton as been of $\approx \hbar c/a = 10^{-26} J$ (with the scale factor $a = 1$ at horizon entry-at Reheating), the merging of the primordial Micro-black-holes generated as due of Quantum fluctuation at Electroweak epoch, it conducts to a number ≈ $10^{96}$ gravitons which determines the mass ($10^{70} J$) and the Universe expansion, and these becomes $e^+e^-$ quarks pairs which can generate at Confinement epoch inside the nucleons an electrical field ($E$) that condensate the free photons, also as resulting from the radiating of the Micro-black-holes, but as gluons of near the same number (≈ $10^{96}$), that representing the component of the magnetic field ($B$).

The further Universe expansion is assured by the gravitons released during $\mu$BHs merging with the formation of BH's nuclei of galaxies, that fraction deriving from the same number of the initially $\mu$BH.

In this way it is entirely confirmed the timeline of Universe.

A correspondence of our unifying model with the recent SM*A*S*H it was done, with the difference that the gravitons were proved in LIGO experiment, despite the axions particles not yet.

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