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


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


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



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 $b(T<T_c)$, the blue line is $b(T=T_c)$ and the green is $b(T>T_c)$. $K$ is given by "blue line $= 0$".

Phase transition:

enter image description here


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