# 物理统计学 Statistical Physics II|PHYS7722 Boston College Assignment

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## Instructions:

Statistical physics is a branch of physics that studies the behavior of large collections of particles, such as atoms or molecules, using statistical methods. It aims to explain macroscopic properties of matter, such as temperature, pressure, and entropy, in terms of the microscopic properties of the constituent particles and the interactions between them.

Statistical physics is based on the principles of probability and statistics, and it provides a framework for understanding complex systems that cannot be analyzed through classical physics. It also provides a link between the microscopic world of quantum mechanics and the macroscopic world of classical mechanics.

The partition function for an ideal gas of $N$ distinguishable particles where particle $k$ has mass $m_k$ can be written as
$$Z=\frac{1}{h^{3 N}} \int_V d^3 \mathbf{q}1 \int{\text {all } \mathbf{p}} d^3 \mathbf{p}1 \cdots \int_V d^3 \mathbf{q}_N \int{\text {all } \mathbf{p}} d^3 \mathbf{p}N \exp \left(-\frac{1}{k_B T} \sum{k=1}^N \frac{\mathbf{p}_k^2}{2 m_k}\right) .,$$
where $V$ is the volume of the system.
(a) Evaluate all the integrals in Eq.(40) and write the final result in terms of $T, V, N$ and the set of masses $m_1, m_2, \ldots, m_N$ (and physical constants).

We can evaluate the integrals by using the fact that they represent the phase space integral for each particle. The integral $\int{d^3\mathbf{q}_k} \int{\text{all } \mathbf{p}_k}d^3\mathbf{p}_k$ represents the phase space integral for the $k$th particle. We can perform this integral by changing to spherical coordinates for the momentum integral and making use of the fact that the volume element $d^3 \mathbf{p}$ is $4\pi p^2dp$, where $p$ is the magnitude of the momentum vector. Thus, we have

$\int d^3 \mathbf{q}_k \int$ all $\mathbf{p}_k d^3 \mathbf{p}_k=V \int_0^{\infty} 4 \pi p_k^2 d p_k=\frac{4}{3} \pi V\left(\frac{2 \pi m_k k_B T}{h^2}\right)^{3 / 2}$

Using this result, we can evaluate the integral over all particles in the partition function: \begin{align*} Z &= \frac{1}{h^{3N}} \prod_{k=1}^N \int{d^3\mathbf{q}_k} \int{\text{all } \mathbf{p}k}d^3\mathbf{p}k \exp \left(-\frac{1}{k_B T} \sum{k=1}^N \frac{\mathbf{p}k^2}{2m_k}\right)\ &= \frac{1}{h^{3N}} \prod{k=1}^N \frac{4}{3} \pi V \left(\frac{2\pi m_k k_B T}{h^2}\right)^{3/2} \exp \left(-\frac{1}{k_B T} \frac{\mathbf{p}k^2}{2m_k}\right)\ &= \frac{1}{N! h^{3N}} V^N \prod{k=1}^N \left(\frac{2\pi m_k k_B T}{h^2}\right)^{3/2} \exp \left(-\frac{1}{k_B T} \frac{\mathbf{p}k^2}{2m_k}\right)\ &= \frac{1}{N!} \left(\frac{V}{v_N}\right)^N \prod{k=1}^N \left(\frac{mk_B T}{2\pi\hbar^2}\right)^{3/2}, \end{align*} where $v_N$ is the volume occupied by a single particle, defined by $v_N = \frac{4}{3}\pi\left(\frac{\hbar}{\sqrt{2\pi mk_B T}}\right)^3$ and $m$ is the average mass of a particle, defined by $m = \frac{1}{N}\sum{k=1}^N m_k$. The result is written in terms of physical constants and the set of masses $m_1, m_2, \ldots, m_N$.

(b) Given the partition function $Z$ of a system, the free energy is
$$F=-k_B T \ln Z$$
Therefore, the free energy of this ideal gas system is
\begin{aligned} F & =-k_B T \ln \left[\frac{V^N}{h^{3 N}}\left(2 \pi k_B T\right)^{3 N / 2} \prod_{k=1}^N \sqrt{m_k^3}\right] \ & =-k_B T \ln \left[V^N\left(\frac{2 \pi k_B T}{h^2}\right)^{3 N / 2} \prod_{k=1}^N m_k^{3 / 2}\right] \ & =-k_B T\left[N \ln V+\frac{3 N}{2} \ln \frac{2 \pi k_B T}{h^2}+\frac{3}{2} \sum_{k=1}^N \ln m_k\right] . \end{aligned}

(c) Given that pressure is the negative of the volume partial derivative of the free energy, derive the relationship between pressure $P$, number of particles $N$, volume $V$ and temperature $T$ for this system.

(c) The pressure of an ideal gas system is given by
$$P=-\frac{\partial F}{\partial V}$$
Computing the pressure from Eq.(45), we find
\begin{aligned} P & =-\frac{\partial}{\partial V}\left{-k_B T\left[N \ln V+\frac{3 N}{2} \ln \frac{2 \pi k_B T}{h^2}+\frac{3}{2} \sum_{k=1}^N \ln m_k\right]\right} \ & =k_B T \frac{\partial}{\partial V} N \ln V=k_B T N \frac{1}{V},\end{aligned}
where in the second line we dropped all terms which were independent of $V$. Therefore, we have
$$P=\frac{N k_B T}{V}$$
which is the standard ideal gas law for identical particles. Therefore, we see that the ideal gas law does not change when the particles are not identical.