Mean squared displacement

In statistical mechanics, the mean squared displacement (MSD, also mean square displacement, average squared displacement, or mean square fluctuation) is a measure of the deviation over time between the position of a particle and some reference position. It is the most common measure of the spatial extent of random motion, and can be thought of as measuring the portion of the system "explored" by the random walker. It prominently appears in the Debye–Waller factor (describing vibrations within the solid state) and in the Langevin equation (describing diffusion of a Brownian particle). The MSD is defined as

${\rm MSD}\equiv\langle (x-x_0)^2\rangle=\frac{1}{T}\sum_{t=1}^T (x(t) - x_0)^2$

where T is the time over which one wants to average, and $x_0$ is the reference position of the particle. Typically this reference position will be the time-averaged position of the same particle, i.e. $\overline{x}$.

Derivation of the MSD for a Brownian particle in 1D

The probability density function (PDF) for a particle in one dimension is found by solving the one-dimensional diffusion equation. (This equation states that the position probability density diffuses out over time - this is the method used by Einstein to describe a Brownian particle. Another method to describe the motion of a Brownian particle was described by Langevin, now known for its namesake as the Langevin equation.)

$\frac{\partial p(x,t \mid x_0)}{\partial t}=D\frac{\partial^2p(x,t \mid x_0)}{\partial x^2},$

given the initial condition $p(x_{0},t={0} \mid x_{0})=\delta(x-x_{0})$; where $x(t)$ is the position of the particle at some given time, $x_0$ is the tagged particle's initial position, and $D$ is the diffusion constant with the S.I. units $m^2s^{-1}$ (an indirect measure of the particle's speed). The bar in the argument of the instantaneous probability refers to the conditional probability. The diffusion equation states that the speed at which the probability for finding the particle at $x(t)$ is position dependent.

It can be shown that the one-dimensional PDF is

$P(x,t)=\frac{1}{\sqrt{4\pi Dt}}\exp\left(-\frac{(x-x_0)^2}{4Dt}\right).$

This states that the probability of finding the particle at $x(t)$ is Gaussian, and the width of the Gaussian is time dependent. More specifically the full width at half maximum (FWHM)(technically/pedantically, this is actually the Full duration at half maximum as the independent variable is time) scales like

${\rm FWHM}\sim\sqrt{t}.$

Using the PDF one is able to derive the average of a given function, $L$, at time $t$:

$\langle L(t) \rangle\equiv \int^\infty_{-\infty} L(x,t) P(x,t) \, dx,$

where the average is taken over all space (or any applicable variable).

The Mean squared displacement is defined as

${\rm MSD}\equiv\langle ( x(t)-x_0)^2\rangle,$

expanding out the ensemble average

$\langle (x-x_0)^2 \rangle =\langle x^2\rangle+x_0^2 - 2x_0\langle x\rangle,$

dropping the explicit time dependence notation for clarity. To find the MSD, one can take one of two paths: one can explicitly calculate $\langle x^2\rangle$ and $\langle x\rangle$, then plug the result back into the definition of the MSD; or one could find the moment-generating function, an extremely useful, and general function when dealing with probability densities. The moment-generating function describes the $k^{\textrm{th}}$ moment of the PDF. The first moment of the displacement PDF shown above is simply the mean: $\langle x\rangle$. The second moment is given as $\langle x^2\rangle$.

So then, to find the moment-generating function it is convenient to introduce the characteristic function:

$G(k)=\langle e^{ikx}\rangle\equiv \int_I e^{ikx}P(x,t\mid x_0) \, dx,$

one can expand out the exponential in the above equation to give

$G(k) = \sum^\infty_{m=0}\frac{(ik)^m}{m!}\mu_m.$

By taking the natural log of the characteristic function, a new function is produced, the cumulant generating function,

$\ln(G(k)) = \sum^\infty_{m=1}\frac{(ik)^m}{m!}\kappa_m,$

where $\kappa_m$ is the $m \textrm{th}$ cumulant of $x$. The first two cumulants are related to the first two moments, $\mu$, via $\kappa_1 =\mu_1;$ and $\kappa_2 =\mu_2-\mu_1^2,$ where the second cumulant is the so-called the variance, $\sigma^2$. With these definitions accounted for one can investigate the moments of the Brownian particle PDF,

$G(k)=\frac{1}{\sqrt{4\pi Dt}}\int_I \exp(ikx)\exp\left(-\frac{(x-x_0)^2}{4Dt}\right) \, dx;$

by completing the square and knowing the total area under a Gaussian one arrives at

$G(k)=\exp(ikx_0-k^2Dt).$

Taking the natural log, and comparing powers of $ik$ to the cumulant generating function, the first cumulant is

$\kappa_1=x_0,$

which is as expected, namely that the mean position is the Gaussian centre. The second cumulant is

$\kappa_2=2Dt, \,$

the factor 2 comes from the factorial factor in the denominator of the cumulant generating function. From this, the second moment is calculated,

$\mu_2=\kappa_2+\mu_1^2=2Dt+x_0^2.$

Plugging the results for the first and second moments back, one finds the MSD,

$\langle (x(t)-x_0)^2 \rangle = 2Dt.$

MSD in experiments

Experimental methods to determine MSDs include neutron scattering and photon correlation spectroscopy.

See also

• Root-mean-square deviation of atomic positions: the average is taken over a group of particles at a single time, where the MSD is taken for a single particle over an interval of time