I already saw this Phys.SE post and it seems perfectly reasonable that the free energy describing a system must be a non-analytic function in order to display a phase transition.
An analytic function can be Taylor expanded around the critical point. In a phase transition the derivatives of the free energy are discontinous, so the function can't be Taylor expanded. Hence free energy must be non-analytic.
My question arises because in the Landau theory of phase transitions, he considers that every free energy must be indeed analytic.
So why Landau states such a thing? Also in the paramagnet-ferromagnet phase transition, the discontinuous function is the magnetization, which is not a derivative of $F(M,T)$, but it is said to be a second-order phase transition, why?
Best Answer
Landau free energy is just an approximation to the real free energy in the thermodynamic limit. For that reason, Landau free energy can be analytic, while the real one is not. Let me show how the approximation works.
As you may know, the Landau free energy is defined in the following way, assuming the Ising model:
$$Z\left(h,T\right)=\sum_{\left\{ s_{i}\right\} }\exp\left(-\beta H\left(\left\{ s_{i}\right\} \right)\right)=\sum_{m}\exp\left(-\beta F_{L}\left(m,h,T\right)\right)$$
where $Z$ is the partition function, $\{s_i\}$ stands for all possible spin configurations, and the sum in $m$ stands for every possible magnetization.
In Landau theory, the critical point the point $m^*$ such that $F_L(m^*,h,T)$ is minimum. Then, note that $ \exp\left(-\beta F_{L}\left(m^*,h,T\right)\right) $ is a maximum because of the minus sign. Then, we write
$$\log Z= \log \left [\sum_{m}\exp\left(-\beta F_{L}\left(m,h,T\right)\right) \right ] \geq \log \left [\sum_{m}\exp\left(-\beta F_{L}\left(m^*,h,T\right)\right) \right ],$$
In addition to this inequality, we can get an upper bound to this expression. The sum includes many different values for the magnetization, from -1 to +1; you can convince yourself (this is the hardest part of the demonstration) that if we replace the sum by a product of the $N$ at the minimum configuration, this quantity will be bigger than the original one:
$$\log Z \leq \log \left [N \exp\left(-\beta F_{L}\left(m^*,h,T\right)\right) \right ]$$
Now we almost have it. The quantity is bounded,
$$\log\left[N\exp\left(-\beta NF_{L}\left(m^{*},h,T\right)\right)\right]\geq\log Z\geq\log\left[\exp\left(-\beta F_{L}\left(m^{*},h,T\right)\right)\right]$$
$$\log N-\beta F_{L}\left(m^{*},h,T\right)\geq \log Z \geq-\beta F_{L}\left(m^{*},h,T\right).$$
Now we use the definition of the real, free energy, $F=-kT\log Z$. Multiplying by $1/\beta$ the expression above, we get that the real, non-analytic free energy, is bounded by the Landau free energy at the minimum:
$$k T\log N-F_{L}\left(m^{*},h,T\right)\geq -F\left(h,T\right)\geq-\beta F_{L}\left(m^{*},h,T\right).$$
Then, we change into intensive variables $F_L=Nf_L$ and divide by the number of spins $N$, to get:
$$\frac{\log N}{N}-f_{L}\left(m^{*},h,T\right)\geq f\left(h,T\right)\geq- f_{L}\left(m^{*},h,T\right)$$.
Notice that once we do the thermodynamic limit, the term $\log(N)/N \rightarrow 0$ and then we have that $f_{L}\left(m^{*},h,T\right) = f\left(m^{*},h,T\right)$, so in the thermodynamic limit, the analytic Landau free energy is the same as the real one. However, real system have real not an infinite number of spins, meaning that this is only an approximation. In experiments, there is an huge number of spins, and this is why Landau theory works very well, but if you work with little $N$ the difference between the two is noticeable.
About your second question, I think you are confusing things a bit: the transition is second order because the magnetization, which is the first derivative of the free energy with respect to the external magnetic field, is continuous at the critical point. However, susceptibility, which is the second derivative, is not continuous, meaning it is a second order phase transition.
Maybe what confuses you is that Landau theory gives you a piecewise defined function for magnetization. However, you precisely compute $T_c$ by the constraint that magnetization has to be continuous.
Edit: the non-analyticity of the free energy, if I am not mistaken, is respect to the magnetization -but not with respect to the external field $h$.