3. Results

The main results of applying the Navier Stokes equation to the atomic nucleus, which behaves like an incompressible nuclear fluid, are:


#### 3.1 The nuclear force and the Navier Stokes force are related

Firstly, we will use the concepts of Classic Mechanics and the formulation of the Yukawa potential, <sup>Φ</sup>ð Þ¼ <sup>r</sup> <sup>g</sup> <sup>4</sup>π<sup>r</sup> ð Þ <sup>A</sup> � <sup>1</sup> <sup>e</sup>�μ<sup>r</sup> to find the nuclear force exerted on each nucleon at interior of the atomic core F<sup>N</sup> ¼ �∇Φð Þr . Also, replace the terms of the potential <sup>e</sup>�μ<sup>r</sup> <sup>¼</sup> <sup>1</sup>�<sup>P</sup> <sup>P</sup> and <sup>1</sup> <sup>r</sup> <sup>¼</sup> <sup>μ</sup> <sup>2</sup> P by the respective terms already obtained in Eq. (10).

$$\Phi(r) = \frac{\mathbf{g}(A - \mathbf{1})}{4\pi r} e^{-\mu r} = \frac{\mathbf{g}\mu(A - \mathbf{1})}{8\pi} (\mathbf{1} - P) \tag{32}$$

The general form of the Eq. (32), is a function of ð Þ x, y, z, t .

$$\Phi(r,t) = \frac{\mathbf{g}(A-\mathbf{1})}{4\pi r} \mathbf{e}^{kt-\mu r} = \frac{\mathbf{g}\mu(A-\mathbf{1})}{8\pi} \left(\mathbf{1} - P(r,t)\right) \tag{33}$$

Secondly, we will obtain the Navier Stokes force equation given by:

$$\frac{d\mathbf{u}}{dt} = \frac{\partial \mathbf{u}}{\partial t} + (\mathbf{u} \cdot \nabla)\mathbf{u} = \frac{\partial \mathbf{u}}{\partial t} + 2\nu^2 \nabla \left(\frac{|\nabla P|^2}{P^2}\right) = -2\mu\nu k P (1 - P) \nabla \mathbf{r} \tag{34}$$

Theorem 9 The Nuclear Force and Navier Stokes Force are proportional inside the atomic nucleus F<sup>N</sup> ¼ CFNS.

Proof. Eq. (34) rigorously demonstrated by theorems and propositions 1 through 8, represent the acceleration of a particle within the atomic nucleus. According to Classical Mechanics the force of Navier Stokes applied to a particle of mass m, would have the form:

Exergy: Mechanical Nuclear Physics Measures Pressure, Viscosity and X-Ray Resonance… DOI: http://dx.doi.org/10.5772/intechopen.95405

$$\mathbf{F}\_{\rm NS} = m \frac{d\mathbf{u}}{dt} = -2m\mu\nu k P(\mathbf{1} - P)\nabla \mathbf{r}.\tag{35}$$

▪ Proof. The nuclear force on its part would be calculated as follows F<sup>N</sup> ¼ �∇Φð Þr :

$$\mathbf{F}\_N = -\nabla \Phi(r) = -\frac{\mathbf{g}\mu}{8\pi}(A-\mathbf{1})\nabla P. \tag{36}$$

Replacing the term <sup>∇</sup><sup>P</sup> <sup>¼</sup> <sup>μ</sup> <sup>P</sup> � <sup>P</sup><sup>2</sup> <sup>∇</sup><sup>r</sup> of Eq. (7), we obtain

$$\mathbf{F}\_N = -\nabla\Phi(r) = -\frac{g\mu^2}{8\pi}(A-1)P(\mathbf{1}-P)\nabla\mathbf{r}.\tag{37}$$

It is possible to write nuclear force as a function of speed.

$$\mathbf{F}\_N = -\nabla \Phi(r) = -\frac{g\mu^2}{8\pi}(A-1)P(1-P)\nabla \mathbf{r}.$$

Finally, we can show that the nuclear force and force of Navier Stokes differ at most in a constant C. Equating (35) and (37), we find the value g as a function of the parameters nuclear viscosity ν, attenuation μ and growth coefficient of the nuclear reaction k, nucleon mass m and C 6¼ 1.

$$\mathbf{g} = \frac{1 \mathfrak{Im}\pi\nu k}{\mu(A-1)} \mathbf{C} \tag{38}$$

▪

#### 3.2 Cross section and Golden ratio <sup>σ</sup><sup>1</sup> σ2 are important elements of the femtoscope

According to NIST and GEANT4 [17], current tabulations of <sup>μ</sup> <sup>ρ</sup> rely heavily on theoretical values for the total cross section per atom, σtot, which is related to <sup>μ</sup> <sup>ρ</sup> by the following equation:

$$\frac{\mu}{\rho} = \frac{\sigma\_{\text{tot}}}{uA} \tag{39}$$

In (Eq. 39), <sup>u</sup> <sup>¼</sup> <sup>1</sup>:<sup>6605402</sup> � <sup>10</sup>�<sup>24</sup>gr is the atomic mass unit (1/12 of the mass of an atom of the nuclide 12C)4.

The attenuation coefficient, photon interaction cross sections and related quantities are functions of the photon energy. The total cross section can be written as the sum over contributions from the principal photon interactions

$$
\sigma\_{\rm tot} = \sigma\_{\rm pt} + \sigma\_{\rm coh} + \sigma\_{\rm incoh} + \sigma\_{\rm trip} + \sigma\_{\rm ph.n} \tag{40}
$$

Where σpe is the atomic photo effect cross section, σcoh and σincoh are the coherent (Rayleigh) and the incoherent (Compton) scattering cross sections, respectively, σpair and σtrip are the cross sections for electron-positron production in the fields of the nucleus and of the atomic electrons, respectively, and σph:<sup>n</sup>. is the photonuclear cross section 3,4.

P rð Þ¼ <sup>≥</sup><sup>0</sup> exp kt

<sup>μ</sup> log 2ð Þ<sup>≤</sup>

where the last equality follows from an application of the L'Hopital's rule

The main results of applying the Navier Stokes equation to the atomic nucleus,

• The Cross Sections in Low energy X ray can explain the Golden Ratio <sup>σ</sup><sup>1</sup>

• The principle of the femtoscope explains that low energy X-rays produce

Firstly, we will use the concepts of Classic Mechanics and the formulation of the

�μ<sup>r</sup> <sup>¼</sup> <sup>g</sup>μð Þ <sup>A</sup> � <sup>1</sup>

kt�μ<sup>r</sup> <sup>¼</sup> <sup>g</sup>μð Þ <sup>A</sup> � <sup>1</sup>

nucleon at interior of the atomic core F<sup>N</sup> ¼ �∇Φð Þr . Also, replace the terms of the

e

e

Secondly, we will obtain the Navier Stokes force equation given by:

<sup>∇</sup> j j <sup>∇</sup><sup>P</sup> <sup>2</sup> P2 !

Theorem 9 The Nuclear Force and Navier Stokes Force are proportional inside the

Proof. Eq. (34) rigorously demonstrated by theorems and propositions 1 through 8, represent the acceleration of a particle within the atomic nucleus. According to Classical Mechanics the force of Navier Stokes applied to a particle of

<sup>4</sup>π<sup>r</sup> ð Þ <sup>A</sup> � <sup>1</sup> <sup>e</sup>�μ<sup>r</sup> to find the nuclear force exerted on each

<sup>2</sup> P by the respective terms already obtained in Eq. (10).

<sup>8</sup><sup>π</sup> ð Þ <sup>1</sup> � <sup>P</sup> (32)

<sup>8</sup><sup>π</sup> ð Þ <sup>1</sup> � P rð Þ , <sup>t</sup> (33)

¼ �2μνkPð Þ 1 � P ∇r (34)

E rð Þ¼ jr≥ 0

appears in the femtoscope.

resonance in layer K.

Yukawa potential, <sup>Φ</sup>ð Þ¼ <sup>r</sup> <sup>g</sup>

<sup>P</sup> and <sup>1</sup>

<sup>r</sup> <sup>¼</sup> <sup>μ</sup>

<sup>Φ</sup>ð Þ¼ <sup>r</sup>, <sup>t</sup> g Að Þ � <sup>1</sup>

<sup>þ</sup> ð Þ <sup>u</sup>:<sup>∇</sup> <sup>u</sup> <sup>¼</sup> <sup>∂</sup><sup>u</sup>

<sup>Φ</sup>ð Þ¼ <sup>r</sup> g Að Þ � <sup>1</sup> 4πr

The general form of the Eq. (32), is a function of ð Þ x, y, z, t .

4πr

∂t <sup>þ</sup> <sup>2</sup>ν<sup>2</sup>

potential <sup>e</sup>�μ<sup>r</sup> <sup>¼</sup> <sup>1</sup>�<sup>P</sup>

du dt <sup>¼</sup> <sup>∂</sup><sup>u</sup> ∂t

34

atomic nucleus F<sup>N</sup> ¼ CFNS.

mass m, would have the form:

P rð Þ¼ <sup>≥</sup><sup>0</sup> lim <sup>t</sup>!<sup>∞</sup> exp kt

Nuclear Materials

3. Results

1 P rð Þ ≥0

which behaves like an incompressible nuclear fluid, are:

• The nuclear force and the Navier Stokes force are related.

• Navier Stokes Equation and Cross Section in Nuclear Physics.

3.1 The nuclear force and the Navier Stokes force are related

2

1 þ exp kt

2

<sup>1</sup><sup>þ</sup> exp kt <sup>¼</sup> 1. ▪

<sup>μ</sup> log 2ð Þ for all <sup>t</sup><sup>≥</sup> 0 (31)

σ2 � �

that

We use data of NIST and simulations with GEANT4 for elements Z ¼ 11 to <sup>Z</sup> <sup>¼</sup> 92 and photon energies 1:<sup>0721</sup> � <sup>10</sup>�<sup>3</sup> MeV to 1:<sup>16</sup> � <sup>10</sup>�<sup>1</sup> MeV, and have been calculated according to:

$$\frac{\mu}{\rho} = \left(\sigma\_{pe} + \sigma\_{coh} + \sigma\_{incoh} + \sigma\_{trip} + \sigma\_{ph.n}\right) / \mu A \tag{41}$$

The attenuation coefficient μ of a low energy electron beam 10, 100 ½ �eV will essentially have the elastic and inelastic components. It despises Bremsstrahlung emission and Positron annihilation.

$$
\sigma\_{\text{tot}} = \sigma\_{\text{coh}} + \sigma\_{\text{incoh}} \tag{42}
$$

We note that left hand side of Eqs. (46) and (47) should have a factor larger than

<sup>¼</sup> <sup>a</sup> <sup>σ</sup><sup>1</sup> σ2 <sup>b</sup>

> σ1 σ2 <sup>2</sup>:<sup>5031</sup>

After performing some simulations it is shown that the thermal a represents the

¼ R<sup>∞</sup>

We use σ1ð Þ Z , σ2ð Þ Z for represent cross section in resonance, R<sup>∞</sup> is the generalized Rydberg constant for all elements of periodic table, and Z atomic number. Last equation with a <sup>R</sup><sup>2</sup> <sup>¼</sup> <sup>0</sup>:9935 was demonstrated and constructed using the

¼ 0:0021Z þ 0:0696

<sup>Z</sup><sup>2</sup> � <sup>0</sup>:0003<sup>Z</sup> <sup>þ</sup> <sup>0</sup>:<sup>004</sup>

This equation was obtained with a <sup>R</sup><sup>2</sup> <sup>¼</sup> <sup>0</sup>:9939, and indicates that the ratio of

This equation obtained for a <sup>R</sup><sup>2</sup> <sup>¼</sup> <sup>0</sup>:9996, complements the system of equations

The radius of the neutron can be obtained using Eq. (49) in the following way.

<sup>N</sup> <sup>r</sup> <sup>þ</sup>

, t≥0 This P is the general solution of the Navier Stokes 3D equations, which satisfies the conditions (50) and (51), allowing to analyze the dynamics of an

Where, u∈ <sup>3</sup> an known velocity vector, ρ<sup>0</sup> constant density of fluid and

Z <sup>N</sup> rp

<sup>P</sup> , where P xð Þ , y, z, t is the logistic

, t≥0 (50)

<sup>1</sup>þekt�μ<sup>r</sup> , <sup>r</sup> <sup>¼</sup> <sup>x</sup><sup>2</sup> <sup>þ</sup> <sup>y</sup><sup>2</sup> <sup>þ</sup> <sup>z</sup><sup>2</sup> ð Þ<sup>1</sup>=<sup>2</sup> defined in

ð Þ <sup>x</sup>, <sup>y</sup>, <sup>z</sup> <sup>∈</sup> <sup>3</sup>

that allow to know the simulation values as a function of Z for σ1ð Þ Z , σ2ð Þ Z and <sup>E</sup><sup>∗</sup> ð Þ <sup>Z</sup> , where <sup>σ</sup>1ð Þ <sup>Z</sup> <sup>&</sup>lt;σ2ð Þ <sup>Z</sup> : The Femtoscope equations further demonstrate that energy <sup>E</sup><sup>∗</sup> ð Þ <sup>¼</sup> min <sup>E</sup> is minimum and Shannon entropy <sup>S</sup> <sup>∗</sup> ð Þ <sup>¼</sup> max <sup>S</sup> is maximum

> <sup>r</sup> <sup>¼</sup> <sup>R</sup><sup>∞</sup> 8000πa

rN <sup>¼</sup> <sup>A</sup>

3.4 Navier Stokes equation and cross section in nuclear physics

ρ0

σ2 <sup>b</sup>

.

Where

(48)

(49)

one due to resonance. The unique factor that holds this requirement is <sup>σ</sup><sup>1</sup>

Exergy: Mechanical Nuclear Physics Measures Pressure, Viscosity and X-Ray Resonance…

8πrλ ð Þ σ<sup>2</sup> � σ<sup>1</sup>

dimensionless Rydberg constant <sup>a</sup> <sup>¼</sup> <sup>R</sup><sup>∞</sup> <sup>¼</sup> <sup>1</sup>:<sup>0973731568539</sup> <sup>∗</sup> 107

elements of the solution of the Navier Stokes equations.

σ1 σ2 

<sup>E</sup><sup>∗</sup> <sup>¼</sup> <sup>2</sup> <sup>∗</sup> <sup>10</sup>�<sup>5</sup>

in resonance, because in equilibrium σ1ð Þ¼ Zi σ2ð Þ Zi .

The speed needs to be defined as <sup>u</sup> ¼ �2<sup>ν</sup> <sup>∇</sup><sup>P</sup>

<sup>þ</sup> ð Þ <sup>u</sup>:<sup>∇</sup> <sup>u</sup> ¼ � <sup>∇</sup><sup>p</sup>

probability function P xð Þ¼ , <sup>y</sup>, <sup>x</sup>, <sup>t</sup> <sup>1</sup>

∂u ∂t

ð Þ <sup>x</sup>, <sup>y</sup>, <sup>z</sup> <sup>∈</sup> <sup>3</sup>

incompressible fluid.

pressure p ¼ p0P∈ .

37

the effective sections fully explain each element of the periodic table.

8000πrλ ð Þ σ<sup>2</sup> � σ<sup>1</sup>

a, b constants.

DOI: http://dx.doi.org/10.5772/intechopen.95405
