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

$\langle \eta (r,t)\rangle =0,\;\langle \eta (r',t')\eta (r,t)\rangle =2dD\delta ^{d}(r-r')\delta (t-t')$

The symbol $\langle \ldots \rangle$ 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:

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

Hence, we have the Edwards Wilkinson equation:

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

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:

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

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

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

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

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

It is useful to introduce the Fourier modes:

${\hat {h}}_{q}(t)={\frac {1}{L}}\int _{0}^{L}e^{iqr}h(r,t),\quad h(r,t)=\sum _{q}e^{-iqr}{\hat {h}}_{q}(t)$

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

$w_{2}(t)=\sum _{q\neq 0}|{\hat {h}}_{q}(t)|^{2}=\sum _{q\neq 0}\left({\hat {h}}_{q}(t){\hat {h}}_{-q}(t)\right)^{2}$

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

$\partial _{t}{\hat {h}}_{q}(t)=-\nu q^{2}{\hat {h}}_{q}(t)+\eta _{q}(t),\quad {\text{with}}\;\langle \eta _{q_{1}}(t')\eta _{q_{2}}(t)\rangle ={\frac {2T}{L}}\delta _{q_{1},-q_{2}}\delta (t-t')$

The solution of this first order linear equation writes

${\hat {h}}_{q}(t)={\hat {h}}_{q}(0)e^{-\nu q^{2}t}+\int _{0}^{t}dse^{-\nu q^{2}(t-s)}\eta _{q}(s)$

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

$\langle {\hat {h}}_{q}(t){\hat {h}}_{-q}(t)\rangle ={\begin{cases}{\dfrac {T(1-e^{-2\nu q^{2}t})}{L\nu q^{2}}},&q\neq 0,\\[1.2em]{\frac {2T}{L}}t,&q=0.\end{cases}}$

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

$\langle w_{2}(t)\rangle ={\dfrac {T}{L\nu }}\sum _{q\neq 0}{\dfrac {1-e^{-2\nu q^{2}t}}{q^{2}}}={\begin{cases}T{\sqrt {\frac {2t}{\pi \nu }}},&t\ll L^{2},\\[1.2em]{\frac {T}{\nu }}{\frac {L}{12}},&t\gg L^{2}.\end{cases}}$