LBan-V

Avalanches at the Depinning Transition

In the previous lesson, we studied the dynamics of an interface in a disordered medium under a uniform external force $F$. In a fully connected model, we derived the force–velocity characteristic and identified the critical depinning force $F_c$.

In this lesson, we focus on the avalanches that occur precisely at the depinning transition. To do so, we introduce a new driving protocol: instead of controlling the external force $F$, we control the position of the interface by coupling it to a parabolic potential. Each block is attracted toward a prescribed position $w$ through a spring of stiffness $k_0$.

For simplicity, we restrict to the fully connected model, where the distance of block $i$ from its local instability threshold is $$x_i=1-(h_{CM}-h_i)-k_0(w-h_i).$$

The first term represents the elastic pull exerted by the rest of the interface, while the second encodes the external driving through the spring.

Here $h_{CM}$ is the center-of-mass position of the interface. Because the sum of all internal elastic forces vanishes, the total external force is balanced solely by the pinning forces. The effective external force can thus be written as a function of $w$: $$F(w)=k_0(w-h_{CM}).$$

As $w$ is increased quasistatically, the force $F(w)$ would increase if $h_{CM}$ were fixed. When an avalanche takes place, $h_{CM}$ jumps forward and $F(w)$ suddenly decreases. However, in the steady state and in the thermodynamic limit $N\to \mathrm{\infty }$, the force recovers a well-defined value. In the limit $k_0\to 0$, this force tends to the critical depinning force $F_c$; at finite $k_0$ it lies slightly below $F_c$.

Quasi-Static Protocol and Avalanche Definition

To study avalanches, the position $w$ is increased quasi-statically so that the block closest to its instability threshold reaches it, $$x_i=0.$$

This block is the epicenter of the avalanche: it becomes unstable and jumps to the next well.

When block $i$ jumps by $\mathrm{\Delta }$, both the elastic contribution and the driving spring relax. This gives $$\{\begin{array}{l}{x}_i=0⟶x_i=\mathrm{\Delta }(1+k_0),\\ x_j⟶x_j-\frac{\mathrm{\Delta }}{N}(j\ne i).\end{array}$$

The key feature of the quasi-static protocol is that $w$ does not evolve during the avalanche: all subsequent destabilizations are triggered exclusively by previously unstable blocks.

It is convenient to organize the avalanche into generations of unstable sites:

• First generation: the epicenter.
• Second generation: sites destabilized by it.
• Third generation: sites destabilized by generation two.
• And so on.

This hierarchical construction allows us to compute avalanche amplification step by step.

Derivation of the Evolution Equation

Our goal is to determine the distribution $P_w(x)$ of distances to threshold at fixed $w$.

We shift the parabola by $w\to w+\mathrm{d}w$. Before the shift: $$x_i(w)=1-k_0(w-h_i(w))-(h_{CM}(w)-h_i(w)).$$

We now follow the dynamics generation by generation.

First generation

During the shift, the center of mass has not yet moved.

• Stable blocks ($x_i>k_{0}dw$):

$$x_i^{t=1}=x_i-k_{0}dw.$$

• Blocks with $0<x_i<k_{0}dw$ are unstable. Since $dw$ is infinitesimal, their fraction is

$$P_w(0)k_{0}dw.$$

Unstable blocks first relax to ($x_i=0$) and are then reinjected with a random kick $\Delta$ drawn from $g(\Delta)$, leading to \[ x_i^{t=1} = (1+k_0)\Delta. \] Let $p(x)dx$ be the probability that an unstable block stabilizes in the interval $(x,x+dx)$. By change of variables we have $$p(x)dx=g(\mathrm{\Delta })d\mathrm{\Delta }.$$ Since $x=(1+k_0)\mathrm{\Delta }$, it follows that $\mathrm{\Delta }=\frac{x}{1+k_0}$ and $$d\mathrm{\Delta }=\frac{dx}{1+k_0}.$$ Therefore, $$p(x)=\frac{1}{1+k_0}g\left(\frac{x}{1+k_0}\right).$$

Second generation

The parabola is now fixed, but the center of mass has advanced: $$h_{CM}\to h_{CM}+\bar{\mathrm{\Delta }}{P}_w(0)k_{0}dw.$$

Thus all sites shift again toward instability.

• Stable sites:

$$x_i^{t=2}=x_i-(1+\bar{\mathrm{\Delta }}{P}_w(0))k_{0}dw.$$

• Newly unstable fraction:

$$P_w(0)k_{0}dw+(\bar{\mathrm{\Delta }}{P}_w(0))P_w(0)k_{0}dw.$$

Higher generations

Iterating produces a geometric amplification: $$1+\bar{\mathrm{\Delta }}{P}_w(0)+(\bar{\mathrm{\Delta }}{P}_w(0){)}^2+\dots =\frac{1}{1-\bar{\mathrm{\Delta }}{P}_w(0)}.$$

The quantity $\bar{\mathrm{\Delta }}{P}_w(0)$ plays the role of a branching ratio: it measures the average number of sites destabilized by one instability.

For stable sites $$x_i(w+dw)=x_i(w)-\frac{k_0}{1-\bar{\mathrm{\Delta }}{P}_w(0)}dw.$$ while unstable sites are reinjected at a random location $\mathrm{\Delta }(1+k_0)$. The fraction of unstable sites is $$\frac{P_w(0)}{1-\bar{\mathrm{\Delta }}{P}_w(0)}{k}_{0}dw$$ is This yields $${\partial }_w{P}_w(x)=\frac{k_0}{1-\bar{\mathrm{\Delta }}{P}_w(0)}[{\partial }_x{P}_w(x)+\frac{P_w(0)}{1+k_0}g\left(\frac{x}{1+k_0}\right)].$$

Stationary solution

At large $w$: $$0={\partial }_x{P}_{\text{stat}}(x)+\frac{P_{\text{stat}}(0)}{1+k_0}g\left(\frac{x}{1+k_0}\right).$$

Solving: $$P_{\text{stat}}(x)=\frac{1}{\bar{\mathrm{\Delta }}(1+k_0)}\underset{x/(1+k_0)}{\overset{\mathrm{\infty }}{\int }}g(z)dz.$$

Critical Force

The average distance from the threshold gives a simple relation for the force acting on the system, namely $$F(k_0)=1-\bar{x}=1-(1+k_0)\frac{\overline{{\mathrm{\Delta }}^2}}{2\bar{\mathrm{\Delta }}}.$$

In the limit $k_0\to 0$ we obtain: $$F_c=F(k_0\to 0)=1-\frac{1}{2}\frac{\overline{{\mathrm{\Delta }}^2}}{\bar{\mathrm{\Delta }}}.$$

Avalanches

We consider an avalanche starting from a single unstable site $x_0=0$.

Ordering sites by stability: $$x_1<x_2<x_3<\dots$$

From order statistics: $$\underset{0}{\overset{x_1}{\int }}{P}_w(t)dt=\frac{1}{N}.$$

Thus $$x_n\sim \frac{n}{N{P}_w(0)}.$$

Each instability gives kicks $\mathrm{\Delta }/N$.

Compare mean kick and mean gap: $$\frac{\bar{\mathrm{\Delta }}}{N}\text{vs}\frac{1}{N{P}_w(0)}.$$

Criticality occurs when $$\bar{\mathrm{\Delta }}{P}_w(0)=1.$$

Using the stationary solution: $$\bar{\mathrm{\Delta }}{P}_w(0)=\frac{1}{1+k_0}.$$

Hence:

• $k_0>0$ → subcritical.
• $k_0=0$ → critical.

Mapping to a Random Walk

Define the random increments $${\eta }_1=\frac{{\mathrm{\Delta }}_1}{N}-x_1,{\eta }_2=\frac{{\mathrm{\Delta }}_2}{N}-(x_2-x_1),{\eta }_3=\frac{{\mathrm{\Delta }}_3}{N}-(x_3-x_2),\dots$$ and the associated random walk $$X_n=\sum _{i=1}^n{\eta }_i.$$

The mean increment is $$\bar{\eta }=\frac{\bar{\mathrm{\Delta }}}{N}-\frac{1}{N{P}_w(0)}.$$

An avalanche remains active as long as $$X_n>0.$$

The avalanche size $S$ is therefore the first-passage time of the walk to zero.

Critical case (k₀ = 0)

At criticality, $$\bar{\eta }=0,\bar{\mathrm{\Delta }}{P}_w(0)=1.$$

The jump distribution is symmetric and has zero drift. We set $X_0=0$.

Let $$Q(n)=\text{Prob}(X_1>0,\dots ,X_n>0)$$ be the survival probability of the walk.

By the Sparre–Andersen theorem, for large $n$, $$Q(n)\sim \frac{1}{\sqrt{\pi n}}.$$

The avalanche-size distribution is the first-passage probability: $$P(S)=Q(S)-Q(S+1).$$

Using the asymptotic form, $$P(S)\sim \frac{1}{\sqrt{\pi S}}-\frac{1}{\sqrt{\pi (S+1)}}\sim \frac{1}{2\sqrt{\pi }}{S}^{-3/2}.$$

Thus, at criticality, $$P(S)=\frac{1}{2\sqrt{\pi }}{S}^{-3/2}(S\gg 1).$$

The universal exponent is $$\tau =\frac{3}{2}.$$

This power law is of Gutenberg–Richter type.

Finite k₀ > 0 (Subcritical case)

Using the stationary solution, $$\bar{\mathrm{\Delta }}{P}_{\text{stat}}(0)=\frac{1}{1+k_0},$$ the mean drift becomes $$\bar{\eta }=-k_0\frac{\bar{\mathrm{\Delta }}}{N}.$$

The random walk is weakly biased toward negative values. For small $k_0$, the walk is only slightly tilted.

In this case the distribution retains the critical form $$P(S)\sim S^{-3/2}$$ up to a cutoff set by the inverse squared drift: $$S_{\max }\sim k_0^{-2}.$$