Viscosity and numerical methods Viggo H. Hansteen October 8, 2018 1 Introduction In any flow, layers move at different velocities and a fluid’s viscosity arises from the shear stress between the layers that ultimately oppose any applied force. These shear stresses are communicated via collisions between the particles comprising the fluid, and the typical scale of viscosity is therefore of order the mean free collisional length λ. This scale is usually much smaller than the scales that can be represented on a numerical grid, λ << ∆x if one desires to model an astrophysical phenomenon of macroscopic size, and an understanding of this limitation is therefore vital to understanding the limitations of numerical modeling. 1.1 The Cauchy{Stokes Decomposition Let us begin by considering the description of the details of the instantanious motion of a fluid. The flow at some small distance ξ from a given point O, moving with velocity uO can to first order be written O j ui(ξ) = ui + ui;jξ; i.e. we are expressing the relative velocity in terms of the velocity gradient tensor ui;j. As with any tensor this can be decomposed into an antisymmetric and a symmetric part giving 1 1 u (ξ) = uO + (u − u )ξj + (u + u )ξj i i 2 i;j j;i 2 i;j j;i This means that we can decompose the motion into a translation, a rigid rotation, and a dilation. The rotation of a rigid body with angular rate !R around an axis n gives an angular velocity around any point ξ uR = !R×ξ j The second term of our decomposed relative velocity is a sum of the form Ωjiξ where Ω is the antisymmetric tensor 1 Ω ≡ (u − u ): ji 2 i;j j;i But a such a sum containing an antisymmetric tensor can always be rewritten ! × ξ where ! is the vector 1 dual of the tensor Ω. 1 Thus we see that our fluid parcel rotates with the angular velocity 1 1 1 !i = eijkΩ = e u = (r × u)i; 2 jk 2 ijk k;j 2 i.e. the local angular velocity ! is equal to one half of the vorticity of the flow. The remaining term term 1 E = (u + u ) ij 2 i;j j;i is the strain tensor, where the diagonal elements are the normal strain while the off-diagonal elements represent the shear deformation. 1.2 The Stress Tensor Figure 1: Stresses acting on an elementary tetrahedron. While the section above described how the motions of a fluid elements in the vicinity of a point O can be decomposed into a translation, a rigid rotation, and a dilation, the stress tensor describes the forces acting on the fluid element. Let us consider the surface forces on the surface dS located at position x with a direction n. If t represents the surface forces such that t(x; n)dS. It is clear that the forces on the other side, away from n, of the surface are given by −t(x; n)dS or t(x; −n)dS which implies that t(x; −n) = −t(x; n) and that t is an antisymmetric function of n. 1 1 The vector dual of an antisymmetric tensor is defined as !i = 2 eijkΩjk and the reciprocal relation Ωij = eijk!k. Note that for such dual pairs the following relation holds for any vector a: Ωjkaj = eijk!iaj = (! × a) . The permutation symbol eijk is defined as 8 < +1 ijk; even permutation of 123 eijk = −1 ijk; odd permutaion of 123 : 0 any two indices are the same 2 Consider now the parallellopeid constructed by dS and the three surfaces defined along the co-ordinate axes i; j; k. The surface areas of these surfaces are dS1 = n1dS, dS2 = n2dS, and dS3 = n3dS, where nl is the lth component of n. In equilibrium we must have that the sum of the forces is zero so t(n)dS + t(−i)dS1 + t(−j)dS2 + t(−k)dS3 ≡ 0; but then by antisymmetry we must have that t(n)dS − t(i)dS1 − t(j)dS2 − t(k)dS3 = 0; or by expressing dSi in terms of ni and dS t(n) = n1t(i) + n2t(j) + n3t(k) Let ti denote the ith component of t(n) and let Tji denote the ith component of t(j), the previous equations then states that i ti = n Tji where Tji is the stress tensor. The diagonal elements of the tensor represent the normal stresses while the off-diagonal elements are the shearing stresses. A fluid at rest can only support normal stresses in which case we write the surface forces as i tj = Tijn = −pnj i.e. t is proportional to n in any arbritary direction, which means that we must write Tij = −pδij: In the case of an ideal fluid no tangetial stresses are supported, for non-ideal fluids we write the total stress tensor as Tij = −pδij + σij: where the latter tensor is the viscous stress tensor. Note that if angular momentum is to be conserved then we have that the stress tensor is symmetric i.e that Tij = Tji. 1.3 Cauchy's equation of motion We need to see how the viscosity appears in the equation of motion for the fluid. A fluid element's momentum is changed by both body forces and surface forces. D Z Z Z ρudV = fdV + tdS Dt V V S Using the equation of mass continuity we can rewrite this in component form as Z Z Z ρ(Dui=Dt)dV = f idV + tidS V V S or using the divergence theorem Z Z Z Z i i ji i ji ρ(Du =Dt)dV = f dV + T njdS = (f + T;j )dV V V S V i where we have written the surface force t as ti = n Tji with the unit vector n representing the normal to the surface. Since the volume V is arbitrary the integrals are only equal if their integrands are equal. Thus we get Cauchy's equation of motion i i ij ρ(Du =Dt) = f + T;j 3 Figure 2: Sketch for illustration of discussion in viscosity: Variations in the velocity shear (dux=dy) are communicated via random (thermal) motions in the fluid with mean free path λ between collisions. This leads to a shear stress τ on area A in the x direction. 1.4 A concrete example To give an idea of what is going on inside the fluid element before drowning in an ocean of mathematics let us consider a simple example which should help tie together ideas of how forces inside fluids are com- municated and our mathematical formulation of these viscous forces through the stress tensor. We follow Wikipedia (http://en.wikipedia.org/wiki/Viscosity) or Mihalas & Mihalas (Foundations of Radiation Hydrodynamics, sections 29 and 30) and consider the relation between the mean free path and the viscosity. For a Newtonian fluid, the shear stress τ on a unit area moving parallel to itself is (empirically) found to be proportional to the rate of change of velocity with distance perpendicular to the unit area parallel to the x − z plane, moving along the x-axis: du τ = µ x : dy Interpreting shear stress as the time rate of change of momentum p per unit area A of an arbitrary surface gives τ =pA _ =m _ huxiA wherem _ is the rate of fluid mass hitting the surface A and huxi is the average velocity along x of fluid molecules hitting the unit area; see figure 2 for a sketch of the geometry. We can write 1 du m_ = ρuA¯ and hu i = λ x x 2 dy where ρ is the density of the fluid andu ¯ = phu2i is the average particle speed. We are and assuming that particles hitting the unit area are equally distributed between distances 0 and λ from A and that their average velocities change linearly with distance (this will always be true with small enough λ). From this 4 we get 1 du µ 1 τ = ρuλ¯ x ) ν = = uλ.¯ (1) 2 dy ρ 2 The quantity µ is the dynamic viscosity while ν ≡ µ/ρ is the kinematic viscosity. We will use later that u¯ ∼ cs, i.e. that the average speed of particles is the speed of sound in the medium. 2 The Stress Tensor in a Newtonian Fluid We have shown the equation of motion to be dependent on the stress tensor Tij. Following Mihalas & Mihalas Foundations of Radiation Hydrodynamics we can derive an explicit expression for the stress tensor in terms of the physical properties of the fluid and its state of motion. Mihalas & Mihalas state the following four physical considerations: 1. Internal frictional forces should only exist when one element of fluid moves relative to another; hence viscous terms must depend on the space derivates of the velocity field, ui;j. This since the velocity at a small distance away from the point (r) can be written du u(r + dr) = u(r) + dr + ::: dr where du=dr is shorthand for the dyadic product of the r operator and the velocity u as discussed in section 1.1. 2. The stress tensor should reduce to its hydrostatic form when the fluid is at rest or translates uniformly. Therefore, we write Tij = −pδij + σij where σij is the viscous stress tensor, which accounts for the internal frictional forces in the flow. 3. For small velocity gradients we expect viscous forces, i.e. σij, to depend only linearly on space deriva- tives of the velocity.