# Brownian dimension

**Classical Mechanics**Level 5

How does the number \(\mathbf{N}\) of circles of radius \(r\) that are required to cover the object change with \(r\)?

For instance, a line of length \(L\) can be covered by \(L/r\) circles of radius \(r\), and \(\mathbf{N}\sim r^{-1}\). A square of side length \(L\) can be covered by \(\sim L^2/r^2\) circles of radius \(r\), and \(\mathbf{N}\sim r^{-2}\). It is for this reason that we call a line 1 dimensional, and we call the square 2 dimensional. Here we'd like to ask, what is the dimensionality of a particle undergoing Brownian motion (diffusion)?

A particle undergoes Brownian motion through the action of a random force so that its displacement in a time interval \(\Delta t\) behaves as \(\sim\sqrt{2D\Delta t}\). Because the particle is 0d, we might expect that its trajectory is 1d, as for particles in projectile motion.

Suppose we have a perfectly sampled trajectory of a particle, \(\gamma_B\). We want to lay down a set of circles, \(C\), of radius \(r\) to cover this path, i.e. so that every section of the particle's path in the 2d plane is overlaid by a portion of a circle \(C_i \in C\). \(\mathbf{N}_\textrm{Brownian}\) changes with \(r\) according to \(\mathbf{N}_\textrm{Brownian} \sim r^d\). What is \(\lvert d \rvert\), i.e. what is the dimensionality of Brownian motion?

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