A magnetic moment circles a field before it settles

A field pushes a magnetic moment at right angles to both the moment and the field, so the moment circles the field at a rate fixed by the field strength: 28.0 GHz per tesla. A weak damping term turns the circle into a slow spiral that ends along the field. With the damping 0.02 of the nanomagnet in this course, the moment goes round about eight times while its tilt shrinks by a factor of e.

Sandbox Physics · 2026-10-29

The moment, the field and the equation

Inside a ferromagnet the electron spins add up to a magnetization $\mathbf{M}=M_s\mathbf{m}$ . Its size, the saturation magnetization $M_s$ , is fixed by the material; micromagnetics follows only its direction, the unit vector $\mathbf{m}$ . A field $\mathbf{H}$ is measured in amperes per metre; we quote it as $\mu_0H$ in millitesla.

Electron spin carries angular momentum, so a field does not rotate it the way it rotates a compass needle in honey. The torque changes the angular momentum, and the angular momentum is the moment itself. The result, without damping, is the first term of the Landau-Lifshitz-Gilbert equation (Abert 2019, Eq. 117), with $\gamma_0=2.211\times10^{5}\,\mathrm{m\,A^{-1}\,s^{-1}}$ , the value the muMAG standard problem uses.

\frac{\mathrm{d}\mathbf{m}}{\mathrm{d}t}=-\gamma_0\,\mathbf{m}\times\mathbf{H}

Why the moment circles

A cross product is perpendicular to both of its factors. The change of $\mathbf{m}$ is therefore perpendicular to $\mathbf{m}$ , so its length stays 1, and perpendicular to $\mathbf{H}$ , so the angle $\theta$ between them stays fixed. The tip of $\mathbf{m}$ can only run around a circle of radius $\sin\theta$ about the field. This motion is precession.

Its speed is $|\mathrm{d}\mathbf{m}/\mathrm{d}t|=\gamma_0H\sin\theta$ . Dividing by the radius gives the angular rate $\omega=\gamma_0H$ , the same at every tilt. Converted to a frequency per tesla of $\mu_0H$ this is 28.0 GHz per tesla.

f=\frac{\gamma_0H}{2\pi}=\frac{\gamma_0}{2\pi\mu_0}\,\mu_0H\approx28.0\,\mathrm{GHz\,T^{-1}}\times\mu_0H

Damping turns the circle into a spiral

Real magnets lose energy to the lattice and to conduction electrons. Gilbert added a term that opposes the motion, $\alpha\,\mathbf{m}\times\dot{\mathbf{m}}$ , with a small dimensionless damping $\alpha$ . Solving for $\dot{\mathbf{m}}$ gives the explicit form below (Abert 2019, Eq. 118). Its second term points from $\mathbf{m}$ toward the field.

The energy density of the moment in the field is $E=-\mu_0M_s\,\mathbf{m}\cdot\mathbf{H}$ . Inserting the explicit form shows that it can only fall, at the rate $\dot{E}=-\alpha\gamma_0\mu_0M_s|\mathbf{m}\times\mathbf{H}|^2/(1+\alpha^2)$ (Abert 2019, Eqs. 121 to 124). With $\alpha=0$ the energy is conserved and the moment circles for ever.

\dot{\mathbf{m}}=-\frac{\gamma_0}{1+\alpha^2}\,\mathbf{m}\times\mathbf{H}-\frac{\alpha\gamma_0}{1+\alpha^2}\,\mathbf{m}\times(\mathbf{m}\times\mathbf{H})

Worked example

How many turns before it settles

Put the field along $z$ and describe $\mathbf{m}$ by its tilt $\theta$ and its azimuth $\varphi$ . The first term of the explicit form moves only $\varphi$ , at $\omega=\gamma_0H/(1+\alpha^2)$ . The second moves only $\theta$ , at $\dot\theta=-\alpha\omega\sin\theta$ .

Separate the variables: $\mathrm{d}\theta/\sin\theta=-\alpha\omega\,\mathrm{d}t$ . The integral of $1/\sin\theta$ is $\ln\tan(\theta/2)$ , so $\tan(\theta/2)$ falls exponentially with time constant $1/(\alpha\omega)$ . In that time the moment makes $\omega/(2\pi)\times1/(\alpha\omega)=1/(2\pi\alpha)$ turns, whatever the field.

For the standard-problem damping $\alpha=0.02$ that is 7.96 turns. In a field of 25 mT the precession frequency is 0.700 GHz and the time constant is 11.4 ns. The macrospin Lab integrates exactly this spiral as its first check, and its fourth-order integrator reproduces the formula at observed order 4.00.

\tan\frac{\theta(t)}{2}=\tan\frac{\theta_0}{2}\,e^{-\alpha\omega t},\qquad N_{e}=\frac{1}{2\pi\alpha}=7.96\quad(\alpha=0.02)

Why the element in the Lab is different

The macrospin Lab treats one nanomagnet, a film 500 nm long, 125 nm wide and 3 nm thick with $M_s=800\,\mathrm{kA/m}$ and $\alpha=0.02$ , the element of muMAG standard problem 4. It follows the element as one rigid moment. That moment does not sit in a uniform external field only: the element makes a strong field of its own, which depends on where the moment points. The next chapter works out that field, and why it makes the moment ring at 4.66 GHz with no external field at all.

Predict, change, explain

Predict how the spiral on the sphere changes if the damping is raised from 0.02 to 0.1: more or fewer turns before the moment settles, and does each turn take longer?

In the macrospin Lab choose “A small kick rings at the Kittel frequency”, watch one run, then set “Gilbert damping” to

0.1

and watch again.

Explain what you see using the number of turns per e-fold,

1/(2\pi\alpha)

, and the factor

1/(1+\alpha^2)

in the precession rate.

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. What is the precession frequency in a field of $\mu_0H=100\,\mathrm{mT}$ ?

Read an explained answer

About 2.80 GHz: 28.0 GHz per tesla times 0.1 T. The damping factor $1/(1+\alpha^2)$ changes this by less than 0.1 per cent for $\alpha=0.02$ .

2. With $\alpha=0.01$ , how many turns does the moment make while $\tan(\theta/2)$ shrinks by a factor of e?

Read an explained answer

About 15.9 turns, from $1/(2\pi\alpha)$ . Halving the damping doubles the number of turns.

3. Show that without damping the energy $E=-\mu_0M_s\,\mathbf{m}\cdot\mathbf{H}$ cannot change in a constant field.

Read an explained answer

Its rate is $\dot E=-\mu_0M_s\,\mathbf{H}\cdot\dot{\mathbf{m}}=\mu_0M_s\gamma_0\,\mathbf{H}\cdot(\mathbf{m}\times\mathbf{H})$ . A cross product is perpendicular to each factor, so the dot product with $\mathbf{H}$ is zero.

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.

muMAG Standard Problem #4, NIST Center for Theoretical and Computational Materials Science

Full specification page: a film 500 nm long, 125 nm wide and 3 nm thick, with saturation magnetization 800 kA/m, exchange 13 pJ/m, damping 0.02, gyromagnetic ratio 221.1 km/(A s) and two reversal fields. The macrospin Lab borrows the geometry and constants only; it does not solve the problem.

The moment, the field and the equation

Why the moment circles

Damping turns the circle into a spiral

How many turns before it settles

Why the element in the Lab is different

Predict, change, explain

Try explaining it yourself

1. What is the precession frequency in a field of μ0H=100 mT\mu_0H=100\,\mathrm{mT}μ0​H=100mT?

2. With α=0.01\alpha=0.01α=0.01, how many turns does the moment make while tan⁡(θ/2)\tan(\theta/2)tan(θ/2) shrinks by a factor of e?

3. Show that without damping the energy E=−μ0Ms m⋅HE=-\mu_0M_s\,\mathbf{m}\cdot\mathbf{H}E=−μ0​Ms​m⋅H cannot change in a constant field.

Sources to read alongside

1. What is the precession frequency in a field of $\mu_0H=100\,\mathrm{mT}$ ?

2. With $\alpha=0.01$ , how many turns does the moment make while $\tan(\theta/2)$ shrinks by a factor of e?

3. Show that without damping the energy $E=-\mu_0M_s\,\mathbf{m}\cdot\mathbf{H}$ cannot change in a constant field.