Let $g=\dot\gamma$ denote the constant rate in the <simple shear flow>
$$
\mathbf u=(gy,0,0).
$$
Then
$$
\dot{\boldsymbol\gamma}
=\begin{pmatrix}0&g&0\\g&0&0\\0&0&0\end{pmatrix},
\qquad
\boldsymbol\Omega
=\frac12\begin{pmatrix}0&g&0\\-g&0&0\\0&0&0\end{pmatrix}.
$$
When $\xi=1$, the structure equation contains the <Jaumann derivative>. In a steady homogeneous flow it becomes
$$
-\boldsymbol\Omega\boldsymbol\alpha
+\boldsymbol\alpha\boldsymbol\Omega+c_1\boldsymbol\alpha
=c_2\dot{\boldsymbol\gamma}.
$$
Solving its component equations gives
$$
\alpha_{xy}=\frac{c_1c_2g}{c_1^2+g^2},
\qquad
\alpha_{xx}=-\alpha_{yy}
=\frac{c_2g^2}{c_1^2+g^2},
\qquad
\alpha_{zz}=0.
$$

The $b_2$ term has no $xy$ component because $\alpha_{xx}+\alpha_{yy}=0$. Thus
$$
\tau_{xy}=\eta_0g+b_1\alpha_{xy},
$$
and the shear viscosity is
$$
\boxed{
\eta(g)=\frac{\tau_{xy}}g
=\eta_0+\frac{b_1c_1c_2}{c_1^2+g^2}}.
$$
It exhibits <shear thinning> whenever $b_1c_1c_2>0$: it decreases from $\eta_0+b_1c_2/c_1$ 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,
$$
\dot{\boldsymbol\gamma}\boldsymbol\alpha+
\boldsymbol\alpha\dot{\boldsymbol\gamma}
-\frac23(\boldsymbol\alpha:\dot{\boldsymbol\gamma})\mathbf I
=\operatorname{diag}\left(\frac23g\alpha_{xy},
\frac23g\alpha_{xy},-\frac43g\alpha_{xy}\right).
$$
The two <normal-stress differences> are therefore
$$
\boxed{
N_1=\tau_{xx}-\tau_{yy}
=\frac{2b_1c_2g^2}{c_1^2+g^2}},
$$
$$
\boxed{
N_2=\tau_{yy}-\tau_{zz}
=\frac{c_2g^2}{c_1^2+g^2}
\left(-b_1+2b_2c_1\right)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 352 / 2 / b / Solution

For the <uniaxial extensional flow>
$$
\mathbf u=\dot\epsilon(-x/2,-y/2,z),
$$
the <spin tensor> vanishes and
$$
\dot{\boldsymbol\gamma}
=\operatorname{diag}(-\dot\epsilon,-\dot\epsilon,2\dot\epsilon).
$$
The flow is steady and homogeneous, so with $\xi=1$ the <Jaumann derivative> of $\boldsymbol\alpha$ vanishes. The structure equation yields
$$
\boldsymbol\alpha=\frac{c_2}{c_1}\dot{\boldsymbol\gamma}.
$$
Writing $k=c_2/c_1$ and using
$\dot{\boldsymbol\gamma}:\dot{\boldsymbol\gamma}=6\dot\epsilon^2$ gives
$$
\dot{\boldsymbol\gamma}\boldsymbol\alpha+
\boldsymbol\alpha\dot{\boldsymbol\gamma}
-\frac23(\boldsymbol\alpha:\dot{\boldsymbol\gamma})\mathbf I
=2k\dot\epsilon\,\dot{\boldsymbol\gamma}.
$$
Consequently
$$
\boldsymbol\tau
=\left(\eta_0+\frac{b_1c_2}{c_1}
+\frac{2b_2c_2}{c_1}\dot\epsilon\right)
\dot{\boldsymbol\gamma}.
$$
The <extensional viscosity> is the tensile stress difference divided by $\dot\epsilon$:
$$
\boxed{
\eta_{\rm ext}
=\frac{\tau_{zz}-\tau_{xx}}{\dot\epsilon}
=3\left(\eta_0+\frac{b_1c_2}{c_1}
+\frac{2b_2c_2}{c_1}\dot\epsilon\right)}.
$$
For $b_2=0$ this is the constant
$$
\boxed{\eta_{\rm ext}=3\left(\eta_0+\frac{b_1c_2}{c_1}\right)}.
$$
The factor three is the <Trouton ratio> associated with the effective zero-rate shear viscosity.

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 352 / 2 / c / Solution

{c}
{wiki}

The Jaumann derivative is the corotational objective derivative $\overset{\circ}{\mathbf C}=D\mathbf C/Dt-\boldsymbol\Omega\mathbf C+\mathbf C\boldsymbol\Omega$, where $\boldsymbol\Omega$ is the <spin tensor>.

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