# How the Gravitons Release Determine the Extension of Spacetime Deformation for the Cosmic Objects (Stars, Galaxies, Planets etc.)

Stefan Mehedinteanu

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# How the Gravitons Release Determines the Extension of Spacetime Deformation for the Cosmic Objects (Stars, Galaxies, Planets etc.).

Stefan Mehedinteanu¹

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

August 18 2016

## Abstract

The Micro-black-holes ($\mu$BHs) particles produced at the horizon entry by quantum fluctuation at Reheating stay at the base of the origin and evolution of Universe. These adjust theirs dimensions by merging when it appears a gravitons release which it deforms the spacetime. Also, it is accompanied by the radiation of the pre-captured photons but kipping someone else, together with the remnant gravitons, finally obtaining particles like electrons, others leptons (quarks), and uncharged particles (neutrinos, dark matter particles, Higgs field).

During the formation of cosmic objects like planets, stars and galaxies as by merging of these particles viewed as $\mu$BHs that means a release of theirs few remnant gravitons it appears a central black-hole (singularity). That gravitons as a gravitational wave (GW), in fact, they deform the spacetime around the objects, where it “fall” others objects, that form systems where the inertia movement of objects maintains the equilibrium.

## 1. Introduction

There are three things which wait to be elucidated of very long time: namely, the black holes (BH) existence and structure, the gravitation, and how is created the Universe. Following the LIGO experiment, partially these things for the first time become more clearly, thus, firstly the existence of BHs, that theirs merging generates the gravitational waves (gravitons), that only these gravitons deform the space-time.

Also, a question persists, namely, how results the dimension of all the cosmic objects, from particles to stars, planets, galaxies, and sure the Universe itself?

In my recently works [1], [2], [3] is proposed that the key of these is the graviton viewed as gravitational charge quantized as $\sqrt{GM}_{Planck} = \hbar c$ and which participate as the inside constituent of any fundamental particle. For example, the electron (quark) 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 $\sqrt{GM}_{Planck}$. In effect, this particular model would states that the electron is in fact just a form of "trapped light", for that a collision $e^+ + e^- \rightarrow 2\gamma$.
The next step it was when we consider the generation of the Virtual Micro-black-holes
pairs (μ BH), which occur with a number density of approximately one per Quantum
bubble, that means one per quantum bubble volume, n<sub>P</sub> = d<sub>H</sub><sup>-3</sup> ≅ 10<sup>61</sup>[m<sup>-3</sup>]; where

Hubble constant $d_{H}^{-} \approx 10^{-20} m$, at Electroweak epoch (EW) when it is considered.

Theoretical arguments suggest that virtual black holes should have mass on the order of
the Planck particles, each one having a graviton number $n_g = \varepsilon_p / (\hbar c \lambda_c^{-1}) = 10^{35}$; where
the energy of the Planck particle is $\varepsilon_p = 10^9 J$; the Compton length is $\lambda_c = \hbar/m_g c \approx 1$;
and the mass of one graviton is $m_g = M_p / 10^{35} = 10^{-43} \text{kg}$. Therefore, per total in

Universe are created $n_g^U = 10^{61} \cdot 10^{35} \cong 10^{96}$, so, the total energy of Universe is

$\epsilon_u = 10^{-26} \cdot 10^{96} = 10^{70} J$, that corresponds with others model of Universe evolution.

Now, my main supposition is that despite the elimination of μBH by instant evaporation
(being of very small size) as has been considered in unanimity, in fact, these μBHs
merge in pairs with release of gravitons as gravitational wave (GW) that deforms the
space-time, after that, by capturing photons these remaining as matter particles [3]. Thus,
in the cited works above, this hypothesis is verified by being calculated the entirely
timeline of Universe, and the LIGO experiment.

Therefore, it seems that there are two fundamental units: the photon and the graviton,

both involving the quantum module $\hbar$, with energies $\varepsilon_{\gamma} = \hbar\nu = \hbar\frac{c}{\lambda}$, and $\varepsilon_g = \hbar c / \kappa a$,

where, $\kappa$ is the wavelength of GW just like $\lambda$ for light wave, see below, and $a$ is the scale factor along the evolution, $\kappa a$ is a kind of decay length of the gravitons after they are released, as for example by a merging mechanism of the primordially micro black holes in quarks tubes of the hadrons. All others ones particles being structures of these two [3]. In our vision the long range interaction between these particles from the structure of an object it means the gravitons release as GW which deforms the space-time, thus establishing the size (the extension) of the object. In the following, one shows how can find the size of cosmic objects, by applying this principle.

## 2. The primordially micro-black holes clustering into central like BH of cosmic objects

In Inflation models [1], 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

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

$d \ln k = -dN(\phi)$ as from [1], and therefore $\ln(k_{\text{end}}/k) = N(\phi)$, or, $k_{\text{end}} = k e^{N[m]}$ where $k_{\text{end}}$ is the scale leaving the horizon at the end of slow-roll inflation, or usually

$k^{-1} \ll k_{\text{end}}^{-1}[m]$, the correct equation being $k = k_{\text{end}} e^{N[m^{-1}]}$. 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 $L = ds \approx a = a_{end}$, so the volume is given by:

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

In reduced-circumference polar coordinates the spatial metric has the form

$$ds_3^2 = \frac{dr^2}{1 - kr^2} + r^2 d\Omega^2; \quad d\Omega^2 = d\theta^2 + \sin^2\theta \cdot d\phi^2$$

$k$ is a constant representing the curvature of the space.

There are two common unit conventions: $k$ may be taken to have units of length$^{-2}$, in which case $r$ has units of length and $a(t)$ is unitless. $k$ is then the Gaussian curvature of the space at the time when $a(t) = 1$. $r$ is sometimes called the reduced circumference because it is equal to the measured circumference of a circle (at that value of $r$), centered at the origin, divided by $2\pi$. Where appropriate, $a(t)$ is often chosen to equal 1 in the present cosmological era, so that measures commoving distance.

Alternatively, $k$ may be taken to belong to the set $\{-1,0,+1\}$ (for negative, zero, and positive curvature respectively). Then $r$ is unitless and $a(t)$ has units of length. When $k = \pm 1$, $a(t)$ is the radius of curvature of the space, and may also be written $R(t)$. (www.astro.yale.edu/vdbosch/astro610_lecture2.pdf)

To estimate the horizon entry we use some derivations done in [1].

During Universe evolution, 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:

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

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

$\rho [kg/m^3]$; 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:

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

During Universe evolution at Electroweak epoch or Reheating due of the quantum fluctuations [1] a huge number of the micro-black holes as Planck particles $n_p$ are generated, $m_{\mu BH} = m_p$; their energy being at horizon leave; $\varepsilon_g = \hbar c/a_{end} = 2 \times 10^{-26} J$; when $a_{leave} = k_{leave} / H_{leave} = 1$, $M_U = 2.2 \times 10^{53} kg$;

$\lambda_C = \hbar c/m_{\mu BH} c^2$; $n_{\mu BH} \approx 1/H_{-1}^3 \cdot V_{available} \approx n_p \approx 10^{61-62}$, or $n_p = E_U / \varepsilon_p = 10^{70} J / 10^9 J$

with $\varepsilon_{EW} = \varepsilon_{\mu BH} / a_{end-EW} = 10^{11} \rightarrow 19J$, $\varepsilon_{\mu BH} = 10^{19} GeV \rightarrow 10^9 J$, it results with

$k_{leave}^{-1} = H_{leave}^{-1} = 10^{-27} [m]$ by iteration for $N = 2.5$; in eq. (2); $k_{end}^{-1} = 1.2 \times 10^{-26} [m]$;

$H^{-1} \approx 10^{-20} [m]$; $t = 3.3 \times 10^{-27} s$; $R = 5 \times 10^{-20} [m]$; $l_C = H_{leave}^{-1} = 10^{-27} [m]$;

$a_{EW} = 8.2 \times 10^7$; 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-black holes created in Universe as the Planck particles during theirs merging (in pairs) that deforms the spacetime.

### Important notice

“At singularity ($10^{-27} \div 10^{-20} m$) as following inflation theory when just are “born” the $\mu$ BHs in pairs and later by the trapping the thermal photons at Electro-Weak Interactions epoch, see [1], [3], here is not consume any graviton (not yet merging- the denominator being zero in eq. (1)), so, the added spacetime deformation is near zero. Just after $\mu$ BHs generation, it appears theirs dividing into a lot of fractions which become as following: a fraction which will becomes matter particles (hadrons) via Quarks-Gluons-Plasma (EWi), see [1], other for bosons ($W^\pm, Z, \text{etc.}$) and neutrinos, other reserved for leptons (electrons), see [1], other reserved for special electrically neutral particles [ (without photons trapping that that determines the electrical charges [3]) (I call these “projectiles”), see [4], the same for dark matter particles, see [1]. The pairs which not merge (so, into perfect equilibrium) go directly into Quantum Vacuum, from where
these could be extracted (mainly by a Schwinger effect) as the real particles, that, by applying different external fields, like in case of experimental collisions (e-e, pp-p (LHC etc.), or instantly as $W^±$ in case of $β$-decay, see [1] as for the free neutrons decay).

The number of gravitons that has been released following $\mu$BHs merging is only

$$n_{at-merging} = \frac{n_g}{(\varepsilon_{EW}/10^{-26})} = \frac{10^{96}}{(19/10^{-26})} \approx 5.1 \times 10^{68}, \text{ and these generate the curvature radius of Universe } R \approx 10^{-20}[m], \text{ or just } d_H = 10^{-20}[m].$$

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

$$n_H^{EW} = 10^{68}/2n_p = 2.6 \times 10^6, \text{ therefore, the available number of gravitons per each } \mu BH \\ \text{is } \approx 3.8 \times 10^{28}. \text{ To note that at singularity } \varepsilon_{EW} = \varepsilon_p = 10^9, \text{ so the consumed is near zero} \\ n_H^{EW} = 10^{61}/2n_p = 0.05.$$

To mention that only this data set match the model.

For the others epochs see the results done in [1]

During the calculation of the Universe timeline based on merging of the primordially micro-black holes ($\mu BH$) with release of theirs inside gravitons which deform the space-time, is obtained the epoch of neutral particles production, where it was obtained the energy for a pure one graviton of $\mu BH$ as from [3],

$$\varepsilon_g = \hbar c / \lambda_c \approx 10^{-26} J; \text{ where } \lambda_c = \hbar / m_g c \approx 3 \text{ m with } m_g = 10^{-43} \text{ kg.}$$

## 2.1 The like BH formation in the center of the cosmic objects

### The particles package-or the mass defect

It is known that the binding energy of nucleons in nuclei is due of short range of nuclear forces, and that it is reduced by Coulomb rejection between protons and others effects like the surface and pairing etc. It is found that the mass defect in fact is due, also, of gravitons release during nucleons packing in nuclei.

For demonstration I will take the example of nuclear fission of $^{235}$U, respectively,

$$^{235}_{92}U + ^{1}_{0}n \rightarrow ^{141}_{56}Ba + ^{92}_{36}Kr + 3^{1}_{0}n, \text{ where the masses are: } ^{235}U = 235.043\text{u}; ^{141}Ba = 140.91\text{u}, \\ ^{92}Kr = 91.92\text{u}; ^{1}n = 1.674927 \times 10^{-27} \text{kg}, \text{lu} = 1.650339 \times 10^{-27} \text{kg}$$

Thus, the difference in the above reaction as the mass defect is

$$\Delta m = 3.896 \times 10^{-25} - (2.3257 \times 10^{-25} + 1.517 \times 10^{-25} + 3 \times 1.674927 \times 10^{-27}) = 3.02789 \times 10^{-28} \text{kg} \rightarrow 169 \text{MeV}$$

Now, we argument that this difference is due of the release of few gravitons during the

$$\text{disruptive nuclei, respectively: } \Delta m_g = 3 \times 10^{-26} \frac{1}{r_n} \rightarrow 215 \text{MeV}; r_n = 0.87 \text{fm}.$$

This demonstrates that the packing of nucleons into nuclei, and in general the packing of particles into objects it means an interaction(merging) of theirs $\mu$BHs with release of GW that propagate in all the Universe.

### Hydrogen made with muons reveals proton size puzzle

A measurement that's off by 7 standard deviations may hint at new physics [5].
But a team of researchers, working at a particle accelerator in Switzerland, has found a different way of measuring the proton's size: put a muon—a heavy, unstable, relative of the electron—in orbit around a proton. The resulting atom, called muonic hydrogen, can be measured during the brief time it exists before the muon decays.

They carry an identical charge but are 207 times heavier than their more familiar cousins. They also typically decay in about $2 \times 10^{-6}$ seconds. Still, if everything is timed perfectly, it's possible to put one into orbit around a proton, creating an atom of muonic hydrogen (also called $\mu$P).

The muon is about 200 times heavier ($m_\mu = 105 MeV$) than the electron, and in an atomic bound state it lies about 200 times closer to the proton than the electron, so proton size effects are greatly magnified.

In the fractions of a second before the muon decays, the properties of these atoms can be measured. The team has focused on tracking the energy involved in shifting the muon between different orbitals around the proton, using that to infer the charge radius of the proton.

Their first attempt showed something strange: the value for the radius they got was significantly smaller than the one obtained when you measure using an electron.

Remember, the muon and the electron should be equivalent, so there should be no difference. Currently, we have no physics that could explain the difference, but, I will try to explain in a different way.

Thus, the Coulomb energy is $E_C = \frac{q^2}{2 \cdot 4\pi \varepsilon_0 r}$, for $q = 1.6 \times 10^{-19} C$, and with the Bohr

radius $r = 5.29 \times 10^{-11} [m]$ is obtained $E_C = 13.4 eV$. Now, if we take into account the new type of interaction as based by a graviton release during the merging between the $\mu$BHs of muon and that of the proton, see in the following, we have the interaction energy $E_g = n_g \cdot \hbar c/r$, for the same Bohr radius we obtain $E_g = 3540 eV$. In order to obtain the same orbit energy the radius it needs to be reduced to $r = 2 \times 10^{-13} m$,

or about of $\frac{3540}{13.4} = 264$, or $\frac{5.29 \times 10^{-11}}{2 \times 10^{-13}} = 265$, that is approximately equally with

$\frac{m_\mu}{m_p} = 210$. In others words the GW generated during the interaction (merging) muon-proton pushes the spacetime, that needs to be balanced by the Coulomb force.

### The electrons production

After interaction of the $\mu$BH of orbit electron with proton are released gravitons as GW of mass $m_{e-g} = 1.3 \times 10^{-40} kg \rightarrow 7 \times 10^{-14} GeV$, or

$\varepsilon_{e-g} = 10^{-23}/a_{end} = 7.3 \times 10^{-14} GeV \rightarrow 1.1 \times 10^{-23} J$, $k_{end} = k_{leave}e^{-N}$; $k_{end}^{-1} = 10^{26} [m]$,

$a_{end} = 0.83$, $\lambda_C = 2.5 \times 10^{-3} [m]$, with eq. (1) $R = 2.6 \times 10^{28} m$, and $H_{end}^{-1} = 10^{26} [m]$,

$t_{end} = H_{end}^{-1}/c = 3 \times 10^{17} s$, $H_{leave}^{-1} = k_{leave}^{-1} = H_{born}^{-1} = 10^{10} [m]$ as has been obtained in [1], we found $N = 37$ 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 of proton and the orbit electron is only $n_g = M_e c^2 / 10^{-26} = 9 \times 10^{12}$. The number of gravitons that has been released following these µBHs merging is only

$$n_{\text{at- merging}} = \frac{n_g}{(\varepsilon_{\mu BH} / 10^{-26})} = \frac{9 \times 10^{12}}{(1.1 \times 10^{-23} / 10^{-26})} \approx 7.7 \times 10^9, \text{ and these generate the above curvature radius of the object R.}$$

### The case of muon

Now, the horizon-entry at EW when is born the muon during Universe evolution from a fraction of µBHs, is of the wave length $k_{end} = k_{leave}e^{-N}$; $k_{end}^{-1} = 1.2[m]$, the scale factor arrives at $a_{end} = k_{end}/H_{end}$, the frequency is $\nu = c/k_{end}^{-1} = 2.9 \times 10^8 \text{Hz}$, and the Compton length $\lambda_C = \hbar/m_{\text{muon\_BH}}c = 1.7 \times 10^{-15}[m]$, $m_\mu = 1.9 \times 10^{-28} \text{kg} \rightarrow 0.1 \text{GeV}$, or $\varepsilon_{\mu-g} = \varepsilon_{\mu BH-EW}/a_{end} = 0.1 \text{GeV} \rightarrow 1.7 \times 10^{-11} \text{J}$, $\varepsilon_{\mu BH-EW} = 8900 \text{GeV}$, $a_{end-e} = 8.3 \times 10^4$, with eq. (2) $R = 10^5 \text{m}$, and $H_{end}^{-1} = 10^5[m]$, $t_{end} = H_{end}^{-1}/c = 3.3 \times 10^{-4} \text{s}$, $H_{leave}^{-1} = k_{leave}^{-1} = H_{EW}^{-1} = 10^{-7}[m]$ as has been obtained in [1], we found $N = 16.3$ 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 had been released following µBHs merging of the proton and the orbit electron is only $n_g = m_\mu c^2 / 10^{-26} = 1.7 \times 10^{15}$. The number of gravitons that had been released following these µBHs merging is only

$$n_{\text{at- merging}} = \frac{n_g}{(\varepsilon_{\mu-g} / 10^{-26})} = \frac{1.7 \times 10^{15}}{(1.7 \times 10^{-11} / 10^{-26})} \approx 1, \text{ and these generate the above curvature radius of the object R.}$$

To note we can obtain the same results if we replace $\varepsilon_{\mu BH-EW} = 8900 \text{GeV}$ with $\varepsilon_\mu = 0.1 \text{GeV}$, and $H_{leave}^{-1} = k_{leave}^{-1} = H_{EW}^{-1} = 10^{-7}[m]$ to $H_{leave}^{-1} = k_{leave}^{-1} = H_{\mu}^{-1} = 10^{-15}[m]$, or that can happen in laboratory conditions as for example by transformations from others particles, similarly is obtained for electrons.

After interaction of the µBHs of orbit muon with proton are released gravitons as GW of mass $m_{\mu-g} = 7.8 \times 10^{-41} \text{kg} \rightarrow 4.4 \times 10^{-14} \text{GeV}$, or

$$\begin{gathered} \varepsilon_{\mu-g} = 10^{-23}/a_{end} = 4.4 \times 10^{-14} \text{GeV} \rightarrow 7 \times 10^{-24} \text{J}, \quad k_{end} = k_{leave}e^{-N}; \quad k_{end}^{-1} = 7 \times 10^{25}[m], \\ a_{end-e} = 1.43, \quad \lambda_C = 4.3 \times 10^{-3}[m], \text{ with eq. (1) } R = 9.5 \times 10^{28} \text{m}, \text{ and } H_{end}^{-1} = 10^{26}[m], \\ t_{end} = H_{end}^{-1}/c = 3.3 \times 10^{17} \text{s}, \quad H_{leave}^{-1} = k_{leave}^{-1} = H_{born}^{-1} = 10^{5}[m] \text{ as obtained above, we found } N = 48 \text{ to match the iterations cycle: } m_g \rightarrow \hbar\nu \rightarrow k_B T \rightarrow \varepsilon \rightarrow R \rightarrow H_{end}^{-1} \rightarrow a_{end} \rightarrow N. \\ \text{The number of gravitons that has been released following } \mu\text{BHs} \text{ merging of proton and the orbit muon, that is only } n_g = m_{\mu-g}c^2/10^{-26} = 2.3 \times 10^{15}. \text{ The number of gravitons that has been released following these } \mu\text{BHs} \text{ merging is only} \end{gathered}$$
$$n_{\text{at- merging}} = \frac{n_g}{(\varepsilon_{\mu-g} / 10^{-26})} = \frac{2.3 \times 10^{15}}{(7 \times 10^{-24} / 10^{-26})} \approx 3.3 \times 10^{12}, \text{ and these generate the above}$$

curvature radius of the object R, that can be assimilated with the influence radius.

We can say that every object after interaction (merging) with another it means a generation of a GW that propagates in the entirely Universe, but, it deforms the spacetime of very neglectable value (≈ $10^{-55}[m]$) as it was obtained if we apply eq. (3).

In other words we are entourne of GW.

## 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

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

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

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

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

The number of gravitons that has been released following $\mu$BHs 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 $n_{\text{at- merging}} = \frac{n_g}{(\varepsilon_{H-g} / 10^{-26})} = \frac{10^{96}}{4.3 \times 10^{-8} / 10^{-26}} \approx 2.3 \times 10^{77}$, and these generate the above curvature radius R.

## The Higgs field as GW

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

$$\varepsilon_{H-g} = \varepsilon_{\mu BH-H} / a_{end} = 5.2 \times 10^{-14} \text{GeV} \rightarrow 8.4 \times 10^{-24} \text{J}, \quad \varepsilon_{\mu BH-g} = 10^{-23} \text{J}; \quad k_{end} = k_{leave} e^{-N};$$

$$k_{end}^{-1} = 8.4 \times 10^{24} [m], \quad a_{end} = 1.2, \quad \lambda_c = 3.6 \times 10^{-3} [m], \text{ with eq. (1) } R = 2 \times 10^{27} m, \text{ and}$$

$$H_{end}^{-1} = 10^{25} [m], \quad t_{end} = H_{end}^{-1}/c = 3.3 \times 10^{16} \text{s}, \quad H_{leave}^{-1} = k_{leave}^{-1} = H_H^{-1} = 10^{-2} [m] \text{ as had been obtained above, we found } N = 62 \text{ to match the iterations cycle:}$$

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

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

$$n_{\text{at- merging}} = \frac{n_g}{(\varepsilon_{H-g} / 10^{-26})} = \frac{4.3 \times 10^{18}}{(8.4 \times 10^{-24} / 10^{-26})} \approx 5 \times 10^{15}, \text{ and these generate the above}$$

curvature radius, or this interaction is propagated everywhere till the Universe limit ≈ $10^{26}[m]$.

## The case of Earth
The number of gravitons in case of Earth with a central like BH (when the mass is entirely outside of its event horizon) is $n_{g\_BH} = U/10^{-26} = 3.2 \times 10^{58}$;

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

The number of gravitons that has been released following µBHs (nucleons) merging (interaction-called nuclear force) is only

$$n_{\text{at-merging}} = \frac{n_g}{(\varepsilon_{\mu BH}/10^{-26})} = \frac{3.2 \times 10^{58}}{(3.6 \times 10^{-10}/10^{-26})} \approx 8.8 \times 10^{41}, \text{ and that generate the curvature radius of the object (} \approx r_{\text{Schw}}\text{).}$$

Now, the horizon-entry is when the wave length $k_{end} = k_{leave}e^{-N}$; $k_{end\_hadrons}^{-1} = 0.0319[m]$, 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 $\lambda_{C-g0} = \hbar/m_{nucleon\_BH}c = 8.2 \times 10^{-17}[m]$, for $\varepsilon_{nucleons} \approx 10^{-10} \text{ J}$ 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 $a_{end\_BH} \approx 1$, and from eq. (2) $H_{end}^{-1} \approx r_{Sch} \approx 8.8 \times 10^{-3}[m]$, $t_{end} = H_{end}^{-1}/c = 2.9 \times 10^{-11} \text{ s}$, 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 $n_{at-merging}$, $R = H_{end}^{-1} = 5.3 \times 10^{-3}[m] \approx r_{Sch}$.

When the wavelength ($k^{-1}[m]$) is large compared to the Hubble length ($H^{-1}[m]$), as $k_{end\_hadrons}^{-1} = 0.0319 > H_{end}^{-1} = 8.8 \times 10^{-3}[m]$, 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, like in the case of CMBR.

## The gravitons release from Earth as a gravitational wave
### Case a)-near the Earth Surface

Therefore, the new horizon leave as gravitational wave (GW) is just when the gravitons escape from the Schwarzschild radius of the central BH. : $a_{leave} = k_{leave}/H_{leave} = 1$, 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 $a_{end\_GW} = 5.3 \times 10^{3}$, $t_{end} = H_{end}^{-1}/c = 0.02 \text{ s}$, we found $N = 12$ to match the iterations cycle: $m_g \rightarrow \hbar\nu \rightarrow k_B T \rightarrow \varepsilon \rightarrow R \rightarrow H_{end}^{-1} \rightarrow a_{end} \rightarrow N$.

For $\varepsilon_{gBH} = \hbar c/a_{end} \cdot \kappa_{end\_hadrons}^{-1} \approx 10^{-23} \text{ J}$, here $a_{end} \approx 1$ and the wave number are chosen when the hadrons there are born $-\kappa_{end\_hadrons}^{-1} = 2.6 \times 10^{-2} [m]$.
The energy of the graviton becomes at an eventually detector (like LIGO) with the above value at merging $\epsilon_{g\,GW} = \frac{\epsilon_{g\,BH}}{a_{end\,GW}} = \frac{10^{-23}}{5.3 \times 10^3} = 1.9 \times 10^{-27} [J] \rightarrow 1.1 \times 10^{-17} \text{GeV}$, 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 $n_g = M_{\text{Earth}}c^2/10^{-26} = 5.3 \times 10^6$, the total energy initially released is $E_{GW} = E_{mBH} = \epsilon_g \cdot n_g = M_{\text{Earth}}c^2 = 5.3 \times 10^{41} [J]$, 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 $R_{\text{Earth}} = 6.5 \times 10^6 [m]$.

### The strain at near the 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:

$$ \left( \frac{r_{Schw}}{R} \right)^2 = \frac{4\pi}{3} \frac{G \cdot \epsilon_g \cdot n_g \cdot r_{Schw}^2}{c^4 \cdot a_{end\,GW}^3} = 1.8 \times 10^{-18} \quad (3) $$

Separately,

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

$$ \frac{r_{Schw}^2}{R_{\text{Earth}}^2} = 1.9 \times 10^{-18} \rightarrow r_{Schw} / R_{\text{Earth}} = 1.3 \times 10^{-9} $$

, so, in both cases the strain is around $\theta = r_{Schw}/R \approx 1.3 \times 10^{-9}$, that is near equally with strain as light bending.

### Case b)-at Moon Orbit

Therefore, the new horizon leave as gravitational wave (GW) is just when the gravitons escape from the Schwarzschild radius (the Object is viewed as an inverse big black hole): $a_{leave} = k_{leave}/H_{leave} = 1$, or $k_{leave}^{-1} = H_{leave}^{-1} = r_{Schw} = 8.8 \times 10^{-3} [m]$.

Now, the new horizon-entry is when the wave length $k_{end} = k_{leave} e^{-N}$;

$k_{end} = 1.9 \times 10^{-4} [m^{-1}]$, $k_{end}^{-1} = 5.3 \times 10^3 [m]$; and the scale arrives factor at $a_{end} = k_{end}/H_{end}$, the Hubble length with Compton length

$\lambda_{c\,GW} = \hbar/m_{g\,GW}c = 217 [m]$; from eq. (1) $H_{end}^{-1} = R = 3.2 \times 10^8 [m]$ equally with Moon Orbit to Earth; it results $a_{end\,GW} = 7.2 \times 10^4$, $t_{end} = H_{end}^{-1}/c = 1.3s$, we found $N = 13.3$ to match the iterations cycle: $m_g \rightarrow \hbar\nu \rightarrow k_B T \rightarrow \epsilon \rightarrow R \rightarrow H_{end}^{-1} \rightarrow a_{end} \rightarrow N$.

For $\epsilon_{g\,BH} = \hbar c/a_{end} \cdot \kappa_{end\_hadrons}^{-1} \approx 10^{-23} J$, here $a_{end} \approx 1$ and the wave number are chosen when the hadrons there are born $-\kappa_{end\_hadrons}^{-1} = 2.6 \times 10^{-2} [m]$.
The energy of the graviton becomes at an eventually detector (like LIGO) with the above value at merging $\epsilon_{g\,GW} = \frac{\epsilon_{gBH}}{a_{end\,GW}} = \frac{10^{-23}}{7.2 \times 10^3} = 1.4 \times 10^{-28} [J] \rightarrow 8.6 \times 10^{-19} \text{GeV}$, and the frequency is $\nu = c/k_{end}^{-1} = 5.6 \times 10^4 \text{Hz}$, the mass is $m_{g\,GW} = 1.5 \times 10^{-45} \text{kg}$; the number of particles released as the gravitational wave (like the photons of the light wave) remains equally with the above value $n_g = M_{\text{Earth}}c^2/10^{-26} = 5.3 \times 10^{67}$, the total energy initially released is $E_{GW} = E_{mBH} = \epsilon_g \cdot n_g = M_{\text{Earth}}c^2 = 5.3 \times 10^{41} [J]$, 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 $R_{\text{Earth\,Moon}} = 3.2 \times 10^8 [m]$. With these data in eq. (3) the strain at Moon orbit becomes $\theta = r_{Schw}/R \approx 2.7 \times 10^{-11}$.

## The case of Sun

The number of gravitons in case of Sun as like BH is $n_{g\,BH} = U/10^{-26} = 3.7 \times 10^{67}$;

$$U = \frac{GM_{\text{Earth}}^2}{R_{\text{Earth}}} = 3.6 \times 10^{41} \text{J}$$

The number of gravitons that has been released following µBHs (nucleons) merging is only $n_{at-merging} = \frac{n_g}{(\epsilon_{µBH}/10^{-26})} = \frac{3.7 \times 10^{67}}{(3.6 \times 10^{-12}/10^{-26})} \approx 10^{53}$, and that generate the curvature radius of the object ($\approx r_{Schw}$).

Now, the horizon-entry is when the wave length $k_{end} = k_{leave}e^{-N}$; $k_{end}^{-1} = 10^2 [m]$, the scale factor arrives at $a_{end} = k_{end}/H_{end}$, the frequency is $\nu = c/k_{end}^{-1} = 2.8 \times 10^6 \text{Hz}$, and the Compton length $\lambda_{C-g0} = \hbar/m_{\text{nucleon\,BH}}c = 8.3 \times 10^{-15} [m]$, for $\epsilon_{\text{nucleons}} \approx 10^{-10} \text{J}$ it results $\epsilon_p = \epsilon_{µBH} = \epsilon_{\text{nucleon}}/a_{end} = 10^{-10}/a_{end} = 3.6 \times 10^{-12} \text{J} \rightarrow 0.022 \text{GeV}$; it results $a_{end\,BH} \approx 28$, and from eq. (2) $H_{end}^{-1} \approx r_{Sch} \approx 2.9 \times 10^3 [m]$, $t_{end} = H_{end}^{-1}/c = 9.8 \times 10^{-6} \text{s}$, we found $N = 43.8$ to match the iterations cycle: $m_g \rightarrow \hbar\nu \rightarrow k_B T \rightarrow \epsilon \rightarrow R \rightarrow H_{end}^{-1} \rightarrow a_{end} \rightarrow N$.

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

When the wavelength ($k^{-1}[m]$) is comparable 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, like in the case of CMBR.

## The gravitons release from Sun as a gravitational wave

Therefore, the new horizon leave as gravitational wave (GW) is just when the gravitons escape from the Schwarzschild radius (the Object is a viewed as an inverse big black hole): $a_{leave} = k_{leave}/H_{leave} = 1$, or $k_{leave}^{-1} = H_{leave}^{-1} = r_{Schw} = 2.9 \times 10^3 [m]$.
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)

$H_{end}^{-1} = R = 4.5 \times 10^{12} [m]$ that corresponds to Neptune orbit; it results $a_{end\_GW} = 1.5 \times 10^{9}$, $t_{end} = H_{end}^{-1}/c = 1.5 \times 10^{4} s$, we found $N = 0.01$ to match the iterations cycle:

$m_g \rightarrow \hbar v \rightarrow k_B T \rightarrow \epsilon \rightarrow R \rightarrow H_{end}^{-1} \rightarrow a_{end} \rightarrow N$, but not greater that it means that this Neptune orbit is a limit.

For $\epsilon_{gBH} = \hbar c / a_{end} \cdot \kappa_{end\_hadrons}^{-1} \approx 10^{-23} J$, here $a_{end} \approx 1$ and the wave number are chosen when the hadrons there are born $-\kappa_{end\_hadrons}^{-1} = 2.6 \times 10^{-2} [m]$.

The energy of the graviton becomes at an eventually spatially located detector (like LIGO) with the above value at merging

$\epsilon_{gGW} = \frac{\epsilon_{gBH}}{a_{end\_GW}} = \frac{10^{-23}}{1.5 \times 10^{9}} = 6.6 \times 10^{-33} [J] \rightarrow 4.1 \times 10^{-23} \text{GeV}$, and the frequency is

$v = c/k_{end}^{-1} = 10^{5} \text{Hz}$, the mass is $m_{g\_GW} = 7.3 \times 10^{-50} \text{kg}$; $\lambda_{C} = 4.5 \times 10^{6} \text{m}$, the number of particles released as the gravitational wave (like the photons of the light wave) remains equally with the above value $n_g = M_{\text{Sun}}c^2 / 10^{-26} = 1.7 \times 10^{73}$, the total energy initially released is $E_{GW} = E_{mBH} = \epsilon_g \cdot n_g = M_{\text{Earth}}c^2 = 1.7 \times 10^{47} [J]$, 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 $R_{\text{Sun\_system}} = 1.7 \times 10^{12} [m]$.

### The strain at Sun-system (Neptune orbit) edge

Now, based on eq. (1) we can derive for the G-wave effect in the deformation (strain) of the space-time between Sun-system and a detector site by using the gravitational pressure due of gravity charges on the area of Schwarzschild radius $r_{Schw}$, we have:

$$ \left( \frac{r_{Schw}}{R} \right)^2 = \frac{4\pi}{3} \frac{G \cdot \epsilon_g \cdot n_g \cdot r_{Schw}^2}{c^4 \cdot a_{end\_GW}^3} = 2.8 \times 10^{-18} $$

Separately,

$$ r_{Schw} = \frac{2G \cdot M_{\text{Sun}}}{c^2} = 2.9 \times 10^3 [m], \text{ we have} $$

$$ \frac{r_{Schw}^2}{R_{\text{Sun-system}}^2} = 4.2 \times 10^{-19} \rightarrow r_{Schw} / R_{\text{Earth}} = 6.5 \times 10^{-10} $$

, so, in both cases the strain is around $\theta = r_{Schw}/R \approx 1.7 \times 10^{-9}$, that is near equally with strain as light bending. To note that at Sun surface it results with eq. (3)

$$ \theta = r_{Schw}/R \approx 4.2 \times 10^{-6} $$

### The case of Milky Way galaxy
The number of gravitons in case of Earth as like BH is $n_{g\_BH} = U/10^{-26} = 1.1 \times 10^{76}$;

$$U = \frac{GM_{\text{Earth}}^2}{R_{\text{Earth}}} = 1.1 \times 10^{50} \text{ J}$$

The number of gravitons that has been released following µBHs (nucleons) merging is only $n_{at-merging} = \frac{n_g}{(\varepsilon_{\mu BH}/10^{-26})} = \frac{1.1 \times 10^{76}}{(2.2 \times 10^{-12}/10^{-26})} \approx 5 \times 10^{61}$, and that generate the curvature radius of the object ($\approx r_{Schw}$).

Now, the horizon-entry is when the wave length $k_{end} = k_{leave}e^{-N}$; $k_{end} = 1.2 \times 10^{-5} [\text{m}^{-1}]$, the scale factor arrives at $a_{end} = k_{end}/H_{end}$, the frequency is $\nu = c/k_{end}^{-1} = 3.5 \times 10^3 \text{ Hz}$, and the Compton length $\lambda_{C-g0} = \hbar/m_{nucleon\_BH}c = 1.3 \times 10^{-14} [\text{m}]$, for $\varepsilon_{nucleons} \approx 10^{-10} \text{ J}$ it results $\varepsilon_P = \varepsilon_{\mu BH} = \varepsilon_{nucleon}/a_{end} = 10^{-10}/a_{end} = 2.2 \times 10^{-12} \rightarrow 0.014 \text{GeV}$; it results $a_{end\_BH} \approx 44$, and from eq. (2) $H_{end}^{-1} \approx r_{Sch} \approx 3.8 \times 10^6 [\text{m}]$,

$$r_{Sch} = \frac{2GM_{MWay\_Sag\_BH}}{c^2} = 3.8 \times 10^6 [\text{m}], \text{ where the mass of CGIRS Be-the BH of}$$

Sagitarus is $M_{MWay\_Sag\_BH} = 1300 \times M_{Sun}$; $t_{end} = H_{end}^{-1}/c = 0.012 \text{ s}$, we found $N = 50.5$ 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 are released gravitons as $n_{at-merging}$, $R = H_{end}^{-1} = 3.8 \times 10^6 [\text{m}] \approx r_{Sch}$. When the wavelength ($k^{-1}[\text{m}]$) is comparable to the Hubble length ($H^{-1}[\text{m}]$), 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, like in the case of CMBR.

## The gravitons release from the BH of Milky Way as a gravitational wave

Therefore, the new horizon leave as gravitational wave (GW) is just when the gravitons escape from the Schwarzschild radius: $a_{leave} = k_{leave}/H_{leave} = 1$, or

$$k_{leave}^{-1} = H_{leave}^{-1} = r_{Schw} = 3.8 \times 10^6 [\text{m}]$$

Now, the new horizon-entry is when the wave length $k_{end} = k_{leave}e^{-N}$; $k_{end} = 2.6 \times 10^{-7} [\text{m}^{-1}]$, $k_{end}^{-1} = 3.8 \times 10^6 [\text{m}]$; and the scale arrives factor at $a_{end} = k_{end}/H_{end}$, $\nu = c/k_{end}^{-1} = 77 \text{Hz}$, the Hubble length with Compton length is $\lambda_{C-GW} = \hbar/m_{g\_Sag\_BH}c = 7.3 \times 10^{11} [\text{m}]$; For $\varepsilon_{gBH} = \hbar c/a_{end} \cdot \kappa_{end\_hadrons}^{-1} \approx 10^{-23} \text{ J}$, here $a_{end} \approx 1$ and the wave number are chosen when the hadrons there are born - $\kappa_{end\_hadrons}^{-1} = 2.6 \times 10^{-2} [\text{m}]$, it results

$$\varepsilon_P = \varepsilon_{\mu BH} = \varepsilon_g / a_{end} = 10^{-23} / a_{end} = 4 \times 10^{-38} \rightarrow 2.5 \times 10^{-28} \text{ GeV}; \text{ in eq. (1) we use the}$$
$$\text{known Milky Way Diameter: } H_{end}^{-1} = 9.4 \times 10^{20} [\text{m}]; \text{ it results } a_{end\_GW} = 2.4 \times 10^{14},$$
$t_{end} = H_{end}^{-1}/c = 3 \times 10^{12} s$, we found $N = 0.01$ to match the iterations cycle:

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

The mass is $m_{g\_GW} = 4.5 \times 10^{-55} \text{kg}$; the number of particles released as the gravitational wave (like the photons of the light wave) remains equally with the above value

$$n_g = M_{\text{MWay\_Sag}} c^2 / 10^{-26} = 2.3 \times 10^{76}, \text{ the total energy initially released is}$$

$$E_{GW} = E_{mBH} = \epsilon_g \cdot n_g = M_{\text{MWay\_Sag}} c^2 = 2.3 \times 10^{50} [\text{J}], \text{ and the curvature radius it results}$$

from eq.(1) with the integral (which the starts at the outside of the $r_{Schw}$) graviton energy with $n_g$, and with $a_{end} = a_{end\_GW}$, it results $R_{Milky\_Way} = 3 \times 10^{18} \approx 9.4 \times 10^{20} [m]$.

### The strain at Milky Way galaxy 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:

$$\left( \frac{r_{Schw}}{R} \right)^2 = \frac{4\pi G \cdot \epsilon_g \cdot n_g \cdot r_{Schw}^2}{3 c^4 \cdot a_{end\_GW}^3} = 1.5 \times 10^{-24}$$

Separately,

$$r_{Schw} = \frac{2G \cdot M_{\text{MWay\_Sag}}}{c^2} = 3.8 \times 10^6 [\text{m}], \text{ we have}$$

$$\frac{r_{Schw}^2}{R_{Milky\_Way}^2} = 1.5 \times 10^{-24} \rightarrow r_{Schw} / R_{Milky\_Way} = 1.2 \times 10^{-12}$$

, so, in both cases the strain is around $\theta = r_{Schw}/R \approx 1.2 \times 10^{-12}$, that is near equally with strain as light bending or a lensing effect from light as coming from others galaxies.

The strains around objects which form this galaxy it need to be higher than $1.2 \times 10^{-12}$, that it is proved, thus: Earth-Moon $2.7 \times 10^{-11}$; the solar system $1.7 \times 10^{-9}$, all that explaining theirs location in this galaxy.

## 3. Conclusions

Again it is proved that the mechanism of merging of $\mu$BHs as applied to the calculation of timeline of Universe an for LIGO experiment is equally applicable to determine the extension of spacetime deformation around cosmic objects. Also, it results that each cosmic object has a central primordial BH due of merging of the matter particles that constitute the objects, releasing gravitons (GW). Since, the wavelength ($k^{-1}[m]$) is greater or comparable 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*, like in the case of CMBR.
It was demonstrated that the packing of nucleons into nuclei, and in general the packing of particles into objects it means an interaction( merging) of theirs µBHs with release of GW that propagate in all the Universe.

1. Stefan Mehedinteanu. The Timeline of Universe Based On the Gravitons Release Due of the Merging of the Primordial Micro-black Holes. 2016. <hal-01316347v4>, https://hal.archives-ouvertes.fr/hal-01316347v4,

2. Stefan Mehedinteanu, The Quantization of the Gravitational Charges At the Origin of the Gravitational Waves From a Binary Black Hole Merger (LIGO), May 2016, <hal-01316349v2>, https://hal.archives-ouvertes.fr/hal-01316349v2

3. Stefan Mehedinteanu. What Is Gravity and How Is Embedded in Mater Particles giving them mass (not the Higgs field)? 2016. <hal-01316929v5>
https://hal.archives-ouvertes.fr/hal-01316929v5.

4. Stefan Mehedinteanu. The Interaction of the Neutral Particles as the Merged Primordial Micro-black Holes With The Quarks of 7Be Can Explains the Cosmological Li Problem. 2016. <hal-01343771>, https://hal.archives-ouvertes.fr/hal-01343771.

5. Carl E. Carlson, The Proton Radius Puzzle, arXiv:1502.05314v1 [hep-ph] Feb 2015.