# 物理代写|统计力学代写Statistical mechanics代考|PHYS3034

## 物理代写|统计力学代写Statistical mechanics代考|Internal energy

For the average internal energy $U$, there is no need to split off the ground state from the sum in Eq. (5.73), because the state with $E=0$ does not contribute to the energy, and we have from Eq. (5.74), $U=\left(3 g V k T /\left(2 \lambda_T^3\right) G_{5 / 2}(z)\right.$. We must still distinguish, however, the two cases for $T>T_B$ and $TT_B \ G_{5 / 2}(1) & TT_B \ \left(\frac{T}{T_B}\right)^{3 / 2} \frac{\zeta(5 / 2)}{\zeta(3 / 2)} & TT_B$ and $n \lambda_{T_B}^3=\zeta(3 / 2)$ for $T \leq T_B$. Note that for $T>T_B$, the energy is that of the classical ideal gas, $\frac{3}{2} N k T$, multiplied by $G_{5 / 2}(z) / G_{3 / 2}(z)$. For $T<T_B$, the energy is in the form $U=\frac{3}{2} N_{\mathrm{ex}} k T \zeta(5 / 2) / \zeta(3 / 2)$, where $N_{\mathrm{ex}}$ is defined in Eq. (5.158). Particles condensed in the ground state have zero energy; only particles in excited states contribute to the energy of the gas for $T<T_B$.

We can use Eq. (B.28) (an expansion of $G_{5 / 2}$ in terms of $G_{3 / 2}$ ) to eliminate reference to the fugacity in Eq. (5.163),
$$\frac{G_{5 / 2}(z)}{G_{3 / 2}(z)}=1-0.1768 G_{3 / 2}(z)-0.0033 G_{3 / 2}^2(z)-0.00011 G_{3 / 2}^3(z)-\cdots,$$ and thus, using $G_{3 / 2}(z)=\zeta(3 / 2)\left(T_B / T\right)^{3 / 2}$, we have from Eq. (5.163),
$$U=\frac{3}{2} N k T\left{\begin{array}{lr} {\left[1-0.4618\left(\frac{T_B}{T}\right)^{3 / 2}-0.0225\left(\frac{T_B}{T}\right)^3-0.0020\left(\frac{T_B}{T}\right)^{9 / 2}-\cdots\right]} & T>T_B \ \frac{\zeta(5 / 2)}{\zeta(3 / 2)}\left(\frac{T}{T_B}\right)^{3 / 2} & T<T_B \end{array}\right.$$

## 物理代写|统计力学代写Statistical mechanics代考|THE MAYER CLUSTER EXPANSION

Consider $N$ identical particles of mass $m$ that interact through two-body interactions, $v_{i j}$, with Hamiltonian
$$H=\frac{1}{2 m} \sum_i p_i^2+\sum_{j>i} v_{i j}, \quad(i, j=1,2, \cdots, N)$$
where $v_{i j} \equiv v\left(\boldsymbol{r}i-\boldsymbol{r}_j\right)$ denotes the potential energy associated with the interaction between particles at positions $\boldsymbol{r}_i, \boldsymbol{r}_j$, and $\sum{j>i}$ indicates a sum over $\left(\begin{array}{c}N \ 2\end{array}\right)$ pairs of particles. ${ }^1$ For central forces, $v_{i j}$ depends only on the magnitude of the distance between particles, $v_{i j}=v\left(\left|\boldsymbol{r}_i-\boldsymbol{r}_j\right|\right)$, which we assume for simplicity. The methods developed here can be extended to quantum systems, but the analysis becomes more complicated; we won’t consider interacting quantum gases. ${ }^2$

For our purposes, the precise nature of the interactions underlying the potential energy function $v(r)$ is not important as long as there is a long-range attractive component together with a shortrange repulsive force. To be definite, we mention the Lennard-Jones potential (shown in Fig. 6.1) for the interaction between closed-shell atoms, which has the parameterized form,
$$v(r)=4 \epsilon\left[\left(\frac{\sigma}{r}\right)^{12}-\left(\frac{\sigma}{r}\right)^6\right],$$
where $\epsilon$ is the depth of the potential well, and $\sigma$ is the distance at which the potential is zero. The $r^{-6}$ term describes an attractive interaction between neutral molecules that arises from the energy of interaction between fluctuating multipoles of the molecular charge distributions. ${ }^3$ The $r^{-12}$ term models the repulsive force at short distances that arises from the Pauli exclusion effect of overlapping electronic orbitals. There’s no science behind the $r^{-12}$ form; it’s analytically convenient, and it provides a good approximation of the interactions between atoms. For the noble gases, $\epsilon$ ranges from $0.003 \mathrm{eV}$ for Ne to $0.02 \mathrm{eV}$ for Xe [18, $\mathrm{p} 398]$. The parameter $\sigma$ is approximately $0.3 \mathrm{~nm}$.

## 物理代写|统计力学代写Statistical mechanics代考|Internal energy

G5/2(和)G3/2(和)=1−0.1768G3/2(和)−0.0033G3/22(和)−0.00011G3/23(和)−⋯,因此，使用G3/2(和)=G(3/2)(吨乙/吨)3/2，我们有从方程式。(5.163),
$$U=\frac{3}{2} N k T\left{ [1−0.4618(吨乙吨)3/2−0.0225(吨乙吨)3−0.0020(吨乙吨)9/2−⋯]吨>吨乙 G(5/2)G(3/2)(吨吨乙)3/2吨<吨乙\正确的。$$

## 物理代写|统计力学代写Statistical mechanics代考|THE MAYER CLUSTER EXPANSION

$$H=\frac{1}{2 m} \sum_i p_i^2+\sum_{j>i} v_{i j}, \quad(i, j=1,2, \cdots, N)$$

$$v(r)=4 \epsilon\left[\left(\frac{\sigma}{r}\right)^{12}-\left(\frac{\sigma}{r}\right)^6\right]$$

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