I have a doubt on systems for which the entropy reduces to configurational entropy (no momentum contribution): for example, hard rods (http://www.sklogwiki.org/SklogWiki/index.php/1-dimensional_hard_rods).

For a hard rod gas, the energy is identically $0$ and the entropy does not depend on the energy. The partition function is $$Z = \frac{(L - N \sigma)^N}{N!} $$ $L$ being the length of the available space, $\sigma$ the rod size and $N$ the number of rods. The free energy is then computed as $$ A = -kT \log Z$$.

Ok, but what does it mean for a hard rod gas to be in equilibrium with a heat bath?

Does not thermal equilibrium at a temperature $T$ imposes the condition $$\frac{1}{T} = \Big(\frac{\partial S}{\partial E} \Big) _{L,N}$$ ? For a hard rod gas the entropy $S$ does not depend on the energy $E$, so the inverse of the temperature must equal $0$.

In other words, if I provided heat to a hard rod gas, where would it go? There is nothing in the system that can "warm up", as the total energy equals identically $0$.

What am I missing?


1 Answer 1


You are missing the subtle and not explained difference between $Z$ and $Q$ in the sklogwiki.org page you are referring to.

Actually, $Z$ is the so-called configurational partition function, i.e., factor $\frac{1}{N!}$ a part, the integral over all the spatial coordinates of the Boltzmann's factor of the potential energy.

The full partition function is $Q=Z/\Lambda^N$, where $\Lambda=h/\sqrt{2 \pi m k_BT}$ is the de Broglie thermal wavelength.

Therefore, is not correct to say that the in the hard rod gas there is no contribution to thermodynamics from the momentum part of the Hamiltonian. It is there, but, since temperature appears only in the kinetic (ideal) part of the partition function and since the partition function can be factorized into a term which is the ideal gas partition function and another depending on $Z$, all thermodynamic functions can be written as the sum of 1D ideal gas contribution, plus an excess term depending on the hard interaction.

Moreover, this observation makes clear how it is possible to speak of temperature and thermal equilibrium.

  • $\begingroup$ Thank you very much. Now it all makes sense. This solved also another doubt of mine, how could dimensional quantities like $V/N$ be the argument of logarithms. $\endgroup$
    – Smerdjakov
    Mar 11, 2019 at 19:53

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