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),


(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.


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