TBan-II

Thermal Interfaces

The dynamics is overdamped, so that we can neglect the inertial term. The Langevin equation of motion is

${\partial }_{t}h(r,t)=-\mu \frac{\delta E_{pot}}{\delta h(r,t)}+\eta (r,t)$

The first term $-\delta E_{pot}/\delta h(r,t)$ is the elastic force trying to smooth the interface, the mobility $\mu$ is the inverse of the viscosity. The second term is the Langevin noise. It is Guassian and defined by

$⟨\eta (r,t)⟩=0,⟨\eta (r^{\prime },t^{\prime })\eta (r,t)⟩=2dD{\delta }^d(r-r^{\prime })\delta (t-t^{\prime })$

The symbol $⟨\dots ⟩$ indicates the average over the thermal noise and the diffusion constant is fixed by the Einstein relation $D=\mu K_{B}T$. We set $\mu =K_B=1$

The potential energy of surface tension ($\nu$ is the stiffness) can be expanded at the lowest order in the gradient:

$E_{pot}\sim \text{const.}+\frac{\nu }{2}\int d^{d}r(\mathrm{\nabla }h{)}^2$

Hence, we have the Edwards Wilkinson equation:

${\partial }_{t}h(r,t)=\nu {\mathrm{\nabla }}^{2}h(r,t)+\eta (r,t)$

Scaling Invariance

The equation enjoys of a continuous symmetry because $h(r,t)$ and $h(r,t)+c$ cannot be distinguished. This is a condition of scale invariance:

$h(br,b^{z}t)\stackrel{inlaw}{\sim }{b}^{\alpha }h(r,t)$

Here $z,\alpha$ are the dynamic and the roughness exponent respectively. From dimensional analysis

$b^{\alpha -z}{\partial }_{t}h(r,t)=b^{\alpha -2}{\mathrm{\nabla }}^{2}h(r,t)+b^{-d/2-z/2}\eta (r,t)$

From which you get $z=2$ in any dimension and a rough interface below $d=2$ with $\alpha =(2-d)/2$.

Width of the interface

Consider a 1-dimensional line of size L with periodic boundary conditions. We consider the width square of the interface

$w_2(t)={\left[\underset{0}{\overset{L}{\int }}\frac{dr}{L}(h(r,t)-\underset{0}{\overset{L}{\int }}\frac{dr}{L}h(r,t))\right]}^2$

It is useful to introduce the Fourier modes:

${\hat{h}}_q(t)=\frac{1}{L}\underset{0}{\overset{L}{\int }}{e}^{iqr}h(r,t),h(r,t)=\sum _q{e}^{-iqr}{\hat{h}}_q(t)$

Here $q=2\pi n/L,n=\dots ,-1,0,1,\dots$ and recall $\underset{0}{\overset{L}{\int }}dr{e}^{iqr}=L{\delta }_{q,0}$. using de Parseval theorem for the Fourier series

$w_2(t)=\sum _{q\ne 0}|{\hat{h}}_q(t){|}^2=\sum _{q\ne 0}{({\hat{h}}_q(t){\hat{h}}_{-q}(t))}^2$

In the last step we used that ${\hat{h}}_q^{*}(t)={\hat{h}}_{-q}(t)$.

Solution in the Fourier space

show that the EW equation writes

${\partial }_t{\hat{h}}_q(t)=-\nu q^2{\hat{h}}_q(t)+{\eta }_q(t),\text{with}⟨{\eta }_{q_1}(t^{\prime }){\eta }_{q_2}(t)⟩=\frac{2T}{L}{\delta }_{q_1,-q_2}\delta (t-t^{\prime })$

The solution of this first order linear equation writes

${\hat{h}}_q(t)={\hat{h}}_q(0)e^{-\nu q^{2}t}+\underset{0}{\overset{t}{\int }}ds{e}^{-\nu q^2(t-s)}{\eta }_q(s)$

• Assume that the interface is initially flat, namely ${\hat{h}}_q(0)=0$. Show that

$⟨{\hat{h}}_q(t){\hat{h}}_{-q}(t)⟩=\{\begin{array}{ll}\frac{T(1-e^{-2\nu q^{2}t})}{L\nu q^2},& q\ne 0,\\ \frac{2T}{L}t,& q=0.\end{array}$

• The mean width square grows at short times and saturates at long times:

$⟨w_2(t)⟩=\frac{T}{L\nu }\sum _{q\ne 0}\frac{1-e^{-2\nu q^{2}t}}{q^2}=\{\begin{array}{ll}T\sqrt{\frac{2t}{\pi \nu }},& t\ll L^2,\\ \frac{T}{\nu }\frac{L}{12},& t\gg L^2.\end{array}$