Kris's Research Notes

November 15, 2010

Rereading Krug

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)),


\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]


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