For ergodicity, what is the significance of the R value and slope?

Question

I am calculating MSDs for the first time and have been struggling to find newbie-level resources on understanding them. If anyone could please suggest any resources or provide guidance on how to interpret MSD results, I would be grateful.

I have been running NPT simulations to collect data for use in fluid dynamics simulations. However, I now realize that I have used low temperatures compared to people who have looked at similar systems. So I want to evaluate MSDs to make sure the simulations are ergodic.

• I don't have much experience with statistics and a basic point I don't understand is whether it is the slope of the MSD or the R value that matters. Or both.

• An example of an MSD I have calculated is below. I get this output (I can't control the time interval used, the software chooses it):

Linear regression interval 41.52 - 83.03 ps.
MSD(t) = -10806.283111 + 1575.888517 * t
R = 0.977891

When I calculate a slope on the interval 1 ps to where the curve goes vertical, I get ~0.8, which I know is not great. I don't know what an acceptable value would be but am guessing >0.9. I do get 0.95 if I calculate the slope not from 1 ps, but from 10 ps. But on a log-log plot, that's a small interval. Is it unacceptably small?

Answer 1

Short introduction to ergodicity

• Ergodicity is when the time-average equals the ensemble-average.
  
• A process is ergodic if the time-average "converges in the square mean" to the ensemble average.
• A sequence $X_t$ converges in the square mean to $X$ if:

$$ \tag{1} \lim_{t\rightarrow \infty}\langle \left|X_t - X\right|^2 \rangle = 0, $$

where $\langle x \rangle$ means the mean (average) of $x$. So if the mean of the square of the absolute difference between the time-average and ensemble-average (i.e. the MSD between them) approaches zero, then the process can be said to be ergodic.

Short introduction to generalized diffusion

If we plot MSD$(t)$ with MSD on the vertical axis and $t$ on the horizontal axis, and fit the data to a power law form:

$$ \tag{2} \textrm{MSD}(t) = Dt^\alpha, $$

where $D$ is the diffusion constant, and $\alpha$ is the generalized diffusion exponent:

• normal diffusion is characterized by $\alpha=1$, meaning MSD$(t)$ is linear.
• sub-diffusion is characterized by $0<\alpha<1$, meaning MSD$(t)$ is sub-linear.
• super-diffusion is characterized by $\alpha>1$, meaning MSD$(t)$ is super-linear.

Short introduction to statistical regression

The coefficient of determination is given by $R^2$ and is a measure of the "goodness of fit". Specifically, when you fit a line or curve through data, how well does that line or curve predict the data? If $R^2 = 1$ then the data is perfectly predicted by your fitted line or curve.

Application to your case

"I don't understand is whether it is the slope of the MSD or the R value that matters. Or both."

You could plot your MSD$(t)$ and fit the data to Eq. 2, which will give you an $\alpha$, which will tell you what type of diffusion you have. But since you chose to do a log-log plot, Eq. 2 has to be modified accordingly:

\begin{align} \tag{3} \log\textrm{MSD}(t) &= \frac{\alpha \log D}{\log 10} \log(t), \\ y &= m x, ~~~ m \equiv\alpha\left(\frac{ \log D}{\log 10}\right). \tag{4} \end{align}

Slope: Eq. 4 tells us that a very large slope may be indicative of super-diffusion, and a very small slope may be indicative of sub-diffusion.

$R$-value: Your $R$ value implies an $R^2$ of 0.956, which means that in the area where you did the fit (41.52 - 83.03 ps) the data is fairly linear (could be more linear but could be a lot worse).