# Kris's Research Notes

## October 6, 2010

### Comparing Marginal Distributions – Part 3

Filed under: Stat. Mech for SOS Model — Kris Reyes @ 4:53 pm

This is a follow up of this post.

Recall we are evolving a height profile $\bar h(t)$ that is initially sinusoidal $\bar h_i(0) = 64 \sin\left(\frac{2\pi}{64}(i-1)\right)$ for $i = 1, \hdots, 64$ and we have fixed  $h_0(t) = 0$ for all $t$. We simulate the evolution with atom hopping rates

$\displaystyle r(i, i\pm1) = \frac{\Omega^\prime}{2}e^{-\beta\gamma n_i}$

where $n_i$ is the number of lateral neighbors of the top atom at site $i$ (always counting half bonds with the wall for $n_M$) and $\Omega^\prime = 5\times 10^7$. We also set the rates $r(1, 0)$ and $r(M, M+1)$ to zero. We noted two things in the previous post. First, we had observed a discontinuity between $h_1(t)$ and $h_0(t) = 0$ for small $t$.  Second, comparing the expected and empirical marginal distributions  $\mathbb P_{emp}(t; h_1)$ and

$\displaystyle \mathbb P_{leq} (t;h_1) = \frac{\sinh \theta - \sinh \beta m_1(t)}{\cosh \theta} e^{\theta|h_1(t)| + m_1(t)h_1(t)},$

we observed that the two distributions did not agree for small $t$.

We had discussed that this was probably due to the discontinuity between $h_0$ and $h_1$.  To avoid this, we decided to use an initial height profile that was very smooth at $x=0$, e.g. $sin^5(x)$. So we repeat the previous experiment with initial profile

$\bar h_i(0) = 64 \sin^5\left(\frac{2\pi}{64}(i-1)\right)$

Here is a movie of the simulated evolution of $\bar h(t)$ with rates specified above and total simulated time of $0.1$ seconds. The predicted average height profile was solved using Euler method with a time step of $\Delta t = 10^{-9}$ seconds.

As we can see, we still get a discontinuity between $h_0 = 0$ and $h_1$. Here are the predicted vs. observed height profiles along with marginals at several times in the interval $[0, 0.1]$.