So the diffusion term in NS in Einstein notation:
$$\upsilon \frac{\partial^2 u_i}{\partial x_j \partial x_j}$$
I saw in some textbooks that the term can also be written as a summation with the rate of strain tensor as such :
$$2 \upsilon \frac{\partial}{\partial x_j} S_{ij}$$
with the rate of strain tensor defined as such: $S_{ij} = \frac{1}{2} \left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right)$
I don't quite see how this can be. This summation relation occurs in other important terms such as the production term of turbulent kinetic energy but for those term the relationship is more evident because it is summed with the Reynold's stress tensor.
The only explanation that I can think of is that
$$2 \frac{\partial}{\partial x_j} S_{ij} = 2 \frac{1}{2} \frac{\partial}{\partial x_j} \left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right) = \frac{\partial^2 u_i}{\partial x_j \partial x_j} + \frac{\partial^2 u_j}{\partial x_j \partial x_i} $$
Then that would mean that the second term $\frac{\partial^2 u_j}{\partial x_j \partial x_i}$ is zero. Could this be because of continuity of the incompressible flow? Am I on the right track or completely wrong?
Derivatives commute, so we have $$\frac{\partial^2 u_j}{\partial x_j \partial x_i} = \frac{\partial}{\partial x_i}\left(\frac{\partial u_j}{\partial x_j }\right),$$ but the term in the parenthesis is just the divergence of $u$, which is $0$ for an incompressible fluid.