Subsections

• Main Idea
• Expansion
• Phase Diagram
• Entropy and Specific Heat
• Susceptibility

Landau Theory

Main Idea

From previous section we learned that:

1.

Free energy A is an analytical function of magnetization M

2.

At T>T_c it has one minimum, at T<T_c--two

Landau idea: let us use only these facts!

Expansion

Let us expand free energy A in powers of M.

First, a symmetric system: H=0. Then M=0 is either a minimum (T>T_c) or a maximum (T<T_c) of the free energy. Due to symmetry only even powers are present:

$$A = A_0 + V(a M^2 + d M^4 + \dots)$$

Assumption:

We neglect all powers higher than 4.

At large $\left\lvert M\right\rvert$ free energy should be large--so d>0. We will see below that at a>0 there is one phase, at a<0--two. So a should change sign at T=T_c. We will expand it:

$$a = \alpha(T-T_c)$$

With the same accuracy:

$$d= \mathit{const}$$

Result:  

$$A = A_0+V\left(\alpha(T-T_c)M^2+dM^4\right)$$ (9)

Phase Diagram

Differentiate (9):

$$2\alpha(T-T_c)M+4dM^3=0$$

Solutions:  

$$M =0\quad \text{or}\quad M^2 = -\frac{\alpha(T-T_c)}{2d}$$ (10)

At T>T_c only the first solution is realized--one phase. At T<T_c we have two solutions--two phases:

Equation of binodal is (10):  

$$2dM^2+\alpha(T-T_c)=0$$ (11)

This is parabola.

Equation of spinodal: $\partial^2 A/\partial M^2=0$ or  

$$6dM^2+\alpha(T-T_c)=0$$ (12)

This is another parabola, inside the binodal

Free energy:  

$$A = \begin{cases} A_0,& T\gt T_c\ A_0 -V\alpha^2(T-T_c)^2/4d, & T<T_c \end{cases}$$ (13)

Entropy and Specific Heat

Entropy is

$$S = -\frac{\partial A}{\partial T}$$

Differentiating (13):

$$S = \begin{cases} S_0, T\gt T_c\ S_0+V\alpha^2(T-T_c)/2d, & T<T_c \end{cases}$$

Specific heat:

$$C = \begin{cases} C_0,& T\gt T_c\ C_0+ V\alpha^2T/2d, & T<T_c \end{cases}$$

There is a jump in specific heat near T_c:

Susceptibility

Now we add a small external field H:  

$$A = A_0+V(\alpha(T-T_c)M^2+dM^4-HM)$$ (14)

and want to calculate

$$\chi = \frac{\partial M}{\partial H}$$

Minimizing (14):  

$$2\alpha(T-T_c)M+4dM^3=H$$ (15)

Two cases:

1.

If T>T_c, then M(H=0)=0. For small H we neglect M³, and

$$M = \frac{H}{2\alpha(T-T_c)}$$

Susceptibility

$$\chi = \frac{1}{2\alpha(T-T_c)}$$

This diverges at $T\to T_c$.

2.

If T<T_c, then $M(H=0)=\pm M_0$ with $M_0 = \pm\sqrt{-\alpha(T-T_c)/2d}$. Let

$$M = M_0+\delta M$$

Expanding (15), we obtain

$$\delta M = - \frac{H}{4\alpha(T-T_c)}$$

and

$$\chi = -\frac{1}{4\alpha(T-T_c)}$$

This also diverges

We obtained a $\lambda$-like curve: