Grain boundaries move by curvature, so a round grain shrinks at a steady rate

Sandbox Physics · 2026-10-30

Heat a metal made of many small crystals and the crystals grow: small ones disappear and boundaries straighten. Nothing pushes the boundaries from outside. They move because boundary length costs energy, and a curved boundary lowers it by moving toward its centre of curvature. For a round grain the result is striking: its area falls at a constant rate,

2\pi M\gamma

, whatever its size. This chapter derives that law from the phase-field model the Lab runs and measures how closely the Lab’s grid follows it.

Boundaries cost energy, and curvature drives them

Where two crystals of different orientation meet, atoms cannot sit in either lattice, so the boundary carries an energy $\gamma$ per unit length in two dimensions. Shortening the boundary lowers the energy. A straight boundary cannot get shorter by moving, but a curved one can: moving toward its centre of curvature shortens it.

The simplest kinetic law makes the speed proportional to the driving force, which for a boundary of energy $\gamma$ and curvature $\kappa$ is $\gamma\kappa$ per unit area. With a mobility $M$ , the speed is $M\gamma\kappa$ . For a round grain of radius $R$ the curvature is $1/R$ , and the radius shrinks ever faster as it gets smaller. Its area, though, falls at a constant rate, because the faster motion acts on a shorter perimeter.

v=M\gamma\,\kappa\quad\Longrightarrow\quad \frac{dR}{dt}=-\frac{M\gamma}{R},\qquad \frac{dA}{dt}=2\pi R\,\frac{dR}{dt}=-2\pi M\gamma

Motion by curvature, and its consequence for a round grain of area

A=\pi R^2

Source: J. K. Mason · A. Kazaryan

One order parameter for each grain

Kazaryan, Wang, Dregia and Patton describe a polycrystal with one field $\eta_i$ per orientation. Inside grain $i$ , $\eta_i=1$ and the others vanish; across a boundary one field falls as the next rises. Each field relaxes down the gradient of a free energy, Eq. (1). The free energy, Eq. (2), adds a gradient term with coefficient $k$ to a bulk density $f$ that has a minimum for each pure grain.

The bulk density of their Eq. (5) with $a_1=a_2=1$ and $a_3=2$ is written out below. The cross term penalises two fields that are both large at one point, so grains do not overlap. The Lab gives every grain its own field, so two grains can never merge by sharing an index, and stores at each cell only the four largest fields.

\frac{\partial\eta_i}{\partial t}=-L\,\frac{\delta F}{\delta\eta_i},\qquad F=\int\Bigl[f(\eta_1,\dots,\eta_p)+\frac{k}{2}\sum_i(\nabla\eta_i)^2\Bigr]d^2r

Allen–Cahn relaxation of each order parameter with mobility

L

(Kazaryan et al., Eqs. 1–2).

f=\sum_i\Bigl(-\frac{\eta_i^2}{2}+\frac{\eta_i^4}{4}\Bigr)+\sum_i\sum_{j>i}\eta_i^2\eta_j^2

The bulk free-energy density used by the Lab, Eq. (5) of Kazaryan et al. with their parameter values.

Source: A. Kazaryan

From the phase field to motion by curvature

Near a gently curved boundary, write the Laplacian in a coordinate $n$ along the normal. It splits into the second derivative across the boundary plus a curvature term. The flat equilibrium profile already balances the bulk force against the second-derivative term, so only the curvature term is left over.

That leftover is $Lk\kappa$ times the slope of the profile. A profile that slides along its normal at speed $v$ changes at exactly that rate times its slope, so the boundary moves with $v=Lk\kappa$ . The bulk density drops out. Comparing with the sharp law gives $M\gamma=Lk$ , and an island grain shrinks as $dA/dt=-2\pi Lk$ , which is Eq. (6) of Kazaryan et al. for an isotropic boundary.

\nabla^2\eta=\partial_n^2\eta+\kappa\,\partial_n\eta\quad\Longrightarrow\quad \partial_t\eta=Lk\kappa\,\partial_n\eta\quad\Longrightarrow\quad v=Lk\,\kappa,\qquad M\gamma=Lk

The curvature term left over near a curved boundary, read as a moving profile.

Source: A. Kazaryan

How close does the grid come?

The route benchmark shrinks round grains on a $128$ by $128$ grid with $L=k=1$ and fits a straight line to the area while it falls from $0.9$ to $0.25$ of its start. Divided by $2\pi Lk$ , the rate is $0.956$ , $0.953$ and $0.952$ for starting radii $20$ , $30$ and $40$ . The rate does not depend on size, as the law says, but is about $5\%$ slow.

Raising $k$ widens the boundary, which then spans more grid cells. With the radius scaled up by the same factor, the runs are the same problem on a finer grid. The ratio rises to $0.978$ at $k=2$ and $0.990$ at $k=4$ . The shortfall halves each time $k$ doubles, which is what a second-order discretisation gives when the boundary width grows as $\sqrt k$ . The gap is a grid error that closes as the boundary is resolved more finely.

Source: A. Kazaryan

Worked example: how long does the Lab’s island last?

Problem. The Lab’s island preset starts a round grain of radius $40$ cells with $L=k=1$ . When does it vanish by the sharp law, and when on the Lab grid?

Sharp law. The area is $\pi\times40^2=5027$ cells and falls at $2\pi=6.283$ cells per unit time, so the grain lasts $R_0^2/(2Lk)=1600/2=800$ time units.

Lab grid. The measured rate is $0.952$ of the law, so the grain lasts $800/0.952=840$ time units.

Check. The radius at half the lifetime is $R_0/\sqrt2=28.3$ cells, not $20$ ; a steady area rate makes the radius fall slowly at first and quickly at the end.

Answer. $800$ time units by the law, about $840$ on the Lab grid.

Source: A. Kazaryan

Try it in the Lab

Time a shrinking round grain

Choose A round grain shrinks at a steady rate. Predict whether the island’s area falls along a straight line or a curve, and how long it lasts.
Reset and double the boundary mobility to 2. Predict the new lifetime, then watch.
Choose A large grain eats its neighbours. Explain why a round grain on its own always shrinks, while a large grain surrounded by small ones can grow.

Exercises with explained answers

With $L=1$ and $k=2$ , how long does a round grain of starting radius $30$ last, by the law and on the grid?

Show the answer

By the law, $R_0^2/(2Lk)=900/4=225$ time units. The benchmark ratio at $k=2$ is $0.978$ , so about $230$ on the grid.

The Lab advances the field with forward Euler steps. Its nine-point Laplacian has its most negative eigenvalue at $-16/3$ , and the bulk term adds a curvature of up to $2$ , so a step is stable only if $L\Delta t\,(16k/3+2)<2$ . Is the default $\Delta t=0.1$ stable for $L=k=2$ ?

Show the answer

The bound is $\Delta t<2/[2\times(32/3+2)]=0.079$ , so $0.1$ is unstable. The Lab takes $0.8$ of the bound, $\Delta t=0.063$ , whenever the default step would exceed it.

Why does a straight boundary between two large grains not move at all in this model, however different their orientations?

Show the answer

Its curvature is zero, so $v=Lk\kappa=0$ . In the isotropic model the driving force comes only from curvature. A flat boundary would need a second driving force, such as a difference in stored energy between the grains, which this model leaves out.

Sources and what we recomputed

A. Kazaryan, Y. Wang, S. A. Dregia and B. R. Patton, Generalized phase field model for computer simulation of grain growth in anisotropic systems, arXiv:cond-mat/9912044v1 (1999)

Version 1, full text. Eq. (1): one Allen–Cahn equation per order parameter. Eq. (2): the free energy with gradient terms. Eq. (5): the bulk free-energy density, used with $a_1=a_2=1$ and $a_3=2$ . Eq. (6) and Fig. 3: an isolated grain shrinks at a constant area rate. The closing paragraphs and Fig. 5: the mean grain area grows linearly in time. This course uses only the isotropic case.

J. K. Mason, E. A. Lazar, R. D. MacPherson and D. J. Srolovitz, Geometric and topological properties of the canonical grain growth microstructure, arXiv:1507.03379v1 (2015)

Version 1. Section II, Eq. (1): the von Neumann–Mullins relation $dA/dt=-2\pi M\gamma(1-n/6)$ , exact for isotropic two-dimensional grain growth. Section III A 1 and Fig. 4(a): the distribution of edges per grain, and Euler’s theorem fixing the mean number of edges at six.