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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 ${0<t<t_{c}}$ 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

(1) 
We will start with writing the Langevin equation for the velocity ${m\dot{v}+\eta v=f(t)}$, we can solve it with Fourier transform:\\
\begin{equation}
(-im\omega+\eta)v_\omega=f(\omega)
\end{equation}
\begin{equation}
v_\omega=\frac{f(\omega)}{\eta-im\omega}
\end{equation}\\
Now we can take square absolute value from both sides and average :\\
\begin{equation}
\langle\left\vert v_\omega\right\vert^2\rangle =\frac{\langle\left\vert f(\omega)\right\vert^2\rangle}{\eta^2+m^2\omega^2}
\end{equation}\\
From the Wiener-Khinchin theorem we get that $\langle\left\vert f(\omega)\right\vert^2\rangle=\tilde{C}_f(\omega)\times t$, so we get:
 \begin{equation}
\tilde{C}_{v}(\omega) =\frac{\tilde{C}_{f}(\omega)}{\eta^2+m^2\omega^2}
 \end{equation}\\
 For case (a) ${\tilde{C}_{f}(\omega)=\nu}$ we get:
 \begin{equation}\tilde{C}_{v}(\omega) =\frac{\nu}{\eta^2+m^2\omega^2}\end{equation}\\
 After inverse Fourier transform:
 \begin{equation}C_{v}(t) =\int \frac{d\omega'}{2\pi}\frac{\nu}{\eta^2+m^2\omega'^2}e^{-i\omega' t}=\frac{\nu}{2m\eta}\int d\omega\frac{1}{\pi}\frac{\frac{\eta}{m}}{\left(\frac{\eta}{m}\right)^2+\omega'^2}e^{-i\omega' t}\end{equation}\\
 This is a Lorentzian, so we get:
 \begin{equation}C_{v}(t)=\langle v(t)v(0)\rangle =\frac{\nu}{2m\eta}e^{-\frac{\left|t\right|}{\left(\frac{m}{\eta}\right)}}=\frac{\nu}{2m\eta}e^{-\frac{\left|t\right|}{t_c}}\end{equation}\\
%%%%%%%%%%%%%%%
\Dn

(2)
Now we use the same equation but in case (b) ${\tilde{C}_{f}(\omega)=c|\omega|}$:
\begin{equation}\tilde{C}_{v}(\omega) =\frac{\tilde{C}_{f}(\omega)}{\eta^2+m^2\omega^2}=\frac{c|\omega|}{\eta^2+m^2\omega^2}\end{equation}\\
we need to do inverse Fourier transform:
\begin{equation}C_{v}(t) =\int \frac{d\omega'}{2\pi}\frac{c|\omega'|}{\eta^2+m^2\omega'^2}e^{-i\omega' t}\end{equation}

We know that the change of $|\omega|$ is slow except near the $\omega=0$, so the shape of $\omega\approx 0$ is determined by the higher $t$ and the shape of $\omega\gg 0$ is determined by the lower $t$. We take the limit $t\gg t_c$ so we can neglect $\omega>\frac{1}{t_c}$ 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}
%%%%%%%%%%%%%%
\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 $T=t'\to 0<T<t,\tau=t'-t''\to 0<\tau<T$.
\begin{equation}\langle\theta^2(t)\rangle=2\int_{0}^{t}dT\int_{0}^{T}d\tau C_v(\tau)=2\int_{0}^{t}dT\int_{0}^{T}d\tau \frac{\nu}{2m\eta}e^{-\frac{\left|\tau\right|}{\left(\frac{m}{\eta}\right)}}\end{equation}
The correlation decay very fast so in the limit $t\gg t_c$ we can take the integral to infinite:
\begin{equation}\langle\theta^2(t)\rangle=2\int_{0}^{t}dT\int_{0}^{\infty}d\tau \frac{\nu}{2m\eta}e^{-\frac{\left|\tau\right|}{\left(\frac{m}{\eta}\right)}}=2t\frac{\nu}{2m\eta}\left(\frac{m}{\eta}\right)=\frac{\nu}{\eta^2}t\end{equation}
%%%%%%%%%%%%%
\Dn

(4)
In the same way:
\begin{equation}\langle\theta^2(t)\rangle=2\int_{0}^{t}dT\int_{0}^{T}d\tau C_v(\tau)\end{equation}
We can neglect the spreading during the transient period ${0<t<t_{c}}$, so we take the limit:
\begin{equation}\langle\theta^2(t)\rangle=2\int_{t_c}^{t}dT\int_{0}^{T}d\tau C_v(\tau)\end{equation}
The solution we found to $C_v(t)$ in case (b) is good just for $t\gg t_c$, so we need to divide the integral to two parts (we assume that the limit $T=t_c$ is the lower limit to our solution):
\begin{equation}\langle\theta^2(t)\rangle=2\int_{t_c}^{t}dT\left(\int_{0}^{\infty}d\tau C_v(\tau)-\int_{T}^{\infty}d\tau C_v(\tau)\right)\end{equation}

The part $\int_{0}^{\infty}d\tau C_v(\tau)=\tilde{C}_v(\omega=0)=0$, so we get:
\begin{equation}\langle\theta^2(t)\rangle=-2\int_{t_c}^{t}dT\int_{T}^{\infty}d\tau C_v(\tau)=2\int_{t_c}^{t}dT\int_{T}^{\infty}d\tau \frac{c}{\eta^2\pi \tau^{2}}=2\int_{t_c}^{t}dT\frac{c}{\eta^2\pi T}=\frac{2c}{\pi \eta^2}\ln{\frac{\left|t\right|}{t_c}}\end{equation}
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\Dn

(5)
We defined $x=\sin{\theta}$:
\begin{equation}\langle x^2(t)\rangle=\langle \sin^2{\theta(t)}\rangle=\langle \frac{\left(e^{i\theta}-e^{-i\theta}\right)^2}{-4}\rangle=\frac{1}{4}\langle \left(2-e^{i2\theta}-e^{-i2\theta}\right)\rangle=\frac{1}{2}-\frac{1}{4}\langle e^{i2\theta}\rangle-\frac{1}{4}\langle e^{-i2\theta}\rangle\end{equation}
We get a tip that for a Gaussian variable that has zero average $\langle e^{i\varphi}\rangle=e^{-(1/2)\langle\varphi^{2}\rangle}$.
Because $\theta$ is a Gaussian variable and it has zero average we get:

\begin{equation}\langle x^2(t)\rangle=\frac{1}{2}-\frac{1}{4}\langle e^{i2\theta}\rangle-\frac{1}{4}\langle e^{-i2\theta}\rangle=\frac{1}{2}-\frac{1}{4}e^{-2\langle\theta^2\rangle-}-\frac{1}{4}e^{-2\langle\theta^2\rangle}=\frac{1}{2}\left(1-e^{-2\langle\theta^2\rangle}\right)=\frac{1}{2}\left(1-e^{-2S(t)}\right) \end{equation}
Note that $\langle\left(2\theta\right)^2\rangle=\langle\left(-2\theta\right)^2\rangle=4\langle\theta^2\rangle$
%%%%%%%%%%%%%
\Dn

(6)
In the previous sections we assumed that $\theta(0)\approx 0$ and we talked about short times, so we could treat $\theta$ like a coordinate and calculate $\langle \theta(t)^2\rangle$. Now there isn't a preferred location on the ring so we can't calculate $S(t)=\langle \theta(t)^2\rangle$, because $\theta$ is not well defined. So we can't calculate $\langle x^2(t)\rangle$ like before, just the correlation between two different times $\langle x(t)x(0) \rangle$.\\
By definition:
\begin{equation}\langle x(t)x(0)\rangle=\int_{0}^{2\pi}\int_{0}^{2\pi}\sin(\theta_t)\sin(\theta_0)\rho(\theta_t,\theta_0)d\theta_td\theta_0\end{equation}
The formula for conditional probability is:
\begin{equation}\rho(A|B)=\frac{\rho(A,B)}{\rho(B)}\to \rho(A,B)=\rho(A|B)\rho(B)\end{equation}
\begin{equation}\langle x(t)x(0)\rangle=\int_{0}^{2\pi}\int_{0}^{2\pi}\sin(\theta_t)\sin(\theta_0)\rho(\theta_t|\theta_0)\rho(\theta_0)d\theta_td\theta_0\end{equation}
We get that in $t=0$, $\theta_0$ has a Uniform distribution, namely $\rho(\theta_0)=\frac{1}{2\pi}$.\\
Additionally $\rho(\theta_t|\theta_0)$ is the probability to find $\theta_t$ when we know where is $\theta_0$, and this is like the previous section, when we assumed that $\theta_0=0$.\\
The probability  $\rho(\theta_t|\theta_0)$ depends only on the difference between $\theta_t$ and $\theta_0$, it doesn't depends on one of them, so let's define $\delta\theta=\theta_t-\theta_0$, when $\rho(\theta_t|\theta_0)d\theta_t=\rho(\delta\theta)d\delta\theta$\\
By using a trigonometric identities we get:
 \begin{equation}\int_{0}^{2\pi}\int_{0}^{2\pi}\sin(\theta_t)\sin(\theta_0)\rho(\theta_t|\theta_0)\rho(\theta_0)d\theta_td\theta_0=
 \frac{1}{4\pi}\int_{0}^{2\pi}\int_{0}^{2\pi}\left(\cos(\delta\theta)-\cos(2\theta_0+\delta\theta)\right)\rho(\delta\theta)d\delta\theta d\theta_0=\end{equation}
 \begin{equation*}=\frac{1}{4\pi}\int_{0}^{2\pi}\int_{0}^{2\pi}\cos(\delta\theta)\rho(\delta\theta)d\delta\theta d\theta_0- \frac{1}{4\pi}\int_{0}^{2\pi}\int_{0}^{2\pi}\cos(2\theta_0+\delta\theta)\rho(\delta\theta)d\delta\theta d\theta_0=\end{equation*}
 The first integral doesn't depend on $\theta_0$, the integral on $\theta_0$ in the second term give $0$, and we get:
  \begin{equation}=\frac{1}{4\pi}\int_{0}^{2\pi}\int_{0}^{2\pi}\cos(\delta\theta)\rho(\delta\theta)d\delta\theta d\theta_0=\frac{1}{2}\int_{0}^{2\pi}\cos(\delta\theta)\rho(\delta\theta)d\delta\theta=\frac{1}{2}\langle\cos(\delta\theta)\rangle\end{equation}
  When we take $\theta_0=0$ we get that $\delta\theta$ is the same $\theta$ we define in the Previous section and get:
  \begin{equation}\frac{1}{2}\langle\cos(\delta\theta)\rangle=\frac{1}{4}\langle e^{i\delta\theta}+e^{-i\delta\theta}\rangle=\frac{1}{4}\left(\langle e^{i\delta\theta}\rangle+\langle e^{-i\delta\theta}\rangle\right)=\frac{1}{2}e^{-\frac{1}{2}\langle\delta\theta^2\rangle}=\frac{1}{2}e^{-\frac{1}{2}S(t)}\end{equation}
%%%%%%%%%%%%%
\Dn

(7)
For case (b):
\begin{equation} S(t)=\frac{2c}{\pi \eta^2}\ln{\frac{t}{t_c}}\end{equation}
\begin{equation}\langle x(t)x(0) \rangle=\frac{1}{2}e^{-\frac{1}{2}S(t)}=\frac{1}{2}\left(\frac{t}{t_c}\right)^{-\frac{c}{\pi \eta^2}}\end{equation}
From the FDT we get the relationship between the correlation function and the response:
\begin{equation}\im \chi\sim\frac{\omega}{2T}\tilde{C}_{xx}(\omega)\end{equation}
In the DC limit, we get:
\begin{equation}\im \chi=\frac{\omega}{2T}\tilde{C}_{xx}(\omega=0)\end{equation}
When:
\begin{equation}\tilde{C}_{xx}(\omega=0)=\int_{-\infty}^{\infty}C_{xx}(t)dt=\int_{-\infty}^{\infty}\frac{1}{2}\left(\frac{t}{t_c}\right)^{-\frac{c}{\pi \eta^2}}dt\end{equation}
We define "phase transition" as When the response diverges. this will happen when $\frac{c}{\pi \eta^2}\le1$, so we get:
\begin{equation}\eta_c=\sqrt{\frac{c}{\pi}}\end{equation}
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