dimensional analysis - Units of the Stokes-Einstein rotational diffusion coefficient

The Stokes-Einstein rotational diffusion relation tells us that we can write down a rotational diffusion coefficient for a sphere as:

$$D_r \approx \frac{k_B T}{\zeta_f} \approx \frac{k_B T}{(8 \pi \eta)(r)^3}$$

Where $k_B$ is Boltzmann's constant, $T$ is the temperature in Kelvin, $\zeta_f \approx (8 \pi \eta)(r)^3$ is the friction, $\eta$ is the viscosity of the medium (e.g. $\approx 1 \space cP$ in pure water), and $r$ is the radius of the sphere.

For an example calculation of $D_r \approx 0.2 \space rad^2/s$ (or $Hz$) of a $\approx 1 \space \mu m$ radius sphere in pure water at room temperature, please see this site (this shortened URL is going to WolframAlpha).

However, as we can directly see in the calculation, the units for $D_r$ are clearly expressible as rad/s or Hz. Why are the units for $D_r$ always reported as rad$^2$/s in the literature?

Answer

The diffusion equation takes the form

$$\frac{\partial f}{\partial t}=D\frac{\partial^2 f}{\partial x^2}$$ where $f$ is some function. For simplicity, let's let $f=\rho$ the mass-density. In this case, then we have, units-wise, $$\frac{{\rm kg/m^3}}{\rm s} = \left[D\right]\frac{\rm kg/m^3}{\rm m^2}$$ Thus, in order to have the correct units on both sides, $\left[D\right]={\rm m^2/s}$. In the case of rotational diffusion, the diffusion equation takes the form $$\frac{\partial f}{\partial t}=D_r\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial f}{\partial\theta}\right)+D_r\frac{1}{\sin^2\theta}\frac{\partial^2f}{\partial\phi^2}$$ Since $\theta$ and $\phi$ are measured in radians and $\sin\theta$ is unitless, then we must have that $\left[D_r\right]={\rm rad^2/s}$.