Statistical Physics =================== Microcanonical Ensemble ----------------------- The entropy is equal to: .. math:: S = k_{\mathrm{B}} \log W = {1\over\beta T} \log W where $W$ is the micronanonical partition function, the number of microstates within the range of energy. Canonical Ensemble ------------------ The partition function is: .. math:: Z_{can} = \sum_n e^{-\beta E_n} The Helmholtz free energy is equal to: .. math:: F = -{1\over\beta} \log Z_{can} = -{1\over\beta} \log \sum_n e^{-\beta E_n} Grand Canonical Ensemble ------------------------ The partition function for fermions is: .. math:: Z_{gr} = \sum_n e^{-\beta(E_n - \mu N_n)} = = \left(\prod_\alpha \sum_{n_\alpha=0}^1 \right) e^{-\beta(E_n - \mu N_n)} = = \left(\prod_\alpha \sum_{n_\alpha=0}^1 \right) e^{-\beta\left(\sum_\alpha n_\alpha \epsilon_\alpha - \mu \sum_\alpha n_\alpha\right)} = = \left(\prod_\alpha \sum_{n_\alpha=0}^1 \right) \prod_\alpha e^{-\beta n_\alpha\left(\epsilon_\alpha - \mu\right)} = = \prod_\alpha \left(\sum_{n_\alpha=0}^1 e^{-\beta n_\alpha\left(\epsilon_\alpha - \mu\right)} \right) = = \prod_\alpha \left(1 + e^{-\beta (\epsilon_\alpha - \mu)} \right) Similarly, for bosons we would get: .. math:: Z_{gr} = \sum_n e^{-\beta(E_n - \mu N_n)} = = \left(\prod_\alpha \sum_{n_\alpha=0}^\infty \right) e^{-\beta(E_n - \mu N_n)} = = \left(\prod_\alpha \sum_{n_\alpha=0}^\infty \right) e^{-\beta\left(\sum_\alpha n_\alpha \epsilon_\alpha - \mu \sum_\alpha n_\alpha\right)} = = \left(\prod_\alpha \sum_{n_\alpha=0}^\infty \right) \prod_\alpha e^{-\beta n_\alpha\left(\epsilon_\alpha - \mu\right)} = = \prod_\alpha \left(\sum_{n_\alpha=0}^\infty e^{-\beta n_\alpha\left(\epsilon_\alpha - \mu\right)} \right) = = \prod_\alpha \left(1 - e^{-\beta (\epsilon_\alpha - \mu)} \right)^{-1} The grand potential for fermions is then equal to: .. math:: \Omega = -{1\over\beta} \log Z_{gr} = = -{1\over\beta} \log\left( \prod_\alpha \left(1 + e^{-\beta (\epsilon_\alpha - \mu)} \right) \right) = = -{1\over\beta} \sum_\alpha \log\left( 1 + e^{-\beta (\epsilon_\alpha - \mu)} \right) Similarly, the grand potential for bosons is equal to: .. math:: \Omega = -{1\over\beta} \log Z_{gr} = = -{1\over\beta} \log\left( \prod_\alpha \left(1 - e^{-\beta (\epsilon_\alpha - \mu)} \right)^{-1} \right) = = -{1\over\beta} \sum_\alpha \log\left( 1 - e^{-\beta (\epsilon_\alpha - \mu)} \right)^{-1} = = {1\over\beta} \sum_\alpha \log\left( 1 - e^{-\beta (\epsilon_\alpha - \mu)} \right) Examples -------- Ideal Gas ~~~~~~~~~ Ideal gas is simply a system of classical particles, where for a given microstate specified by a set of coordinates $\mathbf{x}_i$ and momenta $\mathbf{p}_i$ , the total energy of the microstate is given by the following Hamiltonian: .. math:: H(\mathbf{x}_i, \mathbf{p}_i) = \sum_{i=1}^N {p_i^2 \over 2 m}\,, that is, the particles are non-interacting, each has a mass $m$ and a momentum $\mathbf{p}_i$. The canonical partition function is then equal to: .. math:: Z_{can}(T, V, N) = \sum_n e^{-\beta E_n} = = \int {\d^{3N} x\, \d^{3N} p \over N! (2\pi\hbar)^{3N}} e^{-\beta H(\mathbf{x}_i, \mathbf{p}_i)} = = \int {\d^{3N} x\, \d^{3N} p \over N! (2\pi\hbar)^{3N}} e^{-\beta \sum_{i=1}^N {p_i^2 \over 2 m}} = = {1\over N!}\left( \int {\d^3 x \d^3 p \over (2\pi\hbar)^3} e^{-\beta {p^2 \over 2 m}}\right)^N = = {1\over N!}\left(V \int_0^\infty {4\pi p^2 \d p \over (2\pi\hbar)^3} e^{-\beta {p^2 \over 2 m}}\right)^N = = {1\over N!}\left(V {4\pi \over (2\pi\hbar)^3} {\sqrt\pi (2m)^{3\over 2}\over 4 \beta^{3\over2}} \right)^N = = {1\over N!}\left( \left(m\over2\pi\hbar^2\beta\right)^{3\over2} V\right)^N\,, where we used the following integral: .. math:: \int_0^\infty p^2 e^{-\alpha p^2} \d p = {\sqrt\pi\over4\alpha^{3\over2}}\,. The Helmholtz free energy is then equal to: .. math:: F(T, V, N) = -{1\over\beta} \log Z_{can}(T, V, N) = = -{1\over\beta} \log\left( {1\over N!} \left( \left(m\over2\pi\hbar^2\beta\right)^{3\over2} V\right)^N\right) = = {N\over\beta} \left({\log N!\over N} - \log \left( \left(m\over2\pi\hbar^2\beta\right)^{3\over2} V\right)\right) = = {N\over\beta} \left(\log N - 1 + O\left(\log N\over N\right) - \log \left( \left(m\over2\pi\hbar^2\beta\right)^{3\over2} V\right)\right) = = {N\over\beta} \left( - \log \left( \left(m\over2\pi\hbar^2\beta\right)^{3\over2} {V e \over N}\right) + O\left(\log N\over N\right) \right) = = {N\over\beta} \left({3\over2} - \log \left( \left(m\over2\pi\hbar^2\beta\right)^{3\over2} {V e^{5\over2} \over N}\right) + O\left(\log N\over N\right) \right) = = N k_\mathrm{B} T \left({3\over2} - \log \left( {V T^{3\over2} \over N \left(2\pi\hbar^2\over m k_\mathrm{B}e^{5\over3}\right)^{3\over2} }\right) + O\left(\log N\over N\right) \right)\,, where we used the Stirling's approximation for $N!$. For large $N$ this is equal to the Helmholtz free energy of the ideal gas (see :ref:`ideal_gas`): .. math:: F(T, V, N) = N k_\mathrm{B} T \left(c_V - \log \left({VT^{c_V}\over N\Phi}\right) \right)\,, with $c_V={3\over2}$ and $\Phi=\left(2\pi\hbar^2\over m k_\mathrm{B}e^{5\over3}\right)^{3\over2}$. See that section where all other thermodynamic properties are derived from it. We can also start from the grand canonical partition function: .. math:: Z_{gr}(T, V, \mu) = \sum_{N=0}^\infty e^{\beta \mu N} Z_{can}(T, V, N) = = \sum_{N=0}^\infty e^{\beta \mu N} {1\over N!}\left( \left(m\over2\pi\hbar^2\beta\right)^{3\over2} V\right)^N = = \sum_{N=0}^\infty {1\over N!}\left(e^{\beta \mu} \left(m\over2\pi\hbar^2\beta\right)^{3\over2} V\right)^N = = e^{e^{\beta \mu} \left(m\over2\pi\hbar^2\beta\right)^{3\over2} V} And the grand potential is: .. math:: \Omega(T, V, \mu) = -{1\over\beta} \log Z_{gr}(T, V, \mu) = = -{1\over\beta} {e^{\beta \mu} \left(m\over2\pi\hbar^2\beta\right)^{3\over2} V} = = -{k_\mathrm{B} T} {e^{\mu\over k_\mathrm{B} T} \left(m k_\mathrm{B} T\over2\pi\hbar^2\right)^{3\over2} V} = = - {k_\mathrm{B} V T^{5\over2} \over \left(2\pi\hbar^2\over m k_\mathrm{B}e^{5\over3}\right)^{3\over2} e^{{5\over2}-{\mu\over k_\mathrm{B} T}}} \,. This is equal to the grand potential of an ideal gas: .. math:: \Omega(T, V, \mu) = - {k_\mathrm{B} V T^{c_p} \over \Phi e^{c_p-{\mu\over k_\mathrm{B} T}}} \,, with $c_p={5\over2}$ and $\Phi=\left(2\pi\hbar^2\over m k_\mathrm{B}e^{5\over3}\right)^{3\over2}$. The thermodynamics section then shows that the corresponding Helmholtz free energy is the same as we obtained above from the canonical ensemble. Note that we also obtained the same $\Phi$ as before.