While looking out an airplane window recently, I was thinking about how few rivers run down a mountain. I started writing this out as disordered Plinko—a carnival game that is simply begging to be made—and realized that the solution wasn't too exciting. However, a similar Hamiltonian led to a more surprising phenomenon, which I'll now describe.
Consider ripping a piece of paper of length \(L\). When the tear reaches the bottom of the page, it will have drifted slightly to the side compared to the starting position. This is partly because I'm bad at tearing pieces of paper, so my motion has some deterministic drift. In this experiment, I'm imagining that I'm very good at tearing paper. I only want to know how much drift occurs due to the random material properties of the paper. The well-trained reader may be inclined to mumble \(\sqrt L\) under their breath and close this page—but wait![1]
We tear for distance \(L\) in the \(y\) direction. The paper will rip along a path of weak bonds in the paper; we can imagine these bond strengths as being random. We also introduce a term penalizing sharp turns (\(\kappa > 0\)), so the Hamiltonian describing the dynamics in the \(x\) direction is $$\begin{aligned} \sum_{\ell=1}^L \kappa (x_\ell - x_{\ell-1})^2 + J_{\ell,x_\ell}, \qquad J_{\ell,x} \sim \mathcal{N}(0,1), \end{aligned}$$ where \(J_{\ell,x}\) represents the variance in the bond strength at height \(\ell\) and horizontal position \(x\) on the paper. If at each step in the tear we move horizontally by \(+1\) or \(-1\), we can drop the first term as a constant. The expected displacement we want to compute is thus $$\begin{aligned} \Delta = \left(\mathbb{E} \sum_{S\in\{\pm1\}^L} \left(\sum_{\ell=1}^L S_\ell\right)^2 \frac{e^{-\beta H(S)}}{Z}\right)^{1/2}, \quad H(S) = \sum_{\ell=1}^L J_{\ell,\sum_{k=1}^\ell S_k}, \quad Z = \sum_{S\in\{\pm1\}^L} e^{-\beta H(S)}. \end{aligned}$$ We will compute this with the replica trick \(Z^{-1} = \lim_{n\to 0} Z^{n-1}\), so $$\begin{aligned} \Delta^2 &= \lim_{n\to 0} \mathbb{E}\, Z^{n-1} \sum_{S\in\{\pm1\}^L} \left(\sum_{\ell=1}^L S_\ell\right)^2 \exp\left[-\beta\sum_{m=1}^L J_{m,\sum_{k=1}^m S_k}\right]\\ &= \lim_{n\to 0} \mathbb{E}\, \sum_{S^1,\dots,S^n\in\{\pm1\}^L} \frac{1}{n} \sum_{a=1}^n \left(\sum_{\ell=1}^L S^a_\ell\right)^2 \exp\left[-\beta\sum_{b=1}^n\sum_{m=1}^L J_{m, \sum_{k=1}^m S_k^b}\right]\\ &= \lim_{n\to 0} \frac{1}{n} \sum_{S^1,\dots,S^n\in\{\pm1\}^L} \sum_{a=1}^n \left(\sum_{\ell=1}^L S^a_\ell\right)^2 \exp\left[\frac{\beta^2}{2} \sum_{m=1}^L \left(n + 2\sum_{1\leq a < b \leq n} 1\left\{\sum_{k=1}^m S^a_k = \sum_{k=1}^m S^b_k\right\}\right)\right]\\ &= \lim_{n\to 0} \frac{e^{\beta^2 n L / 2}}{n} \sum_{S^1,\dots,S^n\in\{\pm1\}^L} \sum_{a=1}^n \left(\sum_{\ell=1}^L S^a_\ell\right)^2 \exp\left[\beta^2\sum_{m=1}^L\sum_{1\leq a < b \leq n} 1\left\{\sum_{k=1}^m S^a_k = \sum_{k=1}^m S^b_k\right\}\right] \end{aligned}$$ by the Gaussian moment generating function. The replica trick prescription is to compute the large-\(L\) saddle point for generic \(n\) then take the \(n\to 0\) limit. To evaluate the saddle point, we interpret the exponent as two independent, unbiased random walkers. The sum over \(m\) counts the number of times they bump into each other. If each walker wanders horizontally a distance of \(L^\alpha\), this happens around \(L^{1-\alpha}\) times. We can thus approximate the action as $$\begin{aligned} \Delta^2 &\approx \lim_{n\to 0} \frac{e^{\beta^2nL/2}}{n} \int_0^1 d\alpha \sum_{\substack{S^1, \dots, S^n \in \{\pm1\}^L\\ |\sum_k S_k^a| \approx L^\alpha}} \sum_{a=1}^n L^{2\alpha} \exp\left[\frac{\beta^2 n(n-1)}{2} L^{1-\alpha}\right] \end{aligned}$$ and evaluate the saddle point in \(\alpha\). The number of \(S\) such that \(\sum_{k=1}^L S_k = L^\alpha\) is given by Stirling's approximation as $$\begin{aligned} \binom{L}{(L+L^\alpha)/2} \approx \exp\left[L \log 2 - \frac{1}{2} \log \frac{\pi L}{2} - \frac{L^{2\alpha-1}}{2} - \frac{L^{4\alpha-3}}{12} - \dots\right]. \end{aligned}$$ Taking the leading-order contribution in the above expansion, we solve for the saddle point \(\alpha_*(n)\). Extremizing the \(\alpha\)-dependent action in the \(n\to 0\) limit gives $$\begin{aligned} 0 = \partial_\alpha \left(-\frac{n}{2}L^{2\alpha-1} + \frac{\beta^2 n(n-1)}{2} L^{1-\alpha}\right) \approx \partial_\alpha \left(-\frac{n}{2}L^{2\alpha-1} - \frac{\beta^2 n}{2} L^{1-\alpha}\right) = -\frac{n \log L}{2}\left(\beta^2 L^{1-\alpha} - 2 L^{2\alpha-1}\right). \end{aligned}$$ This is solved at \(\alpha_* \approx 2/3\), giving \(\Delta \approx L^{2/3}\). Is this right? Some physicists in the 1990s were exceptionally good at tearing various pieces of paper they found at the Institute of Paper Research in Budapest. They measured exponents of 0.63-0.72, which seems pretty close.[2]
[1] If you're really well-trained and mumble “KPZ” instead, feel free to close this page.