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This notes explains how thermodynamic potentials $\Phi$ can be used in Boltzmann factor:

$$p = \frac{e^{-\beta\Phi}}{Z}$$

For example, The author claimed that in order to study macrostates instead of microstates, the appropriate potential in this case would be Helmholtz energy $\Phi=E-TS$.

But what I don't understand is what exactly a "macrostate" refers to? Macrostate with particular energy $E-TS$?

Also, if there is Helmholtz energy in the Boltzmann factor, then does this relationship between partition function and Helmholtz energy still hold true?

$$F=-kT\ln Z$$

If so, then how this $F$ can be interpreted? The sum of all possible values of $\Phi$?

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2 Answers 2

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A "macrostate" is a state characterized by macroscopic state variables such as $T$, $V$ and $N$. By contrast, a "microstate" is one that is characterized by a set microscopic variables such as the positions and momenta of each and every particle. Consider now a macrostate $(T,V,N)$, its temperature is fixed by thermal contact to a bath, but its instantaneous energy fluctuates. If the system is found at energy $E'$, its has a corresponding entropy $S'=k \ln\Omega(E',V,N)$ ($\Omega$ is the number of microstates with energy $E'$) and a free energy $\Phi' = E' - T S'$. The probability to find the system at this fluctuation is $$ p = \frac{e^{-\beta \Phi'}}{Z} $$ This probability is very sharply peaked about $E'= \bar E$, or about $\Phi = \bar E - T S(\bar E) \equiv F$, which is the free energy of the system at temperature $T$. This is to say that $\log p \to 0$ when $\Phi'\to F$, which gives $$ F = -\frac{\ln Z}{\beta} $$

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  • $\begingroup$ Thanks for the answer. 1. You make a mistake, $F = -\beta \ln Z$ should be $F = -\frac{1}{\beta} \ln Z$. 2. It seems like free energy calculated from partition function $F = -\frac{1}{\beta} \ln Z$ is the minimum possible free energy a system can have, right? $\endgroup$
    – Ray Siplao
    Commented Apr 1, 2023 at 22:39
  • $\begingroup$ @RaySiplao You're right of course, I corrected the typo. $\endgroup$
    – Themis
    Commented Apr 2, 2023 at 10:39
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I think the notes you refer to are unclear, and they may easily confuse even an expert reader.

The argument starts with an example valid for a system at equilibrium at a constant temperature, volume, and number of particles. By using the standard definition of microstate (the full specification of all the microscopic degrees of freedom) and using the usual meaning of the canonical ensemble probability distribution for the microstates, the author introduces the probability distribution for the energy values, which implies a factor $\Omega(E_A)$, corresponding to the number of microstates with the same energy $E_A$. At this point, $\Omega$ is rewritten in terms of the entropy of a microcanonical ensemble of fixed energy $E_A$.

The final step, which would require some justification, is to conclude that the probability of a macrostate at fixed energy $E_A$ is given by $e^{-\beta F}/Z$, where $F = E_A - ST$ is identified with the Helmholtz free energy. Such identification is possible only at the thermodynamic limit. The equivalence of different ensembles allows us to identify the entropy $S$ of the previous formula, the microcanonical entropy, with the entropy of the canonical ensemble.

Once this result has been obtained, the relation between $F$ and $Z$ is left untouched. The result is just a way to use the canonical probability of the microstates to derive the probability of finding a system at fixed $T$, $V$, and $N$, at a given value of the macroscopic energy.

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