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The FENE-P model replaces infinitely extensible Hookean polymers by finitely extensible nonlinear elastic springs with a mean-field closure. Finite extensibility regularizes the Oldroyd-B extensional catastrophe and produces shear thinning.

The Giesekus model adds the nonlinear anisotropic-drag term $\alpha \lambda {\mathbf{\tau }}^2/\eta$ to the Upper-convected Maxwell model. Positive dimensionless mobility factor $\alpha$ produces shear thinning and a nonzero second normal-stress difference.

The $xy$ component of the stationary constitutive equation isInsert $T_{yy}=-\alpha g{T}_{xy}$ and $T_{xx}=(2\tau -\alpha )g{T}_{xy}$. The coefficient of $g^2{T}_{xy}$ becomes $\tau \alpha +\alpha (\tau -\alpha )=\alpha (2\tau -\alpha )$, givingFor $C=\alpha (2\tau -\alpha )>0$, the small-rate shear stress is $T_{xy}=\mu \dot{\gamma }-\mu C{\dot{\gamma }}^3+O({\dot{\gamma }}^5)$. At large positive rate, $T_{xy}\sim \mu /(C\dot{\gamma })$ and the shear viscosity decreases as $\mu /(C{\dot{\gamma }}^2)$. The shear stress is odd in rate, reaches a positive maximum atand decreases thereafter. The original plot displays this homogeneous constitutive curve. Its decreasing branch can permit shear banding; the local model alone supplies neither an interface-selection rule nor a guaranteed stable homogeneous realization.

This is a Johnson-Segalman fluid without an added Newtonian fluid solvent contribution. To identify conventions explicitly, set $a=1-\alpha /\tau$ and $L=G^T=E+W$, with $W$ the spin tensor. The derivative appearing in the constitutive equation becomesthe Gordon--Schowalter derivative. Its usual slip interval $|a|\le 1$ corresponds to $0\le \alpha \le 2\tau$, while a nonsingular genuinely shear thinning curve requiresAt $\alpha =0$ the model is the Upper-convected Maxwell model; at $\alpha =2\tau$ it is the lower-convected derivative counterpart, and both have constant shear viscosity in this test. At $\alpha =\tau$ the derivative is the Jaumann derivative, giving the corotational Maxwell fluid. Merely requiring $\alpha >0$ is insufficient: for $\alpha =3\tau$ the denominator is $1-3{\tau }^2{\dot{\gamma }}^2$, which vanishes at $|\dot{\gamma }|=1/(\sqrt{3}\tau )$ and then gives a negative steady shear viscosity. There is no solvent-induced large-rate plateau in the stated equation.

Write $q=\dot{\gamma }\tau$ and $f=1+\alpha \mathrm{tr}A$. With $L_{ij}={\partial }_j{v}_i$, the steady homogeneous upper-convected derivative is $-LA-A{L}^T$. Its diagonal and shear equations giveOn the physical branch $f>0$, the other off-diagonal components vanish too. Thus $a_{22}=a_{33}=0$, $a_{11}=2a_{12}^2$ andFor $\alpha \ge 0$ the right side has derivative $1+6\alpha a_{12}^2>0$, so the shear stress determines a unique shear rate. The shear viscosity is $G_0\tau /f$ and decreases with shear magnitude when $\alpha >0$. This is the shear part of shear and pipe flow of an affine linear PTT fluid.

For fully developed pipe flow take $\mathbf{v}=w(r)\hat{\mathbf{z}}$ and the signed axial pressure gradient $dp/dz=\Delta p$. Regularity at the axis and the supplied momentum balance determine the shear stress without using the constitutive equation:Locally the constitutive equations have the same shear form, with axial flow replacing the Cartesian flow direction and $\dot{\gamma }=w^{\prime }(r)$ carrying its sign. HenceIntegrating and imposing no-slip boundary conditions at $r=R$ givesA negative pressure gradient produces positive axial flow. At $\alpha =0$ this is the ordinary Hagen-Poiseuille flow; positive $\alpha$ increases the flux at a fixed pressure drop through shear thinning. The nonzero axial normal stress varies radially, but its axial derivative vanishes in fully developed flow, so it does not change the axial momentum balance used here.

Let the flow direction be $x$, the gradient direction $y$, and the vorticity direction $z$, with simple shear flow $u=(sy,0,0)$ and signed shear rate $s=\dot{\gamma }$. Use the gradient convention established in part (a):At steady homogeneous shear the ordinary time derivative and advection vanish. Put $X={\sigma }_{xx}$, $Y={\sigma }_{yy}$ and $S={\sigma }_{xy}$. The corotational Maxwell fluid law then givesThe $zz$ equation gives ${\sigma }_{zz}=0$, and the unforced $xz,yz$ equations force those components to zero. Solving the three displayed equations gives the steady shear of a corotational Maxwell fluid:The deviatoric stress is indeed traceless. Its isotropic pressure contribution is irrelevant to either shear viscosity or a normal-stress difference.

Consequently the steady shear viscosity isFor $s>0$, $d{\eta }_{\mathrm{sh}}/ds=-2\mu {\lambda }^{2}s/(1+{\lambda }^2{s}^2{)}^2<0$ when $\mu ,\lambda >0$; it decreases with $|s|$ and tends to zero at large rate. This proves shear thinning directly, without confusing a decreasing viscosity with a necessarily decreasing shear stress.

With the standard definitions $N_1={\sigma }_{xx}-{\sigma }_{yy}$ and $N_2={\sigma }_{yy}-{\sigma }_{zz}$,At zero shear both vanish; at small shear they are quadratic, consistently with part (b). The signs and the fact $|N_2|<N_1$ agree qualitatively with many polymeric-fluid experiments. However the fixed ratio $|N_2|/N_1=1/2$ often overestimates the measured second difference and cannot capture variation of that ratio with material and shear rate. For example, measurements on polyethylene show a smaller, rate-dependent relative second normal-stress difference. Thus the qualitative normal-stress signs are reasonable, but the predicted relative magnitude is not generally quantitatively satisfactory; these signs are not a universal statement about every non-Newtonian fluid.

At order ${\alpha }^0$, the preceding equations giveAt order ${\alpha }^1$, the quadratic term is evaluated on ${\mathbf{\tau }}^0$. Solving first the $yy$ equation, then $xy$, then $xx$, givesThus the apparent shear viscosity isso positive $\alpha$ produces shear thinning. The expansion requires $\alpha \ll 1$ and $\alpha (\lambda g{)}^2\ll 1$; it cannot describe arbitrarily high shear rates even when $\alpha$ is numerically small.

Using the normal-stress difference definitionswe findIn the low-rate limit, $-2{\Psi }_2/{\Psi }_1=\alpha +O({\alpha }^2)$. A cone-and-plate or parallel-plate rheometer can measure shear stress and normal thrust over a low-rate range; combining ${\Psi }_1$ and ${\Psi }_2$ then estimates $\alpha$ independently of the viscosity scale.

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

In either half-plane, let $U_{\infty }$ denote its far-field value. Multiplying the ODE by $U^{\prime }$ and integrating from the far field givesSince the profile decreases,Continuity of $U^{\prime }$ at $y=0$ gives

For $n=1$, let $\ell =A^{-1}=\sqrt{hHK/\mu }$ and $\Delta U=U_{-}-U_{+}$. ThenThe Newtonian disturbance therefore has exponential tails.

For $n<1$ and $y>0$, put $W=U-U_{+}$. Integration givesThe corresponding expression on $y<0$ follows by reflection about $(0,U_0)$. A shear thinning power-law material has algebraic rather than exponential margin tails. Measurements of how rapidly an ice-stream speed approaches its far-field values could therefore distinguish an approximately Newtonian rheology from power-law shear thinning and estimate $n$.

The trace isSubstituting it into $F=(L^2-3)/(L^2-T)$ givesFor $\beta \ll 1$, $F=1+\beta +O({\beta }^2)$. For $\beta \gg 1$, $F\sim {\beta }^{1/3}$. The effective shear viscosity is thereforeThe FENE-P curve decreases from ${\eta }_s+{\eta }_p$ toward the solvent plateau ${\eta }_s$, displaying shear thinning. Oldroyd-B has $F=1$ and remains at the constant value ${\eta }_s+{\eta }_p$.

A Couette rheometer measures the torque required to rotate one coaxial cylinder relative to another. With inner radius $R_i$, outer radius $R_o$, wetted length $L$, inner angular velocity $\Omega$ and stationary outer wall, laminar Newtonian flow hasThe shear stress at the inner wall is ${\tau }_i=\mathcal{T}/(2\pi L{R}_i^2)$. In a narrow gap $d=R_o-R_i\ll R_i$, $\dot{\gamma }\simeq \Omega R_i/d$, so $\eta \simeq \mathcal{T}d/(2\pi L\Omega R_i^3)$.

Measuring torque over a range of speeds gives a flow curve: constant viscosity identifies a Newtonian regime; decreasing or increasing apparent viscosity indicates shear thinning or thickening. A yield-stress fit needs a genuine low-rate regime and must not be confused with bearing drag or wall slip. Small oscillatory strain also probes viscoelasticity, with $\tau ={\gamma }_0[G^{\prime }\cos \omega t-G^{\prime \prime }\sin \omega t]$ for imposed $\gamma ={\gamma }_0\cos \omega t$.

Control temperature, calibrate instrument torque and end effects, and check gap dependence or roughened surfaces for slip. Keep inertia small enough to avoid Taylor vortices. The exact Newtonian profile is a calibration model; a non-Newtonian material needs its own stress–rate relation.

A power-law fluid has shear stress $\tau =K|\dot{\gamma }{|}^{n-1}\dot{\gamma }$. It is shear thinning for $n<1$, Newtonian for $n=1$, and shear thickening for $n>1$.

In steady simple shear flow, the tensor equation gives $a_{22}=a_{33}=0$, $a_{11}=2a_{12}^2$ and the equation in the title. A circular pipe with axial pressure gradient $\Delta p$ has $G_0{a}_{rz}=\Delta pr/2$. Thus $w^{\prime }=\Delta pr/(2G_0\tau )+\alpha \Delta p^3{r}^3/(4G_0^3\tau )$, and no-slip boundary conditions givePositive $\alpha$ produces shear thinning; the sign of the pressure gradient determines the direction of the flow.

Shear thinning is the decrease of effective viscosity with increasing shear rate.

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 Gordon--Schowalter parameter $0<a<2$, the steady shear viscosity isIt is shear thinning when ${\lambda }_1>{\lambda }_2$.