L3 · Chapter 5

Reading a relaxation time from the exponential thinning of a polymer filament

Sandbox Physics · 2026-10-29

Stretch a drop between two small plates and surface tension squeezes the bridge in between ever thinner. A Newtonian neck thins at nearly constant speed and soon breaks. A dilute polymer solution instead draws out a thread whose radius falls exponentially and slowly, lasting several relaxation times. The exponential rate depends only on the relaxation time, which makes filament thinning one of the most direct ways to measure extensional rheology.

Surface tension squeezes the bridge

The thinnest part of the thread is close to a cylinder of radius $R(t)$ . Surface tension creates a pressure $\sigma/R$ inside it, pushing liquid out of the neck toward the ends. The axial strain rate at the neck follows from the rate of change of the radius, and the accumulated Hencky strain from the ratio of radii: halving the radius adds $2\ln2$ to the strain.

Nothing external controls the stretch rate in a thinning experiment. The rate is whatever balances surface tension against the stresses inside the liquid, so the same instrument takes different liquids through different stretching histories on its own.

\dot\varepsilon(t)=-\frac{2}{R}\frac{dR}{dt},\qquad \varepsilon(t)=2\ln\frac{R_0}{R(t)}

Eqs. (1) and (2) of McKinley.

Source: G. H. McKinley

A zero-dimensional force balance on the neck

Treat the neck as a uniformly stretched cylinder and neglect inertia and gravity. On the left is the viscous extensional stress of the solvent; the Newtonian extensional viscosity in uniaxial extension is $3\eta_s$ . On the right is the capillary pressure minus the axial polymer tension. The larger the polymer stress, the less driving force is left to deform the solvent, and the slower the thread thins.

The axial polymer stress difference $\tau_{zz}-\tau_{rr}$ comes from the conformation equation in uniaxial extension at the current rate $\dot\varepsilon$ . Together the two equations evolve the radius and the conformation in time.

3\eta_s\left(-\frac{2}{R}\frac{dR}{dt}\right)=\frac{\sigma}{R}-\left(\tau_{zz}-\tau_{rr}\right)

Eq. (15) of McKinley. The polymer stress may come from Oldroyd-B or FENE-P.

Source: G. H. McKinley

A Newtonian thread: the zero-dimensional model is off by more than double

Without polymers the zero-dimensional balance gives $dR/dt=-\sigma/(6\eta_s)$ : the radius falls at constant speed. A real Newtonian neck is not a uniform cylinder; it takes a self-similar shape, and the similarity solution of Papageorgiou, quoted by McKinley in Eq. (16), gives a thinning-speed coefficient of 0.0709 where the zero-dimensional model gives $1/6$ .

Their ratio is $1/(6\times0.0709)=2.35$ , so the zero-dimensional model overestimates the thinning speed by more than a factor of two, as McKinley points out after Eq. (16). Use the zero-dimensional model for trends and for the polymer-dominated stage, not to infer a viscosity from the break-up time of a Newtonian thread.

R(t)=0.0709\,\frac{\sigma}{\eta_s}\left(t_c-t\right),\qquad t_c\approx14.1\,\frac{\eta_sR_0}{\sigma}

Eqs. (16) and (10) of McKinley. The break-up time is the initial radius divided by the similarity thinning speed.

Course-drawn curves, not data, on a logarithmic radius axis. Cyan: the elasto-capillary stage of McKinley’s Eq. (8), a straight line whose slope is

-1/(3\lambda)

in the natural logarithm; its front factor is omitted. Amber: a thread thinning at constant speed, which bends down as it nears break-up. The amber time scale is chosen for display only.

Source: G. H. McKinley

The elasto-capillary stage: exponential thinning

In a dilute polymer solution the solvent viscosity is small, and after a short start the capillary pressure is balanced mainly by polymer tension: $\sigma/R\approx G A_{zz}$ . The smaller the radius, the larger the capillary pressure and the further the polymers must be stretched; and for $A_{zz}$ to grow, the stretch rate must exceed the coil–stretch threshold.

Solve it self-consistently. In logarithms, the growth rate of $\ln A_{zz}$ equals that of $-\ln R$ , which is $\dot\varepsilon/2$ ; the conformation equation requires it to be $2\dot\varepsilon-1/\lambda$ . Setting the two equal gives $\dot\varepsilon=2/(3\lambda)$ . The thread holds its own stretch rate at a Weissenberg number of 2/3, above the threshold of one half found in Chapter 3.

The radius therefore falls as $e^{-t/(3\lambda)}$ . On a logarithmic axis this is a straight line whose slope contains only the relaxation time, not the surface tension or the modulus. The front factor of McKinley’s Eq. (8) contains $(GR_0/\sigma)^{1/3}$ ; it differs from the value listed in Table 2 by a factor of the cube root of two, which Section 5.4 attributes to the one-dimensional similarity solution. The course uses only the slope.

\frac{R(t)}{R_0}\propto\exp\!\left(-\frac{t}{3\lambda}\right),\qquad \dot\varepsilon=\frac{2}{3\lambda},\qquad \mathrm{Wi}=\lambda\dot\varepsilon=\frac23

Eq. (8) of McKinley, with the front factor omitted.

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

Once the chains reach full length

During the exponential stage $A_{zz}$ keeps growing. With FENE-P it eventually approaches $L^2$ , the polymer stress stops increasing with stretch, and the extensional viscosity settles at the ceiling of Chapter 4, $3\eta_s+2\eta_pL^2$ .

The force balance then describes a thread of constant extensional viscosity, whose radius falls linearly in time again and breaks in finite time. The linear spring of Oldroyd-B has no ceiling, and in the zero-dimensional model its thread keeps thinning exponentially forever. So the break-up time of a real thread carries information about chain length, while the slope of the exponential stage carries information about the relaxation time only.

\frac{dR}{dt}=-\frac{\sigma}{2\eta_E},\qquad \eta_E=3\eta_s+2\eta_pL^2

Eq. (11) and Section 5.4 of McKinley. It follows from the zero-dimensional balance with a constant extensional viscosity.

Source: G. H. McKinley

Worked example: a relaxation time from a radius record

Suppose a measurement in which the neck radius falls during the exponential stage from $200\,\mu\mathrm m$ to $50\,\mu\mathrm m$ in $0.83\,\mathrm s$ . The slope gives $\lambda=0.83/(3\ln4)=0.200\,\mathrm s$ . The stretch rate in this stage is $2/(3\lambda)=3.34\,\mathrm{s^{-1}}$ and the accumulated Hencky strain is $2\ln4=2.77$ .

For comparison, take a Newtonian oil with viscosity $1.0\,\mathrm{Pa\,s}$ and surface tension $0.030\,\mathrm{N/m}$ , starting at a radius of $0.3\,\mathrm{mm}$ . The similarity thinning speed is $0.0709\times0.030/1.0=2.13\,\mathrm{mm/s}$ and the break-up time $14.1\times1.0\times3\times10^{-4}/0.030=0.141\,\mathrm s$ . The zero-dimensional model gives $5\,\mathrm{mm/s}$ and $0.06\,\mathrm s$ , breaking the thread in less than half the time.

Choose the time window carefully when reading a relaxation time. Early on the solvent viscosity still matters, and late in the run the chains approach full length; only the straight stretch in between corresponds to a Weissenberg number of 2/3.

Source: G. H. McKinley

Try it in the Lab

Read the relaxation time off the straight stretch

Choose the Polymers hold the thread preset. On the log-radius curve, find the straight stretch, estimate its slope $s$ , compute $\lambda=-1/(3s)$ and compare it with the relaxation time read from the slope by the Lab.
Switch to Solvent only and see how much earlier the thread breaks. Then choose the A thread that never breaks preset and explain why an Oldroyd-B thread does not break in this model.
Choose the Short chains break early preset, find the moment when the straight stretch ends and the radius starts falling linearly, and use the ceiling on extensional viscosity to explain why shorter chains break sooner.

Compare with a Newtonian thread

Choose the Viscosity slows the neck preset and note the minimum radius at several times.
Plot the logarithm of those radii against time. Do they fall on a straight line like the polymer thread? Use the zero-dimensional balance of this chapter to say where the difference comes from.

Exercises with explained answers

During the exponential stage the log radius of a polymer thread falls by one every 0.6 s. What are the relaxation time and the stretch rate?
Show the answer
The slope is $-1/(3\lambda)=-1/0.6\,\mathrm{s^{-1}}$ , so $\lambda=0.2\,\mathrm s$ , and the stretch rate is $2/(3\lambda)=3.33\,\mathrm{s^{-1}}$ .
A solution has $\eta_s=0.01\,\mathrm{Pa\,s}$ , $2\eta_pL^2=100\,\mathrm{Pa\,s}$ and surface tension $0.05\,\mathrm{N/m}$ . Once the chains are fully stretched, how fast does the radius fall?
Show the answer
$\eta_E=0.03+100=100.03\,\mathrm{Pa\,s}$ and $dR/dt=-0.05/(2\times100.03)=-2.50\times10^{-4}\,\mathrm{m/s}$ , about 0.25 mm per second.
Why must the Weissenberg number of the elasto-capillary stage exceed one half?
Show the answer
The capillary pressure grows as the radius shrinks and must be balanced by polymer tension, so that tension must keep growing. Chapter 3 shows that the conformation keeps growing only above the threshold stretch rate; below it the stress settles at a finite value and cannot keep up with the rising capillary pressure.
By what factor do the zero-dimensional model and the similarity solution disagree on the break-up time of a Newtonian thread, and which breaks first?
Show the answer
The zero-dimensional thinning speed is $\sigma/(6\eta_s)$ and the similarity speed $0.0709\,\sigma/\eta_s$ . The first is $2.35$ times faster, so the zero-dimensional thread breaks first, at $1/2.35\approx0.43$ of the similarity break-up time.

Sources and what we recomputed

A. Morozov and S. E. Spagnolie, Introduction to complex fluids, in Complex Fluids in Biological Systems (Springer, 2015), chapter 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.
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)
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.