# 2nd order phase transition trouble deriving coefficient in fluctuations analysis

I can't get one of the coefficients in the equation for $T < T_c$ in the bottom, specifically the equation with the factor of two. any help appreciated.

Consider an ising type expansion of the free energy density in the order parameter $\psi$. (the mean values of $\psi$ corresponding to a uniform configuration.)

\begin{equation} f(T,\psi) = f_0 + G(\nabla\psi)^2+ A\psi^2 + B\psi^4 \end{equation} \begin{equation} \bar{\psi} = \begin{cases} \pm\sqrt{\frac{-A}{2B}} & \qquad T < T_c \\ 0 & \qquad T > T_c \end{cases} \end{equation}

these solutions come from minimizing the free energy density. Expanding the coefficients near the critical temperature, we have \begin{equation} A = at + ... \end{equation} \begin{equation} B = B_0 + ... \end{equation} \begin{equation} G= G_0 + ... \end{equation} Where $t=T-T_c$. Then evaluating the free energy density at $\psi = \bar\psi$ we get \begin{equation} \left. f(T,\psi)\right|_{\psi=\bar{\psi}} = \begin{cases} f_0 - \frac{a^2t^2}{4B_0} & \qquad T < T_c \\ f_0 & \qquad T>T_C \end{cases} \end{equation} To study fluctuations in the order paramater define, $\delta\psi = \psi -\bar{\psi}$.

\begin{equation} \delta\psi = \begin{cases} \psi & \text{in symmetrical phase} \\ \psi - \sqrt{\frac{-at}{2B_0}} & \text{in disordered phase} \end{cases} \end{equation}

We now calculate the change in free energy though I am not sure how exactly to calculate this, for some reason I am missing something simple I'm sure (!).

\begin{equation} \Delta F(T,\delta\psi) = \begin{cases} \int d^{d}r\,\left( G(\nabla\delta \psi)^2 + at\delta \psi^2\right) & \qquad T > T_c \\ \int d^{d}r\,\left( G(\nabla\delta \psi)^2 -2at\delta \psi^2\right) & \qquad T < T_c \end{cases} \end{equation}

I have spent some time trying to calculate the coefficient above for $\delta\psi^2$ for $T < T_c$. I am clearly not understanding something here, since when I calculate $f(\psi) - f(\bar{\psi})$ I do not get the above expression. In fact I am not sure exactly how to go about calculating this. Landau and Lifshitz gives the above coefficient in section 146 "Fluctuations of the Order Parameter". I am also confused because it seems like there are two values of the mean $\bar{\psi} = \pm (\frac{-at}{2B_0})^\frac{1}{2}$ however in the book I only see a reference to the positive square root. This topic is a bit obscure so I'm not sure anybody really is gonna be able to help, but I'd appreciate any insight.

## 1 Answer

To get the coefficient, you just have to expand the free energy up to $\delta\psi^2$. The linear term gives zero (because you're expanding around the minimum) whereas the quadratic term is $A+6B\bar\psi^2$, which gives the expected result. You can see that this result is independent of the minimum choose (independent of the sign).