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

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

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

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