Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 28 April 2020 doi:10.20944/preprints202003.0030.v3 Correct expression of the material derivative in continuum physics |the solution existence condition for the Navier-Stokes equation Bohua Sun1 1 School of Civil Engineering & Institute of Mechanics and Technology Xi'an University of Architecture and Technology, Xi'an 710055, China http://imt.xauat.edu.cn email: [email protected] (Dated: April 27, 2020) The material derivative is important in continuum physics. This Letter shows that the expression d @ dt = @t + (v · r), used in most literature and textbooks, is incorrect. The correct expression d(:) @ dt = @t (:) + v · [r(:)] is formulated. The solution existence condition of Navier-Stokes equation has been proposed from its form-solution, the conclusion is that "The Navier-Stokes equation has solution if and only if the determinant of flow velocity gradient is not zero, namely det(rv) 6= 0." the material derivative with respect to time must be In the latest version of Landau & Lifshitz,[13] the ma- defined. For mass density ρ(x; t), flow velocity v(x; t) = terial derivative of flow velocity is given in another form: vkek, and stress tensor σ = σijei ⊗ ej, the material dv @v derivatives are given by: = + (vr)v: (8) dt @t dρ @ρ @ρ @xk @ρ @ρ @ρ @ρ = + = +v1 1 +v2 2 +v3 3 ; (1) dt @t @xk @t @t @x @x @x Besides the books mentioned above, there is a large amount of literature in fluid mechanics taking the same expressions as Refs. 1 and 13. This makes it difficult dv @v @v @xk @v @v @v @v = + = + v1 1 + v2 2 + v3 3 ; for readers to identify which expression for the materi- dt @t @xk @t @t @x @x @x (2) al derivative is correct, causing great confusion to both and students and scholars. Although some authors, such as Lighthill,[14] dσ @σ @σ @xk @σ @σ @σ @σ Batchelor,[15] Frisch,[16], Nhan & Nam [17], Xie,[18] and = + = + v1 1 + v2 2 + v3 3 ; dt @t @xk @t @t @x @x @x Zhao [19] have used the correct expression, it seems that (3) most fluid mechanics textbooks and academic literature respectively, where t is time, x = xkek is position, ek is have adopted an incorrect expression for the material a base vector and v = @xk is the flow velocity. k @t derivative as in Ref. 1. Therefore, to revive the great In most textbooks, handbooks and encyclopedia, such influence of Landau in physics and fluid mechanics, we as Refs. 1{12, the above material derivatives are ex- attempt to address this issue in this dedicated paper, pressed as: where we revisit the material derivative to show why Eqs. dρ @ρ (7) and (8) are incorrect, and derive a correct expression = + (v · r)ρ, (4) dt @t using standard tensor calculus. In Landau & Lifshitz,[1] the material derivative is given dv @v as: = + (v · r)v; (5) dt @t @v @v @v dx + dy + dz = (dr · grad)v: (9) and @x @y @z dσ @σ = + (v · r)v; (6) Although Landau's fluid mechanics is well known world- dt @t wide, the above expression is incorrect. @ where the gradient operator r = ek . @xk Proof: To simplify Eqs. (1){(3) further, a differential operator is introduced as follows: Since v · grad = v · r = (viei) · (@jej) = vi;jei · ej = vi;jδij = divv, and (v · grad)v = d @ = + (v · r): (7) (v · r)v = (vj;iei · ej)(vkek) = δijvj;ivkek = dt @t vi;ivkek = (divv)v, where the divergence is @v1 @v2 @v3 @v1 This operator is used by most fluid mechanics textbooks, divv = @x1 + @x2 + @x3 , so (divv)v = ( @x1 + @v2 @v3 including well-known graduate textbooks such as Landau @x2 + @x3 )v. Thus: & Lifshitz,[1, 2] Prandtl,[3, 4] Anderson,[5] Pope,[6] Cen- gel & Cimbala,[7], Kundu et al.[8], Woan [9], wikipedia @v @v @v (v · grad)v 6= v + v + v : (10) [10, 11] and britannica [12]. 1 @x1 2 @x2 3 @x3 © 2020 by the author(s). Distributed under a Creative Commons CC BY license. Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 28 April 2020 doi:10.20944/preprints202003.0030.v3 2 @v T In the latest version of Landau & Lifshitz,[13] for rea- For a vector @x = v ⊗ r = vr = (rv) . Strictly sons which are unclear, the material derivative expression speaking, the right gradient of v should be written as dv @v has been changed to dt = @t + (vr)v; this is also incor- v ⊗ r; here, in order to agree as far as possible with rect. conventional presentation, we write v ⊗ r = vr and, similarly for the left gradient, r⊗v = rv. Hence (vr)· Proof: @v @ρ v = v · (rv) and dv = @t dt + (vr) · vdt = @t dt + v · Since vr = (viei)(@jej) = vi;jeiej, (rv)dt, or, dividing both sides by dt, dv @v @v (vr)v = (vj;ieiej)(vkek) = (vj;ieiej)(vkek); = + (vr) · v = + v · (rv): (13) dt @t @t which indicates that vr is a 3rd-order tensor 3. The 2nd order stress tensor σ = σ(x; t) is a tensor- rather than a vector. Thus: valued function of x and t and its differential is dσ = @v @v @v @σ dt + @σ · @x dt. For an arbitrary tensor @σ = σr 6= (vr)v 6= v + v + v : (11) @t @x @t @x 1 @x1 2 @x2 3 @x3 T @σ (rσ) , hence dσ = @t dt + (σr) · vdt, or dividing both dσ @σ sides by dt leads to dt = @t +(σr)·v. Since (σr)·v = To obtain the correct formulation of the material v · (rσ), derivative, we perform some basic tensor calculus [20]. The del operator is a vector differential operator, and dσ @σ @ = + v · (rσ): (14) defined by r = ei . An important note concerning @xi dt @t the del operator r is in order. Two types of gradients If an operator must be introduced for the material are used in continuum physics: left and right gradients. derivative, it should be in the following form: The left gradient is the usual gradient and the right gra- dient is the transpose of the forward gradient operator. d(:) @ = (:) + v · [r(:)]: (15) To see the difference between the left and right of dt @t gradients, consider a vector function A = Ai(x)ei. @ For a scalar function f(x; t), this leads to df = @f + v · The left gradient of a vector A is r ⊗ A ≡ ej ⊗ dt @t @xj @Ai (rf). If f is considered as a distributed function of gas (Aiei) = ej ⊗ ei = Ai;jej ⊗ ei, or written as rA ≡ @xj molecules in their phase space, this derivative is as in Eq. @ @Ai ej (Aiei) = ejei = Ai;jejei. The right gradient @xj @xj (3.2) in Physical Kinetics by Lifshitz and Pitaevskii. [2] @ @Ai du of a vector A is A ⊗ r ≡ (Aiei) ⊗ ej = ei ⊗ For a vector function u(x; t), Eq. (15) yields = @xj @xj dt @ @u + v · (ru). For a tensor function A(x; t), Eq. (15) ej = Ai;jei ⊗ ej, or written as Ar ≡ (Aiei)ej = @t @xj dA @A @Ai @Ai yields = + v · (rA). eiej = Ai;jeiej, where Ai;j = . dt @t @xj xj The Navier-Stokes equations of incompressible flow can The gradient of a scalar function is a vector, the diver- be expressed as follows: the momentum equation: gence of a vector-valued function is a scalar r · A, and the gradient of a vector-valued function is a second-order @v 1 + v · (rv) = − rp + νr2v; (16) tensor rA. Although the del operator has some of the @t ρ properties of a vector, it does not have them all, because it is an operator. For instance, r · A is a scalar (called and mass conservation equation: r · v = 0. the divergence of A) whereas A·r is a scalar differential Applying the divergence operation to both sides of operator, where A is a vector. Thus the del operator r the momentum equation Eq.16 and use the mass con- 2 does not commute in this sense. servation leads to a pressure equation: r · (p1) = 2 It worth to emphasise that the right gradient of a vec- −ρr · (v · rv) = −ρf(rv) + [r(vr)] · vg. In which, tor Ar is a more natural one, is often used in defining v(x; t) is flow velocity field, ρ is constant mass density, the deformation gradient tensor, displacement gradien- p(x; t) is flow pressure, ν is kinematical viscosity, t is k t tensor, and velocity gradient tensor. It is clear that time, x = x ek is position coordinates, ek is a base vec- @ T Ar = (rA) . tor and v is flow velocity, r = ek @xk is gradient operator T 1. Mass density ρ = ρ(x; t) is a scalar-valued function and 1 = ekek is an identity tensor, and vr = (rv) . of x and t and its differential is dρ = @ρ dt + @ρ · @x dt. Both solution existence and smoothness of the Navier- @t @x @t Stokes equations are still an open problem in mathemat- For a scalar @ρ = ρr = rρ, so dρ = @ρ dt+(ρr)·vdt = @x @t ics [21], regardless of numerous abstract studies that have @ρ dt + v · (rρ)dt, or, dividing both sides by dt, @t been done by mathematicians. Now the question is that, dρ @ρ @ρ is it possible to propose an alternative and simple direct = + (ρr) · v = + v · (rρ): (12) approach to express the flow velocity field without using dt @t @t complicated mathematics. Even we can not find the final 2. Flow velocity v = v(x; t) is a vector-valued function solution of the velocity field, it is still plausible if we can @v @v @x of x and t and its differential is dv = @t dt + @x · @t dt.