Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 352 / 2 / d

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.