# What Is Gravity and How Is Embedded in Mater Particles giving them mass (not the Higgs field)?

Stefan Mehedinteanu

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Stefan Mehedinteanu. What Is Gravity and How Is Embedded in Mater Particles giving them mass (not the Higgs field)?. 2017. hal-01316929v6

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# What Is Gravity (Graviton) and How Is Embedded in Mater Particles Giving Them Mass (Not the Higgs Field)?

Stefan Mehedinteanu¹

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

> "When Einstein was asked about the discovery of new particles, the answer it was: firstly to clarify what is with the electron!"

## Abstract

It will be shown how the Micro-black-holes particles produced at the horizon entry into a number of $N \approx 10^{61-62}$ by quantum fluctuation as virtual micro-black holes pairs like $e^+$-$e^-$ creation, stay at the base of: the origin and evolution of Universe, the Black Holes (BH) nuclei of galaxies, of the free photons creation of near mass-less as by radiation decay that condensate later at Confinement into the structure of gauge bosons (gluons). Also, these ex-Micro-black-holes ($\mu$BHs) by merging release gravitons which deform the space-time, and also decay of all free photons as BH's nuclei of galaxies, or if kipping someone else, as electrons, others leptons (quarks) and uncharged particles (neutrinos). After radiation decay, the heritage of gravitational charge (gravitons) quantized as $\sqrt{GM}_{Planck} = \hbar c$ inside (or on event horizon) into equilibrium with the inner field like electrostatic Coulomb field as being generated by the few photons remained, finally give the mass. Thus, is explained how the gravity (only few gravitons remaining at Confinement in nucleons) is embedded in matter particles, and in the BH's nuclei of galaxies of gravitational charge only (gravitons), an old waited answer.

Also, in this context it results that an equal part of the ex-Micro-black-holes are generated at horizon entry that correspond to Electroweak period during Universe evolution ($10^{11}$ GeV; $10^{-29}$ s). A proof of the model is done by applying it to light bending due of Earth. It results that the graviton is first unit (indivisible) of Universe.

A very stranger result is obtained if we divide the gravitational charge $\hbar c$ to the Compton length for every particle, or in other words the gravitational energy of the particle distributed in their Compton length is just the particle energy or the "rest of mass".

**Keywords: Origin of gravity, Planck particle, gravitational charge, virtual black holes, free photons, graviton, color magnetic charges, gluons, photons mass, electrons**

## 1.Introduction

Independently and for the first time, in my work [1c], I got the idea that a nucleon has inside a field in equilibrium with the gravity charge. This match with the idea of the electron which is in fact a sphere with a very small radius and inside this is distributed the momentum of photon energy in equilibrium with gravitational charge $\sqrt{GM}_{Planck}$ as it was presented in [1d;1e]. In effect, this particular model would state that the electron is in fact just a form of "trapped light". In this paper, I have taken the theory further with
numerical examples, not only to reconcile the original idea's that the classical (or
perhaps) semi-classical electron has a mass of electromagnetic nature and that the
gravitational charge (the inertial mass) is in fact intrinsically-related to the same
electromagnetic features. To do this, "I have had to read for many years now, the theories
of times past which involved the photon configuration inside of particles "- Lloyd Motz
[1e]; which he was the first to propose a gravitational charge to a particle and even
speculated on bound photon particles in a type of orbital motion; the original idea's
brought forth in this paper is how to think of the gravitational charge in terms of the
electromagnetic field and we will also study the implications of certain equations under
the same investigative field. To make short, the electron is taken in this paper, as a
fluctuation of either bound or single photons following toroidal or other topological paths
in a dense curved spacetime. To finish off, we will also study what it means to talk about
the spin of an electron.

I will show that, only few free photons of photons dimension could remain embedded in
ex-like-Micro-black-holes which it adjusts its dimensions to electrons, other leptons,
noncharged particles (neutrinos, Higgs) in order to equilibrate the gravitational charge
$\sqrt{GM}_{Planck}$ as it was explained before following electron model [7]. Also we shown how
is generated a huge number ($10^{61-62}$) into each Quantum bubble of one Micro-black-
holes (virtual micro black holes) as by Quantum fluctuations. In [5b] is interpreted in the
topological fluctuations S2 ×S2 bubbles in spacetime foam as virtual
black hole loops. It needs electric or magnetic charges to produce a pair of black holes
from Ernst solution. However the quantum bubbles form even in absence of any field.
The reason for that is that virtual black holes are not classical geometries and hence need
not satisfy Einstein's equations. They are rather solutions of the Wheeler-DeWitt
equation [6] and hence need not have any electric or magnetic charges as would be
required if they satisfied Einstein equations. Virtual black holes lead to the loss of
quantum coherence, which is calculated in this paper [6]. Virtual black holes also lead the
space-time to have an intrinsic entropy. The soft photons radiated during these virtual
micro-black holes inherently decay its being of the same order as the Micro-black-holes,
these are later incorporated as gluons into hadrons. In the same context is considered that
at horizon entry a large part of the Micro-black-holes by merging become BH's nuclei of
galaxies.

After Inflation has ended, scalar perturbations begin the re-enter the horizon and interact
in such a way as to induce gravitational waves. These gravitational waves are due of
gravitons release due of µBHs merging in BHs nuclei of galaxies, and which deform the
space-time.

## 2. The concept of like-Micro-black-holes particle at the origin of Universe

We consider that the firstly Universe beginning is due of Quantum instability fluctuations
when is obtained a lot of Micro-black-holes particles as Planck particles, and which we
will consider its as to be the *primordial seeds* of Universe, in the following we will argue
this hypothesis.

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 $E_U = M_U c^2$):
$$ \frac{GM_U^2}{R} = M_U c^2 \rightarrow G = \frac{c^5}{E_U} t \rightarrow G = \frac{R_U^5}{t^4 E_U} , $$

$R_U = 1.6 \times 10^{26} [m]$, 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 fact as already noted [23] from [1b], a Micro-black-holes mass particle decays via the Bekenstein radiation.

The Maxwell equations in a vacuum with a non zero conductivity coefficient, can be shown to lead to a loss of energy (z-shift) of a photon during its propagation, see [2a]. This is because the dissipating mechanism leads to an extra term in the usual Maxwell's equations proportional to $\partial E / \partial t$, see [1b].

This immense energy $\mu BH$ can constitutes a spectrum for blackbody radiation when photon creation takes place has also been proposed by author in [2a], but now I consider that this radiation is obtained by the decaying of Micro-black-holes ($\mu BH$) viewed as virtual micro black-holes with hair [5a; 5b]; [5c].

## Virtual micro-black holes

The picture of virtual black holes given here also is suggested that macroscopic black holes will evaporate down to the Planck particles ($10^{-35} m$) size and then disappear in the sea of virtual black holes [5b]. However, in his paper Hawking says: "I shall be less concerned with real processes like pair creation, which can occur only when there is an external field to provide the energy, than with virtual processes that should occur even in the vacuum or ground state".

However, in addition to black holes formed by stellar collapse, there might also be much smaller black holes which were formed by density fluctuations in the early universe [5a]. These small black holes, being at a higher temperature, would radiate more than they absorbed. They would therefore presumably decrease in mass. As they got smaller, they would get hotter and so would radiate faster.

In quantum gravity, a **virtual black hole** is a black hole that exists temporarily as a result of a quantum fluctuation of spacetime [5b], [6]. It is an example of **quantum foam** and is the gravitational analog of the **virtual electro-positron pairs** found in quantum electrodynamics. At such small scales of time and space, the Heisenberg uncertainty principle allows energy to briefly decay into particles and antiparticles and then annihilate without violating physical conservation laws. Theoretical arguments suggest that virtual black holes should have mass and the lifetime on the order of the Planck particles, but we have changed to a not so small dimension since not respect the general terms of energy and mass of universe. Therefore, we consider also Micro-black-holes pairs, but that occur with a number density of approximately *one per Quantum bubble*.
[6], initially it was considered per Planck volume [2b], that means one per quantum bubble volume, that means $n_p = d_{H^-}^{-3} \approx 10^{60}[m^{-3}]$; where Hubble constant $d_{H^-} \approx 10^{-20} m$, see below.

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 [2a] to calculate the evolution of Universe. To estimate the horizon entry we use some derivations done in [2a]. 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 $d \ln k = d(\ln(aH)) \approx d \ln a = \frac{\dot{a}dt}{a} = Hdt$. From the definition Eq. (38) of [2] this gives $d \ln k = -dN(\phi)$ as of eq.(46) from [2], 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.

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 $L = ds \approx a = a_{end}$, so the volume is given by:

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

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 [2], 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

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

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

In Newtonian interpretation, the Friedmann equations are equivalent to this pair of equations:
$$ \frac{\dot{a}^2}{2} = \frac{G \frac{4\pi a^3}{3} \rho}{a} $$

$\rho[\text{kg/m}^3]$; energy density $p = -\rho c^2$

If we divide with $a^2$ we obtain for outside the object (BH, Universe, 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:

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

During Universe evolution at Electroweak epoch (EW) 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 $a_{leave} = k_{leave} / H_{leave} = 1$, $a_{end} \approx 1$; $M_U = 2.2 \times 10^{53} \text{ kg}$;

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

First step

$n_p = E_U / \varepsilon_p = 10^{70} \text{ J}/10^9 \text{ J} = 10^{61}$ with $\varepsilon_{\mu BH\_EW} = 10^{19}/a_{end-EW} = 10^{11} \text{ GeV} \rightarrow 17 \text{ J}$, it

results with $k_{leave}^{-1} = H_{leave}^{-1} = 10^{-27} [\text{m}]$ by iteration for $N = 2.5$; in eq. (2);

$k_{end} = 9.2 \times 10^{19} [\text{m}^{-1}]$; $a_{end-EW} = 8.2 \times 10^7$; $H^{-1} = 10^{-18} [\text{m}]$; $t = 3.3 \times 10^{-27} \text{ s}$;

$R = 6.5 \times 10^{-20} [\text{m}]$; $l_C = l_p = 1.5 \times 10^{-27} [\text{m}]$;

Second step

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

$k_{leave}^{-1} = H_{leave}^{-1} = 10^{-27} [\text{m}]$ by iteration for $N = 16.2$; in eq. (2); $k_{end} = 9.2 \times 10^{19} [\text{m}^{-1}]$;

$a_{EW} = 0.92$; $H^{-1} = 10^{-20} [\text{m}]$; $t = 3.3 \times 10^{-29} \text{ s}$; $R = 6.5 \times 10^{-20} [\text{m}]$;

$l_C = l_p = 1.7 \times 10^{-27} [\text{m}]$; $a_{EW} = 0.92$; 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 $n_{at-merging} = \frac{n_g}{(\varepsilon_{\mu BH\_EW}/10^{-26})} = \frac{10^{96}}{(17/10^{-26})} \cong 10^{68}$, 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
$n_H^{EW} = 10^{68}/2n_p = 5 \times 10^5$, for two µBHs merging.

To mention that only this data set match the model.

Electroweak symmetry breaking-quarks epoch

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

$$
\epsilon_{\mu BH\_QGP} = \epsilon_{\mu BH\_EW} / a_{end\_QGP} = 7.9 \times 10^4 \text{GeV} \rightarrow 1.26 \times 10^{-5} \text{J},
$$

$$
\epsilon_{\mu BH\_EW} = 10^9 / a_{end-EW} = 10^{11} \text{GeV} \rightarrow 17 \text{J}; a_{end-EW} = 8.2 \times 10^7; k_{end} = k_{leave} e^{-N};
$$

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

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

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

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

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

$$
n_{at-merging} = \frac{n_g}{\left(\frac{\epsilon_{\mu BH\_QGP}}{a_{end\_QGP}}/10^{-26}\right)} = \frac{10^{96}}{\left(1.26 \times 10^{-5}/10^{-26}\right)} \cong 7.9 \times 10^{74}, \text{ and these generate the curvature radius of the object } R.
$$

In other words the contribution to the space-time deformation is due of
$n_H^{QGP} = 7.9 \times 10^7 / 2n_p = 4 \times 10^{12}$, 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 $n_g = U/10^{-26} = 3.2 \times 10^{58}$;

$$
U = \frac{GM_{\text{Earth}}^2}{R_{\text{Earth}}} = 3.2 \times 10^{32} \text{ J}
$$

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 $k_{end} = k_{leave}e^{-N}$; $k_{end} = 3 \text{ l[m}^{-1}\text{]}$; the
scale factor arrives at $a_{end} = k_{end}/H_{end}$, the frequency is $\nu = c/k_{end}^{-1} = 9.3 \times 10^9 \text{Hz}$, and
the Compton length $\lambda_{C-g0} = \hbar/m_{nucleon\_BH}c = 8.2 \times 10^{-17} [\text{m}]$, for $\varepsilon_{nucleons} \cong 10^{-10} \text{J}$ it
results $\varepsilon_p = \varepsilon_{\mu BH} = \varepsilon_{nucleon}/a_{end} = 10^{-10}/a_{end} = 3.6 \times 10^{-10} \rightarrow 2.2 \text{GeV}$; 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:

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

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

The energy of the graviton becomes at an eventually detector (like LIGO) with the above value at merging $\varepsilon_{g\_GW} = \frac{\varepsilon_{g\_BH}}{a_{end\_GW}} = \frac{10^{-23}}{6.3 \times 10^{17}} = 1.9 \times 10^{-27} [J] \rightarrow 1.1 \times 10^{-17} \text{GeV}$, and the frequency is $\nu = c / k_{end}^{-1} = 2.4 \times 10^{5} \text{Hz}$, the mass is $m_{g\_GW} = 2 \times 10^{-44} \text{kg}$; 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:

$$\left(\frac{r_{Schw}}{R}\right)^2 = \frac{4\pi G \cdot \varepsilon_g \cdot n_g \cdot r_{Schw}^2}{c^4 \cdot a_{end\_GW}^3} = 1.24 \times 10^{-18}$$

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 $\theta = r_{Schw}/R \cong 1.1 \times 10^{-9}$, 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 (at CMBR!) results $R = 10^{21}[m]$,

$t_{end} = H_{end}^{-1}/c \cong R/c = 3.3 \times 10^{12} s$, 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

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

In other words the contribution to the space-time deformation is due of $n_H^{\text{dark}} = 7.4 \times 10^{88}/2n_p = 7.4 \times 10^{26}$, $n_p = 10^{62}$, for two µBHs merging.

Thus, from the overall balance the number of gravitons which are consumed to push the space time is $:= 5 \times 10^5 \cdot 7.4 \times 10^{26} = 3.5 \times 10^{31} \ll n_{\mu BH-p} = 10^{35}$.

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 $3 \times 10^{12} s \cong t_{\text{CMBR}} = 1.13 \times 10^{13} s$, the temperature $\sim 10^4 K$, the wave number of $\sim 8.57 \text{ m}$ (microwave) and the frequency of $10^8 \text{ Hz}$, all correspond to the situation at cosmic microwave background radiation (CMBR) time. Since, at LGO experiment [9], it was observed also a radiation, is advanced the idea that the all-sky map (pattern) of the CMB represents in fact the radiation emitted during µBHs merging (clustering), that is contrary to the thermal radiation at Recombination due of photon decoupling.

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

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

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

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

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

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

$n_g = \frac{10^{35}}{7.4 \times 10^{26}} \cdot \frac{1}{n_{\text{galaxies}}} \cdot \frac{10^{60}}{(\varepsilon_{\text{mgBH}}/10^{-26})} \cong 4.2 \times 10^{51}$, $n_{\text{galaxies}} \cong 10^{11}$, 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 $\frac{10^{62}/10^{11}}{4.2 \times 10^{51}} \cong 4$, or in others words the gravitational potential due of dark matter which deforms the space time is entirely consumed.

The energy released as a gravitational wave which push the space-time is $E_{\text{GW-mgBHs}} = 4.2 \times 10^{51} \times \varepsilon_{\text{mgBH}} = 10^{33} \text{ J} \rightarrow 10^{16} \text{ Kg}$ which is in the range of primordial BH galaxies nuclei $10^{14} \div 10^{23} \text{ kg}$.

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

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

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

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

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:

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

$$\epsilon_g = 2 \times 10^{-26} / a_{end\_BH} = 10^{-26} \text{ J and } n_g = 10^{59}$$

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

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

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

, 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

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.
With $\varepsilon_{\bar{q}q\_g} = \varepsilon_{\mu\bar{B}H\_QGP} / a_{end\_q\bar{q}\_g} = 8.9 \times 10^4 / 3.7 \times 10^3 = 2.1 \text{GeV} \rightarrow 3.4 \times 10^{-10} \text{J}$ at Confinement, where $\varepsilon_{\mu\bar{B}H\_QGP} = \varepsilon_{\mu\bar{B}H\_EW} / a_{end\_QGP} = 7.9 \times 10^4 \text{GeV} \rightarrow 1.26 \times 10^{-5} \text{J}$, and when $k_{end} = k_{leave}e^{-N}$; $k_{end}^{-1} = 0.4[m]$, $a_{end\_\bar{q}q\_g} = 3.7 \times 10^3$, with eq. (2) $H_{end}^{-1} = 150[\text{m}]$, $R = 144[\text{m}]$; $t_{end} = H_{end}^{-1}/c = 5 \times 10^{-6} \text{s}$, and $H_{leave}^{-1} = k_{leave}^{-1} = H_{QGP}^{-1} = 10^{-7}[\text{m}]$ we found $N = 15.2$ to match the iterations cycle:

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

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

$$n_{at-merging} = \frac{n_g}{(\varepsilon_{\bar{q}\bar{q}\_g} / 10^{-26})} = \frac{10^{96}}{(3.4 \times 10^{-10} / 10^{-26})} \approx 2.9 \times 10^{79}, \text{ and these generate the curvature radius of the object } R.$$

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

$$n_H^{q\bar{q}\_g} = 2.9 \times 10^{79} / 2n_p \approx 10^{17}, \text{ for two µBHs merging.}$$

Thus, from the overall balance of µBHs the number of gravitons consumed (via EW $\rightarrow$ QGP $\rightarrow$ QCD) to push the space time is
$$= 1/8 \cdot 5 \times 10^5 \cdot 4 \times 10^{12} \cdot 10^{17} \approx 0.25 \times 10^{35} \approx n_p = 10^{35}.$$

In other words, at Confinement it remains at least few ($\approx 4$) gravitons, so
$$\varepsilon_{nucleon} = 4 \times 10^{-26} / \lambda_c \approx 10^{-10} \text{ J} \approx 0.6 \text{GeV}.$$

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)

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

,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;2.c) at a more deep penetration

$$\lambda_{C-qg} = 3.6 \times 10^{-16} > \lambda_{C-gluon} = 8.86 \times 10^{-17} [\text{m}], \text{ with } m_{gluons\_condensate} = 3.8 \times 10^{-27} \text{Kg} \text{ see below.}$$

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 cromoelectrical flux into a vortex line, followed by forcing an organization into a triangular Abrikosov lattice, see figure 1.

From [2a], we have the lower critical field:

$$B(x) = \frac{2\Phi_0}{2\pi\lambda^2} \log\frac{\lambda}{\xi} = \frac{\pi\hbar c}{\pi\lambda^2 c} \log(\kappa) = 10^{15} \left[ \frac{J}{Am^2} \right], \text{ where } \xi = 0.1114, \text{ and when}$$

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

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

In the case of a homogeneous potential directed along the z-axis [9], the Einstein stress-energy tensor is:
$T^{00} = T^{11} = T^{22} = -T^{33} = \rho_B = \frac{\epsilon_0 c^2 B^2}{8\pi}$; $T^{0i} = 0$, where $\rho_B[J/m^3]$-the magnetic energy density.

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

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

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

$$ \epsilon = \frac{V_{vol} \epsilon_0 c^2 B^2}{8\pi} = \rho_B V_{vol} = V[J], \qquad (2.a) $$

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

$$ E^2 = \frac{(V)}{\epsilon_0 (\lambda_C^{e*})^3} \qquad (2.b) $$

With $V = \epsilon_{gluons}$ as above is obtained $B = E_{q\bar{q}}/c = 2.98 \times 10^{15} [T]$, where $E_{q\bar{q}} = 8.9 \times 10^{23}$ with eq. (2.a), that are identically with the above values, **indubitable** meaning that this force is equilibrated by the gravity charges from quarks tubes, see below.

With equation (4), see below,

$$ \langle E \rangle = \sqrt{\frac{6\hbar c}{\pi \epsilon_0 \lambda^4}} $$

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

### 3. The Quantization of Mass (or Gravitational Charge)

In the classical Abraham-Lorentz theory of the electron, as in references [26, 27, 28] cited in [7], the energy contained in the Coulomb field of a charge $e$ in all space outside its radius $R$ is

$$ U_{elect} = \int_{\vec{r} \ge R} \frac{\epsilon_0}{2} E^2 d\vec{r} = \int_{R}^{\infty} \frac{e^2}{8\pi\epsilon_0 r^2} dr = \frac{e^2}{8\pi\epsilon_0 R} \qquad (3) $$

For a point charge, with $R=0$, the total energy $U_{elect}$ is infinite. The physical mass of the electron, $m_e = U/c^2 = 0.5 \text{MeV}$, then imposes a lower limit on its size of the order of the so-called classical electron radius $r_0 = 2R = 2.82 \times 10^{-15} [m]$.

From [7], our main motivation for the central postulate stated above arises from a consideration of the experimentally well-established (parapositronium) electron-positron annihilation and creation processes as in the here cited reference [36].

$$ e^+ e^- \leftrightarrow \gamma\gamma $$
Otherwise, in all the Feynman diagrams is considered the particles transformation as been accompanied by $\gamma\gamma$ (vortex).

We envisage a quantized solution where, just as is the case for the free photon, we have time varying fields, but where the field distribution is self-confined in space.

The mass of any confined photon will be $m = U/c^2$ where $U = \hbar c/\lambda$ is the energy of the photon of wavelength $\lambda$. From relation $e^+e^- \leftrightarrow \gamma\gamma$ it is clear that for the case where the electron and positron annihilate at rest, the decay photon wavelengths $\lambda$ are just the electron Compton wavelength $\lambda_c = \hbar/m_e c \approx 3.7 \times 10^{-13} [m]$. We therefore, in the first instance, look for a quantized solution defined by periodic boundary conditions of length one Compton wavelength $\lambda_c$, which is confined to some closed path in 3-D space. The magnitude of the apparent charge of our model object is based on the length scales estimated in the previous section. We confine an arbitrary photon with wavelength $\lambda$ to a spherical volume $V = \frac{4}{3}\pi(\frac{\lambda}{2})^3$.

The energy density of the electromagnetic field in the volume is

$$W = \frac{1}{2}(\varepsilon_0|\bar{E}|^2 + \mu_0^{-1}|\bar{B}|^2).$$

For a propagating photon inside the volume, where space is curved, we take $E = cB$ and $c^{-2} = \varepsilon_0\mu_0$ as is the case for a free-space photon.

The electric field energy $U_E$ and the magnetic field energy $U_B$ are then one half of the total confined photon energy $U$ (i.e. $U_E = U_B = 1/2U$). We find for the average energy density of the electric field in the volume $V$, $W_E = U_E/V = \frac{1}{2}U/V$ and also $W = \frac{1}{2}\varepsilon_0 E^2$. The average magnitude of the electric field inside the model electron is then

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

, it results $\langle E \rangle = 5.7 \times 10^{17} [N/C]$, for electrons of $\lambda_c = \hbar/m_e c \approx 3.7 \times 10^{-13} [m]$.

In case of quarks we have

$$\langle E \rangle = 3.7 \times 10^{23} [N/C], \text{ for } m_q = 7.2 \times 10^{-28} \text{ kg} \leftrightarrow 0.4 \text{ GeV},$$

$$\lambda_c \approx \hbar/m_q c \approx 4.6 \times 10^{-16} [m],$$

that corresponds with the value from [2] $\approx 8.8 \times 10^{23} [N/C]$.

To estimate the charge in our model we need to compare the magnitude of the inward directed electric field to that for a point charge at the origin. Making the plausible assumptions that the relevant length scale from where the electric field is effectively inward-directed is the mean radius of energy transport $\bar{r} = \lambda/4\pi$, and that the average electric field of the confined photon, Eq. (3), is a good estimate of the field at this radius, we obtain the effective charge, $q$, by comparing this to the Coulomb field of a point charge at distance $\bar{r} = r$.
$$E = \frac{q}{4\pi\varepsilon_0 r^2}$$

which then yields the charge from our model in terms of the elementary charge $e$

$$q = \frac{1}{2\pi} \sqrt{3\varepsilon_0 \hbar c} \approx 0.91e \quad (5)$$

, where this apparent charge arises from the electric field of the confined photon. Note that $q$ is independent of the energy of the photon (the size of the object) and is a result of the toroidal topology.

The rotational energy of a relativistic object is $U_{rot} = L\omega$, with $L$ the angular momentum, and $\omega$ the angular frequency. For a photon $L = \hbar$, and the total energy of a photon with frequency $\omega$ is $U_{photon} = \hbar\omega$. Thus, the energy of a photon is entirely electromagnetic and contained in its spin. The confined photon in our model has to travel around twice to complete its path of length $\lambda = 2\pi c/\omega$. Consequently, the internal rotational frequency of the model is twice the photon frequency $\omega_s = 2\omega$. The internal rotational energy is equal to the confined photon energy, and we may write

$U_{model} = L\omega_s = \hbar\omega$. Our model must then have an intrinsic angular momentum

$L = \hbar\omega / \omega_s = \frac{1}{2}\hbar$. We see that this describes an object of half-integer spin. If the spin-statistics theorem applies, our self-confined photon should be a fermions. This is again a direct consequence of the topology of our model; the field vectors must rotate through $720^\circ$ before coming back to their starting position with the same orientation. In quantum mechanics, the spin angular momentum has a fixed value $s = \frac{1}{2}$, therefore we cannot take the intrinsic spin to a classical limit by letting $s \to \infty$ and there is no classical correspondence with half-integer spin. In our model this is ensured because, for our topology, we have necessarily one and only one wavelength, and this gives a fixed, length-scale independent value of $s$.

If the electron is indeed constituted by a photon, other elementary particles may also be composed of photon states, but in some other configuration [7]. The possibility that muons and tauons may be formed by electron-like states with a different internal curvature has been discussed in the literature [8]. We speculate that the hadrons may be described by composite confined photon states as gluons [2] together with quarks pairs at the origin of an electrical field and of gravity. If we identify a quark with a confined photon state which is not sufficient in itself to complete a closed loop in space, but transforms a photon going in one spatial direction to one travelling in another, it would then only be possible to build closed three-dimensional loops from these elements with $qqq$ and $q\bar{q}$ combinations.

The gravitational theory here proposed [1e] says that the intense interior gravitational field of the electron is just sufficient to compensate for the repulsive forces in an electron of finite size. It shows that if the electronic charge is pictured as being distributed over a region with a radius equal to the classical radius of electron ($e^2/mc^2$), and the gravitational mass is distributed over a region with a radius of order $\lambda_c = \hbar/mc$, a balance can be achieved between the repulsive force of spin and electric and attractive
gravitational forces. We see from the basic equation of the quantized gravitational charge
see section 3., respectively

$$
G \approx \frac{\hbar c}{M_P^2}, \qquad (6)
$$

that $\sqrt{GM_P^2} \approx (\hbar c)^{1/2}$. Since the gravitational charge is distributed over a region of the order of $\hbar/mc$ it gives a negative energy of the order of

$$
\hbar c / (\hbar / m c) = m c^2
$$

while the electric charge distributed over a region of the order of $e^2/mc^2$ (since

$E = \frac{e^2}{\varepsilon_0 r} = m_e c^2 \rightarrow r = \frac{e^2}{\varepsilon_0 E} = \frac{e^2}{\varepsilon_0 m_e c^2}$) gives a positive self energy of $mc^2$. Since the two self energies are of the same order of magnitude, they be brought into balance. Consider now a photon and suppose that it maintain its identity in virtue of an interior gravitational field. I we picture the gravitational mass $m$ of the photon as being distributed over a volume whose radius is of the order of $\lambda$, the wavelength of the photon, we see that the energy of the photon, in virtue of gravitational mass $m=M_P$, must be of the order of $Gm^2/\lambda$ or $Gm^2v/c \rightarrow 10^9[J]$. But the energy of the photon is also $h\nu \rightarrow \approx 10^9[J]$. Hence we see again that $Gm^2 \approx \hbar c$, that means for $\nu=10^{43} s^{-1}$. The same at confinement in nucleons, $\nu = 6.9 \times 10^{24} Hz$, it results the photons energy $Gm^2v/c \rightarrow 1.8 \times 10^{-10} [J] \rightarrow 2GeV$

**There are Higgs particles?**

At LHC were measured, in fact, two high energy photons, that following my model they
resulting from the disrupturing of quarks (like the electrons) as constituents of the two
collided protons *pp*, respectively *2g* → *2γ*.

Therefore, in the case of Higgs particle at LHC is argued the decay $H \to \gamma\gamma$, but
following us these result as above: $Gm^2\nu/c \to 8.3 \times 10^{-08}[J] \to 112\,GeV$, with
$\nu = 8 \times 10^{26} \,Hz$; and $\hbar\nu = 8 \times 10^{-8}[J]$. In this way we can explain any particles collision
as beginning with $e^+e^- \leftrightarrow \gamma\gamma$ to every Feynman diagrams

A very stranger result is obtained if we divide the gravitational charge to Compton
length of any particle (example Higgs particle) $\lambda_C^H = \hbar/m_H c \rightarrow 2.3 \times 10^{-18} [m]$;

$\epsilon_H = \hbar c / \lambda_C^H \rightarrow 1.25 \times 10^{-8} [J] \rightarrow 125 \, \text{GeV}$, or in other words the energy of particle as to be distributed in their Compton length is just the energy or the "mass".

The neutrino, according to this picture, must be represented as consisting of two photons
bound together gravitationally and revolving around a common center. To account for the
spin of the neutrino, which is a Fermi particle, we need assume that the sum of the orbital
angular momenta of photons equals $\hbar/2$. The two photons must then be revolving such a
way that : 1) their own intrinsic spins (each equal to $\hbar$ are antiparallel , and 2) that the
electromagnetic field of one cancels that of the other. This can always be achieved. A
neutrino consisting of two photons, has, of course, zero rest mass, and therefore this
gravitational of the neutrino accounts for one of its remarkable properties. For other type of neutrinos see [1e].

With above value $T = 2.66 \times 10^{26}$ K ; $t = H^{-1}/c = 3.3 \times 10^{-32}$ s as Beckenstein radiation of $\mu BH$ particle of the like Quantum bubbles (QB), of volume $V = (H^{-1})^3 \approx (10^{-20})^3 [m^{-3}]$ at horizon entry (when the scale factor is $a \approx 2.6$, it results the total number of primordial high energy photons following the “decay” of a like-$\mu BH$ particle and which “fill” the vacuum till are embedded as a Bose Einstein Condensate (B.E.C) in matter (gluons) at symmetries breakings, or when is attained a critical temperature (like in case of superconductors!) as: $n_{\gamma\gamma}^{\text{total}} = 10^{80}$ as above.

The Micro-black-holes decay into photons in $t_p$ is the Hawking radiation where, the gravitational field is so strong that it causes the spontaneous production of photon pairs (with black body energy distribution) and even of particle pairs.

Sure, only few free photons of photons dimension could remain embedded in ex-like-Micro-black-holes which it adjusts its dimensions to electrons, other leptons, noncharged particles (neutrinos, Higgs) in order to equilibrate the gravitational charge $\sqrt{G}m_p$ as it was explained before following electron model ref. [7] cited in [3a].

## The energy of graviton

Very early in 1960 in a seminal paper L.Motz ref. [1d] has elaborated a new theory of the structure of fundamental particles, which introduces gravitational field into the interior of particles such as electrons to account for their stability.

Now, $\sqrt{G}m_p$ is the gravitational charge defined by L. Motz [1d] as resulting from:

$$F_G = \frac{\sqrt{G}m_1 \sqrt{G}m_2}{r^2}, \text{ where it can say that the gravitational charge } q = \sqrt{G}m_1 \text{ is the source of gravitational field } \sqrt{G}m_1/r^2 \text{ at the distance } r, \text{ and that this field is coupled to the gravitational charge } \sqrt{G}m_2 \text{ at the position } r \text{ (relative to the source of the field) via the product of the field strength and the charge. L. Motz derived the quantization condition on gravitational charge in a similar manner by noting that moving particle with velocity } \vec{V} \text{ and with gravitational charge } \sqrt{G}m_p \text{ is coupled not only to the Newtonian gravitational fields of all other particles (the gravielectric field) in the usual way, but also to the Coriolis field } m\omega \times v \text{ divided by gravitational charge } \sqrt{G}m_p \text{ like in gravitoelectromagnetism (GEM) equations (see below), due of the inertial Coriolis force, see figure 2, (defined as } F_C = -2m_p\omega \times v = -mv^2/r \text{), by means of the cross product } (\sqrt{G}m_p\vec{V}/c) \times (2\vec{\omega}\vec{c}/\sqrt{G}) \text{, for } v=c, \text{ see below.}}$$

The gravimagnetic field, in GEM equations, we have: $F_e = q(E+v \times B)$, or equivalently

$F = m(E_g + 4v \times B_g)$; $L = rmv$; the electrical field $E = \frac{q}{4\pi\varepsilon_0 r^2}$; the gravitelectric field
$E_g \cong g = \frac{F}{m} \Leftrightarrow 4\nu \times B_g[ms^{-2}] \rightarrow B_g \equiv \omega[s^{-1}];$ the mass with the charge and the mass

density with the charge density, so, the gravimagnetic field is $E_g = \frac{F_c}{q} = \frac{F_c}{\sqrt{Gm_p}} = \frac{\omega c}{\sqrt{G}}$.

This cross product term give rice to an angular momentum which is proportional to moment of inertia $I$ and angular speed $\omega$, $L = I\omega$. Because $I = r^2m$ for a single particle and $\omega = \nu/r$ for circular motion, angular momentum can be expanded as $L = rmv$, or,

$L = rm_p\nu$ the component in the motion of the particle which is of the order of $L = 2m_p r^2\omega$ and is parallel to the field $\vec{\omega}$.

Therefore, it is obtained the quantization condition as: $L = 2m_p r^2\omega = 2\hbar$.

Since it was interested in the quantization of fundamental gravitational charge, it is found a value for $r$ and $\omega$ that must be associated with such charge. To do this, L.Motz considered the Universe to two such charges in gravitational equilibrium and resolving about each other in the first Bohr orbit. The radius of this orbit is just $r = \hbar^2/Gm_p^3$, which it was taken as to be $r$. In this case (unlike the standard Coriolis force), the force is not fictional, but is due to frame dragging induced by the rotating body when $\omega \leftrightarrow \frac{2GL}{c^2r^3}$.

Since, for a spherical body the Coriolis field is the gravitomagnetic field $B_g$ near a rotating body can be derived from the GEM equations. It is exactly half of the Lense-Thirring precession rate is $B_g = H = \omega = \frac{GL}{2c^2r^3}$, and introducing into Coriolis force

expression $F_c = 2m\omega v = \frac{2mGL}{2c^2r^3}v = \frac{m^2G}{r^2} = F_G$; for $v=c$. He also noted that the resulting Coriolis field $2\omega c/\sqrt{G}$ must, according to Mach's principle, produces the centrifugal force $m\omega^2 c$. "A very general statement of Mach's principle is "*Local physical laws are determined by the large-scale structure of the universe*".

This will be so only if $v = \omega r \cong c$ is of the order of $c$ since the centrifugal force (being charge times field) is just $(\sqrt{Gm}) \cdot (2\omega c/\sqrt{G})$. Introducing these two relationships into the above quantization equation, is obtained $m\frac{v^2}{r}$; with $L = 2m_p c r = 2\hbar$, or, substituting for $r$ as Bohr orbit when the electron is held in a circular orbit by electrostatic attraction, after some manipulations:

$\frac{m \cdot m_e v^2 \cdot r^2}{r^2 \cdot r} = \frac{Z k_e e^2 \cdot m}{r^2}$; $m = m_p$, $k_e = 1/4\pi\epsilon_0$; with

$mvr = n\hbar$, $(mvr)^2 = \hbar^2$; $n=1$; $r = \frac{\hbar^2}{k_e e^2 m}$, $r = \frac{\hbar^2}{k_e e^2 m} = \frac{\hbar^2}{Gm^2 m}$; when $e \rightarrow \sqrt{Gm_p}$ as in GEM. So, introducing in $L$ we have:
$2m_P c \frac{\hbar^2}{Gm_P^3} = \hbar$, or $\frac{Gm_P^2}{c} = \hbar$, the mass quantization (or gravitational charge)

With $m_P = 1.7 \times 10^{-8} \text{ kg}$; $\hbar = 10^{-34} \text{ Hz}$, it results

$r_P = 3.3 \times 10^{-35} m \approx r_{\mu BH} = \lambda_C = \hbar/mc = 2 \times 10^{-35}$, and for the electron

$m_e = 9 \times 10^{-31} \text{ eV}$; $e = 1.6 \times 10^{-19} \text{ C}$ it results $r_e = 5.9 \times 10^{-11} \text{ m}$.

<figure><img src="image_3.png" /><figcaption>Figure 2, Coriolis field H(B)</figcaption></figure>

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 and of electromagnetic between the vortex and the quark flux tube (electric field), see figure 1., becomes

$$ \frac{|F_G|}{|F_{M\_condensate}|} = \frac{\frac{Gm_{Planck}^2}{r^2}}{\frac{ec\pi\hbar c}{\pi ec\lambda^2}} = \frac{G\hbar c}{r^2 G} * \frac{\lambda^2}{\hbar c} = 1 \\ \text{; For } \lambda \approx r \quad (9) $$

$F_{M-\text{condensate}} = ecB \quad (10)$

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

$$ \frac{|F_G|}{|F_{\text{quark}}|} = \frac{\frac{Gm_{\text{Planck}}^2}{r^2}}{\frac{e^2}{4\pi\epsilon_0\lambda^2}} = \frac{G\hbar c}{r^2 G} * \frac{4\pi\epsilon_0\lambda^2}{e^2} = \alpha_s = 137 \quad (11) $$
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 $\alpha_s$:

Therefore, if we consider the string force due to the Coulomb flux tube as given by the colors quarks $\overline{q}q$ pairs, see figure 1., results

$$G = \frac{\alpha_s F_{quarks} \lambda^2}{m_{Planck}^2} \rightarrow \frac{137*1636*3.75e-16^2}{2.2e-08^2} \rightarrow 6.65e-11 \text{m}^3/\text{kg} \cdot \text{s},$$

Or again, $$G = \frac{4\pi\epsilon_0\hbar c\lambda^2}{e^2 m_{Planck}^2} \times \frac{e^2}{4\pi\epsilon_0\lambda^2} = \frac{\hbar c}{m_{Planck}^2}$$

<figure><img src="image_4.png" /><figcaption>Fig.1. The gluons embedded in Giant-Vortex [2] that could be also the arrangement for the nucleon (only illustration). A spin-orbit nonabelian field is shown.</figcaption></figure>

If we accept the quantization condition for the gravitational charge we are led to the value $m \approx 2.2 \times 10^{-8} \text{kg}$ for the inertial mass of the fundamental particle of Universe, which L.Motz shall now call the uniton.

Gamma-gamma interactions can either give up mass in the form of special types of decay processes or, vice versa, the energy can come from antiparticle interactions. This is called the parapositronium decay. Such a phase transition is given as

$$\gamma\gamma \leftrightarrow e^- e^+$$
, and

Now, with data from [2], in case of gluon's condensate of high energy photon

$$\epsilon_{gluon\_condens} \approx 2.18 \text{GeV} \rightarrow 3.48 \times 10^{-10} [\text{J}]$$
, we

have $r_{gluon-condensate} = \lambda_C = \hbar/m_{gluon-condensate}c \rightarrow 8.6 \times 10^{-17} [m]$,

$m_{gluon-condensate} = 3.9 \times 10^{-27} [kg]$ it results the frequency

$\omega = \varepsilon_{gluon\_condens} / \hbar \approx 1.7 \times 10^{24} Hz$.

**It is the graviton the first unity (indivisible) of Universe !**

What it is dimension of a graviton?

To calculate it we use Schwarzschild radius $r_{Schw} = 2Gm_g/c^2 = 1.5 \times 10^{-70} [m]$, for mass of graviton $m_g = 10^{-26}/c^2 = 10^{-43} kg$. Therefore the total dimension of Universe filled by gravitons is $n_g \cdot r_{Schw} = 1.5 \times 10^{26} [m]$ where $n_g = 10^{96}$, which is just the present dimension of Universe, that being an extraordinary result. During all above calculations is used a micro-black hole $\mu BHs$ which contain $n_{g-\mu BH} = \varepsilon_{Planck}/10^{-26} = 10^9/10^{-26} = 10^{35}$, we can verify this as $r_{\mu BH} = 1.5 \times 10^{-70} \times 10^{35} = 1.5 \times 10^{-35}$ or this results to be equally with $r_{Schw-\mu BH} = 2GM_P/c^2 = 2.6 \times 10^{-35} [m]$ with Planck mass $M_P = 1.79 \times 10^{-8} [kg]$.

Therefore, into a Planck particle are still embedded a huge number of such tiny particles ~$10^{35}$, in other words it seems that this *tiny particle graviton* is the fundamental unity (indivisible) of the Universe. Even the graviton have a such infinitely small dimension, then action length (wave length) and the Compton length is much longer.

**4. Cosmological and other consequences of the uniton existence**

If the square of quantum charge is, indeed $\hbar c$, we can account for the stable electrically charged particles such as electrons by balancing the explosive positive electro-static energy with the negative binding energy of the gravitationally charge. Accordingly to this point of view electrons and nucleons are the lowest bound states of two or more *unitons* that collapsed down to the appropriate dimensions gravitationally and radiated anyway most of their energy in the process. It is clear that a gravitational charge of magnitude $\hbar c$ will contribute negative gravitational binding energy of the order of $mc^2$ if this charge is distributed over a region whose dimensions are equal to the Compton wave length of the particle. To note again that each uniton of a total $10^{96}$ it produces by radiation $\approx 10^{96}$ of soft photons as due to hair concept of micro-black-holes [5c], that represent the gluons which again together with quarks condensate at Confinement (V=2.1GeV, ) into nucleons, see more details in [2a].

The binding energy will be sufficient to balance the positive electrostatic energy of the charge $e$ distributed over the classical radius $e^2/4\pi\varepsilon_0m_ec^2$, as from

$$E = \frac{e^2}{4\pi\varepsilon_0 r_e} = m_e c^2 \qquad (14)$$

, it results $r_e = 2.8 \times 10^{-15} [m]$,

**5. Conclusions**
Accounting that the mass of graviton results as been of $\simeq \hbar c/a = 10^{-26} J$ (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 $\simeq 10^{96}$ 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 ($\simeq 10^{96}$), that representing the component of the magnetic field ($B$). The further Universe expansion is assured all the time by the gravitons released during the merging, and finally is accelerated when an another fraction of $\mu$BHs merging (clustering) and later into greater structures as BH's nuclei of galaxies and, that fraction derives from the same number of the initially $\mu$BH.

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

A proof of the model is done by applying it to light bending due of Earth.

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