I calculated the partition function of $N$ classical atoms of identical mass $m$ who all experience a mutual spring forces with identical spring constant $k$. The Hamilton is \begin{align} H = \dfrac{1}{2m} \sum_{j=1}^{N} |\boldsymbol{p}_j|^2 + \dfrac{k}{2} \sum_{j=1}^{N} \sum_{i=1}^{j-1} |\boldsymbol{q}_j-\boldsymbol{q}_i|^2 \end{align}

I calculated the partition function as \begin{align} Z &=& \int_{-\infty}^{\infty} \cdots \int_{-\infty}^{\infty} e^{-\beta H} (d^3p_1 \cdots d^3p_N) (d^3q_1 \cdots d^3q_N) \\&=& V T^{(3/2)N} \left( \left( 2m\pi k_B \right)^N \left( \dfrac{2 \pi}{k} \right)^{N-1} N^{-N} \right)^{3/2} \end{align}

I discovered the equation of state is \begin{align} p = k_B T \dfrac{\partial \ln(Z)}{\partial V} = \dfrac{k_B T}{V} \end{align} which states the pressure $p$, temperature $T$, and volume $V$ do not depend on the number of particles in the gas.

Why would a gas purely coupled by spring forces have an equation of state independent of the amount of gas particles? Also, how would such a gas behave in the real world?


Not sure the calculation was done correctly (on first glance). However, that is not important here, just think about what you are doing: You have N particles all of which are mutually interacting via a super-long-ranged potential (Coulomb-interaction $\sim r^{-1}$ would be considered long-ranged, you are using a parabolic $\sim r^2$ potential). So, what you are defining is by all means a solid (a really, really compact one). So let's have a look: https://en.wikipedia.org/wiki/Equation_of_state#Equations_of_state_for_solids No $N$-dependence there either.

The physical reason is: there is no macroscopic entropy, what you can do with your ball of springs are the 6 rigid-body degrees of freedom (i.e. center of mass translation and rotation).

  • $\begingroup$ Should I include the full derivation? It is possible I might have made a mistake in the math. Thanks. $\endgroup$ – linuxfreebird Jul 17 '15 at 18:01

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