Viscous stress tensor for an incompressible Newtonian fluid I am reading the cengel book and found that the viscous stress tensor for an incompressible Newtonian fluid with constant properties is given by:
$$\tau_{ij} = 2 \mu \epsilon_{ij}$$
where $\epsilon_{ij}$ is the strain rate tensor. My question is if there is a derivation of this property or any reference that can demonstrate it
 A: It's convenient when modeling viscosity to relate the shear stress $\tau$ to the rate of change $\frac{d\gamma}{dt}$ in a corner angle  that was originally 90°, as this parameter is easily accessible in experiments:
$$\tau\sim\frac{d\gamma}{dt}=\dot\gamma;$$
$$\tau=\mu\dot\gamma.$$
We call the constant of proportionality the viscosity $\mu$.
It's also convenient to model strains as
$$\varepsilon_{ij}=\frac{1}{2}\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right)$$
because this compact expression works for both normal and shear strains:
$$\varepsilon_{11}=\frac{1}{2}\left(\frac{\partial u_1}{\partial x_1}+\frac{\partial u_1}{\partial x_1}\right)=\frac{\partial u_1}{\partial x_1};$$
$$\varepsilon_{22}=\frac{1}{2}\left(\frac{\partial u_2}{\partial x_2}+\frac{\partial u_2}{\partial x_2}\right)=\frac{\partial u_2}{\partial x_2};$$
$$\varepsilon_{12}=\frac{1}{2}\left(\frac{\partial u_1}{\partial x_2}+\frac{\partial u_2}{\partial x_1}\right);$$
for example. We actually get all nine components of the strain tensor this way.
You can prove to yourself through diagrams and trigonometry that the tensorial shear strain $\varepsilon_{12}$ ends up being one-half the so-called engineering shear strain $\gamma$. For this reason, we need to insert a factor of two in the original equation to obtain
$$\tau_{ij}=2\mu\dot\varepsilon_{ij}.$$
Does this make sense?
