advanced32 min studyLesson 527 in Physics

Landau–Lifshitz–Gilbert dynamics

Separate precession from dissipation, convert the Gilbert equation, and compute constant-field relaxation with explicit field units.

Before you begin

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The question for today

What you’ll understand

Use the magnetization and gyromagnetic signs consistently.
Convert implicit Gilbert damping to its explicit form.
Check norm conservation, energy decrease, and damping-dependent time scales.

A magnetic moment experiences a torque in a field, but torque alone makes it precess rather than align. Alignment requires energy transfer to other degrees of freedom. The Landau–Lifshitz–Gilbert equation combines these two effects in an equation for the orientation of a nearly fixed magnetization. Its compact form hides several conventions: whether the field is B or H, whether the gyromagnetic ratio is signed, and whether damping is specified before or after eliminating an implicit time derivative.

Define the field and orientation before taking a cross product

Let $\mathbf n=\mathbf M/M_s$ be the unit magnetization direction with positive saturation magnitude Ms. Use a positive gyromagnetic magnitude gamma in radians per second per tesla. Define the effective field in tesla by $\mathbf B_{\mathrm{eff}}=-(1/M_s)\delta E/\delta\mathbf n$ , where the functional derivative is an energy density. This field can include applied, exchange, anisotropy, and demagnetizing contributions. It is a thermodynamic derivative in these units, not automatically the local Maxwell induction inside a material.

With this convention the precession term is minus gamma times n crossed into the effective field. Starting with n along positive x in a positive-z field, the derivative points along positive y. The electron angular momentum points opposite to magnetization; switching between those vectors without changing signs would reverse the predicted motion. Some software writes the equation with H in amperes per metre and a gyromagnetic coefficient containing a factor μ0, so numerical gyromagnetic values cannot be transferred without checking the field units.

\dot{\mathbf n}=-\gamma\mathbf n\times\mathbf B_{\mathrm{eff}}+\alpha\mathbf n\times\dot{\mathbf n}=-\frac{\gamma}{1+\alpha^2}\left[\mathbf n\times\mathbf B_{\mathrm{eff}}+\alpha\mathbf n\times(\mathbf n\times\mathbf B_{\mathrm{eff}})\right]

This is the Gilbert equation and its equivalent explicit form for |n|=1, γ>0 in s−1 T−1, Beff in tesla, and dimensionless damping α≥0. Converting to an H-based equation introduces μ0 and does not change the physical trajectory.↔ Scroll wide equations horizontally.

Check the norm and the energy separately

Both cross-product terms are perpendicular to n, so the exact equation preserves its length. The first term is also perpendicular to the effective field and does no work on a static energy functional. The second term turns the orientation toward lower energy. Substituting the effective-field definition gives a nonpositive energy derivative proportional to the squared transverse torque. This sign check catches either an incorrect damping sign or an inconsistent definition of the field.

Energy decrease assumes the functional has no explicit time dependence and that there are no applied spin-transfer or other driving torques. A changing applied field can supply energy even when damping remains positive. Likewise, the deterministic equation omits thermal noise, which requires a fluctuation-consistent stochastic extension at finite temperature. It is a classical collective-orientation equation; it does not compute a many-body quantum density matrix or the entanglement of neighboring microscopic spins.

\frac{dE}{dt}=-\frac{\alpha\gamma M_s}{1+\alpha^2}\int|\mathbf n\times\mathbf B_{\mathrm{eff}}|^2dV\leq0,\qquad \dot\phi=\frac{\gamma B}{1+\alpha^2},\qquad \tan\frac{\theta(t)}2=\tan\frac{\theta(0)}2e^{-t/\tau},\quad \tau=\frac{1+\alpha^2}{\alpha\gamma B}

The energy identity uses a static functional and constant Ms. The angular solution additionally assumes one uniform magnetization in a constant field B along positive z, with α>0. θ is its polar angle and φ its azimuth. τ is the decay time of tan(θ/2), not exactly of θ at arbitrary angle.↔ Scroll wide equations horizontally.

A damping parameter changes more than an envelope

For small tilt the polar angle itself decays approximately exponentially with the same time constant, while the azimuth advances at the explicit precession frequency. At zero damping the orientation stays at fixed polar angle and precesses forever in this ideal model. At finite damping the trajectory spirals toward the field, except for an exactly antiparallel stationary orientation that requires a perturbation to depart. The exact constant-field solution makes initial conditions as important as the energy minimum.

Keeping the Gilbert gamma fixed, the relaxation rate is proportional to alpha divided by one plus alpha squared. It grows initially, peaks at alpha equal to one, and decreases for still larger damping. Thus the slogan that more damping always means faster relaxation is false in this parametrization. Micromagnetic simulations must also resolve spatial gradients and preserve the unit constraint numerically; satisfying those checks matters independently of whether a plotted trajectory looks like a plausible spiral.

Work it through

Time scales in a uniform static field

A uniform magnetization obeys the Gilbert convention above with α=0.10 and γB=2.0×10¹⁰ s−1. Find its azimuthal angular frequency, precession period, and the decay time of tan(θ/2). Assume an initial tilt strictly between zero and π.

The implicit Gilbert derivative supplies the denominator $1+\alpha^2=1.01$ . Thus $\Omega=2.0\times10^{10}/1.01\simeq1.9802\times10^{10}\,\mathrm{s^{-1}}$ .
A full rotation is two pi radians, so $T_{\mathrm{prec}}=2\pi/\Omega\simeq3.173\times10^{-10}\,\mathrm s=0.3173\,\mathrm{ns}$ . Using inverse angular frequency alone would miss the factor two pi.
The polar-variable decay time is $\tau=1.01/[0.10(2.0\times10^{10})]=5.05\times10^{-10}\,\mathrm s=0.505\,\mathrm{ns}$ . It describes tan of half the tilt exactly and the tilt itself only near alignment.

The result

The angular frequency is about 1.9802×10¹⁰ s−1, the period is 0.3173 ns, and the half-angle variable decays on a 0.505 ns time scale.

The living notebook

Explore the model

Interactive lab

Interactive lab · oscillator

Distinguish an oscillation from relaxation

For each independent run, Reset and set BOTH Natural frequency ω₀=1 and Damping ratio ζ. Compare ζ=0.20 with ζ=1.00. Predict the underdamped and critical regime readouts, respectively, and use Play or Step to inspect their return from x(0)=1 with zero initial velocity. This positive-stiffness scalar oscillator illustrates different relaxation shapes. Its ζ is not the Gilbert α, and it does not solve vector precession, preserve a magnetization norm, or calculate a magnetic field.

Displacement x(t)Current time

Displacement x(t): 1
Velocity v(t): 0
Regime: underdamped

Damping ratio ζ0.2Natural frequency ω₀1.4

t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forDissipation changes the temporal response, but damping conventions and dynamical equations must be identified before comparing rates.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The explicit equation contains 1+α². With the Gilbert gyromagnetic magnitude held fixed, the relaxation rate peaks at α=1 rather than increasing indefinitely.

Your turn

Try it yourself

In a constant field with γB=1.0×10⁹ s−1, compare α=1 and α=2 for an initial polar angle θ(0)=90°. Compute both decay times and the value of tan(θ/2) after 2 ns. Use e−1≈0.3679 and e−0.8≈0.4493.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The initial half-angle tangent is one. Apply the exact half-angle decay rather than assuming the full polar angle decays exponentially from 90 degrees.

Compare with the solution

For α=1, $\tau=(1+1)/(10^9)=2.0\,\mathrm{ns}$ , giving $\tan[\theta(2\,\mathrm{ns})/2]=e^{-1}\simeq0.3679$ . For α=2, $\tau=(1+4)/(2\times10^9)=2.5\,\mathrm{ns}$ , giving the tangent $e^{-0.8}\simeq0.4493$ . The larger remaining half-angle variable means the α=2 magnetization has aligned less by that time. The corresponding polar angles are about 40.4 and 48.4 degrees. This comparison holds γ in the implicit Gilbert form fixed; a different choice of fixed precession coefficient changes how damping parameters are compared.

Keep the idea

Take it with you

The sign and units of the effective field determine the precession convention.
Positive Gilbert damping reduces a static energy while preserving the magnetization magnitude.
At fixed Gilbert γ, the exact constant-field relaxation rate is proportional to α/(1+α²).

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