5.3: Entropy and Counting States
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Suppose we are to partition N particles among J possible distinct single particle states. How many ways Ω are there of accomplishing this task? The answer depends on the statistics of the particles. If the particles are fermions, the answer is easy: \boldsymbol{\ROmega\ns_\ssr{FD}={J\choose N}}. For bosons, the number of possible partitions can be evaluated via the following argument. Imagine that we line up all the N particles in a row, and we place J−1 barriers among the particles, as shown below in Figure [BEcount]. The number of partitions is then the total number of ways of placing the N particles among these N+J−1 objects (particles plus barriers), hence we have \boldsymbol{\ROmega\ns_\ssr{BE}={N+J-1\choose N}}. For Maxwell-Boltzmann statistics, we take \boldsymbol{\ROmega\ns_\ssr{MB}=J^N/N!} Note that \boldsymbol{\ROmega\ns_\ssr{MB}} is not necessarily an integer, so Maxwell-Boltzmann statistics does not represent any actual state counting. Rather, it manifests itself as a common limit of the Bose and Fermi distributions, as we have seen and shall see again shortly.
![[BEcount] Partitioning N bosons into J possible states (N=14 and J=5 shown). The N black dots represent bosons, while the J-1 white dots represent markers separating the different single particle populations. Here n_1=3, n_2=1, n_3=4, n_4=2, and n_5=4.](https://phys.libretexts.org/@api/deki/files/14921/imageedit_4_2123889975.png?revision=1)
The entropy in each case is simply S=kBlnΩ. We assume N≫1 and J≫1, with n≡N/J finite. Then using Stirling’s approximation, ln(K!)=KlnK−K+O(lnK), we have \boldsymbol{\begin{split} S\ns_\ssr{MB}&=-J\kB \, n\ln n \\ S\ns_\ssr{BE}&=-J\kB\big[ n\ln n - (1+n)\ln (1+n)\big] \bvph \\ S\ns_\ssr{FD}&=-J\kB\big[ n\ln n + (1-n)\ln (1-n)\big]\ . \end{split}} In the Maxwell-Boltzmann limit, n≪1, and all three expressions agree. Note thatR \boldsymbol{\begin{split} \pabc{S\ns_\ssr{MB}}{N}{J} &= -\kB \, \big( 1 + \ln n\big) \\ \pabc{S\ns_\ssr{BE}}{N}{J} &= \kB\ln\!\big(n^{-1}+1\big) \bvph \\ \pabc{S\ns_\ssr{FD}}{N}{J} &= \kB\ln\!\big(n^{-1}-1\big)\ . \end{split}}
Now let’s imagine grouping the single particle spectrum into intervals of J consecutive energy states. If J is finite and the spectrum is continuous and we are in the thermodynamic limit, then these states will all be degenerate. Therefore, using α as a label for the energies, we have that the grand potential Ω=E−TS−μN is given in each case by \boldsymbol{\begin{split} \Omega\ns_\ssr{MB} &= J\sum_\alpha \Big[ (\ve\ns_\alpha-\mu)\,n\ns_\alpha+\kT\,n\ns_\alpha\ln n\ns_\alpha\Big] \\ \Omega\ns_\ssr{BE} &= J\sum_\alpha \Big[ (\ve\ns_\alpha-\mu)\,n\ns_\alpha+\kT\,n\ns_\alpha\ln n\ns_\alpha -\kT\,(1+n\ns_\alpha)\ln (1+n\ns_\alpha)\Big] \\ \Omega\ns_\ssr{FD} &= J\sum_\alpha \Big[ (\ve\ns_\alpha-\mu)\,n\ns_\alpha+\kT\,n\ns_\alpha \ln n\ns_\alpha +\kT\,(1-n\ns_\alpha)\ln (1-n\ns_\alpha)\Big] \ . \end{split}} Now - lo and behold! - treating Ω as a function of the distribution {n∗α} and extremizing in each case, subject to the constraint of total particle number N=J∑αn∗α, one obtains the Maxwell-Boltzmann, Bose-Einstein, and Fermi-Dirac distributions, respectively: \boldsymbol{{\delta\over\delta n\ns_\alpha}\Big(\Omega-\lambda \, J\sum_{\alpha'} n\ns_{\alpha'}\Big) = 0 \quad\Rightarrow \quad \begin{cases} n^\ssr{MB}_\alpha=e^{(\mu-\ve\ns_\alpha)/k\ns_\RB T} \\ \\ n^\ssr{BE}_\alpha=\big[e^{(\ve\ns_\alpha-\mu)/k\ns_\RB T}-1\big]^{-1}\\ \\ n^\ssr{FD}_\alpha=\big[e^{(\ve\ns_\alpha-\mu)/k\ns_\RB T}+1\big]^{-1} \ . \end{cases}} As long as J is finite, so the states in each block all remain at the same energy, the results are independent of J.