5
$\begingroup$

In the Landau theory of phase transitions, one typically considers a "free energy" $F$ as a function of the temperature $T$ and the "order parameter" $\psi$: $F(T, \psi)$.

For the sake of clarity, let's consider the liquid-gas transition, in which the order parameter is usually taken to be the difference of densities between the ordered phase and the disordered phase. In this case, it looks that the order parameter $\psi$ is closely related to the the volume $V$, a global variable of state.

What is the correct interpretation of $\psi$ as a thermodynamic variable?

I consider several logical possibilities, but all interpretations look troublesome to me:

  1. $\psi$ is a variable of state replacing the volume. In this case I have a few related concerns:

    • Is the "free energy" $F$ the Helmholtz or the Gibbs free energy?

      • If $F$ is the Gibbs free energy, $F$ should depend on the temperature $T$ and the pressure $p$ (the conjugate variable of the volume), so it shouldn't depend on $\psi$.
      • If it is the Helmholtz free energy, it's ok that it depends on $\psi$, but the experiments are usually carried out at constant pressure, not constant volume.
    • Usually, state variables are something one can select at will [for instance, think of the phase diagram $(p, T)$]. However, the order parameter $\psi$ is not selected at will, but it is selected by the system as a result of the minimization process according to the Landau theory.

  2. $\psi$ is yet another variable of state in addition to volume and temperature.

    • Why the pressure is not usually taken into account as a variable of state, and only the temperature is considered?

    • What is the conjugate variable to $\psi$?

    • Can really $\psi$ and $V$ be taken as independent?

  3. $\psi$ is not a variable of state, is something else.

    • What else can $\psi$ be from the thermodynamic point of view?
    • Same question as above, why is pressure not usually taken into account in Landau theory?
$\endgroup$
3
$\begingroup$

First some preliminaries. There is, in general no requirement that a thermodynamic system should have exactly 2 degrees of freedom. The fundamental thermodynamic equation has the form $$ dU = TdS + \sum_{i=1}^{n-1} F_i\,dX_i $$ Where $F_i$ and $X_i$ are the generalised forces and generalised displacements. Each term in the equation introduces 2 variables and one relation $$ \frac{\partial U}{\partial X_i} = F_i $$ So the total number of degrees of freedom i the system is $n$. This means that we can talk about thermodynamic system with any number of degrees of freedom but we can see that the usual $p-V$ systems in introductory thermo courses only have 2.

So on to your question. The order parameter $\psi$ is a function of state. Exactly which function of state will depend on the system in question, and so it may be either your case 1 or case 2. For example, in a liquid gas transition the order parameter is usually taken to the the difference in density from the density at the phase transition, $\rho - \rho_0$, which would be an alternative to pressure or volume, but for a paramagnet-feromagnet transition it would normally be taken to be the net magnetisation, $m$, which is independent of volume.

As for which free energy to use the answer is simply "the appropriate one which is minimal at equilibrium". It is normally assumed that it is the Helmholtz free energy in textbooks. In any case, if dealing with a non-$p-V$ system or a system with more than 2 degrees of freedom I'm not sure how fixed the naming conventions are, and it is normally better to spell out which variables it depends on explicitly.

$\endgroup$
2
$\begingroup$

This answer may be incomplete, but I hope it will help you anyway.

In Landau's book Statistical Physics (chapter XIV, phase transitions of the second kind and critical phenomena), the author writes the Gibbs free energy $\Phi$ as a function of $T,P,\eta$, where $\eta$ is the order parameter:

$$\Phi=\Phi(T,P,\eta)$$

He then states that $\eta$ is different from $P$ and $T$ because the latter can have arbitrary values, while $\eta$ must satisfy

$$\frac{\partial \Phi}{\partial \eta} =0$$

So I think that we can say that $\eta$ is indeed a thermodynamic variable, but different from the others because the Gibbs free energy $\Phi$ must be minimized with respect to it at equilibrium, so it cannot assume an arbitrary value.

I've also noticed that many authors omit $P$ and just write $F(T,\psi)$ (which in Landau's notation would be $\Phi(T,\eta)$), but this may be simply because, as you correctly wrote, experiments are usually carried out at $P=const$, so maybe there is this "hidden" assumption behind such a notation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.