# Calculating Parameters in Free Energy During Phase Transitions

I have recently started to study statistical mechanics with Pliscke's and Bergersen's book, but there is one example from the book regarding phase transitions I don't understand.

In the example we are given the Landau free energy $G=\frac{b(T)}{2}m^2+\frac{c(T)}{4}m^4+\frac{d(T)}{6}m^6$ and are supposed to calculate the magnetisation at the critical temperature, $m(T_c)$.

At the critical temperature there will be a phase transition, which according to my knowledge is characterised by all minima of the free energy being local rather than one global, i.e. $G(m_0)=G(0)=0$, where $m_0$ is the magnetisation at a free energy minimum. Is this true for both first order and higher order phase transitions? With a second order phase transition there will only be one minimum at zero at the critical temperature?

Calculating the free energy during the phase transition yields

$G(m_0)-G(0)=0=\frac{b(T)}{2}m_0^2+\frac{c(T)}{4}m_0^4+\frac{d(T)}{6}m_0^6$

which gives the solution $m_0=\pm\frac{1}{2}\sqrt{\frac{\pm\sqrt{3(3c^2-16bd)}-3c}{d}}$

Now, according to the book the constant $b$ should be

$b(T_c)=\frac{3c^2}{16d}$

but it doesn't explain why. What is the reason to make the inner root in $m_0$ zero, and thus remove two solutions during the phase transition? With this value of $b$ the solutions $m_0$ become

$m_0^2=-\frac{3c(T_c)}{4d}$

Lets assume $$b(T_c)$$ can have either sign, $$c(T_c)$$ is negative and $$d(T_c)$$ is positive. Then $$G(m)$$ can describe a first order phase transition from $$m=0$$ to $$m=K$$. You would like to find $$K$$ at the transition point.
The equilibrium order parameter will be the global minimum of $$G$$, that is we need $$G'(m) = 0$$. Since we know that $$G(0)=0$$ we have to solve $$G(m) = G'(m) = 0$$ ( and also $$G''(m)<0$$). Solving first $$G(m)=0$$ I get the same as your $$m_0$$ and solving afterwards $$G'(m) = 0$$ and equating the results I find $$b = \frac{3c^2}{16d}$$. Substituting $$b$$ into $$m_0$$ we find the desired result.
Perhaps it is a bit easier to see on a picture. The red line is $$b(T, the blue line is $$b(T=T_c)$$ and the green is $$b(T>T_c)$$. $$K$$ is given by "blue line $$= 0$$".