LBan-II: Difference between revisions

From Disordered Systems Wiki
Jump to navigation Jump to search
Line 67: Line 67:
From which you get <math> z=2 </math> in any dimension and a rough interface below <math> d=2 </math> with <math> \alpha =(2-d)/2 </math>.
From which you get <math> z=2 </math> in any dimension and a rough interface below <math> d=2 </math> with <math> \alpha =(2-d)/2 </math>.


== Explicit Solution ==
== Explicit Solution: the width of the interface ==


For simplicity, consider a 1-dimensional line of size L with periodic boundary conditions. It is useful to introduce the Fourier modes:
Consider a 1-dimensional line of size L with periodic boundary conditions. It is useful to introduce the Fourier modes:
<center> <math>  
<center> <math>  
\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)
\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)
Line 90: Line 90:
\end{cases}
\end{cases}
</math></center>
</math></center>
* Compute the width  <math> \langle h(r,t)^2\rangle = \sum_q \langle \hat h_q(t) \hat h_{-q}(t) \rangle  </math>. Comment about the roughness and the short times growth.
We consider the width square of the interface
<center> <math>
w_2(t) = [\int_0^L d r/L (h(r,t) - \int_0^L dr h(r,t)/L)]^2
</math></center>
using de Parseval theorem for the Fourier series
<center> <math>
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 
</math></center>
* the width square is a random variable and we can compute its mean:
<center> <math>
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 
</math></center>
and the displacement of a point of the interface that we tagg
  <math> \langle h(r,t)^2\rangle = \sum_q \langle \hat h_q(t) \hat h_{-q}(t) \rangle  </math>. Comment about the roughness and the short times growth.

Revision as of 08:47, 25 August 2025

Introduction: Interfaces and Directed Polymers

The physical properties of many materials are governed by manifolds embedded in them. Examples include: dislocations in crystals, domain walls in ferromagnets or vortex lines in superconductors. We fix the following notation: - D: spatial dimension of the embedding medium – d: internal dimension of the manifold – N: dimension of the displacement (or height) field

These satisfy the relation:

D=d+N

We focus on two important cases:

Directed Polymers (d = 1)

The configuration is described by a vector function: x(t), where t is the internal coordinate. The polymer lives in D=1+N dimensions.

Examples: vortex lines, DNA strands, fronts.

Although polymers may form loops, we restrict to directed polymers, i.e., configurations without overhangs or backward turns.

Interfaces (N = 1)

The interface is described by a scalar height field: h(r,t), where rd is the internal coordinate and t represents time.

Examples: domain walls and propagating fronts

Again we neglect overhangs or pinch-off: h(r,t) is single-valued

Note that using our notation the 1D front is both an interface and a directed polymer

Thermal Interfaces

  • The dynamics is overdamped, so that we can neglect the inertial term.

The Langevin equation of motion is

th(r,t)=μδEpotδh(r,t)+η(r,t)

The first term δEpot/δh(r,t) is the elastic force trying to smooth the interface, the mobility μ is the inverse of the viscosity. The second term is the Langevin noise. It is Guassian and defined by

η(r,t)=0,η(r,t)η(r,t)=2dDδd(rr)δ(tt)

The symbol indicates the average over the thermal noise and the diffusion constant is fixed by the Einstein relation D=μKBT. We set μ=KB=1

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

Epot=νddr1+(h)2const.+ν2ddr(h)2

Hence, we have the Edwards Wilkinson equation:

th(r,t)=ν2h(r,t)+η(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,bzt)inlawbαh(r,t)

Here z,α are the dynamic and the roughness exponent respectively. From dimensional analysis

bαzth(r,t)=bα22h(r,t)+bd/2z/2η(r,t)

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

Explicit Solution: the width of the interface

Consider a 1-dimensional line of size L with periodic boundary conditions. It is useful to introduce the Fourier modes:

h^q(t)=1L0Leiqrh(r,t),h(r,t)=qeiqrh^q(t)

Here q=2πn/L,n=,1,0,1, and recall 0Ldreiqr=Lδq,0.

  • Show that the EW equation writes
th^q(t)=νq2h^q(t)+ηq(t),withηq1(t)ηq2(t)=2TLδq1,q2δ(tt)

The solution of this first order linear equation writes

h^q(t)=h^q(0)eνq2t+0tdseνq2(ts)ηq(s)
  • Assume that the interface is initially flat, namely h^q(0)=0. Show that
h^q(t)h^q(t)={T(1e2νq2t)Lνq2,q0,2TLt,q=0.

We consider the width square of the interface

w2(t)=[0Ldr/L(h(r,t)0Ldrh(r,t)/L)]2

using de Parseval theorem for the Fourier series

w2(t)=q0|h^q(t)|2=q0(h^q(t)h^q(t))2
  • the width square is a random variable and we can compute its mean:
w2(t)=q0|h^q(t)|2=q0(h^q(t)h^q(t))2

and the displacement of a point of the interface that we tagg

 h(r,t)2=qh^q(t)h^q(t). Comment about the roughness and the short times growth.