2. Model

layers of the nuclear fluid. For the same radial velocity profile, the required tension is proportional to the viscosity of the nuclear fluid or its composition given by

Obtaining nuclear viscosity and nuclear pressure from the speed of the neutron particles in the disintegration of chemical element. Figure 1a. indicates that the BE=A ratio is proportional to the nuclear pressure, p<sup>0</sup> represented in Figure 1b. Figure 1c is the graph of the viscosity in equilibrium and out of equilibrium, the viscosity in equilibrium is greater than the viscosity at the moment of nuclear decay. Figure 1d is the average

. The half-lives of radioactive atoms do not have a known

3, where it is evident that the volume of the atomic nucleus

Radioactive decay is a stochastic process, at the level of individual atoms. According to quantum theory, it is impossible to predict when a particular atom will decay, regardless of how long the atom has existed. However, for a collection of atoms of the same type [8], the expected decay rate is characterized in terms of its

upper limit, since it covers a time range of more than 55 orders of magnitude, from

Other characteristics of the proton such as its size have been studied in many institutes such as Max Plank, where it has been measured with high precision ranges

The atomic nucleus is an incompressible fluid, justified by the formula of the

changes linearly with A ¼ Z þ N, giving a density constant [11]. All incompressible fluid and especially the atomic nucleus comply with the Navier Stokes equations. We present a rigorous demonstration on the incomprehensibility of the atomic nucleus, which allows to write explicitly the form of the nuclear force F<sup>N</sup> ¼

<sup>8</sup><sup>π</sup> ð Þ A � 1 Pð Þ 1 � P ∇r, which facilitates the understanding of nuclear decay. The Navier Stokes equations are a problem of the millennium [12, 13], that has not been resolved yet in a generalized manner. We present a solution that logically meets all the requirements established by the Clay Foundation [14, 15]. This solution coherently explains the incompressible nuclear fluid and allows calculations of

The alpha particle is one of the most stable. Therefore it is believed that it can exist as such in the heavy core structure. The kinetic energy typical of the alpha

For our demonstrations, we will use strictly the scheme presented by Fefferman in http://www.claymath.org/millennium-problems [13, 14], where six demonstrations

almost instantaneous to much longer than the age of the universe [8, 9].

rp <sup>¼</sup> <sup>0</sup>:84184 67 ð Þfm , providing new research methods [10, 11].

ð Þ Z, N [6–8].

Figure 1.

Nuclear Materials

decay constant <sup>k</sup> <sup>¼</sup> <sup>1</sup>

nuclear radius, <sup>R</sup> <sup>¼</sup> <sup>1</sup>:2A<sup>1</sup>

� <sup>g</sup>μ<sup>2</sup>

24

T1=<sup>2</sup>

half-time of each isotope, 9≤Z ≤92 and 9≤ N ≤200:

the nuclear viscosity and nuclear pressure [1, 2].

particles resulting from the decay is in the order of 5 MeV.

The velocity defined as <sup>u</sup> ¼ �2<sup>ν</sup> <sup>∇</sup><sup>P</sup> <sup>P</sup> , with a radius noted as <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 Þ1=<sup>2</sup> where P xð Þ , <sup>y</sup>, <sup>z</sup>, <sup>t</sup> is the logistic probability function P xð Þ¼ , <sup>y</sup>, <sup>z</sup>, <sup>t</sup> <sup>1</sup> <sup>1</sup>þekt�μ<sup>r</sup> , and the expected value E rð Þ <sup>j</sup>r≥<sup>0</sup> <sup>&</sup>lt;<sup>C</sup> exist. The term <sup>P</sup> is defined in ð Þ <sup>x</sup>, <sup>y</sup>, <sup>z</sup> <sup>∈</sup> <sup>3</sup> , t≥0 � �, where constants k>0, μ>0 and P xð Þ , y, x, t is the general solution of the Navier– Stokes 3D equation, which has to satisfy the conditions (1) and (2), allowing us to analyze the dynamics of an incompressible fluid [12–14].

$$\frac{\partial \mathbf{u}}{\partial t} + (\mathbf{u}, \nabla) \mathbf{u} = \nu \nabla^2 \mathbf{u} - \frac{\nabla p}{\rho\_0} \tag{1} \\ \qquad \qquad \left( (\mathbf{x}, y, z) \in \mathbb{R}^3, t \ge 0 \right) \tag{1}$$

With, u∈ <sup>3</sup> an known velocity vector, ρ<sup>0</sup> constant density of fluid, η dynamic viscosity, <sup>ν</sup> cinematic viscosity, and pressure <sup>p</sup> <sup>¼</sup> <sup>p</sup>0<sup>P</sup> in ð Þ <sup>x</sup>, <sup>y</sup>, <sup>z</sup> <sup>∈</sup> <sup>3</sup> , t≥ 0 � �.

Where velocity and pressure are depending of r and t. We will write the condition of incompressibility.

$$\nabla.\mathbf{u} = \mathbf{0} \tag{2}$$

$$\left(\nabla.\mathbf{u} = \mathbf{0}\right) \qquad \left((x, y, z) \in \mathbb{R}^3, t \ge \mathbf{0}\right) \tag{2}$$

The initial conditions of fluid movement <sup>u</sup><sup>0</sup>ð Þ <sup>x</sup>, <sup>y</sup>, <sup>z</sup> , are determined for <sup>t</sup> <sup>¼</sup> 0. Where speed u<sup>0</sup> must be C<sup>∞</sup> divergence-free vector.

$$\mathbf{u}(\boldsymbol{\kappa}, \boldsymbol{y}, \boldsymbol{z}, \mathbf{0}) = \mathbf{u}^{0}(\boldsymbol{\kappa}, \boldsymbol{y}, \boldsymbol{z}) \tag{3}$$

For physically reasonable solutions, we make sure uð Þ x, y, z, t does not grow large as <sup>r</sup> ! <sup>∞</sup>: We will restrict attention to initial conditions <sup>u</sup><sup>0</sup> that satisfy.

$$\left|\partial\_{\mathbf{x}}^{a}\mathbf{u}^{0}\right| \leq \mathcal{C}\_{aK}(\mathbf{1}+r)^{-K} \qquad\qquad\text{on}\ \mathbb{R}^{3}\ \text{for}\ \mathbf{any}\ a\text{and}K\tag{4}$$

The Clay Institute accepts a physically reasonable solution of (1), (2) and (3), only if it satisfies:

$$(p, \mathbf{u} \in \mathbb{C}^{\approx}(\mathbb{R}^3 \times \mathbf{0}, \infty))\tag{5}$$

and the finite energy condition [14–16].

$$\int\_{\mathbb{R}^3} |\mathbf{u}(x, y, z, t)|^2 dx dy dz \le \mathbb{C} \qquad \text{for all } t \ge 0 \qquad \text{(bounded energy)}.\tag{6}$$

The problems of Mathematical Physics are solved by the Nature, guiding the understanding, the scope, the limitations and the complementary theories. These guidelines of this research were: the probabilistic elements of Quantum Mechanics, the De Broglie equation and the Heisenberg Uncertainty principle.

#### 2.1 Definitions

Nuclear reaction velocity coefficient.

We will use an equation analogous to concentration equation of Physical Chemistry <sup>C</sup> <sup>¼</sup> <sup>C</sup>0ekt, where <sup>k</sup> <sup>¼</sup> <sup>p</sup><sup>0</sup> <sup>2</sup><sup>η</sup> , is velocity coefficient, p<sup>0</sup> is the initial pressure of our fluid, η the dynamic viscosity and C<sup>0</sup> the initial concentration of energetic fluid molecules.

∇:u ¼ �2νμ∇ð Þ ð Þ 1 � P ∇r

.

:

<sup>μ</sup><sup>2</sup> <sup>P</sup> � <sup>P</sup><sup>2</sup> � �<sup>2</sup>

j j ∇r

Nuclear Physics, r<sup>0</sup> <r<1:2A<sup>1</sup>=<sup>3</sup>

<sup>P</sup> <sup>¼</sup> <sup>1</sup>

summarized in: ∇<sup>2</sup>

<sup>∂</sup><sup>t</sup> <sup>þ</sup> ð Þ <sup>u</sup>:<sup>∇</sup> <sup>u</sup> <sup>¼</sup> <sup>ν</sup>∇<sup>2</sup>

(1987).

(1), <sup>∂</sup><sup>u</sup>

1

∇ ∇<sup>2</sup>

derivative <sup>∂</sup><sup>u</sup>

obtain (11).

27

∂t

<sup>2</sup> <sup>¼</sup> <sup>μ</sup><sup>2</sup> <sup>P</sup> � <sup>P</sup><sup>2</sup> � �<sup>2</sup>

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

<sup>1</sup> <sup>þ</sup> ekt�<sup>μ</sup> <sup>x</sup>2þy2þz<sup>2</sup> ð Þ1=<sup>2</sup> <sup>¼</sup> <sup>2</sup>

Laplace equation for the pressure of the fluid ∇<sup>2</sup>

<sup>u</sup> � <sup>∇</sup><sup>p</sup> ρ0

ð Þ <sup>u</sup>:<sup>∇</sup> <sup>u</sup> <sup>¼</sup> <sup>1</sup>

∇2

2

∂u

� <sup>∇</sup><sup>p</sup> ρ0

After replacing the last four results ð Þ <sup>u</sup>:<sup>∇</sup> <sup>u</sup>, <sup>∇</sup><sup>2</sup>

<sup>∇</sup> � <sup>∇</sup><sup>θ</sup> <sup>¼</sup> 0, we have: ð Þ <sup>u</sup>:<sup>∇</sup> <sup>u</sup> <sup>¼</sup> ð Þ <sup>∇</sup>θ:<sup>∇</sup> <sup>∇</sup><sup>θ</sup> <sup>¼</sup> <sup>1</sup>

<sup>P</sup> <sup>¼</sup> <sup>2</sup> <sup>μ</sup> ∇<sup>2</sup> <sup>1</sup> r

¼ �2νμ∇ð Þ ð Þ 1 � P ∇r

¼ �2νμ �<sup>μ</sup> <sup>P</sup> � <sup>P</sup><sup>2</sup> � �j j <sup>∇</sup><sup>r</sup>

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

Where the gradient modulus of <sup>∇</sup><sup>P</sup> <sup>¼</sup> <sup>μ</sup> <sup>P</sup> � <sup>P</sup><sup>2</sup> � �∇r, has the form j j <sup>∇</sup><sup>P</sup> <sup>2</sup> <sup>¼</sup>

∇:u ¼ �2νμð Þ� 1 � P μP þ

Simplifying for 1ð Þ � P 6¼ 0, we obtain the main result of this paper, which represents a fixed point of an implicit function f tð Þ ,<sup>r</sup> where f tð Þ¼ ,<sup>r</sup> <sup>P</sup> � <sup>2</sup>

Eq. (10) has a solution according to the fixed-point theorem of an implicit function, and it is a solution to the Navier Stokes stationary equations, which are

To this point, we need to verify that Eq. (10) is also a solution of requirement

in Eq. (1). Taking into account that θ ¼ �2ν ln ð Þ P , and that ∇θ is irrotational,

<sup>2</sup> ∇ ∇ð Þ <sup>θ</sup>:∇<sup>θ</sup> , and <sup>∇</sup><sup>2</sup><sup>u</sup> <sup>¼</sup> ∇ ∇ð Þ� :<sup>u</sup> <sup>∇</sup> � ð Þ¼ <sup>∇</sup> � <sup>u</sup> ∇ ∇ð Þ� :∇<sup>θ</sup> <sup>∇</sup> � ð Þ¼ <sup>∇</sup> � <sup>∇</sup><sup>θ</sup>

<sup>u</sup> <sup>¼</sup> ∇ ∇ð Þ¼ :<sup>u</sup> ∇ ∇<sup>2</sup>

¼ �2ν<sup>∇</sup> j j <sup>∇</sup><sup>P</sup> <sup>2</sup>

θ � �: Simplifying terms in order to replace these results in Eq. (1) we obtain

∇ ∇ð Þ¼ <sup>θ</sup>:∇<sup>θ</sup> <sup>2</sup>ν<sup>2</sup>

The explicit form of velocity is u ¼ �2μνð Þ 1 � P ∇r: Next, we need the partial

<sup>∂</sup><sup>t</sup> ¼ �2μνkPð Þ <sup>1</sup> � <sup>P</sup> <sup>∇</sup>r,

¼ � <sup>μ</sup>p<sup>0</sup> ρ0

<sup>2</sup> <sup>þ</sup> ð Þ <sup>1</sup> � <sup>P</sup> <sup>∇</sup><sup>2</sup>

2 r

� �

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

� � <sup>¼</sup> <sup>0</sup>: Furthermore, it is the typical solution of the

<sup>p</sup> <sup>¼</sup> <sup>p</sup>0∇<sup>2</sup>

. We will do the equivalence u ¼ ∇θ after we replace

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

¼ 0

<sup>θ</sup> � � <sup>¼</sup> <sup>0</sup>

Pð Þ 1 � P ∇r:

u, ∂u

<sup>∂</sup><sup>t</sup> and � <sup>∇</sup><sup>p</sup> ρ0

in Eq. (1) we

P P

<sup>P</sup><sup>2</sup> � <sup>∇</sup><sup>2</sup>

!

<sup>2</sup> ∇ ∇ð Þ� θ:∇θ ∇θ � ð Þ¼ ∇ � ∇θ

h i

r

¼ 0 (9)

, t≥ 0 � � (10)

P ¼ 0. Kerson Huang

(8)

<sup>μ</sup><sup>r</sup> ¼ 0. In

It is evident that, in equilibrium state we can write μr ¼ kt, however, the Navier–Stokes equation precisely measures the behavior of the fluids out of equilibrium, so that: μr 6¼ kt.

Fortunately, there is a single solution for out-of-equilibrium fluids, using the fixed-point theorem for implicit functions, <sup>1</sup> <sup>1</sup>þekt�μ<sup>r</sup> <sup>¼</sup> <sup>2</sup> <sup>μ</sup><sup>r</sup> , the proof is proved in Theorem 1.

Attenuation coefficient.

We will use the known attenuation formula of an incident flux I0, for which <sup>I</sup> <sup>¼</sup> <sup>I</sup>0e�μ<sup>r</sup> . Where, I<sup>0</sup> initial flux and μ attenuation coefficient of energetic molecules that enter into interaction and/or resonance with the target molecules, transmitting or capturing the maximum amount of energy [5].

Dimensional analysis and fluid elements.

We will define the respective dimensional units of each one of variables and physical constants that appear in the solution of the Navier–Stokes 3D equation [12–14, 16].

Nuclear decay N tðÞ¼ <sup>N</sup>0e�kt: Where <sup>k</sup> <sup>¼</sup> <sup>1</sup> <sup>T</sup>1=<sup>2</sup> [1/s], the velocity coefficient, and T1=<sup>2</sup> ground state half-life.

Kinematic viscosity <sup>ν</sup> <sup>¼</sup> <sup>η</sup> ρ0 , <sup>m</sup><sup>2</sup> s h i.

Dynamic viscosity η, [pa.s], where pa represents pascal pressure unit. Initial Pressure of out of equilibrium. p0, [pa].

Fluid density <sup>ρ</sup>0, kg m<sup>3</sup> h i, where kg is kilogram and <sup>m</sup><sup>3</sup> cubic meters.

Logistic probability function, P xð Þ¼ , <sup>y</sup>, <sup>z</sup>, <sup>t</sup> <sup>1</sup> <sup>1</sup>þekt�μ<sup>r</sup> , it is a real number 0 <sup>≤</sup>P≤1: Equilibrium condition, <sup>r</sup> <sup>¼</sup> <sup>k</sup> <sup>μ</sup> <sup>t</sup> <sup>¼</sup> <sup>p</sup><sup>0</sup> <sup>2</sup>ρ0νμ t ¼ u<sup>e</sup> j jt, [m].

Fluid velocity in equilibrium, u<sup>e</sup> j j, ½ � m=s : Protons, neutrons and alpha particles are the elements of the fluid.

Fluid field velocity out of equilibrium, u ¼ �2νμð Þ 1 � P ∇r: [m/s]. All nuclear decay is a process out of nuclear equilibrium.

Position, <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> , [m]. Attenuation coefficient, μ, [1/m]. Growth coefficient, <sup>k</sup> <sup>¼</sup> <sup>p</sup><sup>0</sup> <sup>2</sup>ρ0<sup>ν</sup> <sup>¼</sup> <sup>p</sup><sup>0</sup> <sup>2</sup><sup>η</sup> , [1/s]. Concentration C ¼ C<sup>0</sup> 1�P <sup>P</sup> :

Theorem 1 The velocity of the fluid is given by <sup>u</sup> ¼ �2<sup>ν</sup> <sup>∇</sup><sup>P</sup> <sup>P</sup> , where P xð Þ , y, z, t is the logistic probability function P xð Þ¼ , <sup>y</sup>, <sup>x</sup>, <sup>t</sup> <sup>1</sup> <sup>1</sup>þekt�μ<sup>r</sup> , and p pressure such that <sup>p</sup> <sup>¼</sup> <sup>p</sup>0P, both defined on xð Þ , <sup>y</sup>, <sup>z</sup> <sup>∈</sup> <sup>3</sup> , t≥0 � �: The function P is the general solution of the Navier Stokes equations, which satisfies conditions (1) and (2).

Proof. To verify condition (2), ∇:u ¼ 0, we must calculate the gradients

and laplacians of the radius. <sup>∇</sup><sup>r</sup> <sup>¼</sup> <sup>x</sup> r , y r , z r � �, and <sup>∇</sup><sup>2</sup><sup>r</sup> <sup>¼</sup> <sup>∇</sup>:∇<sup>r</sup> <sup>¼</sup> <sup>y</sup>2þz<sup>2</sup> ð Þþ <sup>x</sup>2þz<sup>2</sup> ð Þþ <sup>x</sup>2þy<sup>2</sup> ð Þ <sup>x</sup>2þy2þz<sup>2</sup> ð Þ3=<sup>2</sup> <sup>¼</sup> <sup>2</sup> r :

$$\nabla \cdot \boldsymbol{\mu} = -2\nu \nabla \cdot \frac{\nabla P}{P} = -2\nu \mu \nabla ((\mathbf{1} - P)\nabla r) \tag{7}$$

Replacing the respective values for the terms: <sup>∇</sup><sup>2</sup><sup>r</sup> and j j <sup>∇</sup><sup>r</sup> <sup>2</sup> in the Eq. (7). Exergy: Mechanical Nuclear Physics Measures Pressure, Viscosity and X-Ray Resonance… DOI: http://dx.doi.org/10.5772/intechopen.95405

$$\begin{split} \nabla \mathcal{U} &= -2\nu \mu \nabla ( (\mathbf{1} - P) \nabla r ) \\ &= -2\nu \mu \nabla ( (\mathbf{1} - P) \nabla r ) \\ &= -2\nu \mu \left[ -\mu \left( P - P^2 \right) |\nabla r|^2 + (\mathbf{1} - P) \nabla^2 r \right] \end{split} \tag{8}$$

Where the gradient modulus of <sup>∇</sup><sup>P</sup> <sup>¼</sup> <sup>μ</sup> <sup>P</sup> � <sup>P</sup><sup>2</sup> � �∇r, has the form j j <sup>∇</sup><sup>P</sup> <sup>2</sup> <sup>¼</sup> <sup>μ</sup><sup>2</sup> <sup>P</sup> � <sup>P</sup><sup>2</sup> � �<sup>2</sup> j j ∇r <sup>2</sup> <sup>¼</sup> <sup>μ</sup><sup>2</sup> <sup>P</sup> � <sup>P</sup><sup>2</sup> � �<sup>2</sup> .

$$\nabla \cdot \boldsymbol{\mu} = -2\nu \mu (\mathbf{1} - P) \left[ -\mu P + \frac{\mathbf{2}}{r} \right] = \mathbf{0} \tag{9}$$

Simplifying for 1ð Þ � P 6¼ 0, we obtain the main result of this paper, which represents a fixed point of an implicit function f tð Þ ,<sup>r</sup> where f tð Þ¼ ,<sup>r</sup> <sup>P</sup> � <sup>2</sup> <sup>μ</sup><sup>r</sup> ¼ 0. In Nuclear Physics, r<sup>0</sup> <r<1:2A<sup>1</sup>=<sup>3</sup> :

$$P = \frac{1}{\mathbf{1} + e^{\mathbf{k}t - \mu(\mathbf{x}^2 + \mathbf{y}^2 + \mathbf{z}^2)^{1/2}}} = \frac{2}{\mu(\mathbf{x}^2 + \mathbf{y}^2 + \mathbf{z}^2)^{1/2}} \qquad \left( (\mathbf{x}, \mathbf{y}, \mathbf{z}) \in \mathbb{R}^3, t \ge 0 \right) \tag{10}$$

Eq. (10) has a solution according to the fixed-point theorem of an implicit function, and it is a solution to the Navier Stokes stationary equations, which are summarized in: ∇<sup>2</sup> <sup>P</sup> <sup>¼</sup> <sup>2</sup> <sup>μ</sup> ∇<sup>2</sup> <sup>1</sup> r � � <sup>¼</sup> <sup>0</sup>: Furthermore, it is the typical solution of the Laplace equation for the pressure of the fluid ∇<sup>2</sup> <sup>p</sup> <sup>¼</sup> <sup>p</sup>0∇<sup>2</sup> P ¼ 0. Kerson Huang (1987).

To this point, we need to verify that Eq. (10) is also a solution of requirement (1), <sup>∂</sup><sup>u</sup> <sup>∂</sup><sup>t</sup> <sup>þ</sup> ð Þ <sup>u</sup>:<sup>∇</sup> <sup>u</sup> <sup>¼</sup> <sup>ν</sup>∇<sup>2</sup> <sup>u</sup> � <sup>∇</sup><sup>p</sup> ρ0 . We will do the equivalence u ¼ ∇θ after we replace in Eq. (1). Taking into account that θ ¼ �2ν ln ð Þ P , and that ∇θ is irrotational, <sup>∇</sup> � <sup>∇</sup><sup>θ</sup> <sup>¼</sup> 0, we have: ð Þ <sup>u</sup>:<sup>∇</sup> <sup>u</sup> <sup>¼</sup> ð Þ <sup>∇</sup>θ:<sup>∇</sup> <sup>∇</sup><sup>θ</sup> <sup>¼</sup> <sup>1</sup> <sup>2</sup> ∇ ∇ð Þ� θ:∇θ ∇θ � ð Þ¼ ∇ � ∇θ 1 <sup>2</sup> ∇ ∇ð Þ <sup>θ</sup>:∇<sup>θ</sup> , and <sup>∇</sup><sup>2</sup> u ¼ ∇ ∇ð Þ� :u ∇ � ð Þ¼ ∇ � u ∇ ∇ð Þ� :∇θ ∇ � ð Þ¼ ∇ � ∇θ ∇ ∇<sup>2</sup> θ � �: Simplifying terms in order to replace these results in Eq. (1) we obtain

$$(\mathbf{u} \cdot \nabla)\mathbf{u} = \frac{1}{2} \nabla(\nabla \theta . \nabla \theta) = 2\nu^2 \nabla \left(\frac{|\nabla P|^2}{P^2}\right)$$

$$\nabla^2 \mathbf{u} = \nabla(\nabla . \mathbf{u}) = \nabla \left(\nabla^2 \theta\right) = \mathbf{0}$$

$$= -2\nu \nabla \left(\frac{|\nabla P|^2}{P^2} - \frac{\nabla^2 P}{P}\right) = \mathbf{0}$$

The explicit form of velocity is u ¼ �2μνð Þ 1 � P ∇r: Next, we need the partial derivative <sup>∂</sup><sup>u</sup> ∂t

$$\frac{\partial \mathbf{u}}{\partial t} = -2\mu\nu k P(\mathbf{1} - P) \nabla r,$$

$$ -\frac{\nabla p}{\rho\_0} = -\frac{\mu p\_0}{\rho\_0} P(\mathbf{1} - P) \nabla r.$$

After replacing the last four results ð Þ <sup>u</sup>:<sup>∇</sup> <sup>u</sup>, <sup>∇</sup><sup>2</sup> u, ∂u <sup>∂</sup><sup>t</sup> and � <sup>∇</sup><sup>p</sup> ρ0 in Eq. (1) we obtain (11).

We will use an equation analogous to concentration equation of Physical

of our fluid, η the dynamic viscosity and C<sup>0</sup> the initial concentration of energetic

It is evident that, in equilibrium state we can write μr ¼ kt, however, the Navier–Stokes equation precisely measures the behavior of the fluids out of

Fortunately, there is a single solution for out-of-equilibrium fluids, using the

We will use the known attenuation formula of an incident flux I0, for which

that enter into interaction and/or resonance with the target molecules, transmitting

We will define the respective dimensional units of each one of variables and physical constants that appear in the solution of the Navier–Stokes 3D equation

<sup>1</sup>þekt�μ<sup>r</sup> <sup>¼</sup> <sup>2</sup>

. Where, I<sup>0</sup> initial flux and μ attenuation coefficient of energetic molecules

, where kg is kilogram and m<sup>3</sup> cubic meters.

<sup>2</sup>ρ0νμ t ¼ u<sup>e</sup> j jt, [m]. Fluid velocity in equilibrium, u<sup>e</sup> j j, ½ � m=s : Protons, neutrons and alpha particles

Fluid field velocity out of equilibrium, u ¼ �2νμð Þ 1 � P ∇r: [m/s]. All nuclear

<sup>2</sup><sup>η</sup> , is velocity coefficient, p<sup>0</sup> is the initial pressure

<sup>μ</sup><sup>r</sup> , the proof is proved in

<sup>T</sup>1=<sup>2</sup> [1/s], the velocity coefficient, and

<sup>1</sup>þekt�μ<sup>r</sup> , it is a real number 0 <sup>≤</sup>P≤1:

<sup>P</sup> , where P xð Þ , y, z, t

<sup>2</sup> in the Eq. (7).

<sup>1</sup>þekt�μ<sup>r</sup> , and p pressure such that

<sup>P</sup> ¼ �2νμ∇ð Þ ð Þ <sup>1</sup> � <sup>P</sup> <sup>∇</sup><sup>r</sup> (7)

r and j j ∇r

, t≥0 � �: The function P is the general solution of

� �, and <sup>∇</sup><sup>2</sup><sup>r</sup> <sup>¼</sup> <sup>∇</sup>:∇<sup>r</sup> <sup>¼</sup>

Chemistry <sup>C</sup> <sup>¼</sup> <sup>C</sup>0ekt, where <sup>k</sup> <sup>¼</sup> <sup>p</sup><sup>0</sup>

fixed-point theorem for implicit functions, <sup>1</sup>

or capturing the maximum amount of energy [5]. Dimensional analysis and fluid elements.

Nuclear decay N tðÞ¼ <sup>N</sup>0e�kt: Where <sup>k</sup> <sup>¼</sup> <sup>1</sup>

ρ0 , <sup>m</sup><sup>2</sup> s h i . Dynamic viscosity η, [pa.s], where pa represents pascal pressure unit.

Initial Pressure of out of equilibrium. p0, [pa].

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

<sup>μ</sup> <sup>t</sup> <sup>¼</sup> <sup>p</sup><sup>0</sup>

, [m].

<sup>2</sup><sup>η</sup> , [1/s].

Proof. To verify condition (2), ∇:u ¼ 0, we must calculate the gradients

r , y r , z r

∇P

<sup>2</sup>ρ0<sup>ν</sup> <sup>¼</sup> <sup>p</sup><sup>0</sup>

the Navier Stokes equations, which satisfies conditions (1) and (2).

∇:u ¼ �2ν∇:

Replacing the respective values for the terms: ∇<sup>2</sup>

1�P <sup>P</sup> : Theorem 1 The velocity of the fluid is given by <sup>u</sup> ¼ �2<sup>ν</sup> <sup>∇</sup><sup>P</sup>

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

r :

m<sup>3</sup> h i

decay is a process out of nuclear equilibrium.

equilibrium, so that: μr 6¼ kt.

Attenuation coefficient.

T1=<sup>2</sup> ground state half-life.

Fluid density <sup>ρ</sup>0, kg

are the elements of the fluid.

Concentration C ¼ C<sup>0</sup>

Kinematic viscosity <sup>ν</sup> <sup>¼</sup> <sup>η</sup>

Equilibrium condition, <sup>r</sup> <sup>¼</sup> <sup>k</sup>

Position, <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>

Attenuation coefficient, μ, [1/m]. Growth coefficient, <sup>k</sup> <sup>¼</sup> <sup>p</sup><sup>0</sup>

<sup>p</sup> <sup>¼</sup> <sup>p</sup>0P, both defined on xð Þ , <sup>y</sup>, <sup>z</sup> <sup>∈</sup> <sup>3</sup>

and laplacians of the radius. <sup>∇</sup><sup>r</sup> <sup>¼</sup> <sup>x</sup>

<sup>x</sup>2þy2þz<sup>2</sup> ð Þ3=<sup>2</sup> <sup>¼</sup> <sup>2</sup>

<sup>y</sup>2þz<sup>2</sup> ð Þþ <sup>x</sup>2þz<sup>2</sup> ð Þþ <sup>x</sup>2þy<sup>2</sup> ð Þ

26

fluid molecules.

Nuclear Materials

Theorem 1.

<sup>I</sup> <sup>¼</sup> <sup>I</sup>0e�μ<sup>r</sup>

[12–14, 16].

$$-2\mu\nu k P(\mathbf{1} - P)\nabla r = 2\nu^2 \nabla \left(\frac{|\nabla P|^2}{P^2}\right) - \frac{\mu p\_0}{\rho\_0} P(\mathbf{1} - P) \nabla r. \tag{11}$$

with which we can write the following inequality:

Where <sup>α</sup> <sup>¼</sup> <sup>1</sup> � <sup>m</sup>

<sup>u</sup>ð Þ¼� <sup>x</sup>, <sup>y</sup>, <sup>z</sup>, 0 <sup>2</sup><sup>ν</sup> <sup>∇</sup><sup>P</sup>

exists the derivative C<sup>∞</sup>.

∂n x 1 r � �

∂n y 1 r � �

∂n z 1 r � �

29

¼ �ð Þ<sup>1</sup> <sup>n</sup>

¼ �ð Þ<sup>1</sup> <sup>n</sup>

¼ �ð Þ<sup>1</sup> <sup>n</sup>

<sup>u</sup>ð Þ¼ <sup>x</sup>, <sup>y</sup>, <sup>z</sup>, 0 <sup>u</sup><sup>0</sup>ð Þ¼� <sup>x</sup>, <sup>y</sup>, <sup>z</sup> <sup>2</sup>νμð Þ <sup>1</sup> � <sup>P</sup><sup>0</sup> <sup>x</sup>

�≤C<sup>α</sup>Kð Þ <sup>1</sup> <sup>þ</sup> <sup>r</sup> �<sup>K</sup>

∂α xu<sup>0</sup> � � �

derivable.

M

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

<sup>u</sup>ð Þ¼ <sup>x</sup>, <sup>y</sup>, <sup>z</sup>, 0 <sup>u</sup><sup>0</sup>ð Þ¼� <sup>x</sup>, <sup>y</sup>, <sup>z</sup> <sup>2</sup>νμð Þ <sup>1</sup> � <sup>P</sup><sup>0</sup>

1 þ e�μr<sup>0</sup>

Proof. Taking the partial derivatives of ∂<sup>n</sup>

∂n x x r � � <sup>¼</sup> <sup>n</sup>∂<sup>n</sup>�<sup>1</sup> <sup>x</sup>

∂n y x r � � <sup>¼</sup> <sup>n</sup>∂<sup>n</sup>�<sup>1</sup> y

∂n z x r � � <sup>¼</sup> <sup>n</sup>∂<sup>n</sup>�<sup>1</sup> <sup>z</sup>

<sup>P</sup><sup>0</sup> <sup>¼</sup> <sup>1</sup>

j j <sup>T</sup>φðÞ�<sup>t</sup> <sup>T</sup>ψð Þ<sup>t</sup> <sup>¼</sup> j j <sup>φ</sup>ð Þ�<sup>t</sup> <sup>ψ</sup>ð Þ<sup>t</sup> <sup>1</sup> � <sup>m</sup>

is valid for all t of ½ � a, b . Where T is a contraction operator and the proof is

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

complete, since for every contraction operator T : C ! C there exists one and only one continuous function φ in C, such that Tð Þ¼ φ φ. Using Eq. (10), which represents the fundamental solution of the Navier–Stokes 3D equation, we verify Eq. (2), which represents the second of the six requirements of an acceptable

solution. ▪ Proposition 3 Requirement (3). The initial velocity can be obtained from:

M

� �. Since 0< m ≤ M, we have 0≤ α<1. The above inequality

<sup>P</sup> , where each of the components ux, uy and uz are infinitely

x r , y r , z r

x x r � �, ∂<sup>n</sup> y x r � � and ∂<sup>n</sup>

> <sup>þ</sup> <sup>x</sup>∂<sup>n</sup> x 1 r � �

<sup>þ</sup> <sup>y</sup>∂<sup>n</sup> y 1 r � �

<sup>þ</sup> <sup>z</sup>∂<sup>n</sup> z 1 r � �

> x <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>

> y <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>

> z <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>

on <sup>3</sup> foranyα andK

� �, it is evident that

!

!

!

1 r � �

1 r � �

1 r � �

Recalling the derivatives of special functions (Legendre), it is verified that there

<sup>2</sup> Pn

<sup>2</sup> Pn

<sup>2</sup> Pn

Physically, this solution is valid for the initial velocity, indicated by Eq. (4), where the components of the initial velocity are infinitely differentiable, and make it possible to guarantee that the velocity of the fluid is zero when r ! ∞ [6–9]. Proposition 4 Requirement (4). Using the initial velocity of a moving fluid given by

> r , y r , z r

<sup>n</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>n</sup>þ<sup>1</sup>

<sup>n</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>n</sup>þ<sup>1</sup>

<sup>n</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>n</sup>þ<sup>1</sup>

� � <sup>≤</sup>α φk k � <sup>ψ</sup> : (13)

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

z x r � �. (14)

(15)

(16)

▪

The Eq. (11) is equivalent to Eq. (1). After obtaining the term j j <sup>∇</sup><sup>P</sup> <sup>2</sup> <sup>P</sup><sup>2</sup> from the incompressibility equation ∇ ∇<sup>2</sup> <sup>θ</sup> � � ¼ �2ν<sup>∇</sup> � j j <sup>∇</sup><sup>P</sup> <sup>2</sup> <sup>P</sup><sup>2</sup> <sup>þ</sup> <sup>∇</sup>2<sup>P</sup> P � � <sup>¼</sup> 0 and replacing in Eq. (11).

$$-2\mu\nu k P(\mathbf{1} - P)\nabla r = 2\nu^2 \nabla \left(\frac{\nabla^2 P}{P}\right) - \frac{\mu p\_0}{\rho\_0} P(\mathbf{1} - P) \nabla r. \tag{12}$$

Eq. (10) simultaneously fulfills requirements (1) expressed by Eq. (12) and requirement (2) expressed by Eq. (7), for a constant <sup>k</sup> <sup>¼</sup> <sup>p</sup><sup>0</sup> <sup>2</sup>ρ0<sup>ν</sup> <sup>¼</sup> <sup>p</sup><sup>0</sup> <sup>2</sup><sup>η</sup> : Moreover, according to Eq. (10), the probability <sup>P</sup> <sup>¼</sup> <sup>2</sup> <sup>μ</sup><sup>r</sup> which allows the Laplace equation to be satisfied: <sup>∇</sup><sup>2</sup><sup>P</sup> <sup>¼</sup> <sup>2</sup> <sup>μ</sup> ∇<sup>2</sup> <sup>1</sup> r � � <sup>¼</sup> <sup>0</sup>: In other words, the Navier–Stokes 3D equation system is solved. ▪

Implicit Function.

An implicit function defined as (10), f tð Þ¼ ,<sup>r</sup> <sup>1</sup> <sup>1</sup>þekt�μ<sup>r</sup> � <sup>2</sup> <sup>μ</sup><sup>r</sup> ¼ 0 has a fixed point ð Þ t,r of R ¼ f g ð Þj t,r 0<a≤t≤b, 0 <r< þ ∞ , where m and M are constants, such as: m ≤ M: Knowing that the partial derivative exists: <sup>∂</sup>rf tð Þ¼ ,<sup>r</sup> <sup>ν</sup>Pð Þþ <sup>1</sup> � <sup>P</sup> <sup>2</sup> <sup>μ</sup>r<sup>2</sup> we can assume that: 0 <sup>&</sup>lt; <sup>m</sup> <sup>≤</sup> <sup>∂</sup>rf tð Þ ,<sup>r</sup> <sup>≤</sup> <sup>M</sup>: If, in addition, for each continuous function <sup>φ</sup> in ½ � <sup>a</sup>, <sup>b</sup> the composite function g tðÞ¼ f tð Þ , φð Þt is continuous in ½ � a, b , then there is one and only one function:r ¼ φð Þt continuous in ½ � a, b , such that f t½ �¼ , φð Þt 0 for all t in ½ � a, b .

Theorem 2 An implicit function defined as (10) f tð Þ¼ ,<sup>r</sup> <sup>1</sup> <sup>1</sup>þekt�μ<sup>r</sup> � <sup>2</sup> <sup>μ</sup><sup>r</sup> ¼ 0 has a fixed point tð Þ ,r of R ¼ f g ð Þj t,r 0<a≤t≤ b, 0< r< þ ∞ . In this way, the requirements (1) and (2) are fulfilled.

Proof. Let C be the linear space of continuous functions in ½ � a, b , and define an operator T : C ! C by the equation:

$$T\rho(t) = \rho(t) - \frac{1}{M}f[t, \rho(t)].$$

Then we prove that T is a contraction operator, so it has a unique fixed point r ¼ φð Þt in C. Let us construct the following distance.

$$T\rho(t) - T\psi(t) = \rho(t) - \psi(t) - \frac{f[t, \rho(t)] - f[t, \psi(t)]}{M}.$$

Using the mean value theorem for derivation, we have

$$f[t, \rho(t)] - f[t, \psi(t)] = \partial\_{\varnothing} f(t, x(t))[\rho(t) - \psi(t)].$$

Where ϕð Þt is situated between φð Þt and ψð Þt . Therefore, the distance equation can be written as:

$$T\boldsymbol{\rho}(t) - T\boldsymbol{\nu}(t) = [\boldsymbol{\rho}(t) - \boldsymbol{\nu}(t)] \left[ \mathbf{1} - \frac{\partial \boldsymbol{\phi}^{\boldsymbol{f}}(t, \boldsymbol{z}(t))}{M} \right]$$

Using the hypothesis 0<sup>&</sup>lt; <sup>m</sup> <sup>≤</sup>∂rf tð Þ ,<sup>r</sup> <sup>≤</sup> <sup>M</sup> we arrive at the following result:

$$0 \le \mathbf{1} - \frac{\partial\_{\boldsymbol{\Phi}} f(t, \boldsymbol{\phi}(t))}{M} \le \mathbf{1} - \frac{m}{M},$$

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

with which we can write the following inequality:

�2μνkPð Þ <sup>1</sup> � <sup>P</sup> <sup>∇</sup><sup>r</sup> <sup>¼</sup> <sup>2</sup>ν<sup>2</sup>

�2μνkPð Þ <sup>1</sup> � <sup>P</sup> <sup>∇</sup><sup>r</sup> <sup>¼</sup> <sup>2</sup>ν<sup>2</sup>

requirement (2) expressed by Eq. (7), for a constant <sup>k</sup> <sup>¼</sup> <sup>p</sup><sup>0</sup>

An implicit function defined as (10), f tð Þ¼ ,<sup>r</sup> <sup>1</sup>

r ¼ φð Þt in C. Let us construct the following distance.

Using the mean value theorem for derivation, we have

Knowing that the partial derivative exists: <sup>∂</sup>rf tð Þ¼ ,<sup>r</sup> <sup>ν</sup>Pð Þþ <sup>1</sup> � <sup>P</sup> <sup>2</sup>

Theorem 2 An implicit function defined as (10) f tð Þ¼ ,<sup>r</sup> <sup>1</sup>

TφðÞ¼ t φðÞ�t

according to Eq. (10), the probability <sup>P</sup> <sup>¼</sup> <sup>2</sup>

<sup>μ</sup> ∇<sup>2</sup> <sup>1</sup> r

operator T : C ! C by the equation:

incompressibility equation ∇ ∇<sup>2</sup>

Eq. (11).

Nuclear Materials

satisfied: <sup>∇</sup><sup>2</sup><sup>P</sup> <sup>¼</sup> <sup>2</sup>

Implicit Function.

and (2) are fulfilled.

can be written as:

28

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

The Eq. (11) is equivalent to Eq. (1). After obtaining the term j j <sup>∇</sup><sup>P</sup> <sup>2</sup>

<sup>θ</sup> � � ¼ �2ν<sup>∇</sup> � j j <sup>∇</sup><sup>P</sup> <sup>2</sup>

∇

Eq. (10) simultaneously fulfills requirements (1) expressed by Eq. (12) and

is solved. ▪

of R ¼ f g ð Þj t,r 0<a≤t≤b, 0 <r< þ ∞ , where m and M are constants, such as: m ≤ M:

that: 0 <sup>&</sup>lt; <sup>m</sup> <sup>≤</sup> <sup>∂</sup>rf tð Þ ,<sup>r</sup> <sup>≤</sup> <sup>M</sup>: If, in addition, for each continuous function <sup>φ</sup> in ½ � <sup>a</sup>, <sup>b</sup> the composite function g tðÞ¼ f tð Þ , φð Þt is continuous in ½ � a, b , then there is one and only one function:r ¼ φð Þt continuous in ½ � a, b , such that f t½ �¼ , φð Þt 0 for all t in ½ � a, b .

point tð Þ ,r of R ¼ f g ð Þj t,r 0<a≤t≤ b, 0< r< þ ∞ . In this way, the requirements (1)

Proof. Let C be the linear space of continuous functions in ½ � a, b , and define an

Then we prove that T is a contraction operator, so it has a unique fixed point

<sup>T</sup>φðÞ�<sup>t</sup> <sup>T</sup>ψðÞ¼ <sup>t</sup> <sup>φ</sup>ðÞ�<sup>t</sup> <sup>ψ</sup>ðÞ�<sup>t</sup> f t½ �� , <sup>φ</sup>ð Þ<sup>t</sup> f t½ � , <sup>ψ</sup>ð Þ<sup>t</sup>

f t½ �� , <sup>φ</sup>ð Þ<sup>t</sup> f t½ �¼ , <sup>ψ</sup>ð Þ<sup>t</sup> <sup>∂</sup>ϕf tð Þ , z tð Þ ½ � <sup>φ</sup>ðÞ�<sup>t</sup> <sup>ψ</sup>ð Þ<sup>t</sup> :

Where ϕð Þt is situated between φð Þt and ψð Þt . Therefore, the distance equation

<sup>T</sup>φðÞ�<sup>t</sup> <sup>T</sup>ψðÞ¼ <sup>t</sup> ½ � <sup>φ</sup>ðÞ�<sup>t</sup> <sup>ψ</sup>ð Þ<sup>t</sup> <sup>1</sup> � <sup>∂</sup>ϕf tð Þ , z tð Þ

Using the hypothesis 0<sup>&</sup>lt; <sup>m</sup> <sup>≤</sup>∂rf tð Þ ,<sup>r</sup> <sup>≤</sup> <sup>M</sup> we arrive at the following result:

<sup>M</sup> <sup>≤</sup><sup>1</sup> � <sup>m</sup>

<sup>0</sup>≤<sup>1</sup> � <sup>∂</sup>ϕf tð Þ , <sup>ϕ</sup>ð Þ<sup>t</sup>

1

<sup>M</sup> f t½ � , <sup>φ</sup>ð Þ<sup>t</sup> :

∇<sup>2</sup>P P � � � <sup>μ</sup>p<sup>0</sup> ρ0

<sup>P</sup><sup>2</sup> <sup>þ</sup> <sup>∇</sup>2<sup>P</sup> P

> � <sup>μ</sup>p<sup>0</sup> ρ0

� � <sup>¼</sup> <sup>0</sup>: In other words, the Navier–Stokes 3D equation system

<sup>1</sup>þekt�μ<sup>r</sup> � <sup>2</sup>

� �

Pð Þ 1 � P ∇r: (11)

¼ 0 and replacing in

Pð Þ 1 � P ∇r: (12)

<sup>2</sup><sup>η</sup> : Moreover,

<sup>μ</sup><sup>r</sup> ¼ 0 has a fixed point ð Þ t,r

<sup>μ</sup>r<sup>2</sup> we can assume

<sup>μ</sup><sup>r</sup> ¼ 0 has a fixed

<sup>2</sup>ρ0<sup>ν</sup> <sup>¼</sup> <sup>p</sup><sup>0</sup>

<sup>μ</sup><sup>r</sup> which allows the Laplace equation to be

<sup>1</sup>þekt�μ<sup>r</sup> � <sup>2</sup>

<sup>M</sup> :

M � �

M ,

<sup>P</sup><sup>2</sup> from the

$$|T\boldsymbol{\varrho}(t) - T\boldsymbol{\wp}(t)| = |\boldsymbol{\wp}(t) - \boldsymbol{\wp}(t)| \left(1 - \frac{m}{M}\right) \le a||\boldsymbol{\wp} - \boldsymbol{\wp}||.\tag{13}$$

Where <sup>α</sup> <sup>¼</sup> <sup>1</sup> � <sup>m</sup> M � �. Since 0< m ≤ M, we have 0≤ α<1. The above inequality is valid for all t of ½ � a, b . Where T is a contraction operator and the proof is complete, since for every contraction operator T : C ! C there exists one and only one continuous function φ in C, such that Tð Þ¼ φ φ. Using Eq. (10), which represents the fundamental solution of the Navier–Stokes 3D equation, we verify Eq. (2), which represents the second of the six requirements of an acceptable solution. ▪

Proposition 3 Requirement (3). The initial velocity can be obtained from: <sup>u</sup>ð Þ¼� <sup>x</sup>, <sup>y</sup>, <sup>z</sup>, 0 <sup>2</sup><sup>ν</sup> <sup>∇</sup><sup>P</sup> <sup>P</sup> , where each of the components ux, uy and uz are infinitely derivable.

$$\mathbf{u}(\mathbf{x}, \mathbf{y}, \mathbf{z}, \mathbf{0}) = \mathbf{u}^{0}(\mathbf{x}, \mathbf{y}, \mathbf{z}) = -2\nu\mu(\mathbf{1} - P\_{0}) \left(\frac{\mathbf{x}}{r}, \frac{\mathbf{y}}{r}, \frac{\mathbf{z}}{r}\right) \tag{18.15}$$

$$P\_{0} = \frac{\mathbf{1}}{\mathbf{1} + e^{-\mu r\_{0}}} \tag{14.16}$$

Proof. Taking the partial derivatives of ∂<sup>n</sup> x x r � �, ∂<sup>n</sup> y x r � � and ∂<sup>n</sup> z x r � �.

$$\begin{aligned} \partial\_x^n \left( \frac{\varkappa}{r} \right) &= n \partial\_x^{n-1} \left( \frac{1}{r} \right) + x \partial\_x^n \left( \frac{1}{r} \right) \\ \partial\_\nu^n \left( \frac{\varkappa}{r} \right) &= n \partial\_\gamma^{n-1} \left( \frac{1}{r} \right) + y \partial\_\gamma^n \left( \frac{1}{r} \right) \\ \partial\_z^n \left( \frac{\varkappa}{r} \right) &= n \partial\_z^{n-1} \left( \frac{1}{r} \right) + z \partial\_z^n \left( \frac{1}{r} \right) \end{aligned} \tag{15}$$

Recalling the derivatives of special functions (Legendre), it is verified that there exists the derivative C<sup>∞</sup>.

$$\begin{aligned} \partial\_x^n \left( \frac{1}{r} \right) &= (-1)^n n! \left( \mathbf{x}^2 + \mathbf{y}^2 + \mathbf{z}^2 \right)^{-\frac{(n+1)}{2}} P\_n \left( \frac{\mathbf{x}}{\left( \mathbf{x}^2 + \mathbf{y}^2 + \mathbf{z}^2 \right)^{1/2}} \right) \\ \partial\_y^n \left( \frac{1}{r} \right) &= (-1)^n n! \left( \mathbf{x}^2 + \mathbf{y}^2 + \mathbf{z}^2 \right)^{-\frac{(n+1)}{2}} P\_n \left( \frac{\mathbf{y}}{\left( \mathbf{x}^2 + \mathbf{y}^2 + \mathbf{z}^2 \right)^{1/2}} \right) \\ \partial\_z^n \left( \frac{1}{r} \right) &= (-1)^n n! \left( \mathbf{x}^2 + \mathbf{y}^2 + \mathbf{z}^2 \right)^{-\frac{(n+1)}{2}} P\_n \left( \frac{\mathbf{z}}{\left( \mathbf{x}^2 + \mathbf{y}^2 + \mathbf{z}^2 \right)^{1/2}} \right) \end{aligned} \tag{16}$$

▪ Physically, this solution is valid for the initial velocity, indicated by Eq. (4), where the components of the initial velocity are infinitely differentiable, and make it possible to guarantee that the velocity of the fluid is zero when r ! ∞ [6–9].

Proposition 4 Requirement (4). Using the initial velocity of a moving fluid given by <sup>u</sup>ð Þ¼ <sup>x</sup>, <sup>y</sup>, <sup>z</sup>, 0 <sup>u</sup><sup>0</sup>ð Þ¼� <sup>x</sup>, <sup>y</sup>, <sup>z</sup> <sup>2</sup>νμð Þ <sup>1</sup> � <sup>P</sup><sup>0</sup> <sup>x</sup> r , y r , z r � �, it is evident that

$$\left|\partial\_{\mathbf{x}}^{a}\mathbf{u}^{0}\right| \leq \mathbf{C}\_{a\mathbf{K}}(\mathbf{1}+r)^{-K} \qquad\qquad\qquad on\ \mathbb{R}^{3}\ for any a\,\,d\,\mathbf{K}$$

$$\left(\partial\_{\mathbf{x}}^{a}\frac{\mathbf{x}}{r}\right)^{2} = \left(a\partial\_{\mathbf{x}}^{a-1}\left(\frac{\mathbf{1}}{r}\right) + \mathbf{x}\partial\_{\mathbf{x}}^{a}\left(\frac{\mathbf{1}}{r}\right)\right)\left(a\partial\_{\mathbf{x}}^{a-1}\left(\frac{\mathbf{1}}{r}\right) + \mathbf{x}\partial\_{\mathbf{x}}^{a}\left(\frac{\mathbf{1}}{r}\right)\right)$$

$$\left(\partial\_x^{\alpha}\frac{x}{r}\right)^2 = \alpha^2 \left(\partial\_x^{\alpha-1}\frac{1}{r}\right)^2 + 2\alpha x \partial\_x^{\alpha-1}\frac{1}{r}\partial\_x^{\alpha}\frac{1}{r} + x^2 \left(\partial\_x^{\alpha}\frac{1}{r}\right)^2$$

$$\left(\partial\_y^{\alpha}\frac{y}{r}\right)^2 = \alpha^2 \left(\partial\_y^{\alpha-1}\frac{1}{r}\right)^2 + 2\alpha y \partial\_y^{\alpha-1}\frac{1}{r}\partial\_y^{\alpha}\frac{1}{r} + y^2 \left(\partial\_y^{\alpha}\frac{1}{r}\right)^2\tag{17}$$

$$\left(\partial\_x^{\alpha}\frac{x}{r}\right)^2 = \alpha^2 \left(\partial\_x^{\alpha-1}\frac{1}{r}\right)^2 + 2\alpha x \partial\_x^{\alpha-1}\frac{1}{r}\partial\_x^{\alpha}\frac{1}{r} + z^2 \left(\partial\_x^{\alpha}\frac{1}{r}\right)^2$$

$$\begin{aligned} \partial\_{\mathbf{x}}^a \frac{\mathbf{1}}{r} &= (-\mathbf{1})^a a! (\mathbf{x}^2 + \mathbf{y}^2 + \mathbf{z}^2)^{-\frac{(a+1)}{2}} P\_a \left( \frac{\mathbf{x}}{\left(\mathbf{x}^2 + \mathbf{y}^2 + \mathbf{z}^2\right)^{1/2}} \right) \\\\ \partial\_{\mathbf{x}}^{a-1} \frac{\mathbf{1}}{r} &= (-\mathbf{1})^{a-1} (a-1)! \left(\mathbf{x}^2 + \mathbf{y}^2 + \mathbf{z}^2\right)^{-\frac{a}{2}} P\_{a-1} \left( \frac{\mathbf{x}}{\left(\mathbf{x}^2 + \mathbf{y}^2 + \mathbf{z}^2\right)^{1/2}} \right) \end{aligned} \tag{18}$$

$$\begin{aligned} \varkappa^2 \left( \partial\_x^x \left( \frac{1}{r} \right) \right)^2 &\le \varkappa^2 (a!)^2 r^{-2(a+1)} \\\\ 2\varkappa \alpha \partial\_x^{x-1} \frac{1}{r} \partial\_x^x \frac{1}{r} &\le 2\varkappa a (a!) (a-1)! (-1)^{2a-1} r^{-2a-1} \\\\ \varkappa^2 \left( \partial\_x^{x-1} \left( \frac{1}{r} \right) \right)^2 &\le a^2 ((a-1)!)^2 r^{-2a} \end{aligned} \tag{19}$$

$$\left(\partial\_x^a \frac{x}{r}\right)^2 \le r^{-2a} \left[\frac{x^2(a!)^2}{r^2} - \frac{2x(a!)^2}{r} + a^2((a-1)!)^2\right]$$

$$\left(\partial\_y^a \frac{y}{r}\right)^2 \le r^{-2a} \left[\frac{y^2(a!)^2}{r^2} - \frac{2y(a!)^2}{r} + a^2((a-1)!)^2\right] \tag{20}$$

$$\left(\partial\_x^a \frac{x}{r}\right)^2 \le r^{-2a} \left[\frac{x^2(a!)^2}{r^2} - \frac{2x(a!)^2}{r} + a^2((a-1)!)^2\right]$$

$$\left| \partial\_{\mathbf{x}}^{\alpha} \mathbf{u}^{0} \right| \le r^{-2\alpha} \left[ \mathfrak{Z}(a!)^{2} + a^{2} ((a-1)!)^{2} - \frac{2(\varkappa + \wp + z)(a!)^{2}}{r} \right]$$

$$\begin{aligned} \left| \partial\_{\mathbf{x}}^a \mathbf{u}^0 \right| &\leq \frac{2(a!)^2}{r^{2a}} \left[ 2 + \left| \frac{\mathbf{x}}{r} \right| + \left| \frac{\mathcal{Y}}{r} \right| + \left| \frac{\mathbf{z}}{r} \right| \right] \\ \left| \partial\_{\mathbf{x}}^a \mathbf{u}^0 \right| &\leq \frac{10 \left(a!\right)^2}{r^{2a}} \end{aligned}$$

$$\begin{aligned} \mathbf{u}(\mathbf{x}, \mathbf{y}, z, t) &= 2\nu^2 \left( \frac{\mathbf{x}}{r^2}, \frac{\mathbf{y}}{r^2}, \frac{z}{r^2} \right) & \quad & \quad \left( (\mathbf{x}, \mathbf{y}, z) \in \mathbb{R}^3 \right) \\ P(\mathbf{x}, \mathbf{y}, z, t) &= \frac{1}{\mathbf{1} + e^{\frac{\mathbf{p}\_0}{2\eta} - \mu r}} = \frac{2}{\mu r} \end{aligned} \tag{21}$$

$$
\begin{split}
\partial\_x^n \left(\frac{\chi}{r^2}\right) &= n \partial\_x^{n-1} \left(\frac{1}{r^2}\right) + \varkappa \partial\_x^n \left(\frac{1}{r^2}\right) \\
\partial\_y^n \left(\frac{\chi}{r^2}\right) &= n \partial\_y^{n-1} \left(\frac{1}{r^2}\right) + \wp \partial\_y^n \left(\frac{1}{r^2}\right) \\
\partial\_z^n \left(\frac{\chi}{r^2}\right) &= n \partial\_z^{n-1} \left(\frac{1}{r^2}\right) + z \partial\_z^n \left(\frac{1}{r^2}\right)
\end{split} \tag{22}
$$

$$\begin{aligned} \partial\_x^n \left( \frac{1}{r^2} \right) &= (-1)^n n! \left( \mathbf{x}^2 + \mathbf{y}^2 + z^2 \right)^{-(n+1)} P\_n \left( \frac{\mathbf{x}}{(\mathbf{x}^2 + \mathbf{y}^2 + z^2)} \right) \\ \partial\_y^n \left( \frac{1}{r^2} \right) &= (-1)^n n! \left( \mathbf{x}^2 + \mathbf{y}^2 + z^2 \right)^{-(n+1)} P\_n \left( \frac{\mathbf{y}}{(\mathbf{x}^2 + \mathbf{y}^2 + z^2)} \right) \\ \partial\_x^n \left( \frac{1}{r^2} \right) &= (-1)^n n! \left( \mathbf{x}^2 + \mathbf{y}^2 + z^2 \right)^{-(n+1)} P\_n \left( \frac{z}{(\mathbf{x}^2 + \mathbf{y}^2 + z^2)} \right) \end{aligned} \tag{23}$$

Proposition 6 Requirement (5). The pressure is totally defined by the equivalence p xð Þ¼ , y, z, t p0P xð Þ , y, z, t and is infinitely differentiable in each of its components.

$$p(\mathbf{x}, y, z, t) = p\_0 P(\mathbf{x}, y, z, t) \qquad \qquad \left( (\mathbf{x}, y, z) \in \mathbb{R}^3 \right) \tag{24}$$

In this way the value of the constant <sup>C</sup> is <sup>C</sup> <sup>¼</sup> <sup>32</sup>πν<sup>2</sup>

P xð Þ , <sup>y</sup>, <sup>z</sup>, <sup>t</sup> associated with the velocity <sup>u</sup> ¼ �2<sup>ν</sup> <sup>∇</sup><sup>P</sup>

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

orders lower than the minimum experimental value.

of <sup>ξ</sup> <sup>¼</sup> <sup>r</sup> � <sup>ϕ</sup>ð Þ <sup>t</sup>, <sup>k</sup>, <sup>μ</sup> . The function <sup>ϕ</sup>ð Þ¼ <sup>t</sup>, <sup>k</sup>, <sup>μ</sup> <sup>k</sup>

equation, <sup>1</sup>

1þe �<sup>μ</sup> <sup>r</sup>�<sup>k</sup> <sup>μ</sup> ð Þ<sup>t</sup> <sup>¼</sup> <sup>2</sup>

Derivation of E rð Þ jr≥0 :

½ � <sup>1</sup><sup>þ</sup> exp ð Þ �<sup>ξ</sup> <sup>2</sup> , where <sup>1</sup>

calculations we put <sup>ϕ</sup>ðÞ¼ • <sup>ϕ</sup>ð Þ¼ <sup>t</sup>, <sup>k</sup>, <sup>μ</sup> <sup>k</sup>

when <sup>r</sup>≥0 is given by f rð Þ¼ <sup>j</sup>r≥<sup>0</sup> f rð Þ

<sup>1</sup><sup>þ</sup> exp ð ÞÞ <sup>ϕ</sup>ð Þ• <sup>¼</sup> <sup>1</sup>

P rð Þ ≥ 0

P rð Þ ≥0

where we have used the fact that

ð<sup>∞</sup> 0

distribution function for <sup>r</sup> is given by F rð Þ¼ <sup>1</sup>

<sup>h</sup>ð Þ¼ <sup>ξ</sup> <sup>μ</sup> exp ð Þ �<sup>ξ</sup>

<sup>1</sup> � <sup>F</sup>ð Þ¼ <sup>0</sup> exp ð ÞÞ <sup>ϕ</sup>ð Þ•

E rð Þ¼ jr≥0

E rð Þ¼ <sup>j</sup>r≥<sup>0</sup> <sup>1</sup>

E rð Þ¼ <sup>j</sup>r<sup>≥</sup> <sup>0</sup> <sup>1</sup>

follows:

33

completely. In general, Eq. (10) can be written f tð Þ¼ ,r þ r<sup>0</sup>

infinities ð Þ <sup>x</sup>, <sup>y</sup>, <sup>z</sup> that hold the relationship <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>

Stokes 3D equation, where all fluid is in accelerated motion <sup>∂</sup><sup>u</sup>

<sup>μ</sup> <sup>¼</sup> <sup>σ</sup> ffiffiffi

<sup>ϕ</sup> <sup>t</sup>, <sup>p</sup>0, <sup>η</sup>, <sup>μ</sup> � � <sup>þ</sup> <sup>ξ</sup> function for <sup>r</sup> is then f rð Þ¼ exp ½ � �r�ϕð Þ• <sup>=</sup><sup>τ</sup>

<sup>μ</sup>rf rð Þ <sup>j</sup>r≥<sup>0</sup> dr <sup>¼</sup> <sup>1</sup>

2

We replaced in Eq. (29) dP <sup>¼</sup> <sup>μ</sup>Pð Þ <sup>1</sup> � <sup>P</sup> dr and <sup>r</sup><sup>2</sup> <sup>¼</sup> <sup>2</sup>

2

ð1 1=2

ð1 1=2 P rð Þ ≥0

<sup>μ</sup><sup>P</sup> ð Þ <sup>μ</sup>Pð Þ <sup>1</sup> � <sup>P</sup> dP

<sup>μ</sup><sup>P</sup> ð Þ <sup>μ</sup>Pð Þ <sup>1</sup> � <sup>P</sup> dP

ð<sup>∞</sup> 0

Pð Þ 1 � P

and in this way discontinuities are avoided when r ! 0, but this problem does not occur since in the atomic nucleus <sup>r</sup><sup>0</sup> <sup>&</sup>lt;r<1:2A1=<sup>3</sup> is satisfied. ▪ Lemma 8 The irrotational field represented by the logistic probability function

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

stochastic behavior of the physical variables p0, η, μ: These stochastic variations are in

attenuation coefficient μ. Due to Heisenberg uncertainty principle, these parameters have a variation when we measure and use them, as is the case of the estimate

results, when �∞ <ξ< þ ∞. The physical and mathematical realities are mutually conditioned and allow for these surprising results. For a definite t there exist

are trajectories in the sphere as long as it is probabilistically possible, this is reduced to

The logistic density function for <sup>ξ</sup> when <sup>E</sup>ð Þ¼ <sup>ξ</sup> 0 and Varð Þ¼ <sup>ξ</sup> <sup>σ</sup><sup>2</sup> is defined by.

<sup>3</sup> <sup>p</sup> <sup>=</sup><sup>π</sup> is a scale parameter. Given that <sup>r</sup> <sup>¼</sup>

P rð Þ <sup>≥</sup><sup>0</sup> for <sup>r</sup>≥0: Given that the cumulative

<sup>μ</sup><sup>r</sup> exp ½ � kt � <sup>μ</sup><sup>r</sup> f g 1 þ exp ½ � kt � μr

> μP � �<sup>2</sup>

> > P rð Þ ≥ 0

2

<sup>P</sup>ð Þ <sup>1</sup> � <sup>P</sup> <sup>¼</sup> <sup>1</sup>

<sup>1</sup><sup>þ</sup> exp ð ÞÞ �ϕð Þ• . The derivation of E rð Þ <sup>j</sup>r≥<sup>0</sup> then proceeds as

<sup>τ</sup> <sup>1</sup><sup>þ</sup> exp ð � �ð Þ <sup>r</sup>�ϕð Þ• <sup>=</sup><sup>τ</sup> <sup>2</sup> <sup>½</sup> , to facilitate the

<sup>1</sup><sup>þ</sup> exp ð Þ kt�μ<sup>r</sup> , it follows that P rð Þ¼ <sup>≥</sup><sup>0</sup>

<sup>2</sup> dr

of this manner we obtain

<sup>μ</sup> log 2ð Þ (30)

(29)

<sup>μ</sup> t. By definition, the truncated density for r

definite r there are infinities t that respect the fixed-point theorem and create spherical trajectories. When the physical variables k, μ vary, even at levels of 1/100 or 1/1000, they remain below the minimum variation of the experimental value. We could try to avoid the existence of trajectories on the spherical surface, for which we must assume that the fluid is at rest or it is stationary, which contradicts the Navier–

showing that the expected value of the radius E r½ � jr≥0 exists and it is finite.

Proof. The implicit function representing the solution of the Navier–Stokes 3D

<sup>μ</sup><sup>r</sup> depends on the values of initial pressure p0, viscosity ν and

<sup>μ</sup> : Verifying the proposition (6)

kt�<sup>μ</sup> <sup>r</sup>þ<sup>r</sup> ð Þ <sup>0</sup> � <sup>2</sup>

. Moreover, for a

<sup>∂</sup><sup>t</sup> 6¼ 0. In short, if there

μμð Þ <sup>r</sup>þr<sup>0</sup> <sup>¼</sup> <sup>0</sup>

1 1þe

<sup>P</sup> , can produce vortices, due to the

<sup>μ</sup> t, expressly incorporates these

Proof. Taking partial derivatives for ∂<sup>n</sup> x 1 r � �, ∂<sup>n</sup> y 1 r � � and ∂<sup>n</sup> z 1 r � �, recalling the derivatives of special functions of Eq. (16), it is shown that the derivative C<sup>∞</sup>. We only have to find the time derivatives: ∂<sup>n</sup> <sup>t</sup> <sup>p</sup>0<sup>P</sup> � � <sup>¼</sup> <sup>p</sup>0∂<sup>n</sup> <sup>t</sup> ð Þ P . Using Eq. (21) for P, we have.

$$\begin{aligned} \partial\_t P &= (-k)P(1-P) \\ \partial\_t^2 P &= (-k)^2 (1-2P)P(1-P) \\ \partial\_t^3 P &= (-k)^3 \left(1 - 6P + 6P^2\right)P(1-P) \\ \partial\_t^4 P &= (-k)^4 \left(1 - 14P + 36P^2 - 24P^3\right)P(1-P) \\ \partial\_t^5 P &= (-k)^5 \left(1 - 30P + 150P^2 - 240P^3 + 120P^4\right)P(1-P) \\ \partial\_t^6 (P) &= \partial\_t \left(\partial\_t^{n-1}(P)\right) \end{aligned} \tag{25}$$

It is always possible to find the derivative ∂<sup>n</sup> <sup>t</sup> ð Þ P as a function of the previous

derivative, since the resulting polynomial of each derivative <sup>n</sup> � 1 is of degree <sup>n</sup>. ▪ Proposition 7 Requirement (6). The energy must be limited in a defined volume and fundamentally it must converge at any time, such that t≥0.

$$\int\_{\mathbb{R}^3} |\mathbf{u}(x, y, z, t)|^2 dx dy dz \le \mathbb{C} \qquad for all \ t \ge 0 \qquad (bounded \ energy).$$

Proof. We will use the explicit form of velocity given in Eq. (21) uð Þ¼ x, y, z, t 2νμð Þ 1 � P ∇r, to obtain the vector module: j j u <sup>2</sup> <sup>¼</sup> <sup>4</sup>ν<sup>2</sup>μ<sup>2</sup>ð Þ <sup>1</sup> � <sup>P</sup> <sup>2</sup> . Rewriting Eq. (21), and applying a change of variable in: dxdydx <sup>¼</sup> <sup>4</sup>πr<sup>2</sup>dr.

$$\int\_{\mathbb{R}^3} \left| \mathbf{u}(\mathbf{x}, y, z, t) \right|^2 dx dy dz = \mathbf{1} \mathbf{6} \pi \nu^2 \mu^2 \int\_{r\_0}^{\infty} r^2 (\mathbf{1} - P)^2 dr \tag{26}$$

Making another change of variable dP ¼ μPð Þ 1 � P dr. Using (10), replacing <sup>r</sup><sup>2</sup> <sup>¼</sup> <sup>2</sup> μP � �<sup>2</sup> we have

$$\begin{split} \int\_{\mathbb{R}^3} |\mathbf{u}(\mathbf{x}, \mathbf{y}, z, t)|^2 d\mathbf{x} d\mathbf{y} dz &= 16\pi\nu^2 \mu^2 \int\_{P\_0}^{P\_\infty} \left(\frac{2}{\mu P}\right)^2 (1 - P)^2 \frac{dP}{\mu P(1 - P)} \\ &= \frac{64\pi\nu^2}{\mu} \int\_{P\_0}^{P\_\infty} \frac{1 - P}{P^3} dP \end{split} \tag{27}$$

Where radius <sup>r</sup> ! <sup>∞</sup>, when <sup>t</sup>≥0, we have lim <sup>r</sup>!<sup>∞</sup><sup>P</sup> <sup>¼</sup> lim <sup>r</sup>!<sup>∞</sup> <sup>1</sup> <sup>1</sup><sup>þ</sup> exp kt ð Þ exp ð Þ μr ¼ P<sup>∞</sup> ¼ 1: Moreover, physically if r ! r<sup>0</sup> ≈ 0 then t ! 0 we have lim <sup>r</sup>!<sup>0</sup>P ¼ lim <sup>r</sup>!<sup>0</sup> 1 <sup>1</sup><sup>þ</sup> exp kt ð Þ exp ð Þ μr ¼ <sup>P</sup><sup>0</sup> <sup>¼</sup> <sup>1</sup> <sup>2</sup> : Here, a probability <sup>1</sup> <sup>2</sup> represents maximum entropy.

$$\int\_{\mathbb{R}^3} |\mathbf{u}(\mathbf{x}, \mathbf{y}, \mathbf{z}, t)|^2 d\mathbf{x} d\mathbf{y} d\mathbf{z} = \frac{64\pi\nu^2}{\mu} \int\_{1/2}^1 \frac{1 - P}{P^3} dP = \frac{64\pi\nu^2}{\mu} \left[\frac{2P - 1}{2P^2}\right]\_{1/2}^1 \tag{28}$$

$$\int\_{\mathbb{R}^3} |\mathbf{u}|^2 d\mathbf{x} d\mathbf{y} dz \le \frac{32\pi\nu^2}{\mu} \tag{29}$$

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

In this way the value of the constant <sup>C</sup> is <sup>C</sup> <sup>¼</sup> <sup>32</sup>πν<sup>2</sup> <sup>μ</sup> : Verifying the proposition (6) completely. In general, Eq. (10) can be written f tð Þ¼ ,r þ r<sup>0</sup> 1 1þe kt�<sup>μ</sup> <sup>r</sup>þ<sup>r</sup> ð Þ <sup>0</sup> � <sup>2</sup> μμð Þ <sup>r</sup>þr<sup>0</sup> <sup>¼</sup> <sup>0</sup> and in this way discontinuities are avoided when r ! 0, but this problem does not occur since in the atomic nucleus <sup>r</sup><sup>0</sup> <sup>&</sup>lt;r<1:2A1=<sup>3</sup> is satisfied. ▪

Lemma 8 The irrotational field represented by the logistic probability function P xð Þ , <sup>y</sup>, <sup>z</sup>, <sup>t</sup> associated with the velocity <sup>u</sup> ¼ �2<sup>ν</sup> <sup>∇</sup><sup>P</sup> <sup>P</sup> , can produce vortices, due to the stochastic behavior of the physical variables p0, η, μ: These stochastic variations are in orders lower than the minimum experimental value.

Proof. The implicit function representing the solution of the Navier–Stokes 3D equation, <sup>1</sup> 1þe �<sup>μ</sup> <sup>r</sup>�<sup>k</sup> <sup>μ</sup> ð Þ<sup>t</sup> <sup>¼</sup> <sup>2</sup> <sup>μ</sup><sup>r</sup> depends on the values of initial pressure p0, viscosity ν and attenuation coefficient μ. Due to Heisenberg uncertainty principle, these parameters have a variation when we measure and use them, as is the case of the estimate of <sup>ξ</sup> <sup>¼</sup> <sup>r</sup> � <sup>ϕ</sup>ð Þ <sup>t</sup>, <sup>k</sup>, <sup>μ</sup> . The function <sup>ϕ</sup>ð Þ¼ <sup>t</sup>, <sup>k</sup>, <sup>μ</sup> <sup>k</sup> <sup>μ</sup> t, expressly incorporates these results, when �∞ <ξ< þ ∞. The physical and mathematical realities are mutually conditioned and allow for these surprising results. For a definite t there exist infinities ð Þ <sup>x</sup>, <sup>y</sup>, <sup>z</sup> that hold the relationship <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> . Moreover, for a definite r there are infinities t that respect the fixed-point theorem and create spherical trajectories. When the physical variables k, μ vary, even at levels of 1/100 or 1/1000, they remain below the minimum variation of the experimental value. We could try to avoid the existence of trajectories on the spherical surface, for which we must assume that the fluid is at rest or it is stationary, which contradicts the Navier– Stokes 3D equation, where all fluid is in accelerated motion <sup>∂</sup><sup>u</sup> <sup>∂</sup><sup>t</sup> 6¼ 0. In short, if there are trajectories in the sphere as long as it is probabilistically possible, this is reduced to showing that the expected value of the radius E r½ � jr≥0 exists and it is finite.

Derivation of E rð Þ jr≥0 :

Proposition 6 Requirement (5). The pressure is totally defined by the equivalence p xð Þ¼ , y, z, t p0P xð Þ , y, z, t and is infinitely differentiable in each of its components.

> x 1 r � �, ∂<sup>n</sup> y 1 r � � and ∂<sup>n</sup>

tives of special functions of Eq. (16), it is shown that the derivative C<sup>∞</sup>. We only

<sup>t</sup> <sup>P</sup> ¼ �ð Þ<sup>k</sup> <sup>5</sup> <sup>1</sup> � <sup>30</sup><sup>P</sup> <sup>þ</sup> <sup>150</sup>P<sup>2</sup> � <sup>240</sup>P<sup>3</sup> <sup>þ</sup> <sup>120</sup>P<sup>4</sup> � �Pð Þ <sup>1</sup> � <sup>P</sup>

derivative, since the resulting polynomial of each derivative <sup>n</sup> � 1 is of degree <sup>n</sup>. ▪ Proposition 7 Requirement (6). The energy must be limited in a defined volume and

Proof. We will use the explicit form of velocity given in Eq. (21) uð Þ¼ x, y, z, t

dxdydz <sup>¼</sup> <sup>16</sup>πν<sup>2</sup>

Making another change of variable dP ¼ μPð Þ 1 � P dr. Using (10), replacing

μ2 ð<sup>P</sup><sup>∞</sup> P0

> ð<sup>P</sup><sup>∞</sup> P0

<sup>2</sup> represents maximum entropy.

1 � P

32πν<sup>2</sup> μ

<sup>P</sup><sup>3</sup> dP <sup>¼</sup> <sup>64</sup>πν<sup>2</sup>

μ

dxdydz ≤C for all t≥0 ð Þ bounded energy :

μ2 ð<sup>∞</sup> r0 r 2 ð Þ <sup>1</sup> � <sup>P</sup> <sup>2</sup>

2 μP � �<sup>2</sup>

1 � P <sup>P</sup><sup>3</sup> dP

<sup>2</sup> <sup>¼</sup> <sup>4</sup>ν<sup>2</sup>μ<sup>2</sup>ð Þ <sup>1</sup> � <sup>P</sup> <sup>2</sup>

ð Þ <sup>1</sup> � <sup>P</sup> <sup>2</sup> dP

μPð Þ 1 � P

2P � 1 2P<sup>2</sup> � �<sup>1</sup>

forall t≥0

<sup>1</sup><sup>þ</sup> exp kt ð Þ exp ð Þ μr

1=2

<sup>t</sup> <sup>p</sup>0<sup>P</sup> � � <sup>¼</sup> <sup>p</sup>0∂<sup>n</sup>

Proof. Taking partial derivatives for ∂<sup>n</sup>

<sup>∂</sup>tP ¼ �ð Þ<sup>k</sup> <sup>P</sup>ð Þ <sup>1</sup> � <sup>P</sup>

ð Þ 1 � 2P Pð Þ 1 � P

<sup>t</sup> <sup>P</sup> ¼ �ð Þ<sup>k</sup> <sup>4</sup> <sup>1</sup> � <sup>14</sup><sup>P</sup> <sup>þ</sup> <sup>36</sup>P<sup>2</sup> � <sup>24</sup>P<sup>3</sup> � �Pð Þ <sup>1</sup> � <sup>P</sup>

<sup>t</sup> <sup>P</sup> ¼ �ð Þ<sup>k</sup> <sup>3</sup> <sup>1</sup> � <sup>6</sup><sup>P</sup> <sup>þ</sup> <sup>6</sup>P<sup>2</sup> � �Pð Þ <sup>1</sup> � <sup>P</sup>

have to find the time derivatives: ∂<sup>n</sup>

<sup>t</sup> <sup>P</sup> ¼ �ð Þ<sup>k</sup> <sup>2</sup>

<sup>t</sup> ð Þ¼ <sup>P</sup> <sup>∂</sup><sup>t</sup> <sup>∂</sup><sup>n</sup>�<sup>1</sup> <sup>t</sup> ð Þ <sup>P</sup> � �

j j <sup>u</sup>ð Þ <sup>x</sup>, <sup>y</sup>, <sup>z</sup>, <sup>t</sup> <sup>2</sup>

ð 3

we have

j j <sup>u</sup>ð Þ <sup>x</sup>, <sup>y</sup>, <sup>z</sup>, <sup>t</sup> <sup>2</sup>

<sup>2</sup> : Here, a probability <sup>1</sup>

j j <sup>u</sup>ð Þ <sup>x</sup>, <sup>y</sup>, <sup>z</sup>, <sup>t</sup> <sup>2</sup>

2νμð Þ 1 � P ∇r, to obtain the vector module: j j u

It is always possible to find the derivative ∂<sup>n</sup>

fundamentally it must converge at any time, such that t≥0.

and applying a change of variable in: dxdydx <sup>¼</sup> <sup>4</sup>πr<sup>2</sup>dr.

j j <sup>u</sup>ð Þ <sup>x</sup>, <sup>y</sup>, <sup>z</sup>, <sup>t</sup> <sup>2</sup>

dxdydz <sup>¼</sup> <sup>16</sup>πν<sup>2</sup>

dxdydz <sup>¼</sup> <sup>64</sup>πν<sup>2</sup>

ð 3 j j u 2

<sup>¼</sup> <sup>64</sup>πν<sup>2</sup> μ

Where radius <sup>r</sup> ! <sup>∞</sup>, when <sup>t</sup>≥0, we have lim <sup>r</sup>!<sup>∞</sup><sup>P</sup> <sup>¼</sup> lim <sup>r</sup>!<sup>∞</sup> <sup>1</sup>

Moreover, physically if r ! r<sup>0</sup> ≈ 0 then t ! 0 we have lim <sup>r</sup>!<sup>0</sup>P ¼ lim <sup>r</sup>!<sup>0</sup>

ð1 1=2

dxdydz≤

μ

∂2

Nuclear Materials

∂3

∂4

∂5

∂n

ð 3

<sup>r</sup><sup>2</sup> <sup>¼</sup> <sup>2</sup> μP � �<sup>2</sup>

<sup>P</sup><sup>0</sup> <sup>¼</sup> <sup>1</sup>

32

ð 3

ð 3 p xð Þ¼ , <sup>y</sup>, <sup>z</sup>, <sup>t</sup> <sup>p</sup>0P xð Þ , <sup>y</sup>, <sup>z</sup>, <sup>t</sup> ð Þ <sup>x</sup>, <sup>y</sup>, <sup>z</sup> <sup>∈</sup> <sup>3</sup> � � (24)

z 1 r

� �, recalling the deriva-

(25)

. Rewriting Eq. (21),

dr (26)

(27)

¼ P<sup>∞</sup> ¼ 1:

(28)

1 <sup>1</sup><sup>þ</sup> exp kt ð Þ exp ð Þ μr ¼

<sup>t</sup> ð Þ P . Using Eq. (21) for P, we have.

<sup>t</sup> ð Þ P as a function of the previous

The logistic density function for <sup>ξ</sup> when <sup>E</sup>ð Þ¼ <sup>ξ</sup> 0 and Varð Þ¼ <sup>ξ</sup> <sup>σ</sup><sup>2</sup> is defined by. <sup>h</sup>ð Þ¼ <sup>ξ</sup> <sup>μ</sup> exp ð Þ �<sup>ξ</sup> ½ � <sup>1</sup><sup>þ</sup> exp ð Þ �<sup>ξ</sup> <sup>2</sup> , where <sup>1</sup> <sup>μ</sup> <sup>¼</sup> <sup>σ</sup> ffiffiffi 3 <sup>p</sup> <sup>=</sup><sup>π</sup> is a scale parameter. Given that <sup>r</sup> <sup>¼</sup> <sup>ϕ</sup> <sup>t</sup>, <sup>p</sup>0, <sup>η</sup>, <sup>μ</sup> � � <sup>þ</sup> <sup>ξ</sup> function for <sup>r</sup> is then f rð Þ¼ exp ½ � �r�ϕð Þ• <sup>=</sup><sup>τ</sup> <sup>τ</sup> <sup>1</sup><sup>þ</sup> exp ð � �ð Þ <sup>r</sup>�ϕð Þ• <sup>=</sup><sup>τ</sup> <sup>2</sup> <sup>½</sup> , to facilitate the calculations we put <sup>ϕ</sup>ðÞ¼ • <sup>ϕ</sup>ð Þ¼ <sup>t</sup>, <sup>k</sup>, <sup>μ</sup> <sup>k</sup> <sup>μ</sup> t. By definition, the truncated density for r when <sup>r</sup>≥0 is given by f rð Þ¼ <sup>j</sup>r≥<sup>0</sup> f rð Þ P rð Þ <sup>≥</sup><sup>0</sup> for <sup>r</sup>≥0: Given that the cumulative distribution function for <sup>r</sup> is given by F rð Þ¼ <sup>1</sup> <sup>1</sup><sup>þ</sup> exp ð Þ kt�μ<sup>r</sup> , it follows that P rð Þ¼ <sup>≥</sup><sup>0</sup> <sup>1</sup> � <sup>F</sup>ð Þ¼ <sup>0</sup> exp ð ÞÞ <sup>ϕ</sup>ð Þ• <sup>1</sup><sup>þ</sup> exp ð ÞÞ <sup>ϕ</sup>ð Þ• <sup>¼</sup> <sup>1</sup> <sup>1</sup><sup>þ</sup> exp ð ÞÞ �ϕð Þ• . The derivation of E rð Þ <sup>j</sup>r≥<sup>0</sup> then proceeds as follows:

$$\begin{split} E(r|r \ge 0) &= \int\_{0}^{\infty} \mu r f(r|r \ge 0) dr = \frac{1}{P(r \ge 0)} \int\_{0}^{\infty} \mu r \frac{\exp\left[kt - \mu r\right]}{\left\{1 + \exp\left[kt - \mu r\right]\right\}^2} dr\\ E(r|r \ge 0) &= \frac{1}{P(r \ge 0)} \int\_{1/2}^{1} \frac{2}{\mu P} (\mu P(1 - P)) \frac{dP}{P(1 - P)} \end{split} \tag{29}$$

We replaced in Eq. (29) dP <sup>¼</sup> <sup>μ</sup>Pð Þ <sup>1</sup> � <sup>P</sup> dr and <sup>r</sup><sup>2</sup> <sup>¼</sup> <sup>2</sup> μP � �<sup>2</sup> of this manner we obtain

$$E(r|r\geq 0) = \frac{1}{P(r\geq 0)} \int\_{1/2}^{1} \frac{2}{\mu P} (\mu P(1-P)) \frac{dP}{P(1-P)} = \frac{1}{P(r\geq 0)} \frac{2}{\mu} \log(2) \tag{30}$$

where we have used the fact that

$$P(r \ge 0) = \frac{\exp kt}{1 + \exp kt}$$

$$E(r|r \ge 0) = \frac{1}{P(r \ge 0)} \frac{2}{\mu} \log \left(2\right) \le \frac{2}{\mu} \log \left(2\right) \qquad \text{for all } t \ge 0 \tag{31}$$

FNS ¼ m

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

du

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

<sup>F</sup><sup>N</sup> ¼ �∇Φð Þ¼� <sup>r</sup> <sup>g</sup><sup>μ</sup>

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

<sup>F</sup><sup>N</sup> ¼ �∇Φð Þ¼� <sup>r</sup> <sup>g</sup>μ<sup>2</sup>

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

<sup>F</sup><sup>N</sup> ¼ �∇Φð Þ¼� <sup>r</sup> <sup>g</sup>μ<sup>2</sup>

reaction k, nucleon mass m and C 6¼ 1.

3.2 Cross section and Golden ratio <sup>σ</sup><sup>1</sup>

femtoscope

the following equation:

cross section 3,4.

35

of an atom of the nuclide 12C)4.

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

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

> <sup>g</sup> <sup>¼</sup> <sup>16</sup>mπν<sup>k</sup> μð Þ A � 1

> > σ2

theoretical values for the total cross section per atom, σtot, which is related to <sup>μ</sup>

μ <sup>ρ</sup> <sup>¼</sup> <sup>σ</sup>tot

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

The attenuation coefficient, photon interaction cross sections and related quantities are functions of the photon energy. The total cross section can be written as

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

σtot ¼ σpe þ σcoh þ σincoh þ σtrip þ σph:<sup>n</sup> (40)

According to NIST and GEANT4 [17], current tabulations of <sup>μ</sup>

the sum over contributions from the principal photon interactions

dt ¼ �2mμνkPð Þ <sup>1</sup> � <sup>P</sup> <sup>∇</sup>r: (35)

<sup>8</sup><sup>π</sup> ð Þ <sup>A</sup> � <sup>1</sup> <sup>∇</sup>P: (36)

<sup>8</sup><sup>π</sup> ð Þ <sup>A</sup> � <sup>1</sup> <sup>P</sup>ð Þ <sup>1</sup> � <sup>P</sup> <sup>∇</sup>r: (37)

C (38)

uA (39)

<sup>8</sup><sup>π</sup> ð Þ <sup>A</sup> � <sup>1</sup> <sup>P</sup>ð Þ <sup>1</sup> � <sup>P</sup> <sup>∇</sup>r:

are important elements of the

▪

▪

<sup>ρ</sup> by

<sup>ρ</sup> rely heavily on

where the last equality follows from an application of the L'Hopital's rule P rð Þ¼ <sup>≥</sup><sup>0</sup> lim <sup>t</sup>!<sup>∞</sup> exp kt <sup>1</sup><sup>þ</sup> exp kt <sup>¼</sup> 1. ▪
