Subsections

• Euler-Lagrange Equation
• Solution of Euler-Lagrange Equation
• Surface Tension
• Critical Behavior

Interface Layer in Landau Theory

Euler-Lagrange Equation

We need to minimize the functional

$$\begin{split} &A = A_0+ \\ & W\int\left[aM^2+dM^4-HM+g\left(\frac{\partial M}{\partial x}\right)^2\right]\,dx \end{split}$$

with

$$M(-\infty) = M_0, \quad M(\infty) = -M_0, \quad M_0 = \sqrt{-a/2d},$$

This is a Lagrange problem with

$$\mathcal{L} = aM^2+dM^4-HM+g{M'}^2$$

On the coexistence line H=0 (Why?). We have:

$$\begin{aligned} \frac{\partial\mathcal{L}}{\partial M} &= 2... ...\frac{\partial\mathcal{L}}{\partial M'} & = 2gM' \end{aligned}$$

and Euler-Lagrange equation becomes

 

gM''=aM+2dM³

Solution of Euler-Lagrange Equation

Multiply (6) by M'. Then

$$gM''M' = \frac{g}{2}\frac{d}{dx}\left(\frac{dM}{dx}\right)^2$$

and

$$(aM+2dM^3)M' = \frac12\frac{d}{dx}(aM^2+dM^4)$$

Therefore

$$\frac{g}{2}\frac{d}{dx}\left(\frac{dM}{dx}\right)^2 = \frac12\frac{d}{dx}(aM^2+dM^4)$$

and

$$g \left(\frac{dM}{dx}\right)^2 = aM^2+dM^4+C$$

To find C note that at $x\to-\infty$ we have

$$M'=0,\quad M^2=-a/2d, \quad aM^2+dM^4=-a^2/4d$$

We obtain:

$$C = \frac{a^2}{4d},\quad aM^2+dM^4+C = d\left(M^2-M_0^2\right)^2$$

This gives

$$g \left(\frac{dM}{dx}\right)^2 = \left(M^2-M_0^2\right)^2d$$

or  

$$\frac{dM}{dx} = -\sqrt{\frac{d}{g}}\left(M^2-M_0^2\right)$$ (7)

and

$$\sqrt{\frac{g}{d}}\int\frac{dM}{M^2-M_0^2} = -x$$

Solution:  

$$x = \sqrt{-\frac{g}{2a}}\ln\left(\frac{M_0-M}{M+M_0}\right)$$ (8)

This gives x(M). Inverting this function, we obtain density profile M(x)

Width of the interface  

$$L\sim\sqrt{-\frac{g}{2a}}\sim\xi$$ (9)

The length of the interface is proportional to the correlation length!

Surface Tension

In the bulk the energy per unit volume is

$$f_0=\frac{A_0}{V}+aM_0^2+dM_0^4 = \frac{A_0}{V} - \frac{a^2}{4d}$$

In the interface layer we have free energy per unit volume

$$f = \frac{A_0}{V} +aM^2+dM^4 + g(M')^2$$

Additional free energy per unit volume

$$\Delta f = f-f_0=2d(M-M_0)^2$$

Surface tension is surface energy per unit area, i.e.

$$\gamma = \int_{-\infty}^{\infty} \Delta f\,dx$$

Change of variables:

$$\gamma = \int_{M_0}^{-M_0} \Delta f\frac{dx}{dM}\,dM$$

or, substituting $\Delta f$ and M'  

$$\gamma=2\sqrt{gd}\int_{-M_0}^{M_0}(M_0^2-M^2)\,dM= -\frac{4a}{3}\sqrt{\frac{g}{d}}$$ (10)

Critical Behavior

What happens at $T\to T_c$? Coefficient a tends to zero as $a=\alpha(T-T_c)$.

From (9) and (10) we obtain:

$$L\sim (T_c-T)^{-1/2},\quad \gamma\sim(T-T_c)$$

As we are closer to the critical point, the interface layer becomes thicker, and the surface tension drops. Since $L\sim\xi$, correlation length diverges, and $\gamma\to0$ means that the difference between the phases disappears. The penalty for forming interface becomes lower, and fluctuations grow.