Blundell Thermal Physics · Problem 29.6

Bose–Einstein and Fermi–Dirac distributions

Physics — Statistical Mechanics · fermi-dirac-bose-einstein-microcanonical

Problem

In this exercise, we derive Fermi--Dirac and Bose--Einstein statistics using the microcanonical ensemble. (a) Show that the number of ways of distributing njn_j fermions among gjg_j states with not more than one particle in each is Ωj=gj!nj!(gj−nj)!.(29.40)\Omega_j = \frac{g_j!}{n_j!(g_j - n_j)!}. \tag{29.40} Here, jj labels a particular group of states. Hence the entropy SS is given by S=kBln⁡[∏jgj!nj!(gj−nj)!].(29.41)S = k_{\mathrm{B}} \ln\left[\prod_j \frac{g_j!}{n_j!(g_j - n_j)!}\right]. \tag{29.41} Hence show (using Stirling's approximation) that S=−kB∑jgj[nˉjln⁡nˉj+(1−nˉj)ln⁡(1−nˉj)],(29.42)S = -k_{\mathrm{B}} \sum_j g_j[\bar{n}_j \ln \bar{n}_j + (1 - \bar{n}_j)\ln(1 - \bar{n}_j)], \tag{29.42} where nˉj=nj/gj\bar{n}_j = n_j/g_j are the mean occupation numbers of the quantum states. Maximize this expression subject to the constraint that the total energy EE and number of particles NN are constant, and hence show that nˉj=1eα+βEj+1.(29.43)\bar{n}_j = \frac{1}{\mathrm{e}^{\alpha + \beta E_j} + 1}. \tag{29.43} (b) Show that the number of ways of distributing njn_j bosons among gjg_j states with any number of particles in each is Ωj=(gj+nj−1)!nj!(gj−1)!.(29.44)\Omega_j = \frac{(g_j + n_j - 1)!}{n_j!(g_j - 1)!}. \tag{29.44} Hence show that S=kB∑jgj[(1+nˉj)ln⁡(1+nˉj)−nˉjln⁡nˉj].(29.45)S = k_{\mathrm{B}} \sum_j g_j[(1 + \bar{n}_j)\ln(1 + \bar{n}_j) - \bar{n}_j \ln \bar{n}_j]. \tag{29.45} Maximize this expression subject to the constraint that the total energy EE and number of particles NN are constant, and hence show that nˉj=1eα+βEj−1.(29.46)\bar{n}_j = \frac{1}{\mathrm{e}^{\alpha + \beta E_j} - 1}. \tag{29.46}

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