The Physics of Percolation

Why power laws?

Isn’t it remarkable that out of all the functions nature as to offer, simple power laws emerge contrasting the many-body complexity of the system? These power laws have established themselves as ubiquitous with critical phenomena and hence critical exponents. The following is an optional toy calculation to understand the origin of power laws using something called coarse-graining. It does not correspond to real 3D percolation system you’re working with as it does not produce the correct critical threshold $p_{\rm, c}$, but the ideas and the implications are the same. It is also not required for your notebook nor report, so don’t spend time on this during lab hours but If you work through it, I guarantee a highly rewarding conclusion. This exercise is divided into two parts: coarse-graining and the emergence of power laws.

Coarse-graining

Let us start with a toy model on a square lattice of size $N\times N$ with $N^2$ sites. The sites can occupy two states, which we will interpret as metal or plastic. Let $p$ be the fraction of metal beads in the system. Consider a window of size $w\times w$ that contains $w^2$ sites which divide the whole lattice into $L^2/w^2$ blocks (you may assume periodic boundary conditions if it makes the following calculation easier). For a fixed fraction $p$ and a configuration sampled from this fraction, there’s going to be some number of metal beads and plastic beads that end up in this window. You maybe tempted to conclude that the fraction of metal beads is $p$, however it turns out that is not the case. I invite you to think about why but if you don’t have a solid answer, come back to this question after you’ve finished your calculation.

We seek to understand what the effective phase of our system–is it metallic or plastic? To answer that question, we want understand how the system behaves as $w\to \infty$ for the infinite $N\to \infty$ lattice. Note that the order of limits matter here, we first take $N\to \infty$ to obtain our large system and then consider how coarse-graining affects our system. For each $w$, we coarse grain our system by assigning a simple majority rule:

Within the window $w$, if the number of metal beads forms a majority, then we replace the window with a single metal bead. Conversely, if the plastic beads form a majority, then we replace the window with a single plastic bead. Such a coarse graining procedure essentially allows us to zoom out and filter out the minutia and in essence, get a bigger picture.

Take a moment to pause and wonder what the system effectively looks like visually after coarse graining for

1. $p\approx 0$
2. $p \approx 1$

Now imagine slowing moving away from the extreme bounds, what do you expect to see? Make a prediction before we go ahead and attempt to compute this. For a fantastic animation of coarse-graining to help you visualize this, check out How Fractals Make the Best Coffee (11:12).

Now fixed a $p$ which implies that $p N^2$ is the total number of metal beads in the system. Let $P_{N,w}(k)$ be the probability of finding $k$ number of metal beads inside a window $w^2$ on the lattice of size $N^2$. For the majority rule, we are ultimately interested in finding \(\lim_{w\to \infty} \lim_{N\to \infty}\sum_{k>w^2/2}^{w^2}P_{N,w}(k)\) which is the probability that the window coarse grains into a metal bead if it forms a majority ⁵. Let’s start with $P_{N,w}(k)$ which requires us to answer the following questions

1. What are the total number of ways of finding the metal beads for a fixed p on the grid?(This is your normalization factor!)
2. How many ways can you populate the window of size $w^2$ with $k$ metal beads?
3. How many ways can you populate the remaining sites with the remaining metal beads?
4. Put 2 and 3 together, how many ways are there to find $k$ metal beads in this window?
5. Finally, use 1-5 to find $P_{N,w}(k)$.

Now, we want to take the large $N$ limit and deduce \(\lim_{N\to \infty} P_{N,w}:= P_w(k).\) However, care must be exercised when taking this limit else we’ll have infinities over infinities!

1. Using the definition of the binomial coefficient, write out the expression for $P_w(k)$ in terms of factorials.
2. Simplify the factorials in terms of products. Use \(\frac{X!}{(X-Y)!}= (X-Y+1)(X-Y+2)\dots (X-Y+Y)=\prod_{j=1}^Y(X-Y+j)\) to make your lives easier : ).
3. Factor all $N^2$ terms and perform the limit $N\to \infty$, you will find an elegant, simple sum that involves the cherished Bernoulli distribution, reminiscent of flipping coins.

This relation actual goes further: Indeed, there is an exact mapping of the $w\to \infty$ limit to the coin problem that allows us to evaluate this limit:

• What is the probability to end up with greater than $50\%$ heads as the number of flips, $n=w^2$ becomes infinite? i.,e \(P(\text{number of heads after } n \text{ flips} > n/2)\) in the limit $n\to\infty$.*

There are two different cases for you to consider

After computing these, take a moment to digest your incredible result, what does it mean? What does the system look like when you zoom out when $p<1/2$? what about when $p>1/2$? and most critically, what about when $p=1/2$? Is there anything geometric that sticks out to you about the system when $p=1/2$?

Power laws

Having set our background on criticality, we are ready to answer why Power laws emerge. The key fact that we will use is the scale-invariance at criticality that you observed in the previous toy example. In particular, we will show that

I will be referencing Brain Criticality - Optimizing Neural Computations (13:12), A. Kirsanov for the following argument.

So how do we quantify scale-invariance? Let us suppose that we are at criticality to begin with and define $P(x)$ as the probability of finding a cluster of size $x$. By cluster, we make the metal-bead network (or you’re familiar with the Ising model, we mean a connected group of the same spin) and by size we mean the number of beads that form a connected network. We will show that this probability follows a power law i.e

at criticality and consequently any observables that can be computed from it using expectation values viz. the conductance or capacitance also follows a power law ⁸.

Imagine for a brief moment that you were able to diligently compute the probabilities of observing different clusters of sizes $P(x)$ for the entire lattice. Now suppose you zoom out so that you compute all clusters of size $2x$. A connected cluster of size $2x$ is going to be less likely than a cluster of size $x$ because there are fewer ways for a huge cluster to form than a small cluster which can be present at many locations. Therefore, $P(2x)<P(x)$ and so far this still does not give us any information about what these could be. It might be tempting to assume that scale invariance implies that $P(2x)=P(x)$, but this is not true, as the fraction of bigger clusters are necessarily going to be rarer than smaller clusters. Consider now instead the ratio or relative probability $P(2x)/P(x)$. This ratio clearly depends on the size of the cluster $x$ in normal scenarios, but at criticality, something different happens due to scale invariance. Although the individual $P(x)$ and $P(2x)$ depend on $x$, scale invariance implies that the ratio should not depend on $x$! This is because here we are free to zoom out by a factor of $2$ and since we get an identical “looking” system, our statistics is identical, but all our probabilities are shrunken by a factor (function) that depends only on your zoom factor. More precisely, we have that for any scale $k>0$ that we zoom out,

for a “nice” (smooth) function $g(k)$ which, importantly only depends on $k$!

Another way to see this is through the observation that our statistics when zooming out by a factor of $k$, is identical to what we had (because we see similar looking clusters) except that all our probabilities are scaled by some function that depends on how much we’ve zoomed out, i.e

\(P(kx)= g(k) P(x). \tag{1}\)

Infact, the above two equations are the definition of scale invariance and can be used if one has scale invariance. You can easily see power laws $\implies$ scale invariance by plugging in a generic power law to obtain the above equation. The converse however is non-trivial, but the fantastic result above and some calculus is all we need to show power laws–which I will walk you through with the following exercise.

1. Assume that $P$ is a smooth function, differentiate Eq. 1 with respect to $k$ and set $k=1$.
2. Solve the resultant ODE for $P(x)$ ⁹.
3. Call $g^\prime(1)=-\gamma$, what functional form do you find for $P(x)$?

After an interesting cocktail of combinatorics and calculus, and the simple idea of scale invariance at criticality, we have derived the emergence of power laws. This is why power laws are ubiquitous with critical phenomena.

Scaling theory and predicting $u$

Below is an optional exercise to get a taste of scaling theory which predicts what the exponent $u$. We use observation that the admittance is non-zero and finite at $p=p_{\rm c}$ and depends on $\omega$. I will be referencing 6. The Scaling Hypothesis Part 1, M. Kardar.

We begin by considering the frequency dependent complex admittance $Y(\omega,p)$ which contains both the conductance and capacitance defined by

Electronically, the metal-plastic bead system is equivalent to an RC circuit with effective resistance $R$ and capacitance $C$ (both of which you measure) that sets a characteristic frequency scale $\omega_{0}$.

1. Recall and state the characteristic frequency scale $\omega_0$ for an RC circuit. Using $G=1/R$, express $\omega_0$ in terms of the conductance $G$.

Now, the admittance also depends strongly on the fraction $p$. If you look the graphs for $G(p)$ and $C(p)$, we notice that just above the percolation threshold, the capacitance decays while the conductance grows rapidly (limited only by the fact that the beads have some contact resistance). Conversely, just below the threshold, the capacitance grows rapidly whereas the conductance is tiny. We can reframe our observation as

where the factors $A$ and $B$ are used to ensure that the dimensions work out ($A$ has units of conductance, siemens). Now we may imagine trying to write this as a single expression and imagine we factor the sign independent $(p-p_{\rm c})^t$ from it to get

which has the same scaling when $p>p_{\rm c}$. Note the dimensionless nature of this new function. This new function should have as an input

1. something that depends on the frequency $\omega$ but dimensionless
2. something that depends on $p-p_{\rm c}$ with some arbitrary exponent

Denote $\omega_0$ the natural frequency scale set by our system and let $\phi$ be this arbitrary exponent. This gives us that the scaling function $g_{\pm}$ should assume the form

where $g_{+}$ is the scaling function for $p-p_{\rm c}>0$ and $g_{-}$ for $p-p_{\rm c}<0$ both depending on the dimensionless variable $x=\frac{\omega}{\omega_0} |p-p_{\rm c}|^{\phi}$.

1. Assume that $p>p_{\rm c}$, in the small frequency limit $\omega\to 0$ (remember that by “small” $\omega$ we mean $\omega \ll \omega_0 $), we should have $Y(\omega,p)\sim A |p-p_{\rm c}|^t$. Given this, what should the $\lim_{x\to 0} g_{+}(x)$ become in order to match what we should have? (you don’t need to find the exact value of $g(0)$, but rather only identify and choose if it should be a power law, a constant, a divergence or an exponential.)

2. Conversely, given that $p<p_{\rm c}$, what should $\lim_{x\to 0}g_{-}$ grow like to match $\omega |p-p_{\rm c}|^{-s}?$ Use this matching condition to derive $\phi=-(s+t)$.

Now that we have $\phi$, the general form reads

Since $s,t>0$, at criticality, $|p-p_{\rm c}|^{-(s+t)}\to \infty$ when $|p-p_{\rm c}|\to 0$. But experimentally we know (or you will find) that $Y(\omega,p)\sim (i \omega)^u$ for some exponent $u$. The key fact to keep in mind is that $Y(\omega,p_{\rm c})$ is neither zero nor infinite. Since $|p-p_{\rm c}|^t \to 0$ but $\frac{1}{|p-p_{\rm c}|^{s+t}}\to \infty$ as $|p-p_{\rm c}|\to 0$, we have two competing limits that requires some care.

4. To obtain a finite limit for $Y$, we need $g_{\pm} (x)$ to grow. Suppose that it grows in the following manner $\lim_{x\to \infty} g(x)\sim x^q$ for some exponent $q$. Match the limit with $Y(\omega,p_{\rm c})\sim i \omega^u$. What do you get for $q$? Using the fact that $Y$ is neither infinite nor zero, obtain the scaling relation \(u= \frac{t}{s+t}\)

Finally, now we deduce the critical angle $\delta_{\rm c}$. The loss tangent is defined as

5. Using $Y= G + i \omega C= (i\omega)^u$ and Euler’s identity, find $G$ and $C$ in terms of $u$ and express $\tan(\delta)$ in terms of $u$. The resultant angle is our critical angle!

With essentially what’s Dimensional Analysis part II, you have derived your first scaling relation! Congratulations! Now go forth and test your prediction. If you liked this calculation, you invite you to check out M. Kardar’s lecture video that I cited.