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In Equilibrium Statistical Physics by Plischke and Bergersen the canonical partition function is defined (on page 37, eq. (2.33)) as $$Z_C = \int \frac{dE}{\delta E} \Omega(E) \exp\{-\beta E\}\tag{1},$$ where $\Omega(E)$ is the accessible phase space volume at energy $E$ for the system under consideration and $\delta E$ is a "tolerance" term which is later set to be proportional to $\sqrt{\langle E \rangle}$ to obtain agreement with classial thermodynamics.

But then in the end of chapter problems, if one looks at the official solutions manual for example, one instead uses $$Z_C = \frac{1}{h^{3N}} \int d^{3N}q \int d^{3N}p \exp\{-\beta H\}\tag{2},$$ with $H$ the Hamiltonian of the system. But nowhere does the text show the equivalence of these two integrals.

So how are (1) and (2) equivalent? How does one go from the definition (1) to (2)?

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(2) is the canonical partition function of a classical system of $N$ particles with Hamiltonian H. The Boltzmann weight is integrated over the $6N$-dimensional phase space $(q_i,p_i)$. This integral can be cast in the form of (1) by introducing $$\Omega(E)={1\over h_0^3}\int d^{3N}q\int d^{3N}p\ \!\delta(E-H(q,p))$$ which counts (up to the factor h_0^3) the volume of phase space which is accessible to the system with an energy $E$. It follows that $$Z_C=\int \Omega(E)e^{-\beta E}dE$$

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  • $\begingroup$ Thank you, I think that pretty much resolves it! $\endgroup$
    – ummg
    Sep 7 at 15:32
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    $\begingroup$ Maybe we can take care of the tolerance $\delta E$ also. The book takes $E<H(\mathbf q,\mathbf p)<E+\delta E$ as the accessible phase space volume, so maybe we could write $\Omega=(\delta E/h^{3N}) \int d^{3N}q \int d^{3N}p \, \delta(H(\mathbf q,\mathbf p)-E)$. Then, when inserting into (1), the $\delta E$ factors cancel nicely and one obtains (2). Would that be correct? $\endgroup$
    – ummg
    Sep 7 at 15:48
  • $\begingroup$ Hmm, I suppose that would be dimensionally incorrect, if nothing else. Is there some other way that we can also account for the tolerance? $\endgroup$
    – ummg
    Sep 7 at 15:57
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    $\begingroup$ Your expression is correct in the limit $\delta E\ll E$. $\Omega(E)$ is not the volume of phase space accessible at energy $E$ anymore but a density of states per unit of energy. I do not see any problem. At the end, the two $\delta E$ cancel and you get the correct partition function. $\endgroup$
    – Christophe
    Sep 7 at 19:21

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