How and why is the Landau free energy any different from thermodynamic free energies?
It is written on page 140 of Nigel Goldenfeld's book Lectures on Phase Transitions and The Renormalization Group that
The Landau free energy has dimensions of energy, and is related to, but, as we will see, is not identical with the Gibbs free energy of the system.
The explanation in section 5.6 is quite elaborate and too complicated. Please help me with a simple understanding of why the Landau free energy is not the Helmholtz free energy or the Gibbs free energy, and how it is related to the thermodynamic free energies.
Answer
The Landau free energy, also called the Landau-Ginzburg Hamiltonian, is treated in an adhoc and rather confusing manner in a lot of textbooks. But in the modern view, it has a simple interpretation as an effective Hamiltonian attained by integrating out degrees of freedom.
Suppose we have a spin system, such as an Ising magnet. We can describe the state of the system by a magnetization field $\phi(x)$, noting that this field doesn't make sense if we examine length scales smaller than the lattice spacing $a$. We can write a sum over all spin states by an integral over field configurations, as long as the integral is cut off at the distance scale $a$.
If the Hamiltonian is $H[\phi]$, then the thermodynamic free energy $F$ obeys $$Z = e^{-\beta F} = \int_{\Delta x\, >\, a} \mathcal{D}\phi \, e^{-\beta H[\phi]}$$ which is just a rewording of the standard identity $F = - k_B T \log Z$. In the Wilsonian view, the thermodynamic free energy is acquired by integrating out all microscopic degrees of freedom. The result only depends on macroscopic quantities like temperature, pressure, and the external field. This is useful because the entire point of thermodynamics is to ignore the microscopic details and focus on macroscopic quantities that are easy to measure. For example, using just the function $F$, we can determine the equilibrium magnetization by minimizing it.
Now, the Landau free energy $H_L$ satisfies $$Z = \int_{\Delta x \, > \, b} \mathcal{D}\phi \, e^{-\beta H_L[\phi]}$$ where $b$ is a mesoscopic distance scale, larger than $a$ but still much smaller than a macroscopic length. In the Wilsonian view, the Landau free energy is the effective Hamiltonian acquired by integrating out degrees of freedom on lengthscales $a < x < b$. The point of the Landau free energy is that it represents a compromise between the completely microscopic $H$, which has too much detail to be useful, and the completely macroscopic $F$, which tells us nothing about, e.g. position dependence. Like $H$, $H_L$ is a functional, but it's a functional of "fewer variables".
The above explains why $H_L$ can be called a Hamiltonian, but why is it also called a free energy? Usually, the starting point for applying Landau theory is the saddle point approximation, which states that typical equilibrium field configurations minimize $H_L$. Since we're minimizing $H_L$, we're treating it like we would a free energy, which is why it's sometimes called the Landau free energy.
But why is this valid? You definitely can't get the right answer to any thermodynamic question by minimizing $H$, because it doesn't take into account thermal effects; you instead have to minimize $F$. Minimizing $H_L$ gives the right answer precisely when thermal effects are negligible on distance scales greater than $b$. This is true when $b$ is much greater than the system's correlation length $\xi$, which is why Landau theory does such a good job, and usually not true at a critical point where $\xi$ diverges, which is why Landau theory fails to describe continuous phase transitions.
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