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\exnumber{8027}
\heading{Brownian motion of a diatomic particle} 
\auname{Elad Benjamin and Ido Michealovich}


{\bf The problem:}

\Dn

Consider the 1D motion of two beads of mass m, attached by a very
flexible spring with constant k (the length of the spring at rest
is much smaller than the fluctuations caused by random forces).The
beads are immersed in viscous liquid with friction coefficients $\gamma_1,\gamma_2$ and temperature T. Disregard the hydrodynamic interactions
between the beads and the direct collisions of the beads.

\Dn

(1) Write down Langevin equations for the beads. Neglect accelerations.

\Dn

(2) For $\gamma_1=\gamma_2\equiv\gamma$, Define $R=\frac{x_{1}+x_{2}}{2}$, $r=x_{1}-x_{2}$ and find $\langle(r(t)-r_0)^2\rangle$ and $\langle(R(t)-R_0)^2\rangle$

(where $r_0$ is half the initial distance between the "atoms" and $R_0$ is the initial location of the "molecule").

\Dn

(3) Solve the equations for $\gamma_1\neq\gamma_2$ and show that the same solution as in (2) is obtained by setting $\gamma_1=\gamma_2\equiv\gamma$.

\Dn

(4) Generalize the results to 3D.

\Dn\Dn


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

\Dn

(1)
The equations of motion:
\begin{equation}
\begin{cases}
m\ddot{x}_{1}=-k(x_{1}-x_{2})-\gamma_1\dot{x}_{1}+F_{1}(t)\\
m\ddot{x}_{2}=k(x_{1}-x_{2})-\gamma_2\dot{x}_{2}+F_{2}(t)
\end{cases}
\end{equation}

Where $x_{i}$ is the displacement of the $i$th bead, and $F_{i}(t)$
is the random force acting on the $i$th bead at time t. The correlation
of the random forces is:
\begin{equation}
\langle F_{i}(t)F_{j}(t')\rangle=2\gamma_i T\delta_{ij}\delta(t-t')
\end{equation}

Neglecting the acceleration term and defining $f_{i}=\frac{1}{\gamma_i}F_{i}$,
$\alpha_i=\frac{k}{\gamma_i}$ we get:
\begin{equation}
\begin{cases}
\dot{x}_{1}=-\alpha_1 (x_{1}-x_{2})+f_{1}(t)\\
\dot{x}_{2}=\alpha_2 (x_{1}-x_{2})+f_{2}(t)
\end{cases}
\end{equation}

%%%%%%%%%%%%%%
\Dn

(2)
The Langevin eqs. are coupled so we move to the coordinates of
the center of mass $r_{c.m}\equiv R$ and relative motion
$r_{rel}\equiv r$:
\begin{equation}
\begin{cases}
R=\frac{x_{1}+x_{2}}{2}\\
r=x_{1}-x_{2}
\end{cases}\Rightarrow\begin{cases}
x_{1}=R+\frac{1}{2}r\\
x_{2}=R-\frac{1}{2}r
\end{cases}
\end{equation}

Substituting (4) into eqs. (3) gives:
\begin{equation}
\dot{R}+\frac{1}{2}\dot{r}=-\alpha r+f_{1}(t)
\end{equation}
\begin{equation}
\dot{R}-\frac{1}{2}\dot{r}=\alpha r+f_{2}(t)
\end{equation}

Adding (5) and (6) gives a Langevin type eq. for the center of mass:
\begin{equation}
\label{eq:sec2langevin}
\dot{R}=\frac{1}{2}\big[f_{1}(t)+f_{2}(t)\big]
\end{equation}

which is the same as the Langevin eq. for a Brownian particle.

Taking the same step as done in class, we get:

\begin{equation}
\langle\Delta R^{2}(t)\rangle=\frac{T}{\gamma}\cdot t=2\big(\frac{D}{2}\big)t
\end{equation}

Where D is the diffusion coefficient for Brownian particle. We see
that the center of mass diffuses half as quickly as one free bead,
which makes sense, since effectively the drag is twice as big.

Now for the relative motion. Subtracting (6) from (5) we get:
\begin{equation}
\label{eq:sec2langevinB}
\dot{r}=-2\alpha r+\Big[f_{1}(t)-f_{2}(t)\Big]\equiv-2\alpha r+\phi(t)
\end{equation}

To solve the differential eqn., lets make the substitution $\{r=y e^{-2\alpha t}\Rightarrow\dot{r}=\dot{y}e^{-2\alpha t}-2\alpha y e^{-2\alpha t}\}$
to get:
\[
\dot{y}e^{-2\alpha t}-2\alpha y e^{-2\alpha t}=-2\alpha y e^{-2\alpha t}+\phi(t)\Rightarrow\dot{y}=\phi(t)e^{2\alpha t}
\]
\[
\dot{y}(t)=\dot{y}(0)+\int_{0}^{t}dt'\phi(t')e^{2\alpha t'}
\]

Going back to $r$:
\begin{equation}
r(t)=r_{0}e^{-2\alpha(t)}+\int_{0}^{t}dt'\phi(t')e^{-2\alpha(t-t')}
\end{equation}
So:
\[
\langle\Delta r^{2}(t)\rangle=\langle(r-r_{0})^{2}\rangle=\langle\Big[r_{0}(e^{-2\alpha t}-1)+\int_{0}^{t}dt'\phi(t')e^{-2\alpha(t-t')}\Big]^{2}\rangle
\]
\[
\langle\Delta r^{2}(t)\rangle=r_{0}^{2}(e^{-2\alpha t}-1)^{2}+2(e^{-2\alpha t}-1)\int_{0}^{t}dt'\langle r_{0}\phi(t')\rangle e^{-2\alpha(t-t')}+\int_{0}^{t}\int_{0}^{t}dt'dt''\langle\phi(t')\phi(t'')\rangle e^{-2\alpha(2t-t'-t'')}
\]

The initial relative displacement and the random forces on the beads
at some future time $t'>0$ are obviously uncorrelated, so the middle
term vanishes. Substituting $\phi$ and $\alpha$ back into
the eqn. we get:
\[
\langle\Delta r^{2}(t)\rangle=r_{0}^{2}(e^{-2\alpha t}-1)^{2}+\frac{1}{\gamma^{2}}\int_{0}^{t}\int_{0}^{t}dt'dt''\langle\big[F_{1}(t')-F_{2}(t')\big]\big[F_{1}(t'')-F_{2}(t'')\big]\rangle e^{-2\alpha(2t-t'-t'')}
\]
\[
\langle\Delta r^{2}(t)\rangle=r_{0}^{2}(e^{-2\alpha t}-1)^{2}+\frac{4T}{\gamma}\int_{0}^{t}\int_{0}^{t}dt'dt''\delta(t'-t'')e^{-2\alpha(2t-t'-t'')}=r_{0}^{2}(e^{-2\alpha t}-1)^{2}+\frac{4T}{\gamma}\int_{0}^{t}dt'e^{-2\alpha(2t-2t')}
\]
\begin{equation}
\langle\Delta r^{2}(t)\rangle=r_{0}^{2}(e^{-2\alpha t}-1)^{2}+\frac{T}{\alpha\gamma}e^{-4\alpha t}(e^{4\alpha t}-1)=r_{0}^{2}(e^{-\frac{2kt}{\gamma}}-1)^{2}+\frac{T}{k}(1-e^{-\frac{4kt}{\gamma}})
\end{equation}

Looking at Eq(10) we see that it is the sum of noise-free solution that depends on $r_0$
and a second "random walk" stochastic component (integrating the noise over duration $t$).
Therefore the expression for the spreading in Eq(11) is the sum of two terms: a noise-free solution that depends on $r_0$,
and a second term which is the same as for "random walk", namely $2[C(0)-C(t)]$,
where $C$ is the correlation function of $r$ (formally identical to $C_{vv}$ whose calculation has been worked out in the lecture).

\Dn

(3)
Again the Langevin eqs. are:
\begin{equation}
\begin{cases}
\dot{x}_{1}=-\alpha_1(x_{1}-x_{2})+f_{1}(t)\\
\dot{x}_{2}=\alpha_2(x_{1}-x_{2})+f_{2}(t)
\end{cases}
\end{equation}

Or in matrix notation:
\begin{equation}
\label{eq:sec3langevin}
\frac{d}{dt}\begin{pmatrix}x_{1}\\
x_{2}
\end{pmatrix}=M\begin{pmatrix}x_{1}\\
x_{2}
\end{pmatrix}+\begin{pmatrix}f_{1}\\
f_{2}
\end{pmatrix}
\end{equation}
Where 
\begin{equation}
M=\begin{pmatrix}-\alpha_{1} & \alpha_{1}\\
\alpha_{2} & -\alpha_{2}
\end{pmatrix}
\end{equation}

To de-couple the eqs., we must move to the basis of the modes of motion
- $\begin{pmatrix}\xi_{1}\\
\xi_{2}
\end{pmatrix}$. We diagonalize M so $M=UDU^{-1}$:
\[
D=\begin{pmatrix}0 & 0\\
0 & -(\alpha_{1}+\alpha_{2})
\end{pmatrix}\,\,\,\,\,\,\,U=\begin{pmatrix}1 & 1\\
1 & -\frac{\gamma_{1}}{\gamma_{2}}
\end{pmatrix}\,\,\,\,\,\,\,U^{-1}=\begin{pmatrix}\frac{\gamma_{1}}{\gamma_{1}+\gamma_{2}} & \frac{\gamma_{2}}{\gamma_{1}+\gamma_{2}}\\
\frac{\gamma_{2}}{\gamma_{1}+\gamma_{2}} & \frac{-\gamma_{2}}{\gamma_{1}+\gamma_{2}}
\end{pmatrix}
\]

The transformation matrix U takes us to the new basis:~~~$\begin{pmatrix}x_{1}\\
x_{2}
\end{pmatrix}=U\begin{pmatrix}\xi_{1}\\
\xi_{2}
\end{pmatrix}$. ~~~We can see that $\xi_{1}$ is the coordinate of the center
of mass up to a factor, and $\xi_{2}$ plays the part of the \emph{effective}
relative motion.

So now eq.~(\ref{eq:sec3langevin}) becomes:
\[
\frac{d}{dt}U\begin{pmatrix}\xi_{1}\\
\xi_{2}
\end{pmatrix}=MU\begin{pmatrix}\xi_{1}\\
\xi_{2}
\end{pmatrix}+\begin{pmatrix}f_{1}\\
f_{2}
\end{pmatrix}\Rightarrow U^{-1}U\frac{d}{dt}\begin{pmatrix}\xi_{1}\\
\xi_{2}
\end{pmatrix}=U^{-1}MU\begin{pmatrix}\xi_{1}\\
\xi_{2}
\end{pmatrix}+U^{-1}\begin{pmatrix}f_{1}\\
f_{2}
\end{pmatrix}
\]
\begin{equation}
\frac{d}{dt}\begin{pmatrix}\xi_{1}\\
\xi_{2}
\end{pmatrix}=D\begin{pmatrix}\xi_{1}\\
\xi_{2}
\end{pmatrix}+U^{-1}\begin{pmatrix}f_{1}\\
f_{2}
\end{pmatrix}
\end{equation}
So with plugging $\alpha_{1,2}$ back in, we arrive at the eqs:
\begin{equation}
\begin{cases}
\dot{\xi_{1}}=\frac{1}{\gamma_{1}+\gamma_{2}}(F_{1}+F_{2})\\
\dot{\xi}_{2}=-k\big(\frac{1}{\gamma_{1}}+\frac{1}{\gamma_{2}}\big)\xi_{2}+\frac{1}{\gamma_{1}+\gamma_{2}}\big(\frac{\gamma_{2}}{\gamma_{1}}F_{1}-F_{2}\big)
\end{cases}
\end{equation}

Indeed we see that for~~~ $\gamma_{1}=\gamma_{2}=\gamma$~~~
we get eqs.~(\ref{eq:sec2langevin}) and~(\ref{eq:sec2langevinB}) from the previous section up to a factor of $\frac{1}{2}$, which can be mended with multiplying one of the eigenvectors by 2.

The solution for the center of mass is exactly the same as before
(with the appropriate factor change):
\begin{equation}
\langle\Delta\xi_{1}^{2}(t)\rangle=\frac{2T}{\gamma_{1}+\gamma_{2}}\cdot t
\end{equation}


With the relative motion we need to be a bit more careful because
of the factor in front of $F_{1}$. We can apply the same method of
solution to finally get:
\begin{equation}
\langle\Delta\xi_{2}^{2}(t)\rangle=\xi_{2}^{2}(0)\Big(e^{-k\big(\frac{1}{\gamma_{1}}+\frac{1}{\gamma_{2}}\big)t}-1\Big)^{2}+\frac{2T}{k\big(\frac{\gamma_{1}}{\gamma_{2}}+1\big)^{3}}\Big(1-e^{-2k\big(\frac{1}{\gamma_{1}}+\frac{1}{\gamma_{2}}\big)t}\Big)
\end{equation}


Again we see that for~~~ $\gamma_{1}=\gamma_{2}=\gamma$~~~
we get the same results as in the previous section up to a factor of $\frac{1}{4}$ which can be fixed as explained before.

\Dn

(4)
Generalizing this result to 3D is simple enough- Since the forces
in different directions are independent of each other, we would simply
get the same answer as in section (3) for each direction separately.
So:
\begin{equation}
\langle\Delta\vec{\xi}_{1}^{2}(t)\rangle=\langle\Delta\xi_{1_{x}}^{2}(t)\rangle+\langle\Delta\xi_{1_{y}}^{2}(t)\rangle+\langle\Delta\xi_{1_{z}}^{2}(t)\rangle=\frac{6T}{\gamma_{1}+\gamma_{2}}\cdot t
\end{equation}
\begin{equation}
\langle\Delta\vec{\xi}_{2}^{2}(t)\rangle=\vec{\xi}_{2}^{2}(0)\Big(e^{-k\big(\frac{1}{\gamma_{1}}+\frac{1}{\gamma_{2}}\big)t}-1\Big)^{2}+\frac{6T}{k\big(\frac{\gamma_{1}}{\gamma_{2}}+1\big)^{3}}\Big(1-e^{-2k\big(\frac{1}{\gamma_{1}}+\frac{1}{\gamma_{2}}\big)t}\Big)
\end{equation}

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\end{document} 
