Q:

Why is the stress tensor simplifying like this

\begin{eqnarray}
\nabla \cdot \boldsymbol \tau &=& 2 \mu \nabla \cdot \boldsymbol \varepsilon\\
&=& \mu \nabla \cdot \left( \nabla\mathbf{u} + (\nabla\mathbf{u}) ^\mathrm{T} \right)\\
&=& \mu \, \nabla^2 \mathbf{u}
\end{eqnarray}
Given $$ \nabla \cdot \boldsymbol u =0 \ $$
How do you go from step 3 to step 4 and get rid of the transpose? Isn't the divergence operator distributive

A:

The divergence of the transpose of a vector gradient is equal to the gradient of the divergence of the vector. You can prove it quickly using some index notation:
$$\nabla \cdot (\nabla \vec{u})^T = \frac{\partial^2 u_j}{\partial x_i \partial x_j} = \frac{\partial}{\partial x_i}\frac{\partial u_j}{\partial x_j} = \nabla(\nabla \cdot \vec{u})$$
I assumed everything is in orthonormal coordinates so no need to raise or lower indices. Since the flow is incompressible by your assumption, this term cancels out, and you're only left with the divergence of the vector gradient (which gives you the vector Laplacian).

