# Kris's Research Notes

## September 29, 2010

### How the average height profile evolves over time.

Filed under: Stat. Mech for SOS Model — Kris Reyes @ 6:47 am

Recall we defined the average height profile

$\bar h_i(t) = \mathbb E_{P(t)} [h_i] = \displaystyle \sum_{h \in \mathcal X} h_i P(t;h)$.

If we approximated $P(t)$ with the local equilibrium distribution

$\displaystyle \pi_{leq}(t;h) = \frac{1}{\Xi} \exp\left[\beta (H(h) - \sum_{i=1}^N \mu_i h_i\right],$

we showed that $\bar h$ evolves approximately like

$\displaystyle \frac{d\bar h_k}{dt} =\frac{\Omega}{2} \left(e^{\beta\mu_{k-1}} - 2 e^{\beta\mu_k} + e^{\beta \mu_{k+1}}\right),$

for $k=2, \hdots, N-1$ and

$(1) \displaystyle \frac{d\bar h_1}{dt} =\frac{\Omega}{2} \left( - e^{\beta\mu_1} + e^{\beta \mu_2}\right),$

and

$(2) \displaystyle \frac{d\bar h_N}{dt} = \frac{\Omega}{2} \left( e^{\beta\mu_{N-1}} - e^{\beta\mu_N}\right).$

Here, $\Omega$ is the prefactor used in the hopping rate

$(3) \displaystyle r (h, h-e_i+e_{i\pm 1}) = \Omega e^{\beta[H(h) - H(h(i))]} = \Omega e^{-\beta\gamma(n_i(h)-2)},$

where $n_i(h)$ is the number of bonds the top atom at site $i$ has. If instead we wished to use the number of lateral neighbors, $\ell_i(h)$, then $n_i(h) = \ell_i(h)+1$ and the rates are given by

$\displaystyle r (h, h-e_i+e_{i\pm 1}) = \Omega e^{-\beta\gamma(\ell_i(h)-1)} = \Omega e^{\beta\gamma}e^{-\beta\gamma \ell_i(h)}.$

We make this point because in KMC simulations, we often define rates in terms of $\ell_i(h)$. For example, in our current simulation we set our rates such that $\Omega e^{\beta\gamma} = 5\times 10^7$, hence in our analysis we must fix $\Omega = e^{-\beta\gamma}5\times 10^7$. With $\gamma = 0.25$, this means $\Omega = 2.748\times 10^6.$ Note this is a large number.

Consider the average profile near equilibrium. Here the average profile does not change much in space ($\bar h_i - \bar h_{i-1} \approx 0$) or time ($\frac{d\bar h}{dt} \approx 0$). Then

$m_i = kT \mathcal M(\bar h_i - \bar h_{i-1}) = kt \mathcal M(\epsilon).$

Consider $kT \mathcal M(x)$ near $x = 0$:

Observe the $m_i$ are small even for (relatively) large difference in height profile. Then the $mu_i = -(m_{i+1} - m_i)$  are also small. Using the above graph as an example,  we see that near equilibrium $\mu_i \in [-0.3,0.3]$ with high probability.  In equilibrium all the $\mu_i$ are equal and so, by examining equations (1), (2), and (3) we see that $\frac{d\bar{h_i}}{dt} = 0$ in this case.

Now consider the system not near equilibrium. In particular, suppose we had the following height profile:

We wish to consider how our model predicts it will evolve with respect to the equations (1), (2), and (3) above. To that end, consider the plot of $e^{\beta \mu_i}$ where the $\mu_i$ are calculated from this average profileAs we see, the values are somewhat close together — but not close enough! That is, consider $\displaystyle \frac{1}{\Omega/2} \frac{d\bar h}{dt}$, the unnormalized rates:

When we scale by $\Omega/2 \approx 10^6$ we see that $\frac{d\bar h}{dt}$ is very large! This could leads to some unstable behavior if we try to evolve $\frac{d\bar h}{dt}$ using e.g Euler’s Method.