Monday, July 20, 2009

[1] MSD of random walks with given SWD and directional correlations

We consider a stationary random walk of a particle in 2D. The position of the particle is measured at regular time intervalls
t_n=n\Delta t \; \mbox{with} \; n=0,1,\ldots
After each intervall, the particle has moved by one step, characterized by a positive step width and a direction (unit) vector:
\Delta\vec{R}_n = s_n\;\vec{e}_n = s_n \; \left( \cos(\phi_n) , \sin(\phi_n) \right) .
Let the probability distribution function of the step widths (abbreviated by SWD) be given by
P(s) = \mbox{Prob}(s_n=s)
with finite mean and variance and assume that the successive step widths are statistically independent random variables.
\left\langle (s_m-\overline{s})(s_n-\overline{s})\right\rangle = \overline{s^2}\delta_{mn}.
Let the correlation function of the direction vectors be defined as
C_{mn}=\left\langle \vec{e}_m  \vec{e}_n \right\rangle = \left\langle \cos(\phi_m-\phi_n) \right\rangle,
where the brackets denote the ensemble average. Assume that the width and direction of each step are statistically uncorrelated.
\left\langle (s_m-\overline{s})\vec{e}_n\right\rangle = \vec{0}.
The position of the particle at time step N is given by
\vec{R}_N = \sum_n \Delta\vec{R}_n.

What is the ensemble averaged mean squared displacement (MSD) of the particle ?

We have
\left\langle \vec{R}^2_N  \right\rangle = \left\langle  \sum_m \sum_n \Delta\vec{R}_m \Delta\vec{R}_n \right\rangle 

=\sum_{m,n} \left\langle s_m\vec{e}_m\;s_n\vec{e}_n \right\rangle 

Due to the independence of step lengths and directions this factorizes to
=\sum_{m,n}  \left\langle s_m s_n \right\rangle \; \left\langle \vec{e}_m \vec{e}_n  \right\rangle

=\sum_{m,n} \left\langle s_m s_n \right\rangle \; C_{mn}.

We next write the step widths in the form of average plus fluctuation:
s_m = \overline{s} + \Delta s_m,
understanding that the fluctuations are zero mean:
\left\langle \Delta s_m \right\rangle = 0.
This yields
\left\langle \vec{R}^2_N  \right\rangle = \sum_{m,n} C_{mn} \left\langle (\overline{s} + \Delta s_m) (\overline{s} + \Delta s_n) \right\rangle.
Under our statistical assumptions, we have
\left\langle (\overline{s} + \Delta s_m) (\overline{s} + \Delta s_n) \right\rangle = \overline{s}^2 + \left\langle \Delta s_m \Delta s_n \right\rangle =   \overline{s}^2 + \overline{s^2}\;\delta_{mn},
Note that here
stands for the variance of the fluctuation of the step widths. We obtain
\left\langle \vec{R}^2_N  \right\rangle = \overline{s}^2\sum_{mn}C_{mn}+\overline{s^2}\sum_m C_{mm}.

 C_{mm}=\left\langle \vec{e}^2_m \right\rangle = 1,

we obtain
 \left\langle \vec{R}^2_N  \right\rangle = \overline{s}^2\sum_{mn}C_{mn}+N\;\overline{s^2}.
Because our random walk was assumed to be a stationary random process, the correlation functions can only depend on time differences,
, and thus we have
 \left\langle \vec{R}^2_N  \right\rangle = N\;\overline{s^2} + \overline{s}^2\sum_{mn}C_{m-n}.

It is easy to show that
 \sum_{m=1}^N \sum_{n=1}^N f_{(m-n)}=\sum_{k=-N+1}^{N-1}(N-k)  f_k.

Using this, we arrive at
 \left\langle \vec{R}^2_N  \right\rangle = \overline{s^2}\;N + \overline{s}^2\sum_{k=-N+1}^{N-1}(N-k)C_k
 =\overline{s^2}\;N + N\overline{s}^2\sum_{k=-N+1}^{N-1}C_k - \overline{s}^2\sum_{k=-N+1}^{N-1}k C_k.
Since the correlation function is symmetric with respect to k, while k is asymmetric, the last term vanishes, yielding
 \left\langle \vec{R}^2_N  \right\rangle = \overline{s^2}\;N + \overline{s}^2\; N\;\sum_{k=-N+1}^{N-1}C_k.

In the sum, we can further extract the k=0 term
and make use of the symmetry
to obtain
\left\langle \vec{R}^2_N  \right\rangle = \overline{s^2}\;N + \overline{s}^2\;N + \overline{s}^2\; 2N\;\sum_{k=1}^{N-1}C_k.
Thus we finally arrive at
\left\langle \vec{R}^2_N  \right\rangle = \left( \overline{s^2}+ \overline{s}^2+ 2\overline{s}^2\sum_{k=1}^{N-1}C_k\right) N.

We see that the MSD of our particle is completely determined by the mean and variance of the step width distribution and by the directional correlation function. All step width distributions with the same mean(s) and var(s) produce the same MSD. Moreover, interesting (non-diffusive) MSD-versus-lagtime relations can only be generated via the directional correlations. We have excluded the possibility of Levi flights and the like by demanding that the SWD is not heavy-tailed.

Note to self: There are at least 4 possibilities how fractional powerlaw MSDs can arise: (1) heavy-tailed step width, (2) heavy-tailed waiting times, (3) persistent directions, or (4) long-time correlated statistical dependencies between the widths and directions of the steps. It might be interesting to follow possibility (4), if I should ever find the time.

We can check a few limiting cases:

(a) Brownian diffusion

In this case
so that
 \sum_{k=1}^{N-1}C_k = 0 
and, hence,
 \left\langle \vec{R}^2_N  \right\rangle = (\overline{s}^2 + \overline{s^2})\;N,
as it should be for a diffusive process. Note that for an exponential step width distribution,
\overline{s}^2 = \overline{s^2},
so that
 \left\langle \vec{R}^2_N  \right\rangle = 2 \overline{s}^2 N.

(b) Ballistic motion

In this case
so that we get
and so
and so
 \sum_{1}^{N-1}C_k = (N-1) 
and so
\left\langle \vec{R}^2_N  \right\rangle = \left( \overline{s^2}+ \overline{s}^2+ 2\overline{s}^2 (N-1)\right) N.
In the limit of large N we obtain
\left\langle \vec{R}^2_N  \right\rangle = 2\overline{s}^2 N^2,
i.e. the MSD grows quadratically with time, as expected for ballistic motion.

(c) Exponentially decaying directional correlations

In this case,
For a measuring time intervall dt and a physical decorrelation time \tau, we would have to set
 q=e^{\Delta t/\tau}.
The sum can be written as
\sum_{k=1}^{N-1}C_k = -1 + \sum_{k=0}^{N-1} (q^{-1})^k = -1 +\frac{1-q^{-N}}{1-q}, 
so that
\left\langle \vec{R}^2_N  \right\rangle = \left( \overline{s^2}+ \overline{s}^2+ 2\overline{s}^2 (-1 +\frac{1-q^{-N}}{1-q}) \right) N.
=\left( \overline{s^2}- \overline{s}^2+ 2\overline{s}^2 \frac{1-q^{-N}}{1-q} \right) N. 
Note that for
N\Delta t/\tau<<1
we have
 1-q^{-N}=1-e^{-N\Delta t/\tau}\approx 1-1+N\Delta t/\tau=N\Delta t/\tau=N \ln(q).
So for lagtimes smaller than the decorrelation time, the MSD is ballistic, for larger lagtimes diffusive, as it should be.

(c) Long-time correlated directional correlations

We next assume that for large lagtimes, the correlations decac according to a power law
C_k\rightarrow c_0 N^{-\gamma} = c_0 (t/\Delta t)^{-\gamma}.

To treat this case, we change from discrete time steps to a continuous time by replacing
 N\rightarrow t/\Delta t
\sum_{k=1}^{N-1}C_k \rightarrow \frac{1}{\Delta t}\int_{t0}^t C(t^{\prime})dt^{\prime},
to obtain
\left\langle \vec{R}^2(t)  \right\rangle = \left( \overline{s^2}+ \overline{s}^2+ \frac{2\overline{s}^2}{\Delta t}\int_{t0}^t C(t^{\prime})dt^{\prime}\right)t/\Delta t .
In the limit of large lagtimes we have
\int_{t0}^t C(t^{\prime})dt^{\prime} = \int_{t0}^t c_0(t^{\prime}/\Delta t)^{-\gamma}dt^{\prime} \rightarrow (t/\Delta t)^{1-\gamma},
so that
\left\langle \vec{R}^2(t)  \right\rangle \rightarrow R_0^2\;(t/\Delta t)^{2-\gamma} .
For example, directional correlations decaying with lagtime as an inverse square root will produce a MSD growing with a fractional powerlaw exponent of 1.5.

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