Shear stress

Example

Considering a 2D space in Cartesian coordinates (x,y) (the flow velocity components are respectively (u,v)), the shear stress matrix given by$${\displaystyle {\begin{pmatrix}\tau _{xx}&\tau _{xy}\\\tau _{yx}&\tau _{yy}\end{pmatrix}}={\begin{pmatrix}x{\frac {\partial u}{\partial x}}&0\\0&-t{\frac {\partial v}{\partial y}}\end{pmatrix}}}$$represents a Newtonian flow; in fact it can be expressed as$${\displaystyle {\begin{pmatrix}\tau _{xx}&\tau _{xy}\\\tau _{yx}&\tau _{yy}\end{pmatrix}}={\begin{pmatrix}x&0\\0&-t\end{pmatrix}}\cdot {\begin{pmatrix}{\frac {\partial u}{\partial x}}&{\frac {\partial u}{\partial y}}\\{\frac {\partial v}{\partial x}}&{\frac {\partial v}{\partial y}}\end{pmatrix}},}$$i.e., an anisotropic flow with the viscosity tensor$${\displaystyle {\begin{pmatrix}\mu _{xx}&\mu _{xy}\\\mu _{yx}&\mu _{yy}\end{pmatrix}}={\begin{pmatrix}x&0\\0&-t\end{pmatrix}},}$$which is nonuniform (depends on space coordinates) and transient, but is independent of the flow velocity:$${\displaystyle {\boldsymbol {\mu }}(x,t)={\begin{pmatrix}x&0\\0&-t\end{pmatrix}}.}$$This flow is therefore Newtonian. On the other hand, a flow in which the viscosity was$${\displaystyle {\begin{pmatrix}\mu _{xx}&\mu _{xy}\\\mu _{yx}&\mu _{yy}\end{pmatrix}}={\begin{pmatrix}{\frac {1}{u}}&0\\0&{\frac {1}{u}}\end{pmatrix}}}$$is non-Newtonian since the viscosity depends on flow velocity. This non-Newtonian flow is isotropic (the matrix is proportional to the identity matrix), so the viscosity is simply a scalar:$${\displaystyle \mu (u)={\frac {1}{u}}.}$$