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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.

Multiply the Hookean dumbbell model configuration equation by $R_i{R}_j$ and integrate twice by parts, with sufficient decay at large separation. The linear velocity field contributes $KC+C{K}^T$, the spring drift contributes $-4HC/\zeta$, and diffusion contributes $4k_{B}TI/\zeta$. Thus the dimensionless conformation tensor $A=HC/(k_{B}T)$ obeys $\dot{A}-KA-A{K}^T=-(A-I)/\lambda$, with polymer relaxation time $\lambda =\zeta /(4H)$. Its left side is an upper-convected derivative for this gradient convention.

Let $C=⟨\mathbf{R}{\mathbf{R}}^T⟩$ be the unnormalized conformation tensor, $G_{ij}={\partial }_j{u}_i$ the velocity gradient, and $M/\zeta$ a symmetric positive hydrodynamic mobility matrix independent of the individual connector. The connector Fokker-Planck equation has drift $G\mathbf{R}-(2H/\zeta )M\mathbf{R}$ and diffusion matrix $(2k_{B}T/\zeta )M$. Integrating its second moment twice by parts gives the displayed upper-convected derivative, $C^{\nabla }=D_{t}C-GC-C{G}^T$. If $M$ is a polynomial in $C$, the two matrices commute and the relaxation term simplifies to $-(4H/\zeta )M(C-k_{B}TI/H)$. No Gaussian distribution assumption is required for this moment closure.

For component-first velocity gradient $L_{ij}={\partial }_j{u}_i$, ideal incompressible magnetic field stretching gives the displayed evolution of $M_{ij}=B_i{B}_j$. It has the same upper-convected derivative as a polymer conformation tensor with negligible relaxation. Finite polymer relaxation time introduces a local return toward thermal equilibrium, whereas magnetic diffusion acts through spatial gradients and preserves the solenoidal magnetic-field constraint. The analogy does not identify those dissipative mechanisms.

Label a trajectory by $\mathbf{a}$ and write its flow map as $\mathbf{X}(\mathbf{a},t)$, with ${\partial }_t\mathbf{X}=\mathbf{u}(\mathbf{X},t)$. An infinitesimal material line joining neighboring labels has tangent ${\ell }_i=F_{ij}d{a}_j$, where $F_{ij}=\partial X_i/\partial a_j$ is the deformation gradient. Differentiating at fixed labels givesThis identity holds for an unnormalized material line tangent even in a compressible flow. Incompressibility also makes an advected magnetic field obey the same stretching law. A unit tangent would instead require a projection removing its change of length.

Use the source's velocity gradient convention $G_{ij}={\partial }_i{u}_j$. For a outer product $A_{ij}={\ell }_i{\ell }_j$, the product rule yieldsThe same law holds for a sum or locally transported average of such outer products. Thus the dyadic transport of material line elements motivatesThis upper-convected derivative vanishes for a symmetric tensor transported solely by stretching of material lines. It subtracts both advection and geometric deformation, allowing a constitutive equation to isolate intrinsic relaxation or production. More generally $\mathsf{A}=F{\mathsf{A}}_{\mathrm{ref}}{F}^T$ gives $\stackrel{▽}{\mathsf{A}}=F{\dot{\mathsf{A}}}_{\mathrm{ref}}{F}^T$. Under a time-dependent rigid rotation $Q$, its value transforms as $Q\stackrel{▽}{\mathsf{A}}{Q}^T$, making it an objective time derivative. This is a suitable derivative for this transport type; symmetry alone does not make it the unique possible objective time derivative.

Let $p_T=n{k}_{B}T$, $C=⟨\mathbf{R}{\mathbf{R}}^T⟩$, and $M=I+\alpha \sigma$, so the single-bead mobility is $M/\zeta$. Assume an incompressible solvent, constant $n,H,T,\zeta$, negligible bead inertia, and a symmetric positive-definite $M$. The last requirement makes the resistance invertible and Brownian diffusion physically admissible. At a given position and time, $M$ depends on the moment $C$ but not on the individual configuration $\mathbf{R}$.

Use the source's gradient convention $K_{ij}={\partial }_i{u}_j$, and write $G=K^T$ for the component-first gradient. Then $\mathbf{u}({\mathbf{r}}_2)-\mathbf{u}({\mathbf{r}}_1)\simeq G\mathbf{R}$. The spring forces on the two beads are ${\mathbf{F}}_1=H\mathbf{R}$ and ${\mathbf{F}}_2=-H\mathbf{R}$. Their overdamped probability currents areHere the Brownian term written as $-k_{B}T\nabla \psi$ is a force weighted by probability density. Its corresponding conditional osmotic force is $-k_{B}T\nabla \log \psi$; treating the given term as an unweighted deterministic force per bead would not produce the stated Fokker–Planck equation.

With midpoint $\mathbf{x}=({\mathbf{r}}_1+{\mathbf{r}}_2)/2$, the derivatives are ${\nabla }_{r_1}=\frac{1}{2}{\nabla }_x-{\nabla }_R$ and ${\nabla }_{r_2}=\frac{1}{2}{\nabla }_x+{\nabla }_R$. Subtracting the bead currents gives the configuration currentThe midpoint current is $\mathbf{u}\psi$ after neglecting the specified spatial Brownian diffusion. Configuration probability conservation therefore givesSince $G\mathbf{R}$ has components $R_j{\partial }_j{u}_i$, this agrees with the source's $\mathbf{R}\cdot \nabla \mathbf{u}$. The factor two in both spring relaxation and configuration diffusivity comes from relative motion of both beads.

The second moment $C$ is the unnormalized conformation tensor. Now multiply by $R_i{R}_j$ and integrate over configuration space. Assume sufficiently decaying tails so every boundary term at infinite separation vanishes. The affine current gives $GC+C{G}^T$; spring drift gives $-(2H/\zeta )(MC+CM)$; two integrations by parts of diffusion give $(4k_{B}T/\zeta )M$. HenceBecause $\sigma =nHC-p_{T}I$, $M$ is a polynomial in $C$ and commutes with it. ThusThis is the Hookean dumbbell conformation equation with anisotropic mobility, an exact second-moment closure within the stated model; no Gaussian approximation to $\psi$ was needed.

The upper-convected derivative of the identity is $I^{\nabla }=-(G+G^T)$. Applying it to the polymer stress therefore givesConsequently the coefficients and the closed stress equation arewhere ${\sigma }^{\nabla }=D_t\sigma -\sigma K-K^T\sigma$ in the printed convention. Here $\lambda$ is a polymer relaxation time, unrelated to the inverse-length parameter of question c; $\mu$ is the polymer viscosity scale. The coefficient $\alpha$ has inverse-stress units. In the dimensional mobility parameter in the Giesekus model, the dimensionless factor is $b=\alpha p_T$, not the dimensional $\alpha$ printed in this problem.

The upper-convected rate removes the affine stretching and rotation of material connectors. If $\dot{F}=GF$, then a tensor transported purely as $C=F{C}_0{F}^T$ has $C^{\nabla }=0$. Under a time-dependent rigid rotation $Q$, the rate transforms as $Q{C}^{\nabla }{Q}^T$: this is material frame indifference, not an ordinary componentwise time derivative. The resulting equation is the Giesekus model in these dimensional conventions.

The isotropic subtraction is fixed by zero polymer extra stress at thermal equilibrium. With no flow and positive mobility, the stationary distribution isso $\boxed{{\sigma }_{\mathrm{eq}}=0}$. The constitutive equation correspondingly has a zero-stress quiescent state and a Newtonian low-rate limit $\sigma \simeq \mu (K+K^T)$. Without subtracting exactly $p_{T}I$, the spring contribution would retain an equilibrium isotropic stress and the equation would acquire corresponding reference terms. Because mobility itself depends on $\sigma$, that reference cannot simply be shifted while retaining the same model. The separate fluid pressure enforcing incompressibility is not included in this polymer extra stress.

For steady uniaxial extension of the Giesekus model, the component-first velocity gradient is diagonal, $G=\mathrm{diag}(\gamma ,-\gamma /2,-\gamma /2)$. For a steady spatially uniform diagonal stress, the material derivative vanishes and the convected terms are $-2\gamma {\sigma }_{11}$ and $+\gamma {\sigma }_{22}$. The off-diagonal equations vanish identically, and symmetry permits equal transverse components. The remaining equations areSet $W=\lambda \gamma \ge 0$, $b=\alpha p_T$ and $s_i={\sigma }_{ii}/p_T$, with $\mu =\lambda p_T$. For $b>0$, the branches continuously connected to zero stress at zero rate areA sufficient nondegenerate range for physical solutions at every nonnegative rate is $0<b<1$. Indeed $M=(1-b)I+bA$ with $A=HC/(k_{B}T)$, so positive conformation implies positive mobility. The longitudinal quadratic equation is nonpositive at $s_1=0$ and has a nonnegative root of a polynomial. The transverse quadratic equation has values $b-1<0$ at $s_2=-1$ and $W\ge 0$ at $s_2=0$, so its larger root of a polynomial lies in $(-1,0]$. Thus the displayed branches satisfy $s_i>-1$, giving $A_i=1+s_i>0$. The other algebraic roots do not provide a physical branch connected to the equilibrium distribution. For constant positive diagonal $A$, the associated stationary configuration distribution is a Gaussian with covariance $C=(k_{B}T/H)A$.

For positive $\alpha$ on the physical branch, the stretching stress remains finite through $W=1/2$. At that rate,At large extension rate,Thus the stress continues to grow, but the uniaxial extensional viscosity ${\eta }_E=({\sigma }_{11}-{\sigma }_{22})/\gamma$ approaches the finite plateau $2\lambda /\alpha$ rather than developing a finite-rate pole. The transverse stress approaches $-p_T$ as its conformation tensor variance becomes small. At small rate the two stresses are $2\mu \gamma$ and $-\mu \gamma$, so ${\eta }_E\to 3\mu$.

When $\alpha =0$, the physical steady branch instead isThe stretching stress diverges as $W\to 1/2$ from below. Above that threshold its algebraic continuation has $1+s_1=1/(1-2W)<0$ and cannot be a conformation covariance. There is no physical finite steady stretching state. The transient equationgives linear growth at $W=1/2$ and exponential growth at larger rates, the Hookean coil-stretch failure of the upper-convected Maxwell limit.

Parameter qualification. Positivity of $\alpha$ alone, without a restriction on $b=\alpha n{k}_{B}T$, does not ensure real physical steady solutions for the entire tensor. For example $b=2$, $W=1$ makes the transverse equation $2s_2^2+2s_2+1=0$, which has no real root. Its diffusion/resistance interpretation is already inadmissible for some conformations. The physically positive-mobility regime stated above avoids that counterexample; the source's positive-$\alpha$ case must not be taken as a guarantee for arbitrary parameters.

The common part of the magnetohydrodynamic-viscoelastic analogy is stretching and advection of a line element. With the component-first velocity gradient $L_{ij}={\partial }_j{u}_i$, the ideal tensor equation isIts left side is an upper-convected derivative, the same deformation law that appears for a polymer conformation tensor when its relaxation is neglected.

There are two important physical differences. First, the solenoidal magnetic-field constraint prevents a magnetic field line from having an isolated source or sink inside the fluid. A polymer strand can have molecular endpoints, so there is no corresponding no-endpoint constraint on the strands. Second, a stretched polymer has a preferred relaxed distribution of configurations and a finite polymer relaxation time; it can recoil relative to the solvent through molecular drag and thermal motion. An ideal magnetic field has no analogous local tendency to return to a prescribed coil size. Real polymers also have a finite contour length, a limitation absent from the ideal field-stretching law.

For example, a normalized Hookean dumbbell conformation equation has the dissipative extensionThis is local relaxation toward an isotropic thermal configuration. It remains nonzero for a spatially uniform stretched $C$. By contrast, constant magnetic diffusivity ${\eta }_m$ adds ${\eta }_m{\nabla }^2\mathbf{B}$ to the resistive induction equation. The product rule gives the resistive evolution of the magnetic second-moment tensor:The magnetic correction involves spatial gradients, vanishes for a uniform magnetic field, and does not create field-line endpoints: ${\partial }_t(\nabla \cdot \mathbf{B})={\eta }_m{\nabla }^2(\nabla \cdot \mathbf{B})$. It permits slippage and magnetic reconnection while preserving the no-monopole constraint. After averaging, its gradient-product term generally needs additional information beyond the averaged $M$. One cannot replace it indiscriminately by the local polymer relaxation term, or even by ${\eta }_m{\nabla }^{2}M$ alone.

Let $K_{ij}={\partial }_j{u}_i$, $C_{ij}=⟨R_i{R}_j⟩$, $\alpha =2H/\zeta$ and $D_R=2k_{B}T/\zeta$. The configuration equation of the Hookean dumbbell model becomesHere the paper's $k$ denotes the Boltzmann constant $k_B$. Integrating the second moment equation against $R_i{R}_j$ and discarding boundary terms at infinite separation givesFor example the spring drift differentiates each factor $R_i,R_j$, while the Laplacian gives ${\nabla }_{\mathbf{R}}^2(R_i{R}_j)=2{\delta }_{ij}$. The Hookean dumbbell conformation equation isIn the row-vector convention of $\mathbf{R}\cdot \nabla \mathbf{u}$, the displayed gradient tensor is $K^T$, giving exactly the transposes in the requested equation. The dimensionless conformation tensor $A=HC/(k_{B}T)$ consequently has upper-convected derivative $-(A-I)/\lambda$, with polymer relaxation time $\lambda =\zeta /(4H)$. This normalization matters: the unscaled $C$ has equilibrium value $(k_{B}T/H)I$, rather than $I$.

Write $q=\dot{\gamma }\tau$ and $f=1+\alpha \mathrm{tr}A$. With $L_{ij}={\partial }_j{v}_i$, the steady homogeneous upper-convected derivative is $-LA-A{L}^T$. Its diagonal and shear equations giveOn the physical branch $f>0$, the other off-diagonal components vanish too. Thus $a_{22}=a_{33}=0$, $a_{11}=2a_{12}^2$ andFor $\alpha \ge 0$ the right side has derivative $1+6\alpha a_{12}^2>0$, so the shear stress determines a unique shear rate. The shear viscosity is $G_0\tau /f$ and decreases with shear magnitude when $\alpha >0$. This is the shear part of shear and pipe flow of an affine linear PTT fluid.

For fully developed pipe flow take $\mathbf{v}=w(r)\hat{\mathbf{z}}$ and the signed axial pressure gradient $dp/dz=\Delta p$. Regularity at the axis and the supplied momentum balance determine the shear stress without using the constitutive equation:Locally the constitutive equations have the same shear form, with axial flow replacing the Cartesian flow direction and $\dot{\gamma }=w^{\prime }(r)$ carrying its sign. HenceIntegrating and imposing no-slip boundary conditions at $r=R$ givesA negative pressure gradient produces positive axial flow. At $\alpha =0$ this is the ordinary Hagen-Poiseuille flow; positive $\alpha$ increases the flux at a fixed pressure drop through shear thinning. The nonzero axial normal stress varies radially, but its axial derivative vanishes in fully developed flow, so it does not change the axial momentum balance used here.

With its material state fixed by a prescribed deformation history, the differential pom-pom model is a simple-fluid constitutive law. The stress depends on the local history of a material element through its molecular stretch and orientation, without gradients of that history between neighbouring elements. A differential representation does not prevent a material from being a simple fluid.

To see the history dependence explicitly, use the component-first velocity gradient $K$, so the tensor evolution is $D_{t}A=KA+A{K}^T-(A-I)/{\tau }_1$. Let $F(t,s)$ be the relative deformation gradient, satisfying ${\partial }_{t}F=K(t)F$ and $F(s,s)=I$. Along a material trajectory, an integrating factor givesThus $A$ is a functional of the local deformation history and its prepared initial state. If the initial $A$ is a positive-definite matrix, it retains that property, so $B=A/\mathrm{tr}A$ is well defined. Given this $B(t)$, the scalar stretch solves $\dot{\lambda }=r(t)\lambda +1/{\tau }_2$, where $r=B:K-1/{\tau }_2$. Its variation of constants formula again depends only on that local history:The differential pom-pom model also satisfies material frame indifference. Under an observer rotation, $A$ and $B$ transform as objective tensors, while $\lambda$ is a scalar. The extra skew part of the transformed velocity gradient has zero contraction with symmetric $B$, so the stretch equation is unchanged. The tensor evolution uses an upper-convected derivative, and the final stress transforms objectively. Initial values cannot be independently forgotten when specifying a finite-past experiment: a fully specified prepared state, such as rest, is part of the history specification.

Write $\mathbf{\tau }$ for the Kirchhoff stress tensor and $S$ for the second Piola-Kirchhoff stress tensor. Since $\mathbf{\tau }=FS{F}^T$ and $\dot{F}=LF$ for the velocity gradient,Therefore the upper-convected rate as a stress push-forward is

For the simple shear flow, put $\gamma =f_{,2}$ and $g=\dot{\gamma }$. The velocity gradient and rate-of-strain tensor areTheir upper-convected derivative givesDenote the symmetric extra stress by $E={\sigma }^d$, and the relaxation time by ${\tau }_r$ to distinguish it from $\mathbf{\tau }$. The Oldroyd-B model in the question becomesFor the steady constitutive response, $E_{22}=E_{33}=E_{13}=E_{23}=0$, $E_{12}=\mu g$ and $E_{11}=2(\mu -{\mu }_r){\tau }_r{g}^2$. HenceA constant imposed shear rate can still have startup transients; these values describe the steady response.

For circular Couette flow $v(r)e_{\theta }$, the velocity gradient in the orthonormal polar basis $(e_r,e_{\theta })$ isFor completeness, in fixed Cartesian coordinates let $\omega (r)=v(r)/r$ and $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. Then $v=\omega Jx$ and $L=\omega J+({\omega }^{\prime }/r)(Jx)\otimes x$. Along a particle, the polar basis rotates at angular speed $\omega$. Its contribution to the material derivative cancels the rigid-rotation part of $L$ in the upper-convected derivative. The rotating-frame reduction of circular viscoelastic shear therefore leavesThe flow direction is now $e_{\theta }$ and the gradient direction is $e_r$. With $g=v^{\prime }-v/r$, the same simple-shear constitutive calculation gives

Angular force balance implies $r^2\mu g$ is constant. Integrating $v^{\prime }-v/r\propto r^{-2}$ gives $v=Ar+B/r$. With $v(a)=V_a$ and $v(b)=0$,The radial force balance is $-p^{\prime }-2(\mu -{\mu }_r){\tau }_r{g}^2/r=0$. Thus the inertialess Oldroyd-B circular Couette pressure isThe constant $p_0$ is fixed by a pressure or normal-traction boundary condition.

An objective time derivative transforms as a physical tensor under a time-dependent rigid change of observer. If $x^{\ast }=Q(t)x+c(t)$, with $Q$ orthogonal and $A^{\ast }=QA{Q}^T$, objectivity requiresThis means the constitutive equation makes the same physical prediction in translating and rotating frames, rather than requiring individual tensor components to be invariant.

The printed plus commutator uses the derivative-index-first velocity gradient, $G_{ij}={\partial }_i{u}_j$. This is the transpose of the component-index-first convention $L_{ij}={\partial }_j{u}_i$ often used for the velocity gradient. In this notation $\omega =G-G^T$, and the rate is the Jaumann derivativeTo check the covariance of the Jaumann derivative, let $R=\dot{Q}{Q}^T$. Direct transformation givesThe observer-rotation terms cancel in the displayed corotational rate. Equivalently, in the $L$ convention the physical spin tensor is $W=(L-L^T)/2=-\omega /2$, and the same rate is $D_{t}A-WA+AW$. One must transpose the gradient and reverse the commutator together; using $L$ in the printed plus-sign expression would not be objective.

Another example is the upper-convected derivative, written consistently in the $L$ convention asThe extra terms in $L^{\ast }=QL{Q}^T+R$ cancel those in $D_t^{\ast }{A}^{\ast }$, so this also transforms covariantly. In contrast, the uncorrected material derivative $D_{t}A$ of a second-rank tensor is not objective: its transformation contains $R{A}^{\ast }-A^{\ast }R$. For instance a constant anisotropic tensor in a stationary fluid has zero material derivative in its original frame, but its components have a nonzero commutator derivative in a rotating observer's frame. The ordinary partial time derivative is likewise not an objective tensor rate.

In the Giesekus model, $\lambda$ is the viscoelastic relaxation time: after deformation stops, polymeric stress relaxes on that timescale. The parameter $\eta$ has dimensions of dynamic viscosity and is the model's zero-rate viscosity scale. Since $\lambda {\mathbf{\tau }}^2/\eta$ already has dimensions of stress, $\alpha$ is dimensionless.

The symbol $\stackrel{▽}{\tau }$ is the upper-convected derivativeThe convective and velocity-gradient terms account for translation, rotation, and affine stretching of material elements. They make the constitutive law objective under time-dependent rigid changes of observer.

An objective time derivative transforms as a tensor under every superposed time-dependent rigid translation and rotation of the observer. A constitutive prediction using it is therefore frame-indifferent. The upper-convected derivative, lower-convected derivative, and Jaumann derivative are objective; the componentwise partial time derivative, and generally the uncorrected material derivative of a tensor, are not objective under rotating observers.

The Johnson--Segalman--Oldroyd model uses the Gordon--Schowalter derivativeHere $\eta$ is the zero-shear viscosity, ${\lambda }_1$ is the stress-relaxation time, and ${\lambda }_2$ is the retardation time associated with the more rapidly responding viscous contribution. Passive Oldroyd-type materials normally have $0\le {\lambda }_2\le {\lambda }_1$.

A low-rate steady-shear experiment measures $\eta$. A step strain followed by stress relaxation determines ${\lambda }_1$ from the exponential decay. Creep and recovery, or a frequency sweep in small-amplitude oscillatory shear, then separates ${\lambda }_2$ from ${\lambda }_1$; fitting both storage modulus and loss modulus is especially direct.

Write the velocity gradient aswhere $\mathbf{\Omega }$ is the spin tensor. Expanding the upper-convected derivative in the structure equation givesFor $\xi =0$, the first four terms reproduce the upper-convected derivative. Settingtherefore givesWith polymeric stress ${\mathbf{\tau }}_p=b_1\mathbf{\alpha }$, this is the Oldroyd-B model. For $c_1>0$, its relaxation time and polymer viscosity areIndeed the total stress obeys

For $\xi =2$, the coefficient of
$\dot{\mathbf{\gamma }}\mathbf{\alpha }+\mathbf{\alpha }\dot{\mathbf{\gamma }}$
is $+1/2$, so the objective derivative becomes the lower-convected derivative. Hencerecovers the Oldroyd-A model, again with $c_1>0$ for a finite positive relaxation time. Parameter choices such as $b_1{c}_2=0$ give the degenerate Newtonian limit.

A plain material derivative of the conformation tensor is not an objective time derivative: an observer undergoing a time-dependent rigid rotation would infer a different constitutive response, and even rigid-body rotation could appear to change polymer deformation. The upper-convected derivative subtracts deformation and rotation carried by the velocity gradient and is frame indifferent.

The Oldroyd-B model is often inadequate because its Hookean dumbbells are infinitely extensible. It predicts constant shear viscosity rather than shear thinning, zero second normal-stress difference, and an unbounded extensional viscosity at a finite extension rate. Real polymer chains have finite extensibility and commonly exhibit shear thinning, bounded extensional stress, multiple relaxation times, and nonlinear solvent or concentration effects.

The affine linear PTT fluid considered here uses a symmetric dimensionless tensor $A$ with $\stackrel{\nabla }{A}+(1+\alpha \mathrm{tr}A)A/\tau =2E$ and polymer stress $G_{0}A$. Here $\stackrel{\nabla }{A}$ is the upper-convected derivative, $G_0,\tau >0$ and the nonlinear relaxation parameter $\alpha$ is usually positive. Linearizing about rest gives the Maxwell fluid relaxation modulus $G(s)=G_0{e}^{-s/\tau }$. The PTT family also admits other relaxation functions and nonaffine derivatives; their nonlinear predictions need not equal those of this specified version.

Using the component-first velocity gradient $K_{ij}=\partial v_i/\partial x_j$, the Rivlin-Ericksen tensors are defined by the displayed recurrence. In particular $A_1$ is twice the rate-of-strain tensor. These tensors transform objectively under time-dependent rigid observer changes. With this convention the upper-convected derivative of $A_1$ is $A_2-2A_1^2$, an identity useful in the slow-flow expansion of a second-order fluid.

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.