 Solution

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.