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: $$h \nabla^2\mathbf{u}$$.
Best Answer
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 $\nabla\cdot(\eta \underline{\underline{D}} + \eta_v \underline{\underline{I}}\nabla\cdot u )$