= Chiral active Brownian displacement
{title2=$M(t)=4Dt+2v^2\operatorname{Re}[t/z-(1-e^{-zt})/z^2],\quad z=D_r-i\omega$}
A planar self-propelled particle with translational <diffusion coefficient> $D$, propulsion speed $v$, rotational diffusion $D_r$ and constant angular drift $\omega$ has orientation correlation $e^{-D_r|t|}\cos(\omega t)$. Integrating this correlation twice gives the displayed mean-square displacement. For $D_r>0$, its long-time <diffusion coefficient> is $D+v^2D_r/[2(D_r^2+\omega^2)]$. The full position-angle process has the <Markov property>, whereas position alone generally retains orientational memory. The angular second moment refers to an unwrapped angle.
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