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Linearize about a motionless, isotropic, relaxed state by taking velocity, deformation and extra stress to be first order in a small amplitude. The advection $u\cdot \nabla \sigma$ and rotational commutator $\omega \sigma -\sigma \omega$ are then second order. The corotational Maxwell fluid therefore reduces to the linear Maxwell fluid:where $e$ is the rate-of-strain tensor. This is the linear Maxwell viscoelasticity model, mechanically a spring and dashpot in series.

For one shear component, dividing by $\mu$ gives $\dot{\gamma }=\sigma /\mu +(\lambda /\mu )\dot{\sigma }$. Comparing with a dashpot of viscosity $\mu$ in series with a spring of shear modulus $G$ identifiesIn particular a held strain with zero subsequent shear rate has $\sigma (t)=\sigma (0)e^{-t/\lambda }$. At long times under steady small shear, $\sigma =\mu \dot{\gamma }$, whereas a sufficiently rapid deformation probes the elastic modulus $G$.

In linear simple shear, $\dot{\gamma }$ has only off-diagonal $xy$ components. The diagonal stresses have no forcing and, from a relaxed initial state, stay zero. Any pre-existing diagonal stress decays homogeneously. Thus both normal-stress differences vanish in the linear response to shear. The finite-shear normal-stress differences generated by products of shear rate and stress are quadratic in deformation amplitude and have been discarded by this linearization. This conclusion is about simple shear, not about every possible extensional deformation.

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.

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.