Does ∇⋅τ = μΔv in the Cauchy Momentum Equation?

Question

I find two versions of the Cauchy momentum equation (1, 2): $$ \rho \frac{D\vec{v}}{Dt}=\rho\vec{g} - \nabla{p} + \mu\nabla^2\vec{v} $$ $$ \rho \frac{D\vec{v}}{Dt}=\rho\vec{g} - \nabla{p} + \nabla \cdot \bf\tau $$ where $\tau$ is the stress tensor and $\vec{v}$ is the fluid velocity.

I'm tempted to conclude that $\mu\nabla^2\vec{v} = \nabla \cdot \bf\tau$. However, when I expand and compare terms on both sides of the equation they look widely different.

Does this equality actually hold? If so, what is the physical relationship?

Answer 1

The stress tensor is related to velocity in a Newtonian fluid as \begin{equation}\tag{1} \tau_{ij} = \mu\left(\frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i}\right) \end{equation} If we assume that the viscosity of the fluid is independent of its position then the divergence of stress is \begin{equation}\tag{2} \frac{\partial\tau_{ij}}{\partial x_j} = \mu\frac{\partial^2 v_i}{\partial x_j^2} + \mu\frac{\partial}{\partial x_i}\left(\frac{\partial v_j}{\partial x_j}\right). \end{equation}

If we further assume that the fluid is incompressible then $\nabla\cdot\vec{v} = 0$ so that equation (2) simplifies to \begin{equation}\tag{3} \frac{\partial\tau_{ij}}{\partial x_j} = \mu\frac{\partial^2 v_i}{\partial x_j^2}. \end{equation}

It is under these assumptions (marked bold) that $\nabla\cdot\tau = \mu\nabla^2\vec{v}$.

You may find the third chapter of Batchelor's book interesting if you wish to know more about this topic.