Tuesday, April 3, 2018

Diffusion and Random Walks

In the previous post, we saw how the probability distribution \(p(x,N)\) after \(N\) random steps on a unit lattice is given by, \[p(x, N) = \dfrac{1}{\sqrt{2 \pi N}} \exp\left(-\dfrac{x^2}{2N}\right)\] If the average step size is \(b\) instead of \(b=1\), then we can generalize, and write the formula as:
\[p(x, N) = \dfrac{1}{\sqrt{2 \pi Nb^2}} \exp\left(-\dfrac{x^2}{2Nb^2}\right)\] Now consider a Gaussian random walk in 1D. Suppose the stepsize at each step is drawn from a normal distribution \(\mathcal{N}(0, 1)\). While it has the same average stepsize as a walk on the lattice, an individual step may be shorter or longer than b=1.

In polymer physics, where a Gaussian coil is often used as a model for polymer conformations, \(b\) is called the Kuhn length, and \(N\) is proportional to the molecular weight.

Due to the connection between Brownian motion and random walks, the mean squared distance travelled by a particle in 1D with self-diffusivity \(D\) is \(\langle x^2(t) \rangle = 2Dt\). Similarly, the mean end-to-end squared distance of a Gaussian random walk is given by, \[\langle x^2(N) \rangle = \int_{-\infty}^{\infty} x^2 p(x, N) dx = Nb^2 \equiv 2Dt = \langle x^2(t) \rangle.\] This allows us to re-parameterize the equation for the position of a Brownian diffusors. \[p(x, t) = \dfrac{1}{\sqrt{4 \pi Dt}} \exp\left(-\dfrac{x^2}{4Dt}\right)\] Look at the correspondence between \(t\) and \(N\), and \(b\) and \(\sqrt{2D}\).

No comments: