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For simple shear flow $u=(\dot{\gamma }y,0,0)$, put $s=\dot{\gamma }$ and $\ell =\lambda s$. The steady deviatoric stress isThis follows by setting ${\omega }_{xy}=-s$, ${\omega }_{yx}=s$ in the corotational Maxwell fluid equation: ${\sigma }_{xx}=\lambda s{\sigma }_{xy}$, ${\sigma }_{yy}=-\lambda s{\sigma }_{xy}$, and ${\sigma }_{xy}+\lambda s({\sigma }_{xx}-{\sigma }_{yy})/2=\mu s$. Hence the shear viscosity is $\mu /(1+{\lambda }^2{s}^2)$, which exhibits shear thinning, while $N_1=2\mu \lambda s^2/(1+{\lambda }^2{s}^2)$ and $N_2=-N_1/2$. These differences are independent of the isotropic pressure.

For constant simple shear flow rate $s$, the nonpressure stress in the component-first gradient convention $L_{12}=s$ isThus the first normal-stress difference is $N_1=2(\mu -{\mu }_r)\tau s^2$, while the second is $N_2=-N_1$. Adding the incompressibility pressure changes neither difference.

In simple shear flow, a persistent polar order parameter with $|\xi |<1$ has no fixed angle and rotates continuously. Integrating $\dot{\theta }=g(\xi \cos 2\theta -1)/2$ through $2\pi$ gives the displayed period. An unoriented nematic director repeats after half that time. At $\xi =0$, the magnitude relaxes independently while the angular speed is $-g/2$.

A viscometric flow has the local relative deformation history of steady simple shear flow, up to rigid changes of frame. For an isotropic simple fluid, symmetry leaves one shear stress and two independent normal-stress differences. The corresponding viscometric functions depend on the magnitude of the shear rate, while pressure supplies the isotropic part.

The three viscometric functions describe steady homogeneous simple shear flow: its shear viscosity and the first and second normal-stress differences per squared shear rate. Take x as the flow direction, y as the gradient direction and z as the vorticity direction, so $N_1={\sigma }_{xx}-{\sigma }_{yy}$ and $N_2={\sigma }_{yy}-{\sigma }_{zz}$. Isotropic pressure drops out of these differences.

Use a linear shear flow to see why straight pathlines can still carry local spin. Faster layers pull past slower ones, and a small symmetric paddle can turn clockwise while its center travels straight ahead.

For the planar simple shear flow $\mathbf{u}=(\gamma y,0,0)$ with constant signed shear rate $\gamma$, direct differentiation givesWith $\gamma =1{\mathrm{s}}^{-1}$, these are $-1{\mathrm{s}}^{-1}$ and $-0.5{\mathrm{s}}^{-1}$: both signs indicate clockwise orientation under the right-hand rule. Yet a particle initially at $(x_0,y_0)$ follows the straight pathline $x(t)=x_0+\gamma y_{0}t$, $y(t)=y_0$. The curl measures differences of velocity between neighboring particles, rather than bending of one particle's pathline.

The velocity gradient has $L_{xy}=\gamma$ as its only nonzero entry. Its spin tensor supplies clockwise rotation, while its rate-of-strain tensor supplies simultaneous deformation. For a material line making angle $\alpha$ to the horizontal, the equation $\dot{\delta \mathbf{x}}=L\delta \mathbf{x}$ gives $\dot{\alpha }=-\gamma {\sin }^2\alpha$. Initially horizontal and vertical material lines therefore have angular velocities $0$ and $-\gamma$; their mean is $-\gamma /2$. At $t=0.5\mathrm{s}$ with $\gamma =1{\mathrm{s}}^{-1}$, an initially vertical displacement becomes $(0.5,1)$ times its original length and has angle $\arctan 2\approx 63.435^{\circ }$, a clockwise change of $26.565^{\circ }$. The plotted cross deforms; an actual paddle's response also depends on its shape and interaction with the fluid. The exact local spin statement concerns the spin tensor, rather than every tracer's orientation.