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The conformation tensor describes the average stretch and orientation of polymer molecules. Its equilibrium value is the identity tensor under a common normalization.

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.

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.

Represent one polymer by two beads joined by an ideal linear entropic spring, and stipulate a dimensionless conformation tensor $C$ with equilibrium value $I$. The displayed model uses velocity gradient $A=\mathrm{diag}(\dot{ϵ},-\dot{ϵ},0)$ and positive polymer relaxation time $\lambda$. It omits finite chain extensibility, interactions with other chains, and feedback from the polymer onto the flow. It is a teaching model, not a fitted description of Keller's apparatus.

Define the Weissenberg number for this model as $\mathrm{Wi}=\lambda \dot{ϵ}$ and dimensionless time $u=t/\lambda$. With an initially isotropic $C=I$, the diagonal entries obeyBelow $\mathrm{Wi}=1/2$, both transverse and longitudinal responses approach positive finite steady values:For a chosen $\mathrm{Wi}=0.4$, $C_{xx}=5$ and $C_{yy}=5/9$: the longitudinal mean-square separation is five times its quiescent value, while the transverse separation contracts. A root-mean-square length grows as the square root of a variance, so the longitudinal length factor is $\sqrt{5}$, not five. The anisotropy $C_{xx}-C_{yy}$ is an orientational indicator; converting it to measured birefringence requires an additional calibrated optical relation.

At $\mathrm{Wi}=1/2$, $C_{xx}=1+u$. Above that value, the growing homogeneous solution prevents a positive finite steady state. For instance, $\mathrm{Wi}=0.6$ gives $C_{xx}=6e^{u/5}-5$. Real chains have finite contour lengths, so this growth identifies the failure of the linear spring model. It does not establish a universal threshold for arbitrary flow histories, finite chains or other definitions of polymer relaxation time. Combining the growth equation with residence time in planar extension also explains why a large instantaneous Weissenberg number alone does not guarantee a strongly stretched exiting chain.

Represent a dilute polymer by two beads connected by a spring obeying Hooke's law, with stiffness $H>0$ and bead drag coefficient $\zeta >0$. Neglect bead inertia and hydrodynamic interactions. For component-first velocity gradient $K_{ij}={\partial }_j{u}_i$, the configuration Fokker-Planck probability current isThe factor two comes from relative motion of both beads. At rest its normalized probability density is ${\psi }_{\mathrm{eq}}=(H/(2\pi k_{B}T){)}^{3/2}{e}^{-H{R}^2/(2k_{B}T)}$. The conformation tensor $C=⟨\mathbf{R}{\mathbf{R}}^T⟩$ has dimensions of length squared; $(H/(k_{B}T))C$ is its dimensionless version. The Hookean dumbbell in planar extension is a particular imposed flow in this model.

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.

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 $I$.

The claimed equality does not hold for the printed joint distribution. The original PDF specifies a centered Maxwell-Boltzmann distribution of $\mathbf{v}=\dot{\mathbf{R}}$ independent of $\mathbf{R}$, withSince the double average includes number density $n$, evaluating the Gaussian integrals givesThe cross term vanishes because the conditional mean of $\mathbf{v}$ is zero. In particular, set $K=0$ and take the equilibrium configuration probability density ${\psi }_{\mathrm{eq}}\propto e^{-H{R}^2/(2k_{B}T)}$. The left side is $3n\zeta k_{B}T/m>0$, while $nHC:K=0$. This is a counterexample even at thermal equilibrium, so no rearrangement of the earlier conformation tensor equation can prove the printed statement.

The discrepancy is thermal bead drag and heat-bath power: gross viscous loss of microscopic thermal motion is replenished by the thermal bath. It is not identical to mechanical polymer stress power. Moreover, $\dot{\mathbf{R}}$ in the earlier configuration continuity equation is a probability current velocity, including a distribution-dependent osmotic drift; it cannot simultaneously be identified with an independently sampled centered Maxwellian velocity.

A consistent overdamped result is the free-energy balance of a Hookean dumbbell. Define $\mathbf{q}=H\mathbf{R}+k_{B}T{\nabla }_{\mathbf{R}}\log \psi$ andDifferentiate the relative entropy and use the configuration Fokker-Planck probability current $\psi \mathbf{V}$. After integration by parts,The affine osmotic term is $n{k}_{B}T\int \nabla \psi \cdot K\mathbf{R}d\mathbf{R}=-n{k}_{B}T\mathrm{tr}K=0$. HenceThe polymer stress tensor ${\sigma }_p=n(HC-k_{B}TI)$ has ${\sigma }_p:K=nHC:K$ in an incompressible flow. It stores or releases free energy as well as dissipating it, so the total stress-work term in part a need not itself be a purely dissipative quantity for this viscoelastic material. Stress power equals $\mathcal{D}$ only for stationary free energy, and $\mathcal{D}$ uses the probability-current slip. Neither condition turns the printed independent-Maxwellian average into that expression.

Choose the extending direction as $x$ and write $s=\dot{\gamma }>0$. The uniaxial extensional flow, satisfying incompressibility, hasStarting from rest, symmetry preserves $A=\mathrm{diag}(a,b,b)$. The differential pom-pom model tensor equation becomesPut $x=s{\tau }_1$. For $x\ne 1/2$, the solutions areIf $x<1/2$, both transients decay, givingIf $x>1/2$, the axial component is instead $a=(2x{e}^{(2x-1)t/{\tau }_1}-1)/(2x-1)$ and grows exponentially without bound, while $b\to 1/(1+x)$. At the boundary $x=1/2$, which lies between the two stated regimes, $a=1+t/{\tau }_1$ grows linearly and $b\to 2/3$.

Normalization makes the molecular orientation finite even when the unnormalized conformation tensor diverges. For $0\le x<1/2$, direct division by $a+2b$ givesFor $x>1/2$, the divergent axial component dominates the matrix trace, soThe same normalized limit holds at $x=1/2$, agreeing continuously with the limit of the subcritical expression. The limiting orientations have nonnegative eigenvalues and unit matrix trace, as required for the normalized molecular orientation.

Now set $S(t)=B(t):K=s(a-b)/(a+2b)$. Its limiting value isThe scalar stretch equation is $\dot{\lambda }=[S(t)-1/{\tau }_2]\lambda +1/{\tau }_2$. If $S_{\infty }<1/{\tau }_2$, its coefficient is eventually bounded above by a negative constant. The variation of constants formula gives a bounded solution, converging toIf $S_{\infty }>1/{\tau }_2$, the coefficient is eventually positive, and the positive stretch grows without bound. At equality a bounded stretch would imply $\dot{\lambda }\to 1/{\tau }_2>0$, which contradicts boundedness. Thus equality is not an additional bounded steady regime.

The function $S_{\infty }(s)$ is continuous and strictly increasing from zero to infinity. Consequently there is a unique critical extension rate defined byand the molecular stretch remains bounded precisely for $\dot{\gamma }<{\dot{\gamma }}_c$. This is distinct from the tensor threshold $\dot{\gamma }{\tau }_1=1/2$. If ${\tau }_2<2{\tau }_1$, for example, the normalized orientation can remain well defined and the stretch remain bounded for some rates above the tensor threshold; divergence of $A$ alone does not prove divergence of the physical stress.

In the requested doubly subcritical regime, write $D=1-s{\tau }_1-s^2{\tau }_1{\tau }_2>0$. The steady uniaxial extension of the differential pom-pom model hasThe isotropic pressure cancels from the tensile normal-stress difference. Hence the extensional viscosity isorIts zero-rate limit is $G{\tau }_1=3{\mu }_0$, recovering the Trouton ratio of three. The restrictions exclude both loss of a finite tensor steady state and loss of a finite stretch steady state; neither denominator may be continued through its singularity as a physical steady constitutive response.

Let two observers be related by ${\mathbf{x}}^{\ast }=\mathbf{c}(t)+Q(t)\mathbf{x}$, with Q a proper rotation matrix. A spatial second-rank tensor has material frame indifference if its values at corresponding events satisfy $\boxed{J^{\ast }=QJ{Q}^T}$. This defines an objective second-rank tensor, and allows the observer's rotation to depend on time.

For Cauchy stress tensor, consider a material surface element with unit normal n and contact traction $\mathbf{t}=\sigma \mathbf{n}$. A rigid observer change preserves its area and rotates both its normal and actual contact force: ${\mathbf{n}}^{\ast }=Q\mathbf{n}$, ${\mathbf{t}}^{\ast }=Q\mathbf{t}$. Thus ${\sigma }^{\ast }Q\mathbf{n}=Q\sigma \mathbf{n}$ for every n, proving ${\sigma }^{\ast }=Q\sigma Q^T$. Fictitious body forces of an accelerated observer do not alter this transformation of physical contact traction.

To keep the vorticity convention explicit, write $L_{ij}={\partial }_j{v}_i$, $E=(L+L^T)/2$ and $\Omega =(L^T-L)/2$, the negative of the usual component-first spin tensor. Equivalently, with the derivative-index-first $\nabla v$ used in the paper, $\Omega =[\nabla v-(\nabla v{)}^T]/2$. Put $R=\dot{Q}{Q}^T$, which is antisymmetric. The transformed velocity is ${\mathbf{v}}^{\ast }=\dot{\mathbf{c}}+\dot{Q}\mathbf{x}+Q\mathbf{v}$, and direct differentiation givesThe additional observer spin cancels from the symmetric part, proving that the rate-of-strain tensor is objective.

Along the same material particle, differentiating $J^{\ast }=QJ{Q}^T$ givesThe two extra terms cancel those from ${\Omega }^{\ast }{J}^{\ast }-J^{\ast }{\Omega }^{\ast }$. Moreover $E^{\ast }{J}^{\ast }+J^{\ast }{E}^{\ast }=Q(EJ+JE)Q^T$. Hence the complete objective time derivative satisfiesThis proves covariance for every constant scalar $\alpha$; changing the spin convention without changing the commutator sign would not. It is a Gordon--Schowalter derivative with the paper's slip-parameter convention.

For the Johnson-Segalman fluid, first use the standard uniaxial extensional flow $\mathbf{v}=(\dot{ϵ}x,-\dot{ϵ}y/2,-\dot{ϵ}z/2)$, with $\dot{ϵ}>0$. Here $\Omega =0$ and E is diagonal. The steady conformation equation giveswith vanishing off-diagonal components on the stable steady branch. The tensile normal-stress difference is $\alpha G_0(A_{xx}-A_{yy})+3{\mu }_0\dot{ϵ}$. Dividing by the extension rate gives the Johnson-Segalman extensional response:The conformation tensor evolution in each principal direction is ${\dot{A}}_{ii}+(1/\tau -2\alpha E_{ii})A_{ii}=1/\tau$. Therefore a positive, attracting steady conformation requires both displayed denominators positive. For $\alpha >0$ the first pole is at $\dot{ϵ}=1/(2\alpha \tau )$; for $\alpha <0$ it is at $\dot{ϵ}=1/(|\alpha |\tau )$ in the transverse components. At a pole the corresponding component grows linearly in time, and beyond it generically grows exponentially. The model has a finite-rate extensional catastrophe; the algebraic viscosity cannot be continued as a physical steady response past the first pole. At zero rate, ${\eta }_E=3({\mu }_0+{\alpha }^2{G}_0\tau )$, giving Trouton ratio three. For $\alpha =0$ the polymer contribution to stress disappears and the response is simply Newtonian.

The question does not specify extension geometry. If a planar convention $\mathbf{v}=(\dot{ϵ}x,-\dot{ϵ}y,0)$ is intended instead, the same diagonal calculation gives $({\sigma }_{xx}-{\sigma }_{yy})/\dot{ϵ}=4{\mu }_0+4{\alpha }^2{G}_0\tau /[1-4{\alpha }^2{\tau }^2{\dot{ϵ}}^2]$, valid on the stable branch $|2\alpha \tau \dot{ϵ}|<1$. The preceding boxed formula uses the standard uniaxial definition of extensional viscosity.

For simple shear flow $v_x=\dot{\gamma }y$, write $A_{xx}=X$, $A_{yy}=Y$, $A_{xy}=A_{yx}=Z$, and $A_{zz}=1$. The xx, yy and xy components of the steady constitutive equation giveSubstituting the first two relations into the third yields $Z=\alpha \tau \dot{\gamma }/D$, with $D=1+(1-{\alpha }^2){\tau }^2{\dot{\gamma }}^2$. Define the viscometric functions by ${\sigma }_{xy}=\mu \dot{\gamma }$, $N_1={\psi }_1{\dot{\gamma }}^2$ and $N_2={\psi }_2{\dot{\gamma }}^2$, using x for flow, y for the shear gradient and z for vorticity. The Johnson-Segalman steady viscometric functions are thereforeThe solvent creates no normal-stress differences in this flow. For nonzero $\alpha$, the fluid is shear thinning with positive first and negative second normal-stress differences.

To classify the stress curves, set $x=\tau \sqrt{1-{\alpha }^2}\dot{\gamma }$ and $b={\mu }_0/({\alpha }^2{G}_0\tau )$. The scaled shear stress is $S=x[b+(1+x^2{)}^{-1}]$, andThe second term has minimum $-1/8$ at $x^2=3$. Thus the shear stress is monotone for $b\ge 1/8$ and has a decreasing segment for $\boxed{0<b<1/8}$. At equality it has a horizontal tangent but no decreasing interval. Below the threshold, its maximum and minimum occur atwith the smaller root giving the maximum. Negative-rate curves are obtained by odd reflection.

For small b, polymer stress dominates the low-rate response, while the solvent restores a positive high-rate slope. The Johnson-Segalman negative-slope shear threshold signals a constitutive instability of homogeneous shear: different local shear rates can carry the same shear stress, permitting shear banding. A stress plateau or a selected interface position cannot be inferred from the local curve alone; spatial regularization and boundary/history information matter. This limitation is illustrated in a primary diffusive Johnson-Segalman study. If $\alpha =0$, the scaling by b is inapplicable and the stress is just ${\mu }_0\dot{\gamma }$.

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.

For extension rate $s>0$ and $x=s{\tau }_1<1/2$, the steady conformation tensor is $\mathrm{diag}((1-2x{)}^{-1},(1+x{)}^{-1},(1+x{)}^{-1})$. Its normalized orientation has axial component $(1+x)/[3(1-x)]$ and transverse components $(1-2x)/[3(1-x)]$. Stretch is finite only if $s^2{\tau }_1{\tau }_2<1-x$, when ${\lambda }_{\infty }=(1-x)/(1-x-s^2{\tau }_1{\tau }_2)$ and the displayed extensional viscosity follows. Above $x=1/2$, the unnormalized axial tensor grows exponentially but $B\to \mathrm{diag}(1,0,0)$; then stretch remains bounded only if $s{\tau }_2<1$. Orientation normalization and stretch relaxation therefore impose different thresholds.