### 物理代写|统计力学代写Statistical mechanics代考|PHYC30017

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 物理代写|统计力学代写Statistical mechanics代考|High-temperature form

As $\beta \rightarrow 0$ there are contributions to Eq. (5.33) from large values of the quantum number $l$, which suggests we approximate the sum in Eq. (5.33) with an integral, using the form of $Z$ in Eq. (4.15). That route requires the density-of-states function, $\Omega(E)$, the derivative with respect to energy of the total number of energy states up to and including $E$. Energy at a specified value $E$ implies a maximum value of $l$ determined by $E=\hbar^2 l_{\max }\left(l_{\max }+1\right) /(2 I) \approx \hbar^2 l_{\max }^2 /(2 I)$ because $l_{\max } \gg 1$. How many states are there for $0 \leq l \leq l_{\max }$ ? It can be shown that
$$\sum_{l=0}^{l_{\max }}(2 l+1)=\left(l_{\max }+1\right)^2 \approx l_{\max }^2 \approx \frac{2 I}{\hbar^2} E .$$
The density of states is therefore $\Omega(E)=2 I / \hbar^2$. Thus, we can approximate Eq. (5.33),
$$Z_{1, \text { rol }}(T)=\frac{2 I}{\hbar^2} \int_0^{\infty} \mathrm{e}^{-\beta E} \mathrm{~d} E=\frac{2 I}{\beta \hbar^2} \equiv \frac{T}{\Theta_r}, \quad\left(T \gg \Theta_r\right)$$
where $\Theta_r=\hbar^2 /(2 I k)$ sets a characteristic temperature for rotational motions. ${ }^{23}$ Using equations that we’ve now used several times (Eqs. (4.40) and (P4.1)), with $Z=\left(Z_1\right)^N$,
\begin{aligned} \langle E\rangle_{\mathrm{rot}} &=N k T \ \left(C_V\right){\mathrm{rot}} &=N k, \quad(T \rightarrow \infty) \end{aligned} the same as what we obtain from the equipartition theorem. A more accurate high-temperature form can be obtained using the result of Exercise 5.11: $$Z{1, \mathrm{rot}}(T)=\frac{T}{\Theta_r}+\frac{1}{3}+\frac{1}{15} \frac{\Theta_r}{T}+\frac{4}{315}\left(\frac{\Theta_r}{T}\right)^2+\cdots . \quad\left(T \gg \Theta_r\right)$$
From Eq. (5.37) we obtain an expression for the heat capacity more general than Eq. (5.36) (see Exercise 5.12),
$$\left(C_V(T)\right){\mathrm{rot}}=N k\left[1+\frac{1}{45}\left(\frac{\Theta_r}{T}\right)^2+\frac{16}{945}\left(\frac{\Theta_r}{T}\right)^3+\cdots\right] .$$ We see that $\left(C_V(T)\right){\text {rot }}$ exceeds the classical value $N k$, a value that it tends to as $T \rightarrow \infty$.

## 物理代写|统计力学代写Statistical mechanics代考|Low-temperature form

In the low-temperature regime, $T \ll \Theta_r$, we have, from Eq. (5.33),
$$Z(T){1, \mathrm{rot}}=1+3 \mathrm{e}^{-2 \Theta_r / T}+5 \mathrm{e}^{-6 \Theta_r / T}+\cdots .$$ In this case, the variable $\mathrm{e}^{-\Theta_r / T}$ is exponentially small as $T \rightarrow 0$. From Eq. (5.39), we find to lowest order $$\left(C_V(T)\right){\mathrm{rot}} \approx 12 N k\left(\frac{\Theta_r}{T}\right)^2 \mathrm{e}^{-2 \Theta_r / T} . \quad\left(T \ll \Theta_r\right)$$

As $T \rightarrow 0,\left(C_V(T)\right)_{\text {rot }}$ drops to zero exponentially fast; rotational degrees of freedom can’t be excited at sufficiently low temperature – they become “frozen out.”

The two equations, (5.38) and (5.40), are limiting forms of $\left(C_V(T)\right){\mathrm{rot}}$ in the high- and lowtemperature regimes. They each show that the heat capacity is temperature dependent. To obtain the complete temperature dependence of $\left(C_V(T)\right){\text {rot }}$ requires the use of a computer to evaluate the sum in Eq. (5.33) at each temperature. A detailed analysis shows there is a maximum value of $\left(C_V(T)\right)_{\mathrm{rot}} \approx 1.1 \mathrm{Nk}$ at $T \approx 0.81 \Theta_r$. Given that $\Theta_r \approx 10 \mathrm{~K}$, measurements of $C_V$ on diatomic gases at room temperature are consistent with the prediction of the equipartition theorem.

## 物理代写|统计力学代写Statistical mechanics代考|High-temperature form

$$\sum_{l=0}^{l_{\max }}(2 l+1)=\left(l_{\max }+1\right)^2 \approx l_{\max }^2 \approx \frac{2 I}{\hbar^2} E$$

$$Z_{1, \text { rol }}(T)=\frac{2 I}{\hbar^2} \int_0^{\infty} \mathrm{e}^{-\beta E} \mathrm{~d} E=\frac{2 I}{\beta \hbar^2} \equiv \frac{T}{\Theta_r}, \quad\left(T \gg \Theta_r\right)$$

$$\langle E\rangle_{\mathrm{rot}}=N k T\left(C_V\right) \text { rot }=N k, \quad(T \rightarrow \infty)$$

$$Z 1, \operatorname{rot}(T)=\frac{T}{\Theta_r}+\frac{1}{3}+\frac{1}{15} \frac{\Theta_r}{T}+\frac{4}{315}\left(\frac{\Theta_r}{T}\right)^2+\cdots . \quad\left(T \gg \Theta_r\right)$$

$$\left(C_V(T)\right) \operatorname{rot}=N k\left[1+\frac{1}{45}\left(\frac{\Theta_r}{T}\right)^2+\frac{16}{945}\left(\frac{\Theta_r}{T}\right)^3+\cdots\right] .$$

## 物理代写|统计力学代写Statistical mechanics代考|Low-temperature form

$$Z(T) 1, \text { rot }=1+3 \mathrm{e}^{-2 \Theta_r / T}+5 \mathrm{e}^{-6 \Theta_r / T}+\cdots .$$

$$\left(C_V(T)\right) \operatorname{rot} \approx 12 N k\left(\frac{\Theta_r}{T}\right)^2 \mathrm{e}^{-2 \Theta_r / T} . \quad\left(T \ll \Theta_r\right)$$

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