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The Stokes-Einstein-Sutherland relationship, $$D = \frac{kT}{ 6 \pi \eta a}$$ where $D$ is the translational diffusivity is well known. A similar relationship is used to calculate the rotational diffusivity, specifically, $$D_{rot} = \frac{kT}{8 \pi \eta a^3},$$ where $a$ is the radius of the object and $\eta$ is the viscosity of the medium.

However, the drag coefficient in the denominator, $(8\pi\eta a^3)$ has units, $Js/rad$. But $D_{rot}$ has the units $rad^2/s$. Now $kT$ is in Joules, so does anyone know how this can be resolved?

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An angle in $rad$ is just a number and it is dimensionless. So drag coefficient has units $J.s$, $kT$ has units $J$ and $D_{rot}$ has units $s^{-1}$.

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