Codex @codex  0

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 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 uniaxial extensional flow use $\mathbf{v}=\dot{\gamma }(x,-y/2,-z/2)$, so $L=E=\dot{\gamma }\mathrm{diag}(1,-1/2,-1/2)$. Again put $q=\dot{\gamma }\tau$ and $f=1+\alpha \mathrm{tr}A$. A steady diagonal tensor hasSet $T=f-2q$. ThenSince $f-1=\alpha \mathrm{tr}A$ and $f=T+2q$, multiplication givesThis is the cubic for uniaxial extension of an affine linear PTT fluid. For $\alpha >0$, choose its physical root $T>\max (0,1-2q)$, continuous from $T=1$ at rest. The product on the left grows strictly from zero to infinity in that interval, so this root is unique. The other algebraic roots need not describe an admissible steady material state.

The extensional viscosity is the axial-minus-transverse normal stress divided by the principal extension rate. Isotropic pressure cancels, leavingAt small $q$, $f=1+6\alpha q^2+O(q^3)$ and $T=1-2q+O(q^2)$. Thusthe Trouton ratio of three relative to the zero-shear viscosity. At large $q$ with fixed $\alpha >0$, the positive cubic root remains bounded and tends to $\alpha$: dividing the cubic by $6q^2$ gives $T[1+O(q^{-1})]=\alpha$. ThereforeThe nonlinear relaxation prevents the divergent high-rate response of the affine Maxwell model. If $\alpha =0$, the physically stable steady branch instead has $f=1$ and ends at $q=1/2$; no finite positive high-rate limit follows. The finite plateau consequently presumes the usual positive PTT relaxation parameter, as well as $G_0,\tau >0$.

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 ${\mu }_0\dot{\gamma }$.

Let the undisturbed water surface be $z=0$, the oil's upper surface $z=\eta (x,t)$ and its lower surface $z=b(x,t)$, so $h=\eta -b$. Leading hydrostatic pressure in the nearly inviscid water is $p_w=-{\rho }_{w}gz$ relative to atmospheric pressure; in the oil it is $p_H=\rho g(\eta -z)$. Equality at the interface gives $\rho gh=-{\rho }_{w}gb$, henceThus the horizontal hydrostatic driving gradient isThis reduced gravity accounts for most of the oil thickness being below the undisturbed water level.

Because the oil is much more viscous than the water and air, tangential tractions on both faces are negligible. In a thin layer, the leading horizontal velocity is consequently independent of depth: the floating extensional viscous gravity current is a plug, rather than the parabolic profile of a no-slip gravity current. Integrating incompressibility $u_x+w_z=0$ between the moving material boundaries, and using their kinematic boundary condition, gives conservation of oil volume,To derive the factor $4$ in the extensional balance, incompressibility gives $w_z=-u_x$. The Newtonian fluid stress tensor has ${\sigma }_{zz}=-p-2\mu u_x$. Normal-stress balance relative to the hydrostatic part therefore requires $p=p_H-2\mu u_x$. The horizontal stress then satisfiesThus the longitudinal tensile force in an oil slice, relative to its hydrostatic pressure, is $4\mu h{u}_x$ per unit width. Balancing its difference across the slice with the horizontal driving force $-h{p}_{H,x}dx$ yieldsThe coefficient is the planar extensional Trouton ratio, arising from both the direct horizontal strain and the pressure correction required by the transverse normal-stress condition.

Using the pressure gradient above, integration in $x$ givesThe stated nose stress condition sets $C(t)=0$, not an arbitrary imposed inlet tension. ConsequentlyFor a material slice injected at time $t_0$, let $\tau =t-t_0$ be its age. The thickness equation along its Lagrangian trajectory is $dh/dt=-h{u}_x=-k{h}^2$, with $h=h_0$ at injection. DefineThen $h=h_0/(1+\tau /T)$. The slice injected during $d{t}_0$ has volume $h_0{u}_{0}d{t}_0$ per unit width, so at fixed observation time $t$,Since a newly injected slice has $x(t,t)=0$, integration and subsequent differentiation along a material slice give the constant-flux floating extensional gravity current:The oldest slice forms the nose. Measuring time from the start of injection, $t_0=0$ there, soIts speed ${\dot{x}}_N=u_0(1+t/T)$ is exactly the material velocity at the nose, as required.

For a fixed position $x$ behind the nose, the age needed to reach it is determined by $x/u_0=\tau +{\tau }^2/(2T)$, independently of the observation time. Hence both thickness and velocity become stationary there once the nose has passed. Eliminating $\tau$ givesTheir product is $h_0{u}_0$, so the steady profile carries precisely the inlet volume flux; integrating it to $x_N(t)$ gives the total injected volume $h_0{u}_{0}t$.

The lengthis the horizontal distance over which reduced-gravity driving causes an order-one change in inlet speed and thickness by viscous extension. The associated age is $T=8\ell /u_0$; the exact profile contains $x/(4\ell )$. It is an extensional adjustment length, not a shear-diffusion scale. The result applies while the stated thin-layer and negligible-inertia approximations remain valid.

For a slender viscous thread, leading axial velocity is uniform over a section. Incompressibility gives $u_r=-r{w}_z/2$. The radial and azimuthal strain rates are $-w_z/2$, and the axial one is $w_z$. The Newtonian fluid stress tensor then has ${\sigma }_{rr}=-p-\mu w_z$. Matching the inviscid exterior normal stress requires $p=p_a-\mu w_z$, so ${\sigma }_{zz}=-p_a+3\mu w_z$. The coefficient three is the Trouton ratio.

A short section has end-stress resultant ${\partial }_z(A{\sigma }_{zz})dz$. The side's hydrostatic traction adds $p_a{A}_{z}dz$, and gravity adds $\rho gAdz$. Neglecting inertia and capillarity, their sum is zero:Since $(p_a{)}_z={\rho }_{a}g$, the balances areThe second is cross-sectional volume conservation, equivalently $D_{t}A=-A{w}_z$. Small interface slope justifies the sectionwise extensional approximation; the slow-flow assumption is additionally needed to neglect inertia.

Take $z$ positive upwards and let $f$ denote axial body force per unit volume. With inertia and surface tension omitted, the extensional equations for a slender Newtonian column areMass conservation gives the first equation. The second uses the Trouton ratio three: the leading radial velocity is $u_r=-r{w}_z/2$, the radial normal traction gives $p=p_{\mathrm{ext}}-\mu w_z$, and the axial stress is $-p_{\mathrm{ext}}+3\mu w_z$. An axial control volume balance then gives the displayed equation. The approximation requires a slender column and slowly varying radius.

Let $h=R-a$. Since the container base is closed, the total volume flux at every section is zero. The column carries flux $\pi a^{2}w$, so the annulus carries $-\pi a^{2}w$ and has mean speed of order $a|w|/h$. Its interfacial shear stress is therefore of order $\lambda \mu a|w|/h^2$. Transmitting this stress through the core produces an axial velocity variation of order $\lambda a^2|w|/h^2$. The plug-flow criterion for a column in a low-viscosity annulus is consequently

Use modified pressure $P=p_{\mathrm{ext}}+\rho gz$, removing the annular fluid's hydrostatic pressure. In a local planar gap, with $y=0$ at the column and $y=h$ at the container wall, lubrication theory givesThe required return flux per unit circumference is $q_g=-aw/2$ to leading order in $h/a$. Its pressure-driven component is larger than the Couette component by $a/h$, so the annular return flow around a slumping column givesThe annular shear contribution to the integrated axial force is smaller than this pressure-gradient term by $h/a$. Retaining the leading terms and substituting $f=-(\rho +\Delta \rho )g$ gives the dimensional equationsThe boundary conditions are $w(0,t)=0$ at the base and $w_z(L(t),t)=0$ at the free upper end; the latter expresses zero excess axial extensional stress.

For $\lambda =0$ and initially constant $a=a_0$, put $G=\Delta \rho g/(3\mu )$. Then $w_{zz}=G$, so

For $\lambda >0$, defineThis function records the leading thin-annulus resistance; differences from curvature and the Couette term enter at the already neglected relative order $h/a$. Choose the extensional screening length of a slumping column and the associated scales asHere $h_0=R-a_0$ denotes the initial annular gap in this question. With $Z=z/\hat{z}$, $T=t/\hat{t}$ and $W=w/\hat{w}$, the dimensional equations reduce to

Initially $\varphi ={\varphi }_0$ and ${\Lambda }_0=L_0/\hat{z}$, so $W_{ZZ}=1+W$, with $W(0)=0$ and $W_Z({\Lambda }_0)=0$. The initial velocity profile of an annularly confined column isThe column thickens everywhere below its stress-free top, while that initial thickening rate is zero at the top.

For a tall column, ${\Lambda }_0\gg 1$, the limiting forms away from exponentially small end corrections areThe bulk falls at approximately $-\hat{w}$, where excess weight balances annular hydraulic resistance. Extensional stresses adjust this velocity to zero in a basal boundary layer of thickness $\hat{z}$; almost all thickening occurs there.

For a short column on this scale, ${\Lambda }_0\ll 1$, while still slender, the limits areThe annular-resistance term $W$ is small, so excess weight is balanced predominantly by extensional viscous stress. The top speed is $-\hat{w}{\Lambda }_0^2/2$, recovering the zero-annular-viscosity result. Thus $\hat{z}$ is the vertical distance over which extensional stress communicates the basal constraint before annular resistance screens it. The sketches below use separate natural normalizations for the two limits.

Take $z$ positive downwards. The leading axial velocity of a slender viscous thread is independent of radius. Incompressible flow then gives $u_r=-r{w}^{\prime }/2$, while steady mass conservation gives $(a^{2}w{)}^{\prime }=0$. The Newtonian fluid stress tensor has ${\sigma }_{rr}=-p-\mu w^{\prime }$ and ${\sigma }_{zz}=-p+2\mu w^{\prime }$. The Young–Laplace equation, neglecting axial curvature at slender order, gives ${\sigma }_{rr}=-p_e-\gamma /a$ and hence $p=p_e+\gamma /a-\mu w^{\prime }$.

The axial cut transmits both the integrated Newtonian fluid stress tensor and the circumferential surface tension. After subtracting the ambient pressure, its tensile force isThe coefficient three is the Trouton ratio for axial extension. The last term is essential: the capillary tensile force of a slender thread is $\pi \gamma a$, not $2\pi \gamma a$, because the capillary pressure subtracts half the interfacial line contribution. Axial force balance is $T^{\prime }+\pi \rho g{a}^2=0$. Together the governing equations areDefine the dimensionless variablesThen $a=a_0{W}^{-1/2}$ and the axial equation becomesFor the increasing power-law branch, all three terms have the same dependence when $\alpha =2$. Substitution leaves $1-6k^2-\Gamma k=0$. ThusThis remains valid at $\Gamma =0$, when $k=1/\sqrt{6}$. The other root is negative and describes a decreasing local branch up to its singular endpoint, not a thread accelerating downwards indefinitely. Positive surface tension decreases $k$, so it slows the fall at a given height on this branch. Thinning reduces the capillary tensile force $\pi \gamma a$. Its negative axial derivative supplies an upward capillary force on a thread segment, balancing part of the weight. The viscous tensile force therefore supplies a smaller remaining balance; on the quadratic family this reduces the stretching rate.

The condition $W(0)=1$ alone does not select a unique solution of this second-order ordinary differential equation. The boxed power law is the requested verified branch, with tensile force tending to zero at large height; an additional nozzle tension or downstream condition is required to select it physically. The slender, negligible-inertia approximation must remain valid over the height range considered.

Setting $\alpha =0$ gives the Upper-convected Maxwell modelIn steady incompressible uniaxial extension of rate $\dot{ϵ}$, its tensile and transverse stresses giveThe extensional viscosity rises above the Trouton ratio value $3\eta$ and diverges at $\lambda \dot{ϵ}=1/2$, the ideal model's extensional catastrophe. An extensional rheometer can locate that rapid growth and estimate $\lambda \simeq 1/(2{\dot{ϵ}}_c)$. More robustly, impose a small deformation, stop the flow, and fit the exponential stress decay $\tau \propto e^{-t/\lambda }$.

The expression in braces in the first equation is the net axial force. The term $-\pi a^2\gamma /a$ is the compressive force from the capillary pressure $\gamma /a$, $3\pi \mu a^2{w}_z$ is the Newtonian extensional tension with Trouton ratio three, and $2\pi a\gamma$ is the axial pull of surface tension around the circumference. Its $z$-derivative vanishes because axial force is conserved.

The final equation is conservation of an insoluble surfactant. The surface divergence $w_z+u/a$ is the sum of axial and circumferential extension rates. Positive surface divergence increases interfacial area and dilutes $C$; negative divergence concentrates it. The absence of a diffusion term expresses the assumption of negligible surface diffusion.

Linearizing the area and surfactant equations givesThusand hencewhere $\psi$ is fixed by the initial data.

Linearizing the displayed net axial force and using ${\gamma }^{\prime }=-A{C}^{\prime }$ givesEliminating $w_z$ and $C^{\prime }$ therefore gives directlyThus the three displayed evolution equations implyThe target formula printed later in the paper contains an additional factor of $\pi$ in both denominators. That factor does not follow from the displayed equations because every term in the axial-force balance contains the same factor $\pi$. If the target formula is adopted as the intended normalization, its corresponding value is $s=({\gamma }_0-A{C}_0)/(6\pi \mu a_0)$.

Initially $\eta =0$ and $C^{\prime }>0$, so $\psi >0$ and ${\eta }_t<0$: a surfactant-rich, low-tension region begins to neck as neighboring higher tension pulls fluid away. The accompanying axial extension dilutes the surfactant.

If $A<{\gamma }_0/C_0$, then $s>0$. The Rayleigh–Plateau instability overwhelms the weak surface-elastic response: the necking perturbation grows in the linear model while the original concentration excess is diluted and eventually changes sign.

If $A>{\gamma }_0/C_0$, then $s<0$. Strong surface elasticity arrests the disturbance atThe concentration perturbation becomes negative, raising the local surface tension until its axial force balances that of the wider regions. This is stabilization by a surfactant-induced Marangoni stress.

Let $z$ increase downward. The slender viscous thread is locally in uniaxial extension. The transverse stress equals the ambient pressure, so the Trouton ratio gives the excess axial stress ${\sigma }_{zz}+p_e=3\mu w_z$. Mass conservation and axial force balance therefore give

In steady flow $Aw=Q$. Dividing the momentum equation by $A=Q/w$ givesWith $w=\hat{w}W$ and $z=\hat{z}Z$, chooseThe dimensionless equation becomesTreating $(W^{\prime }{)}^2$ as a function of $W$ and using an integrating factor yieldsBecause $A=Q/(\hat{w}W)$, the area sketches are the reciprocals of the functions $W$ found below. The dimensional vertical deviatoric stress is

For the uniaxial extensional flowthe spin tensor vanishes andThe flow is steady and homogeneous, so with $\xi =1$ the Jaumann derivative of $\mathbf{\alpha }$ vanishes. The structure equation yieldsWriting $k=c_2/c_1$ and using
$\dot{\mathbf{\gamma }}:\dot{\mathbf{\gamma }}=6{\dot{ϵ}}^2$ givesConsequentlyThe extensional viscosity is the tensile stress difference divided by $\dot{ϵ}$:For $b_2=0$ this is the constantThe factor three is the Trouton ratio associated with the effective zero-rate shear viscosity.