{c}
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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>.

Jaumann derivative

Write the <velocity gradient> as
$$
\mathbf L=\nabla\mathbf u
=\frac12\dot{\boldsymbol\gamma}+\boldsymbol\Omega,
$$
where $\boldsymbol\Omega$ is the <spin tensor>. Expanding the <upper-convected derivative> in the structure equation gives
$$
\frac{D\boldsymbol\alpha}{Dt}
-\boldsymbol\Omega\boldsymbol\alpha
+\boldsymbol\alpha\boldsymbol\Omega
+\frac{\xi-1}{2}
\left(\dot{\boldsymbol\gamma}\boldsymbol\alpha+
\boldsymbol\alpha\dot{\boldsymbol\gamma}\right)
+c_1\boldsymbol\alpha
=c_2\dot{\boldsymbol\gamma}.
$$
For $\xi=0$, the first four terms reproduce the upper-convected derivative. Setting
$$
\boxed{\xi=0,\qquad b_2=0}
$$
therefore gives
$$
\boldsymbol\tau=\eta_0\dot{\boldsymbol\gamma}+b_1\boldsymbol\alpha,
\qquad
\boldsymbol\alpha^{\triangledown}+c_1\boldsymbol\alpha
=c_2\dot{\boldsymbol\gamma}.
$$
With polymeric stress $\boldsymbol\tau_p=b_1\boldsymbol\alpha$, this is the <Oldroyd-B model>. For $c_1>0$, its relaxation time and polymer viscosity are
$$
\lambda=\frac1{c_1},
\qquad
\eta_p=\frac{b_1c_2}{c_1}.
$$
Indeed the total stress obeys
$$
\boxed{
\boldsymbol\tau+\lambda\boldsymbol\tau^{\triangledown}
=(\eta_0+\eta_p)\dot{\boldsymbol\gamma}
+\lambda\eta_0\dot{\boldsymbol\gamma}^{\triangledown}}.
$$

For $\xi=2$, the coefficient of
$\dot{\boldsymbol\gamma}\boldsymbol\alpha+
\boldsymbol\alpha\dot{\boldsymbol\gamma}$
is $+1/2$, so the objective derivative becomes the <lower-convected derivative>. Hence
$$
\boxed{\xi=2,\qquad b_2=0}
$$
recovers the <Oldroyd-A model>, again with $c_1>0$ for a finite positive relaxation time. Parameter choices such as $b_1c_2=0$ give the degenerate Newtonian limit.

Solution

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

{title2=$\boldsymbol\Omega$}
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The spin tensor is the antisymmetric part of the velocity gradient, $\boldsymbol\Omega=(\nabla\mathbf u-(\nabla\mathbf u)^T)/2$. It describes local rigid-body rotation.


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 Incoming links: Spin tensor