Why does the free energy depend on the configuration, in the Peierls argument?

Question

Kendrick Mendez · Accepted Answer

As defined in standard thermodynamics, the Helmholtz free energy F is a function of the macroscopic variables T, V, and N. It does not depend on the configuration.
However, there are related "free energies" that do depend on some features of the configuration. Note that F is defined in statistical mechanics as

e^{- β F} = \sum_{{s_{i}}} e^{- β H [s_{i}]}

where we sum over all configurations. But then too little information remains in F to do the Peierls argument, which involves counting domain walls in particular configurations.
Instead, we can perform the sum "halfway". We let

e^{- β F} = \sum_{n} e^{- β F^{'} (n)}, e^{- β F^{'} (n)} = \sum_{n domain walls} e^{- β H [s_{i}]} .

The notation is a bit clunky, but the point is that while the thermodynamic free energy F sums over all configurations, the quantity

F^{'} (n)

sums over only configurations with n domain walls. Your quantities

F_{1}

and

F_{2}

are

F^{'} (0)

and

F^{'} (1)

The probability of having n domain walls is proportional to

e^{- β F^{'} (n)}

, so your calculation shows that having one domain wall is much more likely than having none. That's the Peierls argument.

Why does the free energy depend on the configuration, in the Peierls argument?

Open question

Answer & Explanation

New Questions in Thermal Physics