# Kris's Research Notes

## November 15, 2010

Filed under: Stat. Mech for SOS Model — Kris Reyes @ 8:54 am

I am currently rereading J. Krug’s paper “Atom Mobility for the Solid-on-Solid Model.” I think I understand more of it than last time, so I will now attempt to discuss the continuum limit for the evolution of a height profile.

Recall, in the solid on solid model, we move from one height profile $h = (h_0, \hdots, h_{N-1})$ to another one $h^\prime$ via atom hops:

$\displaystyle h^{\prime} = h - e_i + e_{i\pm 1}$,

where the indexing is done modulo $N$.

In the continuum limit, we describe a height profile as $h(x,t) : [0,N] \times \mathbb R \rightarrow \mathbb R$.  We have a conservation law:

$\displaystyle \frac{\partial h}{\partial t} = - \frac{\partial J}{\partial x}$

where the $\frac{\partial J}{\partial x}$ is the mass transport term. I’m not entirely sure about the origin of this term but, we may define $J$ by

$\displaystyle J = -\sigma \frac{\partial \mu}{\partial x}$

where $\sigma$ describes adatom mobility and $\mu(x,t)$  is the chemical potential at $x$. While I don’t understand this completely, I believe this says that atoms hop because of differences in chemical potential.

Now as we have pointed out (see e.g. my thesis proposal), we may write the chemical potential $\mu = - \frac{\partial m}{\partial x}$, where the quantity $m$ has a functional dependence on $u(x, t) = \mathbb E_t\left[\frac{\partial h}{\partial x}\right]$, the expected slope of the height at time $t$.  Explicitly we have

$\displaystyle u = u(m(x,t)) = \frac{\sinh(m(x,t))}{\cosh \theta - \cosh m(x,t)},$

and upon inverting we have:

$m = m(u(x,t)) = kt G(u(x,t)),$

where

$\displaystyle G(x) = sgn(x)\cosh^{-1}\left[\frac{x^2\cosh\theta - \sqrt{1+x^2\sinh^2\theta}}{x^1-1}\right],$

where $\theta = \frac{\beta \gamma}{2}$ and $\gamma$ is the bond strength.

The time dependence here is because the underlying probability distribution changes in time according to the master equation). Note this is different from Krug, where he defines $u(x,t) = \frac{\partial h}{\partial x}$ instead of the expectation. Either way, we may write $m = m(u(x,t))$ and hence

$\displaystyle \mu(x,t) = - \frac{\partial m}{\partial x} = - \frac{dm}{du} \frac{\partial u}{\partial x}$

Here $\frac{dm}{du}$ corresponds to the effective stiffness of the surface, and we denote it by $\hat \gamma(x,t)$. The inverse of this, $\frac{du}{dm}$, turns out to be the variance $V(x,t)$ of $\frac{\partial h}{\partial x}$. Therefore, we may write the chemical potential

$\displaystyle \mu(x,t) = - \hat\gamma(x,t) \frac{\partial u}{\partial x} = - \hat \gamma(x,t) \mathbb E_t\left[\frac{\partial^2 h}{\partial x^2}\right]$

where I have assumed (but not proved that) $\frac{\partial}{\partial x} \mathbb E_t\left[\frac{\partial h}{\partial x}\right] =\mathbb E_t\left[ \frac{\partial^2 h}{\partial x^2}\right]$. If instead we use Krug’s definition for $u$, we have

$\displaystyle \mu(x,t) = - \hat \gamma(x,t) \frac{\partial^2 h}{\partial x^2}.$

Then the conservation law above may be written as:

$\displaystyle \frac{\partial h}{\partial t} = - \frac{\partial }{\partial x}\left[ - \sigma \frac{\partial \mu}{\partial x}\right] = - \frac{\partial }{\partial x}\left[ \sigma \frac{\partial}{\partial x}\hat \gamma(x,t) \mathbb E_t\left[\frac{\partial^2 h}{\partial x^2}\right]\right]$

or with Krug’s definition

$\displaystyle \frac{\partial h}{\partial t} = - \frac{\partial }{\partial x}\left[ - \sigma \frac{\partial \mu}{\partial x}\right] = - \frac{\partial }{\partial x}\left[ \sigma \frac{\partial}{\partial x}\hat \gamma(x,t) \frac{\partial^2 h}{\partial x^2} \right]$