L-2

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Goal: The physical properties of many materials are controlled by the interfaces embedded in it. This is the case of the dislocations in a crystal, the domain walls in a ferromagnet or the vortices in a supercoductors. In the next lecture we will discuss how impurities affect the behviour of these interfaces. Today we focus on thermal fluctuations and introduce two important equations for the interface dynamics: the Edward Wilkinson euqation and the Kardar Parisi Zhang equation.

An interface at Equilibrium: the Edward Wilkinson equation

Consider domain wall h(r,t) fluctuating at equilibrium at the temparature T. Here t is time, r defines the d-dimensional coordinate of the interface and h is the scalar height field. Hence, the domain wall separating two phases in a film has d=1,r∈ℛ, in a solid instead d=2,r∈ℛ𝟐.

Two assumptions are done:

  • Overhangs, pinch-off are neglected, so that h(r,t) is a scalar univalued function.
  • The dynamics is overdamped, so that we can neglect the inertial term.

Derivation

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 inversily proportional to the viscosity. The second term is the Langevin Gaussian noise defined by the correlations

⟨η(r,t)⟩=0,⟨η(r′,t′)η(r,t)⟩=2Dδd(r−r′)δ(t−t′)

The symbol ⟨…⟩ indicates the average over the thermal noise. The diffusion constant is fixed by the Eistein relation (fluctuation-dissipation theorem):

D=μKBT

The potential energy of surface tension can be expanded at the lowest order in the gradient:

Epot=σ∫ddr1+(∇h)2∼const.+σ2∫ddr(∇h)2

Setting μ=1,σ=1/2 we have the Edward Wilkinson equation:

∂th(r,t)=∇2h(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 conndition os scale invariance:

h(br,bzt)∼inlawbαh(r,t)

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

bα−z∂th(r,t)=bα−2∇2h(r,t)+b−d/2−z/2η(r,t)

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

Exercise L2-A: Solve Edward-Wilkinson