---
title: "From a bead-spring dumbbell to the conformation tensor and the Oldroyd-B equation"
description: "Treat a polymer as two beads joined by a spring, derive the conformation tensor and Oldroyd-B stress, and solve startup shear with normal stress."
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[Complex fluids](https://sandboxphysics.com/stories/complex-fluids/)

L2 · Chapter 2

# From a bead-spring dumbbell to the conformation tensor and the Oldroyd-B equation

Sandbox Physics · 2026-10-29

The Maxwell equation of the previous chapter has a single scalar stress and cannot say in which direction a polymer is stretched. Replace a polymer chain by two beads joined by a spring and average over thermal motion, and the result is a tensor that records how far the coil is stretched in each direction. The equation for this conformation tensor, plus the viscous stress of the solvent, is the Oldroyd-B model. In shear it gives one thing Maxwell cannot: a normal stress along the flow.

## Before you begin

The Maxwell liquid and startup shear of Chapter 1; matrix products, the components of the velocity-gradient tensor, and systems of first-order linear ordinary differential equations.

- Derive the upper-convected equation for the conformation tensor and the polymer stress from a Hookean dumbbell.
- Solve the Oldroyd-B response to startup shear and to an imposed stress, and compute the first normal-stress difference.

## Two beads and a spring

Replace a polymer chain by two beads. Each feels solvent drag with coefficient $\zeta$, the beads are joined by a spring of stiffness $K$, and thermal kicks act on both. The end-to-end vector $\mathbf R$ jumps about, but its second moment $\langle\mathbf R\mathbf R\rangle$ obeys a deterministic equation.

At equilibrium the mean energy stored in the spring is set by the thermal energy, $K\langle\mathbf R\mathbf R\rangle=k_BT\,\mathbf I$. Measure $\langle\mathbf R\mathbf R\rangle$ in units of $k_BT/K$ and you get the conformation tensor $\mathbf A$, which is the identity at rest. The square root of its trace over three is the mean coil size relative to equilibrium, called the stretch below.

The polymer stress comes from the spring force. With $n$ springs per unit volume it is $n$ times the mean product of spring tension and end-to-end vector, minus the equilibrium part. The result is proportional to $\mathbf A-\mathbf I$, and the prefactor $G=nk_BT$ is the modulus of Chapter 1.

$$
\mathbf A=\frac{K\langle\mathbf R\mathbf R\rangle}{k_BT},\qquad \boldsymbol\tau_p=G\left(\mathbf A-\mathbf I\right),\qquad G=nk_BT=\frac{\eta_p}{\lambda},\qquad \lambda=\frac{\zeta}{4K}
$$

Eqs. (1.98), (1.104), (1.109) and (1.110) of Morozov and Spagnolie. The relaxation time is set by the bead drag and the spring stiffness.

Source: A. Morozov and S. E. Spagnolie

## How the conformation tensor follows the flow

The flow stretches and turns a polymer in the same way it stretches and turns a material line moving with the fluid. A polymer that never relaxed would deform exactly like those lines, which mathematically means its upper-convected derivative is zero. The spring and the thermal motion pull it back toward $\mathbf I$ at the rate $1/\lambda$.

In the upper-convected derivative the terms $\mathbf L\mathbf A+\mathbf A\mathbf L^{\mathsf T}$ are the stretching and turning by the velocity gradient. Add the viscous solvent stress $2\eta_s\mathbf D$ to the polymer stress and you have the Oldroyd-B model. It has only three parameters: solvent viscosity, polymer viscosity and relaxation time.

The conformation tensor is an average coil shape and must stay positive definite: its three eigenvalues are squared coil sizes along three principal directions and cannot be negative. A numerical solution that produces a negative eigenvalue has lost its meaning. The review by Datta and co-authors writes the common models in conformation-tensor form partly because this makes the property easy to check and to preserve.

$$
\overset{\nabla}{\mathbf A}\equiv\frac{D\mathbf A}{Dt}-\mathbf L\mathbf A-\mathbf A\mathbf L^{\mathsf T}=-\frac{1}{\lambda}\left(\mathbf A-\mathbf I\right),\qquad L_{ij}=\frac{\partial u_i}{\partial x_j}
$$

Eq. (1.72) of Morozov and Spagnolie defines the upper-convected derivative; Eqs. (1.103) and (1.110) give the dumbbell result, and Eq. (1.80) is Oldroyd-B with the solvent stress added. Here the velocity gradient is written $\mathbf L$, the transpose of the gradient tensor used in that chapter.

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

## Startup shear: shear stress and normal stress

In simple shear $u_x=\dot\gamma y$ the tensor $\mathbf L$ has one component, $L_{xy}=\dot\gamma$. The component equations then separate. $A_{yy}$ has no stretching term and stays equal to one; $A_{xy}$ is driven by $\dot\gamma A_{yy}$ and approaches $\mathrm{Wi}$ exponentially; $A_{xx}$ is in turn driven by $2\dot\gamma A_{xy}$.

Solving in that order, the shear stress is exactly that of Chapter 1: the startup viscosity $\eta^+$ grows from $\eta_s$ to $\eta_0$, and the steady viscosity does not depend on shear rate. The new quantity is the first normal-stress difference $N_1$: the tension along the flow exceeds the tension across it, and its steady value grows as the square of the shear rate. It rises more slowly than the shear stress and starts with zero slope, because $A_{xx}$ must wait for $A_{xy}$ to grow first.

In steady shear the ratio of $N_1$ to the polymer shear stress is $2\mathrm{Wi}$. The larger the Weissenberg number, the more the normal stress matters relative to the shear stress. Oldroyd-B has no second normal-stress difference.

$$
\eta^{+}(t)=\eta_s+\eta_p\left(1-e^{-t/\lambda}\right),\qquad N_1^{+}(t)=2\eta_p\lambda\dot\gamma^2\left[1-\left(1+\frac{t}{\lambda}\right)e^{-t/\lambda}\right]
$$

Obtained by integrating the components in turn; as $t\to\infty$ it agrees with Eq. (1.116) of Morozov and Spagnolie.

Oldroyd-B startup shear with solvent share $\beta=0.5$, time in relaxation times. Cyan: shear viscosity over the zero-shear viscosity, starting at the solvent share. Amber: first normal-stress difference over its steady value, starting with zero slope. Analytic curves computed for this course.

Source: A. Morozov and S. E. Spagnolie

## Creep under an imposed stress

Impose the total shear stress $\tau_0$ instead. At first the polymer stress is zero, the solvent carries everything, and the shear rate is $\tau_0/\eta_s$. As the polymer stress grows it takes over part of the load and the shear rate falls to $\tau_0/\eta_0$.

Write the solvent stress as $\tau_0$ minus the polymer stress and insert it into the polymer shear-stress equation. The result is an exponential with time constant $\beta\lambda$: the smaller the solvent share, the faster the start and the shorter the transition. Write $\mathrm{Wi}_N=\lambda\tau_0/\eta_0$ for the final Weissenberg number.

$$
\lambda\dot\gamma(t)=\mathrm{Wi}_N\left[1+\frac{\eta_p}{\eta_s}\,e^{-t/(\beta\lambda)}\right],\qquad \mathrm{Wi}_N=\frac{\lambda\tau_0}{\eta_0}
$$

Obtained by solving the Oldroyd-B shear component together with the condition of constant total stress. At the start the bracket equals the inverse solvent share.

Source: A. Morozov and S. E. Spagnolie

## Worked example: startup shear at a Weissenberg number of 2

Take $\eta_0=1\,\mathrm{Pa\,s}$, $\beta=0.5$ and $\lambda=1\,\mathrm{s}$, so $\eta_s=\eta_p=0.5\,\mathrm{Pa\,s}$ and $G=0.5\,\mathrm{Pa}$. Start shearing at $\dot\gamma=2\,\mathrm{s^{-1}}$, so $\mathrm{Wi}=2$.

At $t=\lambda$, $\eta^+=0.5+0.5\,(1-e^{-1})=0.816\,\mathrm{Pa\,s}$: 63% of the shear-stress growth is done. The steady normal stress is $2\times0.5\times1\times4=4\,\mathrm{Pa}$; at $t=\lambda$ it is only $4\,(1-2e^{-1})=1.06\,\mathrm{Pa}$, or 26%, and at $t=3\lambda$ it is $4\,(1-4e^{-3})=3.20\,\mathrm{Pa}$.

The steady conformation is $A_{xx}=1+2\mathrm{Wi}^2=9$, $A_{xy}=\mathrm{Wi}=2$, $A_{yy}=A_{zz}=1$. The trace is 11 and the stretch $\sqrt{11/3}=1.91$. The coil is elongated along the flow and tilted in the shear plane, but it is less than twice its equilibrium size.

Impose a stress instead, chosen so that $\mathrm{Wi}_N=2$. At first $\lambda\dot\gamma=2/\beta=4$; at $t=1\,\mathrm{s}$, two times $\beta\lambda$ later, it has fallen to $2\,(1+e^{-2})=2.27$.

Source: A. Morozov and S. E. Spagnolie

## Try it in the Lab

### [Watch the conformation tensor grow in shear](https://sandboxphysics.com/labs/polymer-stretch-startup/)

1. Switch the spring model to Oldroyd-B, keep the imposed rate, set the solvent share to 0.5 and the Weissenberg number to 2.
2. Look at the shear side only. The coil stretch should approach $\sqrt{11/3}=1.91$. In the detailed readouts, find the ratio of the normal-stress difference to the shear stress in shear; for the polymer part it is $2\mathrm{Wi}$, that is 4.
3. Switch to the imposed stress with the same settings and watch the shear-side Weissenberg number fall from about $2/\beta$ to 2. Set the solvent share to 0.2 and predict whether the transition becomes faster or slower.

## Exercises with explained answers

1. Why does $A_{yy}$ stay equal to one in startup simple shear?

   Show the answer

   The only velocity gradient in simple shear is $L_{xy}$. In the upper-convected derivative the stretching terms of the $A_{yy}$ equation contain $L_{yx}$ or $L_{yy}$, both zero, so only relaxation toward one remains. It starts at one and stays there.

2. In startup shear, what fraction of its steady value has the first normal-stress difference reached at $t=2\lambda$? And the polymer shear stress?

   Show the answer

   The normal-stress difference is at $1-3e^{-2}=0.594$, about 60%; the polymer shear stress is at $1-e^{-2}=0.865$. The normal stress starts slowly and needs more time.

3. Under an imposed stress with solvent share $\beta=0.1$, how many times the final shear rate is the initial one, and what is the time constant of the transition?

   Show the answer

   The initial shear rate is $1/\beta=10$ times the final one, and the time constant is $\beta\lambda=0.1\lambda$. The thinner the solvent, the harder the start and the quicker the transition.

4. Reverse the direction of shear. What happens to the signs of the shear stress and of the first normal-stress difference?

   Show the answer

   The shear stress is proportional to $\dot\gamma$ and changes sign. $N_1$ is proportional to $\dot\gamma^2$ and keeps its sign, so whichever way you turn, the normal stress puts tension along the streamlines.

[Previous: Stress relaxation, creep and the Weissenberg number in a Maxwell liquid](https://sandboxphysics.com/stories/complex-fluids/stress-relaxation-and-creep/) [Next: Why extensional flow unravels polymer coils while shear only tumbles them](https://sandboxphysics.com/stories/complex-fluids/coil-stretch-in-extension/)

## 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. [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.