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.
\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).
p_{dD}(r, t) = \left(\dfrac{1}{4 \pi Dt}\right)^{d/2} \exp\left(-\dfrac{r^2}{4Dt}\right).
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).
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