David Izabel
email: d.izabel@aliceadsl.fr

Can we understand the concepts of the General Relativity and of the Quantum Mechanics based on the principles of the strength of materials? Reflections and proposals

The quantum mechanics, in the infinitesimal world, is the reign of the quantification of energies, and of the probability of presence of particles with the famous wave function ψ associated at the Schrodinger equation. The General relativity, is for it, reserved at the infinitely large world where the mater, the energy density 𝑇 𝜇𝜈 influence the curvature 𝐺 𝜇𝜈 of the space time at 4 dimensions and reciprocally. Finally, the strength of materials is the mechanic of objects at the human scale (beams, columns, plates, shells) used for the design of structures, without common measures with this two pillars of the of the physics cited earlier. We are going to show in first part of this paper that the results of the quantum mechanics (quantification of energy, shape of the wave function) are similar at the eigen frequencies and eigen modes of a beam. In second part, we will show on simples cases, that the concepts of curvature linked with energy density present in general relativity are equally at the bases of the fundamental equations of the elasticity and of the strength of material. We will demonstrate finally that the stress energy tensor written for small speeds non relativistic is an extension in 4 dimensions of stress tensor of the elasticity theory.

st part) -Quantum mechanics and beam vibration 1.1) Free vibration of a beam on two supports -natural frequencies and mode of vibration

We consider a beam on two supports, with a span L (m), carried out with a material of elastic modulus E (MPa), that have a section with an inertia I (m 4 ) and an area S (m²) and that have a mass by unit of length (m= S) in kg/m (cf. figure 1). [START_REF] Ménad | Compléments technique relatif a l'effet de surpression Recommandations et précautions en vue de réduire les risques -rapport d'étude For the general relativity[END_REF] The beam is in free vibration and as a deflection along the time 𝑦 (𝑥,𝑡) : In addition, we have by connecting the radius of curvature R (m) at the bending moment 𝑀 (𝑥,𝑡) :

1 𝑅 = 𝜕 2 𝑦 (𝑥,𝑡) 𝜕𝑥 2 = 𝑀 (𝑥,𝑡) 𝐸𝐼 (2)
Knowing that:

𝜕𝑀 (𝑥,𝑡) 𝜕𝑥 = 𝑉 (𝑥,𝑡) (3)
𝜕𝑉 (𝑥,𝑡) 𝜕𝑥 = -𝑞 (𝑥,𝑡) [START_REF] Ménad | Compléments technique relatif a l'effet de surpression Recommandations et précautions en vue de réduire les risques -rapport d'étude For the general relativity[END_REF] We obtain by deriving twice respect to 𝑥 the equation (2): 𝜕𝑥 4 = -𝑞 (𝑥,𝑡) [START_REF] Beau | Théorie des champs des contraintes et des déformations en relativité générale et expansion cosmologique[END_REF] By equating equations ( 1) and [START_REF] Beau | Théorie des champs des contraintes et des déformations en relativité générale et expansion cosmologique[END_REF] in the case of auto-vibration, we obtain the well-known differential equation which controls the self-vibration of a beam on 2 supports: If we put 𝑦 (𝑥,𝑡) = 𝑞 (𝑡)  (𝑥) [START_REF] Beau | Sur l'hypothèse concernant l'existence d'un champ de gravité de type vectoriel couplant avec l'accélération des masses[END_REF] By replacing in equation ( 8) above, we obtains:

𝐸𝐼𝑞 (𝑡)  (𝑥) 𝑖𝑉 + 𝑆𝑞̈( 𝑡) 𝜙 (𝑥) = 0 [START_REF] Collective | Observation of Gravitational Waves from a Binary Black Hole Merger physical review letter[END_REF] We can therefore separate the variables: [START_REF]Cours de physique « tenseur energie impulsion[END_REF] As this equation has to be verified for all 𝑥 or t, the above ration have to be constant.

 𝐼𝑉  = - 𝑆 𝐸𝐼 × 𝑞q
We therefore define:

 𝐼𝑉  = - 𝑆 𝐸𝐼 × 𝑞q = 𝛼 4 𝐿 4 (12)
And we define the pulsation (circular natural frequency) ω thus:

 2 = 𝛼 4 𝐿 4 × 𝐸𝐼 𝑆 (13)
We can cut in two parts the differential equation [START_REF]Cours en ligne de richard Taillet univ grenoble-initiation à la relativité générale et restreinte For the quantum mechanics[END_REF] in two equations (separation of variables):

 (𝑥) 𝐼𝑉 -𝛼 4 𝐿 4  (𝑥) = 0 [START_REF] Fowler | Schrödinger's Equation in 1-D: Some Examples Curvature of Wave Functions[END_REF] 𝑞̈( 𝑡) +  2 𝑞 (𝑡) = 0 [START_REF] Errede | [END_REF] The study of the first equation gives the following general equation:

 (𝑥) =
The 4 boundary conditions allows to write:

0 ) 0 (   x  0 ) (  L x  0 ) 0 (   x II  0 ) (  L x II   (𝑥=0) = 𝐴𝑠𝑖𝑛 (𝛼 0 𝐿 ) + 𝐵𝑐𝑜𝑠 (𝛼 0 𝐿 ) + 𝐶𝑠𝑖𝑛ℎ (𝛼 0 𝐿 ) + 𝐷𝑐𝑜𝑠ℎ (𝛼 0 𝐿 ) (21)
 (𝑥=0) = 0𝐴 + 𝐵 + 0𝐶 + 𝐷 = 0 (22)  (𝑥=𝐿) = 𝐴𝑠𝑖𝑛 (𝛼 𝐿 𝐿 ) + 𝐵𝑐𝑜𝑠 (𝛼 𝐿 𝐿 ) + 𝐶𝑠𝑖𝑛ℎ (𝛼 𝐿 𝐿 ) + 𝐷𝑐𝑜𝑠ℎ (𝛼 𝐿 𝐿 ) (23)
 (𝑥=𝐿) = 𝐴𝑠𝑖𝑛(𝛼) + 𝐵𝑐𝑜𝑠(𝛼) + 𝐶𝑠𝑖𝑛ℎ(𝛼) + 𝐷𝑐𝑜𝑠ℎ(𝛼) = 0 (24)

′′ (𝑥=0) = -𝐴 𝛼 2 𝐿 2 𝑠𝑖𝑛 (𝛼 0 𝐿 ) -𝐵 𝛼 2 𝐿 2 𝑐𝑜𝑠 (𝛼 0 𝐿 ) + 𝐶 𝛼 2 𝐿 2 𝑠𝑖𝑛ℎ (𝛼 0 𝐿 ) + 𝐷 𝛼 2 𝐿 2 𝑐𝑜𝑠ℎ (𝛼 0 𝐿 ) (25) ′′ (𝑥=0) = 0𝐴 -𝐵 𝛼 2 𝐿 2 + 0𝐶 + 𝐷 𝛼 2 𝐿 2 = 0 (26) ′′ (𝑥=𝐿) = -𝐴 𝛼 2 𝐿 2 𝑠𝑖𝑛 (𝛼 𝐿 𝐿 ) -𝐵 𝛼 2 𝐿 2 𝑐𝑜𝑠 (𝛼 𝐿 𝐿 ) + 𝐶 𝛼 2 𝐿 2 𝑠𝑖𝑛ℎ (𝛼 𝐿 𝐿 ) + 𝐷 𝛼 2 𝐿 2 𝑐𝑜𝑠ℎ (𝛼 𝐿 𝐿 ) (27)
′′ (𝑥=𝐿) = -𝐴 𝛼 2 𝐿 2 𝑠𝑖𝑛(𝛼) -𝐵 𝛼 2 𝐿 2 𝑐𝑜𝑠(𝛼) + 𝐶 𝛼 2 𝐿 2 𝑠𝑖𝑛ℎ(𝛼) + 𝐷 𝛼 2
𝐿 2 𝑐𝑜𝑠ℎ(𝛼) = 0 (28)

We can put these equations on the form of a matrix:

[ 0 𝐵 0 𝐷 𝐴𝑠𝑖𝑛𝛼 𝐵𝑐𝑜𝑠𝛼 𝐶𝑠𝑖𝑛ℎ𝛼 𝐷𝑐𝑜𝑠ℎ𝛼 0 -𝐴 𝛼²𝑠𝑖𝑛𝛼 𝐿² -𝐵 𝛼² 𝐿² -𝐵 𝛼²𝑐𝑜𝑠𝛼 𝐿² 0 𝐶 𝛼²𝑠𝑖𝑛ℎ𝛼 𝐿² 𝐷 𝛼² 𝐿² 𝐷 𝛼²𝑐𝑜𝑠ℎ𝛼 𝐿² ] (29)
The determinant of this matrix have to be null.

We notice that from the equations ( 1) and ( 3) that B = D = 0

 (𝑥) = 𝐴𝑠𝑖𝑛 (𝛼 𝑥 𝐿 ) + 𝐵𝑐𝑜𝑠 (𝛼 𝑥 𝐿 ) + 𝐶𝑠𝑖𝑛ℎ (𝛼 𝑥 𝐿 ) + 𝐷𝑐𝑜𝑠ℎ (𝛼 𝑥 𝐿 ) (30)
 (𝑥) = 𝐴𝑠𝑖𝑛 (𝛼 𝑥 𝐿 ) + 𝐶𝑠𝑖𝑛ℎ (𝛼 𝑥 𝐿 ) (31)
And we have: -

𝑆 𝐸𝐼 × 𝑞q = 𝛼 4 𝐿 4 (32)
The two equations ( 2) and (4) gives:

 (𝑥=𝐿) = 𝐴𝑠𝑖𝑛(𝛼) + 𝐶𝑠𝑖𝑛ℎ(𝛼) = 0 (33) ′′ (𝑥=𝐿) = -𝐴 𝛼 2 𝐿 2 𝑠𝑖𝑛(𝛼) + 𝐶 𝛼 2
𝐿 2 𝑠𝑖𝑛ℎ(𝛼) = 0 (34) So:

 (𝑥=𝐿) = 𝐴𝑠𝑖𝑛(𝛼) + 𝐶𝑠𝑖𝑛ℎ(𝛼) = 0 (35) ′′ (𝑥=𝐿) = -𝐴𝑠𝑖𝑛(𝛼) + 𝐶𝑠𝑖𝑛ℎ(𝛼) = 0 (36)
The sum of these two equations gives:

2𝐶𝑠𝑖𝑛ℎ(𝛼) = 0 => 𝛼 = 0 𝑜𝑢 𝐶 = 0 (37)
The subtraction of these two equations gives:

2𝐴𝑠𝑖𝑛(𝛼) = 0 => 𝛼 = 𝑛 And so C = 0 (38)
Is the Eigen value of the system.

We obtain the following equation of the Eigen mode:

 (𝑥) = 𝐴𝑠𝑖𝑛 ( 𝑛𝑥 𝐿 ) (39)
That we write in function of n (quantification):

The Eigen mode is so written:

 𝑛(𝑥) = 𝐴 𝑛 𝑠𝑖𝑛 ( 𝑛𝑥 𝐿 ) (40)
The norm of the Eigen mode is so written:

∫  𝑛 (𝑥) 𝐿 0  𝑛 (𝑥) = ∫ 𝐴 𝑛 2 𝑠𝑖𝑛 2( 𝑛𝑥 𝐿 ) 𝑑𝑥 (41) 𝐿 0
We replace du sin²(ax) by (1 -cos (2𝑎𝑥)) (42)

∫  𝑛 (𝑥) 𝐿 0  𝑛 (𝑥) = 𝐴 𝑛 2 ∫ 1 2 𝐿 0 (1 -𝑐𝑜𝑠 ( 2𝑛𝑥 𝐿 )) 𝑑𝑥 (43) We define 𝑢 = 2𝑛𝑥 𝐿 𝑑𝑢 = 2𝑛 𝐿 𝑑𝑥 𝑑𝑥 = 𝐿 2𝑛 𝑑𝑢 ∫  𝑛 (𝑥) 𝐿 0  𝑛 (𝑥) = 𝐴 𝑛 2 ∫ 1 2 𝑑𝑥 + 𝐴 𝑛 2 1 2 ∫ 𝐿 2𝑛 𝑐𝑜𝑠𝑢 𝑑𝑢 (44) 𝐿 0 𝐿 0
The antiderivative of cosu is sinu :

∫  𝑛 (𝑥) 𝐿 0  𝑛 (𝑥) = 𝐴 𝑛 2 [( 𝑥 2 - 𝐿 4𝑛 𝑠𝑖𝑛 ( 2𝑛𝑥 𝐿 ))] 0 𝐿 = 𝐴 𝑛 2 ( 𝐿 2 -0 -(0 -0)) = 𝐴 𝑛 2 𝐿 2 (45)
We note that if we chose𝐴 𝑛 = √ 2 𝐿 (46)) the modes  𝑛 (𝑥) would form an orthonormal basis for the canonical scalar product.

In the case of normed Eigen vector we have so:

 𝑛(𝑥) = √ 2 𝐿 𝑠𝑖𝑛 ( 𝑛𝑥 𝐿 ) (47)
We have to study the second equation now :

𝑞̈( 𝑡) +  2 𝑞 (𝑡) = 0
We fix 𝑞 (𝑡) as following:

𝑞 (𝑡) = 𝑎𝑐𝑜𝑠𝑡 + 𝑏𝑠𝑖𝑛𝑡 𝑞̇( 𝑡) = -𝑎𝑠𝑖𝑛𝑡 + 𝑏𝑐𝑜𝑠𝑡 𝑞̈( 𝑡) = -𝑎²𝑐𝑜𝑠𝑡 -𝑏 2 𝑠𝑖𝑛𝑡
At the time t = 0, the system is in static, and so the speed is null:

𝑞̇( 𝑡=0) = -𝑎𝑠𝑖𝑛0 + 𝑏𝑐𝑜𝑠0
That imply b = 0:

And so the final deflection of the beam is:

𝑦 (𝑥,𝑡) = 𝑞 (𝑡)  (𝑥) (9) 𝑦 𝑛(𝑥,𝑡) = 𝑎 √ 2 𝐿 𝑠𝑖𝑛 ( 𝑛𝑥 𝐿 ) cos(𝜔𝑡) (9𝑏𝑖𝑠)

Note

We find this result by writing that the determinant of the matrix of the components must be zero.

Determination of circular natural frequency:

 2 = 𝛼 4 𝐿 4 × 𝐸𝐼 𝑆 (48)
With α = n (49)

 2 = 𝑛 4  4 𝐿 4 × 𝐸𝐼 𝑆 (50) =>  = 𝑛²² 𝐿² √ 𝐸𝐼 𝑆 ( 51
)
With these equations, we can define the different Eigen mode of vibration of the beam (cf. figure 2).

1 st mode :

 1 = ² 𝐿² √ 𝐸𝐼 𝑆  1(𝑥) = √ 2 𝐿 𝑠𝑖𝑛 ( 𝑥 𝐿 )
2 nd mode :

3 rd mode : We consider a particle in a potential well as defined below (cf. figure 3).

Figure 3 -Particle in movement in a potential well -

The Schrodinger equation independent of the time can be written as following:

( ħ 2 2𝑚 ) 𝜕 2  (𝑥)
𝜕𝑥 2 + 𝐸 𝑚  (𝑥) = 0 (52)

 2 = 4² 𝐿² √ 𝐸𝐼 𝑆  2(𝑥) = √ 2 𝐿 𝑠𝑖𝑛 ( 2𝑥 𝐿 )  2 = 9² 𝐿² √ 𝐸𝐼 𝑆  3(𝑥) = √ 2 𝐿 𝑠𝑖𝑛 ( 3𝑥 𝐿 ) m m v v 𝑥 L 0
This equation can be also written on the form curvature = k energy

𝑑 2  (𝑥) 𝑑𝑥 2 = -( 2𝑚 ħ 2  (𝑥) ) 𝐸 𝑚 (52 𝑏𝑖𝑠)
The dimensional equation is so the following:

1 𝑚² = 𝑘𝑔 ( 𝑘𝑔𝑚²𝑠 𝑠² ) 2 × 𝑘𝑔𝑚² 𝑠² 1 𝑚² = 𝑠² 𝑘𝑔𝑚 4 × 𝑘𝑔𝑚² 𝑠² 1 𝑚² = 𝑠² 𝑘𝑔𝑚 × 𝑈 𝑚 3
The boundary conditions are therefore the following:

The particle is present in the potential well:

The particle is not present in the potential well:

0 ≤ 𝑥 ≤ 𝐿 𝑥 < 0 ou 𝑥 > 𝐿 ( ħ 2 2𝑚 ) 𝜕 2  (𝑥) 𝜕𝑥 2 + 𝐸 𝑚  (𝑥) = 0  (𝑥) = 0
The particle follows the Schrodinger equation there is no particle Otherwise,  (𝑥) is a continuous function. The consequence is so:

 (0) =  (𝐿) = 0 So: 𝑑 2  (𝑥) 𝑑𝑥 2 + 𝑘 2  (𝑥) = 0 (53)
With:

𝑘 2 = 2𝑚𝐸 𝑚 ħ 2 (54)
The solution of the wave function for this differential equation is of the following shape:

 (𝑥) = 𝐴𝑠𝑖𝑛(𝑘𝑥) + 𝐵𝑐𝑜𝑠(𝑘𝑥) (55)
We suppose that the energy Em is positive.

By use of the boundary condition we can find the constant A and B:

In 𝑥 = 0,  (0) = 0 :

That this imply that 𝐵 = 0

 (𝑥) = 𝐴𝑠𝑖𝑛(𝑘𝑥) In 𝑥 = 𝐿,  (𝐿) = 0 :  (𝐿) = 𝐴𝑠𝑖𝑛(𝑘𝐿) = 0
It is so necessary that with A and k different from 0 that 𝑘𝐿 = 𝑛 That imply that:

𝑘 = 𝑛 𝐿 ( 56
)
With n an integer =1,2,3……

As we have posed:

𝑘 2 = 2𝑚𝐸 𝑚 ħ 2
The result is :

𝑘 2 = 2𝑚𝐸 𝑚 ħ 2 = 𝑛 2  2 𝐿 2 (57)
We finally obtain the quantified values of the energy:

𝐸 𝑚,𝑛 = 𝑛 2  2 ħ 2 2𝑚𝐿 2 (58) With n = 1,2,3,…
So, the Eigen values of the energy are quantified.

We can finally write the expression of the wave function:

 (𝑥) = 𝐴𝑠𝑖𝑛 ( 𝑛𝑥 𝐿 𝑥)
We are looking for now the value of the constant A:

As the particle has to be somewhere in the quantum well

The function ‖ (𝑥) ‖ 2 = 1 to satisfy the density of probability (the particle must be somewhere in the box between 0 and L).

∫ ‖ (𝑥) ‖ 2 𝑑𝑥 = 1 (59) 𝐿 0 ∫ 𝐴 2 𝑠𝑖𝑛² ( 𝑛𝑥 𝐿 ) 𝑑𝑥 𝐿 0 = 𝐴 2 2 ∫ {1 -𝑐𝑜𝑠 ( 2𝑛𝑥 𝐿 )} 𝑑𝑥 𝐿 0 = 𝐴 2 2 [𝑥 - 𝐿 2𝑛 𝑠𝑖𝑛 ( 2𝑛𝑥 𝐿 )] 0 𝐿 = 𝐴 2 2 (𝐿 - 𝐿 2𝑛 𝑠𝑖𝑛 ( 2𝑛𝐿 𝐿 )) = 𝐴 2 𝐿 2 = 1 So 𝐴 = √ 2 𝐿 (60)
And the solution of the problem in the quantum well, the wave function, is so:

 (𝑥) = √ 2 𝐿 𝑠𝑖𝑛 ( 𝑛𝑥 𝐿 𝑥) (61)
Wave function probability to be present

 (𝑥) = √ 2 𝐿 𝑠𝑖𝑛 ( 𝑛𝑥 𝐿 𝑥) ‖ (𝑥) ‖ 2 = 2 𝐿 𝑠𝑖𝑛² ( 𝑛𝑥 𝐿 𝑥) (62)
Not about the Energy: By the following classic notation:

𝐸 = ℎ𝑣 = ℎ 𝑇 And with 𝑇 = 2  𝐸 = ℎ 2 = ħ ħ = 𝑛²² 𝐿² ( ħ 2 2m )  = 𝑛²² 𝐿² ( ħ 2m ) (63)

Conclusion of this first part

Table 1 below shows the perfect analogy between the vibrations of a beam on two supports (resulting from the strength of the materials) and the quantification of the energy of a particle in a quantum well (derived from quantum mechanics).

Case studied

Eigen value

Eigen Modes Natural circular Frequencies Quantified energy

Eigen mode of oscillation of a beam Expression of the Wave function Beam on two supports

 = 𝑛²² 𝐿² (√ 𝐸𝐼 𝑆 )  𝐼𝑉  = - 𝑆 𝐸𝐼 × 𝑞q = 𝛼 4 𝐿 4  (𝑥) 𝐼𝑉 - 𝛼 4 𝐿 4  (𝑥) = 0 𝑦 (𝑥,𝑡) = 𝑞 (𝑡)  (𝑥) Without interests  𝑛(𝑥) = √ 2 𝐿 𝑠𝑖𝑛 ( 𝑛𝑥 𝐿 )  𝐼𝑉  = - 𝑆 𝐸𝐼 × 𝑞q = 𝛼 4 𝐿 4 𝑑 2 𝑞 (𝑡) 𝑑𝑡² + ( 𝛼 4 𝐸𝐼 𝐿 4 𝑆 ) 𝑞 (𝑡) = 0 𝑦 (𝑥,𝑡) = 𝑞 (𝑡)  (𝑥)

Without interests

Particle in the potential well

With the notation :

𝐸 = ℎ𝑣 = ℎ 𝑇  = 𝑛²² 𝐿² ( ħ 2m ) 𝑑 2  (𝑥) 𝑑𝑥 2 + ( 2𝑚𝐸 𝑚 ħ 2 )  (𝑥) = 0 E = 𝑛²² 𝐿² ( ħ 2 2m ) Without interests  𝑛(𝑥) = √ 2 𝐿 𝑠𝑖𝑛 ( 𝑛𝑥 𝐿 ) 𝑑 2  (𝑥) 𝑑𝑥 2 + ( 2𝑚𝐸 𝑚 ħ 2 )  (𝑥) = 0

2.1) Case of the beam on two simple supports

Considering the same beam as that defined in figure 1.

The fundamental relation connecting the curvature (1/R) at the bending moment 𝑀 (𝑥) and at the second derivative of the deflection 𝑦 (𝑥) writes:

𝑑 2 𝑦 𝑑𝑥 2 = - 𝑀 (𝑥) 𝐸𝐼 = 1 𝑅 (64)
In this expression, 𝑦 (𝑥) is the deflection of the beam (m), 𝑀 (𝑥) the bending moment (N.m), E the young modulus of the material that constitute the beam (N/m²), I the inertia in(m 4 ) and R the curvature radius in (m).

The exact expression of the curvature is given in the expression (65). The part in cube root is negligible.

1 𝑅 = 𝑑²𝑦 𝑑𝑥² ( √ 1 + ( 𝑑𝑦 𝑑𝑥 ) 2 ) 3 (65)
In addition the elastic bending energy of a beam can be written as:

𝑈 = 1 2 ∫ 𝑀 (𝑥) 2 𝐸𝐼 𝑑𝑥 (66) 𝐿 0
Considering to simplify a beam under a constant bending moment M at each extremity (cf. figure 4).

Figure 4 -Beam on two supports under a bending moment M at each extremity -

The bending moment equation (pure bending) is the following:

𝑀 (𝑥) = 𝑀 (67)
From the equation (64) w obtain:

𝑑 2 𝑦 𝑑𝑥 2 = - 𝑀 𝐸𝐼 = 1 𝑅 (68) 𝑀 (𝑥) = 𝑀 = - 𝐸𝐼 𝑅 ( 69
)
By introducing the expression (69) in the expression of bending energy (66):

𝑈 = 1 2 ∫ (𝐸𝐼) 2 𝑅 2 𝐸𝐼 𝑑𝑥 = 1 2 ∫ 𝐸𝐼 𝑅 2 𝑑𝑥 𝐿 0 𝐿 0
With the curvature that is constant, we obtain:

𝑈 = 1 2 𝐸𝐼𝐿 𝑅 2
So in pure bending:

1 𝑅 2 = 2 𝐸𝐼 ( 𝑈 𝐿 ) (70) 𝑀, 𝐸, 𝐼 𝑥 𝑦 (𝑥,𝑡) M R 𝑦 𝐿 M
We obtain thus a relation between the curvature and the strain energy of a beam Note: By integral two times from 𝑥 the expression (68) and considering that the deflection have to be null on each support, we obtain the expression of the deflection.

𝑦 (𝑥) = - 𝑀 2𝐸𝐼 𝑥 2 + 𝑀𝐿 2𝐸𝐼 𝑥 (71)
And we find the well-known result of the strength of materials:

𝑦 (𝐿/2) = - 𝑀𝐿 2 8𝐸𝐼 + 𝑀𝐿 2 4𝐸𝐼 = 𝑀𝐿 2 8𝐸𝐼 (72) 𝑑 2 𝑦 (𝑥) 𝑑𝑥 2 - 𝑀 𝐸𝐼 = 1 𝑅

2.2) Case of a thin plate of thickness h

Considering a thin plane h, of sides 𝑥 et 𝑦 sunder a bending moments Mx (cf. figure 5) : The elastic bending energy of the plate can be written so:

𝑈 = 1 2
𝑀 𝑥 𝑦𝜃 (76)

𝑈 𝑥 = - 1 2 𝑀 𝑥 𝜕 2 𝑤 (𝑥,𝑦) 𝜕𝑥 2 𝑥𝑦 (77)
The contribution of energy following 𝑦 can be written following the same approach if a bending 𝑀 𝑦 is now applied on the side 𝑥 :

𝑈 𝑦 = - 1 2 𝑀 𝑦 𝜕 2 𝑤 (𝑥,𝑦) 𝜕𝑦 2 𝑥𝑦 (78)
The contribution due at the torsion can be also written:

𝑈 𝑥𝑦 = 1 2 2𝑀 𝑥𝑦 𝜕 2 𝑤 (𝑥,𝑦) 𝜕𝑦𝜕𝑥 𝑥𝑦 (79)
Finally the strain energy of the plate can be defined: We obtain a relation between the curvature and the density of strain energy of the plate:

𝑈 = 𝑈 𝑥 + 𝑈 𝑦 + 𝑈 𝑥𝑦 (80) 𝑈 = - 1 2
[( 1 𝑅 𝑥 ) 2 + ( 1 𝑅 𝑦 ) 2 + 2(1 -𝜈) {( 1 𝑅 𝑥𝑦 ) 2 } + 2𝜈 { 1 𝑅 𝑥 1 𝑅 𝑦 }] = 24(1 -𝜈 2 ) 𝐸ℎ 3 × 𝑈 𝑥𝑦 (90) Note:
The expression below allow to reformulate and to simplify the expression (89): Therefore the expression (89) can be written from the two expression (91) and (92):

-2(1 -𝑣) [ 𝜕 2 𝑤 𝜕𝑥 2 𝜕 2 𝑤 𝜕𝑦 2 -( 𝜕 2 𝑤 𝜕𝑥𝜕𝑦 ) 2 ] = -2 𝜕 2 𝑤 𝜕𝑥 2 𝜕 2 𝑤 𝜕𝑦 2 + 2 ( 𝜕 2 𝑤 𝜕𝑥𝜕𝑦 ) 2 + 2𝑣 𝜕 2 𝑤 𝜕𝑥 2 𝜕 2 𝑤 𝜕𝑦 2 -2𝑣 ( 𝜕 2 𝑤 𝜕𝑥𝜕𝑦 ) 2 (91)
𝑈 = 𝐷 2 [( 𝜕 2 𝑤 (𝑥,𝑦) 𝜕𝑥 2 + 𝜕 2 𝑤 (𝑥,𝑦) 𝜕𝑦 2 ) 2 -2(1 -𝑣) [ 𝜕 2 𝑤 𝜕𝑥 2 𝜕 2 𝑤 𝜕𝑦 2 -( 𝜕 2 𝑤 𝜕𝑥𝜕𝑦 ) 2 ]] 𝑥𝑦 (93)
With for the bending rigidity D:

𝐷 = 𝐸ℎ 3 12(1 -𝑣²) (94) 𝑈 = 𝐷 2 [(- 1 𝑅 𝑥 - 1 𝑅 𝑦 ) 2 -2(1 -𝑣) [ 1 𝑅 𝑥 1 𝑅 𝑦 -( 1 𝑅 𝑥𝑦 ) 2 ]] 𝑥𝑦 (95)
If we present the expression to put in first plane the curvature in one side and the energy of the other side:

[(- 1 𝑅 𝑥 - 1 𝑅 𝑦 ) 2 -2(1 -𝑣) [ 1 𝑅 𝑥 1 𝑅 𝑦 -( 1 𝑅 𝑥𝑦 ) 2 ]] = ( 2 𝐷 ) 𝑈 𝑥𝑦 (96)
We are developing now this expression in the case of the pure flexion:

The radius of curvature are all equal. The radius of curvature due to the torsion is null.

𝜕 2 𝑤 𝜕𝑥𝜕𝑦 = 0, 1 𝑅 𝑥 = 1 𝑅 𝑦 = 1 𝑅 (97)
By introducing these expressions in the equation ( 96) we obtain:

𝑈 = 𝐷 2 [(- 2 𝑅 ) 2 -2(1 -𝑣) [ 1 𝑅 2 ]] 𝑥𝑦 𝑈 = 𝐷 2 [ 4 𝑅 2 - 2 𝑅 2 + 2𝑣 1 𝑅 2 ] 𝑥𝑦 𝑈 = 𝐷 2 [ 2 𝑅 2 + 2𝑣 1 𝑅 2 ] 𝑥𝑦 𝑈 = 𝐷 𝑅 2 [1 + 𝑣]𝑥𝑦 So: 1 𝑅 2 = 1 𝐷(1 + 𝑣) 𝑈 𝑥𝑦 (98)
If we take into account of the bending rigidity (94) and if we define fix 𝑥. 𝑦 = 𝐴

1 𝑅 2 = 12(1 -𝑣²) 𝐸ℎ 3 (1 + 𝑣) 𝑈 𝐴 1 𝑅 2 = 12(1 -𝑣²) 𝐸ℎ 2 (1 + 𝑣) × 𝑈 𝐴ℎ 1 𝑅 2 = 12(1 + 𝑣)(1 -𝑣) 𝐸ℎ 2 (1 + 𝑣) × 𝑈 𝐴ℎ
We obtain in the case of the pure bending:

1 𝑅 2 = 12(1 -𝑣) 𝐸ℎ 2 × 𝑈 𝐴ℎ ( 99
)
And the dimensional equation is:

1 𝑚² = 1 𝑘𝑔𝑚 𝑠²𝑚² × 𝑚² × 𝑈 𝑚 3 1 𝑚² = 𝑠² 𝑘𝑔𝑚 × 𝑈 𝑉

Energy density Curvature

Proportionality factor proportionnalité

2.3) Case of the General relativity

2.3.1) Expression Einstein Equation

The Einstein equation has to be written as following:

𝐺 𝜇𝑣 = 𝜅𝑇 𝜇𝑣 (100)
In this equation, the curvature and the energy of the space time are linked (cf. figure 7). In addition the dimensional equation of κ is the following:

𝜅 = 𝐿 3 𝑇 4 𝑀𝑇²𝐿 4 = 𝑇² 𝑀𝐿 = 𝑠² 𝑘𝑔𝑚 (102)
The developed equation of Einstein becomes:

𝑅 𝜇𝜈 - 1 2 𝑔 𝜇𝜈 𝑅 + 𝛬𝑔 𝜇𝜈 = 8𝜋𝐺 𝑐 4 𝑇 𝜇𝜈 (103) 1 𝑚² = 𝑠² 𝑘𝑔𝑚 × 𝑈 𝑉

2.3.2) Details on the curvature tensor

The Ricci tensor is obtained by contraction of the Riemann tensor on the indices λ:

𝑅 𝜇𝑣 = 𝑅 𝜇𝑣  R is a contraction of the tensor 𝑅 𝜇𝑣 .
The curvature tensor or Riemann tensor is written as following:

𝑅 𝜇𝑣𝛼  = 𝛤 𝜇𝛼,𝑣  -𝛤 𝜇𝑣,𝛼  + 𝛤 𝑣  𝛤 𝜇𝛼  -𝛤 𝛼  𝛤 𝜇𝑣  (104)
Or in another way:

𝑅 𝜇𝑣𝛼  = 𝜕𝛤 𝜇𝛼  𝜕𝑥 𝑣 - 𝜕𝛤 𝜇𝑣  𝜕𝑥 𝛼 + 𝛤 𝑣  𝛤 𝜇𝛼  -𝛤 𝛼  𝛤 𝜇𝑣  (104 𝑏𝑖𝑠)
By definition the definition of the Christoffel symbols is the following:

𝛤 𝜇𝑣  = 1 2 𝑔  (𝑔 𝜇,𝑣 + 𝑔 𝑣,𝜇 -𝑔 𝜇𝑣, ) = 1 2 𝑔  ( 𝜕𝑔 𝜇 𝜕𝑥 𝑣 + 𝜕𝑔 𝑣 𝜕𝑥 𝜇 - 𝜕𝑔 𝜇𝑣 𝜕𝑥  ) (105)
gμν is the metric,

The coefficient of the metric are issued of the differential distance (special relativity):

𝑑𝑠 2 = 𝑐 2 𝑑𝑡 2 -𝑑𝑥 2 -𝑑𝑦 2 -𝑑𝑧 2 (106)
So, considering the Einstein convention of summation:

𝑑𝑠 2 = 𝑔 𝜇𝑣 𝑑𝑥 𝜇 𝑑𝑥 𝑣 (107) 𝑔 𝜇𝑣 =  𝜇𝑣 𝜕𝜉 𝛼 𝜕𝑥 𝜇 𝜕𝜉 𝛽 𝜕𝑥 𝑣 (108)
𝜉 0 = 𝑐𝑡; 𝜉 1 = 𝑥; 𝜉 2 = 𝑦; 𝜉 3 = 𝑧 (109)

 𝜇𝑣 = [ 1 -0 -0 -0 0 -1 -0 -0 0 0 -0 -0 -1 -0 -0 -1 ] (110)
Λ is the cosmologic constant (possible candidate to explain the dark energy and dark matter)

The    are the first derivative of the metric gμν :

For example in coordinates (t,r,θ,φ) :

𝛤 𝑟,𝑟  = 𝛤 𝑟𝑡,𝑟 𝑡 + 𝛤 𝑟𝑟,𝑟 𝑟 + 𝛤 𝑟𝜃,𝑟 𝜃 + 𝛤 𝑟𝜑,𝑟 𝜑 = 𝜕𝛤 𝑟𝑡 𝑡 𝜕𝑟 + 𝜕𝛤 𝑟𝑟 𝑟 𝜕𝑟 + 𝜕𝛤 𝑟𝜃 𝜃 𝜕𝑟 + 𝜕𝛤 𝑟𝜑 𝜑 𝜕𝑟 (111)
So the 𝛤 𝑟,𝑟  are the second derivatives of the metrics 𝑔 𝜇𝑣 that we can by analogy compare in one and two dimensions with the expressions of curvatures given in the expression (70) , (90), ( 96) and (99).

So 𝐺 𝜇𝑣 has the dimensional value of a curvature in 1/m² For example the first terms 𝑅 𝜇𝑣𝛼  (see 104 bis)

𝜕𝛤 𝜇𝛼  𝜕𝑥 𝑣 = 𝜕 { 1 2 𝑔  ( 𝜕𝑔 𝜇 𝜕𝑥 𝑣 + 𝜕𝑔 𝑣 𝜕𝑥 𝜇 - 𝜕𝑔 𝜇𝑣 𝜕𝑥  )} 𝜕𝑥 𝑣

2.3.3) Detail on the stress energy tensor

We have showing below that the component of the stress energy tensor 𝑇 𝜇𝜈 have the dimension of an energy density G = 6.6726x10 -11 m 3 kg -1 s -2 1J = 1W.s = 1 N.m = 1kg.m².s -2 kg

𝐶𝑢𝑟𝑣𝑎𝑡𝑢𝑟𝑒 = 8𝜋𝐺 𝑐 4 × 𝐸𝑛𝑒𝑟𝑔𝑦 𝑉𝑜𝑙𝑢𝑚𝑒 = 8𝜋𝐺 𝑐 2 𝑀𝑎𝑠𝑠 𝑉𝑜𝑙𝑢𝑚𝑒 (112) 1/L² (L/T) 4 (L 3 )
The stress energy tensor is a matrix where the component are given below (113): with m the mass and V unitary volume, c the speed of light, vi a speed in the direction i.

𝑇 𝜇𝑣 = [ 𝑚𝛾
The factor of Lorentz becomes from the Lorentz transformation that imply that the speed of light stay constant in all the referential.

𝛾 = 1 √ 1 - 𝑣 2 𝑐 2 = 1 √1 -𝛽 2 (114)
We show that at low velocity (in a non-relativistic situation) and in 3 dimensions, this tensor with 16 components in four-dimensional space-time actually includes the stress tensor (9 components) of the continuum mechanics from which derive the strength of materials [START_REF]Cours de physique « tenseur energie impulsion[END_REF].

The stress tensor in two dimensions can be written as following:

𝜎 𝑖𝑗 = [ 𝜎 𝑥𝑥  𝑥𝑦  𝑥𝑦 𝜎 𝑦𝑦 ] (115)
And in 3 dimensions:

𝜎 𝑖𝑗 = [ 𝜎 𝑥𝑥 𝜏 𝑥𝑦 𝜏 𝑥𝑧 𝜏 𝑦𝑥 𝜎 𝑦𝑦 𝜏 𝑦𝑧 𝜏 𝑧𝑥 𝜏 𝑧𝑦 𝜎 𝑧𝑧 ] (116)
In theory of elasticity, that come from the continuum mechanics, the relation between the stress tensor 𝜎 𝑖𝑗 and the applied force Qi on a surface of normal nj can be written as following:

𝑄 𝑖 = 𝜎 𝑖𝑗 𝑛 𝑗 (117)
In a field of a variational approach, the stress tensor can be written as follow:

𝑢 𝜇 =
The Einstein equation build a link in 4 dimensions (space time) with the curvature tensor 𝐺 𝜇𝑣 (dimension 1/m²) and the stress energy tensor 𝑇 𝜇𝑣 (dimension energy/m 3 ) that is itself a generalization in 4 dimensions of the stress tensor of the continuum mechanics.

2.4) Conclusion of this second part

The table 2 below makes a synthesis of the results obtained. We can see that the general relativity is a generalization of the elastic theory in 4 dimensions of the space time.

Theory considered

Number of dimensions

Formulation between energy and curvature Example of application (pure bending) or expression at low speed

Beam on two simple supports in elasticity

1 1 𝑅 = 𝑑²𝑦 𝑑𝑥² ( √ 1 + ( 𝑑𝑦 𝑑𝑥 ) 2 ) 3 𝑈 = 1 2 ∫ 𝑀 (𝑥) 2 𝐸𝐼 𝑑𝑥 𝐿 0 𝑑 2 𝑦 𝑑𝑥 2 = - 𝑀 (𝑥) 𝐸𝐼 = 1 𝑅 1 𝑅 2 = 2 𝐸𝐼 ( 𝑈 𝐿 ) Thin plate 2 [(- 1 𝑅 𝑥 - 1 𝑅 𝑦 ) 2 -2(1 -𝑣) [ 1 𝑅 𝑥 1 𝑅 𝑦 -( 1 𝑅 𝑥𝑦 ) 2 ]] = ( 2

Conclusions

We have showed on several examples that the strength of materials allows to understand at our scale certain fundamental principles of the quantum mechanics:

-The eigen mode of a beam on two simple supports correspond at the different shape of the wave function connected with the jump energy of a particle in a quantum well, -The natural circular frequencies of a beam on two simple supports correspond at the quantification of the energy of a particle in a potential well.

We have showed that the general relativity is a generalization in 4 dimensions (space time) of the relation curvature/energy equally present in strength of materials for the beam and for the thin plates Finally, we have showed the stress energy tensor written at low speed give the stress tensor of the elasticity theory.

Extension of this article, next steps:

We have equally shown that in strength of material 2 main equations are necessary. One in static (curvature = K energy density) and one in dynamic (natural frequencies and eigen modes). So if the analogy between the beam (or plate) in strength of material and the space time is total; two equations should be also necessary at large scale. The first is the Einstein Equation connecting the curvature and the energy. The second remain to find.

If the analogy with the strength of the materials is exact we have some clues about this second equation.

This second equation would allow to quantify the space time (analogy with the natural frequency and eigen mode of the beam that represent also the energy quantification and shape of the wave function ψ in quantum mechanics).see table 3 in annex A.

Notice that this equation should be in 4th derivative of the space metric and 2th derivative of the time metric, and of course tensorial written to be valid in all the referential (covariant derivative).

Figure 1 -

 1 Figure 1 -Beam on 2 supports under its self-weight -

Figure 2 - 1 . 2 )

 212 Figure 2 -Eigen mode and circular natural frequencies of a beam on two supports -1.2) Particle in a potential well with an infinite dimension

Tableau 1 -

 1 Analogy between the natural circular frequencies of vibration of a beam on 2 supports and the quantified energy of a particle in a quantum wellThe circular natural frequencies  of a beam in strength of materials have an analogy with the jump of quantified energy Em of a particle in a quantum well. modes s  of vibration of a beam in strength of materials have an analogy with the Eigen mode of the wave function in the quantum well.2 nd part) Relation between curvature and energy in 1 and 2 dimensions (beam and slab in strength of materials) and in 4 dimensions (General relativity)

Figure 5 -Figure 6 -

 56 Figure 5 -Plate under a bending moment on its sides -𝑀 𝑥 Represent a bending moment by unit of length:

Figure 7 -

 7 Figure 7 -Symbolic view of the curvature of the space time by projection in a plane space -With: 𝐺 𝜇𝑣 is the Einstein tensor.

Annex A -Synthesis of the results - Table 3 -

 3 Basic evidence of one equation is missing at the space time scale by analogy with the 2 main equations in strength of material -function 𝜓 (𝑥)Deflection of the beam 𝑦 (𝑥,𝑡) Metric 𝑔 𝜇𝜈 with𝜇𝜈 = 0(𝑡) 𝑡𝑜 3(𝑥, 𝑦, 𝑧)

  2 𝑐² 𝑉 𝜌𝛾 2 𝑐𝑣 𝑥 𝜌𝛾 2 𝑐𝑣 𝑦 𝜌𝛾 2 𝑐𝑣 𝑧 𝜌𝛾 2 𝑐𝑣 𝑥 𝜌𝛾 2 𝑣 𝑥 𝑣 𝑥 𝜌𝛾 2 𝑣 𝑥 𝑣 𝑦 𝜌𝛾 2 𝑣 𝑥 𝑣 𝑧

				(113)
	𝜌𝛾 2 𝑐𝑣 𝑦	𝜌𝛾 2 𝑣 𝑦 𝑣 𝑥	𝜌𝛾 2 𝑣 𝑦 𝑣 𝑦	𝜌𝛾 2 𝑣 𝑦 𝑣 𝑧
	𝜌𝛾 2 𝑐𝑣 𝑧	𝜌𝛾 2 𝑣 𝑧 𝑣 𝑥	𝜌𝛾 2 𝑣 𝑧 𝑣 𝑦	𝜌𝛾 2 𝑣 𝑧 𝑣 𝑧 ]
	Where γ is le factor of Lorentz (114) (cf. special relativity), the energy density 𝜌 =	𝑚 𝑉

Comparison of the different relations between energy and curvature in function of the number of dimensions considered -

						𝐷	)	𝑈 𝑥𝑦	1 𝑅 2 =	12(1 -𝑣) 𝐸ℎ 3	×	𝑈 𝐴
	General relativity	4	𝑅 𝜇𝜈 -	1 2	𝑔 𝜇𝜈 𝑅 + 𝛬𝑔 𝜇𝜈 =	8𝜋𝐺 𝑐 4 𝑇 𝜇𝜈		𝑇 𝜇𝑣 =	𝐺 𝜇𝑣 = 𝜅𝑇 𝜇𝑣 𝜌𝑐𝑣 𝑥 𝜌𝑐𝑣 𝑦 𝜌𝑐𝑣 𝑧 𝜏 𝑥𝑦 𝜏 𝑥𝑧 𝜌𝑐𝑣 𝑥 𝜎 𝑥𝑥 𝑚𝑐² 𝑉 𝜌𝑐𝑣 𝑦 [ 𝜌𝑐𝑣 𝑧 𝜏 𝑦𝑥 𝜏 𝑧𝑥 𝜎 𝑦𝑦 𝜏 𝑦𝑧 𝜏 𝑧𝑦 𝜎 𝑧𝑧 ]
		Tableau 2 -

  To be developed… In 4 dimensions In 4 th derivative of the space metric and 2 th derivative of the time metric, and of course tensorial written to be valid in all the referential (covariant derivative)?

										Curvature and energy density
										Example in 4 dimensions
										𝑹 𝝁𝝂 -	𝟏 𝟐	𝒈 𝝁𝝂 𝑹 + 𝜦𝒈 𝝁𝝂 =	𝟖𝝅𝑮 𝒄 𝟒 𝑻 𝝁𝝂
										1 𝑅 2 =	2 𝐸𝐼 ( 𝑈 𝐿 )
	Eigen value	Example of a particle in a	Example of a beam simply supported in
	(natural	well potential in 1 dimension	1 dimension (natural frequencies)
	frequency) and eigen mode									 𝟒 𝒚 (𝒙,𝒕) 𝝏𝒙 𝟒 + ( 𝑬𝑰 𝒎 ) 𝑦 (𝑥,𝑡) = 𝑞 (𝑡)  (𝑥) 𝝏²𝒚 (𝒙,𝒕) 𝝏𝒕²	= 𝟎
										𝒚 𝒏(𝒙,𝒕) = 𝒂 √ 𝟐 𝑳	𝒔𝒊𝒏 ( 𝒏𝒙 𝑳 ) 𝐜𝐨𝐬(𝝎𝒕)
		 =	𝑛²² 𝐿²	(	ħ 2m	)	 =	𝑛²² 𝐿²	(√ 𝑆 𝐸𝐼	)
		E =	𝑛²² 𝐿²	(	ħ 2 2m	)	 𝐼𝑉   (𝑥) = -𝐼𝑉 -In space :	𝑆 𝐸𝐼 𝛼 4 𝐿 4  (𝑥) = 0 × 𝛼 4 𝑞q = 𝐿 4
										𝑑 4  (𝑥) 𝑑𝑥 4 -	𝛼 4 𝐿 4  (𝑥) = 0
		 𝑛(𝑥) = √	2 𝐿	𝑠𝑖𝑛 ( 𝑛𝑥 𝐿	)	 𝑛(𝑥) = √ 2 𝐿	𝑠𝑖𝑛 ( 𝑛𝑥 𝐿	)
		𝑑 2  (𝑥) 𝑑𝑥 2 + ( 2𝑚𝐸 𝑚 ħ 2 )  (𝑥) = 0	In time : 𝑑 2 𝑞 (𝑡) 𝑑𝑡²	+ (	𝛼 4 𝐸𝐼 𝐿 4 𝑆	) 𝑞 (𝑡) = 0

𝜎 𝑖𝑗 =

𝛥𝑄 𝑖 𝛥𝑛 𝑗 𝑤𝑖𝑡ℎ 𝛥 𝑛𝑗 → 0 (118)

So, with m the mass, ρ the density of mass energy, V the volume and ai the acceleration, we have:

If we make the hypothesis that the variation of the force is due only to the variation of the Volume V in function of the time t we obtain with:

We can of course define the volume V:

Thus we obtain:

We can replace the nj by its value: After simplification we obtain:

By definition of a speed, we have:

We obtain finally the expression of the stress tensor at low speed in function of the energy density ρ and based on the multiplication of the velocity vi and vj:

The stress energy tensor becomes from the product of the density of energy and the multiplication of the four velocity (4 dimension of the space time) issued from the general relativity.

𝑇 𝜇𝑣 = 𝜌𝑢 𝜇 𝑢 𝑣 (129)

With for the four velocity:


David Izabel Ingénieur
email: d.izabel@aliceadsl.fr

Can we estimate the Young's modulus of the space-time from simple analogies based on the concepts of the strength of materials? Reflection and proposals

Einstein has shown that the matter curves the space-time. In its model, the space-time is a mathematical conceptualization unifying the space and the time that accounts for measured physical phenomena (delay of the perihelion of Mercury, deviation, curvature of light by a large mass, relativity of length and time, propagation of gravitational waves, caused in particular by coalescence of black holes, which materialize by infinitesimal deformations L / L of space-time (10 -21 )), etc. The question that then comes to mind is whether the mathematical conceptualization of space-time has a physical reality? In this case, what is the material of the fabric of the space-time that undulates or bends under gigantic masses in motion? Where, from a mechanical point of view, can we find an equivalent Young's modulus E to the fabric of the space-time? A first clue is the Casimir's force, which tells us that the vacuum of space is not empty ... From recent publications on the subject and simple analogies based on the strength of materials, we will show that if the E-module of space-time exists, it seems incommensurably large whatever the approach considered (elastic plates in 2 dimensions, plates analyzed by a tensor calculation in 4 dimensions, strings, Casimir force). In any case, only a measure of the latter would make it possible to decide definitively.

1) Determination of the young modulus of the fabric of the space time by analogy with the classical mechanic then with a tensorial calculation in 4 dimensions

1.1) Determination of the base equations

We have shown in previous publications that the relation between the curvature and the energy can be written in 2 dimensions (plate theory) and in 4 dimensions (General relativity theory) [START_REF] Izabel | HAL Peut-on comprendre et unifier les fondamentaux de la relativité générale et de la mécanique quantique à partir des concepts de la résistance des matériaux ? Réflexions et propositions[END_REF].

We have so:

-In plate theory in two dimensions: ] 𝑥𝑦 (1)

𝑈 = 𝐷
Or again, if we put following a more clearly way the curvature in one side and the energy on the other side of the equation, we obtain:

[( 1 𝑅 𝑥 ) 2 + ( 1 𝑅 𝑦 ) 2 + 2(1 -𝜈) {( 1 𝑅 𝑥𝑦 ) 2 } + 2𝜈 { 1 𝑅 𝑥 1 𝑅 𝑦 }] = 24(1 -𝜈 2 ) 𝐸ℎ 3 × 𝑈 𝑥𝑦 (2)
-Following the general relativity, if we put following a more clearly way the curvature of the space time in one side and the stress energy tensor on the other side of the equation, we obtain:

𝑅 𝜇𝜈 - 1 2 𝑔 𝜇𝜈 𝑅 + 𝛬𝑔 𝜇𝜈 = 8𝜋𝐺 𝑐 4 𝑇 𝜇𝜈 (3)
With for memory for the stress energy tensor:

𝑇 𝜇𝑣 = [ 𝑚𝑐² 𝑉 𝜌𝑐𝑣 𝑥 𝜌𝑐𝑣 𝑦 𝜌𝑐𝑣 𝑧 𝜌𝑐𝑣 𝑥 𝜎 𝑥𝑥 𝜏 𝑥𝑦 𝜏 𝑥𝑧 𝜌𝑐𝑣 𝑦 𝜌𝑐𝑣 𝑧 𝜏 𝑦𝑥 𝜏 𝑧𝑥 𝜎 𝑦𝑦 𝜏 𝑧𝑦 𝜏 𝑦𝑧 𝜎 𝑧𝑧 ] (4)
If we suppose that the space-time fabric is a superposition of thin sheets of thickness Lp (with Lp the length of Planck), we obtain a multi sandwich universe. It is then possible by analogy between the two models (Einstein's general relativity and the plate theory in strength of materials) to find an evaluation of the Young's modulus E of the space-time E (N / m²).

To do this, it is sufficient, to compare the curvature of the two equations ( 2) and (3). We obtain then:

24(1 -𝜈 2 ) 𝐸ℎ² × 𝑈 𝑎𝑏ℎ ≈ 8𝜋𝐺 𝑐 4 × 𝑈 𝑉 ( 5
)
By simplifying by U / V on each side of the equation, we obtain:

8𝜋𝐺 𝑐 4 ≈ 24(1 -𝜈 2 ) 𝐸ℎ² (6)
The dimensional equation allows us to verify that this equivalence is reasonable:

𝐿 3 𝑀𝑇 2 𝐿 4 𝑇 4 => 1 𝑀 𝐿 𝑇 2 = 1 𝑀 𝐿² 𝐿 𝑇² 𝐿² => 1 𝑀 𝐿 𝑇² = 𝑠² 𝑘𝑔𝑚 (7)
We can extract from the equation ( 6) an estimation of the Young's modulus E of the space-time fabric..

𝐸 𝑓𝑎𝑏𝑟𝑖𝑐 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑝𝑎𝑐𝑒 𝑡𝑖𝑚𝑒 = 24(1 -𝜈 2 )𝑐 4 ℎ²8𝜋𝐺 (8)
By introducing 𝜅 = 8𝜋𝐺 𝑐 4 [START_REF] Beau | Sur l'hypothèse concernant l'existence d'un champ de gravité de type vectoriel couplant avec l'accélération des masses[END_REF] We obtain the Young modulus E of the space-time fabric by analogy with a thin sheet of thickness h of the space-time in bending:

𝐸 𝑓𝑎𝑏𝑟𝑖𝑐 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑝𝑎𝑐𝑒 𝑡𝑖𝑚𝑒 = 24(1 -𝜈 2 ) ℎ²𝜅 (10)
v is the Poisson's ratio of the space-time .

If we assume that the thickness of the thin sheet is equal to the length of Planck Lp (smaller than it is possible to reach), we obtain:

𝐸 𝑓𝑎𝑏𝑟𝑖𝑐 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑝𝑎𝑐𝑒 𝑡𝑖𝑚𝑒 = 24(1 -𝜈 2 ) 𝐿 𝑝 ²𝜅 (11) 𝑤𝑖𝑡ℎ 𝐿 𝑝 = √ ћ𝐺 𝑐 3 (12) 𝑎𝑛𝑑 𝜅 = 8𝜋𝐺 𝑐 4 𝐸 𝑓𝑎𝑏𝑟𝑖𝑐 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑝𝑎𝑐𝑒 𝑡𝑖𝑚𝑒 = 6(1 -𝜈 2 )𝑐 7 𝐺 2 ℎ = 3(1 -𝜈 2 )𝑐 7 𝐺 2 ћ𝜋 ( 13
)
It will be noted that this last expression associates the 3 fundamental constants of physics, c, G and ℏ.

1.2) Numerical application

We apply the formula [START_REF]Cours de physique « tenseur énergie impulsion[END_REF] from the following data:

v = 0 (hypothesis) G = 6.6726x10 -11 m 3 /kg/s² c = 299792458 m/s h =6.62607004x10 -34 m² kg/s κ =2.07651x10 -43 s²/(kg . m) Lp = 1.62 x10 -35 m
This results for the Young's modulus E is so:

Espace time fabric = 4.4247E +113 N/m²
We can see the extremely high value of this young modulus following this approach.

1.3) Comparison of the obtained results with the state of the art-tensorial calculation of a plate in 4 dimensions

In the publication [START_REF] Tenev | The Mechanics of Spacetime { A Solid Mechanics Perspective on the Theory of General[END_REF]: "The Mechanics of Spacetime {A Solid Mechanics Perspective on the Theory of General Relativity} T G Tenev and M F Horstemeyer" we have an expression of young modulus.

The same approach was developed, but this time considering a plate in 4 dimensions and performing the tensor calculations. The authors arrive at the following expression: See. Formula 3.12 or 4.45 following the date of publication [START_REF] Tenev | The Mechanics of Spacetime { A Solid Mechanics Perspective on the Theory of General[END_REF] considered with v = 1:

𝑌 = 2(1 + 𝑣)𝜇 With 𝜇 =
𝑌 = 𝐸 𝑠𝑝𝑎𝑐𝑒 𝑡𝑖𝑚𝑒 𝑓𝑎𝑏𝑟𝑖𝑐 = 48 𝐿 𝑝 ²𝜅 = 24(1 + 𝑣) 𝐿 𝑝 ²𝜅 (14)
Espace time fabric (v=1) = 8.8495E +113 N /m² By taking v = 0 (Poisson's ratio) to compare with our results, we obtain:

𝐸 𝑠𝑝𝑎𝑐𝑒 𝑡𝑖𝑚𝑒 𝑓𝑎𝑏𝑟𝑖𝑐 = 24 𝐿 𝑝 ²𝜅 = 24(1 + 𝑣) 𝐿 𝑝 ²𝜅 (15)
Espace time fabric (v=0) = 4.43E +113 N /m² 2) Determination of the Young modulus of the space time by an approach of wave transmit by the elastic material constituting the fabric of the universe

2.1) Determination of the base equations

We can consider the space-time fabric as an assembly of extremely fine and short strings (see Figure 1).

We also consider that it is this fabric (invisible to our eyes since it is made up of elements of the size of Planck's length) which transmits the gravitational waves and bends in the presence of matter. By denoting by S the section of the string and ρ the density of the material constituting it, we denote by u(x,t) the displacement of a point of a string :

R E,S E,S

𝜌𝑆 × 𝛥𝑥 × 𝜕²𝑢(𝑥, 𝑡) 𝜕𝑡² = 𝐹(𝑥 + 𝑑𝑥) -𝐹(𝑥) (17)
In addition:

𝐹(𝑥 + 𝛥𝑥) = 𝐹(𝑥) + 𝜕𝐹(𝑥) 𝜕𝑥 𝛥𝑥 (18)
By introducing the expression (18) into the expression ( 17) we obtain:

𝜌𝑆 × 𝜕²𝑢(𝑥, 𝑡) 𝜕𝑡² = 𝜕𝐹(𝑥) 𝜕𝑥 ( 19
)
In addition, the Hooke law allows to write:

𝐹(𝑥) = 𝜎𝑆 = 𝐸𝜀𝑆 ( 20
)
By definition of a strain we have:

𝜀 = 𝜕𝑢(𝑥) 𝜕𝑥 (21)
𝐹(𝑥) = 𝜎𝑆 = 𝐸𝜀𝑆 = 𝐸𝑆 𝜕𝑢(𝑥) 𝜕𝑥 (22)
So for a plane wave by introducing the equation [START_REF] Guy | Relier la mécanique et la relativité générale ? Réflexions et propositions[END_REF] in the equation ( 19) we obtain:

𝜌𝑆 × 𝜕²𝑢(𝑥, 𝑡) 𝜕𝑡² = 𝐸𝑆 𝜕²𝑢(𝑥) 𝜕𝑥² ( 23
)
We obtain the d'Alembert's equation:

𝜌 𝐸 × 𝜕²𝑢(𝑥, 𝑡) 𝜕𝑡² = 𝜕²𝑢(𝑥) 𝜕𝑥² ( 24
)
By definition of a strain, we obtain for a plane wave:

𝜕𝑢(𝑥, 𝑡) 𝜕𝑥 = 𝜀(𝑥, 𝑡) = - 𝑎 𝐶 𝑓′ (𝑡 - 𝑥 𝐶 ) (25)
And by the Hook law we obtain also:

𝐹(𝑥) = 𝜎𝑆 = 𝐸𝜀𝑆 𝜎(𝑥, 𝑡) = - 𝑎𝐸 𝐶 𝑓′ (𝑡 - 𝑥 𝐶 ) (26) 𝜕²𝑢(𝑥, 𝑡) 𝜕𝑥² = 𝑎 𝐶² 𝑓′′ (𝑡 - 𝑥 𝐶 ) (27)
In addition we have:

𝜕𝑢(𝑥, 𝑡) 𝜕𝑡 = 𝑎 × 𝑓′ (𝑡 - 𝑥 𝐶 ) (28) 𝜕²𝑢(𝑥, 𝑡) 𝜕𝑡² = 𝑎 × 𝑓′′ (𝑡 - 𝑥 𝐶 ) (29)
If we introduce these two equations ( 27) and ( 29) in the d'Alembert's formula we obtain:

𝜌 𝐸 𝑎 × 𝑓′′ (𝑡 - 𝑥 𝑐 ) = 𝑎 𝐶 2 𝑓′′ (𝑡 - 𝑥 𝑐 ) (30)
And finally we obtain:

𝜌 𝐸 = 1 𝑐² (31)
We can so deduct from the expression (31) the equivalent young modulus of the fibers that constitute the fabric of the space time.

𝜌 𝐸 = 1 𝑐² => 𝐸 = 𝜌𝑐 2 (32)
If we consider that the space-time fabric consists of an infinity of these fibers / cords and that the gravitational waves propagate at the speed of light, we can evaluate the Young-E modulus of the spacetime.

2.2) Numerical application (cf. [30])

We envisage two hypotheses, because the state of the art of current physics gives two values of the energy density of the vacuum.

a) Hypothesis n°1 : the vacuum energy comes of the cosmologic constant

The vacuum energy density: 𝜌 𝑐𝑜𝑠𝑚𝑜𝑙𝑜𝑔𝑖𝑐 is 1E -9 J/m 3

The vacuum density from Ev =mc² => Ev/c² = m that imply 𝜌 𝑐𝑜𝑠𝑚𝑜𝑙𝑜𝑔𝑖𝑐 = 1.11E -26 kg/m 3 or 10 -29 g/cm 3 The equivalent Young modulus E is so 1.11E -26 x 299792458² = 1E -9 N/m² Efabric of the space time = 1E -9 N/m² b) Hypothesis n°2 : the vacuum energy comes of the quantum mechanics (field theory)

The vacuum energy density:

𝜌 𝑞𝑢𝑎𝑛𝑡𝑖𝑞𝑢𝑒 is 1E +113 J/m 3
The vacuum density from Ev =mc² => Ev/c² =m that imply 𝜌 𝑞𝑢𝑎𝑛𝑡𝑖𝑐 = 1.11E +96 kg/m 3

The equivalent Young modulus E is so 1.11E 96 x 299792458 ²= 1E 113 N/m²

Efabric of the space time = 1E 113 N/m²

The hypothesis of the quantum mechanics seems more realistic than the cosmologic approach by comparison with the previous results obtained.

3) Determination of the young modulus of the space time by a quantic approach-field theory and Casimir force

3.1) Determination of the base equations

The principle of the Casimir force is given figure 2 below. The vacuum energy of each side of the two plates create a force F which brings them closer together:

Casimir force (vacuum energy)

Figure 2 -Definition of the Casimir force confirmed experimentally -

In the case of an approach in one dimension, the energy due to the Casimir force takes the following expression (see [START_REF] Ingold | Casimir effect from a scattering approach Gert-Ludwig Ingold[END_REF] and [START_REF] Joseph | Casimir effect due to a single boundary as a manifestation of the Weyl problem[END_REF]):

𝑈 = 𝜋ћ𝑐 24𝐿 (33)
To obtain the associated force, we use the fact that a force derives of a potential:

𝐹 = 𝑔𝑟𝑎𝑑 (𝑈) (34)
We obtain by deriving the energy by L:

𝐹 = - 𝜋ћ𝑐 24𝐿 2 (35)
Moreover, we can assimilate the displacement of the two plates due to the Casimir force F to an application of Hooke's law to the space-time fabric located between these two plates (see Figure 3).

In this case, it is possible to consider, between the two surface plates S of a length L, an elastic material of Young's modulus E. The strain energy of this type of system loaded with a normal effort F supposed constant for L fixed is the following:

𝑈 𝑒𝑙𝑎𝑠𝑡𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 = 1 2 ∫ (𝑁 (𝑥) ) 2 𝐸𝑆 𝐿 0 𝑑𝑥 = 1 2 × 𝐹 2 𝐿 𝐸𝑆 (36)
Assuming that U = Ustrain energy and considering the force F of Casimir we obtain for the elastic deformation energy of the fabric of the space time located between the two plates:

𝑈 𝑒𝑙𝑎𝑠𝑡𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝑈 𝑐𝑎𝑠 = ћ𝑐𝜋 24𝐿 = 1 2 × (- ћ𝑐𝜋 24𝐿 2 ) 2 𝐿 𝐸𝑆 (37)
So:

ћ𝑐𝜋 24𝐿 = 1 2 × ћ 2 𝑐²𝜋 2 𝐿 24²𝐸𝑆𝐿 4 (38)
Or again:

ћ𝑐𝜋 24𝐿 = 1 𝐸𝑆 × 1 2 × ћ 2 𝑐²𝜋 2 24² × 1 𝐿 3
After simplification we obtain:

1 𝐿 = 1 𝐸𝑆 × ћ 𝑐𝜋 24𝐿 × 1 2𝐿² (39)
With:

𝑈 = 𝜋ћ𝑐 24𝐿 (40)
So we obtain:

1 𝐿 = 1 2 × 1 𝐸𝑆 × 𝑈 𝐿² (41)
The dimensional equation of 1/ES is:

1 𝐸𝑆 => 1 𝑀𝑃𝑎. 𝑚² = 1 𝑀𝑁 𝑚² 𝑚 2 = 1 𝑀𝑁 = 1 𝑘𝑔 𝑚 𝑠² = 𝑠² 𝑘𝑔𝑚 = 𝑇² 𝑀𝐿 (42)
From the expression (39), we can extract the equivalent Young's modulus E from the space-time fabric, which is also here that of the energy of the vacuum created by the disintegrating creations of the particles continuously and instantaneously (see quantum field theory). The results obtained is completely in connection with the previous results obtained.

4) Conclusions

The results are therefore similar regardless of the simplified approaches envisaged (see Table 1

below).

The Young-E modulus of space-time fabric therefore seems extremely large. E seems to be of the order 1E + 113 N / m². This remains to be confirmed by experience and measurement.

Hypothesis

Figure 1 -

 1 Figure 1 -Fabric of the space time made of an assembly of thin strings that have a young modulus E and a section S -

Figure 3 -

 3 Figure 3 -Equivalent material (fabric of the space time) located between the two plates loaded by theCasimir Force F-

  By hypothesis we consider for L the Planck distance Lp in view to have the possibility to compare the results with the previous results obtained (S = L² = Lp ² ). ћ =1.05457E -34 J.s Lp = 1.616252E -35 m c = 299792458 m/s Etissu de l'espace-temps = 3.03 E 112 N/m²

Table 1 -Comparison of different approaches and values obtained from the young modulus E of the space time fabric -

 1

		Theory followed	Equivalent Young	Numerical application
			modulus E obtained	at the Planck scale (E in
								N/m²)
	Fabric of the space time in compression	Casimir force developed in one direction and use		𝐸 =	ћ 𝑐𝜋 48𝑆𝐿 2	L =Lp S = Lp²
	between the two	of the Hook law						E = 3.03 E 112
	plates
	Waves propagation	d'Alembert's			𝐸 = 𝜌𝑐²	𝜌 issued of the
	in the fabric of the	Equation						quantum field theory
	space time							E = 1E 113
	The space time is a multi-sandwich constituted of a	By analogy with the relation between the	𝐸 =	24(1 -𝜈 2 ) ℎ²𝜅	h = Lp if v = 0
	multitude of thin	curvature and
	plates of Planck	energy issued of
	thickness Lp	the plate theory
		and the formula of						E = 4.43E +113
		Einstein general
		relativity
	Elastic theory and tensor approach of the plates theory in	See [5]	𝐸 =	48 𝐿 𝑝 ²𝜅	=	24(1 + 𝑣) 𝐿 𝑝 ²𝜅	h =Lp If v=1 E = 8.8495E +113
	4 dimensions							If v=0 (added by DI)
								E = 4.43E +113

Bibliographic references

For the strength of materials (frame mechanics):