## Landau theory of a first order phase transition

The Landau theory of phase transitions is based on the idea that the free energy can be expanded as a power series in the order parameter $m$. For a second order phase transition, the order parameter grows continuously from zero at the phase transition so the first few terms of the power series will dominate. If the free energy is expanded to sixth order in the order parameter, the system will undergo a first order phase transition if $\alpha_0 > 0$, $\beta < 0$, and $\gamma > 0$.

$$$\label{eq:f} f\left(T\right) = f_0\left(T\right)+\alpha_0\left(T-T_c\right)m^2+\frac{1}{2}\beta m^4+ \frac{1}{3}\gamma m^6 \hspace{1cm} \alpha_0 > 0, \hspace{1cm} \beta < 0, \hspace{1cm} \gamma > 0.$$$

Here $f_0\left(T\right)$ describes the temperature dependence of the high temperature phase near the phase transition. The form below can be used to plot the free energy for different temperatures. The parameters that are input into the form are also used to plot the temperature dependence of the order parameter, the free energy, the entropy, and the specific heat. The temperature dependence of the high temperture phase $f_0(T)$ can be input in the form. For a metal at low temperature, the electron contribution to the free energy will dominate and it will be $-\gamma T^2/2$ where $\gamma$ describes the linear specific heat of a metal at low temperatures, $c_v=\gamma T$. For an insulator at low temperature, the phonon contribution will dominate and it will be proportional to $-T^4$.

 $f-f_0$ $m$

 $\alpha_0$ = $\beta$ = $\gamma$ = $T$ = $T_c$ = $f_0\left( T\right)$ =

Ferroelectricity
Determining α0, β and γ

Order parameter
The order parameter that minimizes the free energy will be the one that is observed.

$$$\label{eq:dfdm} \frac{df}{dm} = 0=2\alpha_0\left(T-T_c\right)m+2\beta m^3+ 2\gamma m^5$$$

Clearly one solution is $m=0$. Canceling one factor of $m$ from the equation yields,

$$$0=2\alpha_0\left(T-T_c\right)+2\beta m^2+ 2\gamma m^4.$$$

This can be solved for $m^2$,

$$$\label{eq:m2} m^2=0,\frac{-\beta\pm\sqrt{\beta^2-4\alpha_0(T-T_c)\gamma}}{2\gamma}.$$$

Of the five solutions to \eqref{eq:dfdm} for $m$, at most 3 are real and stable. These solutions are,

$$$\begin{matrix} \label{eq:m} m=\pm\sqrt{\frac{-\beta+\sqrt{\beta^2-4\alpha_0(T-T_c)\gamma}}{2\gamma}} & \hspace{1cm}\text{for}\hspace{0.5cm}T<T_c,\hspace{1cm}\\ m=0,\pm\sqrt{\frac{-\beta+\sqrt{\beta^2-4\alpha_0(T-T_c)\gamma}}{2\gamma}} & \hspace{1cm}\text{for}\hspace{0.5cm}T_c<T<T_1,\\ m=0 & \hspace{1cm}\text{for}\hspace{0.5cm}T_1<T,\hspace{1cm} \end{matrix}$$$

where the temperature $T_1$ is where the two non-zero solutions become unstable,

$$$T_1=\frac{\beta^2}{4 \alpha_0\gamma} + T_c.$$$
 $m$ $T$

The jump in the order parameter at $T_c$ is,

\begin{align} \Delta m &=\sqrt \frac{-\beta}{\gamma}. \end{align}

When evaluating the jump in the order parameter, note that $\beta$ is negative so $-\beta+\sqrt{\beta^2} = -2\beta$. The jump in the order parameter at $T_1$ is,

\begin{align} \Delta m =\sqrt \frac{-\beta}{2 \gamma}. \end{align}

Cooling from a high temperature ($T>T_1$) the system will follow the red curve until $T=T_c$ where it becomes unstable. There is a symmetry breaking at this point and the system has to jump to one of the two blue curves. If the system is then heated back up, it remains on the blue curve until $T=T_1$ where it jumps back to the red curve.

The jumps in the order parameter puts the power series expansion in question. For a second order phase transition, we could argue that since the order parameter goes continuously to zero, there must be a range close to the phase transition where only the first few terms of the power series are relevant. With a jump in the order parameter, we have to consider the possibility that higher order terms should be included in the power series expansion. We continue with the analysis here assuming that the free energy can be described by a power series including only even terms up to $m^6$.

Free energy
The temperature dependence of the free energy can be found by substituting the expression for the order parameter \eqref{eq:m} back into the Eq. \eqref{eq:f} for the free energy. After some algebra, the two branches of the solution can be written as,

$$$\begin{matrix} f(T)=f_0(T) & \hspace{1cm}\text{for}\hspace{0.5cm}T>T_c,\\ f(T)=f_0(T) - \frac{\left|\left( \beta^2-4\alpha_0(T-T_c)\gamma\right)^{3/2}\right|}{12\gamma^2} -\frac{\alpha_0 \beta (T-T_c)}{2\gamma} + \frac{\beta^3}{12 \gamma^2} & \hspace{1cm}\text{for}\hspace{0.5cm}T<T_1. \end{matrix}$$$

There is a subtle issue with the sign here. The factor $\beta$ is negative and $\beta^2$ is positive. The factor $(\beta^2)^{3/2}=\beta^3$ which could be interpreted as positive or negative depending on which sign is taken for the square root. To avoid the ambiguity, the absolute value was taken. At $T=T_c$,

$$f(T)-f_0(T) = - \frac{\left|\left( \beta^2\right)^{3/2}\right|}{12\gamma^2} + \frac{\beta^3}{12 \gamma^2} = -\frac{|\beta^3|}{6 \gamma^2}.$$

The high temperature (red) solution is valid for temperatures above $T_c$ and the low temperature (blue) solution is valid for temperatures below $T_1$.
 $f$ $T$

Apart from $T_c=$  K and $T_1=$ K, there is another important temperature in this problem, namely where the free energies of the two phases are equal. This temperature is defined to be $T^*=$ K. At $T^*$,

$$- \frac{\left|\left( \beta^2-4\alpha_0(T^*-T_c)\gamma\right)^{3/2}\right|}{12\gamma^2} -\frac{\alpha_0 \beta (T^*-T_c)}{2\gamma} + \frac{\beta^3}{12 \gamma^2} =0.$$

The first term is negative. The second term is positive because of the minus sign in front and the factor of $\beta$. The third term is negative because of the factor of $\beta^3$. All of the terms are divided by the negative factor $\frac{\beta^3}{12 \gamma^2}$ and they all change sign.

$$\left|\left( 1-\frac{4\alpha_0(T^*-T_c)\gamma}{\beta^2}\right)^{3/2}\right| -\frac{6\alpha_0 (T^*-T_c)\gamma}{\beta^2} + 1 =0.$$

Making the substitution, $r = \frac{\alpha_0 (T^*-T_c)\gamma}{\beta^2}$, this can be written as,

$$\left|\left( 1-4r\right)^{3/2}\right| -6r + 1 =0.$$

The solution to this equation is $r=\frac{3}{16}$ thus,

$$T^*-T_c = \frac{3\beta^2}{16\alpha_0\gamma}.$$

From this expression, it can be shown that $3(T_1 - T^*) = T^* -T_c$. This provides another expression for the free energy that is useful to fit $\alpha$, $\beta$, and $\gamma$ if $f(T) - f_0(T)$ is known,

$$$f(T)-f_0(T) = - \frac{\left|\left( \frac{\beta^2}{4}-4\alpha_0(T-T^*)\gamma\right)^{3/2}\right|}{12\gamma^2} -\frac{\alpha_0 \beta (T-T^*)}{2\gamma} - \frac{\beta^3}{96 \gamma^2} \hspace{1cm}\text{for}\hspace{0.5cm}T < T_1.$$$

Entropy
The entropy is the derivative of the free energy with repect to temperature,

$\begin{equation*} s(T) = -\frac{df}{dT}. \end{equation*}$

The two branches of the solution are,

$$$\begin{matrix} s(T)=s_0(T) & \hspace{1cm}\text{for}\hspace{0.5cm}T>T_c,\\ s(T)= s_0(T) - \frac{\alpha_0}{2\gamma} \left (-\beta+ \sqrt{\beta^2-4\alpha_0(T-T_c)\gamma} \right ) & \hspace{1cm}\text{for}\hspace{0.5cm}T<T_1. \end{matrix}$$$
 $s$ $T$

Latent heat
A first order transition has a finite latent heat, $L=T\Delta s$. The latent heat at $T_c$ is,

$$$L = \frac{\alpha_0\beta T_c}{\gamma}.$$$

At $T_1$, the latent heat is,

$$$L = \frac{\alpha_0\beta T_1}{2\gamma}.$$$

Specific heat
The specific heat can be determined from the entropy using the relation,

$$$c_v(T) = T\frac{ds}{dT}.$$$

The two branches of the solution are,

$$$\begin{matrix} c_v(T)=c_v0(T) & \hspace{1cm}\text{for}\hspace{0.5cm}T>T_c,\\ c_v(T)= c_{v0} (T) + \frac{\alpha_0^2 T }{\sqrt{\beta^2-4\alpha_0(T-T_c)\gamma}} & \hspace{1cm}\text{for}\hspace{0.5cm}T<T_1. \end{matrix}$$$
 $c_v$ $T$

The jump in the specific heat at $T_c$ is,

$$$\Delta c_v = \frac{\alpha_0^2T_c}{\beta}.$$$

The specific heat diverges at $T_1$ so there is no clearly defined jump at that temperature.

Susceptibility
The susceptibility is the response of the system to a field,

$$$\chi = \frac{dm}{dB}.$$$

In order to calculate the susceptibility, the field must be included in the expression for the free energy,

$$$f\left(T\right) = f_0\left(T\right)+\alpha_0\left(T-T_c\right)m^2+\frac{1}{2}\beta m^4+ \frac{1}{3}\gamma m^6 - mB .$$$

Take the derivative with respect to $m$ and set it to zero,

$$$\frac{df}{dm} = 2\alpha_0\left(T-T_c\right)m+2\beta m^3+ 2\gamma m^5 - B = 0.$$$

At the minima of the free energy we get

$$$B = 2\alpha_0\left(T-T_c\right)m+2\beta m^3+ 2\gamma m^5.$$$

Taking the derivative again, yields

$$$dB = \left ( 2\alpha_0\left(T-T_c\right)+6\beta m^2+ 10\gamma m^4 \right ) dm.$$$

At the critical temperatures $m = 0, \sqrt{ \frac{-\beta}{2 \gamma} }$. So we get the following result for the susceptibility in the vicinity of the phase transition,

\begin{align} \mathrm{at}~ T_c \hspace{2cm} \chi & = \frac{dm}{dB} \Bigg \vert_{m=0} & = \frac{1}{2 \alpha_0 \left ( T-T_c \right )}, \\ \mathrm{at}~ T_1 \hspace{2cm} \chi & = \frac{dm}{dB} \Bigg \vert_{m=\sqrt{ \frac{- \beta}{2 \gamma} }} & = \frac{1}{2 \alpha_0 \left ( T_1-T \right )}. \\ \end{align}

The susceptibility obeys a Curie - Weiss law, as it does for second order transitions.

 $\chi$ $T$