# Viscosity tensor in Navier-Stokes equation?

In my Hydrodynamics notes the viscosity term in the Navier-Stokes equation is of the form:

$$\nabla\cdot(\underline{\underline{h}}\cdot\nabla)\mathbf{u}$$

where $\underline{\underline{h}}$ is the viscosity tensor and $\mathbf{u}$ is the velocity of the fluid.

(The left hand side of the N-S equation is $\rho\frac{\partial \mathbf{u}}{\partial t}$.)

Why does this term have this form? Is there a physical meaning?

I have tried to look this up online but most pages consider the viscosity to be homogeneous and isotropic so that it can be taken outside the brackets: $$h \nabla^2\mathbf{u}$$.

## 2 Answers

There are three distinct things:

1) The rate-of-strain tensor is $2\underline{\underline{D}} = \nabla u + \nabla u^T$. Thus viscous stresses of homogeneous viscosity will be $\eta\nabla\cdot D$, which is equal to $\eta \nabla^2 u$ only if $\nabla\cdot u =0$ (incompressible and constant density).

2) The viscosity may be heterogeneous in space, in case of mixtures, temperature gradients or for some simple non-Newtonian fluids. Thus, in general, you have $\nabla\cdot(\eta D)$.

3) The viscosity tensor above can include in just one term the first (shear) and second (volume) viscosity, which can be relevant in compressible flows : in that case, the viscous stresses write $\nabla\cdot(\eta \underline{\underline{D}} + \eta_v \underline{\underline{I}}\nabla\cdot u )$

• why does (1) have that form? Apr 11, 2014 at 14:52
• In index notation $\nabla u + \nabla u^T = v_{i,j} + v_{j,i} =\sigma_{ij}$. In a fluid in uniform rotation there is no strain, so the rate-of-strain tensor must be symmetric. In uniform motion, there is also no strain, so the tensor can depend only on derivatives of $v_i$. $\sigma_{ij}$ is the symmetrization of $v_{i,j}$. Apr 11, 2014 at 15:06

The form of the viscosity term can be derived on physical grounds. You can most likely do no better than to consult Landau and Lifshitz, volume 6, Chapter II. The Euler equation is the conservation law for momentum, it can be written in the form $$\frac{\partial}{\partial t} (\rho v_i) = -\partial_i \Pi_{ij}$$ that is the change of momentum density is the divergence of something. Since momentum is a vector this "something" has to be a tensor. From the ideal fluid equations, $\Pi_{ij}$ includes at least the advection term and the pressure, $$\Pi_{ij} = p\delta_{ij} + \rho v_i v_j - \sigma'_{ij} = \rho v_i v_j + \sigma_{ij}$$ and $\sigma_{ij}$ is the viscous stress tensor.

To find the form of $\sigma_{ij}'$ the argument starts with that it must vanish for the two cases

• uniform linear motion
• uniform rotation

since it is supposed to represent internal friction in the fluid, friction is clearly absent in uniform motion. The first condition says that $\sigma'_{ij}$ depends only on derivatives of $v_i$, the second implies that it is a symmetric tensor. Assuming isotropy, the most general tensor satisfying these conditions is $$\sigma'_{ij} = \eta_0(v_{i,j} + v_{j,i}) + \xi_0 \delta_{ij} v_{k,k}.$$ You can take the divergence of this without assuming that $\eta_0$ or $\xi_0$ are homogenous.