 Planar rotational diffusion

For free planar turning with constant $D_r$, no imposed torque and negligible inertia, model an unwrapped angle by the stochastic differential equationHere $W$ is standard one-dimensional Brownian motion, with increment variance $dt$, and brackets denote expected values over repeated realizations. The angular variance is conventionally reported in $\mathrm{ra}{\mathrm{d}}^2$. “Unwrapped” means recording accumulated turns continuously, rather than resetting the angle at each revolution. Directly subtracting wrapped camera angles creates artificial jumps; for an unmarked ellipsoid the wrapping period is $\pi$, and resolving successive turns additionally requires sufficiently frequent images.

For a planar arrow $\mathbf{u}=(\cos \varphi ,\sin \varphi )$, the orientational autocorrelation isFor an unoriented planar axis, use an autocorrelation unchanged by replacing either endpoint angle by that angle plus $\pi$:The same $D_r$ thus gives arrow and axis correlation times $1/D_r$ and $1/(4D_r)$, respectively, where a correlation time here is the lag at which the relevant curve reaches $1/e$. A short explanation is enough: $\Delta \varphi$ has a Gaussian distribution of variance $2D_{r}t$, so its characteristic function gives $⟨\cos (n\Delta \varphi )⟩=e^{-n^2{D}_{r}t}$. The integer $n$ selects what the observable can distinguish; it does not change the underlying angular spreading.

For an original numerical example, choose $D_r=0.25{\mathrm{s}}^{-1}$. After $2\mathrm{s}$, the angular variance is $1\mathrm{ra}{\mathrm{d}}^2$ and its root mean square displacement is $1\mathrm{rad}$, although the mean displacement is zero. The planar arrow autocorrelation is $e^{-0.5}\simeq 0.607$, while the planar axis autocorrelation is $e^{-2}\simeq 0.135$. Calling both “the orientation correlation” without specifying the observable would hide a factor of four in decay rate.