---
title: "The Peach–Koehler force makes like dislocations stack into walls"
description: "Turn stress into a force on a dislocation, let dislocations glide in each other’s fields, and see why like signs stack and opposite signs settle at 45 degrees."
canonical_url: "https://sandboxphysics.com/stories/crystal-defects-microstructure/peach-koehler-force-and-glide/"
language: "en"
page_type: "story"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/stories/crystal-defects-microstructure/peach-koehler-force-and-glide/"
last_modified: "2026-10-30"
---

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

L2 · Chapter 2

# The Peach–Koehler force makes like dislocations stack into walls

Sandbox Physics · 2026-10-30

A dislocation feels a force wherever the crystal around it is stressed, whether the stress comes from the testing machine or from another dislocation. One formula, due to Peach and Koehler, gives that force from the local stress and the Burgers vector. Applied to two dislocations on parallel planes, it explains why dislocations of one sign stack on top of each other into a wall, and why a pair of opposite signs settles at 45 degrees instead of annihilating.

## Before you begin

Chapter 1: the Burgers vector and the shear stress of an edge dislocation, including where it changes sign. Cross products, and the idea that a force does work equal to force times displacement.

- Derive the glide force on an edge dislocation from the work done when it moves, and say why the Lab lets dislocations glide but not climb.
- Find where the glide force between two parallel edge dislocations vanishes, and decide which of those positions are stable for like and for opposite signs.

## A force from the work done by slip

Let a straight edge dislocation of unit length glide a distance $\delta x$. The crystal above its glide plane has now slipped by $b$ over an extra area $\delta x$. If the shear stress on that plane is $\tau$, the stress does work $\tau\,b\,\delta x$. Work divided by distance is force, so the line is pushed with $b\tau$ per unit length. The KIT lecture states this on slide 23.

The general version, written by Peach and Koehler, takes the stress tensor $\boldsymbol\sigma$, the Burgers vector $\mathbf b$ and the unit vector $\boldsymbol\xi$ along the line. For our edge dislocation along $z$ with $\mathbf b$ along $x$, the force has a glide part along $x$ set by the shear stress, and a climb part along $y$ set by the normal stress along the Burgers vector.

$$
\mathbf f=(\boldsymbol\sigma\cdot\mathbf b)\times\boldsymbol\xi,\qquad f_{\mathrm g}=b\,\sigma_{xy},\qquad f_{\mathrm c}=-b\,\sigma_{xx}
$$

The Peach–Koehler force per unit length, and its glide and climb components for an edge dislocation (Niu et al., Eq. 2.1).

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

## Glide, but not climb

To glide, a dislocation only rearranges bonds next to its core. To climb, the extra half-plane must grow or shrink, which needs vacancies or interstitial atoms to arrive by diffusion. At low temperature that is usually much slower than glide, so the Lab freezes climb and lets each dislocation move only along its own glide plane.

The Lab treats the motion as overdamped, as in Eq. (2.8) of Niu and co-workers: the velocity is proportional to the force, with a mobility $M$. Each dislocation feels the applied shear plus the shear stress of every other dislocation at its position. The sign $s_i=\pm1$ records which way its Burgers vector points, so an applied stress pushes opposite dislocations in opposite directions. The Lab measures stress in $\mu$, length in $b$ and time in $1/(\mu M)$.

$$
\frac{dx_i}{dt}=M\,s_i\,b\Bigl[\tau+\sum_{j\ne i}\sigma_{xy}^{(j)}(x_i-x_j,\;y_i-y_j)\Bigr]
$$

Overdamped glide of dislocation $i$ under the applied shear $\tau$ and the fields of the others. The Lab removes an opposite pair that meets within $b$ on one glide plane.

Source: X. Niu

## Two dislocations on parallel planes

Put one dislocation at the origin and a second on a parallel glide plane at height $h$, offset by $x$. The glide force on the second is its Burgers vector times the first one’s shear stress at that point. The factor $x(x^2-h^2)$ vanishes at three offsets: directly above, $x=0$, and at 45 degrees, $x=\pm h$.

For like signs, $s_1s_2=1$, the force is negative for $0<x<h$ and positive for $x>h$. A dislocation slightly off the vertical is pulled back to $x=0$, and one slightly past 45 degrees is pushed away. So like dislocations are stable stacked vertically and unstable at 45 degrees. For opposite signs every force reverses: the vertical position becomes unstable, and the pair settles at 45 degrees.

The forces on the two dislocations are equal and opposite, because moving the pair together does not change their relative position. Both glide toward the stable arrangement at once.

$$
f_x=s_1s_2\,\frac{\mu b^2}{2\pi(1-\nu)}\,\frac{x\,(x^2-h^2)}{(x^2+h^2)^2}
$$

Glide force on a dislocation at offset $x$ and height $h$ from another, with $s_1s_2=+1$ for like and $-1$ for opposite signs.

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

## Like signs build a wall

Several dislocations of one sign on equally spaced planes slide until they sit one above the other. Each brings one extra half-plane, so the lattice on one side of the column is tilted relative to the other. With planes $h$ apart, the tilt is about $b/h$ radians. This is a low-angle tilt boundary, the simplest grain boundary, and the next chapters return to boundaries of larger angle.

The route benchmark relaxes seeded pairs with no applied stress. Like pairs end at $90.000^{\circ}$ and opposite pairs at $45.000^{\circ}$, with standard errors near $10^{-15}$ degrees.

Source: A. Kauffmann and D. Schliephake

## Worked example: which way does each dislocation go?

Problem. Two like edge dislocations sit on planes $4b$ apart; the upper one is offset by $2b$ from the lower. Take $\nu=0.3$. Find the glide and climb forces on each and say what the pair does with no applied stress.

Upper dislocation. Chapter 1 gave the shear stress of the lower one at $(2b,4b)$: $\sigma_{xy}=-0.0136\,\mu$. The glide force is $f_{\mathrm g}=b\sigma_{xy}=-0.0136\,\mu b^2$, toward smaller $x$, that is, toward the vertical.

Lower dislocation. Seen from the upper one it sits at $(-2b,-4b)$. Both factors of $x$ change sign, so $\sigma_{xy}=+0.0136\,\mu$ and $f_{\mathrm g}=+0.0136\,\mu b^2$: equal and opposite, also toward the vertical.

Climb force on the upper one. $\sigma_{xx}=-0.2274\,\mu b\,y(3x^2+y^2)/(x^2+y^2)^2=-0.2274\times4\times28/400\,\mu=-0.0637\,\mu$, so $f_{\mathrm c}=+0.0637\,\mu b^2$. It is nearly five times the glide force, but the Lab ignores it because climb is frozen.

Answer. The pair slides together until the two are stacked vertically, $x=0$, which is the stable arrangement for like signs. The offset is inside $h$, so the pair stays on the stable side of 45 degrees the whole way.

Source: X. Niu

## Try it in the Lab

### [Watch five like dislocations build a wall](https://sandboxphysics.com/labs/edge-dislocation-glide/)

1. Choose Dislocations line up into a wall. Before pressing play, predict the angle the pair-angle readout will settle at.
2. Let it run, then switch the arrangement to Dipole of opposite signs with the same spacing and drag one dislocation sideways. Predict where it returns to.
3. Explain both end states from the sign of the shear-stress lobes: why like signs stack vertically and opposite signs stop at 45 degrees.

## Exercises with explained answers

1. Show that the 45-degree position $x=h$ is unstable for two like dislocations and stable for two opposite ones.

   Show the answer

   Write $u=x/h$ and $g(u)=u(u^2-1)/(u^2+1)^2$. Near $u=1$, $g\approx(u-1)/2$, which grows through zero. For like signs the force is proportional to $+g$, so a small step past $u=1$ is pushed further: unstable. For opposite signs it is $-g$, which pushes back: stable.

2. In the Lab’s units a lone dislocation glides at $dx/dt=s\,\tau$. How long does it take to cross $20\,b$ under $\tau=0.01\,\mu$?

   Show the answer

   The speed is $0.01\,b$ per unit of $1/(\mu M)$, so the crossing takes $2000/(\mu M)$, the length of one default loading period. The Lab model has no lattice friction, so any nonzero stress moves a lone dislocation.

3. A symmetric tilt boundary has like edge dislocations stacked every $h$. Geometry gives $b=2h\sin(\theta/2)$. Find the tilt for $h=10\,b$ and for the Lab’s $h=4\,b$, and compare with $\theta\approx b/h$.

   Show the answer

   For $h=10\,b$, $\theta=2\arcsin0.05=0.1000$ rad, or $5.73^{\circ}$, the same as $b/h$ to four digits. For $h=4\,b$, $\theta=2\arcsin0.125=0.2507$ rad, or $14.4^{\circ}$, against $0.25$ rad from the small-angle formula.

[Previous: The Burgers vector and the stress field of an edge dislocation](https://sandboxphysics.com/stories/crystal-defects-microstructure/burgers-vector-and-edge-dislocation-field/) [Next: Dislocation dipoles and the stress that tears them apart](https://sandboxphysics.com/stories/crystal-defects-microstructure/dislocation-dipoles-and-passing-stress/)

## 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.