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 David Izabel To cite this version: David Izabel. Can we understand the concepts of the General Relativity and of the Quantum Me- chanics based on the principles of the strength of materials? Reflections and proposals. 2017. hal- 01562660 HAL Id: hal-01562660 https://hal.archives-ouvertes.fr/hal-01562660 Preprint submitted on 16 Jul 2017 HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. 1 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 David Izabel Engineer and professor [email protected] July 2017 Summary 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 푇휇휈influence the curvature 퐺휇휈 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. 2 1st 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 (m4) and an area S (m²) and that have a mass by unit of length (m= S) in kg/m (cf. figure 1). [4] The beam is in free vibration and as a deflection along the time 푦(푥,푡) : 푀, 퐸, 퐼 푥 푦(푥,푡) 퐿 푦 Figure 1 – Beam on 2 supports under its self-weight – We can write: 휕²푦(푥,푡) 퐹 = 푚푎 = 푚 = 푚푦̈ (1) 휕푡² (푥,푡) In addition, we have by connecting the radius of curvature R (m) at the bending moment 푀(푥,푡) : 2 1 휕 푦(푥,푡) 푀(푥,푡) = = (2) 푅 휕푥2 퐸퐼 Knowing that: 휕푀(푥,푡) = 푉 (3) 휕푥 (푥,푡) 휕푉(푥,푡) = −푞 (4) 휕푥 (푥,푡) We obtain by deriving twice respect to 푥 the equation (2): 2 4 2 1 푦(푥,푡) 푀(푥,푡) −푞(푥,푡) ( ) = = ( ) = (5) 푥² 푅 휕푥4 푥² 퐸퐼 퐸퐼 So: 4 푦(푥,푡) 퐸퐼 = −푞 (6) 휕푥4 (푥,푡) By equating equations (1) and (6) in the case of auto-vibration, we obtain the well-known differential equation which controls the self-vibration of a beam on 2 supports: 3 4 휕²푦(푥,푡) 푦(푥,푡) 푚 = −퐸퐼 (7) 휕푡² 휕푥4 So : 4 푦(푥,푡) 푚 휕²푦(푥,푡) + ( ) = 0 (8) 휕푥4 퐸퐼 휕푡² If we put 푦(푥,푡) = 푞(푡)(푥) (9) By replacing in equation (8) above, we obtains: 푖푉 퐸퐼푞(푡)(푥) + 푆푞̈(푡)휙(푥) = 0 (10) We can therefore separate the variables: 퐼푉 푆 푞̈ = − × (11) 퐸퐼 푞 As this equation has to be verified for all 푥 or t, the above ration have to be constant. We therefore define: 퐼푉 푆 푞̈ 훼4 = − × = (12) 퐸퐼 푞 퐿4 And we define the pulsation (circular natural frequency) ω thus: 훼4 퐸퐼 2 = × (13) 퐿4 푆 We can cut in two parts the differential equation (12) in two equations (separation of variables): 훼4 퐼푉 − = 0 (14) (푥) 퐿4 (푥) 2 푞̈(푡) + 푞(푡) = 0 (15) The study of the first equation gives the following general equation: 푥 푥 푥 푥 = 퐴푠푛 (훼 ) + 퐵푐표푠 (훼 ) + 퐶푠푛ℎ (훼 ) + 퐷푐표푠ℎ (훼 ) (16) (푥) 퐿 퐿 퐿 퐿 And the successive derivative gives: 훼 푥 훼 푥 훼 푥 훼 푥 ′ = 퐴 푐표푠 (훼 ) − 퐵 푠푛 (훼 ) + 퐶 푐표푠ℎ (훼 ) + 퐷 푠푛ℎ (훼 ) (17) (푥) 퐿 퐿 퐿 퐿 퐿 퐿 퐿 퐿 훼2 푥 훼2 푥 훼2 푥 훼2 푥 ′′ = −퐴 푠푛 (훼 ) − 퐵 푐표푠 (훼 ) + 퐶 푠푛ℎ (훼 ) + 퐷 푐표푠ℎ (훼 ) (18) (푥) 퐿2 퐿 퐿2 퐿 퐿2 퐿 퐿2 퐿 훼3 푥 훼3 푥 훼3 푥 훼3 푥 ′′′ = −퐴 푐표푠 (훼 ) + 퐵 푠푛 (훼 ) + 퐶 푐표푠ℎ (훼 ) + 퐷 푠푛ℎ (훼 ) (19) (푥) 퐿3 퐿 퐿3 퐿 퐿3 퐿 퐿3 퐿 훼4 푥 훼4 푥 훼4 푥 훼4 푥 퐼푉 = 퐴 푠푛 (훼 ) + 퐵 푐표푠 (훼 ) + 퐶 푠푛ℎ (훼 ) + 퐷 푐표푠ℎ (훼 ) (20) (푥) 퐿4 퐿 퐿4 퐿 퐿4 퐿 퐿4 퐿 The 4 boundary conditions allows to write: 4 (x0) 0 (xL) 0 II (x0) 0 II 0 (xL) 0 0 0 0 = 퐴푠푛 (훼 ) + 퐵푐표푠 (훼 ) + 퐶푠푛ℎ (훼 ) + 퐷푐표푠ℎ (훼 ) (21) (푥=0) 퐿 퐿 퐿 퐿 (푥=0) = 0퐴 + 퐵 + 0퐶 + 퐷 = 0 (22) 퐿 퐿 퐿 퐿 = 퐴푠푛 (훼 ) + 퐵푐표푠 (훼 ) + 퐶푠푛ℎ (훼 ) + 퐷푐표푠ℎ (훼 ) (23) (푥=퐿) 퐿 퐿 퐿 퐿 (푥=퐿) = 퐴푠푛(훼) + 퐵푐표푠(훼) + 퐶푠푛ℎ(훼) + 퐷푐표푠ℎ(훼) = 0 (24) 훼2 0 훼2 0 훼2 0 훼2 0 ′′ = −퐴 푠푛 (훼 ) − 퐵 푐표푠 (훼 ) + 퐶 푠푛ℎ (훼 ) + 퐷 푐표푠ℎ (훼 ) (25) (푥=0) 퐿2 퐿 퐿2 퐿 퐿2 퐿 퐿2 퐿 훼2 훼2 ′′ = 0퐴 − 퐵 + 0퐶 + 퐷 = 0 (26) (푥=0) 퐿2 퐿2 훼2 퐿 훼2 퐿 훼2 퐿 훼2 퐿 ′′ = −퐴 푠푛 (훼 ) − 퐵 푐표푠 (훼 ) + 퐶 푠푛ℎ (훼 ) + 퐷 푐표푠ℎ (훼 ) (27) (푥=퐿) 퐿2 퐿 퐿2 퐿 퐿2 퐿 퐿2 퐿 훼2 훼2 훼2 훼2 ′′ = −퐴 푠푛(훼) − 퐵 푐표푠(훼) + 퐶 푠푛ℎ(훼) + 퐷 푐표푠ℎ(훼) = 0 (28) (푥=퐿) 퐿2 퐿2 퐿2 퐿2 We can put these equations on the form of a matrix: 0 퐵 0 퐷 퐴푠푛훼 퐵푐표푠훼 퐶푠푛ℎ훼 퐷푐표푠ℎ훼 훼² 훼² 0 −퐵 0 퐷 (29) 퐿² 퐿² 훼²푠푛훼 훼²푐표푠훼 훼²푠푛ℎ훼 훼²푐표푠ℎ훼 −퐴 −퐵 퐶 퐷 [ 퐿² 퐿² 퐿² 퐿² ] The determinant of this matrix have to be null. We notice that from the equations (1) and (3) that B = D = 0 푥 푥 푥 푥 = 퐴푠푛 (훼 ) + 퐵푐표푠 (훼 ) + 퐶푠푛ℎ (훼 ) + 퐷푐표푠ℎ (훼 ) (30) (푥) 퐿 퐿 퐿 퐿 푥 푥 = 퐴푠푛 (훼 ) + 퐶푠푛ℎ (훼 ) (31) (푥) 퐿 퐿 푆 푞̈ 훼4 And we have: − × = (32) 퐸퐼 푞 퐿4 The two equations (2) and (4) gives: (푥=퐿) = 퐴푠푛(훼) + 퐶푠푛ℎ(훼) = 0 (33) 훼2 훼2 ′′ = −퐴 푠푛(훼) + 퐶 푠푛ℎ(훼) = 0 (34) (푥=퐿) 퐿2 퐿2 5 So: (푥=퐿) = 퐴푠푛(훼) + 퐶푠푛ℎ(훼) = 0 (35) ′′(푥=퐿) = −퐴푠푛(훼) + 퐶푠푛ℎ(훼) = 0 (36) The sum of these two equations gives: 2퐶푠푛ℎ(훼) = 0 => 훼 = 0 표푢 퐶 = 0 (37) The subtraction of these two equations gives: 2퐴푠푛(훼) = 0 => 훼 = 푛 And so C = 0 (38) Is the Eigen value of the system. We obtain the following equation of the Eigen mode: 푛푥 = 퐴푠푛 ( ) (39) (푥) 퐿 That we write in function of n (quantification): The Eigen mode is so written: 푛푥 = 퐴 푠푛 ( ) (40) 푛(푥) 푛 퐿 The norm of the Eigen mode is so written: 퐿 퐿 푛푥 2( ) 2 퐿 ∫ 푛(푥) 푛(푥) = ∫ 퐴푛푠푛 푑푥 (41) 0 0 1 We replace du sin²(ax) by (1 − cos (2푎푥)) (42) 2 퐿 퐿 2 1 2푛푥 ∫ 푛(푥) 푛(푥) = 퐴푛 ∫ (1 − 푐표푠 ( )) 푑푥 (43) 0 0 2 퐿 2푛푥 We define 푢 = 퐿 2푛 푑푢 = 푑푥 퐿 퐿 푑푥 = 푑푢 2푛 퐿 퐿 퐿 2 1 2 1 퐿 ∫ 푛(푥) 푛(푥) = 퐴푛 ∫ 푑푥 + 퐴푛 ∫ 푐표푠푢 푑푢 (44) 0 0 2 2 0 2푛 The antiderivative of cosu is sinu : 퐿 퐿 2 푥 퐿 2푛푥 퐿 퐴푛퐿 ∫ (푥) (푥) = 퐴2 [( − 푠푛 ( ))] = 퐴2 ( − 0 − (0 − 0)) = (45) 푛 푛 푛 2 4푛 퐿 푛 2 2 0 0 2 We note that if we chose퐴 = √ (46)) the modes (푥) would form an orthonormal basis for the 푛 퐿 푛 canonical scalar product. In the case of normed Eigen vector we have so: 6 2 푛푥 = √ 푠푛 ( ) (47) 푛(푥) 퐿 퐿 We have to study the second equation now : 2 푞̈(푡) + 푞(푡) = 0 We fix 푞(푡)as following: 푞(푡) = 푎푐표푠푡 + 푏푠푛푡 푞̇(푡) = −푎푠푛푡 + 푏푐표푠푡 2 푞̈(푡) = −푎²푐표푠푡 − 푏 푠푛푡 At the time t = 0, the system is in static, and so the speed is null: 푞̇(푡=0) = −푎푠푛0 + 푏푐표푠0 That imply b = 0: And so the final deflection of the beam is: 푦(푥,푡) = 푞(푡)(푥) (9) 2 푛푥 푦 = 푎√ 푠푛 ( ) cos(휔푡) (9푏푠) 푛(푥,푡) 퐿 퐿 Note We find this result by writing that the determinant of the matrix of the components must be zero.