 Solution

Use the unwrapped angle on the real line. Integrating its Langevin equation gives $\theta (t)-\theta (0)=\omega t+\eta (t)$, where $\eta ={\int }_0^t\beta (s)ds$ is a centered Gaussian random variable with variance $2D_{r}t$. ThereforeA wrapped angular difference is bounded and would not satisfy this formula.

For the orientation correlation, $\mathbf{n}(t)\cdot \mathbf{n}(0)=\cos [\theta (t)-\theta (0)]$. The characteristic function gives $⟨e^{i\eta }⟩=e^{-⟨{\eta }^2⟩/2}=e^{-D_{r}t}$, soThe translational noise is independent of the orientation, hence the cross terms vanish. Summing its two component correlations gives $⟨\mathbf{u}(t)\cdot \mathbf{u}(0)⟩=4D\delta (t)$. Consequently, for $t\ge 0$,The extension to negative time uses $e^{-D_r|t|}$. Because the position contains Brownian motion, the instantaneous noise velocity and its correlation are distributions; the delta contribution is not a finite ordinary velocity at $t=0$.