Viscoelasticity

Viscoelasticity combines viscous dissipation with elastic storage and relaxation of deformation.

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.

For uniaxial extensional flow put $q=\tau \dot{\gamma }>0$, $f=1+\alpha \mathrm{tr}A$ and $T=f-2q$. The diagonal components are $a_{11}=2q/T$, $a_{22}=a_{33}=-q/(T+3q)$. The identity $f-1=\alpha \mathrm{tr}A$ gives the cubic equation in the title. For $\alpha >0$ its physical root has $T>\max (0,1-2q)$ and connects to $T=1$ at rest. Its extensional viscosity is ${\eta }_E=G_0\tau [2/T+1/(T+3q)]$, tending to $3G_0\tau$ at small $q$ and $2G_0\tau /\alpha$ at large $q$. At $\alpha =0$ the stable steady Maxwell extension instead ends at $q=1/2$; the finite high-rate limit must not be extended to that degenerate case.

In steady simple shear flow, the tensor equation gives $a_{22}=a_{33}=0$, $a_{11}=2a_{12}^2$ and the equation in the title. A circular pipe with axial pressure gradient $\Delta p$ has $G_0{a}_{rz}=\Delta pr/2$. Thus $w^{\prime }=\Delta pr/(2G_0\tau )+\alpha \Delta p^3{r}^3/(4G_0^3\tau )$, and no-slip boundary conditions givePositive $\alpha$ produces shear thinning; the sign of the pressure gradient determines the direction of the flow.

The Weissenberg number measures deformation accumulated over a viscoelastic relaxation time, often as the product of that time and a shear rate or an extension rate. Its precise rate convention should be stated. Unlike a Deborah number based on externally imposed temporal variation, it can remain nonzero in a steady flow. A weak-flow constitutive expansion generally requires both slow temporal change and small deformation over the memory time.

The Deborah number compares a material's viscoelastic relaxation time with the time scale of an imposed process. A small value allows substantial stress relaxation during that process; a large value can reveal stored elasticity. In small-amplitude oscillatory shear, one commonly uses $\mathrm{De}=\omega \tau$. Large frequency does not by itself invalidate a linear viscoelastic fluid approximation if the deformation amplitude remains sufficiently small.

This differential model separates normalized molecular orientation $B=A/\mathrm{tr}A$ from scalar stretch $\lambda$. With component-first velocity gradient $K$, it evolves by $D_{t}A=KA+A{K}^T-(A-I)/{\tau }_1$ and $D_t\lambda =\lambda B:K-(\lambda -1)/{\tau }_2$. Its relaxed state is $A=I$, $B=I/3$, $\lambda =1$. The displayed constant-relaxation stretch law is an idealization; variants can add saturation or nonlinear relaxation.

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.

Linearizing the differential pom-pom model about its relaxed state leaves no first-order stretch perturbation in incompressible flow. Its anisotropic orientation obeys $\dot{a}+a/{\tau }_1=A_1$, producing the displayed relaxation modulus. In a slow-flow expansion, $A=I+{\tau }_1{A}_1+{\tau }_1^2(2A_1^2-A_2)+O(3)$. Trace normalization and the first stretch correction affect only isotropic stress at this order, giving the displayed second-order fluid coefficients. Linear small-amplitude response allows arbitrary frequency, whereas this local second-order expansion also requires slow time dependence.

An ideal dashpot resists relative motion with force proportional to velocity. Maintaining a displacement rate $\dot{x}$ requires force $\mu \dot{x}$ and dissipates power $\mu {\dot{x}}^2$ for $\mu >0$. Connecting it in parallel or series with a spring gives different constitutive laws.

The Kelvin-Voigt model puts a spring and dashpot in parallel. Their common displacement makes their forces add. With the convention $e^{i\omega t}$, the complex stiffness is $k+i\omega \mu$, giving storage stiffness $k$ and loss stiffness $\omega \mu$. Under a force step, $x=(F_0/k)(1-e^{-t/\tau })$ with $\tau =\mu /k$. The series spring–dashpot is a Maxwell fluid analogue and has different relaxation and creep behavior.

For a quasistatic cantilever of Kelvin-Voigt material, the bending moment is $M=EI{y}_{xx}+{\eta }_{s}I{y}_{xxt}$. A tip load gives $M=F(t)(\ell -x)$, so the free-tip displacement obeys the displayed equation with $\tau ={\eta }_s/E$. A step load gives a finite asymptotic deflection. If inertia is retained, the interior equation is $\rho A{y}_{tt}+EI{y}_{xxxx}+{\eta }_{s}I{y}_{xxxxt}=0$, with the corresponding moment and shear conditions at the tip.

Stress relaxation is decay of stored stress after a deformation is imposed and then held fixed. It probes a material's memory through its relaxation modulus. An instantaneous generalized Newtonian fluid has no stored shear stress once the shear rate is zero and cannot describe a nontrivial relaxation tail.

A linear viscoelastic fluid has causal stress response linear in the deformation history, often expressed as ${\sigma }^d(t)=2{\int }_0^{\infty }G(s)E(t-s)ds$. The approximation requires small deformation and rotation over the relevant memory times. High frequency is allowed at sufficiently small amplitude. An integrable decaying relaxation modulus gives a finite zero-shear viscosity; an instantaneous solvent term may be written separately.

A small step of shear stress $s_{0}H(t)$ produces shear strain $\gamma (t)=s_{0}J(t)$ from a relaxed state. This time-dependent deformation under held stress is creep. The causal constitutive convolution gives $\hat{s}=p\hat{G}\hat{\gamma }$, proving the displayed Laplace transform relation. For a Maxwell fluid, $J(t)=1/G_0+t/(G_0\tau )$: an elastic jump followed by viscous flow.

For the convention $e^{i\omega t}$, complex viscosity is the ratio of oscillatory shear stress amplitude to shear rate amplitude. A causal relaxation modulus gives the displayed transform. Its associated complex shear modulus is $G^{\ast }=i\omega {\eta }^{\ast }=G^{\prime }+i{G}^{\prime \prime }$, where $G^{\prime }$ is the storage modulus and $G^{\prime \prime }$ the loss modulus.

The relaxation modulus is the shear-stress response at lag s to a unit small step of shear strain. By linear superposition it is the causal memory kernel multiplying shear rate. Its decaying part represents stored stress that relaxes after deformation ceases.

For a linear viscoelastic fluid, write the solvent stress explicitly as $2{\mu }_{0}E(t)$, or use the displayed kernel with a causal endpoint delta satisfying ${\int }_0^{\infty }{\delta }_{+}(s)f(s)ds=f(0)$. If a symmetric delta is integrated with half its weight at the endpoint, its coefficient must instead be $2{\mu }_0$. Mixing those conventions loses a factor of two. The Oldroyd-B model has the displayed exponentially decaying polymer part after linearization.

The three viscometric functions describe steady homogeneous simple shear flow: its shear viscosity and the first and second normal-stress differences per squared shear rate. Take x as the flow direction, y as the gradient direction and z as the vorticity direction, so $N_1={\sigma }_{xx}-{\sigma }_{yy}$ and $N_2={\sigma }_{yy}-{\sigma }_{zz}$. Isotropic pressure drops out of these differences.

A corotational Jeffreys fluid obeys $\stackrel{\circ }{S}+S/\tau =2{\mu }_0\stackrel{\circ }{D}+2{\mu }_{1}D/\tau$, with deviatoric stress $S$, rate-of-strain tensor $D$, Jaumann derivative and relaxation time $\tau >0$. Put $A=S-2{\mu }_{0}D$. In the corotating-frame representation of a Jaumann derivative, ${\tilde{A}}^{\prime }+\tilde{A}/\tau =2({\mu }_1-{\mu }_0)\tilde{D}/\tau$. An integrating factor gives an exponentially weighted history transported between the rotating frames. The infinite-past formula selects histories with $e^{t/\tau }\tilde{A}(t)\to 0$ as $t\to -\infty$; finite initial data also contribute an exponentially decaying homogeneous term.

A linear Maxwell fluid has constitutive equation $\mathbf{\sigma }+\lambda {\partial }_t\mathbf{\sigma }=\mu \dot{\mathbf{\gamma }}$, with $\dot{\mathbf{\gamma }}$ twice the rate-of-strain tensor. Here $\mu$ is the zero-frequency shear viscosity and $\lambda >0$ is the viscoelastic relaxation time. Stress at fixed strain decays as $e^{-t/\lambda }$; the corresponding series spring and dashpot have shear modulus $G=\mu /\lambda$. In small-amplitude simple shear from an isotropic relaxed state, the first-order forcing is off-diagonal, so both normal-stress differences vanish at linear order. This is not a statement about the nonlinear objective model at second order in shear rate.

For a no-slip channel with half-width $h$, a harmonic pressure gradient $\Delta p{e}^{i\omega t}$ and complex viscosity $\hat{\mu }=G_0\tau /(1+i\omega \tau )$, put $z^2=i\omega \rho h^2/\hat{\mu }$. Solving $\hat{\mu }{U}^{\prime \prime }-i\omega \rho U=\Delta p$ gives $U=-\Delta p[1-\cosh (zy/h)/\cosh z]/(i\omega \rho )$; integration gives the displayed flux per unit width. At low frequency it approaches $-2\Delta p{h}^3/(3G_0\tau )$. At fixed positive $h,\tau ,G_0,\rho$, the high-frequency leading flux is $-2\Delta ph/(i\omega \rho )$, although shear waves can persist in the local profile. If instead $\omega \tau \gg 1$ while $\rho {\omega }^2{h}^2/G_0\ll 1$, the negligible-inertia elastic limit is $Q\simeq -2i\omega \Delta p{h}^3/(3G_0)$. The last two limits impose different scale orderings.

For a half-space initially at rest and a wall given a step velocity $U$, momentum conservation and the Maxwell fluid stress law give $\tau u_{tt}+u_t=(G_0\tau /\rho )u_{yy}$. The displayed decaying Laplace transform solution obeys the wall condition. Inverting on a right-half-plane Bromwich contour yields a shear signal of speed $\sqrt{G_0/\rho }$. The limit $\tau \to 0$ at fixed $G_0\tau =\eta$ gives viscous error-function penetration; $\tau \to \infty$ at fixed $G_0$ gives a sharp elastic front.

For a quasistatic beam made of a two-element series Maxwell material, $\dot{M}+M/\tau =EI{\partial }_t{y}_{xx}$. A force step therefore produces an instantaneous elastic deflection followed by unbounded linear creep, as displayed. It cannot have the finite long-time plateau of a Kelvin-Voigt cantilever creep response; a solid with finite relaxed modulus requires an additional elastic element.

A prescribed local linear Maxwell force law filters a harmonic drag amplitude by $(1+i\omega \lambda {)}^{-1}$. For periodic states its time derivative has zero mean, so the mean force equals the mean driving force. Leading harmonic power is reduced by $[1+(\omega \lambda {)}^2{]}^{-1}$, as elastic storage changes the in-phase dissipative response.

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

An objective time derivative transforms covariantly under time-dependent rigid changes of observer, so constitutive predictions do not depend on the observer's rotation.

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.

With velocity gradient $A_{ij}={\partial }_i{v}_j$, set $\Omega =(A-A^T)/2$. A vector rigidly rotating with the material has material derivative $D_t\mathbf{p}=-\Omega \mathbf{p}$, so its corotational derivative vanishes. Its unit rotational coefficient is fixed by an objective time derivative; adding $-\xi D\mathbf{p}$ describes the material's flow alignment of a polar order parameter.

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 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.

The lower-convected derivative of a covariant tensor is $\stackrel{△}{\mathbf{C}}=D\mathbf{C}/Dt+(\nabla \mathbf{u}{)}^T\mathbf{C}+\mathbf{C}\nabla \mathbf{u}$.

For $L=\dot{F}{F}^{-1}$, the product rule proves the displayed lower-convected derivative identity. Under a superposed rotation $Q(t)$, ${\tau }^{\ast }=Q\tau Q^T$ and $L^{\ast }=QL{Q}^T+\dot{Q}{Q}^T$. The two rotation terms in ${\dot{\tau }}^{\ast }$ cancel those in $(L^{\ast }{)}^T{\tau }^{\ast }+{\tau }^{\ast }{L}^{\ast }$, proving that the lower-convected rate is an objective time derivative.

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

If an orthogonal matrix $Q$ satisfies $\dot{Q}=WQ$ for the spin tensor $W^T=-W$, the Jaumann derivative becomes an ordinary derivative in that frame:This follows by differentiating all three factors and using ${\dot{Q}}^T=-Q^{T}W$. It converts a tensor relaxation law using the Jaumann derivative into a componentwise ordinary differential equation.

The corotational Maxwell fluid replaces the ordinary derivative of the linear Maxwell fluid by the Jaumann derivative:Here $\omega =G-G^T$ uses $G_{ij}={\partial }_i{u}_j$, or equivalently the commutator is $-W\sigma +\sigma W$ with the standard spin tensor. The stress has relaxation time $\lambda$ and small-rate viscosity $\mu$. The commutator makes the constitutive equation nonlinear while maintaining an objective time derivative.

For simple shear flow $u=(\dot{\gamma }y,0,0)$, put $s=\dot{\gamma }$ and $\ell =\lambda s$. The steady deviatoric stress isThis follows by setting ${\omega }_{xy}=-s$, ${\omega }_{yx}=s$ in the corotational Maxwell fluid equation: ${\sigma }_{xx}=\lambda s{\sigma }_{xy}$, ${\sigma }_{yy}=-\lambda s{\sigma }_{xy}$, and ${\sigma }_{xy}+\lambda s({\sigma }_{xx}-{\sigma }_{yy})/2=\mu s$. Hence the shear viscosity is $\mu /(1+{\lambda }^2{s}^2)$, which exhibits shear thinning, while $N_1=2\mu \lambda s^2/(1+{\lambda }^2{s}^2)$ and $N_2=-N_1/2$. These differences are independent of the isotropic pressure.

Use the derivative-index-first velocity gradient $G_{ij}={\partial }_i{u}_j$ and $\omega =G-G^T$. For an objective time derivative the tensor transformation $A^{\ast }=QA{Q}^T$ under $x^{\ast }=Q(t)x+c(t)$ must imply $\mathcal{D}{A}^{\ast }/\mathcal{D}t=Q(\mathcal{D}A/\mathcal{D}t)Q^T$. If $R=\dot{Q}{Q}^T$, then $G^{\ast }=QG{Q}^T-R$ andConsequently the extra terms cancel in $D_{t}A+\frac{1}{2}(\omega A-A\omega )$, proving covariance. With the component-index-first gradient $L_{ij}={\partial }_j{u}_i$, the same physical rate is $D_{t}A-WA+AW$, with spin tensor $W=(L-L^T)/2=-\omega /2$. Changing the gradient convention without changing the commutator sign destroys objectivity.

The Oldroyd-B model combines a Newtonian solvent with infinitely extensible Hookean polymer dumbbells. It predicts constant shear viscosity, a positive first normal-stress difference, and an extensional catastrophe at a finite extension rate.

In an inertialess cone-and-plate rheometer, reverse a steady torque from T to -T at time zero. The polymer shear stress remains continuous while the positive solvent viscosity allows the rate to jump. The linear stress equation gives ${\lambda }_2={\mu }_0\tau /({\mu }_0+G_0\tau )$ and the displayed velocity for t>0, where ${\Omega }_{+}=3\alpha T/[2\pi a^3({\mu }_0+G_0\tau )]$. The transient backward speed exceeds the final backward speed because stored polymer stress initially retains its former sign.

For circular Couette flow with $v=Ar+B/r$, the rotating-frame reduction of circular viscoelastic shear gives shear rate $g=-2B/r^2$. The total shear stress is $\mu g$, and the azimuthal minus radial normal-stress difference is $2(\mu -{\mu }_r)\tau g^2$ in the stress convention where ${\sigma }_{rr}=-p$. Radial force balance without inertia is $p^{\prime }=-2(\mu -{\mu }_r)\tau g^2/r$, which integrates to the displayed pressure. The pressure constant is undetermined by these equations.

The FENE-P model replaces infinitely extensible Hookean polymers by finitely extensible nonlinear elastic springs with a mean-field closure. Finite extensibility regularizes the Oldroyd-B extensional catastrophe and produces shear thinning.

The Oldroyd-A model is the lower-convected counterpart of the Oldroyd-B model: its polymeric stress evolves using a lower-convected derivative.

For constant simple shear flow rate $s$, the nonpressure stress in the component-first gradient convention $L_{12}=s$ isThus the first normal-stress difference is $N_1=2(\mu -{\mu }_r)\tau s^2$, while the second is $N_2=-N_1$. Adding the incompressibility pressure changes neither difference.

Put $S={\sigma }^d-2{\mu }_{r}D$ in the Oldroyd-A model. It obeys ${\mathcal{D}}_{l}S+S/\tau =2(\mu -{\mu }_r)D/\tau$. Pulling back as $F^{T}SF$ converts this to an ordinary linear relaxation equation. With decayed remote-past initial memory, its solution isAt a finite initial time, include the homogeneous transported term $e^{-(t-t_0)/\tau }F(t{)}^{-T}F(t_0{)}^{T}S(t_0)F(t_0)F(t{)}^{-1}$. The integral-only law is not equivalent to all arbitrary initial-value solutions without a memory condition.

A second-order fluid is the quadratic, weak-flow expansion of an isotropic simple-fluid constitutive equation. In steady simple shear it predicts first and second normal-stress differences $N_1={\Psi }_1{\dot{\gamma }}^2$ and $N_2={\Psi }_2{\dot{\gamma }}^2$.

For a sphere of radius $a$ rotating with angular velocity $\Omega$, the primary Stokes flow has $v_{\varphi }=K\sin \theta /r^2$, $K=\Omega a^3$. In the weak, inertialess second-order fluid correction, put $c={\psi }_1+2{\psi }_2$. The stream function in the title hasBoth meridional components vanish at $r=a$ and decay at infinity. Indeed, $E_{r\varphi }=-3K\sin \theta /(2r^3)$ gives $[\nabla \times \nabla \cdot (2c{E}^2){]}_{\varphi }=72c{K}^2\sin \theta \cos \theta /r^8$, and the curl of the momentum equation reduces to $(d^2/d{r}^2-6/r^2{)}^{2}H=72c{K}^2/(\mu r^7)$. For $c>0$ the secondary flow enters towards the equator and leaves near the poles; $c<0$ reverses it and $c=0$ removes this correction. The meridional circulation is even in the sign of sphere rotation. Inertial secondary flow is excluded from this calculation.

Write the non-Newtonian extra stress as $S=2({\psi }_1+2{\psi }_2)E^2-{\psi }_1\stackrel{\circ }{E}$, with the Jaumann derivative. For a steady Stokes flow, $\nabla \times \nabla \cdot \stackrel{\circ }{E}=0$. If ${\psi }_1+2{\psi }_2=0$, the extra force is a gradient and only alters pressure. In a planar incompressible flow, $E^2=q\mathrm{diag}(1,1,0)$, so its divergence is also a gradient; no coefficient restriction is needed. A Newtonian velocity with prescribed boundary velocities consequently solves the second-order model with an altered pressure. On the regular weak-flow branch, Uniqueness of Stokes flow makes each velocity correction vanish. This does not establish uniqueness of arbitrary large nonlinear branches or the absence of higher-order effects in a real material.

The Johnson--Segalman--Oldroyd model is a linear relaxation-retardation equation made nonlinear through a Gordon--Schowalter objective derivative. Its slip parameter can produce a rate-dependent steady shear viscosity.

In uniaxial extensional flow, the conformation equations give $A_{xx}=1/(1-2\alpha \tau \dot{ϵ})$ and $A_{yy}=A_{zz}=1/(1+\alpha \tau \dot{ϵ})$. Their stress difference gives the displayed extensional viscosity. A stable positive steady conformation requires both denominators positive; a first pole marks loss of a bounded steady response. At zero extension rate the Trouton ratio is three relative to the zero-shear viscosity.

For stress $-pI+\alpha G_{0}A+2{\mu }_{0}E$ and slip derivative $D_{t}A+\Omega A-A\Omega -\alpha (EA+AE)=-(A-I)/\tau$, the steady viscometric functions are $\mu ={\mu }_0+{\alpha }^2{G}_0\tau /D$, ${\psi }_1=2{\alpha }^2{G}_0{\tau }^2/D$ and ${\psi }_2=-(1-\alpha ){\psi }_1/2$. Solving the three symmetric xy conformation equations gives $A_{xy}=\alpha \tau \dot{\gamma }/D$, $A_{xx}=1+(1+\alpha )\tau \dot{\gamma }{A}_{xy}$, $A_{yy}=1-(1-\alpha )\tau \dot{\gamma }{A}_{xy}$ and $A_{zz}=1$.

For nonzero slip parameter with $|\alpha |<1$, scale shear rate as $x=\tau \sqrt{1-{\alpha }^2}\dot{\gamma }$ and stress as $S=x[b+1/(1+x^2)]$, where $b={\mu }_0/({\alpha }^2{G}_0\tau )$. The derivative is $b+(1-x^2)/(1+x^2{)}^2$ and its least value is $b-1/8$, attained at $x^2=3$. A decreasing homogeneous constitutive segment therefore exists exactly under the displayed condition. It permits different shear rates at the same stress and motivates shear banding; the local curve alone does not select a band-interface stress.

For a tensor $\mathbf{\sigma }$, the Gordon--Schowalter derivative used here iswhich interpolates among objective convected derivatives as the dimensionless slip parameter $a$ changes.

The retardation time controls the delayed strain-rate term in an Oldroyd constitutive equation. Together with the relaxation time it determines the solvent-like high-frequency or high-rate response.

In small-amplitude oscillatory shear,

For Gordon--Schowalter parameter $0<a<2$, the steady shear viscosity isIt is shear thinning when ${\lambda }_1>{\lambda }_2$.

In simple shear, the first and second normal-stress differences are $N_1={\tau }_{xx}-{\tau }_{yy}$ and $N_2={\tau }_{yy}-{\tau }_{zz}$.

A polymeric extrudate can expand sideways after leaving a die as the normal-stress differences and stored deformation generated upstream relax. Axial elastic recovery contributes to transverse swelling in an incompressible flow. A Newtonian extrudate can also swell through rearrangement of its velocity profile, so observing swelling alone does not isolate the elastic contribution.

A rotating rod can draw a liquid with appreciable viscoelasticity inward and raise its free surface around the rod. Tensile normal stresses along curved flow lines produce an inward hoop force, competing with inertia and gravity. It is a manifestation of elastic normal-stress differences, rather than merely a changed scalar shear viscosity.

The uniaxial extensional viscosity is the tensile normal-stress difference divided by the imposed extension rate.

An incompressible uniaxial extensional flow has velocity $\mathbf{u}=\dot{ϵ}(-x/2,-y/2,z)$. Its principal deformation rates are $-\dot{ϵ}/2,-\dot{ϵ}/2,\dot{ϵ}$.

A slender axisymmetric column of a Newtonian fluid, with radius $a(z,t)$, axial velocity $w$, external pressure $p_e$ and axial body-force density $f$, obeys $(a^2{)}_t+(a^{2}w{)}_z=0$ and $3\mu (a^2{w}_z{)}_z/a^2=p_{e,z}-f$ when inertia, surface tension and leading external shear are negligible. The factor three is the Trouton ratio. Radial mass conservation gives $u_r=-r{w}_z/2$; the Newtonian fluid stress tensor gives $p=p_e-\mu w_z$ and axial excess stress $3\mu w_z$. This reduces a three-dimensional Stokes flow to one-dimensional evolving geometry.

In a closed-base cylindrical container, mass conservation makes the annular volume flux opposite to the core flux. With $h=R-a\ll a$ and annular viscosity $\lambda \mu$, Couette-Poiseuille flow in a thin gap gives $q_g=wh/2-h^3{P}_z/(12\lambda \mu )\simeq -aw/2$, where $P$ removes the annular hydrostatic pressure. Hence $P_z=6\lambda \mu aw/h^3[1+O(h/a)]$. The resulting column equation is $(a^2{w}_z{)}_z/a^2=\Delta \rho g/(3\mu )+2\lambda aw/h^3$. Curvature and the Couette flux change only the neglected relative order $h/a$.

For a uniform initial column, annular return flow around a slumping column competes with axial extensional stress over $\hat{z}=\sqrt{h_0^3/(2\lambda a_0)}$. With $G=\Delta \rho g/(3\mu )$, choose velocity $\hat{w}=G{\hat{z}}^2$ and time $\hat{t}=\hat{z}/\hat{w}$. Equivalently $\hat{z}=R/\sqrt{\lambda {\alpha }_0}$, where $\varphi =a^2/R^2$ and $\alpha (\varphi )=2\sqrt{\varphi }/(1-\sqrt{\varphi }{)}^3$ in the leading thin-annulus model. This length describes the vertical range of the basal constraint: a much taller column has a localized basal boundary layer, while a much shorter column responds predominantly through extension.

Using the extensional screening length of a slumping column, an initially uniform column of length $\Lambda$ obeys $W_{ZZ}=1+W$, $W(0)=0$ and $W_Z(\Lambda )=0$. Thus $W=\cosh (\Lambda -Z)/\cosh \Lambda -1$ and its initial area-fraction growth is ${\varphi }_T={\varphi }_0\sinh (\Lambda -Z)/\cosh \Lambda$. For $\Lambda \gg 1$, the velocity is $e^{-Z}-1$ outside exponentially small end corrections; for $\Lambda \ll 1$, it is $Z^2/2-\Lambda Z$. The first limit balances excess weight with annular hydraulic drag, the second with extensional viscous stress.

A column of viscosity $\mu$ in an annulus of viscosity $\lambda \mu$ and gap $h\ll a$ drives return flow of speed $O(a|w|/h)$ in a closed container. Annular shear scales as $\lambda \mu a|w|/h^2$, inducing relative core-velocity variation $O(\lambda a^2/h^2)$. Thus the extensional equations for a slender Newtonian column admit a leading plug-like axial velocity when $\lambda \ll h^2/a^2$. The small viscosity ratio alone is insufficient when the gap is very narrow.

The axial strain rate is the axial component of the rate-of-strain tensor. For axial velocity $w(z)$ it is $E=\partial w/\partial z$. In a uniform uniaxial extensional flow, $w=Ez$ and incompressible flow requires $u_r=-Er/2$; positive $E$ stretches the axial direction and contracts the transverse directions.