分析力学第12次作业

Chasse_neige

7.6 用两种方法证明以下变换是正则变换

Q = \sqrt{2 p} \sin q, P = \sqrt{2 p} \cos q

(2) 计算泊松括号。

计算泊松括号为

\begin{matrix} [Q, P]_{p q} = \frac{\partial Q}{\partial q} \frac{\partial P}{\partial p} - \frac{\partial Q}{\partial p} \frac{\partial P}{\partial q} \\ = \sqrt{2 p} \cos q \frac{\cos q}{\sqrt{2 p}} + \frac{\sin q}{\sqrt{2 p}} \sqrt{2 p} \sin q = 1 \end{matrix}

并且显然有

[Q, Q]_{p q} = [P, P]_{p q} = 0

所以是正则变换。

7.8 用两种方法证明以下变换是正则变换

Q_{i} = q_{i} \cos θ_{i} - p_{i} \sin θ_{i}, P_{i} = q_{i} \sin θ_{i} + p_{i} \cos θ_{i}, i = 1, 2, 3, \dots, s

其中 $θ_{i}$ 是适当的常量， $s$ 是系统自由度个数。

(2) 计算泊松括号。

计算泊松括号

\begin{matrix} [Q_{i}, P_{j}]_{p q} = \sum_{k} \frac{\partial Q_{i}}{\partial q_{k}} \frac{\partial P_{j}}{\partial p_{k}} - \frac{\partial Q_{i}}{\partial p_{k}} \frac{\partial P_{j}}{\partial q_{k}} \\ = \sum_{k} δ_{i k} \cos θ_{i} δ_{j k} \cos θ_{j} - δ_{i k} \sin θ_{k} δ_{j k} \sin θ_{k} \\ = δ_{i j} \end{matrix}

\begin{matrix} [Q_{i}, Q_{j}]_{p q} = \sum_{k} \frac{\partial Q_{i}}{\partial q_{k}} \frac{\partial Q_{j}}{\partial p_{k}} - \frac{\partial Q_{i}}{\partial p_{k}} \frac{\partial Q_{j}}{\partial q_{k}} \\ = \sum_{k} δ_{i k} \cos θ_{i} δ_{j k} \sin θ_{j} - δ_{i k} \sin θ_{k} δ_{j k} \cos θ_{k} \\ = 0 \end{matrix}

\begin{matrix} [P_{i}, P_{j}]_{p q} = \sum_{k} \frac{\partial P_{i}}{\partial q_{k}} \frac{\partial P_{j}}{\partial p_{k}} - \frac{\partial P_{i}}{\partial p_{k}} \frac{\partial P_{j}}{\partial q_{k}} \\ = \sum_{k} δ_{i k} \sin θ_{i} δ_{j k} \cos θ_{j} - δ_{i k} \cos θ_{k} δ_{j k} \sin θ_{k} \\ = 0 \end{matrix}

所以是正则变换。

7.11 求证: 角动量 $J = (J_{1}, J_{2}, J_{3})$ 分量之间满足 $[J_{1}, J_{2}] = J_{3}$ 。

证明：

写出角动量的分量式子

J_{k} = ϵ_{i j k} r_{i} p_{k}

所以

\begin{matrix} [J_{i}, J_{j}] = [ϵ_{k l i} r_{k} p_{l}, ϵ_{m n j} r_{m} p_{n}] = ϵ_{k l i} ϵ_{m n j} [r_{k} p_{l}, r_{m} p_{n}] \\ = ϵ_{k l i} ϵ_{m n j} (δ_{k n} p_{l} r_{m} - δ_{l m} r_{k} p_{n}) \\ = ϵ_{n l i} ϵ_{m n j} p_{l} r_{m} - ϵ_{k m i} ϵ_{m n j} r_{k} p_{n} \\ = (δ_{l j} δ_{i m} - δ_{l m} δ_{i j}) p_{l} r_{m} - (δ_{i n} δ_{k j} - δ_{k n} δ_{i j}) r_{k} p_{n} \\ = p_{j} r_{i} - δ_{i j} p_{l} r_{l} - p_{i} r_{j} + δ_{i j} r_{k} p_{k} = r_{i} p_{j} - r_{j} p_{i} \end{matrix}

所以

[J_{1}, J_{2}] = r_{1} p_{2} - r_{2} p_{1} = J_{3}

7.13 在中心力场 $V (r) = - \frac{k}{r}$ 运动的质点，质量 $m$ ，动量 $p$ ，角动量 $J$ ，还有一个守恒量，拉普拉斯-龙格-楞次 (Laplace-Runge-Lenz) 矢量

\vec{A} = \vec{p} \times \vec{J} - \frac{m k \vec{r}}{r}

利用泊松括号证明该矢量是守恒量。

由于 $A$ 不显含 $t$ ，所以

\frac{d \vec{A}}{d t} = \frac{\partial \vec{A}}{\partial t} + [\vec{A}, H] = [\vec{A}, H]

由于

H = \frac{p^{2}}{2 m} - \frac{k}{r}

所以

[\vec{A}, H] = [\vec{p} \times \vec{J} - \frac{m k \vec{r}}{r}, \frac{p^{2}}{2 m} - \frac{k}{r}]

展开上述式子

\begin{matrix} [p^{2} \vec{r} - (\vec{p} \cdot \vec{r}) \vec{p} - \frac{m k \vec{r}}{r}, \frac{p^{2}}{2 m} - \frac{k}{r}] = [p^{2}, \frac{p^{2}}{2 m}] \vec{r} + [\vec{r}, \frac{p^{2}}{2 m}] p^{2} - [p^{2} \vec{r}, \frac{k}{r}] - [(\vec{p} \cdot \vec{r}) \vec{p}, \frac{p^{2}}{2 m}] + [(\vec{p} \cdot \vec{r}) \vec{p}, \frac{k}{r}] \\ - [\frac{m k \vec{r}}{r}, \frac{p^{2}}{2 m}] + [\frac{m k \vec{r}}{r}, \frac{k}{r}] \end{matrix}

做等价变换

[, \frac{p^{2}}{2 m}] = [, \frac{\vec{p} \cdot \vec{p}}{2 m}] = [, \vec{p}] \cdot \frac{\vec{p}}{m}

带入泊松括号的表达式

[p^{2}, \frac{p^{2}}{2 m}] = 0

[r_{i}, \frac{p^{2}}{2 m}] = [r_{i}, p_{j}] \frac{p_{j}}{m} = δ_{i j} \frac{p_{j}}{m} = \frac{p_{i}}{m}

\begin{matrix} [p^{2} r_{i}, \frac{k}{r}] = [r_{i}, \frac{k}{r}] p^{2} + 2 p_{j} [p_{j}, \frac{k}{r}] r_{i} \\ = 0 + \frac{2 p_{j} r_{j}}{r^{3}} k r_{i} \end{matrix}

[(p_{j} r_{j}) p_{i}, \frac{p^{2}}{2 m}] = [p_{j} r_{j} p_{i}, p_{k}] \frac{p_{k}}{m} = p_{j} p_{i} δ_{j k} \frac{p_{k}}{m} = \frac{p^{2}}{m} p_{i}

[p_{j} r_{j} p_{i}, \frac{k}{r}] = \frac{k}{r^{2}} (r_{j} p_{i} \frac{r_{j}}{r} + p_{j} r_{j} \frac{r_{i}}{r})

[\frac{m k r_{i}}{r}, \frac{p^{2}}{2 m}] = p_{j} k [\frac{r_{i}}{r}, p_{j}] = \frac{p_{i} k}{r} - \frac{k p_{j} r_{j} r_{i}}{r^{3}}

[\frac{m k r_{i}}{r}, \frac{k}{r}] = 0

所以带入得到

[A_{i}, H] = p^{2} \frac{p_{i}}{m} - \frac{2 p_{j} r_{j}}{r^{3}} k r_{i} - \frac{p^{2}}{m} p_{i} + \frac{k}{r^{2}} (r p_{i} + p_{j} r_{j} \frac{r_{i}}{r}) - \frac{p_{i} k}{r} + \frac{k p_{j} r_{j} r_{i}}{r^{3}} = 0

所以是守恒量

7.14 计算拉普拉斯-龙格-楞次矢量 $A$ 各分量之间的泊松括号。

\begin{matrix} [A_{i}, A_{j}] = [ϵ_{l m i} p_{l} J_{m} - \frac{m k r_{i}}{r}, ϵ_{k n j} p_{k} J_{n} - \frac{m k r_{j}}{r}] \\ = ϵ_{l m i} ϵ_{k n j} [p_{l} J_{m}, p_{k} J_{n}] - m k ϵ_{k n j} [\frac{r_{i}}{r}, p_{k} J_{n}] - m k ϵ_{l m i} [p_{l} J_{m}, \frac{r_{j}}{r}] \end{matrix}

第一项

ϵ_{l m i} ϵ_{k n j} [p_{l} J_{m}, p_{k} J_{n}] = - 2 m H ϵ_{i j k} J_{k} - 2 m k \frac{1}{r} ϵ_{i j k} J_{k} + 2 m k (\frac{r_{j}}{r} p_{i} - \frac{r_{i}}{r} p_{j})

第二项

- m k ϵ_{k n j} [\frac{r_{i}}{r}, p_{k} J_{n}] = - 2 m k \frac{1}{r} ϵ_{i j k} J_{k} - 2 m k \frac{r_{j}}{r} p_{i} + 2 m k \frac{r_{i}}{r} p_{j}

第三项

- m k ϵ_{l m i} [p_{l} J_{m}, \frac{r_{j}}{r}] = 2 m k \frac{1}{r} ϵ_{i j k} J_{k} + 2 m k \frac{r_{j}}{r} p_{i} - 2 m k \frac{r_{i}}{r} p_{j}

将三项相加，所有含 $\frac{m k}{r}$ 和 $r_{i} p_{j}$ 的项相互抵消，最终结果为

[A_{i}, A_{j}] = - 2 m H ϵ_{i j k} J_{k}

其中 $H = \frac{p^{2}}{2 m} - \frac{k}{r}$ 是系统的哈密顿量。

分析力学 第12次作业 ​