Grand Partition Function

Gas here means the particles are not intereacting with each other.

Gibbs Factor

Consider a system connected to an isolated reservoir, in which the system and the reservoir as a whole is isolated. The multiplicity of the system at state $s$ is $\Omega_S = 1$ so the total multiplicity is $\Omega_S \cdot \Omega_R = \Omega_R$. The probability ratio of a system between two state $s_1$ and $s_2$ is $$\begin{align} \frac{P(s_1)}{P(s_2)} &= \frac{\Omega_R(s_1)}{\Omega_R(s_2)} \\ &= \frac{e^{S_R(s_1)/k}}{e^{S_R(s_2)/k}} \\ &= e^{(S_R(s_1) - S_R(s_2))/k} \end{align}$$ where $\Omega_R(s)$ and $S_R(s)$ are the multiplicity and entropy of the reservoir at state $s$. If the reservoir volume is fixed but the system is allowed to exchange energy and particles with the reservoir, then $$dS_R = \frac{1}{T}(dU_R + PdV -\mu dN) = \frac{1}{T}(dU - \mu dN)$$ So, $$\frac{P(s_1)}{P(s_2)} = e^{(U_R(s_1) - U_R(s_2))/kT}$$ Let $U_{tot}$ be the total energy, i.e. energy of the reservoir plus energy of the system. As the reservoir and the system as a whole is isolated, $U_{tot}$ is a constant and $$U_R = U_{tot} - E$$ where $E$ is the energy of the system. Let $N_{tot}$ be the total number of particles, i.e. number of particles of the reservoir plus number of particles of the system. As the reservoir and the system as a whole is isolated, $N_{tot}$ is a constant and $$N_R = N_{tot} - N_S$$ where $N_S$ is the number of particles of the system. Hence, $$\frac{P(s_1)}{P(s_2)} = e^{(\mu N_S(s_1) -E(s_1)) - (\mu N_S(s_2) - E(s_2)))/kT}$$ And the probability at a state $s$ is proportional to the term $$\bbox[5px,border:2px solid #666] { P(s) \propto e^{\left[\mu N_S(s)-E(s)\right]/kT} }$$ which is called the Gibbs factor.

Grand Partition Function

Let $C$ be a constant such that $$P(s) = C e^{\left[\mu N(s)-E(s)\right]/kT}$$ As the sum of the probability of all states must be equal to 1, $$\sum_sP(s) = C \sum_s e^{\left[\mu N(s)-E(s)\right]/kT} = 1$$ Define $$\bbox[5px,border:2px solid #666] { Z = \sum_s e^{\left[\mu N(s)-E(s)\right]/kT} }$$ Then, substitute $C = \frac{1}{Z}$ into probability equation, $$P(s) = \frac{1}{Z}e^{\left[\mu N(s)-E(s)\right]/kT}$$

Distribution

Consider a box of volume $V$ with $N$ particles at temperature $T$. Let $\epsilon_s$ and $n_s$ be the energy and number of particles at state $s$ respectively. For the whole system at a state $S$, the total energy is $$E_S = n_1\epsilon_1 + n_2\epsilon_2 + ... = \sum_s n_s\epsilon_s$$ where $$\sum_s n_s = N$$ The partition function of the whole system is $$Z = \sum_S e^{-\beta E_S}$$ The probability to be at state $S$ is given by $$P_S = \frac{1}{Z}(n_1\epsilon_1 + n_2\epsilon_2 + ...)$$ The mean number of particle at state $s$ is $$\begin{align} \bar{n}_s &= \sum_S n_s P_S \\ &= \frac{1}{Z}\left(\sum_S n_s e^{-\beta(n_1\epsilon_1+n_2\epsilon_2+...)}\right)\\ &= \frac{1}{Z}\sum_S\left(-\frac{1}{\beta}\frac{\partial}{\partial \epsilon_s}\right)e^{-\beta(n_1\epsilon_1+n_2\epsilon_2+...)}=-\frac{1}{\beta Z}\frac{\partial Z}{\partial \epsilon_s}\\ &= -\frac{1}{\beta}\frac{\partial \ln Z}{\partial \epsilon_s} \end{align}$$

Fermi-Dirac Distribution

Let there be N fermions in a system. The fermions are indistinguishable and cannot occupy the same state. The grand partition function is given by $$\begin{align} Z &= \sum_S e^{-\beta(n_1\epsilon_1 + n_2\epsilon_2+...)}e^{\beta(\mu n_1 + \mu n_2 + ...)}\\ &= \sum_{n_1,n_2,...}e^{-\beta(\epsilon_1-\mu)n_1}e^{-\beta(\epsilon_2-\mu)n_2}...\\ &= \left(\sum^1_{n_1=0}e^{-\beta(\epsilon_1-\mu)n_1}\right)\left(\sum^1_{n_2=0}e^{-\beta(\epsilon_2-\mu)n_2}\right)...\\ &=(1+e^{-\beta(\epsilon_1-\mu)})(1+e^{-\beta(\epsilon_2\mu)})...\\ &=\prod_s(1+e^{-\beta(\epsilon_s-\mu)})\\ \ln Z &= \sum_s\ln(1+e^{-\beta(\epsilon_s-\mu)}) \end{align}$$ Then the average number of fermions in state $s$ is $$\bar{n}_s=-\frac{1}{\beta}\frac{\partial}{\partial \epsilon_s}\ln Z = -\frac{1}{\beta}\frac{(-\beta)e^{-\beta(\epsilon_s-\mu)}}{1+e^{-\beta(\epsilon_s-\mu)}}$$ So, $$\bbox[5px,border:2px solid #666] { \bar{n}_s = \frac{1}{e^{\beta(\epsilon_s-\mu)}+1} }$$

Fermi Gas

Consider a fermi gas inside a cube with length $L$. The allowed quantum states are $$p_i = \frac{h}{\lambda} = n_i\frac{h}{2L}$$ where $i=x,y,z$ and $n_x,n_y,n_z=0,1,2,3...$. Energy at each states is given by $$E(n_x,n_y,n_z) = \frac{1}{2m}(p_x^2+p_y^2+p_z^2) = \frac{h^2}{8mL^2}(n_x^2+n_y^2+n_z^2)$$ Define $n^2 = n_x^2 + n_y^2 + n_z^2$. The maximum $n$ can be found by $$n_{\text{max}}^2 = \frac{8mL^2}{h^2}E_{\text{max}}$$ Consider $n$ as a 3-dimensional vector. The allowed states must lie within the sphere of radius $n_{\text{max}}$ $$n_{\text{max}}^2 = \geq n_x^2 + n_y^2 + n_z^2$$ As the $n_x,n_y,n_z$ are positive integers, the allowed states must be in the first octant. So, the number of allowed states are $$N = \frac{\text{volume of first octant of the sphere with radius }n_{\text{max}}}{\text{volume per state }n} \times \text{the number of spin states}$$ As $n_x,n_y,n_z$ are positive integers, $\Delta n_x = \Delta n_y = \Delta n_z = 1$, volume per state is 1. Let $s$ be the number of spin states. $$N = \frac{1}{8}\left(\frac{3}{4}\pi n_{\text{max}}^3\right)s = \frac{2\pi s}{3}\left(\frac{8mL^2}{h^2}E_{\text{max}}\right)^{3/2}$$ The density of states, i.e. the number of single-particle state per unit energy, is $$g(\epsilon) = \frac{\partial N}{\partial \epsilon} = \frac{\partial}{\partial \epsilon} \frac{2\pi sV}{3}\left(\frac{8m}{h^2}\epsilon\right)^{3/2}$$ where $V$ is the volume. So, $$\bbox[5px,border:2px solid #666] { g(\epsilon) = \pi sV\left(\frac{8m}{h^2}\right)^{3/2}\epsilon^{1/2} }$$

Bose-Einstein Distribution

Let there be N bosons in a system. The bosons are indistinguishable and can occupy the same state. The grand partition function is given by $$\begin{align} Z &= \sum_S e^{-\beta(n_1\epsilon_1 + n_2\epsilon_2+...)}e^{\beta(\mu n_1 + \mu n_2 + ...)}\\ &= \sum_{n_1,n_2,...}e^{\beta(\mu-\epsilon_1)n_1+\beta(\mu-\epsilon_2)n_2+...}\\ &= \left(\sum^{\infty}_{n_1=0}e^{\beta(\mu-\epsilon_1)n_1}\right)\left(\sum^{\infty}_{n_2=0}e^{\beta(\mu-\epsilon_2)n_2}\right)...\\ &= \left(\frac{1}{1-e^{\beta(\mu-\epsilon_1)}}\right)\left(\frac{1}{1-e^{\beta(\mu-\epsilon_2)}}\right)...\\ \ln Z &= -\sum_s \ln \left(1-e^{\beta(\mu-\epsilon_s)}\right) \end{align}$$ Then the average number of bosons in state $s$ is $$\begin{align} \bar{n}_s &= \sum_S n_s\cdot P(S) = \frac{1}{Z}\sum_S n_s e^{-\beta(n_1\epsilon_1 + n_2\epsilon_2+...)}e^{\beta(\mu n_1 + \mu n_2 + ...)} \\ &= \frac{1}{Z}\left(-\frac{1}{\beta}\frac{\partial}{\partial \epsilon_s}\right)\sum_S e^{-\beta(n_1\epsilon_1 + n_2\epsilon_2+...)}e^{\beta(\mu n_1 + \mu n_2 + ...)} \\ &= -\frac{1}{\beta}\frac{\partial \ln Z}{\partial \epsilon_s}\\ &= -\frac{1}{\beta}\frac{\partial}{\partial \epsilon_s}\left[-\sum_s\ln(1-e^{\beta(\mu-\epsilon_s)})\right]\\ &= -\frac{1}{\beta}\frac{-\beta e^{-\beta(\epsilon_s-\mu)}}{1-e^{-\beta(\epsilon_s-\mu)}}\\ \end{align}$$ So, $$\bbox[5px,border:2px solid #666] { \bar{n}_s = \frac{1}{e^{\beta(\epsilon_s-\mu)}-1} }$$

Bose Gas

Consider a bose gas in a box with length $L$. The allowed quantum states are $$p_i = \frac{h}{\lambda} = n_i\frac{h}{2L}$$ where $i=x,y,z$ and $n_x,n_y,n_z=0,1,2,3...$. For photons, energy at each state is given by $$\epsilon_n = pc = \frac{hcn}{2L}$$ where $n=\sqrt{n_x^2+n_y^2+n_z^2}$. As the bose gas is contained in the box, $\mu =0$ $$\bar{n} = \frac{1}{e^{(\epsilon-(0))/kT}-1} = \frac{1}{e^{\epsilon/kT}-1}$$ For photons, the average energy is $$\bar{\epsilon} = \bar{n}hf = \frac{hf}{e^{hf/kT}-1}$$ The total energy is $$U=2\int^{\infty}_0 \epsilon_n dV$$ where the factor 2 is to account for the two independent polarizations of a photon. Consider $n$ as a 3-dimensional vector. Volume is the first octant of the sphere. $$dV = \frac{1}{8}(4\pi n^2)dn$$ So, $$U=2 \int^{\infty}_0 \left(\frac{hcn}{2L}\right)\frac{1}{e^{\epsilon/kT}-1}\left(\frac{1}{8}4\pi n^2\right)dn$$ The density of states is $$\begin{align} g(\epsilon) &= 2 \left(\frac{1}{8}\right)(4\pi n^2)\frac{dn}{d\epsilon}\\ &= \pi\left(\frac{2L\epsilon}{hc}\right)^2\frac{2L}{hc} \\ &= \frac{8\pi L^3}{(hc)^3}\epsilon^2 \\ &= \frac{8\pi V}{(hc)^3}\epsilon^2 \\ \end{align}$$ The energy density is given by $$\begin{align} u(\epsilon) &= \frac{1}{V}g(\epsilon)\bar{n}(\epsilon)\epsilon \\ &= \frac{1}{V}\left(\frac{8\pi V}{(hc)^3}\epsilon^2\right)\left(\frac{1}{e^{\beta\epsilon}-1}\right)\epsilon \\ &= \frac{8\pi}{(hc)^3}\frac{\epsilon^3}{e^{\beta\epsilon}-1} \end{align}$$