The Tensor of the Viscous Stresses

Chapter 9

The tensor of the viscous stresses

9.1 The tensor of the viscous stresses

We have already seen the possibility of representing the surface stresses be- tween parcels in a perfect fluid by means of a scalar field, the pressure. This is due to the fact that the stresses are always normal to the surfaces to which they refer, and have a magnitude independent of their orientation. But it was clear since then that in such a definition two different vector quantities were implicitly involved, i.e., the normal to the surface and the pressure force. The fact that their directions are coincident and that the magnitude of the pressure stress is independent of them allows us to represent the stress by a simple scalar quantity. However, the example of viscous flow analyzed in the previous chapter shows that the stress can also be tangent to a surface, and that the stress on a horizontal surface can be different from the stress on a vertical surface. Thus, if we want to deal with the most general case, we must take into account the possibility that the directions of the normal to the surface and of the stress can be different, and that the magnitude of this last quantity can depend on the orientation of the surface.

61 62 Chapter 9. The tensor of the viscous stresses

Let us consider an infinitesimal parallelepiped parcel. For each of its faces we allow the stress to be represented by a vector of different magnitude and direction. For the sake of simplicity, in (Fig. 9.1) we represent the total stress subtracted of the pressure component, as defined for an ideal fluid.

...... τxy ...... τ ...... x ...... τ ...... xz ...... • ...... (x,y,z). .τ . . . . xx ...... δz ...... δy ...... δx

Fig. 9.1: The stress vector due to the molecular viscosity of the fluid is represented for one of the six faces of the infinitesimal parallelepiped parcel. Here, τxx,τxy and τxz denote the components of the stress vector τ x relative to the positively oriented surface x = cost. Similar vectors can be defined over the other faces of the parcel.

These six forces must be subject to some conditions requested by the basic principles of statics. Fist of all, if two parcels are adjacent, according to the principle of action and reaction the two forces acting along their separation surface must cancel each other. This means that if we consider two adjacent parcels (1) and (2), the stress τ x1 exerted by the second parcel on the first one must be of equal magnitude and have the opposite direction with respect to the stress τ x2 exerted by the first parcel on the second one (Fig. 9.2). Since the parcel is infinitesimal, we expect that the stress passing from one face to the other parallel face of the parcel can only change by an infinitesimal amount. Thus, to a first approximation, we must have τ τ x+δx/2 = − x−δx/2 + O(δx). (9.1) The change in sign is due to the fact that the normal to the two parallel sides x − δx/2 = cost and x + δx/2 = cost are oriented in opposite directions. Thus, the stresses acting upon any pair of parallel faces of the parcel form a not exactly equilibrated torque. 9.1. The tensor of the viscous stresses 63

...... τ . τ . . . x21 . x32 ...... (1)...... (2). (3) ...... τ . . . . x12 ...... x−δx/2 x+δx/2

Fig. 9.2: The stress τ x32 exerted by parcel (3) on parcel (2) differs from the stress τ x21 exerted by parcel (2) on parcel (1) only for an infinitesimal quantity. For the principle of action and reaction the stress τ x12 exerted by parcel (1) on parcel (2) must be the opposite of the stress τ x21. Thus, τ x32 differs from −τ x12 only for an infinitesimal quantity. Setting τ x32 = τ x+δx/2 and τ x12 = τ x−δx/2, (9.1) follows.

Therefore, the stresses in the interior of the ﬂuid can be represented by three vectors relative to three surfaces normal to the coordinate directions. This means that the representation of the stresses τ x, τ y and τ z requires the use of a tensor, whose structure is τxx τxy τxz T = τ τ τ . (9.2) yx yy yz

τzx τzy τzz

The quantity T is called the tensor of the viscous stresses. The ﬁrst index of each component denotes the surface to which it refers. More exactly, it is the index of the axis to which the surface is normal. The second index refers to the component of the stress. Since each row of the tensor represents the stress along one of the faces of the parcel, we can write

τ x = τxxi + τxyj + τxzk, and analogous expressions for τ y and τ z. After these preliminary remarks, we can immediately see that these nine components are not independent of each other. Let us consider the components able to give rise to a rotation around the z-axis (Fig. 9.3). The torque δM applied to a parcel in the z-direction is given by

δM =(τxy δyδz)δx − (τyx δxδz)δy =(τxy − τyx)δV, 64 Chapter 9. The tensor of the viscous stresses

i τ ...... yx...... −j τxy . . • . . τxy j ...... −τyx i

Fig. 9.3: The global torque in the z-direction of the two couples of forces must vanish, so that τxy = τyx.

while the moment of inertia δI of the parcel is given by

1 δI = ρδV (δx)2 +(δy)2 . 12 h i Since the equation of motion is dΩ δM = δI , dt where Ω is the angular velocity of the parcel along the z-direction, it follows that

1 dΩ 2 2 τxy − τyx = ρ (δx) +(δy) . 12 dt h i Hence, the assumption that the angular acceleration of the parcel is always ﬁnite as the horizontal dimensions of the parcel tend to zero implies

τxy = τyx.

Clearly, it is possible for the parcels to change their rotation speed, but this can only be the result of the infinitesimal torque related to infinitesimal varia- tions of the stress passing from one side of the parcel to the other. This torque applied to the infinitesimal parcel can give rise to an angular acceleration of finite magnitude. Furthermore, since the stress is assumed to vary with continuity from one parcel to the other, the angular acceleration must involve, with a stronger or weaker magnitude, also the adjacent parcels, i.e., a finite volume of fluid. 9.1. The tensor of the viscous stresses 65

Thus, the torque of the two pairs of forces must be equal in magnitude and opposite in direction, with a possible unbalance of infinitesimal order. This in turn implies that the magnitude of the stress along the orthogonal faces of the parcel must be equal. In other words, the components of the stress tensor with the indices interchanged must be equal, so that the tensor is symmetric. Therefore, only six out of its nine components are independent quantities. Furthermore, the stress tensor is responsible for the deformation of the parcels. It is evident that stresses oriented as in Fig. 9.3 give rise to a deformation of the parcel as in Fig. 9.4. Clearly, the deformation of a parcel is hampered by the deformation of the adjacent parcels, so that it is infinitesimal for infinitesimal time intervals.

τ y...... τ ...... x ...... • ...... −τ ...... x...... −τ y

Fig. 9.4: This is an example of the way in which the tangential stresses can give rise to a deformation of the parcel. In the three-dimensional case the deformation is still more complicated.

In the example used in the previous chapter to introduce viscosity, the only component diﬀerent from zero are τzx, and consequently τxz. There, we have set T = τzx, so that the complete stress tensor is

0 0 τxz 0 0 T T = 0 0 0 = 0 0 0 .

τ 0 0 T 0 0 zx

The component τzx derives from the forces of constraint generated by the two horizontal plates. Once the stress tensor is known, it is possible to evaluate the stress exerted on a surface of arbitrary inclination.

The component of the tensor τij represents the value of the viscous stress in the direction xj over a surface at xi constant and oriented along the positive direction of xi. 66 Chapter 9. The tensor of the viscous stresses

Hence, τxx is the stress in the x-direction over a surface at x constant, whose unit vector is i. Conversely, over a negatively oriented surface the stress in the x-direction is equal to −τxx i. Now that this has been established, let us consider the prism of Fig. 9.5. In order that this be in equilibrium, the sum of the forces applied to it must vanish. In particular, the component of this force in the x-direction must be zero

−τxx δSx − τyx δSy + τnx δS =0.

...... δS ...... n.... . − ...... τxx ...... i...... θ ...... τnx = nxτxx + nyτyx ...... δSx ...... θ ...... δS ...... y −τyx i

Fig. 9.5: The stresses in the x-direction over the three faces of the prism must combine in such a way so as to guarantee the static equilibrium of the parcel.

Observe that the forces along the horizontal and vertical surfaces of the prism are negative, because the unit vectors of the surfaces are in the direction opposite with respect to that of the coordinate axes. When the two forces are positive, then both τxx and τyx are negative. From the equation above one obtains

τnx = nxτxx + nyτyx, which, for a generic tetrahedron with three orthogonal faces, becomes 3 τni = nj τji. =1 Xj Adopting a matrix notation, we ﬁnd that the stress τ over an arbitrary surface oriented in the direction of its unit vector n is given by

τxx τxy τxz τ = τ i + τ j + τ k = | n n n | τ τ τ = n ·T = T· n. (9.3) n nx ny nz x y z yx yy yz

τzx τzy τzz

The last step is justiﬁed by the symmetry of the stress tensor .

9.2. Forces associated to the viscous stress tensor 67

9.2 Forces associated to the viscous stress tensor

Let us denote by S the tensor of the total stresses, inclusive of the pressure stresses. We can write σxx σxy σxz −p + τxx τxy τxz S = σ σ σ = τ −p + τ τ = −pI + T , (9.4) yx yy yz yx yy yz

σzx σzy σzz τzx τzy −p + τzz where I is the unit tensor and T represents the tensor to add to the pressure terms in order to obtain the tensor of the global stresses. Hereafter, since the pressure term has been already worked out, we will focus our attention on T , which we will refer to as the stress tensor without any other attribute. In the case of an ideal ﬂuid, the total stress tensor reduces to the diagonal form −p 0 0 S = −pI = 0 −p 0 , (9.5)

0 0 −p which expresses the fact that no tangential stresses exist, but only normal stresses, directed toward the interior of the parcel. Normal stresses are therefore represented by a tensor, whose diagonal terms are all equal to the scalar function pressure. In order to evaluate the forces associated to the tensor T , we may adopt the same procedure followed in the derivation of the pressure forces. The force in the x-direction must depend on the components in this direction τxx,τyx and τzx of the stress vectors over the three orthogonal faces of the parcel δx δx ρ Fv,x δxδyδz = τxx x + ,y,z − τxx x − ,y,z δyδz + " 2 ! 2 !# δy δy + τyx x, y + , z − τyx x, y − , z δxδz + " 2 ! 2 !# δz δz + τzx x,y,z + − τzx x,y,z − δxδy = " 2 ! 2 !# ∂τ ∂τ ∂τ = xx + yx + zx δxδyδz. (9.6) ∂x ∂y ∂z ! Similar expressions can be found for the components in the y and z-directions. Gathering together the three components of the equation, we obtain 1 Fv = ∇·T , ρ 68 Chapter 9. The tensor of the viscous stresses where it is understood that the divergence operator applies to the first index of the tensor (left divergence), as indicated by the position of the del-operator. The divergence of a vector field yields a scalar field, but the divergence of a tensor field yields a vector field. If the tensor is given by the product of a scalar field by a unit tensor, then its divergence is also the gradient of the scalar field. Thus, the divergence of the diagonal tensor (9.5) is also the opposite of the pressure gradient

∇· (−pI) = −∇p.

These contributions to the total force applied to the parcel must be added at the second member of the Euler equations (7.3). We have

du ∇p 1 = − + ∇·T + g. (9.7) dt ρ ρ