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\heading{E8034: Brownian particle on a ring} 
\auname{Yossi Perl}


{\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 ${0<t<t_{c}}$ is assumed to be negligible.
In item~(6) assume that the particle had been launched in the far
past (${t=-\infty}$): accordingly there is no preferred location
on the ring.
\begin{enumerate}
\item Find the exact correlation function $\langle v(t)v(0)\rangle$ in
case (a).
\item Find the correlation function $\langle v(t)v(0)\rangle$ for $t\gg t_{c}$
in case (b).
\item Find the spreading $S(t)\equiv\langle\theta(t)^{2}\rangle$ for $t\gg t_{c}$
in case (a).
\item Find the spreading $S(t)\equiv\langle\theta(t)^{2}\rangle$ for $t\gg t_{c}$
in case (b).
\item Express $\langle x(t)^{2}\rangle$ for a spreading scenario given
$S(t)$.
\item Express the correlation function $\langle x(t)x(0)\rangle$ given
$S(t)$.
\item Write the explicit long time expression for $\langle x(t)x(0)\rangle$
in case (b), and deduce what is the critical value $\eta_{c}$ above
which a "phase transition" is expected in the response characteristics
of the system.
\end{enumerate}
Tips: For a Gaussian variable that has zero average $\langle e^{i\varphi}\rangle=e^{-(1/2)\langle\varphi^{2}\rangle}$.
\\
 The Fourier transform of $|\omega|$ has zero area, with negative
tails $-1/(\pi t^{2})$. \\
 If you fail to solve (6), assume that the answer is the same as in
(5), and proceed to (7). 



\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn

\begin{enumerate}
\item Rewriting the equation as $m\frac{d}{dt'}[v(t')]+\eta v(t')=f(t')$, and multiplying
by the equation $m\frac{d}{dt''}[v(t'')]+\eta v(t'')=f(t'')$, we get the
expression: 
\[
m^{2}\frac{d}{dt'}\frac{d}{dt''}[v(t')v(t'')]+m\eta\frac{d}{dt'}[v(t')v(t'')]+m\eta\frac{d}{dt''}[v(t')v(t'')]+\eta^{2}[v(t')v(t'')]=f(t')f(t'')
\]
 now we shall take the expectation values of both sides of the equation.
$f(t)$ is a stationary signal, i.e. $<f(t')f(t'')>=C_{ff}(t'-t'')$,
we shall assume that $v(t)$ will also be a stationary signal: $<v(t')v(t'')>=C_{vv}(t'-t'')$. This is true when the initial velocity is "forgotten" and can be assumed to be a random variable.
Now we can write the equation with a single variable $t=t'-t''$,
and replace $\frac{d}{dt'}$ with $\frac{d}{dt}$ and $\frac{d}{dt''}$
with $-\frac{d}{dt}$. We get the much simpler equation: 
\[
-m^{2}\frac{d^{2}}{dt^{2}}C_{vv}(t)+\eta^{2}C_{vv}(t)=C_{ff}(t)
\]
 in case a, $C_{ff}(t)$ is a delta function. The simplest way to
solve the equation is using a Fourier transform: 
\[
m^{2}\omega^{2}\tilde{C}_{vv}(\omega)+\eta^{2}\tilde{C}_{vv}(\omega)=\tilde{C}_{ff}(\omega)=\nu
\]
 
\[
\tilde{C}_{vv}(\omega)=\frac{\nu}{(m\omega)^{2}+\eta^{2}}\Rightarrow C_{vv}(t)=\int_{-\infty}^{\infty}\frac{\nu}{(m\omega)^{2}+\eta^{2}}e^{i\omega t}\frac{d\omega}{2\pi}=\int_{-\infty}^{\infty}\frac{\nu}{(m\omega)^{2}+\eta^{2}}e^{i\omega t}\frac{d\omega}{2\pi}=
\]
 
\[
=-\oint_{r\rightarrow\infty}\frac{\nu}{2i\eta m}\frac{e^{i\omega t}}{\omega+\frac{i\eta}{m}}\frac{d\omega}{2\pi}+\oint_{r\rightarrow\infty}\frac{\nu}{2i\eta m}\frac{e^{i\omega t}-e^{-\frac{\eta t}{m}}}{\omega-\frac{i\eta}{m}}\frac{d\omega}{2\pi}+\frac{\nu}{2\eta m}e^{-\frac{\eta t}{m}}\oint_{r\rightarrow\infty}\frac{1}{\omega-\frac{i\eta}{m}}\frac{d\omega}{2\pi i}=
\]
 
$=\frac{\nu}{2\eta m}e^{-\frac{\eta t}{m}}$




Where the orbital integral is over a semi-circle with infinite radius
in the upper semi-plain $Im[\omega]\geq0$, and the last equality
is due to the Residue Theorem. 

\item In case b, the previous method would fail, since $\tilde{C}_{ff}(\omega)=C|\omega|$
is non-analytical at $\omega=0$, the Residue Theorem will not work,
and the inverse Fourier transform is hard to calculate. However, since
we are only asked for the solution at very long times, much longer
than the system's decay time $t_{c}=m/\eta$, we can assume that the
solution is now slowly-varying, i.e. $\frac{d^{2}}{dt^{2}}C_{vv}(t)\ll\frac{1}{t_{c}^{2}}C_{vv}(t)$.
Therefore, the solution for long times, is 
\[
C_{vv}(t)\simeq\frac{1}{\eta^{2}}C_{ff}(t)=-\frac{C}{\eta^{2}\pi t^{2}}
\]

\item  Again, we are only asked for the solution at long times, $t\gg t_{c}$.
Therefore we can go back to the equation $C_{vv}(t)\simeq\frac{1}{\eta^{2}}C_{ff}(t)$.
Now we can write: $S(t)=<(\theta(t))^{2}>=<\int_{0}^{t}v(t')dt'\int_{0}^{t}v(t'')dt''>=\int_{0}^{t}d\tilde{t}\int_{-\widetilde{t}}^{\widetilde{t}}\frac{1}{\eta^{2}}C_{ff}(\tau)d\tau$.
In the last equality the variables $\tilde{t}=t'+t''$ and $\tau=t'-t''$
were used. In case a, $S(t)=\int_{0}^{t}d\tilde{t}\int_{-\widetilde{t}}^{\widetilde{t}}\frac{1}{\eta^{2}}\nu\delta(\tau)d\tau=\frac{\nu}{\eta^{2}}t$
\item $S(t)=\int_{0}^{t}d\tilde{t}\int_{-\widetilde{t}}^{\widetilde{t}}\frac{1}{\eta^{2}}C_{ff}(\tau)d\tau$.
We know how the function $C_{ff}(\tau)$ looks like for a large $\tau$.
However, since the function is symmetric and has zero area $(\tilde{C}_{ff}(\omega=0)=0)$
, $\int_{-\widetilde{t}}^{\widetilde{t}}\frac{1}{\eta^{2}}C_{ff}(\tau)d\tau\equiv f(\tilde{t})=\frac{2}{\eta^{2}}\int_{0}^{\widetilde{t}}C_{ff}(\tau)d\tau=\frac{2}{\eta^{2}}\int_{0}^{\infty}C_{ff}(\tau)d\tau-\frac{2}{\eta^{2}}\int_{\tilde{t}}^{\infty}C_{ff}(\tau)d\tau=-\frac{2}{\eta^{2}}\int_{\tilde{t}}^{\infty}(-\frac{C}{\pi\tau^{2}})d\tau=\frac{2C}{\eta^{2}\pi\tilde{t}}$.
Now we need to calculate $\int_{0}^{t}f(\tilde{t})d\tilde{t}$ for
large $t$. Since we only know the form of $f(\tilde{t})$ for large
$\tilde{t}$, we can write $S(t)=S(t_{c})+\int_{t_{c}}^{t}\frac{2C}{\eta^{2}\pi\tilde{t}}d\tilde{t}=S(t_{c})+\frac{2C}{\eta^{2}\pi}\ln(\frac{t}{t_{c}})$,
where $S(t_{c})$ is some constant. However, if we assume, as instructed, that the spreading for $0<t<t_{c}$ is neglible, then $S(t_{c})=0$, and $S(t)=\frac{2C}{\eta^{2}\pi}\ln(\frac{t}{t_{c}})$
\item For a brownian motion, the noise $f(t)$ is a gaussian random variable, and so $\theta(t)$ is also a gaussian random variable.
Therefore: 
\[
<(x(t))^{2}>=<sin^{2}(\theta(t))>=<\left(\frac{e^{i\theta(t)}-e^{-i\theta(t)}}{2i}\right)^{2}>=\frac{1}{4}\left[2-<e^{2i\theta(t)}>-<e^{-2i\theta(t)}>\right]=
\]
\[
=\frac{1}{4}\left[2-<e^{2i\theta(t)}>-<e^{-2i\theta(t)}>\right]=\frac{1}{4}\left[2-e^{-2\langle(\theta(t))^{2}\rangle}-e^{-2\langle(\theta(t))^{2}\rangle}\right]=\frac{1}{2}\left[1-e^{-2S(t)}\right]
\]

\item Now we are no longer dealing with a spreading scenario. We have to
calculate $<x(t)x(0)>$where $x(0)$ is also a random variable. 
\[
<x(t)x(0)>=<sin(\theta(t))sin(\theta(0))>=<\left(\frac{e^{i\theta(t)}-e^{-i\theta(t)}}{2i}\right)\left(\frac{e^{i\theta(0)}-e^{-i\theta(0)}}{2i}\right)>=
\]
\[
=-\frac{1}{4}<\left(e^{i(\theta(t)+\theta(0))}+e^{i(\theta(t)+\theta(0))}-e^{i(\theta(t)-\theta(0))}-e^{i(\theta(t)-\theta(0))}\right)>=
\]
\[
=\frac{1}{2}\left[e^{-\frac{1}{2}<(\theta(t)-\theta(0))^{2}>}-e^{-\frac{1}{2}<(\theta(t)+\theta(0))^{2}>}\right]
\]
 Using the law of total expectation:
\[
<(\theta(t)-\theta(0))^{2}>=<<(\theta(t)-\theta(0))^{2}|\theta(0)>>=<S(t)>=S(t)
\]
\[
<(\theta(t)+\theta(0))^{2}>=<<(\theta(t)-\theta(0))^{2}|\theta(0)>+4<\theta(t)\theta(0)|\theta(0)>=S(t)+4<\theta^{2}(0)>
\]
 Since the particle was lounched at $t\longrightarrow-\infty$, we
can write the equality as: 
\[
<(\theta(t)+\theta(0))^{2}>=S(t)+4<\theta^{2}(0)-\theta^{2}(-\infty)>+4\theta^{2}(-\infty)=S(t)+4S(\infty)+Const\rightarrow\infty
\]
where the expression $S(\infty)=\lim_{t\rightarrow\infty}(s(t))$
diverges. Therefore:
\[
<x(t)x(0)>=\frac{1}{2}\left[e^{-\frac{1}{2}<(\theta(t)-\theta(0))^{2}>}-e^{-\frac{1}{2}<(\theta(t)+\theta(0))^{2}>}\right]=\frac{1}{2}e^{-\frac{1}{2}S(t)}
\]

\item $<x(t)x(0)>=\frac{1}{2}e^{-\frac{1}{2}S(t)}=\frac{1}{2}e^{-\frac{1}{2}\frac{2C}{\eta^{2}\pi}\ln(\frac{t}{t_{c}})}=\frac{1}{2}(\frac{t_{c}}{t})^{\frac{C}{\eta^{2}\pi}}$.
For $\frac{C}{\eta^{2}\pi}<1\rightarrow\eta>\eta_{c}=\sqrt{\frac{C}{\pi}}$,
the correlation between $X(t)$ and $X(0)$ is strong, which means that the
particle can be thought of as "localized''. In the oposite case, when $\frac{C}{\eta^{2}\pi}>1\rightarrow\eta<\sqrt{\frac{C}{\pi}}$, the correlation is weak. One response charactaristic that experiences a "phase transition" is the integral $\int_{t_c}^{\infty}{<x(t)x(0)>dt}$ which converges only for $\eta<\eta_{c}$.\end{enumerate}



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