---
title: "How finite chain length caps stretch and extensional viscosity in the FENE-P model"
description: "Give the dumbbell spring a maximum length and find the FENE-P ceiling on extensional viscosity, shear thinning and the energy balance."
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language: "en"
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last_modified: "2026-10-29"
---

[Complex fluids](https://sandboxphysics.com/stories/complex-fluids/)

L3 · Chapter 4

# How finite chain length caps stretch and extensional viscosity in the FENE-P model

Sandbox Physics · 2026-10-29

A linear spring can stretch forever, which is why the Oldroyd-B extensional stress grows without bound above the threshold. A real chain has a contour length, and its spring becomes very stiff close to full extension. The FENE-P model writes this into the conformation equation: the closer the stretch is to the limit, the faster the recovery. The extensional viscosity then stops at a finite plateau, the shear viscosity falls with shear rate, and an exact balance still links elastic energy, work and dissipation.

## Before you begin

The conformation equation and polymer stress of Chapter 2 and the planar-extension solution of Chapter 3; solving a one-variable equation by bisection; trace, determinant and eigenvalues of a matrix.

- Write the FENE-P conformation equation and find the steady planar-extension solution and the ceiling on extensional viscosity.
- Derive the FENE-P elastic free energy and dissipation, and use work equals storage plus dissipation to check a numerical solution.

## A spring with a maximum length

The end-to-end distance of a real chain cannot exceed its contour length, and the tension of a FENE spring diverges near that length. The Peterlin approximation replaces the length-dependent factor in the tension by its value at the mean conformation, giving a factor $f$ that depends only on $\operatorname{tr}\mathbf A$. $L^2$ is the upper limit of $\operatorname{tr}\mathbf A$, the squared maximum extension.

At equilibrium $\operatorname{tr}\mathbf A=3$ and $f=1$, so the model coincides with Oldroyd-B. As $\operatorname{tr}\mathbf A$ approaches $L^2$, $f$ diverges and the recovery rate $f/\lambda$ can match any stretch rate. For $L^2\to\infty$ the model returns to Oldroyd-B.

$$
\overset{\nabla}{\mathbf A}=-\frac{1}{\lambda}\left(f\mathbf A-\mathbf I\right),\qquad \boldsymbol\tau_p=G\left(f\mathbf A-\mathbf I\right),\qquad f=\frac{L^2-3}{L^2-\operatorname{tr}\mathbf A}
$$

Eqs. (1.86)–(1.87) of Morozov and Spagnolie write the same model in terms of stress, with the factor $1+\lambda\operatorname{tr}\boldsymbol\tau_p/(\eta_pL^2)$; inserting the stress–conformation relation gives this form. Table II of Datta and co-authors uses the form shown here.

Source: A. Morozov and S. E. Spagnolie · S. S. Datta et al.

## The extensional viscosity reaches a plateau

In steady planar extension the components separate as in Chapter 3, except that the relaxation rate is multiplied by $f$. Each diagonal component is written in terms of $f$, but $f$ depends on their sum, so one self-consistent equation remains; bisection over $f>2\mathrm{Wi}$ finds the root.

Under strong stretching $A_{xx}$ is large, so the denominator $f-2\mathrm{Wi}$ must be small and $f$ sits just above $2\mathrm{Wi}$: the spring stiffens until its recovery rate exactly cancels the stretch rate. With the identity $f(A_{xx}-A_{yy})=2\mathrm{Wi}(A_{xx}+A_{yy})$ the extensional viscosity takes the form below. At large $\mathrm{Wi}$, $A_{xx}$ approaches $L^2$ and the extensional viscosity approaches its ceiling.

Eq. (7) of McKinley gives the ceiling on the tensile stress in uniaxial extension, $2\eta_p\dot\varepsilon L^2$, and Section 5.4 the matching ceiling on the extensional viscosity, $3\eta_s+2\eta_pL^2$. In planar extension the 3 becomes 4.

$$
A_{xx}=\frac{1}{f-2\mathrm{Wi}},\quad A_{yy}=\frac{1}{f+2\mathrm{Wi}},\quad A_{zz}=\frac1f,\qquad \eta_P=4\eta_s+2\eta_p\left(A_{xx}+A_{yy}\right)\xrightarrow{\ \mathrm{Wi}\to\infty\ }4\eta_s+2\eta_pL^2
$$

In the steady state $f$ is fixed by $f\left(L^2-\operatorname{tr}\mathbf A\right)=L^2-3$.

Source: G. H. McKinley · A. Morozov and S. E. Spagnolie

## Shear thinning

In steady shear $A_{yy}=1/f$ and the polymer shear stress is $G\,\mathrm{Wi}/f$, so the shear viscosity is $\eta_s+\eta_p/f$. Faster shear means longer coils, a larger $f$ and a lower viscosity. The Oldroyd-B shear viscosity is constant; the FENE-P shear viscosity falls with shear rate.

Inserting the steady shear components into the self-consistency condition gives a cubic for $f$. The larger $L^2$, the closer $f$ stays to one at a given $\mathrm{Wi}$, and the weaker the thinning.

$$
\eta=\eta_s+\frac{\eta_p}{f},\qquad f^2\left(f-1\right)=\frac{2\,\mathrm{Wi}^2}{L^2}
$$

Obtained by inserting the steady shear values $A_{yy}=1/f$, $A_{xy}=\mathrm{Wi}/f^2$ and $A_{xx}=1/f+2\mathrm{Wi}^2/f^3$ into the self-consistency condition.

Source: A. Morozov and S. E. Spagnolie

## Elastic energy, work and dissipation

FENE-P has an elastic free energy $\psi$ that is zero at equilibrium and positive for every other conformation. Following the fluid, its rate of change equals the power the flow delivers to the polymers, $\boldsymbol\tau_p:\mathbf L$, minus a dissipation $\mathcal D$ that is a sum of squares over the eigenvalues $a_i$ and never negative.

This balance gives an independent check on a numerical solution: accumulate the work, the stored energy and the dissipation over a run, and their mismatch should be close to zero. Eqs. (4)–(5) of Thomases, Shelley and Thiffeault write the same kind of balance for Oldroyd-B, using the integral of $\tfrac12\operatorname{tr}(\mathbf S-\mathbf I)$ as the energy.

$$
\psi=\frac{G}{2}\left[-(L^2-3)\ln\frac{L^2-\operatorname{tr}\mathbf A}{L^2-3}-\ln\det\mathbf A\right],\qquad \frac{D\psi}{Dt}=\boldsymbol\tau_p:\mathbf L-\mathcal D,\qquad \mathcal D=\frac{G}{2\lambda}\sum_i\frac{\left(fa_i-1\right)^2}{a_i}
$$

Derived for this course: differentiate $\psi$, insert the conformation equation and use incompressibility, $\operatorname{tr}\mathbf L=0$. For $L^2\to\infty$, $\psi$ becomes $\tfrac G2\left(\operatorname{tr}\mathbf A-3-\ln\det\mathbf A\right)$.

Source: B. Thomases

## What a numerical solution must preserve

The conformation tensor must stay positive definite, and FENE-P also requires $\operatorname{tr}\mathbf A<L^2$. Discretization error can push the smallest eigenvalue below zero or the trace past the limit, and from then on the stress has no physical meaning. Section 6 of Morozov and Spagnolie warns that Oldroyd-B simulations often fail at high Weissenberg number, known in the literature as the high-Weissenberg-number problem.

Three remedies are common. Solve for the matrix logarithm of the conformation tensor, the log-conformation scheme, which keeps it positive definite automatically. Add a little diffusion to the stress equation to smooth gradients that are too steep. Or switch to a bounded model such as FENE-P. The review by Datta and co-authors discusses the first and states that Oldroyd-B is almost never quantitative at high Weissenberg number. Chapter 7 shows what the second remedy can cost.

Source: A. Morozov and S. E. Spagnolie · S. S. Datta et al.

## Worked example: two chain lengths in planar extension

Take $\beta=0.5$ and $\mathrm{Wi}=3$. Oldroyd-B has no steady state here; the stretch grows exponentially. FENE-P with $L^2=400$ gives, by bisection, $f=6.003$, only 0.003 above $2\mathrm{Wi}=6$. Then $A_{xx}=333.6$, $A_{yy}=0.0833$, $A_{zz}=0.167$, and the trace is 333.9, 83.5% of the limit. The stretch is $\sqrt{333.9/3}=10.5$, against a largest possible value of $\sqrt{400/3}=11.5$.

The extensional viscosity is $\eta_P=4\times0.5+2\times0.5\times(333.6+0.083)=335.7\,\eta_0$, with a ceiling of $4\times0.5+400=402\,\eta_0$.

With $L^2=100$ instead: $f=6.012$, $A_{xx}=83.6$, a trace of 83.9 or 83.9% of the limit, a stretch of 5.29, and an extensional viscosity of $85.7\,\eta_0$ against a ceiling of $102\,\eta_0$. A quarter of the squared length gives roughly a quarter of the extensional viscosity, while the fraction of the limit in use barely changes.

Energy check: with $\lambda=1\,\mathrm{s}$ and $G=0.5\,\mathrm{Pa}$, at $L^2=400$ the flow delivers $(\tau_{xx}-\tau_{yy})\dot\varepsilon=3.00\times10^{3}\,\mathrm{W/m^3}$ to the polymers, and the dissipation computed from the eigenvalues is the same. In the steady state the elastic energy no longer changes and all of the work is dissipated.

Source: A. Morozov and S. E. Spagnolie

## Try it in the Lab

### [Pull the chains to full length](https://sandboxphysics.com/labs/polymer-stretch-startup/)

1. Choose the Chains hit full length preset (FENE-P, Weissenberg number 3, $L^2=100$). At the end of the run the coil stretch in extension should be close to the 5.29 of the worked example, and the largest share of the finite-extension bound used close to 0.84.
2. Change the maximum extension squared to 400 and predict the new stretch and extensional viscosity before comparing with the worked example. In the detailed readouts, check that the largest mismatch of work minus stored minus dissipated is far smaller than the work itself.
3. Choose the Stress puts on the brakes preset. Under an imposed stress, once the extensional viscosity grows, the same stress can only sustain a smaller stretch rate. Watch the extension-side Weissenberg number fall with time.

## Exercises with explained answers

1. A FENE-P solution has $\beta=0.1$ and $L^2=1000$. What is the ceiling on the planar extensional viscosity in units of the zero-shear viscosity, and how many times the Newtonian value is that?

   Show the answer

   The ceiling is $4\times0.1+2\times0.9\times1000=1800.4$ times the zero-shear viscosity. The Newtonian value is $4\eta_0$, so the ceiling is about 450 times larger.

2. Take the elastic energy in the Oldroyd-B limit and let $\mathbf A=\mathrm{diag}(a,1/a,1)$. Write $\psi$ and show that it is never negative.

   Show the answer

   $\operatorname{tr}\mathbf A-3=a+1/a-2$ and $\det\mathbf A=1$, so $\psi=\tfrac G2\left(a+1/a-2\right)$. By the inequality of arithmetic and geometric means, $a+1/a\ge2$, with equality only at $a=1$.

3. FENE-P with $\beta=0.5$ and $L^2=100$ in steady shear at $\mathrm{Wi}=3$. Estimate the shear viscosity.

   Show the answer

   Solve $f^2(f-1)=2\times9/100=0.18$ to get $f\approx1.14$. The shear viscosity is $0.5+0.5/1.14=0.94\,\eta_0$, only about 6% lower, while at the same Weissenberg number the extensional viscosity is 86 times the zero-shear viscosity.

4. Why is $f$ always just above $2\mathrm{Wi}$ under strong stretching?

   Show the answer

   In the steady state $A_{xx}=1/(f-2\mathrm{Wi})$. Under strong stretching $A_{xx}$ is close to $L^2$, a large number, so the denominator must be small. Physically, the spring keeps stiffening until its recovery rate $f/\lambda$ cancels the amplification $2\dot\varepsilon$ by the flow.

[Previous: Why extensional flow unravels polymer coils while shear only tumbles them](https://sandboxphysics.com/stories/complex-fluids/coil-stretch-in-extension/) [Next: Reading a relaxation time from the exponential thinning of a polymer filament](https://sandboxphysics.com/stories/complex-fluids/filament-thinning-and-extensional-rheometry/)

## Sources and what we recomputed

1. [A. Morozov and S. E. Spagnolie, Introduction to complex fluids, in Complex Fluids in Biological Systems (Springer, 2015), chapter 1](https://doi.org/10.1007/978-1-4939-2065-5_1)

   Section 3: the Weissenberg number defined as the relaxation time times the shear rate, and Eq. (1.27), the generalized Newtonian liquid. Section 4.1: Eqs. (1.47)–(1.51), the spring, the dashpot, the Maxwell liquid and its memory integral, with a relaxation time equal to viscosity over modulus. Eq. (1.72): the upper-convected derivative; Eq. (1.80): Oldroyd-B; Eqs. (1.86)–(1.87): FENE-P. Section 4.4: the Hookean dumbbell, Eqs. (1.98)–(1.110). Eq. (1.116): the steady-shear normal stress of Oldroyd-B. Eqs. (1.117)–(1.125): hoop stress and the cone-and-plate normal force. Eqs. (1.128)–(1.131): uniaxial extension and the Oldroyd-B divergence at half the inverse relaxation time. The chapter has no startup solution and no planar extension; the course derives both.

2. [G. H. McKinley, Visco-elasto-capillary thinning and break-up of complex fluids, MIT Hatsopoulos Microfluids Laboratory report 05-P-04, submitted to Annual Rheology Reviews (2005)](https://dspace.mit.edu/handle/1721.1/18085)

   Eqs. (1)–(2): strain rate and Hencky strain of a thinning thread. Eqs. (4)–(7): Oldroyd-B stress growth in uniaxial extension and the finite-extensibility cap. Eq. (8): elasto-capillary thinning, in which the radius falls exponentially with a time constant of three relaxation times, so the thread runs at a Weissenberg number of 2/3. Eq. (11): linear thinning once the extensional viscosity is constant. Eq. (15): the zero-dimensional force balance on the neck. Eq. (16): the Newtonian similarity solution with prefactor 0.0709; the text after it says the simple zero-dimensional estimate is off by more than 100 %. Section 5.4: the terminal extensional viscosity of a finitely extensible dumbbell. The front factor of Eq. (8) and the one listed in Table 2 differ by the cube root of two; Section 5.4 attributes this to the one-dimensional similarity solution. The course uses only the exponent.

3. [S. S. Datta et al., Perspectives on viscoelastic flow instabilities and elastic turbulence, arXiv:2108.09841v1 (2021); Physical Review Fluids 7, 080701 (2022)](https://arxiv.org/abs/2108.09841v1)

   arXiv version 1, 22 August 2021; the journal version was not checked. Section II, Eq. (1): the Pakdel–McKinley criterion, which combines streamline curvature with the ratio of normal stress to shear stress. No critical value is given for Taylor–Couette or cone-and-plate flow. Section III: simulations of cross-slot flows lose symmetry or become time dependent when the largest local value of that criterion exceeds about 4 to 5; Eq. (2) and Table II write Oldroyd-B, FENE-P and other models with a conformation tensor, which allows a positive-definiteness check and log-conformation schemes; a two-dimensional Taylor–Couette simulation became chaotic near a Weissenberg number of 10. The workshop summary states that Oldroyd-B is almost never quantitative at high Weissenberg number.

4. [B. Thomases, M. Shelley and J.-L. Thiffeault, A Stokesian viscoelastic flow: transition to oscillations and mixing, Physica D 240, 1602 (2011)](https://doi.org/10.1016/j.physd.2011.06.011)

   Eqs. (1)–(2) and (6): the same model with stress diffusion 1e-3 for Weissenberg numbers up to 10. Eqs. (4)–(5): the polymer strain energy balance. Section 3.1, Fig. 2: a small perturbation of the symmetric steady state decays at 4.8 and grows at 4.9; 256 by 256 grid, and doubling the resolution does not change the dynamics. Fig. 6(d): at 10 the velocity spectrum has periods of about 330 and 55. Section 5: between about 4.9 and 9 the asymmetric flow settles down again; above about 9 a vortex unpins and persistent oscillations mix the fluid; periodic at 12 and 15, aperiodic above about 20. Appendix B: stress diffusion is calibrated against FENE-P; its effect on the threshold is not reported.