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Let $g=\dot{\gamma }$ denote the constant rate in the simple shear flowThenWhen $\xi =1$, the structure equation contains the Jaumann derivative. In a steady homogeneous flow it becomesSolving its component equations gives

The $b_2$ term has no $xy$ component because ${\alpha }_{xx}+{\alpha }_{yy}=0$. Thusand the shear viscosity isIt exhibits shear thinning whenever $b_1{c}_1{c}_2>0$: it decreases from ${\eta }_0+b_1{c}_2/c_1$ at zero shear rate to the solvent plateau ${\eta }_0$ at large shear rate. If the product vanishes, the viscosity is constant.

For the diagonal stresses,The two normal-stress differences are therefore

A plain material derivative of the conformation tensor is not an objective time derivative: an observer undergoing a time-dependent rigid rotation would infer a different constitutive response, and even rigid-body rotation could appear to change polymer deformation. The upper-convected derivative subtracts deformation and rotation carried by the velocity gradient and is frame indifferent.

The Oldroyd-B model is often inadequate because its Hookean dumbbells are infinitely extensible. It predicts constant shear viscosity rather than shear thinning, zero second normal-stress difference, and an unbounded extensional viscosity at a finite extension rate. Real polymer chains have finite extensibility and commonly exhibit shear thinning, bounded extensional stress, multiple relaxation times, and nonlinear solvent or concentration effects.

For azimuthal velocity $v(r)e_{\theta }$, the velocity gradient in the rotating polar basis is $\left(\begin{array}{cc}0& -v/r\\ v^{\prime }& 0\end{array}\right)$. The basis rotates along a particle at angular speed $v/r$. Accounting for that rotation in the material derivative replaces $L$ by $L-(v/r)\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ in the upper-convected derivative. Its remaining shear rate is $v^{\prime }-v/r$, so the local viscoelastic normal-stress difference follows the same constitutive algebra as simple shear flow.

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.

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.

A rotating rod can draw a liquid with appreciable viscoelasticity inward and raise its free surface around the rod. Tensile normal stresses along curved flow lines produce an inward hoop force, competing with inertia and gravity. It is a manifestation of elastic normal-stress differences, rather than merely a changed scalar shear viscosity.