# 能量均分定理

## 基本概念及簡易例子

### 平移能量與理想氣體

$H^{\mathrm{kin}} = \tfrac12 m |\mathbf{v}|^2 = \tfrac{1}{2} m\left( v_{x}^{2} + v_{y}^{2} + v_{z}^{2} \right),$

$v_{\mathrm{rms}} = \sqrt{\langle v^{2} \rangle} = \sqrt{\frac{3 k_{B} T}{m}} = \sqrt{\frac{3 R T}{M}},$

### 旋轉能量與溶液中的分子滾翻

$H^{\mathrm{rot}} = \tfrac{1}{2} ( I_{1} \omega_{1}^{2} + I_{2} \omega_{2}^{2} + I_{3} \omega_{3}^{2} ),$

### 勢能與諧波振蕩器

$H^{\mathrm{pot}} = \tfrac 12 a q^{2},\,$

$H = H^{\mathrm{kin}} + H^{\mathrm{pot}} = \frac{p^{2}}{2m} + \frac{1}{2} a q^{2}.$

$\langle H \rangle = \langle H^{\mathrm{kin}} \rangle + \langle H^{\mathrm{pot}} \rangle = \tfrac{1}{2} k_{B} T + \tfrac{1}{2} k_{B} T = k_{B} T,$

### 粒子的澱積

$H^{\mathrm{grav}} = m_{b} g z\,,$

## 能量均分定理的通用公式化

$\! \Bigl\langle x_{m} \frac{\partial H}{\partial x_{n}} \Bigr\rangle = \delta_{mn} k_{B} T.$

1. 對所有n$\Bigl\langle x_{n} \frac{\partial H}{\partial x_{n}} \Bigr\rangle = k_{B} T$
2. 對所有mn$\Bigl\langle x_{m} \frac{\partial H}{\partial x_{n}} \Bigr\rangle = 0$

$k_{B} T = \Bigl\langle x_{n} \frac{\partial H}{\partial x_{n}}\Bigr\rangle = 2\langle a_n x_n^2 \rangle,$

$\Bigl\langle p_{k} \frac{\partial H}{\partial p_{k}} \Bigr\rangle = \Bigl\langle q_{k} \frac{\partial H}{\partial q_{k}} \Bigr\rangle = k_{B} T.$

$\Bigl\langle p_{k} \frac{dq_{k}}{dt} \Bigr\rangle = -\Bigl\langle q_{k} \frac{dp_{k}}{dt} \Bigr\rangle = k_{B} T.$

$\Bigl\langle q_{j} \frac{\partial H}{\partial q_{k}} \Bigr\rangle, \quad \Bigl\langle q_{j} \frac{\partial H}{\partial p_{k}} \Bigr\rangle, \quad \Bigl\langle p_{j} \frac{\partial H}{\partial p_{k}} \Bigr\rangle, \quad \Bigl\langle p_{j} \frac{\partial H}{\partial q_{k}} \Bigr\rangle, \quad \Bigl\langle q_{k} \frac{\partial H}{\partial p_{k}} \Bigr\rangle,$   及   $\Bigl\langle p_{k} \frac{\partial H}{\partial q_{k}} \Bigr\rangle$

j≠k則皆為零。

### 與均功定理的關係

$\Bigl\langle \sum_{k} q_{k} \frac{\partial H}{\partial q_{k}} \Bigr\rangle = \Bigl\langle \sum_{k} p_{k} \frac{\partial H}{\partial p_{k}} \Bigr\rangle = \Bigl\langle \sum_{k} p_{k} \frac{dq_{k}}{dt} \Bigr\rangle = -\Bigl\langle \sum_{k} q_{k} \frac{dp_{k}}{dt} \Bigr\rangle,$

## 應用

### 理想氣體定律

\begin{align} \langle H^{\mathrm{kin}} \rangle &= \frac{1}{2m} \langle p_{x}^{2} + p_{y}^{2} + p_{z}^{2} \rangle\\ &= \frac{1}{2} \biggl( \Bigl\langle p_{x} \frac{\partial H^{\mathrm{kin}}}{\partial p_{x}} \Bigr\rangle + \Bigl\langle p_{y} \frac{\partial H^{\mathrm{kin}}}{\partial p_{y}} \Bigr\rangle + \Bigl\langle p_{z} \frac{\partial H^{\mathrm{kin}}}{\partial p_{z}} \Bigr\rangle \biggr) = \frac{3}{2} k_{B} T \end{align}

\begin{align} \langle \mathbf{q} \cdot \mathbf{F} \rangle &= \Bigl\langle q_{x} \frac{dp_{x}}{dt} \Bigr\rangle + \Bigl\langle q_{y} \frac{dp_{y}}{dt} \Bigr\rangle + \Bigl\langle q_{z} \frac{dp_{z}}{dt} \Bigr\rangle\\ &=-\Bigl\langle q_{x} \frac{\partial H}{\partial q_x} \Bigr\rangle - \Bigl\langle q_{y} \frac{\partial H}{\partial q_y} \Bigr\rangle - \Bigl\langle q_{z} \frac{\partial H}{\partial q_z} \Bigr\rangle = -3k_{B} T, \end{align}

$3Nk_{B} T = - \biggl\langle \sum_{k=1}^{N} \mathbf{q}_{k} \cdot \mathbf{F}_{k} \biggr\rangle$

$-\biggl\langle\sum_{k=1}^{N} \mathbf{q}_{k} \cdot \mathbf{F}_{k}\biggr\rangle = P \oint_{\mathrm{surface}} \mathbf{q} \cdot \mathbf{dS},$

$\boldsymbol\nabla \cdot \mathbf{q} = \frac{\partial q_{x}}{\partial q_{x}} + \frac{\partial q_{y}}{\partial q_{y}} + \frac{\partial q_{z}}{\partial q_{z}} = 3,$

$P \oint_{\mathrm{surface}} \mathbf{q} \cdot \mathbf{dS} = P \int_{\mathrm{volume}} \left( \boldsymbol\nabla \cdot \mathbf{q} \right) dV = 3PV,$

$3Nk_{B} T = -\biggl\langle \sum_{k=1}^{N} \mathbf{q}_{k} \cdot \mathbf{F}_{k} \biggr\rangle = 3PV,$

$PV = Nk_{B} T = nRT,\,$

### 雙原子氣體

$H = \frac{\left| \mathbf{p}_{1} \right|^{2}}{2m_{1}} + \frac{\left| \mathbf{p}_{2} \right|^{2}}{2m_{2}} + \frac{1}{2} a q^{2},$

### 極度相對性理想氣體

$H^{\mathrm{kin}} \approx cp = c \sqrt{p_{x}^{2} + p_{y}^{2} + p_{z}^{2}}.$

$p_{x} \frac{\partial H^{\mathrm{kin}}}{\partial p_{x}} = c \frac{p_{x}^{2}}{\sqrt{p_{x}^{2} + p_{y}^{2} + p_{z}^{2}}}$

\begin{align} \langle H^{\mathrm{kin}} \rangle &= \biggl\langle c \frac{p_{x}^{2} + p_{y}^{2} + p_{z}^{2}}{\sqrt{p_{x}^{2} + p_{y}^{2} + p_{z}^{2}}} \biggr\rangle\\ &= \Bigl\langle p_{x} \frac{\partial H^{\mathrm{kin}}}{\partial p_{x}} \Bigr\rangle + \Bigl\langle p_{y} \frac{\partial H^{\mathrm{kin}}}{\partial p_{y}} \Bigr\rangle + \Bigl\langle p_{z} \frac{\partial H^{\mathrm{kin}}}{\partial p_{z}} \Bigr\rangle\\ &= 3 k_{B} T \end{align}

### 非理想氣體

$\langle h^{\mathrm{pot}} \rangle = \int_{0}^{\infty} 4\pi r^{2} \rho U(r) g(r)\, dr$

$H = \langle H^{\mathrm{kin}} \rangle + \langle H^{\mathrm{pot}} \rangle = \frac{3}{2} Nk_{B}T + 2\pi N \rho \int_{0}^{\infty} r^{2} U(r) g(r) \, dr$

$3Nk_{B}T = 3PV + 2\pi N \rho \int_{0}^{\infty} r^{3} U'(r) g(r)\, dr$

### 非諧振器

$H^{\mathrm{pot}} = C q^{s}$

$k_{B} T = \Bigl\langle q \frac{\partial H^{\mathrm{pot}}}{\partial q} \Bigr\rangle = \langle q \cdot s C q^{s-1} \rangle = \langle s C q^{s} \rangle = s \langle H^{\mathrm{pot}} \rangle$

$H^{\mathrm{pot}} = \sum_{n=2}^{\infty} C_{n} q^{n}$

$k_{B} T = \Bigl\langle q \frac{\partial H^{\mathrm{pot}}}{\partial q} \Bigr\rangle = \sum_{n=2}^{\infty} \langle q \cdot n C_{n} q^{n-1} \rangle = \sum_{n=2}^{\infty} n C_{n} \langle q^{n} \rangle$

$\langle H^{\mathrm{pot}} \rangle = \frac{1}{2} k_{B} T - \sum_{n=3}^{\infty} \left( \frac{n - 2}{2} \right) C_{n} \langle q^{n} \rangle$

### 布朗運動

$\frac{d\mathbf{v}}{dt} = \frac{1}{m} \mathbf{F} = -\frac{\mathbf{v}}{\tau} + \frac{1}{m} \mathbf{F}^{\mathrm{rnd}},$

$\Bigl\langle \mathbf{r} \cdot \frac{d\mathbf{v}}{dt} \Bigr\rangle + \frac{1}{\tau} \langle \mathbf{r} \cdot \mathbf{v} \rangle = 0$

（由於Frnd跟位置向量r不相關）。使用數學恒等式

$\frac{d}{dt} \left( \mathbf{r} \cdot \mathbf{r} \right) = \frac{d}{dt} \left( r^{2} \right) = 2 \left( \mathbf{r} \cdot \mathbf{v} \right)$

$\frac{d}{dt} \left( \mathbf{r} \cdot \mathbf{v} \right) = v^{2} + \mathbf{r} \cdot \frac{d\mathbf{v}}{dt},$

$\frac{d^{2}}{dt^{2}} \langle r^{2} \rangle + \frac{1}{\tau} \frac{d}{dt} \langle r^{2} \rangle = 2 \langle v^{2} \rangle = \frac{6}{m} k_{B} T,$

$\langle H^{\mathrm{kin}} \rangle = \Bigl\langle \frac{p^{2}}{2m} \Bigr\rangle = \langle \tfrac{1}{2} m v^{2} \rangle = \tfrac{3}{2} k_{B} T$

$\langle r^{2} \rangle = \frac{6k_{B} T \tau^{2}}{m} \left( e^{-t/\tau} - 1 + \frac{t}{\tau} \right)$

$\langle r^{2} \rangle \approx \frac{3k_{B} T}{m} t^{2} = \langle v^{2} \rangle t^{2}$

$\langle r^{2} \rangle \approx \frac{6k_{B} T\tau}{m} t = 6\gamma k_{B} T t$

### 恒星物理學

$H^{\mathrm{grav}}_{\mathrm{tot}} = -\int_{0}^{R} \frac{4\pi r^{2} G}{r} M(r)\, \rho(r)\, dr,$

$H^{\mathrm{grav}}_{\mathrm{tot}} = - \frac{3G M^{2}}{5R},$

$H^{\mathrm{grav}}_{\mathrm{tot}} = - \frac{3G M^{2}}{5R},$

$\Bigl\langle r \frac{\partial H^{\mathrm{grav}}}{\partial r} \Bigr\rangle = \langle -H^{\mathrm{grav}} \rangle = k_{B} T = \frac{3G M^{2}}{5RN}.$

### 恒星的形成

$\frac{3G M^{2}}{5R} > 3 N k_{B} T$

$M = \frac{4}{3} \pi R^{3} \rho$

$M_{J}^{2} = \left( \frac{5k_{B}T}{G m_{p}} \right)^{3} \left( \frac{3}{4\pi \rho} \right)$

## 推導

### 動能與麥克斯韋-波茲曼分布

$f (v) = 4 \pi \left( \frac{m}{2 \pi k_B T}\right)^{3/2}\!\!v^2 \exp \Bigl( \frac{-mv^2}{2k_B T} \Bigr)$

$\langle H^{\mathrm{kin}} \rangle = \langle \tfrac{1}{2} m v^{2} \rangle = \int _{0}^{\infty} \tfrac{1}{2} m v^{2}\ f(v)\ dv = \tfrac{3}{2} k_{B} T$

### 二次能量與配分函數

$Z_{x} = \int_{-\infty}^{\infty} dx \ e^{-\beta A x^{2}} = \sqrt{\frac{\pi}{\beta A}}$

$\langle H_{x} \rangle = - \frac{\partial \log Z_{x}}{\partial \beta} = \frac{1}{2\beta} = \frac{1}{2} k_{B} T$

### 一般證明

$d\Gamma = \prod_{i} dq_{i} dp_{i}$

$\Gamma (E, \Delta E) = \int_{H \in \left[E, E+\Delta E \right]} d\Gamma .$

$\Sigma (E) = \int_{H < E} d\Gamma .$

$\int_{H \in \left[ E, E+\Delta E \right]} \ldots d\Gamma = \Delta E \frac{\partial}{\partial E} \int_{H < E} \ldots d\Gamma,$

$\Gamma = \Delta E \ \frac{\partial \Sigma}{\partial E} = \Delta E \ \rho(E),$

$\frac{1}{T} = \frac{\partial S}{\partial E} = k_{b} \frac{\partial \log \Sigma}{\partial E} = k_{b} \frac{1}{\Sigma}\,\frac{\partial \Sigma}{\partial E}$

#### 正則系綜

$\mathcal{N} \int e^{-\beta H(p, q)} d\Gamma = 1,$

$\mathcal{N} \int \left[ e^{-\beta H(p, q)} x_{k} \right]_{x_{k}=a}^{x_{k}=b} d\Gamma_{k}+ \mathcal{N} \int e^{-\beta H(p, q)} x_{k} \beta \frac{\partial H}{\partial x_{k}} d\Gamma = 1,$

$\mathcal{N} \int e^{-\beta H(p, q)} x_{k} \frac{\partial H}{\partial x_{k}} \,d\Gamma = \Bigl\langle x_{k} \frac{\partial H}{\partial x_{k}} \Bigr\rangle = \frac{1}{\beta} = k_{B} T.$

#### 微正則系綜

\begin{align} \Bigl\langle x_{m} \frac{\partial H}{\partial x_{n}} \Bigr \rangle &= \frac{1}{\Gamma} \, \int_{H \in \left[ E, E+\Delta E \right]} x_{m} \frac{\partial H}{\partial x_{n}} \,d\Gamma\\ &=\frac{\Delta E}{\Gamma}\, \frac{\partial}{\partial E} \int_{H < E} x_{m} \frac{\partial H}{\partial x_{n}} \,d\Gamma\\ &= \frac{1}{\rho} \,\frac{\partial}{\partial E} \int_{H < E} x_{m} \frac{\partial \left( H - E \right)}{\partial x_{n}} \,d\Gamma, \end{align}

\begin{align} \int_{H < E} x_{m} \frac{\partial ( H - E )}{\partial x_{n}} \,d\Gamma &= \int_{H < E} \frac{\partial}{\partial x_{n}} \bigl( x_{m} ( H - E ) \bigr) \,d\Gamma - \int_{H < E} \delta_{mn} ( H - E ) d\Gamma\\ &= \delta_{mn} \int_{H < E} ( E - H ) \,d\Gamma, \end{align}

$\Bigl\langle x_{m} \frac{\partial H}{\partial x_{n}} \Bigr\rangle = \delta_{mn} \frac{1}{\rho} \, \frac{\partial}{\partial E} \int_{H < E}\left( E - H \right)\,d\Gamma = \delta_{mn} \frac{1}{\rho} \, \int_{H < E} \,d\Gamma = \delta_{mn} \frac{\Sigma}{\rho}$

$\Bigl\langle x_{m} \frac{\partial H}{\partial x_{n}} \Bigr\rangle = \delta_{mn} \Bigl(\frac{1}{\Sigma} \frac{\partial \Sigma}{\partial E}\Bigr)^{-1} = \delta_{mn} \Bigl(\frac{\partial \log \Sigma} {\partial E}\Bigr)^{-1} = \delta_{mn} k_{B} T$

$\! \Bigl\langle x_{m} \frac{\partial H}{\partial x_{n}} \Bigr\rangle = \delta_{mn} k_{B} T,$

## 限制

### 量子效應引起的失敗

$P(E_{n}) = \frac{e^{-n\beta h\nu}}{Z}$

$Z = \sum_{n=0}^{\infty} e^{-n\beta h\nu} = \frac{1}{1 - e^{-\beta h\nu}}$

$\langle H \rangle = \sum_{n=0}^{\infty} E_{n} P(E_{n}) = \frac{1}{Z} \sum_{n=0}^{\infty} nh\nu \ e^{-n\beta h\nu} = -\frac{1}{Z} \frac{\partial Z}{\partial \beta} = -\frac{\partial \log Z}{\partial \beta}$

Z的式子代入得最後結果[7]

$\langle H \rangle = h\nu \frac{e^{-\beta h\nu}}{1 - e^{-\beta h\nu}}$

## 註釋及參考資料

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## 延伸閱讀

• Huang, K. Statistical Mechanics 2nd ed. John Wiley and Sons. 1987: pp. 136–138. ISBN 0-471-81518-7.
• Khinchin, AI. Mathematical Foundations of Statistical Mechanics (G. Gamow, translator). New York: Dover Publications. 1949: pp. 93–98. ISBN 0-486-63896-0.
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