Consider Einstein solid model ($N$ oscillators of same frequency $\omega$, where $n=\sum k_i $ with $k_i$ being the occupation number of single oscillators)

In microcanonical ensemble entropy is $$S=k \ln (\frac{(N+n-1)!}{n! (N-1)!}) \sim_{\mathrm{thermodynamic \,\, lim. \,\, (N\to \infty)\,\,\,\,\,}} k \ln (\frac{(N+n)^{N+n}}{n^n (N)^N})\tag{1}$$

On the other hand in canonical ensemble, since the partition function if $Z=(\frac{1}{2\sinh(\beta\omega\hbar/2)})^N$ one finds that $$S=\frac{kN}{2} [\beta\hbar\omega \coth(\beta\omega\hbar/2)-2 \ln(2\sinh(\beta\omega\hbar/2))]\tag{2}$$

Where $\beta=1/kT$.

I would like to show that two expressions are the same in the thermodynamic limit which was not taken for $(2)$.

Firstly I also got this relation between $\beta $ and $n$ $$\beta(n)=\frac{1}{\hbar \omega} \ln (\frac{N}{n}+1) \tag{3}$$

I tried to put $\beta(n)$ in $(2)$ and make an approximation for $N\to \infty $ in $(2)$ but I could not get to $(1)$ in any way.

What is the way to correctly approximate $(2)$ (with $\beta (n)$) to get to $(1)$?

  • $\begingroup$ Maybe I misunderstand what you're asking, but why would the partition functions for the micro canonical and canonical ensembles have to be equal? The former is not coupled to a heat bath, the latter is. $\endgroup$ Commented Nov 9, 2017 at 16:17

1 Answer 1


Consider first the microcanonical ensemble. The total energy of the $N$ oscillators is denoted $$E=\hbar\omega\left(n+{N\over 2}\right) \ \Leftrightarrow\ n={E\over\hbar\omega}-{N\over 2}$$ i.e. $n$ is the sum of the $N$ quantum numbers $n_i$ of each oscillators. $n$ can be interpreted as the number of energy quanta in the system. The number of stationary states with energy $E$ is, as you wrote, $$\Omega(E)={(N+n-1)!\over n!(N-1)!}$$ Therefore the microcanonical energy is $$S(E)=k_B\ln\Omega(E)=k_B\big[\ln(N+n-1)!-\ln n!-\ln (N-1)!\big]$$ Using Stirling approximation $\ln M!\sim M\ln M-M$, we get after cancellation of the three linear terms $$S(E)\simeq k_B\big[(N+n-1)\ln(N+n-1)-n\ln n-(N-1)\ln(N-1)\big)$$ The microcanonical temperature is defined as $${1\over T}={\partial S\over\partial E}={\partial S\over\partial n}{dn\over dE}={k_B\over\hbar\omega}\big[\ln (N+n-1)-\ln n\big]$$ Invert this relation to write $n$ as a function of $\beta=1/k_BT$ $$\beta={1\over\hbar\omega}\ln{N+n-1\over n}\ \Leftrightarrow\ n={N-1\over e^{\beta\hbar\omega}-1}$$ i.e. the Bose-Einstein distribution ! Noting that $$N-1+n=(N-1){e^{\beta\hbar\omega}\over e^{\beta\hbar\omega}-1}$$ the microcanonical entropy now reads $$\eqalign{ S(E)&\simeq k_B\big[(N+n-1)\ln(N+n-1)-n\ln n-(N-1)\ln(N-1)\big)\cr &=k_B\left[(N-1){e^{\beta\hbar\omega}\over e^{\beta\hbar\omega}-1} \ln{(N-1)e^{\beta\hbar\omega}\over e^{\beta\hbar\omega}-1}-{N-1\over e^{\beta\hbar\omega}-1}\ln{N-1\over e^{\beta\hbar\omega}-1}-(N-1)\ln(N-1)\right]\cr &=(N-1)k_B\Big[\Big(\underbrace{{e^{\beta\hbar\omega}\over e^{\beta\hbar\omega}-1}-{1\over e^{\beta\hbar\omega}-1}-1}_{=0}\Big)\ln(N-1)\Big.\cr &\quad\quad\quad\quad\quad\quad\Big. +{e^{\beta\hbar\omega}\over e^{\beta\hbar\omega}-1}\ln{e^{\beta\hbar\omega}\over e^{\beta\hbar\omega}-1}-{1\over e^{\beta\hbar\omega}-1}\ln{1\over e^{\beta\hbar\omega}-1}\Big]\cr &={(N-1)k_B\over e^{\beta\hbar\omega}-1}\Big[\beta\hbar\omega e^{\beta\hbar\omega}+\Big(e^{\beta\hbar\omega}-1\Big)\ln{1\over e^{\beta\hbar\omega}-1}\Big]\cr }$$ The logarithm can be written as $$\ln{1\over e^{\beta\hbar\omega}-1} =\ln{e^{-\beta\hbar\omega/2}\over e^{\beta\hbar\omega/2}-e^{\beta\hbar\omega/2}}=-{\beta\hbar\omega\over 2}-\ln 2\sinh{\beta\hbar\omega\over 2}$$ The microcanonical entropy is now $$\eqalign{ S&=(N-1)k_B\Big[\beta\hbar\omega {e^{\beta\hbar\omega}\over e^{\beta\hbar\omega}-1}-{\beta\hbar\omega\over 2}-\ln 2\sinh{\beta\hbar\omega\over 2}\Big]\cr &=(N-1)k_B\Big[{\beta\hbar\omega\over 2}{e^{\beta\hbar\omega}+1\over e^{\beta\hbar\omega}-1}-\ln 2\sinh{\beta\hbar\omega\over 2}\Big]\cr &=(N-1)k_B\Big[{\beta\hbar\omega\over 2}{e^{\beta\hbar\omega/2}+e^{-\beta\hbar\omega/2}\over e^{\beta\hbar\omega/2}-e^{-\beta\hbar\omega/2}}-\ln 2\sinh{\beta\hbar\omega\over 2}\Big]\cr &=(N-1)k_B\Big[{\beta\hbar\omega\over 2}{\cosh\beta\hbar\omega/2\over\sinh\beta\hbar/2}-\ln 2\sinh{\beta\hbar\omega\over 2}\Big]\cr &=(N-1)k_B\Big[{\beta\hbar\omega\over 2}{\rm cotanh}\ \!{\beta\hbar\omega\over 2}-\ln 2\sinh{\beta\hbar\omega\over 2}\Big]\cr }$$ which is precisely the expression of the entropy in the canonical ensemble with the assumption $N-1\simeq N$.


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