---
title: "Dislocation dipoles and the stress that tears them apart"
description: "A pair of opposite dislocations on nearby planes resists shear up to a passing stress that falls as one over their spacing. Derive it and test it."
canonical_url: "https://sandboxphysics.com/stories/crystal-defects-microstructure/dislocation-dipoles-and-passing-stress/"
language: "en"
page_type: "story"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/stories/crystal-defects-microstructure/dislocation-dipoles-and-passing-stress/"
last_modified: "2026-10-30"
---

[Crystal defects and microstructure](https://sandboxphysics.com/stories/crystal-defects-microstructure/)

L2 · Chapter 3

# Dislocation dipoles and the stress that tears them apart

Sandbox Physics · 2026-10-30

Two dislocations of opposite sign on glide planes a few atoms apart hold each other at 45 degrees. A shear stress pushes them in opposite directions, and up to a point their mutual attraction pushes back. The largest stress the pair can resist is the passing stress, $\tau_{\mathrm p}=\mu b/[8\pi(1-\nu)h]$. This chapter derives it from the pair force of chapter 2 and checks it against seeded runs of the Lab model.

## Before you begin

Chapter 2: the glide force between two parallel edge dislocations and its zeros. Finding the maximum of a function of one variable by setting its derivative to zero.

- Derive the passing stress of an edge dipole and the angle at which the pair breaks.
- Predict the angle a bound dipole swings through under a cyclic shear below the passing stress, and say why narrow dipoles are stronger.

## The restoring force has a maximum

Hold a dipole with its planes $h$ apart at offset $x$. The applied shear pushes the two dislocations apart with $b\tau$ each; the pair force of chapter 2, with its sign reversed for opposite dislocations, pulls them back. Equilibrium needs the two to balance, which fixes the stress needed to hold the pair at each offset.

The function $g(u)$ is zero at $u=1$, the unloaded 45-degree position. As the stress rises, the pair opens to larger $u$ and $g$ rises with it, but only up to a maximum. Setting the derivative to zero gives $u=1+\sqrt2$, where the line joining the pair makes $22.5^{\circ}$ with the glide plane and $g=1/4$. Beyond that offset the attraction weakens as the pair separates, so a stress large enough to reach it carries the pair all the way apart.

$$
\tau(u)=\frac{\mu b}{2\pi(1-\nu)h}\,g(u),\qquad g(u)=\frac{u\,(u^2-1)}{(u^2+1)^2},\qquad u=\frac{x}{h}
$$

The applied shear that holds an opposite pair at offset $x$ on planes $h$ apart.

$$
g'(u)=0\;\Longleftrightarrow\;u^4-6u^2+1=0\;\Longrightarrow\;u=1+\sqrt2,\quad g(1+\sqrt2)=\tfrac14
$$

Where the holding stress peaks, and its value there.

The shear stress that holds an edge dipole at offset $x$ on planes $h$ apart, in units of $\mu b/[2\pi(1-\nu)h]$. The unloaded pair sits at $x=h$, the 45-degree position. The curve peaks at $1/4$ when $x=(1+\sqrt2)h$, where the pair makes 22.5 degrees with the glide plane; a larger stress has no equilibrium, and the pair passes. The dashed lines mark the passing stress.

Source: A. Kauffmann and D. Schliephake · X. Niu

## The passing stress

Putting $g=1/4$ into the holding stress gives the passing stress below. It falls as one over the plane spacing: a dipole on planes $2b$ apart is twice as strong as one on planes $4b$ apart. With $\nu=0.3$ the values are $0.0284\,\mu$, $0.0142\,\mu$ and $0.00568\,\mu$ for $h=2b$, $4b$ and $10b$.

The same stress pulls the pair the other way when its sign reverses. Then the offset shrinks, the pair swings through the vertical, where it is unstable, to the mirror-image 45-degree position, and breaks beyond $u=-(1+\sqrt2)$ at the same $\tau_{\mathrm p}$. The function obeys $g(1/u)=-g(u)$, so a loading that tilts the pair to an angle $\varphi$ in one direction tilts it to $90^{\circ}-\varphi$ in the other.

The same symmetry puts a second peak of the same height at $u=\sqrt2-1$, so swinging through the vertical also takes $\tau_{\mathrm p}$. A cyclic load just above $\tau_{\mathrm p}$ flips the pair to its mirror image and falls again before the pair reaches the far peak. A larger load flips the pair and then pulls it apart. The Lab shows both.

$$
\tau_{\mathrm p}=\frac{\mu b}{8\pi(1-\nu)\,h}
$$

The passing stress of an edge dipole in an isotropic crystal. The Lab shows it as a readout and as dashed lines on the loading chart.

Source: A. Kauffmann and D. Schliephake

## Checking the threshold numerically

The route benchmark finds the threshold without using the formula. For $16$ seeded combinations of spacing and Poisson ratio, it holds a constant shear on a 45-degree dipole and asks whether the pair separates by $12h$ within $400/\tau$ time units. Eighteen halvings of the stress interval locate the threshold.

The measured threshold divided by the formula is $1.0061\pm0.0012$. The small excess is expected from the method: just above $\tau_{\mathrm p}$ the pair crawls through the region near $u=1+\sqrt2$ where the net force almost vanishes, so a finite time limit records a slightly higher threshold than the static one.

Source: X. Niu

## Worked example: how far does a bound dipole swing?

Problem. A dipole on planes $4b$ apart, with $\nu=0.3$, is loaded by $\tau(t)=0.009\,\mu\,\sin(2\pi t/T)$. If the load changes slowly enough for the pair to follow it, between which angles does the pair swing?

Passing stress. $\tau_{\mathrm p}=1/(8\pi\times0.7\times4)\,\mu=0.01421\,\mu$. The peak load is $0.009/0.01421=0.633$ of it, so the pair holds.

Peak offset. At peak load $g(u)=0.633/4=0.158$. Solving $u(u^2-1)/(u^2+1)^2=0.158$ on the stable branch $1<u<1+\sqrt2$ by bisection gives $u=1.419$, an angle $\arctan(1/1.419)=35.2^{\circ}$. At the opposite peak, $u=1/1.419=0.705$ and the angle is $54.8^{\circ}$.

Check. The Lab preset loads with period $T=8000/(\mu M)$, and its pair swings between $35.2^{\circ}$ and $54.8^{\circ}$, as estimated. Load four times faster, with $T=2000/(\mu M)$, and the swing shrinks to $35.8^{\circ}$ to $54.6^{\circ}$: the overdamped pair lags behind the load. The lag also raises the stress needed to break the pair, which is why the Lab loads slowly.

Answer. About $35^{\circ}$ to $55^{\circ}$, symmetric about $45^{\circ}$.

Source: A. Kauffmann and D. Schliephake

## Try it in the Lab

### [Find the stress that breaks a dipole](https://sandboxphysics.com/labs/edge-dislocation-glide/)

1. Choose A bound dipole rocks. Predict the smallest value the pair-angle readout reaches during a cycle.
2. Raise the amplitude of the cyclic shear to 0.015, just above the passing-stress readout. Predict what happens at the next peak, then watch.
3. Choose A narrow dipole holds, which uses the amplitude that tore the first preset apart. Explain from the passing stress why this pair survives.

## Exercises with explained answers

1. Show that $g'(u)=0$ leads to $u^4-6u^2+1=0$, and check that $g(1+\sqrt2)=1/4$.

   Show the answer

   Differentiating, $g'=[(3u^2-1)(u^2+1)-4u^2(u^2-1)]/(u^2+1)^3=-(u^4-6u^2+1)/(u^2+1)^3$. The roots are $u^2=3\pm2\sqrt2$, so $u=\sqrt2\pm1$. At $u=1+\sqrt2$, $u^2-1=2(1+\sqrt2)$ and $u^2+1=2\sqrt2(1+\sqrt2)$, so $g=2(1+\sqrt2)^2/[8(1+\sqrt2)^2]=1/4$.

2. With $\nu=0.3$, what is the widest dipole that survives a shear of $0.025\,\mu$?

   Show the answer

   Solve $\mu b/[8\pi(0.7)h]=0.025\,\mu$: $h=1/(8\pi\times0.7\times0.025)\,b=2.27\,b$. On the Lab’s whole-number spacings, only $h=2b$ holds, which is why that preset survives the load that breaks $h=4b$.

3. By how much does the passing stress of a dipole with $h=4b$ change between $\nu=0$ and $\nu=0.45$?

   Show the answer

   It scales as $1/(1-\nu)$: $0.00995\,\mu$ at $\nu=0$ and $0.0181\,\mu$ at $\nu=0.45$, a factor of $1.82$.

4. Why does a bisection that asks whether the pair separates within a fixed time give a threshold slightly above the passing stress, and how would you reduce the bias?

   Show the answer

   At exactly $\tau_{\mathrm p}$ the pair stops at $u=1+\sqrt2$, and just above it the net force there is tiny, so crossing takes a time that grows without bound as the stress approaches the threshold. A finite time limit misses those slow crossings. A longer limit, or a test for whether any equilibrium exists, would reduce the bias.

[Previous: The Peach–Koehler force makes like dislocations stack into walls](https://sandboxphysics.com/stories/crystal-defects-microstructure/peach-koehler-force-and-glide/) [Next: Grain boundaries move by curvature, so a round grain shrinks at a steady rate](https://sandboxphysics.com/stories/crystal-defects-microstructure/grain-boundaries-move-by-curvature/)

## Sources and what we recomputed

1. [A. Kauffmann and D. Schliephake, Plasticity, chapter 4c: Elastic theory of dislocations, lecture slides, Karlsruhe Institute of Technology, version 24-05-02 (2024)](https://www.iam.kit.edu/wk/downloads/PoMI/4c-Dislocations-Properties.pdf)

   Version 24-05-02, all slides. Slides 9 and 10: displacement and stress field of a straight edge dislocation with line along $z$ and Burgers vector along $x$, after Hirth and Lothe. Slide 11: maps of the stress components. Slide 13: interactions follow from superposing the fields. Slide 15: the field is cut off near the core, at about $1.5\,b$ for an edge dislocation. Slide 23: the glide force per unit length is $b\tau$.

2. [X. Niu, Y. Zhu, S. Dai and Y. Xiang, A continuum model for distributions of dislocations incorporating short-range interactions, arXiv:1707.05161v2 (2018)](https://arxiv.org/abs/1707.05161v2)

   Version 2, Section 2, the discrete dislocation model. Eq. (2.1): the Peach–Koehler force $\mathbf f=(\boldsymbol\sigma\cdot\mathbf b)\times\boldsymbol\tau$, with glide component $b\sigma_{xy}$ and climb component $-b\sigma_{xx}$. Eqs. (2.2)–(2.3): the isotropic stress field of an edge dislocation. Eqs. (2.4)–(2.7): forces summed over all other dislocations plus the applied stress. Eq. (2.8): the mobility law $v_{\mathrm g}=m_{\mathrm g}f_{\mathrm g}$. Section 4 discusses dipolar walls; its continuum results are not used here.