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}
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