--- title: "The von Neumann–Mullins law says grains with fewer than six sides shrink" description: "In a two-dimensional polycrystal, a grain’s area rate depends only on how many sides it has. Derive the law, count sides with Euler, and measure both." canonical_url: "https://sandboxphysics.com/stories/crystal-defects-microstructure/von-neumann-mullins-and-coarsening/" language: "en" page_type: "story" alternate_languages: zh-CN: "https://sandboxphysics.com/zh/stories/crystal-defects-microstructure/von-neumann-mullins-and-coarsening/" last_modified: "2026-10-30" --- [Crystal defects and microstructure](https://sandboxphysics.com/stories/crystal-defects-microstructure/) L3 · Chapter 5 # The von Neumann–Mullins law says grains with fewer than six sides shrink Sandbox Physics · 2026-10-30 In a flat polycrystal, whether a grain grows or shrinks does not depend on its size or shape, only on how many neighbours it has. Grains with fewer than six sides shrink, grains with more grow, and six-sided grains hold their area. This is the von Neumann–Mullins law. Together with Euler’s count, which fixes the average number of sides at six, it explains why a polycrystal coarsens. This chapter derives both and measures them in seeded simulations, where one of the two numbers comes out $2\%$ off. ## Before you begin Chapter 4: motion by curvature and the identity of the boundary speed with the mobility times the gradient coefficient. The total turning of a closed curve. Least-squares fitting of a line. - Derive the von Neumann–Mullins law from motion by curvature and 120-degree triple junctions. - Show with Euler’s formula that the mean number of sides is six, and use it to explain coarsening. - Compare the measured law with the prediction and say what the remaining discrepancy could and could not mean. ## Counting the turns around a grain A grain’s area changes at the rate its boundary sweeps inward, the integral of the normal speed around it. With motion by curvature that integral becomes $-M\gamma$ times the total curvature of the boundary. Walk once around the grain. Your direction turns by $2\pi$ in total. Part of the turning happens smoothly along the sides; the rest happens at the corners, the triple junctions. Where three boundaries of equal energy meet, their tensions balance only at $120^{\circ}$ angles, so at each corner your direction jumps by $\pi/3$. An $n$-sided grain turns by $n\pi/3$ at its corners, leaving $2\pi-n\pi/3$ for the smooth parts. The result, Eq. (1) of Mason, Lazar, MacPherson and Srolovitz, holds for every grain whatever its size or shape. Six-sided grains have straight sides on average and keep their area. With $M\gamma=Lk$ from chapter 4, the slope of the area rate against $n-6$ is $\pi/3=1.0472$ in units of $Lk$. $$ \frac{dA}{dt}=\oint v_n\,ds=-M\gamma\oint\kappa\,ds,\qquad \oint\kappa\,ds+n\,\frac{\pi}{3}=2\pi $$ The area rate as total curvature, and the turning of the boundary split between the smooth sides and the $n$ corners. $$ \frac{dA_n}{dt}=-2\pi M\gamma\Bigl(1-\frac n6\Bigr)=\frac{\pi}{3}\,M\gamma\,(n-6) $$ The von Neumann–Mullins law for an $n$-sided grain in an isotropic two-dimensional polycrystal. Source: J. K. Mason ## Six is the dividing line, and also the mean A two-dimensional grain network in a periodic box is a map on a torus with $V$ corners, $E$ sides and $F$ grains, and Euler’s formula gives $V-E+F=0$. Three sides meet at every corner and every side has two ends, so $3V=2E$. Each side borders two grains, so the side counts add to $2E$. Together these fix the mean number of sides at exactly six, as Section III A 1 of Mason et al. notes. The two facts fit together. Adding the law over all grains gives a total rate proportional to the sum of $n-6$, which vanishes, so the total area is conserved. Grains with few sides shrink and vanish; their area goes to grains with many sides. Each disappearance removes a grain, so the mean grain area grows. Kazaryan et al. report that the mean area grows linearly in time once the structure looks the same at every scale. $$ V-E+F=0,\quad 3V=2E\;\Longrightarrow\;F=\frac E3,\qquad \langle n\rangle=\frac{2E}{F}=6 $$ Euler’s formula on a torus for a network with three sides at every corner. Source: J. K. Mason · A. Kazaryan ## Measuring the law in a simulated polycrystal The route benchmark grows three seeded polycrystals of $80$ grains on a $160$ by $160$ grid with $L=k=1$ up to time $1000$. It counts each grain’s neighbours, records the area change of every grain over windows in which its side count stays fixed, and fits a line through the origin of rate against $n-6$. That gives $3620$, $3710$ and $4000$ samples per seed. The fitted slope is $1.0704\pm0.0007$ in units of $Lk$ over the three seeds, against $\pi/3=1.0472$. The figure shows where the excess comes from. Five-sided grains shrink at $-1.045$, on the law. Grains with seven or more sides grow $3\%$ to $6\%$ faster than the law, and six-sided grains grow at $+0.035$ on average instead of holding still. The mean number of sides is $5.979\pm0.004$, not exactly six. Those last two numbers point to one likely cause, which we have not tested. Two grains count as neighbours only if they share at least two cell faces, so a very short side can be missed. A seven-sided grain counted as six then raises the six-sided average, and since grains with more sides have more chances to hide a short one, the bias grows with $n$ and steepens the slope. Counting triple junctions directly would test this. Until then the $2\%$ excess is an unexplained systematic error of the measurement. Mean area rate of grains with $n$ sides, in units of $Lk$, pooled over three seeded $160$ by $160$ polycrystals (dots, sample counts from $17$ to $3731$). The dashed line is the von Neumann–Mullins law $(\pi/3)(n-6)$. Five-sided grains sit on it; six-sided grains grow slightly, and grains with more sides grow $3\%$ to $6\%$ faster than the law. Source: J. K. Mason ## Worked example: which grains of a small cluster survive? Problem. In a polycrystal with $L=k=1$, a four-sided grain of area $60$ cells sits next to an eight-sided grain of area $400$ cells. Assuming both keep their side counts, how fast does each change, and when does the small one vanish? Rates. For $n=4$, $dA/dt=(\pi/3)(4-6)=-2.09$ cells per unit time. For $n=8$, $+2.09$. Neither depends on the area. Lifetime. $60/2.09=29$ time units. As it shrinks it is likely to lose sides, which would make it shrink faster, so this is an upper estimate. Check. The pooled benchmark rates are $-2.12$ for four-sided and $+2.16$ for eight-sided grains, $1\%$ and $3\%$ faster than the law. Answer. The four-sided grain loses about $2.1$ cells per unit time and is gone within about $29$ time units; the eight-sided one gains at the same rate as long as it keeps eight sides. Source: J. K. Mason ## What the coarsening run does not show Fitting the mean area against time from $t=333$ to $1000$ gives an exponent of $0.81\pm0.05$, not the $1$ of linear growth. By the end of that window only $18$ to $25$ grains are left in each box, down from $43$ to $55$ at its start. That is too few for the self-similar state that the linear law assumes, and the run is short. The benchmark therefore does not test linear coarsening; a larger box and a longer run would be needed. The free energy fell or stayed level in all $150$ checks. Source: A. Kazaryan ## Try it in the Lab ### [Watch six sides divide the grains](https://sandboxphysics.com/labs/curvature-grain-growth/) 1. Choose One defect in a honeycomb with grains coloured by their number of sides. Predict which grains will change area first and which will stay as they are. 2. Choose A polycrystal coarsens and let it run. Predict the value the measured slope readout should approach. 3. Watch the grain count and the mean number of sides together. Explain why the number of grains falls while the mean number of sides stays close to six. ## Exercises with explained answers 1. Use the law to show that the total area of a periodic polycrystal does not change. Show the answer Adding over all $F$ grains, the total rate is $(\pi/3)M\gamma\sum_i(n_i-6)=(\pi/3)M\gamma F(\langle n\rangle-6)$, which is zero because $\langle n\rangle=6$. 2. If triple junctions met at an angle $\theta$ instead of $120^{\circ}$, which number of sides would hold its area? Show the answer Each corner turns the boundary by $\pi-\theta$, so the smooth parts turn by $2\pi-n(\pi-\theta)$. The rate vanishes at $n=2\pi/(\pi-\theta)$, which is six for $\theta=2\pi/3$. Junction angles are fixed by the balance of three equal tensions, which is why the law needs equal boundary energies. 3. A three-sided grain of area $30$ cells shrinks with $L=k=1$. Estimate its lifetime and compare with the pooled benchmark rate for three-sided grains. Show the answer The law gives $dA/dt=-\pi=-3.14$, so about $9.5$ time units. The benchmark pools $30$ samples with a mean rate of $-3.18$, $1.3\%$ faster. 4. Why would a fit of the mean area against time over a short run with about twenty grains left give an exponent below one, even if the law of linear growth were correct? Show the answer Linear growth of the mean area holds once the grain structure is statistically the same at every scale. Starting from a random arrangement of seeds, the side and size distributions take time to reach that state, and with only about twenty grains each disappearance changes the mean area in a large step. A short window over few grains measures the approach to the scaling state. [Previous: Grain boundaries move by curvature, so a round grain shrinks at a steady rate](https://sandboxphysics.com/stories/crystal-defects-microstructure/grain-boundaries-move-by-curvature/) [Next: Solute trapping in alloy phase-field models and the anti-trapping current](https://sandboxphysics.com/stories/crystal-defects-microstructure/thin-interface-alloy-solidification-and-antitrapping/) ## Sources and what we recomputed 1. [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)](https://arxiv.org/abs/cond-mat/9912044v1) 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. 2. [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)](https://arxiv.org/abs/1507.03379v1) 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.