L1 · Chapter 1

Stress relaxation, creep and the Weissenberg number in a Maxwell liquid

Sandbox Physics · 2026-10-29

Stir honey and the stress depends only on the shear rate at that moment; stop, and the stress is gone. A polymer solution is different. After the deformation stops, its stress takes a relaxation time to fade, and under a steady load it first jumps like a spring and then flows like a liquid. A spring in series with a dashpot turns this memory into an equation and gives the result of three standard experiments.

A spring in series with a dashpot

In an elastic solid the shear stress is proportional to the strain, $\tau=G\gamma$ ; in a Newtonian liquid it is proportional to the shear rate, $\tau=\eta\dot\gamma$ . Put a spring and a dashpot in series. Both carry the same force, and the total strain is the sum of the two. Differentiate in time, eliminate the separate strains, and you have the Maxwell liquid.

A time scale appears, $\lambda=\eta/G$ , the relaxation time. For processes much slower than $\lambda$ the term $\lambda\dot\tau$ is negligible and the material is a liquid of viscosity $\eta$ . For processes much faster than $\lambda$ the dashpot has no time to move and the material is a solid of modulus $G$ .

\tau+\lambda\dot\tau=\eta\dot\gamma,\qquad \lambda=\frac{\eta}{G}

Eq. (1.49) of Morozov and Spagnolie. Their Eq. (1.51) rewrites it as an integral over past shear rates: the present stress is a weighted sum of earlier shear rates, and a rate from a time

s

ago carries the weight

e^{-s/\lambda}

Source: A. Morozov and S. E. Spagnolie

Stress relaxation and creep

Stress relaxation: at $t=0$ impose a shear strain $\gamma_0$ suddenly and hold it. During the jump the dashpot cannot move, so the spring takes all of the strain and the stress is $G\gamma_0$ . Afterwards the shear rate is zero, the equation becomes $\tau+\lambda\dot\tau=0$ , and the stress decays exponentially, to $1/e$ after one relaxation time.

Creep: impose a shear stress $\tau_0$ suddenly and hold it. The spring stretches by $\tau_0/G$ at once and the dashpot keeps flowing at the rate $\tau_0/\eta$ , so the strain grows linearly in time. Remove the stress and the spring part springs back; the part that flowed through the dashpot stays.

\tau(t)=G\gamma_0\,e^{-t/\lambda},\qquad \gamma(t)=\frac{\tau_0}{G}+\frac{\tau_0}{\eta}\,t

Left: the stress after a step strain. Right: the strain under a constant stress. Both follow from integrating Eq. (1.49) directly.

Source: A. Morozov and S. E. Spagnolie

Startup shear and the solvent contribution

The stress in a polymer solution has two parts. The solvent part, with viscosity $\eta_s$ , follows the shear rate instantly. The polymer part has viscosity $\eta_p$ and follows a Maxwell equation with relaxation time $\lambda$ . The zero-shear viscosity $\eta_0$ is their sum, and the solvent share is written $\beta$ .

Startup shear: the liquid is at rest until $t=0$ and is then sheared at a constant rate $\dot\gamma$ . The solvent stress reaches $\eta_s\dot\gamma$ at once; the polymer stress grows to $\eta_p\dot\gamma$ over a few relaxation times. The next chapter shows that this is exactly the shear stress of the Oldroyd-B model in simple shear. The same model also produces a normal stress along the flow, which the scalar Maxwell equation cannot.

\tau(t)=\eta_s\dot\gamma+\eta_p\dot\gamma\left(1-e^{-t/\lambda}\right),\qquad \eta_0=\eta_s+\eta_p,\quad \beta=\frac{\eta_s}{\eta_0}

The Oldroyd-B model, Eq. (1.80) of Morozov and Spagnolie, gives the same shear stress in simple shear.

Source: A. Morozov and S. E. Spagnolie

How much elasticity, how much inertia

The Weissenberg number $\mathrm{Wi}=\lambda\dot\gamma$ compares the relaxation time with the time the flow takes to stretch the material. For $\mathrm{Wi}\ll1$ the polymers relax in time and the solution behaves as an ordinary liquid of viscosity $\eta_0$ . Once $\mathrm{Wi}$ reaches order one, elastic stresses are as large as viscous ones. The Deborah number compares the relaxation time with the duration of the observation instead, and decides whether a material looks solid or liquid.

The Reynolds number compares inertia with viscosity. Purcell writes it as size times speed times density over viscosity; for swimming microorganisms it is only $10^{-4}$ to $10^{-5}$ . Divide one number by the other and the speed cancels: the ratio depends only on the fluid and the size of the apparatus, and the smaller the apparatus, the more elasticity dominates.

Groisman and Steinberg shrank their whole swirling-flow apparatus by a factor of four and measured the same resistance curve at the same Weissenberg number. At equal shear rate the speeds were a quarter of the original and the Reynolds number a sixteenth, yet nothing changed, so the instability they saw does not depend on inertia.

\mathrm{Wi}=\lambda\dot\gamma,\qquad \mathrm{Re}=\frac{\rho UL}{\eta},\qquad \frac{\mathrm{Wi}}{\mathrm{Re}}=\frac{\lambda\eta}{\rho L^2}

The Weissenberg number as defined in Section 3 of Morozov and Spagnolie, the Reynolds number as Purcell writes it. The last expression takes the shear rate as speed over size.

Source: A. Morozov and S. E. Spagnolie · E. M. Purcell · A. Groisman and V. Steinberg

Worked example: startup shear of the elastic-turbulence solution

The solution of Groisman and Steinberg (2000): solvent viscosity $\eta_s=0.324\,\mathrm{Pa\,s}$ , solution viscosity $0.424\,\mathrm{Pa\,s}$ at a shear rate of one per second, and relaxation time $\lambda=3.4\,\mathrm{s}$ , measured from the phase shift between stress and strain in oscillation. Treat the polymer part as one Maxwell element: $\eta_p=0.100\,\mathrm{Pa\,s}$ and modulus $G=\eta_p/\lambda=0.0294\,\mathrm{Pa}$ .

Start shearing at $\dot\gamma=1\,\mathrm{s^{-1}}$ , so $\mathrm{Wi}=3.4$ . The shear stress begins at $0.324\,\mathrm{Pa}$ , is $0.324+0.100\,(1-e^{-1})=0.387\,\mathrm{Pa}$ at $t=\lambda$ , and ends at $0.424\,\mathrm{Pa}$ . The polymer stress takes $\lambda\ln100=15.7\,\mathrm{s}$ to reach 99% of its steady value, so measuring a steady viscosity in this solution means waiting more than ten seconds after every change of speed.

Stop the rotation and the solvent stress vanishes at once while the polymer stress decays as $e^{-t/\lambda}$ . After 10 s it is $0.100\,e^{-10/3.4}=5.28\times10^{-3}\,\mathrm{Pa}$ , about 5% of where it started.

Two approximations sit in these numbers. The paper gives the solution viscosity at one shear rate only, and a real solution’s viscosity changes with shear rate. A single relaxation time also stands in for the many relaxation modes of real polymers. Use the numbers for orders of magnitude, not as the exact flow curve of this solution.

Source: A. Groisman and V. Steinberg · A. Morozov and S. E. Spagnolie

Try it in the Lab

Hold a strain and watch the stress relax

Choose the Viscoelastic hold preset and watch the stress fall from its peak during each hold at fixed strain.
The material here is an equilibrium spring in parallel with one Maxwell arm, so the stress does not fall to zero, only to the part carried by the equilibrium spring. Estimate how long the excess stress takes to fall to $1/e$ and compare it with the Maxwell relaxation time setting.
Predict, then test: change the Maxwell relaxation time from 3 s to 0.3 s and then to 20 s. How does the apparent slope of the loading ramp change?

Where a steady viscosity comes from

With the Newtonian preset, read the torque, shear rate and viscosity; change the rotation speed and confirm that the viscosity stays the same.
Switch to the shear-thinning preset and watch the viscosity fall as the speed rises. This rheometer shows steady values: the viscosity depends only on the present shear rate, with no relaxation time and no normal stress.
Using the solution of the worked example, estimate how long you would wait for a steady reading after stepping the shear rate from zero to one per second. The answer is about $4.6\lambda$ .

Exercises with explained answers

A Maxwell liquid with $G=10\,\mathrm{Pa}$ and $\lambda=2\,\mathrm{s}$ receives a sudden strain of 0.5 that is then held. What is the stress at once and after 4 s?
Show the answer
At once $G\gamma_0=5\,\mathrm{Pa}$ ; after 4 s, $5\,e^{-2}=0.677\,\mathrm{Pa}$ . Two relaxation times leave about 14%.
The same liquid receives a sudden shear stress of $2\,\mathrm{Pa}$ , held for 6 s and then removed. What is the strain just before removal, and how much comes back?
Show the answer
The viscosity is $\eta=G\lambda=20\,\mathrm{Pa\,s}$ . The strain is $2/10+2\times6/20=0.2+0.6=0.8$ . After removal the spring part, 0.2, comes back; the 0.6 that flowed through the dashpot stays.
If the solution of the worked example is sheared at $\dot\gamma=0.1\,\mathrm{s^{-1}}$ , what is the Weissenberg number? Does elasticity matter?
Show the answer
$\mathrm{Wi}=3.4\times0.1=0.34$ , below one. The polymers relax in time and in steady flow the solution behaves mostly as a viscous liquid, although after a start it still takes a few relaxation times, tens of seconds, for the stress to settle.
Shrink every dimension of an apparatus tenfold, keep the fluid, and keep the shear rate. How do the Reynolds and Weissenberg numbers change?
Show the answer
The shear rate is unchanged, so the Weissenberg number is unchanged. Speed and size both fall tenfold, so the Reynolds number falls a hundredfold. Their ratio $\lambda\eta/(\rho L^2)$ grows a hundredfold, and inertia matters even less.

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.
E. M. Purcell, Life at low Reynolds number, American Journal of Physics 45, 3–11 (1977)
The Reynolds number written as size times speed times density over viscosity; for swimming microorganisms it is about 1e-4 to 1e-5, and a bacterium that stops pushing coasts about 0.1 ångström. The scallop theorem: at low Reynolds number only the sequence of shapes matters, not how fast it is run. Cited for the low-inertia framing only. Read from a scanned reprint: journal pages 3–6 only.
A. Groisman and V. Steinberg, Elastic turbulence in a polymer solution flow, Nature 405, 53–55 (2000); arXiv:nlin/0104052v1
arXiv version 1, read in full. Fig. 1: swirling flow between a rotating plate of radius 38 mm and a cup of radius 43.6 mm, liquid depth 10 mm, 12 °C. Methods: 65 % saccharose and 1 % NaCl solvent of viscosity 0.324 Pa s with 80 ppm polyacrylamide of molecular weight 1.8e7; solution viscosity 0.424 Pa s at a shear rate of 1 per second; relaxation time 3.4 s. Fig. 2: transition near a shear rate of 1 per second (Weissenberg number 3.5, Reynolds number about 0.3), with hysteresis; resistance up to about 12 times laminar for a 10 mm gap and 19 for 20 mm. Fig. 3: snapshots at Weissenberg number 13. Fig. 4: brightness spectra. Fig. 5: velocity spectra, a power law with exponent −3.5 at 4 per second over about one decade. Text: Reynolds stress below 0.5 % of the resistance.