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There are some discrepancies on how to introduce entropy in classical kinetic theory. In what follows $f(r,p,t)$ is the usual one-particle distribution function of a monatomic gas, normalised to the number of particles $N$.

The negative of Boltzmann's $H$ function is often taken as the entropy:

$$ S = - k_B \int f \ln f \, d^3 r \, d^3 p, \quad N = \int f \, d^3 r \, d^3 p. $$

On the other hand, in Landau and Lifshitz it is $$ S = k_B \int f \ln \frac{e}{f} \, d^3r \, d^3 p, \quad N = \int f \, d^3 r \, d^3 p $$ This is actually derived and I remember to follow the derivation and find it completely satisfactory, though I forgot it by now.

Another book, "Beyond equilibrium thermodynamics" by Hans Öttinger, arrives via the projection-operator technique to

$$ S = - k_B \int f \ln \frac{f}{N} \, d^3r \, d^3 p, \quad N = \int f \, d^3 r \, d^3 p. $$

All of these expressions differ by a constant depending on $N$. I wonder what would be the most general/universal expression for entropy as a functional of a one-particle distribution function, that would be valid even outside the field of classical kinetic theory. For example, I suppose it would be nice to reproduce the Sackur–Tetrode formula in equilibrium: $$ S = k_B N \left(\frac{5}{2} + \ln \left[\frac{V}{N} \, \biggl(\frac{2\pi \, m \, k_B T}{h^2}\biggr)^{3/2} \right] \right). $$

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  • $\begingroup$ good question... $\endgroup$
    – user71065
    Mar 16, 2015 at 23:12

2 Answers 2

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I think I will go with the version from Landau and Lifshitz with quantisation of phase space:

$$ N = \int f \ln \frac{e}{f} \, \frac{dr^3 \,dp^3}{(2 \pi \hbar)^3}, \quad N = \int f \, \frac{dr^3 \,dp^3}{(2 \pi \hbar)^3}. $$

With the equilibrium function of $$ f_{\text{eq}} = (2 \pi \hbar)^3 \, \frac{N}{V} \Bigl(\frac{1}{2 \pi m k_B T}\Bigl)^{3/2}\exp\biggl[-\frac{p_x^2+p_y^2+p_z^2}{2mk_BT} \biggr] $$

one reproduces the Sackur–Tetrode formula (note $h = 2\pi\hbar$).

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The negative of Boltzmann's H function is often taken as the entropy. When written in a grand-canonical ensemble (N unknown and variable but chemical potential µ known) it is

$$S=−kB∫(f\ln f - f )\,\mathrm d^3r\,\mathrm d^3p,\quad N=∫f\,\mathrm d^3r\,\mathrm d^3p.$$

So, Boltzmann and Landau-Lifchitz expressions match.

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