GENERAL EXACT TETRAHEDRON ARGUMENT FOR THE FUNDAMENTAL LAWS OF CONTINUUM MECHANICS

Transcription

1 GENERAL EXACT TETRAHEDRON ARGUENT FOR THE FUNDAENTAL LAWS OF CONTINUU ECHANICS EHSAN AZADI Abstract. In this article, we give a general exact mathematical framework that all the fundamental relations and conservation equations of continuum mechanics can be derived based on it. We consider a general integral equation contains the parameters that act on the volume and the surface of the integral s domain. The idea is to determine how many local relations can be derived from this general integral equation and what these local relations are. After obtaining the general Cauchy lemma, we derive two other local relations by a new general exact tetrahedron argument. So, there are three local relations that can be derived from the general integral equation. Then we show that all the fundamental laws of continuum mechanics, including the conservation of mass, linear momentum, angular momentum, energy, and the entropy law, can be considered in this general framework. Applying the general three local relations to the integral form of the fundamental laws of continuum mechanics in this new framework leads to exact derivation of the mass flow, continuity equation, Cauchy lemma for traction vectors, existence of stress tensor, general equation of motion, symmetry of stress tensor, existence of heat flux vector, differential energy equation, and differential form of the Clausius-Duhem inequality for entropy law. The general exact tetrahedron argument is an exact proof that removes all the challenges on derivation of the fundamental relations of continuum mechanics. In this proof, there is no approximate or limited process and all the parameters are exact point-based functions. Also, it gives a new understanding and a deep insight into the origins and the physics and mathematics of the fundamental relations and conservation equations of continuum mechanics. This general mathematical framework can be used in many branches of continuum physics and the other sciences.. Introduction Is there a general exact framework that all of the fundamental relations and conservation equations of continuum mechanics can be derived in it? Continuum mechanics is a subject that is the base of a wide range of phenomena and physical behaviors of the nature and industry such as fluid mechanics, solid mechanics, continuum thermodynamics, heat transfer, etc. The birth of modern continuum mechanics is the introduction of the traction vector in 822 by Cauchy that describes the nature of forces on the internal surfaces of the substance [6]. He gave a proof that is called Cauchy tetrahedron argument for the existence of stress tensor. The other important Cauchy s achievements in the foundations of continuum mechanics include the Department of echanical Engineering, Sharif University of Technology, Iran. address: eazadi65@gmail.com, Submitted to vixra.org 7 July 207.

2 2 E. AZADI symmetry of stress tensor and the general equation of motion [6, 7, 2]. During about two centuries, the scientists and authors in continuum mechanics presented some different proofs and processes to derive the fundamental relations and conservation equations of continuum mechanics, more generally and precisely [], [0], [3], [9], [], [4], [8], [5]. We already gave two articles on this subject. In the first one (207, [2]), we provided a comprehensive review of the different tetrahedron arguments and the proofs of the existence of stress tensor, and discussed the challenges and improvements of each one. In the second article (207, [3]), for the first time, we presented and proved the exact tetrahedron argument that removes all of the challenges on the previous tetrahedron arguments and the proofs of the existence of stress tensor. Exact tetrahedron argument led to derivation of both the relations for the existence of stress tensor and general equation of motion, simultaneously. Also, we compared the exact tetrahedron argument with the previous proofs of the existence of stress tensor. Exact tetrahedron argument gave us a new understanding and a deep insight into the physics and mathematics of the stress tensor, general equation of motion, and their origins [3]. In this article, we generalize the exact tetrahedron argument for all of the fundamental relations and conservation laws of continuum mechanics. We prove a general exact mathematical framework and consider the different fundamental laws of continuum mechanics in this framework. Then, we will show that this leads to the exact derivation of the relations for the mass flux, existence of stress tensor, symmetry of stress tensor, surface heat flux, entropy flux, and the differential form of fundamental conservation laws of mass, linear momentum, angular momentum, energy, and the entropy law. Here we consider an integral equation over the control volume as the form: dv φ ds (.) In general, pr, tq is called body term and acts over the volume of, and φ φpr, t, nq is called surface term and acts on the surface of, i.e.,. Where r is the position vector, t is time, and n is the outward unit normal vector on the surface of the control volume. If is scalar then φ must be scalar, and if is vector then φ must be vector. These two functions are continuous over their domains. In the later, we will show that the integral form of all the fundamental laws of continuum mechanics can be written in the form of the integral equation (.). We want to find how many local relations can be derived from this integral equation and what they are. We use the Eulerian approach in the entire of this article, where a control volume is utilized and the changes of quantities are recorded as the effective parameters on the surface or volume of the control volume and the fluxes that pass through control volume surface.

3 GENERAL EXACT TETRAHEDRON ARGUENT, THIRD PAPER 3 Figure. The control volumes and 2, where S Y S m and 2 S 2 Y S m, and the control volume such that V V Y V 2 and S Y S General Cauchy lemma Suppose the control volume splits into and 2 by the surface S m. So, V V Y V 2, S Y S m, 2 S 2 Y S m, and S Y S 2, see Figure. If the integral equation (.) applies to and 2, then the sum of these equations is: dv 2 dv 2 φ ds φ 2 ds 2 y V V Y V 2, the sum of the body term integrals is equal to the integral of the body term on. In addition, by S Y S m and 2 S 2 Y S m, the surface integrals split as: dv φ ds S φ ds S m φ 2 ds S 2 φ 2 ds S m y S Y S 2, the sum of the surface integrals on S and S 2 is equal to the surface integral of φ on, so: dv φ ds φ ds S m φ 2 ds S m Comparing this integral equation with the general integral equation (.), implies that: φ ds φ 2 ds 0 S m S m ut φ on S m is φpr, t, nq, and φ 2 on S m is φpr, t, nq, so: ( φpr, t, nq φpr, t, nq ds 0 S m therefore, we have φpr, t, nq φpr, t, nq (2.) This is the first local relation that is derived from the integral equation (.), and is called general Cauchy lemma. It states the surface terms acting on opposite sides of the same surface at a given point and time are equal in magnitude but opposite in sign. It means that if we have the surface term on one side of a surface at a given point and time, then we can get the surface term on the other side of this surface at that point and time by the equation (2.).

4 4 E. AZADI Figure 2. The geometry of tetrahedron control volume and the exact surface term vectors on the faces. Note that in this figure we suppose that the surface term is a vector, but in general it can be a scalar or a vector, depending on the type of the body term. 3. General exact tetrahedron argument Consider a tetrahedron control volume in continuum media that its vortex is at the point o and its three orthogonal faces are parallel to the three orthogonal planes of the Cartesian coordinate system. The fourth surface of the tetrahedron, i.e., its base, has the outward unit normal vector n 4. For simplicity, the vortex point is at the origin of the coordinate system. The geometrical parameters are shown in Figure 2. The vector r xe x ye y ze z is the position vector from the origin of the coordinate system. Applying the general integral equation (.) to the tetrahedron control volume leads to: φ 4 ds s 4 φ ds s φ 2 ds s 2 φ 3 ds s 3 dv (3.) The key idea of this proof is to write the variables of this equation in terms of the exact Taylor series about a point in the domain. Here, we derive these series about the vortex point of tetrahedron (point o), where the three orthogonal faces pass through it. Note that time (t) is the same in the all terms, so it does not exist in the Taylor series. For pr, tq at any point in the domain of the control volume, we have: o o x x o y y o z z 2 o 2! x 2 x2 2 o y 2 y2 2 o z 2 z2 2 2 o xy xy o 22 xz xz yz o 22 yz 8ÿ 8ÿ 8ÿ pmnkq... x m!n!k! x m y n z ˇˇˇo m y n z k k m 0 n 0 k 0 (3.2) Here o and o {x are the exact values of and {x at the point o, respectively. Similarly, the other derivatives are the exact values of the corresponding derivatives of

5 GENERAL EXACT TETRAHEDRON ARGUENT, THIRD PAPER 5 at the point o. On the surface s, x 0 and n does not change, so: φ φ o φ o y y φ o z z 2! 8ÿ 8ÿ... m 0 k 0 2 φ o y 2 y2 2 φ o z 2 z2 2 2 φ yz o yz pmkq (3.3) φ y m!k! y m z ˇˇˇo m z k k where φ o is the exact value of φ on s at the point o. On the surface s 2, y 0 and n 2 does not change, and on the surface s 3, z 0 and n 3 does not change, so: φ 2 φ 2o φ 2 o x x φ 2 o z z 2 φ 2o 2! x 2 8ÿ 8ÿ... m!k! m 0 k 0 x2 2 φ 2o z 2 z2 2 2 φ xz 2o xz pmkq (3.4) φ 2 x x m z ˇˇˇo m z k k φ 3 φ 3o φ 3 o x x φ 3 o y y 2! 8ÿ 8ÿ... m 0 k 0 2 φ 3o x 2 x2 2 φ 3o y 2 y2 2 2 φ xy 3o xy pmkq (3.5) φ 3 x m!k! x m y ˇˇˇo m y k k Similarly, φ 2o and φ 3o are the exact values of φ 2 and φ 3 at the point o on s 2 and s 3, respectively. For the surface term on s 4 a more explanation is needed. The surface term on s 4 expands based on the surface term on the inclined surface that is parallel to s 4 and passes through the vortex point of tetrahedron (point o). ecause the unit normal vectors of these two surfaces are the same, see Figure 3. Therefore: Figure 3. Inclined surface that is parallel to s 4 and passes through point o.

6 6 E. AZADI φ 4 φ 4o φ 4 o x x φ 4 o y y φ 4 o z z 2 φ 4o 2! x 2 x2 2 φ 4o y 2 8ÿ 8ÿ 8ÿ pmnkq φ 4... x m!n!k! x m y n z ˇˇˇo m y n z k k m 0 n 0 k 0 y2 2 φ 4o z 2 z2 2 2 φ 4o xy xy φ 4o 22 xz xz φ yz 4o 22 yz (3.6) where φ 4o is the exact surface term at the point o on the inclined surface with unit normal vector n 4 that this surface passes exactly through point o, the vertex point of tetrahedron control volume. Here x, y, and z are the components of the position vector r on the surface s 4. Note that φ o, φ 2o, φ 3o, and φ 4o are the exact surface terms at the point o but on different surfaces with unit normal vectors n, n 2, n 3, and n 4, respectively. The body term o is exactly defined at the point o. Therefore, all the surface terms and the body term with subscript o and all their derivatives, such as 2 φ 4o {xy, are exactly defined at the point o and are bounded. As a result, for the convergence of the above Taylor series it is enough that we have r ď in the domain of the control volume. ut the scale of the coordinate system is arbitrary and we can define this scale such that the greatest distance in the domain of the control volume from the origin, is equal to one, i.e., r max. y this scale, in the entire of the tetrahedron control volume we have r ď, that leads to the convergence condition for the above Taylor series. Now all of the variables are prepared for integration in the integral equation (3.). The integration of on the volume of : c bp z c q ap y b z c q " dv o o x x o y y o z z... dx dy dz! 6 abc o o 4 x a o y b c ) (3.7) o... z The integration of φ 4 on s 4 : b ap y b q "c c 2 c 2 φ 4 ds ˆφ 4o φ 4 o s a b x x φ 4 o y y φ 4 o x cp z a y q 2 φ 4o b 2! x 2 x2 2 φ 4o y 2 y2 2 φ 4o x cp z 2 a y q 2 b 2 2 φ 4o xy xy φ 4o 22 xz xcp x a y q 2 2 φ 4o b yz ycp x a y q b... dx dy? a2 b 2 2 a 2 c 2 b 2 c!φ 2 4o 3 2 φ 4o 2 x 2 a2 2 φ 4o y 2 φ4o x a φ 4 o y b φ 4 o z c b2 2 φ 4o z 2 c2 2 φ 4o xy ab 2 φ 4o xz ac 2 φ bc 4o... yz ) (3.8)

7 GENERAL EXACT TETRAHEDRON ARGUENT, THIRD PAPER 7 The integration of φ on s : c bp z c q φ ds s φ o 2! y 2! 2 bc φ o " φ o φ o y y φ o z z y2 2 φ o z 2 z2 2 2 φ yz o... yz 3 φo y b φ o z c 2 2 φ o y 2 dy dz b2 2 φ o z 2 c2 2 φ bc ) o... yz (3.9) The integration of φ 2 on s 2 and φ 3 on s 3 can be done, similarly. The geometrical relations for the area of faces and the volume of the tetrahedron are: s 2 bc, s 2 2 ac, s 3 2 ab s 4? a2 b 2 2 a 2 c 2 b 2 c 2, V (3.0) 6 abc y substituting the obtained relations for the surface terms and the body term into the equation (3.) and using the above geometrical relations, we have:! s 4 φ 4o φ4o 3 x a φ 4 o y b φ c 4 o z 2 φ 4o 2 x 2 a2 2 φ 4o y 2 b2 2 φ 4o z 2 c2 2 φ 4o xy ab 2 φ 4o xz ac 2 φ bc ) 4o... yz! s φ o φo 3 y b φ c o 2 φ o z 2 y 2 b2 2 φ o z 2 c2 2 φ bc ) o... yz! s 2 φ 2o φ2o 3 x a φ c 2 o 2 φ 2o z 2 x 2 a2 2 φ 2o z 2 c2 2 φ ac ) 2o... xz! s 3 φ 3o φ3o 3 x a φ b 3 o 2 φ 3o y 2 x 2 a2 2 φ 3o y 2 b2 2 φ ab ) 3o... xy! V o o 4 x a o y b c ) o... 0 z (3.) In the geometry of tetrahedron, h is the height of the vertex o from the base face, i.e., s 4. So, we have the following geometrical relations for a tetrahedron with n 4 n x e x n y e y n z e z, where a, b, and c are greater than zero, see Figure 2. h n x a, h n y b, h n z c h 2 a 2 b 2 c, s 2 4 abc 2h s n x s 4, s 2 n y s 4, s 3 n z s 4 (3.2) V 6 abc 3 h s 4 If we divide the equation (3.) by s 4 and use the relations (3.2) for the faces and volume of the tetrahedron, then substitute the relations a h{n x, b h{n y, and

8 8 E. AZADI c h{n z into the equation and rearrange it based on the powers of h, we have:!φ 4o n x φ o n y φ 2o n z φ 3o ) " φ4o x n y φ2o x " 2 φ 4o φ 4 o n x y φ 2 o n x z 2 φ 4o x 2 n 2 x y 2 n 2 y n x 2 φ o y 2 n 2 y φ 4 o n y z 2 φ o z 2 n 2 z n z 2 φ 3o 2 φ 3o x 2 n 2 x y 2 n 2 y... 0 n z φ3o n z x n x φo n z y 2 φ 4o z 2 n 2 z 2 φ o yz 2 φ 3o xy φ 3 o n x y 2 φ 4o xy φ o n y z n z o n y 3 h 2 φ 4o n x n y xz n y 2 φ 2o n y n z x 2 o n x n y x n 2 x o n x y 2 φ 4o n x n z yz 2 φ 2o z 2 n 2 z o n y z n y n z 2 φ 2o xz n x n z n z 2 h2 (3.3) Note that by the coordinate system here and by V 0, no one of n x, n y, and n z is exactly zero. So, all of the expressions in the braces tu of the equation (3.3) exist. We can rename the expressions in the braces and rewrite the equation as: E 0 E 3 h E 2 2 h (3.4) If we continue to integrate the higher order derivatives of all terms based on their Taylor series, we have the following equation: E 0 E 3 h E h2 E 3 60 h3... E m pm 2q! hm... 0 (3.5) or 8ÿ 2 E m pm 2q! hm 0 (3.6) m 0 This is a great equation in the foundation of continuum mechanics. E 0, E, and E 2 are shown in the braces of the equation (3.3) and E 3 and other E m s will be presented. We now discuss some aspects of the equation (3.5): E m s are formed by the expressions of surface terms, body term and their derivatives, and the components of unit normal vector of the inclined surface. Each of the E m s exists, because the surface terms, body term and their derivatives are defined as continuous functions in continuum media and by the coordinate system here and by V 0, no one of n x, n y, and n z is exactly zero. Each of the E m s depends on the variables at the point o and the components of unit normal vector of the inclined surface that is parallel to s 4 and passes through point o. ecause the surface terms, body term, and their derivatives are defined at the point o.

9 GENERAL EXACT TETRAHEDRON ARGUENT, THIRD PAPER 9 E m s do not depend on the volume of tetrahedron. h is a geometrical variable and by the scale of the coordinate system on the tetrahedron control volume such that r max ď, the altitude of the tetrahedron (h) is not greater than one. Note that h 0 is not valid, because the general integral equation (.) is defined for the control volumes with nonzero volume. y these properties, we return to the equation (3.5). E 0 E 3 h E h2 E 3 60 h3... E m pm 2q! hm... 0 We must find E m s. Since E m s are independent of h, the series on the left hand side is a power series. A power series is identically equal to zero if and only if all of its coefficients are equal to zero. Therefore: E m 0, m 0,, 2,..., 8 (3.7) Note that these results are valid not only for h Ñ 0 but also for all values of h in the domain. In other words, the results (3.7) are valid not only for an infinitesimal tetrahedron but also for any tetrahedron in the scaled coordinate system in continuum media. In addition, we have not done any approximate process during derivation of the equations (3.5) and (3.7). So, the results (3.7) hold exactly, not approximately. Furthermore, the subscript o in the expressions of E m s in the equation (3.3) indicates the vortex point of the tetrahedron. ut any point in the domain in continuum media can be regarded as the vertex point of a tetrahedron and we could consider that tetrahedron. So, the point o can be any point in continuum media. We conclude that E m s are equal to zero at any point in continuum media. This implies that all their derivatives are equal to zero, as well. For example, we have for E 0 : E 0 x E 0 y E 0 z 0 (3.8) and the other higher derivatives of E 0 are equal to zero. This trend holds for other E m s. ut what are E m s? For E 0 0, from the equation (3.3): E 0 φ 4o n x φ o n y φ 2o n z φ 3o 0 (3.9) In this equation, the four surface terms are exactly defined at the point o on the surfaces that pass exactly through this point. The surface term φ o is defined on the negative side of coordinate plane yz, i.e., n e x, at the point o. If φ xo is the surface term on the positive side of coordinate plane yz at the point o, then by the equation (2.), i.e., φpr, t, nq φpr, t, nq, we have: Similarly, for φ 2o and φ 3o : φ o φ xo (3.20) φ 2o φ yo, φ 3o φ zo (3.2)

10 0 E. AZADI y substituting these relations into (3.9) and rearranging it, we have: φ 4o n x4 φ xo n y4 φ yo n z4 φ zo (3.22) where n x4 n x, n y4 n y, and n z4 n z. So, the surface term φ 4o can be obtained by a linear relation between the surface terms on the three orthogonal planes and the components of its unit normal vector. ut can we use the equation (3.22) for any unit normal vector rather than n 4o? y considering the equations (3.) and (3.3), we find that the equation (3.22) is really the following equation: and this equation is: φ 4o s s 4 φ xo s 2 s 4 φ yo s 3 s 4 φ zo (3.23) φ 4o n x4 φ xo n y4 φ yo n z4 φ zo (3.24) In Figure 2, by a ą 0, b ą 0, and c ą 0, the components of unit normal vector on the inclined surface are greater than zero. So, the equation (3.22) is valid for these cases. For the surfaces that their unit normal vector components are negative and are not zero, consider a tetrahedron control volume by the unit normal vector of its inclined surface (base face), n 4, that all of its components are negative. Therefore, we have n 4o n x 4 e xn y 4 e y n z 4 e z n x e x n y e y n z e z, where n 4o is the outward unit normal vector of the surface that is parallel to the inclined surface and passes through the vortex point of this tetrahedron (point o), and n x, n y, and n z are positive values. Applying the process of exact tetrahedron argument to this new tetrahedron, leads to the following equation similar to the equation (3.9): E 0 φ 4o n x 4 φ xo n y 4 φ yo n z 4 φ zo 0 (3.25) As compared with the equation (3.9), in this equation we have φ xo, φ yo, and φ zo rather than φ o, φ 2o, and φ 3o, respectively. ecause the outward sides of orthogonal faces of this new tetrahedron are in the positive directions of coordinate system. y the equation (3.25) and the components of n 4o, we have: φ 4o n x 4 φ xo n y 4 φ yo n z 4 φ zo n x φ xo n y φ yo n z φ zo n x φ xo n y φ yo n z φ zo (3.26) n x 4 φ xo n y 4 φ yo n z 4 φ zo So, the surface term φ 4o can be obtained by a linear relation between the surface terms on the three orthogonal planes and the components of its unit normal vector. For the surfaces that one or two components of their unit normal vectors are negative but the other ones are not zero, the same process can be done. For the other surfaces that one or two components of their unit normal vectors are equal to zero, the tetrahedron does not form, but due to the continuous property of the surface term on n and the arbitrary choosing for any orthogonal basis for the coordinate system, the surface terms on these surfaces can be described by the equation (3.22), as well. So, in general, the normal unit vector n 4 can be related to any surface that passes

11 GENERAL EXACT TETRAHEDRON ARGUENT, THIRD PAPER through point o in three-dimensional continuum media. Thus, the subscript 4 removes from the equation (3.22) and we have for every n n x e x n y e y n z e z : φ o n x φ xo n y φ yo n z φ zo (3.27) The subscript o in this equation indicates the vortex point of the tetrahedron. ut any point in the domain in continuum media can be the vertex point of a tetrahedron and we could consider this tetrahedron. So, the point o can be any point in continuum media and the subscript o removes from the equation: φ n x φ x n y φ y n z φ z (3.28) or φpr, t, nq n x φpr, t, e x q n y φpr, t, e y q n z φpr, t, e z q (3.29) This is the second local relation that is derived from the general integral equation (.). It states that the surface term acting on any surface at a given point and time in the continuum domain can be obtained by a linear relation between the surface terms on the three orthogonal surfaces at that point and time and the components of the unit normal vector of the surface. It means that if we have the surface terms on three orthogonal surfaces at a given point and time, then we can get the surface term on any surface that passes through that point at that time by using the unit normal vector of the surface and the linear relation (3.29). In the next section, we will show that if φpr, t, nq is scalar then the equation (3.29) leads to the existence of a flux vector and if φpr, t, nq is vector then the equation (3.29) leads to the existence of a second order tensor. Note that if we do not have the relation (2.), i.e., the general Cauchy lemma, the equation (3.29) cannot be derived for every unit normal vector. Now the equation (3.29) contains the relation (2.). Let us see what E 0 tells. From the equation (3.3): E φ4o x n y φ2o x φ 4 o n x y φ 2 o n x z φ 4 o n y z n z φ3o n z x n x φo n z y φ 3 o n x y φ o n y z n z (3.30) o n y As previously stated, on the tetrahedron control volume with V 0, no one of n x, n y, and n z is exactly zero. Therefore, E exists. Furthermore, the unit normal vector n 4 does not change on s 4, so: n 4 x n 4 y n 4 z 0 (3.3)

12 2 E. AZADI Using the relations (3.3) and the equation (3.9), i.e., φ 4o E 0 n x φ o n y φ 2o n z φ 3o, we have for (3.30): E E 0 n x x E 0 n y y E 0 n z z φ o x φ 2 o y φ 3 o z o If we define E as: therefore, we have E φ o x φ 2 o y φ 3 o z o (3.32) E n x E 0 x n y E 0 y n z E 0 z E (3.33) ut we saw in (3.8) that the derivatives of E 0 were equal to zero. So, from (3.33) and E 0, we have: E E 0 (3.34) y (3.32), E is defined at the vertex point of tetrahedron. ut as previously stated, the vertex point of the tetrahedron can be at any point in continuum media. Therefore, by (3.34), E 0 at any point in continuum media. This implies that all derivatives of E are equal to zero at any point in continuum media. So: E x E y E z 0 (3.35) y using the relations (3.20) and (3.2), i.e., φ o φ xo, φ 2o φ yo, and φ 3o φ zo, the equation (3.32) becomes: but E 0, so E φ x o x φ y o y φ z o z o (3.36) o φ x o x φ y o y φ z o z (3.37) As explained earlier, we can remove the subscript o from the equation and tell that this equation is valid at any point and at any time in the continuum domain. Therefore: or pr, tq φpr, t, e xq x φ x x φ y y φ z z φpr, t, e yq y φpr, t, e zq z (3.38) (3.39) This is the third local relation that is derived from the general integral equation (.). It is a partial differential equation and states that the body term at a given point and time in the continuum domain is equal to to the sum of the first order derivatives of the surface terms acting on the three orthogonal surfaces at that point and time. It means that if we have the first derivatives of the surface terms on the three orthogonal surfaces at a given point and time, then we can get the body term at that point and time by using the equation (3.39).

13 Let us see what E 2 0 tells. From the equation (3.3): 2 φ 4o E 2 x 2 n 2 x n x 2 φ o y 2 n 2 y GENERAL EXACT TETRAHEDRON ARGUENT, THIRD PAPER 3 2 φ 4o y 2 n 2 y 2 φ o z 2 n 2 z n z 2 φ 3o 2 φ 3o x 2 n 2 x y 2 n 2 y 2 φ 4o z 2 n 2 z 2 φ o yz 2 φ 3o xy 2 φ 4o xy 2 φ 4o n x n y xz n y 2 φ 2o n y n z x 2 o n x n y x For E 2, similar to the process for E 0, we have: n 2 x o n x y 2 φ 4o n x n z yz 2 φ 2o z 2 n 2 z o n y z n y n z 2 φ 2o xz n z E 2 2 E 0 n 2 x x 2 E 0 2 n 2 y y 2 E 0 2 n 2 z z 2 E 0 2 n x n y xy 2 E 0 n x n z xz 2 E 0 n y n z yz E n x x E n y y E n z z n x n z (3.40) (3.4) y the previous explanations, all derivatives of E 0 and E were equal to zero. Therefore, the equation (3.4) is a correct result of E 2 0. Similar to the previous processes for E and E 2, we have for E 3 0: E 3 n 3 x n x n 2 y 3 E 0 x 3 E 0 3 n 3 y y 3 E 0 3 n 3 z z 3 E 0 3 n 2 xn y x 2 y 3 E 0 n 2 xn z x 2 z 3 E 0 n 2 yn z y 2 z 3 E 0 xy 2 n x n 2 z 3 E 0 xz 2 n y n 2 z 3 E 0 yz 2 n x n y n z 3 E 0 xyz 2 E n 2 x x 2 E 2 n 2 y y 2 E 2 n 2 z z 2 E 2 n x n y xy 2 E n x n z xz 2 E n y n z yz (3.42) We saw that all derivatives of E 0 and E were equal to zero. So, the equation (3.42) is a correct result of E 3 0. This process for other E m s, leads to the expressions that contain the higher derivatives of E 0 and E and the higher powers of the components of the unit normal vector and the results are equal to zero. Therefore, the general integral equation (.) leads to the three important local relations (2.), (3.29), and (3.39). 4. Fundamental laws of continuum mechanics, integral forms, basic local relations, and differential forms In this section, we show that each of the fundamental laws of continuum mechanics can be written in the form of the general integral equation (.) on control volume, i.e.: dv φ ds (4.)

14 4 E. AZADI In this equation pr, tq and φ φpr, t, nq are continuous over the volume and the surface of, respectively. Where r is the position vector, t is time, and n is the outward unit normal vector on the surface of the control volume. Here if is scalar then φ must be scalar, and if is vector then φ must be vector. In the previous sections, by using the Eulerian approach, we showed that this integral equation leads to the three local equations, as: the first φpr, t, nq φpr, t, nq (4.2) second φpr, t, nq n x φpr, t, e x q n y φpr, t, e y q n z φpr, t, e z q (4.3) and third pr, tq φpr, t, e xq φpr, t, e yq φpr, t, e zq (4.4) x y z In the following, we present some properties of a general integral equation in continuum media. If we have the following relation: t0 t t0 ψ dτ (4.5) then by the definition of integrals it can be written as: t dτ ψ dτ Note that we use the Eulerian approach. This implies: t ψ 0, ď t ď p tq If and t are any time and time interval in the time domain, then the general equation (4.5) leads to below equation that holds for any time: ψ (4.6) t In addition, the following integral equation holds for the control volume in the Eulerian approach: Q Q dv dv (4.7) t t efore considering the fundamental laws of continuum mechanics, let us discuss the flow of a physical quantity into a surface in continuum media. If we have a physical quantity such as U Upr, tq that transfers by the velocity of the substance in continuum media, and u upr, tq is U per unit volume, then the flow of this quantity into a surface with outward unit normal vector n is in the form: φ U u v.n (4.8) where v vpr, tq is the velocity vector of the substance. So, φ U φ U pr, t, nq and it has the dimension of rus{pm 2.sq. The negative sign is used because n is the outward unit normal vector of the surface. Here we suppose the fixed control volumes and for the moving control volumes the relative velocity must be used. Note that by the equation (4.8), φ U satisfies the first and second local relations (4.2) and (4.3), as below: u v.n p u v.p nqq

15 therefore and so GENERAL EXACT TETRAHEDRON ARGUENT, THIRD PAPER 5 φ U pr, t, nq φ U pr, t, nq (4.9) u v.n u v.pn x e x n y e y n z e z q n x p u v.e x q n y p u v.e y q n z p u v.e z q φ U pr, t, nq n x φ U pr, t, e x q n y φ U pr, t, e y q n z φ U pr, t, e z q (4.0) y these general relations, we will consider the fundamental laws of continuum mechanics in the next subsections. 4.. Conservation of mass. The basic law of conservation of mass of a control volume says: The total mass over the control volume at time p tq equals the total mass over at time plus the net of mass flow into from to p tq. So: ρ dv t ρ dv φ m ds dτ (4.) where ρ ρpr, tq is the density (mass per unit volume) and φ m φ m pr, t, nq is the mass flow into the surface that it acts. Using the equation (4.8), we have φ m ρ v.n. y rearranging the equation: ρ dv t ρ dv ρ v.n ds dτ (4.2) This is similar to the general equation (4.5), using (4.6) it becomes: ρ dv ρ v.n ds (4.3) t This is the integral equation of mass conservation law in continuum mechanics. y using (4.7) we have: ρ t dv ρ v.n ds (4.4) This equation is similar to the general integral equation (4.), where ρ{t and φ φ m pr, t, nq ρ v.n, so the three general local relations (4.2), (4.3), and (4.4) hold for it. The first and second local relations (4.2) and (4.3) lead to: and φ m pr, t, nq φ m pr, t, nq (4.5) φ m pr, t, nq n x φ m pr, t, e x q n y φ m pr, t, e y q n z φ m pr, t, e z q (4.6) ut as we showed in (4.9) and (4.0), the mass flow φ m pr, t, nq ρ v.n satisfies the two local relations (4.2) and (4.3), and their meanings. So, the above two relations do not give us new results. The third local relation (4.4) for ρ{t and φ φ m pr, t, nq ρ v.n leads to: ρ t x p ρ v.e xq y p ρ v.e yq z p ρ v.e zq

16 6 E. AZADI for v v x e x v y e y v z e z, we have v.e x v x, v.e y v y, and v.e z v z. Substituting these relations into the equation and rearranging it, yields: ρ t pρv xq pρv yq pρv zq 0 (4.7) x y z or ρ.pρvq 0 (4.8) t This is the differential equation of mass conservation law in continuum mechanics that is called the continuity equation Conservation of linear momentum. The basic law of conservation of linear momentum of a control volume says: The total linear momentum over the control volume at time p tq equals the total linear momentum over at time plus the net of linear momentum flow into from to p tq plus the total surface and body forces over from to p tq. So: ρv dv t ρv dv t ds dτ φ lm ds dτ ρb dv dτ (4.9) where ρv is the linear momentum per unit volume, φ lm φ lm pr, t, nq is the linear momentum flow into the surface that it acts, t tpr, t, nq is the surface force per unit area that is called traction vector, and b bpr, tq is the body force per unit mass. y using the general equation (4.8) for the flow of linear momentum, we have φ lm pρvqv.n, and rearranging the equation yields: ( ρv dv ρv dv t pρvqv.n ds ρb dv dτ t (4.20) This is similar to the general equation (4.5). So, by using (4.6) it becomes: ( ρv dv t pρvqv.n ds ρb dv (4.2) t This is the integral equation of linear momentum conservation law in continuum mechanics. Using (4.7) and rearranging the equation:! pρvq ) ( ρb dv t pρvqv.n ds (4.22) t This is similar to the general integral equation (4.) by the vector forms of and φ, where pρvq{t ρb and φ t pρvqv.n t φ lm. So, the three general local relations (4.2), (4.3), and (4.4) hold for it. The first and second local relations (4.2) and (4.3) lead to: tpr, t, nq φ lm pr, t, nq tpr, t, nq φ lm pr, t, nq (4.23)

17 and GENERAL EXACT TETRAHEDRON ARGUENT, THIRD PAPER 7 tpr, t, nq φ lm pr, t, nq n x tpr, t, ex q φ lm pr, t, e x q ( n y tpr, t, ey q φ lm pr, t, e y q ( n z tpr, t, ez q φ lm pr, t, e z q ( (4.24) ut as we showed in (4.9) and (4.0), the linear momentum flow φ lm satisfies the two local relations (4.2) and (4.3), and their meanings. i.e.: pρvqv.n and φ lm pr, t, nq φ lm pr, t, nq (4.25) φ lm pr, t, nq n x φ lm pr, t, e x q n y φ lm pr, t, e y q n z φ lm pr, t, e z q (4.26) So, these terms remove from the equations (4.23) and (4.24). Thus, we have from (4.23): tpr, t, nq tpr, t, nq (4.27) This is the Cauchy lemma for traction vectors and states that the traction vectors acting on opposite sides of the same surface at a given point and time are equal in magnitude but opposite in direction. And from (4.24): tpr, t, nq n x tpr, t, e x q n y tpr, t, e y q n z tpr, t, e z q (4.28) This means that if we have the traction vectors on the three orthogonal surfaces at a given point and time then we can get the traction vector on any surface that passes through that point at that time by having the unit normal vector of this surface and using this linear relation. So, we must define the traction vectors on the three orthogonal surfaces at any point and at any time. The traction vector on the surface with unit normal vector e x by its components, defines as: tpr, t, e x q T xx pr, tq e x T xy pr, tq e y T xz pr, tq e z (4.29) here T xx pr, tq, T xy pr, tq, and T xz pr, tq are scalars that depend only on r and t. In each case the first subscript indicates the direction of normal unit vector of the surface that this case acts on it, and the second subscript indicates the direction of this component of traction vector. Similarly, the traction vectors on the surfaces with unit normal vectors e y and e z, define as: and tpr, t, e y q T yx pr, tq e x T yy pr, tq e y T yz pr, tq e z (4.30) tpr, t, e z q T zx pr, tq e x T zy pr, tq e y T zz pr, tq e z (4.3) y substituting these equations in (4.28) tpr, t, nq n x Txx pr, tq e x T xy pr, tq e y T xz pr, tq e z ( n y Tyx pr, tq e x T yy pr, tq e y T yz pr, tq e z ( n z Tzx pr, tq e x T zy pr, tq e y T zz pr, tq e z (

18 8 E. AZADI by rearranging the equation tpr, t, nq n x T xx pr, tq n y T yx pr, tq n z T zx pr, tq ( e x n x T xy pr, tq n y T yy pr, tq n z T zy pr, tq ( e y n x T xz pr, tq n y T yz pr, tq n z T zz pr, tq ( e z this can be shown as» tpr, t, nq t fi» xpr, t, nq t y pr, t, nqfl T fi xx T xy T xz T yx T yy T yz fl t z pr, t, nq T zx T zy T zz using the vector relations, this becomes n fi x fl (4.32) T» n y n z t T T.n (4.33) where T T pr, tq is a second order tensor and is called stress tensor. This tensor depends only on the position vector and time. This relation means that for describing the state of stress on any surface at a given point and time we need the 9 components of the stress tensor at that point and time. So, the second local relation (4.3) for linear momentum leads to the existence of stress tensor. Let us apply the third local relation (4.4) for linear momentum, where pρvq{t ρb and φ tpr, t, nq pρvqv.n. Thus: pρvq t ρb ( ( tpr, t, ex q pρvqv.e x tpr, t, ey q pρvqv.e y x y ( tpr, t, ez q pρvqv.e z z Using the relations (4.29), (4.30), and (4.3), we have: tpr, t, e x q pρvqv.e x pt xx ρv x v x qe x pt xy ρv y v x qe y pt xz ρv z v x qe z tpr, t, e y q pρvqv.e y pt yx ρv x v y qe x pt yy ρv y v y qe y pt yz ρv z v y qe z tpr, t, e z q pρvqv.e z pt zx ρv x v z qe x pt zy ρv y v z qe y pt zz ρv z v z qe z Substituting these equations into the equation (4.34) and rearranging it, yields: pρvq t therefore pρvq t (4.34) ρb ( ( Txx e x T xy e y T xz e z Tyx e x T yy e y T yz e z x y ( ( Tzx e x T zy e y T zz e z ρvx v x e x ρv y v x e y ρv z v x e z z x ( ( ρvx v y e x ρv y v y e y ρv z v y e z ρvx v z e x ρv y v z e y ρv z v z e z y z ρb x y x».t.pρvvq T fi xx T xy T xz T yx T yy T yz T zx T zy T zz fl x y x» ρv fi xv x ρv x v y ρv x v z ρv y v x ρv y v y ρv y v z fl ρv z v x ρv z v y ρv z v z

19 GENERAL EXACT TETRAHEDRON ARGUENT, THIRD PAPER 9 where ρvv ρv i v j is the last second order tensor in the first line of the equation. y rearranging the equation: pρvq.pρvvq.t ρb (4.35) t This is the differential equation of linear momentum conservation law in continuum mechanics and is called the general equation of motion or Cauchy equation of motion. Using the mass continuity equation (4.8), it becomes: ρ v t ρpv. qv.t ρb (4.36) 4.3. Conservation of angular momentum. The basic law of conservation of angular momentum of a control volume about point r 0 says: The total angular momentum about point r 0 over the control volume at time p tq equals the total angular momentum about point r 0 over at time plus the net of angular momentum flow about point r 0 into from to p tq plus the total moment of surface and body forces about point r 0 over from to p tq. So: pr ˆ ρvq dv t pr ˆ ρvq dv pr ˆ tq ds dτ φ am ds dτ pr ˆ ρbq dv dτ (4.37) by r r r 0 then r ˆ ρv is the angular momentum about r 0 per unit volume, φ am φ am pr, t, nq is the angular momentum flow about r 0 into the surface that it acts, r ˆ t is the moment of surface force about r 0 per unit area, and r ˆ ρb is the moment of body force about r 0 per unit volume. y using the general equation (4.8) for the flow of angular momentum about r 0, we have φ am pr ˆ ρvqv.n r ˆ φ lm, where φ lm pρvqv.n is the linear momentum flow into. y rearranging the equation: pr ˆ ρvq dv t pr ˆ ρvq dv r ˆ t pρvqv.n ( ds pr ˆ ρbq dv dτ (4.38) This is similar to the general equation (4.5). So, by using (4.6) it becomes: pr ˆ ρvq dv r ˆ t pρvqv.n ( ds pr ˆ ρbq dv (4.39) t This is the integral equation of angular momentum conservation law in continuum mechanics. Using (4.7) and rearranging the equation:! t pr ˆ ρvq pr ˆ ρbq) dv r ˆ t pρvqv.n ( ds (4.40)

20 20 E. AZADI This is similar to the integral equation (4.) by the vector forms of and φ, where pr ˆ ρvq{t pr ˆρbq and φ r ˆt pρvqv.n r ˆtpr, t, nqφ lm pr, t, nq. So, the three general local relations (4.2), (4.3), and (4.4) hold for it. The first and second local relations (4.2) and (4.3) lead to: and r ˆ tpr, t, nq φ lm pr, t, nq r ˆ tpr, t, nq φ lm pr, t, nq r ˆ tpr, t, nq φ lm pr, t, nq n x r ˆ tpr, t, e x q φ lm pr, t, e x q ( n y r ˆ tpr, t, e y q φ lm pr, t, e y q ( n z r ˆ tpr, t, e z q φ lm pr, t, e z q ( (4.4) (4.42) ut these equations are the cross product of r and the equations (4.25) and (4.26), respectively, that already were obtained in the subsection of the linear momentum. So, these equations do not give us new results. The third local relation (4.4) for pr ˆ ρvq{t pr ˆ ρbq and φ r ˆ t pρvqv.n, leads to: t pr ˆ ρvq pr ˆ ρbq x y r ˆ tpr, t, e x q pρvqv.e x ( r ˆ tpr, t, e y q pρvqv.e y ( z In the Eulerian approach for r r r 0, we have: r ˆ tpr, t, e z q pρvqv.e z ( (4.43) r t 0, r x e x, r y e y, r z e z (4.44) y using these relations, the equation (4.43) becomes:! pρvq r ˆ t )! ρb r ˆ x tpr, t, ex q pρvqv.e x tpr, t, ey q pρvqv.e y y tpr, t, ez q pρvqv.e z ) z!e x ˆ tpr, t, e x q pρvqv.e x ey ˆ tpr, t, e y q pρvqv.e y e z ˆ tpr, t, e z q pρvqv.e z ) (4.45) ut the first two lines of this equation is the cross product of r and the equation (4.34) that already was obtained in the subsection of the linear momentum. So, these parts remove from the equation and we have: e x ˆ tpr, t, e x q pρvqv.e x ey ˆ tpr, t, e y q pρvqv.e y e z ˆ tpr, t, e z q pρvqv.e z 0 (4.46)

21 GENERAL EXACT TETRAHEDRON ARGUENT, THIRD PAPER 2 thus e x ˆ tpr, t, e x q e y ˆ tpr, t, e y q e z ˆ tpr, t, e z q e x ˆ pρvqv.e x ey ˆ pρvqv.e y ez ˆ pρvqv.e z ρv x pe x ˆ vq ρv y pe y ˆ vq ρv z pe z ˆ vq ρv x p v z e y v y e z q ρv y pv z e x v x e z q ρv z p v y e x v x e y q pρv y v z ρv y v z qe x p ρv x v z ρv x v z qe y pρv x v y ρv x v y qe z 0 so e x ˆ tpr, t, e x q e y ˆ tpr, t, e y q e z ˆ tpr, t, e z q 0 (4.47) substituting the components of the traction vectors from (4.29), (4.30), and (4.3) into the equation, yields: this implies e x ˆ pt xx e x T xy e y T xz e z q e y ˆ pt yx e x T yy e y T yz e z q e z ˆ pt zx e x T zy e y T zz e z q 0 p T xz e y T xy e z q pt yz e x T yx e z q p T zy e x T zx e y q pt yz T zy q e x pt zx T xz q e y pt xy T yx q e z 0 So, we have T xy T yx, T xz T zx, T yz T zy (4.48) or T T T (4.49) therefore, the third local relation (4.4) for conservation of angular momentum leads to the symmetry of stress tensor. y (4.49) we can tell for describing the state of stress on any surface at a given point and time we need the 6 components of the symmetric stress tensor at that point and time Conservation of energy. The basic law of conservation of energy of a control volume says: The total energy over the control volume at time p tq equals the total energy over at time plus the net of energy flow into from to p tq plus the total surface heat into from to p tq plus the total heat generation over from to p tq plus the total work done by surface and body forces over from to p tq. So: pρe 2 ρv2 q dv t pρe 2 ρv2 q dv q s ds t.v ds dτ dτ φ en ds ρ 9q g dv dτ dτ pρbq.v dv dτ (4.50) where e epr, tq is the internal energy per unit mass, and ρe{2ρv 2 is the total energy (internal energy + kinetic energy) per unit volume. Here v 2 v i v i v 2 x v 2 y v 2 z. On the right hand side, φ en φ en pr, t, nq is the energy flow into the surface that it acts,

22 22 E. AZADI q s q s pr, t, nq is the rate of surface heat into per unit area, 9q g 9q g pr, tq is the rate of heat generation per unit mass, t.v and pρbq.v are the rates of work done by the surface force per unit area and body force per unit volume, respectively. y using the general equation (4.8) for the flow of energy we have φ en pρe {2ρv 2 qv.n, and by rearranging the equation: pρe 2 ρv2 q dv t pρe 2 ρv2 q dv t.v qs pρe 2 ρv2 qv.n ( ds pρbq.v ρ 9qg ( dv dτ (4.5) This is similar to the general equation (4.5). So, by using (4.6) it becomes: pρe t 2 ρv2 q dv t.v qs pρe 2 ρv2 qv.n ( ( ds pρbq.v ρ 9qg dv (4.52) This is the integral equation of energy conservation law in continuum mechanics. Using (4.7) and rearranging the equation:! t pρe ) 2 ρv2 q pρbq.v ρ 9q g dv t.v qs pρe 2 ρv2 qv.n ( ds (4.53) This is similar to the integral equation (4.), where pρe {2ρv 2 q{t pρbq.v ρ 9q g and φ t.v q s pρe{2ρv 2 qv.n t.v q s φ en. So, the three general local relations (4.2), (4.3), and (4.4) hold for it. The first and second local relations (4.2) and (4.3) lead to: tpr, t, nq.v q s pr, t, nq φ en pr, t, nq tpr, t, nq.v q s pr, t, nq φ en pr, t, nq (4.54) and tpr, t, nq.v q s pr, t, nq φ en pr, t, nq n x tpr, t, ex q.v q s pr, t, e x q φ en pr, t, e x q ( n y tpr, t, ey q.v q s pr, t, e y q φ en pr, t, e y q ( n z tpr, t, ez q.v q s pr, t, e z q φ en pr, t, e z q ( (4.55) ut as we showed in (4.9) and (4.0), the energy flow φ en pρe {2ρv 2 qv.n satisfies the two local relations (4.2) and (4.3), and their meanings. Therefore, the energy flow terms remove from the two above equations. Also, in (4.27) and (4.28), we saw that the traction vector t satisfies the two local relations (4.2) and (4.3). As a result, t.v satisfies that equations, as well. So, these terms remove from the two above equations and we have from (4.54): q s pr, t, nq q s pr, t, nq (4.56) This is the general Cauchy lemma for surface heat and states that the surface heats acting on opposite sides of the same surface at a given point and time are equal in magnitude but opposite in sign. From (4.55), we have: q s pr, t, nq n x q s pr, t, e x q n y q s pr, t, e y q n z q s pr, t, e z q (4.57) This means that if we have the surface heats on the three orthogonal surfaces at a given point and time then we can get the surface heat on any surface that passes through that

23 GENERAL EXACT TETRAHEDRON ARGUENT, THIRD PAPER 23 point at that time by having the unit normal vector of this surface and using this linear relation. So, we must define the scalar surface heats into the three orthogonal surfaces with unit normal vectors e x, e y, and e z, respectively, as: q s pr, t, e x q q x pr, tq, q s pr, t, e y q q y pr, tq, q s pr, t, e z q q z pr, tq (4.58) the negative sign is due to the fact that we suppose for example q x pr, tq is the exit heat from the surface with unit normal vector e x but q s pr, t, e x q is the surface heat into that surface. Here the subscripts in q x, q y, and q z indicate the direction of unit normal vector of the surfaces that they act on them. So, we have from (4.57):» q s pr, t, nq n x q x pr, tq n y q y pr, tq n z q z pr, tq q x q y q z n fi x fl (4.59) thus n y n z q s pr, t, nq qpr, tq.n (4.60) where qpr, tq is a vector that depends only on the position vector and time and is called heat flux vector. So, the first and second local relations (4.2) and (4.3) for the conservation of energy lead to the existence of heat flux vector qpr, tq. This means that for describing the surface heat on any surface at a given point and time we need the 3 components of qpr, tq at that point and time. The third local relation (4.4) for energy conservation is: t pρe 2 ρv2 q pρbq.v ρ 9q g tpr, t, ex q.v q.e x pρe ( x 2 ρv2 qv.e x tpr, t, ey q.v q.e y pρe ( y 2 ρv2 qv.e y (4.6) z tpr, t, ez q.v q.e z pρe 2 ρv2 qv.e z ( y using the equation (4.33), i.e., t T T.n, the above equation can be shown as:» t pρe 2 ρv2 q pρbq.v ρ 9q g T fi xxv x T xy v y T xz v z x y z T yx v x T yy v y T yz v z fl T zx v x T zy v y T zz v z» q fi» x x y z q y fl pρe fi {2ρv2 qv x x y z pρe {2ρv 2 qv y fl pρe {2ρv 2 qv z q z by vector relations, this becomes t pρe 2 ρv2 q pρbq.v ρ 9q g.pt.vq.q.pρe 2 ρv2 qv by rearranging the equation, we have t pρe 2 ρv2 q.pρe qv 2 ρv2.pt.vq.q pρbq.v ρ 9q g (4.62) This is the differential equation of energy conservation law in continuum mechanics. Also, there are some other forms of energy equation that are obtained from the above

24 24 E. AZADI equation. We have: t p 2 ρv2 q " pρv.vq v. 2 t t pρvq.p qv 2 ρv2.pρv.vqv " v..pρvvq 2 ".pt.vq v..t T : v (4.63) where T : v is the following scalar v j v x T : v T ij T xx x i x T v y xy x T v z xz x T v x yx y T v y yy y T v z yz y v x T zx z T v y zy z T v z zz z y using the relations (4.63), the equation (4.62) becomes: " " " pρeq v. t t pρvq.pρevqv..pρvvq v..t T : v.qpρbq.vρ 9q g by rearranging this equation pρeq t.pρevq T : v.q ρ 9q g " pρvq v..pρvvq.t ρb t but due to the differential equation of linear momentum conservation law (4.35), the expression in the braces in the second line of the above equation is equal to zero, therefore this line removes from the equation and we have: pρeq.pρevq T : v.q ρ 9q g (4.64) t this is the differential equation of internal energy balance. Using the mass continuity equation (4.8) it becomes: ρ e t ρv. e T : v.q ρ 9q g (4.65) 4.5. Entropy law. The basic law of entropy of a control volume says: The total entropy over the control volume at time p tq is greater than or equal to the total entropy over at time plus the net of entropy flow into from to p tq plus the total surface heat per temperature into from to p tq plus the total heat generation per temperature over from to p tq. So: ρs dv t ě ρs dv q s T ds dτ φ ent ds dτ ρ 9q g T dv dτ (4.66)

25 GENERAL EXACT TETRAHEDRON ARGUENT, THIRD PAPER 25 where s spr, tq is the entropy per unit mass and ρs is the entropy per unit volume. On the right hand side, φ ent φ ent pr, t, nq is the entropy flow into the surface that it acts, q s q s pr, t, nq is the rate of surface heat into per unit area, 9q g 9q g pr, tq is the rate of heat generation per unit mass, and T T pr, tq is the absolute temperature. In order to convert this inequality to an equation, we may define the rate of entropy generation per unit mass as 9s g 9s g pr, tq, where 9s g ě 0, and add the total entropy generation over from to p tq to the right hand side of the above inequality. So, we have the following equation: ρs dv t ρs dv q s T ds dτ φ ent ds ρ 9q g T dv dτ dτ ρ 9s g dv dτ (4.67) y using the general equation (4.8) for the flow of energy we have φ ent pρsqv.n, and by rearranging the equation: q s ρs dv ρs dv t T pρsqv.n( ds (4.68) ρ 9q g T ρ 9s ( g dv dτ this is similar to the general equation (4.5). So, by using (4.6) it becomes: q s ρs dv t T pρsqv.n( ρ 9q g ds T ρ 9s ( g dv (4.69) This is the integral equation of entropy law in continuum mechanics. Since 9s g ě 0, by removing the integral of ρ 9s g from the equation, we have: t ρs dv ě q s T pρsqv.n( ds ρ 9q g T dv (4.70) This inequality is called the Clausius-Duhem inequality. Using (4.7) and rearranging the equation (4.69), we have:! pρsq ρ 9q ) g t T ρ 9s q s g dv T pρsqv.n( ds (4.7) This is similar to the integral equation (4.), where pρsq{t ρ 9q g {T ρ 9s g and φ q s {T pρsqv.n q s {T φ ent. So, the three general local relations (4.2), (4.3), and (4.4) hold for it. The first and second local relations (4.2) and (4.3) lead to: q s pr, t, nq{t φ ent pr, t, nq q s pr, t, nq{t φ ent pr, t, nq (4.72) and q s pr, t, nq{t φ ent pr, t, nq n x qs pr, t, e x q{t φ ent pr, t, e x q ( n y qs pr, t, e y q{t φ ent pr, t, e y q ( n z qs pr, t, e z q φ ent pr, t, e z q ( (4.73) ut we have from (4.60) that q s pr, t, nq qpr, tq.n, therefore q s {T satisfies the two local relations (4.2) and (4.3). Also, as we showed in (4.9) and (4.0), entropy flow φ ent pρsqv.n satisfies the two local relations (4.2) and (4.3) and their meanings.

26 26 E. AZADI Thus, the two above equations do not give us new results. Applying the third local relation (4.4) to the entropy integral equation (4.7), where pρsq{t ρ 9q g {T ρ 9s g and φ q s {T φ ent qpr, tq.n{t pρsqv.n, leads to: pρsq t so, we have ρ 9q g T pρsq t ρ 9s g q.e x x T q.e z z T ρ 9q g T by rearranging this equation ρ 9s g x pρsqv.e ( x y pρsqv.e ( z y x».p q T q.pρsvq q fi x{t q y {T q z {T q.e y T fl x y pρsqv.e ( y x» ρsv fi x ρsv y fl ρsv z (4.74) pρsq.pρsvq.p q t T q ρ 9q g T ρ 9s g (4.75) This is the differential equation of entropy law in continuum mechanics. Using the mass continuity equation (4.8) it becomes: ρ s t ρv. s.p q T q ρ 9q g T ρ 9s g (4.76) Since 9s g ě 0, by removing ρ 9s g from the equation (4.75) we have: pρsq.pρsvq ě.p q t T q ρ 9q g T This is the differential form of the Clausius-Duhem inequality. (4.77) 5. Conclusion We considered the general integral equation on the control volume, as the form: dv φ ds where pr, tq is called body term and φ φpr, t, nq is called surface term. These functions are continuous over the volume and the surface of, respectively. Here if is scalar then φ must be scalar, and if is vector then φ must be vector. We wanted to determine how many local relations can be derived from this general integral equation and what they are. We first derived the general Cauchy lemma for surface term from the above integral equation as the first local relation. So: The first local relation: φpr, t, nq φpr, t, nq