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\exnumber{8034}
\heading{Brownian particle on a ring}
\auname{Shimon Haver}
{\bf The problem:}
\Dn
The motion of a classical Brownian particle on a 1D ring is described
by the Langevin equation ${m\ddot{\theta}+\eta\dot{\theta}=f(t)}$,
where~$f(t)$ is due to a noisy electromotive force that has a correlation
function ${\langle f(t')f(t'')\rangle=C_{f}(t'-t'')}$. The power
spectrum $\tilde{C}_{f}(\omega)$ is defined as the Fourier transform
of the correlation function. We consider two cases:
\begin{enumerate}
\item [(a)] High temperature white noise ${\tilde{C}_{f}(\omega)=\nu}$.
\item [(b)] Zero temperature noise ${\tilde{C}_{f}(\omega)=c|\omega|}$.
\end{enumerate}
We define the angular velocity of the particle as ${v=\dot{\theta}}$,
and its Cartesian coordinate as ${x=\sin(\theta)}$. In the absence
of noise the dynamics is characterized by the damping time $t_{c}=m/\eta$.\\
In items (3)-(5) you should assume a spreading scenario: the particle
is initially (${t=0}$) located at ${\theta\sim0}$. The spreading
during the transient period ${0\frac{1}{t_c}$ and get:
\begin{equation}C_{v}(t)=\langle v(t)v(0)\rangle =\int \frac{d\omega'}{2\pi}\frac{c|\omega'|}{\eta^2+m^2\omega'^2}e^{-i\omega' t}=\frac{c}{\eta^2}\int \frac{d\omega'}{2\pi}\frac{|\omega'|}{1+t_c^2\omega'^2}e^{-i\omega' t}\approx\frac{c}{\eta^2}\int \frac{d\omega'}{2\pi}|\omega'|e^{-i\omega' t}\end{equation}
We know that the Fourier transform of $|\omega|$ has zero area, with negative tails $-\frac{1}{\pi t^{2}}$, so we get:
\begin{equation}C_{v}(t)=\langle v(t)v(0)\rangle \approx\frac{c}{\eta^2}\int \frac{d\omega'}{2\pi}|\omega'|e^{-i\omega' t}=-\frac{c}{\eta^2\pi t^{2}}\end{equation}
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\Dn
(3)
In the beginning we define $\dot{\theta}=v$, so we get:
\begin{equation}\theta(t)=\int_{0}^{t}dt'v(t')\end{equation}
\begin{equation}\theta^2(t)=\int_{0}^{t}\int_{0}^{t}dt'dt''v(t')v(t'')\end{equation}
\begin{equation}\langle\theta^2(t)\rangle=\int_{0}^{t}\int_{0}^{t}dt'dt''\langle v(t')v(t'')\rangle=\int_{0}^{t}\int_{0}^{t}dt'dt''C_v(t'-t'') \end{equation}
We can see that $t',t''$ are independent variables, so we can choose that $t'>t''$ and double the result. We can do a change of variables to two dependent variables $T=t'\to 0