L4 · Chapter 7

Predicting where a sheared packing slips next

Sandbox Physics · 2026-10-29

Push the top of a jammed packing sideways a little at a time and the shear stress climbs smoothly, then drops all at once as a few disks slip. Before the slip the disk positions show no sign of it, but the vibrations do: as the slip approaches, one collective motion loses its stiffness, falling as the square root of the strain left. The disks that move most in the softest motions are, more often than chance, the ones that slip.

Slow shear: smooth loading, sudden slips

Quasistatic shear moves the top of a periodic cell sideways by a small strain, here $2\times10^{-4}$ , and lets the disks relax to the nearest energy minimum before the next step. There is no inertia and no temperature: the packing always sits in a minimum, and shear slowly changes the shape of that minimum. While the minimum exists, the stress rises along a smooth curve and each step can be undone: shearing back returns every load-bearing disk to where it was.

A slip is the moment that minimum disappears. The disks fall into another minimum, the stress drops, and the step cannot be undone. In our packings of 64 Hertzian disks at $\phi=0.90$ , the smooth stretches last a median strain of 0.0098, and the disk that moves most in a slip jumps a median of about 0.1 small-disk diameters. To locate a slip within $10^{-9}$ in strain, the failed step is halved again and again.

Source: C. E. Maloney and A. Lemaître

Why one mode goes soft first

Expand the energy along the one collective motion that is about to fail, with coordinate $s$ . Near the slip the lowest terms that matter are a cubic in $s$ and a tilt that the shear gradually removes. Keeping only these two terms gives the stiffness of the mode:

U(s)\approx U_0+b\,(\gamma_c-\gamma)\,s-\frac{a}{3}\,s^{3},\qquad \lambda_{\min}=U''(s_*)=2\sqrt{ab}\,(\gamma_c-\gamma)^{1/2}

For

\gamma<\gamma_c

there is a minimum at

s_*=-\sqrt{b(\gamma_c-\gamma)/a}

and a saddle at

+\sqrt{b(\gamma_c-\gamma)/a}

. They meet at

\gamma=\gamma_c

, and the curvature at the minimum, which is the stiffness of this mode, vanishes as a square root. Maloney and Lemaître call this instability a fold.

Source: C. E. Maloney and A. Lemaître

How close to the slip the square root holds

The square root holds only close to the slip. In our Hertzian packings the local slope of $\log\lambda_{\min}$ against $\log(\gamma_c-\gamma)$ between $3\times10^{-8}$ and $3\times10^{-7}$ has a median of 0.475 over 288 slips; between $10^{-5}$ and $10^{-4}$ it is only 0.254, because the other terms of the expansion still matter there.

Harmonic contacts behave differently. Their stiffness does not depend on the overlap, and drops to zero at once when a contact opens. With harmonic contacts the lowest stiffness usually stays flat until the slip and then collapses: only 12% of slips soften gradually, against 98% with Hertzian contacts.

Reading soft spots from the vibrations

The stiffness of each collective motion comes from the dynamical matrix, the second derivatives of the energy with respect to the disk positions. Its eigenvectors $\mathbf e_m$ are the vibration modes and its eigenvalues $\lambda_m$ their stiffnesses. Disks with no load-bearing contact, the rattlers, are removed first; otherwise each rattler adds a mode of zero stiffness.

Manning and Liu (2011) sheared 2500 Hertzian disks at $\phi=0.95$ . From each of the 30 softest modes they took the 20 particles that move most and grouped them into soft spots, about a tenth of all particles. Each rearrangement fell in one spot; the example in their Fig. 2 has an overlap score of 0.64. Soft spots had 6% fewer contacts and 103% more free volume than the rest, yet those local quantities alone did not find them.

H_{i\alpha,j\beta}=\frac{\partial^{2}U}{\partial r_{i\alpha}\,\partial r_{j\beta}},\qquad P_i=\sum_{m=1}^{N_{\mathrm m}}\left|\mathbf e_{m,i}\right|^{2}

P_i

measures how much disk

i

moves in the

N_{\mathrm m}

softest modes. The

k

disks with the largest

P_i

are the prediction.

Source: M. L. Manning and A. J. Liu

A blind test on packings the method never saw

We repeated the idea on smaller packings with a simpler score. At the start of each smooth stretch, the $k=6$ disks with the largest $P_i$ are named. After the slip, the six load-bearing disks with the largest non-affine jump are the movers. The hit fraction $h$ is the share of named disks that are movers. Two simpler guesses are scored the same way: the six disks with the fewest load-bearing contacts, and the six with the lowest local pressure.

The number of modes was chosen on eight training packings (96 slips) only: one and two modes scored 0.276 and 0.278, and more modes did worse, so $N_{\mathrm m}=2$ . Sixteen other packings then gave 192 test slips. The two softest modes named 26% of the movers, the fewest-contacts guess 18%, the lowest-pressure guess 15%, and a random guess 9%. Slip by slip, the soft modes beat the fewest-contacts guess by 0.080 on average, with a standard error of 0.014. In 44 of the 192 slips they named none of the movers.

These packings are small, 64 disks against 2500 in Manning and Liu, and our score is not their overlap score, so the two sets of percentages cannot be compared directly. A denser packing ( $\phi=0.95$ ) gave 28%, and so did harmonic contacts; chance stayed at 9%.

h=\frac{\left|\mathcal P\cap\mathcal R\right|}{k},\qquad \langle h\rangle_{\text{chance}}=\frac{\left|\mathcal R\right|}{N'}

Choosing

k

disks at random among the

N'

load-bearing disks picks each mover with probability

k/N'

, so the average score is

\left|\mathcal R\right|/N'

Source: M. L. Manning and A. J. Liu

Worked example: scoring one slip

Open the Lab’s default preset. The first slip comes at $\gamma_c=0.006934$ . All 64 disks are load-bearing, so $N'=64$ . At the start of the smooth stretch the two softest modes named disks 62, 36, 26, 22, 23 and 12. The six largest jumps were disks 36, 10, 62, 28, 60 and 26, the largest 0.291 diameters. The shear stress fell from $2.49\times10^{-4}$ to $-1.53\times10^{-4}\,\varepsilon/d^{2}$ .

Step one, the hit fraction: disks 62, 36 and 26 are on both lists, so $h=3/6=0.5$ . The fewest-contacts guess (26, 32, 62, 28, 12, 48) also hits three; the lowest-pressure guess (26, 12, 1, 25, 32, 39) hits one.

Step two, chance: $\langle h\rangle=6/64=0.094$ . The probability that six random disks include three or more movers, worked out below, is under 1%. A single slip this good rarely happens by luck, but the fewest-contacts guess did as well here, so one slip cannot rank the methods. That is why the test averages 192 slips.

Step three, the approach: the lowest stiffness was $1.44\times10^{-6}$ at $3\times10^{-8}$ before the slip and $5.60\times10^{-6}$ at $3\times10^{-7}$ , a local slope of $\log_{10}(5.60/1.44)=0.59$ . One decade farther out, at $3\times10^{-6}$ it was $1.73\times10^{-5}$ , a slope of 0.49. Both are close to the one half of a fold.

\Pr\!\left(h\ge\tfrac12\right)=\frac{\sum_{j=3}^{6}\binom{6}{j}\binom{58}{6-j}}{\binom{64}{6}}=\frac{642\,264}{74\,974\,368}\approx 0.0086

A hypergeometric count: of 64 disks, 6 are movers; drawing 6 at random, the number of ways to hit

j

of them is

\binom{6}{j}\binom{58}{6-j}

Try it in the Lab

Score the soft modes yourself

Predict first: will the six green-ringed disks be the ones that slip?
Play Soft spot gives way. At the first slip, count the green-ringed disks that also get an amber ring, and compare the curve in the stiffness chart with the dashed reference line.
Let it run for a dozen slips and compare the soft-mode bar with the chance bar and the white tick recorded on the test packings. Then set the soft modes used for the prediction to 16 and run again.
Choose Spring contacts give no warning and see how the stiffness chart differs before a slip.

Exercises with explained answers

In the worked example the lowest stiffness was $1.73\times10^{-5}$ at $3\times10^{-6}$ before the slip. If it fell exactly as a square root, what would it be at $3\times10^{-8}$ ? Compare with the measured $1.44\times10^{-6}$ .
Show the answer
A hundred times less strain gives ten times less stiffness, so $1.73\times10^{-6}$ . The measured value is about 17% lower, in line with the slightly steeper local slope of 0.59 in the last decade.
A 64-disk packing has 4 rattlers, and all six movers are load-bearing. What hit fraction does a random choice of six load-bearing disks score on average? Why is the baseline not computed over all 64 disks?
Show the answer
6/60 = 0.1. Both the predictions and the movers come from the load-bearing disks, so chance has to be computed over the same set. Counting all 64 disks would lower the baseline to $6/64\approx0.094$ and flatter every guess.
Why does the lowest stiffness give almost no gradual warning with harmonic contacts?
Show the answer
A harmonic contact has $V''=\varepsilon/\sigma^{2}$ whatever the overlap, so a pair keeps its full normal stiffness until the overlap reaches zero and then contributes nothing. When a slip is triggered by a contact opening, the stiffness drops at once instead of following a square root. A Hertzian contact’s stiffness grows as the square root of the overlap, so it fades smoothly as the contact opens.
The number of modes was chosen on training packings and scored on others. What would go wrong if it were chosen on the test slips?
Show the answer
Picking the best of six candidates on the same slips used to report the score favours whichever candidate was lucky on those slips, so the reported score is biased upward. Separate test packings give a fair estimate. Here one and two modes were nearly tied on training, so the choice itself matters little.

Sources and what we recomputed

C. E. Maloney and A. Lemaître, Amorphous systems in athermal, quasistatic shear, arXiv:cond-mat/0510677v2; Physical Review E 74, 016118 (2006)
Version 2, 28 October 2005. Section IV: elastic branches end when the energy minimum merges with a saddle. Eqs. (9) and (10): motion along the critical mode grows as the square root of the remaining strain. Fig. 5(b): the lowest eigenvalue of the dynamical matrix falls as the square root of the strain left before the event. We checked the square-root approach on our own sheared packings.
M. L. Manning and A. J. Liu, Vibrational modes identify soft spots in a sheared disordered packing, arXiv:1012.4822v1; Physical Review Letters 107, 108302 (2011)
Version 1, 21 December 2010. 2500 Hertzian disks at packing fraction 0.95, sheared quasistatically. Fig. 2: soft spots built from the 20 particles that move most in each of the 30 lowest modes. Eq. (1): the overlap score between a soft spot and the next rearrangement; the example in Fig. 2 scores 0.64. Fig. 4(a): the score of the best spot against the strain left before the event. Text: soft spots have 6 % fewer contacts and 103 % more free volume, yet these local quantities alone did not find them. Our blind test uses smaller packings and its own thresholds.