# How is mean square displacement defined in Brownian motion?

I am trying to calculate mean square displacement (MSD) of water in a simulation to calculate its diffusion coefficient. However I am little confused on its exact form. Can anyone please explain it in little detail? My confusion is: from wikipedia first paragraph on MSD page

$$MSD = 1/T * \sum_\text{t=1}^\text{T} (x(t)-x_\text{0})^2$$

My understanding of it was that:

$$MSD =(x(0)-x_\text{0})^2/1 + \{(x(0)-x_\text{0})^2 + (x(1)-x_\text{0})^2\}/2 + \{(x(0)-x_\text{0})^2+(x(1)-x_\text{0})^2+(x(2)-x_\text{0})^2\}/3$$ ... and so on

That is that each term in MSD is cumulative sum of all previous terms devided by time. However most of online programms that i saw (for example matlab programme at the last page of this link) tries to calculate MSD as just simple square of distance between initial position and position at time T. Can some one please explain if it is a cumulative sum of previous displacements or not.

• Try at physics.SE, it belongs there more. Apr 25 '16 at 4:58

I think you misunderstand the $\sum$ symbol and the fact that the term $1/T$ multiply the whole sum.

If you want to sum the integers between 1 and 3 and multiply this sum by 5 $$S = 5 * (1+2+3)$$ you can write $$S = 5 * \sum_{n=1}^3 n.$$

If you now look at your espression to compute the MSD after a total simulation time $T$, you have almost the same expression. You sum the square distances after each time step $t$ and then you divide by the total time T the whole sum. For example for $T=3$ you have $$MSD(3) = \frac{1}{3}[ (x(1)-x_0)^2 + (x(2)-x_0)^2 + (x(3)-x_0)^2].$$ In general this becomes $$MSD(T) = \frac{1}{T} \sum_{t=1}^T (x(t)-x_0)^2.$$

The mean squared distance is a measure of deviation: how far away two things are.

Let's say we want to find the MSD of a particle from it's origin. What we do is find the square distance at specific time intervals, say every second. We then make a table of data:

$$\begin{array}{c|c|} \text{time} & \text{square distance} \\ \hline 1 & 3.6 \\ \hline 2 & 3.8 \\ \hline 3 & 3.4 \\ \hline 4 & 3.2 \\ \hline \end{array}$$

The formula for the average is adding everything up and dividing by the number of things, so in this example:

$$MSD = \frac{3.6+3.8+3.4+3.2}{4}$$

What I've done above is stated more generally, and much more elegantly, as the formula you stated,

$$MSD = 1/T * \sum_\text{t=1}^\text{T} (x(t)-x_\text{0})^2$$

Your cumulative sum formula doesn't make intuitive sense, it instead finds the sum of the mean square distance at every time increment (there is no reason why you'd want to do that)

• Thank you for your clear answer. Forgive me if my question seems too general or silly; but I want to know now that we have the MSD of "one single" particle in a random-walk environment, what is it useful for? Jul 3 '18 at 23:12

At least in the software I use, MSD is based only on the displacement vector from the initial position. It's computationally cheap and very good at differentiating between diffusion and Brownian motion. In the former you'll get almost always a smooth positive gradient whereas the latter would give random gradients.