How does one measure the chemical potential of a substance/ thermodynamic system? I am asking this question for two reasons:

(1) The measure on phase space: Textbooks typically state that one should take the integral measure on phase space to be $\frac{dpdq}{2\pi\hbar}$. Typically, one "justifies" the presence of the $\hbar$ by arguing that the phase space measure should be dimensionless or that quantum mechanics tells us that phase space is quantized in units of $\hbar$. It seems to me that the presence of the $\hbar$ is unnecessary and has no physical affect other than shifting the chemical potential: \begin{align} d^{3N}pd^{3N}q\rightarrow \frac{d^{3N}pd^{3N}q}{(\hbar)^{3N}}&\implies S\rightarrow S-3Nk_B\ln(\hbar)\\ &\implies\mu\rightarrow\mu+3k_B\ln(\hbar)\tag{1} \end{align} The last implication following from the fact the chemical is $\mu=\frac{\partial S}{\partial N}\vert_{E,V}$. i.e. adding the $\hbar$ term to the measure only affects the $N$ dependence of the entropy and therefore only affects quantities that are related to the entropy by a derivative with respect to $N$, which is only the chemical potential. By Eq.($1$) one can see that the only affect of the $\hbar$ is a constant shift of the chemical potential, hence I am wondering: does a constant shift in the chemical potential have any physical significance? Is the choice of the constant in the phase space measure arbitrary and unnecessary?

(2) Gibbs "paradox": In the famous Gibbs paradox for the mixing of two ideal gases, one figures out that if you want the entropy of an ideal gas to be an extensive quantity then one should add a $\frac{1}{N!}$ to the phase space measure. Again, by the exact same reasoning as above, this will only affect the entropy and the chemical potential: \begin{align} d^{3N}pd^{3N}q\rightarrow \frac{d^{3N}pd^{3N}q}{N!}&\implies S\rightarrow S-k_B\ln(N!)\\ &\implies \mu\rightarrow \mu+k_Bln(N) \end{align} Here we do not have a constant shift of the chemical potential as before, hence I am wondering: how would one experimentally measure whether the $\frac{1}{N!}$ should be there or not by measuring the chemical potential? As a corollary, is the extensivity of the entropy of a gas an empirical question? I know that the justification for the $\frac{1}{N!}$ comes from the quantum indistinguishably of identical particles, but I would also like to understand how you could also use a measurement of the chemical potential to justify its presence in another way.


1 Answer 1


To address your questions:

  1. By its definition the volume of phase space is dimensionless – it is the number of ways to partition extensive properties into discrete microstates. A factor with the dimensions of $\hbar$ is necessary for dimensional homogeneity. A constant shift in $\mu$ has no physical significance. All thermodynamic predictions are based on differences ($\Delta \mu$, $\Delta S$, etc). In fact, for tabulation purposes we often fix the zero of the chemical potential at some arbitrary state and calculate all other values relative to this reference.

  2. We cannot experimentally measure $N!$, just try to calculate the factorial of $N\sim 10^{23}$: it cannot be computed, and it cannot be measured! Instead we measure its logarithm, which is both computable and measurable ($\log (10^{23}!) \approx 53$). The standard experiment is the work needed to separate an ideal mixture into its pure components at constant temperature and constant pressure. According to Gibbs this work is $$ \frac{W}{n_1+n_2} = -RT \log\frac{(n_1+n_2)!}{n_1!n_2!} = RT \log\left(n_1\log\frac{n_1}{n_1+n_2}+n_2\log\frac{n_2}{n_2+n_2}\right) $$ and can be measured experimentally.

  • $\begingroup$ Note that $S$ does have intrinsic significance due to the 3rd law, but it is extremely unique among the thermodynamic potentials for that reason. $\endgroup$ Commented May 12 at 13:56
  • $\begingroup$ Yes, absolute zero $T$ sets the absolute zero of $S$. But it is in how $S$ increases with $T$ where the real problem lies. In principle all degrees of freedom contribute to $S$. In practice it is only those we know about today. It is hard to escape the conclusion that entropy characterizes the observer’s knowledge of the system and that it is not an intrinsic property of the system. After all, there is no observation without an observer. $\endgroup$
    – Themis
    Commented May 14 at 19:06

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