Micromagnetics · A course with experiments

L3 · Chapter 6

Why a skyrmion has a set size and a whole-number charge

An isolated skyrmion in this film is a small disc of down-pointing moments in an up film, ringed by a Néel wall whose handedness DMI fixes. Its radius is a balance: DMI lowers the energy of the wall, and bending the wall into a circle costs exchange. At $3.0\,\mathrm{mJ/m^2}$ the radius is 9.4 nm. Its topological charge counts how many times the moments cover the sphere of directions, and it is −1. Next to an open edge the computed charge drifts slightly away from −1 even when the simulation is correct, because the moments at the edge tilt.

Sandbox Physics · 2026-10-29

DMI makes the wall cheap but keeps it positive

Chapter 4 gave the wall energy per area. Without the small hard-axis term, a Néel wall of the right handedness costs $\sigma=4\sqrt{AK_0}-\pi D$ (Rohart and Thiaville 2013, Eq. 10). For this film $4\sqrt{AK_0}=11.4\,\mathrm{mJ/m^2}$ , so at $D=3.0\,\mathrm{mJ/m^2}$ the wall costs only $2.0\,\mathrm{mJ/m^2}$ .

The wall energy reaches zero at $D_c=4\sqrt{AK_0}/\pi=3.64\,\mathrm{mJ/m^2}$ . Above it, walls lower the energy and the film breaks up into cycloids. Below it the uniform film is the ground state. A skyrmion is then a stable texture of higher energy: small changes do not remove it, but it is not the lowest state. Rohart and Thiaville note that such skyrmions are known as isolated skyrmions. The Lab keeps $D$ between 2.8 and $3.4\,\mathrm{mJ/m^2}$ , below $D_c$ .

Worked example

Closing the wall into a ring

Picture a skyrmion of radius $R_s$ as a straight wall bent into a circle, in a film of thickness $t$ . The wall costs $2\pi R_st\sigma$ . Bending it costs exchange in proportion to $1/R_s$ ; Rohart and Thiaville estimate this term as $4\pi tA\Delta/R_s$ (their Eq. 17). The first term wants a small ring, the second a large one.

Set the derivative with respect to $R_s$ to zero: $R_s^2=2A\Delta/\sigma$ . With $\sigma=4(A/\Delta)(1-D/D_c)$ this gives their Eq. 18. At $D=3.0\,\mathrm{mJ/m^2}$ it predicts 9.47 nm.

The estimate assumes the ring is much wider than the wall, $R_s\gg\Delta$ . The full radial profile, their Eq. 16a solved numerically, tests it: 9.40 nm at $3.0\,\mathrm{mJ/m^2}$ , close to the estimate, but 5.77 nm against 7.42 nm at $2.6\,\mathrm{mJ/m^2}$ , where the ring is barely wider than the wall, and 17.3 nm against 15.5 nm at $3.4\,\mathrm{mJ/m^2}$ . The Lab uses the full profile.

E_s\approx2\pi R_st\sigma+\frac{4\pi tA\Delta}{R_s}\quad\Longrightarrow\quad R_s\approx\frac{\Delta}{\sqrt{2(1-D/D_c)}}

What the topological charge counts

Map every point of the film to the direction of its moment, a point on the unit sphere. The uniform film maps to the north pole. A skyrmion’s moments run from the south pole at the core, through every direction around the equator in the wall, back to the north pole far away: they cover the sphere once. The topological charge counts the coverings, with a sign for orientation (Kim and Mulkers 2020, Eq. 1).

For a radial profile $\theta(r)$ that turns once around the azimuth, the integral reduces to $Q=\tfrac12[\cos\theta(0)-\cos\theta(\infty)]$ . A down core in an up film gives $Q=-1$ ; an up core in a down film gives $+1$ .

Abert describes the consequence: a skyrmion cannot be turned into the uniform state by a continuous change of the magnetization. On a grid of cells that protection is only approximate, which is what the rest of this chapter is about.

Q=\frac{1}{4\pi}\int\mathbf m\cdot\left(\frac{\partial\mathbf m}{\partial x}\times\frac{\partial\mathbf m}{\partial y}\right)\mathrm{d}^2r

Worked example

Two ways to count on a grid

The direct way replaces the derivatives in Eq. 1 by finite differences. It works while neighbouring moments differ little, and its error falls with the square of the cell size.

The lattice way, which Kim and Mulkers take from Berg and Lüscher, uses no derivatives. Split each square of four neighbouring cells into two triangles, and add up the signed solid angle that each triangle of three moments spans on the sphere. Kim and Mulkers average the two ways of splitting a square, giving each triangle weight one half (their Eqs. 2 and 3).

Try it on three moments along $\hat x$ , $\hat y$ and $\hat z$ . The numerator is $\hat x\cdot(\hat y\times\hat z)=1$ and the denominator is 1, so $\tan(q/2)=1$ and $q=\pi/2$ , one eighth of the sphere’s $4\pi$ . Replace $\hat z$ by $-\hat z$ and the sign flips to $-\pi/2$ .

In an unbounded film, a periodic box, the skyrmion in the Lab gives a lattice charge of −1 and a finite-difference charge of −0.956 with 2 nm cells and −0.989 with 1 nm cells. Halving the cells cuts the finite-difference error by about four, as a second-order method should. Kim and Mulkers show the same contrast in their Fig. 1: −0.99978 for a smooth skyrmion and −0.971 once random disorder is added, though the covering is unchanged.

Q=\frac{1}{4\pi}\sum_{\langle ijk\rangle}q_{ijk},\qquad \tan\frac{q_{ijk}}{2}=\frac{\mathbf m_i\cdot(\mathbf m_j\times\mathbf m_k)}{1+\mathbf m_i\cdot\mathbf m_j+\mathbf m_i\cdot\mathbf m_k+\mathbf m_j\cdot\mathbf m_k}

An open edge shifts the charge without anything going wrong

With DMI, the moments at a free edge cannot stay perpendicular. Rohart and Thiaville derive the edge condition (their Eqs. 4b and 6) and, for a straight edge, the tilt $\sin\theta_{\mathrm{edge}}=\Delta/\xi$ with $\xi=2A/D$ (their Eq. 8). At $D=3.0\,\mathrm{mJ/m^2}$ , $\xi=10.7\,\mathrm{nm}$ and the edge moments tilt by $\sin\theta_{\mathrm{edge}}=0.525$ . The tilt fades into the film over about $\Delta$ .

Tilted edge moments cover part of the sphere, so with open edges neither definition has to give a whole number. In the 60 nm track of the skyrmion Lab the lattice charge is −0.999935 with the skyrmion centred and −0.999631 when it sits 23.6 nm from the lower edge. With 1 nm cells the values are −0.999937 and −0.999636. Halving the cells leaves the deficit unchanged, so it is set by the edge, not by the grid. A track without a skyrmion gives 0 in both definitions, because a straight edge tilts the same way all along its length.

Kim and Mulkers report a larger shift in a 200 nm disc with a curved edge: −0.9305 from finite differences and −0.9468 from the lattice at zero temperature, which they attribute to the edge tilt. They conclude that non-integer charges in confined structures with chiral edges need careful reading. This course turns that into a rule: a charge that departs from an integer near an edge is first a statement about the edge, and only a mesh test shows whether part of it is numerical error.

An experiment to think through

Predict, change, explain

Predict both charge readouts for a skyrmion in an unbounded film, then for one running along the centre of the track.
In the skyrmion Lab choose “No edges: the Hall angle stays”, then “No sideways drift when beta equals alpha”, and read the two charge readouts beside the stage each time.
The lattice readout shows −1.000 both times, while the finite-difference readout differs between the two. Explain what each number measures, and which test would show whether the difference is numerical. The exported record gives the lattice charge to more digits.

Drive a skyrmion along a track

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. Use the estimate of Eq. 18 to find the radius at $D=3.2\,\mathrm{mJ/m^2}$ . Is it larger or smaller than the full profile’s 12.3 nm?

Read an explained answer

$R_s\approx5.60\,\mathrm{nm}/\sqrt{2(1-3.2/3.64)}=11.4\,\mathrm{nm}$ , about 7 per cent smaller than the full profile.

2. What is the edge tilt at $D=2.8\,\mathrm{mJ/m^2}$ ?

Read an explained answer

$\xi=2A/D=11.4\,\mathrm{nm}$ , so $\sin\theta_{\mathrm{edge}}=5.60/11.4=0.490$ .

3. A single-precision simulation reports a lattice charge of −1.0000004 for a skyrmion in a periodic film. Is that a mesh error to fix by halving the cells?

Read an explained answer

No. For a smooth texture in a periodic film the lattice charge is an integer apart from rounding, and Kim and Mulkers trace fluctuations of this size (−1.0000008 to −0.9999996 at 400 K) to single-precision arithmetic. A finite-difference charge of −0.956 in the same film is a mesh error, and halving the cells is the test for it.

Sources to read alongside

Abert, Micromagnetics and spintronics: models and numerical methods, European Physical Journal B 92, 120 (2019), arXiv:1810.12365v2
Read: the exchange energy (Eq. 10), the demagnetizing field (Sec. 2.3), uniaxial anisotropy (Sec. 2.4), interfacial DMI (Eq. 33) and its boundary term (Sec. 3.5), the Landau-Lifshitz-Gilbert equation in Gilbert and explicit form (Eqs. 117 and 118), the conservation of the magnetization length and the energy dissipation (Eqs. 120 to 125), the Zhang-Li torque (Eq. 134) and the finite-difference method (Sec. 6.1). Other sections were not read.
Rohart & Thiaville, Skyrmion confinement in ultrathin film nanostructures in the presence of Dzyaloshinskii-Moriya interaction, Physical Review B 88, 184422 (2013), arXiv:1310.0666
Read in full: the Pt/Co/AlOx parameters and the local treatment of the stray field in ultrathin films, the free-edge boundary condition and edge tilt (Eqs. 3 to 8, with the full-dipolar check of Fig. 1), the isolated-skyrmion equation (Eqs. 15 and 16) and the approximate radius (Eq. 18).
Kim & Mulkers, On quantifying the topological charge in micromagnetics using a lattice-based approach, IOP SciNotes 1, 025211 (2020), arXiv:2006.13336
Read in full: the finite-difference charge (Eq. 1), the lattice charge from signed solid angles (Eqs. 2 and 3, Fig. 2), the three tests of Sec. 3 (Figs. 3 to 5) and the disc example in which edge tilt makes both values non-integer.

DMI makes the wall cheap but keeps it positive

Closing the wall into a ring

What the topological charge counts

Two ways to count on a grid

An open edge shifts the charge without anything going wrong

Predict, change, explain

Try explaining it yourself

1. Use the estimate of Eq. 18 to find the radius at D=3.2 mJ/m2D=3.2\,\mathrm{mJ/m^2}D=3.2mJ/m2. Is it larger or smaller than the full profile’s 12.3 nm?

2. What is the edge tilt at D=2.8 mJ/m2D=2.8\,\mathrm{mJ/m^2}D=2.8mJ/m2?

3. A single-precision simulation reports a lattice charge of −1.0000004 for a skyrmion in a periodic film. Is that a mesh error to fix by halving the cells?

Sources to read alongside

1. Use the estimate of Eq. 18 to find the radius at $D=3.2\,\mathrm{mJ/m^2}$ . Is it larger or smaller than the full profile’s 12.3 nm?

2. What is the edge tilt at $D=2.8\,\mathrm{mJ/m^2}$ ?