Upper-convected derivative

The upper-convected derivative of a contravariant tensor is $\stackrel{▽}{\mathbf{C}}=D\mathbf{C}/Dt-(\nabla \mathbf{u})\mathbf{C}-\mathbf{C}(\nabla \mathbf{u}{)}^T$.

For a material line tangent $\mathbf{\ell }$, $D_t\mathbf{\ell }=L\mathbf{\ell }$ with velocity gradient $L_{ij}={\partial }_j{u}_i$. The product rule gives $D_t(\mathbf{\ell }{\mathbf{\ell }}^T)=L\mathbf{\ell }{\mathbf{\ell }}^T+\mathbf{\ell }{\mathbf{\ell }}^T{L}^T$. Consequently the upper-convected derivative vanishes for any transported sum of these outer products. Under the derivative-index-first convention $G=L^T$ the same formula is $D_{t}A=G^{T}A+AG$. The distinction prevents transposition errors in constitutive equations.

For azimuthal velocity $v(r)e_{\theta }$, the velocity gradient in the rotating polar basis is $\left(\begin{array}{cc}0& -v/r\\ v^{\prime }& 0\end{array}\right)$. The basis rotates along a particle at angular speed $v/r$. Accounting for that rotation in the material derivative replaces $L$ by $L-(v/r)\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ in the upper-convected derivative. Its remaining shear rate is $v^{\prime }-v/r$, so the local viscoelastic normal-stress difference follows the same constitutive algebra as simple shear flow.

Differentiate the Kirchhoff stress tensor relation $\mathbf{\tau }=FS{F}^T$, using $\dot{F}=LF$ for the velocity gradient. Subtracting the two deformation terms gives the displayed upper-convected derivative. This removes the transport generated solely by deformation of a fixed material stress tensor.

The Upper-convected Maxwell model evolves viscoelastic stress according toIt has constant shear viscosity and predicts a divergent extensional viscosity at a finite extension rate.

The Giesekus model adds the nonlinear anisotropic-drag term $\alpha \lambda {\mathbf{\tau }}^2/\eta$ to the Upper-convected Maxwell model. Positive dimensionless mobility factor $\alpha$ produces shear thinning and a nonzero second normal-stress difference.

For uniaxial extensional flow with $G=\mathrm{diag}(\gamma ,-\gamma /2,-\gamma /2)$, write $W=\lambda \gamma$, $b=\alpha p_T$, and the stress as $p_T\mathrm{diag}(s_1,s_2,s_2)$. The displayed longitudinal equation is supplemented by $b{s}_2^2+(1+W)s_2+W=0$. Choose the roots tending to zero at $W=0$. For $0<b<1$ these roots have $1+s_i>0$ at every nonnegative rate. The stretching stress is finite at $W=1/2$ and the extensional viscosity $({\sigma }_{11}-{\sigma }_{22})/\gamma$ tends to $2\lambda /\alpha$ as $\gamma \to \infty$. In the zero-$\alpha$ Upper-convected Maxwell model, the physical steady longitudinal branch terminates at $W=1/2$. Arbitrary positive $b$ is insufficient: $b=2,W=1$ gives the impossible real transverse equation $2s_2^2+2s_2+1=0$.

When $\sigma =nHC-n{k}_{B}TI$ and the bead hydrodynamic mobility matrix is $(I+\alpha \sigma )/\zeta$, $\alpha$ has inverse-stress units. With $p_T=n{k}_{B}T$, $A=HC/(k_{B}T)$ and $b=\alpha p_T$, the dimensionless mobility is $M=(1-b)I+bA$. Thus $0<b<1$ and a positive-definite conformation tensor ensure a positive-definite mobility. The closed constitutive equation is $\lambda {\sigma }^{\nabla }+\sigma +\alpha {\sigma }^2=\mu (G+G^T)$, with $\lambda =\zeta /(4H)$ and $\mu =\lambda p_T$. This dimensional $\alpha$ differs from conventions in which the symbol denotes $b$ itself.