## Sunday, April 15, 2018

### Diffusion in Higher Dimensions

In the previous posts (1 and 2), we wrote down the probability or concentration distribution of a bunch of Brownian diffusors initially at $x = 0$ (delta function), $p_{1D}(x, t) = \dfrac{1}{\sqrt{4 \pi Dt}} \exp\left(-\dfrac{x^2}{4Dt}\right)$
The PDF is normalized on the domain $x \in [-\infty, \infty]$ so that, $\int_{-\infty}^{\infty} p_{1D}(x,t)\, dx = 1.$ In 2D, $\langle r^2(t) \rangle = \langle x^2(t) \rangle + \langle y^2(t) \rangle$. If diffusion is isotropic, then $\langle r^2(t) \rangle = 2Dt + 2Dt = 4Dt$. In this case,
\begin{align}
p_{2D}(r, t) & = p_{1D}(x, t) \, p_{1D}(y, t)\\
& = \dfrac{1}{\sqrt{4 \pi Dt}} \dfrac{1}{\sqrt{4 \pi Dt}} \exp\left(-\dfrac{1}{2} \dfrac{x^2+y^2}{2Dt}\right)\\
& =\dfrac{1}{4 \pi Dt} \exp\left(-\dfrac{r^2}{4Dt}\right)
\end{align}

The PDF is normalized such that, $\int_{0}^{\infty} (2\pi r) \, p_{2D}(r,t)\, dr = 1.$
Finally, for isotropic 3D diffusion, $p_{3D}(r, t) = \left(\dfrac{1}{4 \pi Dt}\right)^{3/2} \exp\left(-\dfrac{r^2}{4Dt}\right).$ The PDF is normalized such that, $\int_{0}^{\infty} (4\pi r^2) \, p_{3D}(r,t)\, dr = 1.$ In summary, for $d$ = 1, 2, or 3 dimensions
$p_{dD}(r, t) = \left(\dfrac{1}{4 \pi Dt}\right)^{d/2} \exp\left(-\dfrac{r^2}{4Dt}\right).$