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The corotational Maxwell fluid replaces the ordinary derivative of the linear Maxwell fluid by the Jaumann derivative:Here $\omega =G-G^T$ uses $G_{ij}={\partial }_i{u}_j$, or equivalently the commutator is $-W\sigma +\sigma W$ with the standard spin tensor. The stress has relaxation time $\lambda$ and small-rate viscosity $\mu$. The commutator makes the constitutive equation nonlinear while maintaining an objective time derivative.

Use the printed Maxwell-filtered local drag law at fixed material arclength, after transients have decayed to a periodic response. Averaging the force equation over a period makes the time-derivative term vanish. Therefore the mean force equals the mean instantaneous resistive-force theory driving force, even though its oscillating part is filtered. In particular,with the same sign convention as before.

At first order the tangent is ${\mathbf{e}}_y$ and both velocity components are single harmonics of frequency $\omega$. The linear Maxwell fluid response multiplies their complex force amplitudes by $(1+i\omega \lambda {)}^{-1}$. Only its in-phase real part contributes to the mean work. ConsequentlyThe leading power is strictly smaller when $\lambda >0$ and $\omega >0$, and equal in the Newtonian limit. The zero-frequency component generating the mean force is unattenuated, whereas elastic storage and phase lag reduce the dissipative harmonic response. This conclusion concerns the specified linear local force law and prescribed kinematics; it is not a claim for every nonlinear viscoelastic constitutive model.

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.