---
title: "A field drives a domain wall only so fast"
description: "A field moves a wall by tilting the moment inside it. Derive the steady speed, the Walker field where steady motion fails, and how interfacial DMI raises it."
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language: "en"
page_type: "story"
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  zh-CN: "https://sandboxphysics.com/zh/stories/micromagnetics/domain-walls-and-walker-breakdown/"
last_modified: "2026-10-29"
---

[Micromagnetics · A course with experiments](https://sandboxphysics.com/stories/micromagnetics/)

L3 · Chapter 4

# A field drives a domain wall only so fast

In a perpendicular film a field moves a domain wall through the moment inside the wall: the field turns that moment, and the turned moment drives the wall forward. Up to the Walker field the moment settles at a fixed angle and the speed grows in proportion to the field. Above it the moment keeps turning and the mean speed falls. In the Pt/Co film used here the Walker field is 7.74 mT without DMI and 65.5 mT with $0.5\,\mathrm{mJ/m^2}$ of interfacial DMI.

Sandbox Physics · 2026-10-29

In this chapter

[All eight chapters](https://sandboxphysics.com/stories/micromagnetics/)

Before you start

Chapters 1 to 3. Solving a quadratic equation and finding the maximum of a function.

## A wall and its two coordinates

Take a strip of Pt/Co film along $x$, magnetized up ($+z$) on the left and down on the right. Between the two domains the magnetization turns over a width set by exchange and anisotropy. The profile is $\theta(x)=2\arctan e^{(x-q)/\Delta}$ with $\Delta=\sqrt{A/K_0}=5.60\,\mathrm{nm}$ (Mougin et al. 2007, Eqs. 1a and 1b). Thiaville et al. quote 5.65 nm in their text; recomputing from their stated parameters gives 5.60 nm.

The moment at the centre of the wall lies in the film plane, at an angle $\Phi$ from $+x$. At $\Phi=0$ it points along the strip, a Néel wall; at $90^{\circ}$ it points across, a Bloch wall. The wall’s own stray field favours the Bloch wall through a small hard-axis energy $K=N_x\mu_0M_s^2/2=17.0\,\mathrm{kJ/m^3}$, with $N_x=0.0224$ from Thiaville et al. Interfacial DMI favours one Néel wall, of fixed handedness.

Two numbers describe the wall well enough for this chapter: the position $q$ and the angle $\Phi$. Thiaville et al. write the wall energy per area as $\sigma=2\Delta K\cos^2\Phi-\pi D\cos\Phi+\mathrm{const}$. The minimum moves from Bloch to Néel at the critical DMI $D_c=4\Delta K/\pi$, $0.12\,\mathrm{mJ/m^2}$ with our numbers.

## Two equations for two coordinates

Inserting the rigid profile into the Landau-Lifshitz-Gilbert equation and integrating across the wall gives two equations (Thiaville et al. 2012, Eqs. 4 and 5). The first says the applied field $H$ turns the wall moment, minus a share that goes into motion. The second says the wall moves when the moment is turned away from the angle its energy prefers.

The two fields on the right are the hard-axis field $H_K=2K/(\mu_0M_s)$, 31.0 mT, and the DMI field $H_D=\pi D/(2\mu_0M_s\Delta)$, 127 mT at $D=0.5\,\mathrm{mJ/m^2}$.

$$
\dot\Phi+\frac{\alpha\dot q}{\Delta}=\gamma_0H,\qquad \frac{\dot q}{\Delta}-\alpha\dot\Phi=\gamma_0\left(H_D\sin\Phi-H_K\sin\Phi\cos\Phi\right)
$$

Worked example

## Steady motion and the Walker field

Look for steady motion, with $\dot\Phi=0$. The first equation gives the speed directly: $v=\dot q=\gamma_0\Delta H/\alpha$. The second then fixes the angle: $H=\alpha\sin\Phi\,(H_D-H_K\cos\Phi)$ (Thiaville et al. Eq. 7). A steady angle exists only while $H$ is below the largest value of the right-hand side. That largest value is the Walker field.

Without DMI the right-hand side is $-\tfrac12\alpha H_K\sin2\Phi$, whose largest size is $\alpha H_K/2$. With $\alpha=0.5$ that is 7.74 mT (Mougin et al. Eq. 13, with the local hard axis).

With DMI, set the derivative of $\sin\Phi\,(H_D-H_K\cos\Phi)$ to zero. Using $\sin^2\Phi=1-\cos^2\Phi$ gives the quadratic $2H_K c^2-H_Dc-H_K=0$ in $c=\cos\Phi$. Its negative root gives the maximum and matches Thiaville et al. For $D=0.5\,\mathrm{mJ/m^2}$ it is $c=-0.220$, so the moment can turn to 102.7 degrees before steady motion fails, and the Walker field is 65.5 mT.

Below that field the speed stays proportional to the field. At 40 mT the formula gives 78.8 m/s and the 1 nm chain in the Lab gives 77.9 m/s. The two agree to about 1 per cent; the formula holds the wall profile rigid, while the chain lets it adjust. Without DMI an exact solution exists, found by Walker, in which the wall narrows as its moment turns (Mougin et al. Eqs. 1b, 11 and 15); the chain follows it to 0.4 per cent with 1 nm cells.

$$
v=\frac{\gamma_0\Delta}{\alpha}H,\qquad \cos\Phi_W=\frac{H_D-\sqrt{H_D^{2}+8H_K^{2}}}{4H_K},\qquad \mu_0H_W=\begin{cases}7.74\,\mathrm{mT}&D=0\\65.5\,\mathrm{mT}&D=0.5\,\mathrm{mJ/m^2}\end{cases}
$$

## Above the Walker field the moment keeps turning

Above $H_W$ no steady angle exists. Eliminate $\dot q$ between the two equations: $(1+\alpha^2)\dot\Phi=\gamma_0(H-\alpha H_D\sin\Phi+\alpha H_K\sin\Phi\cos\Phi)$. The moment turns faster where the bracket is large and slower where it is small, and the wall moves back and forth as it turns.

Without DMI the bracket is $H+H_W\sin2\Phi$. One full turn takes the time below; we derived it by integrating $\mathrm{d}\Phi/(H+H_W\sin2\Phi)$ over a period. Averaging $\dot q$ over that turn gives the mean speed. Far above $H_W$ it becomes $\gamma_0\Delta\alpha H/(1+\alpha^2)$, Mougin et al. Eq. 16.

At 30 mT, nearly four times the Walker field, the formula gives 13.4 m/s and a turn every 1.54 ns. At 5 mT, below the Walker field, the steady formula gives 9.85 m/s. Six times the field buys only 1.4 times the speed. The chain reproduces the turn period to 0.1 per cent and the mean speed to 1.2 per cent.

$$
T=\frac{2\pi(1+\alpha^2)}{\gamma_0\sqrt{H^2-H_W^2}},\qquad \bar v=\frac{\gamma_0\Delta}{1+\alpha^2}\left(\alpha H+\frac{H-\sqrt{H^2-H_W^2}}{\alpha}\right)\quad(D=0)
$$

## What the one-dimensional wall leaves out

The chain in the Lab has no strip edges, and its stray field is the single local term. Thiaville et al. compared the one-dimensional model with two-dimensional simulations of strips up to 500 nm wide with $D$ below $0.5\,\mathrm{mJ/m^2}$. The wall neither tilted nor twisted along its length, and the wall width and Walker field changed only slightly, mainly because of the local stray-field assumption. Above the Walker field small differences appeared from the wall’s freedom across the strip, growing with strip width and D. Mougin et al. note that in thin perpendicular films pinning usually dominates at low field and walls creep; the model leaves pinning out.

An experiment to think through

## Predict, change, explain

1. Without DMI the Walker field is 7.74 mT. Predict the wall speed at 5 mT, and whether it will be six times larger at 30 mT.
2. In the domain-wall Lab choose “Without DMI, steady only below 8 mT”, then “Past the Walker field the moment spins”, which sets $\mu_0H=30\,\mathrm{mT}$.
3. Explain why the wall-angle readout changes from a fixed angle to a turn period, and why the speed curve bends down after the Walker field.

[Push a domain wall and launch spin waves](https://sandboxphysics.com/labs/domain-wall-spin-waves/)

The macrospin Lab treats the strip of muMAG Standard Problem 4 as one moment. The wall and skyrmion Labs use the Pt/Co film of Thiaville et al. with local stray-field terms. All three run at zero temperature.

## Try explaining it yourself

### 1. With $D=0$ and $\alpha=0.25$, what is the Walker field?

Read an explained answer

$\alpha\,\mu_0H_K/2=0.25\times31.0/2\,\mathrm{mT}=3.87\,\mathrm{mT}$. Less damping lowers the Walker field.

### 2. With $D=0.5\,\mathrm{mJ/m^2}$ and $\alpha=0.5$, how fast does the wall move at 20 mT?

Read an explained answer

20 mT is below the 65.5 mT Walker field, so $v=\gamma_0\Delta H/\alpha=39.4\,\mathrm{m/s}$.

### 3. Why does reversing the sign of D reverse the wall moment?

Read an explained answer

The DMI term in the energy is $-\pi D\cos\Phi$. For positive D it is lowest at $\Phi=0$; for negative D at $\Phi=180^{\circ}$. The handedness of the wall follows the sign of D.

## Sources to read alongside

1. [Mougin, Cormier, Adam, Metaxas & Ferré, Domain wall mobility, stability and Walker breakdown in magnetic nanowires, EPL 78, 57007 (2007), arXiv:cond-mat/0702492](https://arxiv.org/abs/cond-mat/0702492)

   Read in full: the wall profile and width (Eqs. 1a and 1b), the steady angle (Eq. 11), the Walker field (Eq. 13), the steady speed (Eq. 15) and the mean speed far above the Walker field (Eq. 16).

2. [Thiaville, Rohart, Jué, Cros & Fert, Dynamics of Dzyaloshinskii domain walls in ultrathin magnetic films, EPL 100, 57002 (2012), arXiv:1211.5970](https://arxiv.org/abs/1211.5970)

   Read in full: the Pt/Co/AlOx parameters, the effective anisotropy, the critical DMI for a Néel wall, the collective-coordinate wall equations with DMI and the steady speed below the Walker field.

[Previous Exchange smooths the magnetization and sets how fine the mesh must be](https://sandboxphysics.com/stories/micromagnetics/exchange-and-the-mesh/) [Next How a small tilt travels through a magnet as a spin wave](https://sandboxphysics.com/stories/micromagnetics/spin-waves-and-dispersion/)